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.

1 Introduction

In the quantization of gauge systems, the so-called BRST complex plays a prominent role [12]. In the Hamiltonian formalism, the theory is called BFV theory and goes back to Batalin, Fradkin, Fradkina and Vilkovisky [1, 2, 10, 11].

In the Hamiltonian formulation of gauge theory, the presence of gauge freedom yields constraints in the phase space M of the system. The gauge group still acts on the resulting constraint surface $M_0 \subset M$. The physical observables are the functions on the quotient $\tilde{M}$ of the constraint surface $M_0$ by this action. One wishes to quantize those observables. In the BRST method, one introduces variables of non-zero degree to the Poisson algebra P of functions on the original phase space. One then constructs the so-called BRST differential on the resulting complex and recovers the functions on the subquotient $\tilde{M}$ as the cohomology of that complex in degree zero. One may then attempt to quantize the system by quantizing the BRST complex instead of the algebra of functions on $\tilde{M}$.

The quantization procedure involves the construction of gauge invariant observables from the cohomology of the BRST complex [3, 19]. Kostant and Sternberg gave a mathematically rigorous description of the theory [14] in the case where the constraints arise from a Hamiltonian group action on phase space. They make certain assumptions that allow the BRST complex to be constructed as a double complex combining a Koszul resolution of the vanishing ideal J of the constraint surface $M_0\subset M$ with the Lie algebra cohomology of the gauge group. In more general cases, the Koszul complex does not yield a resolution and one has to use a much bigger Tate resolution.

More recently, Felder and Kazhdan formalized the corresponding construction in the Lagrangian formulation of the theory [8]. They consider general Tate resolutions. The aim of this note is to perform a similar formalization in the Hamiltonian setting. We consider Poisson algebras P as a starting point, which arise in the Hamiltonian viewpoint as the functions on phase space. We define the notion of a BFV model for a coisotropic ideal $J\subset P$. In the Hamiltonian theory, J is given as the vanishing ideal of the constraint surface $M_0 \subset M$. We use techniques from [8, 17] to prove the existence of the BFV models and show that they model the Poisson algebra $(P/J)^J$ cohomologically. This latter Poisson algebra is a physically interesting one, since, in the case where P are the functions on phase space and J is the vanishing ideal of the constraint surface, it corresponds to the function on the subquotient $\tilde{M}$, which are the physical observables of the system. The statements about the existence of what we call BFV models and their cohomology are known [12]. However, a rigorous treatment of the question of uniqueness is missing. Under certain local regularity assumptions on the constraint functions, which for instance imply that the constraint surface $M_0$ is smooth, a construction for a uniqueness proof for the BRST cohomology was given in [9]. Stasheff considers the problem from the perspective of homological perturbation theory [17] and gives further special cases under which such uniqueness theorems hold. For instance, he considers the case where a proper subset of the constraints satisfy a regularity condition. Using the notion of stable equivalence from [8], we show that, for a symplectic polynomial algebra $P={{\mathrm{\mathbb {R}}}}[V]$ with bracket induced from the symplectic structure on a finite-dimensional vector space V, any two BRST complexes for the same coisotropic ideal $J \subset P$ are quasi-isomorphic. Hence, we rigorously prove the uniqueness of the BRST cohomology for such P. In contrast to previous treatments of the problem, the assumption on P does not force the constraint surface to be smooth. Moreover, we do not assume a subset of the constraints to be regular. Our Tate resolutions are allowed to contain infinitely many generators.

Finally, we quantize the BRST complex. Under a cohomological assumption, we construct a quantum BRST charge and discuss its uniqueness. The obstruction to quantize lies in the second degree of the classical BRST cohomology, while the ambiguity lies in the first degree. We do this analysis in a rigorous fashion. To the best of our knowledge, such a rigorous treatment in our setting for general Tate resolutions is new.

In the smooth setting, Schätz has dealt with the problem in [16]. See also [5] for the case of a Hamiltonian group action with regular moment map. In [4, 13], this regularity assumption is replaced by the weaker assumption that the components of the moment map generate the vanishing ideal J and that the Koszul complex is acyclic. The authors construct a BRST complex and quantize it. In [13], the assumptions are weakened to allow Tate resolutions with finitetly many generators.

2 BFV models

We work over ${{\mathrm{\mathbb {K}}}}= {{\mathrm{\mathbb {R}}}}$, but any field of characteristic zero will be sufficient. Let P be a unital, Noetherian Poisson algebra. Let $J \subset P$ be a multiplicative ideal satisfying $\{J, J\} \subset J$. Such ideals are called coisotropic. Then the Poisson structure on P induces one on $(P/J)^J$. The purpose of the BRST complex is to model this Poisson algebra cohomologically.

Let ${{\mathrm{\mathcal {M}}}}$ be a negatively graded real vector space with finite-dimensional homogeneous components ${{\mathrm{\mathcal {M}}}}^j$. Denote its component-wise dual by ${{\mathrm{\mathcal {M}}}}^* = \bigoplus _{j>0} ({{\mathrm{\mathcal {M}}}}^{-j})^*$. Define a Poisson bracket on ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$ via the natural pairing between ${{\mathrm{\mathcal {M}}}}$ and ${{\mathrm{\mathcal {M}}}}^*$. For details of the construction, we refer to Appendix A.

Form the tensor product $X_0 = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$ of the two Poisson algebras defined above. Let ${{\mathrm{\mathcal {F}}}}^p X_0$ denote the ideal generated by all elements in $X_0$ of degree at least p. Using the filtration defined by the ${{\mathrm{\mathcal {F}}}}^pX_0$, complete the space $X_0$ to a graded commutative algebra X with homogeneous components

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

Extend the bracket on $X_0$ to X, thus turning X into a graded Poisson algebra. Again, we refer to Appendix A for details. Denote the bracket on X by $\{-,-\}$.

Set $I \subset X$ to be the homogeneous ideal with homogeneous components

$$\begin{aligned} I^j = \lim _{\leftarrow p} \frac{{{\mathrm{\mathcal {F}}}}^1 X_0 \cap X_0^j}{{{\mathrm{\mathcal {F}}}}^{p+1} X_0 \cap X_0^j} \subset X^j. \end{aligned}$$

The powers of ideals are denoted with exponents in parentheses, e.g. $I^{(k)}$ refers to the k-th power of the ideal I.

