The BRST complex of homological Poisson reduction

Abstract

BRST complexes are differential graded Poisson algebras. They are associated with a coisotropic ideal J of a Poisson algebra P and provide a description of the Poisson algebra $(P/J)^J$ as their cohomology in degree zero. Using the notion of stable equivalence introduced in Felder and Kazhdan (Contemporary Mathematics 610, Perspectives in representation theory, 2014), we prove that any two BRST complexes associated with the same coisotropic ideal are quasi-isomorphic in the case $P = {{\mathrm{\mathbb {R}}}}[V]$ where V is a finite-dimensional symplectic vector space and the bracket on P is induced by the symplectic structure on V. As a corollary, the cohomology of the BRST complexes is canonically associated with the coisotropic ideal J in the symplectic case. We do not require any regularity assumptions on the constraints generating the ideal J. We finally quantize the BRST complex rigorously in the presence of infinitely many ghost variables and discuss the uniqueness of the quantization procedure.

Appendix A: Graded Poisson algebras

Let $(P, [-,-]_0)$ be a unital Poisson algebra over ${{\mathrm{\mathbb {K}}}}= {{\mathrm{\mathbb {R}}}}$. Let ${{\mathrm{\mathcal {M}}}}$ be a negatively graded vector space with finite-dimensional homogeneous components. Let ${{\mathrm{\mathcal {M}}}}^*$ be the positively graded vector space with homogeneous components $(M^*)^i = (M^{-i})^*$. Define the graded algebra $X_0 = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$.

Lemma 62

The bracket $\{-,-\}_0$ on P naturally extends to a skew-symmetric, bilinear map $\{-,-\}$ on $X_0$ via the natural pairing of ${{\mathrm{\mathcal {M}}}}$ and ${{\mathrm{\mathcal {M}}}}^*$. This map has degree zero. Moreover, it is a derivation for the product on $X_0$ and satisfies the Jacobi identity. Thus, it turns $X_0$ into a graded Poisson algebra.

Proof

First, we define a bracket on ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$. For $x \in {{\mathrm{\mathcal {M}}}}$ and $\alpha \in {{\mathrm{\mathcal {M}}}}^*,$ we set

$$\begin{aligned} \{x,x\}_1&= 0,&\{\alpha , \alpha \}_1&= 0,&\{x, \alpha \}_1&= \alpha (x),&\{\alpha , x\}_1&= -(-1)^{\deg \alpha \deg x} \alpha (x) \end{aligned}$$

and extend this definition as a bi-derivation to all of ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$. It is then a bilinear, skew-symmetric map $\{-,-\}_1 : {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*) \times {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*) \rightarrow {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$ of degree zero which is by definition a derivation for the product. The expression

$$\begin{aligned} \zeta (a,b,c) := (-1)^{\deg a \deg c} \{a,\{b,c\}\} + \text { cyclic permutations} \end{aligned}$$

satisfies $\zeta (a_1a_2, b,c) = (-1)^{\deg a_1 \deg c} a_1 \zeta (a_2,b,c) + (-1)^{\deg a_2 \deg b} \zeta (a_1,b,c)a_2$ and similar derivation-like statements for the other entries. Let $\{e_j\}$ be a homogeneous basis of ${{\mathrm{\mathcal {M}}}}$ and $\{e^*_j\}$ its dual basis. Then the bracket of any two of those generators is a scalar, whence $\zeta$ is zero on generators. By the above derivation-type property, $\zeta$ vanishes identically, proving the graded Jacobi identity.

Now, set $\{-,-\} = \{-,-\}_0 + \{-,-\}_1$ on $X_0$. $\square$

The grading on ${{\mathrm{\mathcal {M}}}}$ induces a grading on $X_0$. We obtain a descending filtration indexed by nonnegative integers n: ${{\mathrm{\mathcal {F}}}}^n X_0$ is defined to be the ideal generated by elements of $X_0$ of degree at least n. Note that ${{\mathrm{\mathcal {F}}}}^0 X_0 = X_0$ is the whole algebra. We set ${{\mathrm{\mathcal {F}}}}^n X_0^m = {{\mathrm{\mathcal {F}}}}^n X_0 \cap X_0^m$. We also define $I_0 := {{\mathrm{\mathcal {F}}}}^1 X_0$ and $I_0^{(n)} := I_0 \cdots I_0$ to be the n-fold product ideal.

1.1 A.1: Compatibility of filtration and bracket on $X_0$

We use the derivation properties of the bracket on $X_0$ to derive the compatibility relations between the filtration and the bracket.

Lemma 63

For $m, n \in {{\mathrm{\mathbb {Z}}}}$ and $p,q \in {{\mathrm{\mathbb {N}}}}_0,$ we have $\{{{\mathrm{\mathcal {F}}}}^p X_0^n ,{{\mathrm{\mathcal {F}}}}^q X_0^m\} \subset {{\mathrm{\mathcal {F}}}}^{r_{n,m}(p,q)} X_0^{m+n}$ where

$$\begin{aligned} r_{n,m}(p,q) = \max \{m+n, \min \{\max \{p, q + n\}, \max \{q, p + m\}\}\}. \end{aligned}$$

(5)

Proof

Let $a, b, u, v \in X_0$ be homogeneous elements with $\deg a + \deg u = n$, $\deg b + \deg v = m$, $\deg u = p$, and $\deg v = q$. Suppose without loss of generality that $p \geqslant n$ and $q \geqslant m$. Using the Leibnitz rule we see that $\{au, bv\}$ is in ${{\mathrm{\mathcal {F}}}}^r X_0,$ where $r = \min \{p + q, \max \{p, q + n\}, \max \{q, p + m\}\} = \min \{\max \{p, q + n\}, \max \{q, p + m\}\}$. $\square$

