$A_\infty$-Algebra from Supermanifolds

Abstract

Inspired by the analogy between different types of differential forms on supermanifolds and string fields in superstring theory, we construct new multilinear non-associative products of forms which yield, for a single fermionic dimension, an $A_\infty$-algebra as in string field theory. For multiple fermionic directions, we give the rules for constructing non-associative products, which are the basis for a full $A_\infty$-algebra structure to be yet discussed.

Appendices

Appendix A: A Nod to Coderivations and $A_\infty$-Algebras

For the sake of readability of the paper and for future reference, we now briefly introduce some basic elements in the theory of $A_\infty$-algebras.

Generally speaking, $A_\infty$-algebras are examples of non-associative algebras, first introduced by Stashef, see [34] in the context of homotopy theory. In what follows, by the way, we will give a different treatment—somewhat more abstract—compared to the original one, based on the notion of cotensor algebra and coderivations (see [33] for an extended and in-depth discussion).

We start recalling that, over a field or a ring k, a $\mathbb {Z}$-graded coassociative coalgebra is a pair $(C, \Delta )$ where is $\mathbb {Z}$-graded k-module and $\Delta : C \rightarrow C \otimes C$ is a coassociative coproduct, that is, it satisfies $(1 \otimes \Delta ) \Delta = (\Delta \otimes 1 ) \Delta$.

So far we have described objects in the category ${\mathsf {CoAlg}}_k (C)$: to complete the categorial description, we need to introduce morphisms between the objects of the category. Given two coalgebras $(C, \Delta _C)$ and $(D, \Delta _D)$, we call a cohomomorphisms degree 0 maps $\mathfrak {F} : C \rightarrow D$ satisfying $\Delta _D \mathfrak {F} = (\mathfrak {F} \otimes \mathfrak {F}) \Delta _C$. Notice that here the degree is a $\mathbb {Z}$-degree and it refers to the $\mathbb {Z}$-grading of the k-modules C and D, that is, for a cohomomorphism $\mathfrak {F}: C \rightarrow D$, if c is a homogeneous element of degree i in C, i.e., if $c \in C^{(i)} \subset C$, then $\mathfrak {F} (c)$ is a homogeneous element of degree i in D, i.e., $\mathfrak {F} (c) \in D^{(i)}$.

A coderivation$\mathfrak {D}: C \rightarrow C$ on C is a degree 1 map that satisfies the coLeibniz rule, $\Delta \mathfrak {D} = (1 \otimes \mathfrak {D} + \mathfrak {D} \otimes 1) \Delta$. In particular, one defines a differentially graded (coassociative) coalgebra a triple $(C, \Delta , \mathfrak {D})$ where the pair $(C, \Delta )$ is a coassociative coalgebra and $\mathfrak {D}$ is a coderivation satisfying $\mathfrak {D}^2 = 0.$

More in general, given a coalgebra $(C, \Delta )$, one can allow cohomomorphisms and coderivations of any degree, each satisfying the defining properties. In this case, one can introduce the $\mathbb {Z}$-graded k-module $,$ where , and its sub-module , where . Defining the graded commutator as

$$\begin{aligned}{}[\mathfrak {F}, \mathfrak {G}] = \mathfrak {F} \circ \mathfrak {G} - (-1)^{|\mathfrak {F}||\mathfrak {G}|} \mathfrak {G}\circ \mathfrak {F} \end{aligned}$$

(A.1)

for $\deg {\mathfrak {F}} = |\mathfrak {F}|$ and $\deg (\mathfrak {G}) = |\mathfrak {G}|$, one can observe the following fundamental fact: $(CoDer^\bullet _k (C), [\cdot , \cdot ])$ is a Lie subalgebra of $CoEnd^{\, \bullet }_k (C).$ The proof is straightforward: it is enough to check that the commutator closes in $CoDer^\bullet _k (C)$. This is a very useful result, which will be constantly exploited in the following, since it allows one to use the various operations in the Lie algebras, e.g., Jacobi identity, when dealing with coderivations. Notice, in any case, that it is not true that the composition of two coderivation yields a coderivation: this mirrors the fact that the composition of two derivative (vector fields) does not yields a derivative (vector field), but a suitable commutator of them does.

