BV equivalence with boundary

Abstract

An extension of the notion of classical equivalence of equivalence in the Batalin–Vilkovisky (BV) and Batalin–Fradkin–Vilkovisky (BFV) frameworks for local Lagrangian field theory on manifolds possibly with boundary is discussed. Equivalence is phrased in both a strict and a lax sense, distinguished by the compatibility between the BV data for a field theory and its boundary BFV data, necessary for quantisation. In this context, the first- and second-order formulations of nonabelian Yang–Mills and of classical mechanics on curved backgrounds, all of which admit a strict BV–BFV description, are shown to be pairwise equivalent as strict BV–BFV theories. This in particular implies that their BV complexes are quasi-isomorphic. Furthermore, Jacobi theory and one-dimensional gravity coupled with scalar matter are compared as classically equivalent reparametrisation-invariant versions of classical mechanics, but such that only the latter admits a strict BV–BFV formulation. They are shown to be equivalent as lax BV–BFV theories and to have isomorphic BV cohomologies. This shows that strict BV–BFV equivalence is a strictly finer notion of equivalence of theories.

1 Introduction

The notion of equivalence of field theories is one that can be found throughout physics. Such a concept is relevant and useful for various reasons. At a classical level, the equations of motions of a given theory might be easier to handle than other “equivalent” ones, even if they ultimately yield the same moduli space of solutions. Such reformulations often result in different and enlightening new interpretations of a given problem. Moreover, one theory might be better suited for quantisation than another, but the question of whether two classically equivalent theories result in the same quantum theory is in general still open. With this work, we attempt to take another step towards the answer.

The classical physical content of a given field theory is encoded in the set $\mathcal {EL}$ of solutions of the Euler–Lagrange equations. In the case where the theory in question also enjoys a local symmetry—encoded by a tangent distribution D—we are interested in the moduli space of inequivalent solutions $\mathcal {EL}/D$. Classical observables are then defined to be suitable functions on $\mathcal {EL}/D$. Such a quotient is typically singular: Defining a sensible space of functions over it becomes challenging, and it is often more convenient to find a replacement; a problem best addressed within the Batalin–Vilkovisky (BV) formalism.

The BV formalism was first introduced in [12,13,14] as an extension of the BRST formalism [10, 63], named after Becchi, Rouet, Stora,and Tyutin, used to quantise Lagrangian gauge theories in a way that preserves covariance. Around the same time, the Batalin–Fradkin–Vilkovisky (BFV) formalism was introduced, which deals with constrained Hamiltonian systems [4, 11]. It was later noticed by various authors [8, 31, 32, 39, 40, 43, 44, 61, 62] that the aforementioned formalisms enjoy a rich cohomological structure. For example, a BV theory associates a chain complex to a spacetime manifold, the BV complex, which aims at a resolution of the desired space of functions over the quotient $\mathcal {EL}/D$. In the case of the BFV formalism, one introduces the BFV complex [55, 56, 61] as a resolution of the space of functions over the reduced phase space of a given constrained Hamiltonian system.

One can then address the question of equivalence of theories in the BV setting. Following the discussion above, a natural way of comparing two classical theories is through their BV cohomologies, also called classical observables, as done for example in [3]. However, a BV theory comes equipped with several pieces of data other than the underlying dg-algebra structure (for example, a symplectic structure and a Hamiltonian function) that one might want an equivalence relation to preserve. Finding the appropriate notion of “BV equivalence” is thus a nontrivial open question. In [28, 29], a stronger notion of BV equivalence is implemented, which requires all data to be preserved by a symplectomorphism. A nontrivial example of such equivalence is found between 3d gravity and (nondegenerate) BF theory. In [17], various alternative weaker notions of BV equivalence have been presented, which apply to higher dimensional formulations of General Relativity.

The BV and BFV approaches were linked by Cattaneo, Mnev and Reshetikhin in [19], where the authors showed that a BV theory on the bulk induces a compatible BFV theory on the boundary, provided that some regularity conditions are met. The presence of a boundary will typically spoil the symmetry invariance of the BV data, encoded in the BV cohomology, but this failure will be controlled by the BFV data associated to the boundary. From this perspective, the regularity conditions can be seen as a compatibility condition between the BV complex on the bulk and the BFV complex on the boundary.

Derived geometry [50] extends the above setting to algebraic geometry, even though currently only in the restricted setting of AKSZ theories. The induced boundary theory is in this case an example of derived intersection, [15, 24]. As derived geometry mainly addresses classical problems, the nondegeneracy of the symplectic form is only required up to homotopy, which yields problems in the direction of quantisation.

On the other hand the BV–BFV approach [19, 20] is especially successful since it allows for a quantisation procedure that is compatible with cutting and gluing. This has already been shown to work in various examples such as BF theory [20, 21], split Chern–Simons theory [22], 2D Yang–Mills theory [42] and AKSZ sigma models [23].

This approach was first tested¹ on General Relativity in [57]. For diffeomorphism invariant theories, the compatibility between bulk and boundary data becomes a nontrivial matter, and there are various cases where the regularity conditions necessary for the BV–BFV description fail to be met. Most notable are the examples of Palatini–Cartan gravity in (3+1) dimensions [27], Plebanski theory [57], the Nambu–Goto string [46], and the Jacobi action for reparametrisation-invariant classical mechanics [26]. On the other hand, the respectively classically equivalent Einstein–Hilbert formulation of gravity [25] in (3+1) dimensions and the Polyakov action [46] fulfil the BV–BFV axioms. The question of how one can go around these problems and construct a sensible BV–BFV theory for Palatini–Cartan gravity was addressed in [16, 17].

As not all field theories are suitable for a BV–BFV description, the lax approach to the BV–BFV formalism was proposed in [47], which gathers the data prior to the step where the regularity conditions become relevant. This setting already allows us to construct the BV–BFV complex [47], which is the adaptation of the BV complex to the case with boundary. Likewise, classical observables are contained in its cohomology. As such, the lax BV–BFV formalism offers a sensible way of comparing two field theories on manifolds with boundary, even if one does not have a strict BV–BFV theory.

In this paper, we provide an explicit method to lift classical equivalence to a (potential) BV equivalence, also in the presence of boundaries. This naturally introduces the notion of lax equivalence of BV theories on manifolds with boundary, which is in principle finer than BV equivalence. Our method is applied to the simple cases of classical mechanics on a curved background as well as to (nonabelian) Yang–Mills theory, where we explicitly show that the first- and second-order formalisms are lax BV–BFV equivalent (and hence BV-quasi-isomorphic).

We then turn our attention to the main objective of this paper: the analysis of the classically equivalent Jacobi theory and one-dimensional gravity coupled to matter (1D GR). These two models can be regarded as the one-dimensional counterparts of the Nambu–Goto and Polyakov string models respectively, and they both represent a reparametrisation-invariant version of classical mechanics. In [26], it was shown that while 1D GR satisfies the regularity conditions of the BV–BFV formalism, Jacobi theory produces a singular theory on the boundary, and a similar result was proven for their 2d string-theoretic analogues [46], which raises the question of the origin of this boundary discrepancy.

By comparing the BV and BV–BFV cohomologies of Jacobi theory and 1D GR, we find that, even though the two theories on manifolds with possibly nonempty boundary are lax-equivalent, and hence their associated BV (and lax BV–BFV) complexes are quasi-isomorphic, the chain maps that connect the two theories do not preserve the regularity condition required by the strictification procedure (Theorem 3.5.7).

In other words, quasi-isomorphisms of lax BV–BFV complexes do not preserve strict BV–BFV theories, which then should be taken as a genuine subclass of BV theories: Even in the best case scenario of two theories that are classically equivalent with quasi-isomorphic lax BV–BFV complexes, an obstruction to their strict BV–BFV compatibility distinguishes the two. Indeed, consider two lax equivalent theories (Definition 2.6.3—see, e.g. the case described in Theorem 3.5.6) such that one of the two models fails to be compatible with the strict BV–BFV axioms (cf. Remark 2.5.6). In this case, only one of the two admits a quantisation in the BV–BFV setting. Even if they both could ultimately admit a sensible quantisation, our result suggests anyway that they might have different quantisations in the presence of boundaries.

Another way of viewing our result is the following. Suppose we are given a lax BV–BFV theory that is not strict (Definition 2.5.5 and Remark 2.5.6). Can we find a quasi-isomorphic lax BV–BFV theory that is strictifiable? If so, we may think of the second theory as a good replacement for the first, suitable for quantisation with boundary.

We should stress that the enrichment of the BV complex by the de Rham complex of the source manifold (in the sense of local forms) has been the object of past research (see among all [3]). The lax BV–BFV complex we consider coincides with their Batalin–Vilkovisky–de Rham complex; however, our notion of lax equivalence is different (Definition 2.6.3), as it requires the existence of chain maps that are quasi-inverse to one another and compatibile with the whole lax BV–BFV structure.

Crucially, our approach diverges from other investigations of local field theory that only look at pre-symplectic data. The strictification step is precisely the pre-symplectic reduction of such data, and where the obstruction lies. We are not aware of a viable quantisation procedure for pre-symplectic structures.

This paper is structured as follows: Sect. 2 is dedicated to a review of local Lagrangian field theory (Sect. 2.1), which is followed by the BV formalism (Sect. 2.2) and the BV–BFV and lax BV–BFV formalisms (Sect. 2.4). We will showcase several notions of equivalence in classical field theory, starting from Lagrangian field theory in Sect. 2.1, while the discussion of equivalence in the BV and lax BV–BFV cases can be found in Sects. 2.3 and 2.6 respectively. Later in Sect. 3 we discuss our general procedure to prove lax equivalence between two theories (Sect. 3.1) and three examples of such equivalence, namely

• first- and second-order formulations of classical mechanics on a curved background (Sect. 3.3);

• first- and second-order formulations of (nonabelian) Yang–Mills theory (Sect. 3.4);

• one-dimensional gravity coupled to matter and Jacobi theory (Sect. 3.5).

Results and outlook: We present our notion of BV equivalence (Definition 2.3.1) for theories over closed manifolds and lax equivalence (Definition 2.6.3) in the case of manifolds with higher strata, and show that the latter implies the former for the respective bulk (codimension-0 stratum) BV theories (Theorem 2.6.9).

We then show lax equivalence for the aforementioned examples, in the sense that their lax BV–BFV data can be interchanged in a way that preserves their cohomological structure. In particular, we show that the respective BV–BFV complexes are quasi-isomorphic

$$\begin{aligned} H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1) \simeq H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2). \end{aligned}$$

Most notably, this means that the boundary discrepancy present in the BV–BFV formulations of Jacobi theory and 1D GR found in [26] does not have a cohomological origin, and is rather to be interpreted as an obstruction in pre-quantisation.

We expect the procedure to be applicable to other relevant examples of BV–BFV obstructions such as the Nambu-Goto and Polyakov actions [46] and, for a more challenging one, Einstein–Hilbert and Palatini–Cartan gravity in (3+1) dimensions, whose extendibility as BV–BFV theories have been shown to differ in [25, 27].

This obstruction, which bars certain theories from being quantisable in the BV formalism with boundary without additional requirements on the fields, suggests that, even assuming that some quantum theory exists for both models, they might differ. Alternatively, it might suggest that among various classically and BV-equivalent models, there is a preferred choice for models which are BV–BFV compatible. Either way, these results call for additional investigations in this direction.

2 Field theories and equivalence

We start by presenting the field theoretical structures and objects used throughout this work, following [1, 30]. Subsequently, we review the BV formalism for closed manifolds² [12,13,14] − see also [37, 40, 45] − and the BV–BFV formalism, its generalisation for manifolds with boundaries and corners [18, 19]. As some theories we consider are not compatible with the BV–BFV axioms, we revise the lax BV–BFV formalism [47], which not only lets us study these cases, but presents a better stage for our discussions in the presence of boundaries and corners.

Moreover, this section is used to develop our notion of equivalence of field theories at every step of the way, first showcasing how we want to compare two classical field theories in Definition 2.1.6 and adapting these considerations to the BV and lax BV–BFV formalisms in Definitions 2.3.1 and 2.6.3, respectively.

2.1 Lagrangian field theories

Let M be a manifold of any dimension. In order to build a classical field theory on M, we need a space of fields ${\mathcal {E}}$, a local functional S called the action functional and local observables. In most cases, we can achieve such a construction by considering a (possibly graded) fibre bundle $E\rightarrow M$ over M and by defining the space of fields as its space of smooth sections ${\mathcal {E}} {:}{=}\Gamma (M,E)$ with coordinates $\varphi ^i$. Local objects can then be regarded as a subcomplex of the de Rham bicomplex $\Omega ^{\bullet ,\bullet }({\mathcal {E}}\times M)$, where “local” essentially means that these objects only depend on the first k derivatives of the fields $\varphi ^i$ (or the kth jet). Let us make this notion precise:

Definition 2.1.1

( (Integrated) local forms [1]) Let $E\rightarrow M$ be a (possibly graded) fibre bundle over M, ${\mathcal {E}} = \Gamma (M,E)$ its space of smooth sections, $J^k (E)$ the kth jet bundle and $\{j^k :{\mathcal {E}} \times M \rightarrow J^k(E)\}$ the evaluation maps. We consider $j^\infty$ as the inverse limit of these maps and construct the infinite jet bundle $J^\infty (E)$ as the inverse limit of the sequence

$$\begin{aligned} E = J^0(E) \leftarrow J^1(E) \leftarrow \dots \leftarrow J^k(E) \leftarrow \dots \end{aligned}$$

The bicomplex.³ of local forms on ${\mathcal {E}}\times M$ is defined as

$$\begin{aligned} (\Omega ^{\bullet ,\bullet }_{\text {loc}}({\mathcal {E}}\times M),\delta ,\mathrm {d}) {:}{=}(j^\infty )^*\left( \Omega ^{\bullet ,\bullet }(J^\infty (E)),\mathrm {d}_V,\mathrm {d}_H\right) , \end{aligned}$$

where $\mathrm {d}_H$ and $\mathrm {d}_V$ are the horizontal and vertical differentials on the variational complex for $J^\infty (E)$, respectively. Let $\alpha \in \Omega ^{\bullet ,\bullet }(J^\infty (E))$. The differentials $\delta , \mathrm {d}$ are defined through

$$\begin{aligned} \mathrm {d}(j^\infty )^* \alpha&= (j^\infty )^* \mathrm {d}_H \alpha , \end{aligned}$$

(1a)

$$\begin{aligned} \delta (j^\infty )^* \alpha&= (j^\infty )^* \mathrm {d}_V \alpha . \end{aligned}$$

(1b)

Elements of $\Omega ^{0,\bullet }_{\text {loc}}({\mathcal {E}}\times M)$ will be called local functionals on ${\mathcal {E}}\times M$.

Whenever the manifold M is compact, one can define the complex of integrated local k forms $\Omega ^\bullet _{\int }(\mathcal {E})$, as the image of $\int _M :\Omega ^{k,\mathrm {top}}_{\mathrm {loc}}(\mathcal {E}\times M) \rightarrow \Omega ^k_{\int }(\mathcal {E})$ with the (variational) differential⁴$\delta$.

Remark 2.1.2

(On various notions of local forms) Notice that, in some field theory literature (see, e.g. [30]), the term “local form” is often used to denote integrals over the manifold M of elements of $\Omega ^{\bullet , \mathrm {top}}_{\mathrm {loc}}(\mathcal {E}\times M)$, which instead we call integrated local forms.

When M is not compact, integration comes with caveats. One can either consider compactly supported sections or adopt the point of view of [33], where the Lagrangian density is tested against a compactly supported function. Alternatively, one can forgo integration and consider the following quotient

$$\begin{aligned} \Omega ^\bullet _\mathrm {loc}({\mathcal {E}}) {:}{=}\Omega ^{\bullet ,\mathrm {top}}_\text {loc}({\mathcal {E}}\times M) / \mathrm {d}\Omega ^{\bullet ,\mathrm {top}-1}_\text {loc}({\mathcal {E}}\times M). \end{aligned}$$

In [1, Page 21], the elements of $\Omega ^\bullet _{\mathrm {loc}}(\mathcal {E})$ are called variational forms when endowed with the induced vertical differential⁵$\delta _V$.

Clearly, if M is closed, $(\Omega ^\bullet _{\int }(\mathcal {E}),\delta )$ is isomorphic (as a complex) to $(\Omega ^\bullet _{\mathrm {loc}}(\mathcal {E}),\delta _V)$. Indeed, let $f,g \in \Omega ^{\bullet ,\mathrm {top}}_\text {loc}({\mathcal {E}}\times M)$ and define $\mathscr {F} {:}{=}\int _M f$, $\mathscr {G} {:}{=}\int _M g$, their respective integrals over M. Then $\mathscr {F} =\mathscr {G}$ iff the difference $f - g$ is $\mathrm {d}$-exact

$$\begin{aligned} \mathscr {F} - \mathscr {G} = \int _M (f-g) = \int _M \mathrm {d}(\dots ) = 0, \end{aligned}$$

where we used that M is closed in the last step. Hence, $\Omega _{\mathrm {loc}}^\bullet (\mathcal {E})$ can be taken as a replacement of integrated local forms in the noncompact case (assuming there still is no boundary).

If M has a nonempty boundary $\partial M \ne \emptyset$, these considerations no longer hold: Boundary terms become relevant. If, on the local densities side, we can work with $\Omega ^{\bullet ,\bullet }_{\mathrm {loc}}({\mathcal {E}}\times M)$, we will see in Sect. 2.4 what the consequences of integrating over boundaries bring about in field theory.

In addition to the previous construction, we will extensively use the following type of vector field:

Definition 2.1.3

An evolutionary vector field [1] $X\in \mathfrak {X}_{\text {evo}}({\mathcal {E}})$ on ${\mathcal {E}}$ is a vector field on $J^\infty (E)$ which is vertical with respect to the projection $J^\infty (E) \rightarrow M$, such that

$$\begin{aligned} {[}\mathcal {L}_X,\mathrm {d}] = 0, \end{aligned}$$

where $\mathcal {L}_X = [\iota _X,\delta ]$ is the variational Lie derivative on local forms on ${\mathcal {E}}\times M$.

We are now ready to define the notion of a classical field theory. We will assume for simplicity that M is compact, possibly with boundary:

Definition 2.1.4

A classical field theory on M is a pair $({\mathcal {E}},S)$, consisting of a space of fields ${\mathcal {E}} = \Gamma (M,E)$⁶ and an action functional $S \in \Omega ^{0}_{\int }({\mathcal {E}})$.

Since $S=\int L$ for some local form L, applying the variational differential on $\mathcal {E}$ to S is the same as applying $\delta$, defined in Eq. 1b, to L and integrating. This yields two terms:

$$\begin{aligned} \delta S = \text {EL} + \text {BT}. \end{aligned}$$

The term $\text {EL}$ is an integrated local 1-form.⁷ on ${\mathcal {E}}$, whose vanishing locus defines the Euler–Lagrange equations $\text {EL} = 0$. The space where these are satisfied is called the critical locus, the zero locus $\mathcal {EL}:=\mathrm {Loc}_0(\mathrm {EL}) \subset F$, and its elements are called classical solutions. The term $\text {BT}$ is a boundary term (i.e. an integral over $\partial M$, when not empty), which will be crucial for the construction of field theories on manifolds with boundary (cf. Sect. 2.5).

A further important aspect of field theories is the notion of (gauge) symmetries, which are transformations that leave the action functional S and the critical locus $\mathcal {EL}$ invariant. Infinitesimally, they can be described as follows:

Definition 2.1.5

An infinitesimal local symmetry of a classical field theory $({\mathcal {E}},S)$ is given by a distribution $D \subseteq T\mathcal {E}$, such that⁸

$$\begin{aligned} \mathcal {L}_X S = 0 \qquad \forall X \in \Gamma ({\mathcal {E}},D). \end{aligned}$$

Furthermore, we require D to be involutive on the critical locus $\mathcal {EL}$, i.e. if $X,Y \in \Gamma ({\mathcal {E}},D)$, then $[X,Y]\big \vert _\mathcal {EL} \in \Gamma ({\mathcal {E}},D \vert _\mathcal {EL})$.

Whenever local symmetries are present, the space of interest is not $\mathcal {EL}$ but rather the space of inequivalent configurations $\mathcal {EL}/D$,⁹ i.e. the space of orbits of D on the critical locus $\mathcal {EL}$. Classical observables are then suitable functions over $\mathcal {EL}/D$, whose space we denote by $C^\infty (\mathcal {EL}/D)$. Note that, as a quotient, $\mathcal {EL}/D$ is often singular and defining $C^\infty (\mathcal {EL}/D)$ is a nontrivial task. One way of handling this is to build a resolution of $C^\infty (\mathcal {EL}/D)$, by means of the Koszul–Tate–Chevalley–Eilenberg complex, also known as the BV complex (see Definition 2.2.4).

We are interested in analysing to what extent two field theories are equivalent. Starting the discussion of equivalence in the setting of classical Lagrangian field theory, we consider the

Definition 2.1.6

Let $({\mathcal {E}}_{i}, S_i)$, $i\in \{1,2\}$, be two classical field theories with symmetry distributions $D_i$. We say that $({\mathcal {E}}_{i}, S_i)$ are classically equivalent if

$$\begin{aligned} \mathcal {EL}_1&\simeq \mathcal {EL}_2,\\ D_1\vert _{\mathcal {EL}_1}&\simeq D_2\vert _{\mathcal {EL}_2}. \end{aligned}$$

Remark 2.1.7

If two theories are classically equivalent, we have $\mathcal {EL}_1/D_1 \simeq \mathcal {EL}_2/D_2$. If we have a model for the respective spaces of classical observables, they are isomorphic:

$$\begin{aligned} C^\infty (\mathcal {EL}_1/D_1) \simeq C^\infty (\mathcal {EL}_2/D_2). \end{aligned}$$

This notion will be central in our discussion, and we will provide a refinement of it within the BV formalism, with and without boundary.

Remark 2.1.8

In certain cases, we can find $C_1\subset {\mathcal {E}}_1$, defined as the set of solutions of some of the equations of motion $\text {EL}_1 = 0$. Then, if we can find an isomorphism $\phi _{cl}: C_1 \rightarrow {\mathcal {E}}_2$ such that

$$\begin{aligned}&S_1\vert _{C_1} = \phi _{cl}^*S_2&\text {and}&D_1\vert _{C_1} {\mathop {\simeq }\limits ^{\phi _{cl}}} D_2, \end{aligned}$$

the theories are classically equivalent. This is a simple example of the situation in which two theories are classically equivalent because they differ only by auxiliary fields (see, e.g. [3]).

2.2 Batalin–Vilkovisky formalism

The BV formalism is a cohomological approach to field theory, that allows one to characterise the space of inequivalent field configurations by means of the cohomology of an appropriate cochain complex. It turns out that it also provides a natural notion of equivalence of field theories, that also takes into account “observables” of the theory.

In this setting, a classical field theory is described through the following data:

Definition 2.2.1

A BV theory is the assignment of a quadruple $\mathfrak {F} = (\mathcal {F},\omega ,\mathcal {S},Q)$ to a closed manifold M where

• ${\mathcal {F}= \Gamma (M,F)}$ is the space of smooth sections of a $\mathbb {Z}$-graded bundle¹⁰${F\rightarrow M}$ (the BV space of fields),

• $\omega \in \Omega ^2_\mathrm {{\int }}(\mathcal {F})$ is an integrated, local, symplectic form on $\mathcal {F}$ of degree $-1$ (the BV form),

• $\mathcal {S}\in \Omega ^0_\mathrm {{\int }}(\mathcal {F})$ is an integrated, local, functional on $\mathcal {F}$ of degree 0 (the BV action functional),

• $Q \in \mathfrak {X}_\mathrm {evo}(\mathcal {F})$ is a cohomological, evolutionary, vector field of degree 1, i.e. $[Q,Q] = 2Q^2 = 0$, and $[\mathcal {L}_Q,\mathrm {d}] = 0$,

such that

$$\begin{aligned} \iota _Q \omega = \delta \mathcal {S}. \end{aligned}$$

(2)

The internal degree of $\mathcal {F}$ is called the ghost number and will be denoted by $\text {gh}(\cdot )$.

Remark 2.2.2

In principle, we only need to consider either $\mathcal {S}$ or Q, as they are related to one another through Eq. (2), apart from the ambiguity of an additive constant in $\mathcal {S}$. We will nonetheless regard them as separate data for later convenience, as we will see that introducing a boundary spoils Eq. (2).

As Q is cohomological, its Lie derivative $\mathcal {L}_{Q}$ is a differential on $\Omega ^{\bullet }_{\text {loc}}(\mathcal {F})$, since $\mathrm {gh}(\mathcal {L}_Q) = 1$ and $2 \mathcal {L}_Q^2 = [\mathcal {L}_Q,\mathcal {L}_Q] = \mathcal {L}_{[Q,Q]} = 0$. In this context, $\mathcal {L}_Q$-cocycles are interpreted as (gauge-)invariant local forms.

Remark 2.2.3

It is easy to gather that both $\omega$ and $\mathcal {S}$ are $\mathcal {L}_Q$-cocycles by applying $\delta$ and $\iota _Q$ to Eq. (2), respectively. We have

$$\begin{aligned}&\mathcal {L}_Q \omega = 0, \end{aligned}$$

(3a)

$$\begin{aligned}&\mathcal {L}_Q \mathcal {S}= (\mathcal {S},\mathcal {S})= 0. \end{aligned}$$

(3b)

where $(\cdot ,\cdot )$ is the Poisson bracket induced by $\omega$. Eq. (3b) is known as the Classical Master Eqns [14, 58], and encodes the property that $\mathcal {S}$ is gauge invariant. In particular, Eq. (3) mean that we have the freedom to perform the transformations $\omega \mapsto \omega + \mathcal {L}_Q(\dots )$ and $\mathcal {S}\mapsto \mathcal {S}+ \mathcal {L}_Q(\dots )$, as long they preserve Eq. (2).

Definition 2.2.4

We define the BV complex of a given BV theory $\mathfrak {F}$ as the space of integrated local forms on $\mathcal {F}$ endowed with the differential $\mathcal {L}_Q$

$$\begin{aligned} \mathfrak {BV}^{\bullet } {:}{=}\left( \Omega ^\bullet _{{\int }}(\mathcal {F}),\mathcal {L}_Q \right) , \end{aligned}$$

where the grading on $\mathfrak {BV}^{\bullet }$ is given by the ghost number. Its cohomology will be denoted by $H^\bullet (\mathfrak {BV}^\bullet )$ and called the BV cohomology.

While the BV complex $\mathfrak {BV}^\bullet$ consists of inhomogeneous local forms (inhomogeneous also in ghost number), its 0-form part¹¹$\mathfrak {BV}^{\bullet }_0 \subset \mathfrak {BV}^\bullet$ is of interest as it is a resolution of D-invariant functionals on $\mathcal {EL}$ or, when the quotient is nonsingular, functionals on $\mathcal {EL}/D$ in the sense that the BV cohomology is given by¹² [33, 40, 62]

$$\begin{aligned} H^{-i}(\mathfrak {BV}^{\bullet }_0)&= 0\quad \quad \text {for } i>0,\nonumber \\ H^0(\mathfrak {BV}^{\bullet }_0)&\simeq {C^\infty (\mathcal {EL}/D)}. \end{aligned}$$

Example 2.2.5

(Lie algebra case [14], see also [40, 45]) In this paper, we will only consider examples which enjoy symmetries that come from a Lie algebra action. Let $({\mathcal {E}},S)$ be a classical field theory over a closed manifold M with a symmetry on ${\mathcal {E}}$ given by the action of a Lie algebra $(\mathfrak {g},[\cdot ,\cdot ])$. We can build a BV theory as follows: Choose the space of fields to be

$$\begin{aligned} \mathcal {F}= T^*[-1]({\mathcal {E}} \times \Omega ^0(M,\mathfrak {g})[1]), \end{aligned}$$

with local coordinates $\Phi ^i = (\varphi ^j,\xi ^a)$ on the base ${\mathcal {E}} \times \Omega ^0(M,\mathfrak {g})[1]$ and $\Phi ^\dagger _i = (\varphi ^\dagger _j,\xi ^\dagger _a)$ on the fibres. Usually, one calls $\varphi ^j$ the fields, $\xi ^a$ the ghosts¹³ and $\Phi ^\dagger _i$ the antifields. Note that the ghost numbers are related by $\mathrm {gh}(\Phi ^i) + \mathrm {gh}(\Phi ^\dagger _i) = - 1$ due to the -1 shift on the fibres. We take the BV form to be the canonical symplectic form on $\mathcal {F}$

$$\begin{aligned} \omega = \int _M \langle \delta \Phi ^\dagger , \delta \Phi \rangle , \end{aligned}$$

where $\langle \cdot ,\cdot \rangle$ is a bilinear map with values in $\Omega ^{\bullet ,\mathrm {top}}_\mathrm {loc}(M)$. In the case of a Lie algebra action, the cohomological vector field Q decomposes into the Chevalley–Eilenberg differential $\gamma$ and the Koszul–Tate differential $\delta _{KT}$

$$\begin{aligned} Q = \gamma + \delta _{KT}. \end{aligned}$$

The action of $\gamma$ is defined on the fields and ghosts as

$$\begin{aligned}&\gamma \varphi ^j = \xi ^a v^j_a,\quad \gamma \xi ^a = \frac{1}{2} [\xi ,\xi ]^a, \end{aligned}$$

where $v^i_a$ are the fundamental vector fields of $\mathfrak {g}$ on F. In turn, $\delta _{KT}$ acts as

$$\begin{aligned} \delta _{KT} \varphi ^i&= 0,\quad ~~~~~~ \delta _{KT} \xi ^a = 0,\nonumber \\ \delta _{KT} \varphi ^\dagger _i&= \frac{\delta S}{\delta \varphi ^i}, \quad \delta _{KT} \xi ^\dagger _a = v^i_a \varphi ^\dagger _i. \end{aligned}$$

(4)

The BV action functional can then be constructed as an extension of the classical action functional

$$\begin{aligned} \mathcal {S}[\Phi ,\Phi ^\dagger ] = S[\varphi ] + \int _M \langle \Phi ^\dagger , Q \Phi \rangle \end{aligned}$$

and $Q(\cdot ) = (\mathcal {S},\cdot )$ can be used to compute the full form of $Q\Phi ^\dagger _i$. The data $(\mathcal {F}, \omega , \mathcal {S}, Q)$ form a BV theory.

2.3 Equivalence in the BV setting

We now have all the necessary tools to develop a notion of equivalence in the BV formalism. We are interested in comparing the BV data and cohomology $H^\bullet (\mathfrak {BV}^\bullet _i)$ of two BV theories $\mathfrak {F}_i$, $i \in \{1,2\}$. We recall that a quasi-isomorphism is a chain map between chain complexes which induces an isomorphism in cohomology. In this spirit, we define:

Definition 2.3.1

Two BV theories $\mathfrak {F}_1$ and $\mathfrak {F}_2$ are BV-equivalent if there is a (degree-preserving) map $\phi :\mathcal {F}_2 \rightarrow \mathcal {F}_1$ that induces a quasi-isomorphism $\phi ^*:\mathfrak {BV}^\bullet _1 \rightarrow \mathfrak {BV}^\bullet _2$ of BV complexes, such that $\phi ^*$ preserves the cohomological classes of the BV form and BV action functional as

$$\begin{aligned}&\phi ^* [\omega _1] = [\omega _2], \quad \phi ^* [\mathcal {S}_1] = [\mathcal {S}_2]. \end{aligned}$$

(5)

A BV equivalence is called strong iff $\phi$ is a symplectomorphism that preserves the BV action functionals.

Remark 2.3.2

If $\mathfrak {F}_1$, $\mathfrak {F}_2$ are BV-equivalent, we can find a morphism $\psi : \mathcal {F}_1 \rightarrow \mathcal {F}_2$, such that its pullback map $\psi ^*$ is the quasi-inverse of $\phi ^*$. In particular, the composition maps

$$\begin{aligned}&\chi ^* = \psi ^* \circ \phi ^*: \mathfrak {BV}^\bullet _1 \rightarrow \mathfrak {BV}^\bullet _1,\quad \lambda ^* = \phi ^* \circ \psi ^*: \mathfrak {BV}^\bullet _2 \rightarrow \mathfrak {BV}^\bullet _2, \end{aligned}$$

are the identity in the respective BV cohomologies $H^\bullet (\mathfrak {BV}^\bullet _1)$, $H^\bullet (\mathfrak {BV}^\bullet _2)$. This is equivalent to the existence of two maps $h_\chi :\mathfrak {BV}^\bullet _{1} \rightarrow \mathfrak {BV}^\bullet _{1}$, $h_\lambda :\mathfrak {BV}^\bullet _{2} \rightarrow \mathfrak {BV}^\bullet _{2}$ of ghost number $-1$ such that [64]

$$\begin{aligned}&\chi ^* - \mathrm {id}_{1} = \mathcal {L}_{Q_1} h_\chi + h_\chi \mathcal {L}_{Q_1},\quad \lambda ^* - \mathrm {id}_2 = \mathcal {L}_{Q_2} h_\lambda + h_\lambda \mathcal {L}_{Q_2}. \end{aligned}$$

Furthermore, note that applying $\psi ^*$ to Eq. (5) yields

$$\begin{aligned}&\psi ^* [\omega _2] = [\omega _1], \quad \psi ^* [\mathcal {S}_2] = [\mathcal {S}_1]. \end{aligned}$$

(6)

Let us now explore some direct implications of Definition 2.3.1, in particular that the transformation of $\omega _i$ and $\mathcal {S}_i$ are not independent:

Proposition 2.3.3

Rewrite Eqs. (5) and (6) as

$$\begin{aligned} \phi ^* \omega _1&= \omega _2 + \mathcal {L}_{Q_2}\rho _2,\quad \psi ^* \omega _2 = \omega _1 + \mathcal {L}_{Q_1}\rho _1,\nonumber \\ \phi ^* \mathcal {S}_1&= \mathcal {S}_2 + \mathcal {L}_{Q_2}\sigma _2,\quad \psi ^* \mathcal {S}_2 = \mathcal {S}_1 + \mathcal {L}_{Q_1}\sigma _1, \end{aligned}$$

with $\rho _i \in \Omega ^2_{{\int }}(\mathcal {F}_i)$, $\sigma _i \in \Omega ^0_{{\int }}(\mathcal {F}_i)$. Then

$$\begin{aligned} \mathcal {L}_{Q_i} (\iota _{Q_i} \rho _i + \delta \sigma _i) = 0. \end{aligned}$$

(7)

Moreover, Eq. (7) is satisfied if

$$\begin{aligned}&\rho _i = - \delta \mu _i,\quad \sigma _i = \iota _{Q_i} \mu _i \end{aligned}$$

(8)

with $\mu _i \in \Omega ^1_{{\int }}(\mathcal {F}_i)$.

Proof

Applying $\phi ^*$ to $\iota _{Q_1} \omega _1 = \delta \mathcal {S}_1$ yields

$$\begin{aligned}&\iota _{Q_2} \omega _2 + \iota _{Q_2}\mathcal {L}_{Q_2}\rho _2 = \delta \mathcal {S}_2 + \delta \mathcal {L}_{Q_2}\sigma _2\\ \Rightarrow \quad&\mathcal {L}_{Q_2}(\iota _{Q_2}\rho _2 + \delta \sigma _2) = 0, \end{aligned}$$

and analogously $\mathcal {L}_{Q_1}(\iota _{Q_1}\rho _1 + \delta \sigma _1) = 0$.

The simplified condition (8) implies Eq. (7) since

$$\begin{aligned} \mathcal {L}_{Q_{i}}(\iota _{Q_{i}}\rho _{i} + \delta \sigma _{i}) = \mathcal {L}_{Q_{i}}(-\iota _{Q_{i}}\delta \mu _i + \delta \iota _{Q_{i}} \mu _i) = -\mathcal {L}^2_{Q_{i}}\mu _i = 0, \end{aligned}$$

where we used $\mathcal {L}_{Q_i} = [\iota _{Q_i},\delta ]$. $\square$

Remark 2.3.4

Let $\mathfrak {F}_1$ and $\mathfrak {F}_2$ be BV-equivalent theories as per Definition 2.3.1, and let $\chi ^*: \mathfrak {BV}^\bullet _1 \rightarrow \mathfrak {BV}^\bullet _1$ and $\lambda ^*: \mathfrak {BV}^\bullet _2 \rightarrow \mathfrak {BV}^\bullet _2$ be the chain maps defined in Remark 2.3.2. Then, the theories $\mathfrak {F}_1$ and $\chi ^*\mathfrak {F}_1 {:}{=}(\mathcal {F}_1, \chi ^*\omega _1, \chi ^*\mathcal {S}_1, Q_1)$ are clearly BV-equivalent, and so are $\mathfrak {F}_2$ and $\lambda ^*\mathfrak {F}_2$.

Remark 2.3.5

In the literature, there exists another notion of equivalence of BV theories, based on what is usually called elimination of (generalised) auxiliary fields or reduction of contractible pairs (see, e.g. [3, 41] and [5] for a review). When two theories differ by auxiliary fields content, they have the same BV cohomology. In Sect. 3.2, we show how the presence of auxiliary fields leads to the process of elimination of cohomologically contractible pairs by explicitly constructing chain maps that are quasi-inverse to one another and homotopic to the identity. Hence, theories that differ by auxiliary fields/contractible pairs are BV-equivalent in the sense of Definition 2.3.1.

2.4 Field theories on manifolds with higher strata

The BV formalism can be extended to the case where the underlying manifold M has a nonempty boundary $\partial M \ne \emptyset$, as presented in [19]. This construction relies on the BFV formalism introduced in [4]; see also [55, 56, 61].

Definition 2.4.1

An exact BFV theory over a manifold $\Sigma$ is a quadruple $\mathfrak {F}^\partial = (\mathcal {F}^\partial ,\omega ^\partial ,\mathcal {S}^\partial ,Q^\partial )$ where

• ${\mathcal {F}^\partial = \Gamma (\Sigma ,F^\partial )}$ is the space of smooth sections of a $\mathbb {Z}$-graded fibre bundle ${F^\partial \rightarrow M^\partial }$,

• $\omega ^\partial = \delta \alpha ^\partial \in \Omega ^2_{{\int }}(\mathcal {F}^\partial )$ is an exact, integrated, local, symplectic form on $\mathcal {F}^\partial$ of degree 0,

• $\mathcal {S}^\partial \in \Omega ^0_{{\int }}(\mathcal {F}^\partial )$ is a degree 1, integrated, local, functional on $\mathcal {F}^\partial$,

• $Q^\partial \in \mathfrak {X}_\mathrm {evo}(\mathcal {F}^\partial )$ is a degree 1, cohomological, evolutionary, vector field, i.e. $[Q^\partial ,Q^\partial ] = 0$, and $[\mathcal {L}_{Q^\partial },\mathrm {d}] = 0$

such that $Q^\partial$ is the Hamiltonian vector field of $\mathcal {S}^\partial$

$$\begin{aligned} \iota _{Q^\partial } \omega ^\partial = \delta \mathcal {S}^\partial . \end{aligned}$$

We call $\omega ^\partial ,S^\partial$ the boundary form and boundary action functional, respectively.

Definition 2.4.2

A BV–BFV theory over a manifold M with boundary $\partial M$ is given by the data

$$\begin{aligned} (\mathcal {F}, \omega , \mathcal {S}, Q, \mathcal {F}^\partial , \omega ^\partial , \mathcal {S}^\partial , Q^\partial , \pi ) \end{aligned}$$

where $(\mathcal {F}^\partial , \omega ^\partial , \mathcal {S}^\partial ,Q^\partial )$ is an exact BFV theory over $\Sigma =\partial M$ and $\pi : \mathcal {F}\rightarrow \mathcal {F}^\partial$ is a surjective submersion such that

$$\begin{aligned} \iota _Q \omega = \delta \mathcal {S}+ \pi ^* \alpha ^\partial \end{aligned}$$

(9)

and $Q \circ \pi ^* = \pi ^* \circ Q^\partial$.