Corollary 64

We obtain for $p,q \in {{\mathrm{\mathbb {N}}}}_0$ and $m,n \in {{\mathrm{\mathbb {Z}}}}$,

• (1) $\{{{\mathrm{\mathcal {F}}}}^p X_0^1, {{\mathrm{\mathcal {F}}}}^q X_0^1 \} \subset {{\mathrm{\mathcal {F}}}}^{l(p,q)} X_0$, where $l(p,q) = {\left\{ \begin{array}{ll} \max \{p, q\}, &{} if p \ne q \\ p + 1, &{} if p=q \end{array}\right. }$.

• (2) $\{{{\mathrm{\mathcal {F}}}}^p X_0, X_0^m\} \subset {{\mathrm{\mathcal {F}}}}^p X_0$ provided $m \geqslant 0$.

• (3) $\{{{\mathrm{\mathcal {F}}}}^p X_0^n, X_0^m\} \cup \{X_0^n, {{\mathrm{\mathcal {F}}}}^p X_0^m\} \subset {{\mathrm{\mathcal {F}}}}^{t_{n,m}(p)} X_0^{n+m}$, where $t_{n,m}(p) = p - \max \{|n|,|m|\}.$

Lemma 65

We have $\{X_0^1 \cap I_0^{(2)}, {{\mathrm{\mathcal {F}}}}^{m} X_0\} \subset {{\mathrm{\mathcal {F}}}}^{m+1} X_0$.

Proof

Let $a, u_1, u_2 \in X_0$ with $\deg (a) = 1-n$, $\deg (u_1)+\deg (u_2)=n$, and $\deg (u_1), \deg (u_2) > 0$. Let $b,v \in X_0$ with $\deg (v) = m$. Expand $\{a u_1 u_2, bv\}$ using the Leibnitz rule. $\square$

Lemma 66

The ideal $I_0$ is closed under the bracket.

Proof

Let $a, u, b, v \in X_0$ with $\deg (u)=\deg (v) = 1$. Expand $\{ au, bv \}$ using the Leibnitz rule. $\square$

1.2 A.2: Completion

For each j, we use the filtration on $X_0^j$ to complete this space to the space

$$\begin{aligned} X^j = \lim _{\leftarrow p} \frac{X_0^j}{{{\mathrm{\mathcal {F}}}}^p X_0 \cap X_0^j}. \end{aligned}$$

The sum and scalar multiplication on $X_0^j$ extend to this space, turning $X = \bigoplus _j X^j$ into a graded vector space. The product of two elements $(x_p + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p \in X^j$ and $(y_p + {{\mathrm{\mathcal {F}}}}^p X_0^k)_p \in X^k$ is defined to be $(x_p y_p + {{\mathrm{\mathcal {F}}}}^p X_0^{j+k})_p \in X^{j+k}$. This definition does not depend on the choice of representatives, since the product is compatible with the filtration. Moreover, it defines an element of $X^{j+k}$ since for $p \leqslant q$ we have $x_p y_p \equiv x_q y_q \pmod {{{\mathrm{\mathcal {F}}}}^p X_0^{j+k}}$, and we may shift the representatives of x and y. The multiplication is compatible with the addition turning X into a graded commutative algebra.

Endow $X_0^j / {{\mathrm{\mathcal {F}}}}^{p} X_0^j$ with the discrete topology and $\prod _p X_0^j / {{\mathrm{\mathcal {F}}}}^{p} X_0^j$ with the product topology. Equip $\lim _{\leftarrow } X_0^j / {{\mathrm{\mathcal {F}}}}^{p} X_0^j \subset \prod _p X_0^j / {{\mathrm{\mathcal {F}}}}^{p} X_0^j$ with the subspace topology. Finally, equip $X = \bigoplus _j X^j$ with the product topology. Hence, a sequence $\{x_l\}_l \subset X^j$, with $x_l = ( x_{l,p} + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p$, converges to an element $x = (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p \in X^j$ if and only if for all $p \in {{\mathrm{\mathbb {N}}}}_0$ there exists a $l_0$ such that for all $l \geqslant l_0$ we have $x_{p,l} \equiv x_p \pmod {{{\mathrm{\mathcal {F}}}}^p X_0^j}$. A sequence $\{x_l\}_l \subset X$ converges to an element $x \in X$ if and only if all homogeneous components converge. Since X is first-countable, continuity is characterized by the convergence of sequences. We immediately obtain:

Lemma 67

The sum $X \times X \rightarrow X$ is continuous.

For the product, only a weaker statement holds in general:

Lemma 68

The product $X \rightarrow X$ is continuous in each entry. For each pair $(j,k) \in {{\mathrm{\mathbb {Z}}}}^2$, the product $X^j \times X^k \rightarrow X^{j+k}$ is continuous.