Possibly the most important example of coalgebra is the cotensor algebra of a $\mathbb {Z}$-graded k-module V. The cotensor algebra of V is the pair $({\mathsf {T}} (V), \Delta )$, where and the coassociative multiplication $\Delta$ is defined as

(A.2)

For example, one has that for the tensor $v_1 \otimes v_2 \otimes _3 \in V^{\otimes 3} \subset {\mathsf {T}} (V)$

$$\begin{aligned} \Delta (v_1 \otimes v_2 \otimes v_3)&= 1 \otimes (v_1 \otimes v_2 \otimes v_3) + v_1 \otimes (v_2 \otimes v_3) \\&\quad + (v_1 \otimes v_3) \otimes v_3 + (v_1 \otimes v_2 \otimes v_3) \otimes 1 \end{aligned}$$

where the first summand belongs to $k \otimes V^{\otimes 3} \subset {\mathsf {T}} (V) \otimes {\mathsf {T}} (V)$, the second to $V \otimes V^{\otimes 2} \subset {\mathsf {T}} (V) \otimes {\mathsf {T}} (V)$ and so on.

We are interested into the coderivations corresponding to this coproduct: these can be characterized by dualizing the construction for the ordinary tensor algebra. In particular, there is a map from the coderivations on ${\mathsf {T}}(V)$ to the multilinear maps ${\mathsf {T}} (V) \rightarrow V$, which is just given by the composition of $\mathfrak {D} : {\mathsf {T}} (V) \rightarrow {\mathsf {T}}(V)$ with the projection $\mathfrak {p}_1 : {\mathsf {T}}(V) \rightarrow V$. Its inverse is constructed via the following steps: first, one takes a collection of multilinear maps $\{m_k : V^{\otimes k} \rightarrow V \}_{k \ge 0}$ such that $\deg (m_k ) = 1$ for any $k \ge 0$ and with $v_1 \otimes \cdots \otimes v_k \mapsto m_k (v_1, \ldots , v_k)$ for any $v_1, \ldots , v_k \in V$ of homogeneous degree so that and where we define $m_0 : k \rightarrow V$ is so that $m_0(1) \in V$ has degree 1. Then, the maps are extended to the whole ${\mathsf {T}} (V)$ as follows: $m_k \mapsto \mathfrak {m}_k : {\mathsf {T}} (V) \rightarrow {\mathsf {T}} (V)$ with

(A.3)

where we notice that the sign is there because of the Koszul rule of commutation of the degree 1 map $m_k$ with the homogeneous tensors $v_1, \ldots , v_{\ell -1} \in V$, with $\deg (v_k) = |v_k|$ and that, in particular, $\mathfrak {m}_k : V^{\otimes n} \subset {\mathsf {T}} (V) \rightarrow V^{\otimes n-k +1} \subset {\mathsf {T}} (V).$ Finally, the coderivation $\mathfrak {m}: {\mathsf {T}} (V) \rightarrow {\mathsf {T}}(V)$ is given summing over all of the maps $\mathfrak {m}_k$, as follows:

(A.4)

Once a coderivation $\mathfrak {m} = \sum _k \mathfrak {m}_k$ on ${\mathsf {T}} (V)$ has been constructed, it is natural to ask whenever it is a codifferential, that is, whenever it is such that $\mathfrak {m}^2 = 0:$, the definition is related to this question. Let $({\mathsf {T}}(V), \Delta )$ be the cotensor algebra of a $\mathbb {Z}$-graded vector space, and let $\mathfrak {m} = \sum _k \mathfrak {m}_k$ be the coderivation on ${\mathsf {T}}(V)$. Then, we call a weak$A_\infty$-algebra the differentially graded coalgebra $({\mathsf {T}}(V), \Delta , \mathfrak {m})$, that is, for $({\mathsf {T}}(V), \Delta , \mathfrak {m})$ to be a weak $A_\infty$-algebra, the coderivation $\mathfrak {m}$ is actually a codifferential, satisfying $\mathfrak {m}^2= 0$. In particular, a weak $A_\infty$-algebra is an $A_\infty$-algebra if $m_0 : k \rightarrow V$ is the zero map.