Remark 2.4.3

Equation (9) implies that in general $\omega$ and $\mathcal {S}$ are no longer $\mathcal {L}_Q$-cocycles in the presence of a boundary. Instead, we have [19]

$$\begin{aligned} \mathcal {L}_{Q} \omega&= \pi ^* \omega ^\partial , \end{aligned}$$

(10a)

$$\begin{aligned} \mathcal {L}_{Q} \mathcal {S}&= \pi ^*(2 \mathcal {S}^\partial - \iota _{Q^\partial }\alpha ^\partial ). \end{aligned}$$

(10b)

Note that the failure of the structural BV forms to be $\mathcal {L}_Q$-cocycles is controlled by (boundary) BFV forms. In particular, Eq. (10b) means that $\mathcal {S}$ fails to be gauge invariant, and the right-hand side can be related to Noether’s generalised charges [53]. Furthermore, the CME no longer holds. Instead, we have the modified Classical Master Equation [19]

$$\begin{aligned} \frac{1}{2} \iota _Q \iota _Q \omega = \pi ^* \mathcal {S}^\partial . \end{aligned}$$

2.5 Inducing boundary BFV from bulk BV data

It is important to emphasise how one can try to construct a boundary theory $\mathfrak {F}^\partial$ from a BV theory $\mathfrak {F}$, since there might be obstructions. The problem we want to address is that of inducing an exact BFV theory on the boundary $\partial M$, starting from the BV data assigned to the bulk manifold M.

Define ${\check{\alpha }}$ as:

$$\begin{aligned} {\check{\alpha }} {:}{=}\iota _Q \omega - \delta \mathcal {S}. \end{aligned}$$

(11)

By restricting the fields of $\mathcal {F}$ (and their normal jets) to the boundary $\partial M$, we can define the space of pre-boundary fields ${\check{\mathcal {F}}}^\partial$ and endow it with a pre-boundary 2-form ${\check{\omega }} = \delta {\check{\alpha }}$. Usually ${\check{\omega }}$ turns out to be degenerate. In order to define a symplectic space of boundary fields, one then has to perform symplectic reduction, see, e.g. [60]. Let $\ker {\check{\omega }} = \{X \in \mathfrak {X}(\mathcal {{\check{F}}^\partial }) \left| \iota _X {\check{\omega }} = 0 \right. \}$ and set

$$\begin{aligned} \mathcal {F}^\partial {:}{=}{\check{\mathcal {F}}}^\partial /\ker {\check{\omega }}. \end{aligned}$$

(12)

Since we are taking a quotient, nothing guarantees that $\mathcal {F}^\partial$ is smooth, but we want to assume that this is the case. However, a necessary condition for smoothness is that $\ker {\check{\omega }}$ has locally constant dimension, i.e. it is a subbundle of $T{\check{\mathcal {F}}}^\partial$. As we will see, this condition is not always satisfied, namely that there is a unique symplectic form $\omega ^\partial$ such that $\pi ^*\omega ^\partial = {\check{\omega }}$, and a unique cohomological vector field $Q^\partial$ such that $Q \circ \pi ^* = \pi ^* \circ Q^\partial$. We assume (although this may not be true in general) that there is a 1-form $\alpha ^\partial$ such that $\pi ^*\alpha ^\partial ={\check{\alpha }}$. Note that, in this case, $\alpha ^\partial$ is unique and $\omega ^\partial =\delta \alpha ^\partial$. See [18] for details. However, for $\mathcal {F}^\partial$ smooth, we have the surjective submersion $\pi :\mathcal {F}\rightarrow \mathcal {F}^\partial$.

Consider now

Definition 2.5.1

The graded Euler vector field $E \in \mathfrak {X}_{\text {evo}}(\mathcal {F})$ is defined as the degree 0 vector field which acts on local forms of homogeneous ghost number as

$$\begin{aligned} \mathcal {L}_E F = \text {gh}(F) F. \end{aligned}$$

Similarly, we have $E^\partial = \pi _* E \in \mathfrak {X}_{\text {evo}}(\mathcal {F}^\partial )$ on the boundary.

The cohomological vector field $Q^\partial$ is actually Hamiltonian and the corresponding boundary action functional can be computed as [52]

$$\begin{aligned} \mathcal {S}^\partial = \iota _{E^\partial }\iota _{Q^\partial } \omega ^\partial . \end{aligned}$$

(13)

The data $\mathfrak {F}^\partial = (\mathcal {F}^\partial , \omega ^\partial , \mathcal {S}^\partial , Q^\partial )$ define an exact BFV manifold over the boundary $\partial M$. For completeness, we also define the pre-boundary action functional ${\check{S}} {:}{=}\pi ^*S^\partial$. Pulling back Eq. (13) via $\pi ^*$ yields

$$\begin{aligned} {\check{\mathcal {S}}} = \iota _{E}\iota _{Q} {\check{\omega }}. \end{aligned}$$

(14)

Note that by taking Eqs. (11) and (14), we ensure that the data $({\check{\alpha }}, {\check{S}})$ can always be defined, even if the quotient in Eq. (12) does not yield a smooth structure.

Remark 2.5.2

The procedure we just presented can be repeated in case that the manifold M not only has a boundary but also corners (higher strata), as presented in [19, 47]. If this is possible up to codimension n, then we call the theory a n-extended (exact) BV–BFV theory.

Remark 2.5.3

The quantisation programme introduced in [20] relies on the BV–BFV structure of a given classical theory. As such, even if two theories are classically equivalent, only one might turn out to have a BV–BFV structure and so be suitable for quantisation, as we now explore in the example of the Jacobi theory and 1D GR.

Remark 2.4.3 and Sect. 2.5 discuss two potential roadblocks for our construction of equivalence in the presence of boundaries and corners (and more generally codimension-k strata). First, to extend the notion of equivalence discussed in Sect. 2.3 to the case with higher strata, we wish to capture the possibility of local forms being $\mathcal {L}_Q$-cocycles up to boundary terms, as is the case with $\omega$ and $\mathcal {S}$. This is the problem of descent, where we enrich the differential $\mathcal {L}_Q$ by the de Rham differential on M. The second big problem one encounters is that not all BV theories satisfy the regularity requirement necessary to induce compatible BV–BFV data. In order to describe such theories as well, we will relax our definitions.

In order to do this, we turn to a “lax” version of the BV–BFV formalism [47]. We will work with local forms on $\mathcal {F} \times M$ with inhomogeneous form degree on M, namely $\kappa ^\bullet \in \Omega ^{p,\bullet }_\mathrm {loc}(\mathcal {F} \times M)$, and use the codimension to enumerate them, as it makes the notation less cumbersome and more intuitive, i.e. $\kappa ^{k}$ denotes the $(\mathrm {top} - k)$-form part of

$$\begin{aligned} \kappa ^\bullet = \sum ^{\dim M}_{k=0} \kappa ^{k},\end{aligned}$$

where $\kappa ^k \in \Omega ^{p,(\mathrm {top} - k)}_\mathrm {loc}(\mathcal {F} \times M)$.

Remark 2.5.4

What we call “lax” BV–BFV formalism is a rewriting of known approaches to local field theory in the BV/BRST formalism such as [2, 3, 9, 38, 59]. We use the term “lax” to contrast it with the “strict” version given by the BV–BFV formalism proper.

This should be compared to the standard BV–BFV formalism (extended to codimension k [19]), which instead looks at $\Omega ^{\bullet }_{\mathrm {loc}}(\mathcal {F}^{(k)})$ with $\mathcal {F}^{(k)}$ an appropriate space of codimension-k fields. In other words, we describe the BV–BFV picture presented above in terms of densities instead of integrals (cf. Remark 2.1.2), and forfeiting the symplectic structure at codimension k. This setting allows us to phrase equivalence with higher strata in a cohomological way, and it collects all the relevant data before performing the quotient in Eq. (12), thus temporarily avoiding potential complications.¹⁴

The definitions that we work with rely on the lax degree¹⁵$\#(\cdot )$, which describes the interplay between the co-form degree on M and the ghost number. Let $\text {fd}_M(\cdot )$ denote the form degree on M. The lax degree is defined as the difference of the ghost number $\text {gh}(\cdot )$ and the co-form degree $\text {cfd}_M(\cdot ) {:}{=}\dim M - \text {fd}_M(\cdot )$

$$\begin{aligned} \#(\cdot ) {:}{=}\text {gh}(\cdot ) - \text {cfd}_M(\cdot ). \end{aligned}$$

In particular, if an inhomoegeneous local form has vanishing lax degree, then the codimension of its homogeneous components corresponds to their respective ghost number. Most notably, this will be the case for the Lagrangian density. We will use the total degree for computations, which for elements in $\Omega ^{\bullet ,\bullet }_{\mathrm {loc}}({\mathcal {E}}\times M)$ is given by $|\cdot | = \text {gh}(\cdot ) + \text {fd}_M(\cdot ) + \text {fd}_{\mathcal {F}}(\cdot )$, where $\text {fd}_{\mathcal {F}}(\cdot )$ is the form degree on $\mathcal {F}$.

Definition 2.5.5

(Lax BV–BFV theory) A lax BV–BFV theory over a manifold M is a quadruple $\mathfrak {F}^\text {lax} = (\mathcal {F}^\mathrm {lax},\theta ^\bullet ,L^\bullet ,Q)$ where

• $\mathcal {F}^\mathrm {lax} = \Gamma (M,F)$ for some $\mathbb {Z}$-graded fibre bundle $F\rightarrow M$,

• $\theta ^\bullet \in \Omega ^{1,\bullet }_{\text {loc}}(\mathcal {F}^\mathrm {lax} \times M)$ is a local form with lax degree -1,

• $L^\bullet \in \Omega ^{0,\bullet }_{\text {loc}}(\mathcal {F}^\mathrm {lax} \times M)$ is a local functional with lax degree 0,

• $Q \in \mathfrak {X}_{\text {evo}}(\mathcal {F}^\mathrm {lax})$ is an evolutionary, cohomological vector field on $\mathcal {F}^\mathrm {lax}$ of degree 1, i.e. $[\mathcal {L}_Q,\mathrm {d}] = [Q,Q] = 0$,

such that

$$\begin{aligned} \iota _Q \varpi ^\bullet&= \delta L^\bullet + \mathrm {d}\theta ^\bullet , \end{aligned}$$

(15a)

$$\begin{aligned} \iota _Q \iota _Q \varpi ^\bullet&= 2\mathrm {d}L^\bullet , \end{aligned}$$

(15b)

where $\varpi ^\bullet {:}{=}\delta \theta ^\bullet$.

Remark 2.5.6

(Strictification of lax data)

Let $M^\circ$ be the interior (bulk) of M.

• (1) If $M = M^\circ$ is a closed manifold, then we can assign a BV theory $\mathfrak {F}$ to $M^\circ$ from a lax BV–BFV theory $\mathfrak {F}^\mathrm {lax}$ by choosing¹⁶

  $$\begin{aligned} \mathcal {F}&= \mathcal {F}^\mathrm {lax}\vert _{M^\circ },\quad \omega = \int _{M^\circ } \varpi ^0,\quad \mathcal {S}= \int _{M^\circ } L^0 \end{aligned}$$

  and restricting Q to $\mathcal {F}$.

• (2) Similarly, if M is a compact manifold with boundary, the pre-BFV data on $\partial M$ may be induced by setting

  $$\begin{aligned}&{\check{\mathcal {F}}}^\partial = \mathcal {F}^\mathrm {lax}\vert _{M^\partial },\quad {\check{\alpha }} = \int _{M^\partial } \theta ^1,\quad {\check{\mathcal {S}}} = \int _{M^\partial } L^1, \end{aligned}$$

  and restricting Q to $\check{\mathcal {F}}^\partial$. When pre-symplectic reduction w.r.t. ${\check{\omega }} = \delta {\check{\alpha }} = \int _{M^\partial } \varpi ^1$ is possible [19], we can define the space of boundary fields $\mathcal {F}^\partial :={\check{\mathcal {F}}}^\partial /\mathrm {ker}({\check{\omega }}^\sharp )$. Together with the bulk data presented above, this produces a BV–BFV theory.

The procedure is analogous for higher codimensions: If $M^{(k)}$ denotes the kth-codimension stratum, we can induce a Hamiltonian dg manifold of fields in codimension k by performing pre-symplectic reduction of

$$\begin{aligned} \left( \check{\mathcal {F}}^{(k)}=\mathcal {F}^{\mathrm {lax}}\vert _{M^{(k)}}, \check{\omega }^{(k)}=\int _{M^{(k)}}\delta \theta ^k\right) \leadsto \mathcal {F}^{(k)}:=\check{\mathcal {F}}^{(k)}/\mathrm {ker}(\check{\omega }^{(k)\sharp }) \end{aligned}$$

Notice that pre-symplectic reduction might fail to be smooth, resulting in an obstruction to strictification. When there are no obstructions to the pre-symplectic reduction, this procedure yields an n-extended BV–BFV theory (cf. Remark 2.5.2), and we have an n-strictification of a lax BV–BFV theory. (For more details we refer to [47].) It is crucial to observe that this step can fail [26, 27, 46].

Unlike the latter, a lax BV–BFV theory does not require working with symplectic structures at higher codimensions $\ge 1$. This means that lax data allow us to extract some information about the higher codimension behaviour of the field theory but, as we will see, the fact that a theory is strictifiable at a given codimension yields a refinement of the notion of BV equivalence.

Remark 2.5.7

At codimension $\ge 1$, it is sufficient to know $\theta ^\bullet$ in order to compute $L^\bullet$. Applying $\iota _E$ to Eq. (15a) yields

$$\begin{aligned} \iota _E \iota _Q \varpi ^\bullet = \mathcal {L}_E L^\bullet + \iota _E \mathrm {d}\theta ^\bullet , \end{aligned}$$

which implies

$$\begin{aligned} \mathcal {L}_E L^\bullet = \iota _E \left( \iota _Q \delta - \mathrm {d}\right) \theta ^\bullet . \end{aligned}$$

(16)

We can then compute $L^k$ at codimension $k \ge 1$ by using $\text {gh}(L^k) = \text {cfd}_M(L^k) = k$:

$$\begin{aligned} L^k = \frac{1}{k}\iota _E \left( \iota _Q \delta \theta ^k - \mathrm {d}\theta ^{k+1}\right) . \end{aligned}$$

Lemma 2.5.8

([19, 47]) The following equations hold for a lax BV–BFV theory:

$$\begin{aligned} \mathcal {L}_Q \varpi ^\bullet&= \mathrm {d}\varpi ^\bullet , \end{aligned}$$

(17a)

$$\begin{aligned} \mathcal {L}_Q L^\bullet&= \mathrm {d}(2L^\bullet - \iota _Q \theta ^\bullet ). \end{aligned}$$

(17b)

Remark 2.5.9

Equations (2.5.8) are the density versions of Eq. (10). Comparing the two versions, we see that boundary terms are now encoded as $\mathrm {d}$-exact terms, instead of objects in the image of $\pi ^*$. Note that $\varpi ^\bullet$ is a cocycle of the differential $(\mathcal {L}_Q - \mathrm {d})$ and that $L^\bullet$ is so whenever $L^\bullet - \iota _Q \theta ^\bullet = 0$.

In the lax BV–BFV formalism, the relevant differential will no longer be $\mathcal {L}_Q$, as we want to take the boundary configurations into account. Instead, we want to consider a cochain complex of local forms on $\mathcal {F}^\mathrm {lax} \times M$ with differential $\mathcal {L}_Q - \mathrm {d}$, which describes the interplay between gauge invariance and boundary terms:

Definition 2.5.10

([3, 47]) The BV–BFV complex of a lax BV–BFV theory $\mathfrak {F}^{\text {lax}}$ is defined as the space of inhomogeneous local forms on $\mathcal {F}^\mathrm {lax} \times M$ endowed with the differential $(\mathcal {L}_Q - \mathrm {d})$

$$\begin{aligned} (\mathfrak {BV}\text {-}\mathfrak {BFV})^\bullet&{:}{=}\left( \left( \bigoplus _k \Omega ^{\bullet ,k}_\text {loc}(\mathcal {F}^\mathrm {lax} \times M)\right) ,(\mathcal {L}_Q - \mathrm {d}) \right) , \end{aligned}$$

where the grading of $(\mathfrak {BV}\text {-}\mathfrak {BFV})^\bullet$ is given by the lax degree. We will denote its cohomology by $H^\bullet ((\mathfrak {BV}\text {-}\mathfrak {BFV})^\bullet )$ and call it the BV–BFV cohomology.

Remark 2.5.11

The cocycle conditions for an inhomogenoeous local form $\mathcal {O}^\bullet \in \Omega ^{p,\bullet }_\mathrm {loc}(\mathcal {F}^\mathrm {lax} \times M)$ are often called the descent equations [47, 48, 65, 66]

$$\begin{aligned} (\mathcal {L}_Q - \mathrm {d}) \mathcal {O}^\bullet = 0 \end{aligned}$$

i.e. $\mathcal {L}_Q \mathcal {O}^k = \mathrm {d}\mathcal {O}^{k+1}$ with homogeneous components $\mathcal {O}^k$. Such equations are of interest since their solutions produce classical observables, i.e. local functionals (i.e. $p=0$) which belong to $H^0(\mathfrak {BV}^\bullet )$. Let $\gamma ^k$ denote a $(\dim M - k)$-dimensional closed submanifold of M. We can then construct a classical observable by integrating $\mathcal {O}^k$ over $\gamma ^k$ since

$$\begin{aligned} \mathcal {L}_Q \int _{\gamma ^k} \mathcal {O}^k = \int _{\gamma ^k} \mathcal {L}_Q \mathcal {O}^k = \int _{\gamma ^k} \mathrm {d}\mathcal {O}^{k+1} = 0. \end{aligned}$$

As such, comparing the BV–BFV cohomologies of two lax theories offers a natural way of comparing their spaces of classical observables.

2.6 Equivalence in the lax setting

Before adapting our notion of equivalence to the case when a boundary and corners are present, let us define f-transformations, which encode the facts that eventually (i) we are interested in the 2-forms $\varpi ^\bullet$ (and not in their potentials $\theta ^\bullet$) and (ii) Lagrangian densities will be integrated (so total derivatives become irrelevant). The two issues are actually related.

Definition 2.6.1

Let $f \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}^{\mathrm {lax}} \times M)$ be a local functional with $\#(f) = -1$. An f-transformation of a lax BV–BFV theory $\mathfrak {F}^\mathrm {lax}$ changes $(\theta ^\bullet ,L^\bullet )$ as

$$\begin{aligned}&\theta ^\bullet \mapsto \theta ^\bullet + \delta f^\bullet ,\quad L^\bullet \mapsto L^\bullet + \mathrm {d}f^\bullet . \end{aligned}$$

Remark 2.6.2

Note that an f-transformation preserves Eq. (15) since $\varpi ^\bullet = \delta \theta ^\bullet$ and $\mathrm {d}L^\bullet$ are unchanged, as is the term $\delta L^\bullet + \mathrm {d}\theta ^\bullet$

$$\begin{aligned} \delta L^\bullet + \mathrm {d}\theta ^\bullet \mapsto \delta L^\bullet + \delta \mathrm {d}f + \mathrm {d}\theta ^\bullet +\mathrm {d}\delta f = \delta L^\bullet + \mathrm {d}\theta ^\bullet , \end{aligned}$$

where we used $[\delta ,\mathrm {d}]=0$. Hence, we will also allow this kind of freedom in our definition of equivalence.

In the following, we will denote the vertical differentials on $\mathcal {F}^\mathrm {lax}_i$ by $\delta$ and the horizontal (de Rham) differentials on $M_i$ by $\mathrm {d}$.

Definition 2.6.3

(Lax equivalence) We say that two lax theories $\mathfrak {F}^\mathrm {lax}_1$ and $\mathfrak {F}^\mathrm {lax}_2$ are lax-equivalent if there are two morphisms of graded manifolds $\phi :\mathcal {F}_2 \rightarrow \mathcal {F}_1$ and $\psi :\mathcal {F}_1 \rightarrow \mathcal {F}_2$, which induce quasi-isomorphisms $\phi ^*:\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2$, $\psi ^*:\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$ between the BV–BFV complexes, such that $\phi ^*$ and $\psi ^*$ are quasi-inverse to each other and transform $(\theta ^\bullet _i, L^\bullet _i)$ as

$$\begin{aligned} \phi ^* \theta ^\bullet _1&= \theta ^\bullet _2 + (\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2 + \delta f^\bullet _2,\quad \psi ^* \theta ^\bullet _2 = \theta ^\bullet _1 + (\mathcal {L}_{Q_1} - \mathrm {d}) \beta ^\bullet _1 + \delta f^\bullet _1 \nonumber ,\\ \phi ^* L^\bullet _1&= L^\bullet _2 + (\mathcal {L}_{Q_2} - \mathrm {d}) \zeta ^\bullet _2 + \mathrm {d} f^{\bullet }_2,\quad \psi ^* L^\bullet _2 = L^\bullet _1 + (\mathcal {L}_{Q_1 } - \mathrm {d}) \zeta ^\bullet _1 + \mathrm {d} f^{\bullet }_1, \end{aligned}$$

(18)

with $\beta ^\bullet _i \in \Omega ^{1,\bullet }_\mathrm {loc}(\mathcal {F}^\mathrm {lax}_i\times M_i)$, $\# (\beta ^\bullet _i) = -2$, $\zeta ^\bullet _i \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}^\mathrm {lax}_i \times M_i)$, $\# (\zeta ^\bullet _i) = -1$ and $f^\bullet _i \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}^\mathrm {lax}_i\times M_i)$, $\# (f^\bullet _i) = -1$.

Remark 2.6.4

Similarly to the bulk case, in order to show that the composition maps

$$\begin{aligned} \chi ^*&= \psi ^* \circ \phi ^*: \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1,\\ \lambda ^*&= \phi ^* \circ \psi ^*: \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2, \end{aligned}$$

are the identity when restricted to the respective cohomologies, one needs to find two maps $h_\chi :\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{1} \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{1}$, $h_\lambda :\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{2} \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{2}$ of lax degree $-1$ such that

$$\begin{aligned} \chi ^* - \mathrm {id}_{1}&= (\mathcal {L}_{Q_1} - \mathrm {d}) h_\chi + h_\chi (\mathcal {L}_{Q_1} - \mathrm {d}),\\ \lambda ^* - \mathrm {id}_2&= (\mathcal {L}_{Q_2} - \mathrm {d}) h_\lambda + h_\lambda (\mathcal {L}_{Q_2} - \mathrm {d}). \end{aligned}$$

Proposition 2.6.5

If $\text {gh}(\phi ) = \text {gh}(\psi ) = 0$,¹⁷ then the transformation of $L^\bullet _i$ is not independent from the transformation of $\theta ^\bullet _i$:

• (1) $\zeta ^k_i = \iota _{Q_i} \beta ^k_i$ at codimension $k \ge 1$,

• (2) $\mathcal {L}_{Q_i} (\iota _Q \beta ^0_i - \zeta ^0_i) = 0$.

Proof

• (1) As $\text {gh}(\phi ) = 0$, $\phi ^*$ commutes with the Euler vector fields $\mathcal {L}_{E_i}$: $\mathcal {L}_{E_2} \phi ^* = \phi ^* \mathcal {L}_{E_1}$. Applying $\phi ^*$ to Eq. (16) then yields

  $$\begin{aligned} \mathcal {L}_{E_2} \phi ^* L^\bullet _1&= \iota _{E_2} \left( \iota _{Q_2} \delta - \mathrm {d}\right) \phi ^* \theta ^\bullet _1\\&= \iota _{E_2} \left( \iota _{Q_2} \delta - \mathrm {d}\right) \theta ^\bullet _2 + \iota _{E_2} \left( \iota _{Q_2} \delta - \mathrm {d}\right) (\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2 + \iota _{E_2} \left( \iota _{Q_2} \delta - \mathrm {d}\right) \delta f^\bullet _2. \end{aligned}$$

  The first term is simply $\mathcal {L}_{E_2} L^\bullet _2$. For the second term, we compute

  $$\begin{aligned} \iota _{E_2} \left( \iota _{Q_2} \delta - \mathrm {d}\right) (\mathcal {L}_{Q} - \mathrm {d}) \beta ^\bullet _2&= \iota _{E_2} \left( \delta \iota _{Q_2} + \mathcal {L}_{Q_2} - \mathrm {d}\right) (\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2\\&= \iota _{E_2} \delta \iota _{Q_2} (\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2 = (\mathcal {L}_{E_2} - \delta \iota _{E_2}) \iota _{Q_2} (\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2\\&= \mathcal {L}_{E_2} (\mathcal {L}_{Q_2} - \mathrm {d}) \iota _{Q_2} \beta ^\bullet _2, \end{aligned}$$

  where we used that $(\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2 \in \Omega ^{1,\bullet }(\mathcal {F}^\mathrm {lax}\times M)$ implies $\iota _{E_2} \iota _{Q_2} (\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2 = 0$. The third term reads

  $$\begin{aligned} \iota _{E_2} \left( \iota _{Q_2} \delta - \mathrm {d}\right) \delta f^\bullet _2 = - \iota _{E_2} \mathrm {d}\delta f^\bullet _2 = \mathrm {d}\iota _{E_2} \delta f^\bullet _2 = \mathrm {d}\mathcal {L}_{E_2} f^\bullet _2 = \mathcal {L}_{E_2} \mathrm {d}f^\bullet _2, \end{aligned}$$

  hence

  $$\begin{aligned} \mathcal {L}_{E_2} \phi ^* L^\bullet _1 = \mathcal {L}_{E_2} (L^\bullet _2 + (\mathcal {L}_{Q_2} - \mathrm {d}) \iota _{Q_2} \beta ^\bullet _2 + \mathrm {d}f^\bullet _2). \end{aligned}$$

  By counting degrees, we see that both sides have ghost number k at codimension k, and therefore for $k\ge 1$, one can use this equation to determine $\phi ^*L^k_1$, in particular we have $\zeta ^k_i = \iota _{Q_i} \beta ^k_i$.

• (2) Applying $\phi ^*$ to $\iota _{Q_1} \varpi ^\bullet _1 = \delta L^\bullet _1 + \mathrm {d}\theta ^\bullet _1$ yields

  $$\begin{aligned}&\iota _{Q_2} \delta (\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2 = \delta (\mathcal {L}_{Q_2} - \mathrm {d}) \zeta ^\bullet _2 + \mathrm {d}(\mathcal {L}_{Q_2} - \mathrm {d}) \beta ^\bullet _2\nonumber \\ \Rightarrow \quad&(\mathcal {L}_{Q_2} - \mathrm {d}) \iota _{Q_2} \delta \beta ^\bullet _2 = (\mathcal {L}_{Q_2} - \mathrm {d}) \delta \zeta ^\bullet _2 + (\mathcal {L}_{Q_2} - \mathrm {d})\mathrm {d}\beta ^\bullet _2\nonumber \\ \Rightarrow \quad&(\mathcal {L}_{Q_2} - \mathrm {d}) [(\iota _{Q_2} \delta -\mathrm {d})\beta _2^\bullet - \delta \zeta ^\bullet _2] = 0, \end{aligned}$$

  (19)

  where we used $\iota _{Q_2} \varpi ^\bullet _2 = \delta L^\bullet _2 + \mathrm {d}\theta ^\bullet _2$, $\delta ^2 = 0$ and the fact that f-transformations preserve Eq. (15a). Note that this condition holds automatically for condimention higher than zero due to $\zeta ^k_2 = \iota _{Q_2} \beta ^k_2$. To see what Eq. (19) implies at codimention zero, first note that

  $$\begin{aligned} \iota _{Q_2} \delta \beta _2^1 - \delta \zeta ^1_2 = \iota _{Q_2} \delta \beta _2^1 - \delta \iota _{Q_2} \beta _2^1 = \mathcal {L}_{Q_2} \beta _2^1. \end{aligned}$$

  Keeping in mind that $[\mathcal {L}_Q,\mathrm {d}] = 0$, Eq. (19) gives

  $$\begin{aligned}&\mathcal {L}_{Q_2}[ \iota _{Q_2} \delta \beta _2^0 - \mathrm {d}\beta _2^1 - \delta \zeta ^0_2 ] - \mathrm {d}[ \iota _{Q_2} \delta \beta _2^1 - \delta \zeta ^1_2 ] = 0\\ \Rightarrow \quad&\mathcal {L}_{Q_2} \delta (\iota _{Q_i} \beta ^0_2 - \zeta _2^0) = 0. \end{aligned}$$

  We now apply $\iota _{E_2}$, but first note that

  $$\begin{aligned} {[}\mathcal {L}_Q,\iota _{E}] = \iota _{[E,Q]} = \iota _{\mathcal {L}_E Q} = \iota _Q, \end{aligned}$$

  therefore

  $$\begin{aligned} \iota _{E_2}\mathcal {L}_{Q_2} \delta (\iota _{Q_i} \beta ^0_2 - \zeta _2^0)&= [\iota _Q - \mathcal {L}_Q \iota _E] \delta (\iota _{Q_i} \beta ^0_2 - \zeta _2^0)\\&= 2 \mathcal {L}_{Q_2} (\iota _{Q_i} \beta ^0_2 - \zeta _2^0)\\ \Rightarrow \quad \mathcal {L}_{Q_2} (\iota _{Q_i} \beta ^0_2 - \zeta _2^0)&= 0 \end{aligned}$$

  where we used $\iota _Q \delta (\iota _{Q_i} \beta ^0_2 - \zeta _2^0) = \mathcal {L}_Q (\iota _{Q_i} \beta ^0_2 - \zeta _2^0)$ and $\text {gh}(\iota _{Q_i} \beta ^0_2 - \zeta _2^0) = \#(\iota _{Q_i} \beta ^0_2 - \zeta _2^0) = -1$. The computations are analogous for $i = 1$.

$\square$

Remark 2.6.6

The previous lemma means that there is a redundancy in our definition of lax equivalence, as the transformation of $L^\bullet _i$ at codimension $\ge 1$ can be determined through the transformation of $\theta ^\bullet _i$. In particular, when computing explicit examples one only needs to check whether we have the right transformation for $\theta ^\bullet _i$ and $L^0_i$. Observe that if we have $H^{-1}(\mathcal {L}_Q) = 0$ we can conclude that $\iota _{Q_i} \beta ^0_2 - \zeta _2^0 = \mathcal {L}_Q(\dots )$.

Remark 2.6.7

Our definition of lax equivalence directly implies that the local 2-forms $\varpi ^\bullet _i$ are interchanged up to $(\mathcal {L}_{Q_i} - \mathrm {d})$-exact and $\delta$-exact terms

$$\begin{aligned} \phi ^* \varpi ^\bullet _1&= \delta \phi ^* \theta ^\bullet _1 = \varpi ^\bullet _2 - (\mathcal {L}_{Q_2} - \mathrm {d}) \delta \beta ^\bullet _2,\\ \psi ^* \varpi ^\bullet _2&= \delta \psi ^* \theta ^\bullet _2 = \varpi ^\bullet _1 - (\mathcal {L}_{Q_1} - \mathrm {d}) \delta \beta ^\bullet _1. \end{aligned}$$

Similar to the bulk case (cf. Proposition 2.3.3), choosing $\zeta ^0_i = \iota _{Q_i} \beta _i^0$ ensures that the second condition from Proposition 2.6.5 is satisfied.

Proposition 2.6.8

The theories $\mathfrak {F}^\mathrm {lax}_1$, $\chi ^*\mathfrak {F}^\mathrm {lax}_1 {:}{=}(\mathcal {F}_1^\mathrm {lax}, \chi ^*\theta ^\bullet _1, \chi ^*L^\bullet _1, Q_1)$ and $\mathfrak {F}^\mathrm {lax}_2$, $\lambda ^*\mathfrak {F}^\mathrm {lax}_2 {:}{=}(\mathcal {F}_2^\mathrm {lax}, \lambda ^*\theta ^\bullet _2, \lambda ^*L^\bullet _2, Q_2)$ are pairwise lax-equivalent.

Proof

The proof is analogous to the proof of Proposition 2.3.4. $\square$

Theorem 2.6.9

Let $\mathfrak {F}^\mathrm {lax}_i$, $i \in \{1,2\}$, be lax equivalent. Then the respective BV theories $\mathfrak {F}_i$ (cf. Remark 2.5.6) are BV-equivalent.

Proof

Let $\kappa ^\bullet \in \Omega ^{p,\bullet }_\mathrm {loc}({\mathcal {E}} \times M)$ and $K {:}{=}\int _{M} \kappa ^0 \in \Omega ^{p}_{{\int }}({\mathcal {E}})$. We need to check if:

• (1) $\phi ^*,\psi ^*$ are chain maps w.r.t. the BV complexes,

• (2) the cohomological classes of $\omega _i$ and $\mathcal {S}_i$ are mapped into one another,

• (3) $\chi ^*, \lambda ^*$ are the identity on $H^\bullet (\mathfrak {BV}^\bullet _i)$.

To prove these, we simply need to integrate the various conditions over the bulk M. For the chain map condition, we have

$$\begin{aligned} \phi ^* \circ (\mathcal {L}_{Q_1}-\mathrm {d}) \kappa ^\bullet&= (\mathcal {L}_{Q_2}-\mathrm {d}) \circ \phi ^* \kappa ^\bullet \\ \Rightarrow \int _{M} \phi ^* (\mathcal {L}_{Q_1} \kappa ^0 -\mathrm {d}\kappa ^1)&= \int _{M} (\mathcal {L}_{Q_2} \phi ^* \kappa ^ 0 -\mathrm {d}\phi ^* \kappa ^1)\\ \Rightarrow \quad \phi ^* \mathcal {L}_{Q_1} K&= \mathcal {L}_{Q_2} \phi ^* K. \end{aligned}$$

We can compute the transformations of $\omega _1$ and $\mathcal {S}_1$ in a similar way

$$\begin{aligned} \phi ^* \varpi ^\bullet _1&= \omega ^\bullet _2 - (\mathcal {L}_{Q_2} - \mathrm {d}) \delta \beta ^\bullet _2, \phi ^* L^\bullet _1 = L^\bullet _2 + (\mathcal {L}_{Q_2} - \mathrm {d}) \zeta ^\bullet _2 + \mathrm {d} f^{\bullet }_2,\\ \Rightarrow \phi ^* \omega _1&= \omega _2 - \mathcal {L}_{Q_2} \int _{M} \delta \beta ^0_2, \phi ^* \mathcal {S}_1 = \mathcal {S}_2 + \mathcal {L}_{Q_2} \int _{M} \zeta ^0_2. \end{aligned}$$

In the same manner

$$\begin{aligned} (\chi ^* - \mathrm {id}_{1}) \kappa ^\bullet&= (\mathcal {L}_{Q_1} - \mathrm {d}) h_\chi \kappa ^\bullet + h_\chi (\mathcal {L}_{Q_1} - \mathrm {d}) \kappa ^\bullet ,\\ \Rightarrow (\chi ^* - \mathrm {id}_{1}) K&= \mathcal {L}_{Q_1} h_\chi K + h_\chi \mathcal {L}_{Q_1} K, \end{aligned}$$

implying that $\chi ^*$, and analogously $\lambda ^*$, are also homotopic to the identity in $\mathfrak {BV}^\bullet _i$ and as such the identity when restricted to $H^\bullet (\mathfrak {BV}^\bullet _i)$, meaning that the BV complexes are quasi-isomorphic. $\square$

3 Examples

This section is dedicated to the explicit computation of lax BV–BFV equivalence in three different examples. We start by presenting the general strategy in Sect. 3.1. We then look at the examples of classical mechanics on a curved background and (nonabelian) Yang–Mills theory in Sects. 3.3 and 3.4 respectively. Subsequently we turn our attention to the classically equivalent Jacobi theory and one-dimensional gravity coupled to matter (1D GR) in Sect. 3.5, and show that they are lax BV–BFV equivalent, despite their different boundary behaviours w.r.t. the BV–BFV procedure. Furthermore, we show that the chain maps used to prove lax BV–BFV equivalence spoil the compatibility with the regularity condition for the BV–BFV procedure in the case of 1D GR.

3.1 Strategy

We shortly demonstrate our strategy to show explicitly that two theories $\mathfrak {F}^\mathrm {lax}_i$ are lax BV–BFV equivalent. In practice, we need two maps $\phi ^*$, $\psi ^*$ between the BV–BFV complexes $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _i$

which (cf. Definition 2.6.3):

• (1) are chain maps,

• (2) transform $\theta ^\bullet _i, L^\bullet _i$ in the desired way (cf. Eq. (18)),

• (3) are quasi-inverse to one another.

In order to check the first property, we note that, as pullback maps, $\phi ^*$ and $\psi ^*$ automatically commute with the de Rham differentials $\delta$ and $\mathrm {d}$. Thus it suffices to show that $\phi ^*, \psi ^*$ are chain maps w.r.t. $Q_i$ on the fields $\varphi ^j_i \in \mathcal {F}^\mathrm {lax}_i$

$$\begin{aligned} \phi ^* {Q_1}\varphi ^j_1&= Q_2 \phi ^* \varphi ^j_1, \quad \psi ^* Q_2 \varphi ^j_2 = {Q_1}\psi ^* \varphi ^j_2, \end{aligned}$$

as this together with the fact that they commute with $\delta$ then implies that they are chain maps w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ on $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet$, resulting in the following Lemma:

Lemma 3.1.1

If the pullback maps $\phi ^*, \psi ^*$ are chain maps w.r.t. $Q_i$ on $\mathcal {F}_i$, then they are also chain maps w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ on $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _i$.

Showing the second property is a matter of computation. For the third property, we shortly present our strategy to show that $\chi ^* = \psi ^* \circ \phi ^*$ is the identity in cohomology. The same procedure can then be applied to $\lambda ^* = \phi ^* \circ \psi ^*$. Recall that we need a map $h_\chi :\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{1} \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{1}$ of lax degree $-1$ such that

$$\begin{aligned}&\chi ^* - \mathrm {id}_{1} = (\mathcal {L}_{{Q_1}} - \mathrm {d}) h_\chi + h_\chi (\mathcal {L}_{{Q_1}} - \mathrm {d}). \end{aligned}$$

(20)

We start by constructing a homotopy between $\chi ^*$ and $\mathrm {id}_1$, by finding an evolutionary vector field ${R_1}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_1)$ with $\text {gh}({R_1}) = \#({R_1}) = -1$ and defining a one-parameter family of morphisms of the form

$$\begin{aligned} \chi ^*_s {:}{=}e^{s[(\mathcal {L}_{{Q_1}} - \mathrm {d}), \mathcal {L}_{{R_1}}]}, \end{aligned}$$

such that $\chi ^*_{s=0} = \mathrm {id}_1$ and $\lim _{s\rightarrow \infty }\chi ^*_{s} = \chi ^*$. Note that choosing $s = -\ln \tau$ gives the usual definition of homotopy with $\tau \in [0,1]$, i.e. a continuous map $F(\tau ) {:}{=}\chi ^*_{(-\ln \tau )}$ satisfying $F(0) = \chi ^*$ and $F(1) = \mathrm {id}_1$. We will nonetheless work with the parameter s to keep the calculations cleaner, changing when necessary. Furthermore, we can simplify the term in the exponent by setting $D_1{:}{=}[{Q_1},{R_1}]$, since

$$\begin{aligned}{}[(\mathcal {L}_{{Q_1}} - \mathrm {d}), \mathcal {L}_{{R_1}}] = [\mathcal {L}_{{Q_1}}, \mathcal {L}_{{R_1}}] = \mathcal {L}_{[{Q_1},{R_1}]} = \mathcal {L}_{D_1}, \end{aligned}$$

which results in

$$\begin{aligned} \chi ^*_s = e^{s\mathcal {L}_{D_1}}. \end{aligned}$$

Lemma 3.1.2

The Lie derivative $\mathcal {L}_{D_1}$ commutes with the differential $(\mathcal {L}_{{Q_1}} - \mathrm {d})$.