An element $R \in X$ of odd degree which solves $\{R,R\}=0$ defines a differential $d_R = \{R, -\}$ on X by the Jacobi identity. If $R \in X^1$, the differential $d_R$ induces a differential on X / I since it preserves I.

Definition 1

A BFV model for P and J is a pair (X, R) where $(X,\{-,-\})$ is a graded Poisson algebra constructed as above and $R \in X^1$ is such that the following conditions hold:

• (1) $\{R,R\}=0$.

• (2) $H^j(X/I, d_R) = 0$ for $j \ne 0$.

• (3) $H^0(X/I, d_R) = P/J$.

The first equation is called the classical master equation and the element R is called a BRST charge.

The aim of this note is to prove

Theorem 2

Let P be a Poisson algebra and $J \subset P$ a coisotropic ideal. BFV models exist and in the case of $P = {{\mathrm{\mathbb {R}}}}[V]$ with bracket induced from the symplectic structure on a finite-dimensional vector space V,  the complexes of any two BFV models for the same ideal J are quasi-isomorphic, whence the cohomology $H(X, d_R)$ is uniquely determined by J up to isomorphism.

The existence of the BFV models is known [9, 12]. The problem of uniqueness has been dealt with under certain regularity assumptions [9, 12]. These assumptions imply that the constraint surface is smooth. The novel part is the statement that any two BRST complexes are quasi-isomorphic, which gives the uniqueness of the BRST cohomology as a corollary. We prove this without assuming that the constraint surface is smooth and for Tate resolutions with possibly infinitely many generators. The Noetherian hypothesis ensures that there are only finitely many generators in each degree. This is necessary for our proofs of our convergence results. For completeness, we also include proofs of the already known facts in our framework.

Finally, we quantize the BRST charge rigorously and discuss the uniqueness of the quantization procedure.

3 Existence

3.1 Tate resolutions

To construct BFV models, we first have to construct a suitable commutative graded algebra X. The odd variables are obtained via Tate resolutions.

Let P be a unital, Noetherian Poisson algebra and $J \subset P$ be a coisotropic ideal. Tate constructed resolutions of Noetherian rings by adding certain odd variables to the ring [18]. Consider a Tate resolution $T = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ of P / J given by a negatively graded vector space ${{\mathrm{\mathcal {M}}}}$ with finite-dimensional homogeneous components together with a differential $\delta$ on T of degree 1. Define the dual ${{\mathrm{\mathcal {M}}}}^*$ degree-wise. Extend $\delta$ to $X_0 := P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*) = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}) \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$ by tensoring with the identity. Endow $X_0$ with the natural extension of the Poisson bracket, define the filtration ${{\mathrm{\mathcal {F}}}}^p X_0$, and extend the bracket to the completion X as described in Appendix A. We will frequently refer to statements from that section.

3.1.1 The differential $\delta$

Since $\delta$ is the identity on ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$, it preserves the filtration on $X_0$. Hence it extends to the completion X by Remark 80. Call this extension $\delta$. The extension has degree 1 and preserves the filtration on X. The extension is still an odd derivation, whose square is zero. Since $\delta$ preserves the filtration, it defines a differential on the associated graded mapping ${{\mathrm{\text {gr}}}}^p X^n$ into ${{\mathrm{\text {gr}}}}^p X^{n+1}$.

Define $B = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$. Then, $X_0 = B \otimes _P T$. Since, by definition, the extension of $\delta$ to X leaves elements in ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$ fixed, we have

Remark 3

The natural isomorphism of Lemma 82 identifies the differential $\delta$ on the associated graded with $1 \otimes \delta$ on $B \otimes _P T$.

3.1.2 Contracting homotopy

From the Tate resolution, construct a contracting homotopy $s: T\rightarrow T$ of degree $-1$. Then there exists a ${{\mathrm{\mathbb {K}}}}$-linear split $P/J \rightarrow P$ and a map $\overline{\pi } : T \rightarrow T$ which is defined as the composition $P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}) \rightarrow P \rightarrow P/J \rightarrow P \rightarrow P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ such that

$$\begin{aligned} \delta s + s \delta = 1 - \overline{\pi }. \end{aligned}$$

(1)

Extend $\delta$, s and $\overline{\pi }$ to $X_0$ by tensoring with the identity on ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$. From the definition of $\overline{\pi },$ we find

Remark 4

$\overline{\pi } : X_0 \rightarrow X_0$ is zero on monomials which contain a factor of negative degree.

The homotopy s does not act on elements in ${{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*)$ and hence preserves the filtration. For the same reason, $\bar{\pi }$ preserves the filtration. Both s and $\bar{\pi }$ hence naturally extend to the extension and Eq. 1 is valid in X too. Moreover,

Remark 5

s preserves $I^{(2)}$.

3.2 Constructing the BRST charge

3.2.1 First approximation

Definition 6

Let $Q_0$ be the differential $\delta$ on X / I considered as an element of X.

Hence, the cohomological conditions for $Q_0$ to be a BRST charge are satisfied. However, $Q_0$ does not in general satisfy the classical master equation. We are going to prove the existence part of Theorem 2 by adding correction terms to $Q_0$.

An explicit description of $Q_0$ is the following. Let $e_i$ be a homogeneous basis of ${{\mathrm{\mathcal {M}}}}$, $e^*_i$ its dual basis. Set $d_i := \deg e_i = - \deg e^*_i \equiv \deg e^*_i \pmod 2.$ Assume that $i \leqslant j$ implies $d_i \geqslant d_j$. Define $Q_0 := \sum _j (-1)^{1+d_j} e^*_j \delta (e_j)$. By Lemma 77, this defines an element of $X^1$. For each p, let $L_p$ be an integer with $\{ j \in {{\mathrm{\mathbb {N}}}}: -d_j \leqslant p-1 \} = \{ 1, \dots , L_p\}$ so that $(q_0)_p := \sum _{j=1}^{L_p} (-1)^{1+d_j} e^*_j \delta (e_j)$ defines a representative of the p-th component of $Q_0$. Of course, the element $Q_0$ is independent of the choice of basis $e_j$ of ${{\mathrm{\mathcal {M}}}}$.

Lemma 7

We have $\delta = \sum _j (-1)^{1+d_j} \delta (e_j) \{ e^*_j, -\}$ on X where the operator on the right hand side is well defined.