We now look forward to unravel the condition $\mathfrak {m}^2 = 0$ in the definition of an $A_\infty$-algebra in order to see which sort of condition it gives on the multilinear maps $m_k$. With an eye to the previous section, first of all let us observe the following fact: $\mathfrak {m}^2 : {\mathsf {T}} (V) \rightarrow {\mathsf {T}} (V)$ is a (degree 2) map having image into $V\oplus V^{\otimes 2} \oplus \cdots \subset {\mathsf {T}}(V)$ (recall that $m_0$= 0), therefore requiring $\mathfrak {m}^2= 0$, is equivalent to write down the corresponding conditions in all of the summand $V^{\otimes k \ge 1}$ of the image separately. Defining $\mathfrak {p}_k : {\mathsf {T}} (V) \rightarrow V^{\otimes k}$ the projection map onto the ith component of ${\mathsf {T}} (V)$, the condition $\mathfrak {m}^2 = 0$ is equivalent to $\mathfrak {p}_k \circ \mathfrak {m}^2 = 0$ for any $k\ge 0$. What is crucial, though, is that due to the anticommutativity of the $\mathfrak {m}_i$’s, one has that $\mathfrak {p}_1 \mathfrak {m}^2 = 0$ is a sufficient condition for $\mathfrak {m}^2 = 0:$; therefore, the only thing we will be concerned will be the projection onto the first factor $V \subset {\mathsf {T}} (V).$

The condition $\mathfrak {p}_1 \mathfrak {m}^2 = 0$ can be rewritten in terms of the multilinear maps $m_k : V^{\otimes k } \rightarrow V$ making up the codifferential in a very compact and elegant fashion. It reads:

$$\begin{aligned} \sum _{\kappa {+} \ell = n + 1} \sum _{i=0}^{\kappa -1} (-1)^{\sum _{j=1}^i |v_j |} m_{\kappa } \left( v_1, \ldots , v_{i}, m_{\ell } (v_{i+1}, \ldots , v_{i+\ell }), v_{i+\ell +1}, \ldots , v_n \right) {=} 0,\nonumber \\ \end{aligned}$$

(A.5)

where any tensor $v_j \in V$ is understood to be homogeneous of degree $|v_j|$. Notice that the map is well defined; indeed, $m_\ell$ acts on a $\ell$-tensor $v_{i} \otimes \cdots \otimes v_{i+\ell } \in V^{\otimes \ell }$ and $m_{\kappa }$ acts on a $i+1 + (n-i-\ell ) = n+1-\ell = \kappa$-tensor $v_1 \otimes \cdots , v_i \otimes m_{\ell } (v_{i+1}, \ldots , v_{i+\ell }) \otimes v_{i+ \ell + 1 } \otimes \cdots \otimes v_{n} \in V^{\otimes \kappa }$.

We are now in the position to write the first relations coming from Eq. (A.5): notice that since $m_0 = 0$, the first non-trivial relation is given by the choice $\kappa + \ell = 2$ with $\kappa = \ell = 1$ and it reads

$$\begin{aligned} m^2_1 (v_1) = 0, \end{aligned}$$

(A.6)

which says that recalling that $m_1$ has degree 1, the pair $(V, m_1)$ is a complex of $\mathbb {Z}$-graded k-modules, having $m_1 : V^{(n)} \rightarrow V^{(n+1)}$ as the differential of the complex.