Proof

Since ${R_1}$ and ${Q_1}$ are evolutionary vector fields, we directly have $[\mathcal {L}_{D_1},\mathrm {d}] = 0$. To see that $\mathcal {L}_{D_1}$ commutes with $\mathcal {L}_{{Q_1}}$ we compute

$$\begin{aligned}&{[}D_1,{Q_1}] = [[{Q_1},{R_1}], {Q_1}] = - [[{R_1},{Q_1}], {Q_1}] -[[{Q_1},{Q_1}], {R_1}]= - [D_1,{Q_1}]\\ \Rightarrow&[D_1, {Q_1}] = 0. \end{aligned}$$

where we have used the graded Jacobi identity and $[{Q_1},{Q_1}] = 0$. Thus

$$\begin{aligned} {[}\mathcal {L}_{D_1}, \mathcal {L}_{{Q_1}}]&= \mathcal {L}_{[D_1, {Q_1}]} = 0, \end{aligned}$$

proving the statement. $\square$

We can then determine the map $h_\chi$ by rewriting the RHS of Eq. (20) as

$$\begin{aligned} \chi ^* -\mathrm {id}_1&= \int ^\infty _0 \frac{\mathrm {d}}{ds}e^{s\mathcal {L}_{D_1}} = \int ^\infty _0 e^{s\mathcal {L}_{D_1}} \mathcal {L}_{D_1} \mathrm {d}s\\&= \int ^\infty _0 e^{s\mathcal {L}_{D_1}} [(\mathcal {L}_{{Q_1}}-\mathrm {d}) \mathcal {L}_{{R_1}} + \mathcal {L}_{{R_1}} (\mathcal {L}_{{Q_1}}-\mathrm {d})] \mathrm {d}s\\&= (\mathcal {L}_{{Q_1}}-\mathrm {d}) \left( \int ^\infty _0 e^{s\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} \mathrm {d}s \right) + \left( \int ^\infty _0 e^{s\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} \mathrm {d}s \right) (\mathcal {L}_{{Q_1}}-\mathrm {d}), \end{aligned}$$

resulting in the following Lemma:

Lemma 3.1.3

The map $h_\chi :\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{1} \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _{1}$ defined through

$$\begin{aligned} h_\chi \kappa = \int ^\infty _0 e^{s\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} \kappa \, \mathrm {d}s \end{aligned}$$

satisfies Eq. (20).

If $h_\chi$ converges on $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$, then $\chi ^*$ will be the identity in the BV–BFV cohomology $H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1)$ as desired. For this last step, the next Lemma will be useful:

Lemma 3.1.4

If $h_\chi$ converges on $\mathcal {F}_1$, then it converges on the whole BV–BFV complex $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$.

Proof

Let $\kappa \in \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$. Start by redefining $s = - \ln \tau$ with $\tau \in [0,1]$, such that we integrate over a compact interval instead of over $\mathbb {R}_{\ge 0}$. Performing this transformation results in

$$\begin{aligned} h_\chi \kappa = \int ^\infty _0 e^{s\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} \kappa \,\mathrm {d}s = \int ^0_1 e^{-\ln (\tau )\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} \kappa \, \mathrm {d}(-\ln \tau ) = \int ^1_0 \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } \mathcal {L}_{{R_1}} \kappa \,\mathrm {d}\tau . \end{aligned}$$

Assuming that

$$\begin{aligned} h_\chi \varphi ^j_1 = \int ^\infty _0 e^{s \mathcal {L}_{D_1}} {R_1}\varphi ^j_1 \, \mathrm {d}s = \int ^1_0 \frac{e^{-\ln (\tau ) \mathcal {L}_{D_1}}}{\tau } {R_1}\varphi ^j_1 \, \mathrm {d}\tau < \infty \quad \forall \varphi ^j_1 \in \mathcal {F}_1, \end{aligned}$$

we can show that $h_\chi \kappa$ converges on $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$. Let first $\kappa = f$ be a local functional. Writing ${R_1}= {R_1}\varphi ^j_1 \frac{\delta }{\delta \varphi ^j_1}$ then gives

$$\begin{aligned} h_\chi f&= \int ^1_0 \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } {R_1}f \,\mathrm {d}\tau = \int ^1_0 \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } \left( {R_1}\varphi ^j_1\frac{\delta f}{\delta \varphi ^j_1}\right) \mathrm {d}\tau \\&= \int ^1_0 \left( \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } {R_1}\varphi ^j_1\right) \left( e^{-\ln (\tau )\mathcal {L}_{D_1}} \frac{\delta f}{\delta \varphi ^j_1}\right) \mathrm {d}\tau , \end{aligned}$$

where we used that $\chi ^*_s = e^{s\mathcal {L}_{D_1}} = e^{-\ln (\tau ) \mathcal {L}_{D_1}}$ is a morphism in the last equality. The integral over the first integrand is finite by assumption and the second integrand $e^{-\ln (\tau ) \mathcal {L}_{D_1}} \frac{\delta f}{\delta \varphi ^j_1} = e^{s \mathcal {L}_{D_1}} \frac{\delta f}{\delta \varphi ^j_1}$ is nowhere divergent $\forall \tau \in [0,1]$, since we assume $\chi ^*_s = e^{s\mathcal {L}_{D_1}}$ to be well-defined on $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$.

Consider now the local form $\kappa = f \delta \varphi _J \otimes \nu \in \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$, where J is a multiindex raging over the fields and their jets, and $\nu \in \Omega ^\bullet (M)$ is a form on M. We then have

$$\begin{aligned} h_\chi \kappa =&\int ^1_0 \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } \mathcal {L}_{{R_1}} [f \delta \varphi _J \otimes \nu ] \mathrm {d}\tau \\ =&\int ^1_0 \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } [\mathcal {L}_{{R_1}} f \delta \varphi _J \pm f \mathcal {L}_{{R_1}}\delta \varphi _J] \otimes \nu \mathrm {d}\tau \\ =&\int ^1_0 \bigg [\left\{ \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } {R_1}f \right\} \delta \left( e^{-\ln (\tau )\mathcal {L}_{D_1}} \varphi _J\right) \\&\mp e^{-\ln (\tau )\mathcal {L}_{D_1}} f \delta \left\{ \frac{e^{-\ln (\tau )\mathcal {L}_{D_1}}}{\tau } {R_1}\varphi _J\right\} \bigg ] \otimes \nu \mathrm {d}\tau . \end{aligned}$$

The terms in the brackets $\{\cdot \}$ are just the integrands of $h_\chi f$ and $h_\chi \varphi _J$, which converge. Since the other terms $e^{-\ln (\tau )\mathcal {L}_{D_1}} \varphi _J = e^{s\mathcal {L}_{D_1}} \varphi _J$, $e^{-\ln (\tau )\mathcal {L}_{D_1}} f= e^{s\mathcal {L}_{D_1}} f$ are well-defined $\forall \tau \in [0,1]$ and we are integrating over a compact interval, the integral converges and $h_\chi$ is well-defined on $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$. $\square$

3.2 Contractible pairs

The simplest example we can discuss is when the cohomologically trivial fields are nicely decoupled from the rest. This follows the procedure of [41], which we explicitly embed in our framework by constructing suitable chain maps to fit Definition 2.3.1 (see also [3] and [5, 6, 38]). Namely, we have an action of the form

$$\begin{aligned} S_1[{\tilde{a}},v,{\tilde{a}}^\dagger ,v^\dagger ]= S_2 [{\tilde{a}},{\tilde{a}}^\dagger ] + \frac{1}{2} (v,v), \end{aligned}$$

where (, ) is some constant nondegenerate bilinear form on the space V of the v fields and $S_2$ is a solution of the master equation (w.r.t. the fields $\tilde{a},\tilde{a}^\dagger$). We want to compare this theory with the one defined by $S_2 [a,a^\dagger ]$, where we remove the tilde for clarity of the notation.

The cohomological vector field $Q_1$ of $S_1$ acts on the fields $\tilde{a},\tilde{a}^\dagger$ as $Q_1$. In addition we have

$$\begin{aligned} Q_1 v^\dagger = v,\qquad Q_1 v = 0. \end{aligned}$$

where we hve identified $V^*$ with V using the bilinear fom. The fields $(v,v^\dagger )$ are called a contractible pair.

Define maps $\phi$, $\psi$.

$$\begin{aligned} \phi ^*a = \tilde{a}, \quad \phi ^*a^\dagger&=\tilde{a}^\dagger , \quad \phi ^*v=0, \quad \phi ^*v^\dagger =0, \end{aligned}$$

(21a)

$$\begin{aligned} \psi ^*\tilde{a}&= a,\quad \psi ^*\tilde{a}^\dagger = a^\dagger . \end{aligned}$$

(21b)

Lemma 3.2.1

The composition map $\lambda ^*=\phi ^*\circ \psi ^*$ is the identity, while the composition map $\chi ^*=\psi ^*\circ \phi ^*$ acts as

$$\begin{aligned} \chi ^*a=a,\quad \chi ^*a^\dagger =a^\dagger ,\quad \chi ^*v=0,\quad \chi ^*v^\dagger =0, \end{aligned}$$

and is homotopic to the identity.

Proof

One can directly check that $\lambda ^*$ is the identity. In order to show that $\chi ^*$ is the identity in cohomology, we define a family of maps $\chi ^*_s = e^{s\mathcal {L}_{D_1}}$, where $D_1= [{Q_1},{R_1}]$, and show $\lim _{s\rightarrow \infty } \chi ^*_s = \chi ^*$.

We choose ${R_1}$ to act as

$$\begin{aligned} {R_1}a=0,\quad {R_1}a^\dagger =0,\quad {R_1}v=-v^\dagger ,\quad {R_1}v^\dagger = 0. \end{aligned}$$

We can then compute

$$\begin{aligned} D_1 v = (Q_1 R_1 + R_1 Q_1) v = Q_1 R_1 v = - v \qquad \Longleftrightarrow \qquad D_1^{k} v = (-1)^k v, \quad \forall k\ge 1, \end{aligned}$$

and

$$\begin{aligned} D_1 v^\dag = R_1 v = -v^\dag , \qquad \Longleftrightarrow \qquad D_1^{k} v^\dag = -v^\dag , \end{aligned}$$

which yield

$$\begin{aligned} \chi _s^* v = e^{s\mathcal {L}_{D_1}} v = v+ \sum _{k=1}^\infty \frac{s^k}{k!} D_1^{k} v= e^{-s}v {\mathop {\longrightarrow }\limits ^{s\rightarrow \infty }} \chi _\infty ^* v = 0, \end{aligned}$$

and similarly $\chi _{\infty }^*v^\dag = 0$.

On the other hand,

$$\begin{aligned} D_1 a = D_1 a^\dag = 0 \Longrightarrow e^{s\mathcal {L}_{D_1}}a = a, \quad e^{s\mathcal {L}_{D_1}}a^\dag = a^\dag , \quad \forall s, \end{aligned}$$

so that $\lim _{s\rightarrow \infty }\chi _s^* \equiv \chi ^*$.

Furthermore, the map $h_\chi$ converges on all the fields, as $h_\chi a = h_\chi a^\dagger = h_\chi v^\dagger$ trivially and

$$\begin{aligned} h_\chi v = \int ^\infty _0 e^{s\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} v \, \mathrm {d}s = - \int ^\infty _0 e^{s\mathcal {L}_{D_1}} v^\dagger \, \mathrm {d}s = - \int ^\infty _0 e^{-s} v^\dagger \, \mathrm {d}s = - v^\dagger . \end{aligned}$$

$\square$

A direct consequence of this Lemma, together with the facts that $\phi ^*S_1=S_2$ and $\phi ^* \omega _1= \omega _2$ for the canonical BV forms, is

Theorem 3.2.2

The BV theories defined by $S_1$ and $S_2$ are BV-equivalent.

3.2.1 More general contractible pairs

As in [41], we may consider a situation where $S_1$ depends on fields $\tilde{a},w,\tilde{a}^\dagger ,\tilde{w}^\dagger$, where $(w,w^\dagger )$ are not a contractible pair on the nose but satisfy the condition that

$$\begin{aligned} (Q_1 w^\dagger )|_{w^\dagger =0} = 0 \end{aligned}$$

has a unique solution $w=w(\tilde{a},\tilde{a}^\dagger )$. We can then get closer to the previously discussed case by defining¹⁸

$$\begin{aligned} v = Q_1 w^\dagger ,\quad v^\dagger = w^\dagger . \end{aligned}$$

We have indeed that $Q_1 v^\dagger = v$ and $Q_1 v = 0$. Moreover, the above condition implies that the change of variables $(w,w^\dagger )\mapsto (v,v^\dagger )$ is invertible (near $w^\dagger =0$, or everywhere if $w^\dagger$ is odd) and that the submanifold defined by the constraints $v=0$ and $v^\dagger =0$ is symplectic.

The above strategy then works, in the absence of boundary, with some modifications. Namely, the fact that now v and $v^\dagger$ are not Darboux coordinates requires modifying the map of Eq. (21). In turn, the transformation ${R_1}$ will get a nontrivial action on the fields $(a,a^\dagger )$.

All the examples we discuss below belong to this class of contractible pairs. As we will see, the modifications required in the case of classical mechanics and Yang–Mills theory are minimal, whereas those required in the case of 1D parametrisation invariant theories are more consistent—due to the fact that pairing of the v fields will depend on the a fields. In addition, in all the examples we show how to extend this construction to lax theories in order to encompass the presence of boundaries.

3.3 Classical mechanics on a curved background

We start by discussing the example of classical mechanics on a curved background as a warm-up exercise. We take the source manifold to be a time interval $I = [a,b] \subset \mathbb {R}$ and some smooth Riemannian manifold $(M,g)$ as target. We will denote time derivatives with a dot and use tildes to distinguish fields between the different formulations of the theory.

We can formulate the theory by considering a “matter” field $\tilde{q}\in F_2 {:}{=}C^\infty (I,M)$, and the metric tensor on the target will depend on the map $\tilde{q}$. We introduce the shorthand notation $\tilde{g}{:}{=}\tilde{g}(\tilde{q})$. The classical action functional is given by

$$\begin{aligned} S_2[\tilde{q}] = \int _I \frac{1}{2} \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) \mathrm {d} t. \end{aligned}$$

This is usually called the second-order formulation.

On the other hand, we can phrase the theory in its first-order formulation, by considering again a map $q:I \rightarrow M$, together with an “auxiliary” field¹⁹$p\in C^\infty (I,q^*T^*M)$ and the classical action functional

$$\begin{aligned} S_1[q, p] = \int _I \left( \langle p, \dot{q}\rangle - \frac{1}{2} h(p,p)\right) \mathrm {d} t, \end{aligned}$$

where $h{:}{=}g^{-1}$ denotes the inverse of the target metric.

For ease of notation, we will introduce the musical isomorphisms

$$\begin{aligned}&g^\flat :TM \rightarrow T^*M, \quad h^\sharp :T^*M \rightarrow TM \\&g^\flat (v)(\cdot ) = g(v,\cdot ) \quad h^\sharp (\alpha )(\cdot ) = g^{-1}(\alpha ,\cdot ), \end{aligned}$$

and clearly $g^\flat \circ h^\sharp = h^\sharp \circ g^\flat = \mathrm {id}$.

We recall the rather obvious and well-known

Proposition 3.3.1

The first-order and second-order formulations of classical mechanics with a background metric are classically equivalent.

Proof

Following Definition 2.1.6, we start by solving the EL equation of the first-order theory corresponding to the auxiliary field $p$, we have

$$\begin{aligned} \delta _{p} {S_1}[q,p] = \int _I \langle \left( \dot{q}- h^\sharp (p)\right) ,\delta p\rangle \, \mathrm {d} t, \end{aligned}$$

which results in

$$\begin{aligned} p= g^\flat (\dot{q}). \end{aligned}$$

Let $C_1$ be the set of such solutions and define the map $\phi _{cl}: C_1 \rightarrow F_2$ through $\phi ^*_{cl} \tilde{q}= q$. Then the restriction of ${S_1}[q,p]$ to $C_1$ coincides with the pullback of ${S_2}$ via $\phi ^*_{cl}$

$$\begin{aligned} {S_1}[q,p]\big \vert _{C_1}&= {S_1}[q,p= g^\flat (\dot{q})] = \int _I \frac{1}{2} \langle g^\flat (\dot{q}), \dot{q}\rangle \mathrm {d} t= \phi ^*_{cl}\int _I \frac{1}{2} \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) \mathrm {d} t= \phi ^*_{cl} {S_2}[\tilde{q}], \end{aligned}$$

hence the two formulations are classical equivalent. $\square$

Both of these theories can be extended to the lax BV–BFV formalism. Note that, as there are no gauge symmetries in these models, there is no need to introduce ghost fields. We start with the second-order formulation:

Proposition/Definition 3.3.2

The data

$$\begin{aligned} \mathfrak {F}^\mathrm {lax}_{2CM} = (\mathcal {F}_{2CM}^{{\mathrm {lax}}},\theta _{2}^\bullet ,L_{2}^\bullet ,{Q_2}) \end{aligned}$$

where

$$\begin{aligned} \mathcal {F}_{2CM}^{{\mathrm {lax}}}= T^*[-1]C^\infty (I,M). \end{aligned}$$

together with $\theta _{2}^\bullet \in \Omega ^{1,\bullet }_\mathrm {loc}(\mathcal {F}_{2CM}^{{\mathrm {lax}}})$ and $L_{2}^\bullet \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}_{2CM}^{{\mathrm {lax}}})$, which are given by

$$\begin{aligned} \theta _{2}^\bullet&= \langle \tilde{q}^\dagger , \delta \tilde{q}\rangle \mathrm {d} t+ \langle \tilde{g}^\flat (\dot{\tilde{q}}), \delta \tilde{q}\rangle ,\\ L_{2}^\bullet&= \frac{1}{2} \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) \mathrm {d} t, \end{aligned}$$

and the cohomological vector field ${Q_2}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_{2CM}^{{\mathrm {lax}}})$

$$\begin{aligned}&{Q_2}\tilde{q}= 0,\quad {Q_2}\tilde{q}^\dagger = +\frac{1}{2}\partial \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) - \frac{d}{dt}(\tilde{g}^\flat )(\dot{\tilde{q}}) - \tilde{g}^\flat (\ddot{\tilde{q}}), \end{aligned}$$

where²⁰$\partial \tilde{g}{:}{=}\frac{\delta \tilde{g}}{\delta \tilde{q}}$, defines a lax BV–BFV theory.

Proof

We need to check Eq. (15) at codimension 0 and 1, namely

$$\begin{aligned} \iota _{Q_2}\varpi _{2}^0&= \delta L_{2}^0 + \mathrm {d}\theta _{2}^1, \quad \iota _{Q_2}\iota _{Q_2}\varpi _{2}^0 = 2\mathrm {d}L_{2}^1,\\ \iota _{Q_2}\varpi _{2}^1&= \delta L_{2}^1,\quad \quad \quad \quad \quad \iota _{Q_2}\iota _{Q_2}\varpi _{2}^1 = 0. \end{aligned}$$

Note that the Lagrangian only has a top-form term $L^\bullet _1 = L^0_1$. The only nontrivial equation is $\iota _Q \varpi _{2}^0 = \delta L_{2}^0 + \mathrm {d}\theta _{2}^1$, since $L_{2}^1 = \iota _{Q_2}\iota _{Q_2}\varpi _{2}^0 = \iota _{Q_2}\varpi _{2}^1 = 0$. We compute

$$\begin{aligned} \iota _{Q_2}\varpi _{2}^0&= \langle {Q_2}\tilde{q}^\dagger , \delta \tilde{q}\rangle \mathrm {d} t= \left\langle \left( \frac{1}{2}\partial \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) - \left( \frac{d}{dt} \tilde{g}^\flat \right) (\dot{\tilde{q}}) - \tilde{g}^\flat (\ddot{\tilde{q}})\right) ,\delta \tilde{q}\right\rangle \mathrm {d} t\\ \delta L_{2}^0&= \frac{1}{2} \delta \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) \mathrm {d} t+ \tilde{g}(\delta \dot{\tilde{q}},\dot{\tilde{q}}) \mathrm {d} t\\&= \frac{1}{2} \langle \partial \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) , \delta \tilde{q}\rangle \mathrm {d} t+ \frac{\mathrm {d}}{\mathrm {d} t}(\tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}})) \mathrm {d} t-\frac{\mathrm {d}}{\mathrm {d} t}(\tilde{g})(\delta \tilde{q},\dot{\tilde{q}}) \mathrm {d} t- \tilde{g}(\delta \tilde{q},\ddot{\tilde{q}})\mathrm {d} t\\&= \frac{1}{2} \langle \partial \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) , \delta \tilde{q}\rangle \mathrm {d} t- \mathrm {d}\langle \tilde{g}^\flat (\dot{\tilde{q}}),\delta \tilde{q}\rangle - \left\langle \frac{d}{dt} \tilde{g}^\flat (\dot{\tilde{q}}),\delta \tilde{q}\right\rangle \mathrm {d} t- \langle \tilde{g}^\flat (\ddot{\tilde{q}})),\delta \tilde{q}\rangle \mathrm {d} t\\&=- \mathrm {d}\langle \tilde{g}^\flat (\dot{\tilde{q}}),\delta \tilde{q}\rangle + \left\langle \frac{1}{2} \partial \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) - \frac{d}{dt} \tilde{g}^\flat (\dot{\tilde{q}}) -\tilde{g}^\flat (\ddot{\tilde{q}})) , \delta \tilde{q}\right\rangle \mathrm {d} t\\&= - \mathrm {d}\theta _{2}^1 + \iota _{Q_2}\varpi _{2}^0, \end{aligned}$$

where we used $|\mathrm {d} t| = -1$ and $|\delta \tilde{q}| = -1$. $\square$

In the case of the first-order theory, we have the lax BV–BFV theory:

Proposition/Definition 3.3.3

The data

$$\begin{aligned} \mathfrak {F}^\mathrm {lax}_{1CM} = (\mathcal {F}_{1CM}^{{\mathrm {lax}}},\theta _1^\bullet ,L_{1}^\bullet ,{Q_1}) \end{aligned}$$

where

$$\begin{aligned} \mathcal {F}_{1CM}^{{\mathrm {lax}}}= T^*[-1](C^\infty (I,M) \times C^\infty (I,M) ), \end{aligned}$$

together with $\theta _1^\bullet \in \Omega ^{1,\bullet }_\mathrm {loc}(\mathcal {F}_{1CM}^{{\mathrm {lax}}})$ and $L_{1}^\bullet \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}_{1CM}^{{\mathrm {lax}}})$, which take the forms

$$\begin{aligned} \theta _1^\bullet&= (\langle q^\dagger , \delta q\rangle + \langle p^\dagger , \delta p\rangle ) \mathrm {d} t+ \langle p, \delta q\rangle ,\\ L_{1}^\bullet&= \left( \langle p, \dot{q}\rangle - \frac{1}{2} h(p,p)\right) \mathrm {d} t, \end{aligned}$$

and the cohomological vector field ${Q_1}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_{1CM}^{{\mathrm {lax}}})$

$$\begin{aligned}&{Q_1}q= {Q_1}p= 0,\quad {Q_1}q^\dagger = - \dot{p}- \frac{1}{2} \partial h(p,p),\quad {Q_1}p^\dagger = \dot{q}- h^\sharp (p), \end{aligned}$$

with $\partial h{:}{=}\delta h/ \delta q$, defines a lax BV–BFV theory.

Proof

Again we need to check Eq. (15) at codimension 0 and 1. Explicitly we have

$$\begin{aligned} \iota _{Q_1}\varpi _1^0&= \delta L_{1}^0 + \mathrm {d}\theta _1^1,\quad \iota _{Q_1}\iota _{Q_1}\varpi _1^0 = 2\mathrm {d}L_{1}^1,\\ \iota _{Q_1}\varpi _1^1&= \delta L_{1}^1,\quad \quad \quad \quad \quad \iota _{Q_1}\iota _{Q_1}\varpi _1^1 = 0. \end{aligned}$$

As in the second-order theory, the Lagrangian only has a top-form component. The only nontrivial equation is $\iota _Q \varpi _1^0 = \delta L_{1}^0 + \mathrm {d}\theta _1^1$, since $L_{1}^1 = \iota _{Q_1}\iota _{Q_1}\varpi _1^0 = \iota _{Q_1}\varpi _1^1 = 0$. We compute

$$\begin{aligned} \iota _{Q_1}\varpi _1^0&= (\langle {Q_1}q^\dagger , \delta q\rangle + \langle {Q_1}p^\dagger , \delta p\rangle ) \mathrm {d} t\\&= -\left\langle \dot{p}+ \frac{1}{2} \partial h(p,p), \delta q\right\rangle \mathrm {d} t, + \langle (\dot{q}- h^\sharp (p)), \delta p\rangle \mathrm {d} t,\\ \delta L_{1}^0&= \left( \langle p,\delta \dot{q}\rangle - \frac{1}{2} \langle \partial h(p,p), \delta q\rangle \right) \mathrm {d} t+ \langle \dot{q}- h^\sharp (p), \delta p\rangle \mathrm {d} t\\&= \frac{\mathrm {d}}{\mathrm {d} t}\langle p, \delta q\rangle \mathrm {d} t-\langle \dot{p}+ \frac{1}{2} \partial h( p,p),\delta q\rangle \mathrm {d} t+ \langle \dot{q}- h^\sharp (p), \delta p\rangle \mathrm {d} t\\&= - \mathrm {d}\theta _1^1 + \iota _{Q_1}\varpi _1^0. \end{aligned}$$

$\square$

We now present the main theorem of this section, together with an outline of its proof.²¹ The computational details and the various lemmata are presented afterwards.

Theorem 3.3.4

The lax BV–BFV theories $\mathfrak {F}^\mathrm {lax}_{1CM}$ and $\mathfrak {F}^\mathrm {lax}_{1CM}$ of the first-order and second-order formulations of classical mechanics with a background metric are lax BV–BFV equivalent.

Proof

We need to check all the conditions from Definition 2.6.3. The existence of two maps $\phi$, $\psi$ with the desired properties is presented in Lemmata 3.3.5 and 3.3.6 respectively, where we also show that the pullback maps $\phi ^*$, $\psi ^*$ are chain maps w.r.t. the BV–BFV complexes $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _i$, and that they map $(\theta ^\bullet _i,L^\bullet _i)$ in the desired way.

Furthermore, we need to show that the respective BV–BFV complexes are quasi-isormophic. The composition map $\lambda ^* = \phi ^* \circ \psi ^*$ is shown to be the identity in Lemma 3.3.7. In Lemma 3.3.8, we prove that the composition map $\chi ^* = \psi ^* \circ \phi ^*$, is homotopic to the identity by following the strategy presented in Sect. 3.1.

In Lemma 3.3.9, we demonstrate that $\chi ^*$ is the identity in cohomology by showing that the map

$$\begin{aligned} h_\chi \varphi ^j_1 = \int _0^\infty e^{s\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} \varphi ^j_1 ds \end{aligned}$$

satisfying $\chi ^* - \mathrm {id}_{1} = (\mathcal {L}_{{Q_1}} - \mathrm {d}) h_\chi + h_\chi (\mathcal {L}_{{Q_1}} - \mathrm {d})$ (cf. Lemma 3.1.3) converges, therefore proving that the two lax BV–BFV theories in question have isomorphic BV–BFV cohomologies

$$\begin{aligned} H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}_{1CM}) \simeq H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}_{2CM}) \end{aligned}$$

and thus that they are lax BV–BFV equivalent. $\square$

Let us now look at the computations in detail. We start with the chain maps:

Lemma 3.3.5

Let $\phi : \mathcal {F}_{2CM}^{{\mathrm {lax}}}\rightarrow \mathcal {F}_{1CM}^{{\mathrm {lax}}}$ be the map defined through

$$\begin{aligned} \phi ^* q&= \tilde{q}, \quad \phi ^* p= {\phi ^* p= \tilde{g}^\flat (\dot{\tilde{q}}),}\\ \phi ^* q^\dagger&= \tilde{q}^\dagger , \quad \phi ^* p^\dagger = 0. \end{aligned}$$

Its pullback map $\phi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ and maps the lax BV–BFV data of the first-order theory as

$$\begin{aligned}&\phi ^* \theta _1^\bullet = \theta _{2}^\bullet ,\quad \phi ^* L_{1}^\bullet = L_{2}^\bullet . \end{aligned}$$

Proof

To check that $\phi$ is a chain map, we compute

$$\begin{aligned} \phi ^* {Q_1}q&= \phi ^*(0) = 0 = {Q_2}\tilde{q}= {Q_2}\phi ^* q,\\ \phi ^* {Q_1}p&= \phi ^*(0) = 0 = {Q_2}(\tilde{g}^\flat (\dot{\tilde{q}})) = {Q_2}\phi ^* p,\\ \phi ^* {Q_1}q^\dagger&= \phi ^*\left( - \dot{p}- \frac{1}{2} \partial h(p,p)\right) = - \frac{\mathrm {d}}{\mathrm {d} t}(\tilde{g}^\flat (\dot{\tilde{q}})) - \frac{1}{2} \partial \tilde{h}(\tilde{g}^\flat (\dot{\tilde{q}}),\tilde{g}^\flat (\dot{\tilde{q}}))\\&= - \frac{d}{dt}(\tilde{g}^\flat )(\dot{\tilde{q}}) - \tilde{g}^\flat (\ddot{\tilde{q}}) + \frac{1}{2} \partial \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) = {Q_2}\tilde{q}^\dagger = {Q_2}\phi ^* q^\dagger ,\\ \phi ^* {Q_1}p^\dagger&= \phi ^*(\dot{q}- h^\sharp (p)) = 0 = {Q_2}\phi ^* p^\dagger \end{aligned}$$

Together with Proposition 3.1.1, this shows that $\phi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i} - \mathrm {d})$. In turn, $\phi ^*$ acts on $\theta _1^\bullet ,$ $L_{1}^\bullet$ as

$$\begin{aligned} \phi ^* \theta _1^\bullet&= \phi ^*\left( (\langle q^\dagger , \delta q\rangle + \langle p^\dagger , \delta p\rangle ) \mathrm {d} t+ \langle p, \delta q\rangle \right) = \langle \tilde{q}^\dagger , \delta \tilde{q}\rangle \mathrm {d} t+ \langle \tilde{g}^\flat (\dot{\tilde{q}}), \delta \tilde{q}\rangle = \theta _{2}^\bullet ,\\ \phi ^*L_{1}^\bullet&= \phi ^* \left( \langle p, \dot{q}\rangle - \frac{1}{2} h(p,p)\right) \mathrm {d} t= \left( \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) -\frac{1}{2} \tilde{h}(\tilde{g}^\flat (\dot{\tilde{q}}),\tilde{g}^\flat (\dot{\tilde{q}}))\right) \mathrm {d} t\\&= \frac{1}{2} \tilde{g}( \dot{\tilde{q}},\dot{\tilde{q}}) \mathrm {d} t= L_{2}^\bullet . \end{aligned}$$

$\square$

Lemma 3.3.6

Let $\psi : \mathcal {F}_{1CM}^{{\mathrm {lax}}}\rightarrow \mathcal {F}_{2CM}^{{\mathrm {lax}}}$ be the map defined through

$$\begin{aligned}&\psi ^* \tilde{q}= q,\quad \psi ^* \tilde{q}^\dagger { = q^\dagger - \frac{1}{2} \partial g(p^\dagger , {Q_1}p^\dagger ) -\frac{d}{dt}(g^\flat (p^\dag )) + \partial g(\dot{q},p^\dagger )} \end{aligned}$$

Its pullback map $\psi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ and maps the lax BV–BFV data of the second-order theory as

$$\begin{aligned}&\psi ^* \theta _{2}^\bullet = \theta _1^\bullet + (\mathcal {L}_{{Q_1}} - \mathrm {d}) \beta _1^\bullet + \delta f_1^\bullet ,\quad \psi ^* L_{2}^\bullet = L_{1}^\bullet + (\mathcal {L}_{{Q_1}} - \mathrm {d}) \iota _{{Q_1}} \beta _1^\bullet + \mathrm {d}f_1^\bullet . \end{aligned}$$

where

$$\begin{aligned} \beta _1^\bullet&= \frac{1}{2} g(p^\dagger , \delta p^\dagger ) \mathrm {d} t+ g(p^\dagger , \delta q),\quad f_1^\bullet = - \frac{1}{2} g(p^\dagger , {Q_1}p^\dagger ) \mathrm {d} t. \end{aligned}$$

Proof

The only nontrivial calculation that is needed to check whether $\psi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ is ${Q_1}\psi ^* \tilde{q}^\dagger = \psi ^* {Q_2}\tilde{q}^\dagger$. We compute

$$\begin{aligned} {Q_1}\psi ^* \tilde{q}^\dagger&= {Q_1}q^\dagger - \frac{1}{2} \partial g' ({Q_1}p^\dagger ,{Q_1}p^\dagger ) - g^\flat ({Q_1}\dot{p}^\dagger ) - (\dot{g}^{\flat })({Q_1}p^\dag ) + \partial g(\dot{q},{Q_1},p^\dag )\\&= -\dot{p}- \frac{1}{2} \partial h(p,p) - \frac{1}{2} g(\dot{q},\dot{q}) - \frac{1}{2} \partial g(h^\sharp (p),h^\sharp (p)) \\&\quad +\partial g(\dot{q},h^\sharp (p)) - g^\flat \left( \ddot{q}- \dot{h}^\sharp (p) - h^\sharp (\dot{p})\right) \\&\quad - \dot{g}^\flat \left( \dot{q}- h^\sharp (p)\right) + \partial g(\dot{q},\dot{q}) - \partial g(\dot{q},h^\sharp (p))\\&= -g^\flat (\ddot{q}) - \dot{g}^\flat (\dot{q}) + \frac{1}{2} \partial g(\dot{q},\dot{q}) = \psi ^* {Q_2}\tilde{q}^\dagger . \end{aligned}$$

where we used $\dot{g}^\flat (h^\sharp (p)) = - g^\flat (\dot{h}^\sharp (p))$, in virtue of the fact that $g^\flat \circ h^\sharp = \mathrm {id}$.

Let $\Delta \theta _{2}^\bullet {:}{=}\psi ^* \theta _{2}^\bullet - \theta _1^\bullet$ and $\Delta L_{2}^\bullet {:}{=}\psi ^* L_{2}^\bullet - L_{1}^\bullet$. We need to check whether

$$\begin{aligned} \Delta \theta _{2}^0&= \mathcal {L}_{{Q_1}} \beta _1^0 - \mathrm {d}\beta _1^1 + \delta f_1^0,\\ \Delta \theta _{2}^1&= \mathcal {L}_{{Q_1}} \beta _1^1 + \delta f_1^0,\\ \Delta L_{2}^0&= \mathcal {L}_{{Q_1}} \iota _{{Q_1}} \beta _1^0 - \mathrm {d}\iota _{{Q_1}} \beta _1^1 + \mathrm {d}f_1^1. \end{aligned}$$

Recall that we only need to compute $\Delta L_{2}^0$, as $\Delta L_{2}^k$ for $k>0$ are determined through $\Delta \theta _{2}^\bullet$ (cf. Proposition 2.6.5, Remark 2.6.6). Furthermore, note that $f_1^1 = 0$.

Computation of $\Delta \theta _{2}^0$: For the first equation, we compute

$$\begin{aligned} \Delta \theta _{2}^0&= \psi ^*(\langle \tilde{q}^\dagger , \delta \tilde{q}\rangle \mathrm {d} t) - (\langle q^\dagger , \delta q\rangle + \langle p^\dagger , \delta p\rangle ) \mathrm {d} t\\&= \left[ \left\langle q^\dagger - \frac{1}{2} \partial g(p^\dagger ,{Q_1}p^\dagger ) - g^\flat (\dot{p}^\dagger ) -\frac{d}{dt}(g^\flat )(p^\dag ) + \partial g(\dot{q},p^\dagger ) - q^\dagger , \delta q\right\rangle \right. \\&\left. - \langle p^\dagger , \delta p\rangle \right] \mathrm {d} t\\&= \left[ \left\langle - \frac{1}{2} \partial g(p^\dagger ,{Q_1}p^\dagger ) - g^\flat (\dot{p}^\dagger ) -\frac{d}{dt}(g^\flat )(p^\dag ) + \partial g(\dot{q},p^\dagger ), \delta q\right\rangle - \langle p^\dagger , \delta p\rangle \right] \mathrm {d} t\\ \mathcal {L}_{{Q_1}} \beta _1^0&= \mathcal {L}_{{Q_1}}\left( \frac{1}{2} g( p^\dagger , \delta p^\dagger ) \mathrm {d} t\right) =\left[ \frac{1}{2} g({Q_1}p^\dagger , \delta p^\dagger ) + \frac{1}{2} g( p^\dagger , \delta {Q_1}p^\dagger ) \right] \mathrm {d} t, \\ \mathrm {d}\beta _1^1&= \mathrm {d} t\frac{\mathrm {d}}{\mathrm {d} t}\left( g(p^\dagger ,\delta q) \right) = \left[ \dot{g}( p^\dagger , \delta q) + g(\dot{p}^\dagger ,\delta q) + g(p^\dagger , \delta \dot{q}) \right] \mathrm {d} t,\\ \delta f_1^0&= \delta \left( - \frac{1}{2} g(p^\dagger , {Q_1}p^\dagger ) \mathrm {d} t\right) \\&= \left[ + \frac{1}{2} \langle \partial g( p^\dagger ,{Q_1}p^\dagger ) , \delta q\rangle - \frac{1}{2} g(\delta p^\dagger , {Q_1}p^\dagger ) + \frac{1}{2} g(p^\dagger ,\delta {Q_1}p^\dagger ) \right] \mathrm {d} t. \end{aligned}$$

Using

$$\begin{aligned} \delta {Q_1}q^\dagger&= \delta (\dot{q}- h^\sharp (p)) = \delta \dot{q}- \langle \partial h^\sharp (p),\delta q\rangle -h^\sharp (\delta p),\\ g(\cdot ,\langle \partial h^\sharp (\cdot ),\delta q\rangle )&= - \langle \partial g(\cdot , h^\sharp (\cdot )),\delta q\rangle , \end{aligned}$$

the last three terms together yield

Computation of $\Delta \theta _{2}^1$: In this case, we have

$$\begin{aligned} \Delta \theta _{2}^1&= \psi ^*(\langle \tilde{g}^\flat (\dot{\tilde{q}}), \delta \tilde{q}\rangle ) - \langle p,\delta q\rangle = \langle g^\flat (\dot{q}- h^\sharp (p)), \delta q\rangle = g({Q_1}p^\dagger ,\delta q),\\ \mathcal {L}_{{Q_1}} \beta _1^1&= \mathcal {L}_{{Q_1}}\left( g( p^\dagger , \delta q)\right) = g( {Q_1}p^\dagger , \delta q),\\ \delta f_1^1&= 0, \end{aligned}$$

thus showing $\Delta \theta _{2}^1 = \mathcal {L}_{{Q_1}} \beta _1^1 + \delta f_1^1$.