Proof

Set $\delta ' = \sum _j(-1)^{1+d_j}\delta (e_j) \{ e^*_j, -\}$. This defines a map on X. For $x \in X^n$, the elements $\{e_j^*, x\}$ are in ${{\mathrm{\mathcal {F}}}}^{-d_j + n} X$. Hence, the sum converges by Lemma 77. By linearity, $\delta '$ is defined on all of X. We claim that $\delta '$ is continuous on each $X^n$. Let $x^j = (x_{p}^j + {{\mathrm{\mathcal {F}}}}^p X_0^n)_p \in X^n$ be a sequence converging to zero. Fix p. Then there exists a K, independent of j, such that a p-th representative of $\delta '(x^{j})$ is given by

$$\begin{aligned} \sum _{k=1}^K(-1)^{1+d_k} \delta (e_k) \{e_k^*, x^j_{s_{-d_k,n}(p)}\}, \end{aligned}$$

since the bracket is in $X_0^{-d_k + n}$. Now, let $j_0$ be such that for $j\geqslant j_0$ and for all $k \in \{1, \dots , K\}$ we have $x^{j}_{s_{-d_k,n}(p)} \in {{\mathrm{\mathcal {F}}}}^{s_{-d_k, n}(p)} X_0^n$. Then the above representative vanishes modulo ${{\mathrm{\mathcal {F}}}}^p X_0^{n+1}$ by Corollary 64. Hence, $\delta '$ is continuous on $X^n$. The map $\delta '$ descends to a map on $X_0,$ since the sum is then effectively finite because $\{e^*_j, x\}$ becomes zero for j large enough, depending on $x \in X_0$. This restriction agrees with $\delta$, which can be checked on generators since both maps are derivations. Hence, $\delta ' = \delta$ on each $X^n$ by continuity. Hence, $\delta = \delta '$. $\square$

Lemma 8

For $L_0 := \{Q_0, - \} - \delta ,$ we have $L_0({{\mathrm{\mathcal {F}}}}^p X) \subset {{\mathrm{\mathcal {F}}}}^{p+1} X$.

Proof

Fix $x \in {{\mathrm{\mathcal {F}}}}^p X^n$. Then, by Lemma 74,

$$\begin{aligned} \{Q_0, x\} = \lim _{m \rightarrow \infty } \bigg ( \sum _{j=1}^m (-1)^{1+d_j} \delta (e_j) \{e_j^*, x\} + \sum _{j=1}^m (-1)^{1+d_j} e_j^* \{\delta (e_j), x\} \bigg ). \end{aligned}$$

The first part converges to $\delta (x)$ by Lemma 7. The second part converges by Lemma 77 and hence equals $L_0$. Fix j. By Lemma 75, it suffices to prove that $e_j^* \{\delta (e_j), x\} \in {{\mathrm{\mathcal {F}}}}^{p+1} X$. By the derivation property it suffices to consider $x=e^*_l$ for some l. The term $\delta (e_j)$ is a sum of monomials whose factors have degrees in $\{d_j + 1, \dots , 0\}$. Hence, all elementary factors $e_r$ in $\delta (e_j)$ that could possibly kill $e^*_l$ have degree $d_l$ and get compensated by a factor $e^*_j$ with $\deg (e^*_j) > \deg (e_l^*)$. $\square$

Moreover, we have

Lemma 9

$\{ Q_0, Q_0 \} \in X^2 \cap I^{(2)}\subset {{\mathrm{\mathcal {F}}}}^2 X \cap I^{(2)}$.

Proof

We compute $\deg \{Q_0, Q_0\} = 2 \deg Q_0 = 2$ and hence $\{Q_0, Q_0\} \in {{\mathrm{\mathcal {F}}}}^2 X$. For the last statement, we need to calculate. By Lemma 74,

$$\begin{aligned} \{Q_0, Q_0\}&= \lim _{m\rightarrow \infty }(-1)^{d_j+d_k} \sum _{j,k= 1}^m \{\delta (e_j)e^*_j, \delta (e_k) e^*_k\}\\&=\lim _{m\rightarrow \infty } \sum _{j,k= 1}^m \bigg ( 2(-1)^{1+ d_k} \Big ( (-1)^{1+d_j} \delta (e_j) \{e^*_j, \delta (e_k)\}\Big ) e^*_k \\&\quad +(-1)^{d_j+d_k}e^*_j \{\delta (e_j), \delta (e_k)\} e^*_k \bigg ). \end{aligned}$$

By Lemma 7, the first term is a sum in k with summands that contain factors $\delta (\delta (e_k))=0$ and hence the first term vanishes. By Lemma 75, $\{Q_0,Q_0\} = \sum _{j,k}(-1)^{d_j + d_k} e_j^* \{\delta (e_j), \delta (e_k)\} e_k^* \in I^{(2)}$. $\square$

Corollary 10

$\delta \{Q_0, Q_0\} \in X^3 \subset {{\mathrm{\mathcal {F}}}}^3 X$.

3.2.2 Recursive construction

We now inductively construct out of $Q_0$ a sequence of elements $R_n \in X^1$ by setting

$$\begin{aligned} R_n&= \sum _{j=0}^n Q_j,&Q_0&\text { as defined above},&Q_{n+1}&= -\frac{1}{2} s\{R_n, R_n\}. \end{aligned}$$

The elements $R_n$ have degree 1, since $Q_0$ has and s is of degree $-1$. The idea for the construction is taken from [17]. Also, the proof of the following theorem is adapted from that paper.

Theorem 11

For all n, $\{R_n,R_n\} \in {{\mathrm{\mathcal {F}}}}^{n+2} X \cap I^{(2)}$ and $\delta \{R_n, R_n\} \in {{\mathrm{\mathcal {F}}}}^{n+3} X$.