Proof

Consider a sequence $\{x_i\}_i$ in X converging to $x \in X$ and fix $y \in X$. Denote the homogeneous components of $x_{i}$ by $x_{i}^j = (x_{i,p}^j + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p$ and similarly for x and y. Fix $l \in {{\mathrm{\mathbb {Z}}}}$ and $p \in {{\mathrm{\mathbb {N}}}}_0$. The l-th homogeneous component of $x_i y$ has a p-th component with representative $\sum _{j \in C} x_{i,p}^{l-j} y_{p}^{j}$ where the $C \subset {{\mathrm{\mathbb {Z}}}}$ is the finite set for which $y^j \ne 0$. It does not depend on i. (Such a finite set which is independent of i only exists in general when one entry of the product remains fixed.) We have

$$\begin{aligned} \sum _{j \in C} x_{i,p}^{l-j} y_{p}^{j}&= \sum _{j \in C} \Big ( (x_{i,p}^{l-j} - x_p^{l-j}) y_p^{j} + x_p^{l-j} y_p^{j} \Big ). \end{aligned}$$

For each $j \in C,$ pick a number $i_{0,j}$ such that for $i_j \geqslant i_{0,j}$ we have $x_{i_j,p}^{l-j} \equiv x_p^{l-j} \pmod { {{\mathrm{\mathcal {F}}}}^p X_0^{l-j} }$ and let $i_0$ be their maximum. Now, for $i \geqslant i_0$, we have $\sum _{j \in C} x_{i,p}^{l-j} y_{p}^{j} \equiv \sum _{j \in C} x_p^{l-j} y_p^{j} \pmod { {{\mathrm{\mathcal {F}}}}^p X_0^l}$. The second statement follows similarly. $\square$

Next, we approximate elements in X by elements in $X_0$.

Lemma 69

The map $\iota : X_0^n \longrightarrow X^n$ sending $x \in X_0^n$ to $(x + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p$ is injective.

Proof

Since $\iota$ is linear, the claim follows from $\bigcap _p {{\mathrm{\mathcal {F}}}}^p X_0^n = 0$. $\square$

Lemma 70

For $x = (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p \in X_0^n,$ we have $\lim _{m \rightarrow \infty } \iota (x_m) = x$.

Proof

Fix p. For $m \geqslant p,$ we have that the p-th component of $\iota (x_m) - x$ is $x_m - x_p \in {{\mathrm{\mathcal {F}}}}^p X_0^n$. $\square$

Corollary 71

$X_0$ can be considered a dense subset of X.

Now, we turn to the extension of the bracket to the completion. Let $x = (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p \in X^j$ and $y = (y_p + {{\mathrm{\mathcal {F}}}}^p X_0^k)_p \in X^k$. We define $\{x,y\} \in X^{j+k}$ to be the element

$$\begin{aligned} ( \{x_{s_{j,k}(p)} , y_{s_{j,k}(p)} \} + {{\mathrm{\mathcal {F}}}}^p X_0^{j+k})_p, \end{aligned}$$

where $s_{j,k}(p) := p + \max \{\vert j\vert ,\vert k\vert \}$. This definition does not depend on the representatives of x and y by Corollary 64, since $t_{j,k}( s_{j,k}(p) ) = s_{j,k}( t_{j,k}(p)) = p$. Moreover, it defines an element of $X^{j+k}$: for $p \leqslant q$ we have $\{x_{s_{j,k}(p)},y_{s_{j,k}(p)}\} \equiv \{x_{s_{j,k}(q)},y_{s_{j,k}(q)}\} \pmod { {{\mathrm{\mathcal {F}}}}^{s_{j,k}(p)}},$ since we may shift the representatives of x and y. We extend this bracket as a bilinear map to $X \times X$.

Lemma 72

The extension of the bracket on $X_0$ is a skew-symmetric, bilinear degree zero map on X that satisfies the graded Jacobi identity (i.e. the bracket is an odd derivation for itself).

Proof

It is trivial that the extended bracket is a skew-symmetric, bilinear degree zero map. These properties follow directly from the definitions.

We prove the graded Jacobi identity. Consider elements

$$\begin{aligned} x&= (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p \in X^j&y&= (y_p + {{\mathrm{\mathcal {F}}}}^p X_0^k)_p \in X^k&z&= (z_p + {{\mathrm{\mathcal {F}}}}^p X_0^l)_p \in X^l. \end{aligned}$$

The p-th element of $\{y,z\}$ has representative $\{y_{s_{k,l}(p)},z_{s_{k,l}(p)}\}$. Hence, the p-th element of $\{x,\{y,z\}\}$ has representative $\{x_{s_{j, k+l}(p)}, \{y_{s_{k,l}(s_{j, k+l}(p))}, z_{s_{k,l}(s_{j,k+l}(p))} \}$. We now want to bound the indices from above by a term which is invariant under cyclic permutations of (j, k, l). The function $r_{j,k,l}(p) := p + 2 (\vert j \vert + \vert k \vert + \vert l \vert )$ does the job. So, $(-1)^{j l} \{x, \{y,z\}\} + (-1)^{k j} \{y,\{z,x\}\} + (-1)^{l k} \{z,\{x,y\}\}$ has representative

$$\begin{aligned} (-1)^{j l} \{x_{r_{j,k,l}(p)}, \{y_{r_{j,k,l}(p)},z_{r_{j,k,l}(p)}\}\} + \text {cyclic permutations} \end{aligned}$$

which vanishes by the graded Jacobi identity on $X_0$. $\square$

Lemma 73

The bracket on X is a derivation for the product.