The second relation comes from the choice $\kappa + \ell = 3$, and it yields the condition

$$\begin{aligned} m_1 (m_2 (v_1, v_2)) + m_2 (m_1 (v_1), v_2) + (-1)^{|v_1|} m_2 (v_1, m_1 (v_2)) = 0, \end{aligned}$$

(A.7)

which is the Leibniz rule for the differential $m_1$ with respect to the product $m_2 : V^{\otimes 2 } \rightarrow V$. The third relations—possibly the most characterizing one for an $A_\infty$-algebra—come from choosing $\kappa + \ell = 4$, so that one has

$$\begin{aligned}&m_2 (m_2 (v_1, v_2), v_3) + (-1)^{|v_1|} m_2 (v_1, m_2 (v_2, v_3)) \nonumber \\&\quad + m_1 (m_3 (v_1, v_2, v_3)) + m_3 (m_1(v_1), v_2, v_3) + (-1)^{|v_1|} m_3 (v_1, m_1(v_2), v_3 ) \nonumber \\&\quad + (-1)^{|v_1| + |v_2|} m_3 (v_1, v_2, m_1 (v_3)) = 0. \end{aligned}$$

(A.8)

This condition means that the associativity for the product $m_2$ is broken by the terms containing the 3-product $m_3: V^{\otimes 3} \rightarrow V$: one says that $m_2$ is associative up to homotopy in $m_3$. Keep going up, one sees that the 3-associativity for $m_3$ is broken by a term in $m_4$ and so on.

In this context, the Lie algebra structure on the coderivations offers a very compact and useful environment to reproduce the above relations, defining an $A_\infty$-algebra. In general, given a coderivation as in (A.4) one has to compute

$$\begin{aligned}{}[\mathfrak {m}, \mathfrak {m}] = \sum _{k, l} \left[ \mathfrak {m}_k, \mathfrak {m}_l \right] . \end{aligned}$$

(A.9)

The right-hand side has to be considered carefully. First of all, we note that we have taken $\deg (m_k) = 1$ for any k; likewise, we define $\deg \mathfrak {m}_k = \deg (m_k) = 1$, so the commutator above is indeed an anticommutator for any k and l, that is,

$$\begin{aligned}{}[\mathfrak {m}_k , \mathfrak {m}_l] = \mathfrak {m}_k \mathfrak {m}_l + \mathfrak {m}_l \mathfrak {m}_k. \end{aligned}$$

(A.10)

Again, it is useful to divide the various cases by letting $k+l = n+1$ for $n\ge 1$ as above: then, we have that in general $[\mathfrak {m}_k , \mathfrak {m}_l] : V^{\otimes \ge (k+l-1)} \rightarrow V^{\otimes \ge 1}.$ To make contact with $A_\infty$-relations, we restrict our attention to the case the image of the commutator is just V and we look at the first instances. For $n=1$, we have

$$\begin{aligned}{}[\mathfrak {m}_1, \mathfrak {m}_1] (v_1)= 2 m_1 (v_1). \end{aligned}$$

(A.11)

For $n=2$, we have

$$\begin{aligned}&[\mathfrak {m}_1, \mathfrak {m}_2] (v_1 , v_2) = m_1 (m_2 (v_1, v_2)) \nonumber \\&\quad + m_2 ({m}_1 (v_1) , v_2 ) + (-1)^{|v_1|} m_2 (v_1 \otimes m_1 (v_2)). \end{aligned}$$

(A.12)

Notice that for $n=3$ we should start considering more than one commutator; indeed, we find $[\mathfrak {m}_1, \mathfrak {m}_3]$ and $[\mathfrak {m}_2, \mathfrak {m}_2].$ Clearly, as n grows, there will be more and more commutators to take into account. Now, the $A_\infty$-algebra relations can be written in a very compact way using these commutators; for example, the first relations read

$$\begin{aligned}{}[\mathfrak {m}_1, \mathfrak {m}_1] = 0, \qquad [\mathfrak {m}_1, \mathfrak {m}_2] = 0, \qquad [\mathfrak {m}_1, \mathfrak {m}_3 ] + \frac{1}{2}[\mathfrak {m}_2, \mathfrak {m}_2] = 0, \end{aligned}$$

(A.13)

where the projection on V is understood.