Computation of $\Delta L_{2}^0$: For the third equality, we compute

$$\begin{aligned} \Delta L_{2}^0&= \psi ^*\left( \frac{1}{2} \tilde{g}(\dot{\tilde{q}},\dot{\tilde{q}}) \mathrm {d} t\right) - \left( \langle p, \dot{q}\rangle - \frac{1}{2}h(p,p) \right) \mathrm {d} t\\&= \left[ \frac{1}{2} g(\dot{q},\dot{q}) - \langle p, \dot{q}\rangle + \frac{1}{2}h(p,p) \right] \mathrm {d} t,\\ \mathcal {L}_{{Q_1}} \iota _{{Q_1}} \beta _1^0&= \mathcal {L}_{{Q_1}} \iota _{{Q_1}} \left( \frac{1}{2} g(p^\dagger , \delta p^\dagger ) \mathrm {d} t\right) \\&= \frac{1}{2} g({Q_1}p^\dagger ,{Q_1}p^\dagger ) \mathrm {d} t= \frac{1}{2} g(\dot{q}- h^\sharp ( p),\dot{q}- h^\sharp (p)), \mathrm {d} t\\&= \left[ \frac{1}{2}g(\dot{q},\dot{q}) - \langle p,\dot{q}\rangle + \frac{1}{2} h( p, p) \right] \mathrm {d} t,\\ \mathrm {d}\iota _{{Q_1}} \beta _1^1&= \mathrm {d}\left( g(p^\dagger ,{Q_1}q)\right) = 0,\\ \mathrm {d}f_1^1&= 0, \end{aligned}$$

as desired. $\square$

Lemma 3.3.7

The composition map $\lambda ^* = \phi ^* \circ \psi ^*: \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2$ is the identity

$$\begin{aligned}&\lambda ^* \tilde{q}= \tilde{q},\quad \lambda ^*\tilde{q}^\dagger = \tilde{q}^\dagger \end{aligned}$$

and as such the identity in cohomology.

Proof

Using $\lambda ^* = \phi ^* \circ \psi ^*$, we have

$$\begin{aligned} \lambda ^* \tilde{q}&= \phi ^* \circ \psi ^* (\tilde{q}) = \phi ^* q= \tilde{q},\\ \lambda ^* \tilde{q}^\dagger&= \phi ^* \circ \psi ^* (\tilde{q}^\dagger ) \\&=\phi ^*\left( q^\dagger - \frac{1}{2} \partial g(p^\dagger , {Q_1}p^\dagger ) - g^\flat (\dot{p}^\dagger )-\frac{d}{dt}(g^\flat )(p^\dag ) + \partial g(\dot{q},p^\dagger ))\right) = \tilde{q}^\dagger , \end{aligned}$$

as $\phi ^* p^\dagger = 0$. $\square$

Lemma 3.3.8

The composition map $\chi ^* = \psi ^* \circ \phi ^*: \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$ acts as

$$\begin{aligned} \chi ^*q&= q,\quad \chi ^*p= {g^\flat (\dot{q})},\\ \chi ^*q^\dagger&= q^\dagger - \frac{1}{2} \partial g(p^\dagger , {Q_1}p^\dagger ) - g^\flat (\dot{p}^\dagger ) -\frac{d}{dt}(g^\flat )(p^\dag ) + \partial g(\dot{q},p^\dagger ),\quad \chi ^*p^\dagger = 0. \end{aligned}$$

and is homotopic to the identity.

Proof

By definition $\chi ^* = \psi ^* \circ \phi ^*$. Then

$$\begin{aligned} \chi ^*q&= \psi ^* \circ \phi ^*(q) = \psi ^* \tilde{q},\\ \chi ^*p&= \psi ^* \circ \phi ^*(p) = \psi ^* (\tilde{g}\dot{\tilde{q}}) = { = \psi ^* (\tilde{g}^\flat ( \dot{\tilde{q}})) = g^\flat ( \dot{q})},\\ \chi ^*q^\dagger&= \psi ^* \circ \phi ^*(q^\dagger ) = \psi ^* \tilde{q}^\dagger \\&= q^\dagger - \frac{1}{2} \partial g(p^\dagger , {Q_1}p^\dagger ) - g^\flat (\dot{p}^\dagger ) -\frac{d}{dt}(g^\flat )(p^\dag ) + \partial g(\dot{q},p^\dagger ),\\ \chi ^*p^\dagger&= \psi ^* \circ \phi ^*(p^\dagger ) = \psi ^* (0) = 0. \end{aligned}$$

In order to prove that $\chi ^*$ is homotopic to the identity, we first compute $\chi ^*_s = e^{s\mathcal {L}_{D_1}}$, where $D_1= [{Q_1},{R_1}]$, and show $\lim _{s\rightarrow \infty } \chi ^*_s = \chi ^*$. We choose ${R_1}$ to act as

$$\begin{aligned}&{R_1}q= 0,\quad {R_1}p= {g^\flat (p^\dagger )},\quad {R_1}q^\dagger =0,\quad {R_1}p^\dagger =0. \end{aligned}$$

For $q$ we have

$$\begin{aligned} D_1q&= [{Q_1},{R_1}] q= 0, \\&\Rightarrow e^{s\mathcal {L}_{D_1}} q= q, \\&\Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} q= q= \chi ^* q. \end{aligned}$$

For $p$:

$$\begin{aligned} D_1p&= {Q_1}{R_1}p= {Q_1}(g^\flat (p^\dagger )) = g^\flat (\dot{q})- p,\\ \Rightarrow D_1^2 p&= D_1(g^\flat (\dot{q}) - p) = - D_1p= - (g^\flat (\dot{q}) - p),\\ \Rightarrow D_1^k p&= -(-1)^{k} (g^\flat (\dot{q}) - p) \qquad \text {for } k \ge 1,\\ \Rightarrow e^{s\mathcal {L}_{D_1}} p&= p+ \sum ^\infty _{k=1} \frac{s^k}{k!} D_1^k p= p- \sum ^\infty _{k=1} \frac{(-s)^k}{k!} (g^\flat (\dot{q}) - p)\\&= p- (e^{-s} - 1) (g^\flat (\dot{q}) - p)\\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} p&= g^\flat (\dot{p}) = \chi ^* p. \end{aligned}$$

For $p^\dagger$:

$$\begin{aligned} D_1p^\dagger&= {R_1}{Q_1}p^\dagger {= {R_1}(\dot{q}- h^\sharp (p)) = -h^\sharp (g^\flat (p^\dagger ))} = -p^\dagger , \\ \Rightarrow D_1^k p^\dagger&= (-1)^{k} p^\dagger \qquad \text {for } k \ge 0,\\ \Rightarrow e^{s\mathcal {L}_{D_1}} p^\dagger&= p^\dagger + \sum ^\infty _{k=1} \frac{s^k}{k!} D_1^k p^\dagger = p+ \sum ^\infty _{k=1} \frac{(-s)^k}{k!} p^\dagger = e^{-s} p^\dagger \\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} p^\dagger&= 0 = \chi ^* p^\dagger . \end{aligned}$$

For $q^\dagger$:

$$\begin{aligned} D_1q^\dagger&= {R_1}{Q_1}q^\dagger = -{R_1}\left( \dot{p}+ \frac{1}{2} \partial h(p,p)\right) = -\frac{\mathrm {d}}{\mathrm {d} t}(g^\flat (p^\dagger )) - \partial h(g^\flat (p^\dagger ), p)\\&= -\dot{g}^\flat (p^\dagger ) - g^\flat (\dot{p}^\dagger ) - \partial h(p,g^\flat (p^\dagger )) \\&= -\dot{g}^\flat (p^\dagger ) - g^\flat (\dot{p}^\dagger ) - \partial g({Q_1}p^\dagger ,p^\dagger ) + \partial g(\dot{q},p^\dagger ), \end{aligned}$$

where we used $p= g^\flat ( \dot{q}- Q p^\dagger )$, and $\partial h(g^\flat (\cdot ),g^\flat (\cdot )) = -\partial g(\cdot ,\cdot )$.

We see that

$$\begin{aligned} D_1(\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger ))&= (\dot{g}^\flat (D_1p^\dagger ) + g^\flat (\frac{d}{dt}(D_1p^\dagger )) - \partial g(\dot{q},D_1p^\dagger )) \\&= - (\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger )), \end{aligned}$$

so that, for all $k\ge 0$

$$\begin{aligned} D_1^k(\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger ))&= (-1)^k (\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger )) \end{aligned}$$

and (recalling that $[D_1,{Q_1}] = 0$)

$$\begin{aligned} D_1(\partial g(p^\dagger , {Q_1}p^\dagger ))&= \partial g(D_1p^\dagger , {Q_1}p^\dagger ) + \partial g(p^\dagger , {Q_1}D_1p^\dagger ) = - 2 \partial g(p^\dagger , {Q_1}p^\dagger ), \end{aligned}$$

we have

$$\begin{aligned} D_1^k(\partial g(p^\dagger , {Q_1}p^\dagger ))&= (-2)^k \partial g(p^\dagger , {Q_1}p^\dagger ) \end{aligned}$$

for $k\ge 0$, which ultimately yields

$$\begin{aligned} D_1^k q^\dagger&= - D_1^{k-1}(\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger )) - D_1^{k-1}(\partial g(p^\dagger , {Q_1}p^\dagger ))\\&= - (-1)^{k-1} (\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger )) - (-2)^{k-1} (\partial g(p^\dagger , {Q_1}p^\dagger ))\\&= (-1)^k (\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger )) + \frac{1}{2} (-2)^{k} (\partial g(p^\dagger , {Q_1}p^\dagger )) \end{aligned}$$

for all $k\ge 0$, so that

$$\begin{aligned} e^{s\mathcal {L}_{D_1}} q^\dagger&= q^\dagger + \frac{1}{2} (e^{-2s}-1) (\partial g(p^\dagger , {Q_1}p^\dagger )) \\&\quad + (e^{-s}-1)(\dot{g}^\flat (p^\dagger ) - g^\flat (\dot{p}^\dagger ) + \partial g(\dot{q},p^\dagger ))\\ \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} q^\dagger&= q^\dagger - \frac{1}{2} (\partial g(p^\dagger , {Q_1}p^\dagger )) - (\dot{g}^\flat (p^\dagger ) + g^\flat (\dot{p}^\dagger ) - \partial g(\dot{q},p^\dagger )) = \chi ^* p^\dagger .&\end{aligned}$$

All in all, the homotopy $\chi ^*_s$ takes the form

$$\begin{aligned} \chi ^*_s q&= q,\\ \chi ^*_s p&= e^{-s}p- (e^{-s} - 1) {g^\flat (\dot{q})},\\ \chi ^*_s q^\dagger&= q^\dagger + {\frac{1}{2} (e^{-2s}-1) \partial g(p^\dagger ,{Q_1}p^\dagger ) + (e^{-s}-1) ({\dot{g}}^\flat ( p^\dagger ) - g^\flat ({\dot{p}}^\dagger ) +\partial g(\dot{q},p^\dagger ))},\\ \chi ^*_s p^\dagger&= e^{-s} p^\dagger , \end{aligned}$$

and clearly satisfies $\lim _{s\rightarrow \infty } \chi ^*_s = \chi ^*$. $\square$

Lemma 3.3.9

The map $\chi ^*$ is the identity in cohomology.

Proof

We have to check whether the map $h_\chi$ converges on $\mathcal {F}_{1CM}^{{\mathrm {lax}}}$, namely

$$\begin{aligned} h_\chi \varphi ^j_1 = \int ^\infty _0 e^{s D_1} {R_1}\varphi ^j_1 \, \mathrm {d}s < \infty \quad \forall \varphi ^j_1 \in \mathcal {F}_{1CM}^{{\mathrm {lax}}}. \end{aligned}$$

As $\{q,q^\dagger , p^\dagger \} \in \ker {R_1}$ we have

$$\begin{aligned}&h_\chi q= 0,\quad h_\chi q^\dagger = 0,\quad h_\chi p^\dagger = 0. \end{aligned}$$

For $h_\chi p^\dagger$ we compute

$$\begin{aligned} h_\chi p= \int ^\infty _0 e^{s\mathcal {L}_{D_1}} {R_1}p\,\mathrm {d}s = {\int ^\infty _0 e^{s\mathcal {L}_{D_1}} (g^\flat ( p^\dagger )) \mathrm {d}s = g^\flat ( p^\dagger ) \int ^\infty _0 e^{-s} \mathrm {d}s = g^\flat ( p^\dagger )}, \end{aligned}$$

thus, by Proposition 3.1.4, $h_\chi$ converges on $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$ and $\chi ^*$ is the identity in cohomology. $\square$

3.4 Yang–Mills theory

We now look at the example of (nonabelian) Yang–Mills theory. Let (M, g) be a d-dimensional (pseudo-)Riemanian manifold and G a connected Lie group with Lie algebra $(\mathfrak {g},[\cdot ,\cdot ])$, endowed with an ad-invariant inner product, which for ease of notation will be denoted by means of an invariant trace operation²²$\mathrm {Tr}[\cdot ]$. As we consider two formulations of Yang–Mills theory, we will use tildes to distinguish the fields between the two. We point out that an alternative proof of the equivalence of first- and second-order formulations of Yang–Mills theory has been given in [54] using homotopy transfer of $A_\infty$-structures. We will give here an argument that is different on the surface, but which is compatible to their results. However, we stress that our analysis also includes a comparison of the boundary data of the first- and second-order formulations.

We can phrase the theory by considering connection 1-forms $\tilde{A}\in \Omega ^1(M,\mathfrak {g})$, with curvature $\tilde{F}_{\tilde{A}}$, and the classical action functional

$$\begin{aligned} {S_2}[\tilde{A}] = \int _M \mathrm {Tr}\left[ \frac{1}{2} \tilde{F}_{\tilde{A}}\star \tilde{F}_{\tilde{A}}\right] . \end{aligned}$$

This is often known as the second-order formulation.

Alternatively, one can phrase the theory in its first-order formulation, by considering an additional “auxiliary” field $B\in \Omega ^{d-2}(M,\mathfrak {g})$ and the classical action functional

$$\begin{aligned} {S_1}[A,B] = \int _M \mathrm {Tr}\left[ BF_{A}- \frac{\varepsilon _s}{2} B\star B\right] , \end{aligned}$$

where $\varepsilon _s=\pm 1$ denotes the signature of g.

Proposition 3.4.1

The first- and second-order formulations of Yang–Mills theory are classically equivalent.

Proof

Solving the EL equations of the first-order theory w.r.t. the auxiliary field $B$ gives

$$\begin{aligned} \delta _{B} {S_1}[A,B]&= \int _M \mathrm {Tr}\left[ \delta BF_{A}- \frac{\varepsilon _s}{2} \delta B\star B- \frac{\varepsilon _s}{2} B\star \delta B\right] \\&=\int _M \mathrm {Tr}\left[ \delta B\left( F_{A}- \varepsilon _s\star B\right) \right] . \end{aligned}$$

Let $C_1$ be the set of solutions of $F_{A}= \varepsilon _s\star B$, or equivalently

$$\begin{aligned} \star F_{A}= \varepsilon _s\star ^2 B= \varepsilon _s^2 (-1)^{2(d-2)} B= B, \end{aligned}$$

where we used $\star ^2 = \varepsilon _s(-1)^{k(d-k)}$ when acting on k-forms, and let $\phi _{cl}: C_1 \rightarrow F_2$ be the map defined through $\phi ^*_{cl}\tilde{A}= A$. Then

$$\begin{aligned} {S_1}[A,B] \Big \vert _{C_1}&= \int _M \mathrm {Tr}\left[ \star F_{A}F_{A}- \frac{\varepsilon _s^2}{2} \star F_{A}F_{A}\right] = \int _M \mathrm {Tr}\left[ \frac{1}{2} F_{A}\star F_{A}\right] \\&= \phi ^*_{cl}\int _M \mathrm {Tr}\left[ \frac{1}{2} \tilde{F}_{\tilde{A}}\star \tilde{F}_{\tilde{A}}\right] = \phi ^*_{cl} {S_2}[\tilde{A}], \end{aligned}$$

showing that the two theories are classically equivalent. $\square$

Both first- and second-order formulations of Yang–Mills theory can be extended to lax BV–BFV theories as follows. As the symmetries of the theory are given by a Lie algebra $\mathfrak {g}$, we can follow the construction of Example 2.2.5.

Proposition/Definition 3.4.2

([19, 47]) The data

$$\begin{aligned} \mathfrak {F}^\mathrm {lax}_{2YM} = (\mathcal {F}_{2YM}^{{\mathrm {lax}}},\theta _{2}^\bullet ,L_{2}^\bullet ,{Q_2}), \end{aligned}$$

where

$$\begin{aligned} \mathcal {F}_{2YM}^{{\mathrm {lax}}}= T^*[-1](\Omega ^{1}(M,\mathfrak {g}) \oplus \Omega ^{0}(M,\mathfrak {g})[1]), \end{aligned}$$

together with $\theta _{2}^\bullet \in \Omega ^{1,\bullet }_\mathrm {loc}(\mathcal {F}_{2YM}^{{\mathrm {lax}}}{{\times M}})$ and $L_{2}^\bullet \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}_{2YM}^{{\mathrm {lax}}}{{\times M}})$, which are given by

$$\begin{aligned} \theta _{2}^\bullet&=\, \mathrm {Tr}\left[ \tilde{A}^\dagger \delta \tilde{A}+ \tilde{c}^\dagger \delta \tilde{c}+ \delta \tilde{A}\star \tilde{F}_{\tilde{A}}+ \tilde{A}^\dagger \delta \tilde{c}+ \star \tilde{F}_{\tilde{A}}\delta \tilde{c}\right] ,\\ L_{2}^\bullet&=\, \mathrm {Tr}\left[ \frac{1}{2} \tilde{F}_{\tilde{A}}\star \tilde{F}_{\tilde{A}}+ \tilde{A}^\dagger \mathrm {d}_{\tilde{A}}\tilde{c}+ \frac{1}{2} \tilde{c}^\dagger [\tilde{c},\tilde{c}] + \star \tilde{F}_{\tilde{A}}\mathrm {d}_{\tilde{A}}\tilde{c}+ \frac{1}{2} \tilde{A}^\dagger [\tilde{c},\tilde{c}] + \frac{1}{2} \star \tilde{F}_{\tilde{A}}[\tilde{c},\tilde{c}] \right] , \end{aligned}$$

and the cohomological vector field ${Q_2}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_{2YM}^{{\mathrm {lax}}})$

$$\begin{aligned} {Q_2}\tilde{A}&=\, \mathrm {d}_{\tilde{A}}\tilde{c},\quad {Q_2}\tilde{c}= \frac{1}{2} [\tilde{c},\tilde{c}],\\ {Q_2}\tilde{A}^\dagger&=\, \mathrm {d}_{\tilde{A}}\star \tilde{F}_{\tilde{A}}+ [\tilde{c}, \tilde{A}^\dagger ],\quad {Q_2}\tilde{c}^\dagger =\, \mathrm {d}_{\tilde{A}}\tilde{A}^\dagger + [\tilde{c},\tilde{c}^\dagger ], \end{aligned}$$

defines a lax BV–BFV theory.

In the first-order formulation we have:

Proposition/Definition 3.4.3

([19, 47]) The data

$$\begin{aligned} \mathfrak {F}^\mathrm {lax}_{1YM} = (\mathcal {F}_{1YM}^{\mathrm {lax}},\theta _1^\bullet ,L_{1}^\bullet ,{Q_1}), \end{aligned}$$

where

$$\begin{aligned} \mathcal {F}_{1YM}^{\mathrm {lax}}= T^*[-1](\Omega ^{1}(M,\mathfrak {g})\oplus \Omega ^{d-2}(M,\mathfrak {g}) \oplus \Omega ^{0}(M,\mathfrak {g})[1]). \end{aligned}$$

together with $\theta _1^\bullet \in \Omega ^{1,\bullet }_\mathrm {loc}(\mathcal {F}_{1YM}^{\mathrm {lax}})$ and $L_{1}^\bullet \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}_{1YM}^{\mathrm {lax}})$, which take the form

$$\begin{aligned} \theta _1^\bullet =&\mathrm {Tr}\left[ A^\dagger \delta A+ B^\dagger \delta B+ c^\dagger \delta c+ B\delta A+ A^\dagger \delta c+ B\delta c\right] ,\\ L_{1}^\bullet =&\mathrm {Tr}\left[ BF_{A}- \frac{\varepsilon _s}{2} B\star B+ A^\dagger \mathrm {d}_{A}c+ B^\dagger [c,B] + \frac{1}{2} c^\dagger [c,c]\right. \\&+ \left. B\mathrm {d}_{A}c+ \frac{1}{2} A^\dagger [c,c] + \frac{1}{2} B[c,c] \right] \end{aligned}$$

and the cohomological vector field ${Q_1}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_{1YM}^{\mathrm {lax}})$

$$\begin{aligned}&{Q_1}A= \mathrm {d}_{A}c,\qquad \quad {Q_1}A^\dagger = \mathrm {d}_{A}B+ [c,A^\dagger ],\\&{Q_1}B= [c,B],\qquad {Q_1}B^\dagger = F_{A}- \varepsilon _s\star B+ [c, B^\dagger ],\\&{Q_1}c= \frac{1}{2} [c,c],\qquad {Q_1}c^\dagger = \mathrm {d}_{A}A^\dagger + [c, c^\dagger ] + [B^\dagger ,B], \end{aligned}$$

defines a lax BV–BFV theory.

We now present the main theorem of this section, together with an outline of the proof. The computational details and the various required Lemmata are presented in “Appendix A”.

Theorem 3.4.4

The lax BV–BFV theories $\mathfrak {F}^\mathrm {lax}_{1YM}$ and $\mathfrak {F}^\mathrm {lax}_{2YM}$ of the first- and second-order formulations of nonabelian Yang–Mills theory are lax BV–BFV equivalent.

Proof

We need to check all the conditions from Definition 2.6.3. The existence of two maps $\phi$, $\psi$ with the desired properties is presented in Lemmata A.1.1 and A.1.2 respectively, where we also show that the pullback maps $\phi ^*$, $\psi ^*$ are chain maps w.r.t. the BV–BFV complexes $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _i$, and that they map $(\theta ^\bullet _i,L^\bullet _i)$ in the desired way. Specifically, $\phi : \mathcal {F}_{2YM}^{{\mathrm {lax}}}\rightarrow \mathcal {F}_{1YM}^{\mathrm {lax}}$ is defined through

$$\begin{aligned} \phi ^* A&= \tilde{A},\quad \phi ^* B= \star \tilde{F}_{\tilde{A}},\quad \phi ^* c= \tilde{c},\\ \phi ^* A^\dagger&= \tilde{A}^\dagger ,\quad \phi ^* B^\dagger = 0,\quad \phi ^* c^\dagger = \tilde{c}^\dagger . \end{aligned}$$

and maps the lax BV–BFV data of the first-order theory as

$$\begin{aligned}&\phi ^* \theta _1^\bullet = \theta _{2}^\bullet ,\quad \phi ^* L_{1}^\bullet = L_{2}^\bullet , \end{aligned}$$

whereas $\psi : \mathcal {F}_{1YM}^{\mathrm {lax}}\rightarrow \mathcal {F}_{2YM}^{{\mathrm {lax}}}$ is given by

$$\begin{aligned} \psi ^* \tilde{A}&= A,\quad \quad \quad ~~~~~~~~~~~ \psi ^* \tilde{c}= c,\\ \psi ^* \tilde{A}^\dagger&= A^\dagger - \mathrm {d}_{A}\star B^\dagger ,\quad \psi ^* \tilde{c}^\dagger = c^\dagger - \frac{1}{2} [B^\dagger ,\star B^\dagger ], \end{aligned}$$

and maps the lax BV–BFV data of the second-order theory as

$$\begin{aligned}&\psi ^* \theta _{2}^\bullet = \theta _1^\bullet + (\mathcal {L}_{{Q_1}} - \mathrm {d}) \beta _1^\bullet + \delta f_1^\bullet ,\quad \psi ^* L_{2}^\bullet = L_{1}^\bullet + (\mathcal {L}_{{Q_1}} - \mathrm {d}) \iota _{{Q_1}} \beta _1^\bullet + \mathrm {d}f_1^\bullet , \end{aligned}$$

where

$$\begin{aligned} \beta _1^\bullet&= \mathrm {Tr}\left[ \frac{1}{2} B^\dagger \star \delta B^\dagger + \star B^\dagger \delta A+ \star B^\dagger \delta c\right] ,\quad f_1^\bullet = \mathrm {Tr}\left[ \frac{1}{2} B^\dagger (B-\star F_{A}) \right] \end{aligned}$$

in accordance with our notion of lax BV–BFV equivalence. Note that $f_1^1=f_1^2=0$.

Furthermore, we need to show that the respective BV–BFV complexes are quasi-isormophic. The composition map $\lambda ^* = \phi ^* \circ \psi ^*$ is shown to be the identity in Lemma A.1.3, which follows directly from $\phi ^* B^\dagger = 0$. In Lemma A.1.4, we prove that the composition map $\chi ^* = \psi ^* \circ \phi ^*$, which has the explicit form

$$\begin{aligned} \chi ^* A&= A,\quad \chi ^* A^\dagger = A^\dagger - \mathrm {d}_{A}\star B^\dagger ,\\ \chi ^* B&= \star F_{A},\quad \chi ^* B^\dagger = 0,\\ \chi ^* c&= c,\quad \chi ^* c^\dagger = \tilde{c}^\dagger - \frac{1}{2} [B^\dagger ,\star B^\dagger ], \end{aligned}$$

is homotopic to the identity by constructing the morphism $\chi ^*_s = e^{s\mathcal {L}_{D_1}}$ with $D_1= [{R_1},{Q_1}]$, where ${R_1}$ is chosen to act as

$$\begin{aligned} {R_1}A&= 0,\quad {R_1}A^\dagger = 0,\\ {R_1}B&= \star B^\dagger ,\quad {R_1}B^\dagger = 0,\\ {R_1}c&= 0, {R_1}c^\dagger = 0. \end{aligned}$$

The homotopy is explicitly given by

$$\begin{aligned} \chi ^*_s A&= A,\quad \chi ^*_s A^\dagger = A^ \dagger + (e^{-s}-1) \mathrm {d}_{A}\star B^\dagger ,\\ \chi ^*_s B&= e^{-s} B- (e^{-s}-1) \star F_{A},\quad \chi ^*_s B^\dagger = e^{-s} B^\dagger ,\\ \chi ^*_s c&= c,\quad \chi ^*_s c^\dagger = c^\dagger + \frac{1}{2} (e^{-2s} - 1) [ B^\dagger , \star B^\dagger ]. \end{aligned}$$

and fulfils $\lim _{s\rightarrow \infty } \chi ^*_s = \chi ^*$. In Lemma A.1.5, we demonstrate that $\chi ^*$ is the identity in cohomology by showing that the map

$$\begin{aligned} h_\chi \varphi ^j_1 = \int _0^\infty e^{s\mathcal {L}_{D_1}} \mathcal {L}_{{R_1}} \varphi ^j_1 \mathrm {d}s \qquad \quad \varphi ^j_1 \in \mathcal {F}_{1YM}^{\mathrm {lax}}, \end{aligned}$$

satisfying $\chi ^* - \mathrm {id}_{1} = (\mathcal {L}_{{Q_1}} - \mathrm {d}) h_\chi + h_\chi (\mathcal {L}_{{Q_1}} - \mathrm {d})$ (cf. Lemma 3.1.3) converges to

$$\begin{aligned} h_\chi A&= h_\chi A^\dagger = h_\chi B^\dagger = h_\chi c= h_\chi c^\dagger = 0,\\ h_\chi B&= \star B^\dagger , \end{aligned}$$

therefore proving that the two lax BV–BFV theories in question have isomorphic BV–BFV cohomologies

$$\begin{aligned} H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}_{1YM}) \simeq H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}_{2YM}) \end{aligned}$$

and thus that they are lax BV–BFV equivalent. $\square$

3.5 1D reparametrisation-invariant theories

In this section, we compare two one-dimensional reparamentrisation invariant theories, namely Jacobi theory, which one can think of as classical mechanics at constant energies, and one-dimensional gravity coupled to matter (1D GR). For an in-depth discussion of these theories we refer to [26]. We recall that the motivation to investigate the equivalence of these two theories is that, even though they are classically equivalent, 1D GR is compatible with the BV–BFV procedure while the Jacobi theory is not and yields a singular boundary structure. Firstly, this raises the question whether this boundary discrepancy is reflected at a cohomological level. Secondly, this discrepancy in the boundary behaviour is also present in the classically equivalent Einstein–Hilbert gravity and Palatini–Cartan gravity in (3+1) dimensions, where the latter is incompatible with the BV–BFV procedure. Our hope is that the comparison and analysis of these toy models might shed light in the question of equivalence of the (3+1) dimensional theories.

We take the base manifold to be a closed interval on the real line $M = I = [a,b] \subset \mathbb {R}$ with coordinate t for both theories, which should be interpreted as a finite time interval.

In the case of Jacobi theory, we consider a matter field $\tilde{q}\in \Gamma (\mathbb {R}^n \times I) = C^\infty (I,\mathbb {R}^n)$ with mass m. The kinetic energy is taken to be $T(\dot{\tilde{q}}) = \frac{m}{2}\Vert \dot{\tilde{q}}\Vert ^2$ where $\Vert \cdot \Vert$ is the Euclidean norm on $\mathbb {R}^n$ and $\dot{\tilde{q}}= \partial _t \tilde{q}$ is the time derivative of $\tilde{q}$. Let E denote a parameter and $V(\tilde{q})$ a potential term. We do not assume $E = T(\dot{\tilde{q}}) + V(\tilde{q})$. The Jacobi action functional takes the form

$$\begin{aligned} S_J[\tilde{q}] = \int _I 2\sqrt{(E-V)T} \,\mathrm {d}t. \end{aligned}$$

To see that $S_J$ is parameterisation invariant, note that writing

$$\begin{aligned} \mathrm {d}s^2 = 2m(E-V)\,\mathrm {d}\tilde{q}^2 \end{aligned}$$

lets us interpret the Jacobi action functional as the length of a path in the target space $\mathbb {R}^n$ with metric $\mathrm {d}s^2$. As such the symmetry group of Jacobi theory is the diffeomorphism group of the interval $\text {Diff}(I)$, i.e. the reparameterisations of I. The critical locus of $S_J$ is then given by the geodesics of the metric $\mathrm {d}s^2$, which are the trajectories of classical mechanics with an arbitrary parameterisation [26]. Imposing $E =T(\dot{\tilde{q}}) + V(\tilde{q})$ allows us to recover the standard parameterisation. We set $V(\tilde{q})=0$ for the rest of the discussion. The EL equations can be shown to have the form

$$\begin{aligned} \partial _t\left( \sqrt{\frac{E}{T}}m\dot{\tilde{q}}\right) = 0, \end{aligned}$$

which are singular for $\dot{\tilde{q}}= 0$. As such, the space of fields for Jacobi theory is not $C^\infty (I,\mathbb {R}^n)$ but rather

$$\begin{aligned} {\mathcal {F}_J} = \left\{ \tilde{q}\in C^\infty (I,\mathbb {R}^n)\,\,|\,\, \dot{\tilde{q}}(t)\not =0\ \forall t \in I\right\} . \end{aligned}$$

We can then interpret Jacobi theory as classical mechanics at constant energies where the solutions do not have turning points, i.e. points in which the first derivative vanishes.

In the case of 1D GR [26], we also consider a metric field $\mathsf {g} \in \Gamma (S^2_+T^*I)$ as a nonvanishing section of the bundle of symmetric nondegenerate rank-(0, 2) tensors over I. For simplicity, we write $\mathsf {g} = g \, \mathrm {d}t^2$ and work with the component $g\in C^\infty (I,\mathbb {R}_{>0})$. The space of fields is given by

$$\begin{aligned} {\mathcal {F}_{GR}} ={\mathcal {F}_J} \oplus C^\infty (I,\mathbb {R}_{>0}). \end{aligned}$$

The condition $\dot{q} \ne 0$ in ${\mathcal {F}_J}$ is strictly speaking not necessary in the 1D GR case, but we are ultimately interested in comparing 1D GR with the Jacobi theory and therefore impose it for consistency. In this picture, we can interpret 1D GR as an extension of Jacobi theory. We consider the action functional

$$\begin{aligned} S_{GR}[q,g] = \int _I \left( \frac{T}{g} + E\right) \sqrt{g}\,\mathrm {d}t = \int _I \left( \frac{T}{\sqrt{g}} + \sqrt{g}E\right) \,\mathrm {d}t. \end{aligned}$$

Note that the Ricci tensor vanishes in 1D and hence the Einstein–Hilbert term is absent. The first term in $S_{GR}$ is simply the matter Lagrangian for vanishing potential in the presence of a metric field and the second is a cosmological term. As such we interpret the parameter E as a cosmological constant. Since we are integrating over the Riemannian density $\sqrt{g}\,\mathrm {d}t$ of the metric $\mathrm {d}s^2 = g \mathrm {d}t^2$, the symmetry group is again $\text {Diff}(I)$.

Proposition 3.5.1

([26]) Jacobi theory and 1D GR are classically equivalent.

Let us now turn to the lax BV–BFV formulation of Jacobi theory. We first need to introduce the ghost field, which in case of diffeomorphims invariance is chosen to be $\tilde{\xi }\partial _t \in \mathfrak {X}(I)[1]$ [49]. In this setting, the Chevalley–Eilenberg operator acts on the fields as the Lie derivative $\gamma _J = \mathcal {L}_{\tilde{\xi }\partial _t}$ and on the ghost as the Lie bracket of vector fields (cf. Example 2.2.5). We work with the component $\tilde{\xi }\in C^\infty (I,\mathbb {R})[1]$ for simplicity.

Proposition/Definition 3.5.2

The data

$$\begin{aligned} \mathfrak {F}^\mathrm {lax}_{J} = (\mathcal {F}_{J}^{{\mathrm {lax}}},\theta _{J}^\bullet ,L_{J}^\bullet ,{Q_{J}}) \end{aligned}$$

where

$$\begin{aligned} \mathcal {F}_{J}^{{\mathrm {lax}}}= T^*[-1]({\mathcal {F}_J} \oplus C^\infty (I,\mathbb {R})[1]). \end{aligned}$$

(22)

together with $\theta _{J}^\bullet \in \Omega ^{1,\bullet }_\mathrm {loc}(\mathcal {F}_{J}^{{\mathrm {lax}}}{\times I})$ and $L_{J}^\bullet \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}_{J}^{{\mathrm {lax}}}{\times I})$, which are given by

$$\begin{aligned} \theta _{J}^\bullet&= \left[ \tilde{q}^+ \cdot \delta \tilde{q}+ \tilde{\xi }^+ \delta \tilde{\xi }\right] \mathrm {d}t + \sqrt{\frac{E}{T}} m \dot{ \tilde{q}} \cdot \delta \tilde{q}+ \tilde{q}^+ \tilde{\xi }\cdot \delta \tilde{q}- \tilde{\xi }^+ \tilde{\xi }\delta \tilde{\xi },\\ L_{J}^\bullet&= \left[ 2\sqrt{E T} + \tilde{q}^+ \cdot \tilde{\xi }\dot{\tilde{q}}+ \tilde{\xi }^+ \tilde{\xi }\dot{\tilde{\xi }}\right] \mathrm {d}t. \end{aligned}$$

and the cohomological vector field ${Q_{J}}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_{J}^{{\mathrm {lax}}})$

$$\begin{aligned} {Q_{J}}\tilde{q}&= \tilde{\xi }\dot{\tilde{q}},\quad {Q_{J}}\tilde{q}^+ = -\partial _t\left( \sqrt{\frac{E}{T}} m \dot{\tilde{q}}+ \tilde{q}^+ \tilde{\xi }\right) ,\\ {Q_{J}}\tilde{\xi }&= \tilde{\xi }\dot{\tilde{\xi }},\quad {Q_{J}}\tilde{\xi }^+ = - \tilde{q}^+ \cdot \dot{\tilde{q}}+ \tilde{\xi }\dot{\tilde{\xi }}^+ + 2 \dot{\tilde{\xi }}\tilde{\xi }^+. \end{aligned}$$

defines a lax BV–BFV theory.

Proof

It is a matter of a straightforward calculation to check that the formulas above satisfy the axioms of Definition 2.5.5. $\square$

Remark 3.5.3

Note that we can explicitly decompose the cohomological vector field ${Q_{J}}$ into its Chevalley–Eilenberg and Koszul–Tate parts as ${Q_{J}}= \gamma _J + \delta _J$ by using Eq. (4) and setting $\gamma _J = {Q_{J}}- \delta _J$ on $\{\tilde{q}^+,\tilde{\xi }^+\}$. We have:

$$\begin{aligned} \gamma _J \tilde{q}^+&= \tilde{\xi }\dot{\tilde{q}}^+ + \dot{\tilde{\xi }} q^+,\quad \delta _J \tilde{q}^+ = -\partial _t\left( \sqrt{\frac{E}{T}} m \dot{\tilde{q}}\tilde{\xi }\right) ,\\ \gamma _J \tilde{\xi }^+&= \tilde{\xi }\dot{\tilde{\xi }}^+ + 2 \dot{\tilde{\xi }} \tilde{\xi }^+,\quad \delta _J \tilde{\xi }^+ = - \tilde{q}^+ \cdot \dot{\tilde{q}}. \end{aligned}$$

As $\tilde{q}$ is a function and $\tilde{\xi }$ is the component of a vector field, defining $\tilde{q}^+$ and $\tilde{\xi }^+$ through Eq. (22) lets us interpret them as components of tensor fields in $\Omega ^{\text {top}}(I) \otimes C^\infty (I,\mathbb {R}^n)[-1]$ and $\Omega ^{\text {top}}(I) \otimes \Omega ^{\text {top}}(I)[-2]$, respectively, or rather as components of a rank-(0, 1) and a rank-(0, 2) tensors over I. As such we see that the Chevalley–Eilenberg differential also acts as $\gamma _J = \mathcal {L}_{\tilde{\xi }\partial _t}$ on $\{\tilde{q}^+,\tilde{\xi }^+\}$.

In the case of 1D GR we have

Proposition 3.5.4

The data

$$\begin{aligned} \mathfrak {F}^\mathrm {lax}_{GR} = (\mathcal {F}_{GR}^{\mathrm {lax}},\theta _{GR}^\bullet ,L_{GR}^\bullet ,{Q_{GR}}) \end{aligned}$$

where

$$\begin{aligned} \mathcal {F}_{GR}^{\mathrm {lax}}= T^*[1]({\mathcal {F}_J} \oplus C^\infty (\mathbb {R}_{>0}) \oplus C^\infty (I)[1]). \end{aligned}$$

together with $\theta _{GR}^\bullet \in \Omega ^{1,\bullet }_\mathrm {loc}(\mathcal {F}_{GR}^{\mathrm {lax}}{\times I})$ and $L_{GR}^\bullet \in \Omega ^{0,\bullet }_\mathrm {loc}(\mathcal {F}_{GR}^{\mathrm {lax}}{\times I})$, which are given by

$$\begin{aligned} \theta _{GR}^\bullet&= \left[ q^+ \cdot \delta q + \xi ^+ \delta \xi + g^+ \delta g\right] \mathrm {d}t + \frac{m\dot{q}}{\sqrt{g}} \cdot \delta q + q^+\xi \cdot \delta q \\&\quad +g^+\xi \delta g - (2g^+ g + \xi ^+ \xi ) \delta \xi ,\\ L_{GR}^\bullet&= \left[ \frac{T}{\sqrt{g}} + \sqrt{g} E + q^+ \cdot \xi \dot{q} + g^+ (\xi \dot{g} + 2 g {\dot{\xi }}) + \xi ^+ \xi \dot{\xi }\right] \mathrm {d}t + \left( \frac{T}{g} - E \right) \sqrt{g}\xi . \end{aligned}$$

and the cohomological vector field ${Q_{GR}}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_{GR}^{\mathrm {lax}})$

$$\begin{aligned} {Q_{GR}}q&= \xi \dot{q},\quad {Q_{GR}}q^+ = -\partial _t\left( \frac{m\dot{q}}{\sqrt{g}} + q^+\xi \right) ,\\ {Q_{GR}}g&= \xi \dot{g} + 2 {\dot{\xi }} g,\quad {Q_{GR}}g^+ = \text {EL}_g + \xi \dot{g}^+ - {\dot{\xi }} g^+,\\ {Q_{GR}}\xi&= \xi \dot{q},\quad {Q_{GR}}\xi ^+ = - q^+ \cdot \dot{q} + g^+ \dot{g} + 2 \dot{g}^+ g + \xi {\dot{\xi }}^+ + 2 {\dot{\xi }} \xi ^+. \end{aligned}$$

with

$$\begin{aligned} \text {EL}_g {:}{=}\frac{\delta S_{GR}}{\delta g} = \frac{E}{2\sqrt{g}} - \frac{T}{2g^{3/2}}, \end{aligned}$$

defines a lax BV–BFV theory.