Proof

Let

$$\begin{aligned} x&= (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p \in X^j&y&= (y_p + {{\mathrm{\mathcal {F}}}}^p X_0^k)_p \in X^k&z&= (z_p + {{\mathrm{\mathcal {F}}}}^p X_0^l)_p \in X^l. \end{aligned}$$

The p-th element of $\{xy,z\} - (x \{y,z\} + (-1)^{jk} y \{x,z\})$ has representative

$$\begin{aligned}&\{ x_{s_{j+k,l}(p)} y_{s_{j+k,l}(p)}, z_{s_{j+k,l}(p)}\} - \bigg ( x_p \{ y_{s_{k,l}(p)}, z_{s_{k,l}(p)} \} + (-1)^{jk} y_p \{ x_{s_{j,l}(p)}, z_{s_{j,l}(p)} \} \bigg )\\&\equiv \; \{x_q y_q , z_q\} - (x_q \{y_q,z_q\} + (-1)^{jk} y_q \{x_q,z_q\}) \pmod { {{\mathrm{\mathcal {F}}}}^p X_0^{j+k+l}}, \end{aligned}$$

where $q := p + \vert m\vert + \vert n\vert + \vert k\vert$ is a common upper bound of all indices appearing in the formula. The last line vanishes by the derivation property of the bracket on $X_0$. $\square$

Lemma 74

For each pair $(j,k) \in {{\mathrm{\mathbb {Z}}}}^2$, the map $\{ - ,-\} : X^j \times X^k \rightarrow X$ is continuous. The map $\{-,-\} : X \times X \rightarrow X$ is continuous in each entry.

Proof

Let $x_n = ( x_{n,p} + {{\mathrm{\mathcal {F}}}}^p X_0^j )_p \in X^j$ and $y_n = (y_{n,p} + {{\mathrm{\mathcal {F}}}}^p X_0^k)_p \in X^k$ define two sequences converging to the respective elements $x = ( x_{p} + {{\mathrm{\mathcal {F}}}}^p X_0^j )_p \in X^j$ and $y = (y_{p} + {{\mathrm{\mathcal {F}}}}^p X_0^k) \in X^k$. Fix p, set $s = s_{j,k}(p)$, and pick $n_0$ such that for $n \geqslant n_0$,

$$\begin{aligned} x_{n,s}&\equiv x_s \pmod {{{\mathrm{\mathcal {F}}}}^s X_0^j}&y_{m,s}&\equiv y_s \pmod {{{\mathrm{\mathcal {F}}}}^s X_0^k}. \end{aligned}$$

Let $n \geqslant n_0$. The p-th element of $\{x_n, y_n\} - \{x,y\}$ has the representative $\{x_{n, s}, y_{n, s}\} - \{x_s, y_s\} \in {{\mathrm{\mathcal {F}}}}^{t_{j,k}(s)} \subset {{\mathrm{\mathcal {F}}}}^p X_0^{j+k}$ by Corollary 64.

Now, consider a sequence $\{x_i\}_i$ in X converging to $x \in X$ and fix $y \in X$. Denote the homogeneous components of $x_{i}$ by $x_{i}^j = (x_{i,p}^j + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p$ and similarly for x and y. Fix $l \in {{\mathrm{\mathbb {Z}}}}$ and $p \in {{\mathrm{\mathbb {N}}}}_0$. Set $C = \{ j \in {{\mathrm{\mathbb {Z}}}}: y^{j} \ne 0 \}$. This is a finite set. The l-th homogeneous component of $\{x_i,y\}$ has a p-th component with representative

$$\begin{aligned} \sum _{j \in C} \{x_{i,s_{l-j,j}(p)}^{l-j}, y^j_{s_{l-j,j}(p)}\}. \end{aligned}$$

Set $s = \max \{ s_{l-j,j}(p) : j \in C\}$ and pick $n_0$ such that for $n \geqslant n_0$ and all $j \in C,$ we have $x^{l-j}_{n, s} \equiv x^{l-j}_s \pmod { {{\mathrm{\mathcal {F}}}}^s X_0^{l-j}}$. For such n,

$$\begin{aligned} \sum _{j \in C} \{x_{n,s_{l-j,j}(p)}^{l-j}, y^j_{s_{l-j,j}(p)}\} \equiv \sum _{j \in C} \{x_{n,s}^{l-j}, y^j_{s}\} \equiv \sum _{j \in C} \{x_{s}^{l-j}, y^j_{s}\} \pmod { {{\mathrm{\mathcal {F}}}}^p X_0^l }. \end{aligned}$$

$\square$

1.3 A.3: The filtration and the bracket on the completion

The filtration on $X_0$ induces a descending filtration on the completion with homogeneous components

$$\begin{aligned} {{\mathrm{\mathcal {F}}}}^p X^n := \lim _{\leftarrow q} \frac{{{\mathrm{\mathcal {F}}}}^p X_0^n}{{{\mathrm{\mathcal {F}}}}^{p+q} X_0^n} = \{ (x_q + {{\mathrm{\mathcal {F}}}}^q X_0^n)_{q \geqslant p} \in X^n : x_q \in {{\mathrm{\mathcal {F}}}}^p X_0^n \}. \end{aligned}$$

This defines a homogeneous ideal ${{\mathrm{\mathcal {F}}}}^p X = \bigoplus _n {{\mathrm{\mathcal {F}}}}^p X^n$ in X. Here, p is a nonnegative integer and again ${{\mathrm{\mathcal {F}}}}^0 X = X$. We set $I = {{\mathrm{\mathcal {F}}}}^1 X$ and

$$\begin{aligned} I^{(n)} = \bigoplus _m\; \lim _{\leftarrow p} \frac{X_0^m \cap I_0^{(n)}}{{{\mathrm{\mathcal {F}}}}^p X_0^m \cap I_0^{(n)}}. \end{aligned}$$

Those are homogeneous ideals in X.