Appendix B: How to Compute with $\Theta (\iota _D)$ and $\delta (\iota _D)$

In order to clarify the action of $\Theta (\iota _D), \delta (\iota _D)$ and $Z_D$, we present some detailed calculations. Let us compute the action of $\Theta (\iota _D)$ on $\delta (\mathrm{d}\theta ^\alpha )$ with $\alpha =1,2$.

$$\begin{aligned} \Theta (\iota _D) \delta (\mathrm{d}\theta ^\alpha )= & {} - i \lim _{\epsilon \rightarrow 0 }\int _{-\infty }^\infty \mathrm{d}t \frac{e^{i t \iota _D}}{t + i \epsilon } \delta (\mathrm{d}\theta ^\alpha ) = -i \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{ \delta (\mathrm{d}\theta ^\alpha + i D^\alpha t)}{t + i \epsilon } \nonumber \\= & {} - \frac{1}{D^\alpha } \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{ \delta (t - i \frac{\mathrm{d}\theta ^\alpha }{D^\alpha })}{t + i \epsilon } = \frac{i}{\mathrm{d}\theta ^\alpha } \in \Omega ^{(-1|0)}_{\mathbb {P}^{1|2}} \end{aligned}$$

(B.1)

where the coefficient $D^\alpha$ drops out from the computation (but it must be different from zero in order to have a meaningful computation). In the same way, we have

$$\begin{aligned} \delta (\iota _D) \delta (\mathrm{d}\theta ^\alpha ) = \int _{-\infty }^\infty \mathrm{d}t {e^{i t \iota _D}} \delta (\mathrm{d}\theta ^\alpha ) = \int _{-\infty }^\infty \mathrm{d}t \delta (\mathrm{d}\theta ^\alpha + i D^\alpha t) = -\frac{i}{ D^\alpha } \in \Omega ^{(0|0)}_{\mathbb {P}^{1|2}},\nonumber \\ \end{aligned}$$

(B.2)

using the distributional properties. Again the requirement that $D^\alpha$ is different from zero is crucial.

Let us compute the action of $\Theta (\iota _D)$ on the product $\mathrm{d}\theta ^\beta \delta (\mathrm{d}\theta ^\alpha )$. We assume that $\alpha \ne \beta$; otherwise, it vanishes. Applying the same rules, we have

$$\begin{aligned} \Theta (\iota _D) \Big ( \mathrm{d}\theta ^\beta \delta (\mathrm{d}\theta ^\alpha ) \Big )= & {} - i \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{e^{i t \iota _D}}{t + i \epsilon }\Big ( \mathrm{d}\theta ^\beta \delta (\mathrm{d}\theta ^\alpha ) \Big ) \nonumber \\= & {} -i \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{ (\mathrm{d}\theta _\beta + i D_\beta t) \delta (\mathrm{d}\theta ^\alpha + i D^\alpha t)}{t + i \epsilon } \nonumber \\= & {} \frac{-i}{i D^\alpha } \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{ (\mathrm{d}\theta ^\beta + i D^\beta t)}{t + i \epsilon } \delta \Big (t - \frac{i \mathrm{d}\theta _\alpha }{D^\alpha }\Big ) \nonumber \\= & {} -\frac{1}{D^\alpha } \Big ( \mathrm{d}\theta _\beta + i D^\beta \frac{i \mathrm{d}\theta ^\alpha }{D_\alpha }\Big ) \frac{D^\alpha }{i \mathrm{d}\theta ^\alpha } \nonumber \\= & {} i \Big ( \frac{\mathrm{d}\theta ^\beta }{\mathrm{d}\theta ^\alpha } - \frac{D^\beta }{D^\alpha }\Big ) \in \Omega ^{(0|0)}_{\mathbb {P}^{1|2}} \end{aligned}$$