Proof

It is straightforward to show that these formulas satisfy the axioms of Definition 2.5.5. $\square$

Remark 3.5.5

As in Jacobi theory, we can decompose the cohomological vector field as ${Q_{GR}}= \gamma _{GR} + \delta _{GR}$. We have

$$\begin{aligned} \gamma _{GR} q^+&= \xi \dot{q}^+ + {\dot{\xi }} q^+,\quad \delta _{GR} q^+ = -\partial _t\left( \frac{m \dot{q}}{\sqrt{g}}\right) ,\\ \gamma _{GR} g^+&= \xi \dot{g}^+ - {\dot{\xi }} g^+\quad \delta _{GR} g^+ = \text {EL}_g,\\ \gamma _{GR} \xi ^+&= \xi {\dot{\xi }}^+ + 2 {\dot{\xi }} \xi ^+,\quad \delta _{GR} \xi ^+ = - q^+ \cdot \dot{q} + g^+ \dot{g} + 2 \dot{g}^+ g. \end{aligned}$$

Similarly to the Jacobi case, $q^+$, $g^+$ and $\xi ^+$ are components of tensors in $\Omega ^{\text {top}}(I) \otimes C^\infty (I,\mathbb {R}^n)[-1]$, $\Gamma [-1](S^2_+TI) \otimes \Omega ^{\text {top}}(I)$ and $\Omega ^{\text {top}}(I) \otimes \Omega ^{\text {top}}(I)[-2]$, respectively, or rather components of a rank-(0, 1), a rank-(2, 1) and a rank-(0, 2) tensors over I. Therefore, we again have $\gamma _{GR} = \mathcal {L}_{\xi \partial _t}$ on $\{q^+, g^+, \xi ^+\}$.

Before presenting the main theorem of this section, we need to introduce some useful notation. Let $v \in C^\infty (I,\mathbb {R}^n)$ be a $\mathbb {R}^n$-valued field and let $u = \dot{q} / \Vert \dot{q}\Vert$ denote the normalised velocity of q. Note that u is always well-defined because we assume $\dot{q} \ne 0$. We can then decompose $v = v_\parallel + v_\perp$ into its parallel $v_\parallel$ and perpendicular $v_\perp$ components with respect to u as

$$\begin{aligned} \begin{aligned} v_\parallel&= (u \cdot v) u = (\dot{q} \cdot v) \frac{\dot{q}}{\Vert \dot{q}\Vert ^2} = (\dot{q} \cdot v) \frac{m\dot{q}}{2T},\\ v_\perp&= v - (u \cdot v) u = v - (\dot{q} \cdot v) \frac{m\dot{q}}{2T}, \end{aligned} \end{aligned}$$

(23)

where we used that $T = \frac{m}{2}\Vert \dot{q}\Vert ^2$.

We now present the main theorem of this section, together with an outline of the proof. The computational details and the various required Lemmata are presented in the “Appendix B”.

Theorem 3.5.6

The lax BV–BFV theories $\mathfrak {F}^\mathrm {lax}_{GR}$ and $\mathfrak {F}^\mathrm {lax}_{J}$ of 1D GR and Jacobi theory are lax BV–BFV equivalent.

Proof

We need to check all the conditions from Definition 2.6.3. The existence of two maps $\phi$, $\psi$ with the desired properties is presented in Lemmata B.2.1 and B.2.2 respectively, where we also show that the pullback maps $\phi ^*$, $\psi ^*$ are chain maps w.r.t. the BV–BFV complexes $\mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _i$, and that they map $(\theta ^\bullet _i,L^\bullet _i)$ in the desired way. Specifically, $\phi : \mathcal {F}_{J}^{{\mathrm {lax}}}\rightarrow \mathcal {F}_{GR}^{\mathrm {lax}}$ is defined through

$$\begin{aligned} \phi ^*q&= \tilde{q},\quad \phi ^*g = \frac{T}{E},\quad \phi ^*\xi = \tilde{\xi },\\ \phi ^*q^+&= \tilde{q}^+,\quad \phi ^*g^+ = 0,\quad \phi ^*\xi ^+ = \tilde{\xi }^+. \end{aligned}$$

and maps the lax BV–BFV data of 1D GR as

$$\begin{aligned}&\phi ^* \theta _{GR}^\bullet = \theta _{J}^\bullet ,\quad \phi ^* L_{GR}^\bullet = L_{J}^\bullet . \end{aligned}$$

On the other hand, $\psi : \mathcal {F}_{GR}^{\mathrm {lax}}\rightarrow \mathcal {F}_{J}^{{\mathrm {lax}}}$ is given by

$$\begin{aligned} \psi ^*\tilde{q}&= q,\\ \psi ^*\tilde{\xi }&= \xi ,\\ \psi ^* \tilde{q}^+_\parallel&= \eta ^{3/2} \left( q^+_\parallel - \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} -\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\right) ,\\ \psi ^*\tilde{q}^+_\perp&= \eta ^{3/2} \left( q^+_\perp +\frac{2m}{E} g^+ \ddot{q}_\perp \right) ,\\ \psi ^*\tilde{\xi }^+&= \eta ^{3/2}\left( \xi ^+ + \frac{g^{3/2}}{E} \dot{g}^+ g^+\right) , \end{aligned}$$

where $\eta {:}{=}\frac{gE}{T}$, and maps the lax BV–BFV data of the Jacobi theory as

$$\begin{aligned} \psi ^* \theta _{J}^\bullet&= \theta _{GR}^\bullet + (\mathcal {L}_{{Q_{GR}}} - \mathrm {d}) \beta _{GR}^\bullet + \delta f_{GR}^\bullet ,\\ \psi ^* L_{J}^\bullet&= L_{GR}^\bullet + (\mathcal {L}_{{Q_{GR}}} - \mathrm {d}) \iota _{{Q_{GR}}} \beta _{GR}^\bullet + \mathrm {d}f_{GR}^\bullet , \end{aligned}$$

where (writing $\kappa ^\bullet = \kappa ^0 \mathrm {d} t+ \kappa ^1$)

$$\begin{aligned} \beta _{GR}^0 =&-\frac{4g^{7/2}}{\Omega ^2}Tg^+\delta g^+ +\left( \frac{2g^2}{\Omega } + \eta ^{3/2}\frac{2\sqrt{g}}{E}\right) g^+q^+_\perp \cdot \delta q\\&+\left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \dot{g}^+ g^+ \frac{m\dot{q}}{2T} \cdot \delta q -(\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T} \cdot \delta q,\\ \beta _{GR}^1 =&\,\,\xi \beta _{GR}^0 + \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q,\\ f_{GR}^0 =&\,\,2 g^+\left( g - \frac{2g^{3/2}}{\Omega } T\right) ,\\ f_{GR}^1 =&\,\,\xi f_{GR}^0, \end{aligned}$$

with $\Omega = \sqrt{g}T + g \sqrt{TE}$, in accordance with our notion of lax BV–BFV equivalence.

Furthermore, we need to show that the respective BV–BFV complexes are quasi-isormophic. The composition map $\lambda ^* = \phi ^* \circ \psi ^*$ is shown to be the identity in Lemma B.2.3. In Lemma B.2.5, we prove that the composition map $\chi ^* = \psi ^* \circ \phi ^*$, which has explicitly form

$$\begin{aligned} \chi ^*q&= q ,\\ \chi ^*\xi&= \xi ,\\ \chi ^*g&= \frac{T}{E},\\ \chi ^* q^+_\parallel&= \eta ^{3/2} \left( q^+_\parallel - \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} -\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\right) , \\ \chi ^* q^+_\perp&= \eta ^{3/2} \left( q^+_\perp +\frac{2m}{E} g^+ \ddot{q}_\perp \right) ,\\ \chi ^* \xi ^+&= \eta ^{3/2}\left( \xi ^+ + \frac{g^{3/2}}{E} \dot{g}^+ g^+\right) ,\\ \chi ^* g^+&= 0, \end{aligned}$$

is homotopic to the identity by constructing the morphism $\chi ^*_s = e^{s\mathcal {L}_{{D_{GR}}}}$ with ${D_{GR}}= [{R_{GR}},{Q_{GR}}]$, where ${R_{GR}}$ is chosen to act as

$$\begin{aligned} {R_{GR}}q&= 0,\quad {R_{GR}}\xi = 0,\quad {R_{GR}}g= \frac{-2g^{3/2}}{E} g^+,\nonumber \\ {R_{GR}}q^+_\parallel&= -\frac{3\sqrt{g}}{E} \text {EL}_g \xi ^+ \frac{m\dot{q}}{2T},\quad {R_{GR}}\xi ^+ =0,\quad {R_{GR}}g^+ = 0,\nonumber \\ {R_{GR}}q^+_\perp&= \frac{3\sqrt{g}}{E} g^+ \dot{q}^+_\perp . \end{aligned}$$

The homotopy is given by

$$\begin{aligned} \chi ^*_sq&= q,\\ \chi ^*_s\xi&= \xi ,\\ \chi ^*_s g&= e^{-s}g + (1-e^{-s}) \frac{T}{E},\\ \chi ^*_s q^+_\parallel&= \left( \frac{g}{\chi ^*_s g}\right) ^{3/2} \left( q^+_\parallel + (e^{-s} - 1) 2g^{-3/2} \sigma \left( \frac{T}{E}\right) \frac{m\dot{q}}{2T}\right. \\&\quad \left. + (e^{-2s} - 1) \frac{3}{E} g^{-3/2} \sigma (g^{3/2}EL_g)\frac{m\dot{q}}{2T},\right) \\ \chi ^*_s q^+_\perp&= \left( \frac{g}{\chi ^*_s g}\right) ^{3/2} \left( q^+_\perp - (e^{-s}-1) \frac{2m}{E}\ddot{q}_\perp g^+\right) ,\\ \chi ^*_s \xi ^+&= \left( \frac{g}{\chi ^*_s g}\right) ^{3/2} \left( \xi ^+ -(e^{-2s} - 1)\frac{g^{3/2}}{E}\dot{g}^+ g^+\right) ,\\ \chi ^*_s g^+&= \left( \frac{g}{\chi ^*_s g}\right) ^{3/2}e^{-s} g^+. \end{aligned}$$

where $\sigma (\varphi ) = \varphi \partial _t{(g^{3/2}g^+)} - {\dot{\varphi }} g^{3/2}g^+$, and fulfils $\lim _{s\rightarrow \infty } \chi ^*_s = \chi ^*$. There are some steps that are important to highlight in this case. First, in Lemma B.2.4, we show that ${R_{GR}}$ commutes with the Chevalley–Eilenberg differential $\gamma _{GR}$²³

$$\begin{aligned} {[}{R_{GR}},\gamma _{GR}] = 0, \end{aligned}$$

by using general arguments and $\gamma _{GR} \sim \mathcal {L}_{\xi \partial _t}$, but it can also be checked through straightforward calculations. We do this as it greatly simplifies the computations since ${D_{GR}}= [{R_{GR}},{Q_{GR}}] = [{R_{GR}},\delta _{GR}]$.

Furthermore, while the computations for the action of $\chi ^*_s = e^{s\mathcal {L}_{D_{GR}}}$ on the fields $\varphi \in \{q,\xi ,g\}$ is analogous to the ones presented for the other examples, it turns out that in this case of the antifields $\varphi ^+ \in \{q^+,\xi ^+,g^+\}$ finding a recursive formula for ${D^k_{GR}}\varphi ^+$ is quite challenging. Instead, it is easier to take a slight detour: We first compute $\chi ^*_s(g^{3/2}\varphi ^+)$ through ${D^k_{GR}}(g^{3/2}\varphi ^+)$ and then use the property that $\chi ^*_s = e^{s\mathcal {L}_{{D_{GR}}}}$ is a morphism in order to recover $\chi ^*_s\varphi ^+$

$$\begin{aligned}&\chi ^*_s(g^{3/2}\varphi ^+) = \left( \chi ^*_sg\right) ^{3/2}\chi ^*_s\varphi ^+ \nonumber \\ \Leftrightarrow \quad&\chi ^*_s\varphi ^+ = \frac{\chi ^*_s(g^{3/2}\varphi ^+)}{(\chi ^*_sg)^{3/2}} = \frac{\chi ^*_s(g^{3/2}\varphi ^+)}{\left[ g + (e^{-s} - 1) \left( g-\frac{T}{E}\right) \right] ^{3/2}}. \end{aligned}$$

(24)

The limit $s\rightarrow \infty$ then reads

$$\begin{aligned} \lim _{s\rightarrow \infty } \chi ^*_s \varphi ^+ = \left( \frac{E}{T}\right) ^{3/2} \lim _{s\rightarrow \infty } \chi ^*_s (g^{3/2}\varphi ^+). \end{aligned}$$

(25)

Since $\chi ^*_sg$ is nowhere vanishing for any $s\in \mathbb {R}_{\ge 0}$, this expression is well-defined $\forall s\in \mathbb {R}_{\ge 0}$ iff $\chi ^*_s(g^{3/2}\varphi ^+)$ is well-defined $\forall s\in \mathbb {R}_{\ge 0}$ as well.

We exemplify this procedure with the computation of $\chi ^*_s g^+$. In order to see where the aforementioned problem arises, we compute

$$\begin{aligned} {D_{GR}}g^+&= \left( \delta _{GR} {R_{GR}}+ {R_{GR}}\delta _{GR}\right) g^+ = {R_{GR}}(\text {EL}_g) = \frac{\delta \text {EL}_g}{\delta g} {R_{GR}}g\\&= \left( -\frac{E}{4g^{3/2}} + \frac{3T}{4g^{5/2}}\right) \frac{-2g^{3/2}}{E} g^+ = \left( \frac{1}{2} - \frac{3}{2}\frac{T}{Eg}\right) g^+. \end{aligned}$$

One can then proceed with the calculation of ${D^k_{GR}}g^+$ for higher k’s and notice that the expressions become quite lengthy as ${D_{GR}}g = T/E-g$ (Eq. (43)). The idea to avoid this complication by considering ${D^k_{GR}}(g^{3/2} g^+)$, where a recursive formula becomes apparent. We have

$$\begin{aligned} {D_{GR}}(g^{3/2} g^+)&= \frac{3}{2} g^{1/2} {D_{GR}}g g^+ + g^{3/2} {D_{GR}}g^+ \\&= \frac{3}{2} g^{1/2}\left( \frac{T}{E}-g\right) g^+ + g^{3/2} \left( \frac{1}{2} - \frac{3}{2}\frac{T}{Eg}\right) g^+ \\&= - g^{3/2} g^+. \end{aligned}$$

It is then straightforward to see that

$$\begin{aligned} {D^k_{GR}}(g^{3/2} g^+)&= (-1)^k g^{3/2}g^+ \quad \text {for } k \ge 0,\\ \Rightarrow e^{s\mathcal {L}_{{D_{GR}}}}(g^{3/2}g^+)&= e^{-s} g^{3/2}g^+. \end{aligned}$$

Using Eq. (24) we then have

$$\begin{aligned} e^{s\mathcal {L}_{{D_{GR}}}} g^+&= \left( \frac{g}{\chi ^*_s g}\right) ^{3/2} e^{-s} g^+,\\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{{D_{GR}}}} g^+&= 0 = \chi ^* g^+. \end{aligned}$$

The computations for $q^+,\xi ^+$ are lengthier and can be found in the “Appendix B.2”.

In Lemma B.2.6, we demonstrate that $\chi ^*$ is the identity in cohomology by showing that the map

$$\begin{aligned} h_\chi \varphi ^j = \int _0^\infty e^{s\mathcal {L}_{{D_{GR}}}} \mathcal {L}_{{R_{GR}}} \varphi ^j \mathrm {d}s, \quad \quad \varphi ^j \in \mathcal {F}_{GR}^{\mathrm {lax}}, \end{aligned}$$

satisfying $\chi ^* - \mathrm {id}_{GR} = (\mathcal {L}_{{Q_{GR}}} - \mathrm {d}) h_\chi + h_\chi (\mathcal {L}_{{Q_{GR}}} - \mathrm {d})$ (cf. Lemma 3.1.3) converges to

$$\begin{aligned} h_\chi q&= h_\chi \xi = h_\chi \xi ^+ = h_\chi g^+ = 0,\\ h_\chi g&= -\frac{2g^{3/2}}{E} g^+,\\ h_\chi q^+_\parallel&= (1-\eta ^{3/2}) \left[ \xi ^+ + \frac{g^{3/2}}{E}\dot{g}^+ g^+\right] \frac{m\dot{q}}{2T} + \frac{3 \eta - 2 \sqrt{\eta } - 1}{(\sqrt{\eta }+1)^2} \frac{g^{3/2}}{E}\dot{g}^+ g^+ \frac{m\dot{q}}{2T}, \\ h_\chi q^+_\perp&= \frac{2}{T} \left( \frac{\eta }{\sqrt{\eta }+1} + 1\right) g^{3/2} g^+q^+_\perp . \end{aligned}$$

therefore proving that the two lax BV–BFV theories in question have isomorphic BV–BFV cohomologies

$$\begin{aligned} H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}_{J}) \simeq H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}_{GR}) \end{aligned}$$

and thus that they are lax BV–BFV equivalent. $\square$

In this example, we are also interested on how the composition maps $\lambda ^*$, $\chi ^*$ affect the boundary structure, namely the strict BV–BFV structure of the Jacobi theory and 1D GR. More specifically, we want to investigate how they change the kernel of the pre-boundary forms ${\check{\omega }}$ and as such the quotient $\mathcal {F}^\partial = {\check{\mathcal {F}}}^\partial /\ker {\check{\omega }}$ (cf. Eq. (12)).

In the case of $\lambda ^*$, this is trivial since it is the identity. Regarding $\chi ^*$, we argue that $\ker \chi ^* {\check{\omega }}_{GR}$ has a singular behaviour and that we cannot construct a BV–BFV theory from the data $\chi ^*\mathfrak {F}^\mathrm {lax}_{GR}{:}{=}(\mathcal {F}^\mathrm {lax}_{GR},\chi ^*\theta ^\bullet _{GR},\chi ^*L^\bullet _{GR},{Q_{GR}})$. Thus, although $\chi ^*$ is the identity in the BV–BFV cohomology $H^\bullet (\mathfrak {BV}\text {-}\mathfrak {BFV}_{GR})$, it spoils the BV–BFV structure of 1D GR.

Theorem 3.5.7

The lax BV–BFV theory $\chi ^* \mathfrak {F}^\mathrm {lax}_{GR} {:}{=}(\mathcal {F}^\mathrm {lax}_{GR},\chi ^*\theta ^\bullet _{GR},\chi ^*L^\bullet _{GR},{Q_{GR}})$ does not yield a BV–BFV theory.

Proof

Recall that pulling back $(\theta ^\bullet _{GR},L^\bullet _{GR})$ with $\phi ^*$ gives $\phi ^*\theta ^\bullet _{GR} = \theta ^\bullet _{J}$ and $\phi ^*L^\bullet _{GR} = L^\bullet _{J}$. Applying the map $\chi ^* = \psi ^*\circ \phi ^*$ to $(\theta ^\bullet _{GR},L^\bullet _{GR})$ then yields

$$\begin{aligned} \chi ^* \theta ^\bullet _{GR}&= (\psi ^* \circ \phi ^*) \theta ^\bullet _{GR} = \psi ^* \theta ^\bullet _{J} = \theta ^\bullet _{J}[\psi ^*\tilde{q},\psi ^*\tilde{q}^+,\psi ^*\tilde{\xi },\psi ^*\tilde{\xi }^+],\\ \chi ^* L^\bullet _{GR}&= (\psi ^* \circ \phi ^*) L^\bullet _{GR} = \psi ^* L^\bullet _{J} = L^\bullet _{J}[\psi ^*\tilde{q},\psi ^*\tilde{q}^+,\psi ^*\tilde{\xi },\psi ^*\tilde{\xi }^+]. \end{aligned}$$

Thus, the lax BV–BFV data of $\chi ^* \mathfrak {F}^\mathrm {lax}_{GR}$ has the same form as the lax BV–BFV data for Jacobi theory presented in Proposition/Definition 3.5.2 on the submanifold of $\mathcal {F}^\mathrm {lax}_{GR}$ with local coordinates $\{\psi ^*\tilde{q},\psi ^*\tilde{q}^+,\psi ^*\tilde{\xi },\psi ^*\tilde{\xi }^+\}$. Furthermore, applying $\chi ^*$ to Eq. (15) for the lax BV–BFV formulation of 1D GR yields

$$\begin{aligned} \iota _{{Q_{GR}}} \psi ^* \varpi ^\bullet _J&= \delta \psi ^*L^\bullet _J + \mathrm {d}\psi ^*\theta ^\bullet _J,\\ \iota _{{Q_{GR}}} \iota _{{Q_{GR}}} \psi ^*\varpi ^\bullet _J&= 2\,\mathrm {d}\psi ^* L^\bullet _J. \end{aligned}$$

This means that the theory $\chi ^* \mathfrak {F}^\mathrm {lax}_{GR}$ is just a version of Jacobi theory which is defined on a submanifold of $\mathcal {F}^\mathrm {lax}_{GR}$ with local coordinates $\{\psi ^*\tilde{q},\psi ^*\tilde{q}^+,\psi ^*\tilde{\xi },\psi ^*\tilde{\xi }^+\}$. This theory will have the same behaviour as the original Jacobi theory and as such the kernel of the pre-boundary 2-form

$$\begin{aligned} \chi ^* {\check{\omega }}_{GR}&= \int _{\partial I} \delta \chi ^* \theta ^1_{GR} \mathrm {d}t = \int _{\partial I} \delta \psi ^* \theta ^1_{J} \mathrm {d}t = \psi ^* {\check{\omega }}_{J}\\&= {\check{\omega }}_{J}[\psi ^*\tilde{q},\psi ^*\tilde{q}^+,\psi ^*\tilde{\xi },\psi ^*\tilde{\xi }^+], \end{aligned}$$

will be singular, just as the kernel of the pre-boundary 2-form ${\check{\omega }}_{J}$ of Jacobi theory [26]. As such, the data $\chi ^* \mathfrak {F}^\mathrm {lax}_{GR}$ does not yield a BV–BFV theory. $\square$

Remark 3.5.8

It is clear that the theories $\mathfrak {F}^\mathrm {lax}_{GR}$ and $\chi ^*\mathfrak {F}^\mathrm {lax}_{GR}$ are also lax BV–BFV equivalent (see Remark 2.3.4). We have thus presented two pairs of theories, $(\mathfrak {F}^\mathrm {lax}_{J}, \mathfrak {F}^\mathrm {lax}_{GR})$ and $(\mathfrak {F}^\mathrm {lax}_{GR}, \chi ^*\mathfrak {F}^\mathrm {lax}_{GR})$, which have isomorphic BV–BFV cohomologies, but differ in terms of their compatibility with the BV–BFV axioms, a behaviour which is not present in the examples of classical mechanics on a curved background and Yang–Mills theory which we considered in Sects. 3.3 and 3.4. Indeed, a remarkable feature of the classical equivalence between Jacobi theory and 1d GR is that it can actually be promoted to a quasi-isomorphism of their BV–BFV complexes (lax equivalence), which in particular implies BV equivalence in the sense of Definition 2.3.1. This is compatible with the process of removal of auxiliary fields outlined in [3]. However, the request that two lax-equivalent theories both admit a strictification in the sense of Remark 2.5.6 is a genuine refinement of the notion of BV (and lax) equivalence of field theories. Since the BV–BFV quantisation programme requires a strict theory, this obstruction marks a roadblock for nonstrictifiable lax BV–BFV theories.

Appendices

Appendix A: Lengthy calculation for Yang–Mills

1.1 A.1. Lemmata used in Theorem 3.4.4

In this Appendix we present the lemmata used in Theorem 3.4.4 and the respective detailed proofs and calculations.

Lemma A.1.1

Let $\phi : \mathcal {F}_{2YM}^{{\mathrm {lax}}}\rightarrow \mathcal {F}_{1YM}^{\mathrm {lax}}$ be defined through

$$\begin{aligned} \phi ^* A&= \tilde{A},\quad \phi ^* B= \star \tilde{F}_{\tilde{A}},\quad \phi ^* c= \tilde{c},\\ \phi ^* A^\dagger&= \tilde{A}^\dagger ,\quad \phi ^* B^\dagger = 0,\quad \phi ^* c^\dagger = \tilde{c}^\dagger . \end{aligned}$$

Its pullback map $\phi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ and maps the lax BV–BFV data of the first-order theory as

$$\begin{aligned}&\phi ^* \theta _1^\bullet = \theta _{2}^\bullet ,\quad \phi ^* L_{1}^\bullet = L_{2}^\bullet . \end{aligned}$$

Proof

The computations are in the same line as the ones presented in the example of Classical Mechanics on a curved background (cf. Lemma 3.3.5). One should keep in mind that ${Q_2}\star \tilde{F}_{\tilde{A}}= [\tilde{c},\star \tilde{F}_{\tilde{A}}]$. $\square$

Lemma A.1.2

Let $\psi : \mathcal {F}_{1YM}^{\mathrm {lax}}\rightarrow \mathcal {F}_{2YM}^{{\mathrm {lax}}}$ be the map defined through

$$\begin{aligned} \psi ^* \tilde{A}&= A,\quad \psi ^* \tilde{c}= c,\\ \psi ^* \tilde{A}^\dagger&= A^\dagger - \mathrm {d}_{A}\star B^\dagger ,\quad \psi ^* \tilde{c}^\dagger = c^\dagger - \frac{1}{2} [B^\dagger ,\star B^\dagger ]. \end{aligned}$$

Its pullback map $\psi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ and maps the lax BV–BFV data of the second-order theory as

$$\begin{aligned}&\psi ^* \theta _{2}^\bullet = \theta _1^\bullet + (\mathcal {L}_{{Q_1}} - \mathrm {d}) \beta _1^\bullet + \delta f_1^\bullet ,\quad \psi ^* L_{2}^\bullet = L_{1}^\bullet + (\mathcal {L}_{{Q_1}} - \mathrm {d}) \iota _{{Q_1}} \beta _1^\bullet + \mathrm {d}f_1^\bullet , \end{aligned}$$

where

$$\begin{aligned} \beta _1^\bullet&= \mathrm {Tr}\left[ \frac{1}{2} B^\dagger \star \delta B^\dagger + \star B^\dagger \delta A+ \star B^\dagger \delta c\right] ,\quad f_1^\bullet = \mathrm {Tr}\left[ \frac{1}{2} B^\dagger (B-\star F_{A}) \right] . \end{aligned}$$

Proof

The chain map conditions in the case of $\tilde{A}$ and $\tilde{c}$ are straightforward to check. In the case of $\tilde{A}^\dagger$, we first note that

$$\begin{aligned} {Q_1}\mathrm {d}_{A}\star B^\dagger&= -\mathrm {d}_{A}{Q_1}\star B^\dagger + [\mathrm {d}_{A}c,\star B^\dagger ] = -\mathrm {d}_{A}(\star F_{A}- B) - \mathrm {d}_{A}[c,\star B^\dagger ]+ [\mathrm {d}_{A}c,\star B^\dagger ]\\&= -\mathrm {d}_{A}(\star F_{A}- B) + [c,\mathrm {d}_{A}\star B^\dagger ], \end{aligned}$$

as such we have

$$\begin{aligned} {Q_1}\psi ^* \tilde{A}^\dagger&= {Q_1}(A^\dagger - \mathrm {d}_{A}\star B^\dagger ) = \mathrm {d}_{A}B+ [c,A^\dagger ] - {Q_1}\mathrm {d}_{A}\star B^\dagger \\&= \mathrm {d}_{A}\star F_{A}+ [c,A^\dagger ] - [c,\mathrm {d}_{A}\star B^\dagger ] = \psi ^*(\mathrm {d}_{\tilde{A}}\star \tilde{F}_{\tilde{A}}+ [\tilde{c},\tilde{A}^\dagger ]) = \psi ^*{Q_2}\tilde{A}. \end{aligned}$$

Before addressing the case of $c^\dagger$, we compute

$$\begin{aligned} {Q_1}[B^\dagger ,\star B^\dagger ] =&[F_{A},\star B^\dagger ] - \varepsilon _s[\star B,\star B^\dagger ] + [[c,B^\dagger ],\star B^\dagger ]\\&- [B^\dagger ,F_{A}] - [B^\dagger , B] + [B^\dagger ,[c,\star B^\dagger ]]. \end{aligned}$$

Using $[\alpha ,\star \beta ] = - [\beta ,\star \alpha ]$ for $\alpha ,\beta \in \Omega ^\bullet (M,\mathfrak {g})$, the graded Jacobi identity for $c,B$, $\star B$ and $[F_{A},\star B^\dagger ] = \mathrm {d}_{A}^2 \star B^\dagger$, this yields

$$\begin{aligned} {Q_1}[B^\dagger ,\star B^\dagger ] = 2 \mathrm {d}_{A}^2 \star B^\dagger + 2 [B^\dagger ,B] + [c,[B^\dagger ,\star B^\dagger ]]. \end{aligned}$$

With this in hand, we have

$$\begin{aligned} {Q_1}\psi ^* \tilde{c}^\dagger&= \mathrm {d}_{A}A^\dagger + [c,c^\dagger ] + [B^\dagger ,B] - \frac{1}{2}{Q_1}[B^\dagger ,\star B^\dagger ]\\&= \mathrm {d}_{A}(A^\dagger - \mathrm {d}_{A}\star B^\dagger ) + \left[ c,c^\dagger -\frac{1}{2}[B^\dagger ,\star B^\dagger ]\right] = \psi ^*(\mathrm {d}_{\tilde{A}}\tilde{A}^\dagger + [\tilde{c},\tilde{c}^\dagger ]) = \psi ^* {Q_2}\tilde{c}^\dagger . \end{aligned}$$

Let now $\Delta \theta _{2}^\bullet {:}{=}\psi ^* \theta _{2}^\bullet - \theta _1^\bullet$ and $\Delta L_{2}^\bullet {:}{=}\psi ^* L_{2}^\bullet - L_{1}^\bullet$. We need to check whether

$$\begin{aligned} \Delta \theta _{2}^0&= \mathcal {L}_{{Q_1}} \beta _1^0 - \mathrm {d}\beta _1^1 + \delta f_1^0,\\ \Delta \theta _{2}^1&= \mathcal {L}_{{Q_1}} \beta _1^1 - \mathrm {d}\beta _1^2 + \delta f_1^2,\\ \Delta \theta _{2}^2&= \mathcal {L}_{{Q_1}} \beta _1^2 + \delta f_1^2,\\ \Delta L_{2}^0&= \mathcal {L}_{{Q_1}} \iota _{{Q_1}} \beta _1^0 - \mathrm {d}\iota _{{Q_1}} \beta _1^1 + \mathrm {d}f_1^1. \end{aligned}$$

Note that $f_1^1 = f_1^2 = 0$. Recall that we only need to compute $\Delta L_{2}^0$, as $\Delta L_{2}^k$ for $k>0$ is determined by $\Delta \theta _{2}^\bullet$, as shown in Proposition 2.6.5.

Computation of $\Delta \theta _{2}^0$: Explicitly computing $\Delta \theta _{2}^0 = \psi ^* \theta _{2}^0 - \theta _1^0$ yields

$$\begin{aligned} \Delta \theta _{2}^0 = \mathrm {Tr}\left[ - \mathrm {d}_{A}\star B^\dagger \delta A - \frac{1}{2} [B^\dagger ,\star B^\dagger ] \delta c - B^\dagger \delta B\right] . \end{aligned}$$

Before tackling $\mathcal {L}_{{Q_1}} \beta _1^0 = \mathrm {Tr}[ \frac{1}{2} \mathcal {L}_{{Q_1}}(B^\dagger \star \delta B^\dagger )]$, we note that

$$\begin{aligned} B^\dagger \delta [c, \star B^\dagger ]&= B^\dagger [\delta c, \star B^\dagger ] - [c,\delta \star B^\dagger ] = -[B^\dagger , \star B^\dagger ] \delta c- [c,\delta \star B^\dagger ]\\&\Rightarrow B^\dagger \delta [c, \star B^\dagger ] + [c,\delta \star B^\dagger ] = -[B^\dagger , \star B^\dagger ] \delta c, \end{aligned}$$

where we ignored the term $[B^\dagger \delta c, \star B^\dagger ]$ since $\mathrm {Tr}\big [[\alpha , \beta \gamma ]\big ] = 0$ for $\alpha ,\beta ,\gamma \in \Omega ^\bullet (M,\mathfrak {g})$. As such

$$\begin{aligned} \mathcal {L}_{{Q_1}}(B^\dagger \star \delta B^\dagger )&= (F_{A}- \varepsilon _s\star B+ [c, B^\dagger ]) \star \delta B^\dagger + B^\dagger \star \delta (F_{A}- \varepsilon _s\star B+ [c, B^\dagger ])\\&= \delta B^\dagger (\star F_{A}- B) + [c, B^\dagger ] \star \delta B^\dagger + B^\dagger \delta (\star F_{A}- B) + B^\dagger \delta [c, B^\dagger ]\\&= \delta B^\dagger (\star F_{A}- B) + B^\dagger \delta (\star F_{A}- B) -[B^\dagger , \star B^\dagger ] \delta c\end{aligned}$$

Therefore, keeping in mind that $\delta F_{A}= -\mathrm {d}_{A}\delta A$,

$$\begin{aligned}&\mathcal {L}_{{Q_1}} \beta _1^0 - \mathrm {d}\beta _1^1 + \delta f_1^0 = \mathrm {Tr}\left[ \mathcal {L}_{{Q_1}} \left( \frac{1}{2} B^\dagger \star \delta B^\dagger \right) - \mathrm {d}(\star B^\dagger \delta A) + \delta \left( \frac{1}{2}B^\dagger (B- \star F_{A})\right) \right] \\&\quad = \mathrm {Tr}\left[ \frac{1}{2}\delta B^\dagger (\star F_{A}- B) + \frac{1}{2} B^\dagger \delta (\star F_{A}- B) -\frac{1}{2} [B^\dagger , \star B^\dagger ] \delta c- \mathrm {d}_{A}+ \frac{1}{2} \delta (B^\dagger (B- F_{A}))\right] \\&\quad = \mathrm {Tr}\left[ \mathrm {Tr}[\frac{1}{2} \delta B^\dagger \star F_{A}- \frac{1}{2} \delta B^\dagger B+ \frac{1}{2} B^\dagger \delta \star F_{A}- \frac{1}{2} B^\dagger \delta B- \frac{1}{2} [B^\dagger , \star B^\dagger ] \delta c\right. \\&\qquad \left. - \mathrm {d}_{A}\star B^\dagger \delta A - B^\dagger \star \delta F_{A}+ \frac{1}{2} \delta B^\dagger B- \frac{1}{2} \delta B^\dagger \star F_{A}- \frac{1}{2} B^\dagger \delta B+ \frac{1}{2} B^\dagger \star \delta F_{A}\right] \\&\quad = \mathrm {Tr}\left[ - \mathrm {d}_{A}\star B^\dagger \delta A - \frac{1}{2} [B^\dagger ,\star B^\dagger ] \delta c - B^\dagger \delta B\right] = \Delta \theta _{2}^0. \end{aligned}$$

Computation of $\Delta \theta _{2}^1$: We have $\Delta \theta _{2}^1 = \psi ^* \theta _{2}^1 - \theta _1^1$ and as such

$$\begin{aligned} \Delta \theta _{2}^1 = \mathrm {Tr}\left[ \delta A\star F_{A}- \mathrm {d}_{A}\star B^\dagger \delta c- B\delta A\right] . \end{aligned}$$

Noting the identities

$$\begin{aligned} \delta \mathrm {d}_{A}c&= - \mathrm {d}_{A}\delta c+ [\delta A,c],\\ [c,\star B^\dagger \delta A]&= - \star B^\dagger [c,\delta A] + [c,\star B^\dagger ] \delta A= \star B^\dagger [\delta A,c] + [c,\star B^\dagger ] \delta A, \end{aligned}$$

we see that

$$\begin{aligned}&\mathcal {L}_{Q_1}\beta _1^1 -\mathrm {d}\beta _1^2 + \delta f_1^1 = \mathrm {Tr}\left[ \mathcal {L}_{Q_1}(\star B^\dagger \delta A) - \mathrm {d}(\star B^\dagger \delta c)\right] \\&\quad = \mathrm {Tr}\left[ (\star F_{A}- B+ [c, \star B^\dagger ])\delta A+ \star B^\dagger \delta \mathrm {d}_{A}c- \mathrm {d}_{A}\star B^\dagger \delta c+ \star B^\dagger \mathrm {d}_{A}\delta c\right] \\&\quad = \mathrm {Tr}\left[ (\star F_{A}- B) \delta A- \mathrm {d}_{A}\star B^\dagger \delta c+ [c,\star B^\dagger ] \delta A+ \star B^\dagger [\delta A,c]\right] = \Delta \theta _{2}^1 \end{aligned}$$

Computation of $\Delta \theta _{2}^2$: $\Delta \theta _{2}^2 = \psi ^* \theta _{2}^2 - \theta _1^2$ takes the form

$$\begin{aligned} \Delta \theta _{2}^2 = (\star F_{A}- B) \delta c. \end{aligned}$$

Furthermore since $[c, \star B^\dagger \delta c] = [c, \star B^\dagger ] \delta c - \star B^\dagger [c,\delta c],$ we have

$$\begin{aligned} \mathcal {L}_{Q_1}\beta _1^2 + \delta f_1^2&= \mathcal {L}_{Q_1}(\star B^\dagger \delta c) = (\star F_{A}- B+ [c,\star B^\dagger ])\delta c + \frac{1}{2} \star B^\dagger \delta [c,c]\\&= \Delta \theta _{2}^2 + [c,\star B^\dagger ] \delta c - \star B^\dagger [c,\delta c] = \Delta \theta _{2}^2 \end{aligned}$$

Computation of $\Delta L_{2}^0$: Explicitly $\Delta L_{2}^0 = \psi ^* L_{2}^0 - L_{1}^0$ yields

$$\begin{aligned} \Delta L_{2}^0 = \mathrm {Tr}\left[ \frac{1}{2} F_{A}\star F_{A}- BF_{A}+ \frac{\varepsilon _s}{2} B\star B- \mathrm {d}_{A}\star B^\dagger \mathrm {d}_{A}c-\frac{1}{4} [B^\dagger ,\star B^\dagger ][c,c] - B^\dagger [c,B] \right] . \end{aligned}$$

Before computing $\mathcal {L}_{Q_1}\iota _{{Q_1}} \beta _1^0 - \mathrm {d}\iota _{{Q_1}} \beta _1^1 + \mathrm {d}f_1^1$ we note that

$$\begin{aligned} {[}\star B^\dagger F_{A}, c]&= \star B^\dagger [F_{A}, c] + [\star B^\dagger , c]F_{A}= \star B^\dagger [F_{A}, c] + F_{A}[c, \star B^\dagger ]\\ [c, B^\dagger B]&= - B^\dagger [c, B] + [c, B^\dagger ] B\end{aligned}$$

and

$$\begin{aligned}&[c,B^\dagger ][c,\star B^\dagger ] = [c,B^\dagger [c, \star B^\dagger ]] + B^\dagger [c,[c,\star B^\dagger ]]\\&\quad = [c,B^\dagger [c, \star B^\dagger ]] - B^\dagger [c,[\star B^\dagger ,c]] - B^\dagger [\star B^\dagger [c,c]]\\&\quad = [c,B^\dagger [c, \star B^\dagger ]] + [c,B^\dagger [\star B^\dagger ,c]] - [c,B^\dagger ][c,\star B^\dagger ] \\&\qquad + [\star B^\dagger ,B^\dagger [c,c]] - [\star B^\dagger ,B^\dagger ][c,c]\\&\quad \Rightarrow \mathrm {Tr}\bigg [[c,B^\dagger ][c,\star B^\dagger ]\bigg ] = \mathrm {Tr}\bigg [- \frac{1}{2} [B^\dagger ,\star B^\dagger ][c,c] \bigg ]. \end{aligned}$$

Since $f_1^1 = 0$ we therefore have

$$\begin{aligned}&\mathcal {L}_{Q_1}\iota _{{Q_1}} \beta _1^0 - \mathrm {d}\iota _{{Q_1}} \beta _1^1 + \mathrm {d}f_1^1 = \mathrm {Tr}\bigg [\frac{1}{2} {Q_1}B^\dagger \star {Q_1}B^\dagger - \mathrm {d}(\star B^\dagger {Q_1}A) \bigg ]\\&\quad = \mathrm {Tr}\bigg [ \frac{1}{2} (F_{A}- \varepsilon _s\star B+ [c,B^\dagger ]) (\star F_{A}- B+ [c,\star B^\dagger ]) - \mathrm {d}_{A}(\star B^\dagger \mathrm {d}_{A}c)\bigg ]\\&\quad = \mathrm {Tr}\bigg [\frac{1}{2} \Big ( F_{A}\star F_{A}- F_{A}B+ F_{A}[c,\star B^\dagger ] - \varepsilon _s\star B\star F_{A}+ \varepsilon _s\star BB- \varepsilon _s\star B[c,\star B^\dagger ]\\&\qquad + [c,B^\dagger ] \star F_{A}- [c,B^\dagger ] B+ [c,B^\dagger ] [c,\star B^\dagger ]\Big ) -\mathrm {d}_{A}\star B^\dagger \mathrm {d}_{A}c+ \star B^\dagger \mathrm {d}_{A}^2 c\bigg ]\\&= \mathrm {Tr}\bigg [ \frac{1}{2} F_{A}\star F_{A}- BF_{A}+ \frac{\varepsilon _s}{2} B-\star B- \mathrm {d}_{A}\star B^\dagger \\&\qquad + F_{A}[c,\star B^\dagger ] + \star B^\dagger [F_{A},c] - [c, B^\dagger ] B+ \frac{1}{2} [c,B^\dagger ][c,\star B^\dagger ] \bigg ]\\&= \mathrm {Tr}\bigg [\frac{1}{2} F_{A}\star F_{A}- BF_{A}+ \frac{\varepsilon _s}{2} B\star B- \mathrm {d}_{A}\star B^\dagger - [c,B^\dagger ] B- \frac{1}{4} [B^\dagger ,\star B^\dagger ][c,c]\bigg ]\\&= \Delta L_{2}^0. \end{aligned}$$

$\square$

Lemma A.1.3

The composition map $\lambda ^* = \phi ^* \circ \psi ^*: \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _2$ is the identity

$$\begin{aligned} \lambda ^* \tilde{A}&= \tilde{A},\quad \lambda ^* \tilde{c}= \tilde{c},\\ \lambda ^* \tilde{A}^\dagger&= \tilde{A}^\dagger ,\quad \lambda ^* \tilde{c}^\dagger = \tilde{c}^\dagger , \end{aligned}$$

and as such the identity in cohomology.