Lemma 75

For each $j \in {{\mathrm{\mathbb {Z}}}}$, the sets ${{\mathrm{\mathcal {F}}}}^p X^j$ and $I^{(2)}\cap X^j$ are closed.

Proof

Consider the first statement. Since X is first-countable, it suffices to consider sequences $x_n = ( x_{n,q} + {{\mathrm{\mathcal {F}}}}^q X_0^j )_q$ converging to an $x = ( x_{q} + {{\mathrm{\mathcal {F}}}}^q X_0^j )_q$ in X with $x_{n,q} \in {{\mathrm{\mathcal {F}}}}^p X_0^j$ and show that $x \in {{\mathrm{\mathcal {F}}}}^p X^j$. So, fix $q \geqslant p$. Let n be an integer with $x_{n,q} \equiv x_q \pmod {{{\mathrm{\mathcal {F}}}}^q X_0^j}$. Then, $x_q \equiv x_{n,q} \equiv 0 \pmod { {{\mathrm{\mathcal {F}}}}^p X_0^j}$.

Now, let $x_n = ( x_{n,p} + {{\mathrm{\mathcal {F}}}}^p X_0^j)_p$ be a sequence converging to $x = ( x_{p} + {{\mathrm{\mathcal {F}}}}^p X_0^j )_p$ in X with $x_{n,p} \in I_0^{(2)}$. Fix p. For n large enough, we may replace $x_p$ by $x_{n,p} \in I_0^{(2)}$. $\square$

Lemma 76

Fix $p \in {{\mathrm{\mathbb {N}}}}_0$.

• (1) $\{I^{(2)},I^{(2)}\} \subset I^{(2)}$.

• (2) $\{I^{(2)}, I^{(p)}\} \subset I^{(p+1)} \subset {{\mathrm{\mathcal {F}}}}^{p+1} X.$

• (3) $\{I^{(p)}, X^1\} \subset I^{(p)}.$

Proof

The first statement: Consider elements $u = (u_p + {{\mathrm{\mathcal {F}}}}^p X_0^{j})_p$ and $v = (v_p + {{\mathrm{\mathcal {F}}}}^p X_0^{k})_p$ of X with $u_p, v_p \in I_0^{(2)}$. Then the p-th element of $\{u, v\}$ has the representative $\{ u_{s_{j,k}(p)}, v_{s_{j,k}(p)}\} \in \{I_0^{(2)}, I_0^{(2)}\} \subset I_0^{(2)}$ by the Leibnitz rule.

Now, the second statement: First consider $p=0$. Then by the Leibnitz rule, $\{I_0^{(2)}, X_0\} \subset I_0 \{I_0, X_0\} \subset I_0$. Now consider $p>0$. By repeated use of the Leibnitz rule,

$$\begin{aligned} \{I_0^{(2)}, I_0^{(p)}\} \subset \{I_0^{(2)}, I_0\} I_0^{(p-1)} \subset I_0 \{I_0,I_0\} I_0^{(p-1)} \subset I_0^{(p+1)} \end{aligned}$$

by Lemma 66. The statement generalizes to the completion, as in the case above.

The third statement follows analogously by picking representatives. $\square$

Lemma 77

Let $l \mapsto q(l)$ define an unbounded non-decreasing function ${{\mathrm{\mathbb {N}}}}\rightarrow {{\mathrm{\mathbb {N}}}}$. Let $x_l = (x_{l,p} + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p \in {{\mathrm{\mathcal {F}}}}^{q(l)} X^n$ define a sequence of elements in $X^n$. Then, $\sum _{l=0}^{\infty } x_l$ converges to an element $x \in X^n$.

Proof

We may suppose $q(l) = l$. Define $x_p := \sum _{l=0}^{p-1} x_{l,p}$. Then, $x := (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p$ defines an element of $X^n$ since, for $p \leqslant q$, we have

$$\begin{aligned} x_q - x_p = \sum _{l=0}^{q-1} x_{l,q} - \sum _{l=0}^{p-1} x_{l,p} = \sum _{l=0}^{p-1} (x_{l,q}-x_{l,p}) + \sum _{l=p}^{q-1} x_{l,q} \in {{\mathrm{\mathcal {F}}}}^p X_0^n. \end{aligned}$$

We claim that $\sum _{l=0}^k x_l$ converges to x as $k \rightarrow \infty$. Fix p. Let $k \geqslant k_0 := p$. Then the p-th element of $\sum _{l=0}^k x_l - x$ has representative $\sum _{l=0}^k x_{l,p} - x_p = \sum _{l=p}^k x_{l,p} \in {{\mathrm{\mathcal {F}}}}^p X_0^n$. $\square$

Lemma 78

Each $H \in X^n$ can be expanded as $H = \sum _{p \ge 0} h_p$ with $h_p \in B^p \otimes _P T^{n-p}$.

Proof

Write $H = (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p$ with $x_0 = 0$. Pick a homogeneous basis $e_i$ of the underlying graded vector space. Redefine $x_p$ such that $x_p$ does not contain a monomial in ${{\mathrm{\mathcal {F}}}}^p X_0^n$. Set $h_p = x_{p+1}-x_p \in {{\mathrm{\mathcal {F}}}}^p X_0^n$. It cannot contain a monomial of degree $(p+1)$ or higher. Hence, $h_p \in B^p \otimes _P T^{n-p}$. Then by Lemmas 70 and 77, $\sum _p h_p = H$. $\square$

Lemma 79

All statements from Sect. A.1 in Appendix are valid for $X_0$ replaced by X.