(B.3)

from which it immediately follows that if $\alpha = \beta$, then both members vanish. Analogously, we have

$$\begin{aligned} \delta (\iota _D) \Big ( \mathrm{d}\theta ^\beta \delta (\mathrm{d}\theta ^\alpha ) \Big )= & {} \int _{-\infty }^\infty \mathrm{d}t e^{i t \iota _D} \Big ( \mathrm{d}\theta ^\beta \delta (\mathrm{d}\theta ^\alpha ) \Big )\nonumber \\= & {} \int _{-\infty }^\infty \mathrm{d}t (\mathrm{d}\theta _\beta + i D_\beta t) \delta (\mathrm{d}\theta ^\alpha + i D^\alpha t) \nonumber \\= & {} \frac{1}{i D^\alpha } \int _{-\infty }^\infty \mathrm{d}t (\mathrm{d}\theta ^\beta + i D^\beta t) \delta \Big (t - \frac{i \mathrm{d}\theta _\alpha }{D^\alpha }\Big ) \nonumber \\= & {} \frac{1}{i D^\alpha } \Big ( \mathrm{d}\theta ^\beta - \frac{D^\alpha }{D^\beta } \mathrm{d}\theta ^\alpha \Big ) \in \Omega ^{(1|0)}_{\mathbb {P}^{1|2}} \end{aligned}$$

(B.4)

which also vanishes if $\alpha =\beta$.

Let us also consider the following expressions:

$$\begin{aligned} \Theta (\iota _D) \Big ( \frac{1}{\mathrm{d}\theta _\beta } \delta (\mathrm{d}\theta _\alpha ) \Big )= & {} -i \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{ \delta (\mathrm{d}\theta _\alpha + i D_\alpha t)}{(\mathrm{d}\theta _\beta + i D_\beta t) (t + i \epsilon )} \nonumber \\= & {} \frac{-i}{i D_\alpha } \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{1}{(\mathrm{d}\theta _\beta + i D_\beta t)(t + i \epsilon )} \delta \Big (t - \frac{i \mathrm{d}\theta _\alpha }{D_\alpha }\Big ) \nonumber \\= & {} -\frac{1}{D_\alpha } \frac{1}{\Big ( \mathrm{d}\theta _\beta + i D_\beta \frac{i \mathrm{d}\theta _\alpha }{D_\alpha }\Big )} \frac{D_\alpha }{i \mathrm{d}\theta _\alpha } \nonumber \\= & {} i \frac{1}{\Big ( \frac{\mathrm{d}\theta _\beta }{\mathrm{d}\theta _\alpha } - \frac{D_\beta }{D_\alpha }\Big )} \frac{1}{\mathrm{d}\theta _\alpha ^2} \in \Omega ^{(-2|0)}_{\mathbb {P}^{1|2}} \end{aligned}$$

(B.5)

which is an inverse form. Notice that if $\alpha =\beta$, the product $\Big ( \frac{1}{\mathrm{d}\theta ^\beta } \delta (\mathrm{d}\theta ^\alpha ) \Big )$ is ill defined, and this is consistent with the fact that also the right-hand side is divergent.

Let us now compute the action of $\Theta (\iota _D)$ on $\Omega ^{(0|2)}_{\mathbb {P}^{1|2}}$. This is done as follows:

$$\begin{aligned} \Theta (\iota _D) \Big ( \delta (\mathrm{d}\theta _1) \delta (\mathrm{d}\theta _2) \Big )= & {} -i \lim _{\epsilon \rightarrow 0 } \int _{-\infty }^\infty \mathrm{d}t \frac{e^{i t \iota _D}}{t + i \epsilon } \delta (\mathrm{d}\theta _1) \delta (\mathrm{d}\theta _2) \nonumber \\= & {} -i \lim _{\epsilon \rightarrow 0 }\int _{-\infty }^\infty \mathrm{d}t \frac{\delta (\mathrm{d}\theta _1 + i t D_1) \delta (\mathrm{d}\theta _2 + i t D_2)}{t + i \epsilon } \nonumber \\= & {} \frac{i}{D_1 D_2} \left( D_1 \frac{\delta \Big (\mathrm{d}\theta _2 - \frac{D_2}{D_1} \mathrm{d}\theta _1\Big )}{\mathrm{d}\theta _1} - D_2 \frac{\delta \Big (\mathrm{d}\theta _1 - \frac{D_1}{D_2} \mathrm{d}\theta _2\Big )}{\mathrm{d}\theta _2} \right) \nonumber \\= & {} - \frac{i}{D_1 D_2} \Big (\frac{D_1}{\mathrm{d}\theta _1} + \frac{D_2}{\mathrm{d}\theta _2} \Big ) \delta \Big ( \frac{\mathrm{d}\theta _1}{D_1} - \frac{\mathrm{d}\theta _2}{D_2} \Big ) \nonumber \\= & {} = -i \Big (\frac{D_1}{\mathrm{d}\theta _1} + \frac{D_2}{\mathrm{d}\theta _2} \Big ) \delta (D\cdot \mathrm{d}\theta ) \in \Omega ^{(-1|1)}_{\mathbb {P}^{1|2}}\,. \end{aligned}$$

(B.6)

where $(D \cdot \mathrm{d}\theta ) = D_\alpha \epsilon ^{\alpha \beta } \mathrm{d}\theta _\beta$.

Notice that the linear combination of $\mathrm{d}\theta _1$ and $\mathrm{d}\theta _2$ appearing in the first factor is linearly independent from the linear combination appearing in the Dirac delta argument. Notice also that the sign between the two Dirac deltas in the second line is due to the fermionic nature of $\mathrm{d}t$ and of the Dirac delta form. This sign is crucial for the left-hand side and the right-hand side of Eq. (B.6) be consistent. Indeed, if we interchange $\delta (\mathrm{d}\theta _1)$ with $\delta (\mathrm{d}\theta _2)$ in the left-hand side, we get an overall minus sign; on the other hand, on the right-hand side of the equation, by exchanging $\mathrm{d}\theta _1$ and $\mathrm{d}\theta _2$ in the Dirac delta argument again a sign emerges.

Finally, we can consider another independent odd vector field $D'$ and the corresponding operator $\Theta (\iota _{D'})$. Acting on (B.6), it yields

$$\begin{aligned} \Theta (\iota _{D'}) \Theta (\iota _D) \Big ( \delta (\mathrm{d}\theta _1) \delta (\mathrm{d}\theta _2) \Big ) = \frac{\det (D', D)}{(D' \cdot \mathrm{d}\theta ) (D \cdot \mathrm{d}\theta )} \in \Omega ^{(-2|0)}_{\mathbb {P}^{1|2}} \end{aligned}$$

(B.7)

where $(D \cdot \mathrm{d}\theta ) = D_\alpha \epsilon ^{\alpha \beta } \mathrm{d}\theta _\beta$ and $\det (D',D) = D'_\alpha \epsilon ^{\alpha \beta } D_\beta = D' \cdot D$. Notice that in this case, by interchanging $\delta (\mathrm{d}\theta _1)$ with $\delta (\mathrm{d}\theta _2)$, we get again an overall minus sign. This is obtained also by exchanging the coefficients of the vectors D and $D'$, and in this way, we get a minus sign from the determinant $\det (D', D)$.

Let us also consider the action of $\delta (\iota _D)$ on the product of $\delta (\mathrm{d}\theta ^1)\delta (\mathrm{d}\theta ^2)$. We have

$$\begin{aligned} \delta (\iota _D) (\delta (\mathrm{d}\theta ^1)\delta (\mathrm{d}\theta ^2)) = - i \delta ( D\cdot \mathrm{d}\theta ) \in \Omega ^{(0|1)}_{\mathbb {P}^{1|2}}, \end{aligned}$$

(B.8)

and finally,

$$\begin{aligned} \delta (\iota _{D^\prime }) \delta (\iota _D)( \delta (\mathrm{d}\theta ^1)\delta (\mathrm{d}\theta ^2) ) = \mathrm{det}(D', D) \in \Omega ^{(0|0)}_{\mathbb {P}^{1|2}}, \end{aligned}$$

(B.9)

which also follows from (B.7) by the identity $\mathrm{d}\theta ^\alpha \Theta (\iota _{D}) = \delta (\mathrm{d}\theta ^\alpha )$.