Proof

Keeping in mind that $\phi ^* B^\dagger =0$, this is a straightforward calculation. $\square$

Lemma A.1.4

The composition map $\chi ^* = \psi ^* \circ \phi ^*: \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1 \rightarrow \mathfrak {BV}\text {-}\mathfrak {BFV}^\bullet _1$ acts as

$$\begin{aligned} \chi ^* A&= A,\quad \chi ^* B= \star F_{A},\quad \chi ^* c= c,\\ \chi ^* A^\dagger&= A^\dagger - \mathrm {d}_{A}\star B^\dagger ,\quad \chi ^* B^\dagger = 0,\quad \chi ^* c^\dagger = \tilde{c}^\dagger - \frac{1}{2} [B^\dagger ,\star B^\dagger ], \end{aligned}$$

and is homotopic to the identity.

Proof

The explicit computation for $\chi ^*$ is again straightforward. To show that it is indeed homotopic to the identity, we choose the vector field ${R_1}\in \mathfrak {X}_{\text {evo}}(\mathcal {F}_{1YM}^{\mathrm {lax}})[-1]$ to act as

$$\begin{aligned} {R_1}A&= 0,\quad {R_1}B= \star B^\dagger ,\quad {R_1}c= 0,\\ {R_1}A^\dagger&= 0,\quad {R_1}B^\dagger = 0,\quad {R_1}c^\dagger = 0. \end{aligned}$$

We now want to compute $\chi ^*_s = e^{s\mathcal {L}_{D_1}}$, with $D_1= [{R_1},{Q_1}]$, and show that $\lim _{s\rightarrow \infty } \chi ^*_s = \chi ^*$.

Computation for $A$ and $c$:

$$\begin{aligned} D_1A&= [{Q_1},{R_1}] A= {R_1}\mathrm {d}_{A}c= 0\\ \Rightarrow e^{s\mathcal {L}_{D_1}} A&= A\\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} A&= A= \chi ^* A,\\ D_1c&= [{Q_1},{R_1}] c= \frac{1}{2} {R_1}[c,c]= 0\\ \Rightarrow e^{s\mathcal {L}_{D_1}} c&= c\\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} c&= c= \chi ^* c. \end{aligned}$$

Computation for B:

$$\begin{aligned} D_1B&= [{Q_1},{R_1}] B= {Q_1}\star B^\dagger + {R_1}[c,B] \\&= \star F_{A}- B+ [c,\star B^\dagger ] - [c,\star B^\dagger ] = \star F_{A}- B. \end{aligned}$$

Noting that $D_1A= 0$ implies $D_1F_{A}= 0$, we have

$$\begin{aligned} D_1^2 B&= - D_1B= - (\star F_{A}- B) \\ \Rightarrow D_1^k B&= (-1)^k (B- \star F_{A}) \qquad \text { for } k \ge 1\\ \Rightarrow e^{s\mathcal {L}_{D_1}} B&= B+ \sum ^\infty _{k=1} \frac{s^k}{k!} (-1)^k(B- \star F_{A})\\&= B+ (e^{-s} - 1) (B- \star F_{A})\\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} B&= \star F_{A}= \chi ^* B. \end{aligned}$$

Computation for $B^\dagger$:

$$\begin{aligned} D_1B^\dagger&= {R_1}(F_{A}- \varepsilon _s\star B+ [c,B^\dagger ])= - B^\dagger \\ \Rightarrow D_1^k B^\dagger&= (-1)^k B^\dagger \qquad \text { for } k \ge 0\\ \Rightarrow e^{s\mathcal {L}_{D_1}} B^\dagger&= e^{-s} B^\dagger \\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} B^\dagger&= 0 = \chi ^* B^\dagger . \end{aligned}$$

Computation for $A^\dagger$:

$$\begin{aligned} D_1A^\dagger&= {R_1}(\mathrm {d}_{A}B+ [c,A^\dagger ]) = - \mathrm {d}_{A}\star B^\dagger \\ \Rightarrow D_1^k A^\dagger&= - \mathrm {d}_{A}\star D_1^{k-1} B^\dagger = (-1)^k \mathrm {d}_{A}\star B^\dagger \qquad \text { for } k \ge 1\\ \Rightarrow e^{s\mathcal {L}_{D_1}} A^\dagger&= A^\dagger + (e^{-s}-1) \mathrm {d}_{A}\star B^\dagger \\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} A^\dagger&= A^\dagger - \mathrm {d}_{A}\star B^\dagger = \chi ^* A^\dagger . \end{aligned}$$

Computation for $c^\dagger$:

$$\begin{aligned} D_1c^\dagger&= {R_1}(\mathrm {d}_{A}A^\dagger + [c,c^\dagger ] + [B^\dagger ,B])= -[B^\dagger , \star B^\dagger ]\\ \Rightarrow D_1^2 c^\dagger&= -[D_1B^\dagger , \star B^\dagger ] - [B^\dagger , \star D_1B^\dagger ] = 2 [ B^\dagger , \star B^\dagger ] = - 2 D_1c^\dagger \\ \Rightarrow D_1^k c^\dagger&= -(-2)^{k-1} [ B^\dagger , \star B^\dagger ] = \frac{1}{2} (-2)^{k} [ B^\dagger , \star B^\dagger ]\qquad \text { for } k \ge 1 \\ \Rightarrow e^{s\mathcal {L}_{D_1}} c^\dagger&= c^\dagger + \frac{1}{2} (e^{-2s} - 1) [ B^\dagger , \star B^\dagger ]\\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{D_1}} c^\dagger&= c^\dagger - \frac{1}{2} [ B^\dagger , \star B^\dagger ] = \chi ^* c^\dagger .&\end{aligned}$$

Thus we have shown that $\chi ^*$ is homotopic to the identity. $\square$

Lemma A.1.5

The map $\chi ^*$ is the identity in cohomology.

Proof

We have to show that the map $h_\chi$ converges on $\mathcal {F}_{1YM}^{\mathrm {lax}}$, namely

$$\begin{aligned} h_\chi \varphi _i = \int ^\infty _0 e^{s D_1} {R_1}\varphi _i \, \mathrm {d}s < \infty \quad \forall \varphi _i \in \mathcal {F}_{1YM}^{\mathrm {lax}}. \end{aligned}$$

As ${R_1}A= {R_1}c= {R_1}A^\dagger = {R_1}B^\dagger = {R_1}c^\dagger = 0$ we have

$$\begin{aligned} h_\chi A= h_\chi A^\dagger = h_\chi B^\dagger = h_\chi c= h_\chi c^\dagger = 0. \end{aligned}$$

In the case of $B$ we compute

$$\begin{aligned} h_\chi B= \int ^\infty _0 e^{s D_1} {R_1}B\mathrm {d}s = \int ^\infty _0 e^{- s} \star B^\dagger \mathrm {d}s = \star B^\dagger . \end{aligned}$$

As such $h_\chi$ converges and $\chi ^*$ is the identity in cohomology. $\square$

Appendix B: Lengthy Calculations for the Jacobi theory/1D GR case

1.1 B.1 Preliminaries for calculations—tensor number

This appendix has two purposes. It serves as a preliminary for the computations, by presenting a straightforward way to compute the action of the Chevalley–Eilenberg differentials $\gamma _J$, $\gamma _{GR}$, and it provides an explaination for why they act as $\mathcal {L}_{\xi \partial _t}$ on the antifields and antighosts (see Remarks 3.5.3 and 3.5.5). We will be using the 1D GR theory in this discussion but all considerations hold for the Jacobi theory as well.

Let M be a manifold of arbitrary dimension and $X = X^\sigma \partial _\sigma \in \mathfrak {X}(M)$. Recall that the Lie derivative $\mathcal {L}_X$ acts on the components of a tensor field $\mathbf {A} \in \mathcal {T}^n_m(M)$ of rank-(n, m) as

$$\begin{aligned} \begin{aligned} \mathcal {L}_X \mathbf {A}^{\mu _1\dots \mu _n}_{\nu _1\dots \nu _m}&= X^\sigma \partial _\sigma \mathbf {A}^{\mu _1\dots \mu _n}_{\nu _1\dots \nu _m} - \partial _\sigma X^{\mu _1}\mathbf {A}^{\sigma \dots \mu _n}_{\nu _1\dots \nu _m} - \dots - \partial _\sigma X^{\mu _n}\mathbf {A}^{\mu _1 \dots \sigma }_{\nu _1\dots \nu _m}\\&\quad + \partial _{\nu _1} X^\sigma \mathbf {A}^{\mu _1\dots \mu _n}_{\sigma \dots \nu _m} + \dots + \partial _{\nu _n} X^\sigma \mathbf {A}^{\mu _1\dots \mu _n}_{\nu _1\dots \sigma }. \end{aligned} \end{aligned}$$

(26)

Let now $M=I \subset \mathbb {R}$ denote an interval and $X = \xi \partial _t \in \mathfrak {X}(I)[I]$ be the ghost field. In this setting, Eq. (26) is greatly simplified since $\mu _i = \nu _i = t$, where t is the coordinate on I. Let $\mathbf {A} \in \mathcal {T}^n_m(I)$ and denote its component by A. We define the tensor number as $t(A) = (m-n)$. We then have

$$\begin{aligned} \begin{aligned} \mathcal {L}_{\xi \partial _t} A&= \xi \partial _t A - n\partial _t \xi A + m\partial _t \xi A\\&=\xi \dot{A} + t(A) {\dot{\xi }} A. \end{aligned} \end{aligned}$$

As an example, we list the tensor number for the fields, ghosts, antifields and antighosts of the 1D GR theory

$$\begin{aligned} t(q)&= 0-0 = 0,\quad t(g) = 2-0 = 2,\quad t(\xi ) = 0-1 = -1, \nonumber \\ t(q^+)&= 1-0 = 1,\quad t(g^+) = 1-2 = -1,\quad t(\xi ^+) = 2-0 = 2, \end{aligned}$$

(27)

which explains why we claimed that the Chevalley–Eilenberg differential $\gamma _{GR}$ acts as $\mathcal {L}_{\xi \partial _t}$ on the antifields and antighosts in Remark 3.5.5. As such we have $\gamma _{GR} = \mathcal {L}_{\xi \partial _t}$ on all the functions on $\{q,g,q^+,g^+,\xi ^+\}$ and $\gamma _{GR} = \frac{1}{2}\mathcal {L}_{\xi \partial _t}$ on the ghost. Since the ghost is a special case, we assume that the tensor fields only depend on $\{q,g,q^+,g^+,\xi ^+\}$ for the rest of the discussion. When computing $\gamma _{GR}(\cdot )$, we then consider the parts with ghosts and without separately.

The discussion until now only holds for tensor fields that only depend on the zeroth jets of $\{q,g,q^+,g^+,\xi ^+\}$. The action of $\gamma _{GR}$ is then naturally extended to all jets since we assume that ${Q_{GR}}$, and as such $\gamma _{GR}$, is evolutionary, i.e. $[\mathcal {L}_{\gamma _{GR}},\mathrm {d}] = 0$. For example, if A only depends on 0th-jets then

$$\begin{aligned} \begin{aligned} \gamma _{GR} \dot{A}&= \partial _t \gamma _{GR} A = \partial _t [\xi \dot{A} + t(A) {\dot{\xi }} A] \\&= \xi {\ddot{A}} + [1+t(A)] {\dot{\xi }} \dot{A} + t(A) \ddot{\xi }A. \end{aligned} \end{aligned}$$

(28)

For a general tensor field $\mathcal {A}$ which depends on arbitrary jets of the fields, we have

$$\begin{aligned} \gamma _{GR} \mathcal {A} = \xi \dot{\mathcal {A}} + \sum _{n \ge 1} t_n(\mathcal {A}) \partial ^n_t \xi a_n, \end{aligned}$$

(29)

for some real scalars $t_n(\mathcal {A})$ and some functions $a_n$ that depend on the jets of $\Phi ,\Phi ^+ \in \mathcal {F}_{GR}$. In order to extend the notion of tensor number to such objects we define

Definition B.1.1

Let $\mathcal {A}$ be a tensor field that depends on arbitrary jets of $\Phi ,\Phi ^+ \in \mathcal {F}_{GR}$. The tensor number $t(\mathcal {A})$ of $\mathcal {A}$ is defined as the scalar $t_1(\mathcal {A})$ in Eq. (29).

Note that in order to compute $\gamma _{GR}$ we only have to find out what the $t_n(\mathcal {A})$ are. For most of the computations, we are only going to encounter tensor fields that depend on the zeroth jets, and they will atmost include second jets. As such we want to find a pragmatic way of computing $t(\mathcal {A})$. If necessary, we then look at higher $t_n(\mathcal {A})$, for example by following Eq. (28). We list some useful properties of $t(\cdot )$, since they immensely simplify the explicit computations of $\gamma _{GR}(\mathcal {A})$.

Proposition B.1.2

Let $\mathcal {A},\mathcal {B}$ be two tensor fields that depend on an arbitrary number of jets of $\Phi ,\Phi ^+ \in \mathcal {F}_{GR}$. The tensor number has the following properties:

• (1) $t(\mathcal {A}\mathcal {B}) = t(\mathcal {A}) + t(\mathcal {B})$,

• (2) $t(\mathcal {A}^n) = n t(\mathcal {A})$,

• (3) $t(\dot{\mathcal {A}}) = 1 + t(\mathcal {A})$.

Proof

Note that the only two terms from Eq. (29) that can contribute to these properties are the first two. Therefore we will only show the computations for two tensor fields A, B that only depend on the zeroth jets, but they extend to the general case in a straightforward way.

• (1) Let $gh(A) = a$. We compute

  $$\begin{aligned}&\gamma _{GR} (AB) = (\mathcal {L}_{\xi \partial _t} A) B + (-1)^{a} A(\mathcal {L}_{\xi \partial _t} B)\\&\quad = (\xi \dot{A} + t(A) {\dot{\xi }} A) B + (-1)^{a}A (\xi \dot{B} + t(B) {\dot{\xi }} B) \\&\quad = \xi \partial _t(AB) + [t(A)+t(B)] {\dot{\xi }}(AB). \end{aligned}$$

• (2) Using that $\gamma _{GR}$ is a derivative we see that

  $$\begin{aligned} \gamma _{GR} A^n = n A^{n-1} \gamma _{GR} A = n A^{n-1} [\xi \dot{A} + t(A){\dot{\xi }} A] = \xi \partial _t A^n + nt(A) {\dot{\xi }} A^n. \end{aligned}$$

• (3) This equality follows directly from Eq. (28). $\square$

We finish this section by presenting the action of $\gamma _{GR}$ tensor numbers for some relevant quantities

$$\begin{aligned} \begin{aligned} t(\dot{q})&= 1 + t(q) = 1,\\ t(T)&= t(\Vert \dot{q}\Vert ^2) = 2 t(\dot{q}) = 2,\\ t(u)&= t\left( \frac{\dot{q}}{\Vert q\Vert }\right) = t(\dot{q}) - t(\Vert \dot{q}\Vert ) = 0,\\ t(\text {EL}_g)&= t\left( \frac{E}{2\sqrt{g}} - \frac{T}{2g^{3/2}}\right) = t\left( \frac{1}{\sqrt{g}}\right) = -\frac{1}{2} t(g) = -1. \end{aligned} \end{aligned}$$

(30)

We exemplify this method of calculating $\gamma _{GR}(\mathcal {A})$ with the computation of $\mathcal {A} = T = \frac{m}{2}\Vert \dot{q}\Vert ^2$. Recall that $\gamma _{GR}q = \xi \dot{q}$ and as such $\gamma _{GR} \dot{q} = \xi \ddot{q} + {\dot{\xi }} \dot{q}$. $\gamma _{GR} T$ could potentially have terms proportional to $\ddot{\xi }$ since it depends on the derivative $\dot{q}$, but as there are no such terms in $\gamma _{GR} \dot{q}$ there will not be any in $\gamma _{GR}T$. As such we have

$$\begin{aligned} \gamma _{GR} T = \xi \dot{T} + t(T) {\dot{\xi }} T = \xi \dot{T} + 2 {\dot{\xi }} T. \end{aligned}$$

1.2 B.2 Lemmata used in Theorem 3.5.6

In this subsection we explicitly present the lemmata used in Theorem 3.5.6 and the detailed calculations.

Lemma B.2.1

Let $\phi : \mathcal {F}_{J}^{{\mathrm {lax}}}\rightarrow \mathcal {F}_{GR}^{\mathrm {lax}}$ be defined through

$$\begin{aligned} \phi ^*q&= \tilde{q},\quad \phi ^*\xi = \tilde{\xi },\quad \phi ^*g = \frac{T}{E},\\ \phi ^*q^+&= \tilde{q}^+,\quad \phi ^*\xi ^+ = \tilde{\xi }^+,\quad \phi ^*g^+ = 0. \end{aligned}$$

Its pullback map $\phi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ and maps the lax BV–BFV data of the first-order theory as

$$\begin{aligned}&\phi ^* \theta _{GR}^\bullet = \theta _{J}^\bullet ,\quad \phi ^* L_{GR}^\bullet = L_{J}^\bullet . \end{aligned}$$

Proof

The proof for the chain map condition $\phi ^* \circ {Q_{GR}}= {Q_{J}}\circ \phi ^*$ is a matter of straightforward computations

$$\begin{aligned} \phi ^*{Q_{GR}}q&= \phi ^*(\xi \dot{q}) = \tilde{\xi }\dot{\tilde{q}}= {Q_{J}}\tilde{q}= {Q_{J}}\phi ^* q,\\ \phi ^*{Q_{GR}}\xi&= \phi ^*(\xi {\dot{\xi }}) = \tilde{\xi }\dot{\tilde{\xi }} = {Q_{J}}\tilde{\xi }= {Q_{J}}\phi ^* \xi ,\\ \phi ^*{Q_{GR}}g&= \phi ^*(\tilde{\xi }\dot{g} + 2 \dot{\tilde{\xi }} g ) = \tilde{\xi }\frac{\dot{T}}{E} + 2 \dot{\tilde{\xi }} \frac{T}{E} = {Q_{J}}\frac{T}{E} = {Q_{J}}\phi ^* g,\\ \phi ^*{Q_{GR}}q^+&= \phi ^*\left( -\partial _t\left( \frac{m\dot{q}}{\sqrt{g}}\right) + \xi \dot{q}^+ + {\dot{\xi }} q^+\right) = - \partial _t \left( \sqrt{\frac{E}{T}} m \dot{\tilde{q}}\right) + \tilde{\xi }\dot{\tilde{q}}^+ + \dot{\tilde{\xi }} \tilde{q}^+\\&= {Q_{J}}\tilde{q}^+ = {Q_{J}}\phi ^* q^+ ,\\ \phi ^*{Q_{GR}}\xi ^+&= \phi ^*(- q^+ \cdot \dot{q} + g^+ \dot{g} +2 \dot{g}^+ g + \xi {\dot{\xi }}^+ + 2 {\dot{\xi }} \xi ^+) = - \tilde{q}^+ \cdot \dot{\tilde{q}}+ \tilde{\xi }\dot{\tilde{\xi }}^+ + 2 \dot{\tilde{\xi }} \tilde{\xi }^+\\&= {Q_{J}}\tilde{\xi }^+ = {Q_{J}}\phi ^* \xi ^+,\\ \phi ^*{Q_{GR}}g^+&= \phi ^*\left( \frac{1}{2\sqrt{g}} \left( E-\frac{T}{g}\right) + \xi \dot{g}^+ - {\dot{\xi }} g^+ \right) = 0 = {Q_{J}}0 = {Q_{J}}\phi ^*g^+. \end{aligned}$$

$\phi ^*$ is then a chain map w.r.t. $\mathcal {L}_{Q_i} - \mathrm {d}$ due to Lemma 3.1.1. Applying $\phi ^*$ to $(\theta _{GR}^\bullet ,L_{GR}^\bullet )$ gives

$$\begin{aligned} \phi ^*\theta _{GR}^0&= \phi ^*\left( q^+ \cdot \delta q + \xi ^+ \delta \xi + g^+ \delta g\right) = \tilde{q}^+ \cdot \delta \tilde{q}+ \tilde{\xi }^+ \delta \tilde{\xi }= \theta _{J}^0,\\ \phi ^*\theta _{GR}^1&= \phi ^*\left( \frac{m\dot{q}}{\sqrt{g}} \cdot \delta q + q^+\xi \delta q+ g^+\xi \delta g - (2g^+ g + \xi ^+ \xi ) \delta \xi \right) \\&= \sqrt{\frac{E}{T}} m \dot{\tilde{q}}\cdot \delta \tilde{q}+ \tilde{q}^+\tilde{\xi }\delta \tilde{q}+ - \tilde{\xi }^+ \tilde{\xi }\delta \tilde{\xi }= \theta _{J}^1,\\ \phi ^* L_{GR}^0&= \phi ^*\left( \frac{T}{\sqrt{g}} + \sqrt{g} E + q^+ \cdot \xi \dot{q} + g^+ (\xi \dot{g} + 2 g {\dot{\xi }}) + \xi ^+ \xi \dot{\xi }\right) \\&= 2\sqrt{ET} + \tilde{q}^+ \cdot \tilde{\xi }\dot{\tilde{q}}+ \tilde{\xi }^+ \tilde{\xi }\dot{\tilde{\xi }} = L_{J}^0,\\ \phi ^*L_{GR}^1&= \phi ^*\left( \left( \frac{T}{g} - E \right) \sqrt{g}\xi \right) = 0 = L_{J}^1. \end{aligned}$$

$\square$

Lemma B.2.2

Let $\psi : \mathcal {F}_{GR}^{\mathrm {lax}}\rightarrow \mathcal {F}_{J}^{{\mathrm {lax}}}$ be the map defined through

$$\begin{aligned} \psi ^*\tilde{q}&= q,\\ \psi ^*\tilde{\xi }&= \xi ,\\ \psi ^* \tilde{q}^+_\parallel&= \eta ^{3/2} \left( q^+_\parallel - \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} -\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\right) ,\\ \psi ^*\tilde{q}^+_\perp&= \eta ^{3/2} \left( q^+_\perp +\frac{2m}{E} g^+ \ddot{q}_\perp \right) ,\\ \psi ^*\tilde{\xi }^+&= \eta ^{3/2}\left( \xi ^+ + \frac{g^{3/2}}{E} \dot{g}^+ g^+\right) , \end{aligned}$$

where $\eta \, {:}{=}\, gE/T$ and the $\tilde{q}^+_\parallel ,\tilde{q}^+_\perp$ notation works as in Eq. 23. Its pullback map $\psi ^*$ is a chain map w.r.t. $(\mathcal {L}_{Q_i}-\mathrm {d})$ and maps the lax BV–BFV data of the first-order theory as

$$\begin{aligned} \psi ^* \theta _{J}^\bullet&= \theta _{GR}^\bullet + (\mathcal {L}_{{Q_{GR}}} - \mathrm {d}) \beta _{GR}^\bullet + \delta f_{GR}^\bullet ,\\ \psi ^* L_{J}^\bullet&= L_{GR}^\bullet + (\mathcal {L}_{{Q_{GR}}} - \mathrm {d}) \iota _{{Q_{GR}}} \beta _{GR}^\bullet + \mathrm {d}f_{GR}^\bullet , \end{aligned}$$

where

$$\begin{aligned} \beta _{GR}^0 =&-\frac{4g^{7/2}}{\Omega ^2}Tg^+\delta g^+ +\left( \frac{2g^2}{\Omega } + \eta ^{3/2}\frac{2\sqrt{g}}{E}\right) g^+q^+_\perp \cdot \delta q\\&+\left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \dot{g}^+ g^+ \frac{m\dot{q}}{2T} \cdot \delta q -(\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T} \cdot \delta q,\\ \beta _{GR}^1 =&\,\,\xi \beta _{GR}^0 + \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q,\\ f_{GR}^0 =&\,\,2 g^+\left( g - \frac{2g^{3/2}}{\Omega } T\right) ,\\ f_{GR}^1 =&\,\,\xi f_{GR}^0, \end{aligned}$$

with $\Omega = \sqrt{g}T + g \sqrt{TE}$.

Proof

We start with the chain map condition $\psi ^*\circ {Q_{J}}= {Q_{GR}}\circ \psi ^*$. In the case of the fields $\{\tilde{q},\tilde{\xi }\}$ we simply compute

$$\begin{aligned}&\psi ^*Q_J \tilde{q}= \psi ^*(\tilde{\xi }\dot{\tilde{q}}) = \xi \dot{q} = {Q_{GR}}q = {Q_{GR}}\psi ^* \tilde{q},\\&\psi ^*Q_J \tilde{\xi }= \psi ^*(\tilde{\xi }\dot{\tilde{\xi }}) = \xi {\dot{\xi }} = {Q_{GR}}\xi = {Q_{GR}}\psi ^* \tilde{\xi }. \end{aligned}$$

When dealing with the antifields $\{\tilde{q}^+,\tilde{\xi }^+\}$ it is useful to first show that $\psi ^*$ is a chain map w.r.t. to the Chevalley–Eilenberg differentials and then proceed to show that is also fulfils this condition w.r.t. the Koszul–Tate differentials.

In the case of the Chevalley–Eilenberg differentials $\gamma _J,\gamma _{GR}$ it is sufficient to investigate how $\psi ^*$ changes the tensorial properties of the fields, i.e. to analyse the tensor number introduced “Section B.1.” Indeed using $\psi ^* \tilde{\xi }= \xi$ we can compute

$$\begin{aligned} \psi ^*\gamma _J {\tilde{\Phi }}^+ =\psi ^*(\tilde{\xi }\dot{{\tilde{\Phi }}}^+ + t({\tilde{\Phi }}^+) \dot{\tilde{\xi }}{\tilde{\Phi }}^+) = \xi \partial _t(\psi ^* {\tilde{\Phi }}^+) + t({\tilde{\Phi }}^+) {\dot{\xi }} \psi ^* {\tilde{\Phi }}^+. \end{aligned}$$

(31)

The most general form of the other side of the chain map condition $\gamma _{GR}\psi ^*{\tilde{\Phi }}^+$ is given by

$$\begin{aligned} \gamma _{GR} (\psi ^* {\tilde{\Phi }}^+) = \xi \partial _t(\psi ^* {\tilde{\Phi }}^+) + t(\psi ^*{\tilde{\Phi }}^+) {\dot{\xi }} \psi ^* {\tilde{\Phi }}^+ + \sum _{n \ge 2} \partial ^n_t \xi a_n, \end{aligned}$$

(32)

where the field dependent coefficients $a_n$ do not vanish trivially since the expressions for $\psi ^*{\tilde{\Phi }}^+$ depend on derivative terms such as $\dot{g}, \dot{g}^+$, and $\dot{\text {EL}_g}$. In order to show that the two sides of the chain map condition given in Eqs. (31) and (32) are equal, we need prove that $\psi ^*$ preserves the tensor number $t(\cdot )$ and that the coefficients $a_n$ vanish

$$\begin{aligned}&t({\tilde{\Phi }}^+) = t(\psi ^*{\tilde{\Phi }}^+),\quad a_n = 0. \end{aligned}$$

Recalling that $t(g) = t(T) = 2$, we see that the rescaling factor $\eta ^{3/2}$ has vanishing tensor number

$$\begin{aligned} t(\eta ) = t\left( \frac{g}{T}\right) = t(g) - t(T) = 0. \end{aligned}$$

Furthermore, since $t_{n\ge 2}(g) = t_{n\ge 2}(T) = 0$, we have $\gamma _{GR}\eta = \xi {\dot{\eta }}$ and thus it can be ignored, since it neither changes $t(\cdot )$ nor $t_{n\ge 2}(\cdot )$.

We start by showing $\psi ^* \circ \gamma _J = \gamma _{GR} \circ \psi ^*$ on the antifield $\tilde{q}^+$. Using Eqs. (27), (28) and (30), it then follows that all the terms in $\psi ^*\tilde{q}^+_\parallel$ have tensor number 1

$$\begin{aligned} t(q^+_\parallel )&= t(u (u\cdot q^+)) = t(q^+) + 2t(u)= 1,\\ t\left( g^+\dot{g}\frac{m\dot{q}}{2T}\right)&= -1 + (2 + 1) + (0+1-2) = 1,\\ t\left( g^+\frac{g^{3/2}}{E}\dot{\text {EL}_g} \frac{m\dot{q}}{2T}\right)&= -1 + \frac{3}{2}\cdot 2 + (-1+1) + (0+1-2) = 1,\\ t\left( \dot{g}^+g\frac{m\dot{q}}{2T}\right)&= (-1+1) + 2 + (0+1-2) = 1,\\ t\left( \dot{g}^+\frac{g^{3/2}}{E}\text {EL}_g \frac{m\dot{q}}{2T}\right)&= (-1+1) + \frac{3}{2} \cdot 2 - 1 + (0+1-2) = 1, \end{aligned}$$

showing that $t(\psi ^* \tilde{q}^+_\parallel ) = 1 = t(\tilde{q}^+_\parallel )$. We still need to check what happens with the terms in $\gamma _{GR} \psi ^*\tilde{q}^+_\parallel$ that are proportional to $\ddot{\xi }$. Using Eq. (28), we can see that they take the form

$$\begin{aligned} \ddot{\xi }\left( -\left[ g^+(2g)-2g^+g\right] \frac{m\dot{q}}{2T}-\frac{g^{3/2}}{E}\left[ -\text {EL}_gg^+ + \text {EL}_gg^+\right] \right) = 0, \end{aligned}$$

and thus $\psi ^* \gamma _{J} \tilde{q}^+_\parallel = \gamma _{GR} \psi ^*\tilde{q}^+_\parallel$.

In the case of $\tilde{q}^+_\perp$ we have $t(q^+_\perp ) = t(q^+) + 2t(u) = t(q^+)$ since $t(u) = 0$. The same reasoning applies to $\ddot{q}_\perp$, but here we need to consider terms which are proportional to higher derivatives of the ghost since

$$\begin{aligned}&\gamma _{GR} \ddot{q} = \partial ^2_t (\gamma _{GR} q) = \partial ^2_t (\xi \dot{q}) = \xi \dddot{q} + 2 {\dot{\xi }} \ddot{q} + \ddot{\xi }\dot{q}. \end{aligned}$$

The terms proportional to $\ddot{\xi }$ in $\gamma _{GR}\psi ^* \tilde{q}_\perp$ come from $\gamma _{GR}\ddot{q}$ and $u(u\cdot \gamma _{GR}\ddot{q})$. As such they take the form

$$\begin{aligned} \ddot{\xi }\left( \dot{q} - u(\dot{q}\cdot u)\right) = \ddot{\xi }\left( \dot{q} - \dot{q} \right) = 0, \end{aligned}$$

hence showing that there are no terms proportional to $\ddot{\xi }$ in $\gamma _{GR} \psi ^*\tilde{q}^+_\perp$. Furthermore, $t(g^+\ddot{q}_\perp ) = -1 + 2 = 1$ and as such $\psi ^*\gamma _{J} \tilde{q}^+_\perp = \gamma _{GR}\psi ^*\tilde{q}^+_\perp$. In order to check the chain map condition for $\tilde{\xi }^+$ first note that

$$\begin{aligned} \gamma _{GR} \dot{g}^+ = \partial _t(\xi \dot{g}^+ - {\dot{\xi }} g^+) = \xi \ddot{g}^+ - \ddot{\xi }g^+, \end{aligned}$$

since $\gamma _{GR} g^+ = \xi \dot{g}^+ - {\dot{\xi }} g^+$. This in turn implies that

$$\begin{aligned} \gamma _{GR} (\dot{g}^+ g^+)&= \gamma _{GR} \dot{g}^+ g^+ - \dot{g}^+ \gamma _{GR} g^+ = \xi \ddot{g}^+ g^+ - {\dot{\xi }} \dot{g}^+ g^+\\&= \xi \partial _t(\dot{g}^+ g^+) - {\dot{\xi }} \dot{g}^+ g^+. \end{aligned}$$

As such $t(\dot{g}^+ g^+) = -1$ and $t(g^{3/2}\dot{g}^+ g^+) = \frac{3}{2} \cdot 2 - 1 = 2$. Furthermore, $t(\xi ^+) = 2$ then means that $t(\psi ^*\tilde{\xi }^+) = t(\tilde{\xi }^+) = 2$ and since there are no other derivative terms in $\psi ^* \tilde{\xi }^+$, we have $a_n = 0$, which completes the proof for

$$\begin{aligned} \psi ^* \circ \gamma _{J} = \gamma _{GR} \circ \psi ^*. \end{aligned}$$

We now move to the Koszul–Tate differentials. In the case of $\tilde{q}^+_\parallel$ we first note that

$$\begin{aligned} \psi ^*\delta _J \tilde{q}^+_\parallel = \psi ^*\left( \delta _J(\tilde{q}^+ \cdot \dot{\tilde{q}}) \frac{m\dot{\tilde{q}}}{2T} \right) = -\psi ^*\left( \delta ^2_J \tilde{\xi }^+ \frac{m\dot{\tilde{q}}}{2T} \right) = \psi ^*(0) = 0, \end{aligned}$$

since $\delta _J \tilde{\xi }^+ = -\tilde{q}^+\cdot \dot{\tilde{q}}$ and $\delta ^2_J = 0$. The term $\delta _{GR} \psi ^*\tilde{q}^+_\parallel$ vanishes for a similar reason

$$\begin{aligned} \delta _{GR}\psi ^*(\tilde{q}^+_\parallel )&=\delta _{GR}\left\{ \eta ^{3/2} \left( q^+_\parallel - \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} -\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\right) \right\} \\&= -\eta ^{3/2} \delta ^2_{GR}\left\{ \xi ^+ + \frac{g^{3/2}}{E} \dot{g}^+ g^+\right\} \frac{m\dot{q}}{2T} = 0, \end{aligned}$$

where we used that $\delta _{GR} \xi ^+ = -q^+\cdot \dot{q} + g^+\dot{g} + 2\dot{g}^+ g$ and $\delta _{GR}(\dot{g}^+ g^+) = \dot{\text {EL}_g}g^+ - \dot{g}^+ \text {EL}_g$.

The computations for the perpendicular part of $\tilde{q}^+$ go as follows

$$\begin{aligned}&\psi ^* \delta _J \tilde{q}^+_\perp = \psi ^*\left( - \partial _t \left( \sqrt{\frac{E}{T}} m \dot{\tilde{q}}\right) + {\tilde{u}} \partial _t \left( \sqrt{\frac{E}{T}} m \dot{\tilde{q}}\right) \cdot {\tilde{u}} \right) \\&\quad =\psi ^*\left( \frac{1}{2} \left( \frac{E}{T}\right) ^{3/2} \frac{\dot{T}}{E} m \dot{\tilde{q}}- \sqrt{\frac{E}{T}} m \ddot{\tilde{q}}+ {\tilde{u}} \sqrt{2mE} \underbrace{\dot{{\tilde{u}}} \cdot {\tilde{u}}}_{=0} \right) \\&\quad = \left( \frac{E}{ T}\right) ^{3/2} \left( \frac{\dot{T}}{2E} m \dot{q} - \frac{T}{E}m \ddot{ q} \right) , \end{aligned}$$

the other side of the equation reads

where we have used that $T = m\Vert \dot{q}\Vert /2$ and $\dot{T} = m \ddot{q} \cdot \dot{q}$. The last term can be expanded to give

$$\begin{aligned} \frac{2m}{E}\ddot{q}_\perp \text {EL}_g&= \frac{2m}{E}\ddot{q} \left( \frac{E}{2\sqrt{g}} - \frac{T}{2g^{3/2}} \right) - \frac{2m}{E} \dot{q} \frac{\ddot{q} \cdot \dot{q}}{\Vert \dot{q}\Vert ^2} \left( \frac{E}{2\sqrt{g}} - \frac{T}{2g^{3/2}} \right) \\&= \frac{m\ddot{q}}{\sqrt{g}} - \frac{mT}{Eg^{3/2}} \ddot{q} - \frac{\dot{T}}{2T\sqrt{g}} m \dot{q} + \frac{\dot{T}}{2Eg^{3/2}} m \dot{q}. \end{aligned}$$

Putting everything together results in

$$\begin{aligned} \delta _{GR} \psi ^* \tilde{q}^+_\perp = \eta ^{3/2}\left( \frac{\dot{T}}{2Eg^{3/2}} m \dot{q} - \frac{T}{Eg^{3/2}} m\ddot{q}\right) = \left( \frac{E}{T}\right) ^{3/2} \left( \frac{\dot{T}}{2E} m \dot{q} - \frac{T}{E}m \ddot{ q} \right) , \end{aligned}$$

which shows that $\psi ^* \delta _{J} \tilde{q}^+_\perp = \delta _{GR} \psi ^* \tilde{q}^+_\perp$. Finally we show that $\psi ^*$ acts as a chain map w.r.t. $\delta _{KT}$ on $\xi ^+$. We have

$$\begin{aligned} \psi ^* \delta _{J} \tilde{\xi }^+&= \psi ^*(-\tilde{q}^+ \cdot \dot{\tilde{q}}) = \psi ^*(-\tilde{q}^+) \cdot \dot{q}\\&= \eta ^{3/2} \left( -q^+ \cdot \dot{q} + \left[ g^+\dot{g} + 2\dot{g}^+ g \right] + \frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \right) ,\\ \delta _{GR} \psi ^* \tilde{\xi }^+&= \eta ^{3/2} \delta _{GR}\left( \xi ^+ + \frac{g^{3/2}}{E} \dot{g}^+ g^+\right) \\&= \eta ^{3/2} \left( -q^+\cdot \dot{q} + g^+ \dot{g} + 2 \dot{g}^+ g + \frac{g^{3/2}}{E} \dot{\text {EL}_g} g^+ - \frac{g^{3/2}}{E} \dot{g}^+ \text {EL}_g \right) \\&= \eta ^{3/2} \left( -q^+ \cdot \dot{q} + \left[ g^+\dot{g} + 2\dot{g}^+ g \right] + \frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \right) , \end{aligned}$$

finally showing that

$$\begin{aligned} \psi ^* \circ \delta _{J} = \delta _{GR} \circ \psi ^*, \end{aligned}$$

which show that $\psi ^*$ is indeed a chain map.