Proof

The bracket on X is defined by acting on representatives with the bracket of $X_0$ where the statements hold.

1.4 A.4: Extension of maps

Next, we consider the problem of extending maps on $X_0$ to X.

Remark 80

A linear map on $X_0$ of a fixed degree preserving the filtration up to a fixed shift naturally extends to a linear map on X preserving the filtration up to the same shift. This extension is continuous.

1.5 A.5: The associated graded

The associated graded ${{\mathrm{\text {gr}}}}X$ of X is defined as the graded algebra with homogeneous components ${{\mathrm{\text {gr}}}}^p X = {{\mathrm{\mathcal {F}}}}^p X \big / {{\mathrm{\mathcal {F}}}}^{p+1} X$. We have ${{\mathrm{\text {gr}}}}^0 X = X / I$.

Lemma 81

X / I is naturally identified with $P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$.

Proof

Let $x = (x_p + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p \in X^n$. Let $u_p \in I_0$ and $z_p \in X_0^n$ such that $x_p = u_p + z_p$ and $z_p$ does not contain a summand in $I_0$ (or is zero), i.e. $z_p \in {{\mathrm{\text {Sym}}}}_P({{\mathrm{\mathcal {M}}}})$. Then, $z_p - z_1 = x_p - x_1 - (u_p - u_1) \in I_0$. Hence, $z_p = z_1$ for all p. Therefore, $z := ( z_1 + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p \in X^n$ and x define the same equivalence class in X / I. It is clear that different values of $z_1$ yield different equivalence classes. $\square$

Lemma 82

There is a natural isomorphism ${{\mathrm{\text {gr}}}}^\bullet X \cong B^\bullet \otimes _P T$ of graded algebras.

Proof

The inclusions $B^p \hookrightarrow {{\mathrm{\mathcal {F}}}}^p X$ induce a P-linear map $B \longrightarrow {{\mathrm{\text {gr}}}}X$. From this, we obtain a map $B \otimes _P T \rightarrow {{\mathrm{\text {gr}}}}X$ via $B^p \otimes _P T \ni b \otimes t \mapsto bt \in {{\mathrm{\text {gr}}}}^p X$. The claim follows since the monomials in $B^p$ span the free T-module ${{\mathrm{\text {gr}}}}^p X$: The image of linearly independent monomials in $B^p$ under the above map is obviously T-linearly independent. Now for a given $x \in {{\mathrm{\mathcal {F}}}}^p X,$ decompose it into homogeneous elements $x^n = (x_{n,q} + {{\mathrm{\mathcal {F}}}}^q X_0^n)_q \in {{\mathrm{\mathcal {F}}}}^p X^n$. Split $x_{n,q} = b_{n,q} + y_{n,q}$ with $y_{n,q} \in {{\mathrm{\mathcal {F}}}}^{p+1} X_0^n$ and $b_{n,q} \in {{\mathrm{\mathcal {F}}}}^p X_0^n$ does not contain a summand in ${{\mathrm{\mathcal {F}}}}^{p+1}X_0^n$. Then $b_{n,q} - b_{n,{p+1}} \in {{\mathrm{\mathcal {F}}}}^{p+1}X_0^n$ for $q > p$, and hence this difference vanishes. Set $b^n = (b_{n,p+1} + {{\mathrm{\mathcal {F}}}}^q X_0^n)_q$. We have that $b^n$ and $x^n$ define the same equivalence class in ${{\mathrm{\text {gr}}}}^p X^n$ and hence $b = \sum _n b^n$ and x define the same equivalence class in ${{\mathrm{\text {gr}}}}^p X$. Each $b^n$ is in the image of $B^p \otimes _P T \rightarrow {{\mathrm{\text {gr}}}}^p X$. $\square$

1.6 A.6: Form degree

We can filter the algebra X by form degree. For $n \in {{\mathrm{\mathbb {Z}}}}$ and $j \in {{\mathrm{\mathbb {N}}}}_0$, we set $X_0^{n,j} = P \otimes {{\mathrm{\text {Sym}}}}^j( {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*) \cap X_0^n$ and define the homogeneous components of $X^{(j)} = \bigoplus _n X^{n,j}$ to be

$$\begin{aligned} X^{n,j} = \lim _{\leftarrow p} \frac{X_0^{n,j}}{ {{\mathrm{\mathcal {F}}}}^p X_0^n \cap X_0^{n,j}}. \end{aligned}$$

We have

Lemma 83

If $x_j \in X^n$ have form degree j,  then $\sum _j x_j$ converges in X.

Proof

Fix n. Let g denote the ghost degree and a the anti-ghost degree. This means that $g = \deg$ on (homogeneous elements in) $P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$ and $g = 0$ on ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$. Similarly, $a = \deg$ on $P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ and $a = 0$ on ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$. Hence $g \geqslant 0$, $a \leqslant 0$ and $a+g = \deg$. We decompose a summand $s \in X_0^{n,j}$ of a representative of $x_j$ as $s = a_j \otimes x_{j,1} \dots x_{j,j} \in X_0^{n,j}$ according to form degree. Let $l_j$ be the number of factors of positive degree in this decomposition. Then, $g(x_j) \geqslant l_j$ and $a(x_j) \leqslant -(j-l_j)$. Hence,

$$\begin{aligned} g(x_j)&= a(x_j) + (g(x_j)-a(x_j)) \geqslant a(x_j) + (l_j + (j - l_j)) = a(x_j) + j \\&= n + j - g(x_j). \end{aligned}$$

So, $g(x_j) \geqslant \frac{1}{2}(n+j)$. Set $p(j) = \max \{ k \in {{\mathrm{\mathbb {Z}}}}: k \leqslant \frac{j+n}{2}\}$. We obtain $x_j \in {{\mathrm{\mathcal {F}}}}^{p(j)} X^n$. Now apply Lemma 77. $\square$

Lemma 84

If $x \in X^n$, then there are $x_j \in X^n$ of form degree j with $x = \sum _j x_j$.