Proof

The base step was done in Lemma 9 and Corollary 10. We assume the statement is true for $0 \leqslant j \leqslant n$ and consider

$$\begin{aligned} \{R_{n+1}, R_{n+1}\} = \{R_n, R_n\} + 2 \{R_n, Q_{n+1}\} + \{Q_{n+1},Q_{n+1}\}. \end{aligned}$$

By construction and assumption, $Q_{n+1} = -\frac{1}{2}s\{R_n, R_n\} \in {{\mathrm{\mathcal {F}}}}^{n+2} X^1$. Hence, by Corollary 64,

$$\begin{aligned} \{R_{n+1},R_{n+1}\} \equiv \{R_n, R_n\} + 2 \{R_n, Q_{n+1}\} \pmod {{{\mathrm{\mathcal {F}}}}^{n+3} X}. \end{aligned}$$

Expand $\{R_n, Q_{n+1}\} = \sum _{j=1}^n \{Q_j,Q_{n+1}\} + \{Q_0, Q_{n+1}\}$. We have, for $j \in \{1, \dots , n+1\}$ by inductive hypothesis, that $Q_j = -\frac{1}{2}s\{R_{j-1},R_{j-1}\} \in {{\mathrm{\mathcal {F}}}}^{j+1} X^1 \cap I^{(2)}$. Hence, by Lemma 65,

$$\begin{aligned} \{R_{n+1},R_{n+1}\} \equiv \{R_n, R_n\} + 2 \{Q_0, Q_{n+1}\} \pmod {{{\mathrm{\mathcal {F}}}}^{n+3} X}. \end{aligned}$$

We split $\{Q_0, Q_{n+1}\} = \delta Q_{n+1} + L_0 Q_{n+1}$ and, by Lemma 8,

$$\begin{aligned} \{R_{n+1},R_{n+1}\} \equiv \{R_n, R_n\} + 2 \delta Q_{n+1} \pmod {{{\mathrm{\mathcal {F}}}}^{n+3} X}. \end{aligned}$$

Commuting $\delta$ and s,

$$\begin{aligned} 2 \delta Q_{n+1} = - \delta s \{R_n, R_n\} = s \delta \{R_n, R_n\} - \{R_n, R_n\} + \overline{\pi } \{R_n, R_n\}. \end{aligned}$$

Since $\{R_n, R_n\} \in {{\mathrm{\mathcal {F}}}}^{n+2} X^2$, we have that $\overline{\pi } \{R_n, R_n\} = 0$ for $n > 0$ by Remark 4. For $n = 0,$ we obtain $\overline{\pi }\{R_0,R_0\} = 0$ from $\{Q_0,Q_0\} = \sum _{j,k} \pm e_j^* \{\delta (e_j), \delta (e_k)\} e_k^*$ and the fact that $\overline{\pi }$ is zero on $\{J,J\} \subset J$. Hence,

$$\begin{aligned} \{R_{n+1},R_{n+1}\} \equiv s \delta \{R_n, R_n\} \pmod {{{\mathrm{\mathcal {F}}}}^{n+3}, X}. \end{aligned}$$

which vanishes modulo ${{\mathrm{\mathcal {F}}}}^{n+3}X$ by the assumption on $\delta \{R_n, R_n\}$.

Next, by the graded Jacobi identity, we have $0 = \{R_{n+1},\{R_{n+1},R_{n+1}\}\}$. From Lemmas 8 and 65, we find that $L_{n+1} := \{R_{n+1}, - \} - \delta = L_0 + \sum _{j=1}^{n+1} \{Q_j, -\}$ increases filtration degree. Hence, $L_{n+1}\{R_{n+1}, R_{n+1}\} \in {{\mathrm{\mathcal {F}}}}^{n+4} X$ and thus $\delta \{R_{n+1},R_{n+1}\} \in {{\mathrm{\mathcal {F}}}}^{n+4} X.$

Finally, we prove that $\{R_{n+1},R_{n+1}\} = \{R_n, R_n\} + 2 \{R_n, Q_{n+1}\} + \{Q_{n+1},Q_{n+1}\} \in I^{(2)}$. By hypothesis, $\{R_n, R_n\} \in I^{(2)}$. Next, $\{Q_{n+1}, Q_{n+1}\} \in \{I^{(2)}, I^{(2)}\} \subset I^{(2)}$ by Lemma 76. Now, by the same lemma, for $j \in \{1, \dots , n\}$, $\{Q_j, Q_{n+1}\} \in \{I^{(2)},I^{(2)}\}\subset I^{(2)}$ and $\{Q_0, Q_{n+1}\} \in \{ I, I^{(2)}\} \subset I^{(2)}$ which concludes the proof. $\square$

From $Q_{n+1} = -\frac{1}{2} s\{R_n, R_n\} \in {{\mathrm{\mathcal {F}}}}^{n+2}X^1,$ it follows that the $R_n = \sum _{j=0}^n Q_j$ converge to an element $R \in X^1$ by Lemma 77. From Lemma 74, we obtain $\{R_n, R_n\} \longrightarrow \{R,R\}$ as $n \longrightarrow \infty$. We obtain

Corollary 12

$\{R,R\} = 0$.

Proof

We have $\{R_{n+l}, R_{n+l}\} \in {{\mathrm{\mathcal {F}}}}^{n+2} X^2$ for all $l \geqslant 0$. Hence, $\{R,R\} \in {{\mathrm{\mathcal {F}}}}^{n+2} X^2$ for all n by Lemma 75. Hence, $\{R,R\} = 0$. $\square$

We also remark that R as defined above satisfies $R \equiv Q_0 \pmod I^{(2)},$ since for $j > 0$ we have $Q_j \in I^{(2)}$. We are left to consider the cohomology of $d_R = \{R,-\}$ on X / I.

Lemma 13

The action of $d_R$ preserves the filtration and hence defines a differential on ${{\mathrm{\text {gr}}}}X$, which is identified with $1 \otimes \delta$ under the natural isomorphism of Lemma 82.

Proof

$R \in X^1$ and Lemma 63 imply that $d_R$ preserves the filtration and hence descends to the associated graded. We have $\{Q_0,-\} = L_0 + \delta$. Since $L_0$ increases filtration degree by Lemma 8, we have that $\{ Q_0, - \}$ and $\delta$ induce the same maps on ${{\mathrm{\text {gr}}}}X$. Moreover, $K := R - Q_0 \in I^{(2)}\cap X^1$ by the remark above. Hence by Lemma  65, $\{K, -\}$ increases the filtration degree and thus $d_R = \{R, - \}$ and $\{Q_0, -\}$ induce the same maps on the associated graded. $\square$

Corollary 14

$H^j( X / I, d_R ) \cong P/J$ if $j = 0$ and zero otherwise.

Proof

$H^j(X/I, d_R) = H^j({{\mathrm{\text {gr}}}}^0 X, d_R) \cong H^j( B^0 \otimes _P T, 1 \otimes \delta ) \cong H^j(T, \delta )$. $\square$

Given a unital, Noetherian Poisson algebra P with a coisotropic ideal J, we thus have constructed a BFV model for (P, J).