We now present the calculations for $(\psi ^*\theta ^\bullet _J,\psi ^*L^\bullet _J )$. Recall that $\theta ^\bullet _J = \theta ^0_J \,\mathrm {d}t + \theta ^1_J$, $L^\bullet _J = L^0_J \,\mathrm {d}t + L^1_J$ with

$$\begin{aligned} \theta ^0_J&= \tilde{q}^+ \cdot \delta \tilde{q}+ \tilde{\xi }^+ \delta \tilde{\xi },\quad \theta ^1_J = \sqrt{\frac{E}{T}} m \dot{\tilde{q}}\cdot \delta \tilde{q}+ \tilde{q}^+ \tilde{\xi }\delta \tilde{q}- \tilde{\xi }^+ \tilde{\xi }\delta \tilde{\xi },\\ L^0_J&= 2\sqrt{E T} + \tilde{q}^+ \cdot \tilde{\xi }\dot{\tilde{q}}+ \tilde{\xi }^+ \tilde{\xi }\dot{\tilde{\xi }},\quad L^1_J = 0. \end{aligned}$$

as in Proposition/Definition 3.5.2. Specifically we want compute

$$\begin{aligned} \Delta \theta ^0 \mathrm {d}t&= \mathcal {L}_{{Q_{GR}}} \beta ^0 \mathrm {d}t- \mathrm {d}\beta ^1 + \delta f^0 \mathrm {d}t,\\ \Delta \theta ^1&= \mathcal {L}_{{Q_{GR}}} \beta ^1 + \delta f^1 \mathrm {d}t,\\ \Delta L^0 \mathrm {d}t&= \mathcal {L}_{{Q_{GR}}} \iota _{{Q_{GR}}} \beta ^0 \mathrm {d}t - \mathrm {d}\iota _{{Q_{GR}}} \beta ^1 + \mathrm {d}f^1, \end{aligned}$$

where we skip $\Delta L^1$ as it is determined by $\Delta \theta ^\bullet$. We will from now on drop the label GR in order to keep the calculations cleaner, we will denote the Koszul–Tate operator by $\delta _{KT}$ to avoid confusion with the de Rham differential $\delta$. The terms on the left hand sides can be computed explicitly by using the form of $\psi ^*$. We get:

$$\begin{aligned} \Delta \theta ^0 =&\,\,(\eta ^{3/2} - 1) q^+ \cdot \delta q + (\eta ^{3/2} - 1) \xi ^+ \delta \xi - g^+ \delta g\\&- \eta ^{3/2} \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} \cdot \delta q -\eta ^{3/2}\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\cdot \delta q\\&+ \eta ^{3/2} \frac{2m}{E} g^+ \ddot{q}_\perp \cdot \delta q + \eta ^{3/2} \frac{g^{3/2}}{E} \dot{g}^+ g^+ \delta \xi ,\\ \Delta \theta ^1 =&\,\,\left[ \sqrt{\frac{E}{T}}m\dot{q} - \frac{m\dot{q}}{\sqrt{g}} \right] \cdot \delta q\\&+ (\eta ^{3/2}-1) q^+\xi \delta q - (\eta ^{3/2}-1) \xi ^+\xi \delta \xi - g^+\xi \delta g + 2g^+ g \delta \xi \\&- \eta ^{3/2} \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} \cdot \xi \delta q -\eta ^{3/2}\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\cdot \xi \delta q\\&+\eta ^{3/2}\frac{2m}{E} g^+ \xi \ddot{q}_\perp \cdot \delta q + \eta ^{3/2} \frac{g^{3/2}}{E} \dot{g}^+ g^+ \xi \delta \xi \\ =&\,\, \left[ \sqrt{\frac{E}{T}}m\dot{q} - \frac{m\dot{q}}{\sqrt{g}} \right] \cdot \delta q + 2g^+g\delta \xi - \xi \Delta \theta ^1,\\ \Delta L^0 =&\,\,2\sqrt{ET} -\frac{T}{\sqrt{g}} - \sqrt{g}E\\&+(\eta ^{3/2} - 1) q^+ \cdot \xi \dot{q} + (\eta ^{3/2} - 1) \xi ^+ \xi {\dot{\xi }} - g^+ (\xi \dot{g} + 2{\dot{\xi }} g)\\&- \eta ^{3/2} \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \xi -\eta ^{3/2}\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \xi \\&+ \eta ^{3/2} \frac{g^{3/2}}{E} \dot{g}^+ g^+ \xi {\dot{\xi }}. \end{aligned}$$

The following identities are going to be used throughout the calculations:

$$\begin{aligned} \sqrt{\frac{E}{T}} - \frac{1}{\sqrt{g}}&= \frac{2g^{3/2}}{\Omega } \text {EL}_g, \end{aligned}$$

(33)

$$\begin{aligned} 2\sqrt{ET} - \frac{T}{\sqrt{g}} - \sqrt{g}E&= - \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g^2, \end{aligned}$$

(34)

where $\Omega = \sqrt{g}T + g\sqrt{TE}$ with $t(\Omega ) = 3$. To see that the first Eq. (33) holds we compute

$$\begin{aligned} \sqrt{\frac{E}{T}} - \frac{1}{\sqrt{g}}&=\frac{1}{\sqrt{gT}} \left( \sqrt{gE}-\sqrt{T}\right) =\frac{gE-T}{\sqrt{gT}\left[ \sqrt{T} + \sqrt{gE}\right] }\\&= \frac{2g^{3/2}}{\Omega } \left( \frac{E}{2\sqrt{g}} - \frac{T}{2g^{3/2}}\right) = \frac{2g^{3/2}}{\Omega } \text {EL}_g. \end{aligned}$$

For the second one (34) we have

$$\begin{aligned} 2\sqrt{ET} - \frac{T}{\sqrt{g}} - \sqrt{g}E= -\left( g^{1/4} \sqrt{E}- \frac{\sqrt{T}}{g^{1/4}}\right) ^2. \end{aligned}$$

The term in the brackets can be changed to

$$\begin{aligned} g^{1/4} \sqrt{E} - \frac{\sqrt{T}}{g^{1/4}}&= g^{1/4}\left( \sqrt{E}- \sqrt{\frac{T}{g}}\right) = \frac{g^{1/4}}{\sqrt{E} + \sqrt{\frac{T}{g}}} \left( E-\frac{T}{g}\right) \\&= \frac{g^{5/4}\sqrt{T}}{\sqrt{g}T + g\sqrt{TE}}(2\sqrt{g} \text {EL}_g) = \frac{2g^{7/4}\sqrt{T}}{\Omega }\text {EL}_g, \end{aligned}$$

and as such

$$\begin{aligned} 2\sqrt{ET} - \frac{T}{\sqrt{g}} - \sqrt{g}E = - \frac{4g^{7/2}}{\Omega ^2}T \,\text {EL}_g^2. \end{aligned}$$

Computation of $\Delta \theta ^0$: We want to show

$$\begin{aligned} \Delta \theta ^0 \mathrm {d}t = \mathcal {L}_{Q} \beta ^0 \mathrm {d}t- \mathrm {d}\beta ^1 + \delta f^0 \mathrm {d}t. \end{aligned}$$

In order to compute $\mathcal {L}_Q\beta ^0$ we decompose the cohomological vector field as $Q = \gamma + \delta _{KT}$. Starting with $\mathcal {L}_\gamma$, we write $\beta ^0 = A_1 \delta g^+ + A_\perp \cdot \delta q + A_2 \cdot \delta q$ where

$$\begin{aligned} \begin{aligned} A_1&= -\frac{4g^{7/2}}{\Omega ^2}Tg^+,\\ A_\perp&= \left( \frac{2g^2}{\Omega } + \eta ^{3/2}\frac{2\sqrt{g}}{E}\right) g^+q^+_\perp ,\\ A_2&= \left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \dot{g}^+ g^+ \frac{m\dot{q}}{2T} -(\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T}, \end{aligned} \end{aligned}$$

(35)

with the following tensor and ghost numbers:

$$\begin{aligned} t(A_1)&= \frac{7}{2} \cdot 2 - 2\cdot 3 + 2 -1 = 2,\quad |A_1| = -1,\\ t(A_\perp )&= 2\cdot 2 - 3 - 1 + 1 = 1,\quad |A_\perp | = 1,\\ t(A_2)&= \frac{7}{2} \cdot 2 - 2\cdot 3 + 2 - 1 + 1 - 2 = 1,\quad |A_2| = -2. \end{aligned}$$

We then have

which shows

$$\begin{aligned} \mathcal {L}_\gamma \beta ^0&= \partial _t(\xi \beta ^0) - \frac{4g^{7/2}}{\Omega ^2}Tg^+ \dot{g}^+ - \left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \dot{g}^+ g^+ \delta \xi + (\eta ^{3/2}-1)\xi ^+ \delta \xi \\&=\partial _t(\xi \beta ^0) + \eta ^{3/2}\frac{g^{3/2}}{E} \dot{g}^+ g^+ \delta \xi + (\eta ^{3/2}-1)\xi ^+ \delta \xi . \end{aligned}$$

Before addressing the computation of $\mathcal {L}_{\delta _{KT}}\beta ^0$ we note that

$$\begin{aligned} \delta \text {EL}_g&= \left( -\frac{E}{4g^{3/2}} + \frac{3T}{4g^{5/2}}\right) \delta g - \frac{\delta T}{2g^{3/2}},\\ \delta \Omega&= \delta \left( \sqrt{g}T +g\sqrt{ET}\right) = \left( \frac{T}{2\sqrt{g}} + \sqrt{ET}\right) \delta g + \left( \sqrt{g} + \frac{g\sqrt{E}}{2\sqrt{T}}\right) \delta T\\&= \left( \Omega - \frac{\sqrt{g}T}{2} \right) \frac{\delta g}{g} + \left( \Omega - \frac{g\sqrt{ET}}{2}\right) \frac{\delta T}{T},\\ \Omega ^2&= gT^2 + 2 g^{3/2} T^{3/2} \sqrt{E} + g^2 ET,\\ \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g&= \frac{2g^2}{\Omega } T \left( \sqrt{\frac{E}{T}} - \frac{1}{\sqrt{g}}\right) = \frac{2}{\Omega } \left( g^2\sqrt{ET} - g^{3/2}T\right) \\&= \frac{2}{\Omega } \left( g\Omega - 2g^{3/2} T\right) = 2g- \frac{4g^{3/2}}{\Omega } T,\\ \delta \left( \frac{g^{3/2}}{\Omega }\right)&= \frac{3\sqrt{g}}{2\Omega }\delta g - \frac{g^{3/2}}{\Omega ^2}\left( \Omega - \frac{\sqrt{g}T}{2} \right) \frac{\delta g}{g} - \frac{g^{3/2}}{\Omega ^2}\left( \Omega - \frac{g\sqrt{ET}}{2}\right) \frac{\delta T}{T}\\&= \frac{1}{2}\left( \frac{\sqrt{g}}{\Omega } + \frac{gT}{\Omega ^2}\right) \delta g - \left( \frac{g^{3/2}}{T\Omega } - \frac{g^{5/2}\sqrt{E}}{2\sqrt{T}\Omega ^2} \right) \delta T. \end{aligned}$$

The calculation for the first term in $\mathcal {L}_{\delta _{KT}}\beta ^0$ goes as follows

$$\begin{aligned}&\mathcal {L}_{\delta _{KT}}\left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+ \delta g^+\right) = -\frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g \delta g^+ -\frac{4g^{7/2}}{\Omega ^2}Tg^+\delta \text {EL}_g\\&\quad = -\delta \left( \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g g^+ \right) + \delta \left( \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g\right) g^+ -\frac{4g^{7/2}}{\Omega ^2}Tg^+\delta \text {EL}_g \\&\quad =\delta \left( \frac{4g^{3/2}}{\Omega } T g^+ - 2gg^+\right) + 2\delta g g^+ - \delta \left( \frac{4g^{3/2}}{\Omega }\right) T g^+ -\frac{4g^{3/2}}{\Omega } \delta T g^+ \\&\qquad -\frac{4g^{7/2}}{\Omega ^2}Tg^+\left( -\frac{E}{4g^{3/2}} + \frac{3T}{4g^{5/2}}\right) \delta g +\frac{4g^{7/2}}{\Omega ^2}Tg^+\frac{\delta T}{2g^{3/2}}. \end{aligned}$$

Gathering everything in front of $\delta g$ we have

$$\begin{aligned}&g^+\left[ -2 +\left( \frac{2\sqrt{g}}{\Omega } + \frac{2gT}{\Omega ^2}\right) T +\frac{g^2ET-3gT^2}{\Omega ^2} \right] \delta g\\&\quad = g^+\left[ -2 +\frac{2\sqrt{g}T}{\Omega } +\frac{g^2ET-gT^2}{\Omega ^2} \right] \delta g\\&\quad = g^+\left[ -2 +\frac{2\Omega - 2g\sqrt{ET}}{\Omega } +\frac{\Omega ^2 - 2 g^{3/2}T^{3/2}\sqrt{E} - 2gT^2}{\Omega ^2} \right] \delta g\\&\quad = g^+\left[ 1 - \frac{2g\sqrt{ET}\Omega +2 g^{3/2}T^{3/2}\sqrt{E}+2gT^2}{\Omega ^2} \right] \delta g\\&\quad = g^+\left[ 1 - \frac{2g^2ET +4 g^{3/2}T^{3/2}\sqrt{E}+2gT^2}{\Omega ^2} \right] \delta g\\&\quad = g^+\left[ 1 - \frac{2\Omega ^2}{\Omega ^2} \right] \delta g = -g^+ \delta g. \end{aligned}$$

The terms in front of $\delta T$ simplify to

and as such:

$$\begin{aligned} \mathcal {L}_{\delta _{KT}}\left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+ \delta g^+\right) = \delta \left( \frac{4g^{3/2}}{\Omega } T g^+ - 2gg^+\right) + g^+ \left( \frac{2g^{3/2}}{\Omega }\delta T - \delta g\right) . \end{aligned}$$

The rest of $\mathcal {L}_{\delta _{KT}}\beta ^0$ yields

$$\begin{aligned} \mathcal {L}_{\delta _{KT}}\left( \frac{2g^2}{\Omega } g^+q^+_\perp \cdot \delta q, \right)&=\frac{2g^2}{\Omega } \left( \text {EL}_gq^+_\perp + g^+\frac{m \ddot{q}_\perp }{\sqrt{g}}\right) \cdot \delta q\\&= \sqrt{g}\left( \sqrt{\frac{E}{T}} - \frac{1}{\sqrt{g}}\right) q^+_\perp \cdot \delta q +g^+ \frac{2g^{3/2}}{\Omega } m \ddot{q}_\perp \cdot \delta q\\&= (\eta ^{1/2} - 1)q^+_\perp \cdot \delta q +g^+ \frac{2g^{3/2}}{\Omega } m \ddot{q}_\perp \cdot \delta q,\\ \mathcal {L}_{\delta _{KT}}\left( \eta ^{3/2}\frac{2\sqrt{g}}{E} g^+q^+_\perp \cdot \delta q \right)&= \eta ^{3/2}\frac{2\sqrt{g}}{E}\left[ \left( \frac{E}{2\sqrt{g}}-\frac{T}{2g^{3/2}}\right) q^+_\perp + g^+\frac{m \ddot{q}_\perp }{\sqrt{g}}\right] \cdot \delta q\\&= (\eta ^{3/2}-\eta ^{1/2}) q^+_\perp \cdot \delta q + \eta ^{3/2}\frac{2}{E} g^+ m\ddot{q}_\perp \cdot \delta q,\\ \mathcal {L}_{\delta _{KT}}\left( \frac{4g^{7/2}}{\Omega ^2}T \dot{g}^+ g^+ \frac{m\dot{q}}{2T} \cdot \delta q \right)&= \frac{4g^{7/2}}{\Omega ^2}T \left[ \dot{\text {EL}_g} g^+ - \dot{g}^+ \text {EL}_g\right] \frac{m\dot{q}}{2T} \cdot \delta q, \\ \mathcal {L}_{\delta _{KT}}\left( - \eta ^{3/2}\frac{g^{3/2}}{E}\dot{g}^+ g^+ \frac{m\dot{q}}{2T} \cdot \delta q \right)&= - \eta ^{3/2}\frac{g^{3/2}}{E}\left[ \dot{\text {EL}_g} g^+ - \dot{g}^+ \text {EL}_g\right] \frac{m\dot{q}}{2T} \cdot \delta q, \\ \mathcal {L}_{\delta _{KT}}\left( -(\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T} \cdot \delta q\right)&= (\eta ^{3/2}-1)\left[ q^+\cdot \dot{q} - g^+ \dot{g} - 2\dot{g}^+ g\right] \frac{m\dot{q}}{2T} \cdot \delta q\\&=(\eta ^{3/2}-1)q^+_\parallel \cdot \delta q -(\eta ^{3/2}-1)\left[ g^+ \dot{g} + 2\dot{g}^+ g\right] \frac{m\dot{q}}{2T} \cdot \delta q. \end{aligned}$$

Gathering everything gives

$$\begin{aligned} \mathcal {L}_Q\beta ^0 =&\,\, \partial _t(\xi \beta ^0) + \eta ^{3/2}\frac{g^{3/2}}{E}\dot{g}^+ g^+ \delta \xi + (\eta ^{3/2}-1)\xi ^+ \delta \xi \\&+\delta \left( \frac{4g^{3/2}}{\Omega } T g^+ - 2gg^+\right) + g^+ \left( \frac{2g^{3/2}}{\Omega }\delta T - \delta g\right) \\&+(\eta ^{1/2} - 1)q^+_\perp \cdot \delta q +g^+ \frac{2g^{3/2}}{\Omega } m \ddot{q}_\perp \cdot \delta q\\&+(\eta ^{3/2}-\eta ^{1/2}) q^+_\perp \cdot \delta q + \eta ^{3/2}\frac{2}{E} g^+ m\ddot{q}_\perp \cdot \delta q\\&+\left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \left[ \dot{\text {EL}_g} g^+ - \dot{g}^+ \text {EL}_g\right] \frac{m\dot{q}}{2T} \cdot \delta q \\&+(\eta ^{3/2}-1)q^+_\parallel \cdot \delta q -(\eta ^{3/2}-1)\left[ g^+ \dot{g} + 2\dot{g}^+ g\right] \frac{m\dot{q}}{2T} \cdot \delta q\\ =&\,\,\Delta \theta ^0 + \partial _t(\xi \beta ^0) -\delta f^0 + \frac{2g^{3/2}}{\Omega }g^+ \delta T +g^+ \frac{2g^{3/2}}{\Omega } m \ddot{q}_\perp \cdot \delta q\\&+\frac{4g^{7/2}}{\Omega ^2}T \left[ \dot{\text {EL}_g} g^+ - \dot{g}^+ \text {EL}_g\right] \frac{m\dot{q}}{2T} \cdot \delta q +\left[ g^+ \dot{g} + 2\dot{g}^+ g\right] \frac{m\dot{q}}{2T} \cdot \delta q, \end{aligned}$$

which in turn implies

$$\begin{aligned} \Delta \theta ^0 =&\,\,\mathcal {L}_Q \beta ^0 - \partial _t\left( \xi \beta ^0+\frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q\right) + \delta f^0 +\partial _t\left( \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q\right) \nonumber \\&- \frac{2g^{3/2}}{\Omega }g^+ \delta T -g^+ \frac{2g^{3/2}}{\Omega } m \ddot{q}_\perp \cdot \delta q\nonumber \\&-\frac{4g^{7/2}}{\Omega ^2}T \left[ \dot{\text {EL}_g} g^+ - \dot{g}^+ \text {EL}_g\right] \frac{m\dot{q}}{2T} \cdot \delta q -\left[ g^+ \dot{g} + 2\dot{g}^+ g\right] \frac{m\dot{q}}{2T} \cdot \delta q\nonumber \\ =&\,\,\mathcal {L}_Q \beta ^0 - \partial _t\beta ^1 + \delta f^0\nonumber \\&+g^+\left[ \partial _t\left( \frac{2g^{3/2}}{\Omega }\right) m\dot{q} \cdot \delta q +\frac{2g^{3/2}}{\Omega } m\ddot{q} \cdot \delta q + \frac{2g^{3/2}}{\Omega } m\dot{q} \cdot \delta \dot{q} - \frac{2g^{3/2}}{\Omega } \delta T \right. \nonumber \\&- \frac{2g^{3/2}}{\Omega } m \ddot{q}_\perp \cdot \delta q - \left. \frac{4g^{7/2}}{\Omega ^2}T \dot{\text {EL}_g}\frac{m\dot{q}}{2T} \cdot \delta q - \dot{g} \frac{m\dot{q}}{2T} \cdot \delta q \right] \nonumber \\&+\dot{g}^+\left[ \left( \frac{4g^{3/2}}{\Omega }T - 2g\right) +\frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g \right] \frac{m\dot{q}}{2T} \cdot \delta q. \end{aligned}$$

(36)

The last term vanishes due to Eq. (34). In order to show that the term proportional to $g^+$ vanishes as well we note:

$$\begin{aligned} \delta T&= m\dot{q} \cdot \delta \dot{q},\\ \ddot{q}_\perp&= \ddot{q} - \dot{q}\frac{\dot{q} \cdot \ddot{q}}{\Vert \dot{q}\Vert ^2} = \ddot{q} - \dot{q} \frac{\dot{T}}{2T},\\ \dot{\text {EL}_g}&= \left( -\frac{E}{4g^{3/2}} + \frac{3T}{4g^{5/2}}\right) \dot{g} - \frac{\dot{T}}{2g^{3/2}},\\ \partial _t\left( \frac{2g^{3/2}}{\Omega }\right)&= \left( \frac{\sqrt{g}}{\Omega } + \frac{gT}{\Omega ^2}\right) \dot{g} - \left( \frac{2g^{3/2}}{T\Omega } - \frac{g^{5/2}\sqrt{E}}{\sqrt{T}\Omega ^2} \right) \dot{T}. \end{aligned}$$

Then

Taking this into account and introducing $\mathrm {d}t$ in v (36) yields

$$\begin{aligned} \Delta \theta ^0 \mathrm {d}t = \mathcal {L}_Q \beta ^0 \mathrm {d}t - \partial _t\beta ^1 \mathrm {d}t + \delta f^0 \mathrm {d}t = \mathcal {L}_Q \beta ^0 \mathrm {d}t - \mathrm {d}\beta ^1 + \delta f^0 \mathrm {d}t, \end{aligned}$$

(37)

since $|\beta ^1|=0$.

Computation of $\Delta \theta ^1$: We want to compute

$$\begin{aligned} \Delta \theta ^1 = \mathcal {L}_Q \beta ^1 + \delta f^1. \end{aligned}$$

First note that Eq. (37) implies $\mathcal {L}_Q \beta ^0 = \Delta \theta ^0 + \partial _t \beta ^1- \delta f^0$, which we use to compute $\mathcal {L}_Q \beta ^1$. Since

$$\begin{aligned} \beta ^1 = \xi \beta ^0 + \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q, \end{aligned}$$

we have

$$\begin{aligned} \mathcal {L}_Q \beta ^1 = \xi {\dot{\xi }} \beta ^0 - \xi \left( \Delta \theta ^0 + \partial _t \beta ^1 - \delta f^0\right) + \mathcal {L}_Q\left( \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q\right) . \end{aligned}$$

Keeping in mind that

$$\begin{aligned} t\left( \frac{2g^{3/2}}{\Omega }g^+m\dot{q}\right) = \frac{3}{2}\cdot 2 - 3 -1 +1 = 0, \end{aligned}$$

the last term reads

$$\begin{aligned} \mathcal {L}_\gamma \left( \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q\right) =&\,\, \xi \partial _t\left( \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \right) \cdot \delta q + \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta (\xi \dot{q})\\ =&\,\,\xi \partial _t\left( \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \right) \cdot \delta q + \frac{4g^{3/2}}{\Omega }g^+ T \delta \xi \\&+ \xi \frac{2g^{3/2}}{\Omega }g^+ m\dot{q} \cdot \delta \dot{q},\\ \mathcal {L}_{\delta _{KT}} \left( \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \delta q\right) =&\,\,\frac{2g^{3/2}}{\Omega }\text {EL}_gm\dot{q} \cdot \delta q =\left[ \sqrt{\frac{E}{T}} - \frac{1}{\sqrt{g}}\right] m\dot{q} \cdot \delta q, \end{aligned}$$

which results in

Showing $\Delta \theta ^1 = \mathcal {L}_Q \beta ^1 + \delta f^1$ as desired.

Computation of $\Delta L^0$: We want to show

$$\begin{aligned} \Delta L^0 = \mathcal {L}_Q \iota _Q \beta ^0 - \mathrm {d}\iota _Q \beta ^1 + \mathrm {d}f^1.\end{aligned}$$

Note that $\beta ^0$ is of the form $\beta ^0 = a_i \delta b_i$, where we sum over i. We have

$$\begin{aligned} \mathcal {L}_Q \iota _Q(a_i \delta b_i)&= \mathcal {L}_Q (a_i Q b_i) = Q a_i Qb_i\\&= \gamma a_i \gamma b_i + \delta _{KT} a_i \gamma b_i + \gamma a_i \delta _{KT} b_i + \delta _{KT} a_i \delta _{KT} b_i. \end{aligned}$$

Starting with the term $A_1 \delta g^+$ (see Eq. (35)) we have

$$\begin{aligned} \gamma \left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+\right) \gamma g^+&= \left[ \xi \partial _t\left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+\right) - 2 {\dot{\xi }} \frac{4g^{7/2}}{\Omega ^2}Tg^+\right] \cdot \left[ \xi \dot{g}^+ - {\dot{\xi }} g^+\right] \\&= \xi \frac{4g^{7/2}}{\Omega ^2}T \dot{g}^+ {\dot{\xi }} g^+ -2 {\dot{\xi }} \frac{4g^{7/2}}{\Omega ^2}Tg^+ \xi \dot{g}^+\\&= \frac{4g^{7/2}}{\Omega ^2}T \dot{g}^+ g^+ \xi {\dot{\xi }},\\ \delta _{KT} \left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+\right) \gamma g^+&= -\frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g (\xi \dot{g}^+ - {\dot{\xi }} g^+),\\ \gamma \left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+\right) \delta _{KT} g^+&= \left[ \xi \partial _t\left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+\right) - 2 {\dot{\xi }} \frac{4g^{7/2}}{\Omega ^2}Tg^+\right] \text {EL}_g,\\ \delta _{KT} \left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+\right) \delta _{KT} g^+&= -\frac{4g^{7/2}}{\Omega ^2}T EL^0_g = 2\sqrt{ET} - \frac{T}{\sqrt{g}} - \sqrt{g}, \end{aligned}$$

where we used Eq. (34). As such

$$\begin{aligned}&\mathcal {L}_Q \iota _Q \left( -\frac{4g^{7/2}}{\Omega ^2}Tg^+ \delta g^+\right) \\&\quad = \frac{4g^{7/2}}{\Omega ^2}T \dot{g}^+ g^+ \xi {\dot{\xi }} - \partial _t \left( \xi \frac{4g^{7/2}}{\Omega ^2}T g^+\right) \text {EL}_g - \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g \xi \dot{g}^+ +2\sqrt{ET} - \frac{T}{\sqrt{g}} - \sqrt{g}\\&\quad =\frac{4g^{7/2}}{\Omega ^2}T \dot{g}^+ g^+ \xi {\dot{\xi }} - \partial _t \left( \xi \frac{4g^{7/2}}{\Omega ^2}T g^+ \text {EL}_g\right) +\xi \frac{4g^{7/2}}{\Omega ^2}T g^+ \dot{\text {EL}_g} - \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g \xi \dot{g}^+\\&\qquad +2\sqrt{ET} - \frac{T}{\sqrt{g}} - \sqrt{g}. \end{aligned}$$

For the term $A_\perp \cdot \delta q$ we have

$$\begin{aligned} \mathcal {L}_Q \iota _Q (A_\perp \delta q)&= QA_\perp Qq = \gamma A_\perp \cdot \xi \dot{q} + \delta _{KT}(A_\perp ) \cdot \xi \dot{q} \\&= {\dot{\xi }} A_\perp \cdot \xi \dot{q} + \delta _{KT}(A_\perp \cdot \xi \dot{q}) = 0. \end{aligned}$$

For the computation w.r.t. the term

$$\begin{aligned} \mathcal {L}_Q\iota _Q \left[ \left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \frac{m\dot{q}}{2T} \dot{g}^+g^+\cdot \delta q\right] \end{aligned}$$

first define

$$\begin{aligned}B {:}{=}\left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \frac{m\dot{q}}{2T},\end{aligned}$$

with $t(B) = \frac{7}{2}\cdot 2- 2\cdot 3 + 2 + 1 - 2 = 2$ and note

$$\begin{aligned} t(B\dot{g}^+ g^+) =2 + 1 + t(g^+) + t(g^+) = 1. \end{aligned}$$

As such

$$\begin{aligned}&\mathcal {L}_Q\iota _Q \left[ B\dot{g}^+ g^+ \cdot \delta q \right] = Q\left[ B\dot{g}^+ g^+\right] \xi \dot{q}\\&\quad = \partial _t\left[ \xi B\dot{g}^+ g^+\right] \cdot \xi \dot{q} + B \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \cdot \xi \dot{q}\\&\quad = {\dot{\xi }} B\dot{g}^+ g^+ \cdot \xi \dot{q} + B \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \cdot \xi \dot{q}\\&\quad = \left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \dot{g}^+ g^+ {\dot{\xi }} \xi + \left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \xi . \end{aligned}$$

For the last term in $\beta ^0$

$$\begin{aligned} -(\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T} \cdot \delta q, \end{aligned}$$

we have

$$\begin{aligned}&\mathcal {L}_Q \iota _Q \left[ -(\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T} \cdot \delta q\right] = -\gamma \left[ (\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T}\right] \cdot \xi \dot{q}\\&\qquad - (\eta ^{3/2}-1)\left[ - q^+\cdot \dot{q} + g^+\dot{g} + 2\dot{g}^+ g\right] \frac{m\dot{q}}{2T} \cdot \xi \dot{q}\\&\quad =(\eta ^{3/2}-1)\xi ^+ \xi {\dot{\xi }} - (\eta ^{3/2}-1)\left[ - q^+\cdot \dot{q} + g^+\dot{g} + 2\dot{g}^+ g\right] \xi . \end{aligned}$$

All in all the expression for $\mathcal {L}_Q \iota _Q \beta ^0$ is

(38)

where we used that

$$\begin{aligned} g^+ \dot{g} \xi + 2\dot{g}^+ g \xi&=g^+ \dot{g} \xi + \partial _t\left( 2g^+g\xi \right) - 2 g^+ \dot{g} \xi - 2 g^+ \dot{g} \xi \\&= \partial _t\left( 2g^+g\xi \right) - 2 g^+ \dot{g} \xi - g^+ \dot{g} \xi . \end{aligned}$$

Recall that we are want to show $\Delta L^0 \,\mathrm {d}t = \mathcal {L}_Q \iota _Q \beta ^0 \,\mathrm {d}t- \mathrm {d}\iota _Q \beta ^1 + \mathrm {d} f^1$. The last two terms read

$$\begin{aligned} \mathrm {d}\iota _Q \beta ^1 - \mathrm {d} f^1 = \partial _t\left[ \xi \iota _Q \beta ^0 + \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \iota _Q \delta q - 2 \xi g^+\left( g - \frac{2g^{3/2}}{\Omega } T\right) \right] \mathrm {d}t. \end{aligned}$$

which can be simplified by noting that

$$\begin{aligned} \xi \iota _Q \beta ^0 =&-\xi \frac{4g^{7/2}}{\Omega ^2}Tg^+ Qg^+ +\xi \left( \frac{2g^2}{\Omega } + \eta ^{3/2}\frac{2\sqrt{g}}{E}\right) g^+q^+_\perp \cdot Q q, \nonumber \\&+\xi \left( \frac{4g^{7/2}}{\Omega ^2}T - \eta ^{3/2}\frac{g^{3/2}}{E}\right) \dot{g}^+ g^+ \frac{m\dot{q}}{2T} \cdot Q q -\xi (\eta ^{3/2}-1)\xi ^+ \frac{m\dot{q}}{2T} \cdot Q q\nonumber \\ =&\,\,\frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g g^+ \xi , \end{aligned}$$

(39)

and

$$\begin{aligned} \frac{2g^{3/2}}{\Omega }g^+m\dot{q} \cdot \iota _Q \delta q = \frac{4g^{3/2}}{\Omega }T g^+ \xi . \end{aligned}$$

(40)

The resulting expression for $\mathrm {d}\iota _Q \beta ^1 - \mathrm {d} f^1$ is then

$$\begin{aligned} \mathrm {d}\iota _Q \beta ^1 - \mathrm {d} f^1&= \partial _t\left[ \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g g^+ \xi + \frac{4g^{3/2}}{\Omega }T g^+ \xi - 2 \xi g^+\left( g - \frac{2g^{3/2}}{\Omega } T\right) \right] \mathrm {d}t\\&= \partial _t\left( 2g^+g\xi + \frac{4g^{7/2}}{\Omega ^2}T \text {EL}_g g^+ \xi \right) \mathrm {d}t, \end{aligned}$$

which is exactly the total derivative in Eq. (38). Hence

$$\begin{aligned} \mathcal {L}_Q \iota _Q \beta ^0 \mathrm {d}t&= \Delta L^0 \mathrm {d}t +\mathrm {d}\iota _Q \beta ^1 - \mathrm {d} f^1 \\ \Rightarrow \Delta L^0 \mathrm {d}t&= \mathcal {L}_Q \iota _Q \beta ^0 \mathrm {d}t -\mathrm {d}\iota _Q \beta ^1 + \mathrm {d} f^1. \end{aligned}$$

finishing the proof. $\square$

Lemma B.2.3

The composition map $\lambda ^*$ is the identity

$$\begin{aligned} \lambda ^*\tilde{q}&= \tilde{q},\quad \lambda ^*\tilde{\xi }= \tilde{\xi },\\ \lambda ^*\tilde{q}^+&= \tilde{q}^+,\quad \lambda ^*\tilde{\xi }^+ = \tilde{\xi }^+. \end{aligned}$$

and as such the identity in cohomology.

Proof

On the matter and ghost fields $\{\tilde{q}, \tilde{\xi }\}$ this is trivial, since at this level both $\psi ^*$ and $\phi ^*$ simply interchange $\tilde{q}$ with q and $\tilde{\xi }$ with $\xi$

$$\begin{aligned} \lambda ^*\tilde{q}&= (\phi ^* \circ \psi ^*) \tilde{q}= \phi ^* q = \tilde{q},\\ \lambda ^*\tilde{\xi }&= (\phi ^* \circ \psi ^*) \tilde{\xi }= \phi ^* \xi = \tilde{\xi }. \end{aligned}$$

In order to compute the action of $\lambda ^*$ on the antifield and antighost, first recall that $\phi ^* g^+ = 0$ and note that $\phi ^*g = T/ E$ implies $\phi ^*(\eta ^{3/2}) = (\phi ^*(g)E/T)^{3/2} = 1$. We then have

$$\begin{aligned} \lambda ^* \tilde{q}^+_\parallel&= (\phi ^* \circ \psi ^*) \tilde{q}^+_\parallel \\&= \phi ^*(\eta ^{3/2}) \phi ^*\left( q^+_\parallel - \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} -\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\right) \\&= \phi ^* q^+_\parallel = \tilde{q}^+_\parallel ,\\ \lambda ^* \tilde{q}^+_\perp&= (\phi ^* \circ \psi ^*) \tilde{q}^+_\perp = \phi ^*(\eta ^{3/2}) \phi ^*\left( q^+_\perp +\frac{2m}{E} g^+ \ddot{q}_\perp \right) = \phi ^*q^+_\perp = \tilde{q}^+_\perp ,\\ \lambda ^* \tilde{\xi }^+&= (\phi ^* \circ \psi ^*) \tilde{\xi }^+ = \phi ^*(\eta ^{3/2})\phi ^*\left( \xi ^+ + \frac{g^{3/2}}{E} \dot{g}^+ g^+\right) = \phi ^* \xi ^+ = \tilde{\xi }^+, \end{aligned}$$

thus showing $\lambda = \mathrm {id}_J$. $\square$

We now prove that ${R_{GR}}$ commutes with the Chevalley–Eilenberg differential $\gamma _{GR}$ if ${R_{GR}}\xi = 0$. Effectively, this means that we can ignore the Chevalley–Eilenberg part of ${Q_{GR}}$ in ${D_{GR}}= [{Q_{GR}},{R_{GR}}]$ and only have to regard the Koszul–Tate differential when explicitly computing the action of ${D_{GR}}$. Recall that the Chevalley–Eilenberg differential acts as $\gamma _{GR} = \mathcal {L}_{\xi \partial _t}$ on $\{q,g,q^+,g^+,\xi ^+\}$ and as $\gamma _{GR} = \frac{1}{2}\mathcal {L}_{\xi \partial _t}$ on $\{\xi \}$.

Lemma B.2.4

Let ${R_{GR}}\in \mathfrak {X}_{\mathrm {evo}}(\mathcal {F}_{GR})$ be an evolutionary vector field on $\mathcal {F}_{GR}$ with the following properties

• ${R_{GR}}$ vanishes on $\mathfrak {X}[1](I)$,

• ${R_{GR}}$ preserves the tensor rank on I,

and let $\gamma _{GR}$ be the Chevalley–Eilenberg differential of the 1D GR theory. Then $[{R_{GR}},\gamma _{GR}] = 0$.

Proof

Recall that all the fields we are considering are components of tensor fields over I. In particular, note that the property that ${R_{GR}}$ vanishes on $\mathfrak {X}[1](I)$ implies ${R_{GR}}\xi = 0$, since $\xi \partial _t \in \mathfrak {X}[1](I)$. As $\gamma _{GR} \sim \mathcal {L}_{\xi \partial _t}$, it suffices to show that $[{R_{GR}},\mathcal {L}_{\xi \partial _t}] = 0$ on functions, 1-forms and vector fields over I. We assume that all objects have an internal grading throughout the proof in order to account for the ghost number.