The action of a second PCO decreases the picture number as to bring elements of $\Omega ^{p|2}_{\mathbb {P}^{1|2}}$ into superforms having picture number equal to zero. Note that since the PCO Z is formally exact as stressed above, it maps cohomology classes into cohomology classes, $H_{dR}^{(p|2)} \rightarrow H_{dR}^{(p|0)}$; therefore, it is natural to expect that it can only properly act on cohomology classes, and indeed, acting on representatives of $H_{dR}^{(p|2)}$ one never gets inverse forms. Nonetheless, it can be shown that acting on generic elements of the space $\Omega ^{(p|2)}_{\mathbb {P}^{1|2}}$, not necessarily closed, one never produces inverse forms. Let us show this first in a very simple example.

Consider a generic integral form in $\Omega ^{1|2}_{\mathbb {P}^{1|2}} \cong \mathcal {B}er (\mathbb {P}^{1|2})$

$$\begin{aligned} \omega ^{(1|2)}= A(z,\theta ) \mathrm{d}z \delta (\mathrm{d}\theta ^1) \delta (\mathrm{d}\theta ^2) \end{aligned}$$

(B.10)

where $A(z,\theta ^\alpha )$ is a superfield in the local coordinates of $\mathbb {P}^{1|2}$. Being (the analog of) a top form, it is naturally closed. Acting with $Z_D$, one gets

$$\begin{aligned} Z_D(\omega ^{(1|2)})= & {} \mathrm{d}\left[ -i\Theta (\iota _D) A \mathrm{d}z \delta (\mathrm{d}\theta ^1) \delta (\mathrm{d}\theta ^2)\right] -i \Theta (\iota _D) \left[ \mathrm{d}\left( A \mathrm{d}z \delta (\mathrm{d}\theta ^1) \delta (\mathrm{d}\theta ^2) \right) \right] \nonumber \\= & {} d \Big [ A \, \Big ( \frac{D^1}{\mathrm{d}\theta ^1} + \frac{D^2}{\mathrm{d}\theta ^2} \Big ) \mathrm{d}z \, \delta (D\cdot \mathrm{d}\theta ) \Big ] \nonumber \\= & {} 2 \Big ( (D^1 \partial _1 A + D^2 \partial _2 A) \, \mathrm{d}z\, \delta (D\cdot \mathrm{d}\theta ) \Big ) \nonumber \\= & {} 2 D^\alpha \partial _\alpha A \, \mathrm{d}z\, \delta (D\cdot \mathrm{d}\theta ) \Big ) \in \Omega ^{(1|1)}_{\mathbb {P}^{1|2}}, \end{aligned}$$

(B.11)

where $\partial _\alpha A$ are the derivatives with respect to $\theta ^\alpha$ of the superfield A. The result is in $\Omega ^{(1|1)}_{\mathbb {P}^{1|2}}$, it is closed and no inverse form is required. However, the form (B.11) is not the most general (1|1)-pseudoform.

Let us act with a second PCO:

$$\begin{aligned} Z_{D'}\Big [ 2 D^\alpha \partial _\alpha A \, \mathrm{d}z\, \delta (D\cdot \mathrm{d}\theta ) \Big ) \Big ] = 2 \epsilon ^{\alpha \beta } \partial _\alpha \partial _\beta A \, \mathrm{d}z \in \Omega ^{(1|0)}_{\mathbb {P}^{1|2}} \end{aligned}$$

(B.12)

which is a superform in $\Omega ^{(1|0)}_{\mathbb {P}^{1|2}}$, it does not contain any inverse form and it is independent of the odd vector fields $D, D'$. Note that this particular expression is closed, since $\partial _1^2 = \partial ^2_2 = \{\partial _1, \partial _2\}=0$. No inverse form is needed in this case.