4 Properties

In this section, we describe the general properties of the BFV models. We postpone the discussion of their cohomology to Sect. 6.

Let (X, R) be a BFV model of (P, J) with X being the completion of $P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$. Since R is of degree one, the differential $d_R$ preserves the filtration and hence descends to ${{\mathrm{\text {gr}}}}X$. Let $\pi : X \rightarrow X/I = T = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ be the canonical projection. Let $j : T \rightarrow X_0 \rightarrow X$ be the inclusion given by $t \mapsto 1 \otimes t \in {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}^*) \otimes T = X_0$. Define $\delta = \pi \circ d_R \circ j : T \rightarrow T$.

Lemma 15

The map $\delta : T \rightarrow T$ is a derivation and a differential of degree 1.

Proof

The derivation property follows immediately. Let $a \in T$. We have $(j \circ \pi - {{\mathrm{\text {id}}}}_X) (d_R(j(a)))\in I$ and hence $d_R( j ( \pi ( d_R(j(a)))) = d_R( (j \circ \pi - {{\mathrm{\text {id}}}}_X ) (d_R(j(a)))) \in I$ is in the kernel of $\pi$. The statement about the degree is obvious. $\square$

Lemma 16

Under the identification of Lemma 82, the differential $d_R$ induced on grX corresponds to the differential $1 \otimes \delta$ on $B \otimes _P T$.

Proof

Let $x \in {{\mathrm{\text {gr}}}}^p X$ and pick a representative $a \otimes b \in B^p \otimes _P T$. (It suffices to consider the case where this is a pure tensor.) Then, $d_R(ab) = d_R(a) b + (-1)^p a d_R(b)$. The first summand is in ${{\mathrm{\mathcal {F}}}}^{p+1}X$ and the second is equivalent to $1 \otimes \delta (a\otimes b)$ modulo ${{\mathrm{\mathcal {F}}}}^{p+1}X$. $\square$

Let $Q_0$ be the differential $\delta$ on X / I as an element of X.

Remark 17

The complex $(X/I, d_R) = (T, \delta )$ is a Tate resolution of P / J. Hence, the results from Appendix A and Sects. 3.1 and 3.2.1 apply.

Lemma 18

We have $R \equiv Q_0 \pmod I^{(2)}$. Moreover, $\{R, -\} \equiv \{Q_0, -\} \pmod I$.

Proof

We have $\{R, - \} \equiv \{Q_0, -\} \pmod I$ by construction. Expand $R-Q_0 = \sum _{j \geqslant 0} h_j$ with $h_j \in B^j \otimes _P T^{1-j}$. Such a decomposition exists by Lemma 78. Decompose $h_j = \alpha _j + \beta _j$ with $\alpha _j \in B^j \otimes _P T^{1-j} \cap I_0^{(2)}$ and $\beta _j \in B^j \otimes _P T^{1-j} \setminus I_0^{(2)}$. Let $\{e_k^{(l)}\}_k$ be a basis of ${{\mathrm{\mathcal {M}}}}^{-l}$ with dual basis $\{ {e_k^{(l)}}^*\}_k$. By the Leibnitz rule, $\sum _j \{\beta _j, e_k^{(l)}\} = \{R-Q_0, e_k^{(l)}\} - \sum _j \{\alpha _j, e_k^{(l)}\} \in I$. Expand each $\beta _j = \sum _s a_{j,s} {e_s^{(j)}}^*$ with $a_{j,s} \in T$. We obtain $a_{l,k} = (-1)^{1+l}\sum _j\{\beta _j, e_k^{(l)}\} \in I$; hence all $a_{l,k}$ vanish. $\square$

5 Uniqueness

Fix a unital, Noetherian Poisson algebra P and a coisotropic ideal J. In a first step, we prove that two BFV models for (P, J) related to the same Tate resolutions have isomorphic cohomologies. This is a known fact [9, 12] and is presented in Sects. 5.1–5.2. The key tool will be the notion of gauge equivalences. In a second step, we prove that BFV models for $(P = {{\mathrm{\mathbb {R}}}}[V],J)$, V a finite-dimensional symplectic vector space, on different spaces X have isomorphic cohomologies too. We present this result in Sects. 5.3–5.5. Here, the key tool will be the notion of stable equivalence, introduced in the corresponding Lagrangian setting in [8]. The novel part is that we do not require regularity assumptions, which would imply that the constraint surface is smooth.

5.1 Gauge equivalences

We adapt the language of [8] and call the elements in ${{\mathrm{\mathfrak {g}}}}= X^0 \cap I^{(2)}$ generators of gauge equivalences. Different BRST charges for the same Tate resolution will be related by these equivalences.

Lemma 19

The set of generators of gauge equivalences $\mathfrak {g}$ is a closed subset which forms a Lie algebra acting nilpotently on $X/{{\mathrm{\mathcal {F}}}}^pX$ via the adjoint representation. The Lie algebra ${{\mathrm{\text {ad}}}}({{\mathrm{\mathfrak {g}}}})$ exponentiates to a group G acting on X by Poisson automorphisms.

Proof

By Lemma 75, the set is closed. By Lemma 76 and the fact that $\{X^0, X^0\} \subset X^0$, this is a Lie algebra. By Corollary 64, ${{\mathrm{\mathfrak {g}}}}$ acts on $X/{{\mathrm{\mathcal {F}}}}^p X$. By Lemma 76, this action is nilpotent. Hence, ${{\mathrm{\text {ad}}}}{{\mathrm{\mathfrak {g}}}}$ exponentiates to a group acting on X by vector space automorphisms. Since ${{\mathrm{\text {ad}}}}{{\mathrm{\mathfrak {g}}}}$ consists of derivations both for the product and the bracket, those automorphisms are Poisson. $\square$

The elements of G are called gauge equivalences.

Lemma 20

For $x \in X^1$ and a gauge equivalence g,  we have $gx \equiv x \pmod { I^{(2)}}$.

Proof

Let $c \in X^0 \cap I^{(2)}$ be a generator. Then, $gx - x = \sum _{j>0} \frac{1}{j!} {{\mathrm{\text {ad}}}}_c^j x \in I^{(2)}$ by Lemmas 75 and 76. $\square$

5.2 Uniqueness for fixed Tate resolution

In this section, we prove that two solutions $R,R'$ of the classical master equation in the same space X which induce the same map on X / I are related by a gauge equivalence. Since by Lemma 19, gauge equivalences are Poisson automorphisms, this implies that they have isomorphic cohomologies. We use known techniques, which are adapted from [8].