We will first show $[{R_{GR}},\gamma _{GR}]=0$ for functions and 1-forms. Since $\mathcal {L}_{\xi \partial _t} = [\iota _{\xi \partial _t}, \mathrm {d}]$ on $\Omega ^\bullet (I)$ we have

$$\begin{aligned} {[}{R_{GR}},\mathcal {L}_{\xi \partial _t}]&= [{R_{GR}},[\iota _{\xi \partial _t}, \mathrm {d}]] = [\mathrm {d}, [{R_{GR}},\iota _{\xi \partial _t}]] + [\iota _{\xi \partial _t}, [\mathrm {d},{R_{GR}}]] \\&= [\mathrm {d}, [{R_{GR}},\iota _{\xi \partial _t}]], \end{aligned}$$

where we used that ${R_{GR}}$ is evolutionary. As such it is sufficient to show that $\Omega ^\bullet (I) \subset \ker [{R_{GR}},\iota _{\xi \partial _t}]$. By definition all functions on I are in the kernel of $\iota _{\xi \partial t}$: $C^\infty (I) = \Omega ^0(I) \subset \ker \iota _{\xi \partial _t}$. Let $f\in \Omega ^0(I)$. Since we assume that ${R_{GR}}$ preserves the tensor rank we also have ${R_{GR}}f\in \Omega ^0(I)$, then

$$\begin{aligned}&{R_{GR}}\,\iota _{\xi \partial _t} f = 0,\quad \iota _{\xi \partial _t} {R_{GR}}f = 0, \end{aligned}$$

and as such $[{R_{GR}},\iota _{\xi \partial _t}]f = 0$. Let now $\varpi = f \mathrm {d}t \in \Omega ^1(I)$ be a 1-form. Taking into account that $|\mathrm {d}t| = 1$, we have

$$\begin{aligned} {R_{GR}}\,\iota _{\xi \partial _t} \varpi = {R_{GR}}\,\iota _{\xi \partial _t} (f \mathrm {d}t)&= {R_{GR}}(f \mathrm {d}t (\xi \partial _t)) = -{R_{GR}}(f\xi ) = - {R_{GR}}f \xi ,\\ \iota _{\xi \partial _t} {R_{GR}}\varpi&= \iota _{\xi \partial _t} {R_{GR}}(f \mathrm {d}t) = \iota _{\xi \partial _t} [{R_{GR}}f \mathrm {d}t] \\&= {R_{GR}}f \mathrm {d}t (\xi \partial _t) = - {R_{GR}}f \xi ,\\ \Rightarrow [{R_{GR}},\iota _{\xi \partial _t}] \varpi&= {R_{GR}}\iota _{\xi \partial _t}\varpi - \iota _{\xi \partial _t} {R_{GR}}\varpi = -{R_{GR}}f \xi + {R_{GR}}f \xi = 0, \end{aligned}$$

thus showing that $\Omega ^0(I) \times \Omega ^1(I) \subset \ker [{R_{GR}},\iota _{\xi \partial _t}]$. This implies

$$\begin{aligned}{}[\mathrm {d},[{R_{GR}},\iota _{\xi \partial _t}]] f = 0,\quad [\mathrm {d} ,[{R_{GR}},\iota _{\xi \partial _t}]] \varpi = \mathrm {d} [{R_{GR}},\iota _{\xi \partial _t}] \varpi = 0, \end{aligned}$$

where we used $f \in \Omega ^0(I) \subset \ker [{R_{GR}},\iota _{\xi \partial _t}]$, $\mathrm {d}f \in \Omega ^1(I) \subset \ker [{R_{GR}},\iota _{\xi \partial _t}]$ and $\mathrm {d}\varpi = 0$, since $\Omega ^1(I)=\Omega ^\text {top}(I)$.

Consider now a vector field $X = f\partial _t \in \mathfrak {X}(I)$ of degree n. In this case we have:

$$\begin{aligned} {R_{GR}}\mathcal {L}_{\xi \partial _t} X&={R_{GR}}[\xi \partial _t,f \partial _t] ={R_{GR}}(\xi \dot{f} - (-1)^nf{\dot{\xi }}) \partial _t\\&=-(\xi {R_{GR}}\dot{f} + (-1)^n {R_{GR}}f {\dot{\xi }}) \partial _t,\\ \mathcal {L}_{\xi \partial _t} {R_{GR}}X&= [\xi \partial _t,{R_{GR}}f \partial _t] = (\xi {R_{GR}}\dot{f} \partial _t - (-1)^{n-1} {R_{GR}}f {\dot{\xi }}) \partial _t,\\ \Rightarrow [{R_{GR}},\mathcal {L}_{\xi \partial _t}]X&= {R_{GR}}\mathcal {L}_{\xi \partial _t}X + \mathcal {L}_{\xi \partial _t}{R_{GR}}X = 0, \end{aligned}$$

where we used $\partial _t({R_{GR}}f) = {R_{GR}}\dot{f}$. Since $[{R_{GR}},\mathcal {L}_{\xi \partial _t}] = 0$ on functions, 1-forms and vector fields it holds for all tensors. As such we have $[{R_{GR}},\gamma _{GR}] = 0$ on $\mathcal {F}_{GR}$. $\square$

A direct implication of Lemma B.2.4 is that the vector field ${D_{GR}}$ reduces to

$$\begin{aligned} {D_{GR}}= [{Q_{GR}},{R_{GR}}] = [\delta _{GR},{R_{GR}}]. \end{aligned}$$

Lemma B.2.5

The composition map $\chi ^*$ acts as

$$\begin{aligned} \begin{aligned} \chi ^*q&= q ,\\ \chi ^*\xi&= \xi ,\\ \chi ^*g&= \frac{T}{E},\\ \chi ^* q^+_\parallel&= \eta ^{3/2} \left( q^+_\parallel - \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} -\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\right) , \\ \chi ^* q^+_\perp&= \eta ^{3/2} \left( q^+_\perp +\frac{2m}{E} g^+ \ddot{q}_\perp \right) ,\\ \chi ^* \xi ^+&= \eta ^{3/2}\left( \xi ^+ + \frac{g^{3/2}}{E} \dot{g}^+ g^+\right) ,\\ \chi ^* g^+&= 0. \end{aligned} \end{aligned}$$

(41a)

and is homotopic to the identity.

Proof

Recall that $\chi ^* = \psi ^* \circ \phi ^*$ and let $\varphi _i \in \{q,q^+,\xi ,\xi ^+\}$. In this case we have $\phi ^*\varphi _i = {\tilde{\varphi }}_i$ and as such $\chi ^* \varphi _i = \psi ^* {\tilde{\varphi }}_i$, which reproduces the expressions above due to the explicit form of $\psi ^*$. For $\{g,g^+\}$ we compute

$$\begin{aligned} \chi ^* g&= (\psi ^* \circ \phi ^*) g = \psi ^*\left( \frac{T}{E}\right) = \frac{T}{E},\\ \chi ^* g^+&= (\psi ^* \circ \phi ^*) g^+ = \psi ^*(0) = 0. \end{aligned}$$

In order to show that $\chi ^*$ is homotopic to the identity we choose the vector field ${R_{GR}}$ to act as

$$\begin{aligned} {R_{GR}}q&= 0,\quad {R_{GR}}\xi = 0,\quad {R_{GR}}g= \frac{-2g^{3/2}}{E} g^+,\nonumber \\ {R_{GR}}q^+_\parallel&= -\frac{3\sqrt{g}}{E} \text {EL}_g \xi ^+ \frac{m\dot{q}}{2T},\quad {R_{GR}}\xi ^+ =0,\quad {R_{GR}}g^+ = 0,\nonumber \\ {R_{GR}}q^+_\perp&= \frac{3\sqrt{g}}{E} g^+ q^+_\perp , \end{aligned}$$

(42)

Since ${R_{GR}}\xi = 0$ we can use Lemma B.2.4. We start with the computation for q and $\xi$. Recalling that they are both in the kernel of $\delta _{GR}$ (cf. Remark 3.5.5) we have

$$\begin{aligned} {D_{GR}}q&= (\delta _{GR} {R_{GR}}+ {R_{GR}}\delta _{GR}) q = R(\xi \dot{q}) = 0,\\&\Rightarrow e^{s\mathcal {L}_{{D_{GR}}}} q = q,\\&\Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{{D_{GR}}}} q = q = \chi ^* q, \\ {D_{GR}}\xi&= (\delta _{GR} {R_{GR}}+ {R_{GR}}\delta _{GR}) \xi = 0,\\&\Rightarrow e^{s\mathcal {L}_{{D_{GR}}}} \xi = \xi ,\\&\Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{{D_{GR}}}} \xi = \xi = \chi ^* \xi . \end{aligned}$$

Keeping in mind that $\delta _{GR}g = 0$ and

$$\begin{aligned} \delta _{GR} g^+ = \text {EL}_g = \frac{E}{2\sqrt{g}}-\frac{T}{2g^{3/2}}, \end{aligned}$$

(cf. Remark 3.5.5), we compute ${D_{GR}}g$ to be

$$\begin{aligned} {D_{GR}}g&= \left( \delta _{GR} {R_{GR}}+ {R_{GR}}\delta _{GR}\right) g = \delta _{GR} \left( \frac{-2g^{3/2}}{E}g^+\right) \nonumber \\&= \frac{-2g^{3/2}}{E} \left( \frac{E}{2\sqrt{g}}-\frac{T}{2g^{3/2}} \right) = -\left( g - \frac{T}{E}\right) . \end{aligned}$$

(43)

We can then show

$$\begin{aligned} {D^k_{GR}}g = (-1)^k \left( g-\frac{T}{E}\right) \quad \text {for } k \ge 1, \end{aligned}$$

(44)

using induction. As we have computed, this holds for $k=1$. Assuming that Eq. (44) holds for an arbitrary k, we then have the following for $k+1$

$$\begin{aligned} {D_{GR}}^{k+1} g = (-1)^k \left( {D_{GR}}g - {D_{GR}}\frac{T}{E}\right) = (-1)^{k+1}\left( g-\frac{T}{E}\right) , \end{aligned}$$

where we used ${D_{GR}}T = m\dot{q} {D_{GR}}\dot{q} = m\dot{q} \partial _t({D_{GR}}q) = 0$. This results in

$$\begin{aligned} e^{s\mathcal {L}_{{D_{GR}}}} g&= g + \sum _{k \ge 1} \frac{s^k}{k!} (-1)^k \left( g-\frac{T}{E}\right) = g + (e^{-s} - 1) \left( g-\frac{T}{E}\right) \\ \Rightarrow \lim _{s\rightarrow \infty } e^{s\mathcal {L}_{{D_{GR}}}} g&= \frac{T}{E} = \chi ^* g. \end{aligned}$$

We now compute $\chi ^*_s\xi ^+$ and its $s\rightarrow \infty$ limit, which follows the same strategy as the analogous computations for $g^+$ presented in the main proof. We start with ${D_{GR}}\xi ^+$

$$\begin{aligned} {D_{GR}}\xi ^+&= {R_{GR}}\delta _{GR} \xi ^+ = {R_{GR}}\left( -q^+\cdot \dot{q} + g^+ \dot{g} + 2 \dot{g}^+ g \right) \\&= -{R_{GR}}(q^+\cdot \dot{q}) - g^+\partial _t\left( \frac{-2g^{3/2}}{E}g^+\right) -\dot{g}^+ 2\frac{(-2)g^{3/2}}{E}g^+\\&= -{R_{GR}}(q^+\cdot \dot{q}) + \frac{2g^{3/2}}{E} g^+ \dot{g}^+ +\frac{4g^{3/2}}{E} \dot{g}^+g^+ \\&= -{R_{GR}}(q^+\cdot \dot{q}) + \frac{2g^{3/2}}{E} \dot{g}^+ g^+, \end{aligned}$$

where we used that $g^+ g^+ = 0$ in the second line and $g^+ \dot{g}^+ = -\dot{g}^+ g^+$ in the third. With this in hand we can proceed with the calculation of ${D_{GR}}(g^{3/2}\xi ^+)$

$$\begin{aligned} {D_{GR}}(g^{3/2} \xi ^+)&= \frac{3}{2} \sqrt{g} {D_{GR}}g \xi ^+ + g^{3/2} {D_{GR}}\xi ^+\\&= - \frac{3g^2}{E} \text {EL}_g \xi ^+ -g^{3/2} {R_{GR}}(q^+\cdot \dot{q}) + \frac{2g^{3}}{E} \dot{g}^+ g^+ \\&= \frac{2}{E} \partial _t(g^{3/2}g^+)g^{3/2}g^+. \end{aligned}$$

It should now be clear why we chose the specific form for ${R_{GR}}q^+\cdot \dot{q}$: the first two terms in the second line cancel and we are left with a term for which we know how to compute ${D^k_{GR}}$. Using induction we can then prove

$$\begin{aligned} {D^k_{GR}}(g^{3/2} \xi ^+) = -\frac{(-2)^k}{E} \partial _t(g^{3/2}g^+)g^{3/2}g^+ \quad \text {for } k\ge 1. \end{aligned}$$

We have already shown that it holds for $k=1$. Assuming that it is true for k, the expression for $k+1$ reads:

$$\begin{aligned} {D_{GR}}^{k+1} (g^{3/2} \xi ^+)&= -\frac{(-2)^k}{E} \partial _t({D_{GR}}g^{3/2}g^+)g^{3/2}g^+ -\frac{(-2)^k}{E} \partial _t(g^{3/2}g^+){D_{GR}}g^{3/2}g^+\\&= 2 \frac{(-2)^k}{E} \partial _t(g^{3/2}g^+)g^{3/2}g^+ = - \frac{(-2)^{k+1}}{E} \partial _t(g^{3/2}g^+)g^{3/2}g^+. \end{aligned}$$

Since $\partial _t(g^{3/2}g^+)g^{3/2}g^+ = g^3 \dot{g}^+ g^+$ due to $g^+g^+ = 0$, we then have

$$\begin{aligned} \chi ^*_s(g^{3/2} \xi ^+)&= e^{s\mathcal {L}_{{D_{GR}}}}(g^{3/2} \xi ^+) = \sum _{k\ge 0} \frac{s^k}{k!} {D^k_{GR}}(g^{3/2} \xi ^+)\\&= g^{3/2} \xi ^+ - \left( \sum _{k\ge 1} \frac{(-2s)^k}{k!}\right) \frac{g^3}{E} \dot{g}^+ g^+ = g^{3/2} \xi ^+ -(e^{-2s} - 1)\frac{g^{3}}{E}\dot{g}^+ g^+, \end{aligned}$$

and

$$\begin{aligned} \lim _{s\rightarrow \infty } \chi ^*_s \xi ^+&= \left( \frac{E}{T}\right) ^{3/2} \lim _{s\rightarrow \infty } \chi ^*_s (g^{3/2}\xi ^+)\\&= \left( \frac{E}{T}\right) ^{3/2} \lim _{s\rightarrow \infty } \left( g^{3/2} \xi ^+ -(e^{-2s} - 1)\frac{g^{3}}{E}\dot{g}^+ g^+\right) \\&= \eta ^{3/2}\left( \xi ^+ +\frac{g^{3/2}}{E}\dot{g}^+ g^+\right) = \chi ^*\xi ^+. \end{aligned}$$

The strategy in the case of $q^+$ is the same as for $g^+$ and $\xi ^+$. Due to

$$\begin{aligned} {D_{GR}}\left( \frac{m \dot{q}}{2T}\right) = 0, \end{aligned}$$

we have

$$\begin{aligned} {D_{GR}}(g^{3/2} q^+_\parallel ) = {D_{GR}}\left( g^{3/2} q^+\cdot \frac{m \dot{q}}{2T}\right) = {D_{GR}}(g^{3/2}q^+ \cdot \dot{q}) \frac{m \dot{q}}{2T}. \end{aligned}$$

We will therefore omit the term $m \dot{q}/(2T)$ from the computations in order to keep them cleaner. We start by calculating

$$\begin{aligned} {D_{GR}}(q^+ \cdot \dot{q})&= \left( \delta _{GR} {R_{GR}}+ {R_{GR}}\delta _{GR}\right) q^+ \cdot \dot{q}\\&= - \delta _{GR} \left( \frac{3\sqrt{g}}{E} \text {EL}_g \xi ^+\right) - {R_{GR}}\left( \partial _t\left( \frac{m\dot{q}}{\sqrt{g}}\right) \right) \cdot \dot{q}\\&= \frac{3\sqrt{g}}{E} \text {EL}_g (q^+ \cdot \dot{q} - g^+ \dot{g} - 2\dot{g}^+ g) - \partial _t\left( \frac{m\dot{q}}{2g^{3/2}} \frac{2g^{3/2}}{E}g^+\right) \cdot \dot{q}\\&=\frac{3}{2}\left( 1 - \frac{T}{Eg}\right) q^+ \cdot \dot{q} - \frac{3\sqrt{g}}{E} \text {EL}_g(g^+ \dot{g} + 2 \dot{g}^+ g) - \frac{m \ddot{q} \cdot q}{E}g^+ - \frac{m\Vert \dot{q}\Vert ^2}{E} \dot{g}^+\\&=\frac{3}{2}\left( 1 - \frac{T}{Eg}\right) q^+ \cdot \dot{q} - \frac{3\sqrt{g}}{E} \text {EL}_g(g^+ \dot{g} + 2 \dot{g}^+ g) - \frac{\dot{T}}{E}g^+ - \frac{2T}{E} \dot{g}^+. \end{aligned}$$

Let $\sigma (\varphi ) = \varphi \partial _t{(g^{3/2}g^+)} - {\dot{\varphi }} g^{3/2}g^+$. With the result for ${D_{GR}}(q^+ \cdot \dot{q})$ we compute the following

$$\begin{aligned} {D_{GR}}(g^{3/2} q^+ \cdot q)&= \frac{3\sqrt{g}}{2} {D_{GR}}g q^+ \cdot \dot{q} + g^{3/2} {D_{GR}}(q^+\cdot \dot{q})\\&= \frac{3\sqrt{g}}{2} \left( \frac{T}{E}-g\right) q^+ \cdot \dot{q} +g^{3/2} {D_{GR}}(q^+\cdot \dot{q})\\&= - \frac{3g^2}{E} \text {EL}_g(g^+ \dot{g} + 2 \dot{g}^+ g) - g^{3/2}\frac{\dot{T}}{E}g^+ - g^{3/2}\frac{2T}{E} \dot{g}^+\\&= \left( \sqrt{g}\frac{3T}{2E} - \frac{3}{2}g^{3/2} \right) (g^+ \dot{g} + 2 \dot{g}^+ g) - g^{3/2}\frac{\dot{T}}{E}g^+ - g^{3/2}\frac{2T}{E} \dot{g}^+\\&= \frac{T}{E}\partial _t(g^{3/2}) g^+ + \frac{3T}{E} g^{3/2}\dot{g}^+ - \frac{3}{2} g^{3/2} \dot{g} g^+ - 3 g g^{3/2}\dot{g}^+\\&\quad -g^{3/2}\frac{\dot{T}}{E}g^+ - g^{3/2}\frac{2T}{E} \dot{g}^+\\&= \frac{T}{E} \partial _t(g^{3/2} g^+) - \frac{\dot{T}}{E}g^{3/2} g^+ - 3\left[ g \partial _t(g^{3/2} g^+) - \dot{g} (g^{3/2} g^+)\right] \\&= \sigma \left( \frac{T}{E}\right) - 3\sigma (g) = -2\sigma \left( \frac{T}{E}\right) -3 \sigma \left( g-\frac{T}{E}\right) \\&= -2\sigma \left( \frac{T}{E}\right) - 2\frac{3}{E}\sigma (g^{3/2}\text {EL}_g). \end{aligned}$$

Reintroducing $m\dot{q}/(2T)$ gives

$$\begin{aligned} {D_{GR}}(g^{3/2} q^+_\parallel )&= -2\sigma \left( \frac{T}{E}\right) \frac{m\dot{q}}{2T} - 2\cdot \frac{3}{E}\sigma (g^{3/2}\text {EL}_g)\frac{m\dot{q}}{2T}. \end{aligned}$$

(45)

Using induction it is then possible to show that

$$\begin{aligned} {D^k_{GR}}(g^{3/2} q^+_\parallel ) = (-1)^k 2 \sigma \left( \frac{T}{E}\right) \frac{m\dot{q}}{2T} + (-2)^k \frac{3}{E} \sigma (g^{3/2}\text {EL}_g) \frac{m\dot{q}}{2T}, \end{aligned}$$

(46)

for $k\ge 1$. The case $k=1$ is presented in Eq. (45). To see how the case $k+1$ follows from the case k, note that

$$\begin{aligned} {D_{GR}}\sigma \left( \frac{T}{E}\right) = {D_{GR}}\left( \frac{T}{E} \partial _t(g^{3/2} g^+) - \frac{\dot{T}}{E}g^{3/2} g^+ \right) = - \sigma \left( \frac{T}{E}\right) , \end{aligned}$$

where we used ${D_{GR}}T = 0$ and ${D_{GR}}(g^{3/2}g^+) = - g^{3/2}g^+$. Before computing ${D_{GR}}\sigma (g^{3/2}\text {EL}_g)$ note that

$$\begin{aligned} g^{3/2}\text {EL}_g&= \frac{E}{2}\left( g-\frac{T}{E}\right) = -\frac{E}{2}{D_{GR}}g\\&\Rightarrow \quad {D_{GR}}(g^{3/2}\text {EL}_g) = - \frac{E}{2} {D_{GR}}^2g = \frac{E}{2} {D_{GR}}g = - g^{3/2}\text {EL}_g. \end{aligned}$$

The action of ${D_{GR}}$ on $\sigma (g^{3/2}\text {EL}_g)$ is then

$$\begin{aligned} {D_{GR}}(\sigma (g^{3/2}\text {EL}_g))&= {D_{GR}}\left( g^{3/2}\text {EL}_g \partial _t(g^{3/2}g^+) - \partial _t(g^{3/2}\text {EL}_g) g^{3/2}g^+\right) \\&= -2\left( g^{3/2}\text {EL}_g \partial _t(g^{3/2}g^+) - \partial _t(g^{3/2}\text {EL}_g) g^{3/2}g^+\right) = -2\sigma (g^{3/2}\text {EL}_g), \end{aligned}$$

which proves Eq. (46). Having ${D^k_{GR}}(g^{3/2} q^+_\parallel )$ we can now write

$$\begin{aligned} \chi ^*_s q^+_\parallel&= e^{s\mathcal {L}_{{D_{GR}}}} (g^{3/2} q^+_\parallel ) = \sum _{k\ge 0} \frac{s^k}{k!}{D^k_{GR}}(g^{3/2} q^+_\parallel )\\&= g^{3/2} q^+_\parallel + \left( \sum _{k\ge 1} \frac{(-s)^k}{k!}\right) 2 \sigma \left( \frac{T}{E}\right) \frac{m\dot{q}}{2T} + \left( \sum _{k\ge 1} \frac{(-2s)^k}{k!}\right) \frac{3}{E} \sigma (g^{3/2}\text {EL}_g) \frac{m\dot{q}}{2T}\\&= g^{3/2} q^+_\parallel + (e^{-s} - 1) 2 \sigma \left( \frac{T}{E}\right) \frac{m\dot{q}}{2T} + (e^{-s} - 1) \frac{3}{E} \sigma (g^{3/2}\text {EL}_g)\frac{m\dot{q}}{2T}, \end{aligned}$$

taking the $s\rightarrow \infty$ limit then yields

$$\begin{aligned} \lim _{s\rightarrow \infty }\chi ^*_s (g^{3/2} q^+_\parallel )&= g^{3/2} q^+_\parallel - 2\sigma \left( \frac{T}{E}\right) \frac{m\dot{q}}{2T} - \frac{3}{E} \sigma (g^{3/2}\text {EL}_g)\frac{m\dot{q}}{2T} \end{aligned}$$

as desired. We can then extract $\lim _{s\rightarrow \infty }\chi ^*_sq^+_\parallel$ from this expression using

$$\begin{aligned} \lim _{s\rightarrow \infty }\chi ^*_sq^+_\parallel = (E/T)^{3/2}\lim _{s\rightarrow \infty }\chi ^*_s(g^{3/2} q^+_\parallel ), \end{aligned}$$

see Eq. (25). We have

$$\begin{aligned} \begin{aligned} \lim _{s\rightarrow \infty }\chi ^*_sq^+_\parallel&= \eta ^{3/2} \left( q^+_\parallel - 2g^{-3/2}\sigma \left( \frac{T}{E}\right) \frac{m\dot{q}}{2T} - \frac{3}{E}g^{-3/2}\sigma (g^{3/2}\text {EL}_g)\frac{m\dot{q}}{2T}\right) \\&= \eta ^{3/2} \left( q^+\cdot \dot{q} - 2g^{-3/2}\sigma \left( \frac{T}{E}\right) - \frac{3}{E}g^{-3/2}\sigma (g^{3/2}\text {EL}_g)\right) \frac{m\dot{q}}{2T}. \end{aligned} \end{aligned}$$

(47)

This expression can be further simplified, but first note that

$$\begin{aligned} \dot{\text {EL}_g}&= -\frac{E}{4g^{3/2}} \dot{g} + \frac{3T}{4g^{5/2}} \dot{g} - \frac{\dot{T}}{2g^{3/2}}\\&\Rightarrow \frac{3T}{Eg} \dot{g} - \frac{2\dot{T}}{E} = \frac{4g^{3/2}}{E} \dot{\text {EL}_g} + \dot{g}, \end{aligned}$$

and

$$\begin{aligned} \text {EL}_g =&\frac{E}{2\sqrt{g}} - \frac{T}{2g^{3/2}}\\ \Rightarrow&\frac{2T}{E} = 2g - \frac{4g^{3/2}}{E} \text {EL}_g. \end{aligned}$$

Using these two identities the last two terms in Eq. (47) yield

$$\begin{aligned}&2g^{-3/2}\sigma \left( \frac{T}{E}\right) + \frac{3}{E}g^{-3/2}\sigma (g^{3/2}\text {EL}_g)\\&\quad = \frac{2T}{E} g^{-3/2} \partial _t(g^{3/2}g^+) - \frac{2 \dot{T}}{E} g^+ + \frac{3}{E} \text {EL}_g \partial _t(g^{3/2}g^+) - \frac{3}{E}\partial _t(g^{3/2}\text {EL}_g) g^+\\&\quad = g^+\left[ \frac{3T}{Eg} \dot{g} - \frac{2\dot{T}}{E} + \frac{3}{E} \text {EL}_g \partial _t(g^{3/2}) - \frac{3}{E} \partial _t(g^{3/2}\text {EL}_g) \right] + \dot{g}^+\left[ \frac{2T}{E} + \frac{3}{E} g^{3/2} \text {EL}_g \right] ,\\&\quad =g^+\left[ \dot{g} + \frac{g^{3/2}}{E} \dot{\text {EL}_g}\right] + \dot{g}^+\left[ 2g - \frac{g^{3/2}}{E} \text {EL}_g\right] \\&\quad =\left[ g^+\dot{g} + 2\dot{g}^+ g \right] +\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] , \end{aligned}$$

and as such

$$\begin{aligned} \lim _{s\rightarrow \infty }\chi ^*_s q^+_\parallel&= \eta ^{3/2} \left( q^+_\parallel - \left[ g^+\dot{g} + 2\dot{g}^+ g \right] \frac{m\dot{q}}{2T} -\frac{g^{3/2}}{E} \left[ \dot{\text {EL}_g}g^+ - \text {EL}_g \dot{g}^+ \right] \frac{m\dot{q}}{2T}\right) \\&= \chi ^*q^+. \end{aligned}$$

We now move to the computation of $\chi ^*_sq^+_\perp$. As before our strategy will be to compute $\chi ^*_sq^+_\perp$ via ${D^k_{GR}}(g^{3/2} q^+_\perp )$. First note that we have

It follows that

$$\begin{aligned} {D_{GR}}q^+_\perp&= (\delta _{GR} {R_{GR}}+ {R_{GR}}\delta _{GR}) q^+_\perp = \delta _{GR}{R_{GR}}q^+_\perp - {R_{GR}}\left( \frac{m \ddot{q}_\perp }{\sqrt{g}}\right) \\&= \delta _{GR}{R_{GR}}q^+_\perp + \frac{m \ddot{q}_\perp }{2g^{3/2}} \frac{(-2)g^{3/2}}{E} g^+ = \delta _{GR}{R_{GR}}q^+_\perp - \frac{m\ddot{q}_\perp }{E}g^+, \end{aligned}$$

and thus

$$\begin{aligned}&{D_{GR}}(g^{3/2} q^+_\perp ) = \frac{3\sqrt{g}}{2} {D_{GR}}g q^+_\perp + g^{3/2} {D_{GR}}q^+_\perp \\&\quad = - \frac{3g^2}{E} \text {EL}_g q^+_\perp + g^{3/2} \delta _{GR} {R_{GR}}q^+_\perp - g^{3/2} \frac{m\ddot{q}_\perp }{E}g^+\\&\quad = - \frac{3g^2}{E} \delta _{GR}g^+ q^+_\perp +\frac{3g^2}{E} \delta _{GR}(g^+ q^+_\perp ) - g^{3/2} \frac{m\ddot{q}_\perp }{E}g^+ \\&\quad = - \frac{3g^2}{E} g^+ \delta _{GR}q^+_\perp - g^{3/2} \frac{m\ddot{q}_\perp }{E}g^+\\&\quad = \frac{3g^2}{E} g^+ \frac{m\ddot{q}_\perp }{\sqrt{g}} - g^{3/2} \frac{m\ddot{q}_\perp }{E}g^+ = \frac{2m}{E}\ddot{q}_\perp g^{3/2} g^+. \end{aligned}$$

Since ${D_{GR}}\ddot{q}_\perp = 0$, the computation of the higher powers of ${D^k_{GR}}(g^{3/2} q^+_\perp )$ becomes quite straightforward. We have

$$\begin{aligned} {D^k_{GR}}(g^{3/2} q^+_\perp ) = - (-1)^k \frac{2m}{E}\ddot{q}_\perp g^{3/2} g^+ \quad \text {for } k\ge 1, \end{aligned}$$

which results in

$$\begin{aligned} \chi ^*_s (g^{3/2} q^+_\perp )&= e^{s\mathcal {L}_{{D_{GR}}}} (g^{3/2} q^+_\perp ) = \sum _{k\ge 0} \frac{s^k}{k!} {D^k_{GR}}(g^{3/2} q^+_\perp )\\&= g^{3/2} q^+_\perp - \left( \sum _{k\ge 1} \frac{(-s)^k}{k!}\right) \frac{2m}{E}\ddot{q}_\perp g^{3/2} g^+\\&= g^{3/2} q^+_\perp - (e^{-s}-1) \frac{2m}{E}\ddot{q}_\perp g^{3/2} g^+, \end{aligned}$$

and as such

$$\begin{aligned} \lim _{s\rightarrow \infty }\chi ^*_s q^+_\perp = \eta ^{3/2} \left( q^+_\perp + \frac{2m}{E} g^+ \ddot{q}_\perp ,\right)&= \chi ^*q^+. \end{aligned}$$

as desired. $\square$

Lemma B.2.6

The map $\chi ^*$ is the identity in cohomology.

Proof

We need to show that

$$\begin{aligned} h_\chi f = \int ^\infty _0 e^{s\mathcal {L}_{{D_{GR}}}} \mathcal {L}_{{R_{GR}}} f \mathrm {d}s \end{aligned}$$

is well-defined on $\mathcal {F}_{GR}$. Note that $\{q,\xi ,\xi ^+,g^+\} \in \ker {R_{GR}}$ and as such

$$\begin{aligned}&h_\chi q = h_\chi \xi = h_\chi \xi ^+ = h_\chi g^+ = 0. \end{aligned}$$

In the case of the metric field g we compute

$$\begin{aligned} h_\chi g&= \int ^\infty _0e^{s\mathcal {L}_{{D_{GR}}}}{R_{GR}}g \,\mathrm {d}s = -\frac{2}{E} \int ^\infty _0e^{s\mathcal {L}_{{D_{GR}}}}(g^{3/2}g^+) \mathrm {d}s\\&= -\frac{2}{E} \int ^\infty _0e^{-s} g^{3/2}g^+ \mathrm {d}s = -\frac{2g^{3/2}}{E} g^+. \end{aligned}$$

We now compute the action of the map $h_\chi$ on $q^+_\parallel$ and $q^+_\perp$. For the perpendicular part of $q^+$ we have

The integral yields

$$\begin{aligned} I_\perp&= \int ^\infty _0 \frac{e^{-s}}{\left[ e^{-s}g + (1-e^{-s})\frac{T}{E}\right] ^{5/2}} \mathrm {d}s = \left. \frac{2}{3} \left( g-\frac{T}{E} \right) ^{-1} \left[ e^{-s}g + (1-e^{-s})\frac{T}{E}\right] ^{-3/2}\right| ^\infty _0\\&= \frac{2}{3} \left( g-\frac{T}{E} \right) ^{-1} \left[ \left( \frac{E}{T}\right) ^{3/2} - \frac{1}{g^{3/2}}\right] = \frac{2}{3} \frac{E}{Tg^{3/2}} \frac{\eta ^{3/2}-1}{\eta -1}\\&= \frac{2}{3} \frac{E}{Tg^{3/2}} \left( \frac{\eta }{\sqrt{\eta }+1} + 1\right) . \end{aligned}$$

Which results in

$$\begin{aligned} h_\chi q^+_\perp = \frac{2}{T} \left( \frac{\eta }{\sqrt{\eta }+1} + 1\right) g^{3/2} g^+q^+_\perp . \end{aligned}$$

Similarly we have

$$\begin{aligned} h_\chi q^+_\parallel&= \int ^\infty _0 e^{s\mathcal {L}_{{D_{GR}}}} {R_{GR}}q^+_\parallel \mathrm {d}s = -\frac{3}{E}\int ^\infty _0 e^{s\mathcal {L}_{{D_{GR}}}} \left( \sqrt{g}\left( \frac{E}{2\sqrt{g}} - \frac{T}{2g^{3/2}}\right) \xi ^+ \frac{m\dot{q}}{2T}\right) \mathrm {d}s\\&= \frac{3}{2} \frac{m\dot{q}}{2T} \int ^\infty _0 e^{s\mathcal {L}_{{D_{GR}}}} \left( \left( \frac{T}{Eg} - 1\right) \xi ^+\right) \mathrm {d}s\\&= \frac{3}{2} \frac{m\dot{q}}{2T} \int ^\infty _0 \left( \frac{T}{E} (e^{s\mathcal {L}_{{D_{GR}}}}g)^{-1} - 1\right) \\&\qquad \times (e^{s\mathcal {L}_{{D_{GR}}}}g)^{-3/2} \left[ g^{3/2} \xi ^+ -(e^{-2s} - 1)\frac{g^{3}}{E}\dot{g}^+ g^+\right] \mathrm {d}s\\&= \frac{3}{2} \frac{m\dot{q}}{2T} \int ^\infty _0 \left( \frac{T}{E} - e^{s\mathcal {L}_{{D_{GR}}}}g \right) \\&\qquad \times (e^{s\mathcal {L}_{{D_{GR}}}}g)^{-5/2} \left[ g^{3/2} \xi ^+ + \frac{g^{3}}{E}\dot{g}^+ g^+ - e^{-2s} \frac{g^{3}}{E}\dot{g}^+ g^+ \right] \mathrm {d}s\\&= \frac{3}{2} \frac{m\dot{q}}{2T} \left[ g^{3/2} \xi ^+ + \frac{g^{3}}{E}\dot{g}^+ g^+\right] I_1 - \frac{3}{2} \frac{m\dot{q}}{2T} \frac{g^{3}}{E}\dot{g}^+ g^+ I_2. \end{aligned}$$

The integrals that we need to consider are

$$\begin{aligned} I_1&= \int ^\infty _0 \left( \frac{T}{E} - e^{-s}g - (1-e^{-s})\frac{T}{E}\right) \left[ e^{-s}g + (1-e^{-s})\frac{T}{E}\right] ^{-5/2} \mathrm {d}s\\&= \left( \frac{T}{E}-g\right) \int ^\infty _0 \frac{e^{-s}}{\left[ e^{-s}g + (1-e^{-s})\frac{T}{E}\right] ^{5/2}} \mathrm {d}s\\&= \frac{T}{E}(1-\eta ) I_\perp = - \frac{2}{3} g^{-3/2} (\eta ^{3/2}-1), \end{aligned}$$

and

$$\begin{aligned} I_2&= \int ^\infty _0 \left( \frac{T}{E} - e^{-s}g - (1-e^{-s})\frac{T}{E}\right) \frac{e^{-2s}}{(e^{s\mathcal {L}_{{D_{GR}}}}g)^{5/2}} \mathrm {d}s\\&= \left( \frac{T}{E}-g\right) \int ^\infty _0 \frac{e^{-3s}}{(e^{s\mathcal {L}_{{D_{GR}}}}g)^{5/2}} \mathrm {d}s = -\frac{2}{3} \int ^\infty _0 e^{-2s} \frac{\mathrm {d}}{\mathrm {d}s} \left[ (e^{s\mathcal {L}_{{D_{GR}}}}g)^{-3/2} \right] \mathrm {d}s\\&= -\frac{2}{3} \left. \frac{e^{-2s}}{(e^{s\mathcal {L}_{{D_{GR}}}}g)^{3/2}} \right| ^\infty _0 -\frac{4}{3} \int ^\infty _0 \frac{e^{-2s}}{(e^{s\mathcal {L}_{{D_{GR}}}}g)^{3/2}} \mathrm {d}s\\&= \frac{2}{3} g^{-3/2} -\frac{4}{3} \int ^\infty _0 e^{-s}\frac{\mathrm {d}}{\mathrm {d}s} \left[ \frac{2}{\left( g-\frac{T}{E}\right) \sqrt{e^{s\mathcal {L}_{{D_{GR}}}}g}}\right] \mathrm {d}s\\&= \frac{2}{3} g^{-3/2} -\frac{8}{3} \left. \frac{e^{-s}}{\left( g-\frac{T}{E}\right) \sqrt{e^{s\mathcal {L}_{{D_{GR}}}}g}}\right| ^\infty _0 - \frac{8}{3} \int ^\infty _0 \frac{e^{-s}}{\left( g-\frac{T}{E}\right) \sqrt{e^{s\mathcal {L}_{{D_{GR}}}}g}} \mathrm {d}s\\&= \frac{2}{3} g^{-3/2} + \frac{8}{3} \left( g-\frac{T}{E}\right) ^{-1} g^{-1/2} + \frac{16}{3} \left( g-\frac{T}{E}\right) ^{-2} \left. \sqrt{e^{s\mathcal {L}_{{D_{GR}}}}g}\right| ^\infty _0\\&= \frac{2}{3} g^{-3/2} + \frac{8}{3} \left( g-\frac{T}{E}\right) ^{-1} g^{-1/2} + \frac{16}{3} \left( g-\frac{T}{E}\right) ^{-2} \left( \sqrt{\frac{T}{E}} - \sqrt{g}\right) \\&=\left( \frac{E}{T}\right) ^{3/2} \left[ \frac{2}{3} \frac{1}{\eta ^{3/2}} + \frac{8}{3} \frac{1}{\left( \eta -1\right) \sqrt{\eta }} + \frac{16}{3} \frac{1 - \sqrt{\eta }}{\left( \eta -1\right) ^{2}} \right] \\&= -\frac{2}{3}g^{-3/2}\frac{3 \eta - 2 \sqrt{\eta } - 1}{(\sqrt{\eta } + 1)^2}. \end{aligned}$$

Where on the third line we used the integral in $I_\perp$ and similarly for the other integrals. Gathering everything results in

$$\begin{aligned} h_\chi q^+_\parallel = (1-\eta ^{3/2}) \left[ \xi ^+ + \frac{g^{3/2}}{E}\dot{g}^+ g^+\right] \frac{m\dot{q}}{2T} + \frac{3 \eta - 2 \sqrt{\eta } - 1}{(\sqrt{\eta }+1)^2} \frac{g^{3/2}}{E}\dot{g}^+ g^+ \frac{m\dot{q}}{2T}. \end{aligned}$$

$\square$