Proof

Write $x = \sum _l x^l$ with $x^l \in {{\mathrm{\mathcal {F}}}}^l X_0^n$. Expand each $x^l = \sum _j x_j^l$ where the sum is finite with $x_j^l$ being of form degree j. By Lemma 77, $x_j = \sum _l x_j^l$ converges to an element of $X^n$ of form degree j. By Lemma 83, the sum $\sum _j x_j$ converges. We have $x = \sum _l \sum _j x_j^l = \sum _j \sum _l x_j^l$, which can be verified evaluating both sides modulo ${{\mathrm{\mathcal {F}}}}^p$ for general p. $\square$

Lemma 85

For $\xi _j \in X^n$ of form degree j with $\sum _j \xi _j = 0,$ we have $\xi _j = 0$.

Proof

Write $\xi _j = (\xi _{j,p} + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p$ with $\xi _{j,p} \in X_0^{n,j}$. The p-th component of $\sum _j \xi _j$ has representative $\sum _{j} \xi _{j,p} \in {{\mathrm{\mathcal {F}}}}^p X_0^n$, where this sum is effectively finite by the proof of Lemma 83. We see that $\xi _{j,p} \in {{\mathrm{\mathcal {F}}}}^p X_0^n$ by expanding in a P-basis of $X_0$ consisting of monomials in basis elements of the underlying vector space ${{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*$. $\square$

1.7 A.7: Symplectic case

Consider the graded commutative algebra $X_0 = {{\mathrm{\mathbb {R}}}}[x_i, y_i, e_{j}^{(l)}, {e_{j}^{(l)}}^*]$ where $x_i, y_i$ are of degree zero and $e_{j}^{(l)}$ and ${e_j^{(l)}}^*$ are of degree $-l$ and l, respectively, and there are only finitely many generators of a given degree. Define a Poisson structure by setting $\{x_i, y_i\} = \delta _{ij}$, $\{e_j^{(l)}, e_m^{(n)}\} = \delta _{jm}\delta _{ln}$ and setting all other brackets between generators to zero. We complete the space $X_0$ to the space X as described above. The partial derivatives

$$\begin{aligned} \frac{\partial }{\partial x_i}&= - \{y_i, -\}&\frac{\partial }{\partial y_i}&= \{x_i, -\}&\frac{\partial }{\partial e_j^{(l)}}&= (-1)^{l+1}\{ {e_j^{(l)}}^*, -\}&\frac{\partial }{\partial {e_j^{(l)}}^*}&= \{ e_j^{(l)}, -\} \end{aligned}$$

are defined via the bracket and hence are all well defined on X.

Lemma 86

Let $X_i, Y_i, E_j^{(l)}, {E_j^{(l)}}^*$ be elements of X such that the assignments

$$\begin{aligned} (x_i, y_i, e_j^{(l)}, {e_j^{(l)}}^*)&\mapsto (x_i, Y_i, e_j^{(l)}, {E_j^{(l)}}^*) \\ (x_i, y_i, e_j^{(l)}, {e_j^{(l)}}^*)&\mapsto (X_i, Y_i, E_j^{(l)}, {E_j^{(l)}}^*) \end{aligned}$$

both define automorphisms $X \rightarrow X$ of graded commutative algebras. Then the latter map is a Poisson automorphism if there exists an element $S(x_i, Y_i, e_j^{(l)}, {E_j^{(l)}}^*) \in X$ such that

$$\begin{aligned} \frac{ \partial S}{ \partial x_i}&= y_i&\frac{ \partial S }{ \partial Y_i}&= X_i&\frac{\partial S}{\partial e_j^{(l)}}&= (-1)^l {e_j^{(l)}}^*&\frac{\partial S}{\partial {E_j^{(l)}}^*}&= E_j^{(l)}. \end{aligned}$$

Here, the partial derivatives with respect to the new variables are defined via the chain rule.

Proof

Set $\xi _i^{(0)} = x_i$ and $\xi _j^{(l)} = e^{(l)}_j$ for $l > 0$ and similarly $\eta ^{(0)}_i = y_i$ and $\eta _j^{(l)} = {e^{(l)}_j}^*$. We use capital Greek letters $\Xi$ and H for the corresponding transformed variables. We express the bracket as

$$\begin{aligned} \{f,g\} =&\sum _{l} (-1)^{l \deg f} \sum _j \left( (-1)^{l} \frac{ \partial f}{ \partial \xi _j^{(l)}}\frac{\partial g}{\partial \eta _j^{(l)}} - \frac{\partial f}{ \partial \eta _j^{(l)}} \frac{\partial g}{\partial \xi _j^{(l)}}\right) , \end{aligned}$$

since both sides define derivations which agree on generators. The sums converge by Lemma 77. We have

$$\begin{aligned} \frac{\partial S}{\partial \xi _j^{(l)}}= (-1)^l \eta _j^{(l)} \,\, \text { and } \,\, \frac{\partial S}{\partial H_j^{(l)}} = \Xi _j^{(l)} \,\, \text { so also } \,\, \quad \frac{ \partial \eta _{p}^{(q)}}{ \partial \xi _{j}^{(l)}} = \frac{ \partial \eta _{j}^{(l)}}{ \partial \xi _{p}^{(q)}} (-1)^{q+l+ql}. \end{aligned}$$