Remark 21

If R solves $\{R,R\} = 0$ and $g \in G$ is a gauge equivalence, then also $\{gR,gR\} = 0$.

We now discriminate elements in the associated graded ${{\mathrm{\text {gr}}}}^p X^n$ according to how many positive factors they contain at least by defining $A^n_{p,q}:= \{ v \in {{\mathrm{\text {gr}}}}^p X^n : v$ has representative in $I^{(q)}$ }. From the proof of Lemma 82, we see that $A^n_{p,q}$ can be identified with $(B^p \cap I_0^{(q)}) \otimes _P T$, where $I_0 = {{\mathrm{\mathcal {F}}}}^1 X_0$. We now use Remark 3 to see that $A^\bullet _{p,q}$ is a subcomplex and bound its cohomology:

Lemma 22

Fix p and q. We have $H^j( A^\bullet _{p,q}, \delta ) = 0$ for $j < p$.

Proof

From Remark 3, we have $H^j( A^\bullet _{p,q}, \delta ) \cong H^j( (B^p \cap I_0^{(q)}) \otimes _P T, 1\otimes \delta )$. Now, we may factor this space into $(B^p \cap I_0^{(q)}) \otimes _P H^{j-p}(T, \delta ),$ since $B^p \cap I_0^{(q)}$ is a free P-module. For $j<p$, the second factor vanishes, since T is a resolution of P / J. $\square$

Lemma 23

Fix $p \geqslant 2$. Let $R, R' \in X^1$ be two solutions of the classical master equation which induce the same maps on X / I. Then, for $2 \leqslant q \leqslant p$, we have that $R \equiv R' \pmod { I^{(q)} \cap {{\mathrm{\mathcal {F}}}}^p X^1 + {{\mathrm{\mathcal {F}}}}^{p+1} X^1}$ implies the existence of a gauge equivalence g with generator $c \in {{\mathrm{\mathcal {F}}}}^p X^0 \cap I^{(2)}$ such that $gR \equiv R' \pmod {I^{(q+1)} \cap {{\mathrm{\mathcal {F}}}}^p X^1 + {{\mathrm{\mathcal {F}}}}^{p+1} X^1}$. Moreover, the element $gR \in X^1$ sill satisfies the classical master equation and induces the same map on X / I as R and $R'$.

Proof

Let $\delta$ be the common differential on X / I and $Q_0$ be the map $\delta$ as an element of X. Hence, $R \equiv Q_0 \equiv R' \pmod I^{(2)}$ by Lemma 18. Define $v := R-R' \in I^{(q)} \cap {{\mathrm{\mathcal {F}}}}^p X^1+ {{\mathrm{\mathcal {F}}}}^{p+1} X^1 \subset {{\mathrm{\mathcal {F}}}}^p X^1$. We have

$$\begin{aligned} 0&= \{R + R', R- R'\} = 2 \{Q_0, v\} + \{R-Q_0, v\} + \{R'-Q_0, v\}\\&\equiv 2 \{Q_0, v\} \pmod {{{\mathrm{\mathcal {F}}}}^{p+1} X^2} \end{aligned}$$

by Lemma 65. By Lemma 16, the maps $d_R$ and $d_{R'}$ also induce the same map on of ${{\mathrm{\text {gr}}}}X$ which we denote by $\delta$ too. Since $\delta v = \{Q_0, v\} - L_0 v \equiv \{Q_0, v\} \pmod {{{\mathrm{\mathcal {F}}}}^{p+1} X^2}$ by Lemma 8, the above implies $\delta v \equiv 0 \pmod { {{\mathrm{\mathcal {F}}}}^{p+1} X^2}$. Hence, v defines a cocycle $\bar{v}$ in ${{\mathrm{\text {gr}}}}^p X^1$. We have $p>1$. By Lemma 22, there exists $\bar{c} \in {{\mathrm{\text {gr}}}}^p X^{0}$ with $\delta \bar{c} = \bar{v}$ and a corresponding representative $c \in {{\mathrm{\mathcal {F}}}}^p X^0 \cap I^{(q)}$, so that $\delta c \equiv v \pmod { {{\mathrm{\mathcal {F}}}}^{p+1} X^1}$. This c will be the generator of the gauge equivalence we seek. Set $g := \exp {{\mathrm{\text {ad}}}}_c$. We have

$$\begin{aligned} g R - R'&= v + \sum _{j=1}^\infty \frac{1}{j!} {{\mathrm{\text {ad}}}}_c^j R \equiv \delta (c) - d_R(c) + \sum _{j=2}^\infty \frac{1}{j!} {{\mathrm{\text {ad}}}}_c^j R \\&\equiv \sum _{j=2}^\infty \frac{1}{j!} {{\mathrm{\text {ad}}}}_c^j R \pmod { {{\mathrm{\mathcal {F}}}}^{p+1} X^1} . \end{aligned}$$

From Lemma 63, we know that this sum is in ${{\mathrm{\mathcal {F}}}}^p X^1$. We are left to show that the sum is in $I^{(q+1)}$. By Lemma 76, we have ${{\mathrm{\text {ad}}}}_c R \in I^{(q)}$. By Lemma 76, we obtain ${{\mathrm{\text {ad}}}}_c^j R \in I^{(q+1)}$ for all $j \geqslant 2$ since $q \geqslant 2$.

By Remark 21, gR still satisfies the classical master equation and $gR \equiv Q_0 \pmod { I^{(2)}}$ by Lemma 20, whence all maps $R, R', gR$ induce the same map on X / I. $\square$

Theorem 24

Let $R, R' \in X^1$ be solutions of the classical master equation with differentials inducing the same maps on X / I. Then there exists a gauge equivalence $g \in G$ with $R' = g R$.

Proof

First, we inductively construct a sequence of gauge equivalences $g_2, g_3, \dots$ such that for all $p \geqslant 2$ we have $g_p \cdots g_2 R \equiv R' \pmod {{{\mathrm{\mathcal {F}}}}^{p+1} X^1 \cap I^{(2)}}$. By Lemma 20, it suffices to ensure that $g_p \cdots g_2 R \equiv R' \pmod {{{\mathrm{\mathcal {F}}}}^{p+1} X^1}.$

For $p = 2,$ note that $R - R' \in I^{(2)} \subset (I^{(2)} \cap {{\mathrm{\mathcal {F}}}}^2 X^1) + {{\mathrm{\mathcal {F}}}}^3 X^1$ by Lemma 18. Now, apply Lemma 23 with $q = p$ to obtain $g_2$.

Next, assume the $g_2, \dots , g_p$ have been constructed to fulfill

$$\begin{aligned} R'' := g_p \cdots g_2 R \equiv R' \pmod { {{\mathrm{\mathcal {F}}}}^{p+1} X^1 \cap I^{(2)}}. \end{aligned}$$

By Remark 21, $R''$ solves the classical master equation. Moreover, $R'' \equiv Q_0 \pmod { I^{(2)}}$ by Lemmas 18 and 20. Hence, the pair $(R'', R')$ satisfies the requirements of Lemma 23 with $q=2$. We obtain a gauge equivalence $g_{p+1,2}$ with generator $c_{p+1, 2} \in {{\mathrm{\mathcal {F}}}}^{p+1} X^0 \cap I^{(2)}$ and

$$\begin{aligned} g_{p+1,2} R'' \equiv R' \pmod { I^{(3)} \cap {{\mathrm{\mathcal {F}}}}^{p+1} X^1 + {{\mathrm{\mathcal {F}}}}^{p+2} X^1}. \end{aligned}$$

If we continue to apply the lemma for $q = 3, \dots , p+1,$ we obtain gauge equivalences $g_{p+1,3}, \dots , g_{p+1,p+1}$ with generators $c_{p+1, 3}, \dots , c_{p+1,p+1} \in {{\mathrm{\mathcal {F}}}}^{p+1} X^0 \cap I^{(2)}$ such that

$$\begin{aligned} g_{p+1, p+1} \cdots g_{p+1, 2} R'' \equiv R' \pmod {{{\mathrm{\mathcal {F}}}}^{p+2} X^1 \cap I^{(2)}}. \end{aligned}$$

Set $g_{p+1} := g_{p+1,p+1} \cdots g_{p+1,2}$. The construction of the sequence is complete.

We claim that $\lim _{m \rightarrow \infty } g_m g_{m-1} \cdots g_2$ converges pointwise to a gauge equivalence g. Since all generators $c_{m, j}$ are in ${{\mathrm{\mathcal {F}}}}^{m} X^0 \cap I^{(2)}$ and this set is closed under the bracket, the Campbell–Baker–Hausdorff formula implies that the generator $c_m$ of $g_m$ is also in ${{\mathrm{\mathcal {F}}}}^{m} X^0 \cap I^{(2)}$. Now, denote the generator of $g_m \cdots g_2$ by $\gamma _m$. We have $\gamma _m \in I^{(2)}$ by the CBH formula. Moreover, the CBH formula implies that the generator $\gamma _{m+1}$ of $g_{m+1} g_m \cdots g_2$ satisfies

$$\begin{aligned} \gamma _{m+1} = c_{m+1} + \gamma _{m} + \text {higher terms}, \end{aligned}$$

where “higher terms” are those involving commutators of $c_{m+1}$ and $\gamma _m$ where each contains at least one instance of $c_{m+1} \in {{\mathrm{\mathcal {F}}}}^{m+1} X^0$. Since $\gamma _m \in X^0,$ all these terms are in ${{\mathrm{\mathcal {F}}}}^{m+1} X^0$. Hence,

$$\begin{aligned} \gamma _{m+1} \equiv \gamma _m \pmod {{{\mathrm{\mathcal {F}}}}^{m+1} X^0}. \end{aligned}$$

Hence, there exists $\gamma \in X^0$ with $\gamma _m \rightarrow \gamma$ as $m \rightarrow \infty$. We set $g := \exp {{\mathrm{\text {ad}}}}_\gamma$. By Lemma 75, this element defines a gauge equivalence. We claim that $\exp {{\mathrm{\text {ad}}}}_{\gamma _m} = g_m \cdots g_2 \rightarrow g$ pointwise. Let $x \in X^n$. Then,

$$\begin{aligned} \exp {{\mathrm{\text {ad}}}}_{\gamma _m} x - \exp {{\mathrm{\text {ad}}}}_{\gamma } x&= \{\gamma _m - \gamma , x\} + \frac{1}{2} \{\gamma _m,\{\gamma _m,x\}\} - \frac{1}{2} \{\gamma , \{\gamma ,x\}\} + \cdots . \end{aligned}$$

Modulo a fixed ${{\mathrm{\mathcal {F}}}}^k X$, this sum is finite and the number of terms does not depend on m,  since all $\gamma _m$ are at least in $I^{(2)}$. Since $\gamma _m \rightarrow \gamma$ and the bracket are continuous in fixed degree by Lemma 74, we obtain the claim.

Finally, $\exp {{\mathrm{\text {ad}}}}_{\gamma _{m+l}} R - R' \in {{\mathrm{\mathcal {F}}}}^{m} X^1$ implies $g R - R' \in {{\mathrm{\mathcal {F}}}}^{m} X^1$ for all m which shows that $gR = R'$. $\square$

5.3 Trivial BFV models

The key construction in the proof of uniqueness for different spaces X in Theorem 2 is the notion of stable equivalence. The idea of adding variables that do not change the cohomology was already present in [12]. It was first explicitly formalized in [8] in a similar situation in the Lagrangian setting. Roughly speaking, one proves that different BRST complexes for the same pair (P, J) are quasi-isomorphic by adding more variables of non-zero degree. This is formalized by taking products with so-called trivial BFV models.

Let $P = {{\mathrm{\mathbb {R}}}}$ with zero bracket and $J = 0$. Then P is a unital, Noetherian Poisson algebra and J is a coisotropic ideal. Let ${{\mathrm{\mathcal {N}}}}$ be a negatively graded vector space and ${{\mathrm{\mathcal {N}}}}[1]$ the same space with degree shifted by $-1$. Define the differential $\delta$ on ${{\mathrm{\mathcal {M}}}}= {{\mathrm{\mathcal {N}}}}\oplus {{\mathrm{\mathcal {N}}}}[1]$ by $\delta ( a \oplus b) = b \oplus 0$. Set $T = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ and extend $\delta$ to an odd, P-linear derivation on T.

Lemma 25

The complex $(T, \delta )$ has trivial cohomology and hence defines a Tate resolution of $P/J = {{\mathrm{\mathbb {R}}}}$.