There are functions $f_{j,l}(\xi , \eta ) = \Xi _{j}^{(l)}$ and $g_{j,l}(\xi ,\eta ) = H_j^{(l)}$ realizing the change of coordinates so that

$$\begin{aligned} f_{j,l}(\xi , \eta ( \xi , H))&= \frac{\partial S}{\partial H_j^{(l)}}&g_{j,l}(\xi , \eta ( \xi , H))&= H_j^{(l)}. \end{aligned}$$

We obtain in the variables $(\xi _j^{(l)}, H_j^{(l)}),$

$$\begin{aligned} \frac{\partial f_{m,n}}{\partial \xi _j^{(l)} } + \sum _{p,q} \frac{\partial \eta _p^{(q)}}{\partial \xi _j^{(l)}} \frac{\partial f_{m,n}}{\partial \eta _{p}^{(q)}}&= \frac{\partial ^2 S}{\partial \xi _j^{(l)} \partial H_m^{(n)}}&&,\\ \frac{ \partial g_{m',n'} }{ \partial \xi _{j}^{(l)}} + \sum _{pq} \frac{\partial \eta _{p}^{(q)}}{\partial \xi _j^{(l)}} \frac{\partial g_{m',n'} }{\partial \eta _p^{(q)}}&= 0&\sum _{pq}\frac{ \partial \eta _p^{(q)}}{\partial H_{m}^{(n)}} \frac{\partial g_{m',n'}}{\partial \eta _{p}^{(q)}}&= \delta _{mm'} \delta _{nn'}. \end{aligned}$$

These expressions make sense in the completion by Lemma 77 since (j, l, n, m) are fixed and the $\eta _p^{(q)}$ derivatives are of non-decreasing and unbounded degree. Using those equalities, we calculate in the variables $(\xi _j^{(l)}, H_j^{(l)})$ the bracket $\{f_{m,n}, g_{m',n'}\}$ as

$$\begin{aligned}&\sum _{l} (-1)^{l n} \sum _j ( (-1)^{l} \frac{ \partial f_{m,n}}{ \partial \xi _j^{(l)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}} - \frac{\partial f_{m,n}}{ \partial \eta _j^{(l)}} \frac{\partial g_{m',n'}}{\partial \xi _j^{(l)}})\\&\quad = \sum _{jl} (-1)^{l(n+1)} \frac{\partial ^2 S}{\partial \xi _j^{(l)} \partial H_m^{(n)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}} \\&\qquad - \sum _{jl} (-1)^{ln} \bigg ( \sum _{pq} (-1)^l \frac{ \partial \eta _p^{(q)}}{ \partial \xi _j^{(l)}} \frac{\partial f_{m,n} }{ \partial \eta _p^{(q)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}} + \frac{\partial f_{m,n}}{ \partial \eta _j^{(l)}} \frac{\partial g_{m',n'}}{\partial \xi _j^{(l)}} \bigg )\\&\quad = \sum _{jl} (-1)^{l} \frac{\partial ^2 S}{ \partial H_m^{(n)} \partial \xi _j^{(l)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}}\\&\qquad - \sum _{jl} (-1)^{ln} \sum _{pq} \bigg ( (-1)^l \frac{ \partial \eta _p^{(q)}}{ \partial \xi _j^{(l)}} \frac{\partial f_{m,n} }{ \partial \eta _p^{(q)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}} - \frac{\partial f_{m,n}}{ \partial \eta _j^{(l)}} \frac{\partial \eta _{p}^{(q)}}{\partial \xi _j^{(l)}} \frac{\partial g_{m',n'} }{\partial \eta _p^{(q)}} \bigg )\\&\quad = \sum _{jl} \frac{\partial \eta _{j}^{(l)}}{\partial H_m^{(n)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}}\\&\qquad - \sum _{jlpq} \bigg ( (-1)^{ln+l} \frac{ \partial \eta _p^{(q)}}{ \partial \xi _j^{(l)}} \frac{\partial f_{m,n} }{ \partial \eta _p^{(q)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}} - (-1)^{ql + qn+l} \frac{\partial \eta _{p}^{(q)}}{\partial \xi _j^{(l)}} \frac{\partial f_{m,n}}{ \partial \eta _j^{(l)}} \frac{\partial g_{m',n'} }{\partial \eta _p^{(q)}} \bigg )\\&\quad = \delta _{mm'}\delta _{nn'} \\&\qquad - \sum _{jlpq} \bigg ( (-1)^{ln+l} \frac{ \partial \eta _p^{(q)}}{ \partial \xi _j^{(l)}} \frac{\partial f_{m,n} }{ \partial \eta _p^{(q)}} \frac{\partial g_{m',n'}}{\partial \eta _j^{(l)}} - (-1)^{qn + q} \frac{\partial \eta _{j}^{(l)}}{\partial \xi _p^{(q)}} \frac{\partial f_{m,n}}{ \partial \eta _j^{(l)}} \frac{\partial g_{m',n'} }{\partial \eta _p^{(q)}} \bigg ) \\&\quad = \delta _{mm'} \delta _{nn'}. \end{aligned}$$

$\square$