Proof

On ${{\mathrm{\mathcal {M}}}},$ there is a map $s( a \oplus b ) = 0 \oplus a$ with $s \delta + \delta s = {{\mathrm{\text {id}}}}_{{{\mathrm{\mathcal {M}}}}}$. Extend s to an odd, P-linear derivation on T. Then, $s \delta + \delta s$ is an even derivation on T which is the identity on ${{\mathrm{\mathcal {M}}}}$ and hence

$$\begin{aligned} s \delta + \delta s = k {{\mathrm{\text {id}}}}\quad \text {on } P \otimes {{\mathrm{\text {Sym}}}}^k({{\mathrm{\mathcal {M}}}}). \end{aligned}$$

Since both s and $\delta$ preserve the k-degree, we have

$$\begin{aligned} H^j( T, \delta )&= \oplus _k H^j( P \otimes {{\mathrm{\text {Sym}}}}^k({{\mathrm{\mathcal {M}}}}), \delta ) = H^j( P \otimes {{\mathrm{\text {Sym}}}}^0({{\mathrm{\mathcal {M}}}}), \delta )\\&= {\left\{ \begin{array}{ll} P, &{} \text {if } j=0\\ 0, &{} \text {otherwise} \end{array}\right. }. \end{aligned}$$

$\square$

Complete the space $Y_0 = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$ to the space Y. Let $e_j$ be a homogeneous basis of ${{\mathrm{\mathcal {M}}}}$ such that $\delta (e_j) = e_k$ for some k depending on j. Define $Q_0 = \sum _j (-1)^{1+d_j} e^*_j \delta (e_j)$ as in Lemma 8. Since $\{Q_0, Q_0\} = \sum _{j,k} \pm e^*_j\{\delta (e_j), \delta (e_k)\}e^*_k = 0$, the construction of Sect. 3 yields the BRST charge $S = Q_0$. Hence, (Y, S) is a BFV model for $(P,J) = ({{\mathrm{\mathbb {R}}}},0)$. BFV models arising from this construction are called trivial.

Lemma 26

For trivial BFV models, $d_S$ equals the induced map of $\Delta = \delta \oplus \delta ^* : {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^* \rightarrow {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*$ on Y, where the dual differential $\delta ^* : {{\mathrm{\mathcal {M}}}}^* \rightarrow {{\mathrm{\mathcal {M}}}}^*$ is given by $\delta ^* ( u ) = (-1)^{\deg u} u \circ \delta$, i.e. $\delta ^*( a \oplus b) = (-1)^{j} 0 \oplus a$ on $({{\mathrm{\mathcal {M}}}}^*)^j$. Conversely, the map $d_S$ induces a differential on $Y_0$ which coincides with the induced map of $\delta \oplus \delta ^*$ on $Y_0$.

Proof

By acting on the generators $e_j$ and $e_j^*$ defined above, one sees that the induced differential $d_S$ on $Y_0$ equals $\Delta = \delta \oplus \delta ^*$. Since $Y_0$ is dense in Y and both maps are continuous, the first claim follows. The second claim follows from the observation that the sum $d_S(x) = \sum _j \pm \{e_j \delta (e_j), x\}$ is effectively finite if $x \in Y_0$. $\square$

Lemma 27

We have $H^j( Y, d_S ) = 0$ for $j \ne 0$ and $H^0( Y, d_S ) = {{\mathrm{\mathbb {R}}}}$. The same statement holds if we replace Y by $Y_0$.

Proof

Since the cohomology of the complex $({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*, \Delta )$ is trivial, there is a map $s : {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^* \rightarrow {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*$ of degree $-1$ with $\Delta s + s \Delta = {{\mathrm{\text {id}}}}$. Its extension to $Y_0$ as a derivation thus satisfies

$$\begin{aligned} s \Delta + \Delta s = j {{\mathrm{\text {id}}}}\quad \text {on } Y_0^{n,j}. \end{aligned}$$

(2)

Since both maps $\Delta$ and s preserve form degree, we obtain the statement about $H^j(Y_0, d_S)$.

Let $x \in Y^n$ with $d_S x = 0$. By Lemma 84, there are $x_j \in Y^n$ of form degree j with $\sum _j x_j = x$. By continuity of the bracket, $0 = \sum _{j} d_S x_{j}$. By Lemma 85, $d_S x_{j} = 0$ for all j, since $d_S$ preserves form degree. Lift $\Delta$ and s to Y. Then, Eq. 2 is still valid on $Y^{n,j}$. For $j >0$, there are $y_{j} \in Y^{n-1, j}$ with $d_S y_{j} = x_{j}$. By Lemma 83, the element $y = \sum _{j> 0} y_{j}$ is well defined and

$$\begin{aligned} x = \sum _{j>0}x_{j} + x_{0} = d_S y + x_{0} \end{aligned}$$

with $x_{0} \in Y^{n,0}$. For $n \ne 0,$ this is the empty set and hence x is exact. For $n = 0,$ this set is ${{\mathrm{\mathbb {R}}}}$. We are left to show that two distinct $d_S$-closed elements of ${{\mathrm{\mathbb {R}}}}$ always define the distinct cohomology classes. This follows from the fact that each summand in $d_s y = \sum _j (\pm \delta (e_j) \{e^*_j, y\} \pm e^*_j \{\delta (e_j), y\})$ is zero or has a factor of nonzero degree since $\delta (e_j) = e_k$ for some k depending on j. $\square$

5.4 Stable equivalence

Let P be a unital, Noetherian Poisson algebra and $J \subset P$ a coisotropic ideal. Let (X, R) be a BFV model for (P, J) and (Y, S) be a trivial BFV model. Let ${{\mathrm{\mathcal {M}}}}$ and ${{\mathrm{\mathcal {N}}}}$ be the corresponding vector spaces. Define Z as the completion of $Z_0 := X_0 \otimes Y_0 = P \otimes {{\mathrm{\text {Sym}}}}( {{\mathrm{\mathcal {U}}}}\oplus {{\mathrm{\mathcal {U}}}}^*),$ where ${{\mathrm{\mathcal {U}}}}= {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {N}}}}$ and $L = R \otimes 1 + 1 \otimes S$. Both X and Y naturally sit inside Z as Poisson subalgebras, since the inclusions $X_0 \rightarrow Z_0$ and $Y_0 \rightarrow Z_0$ preserve the respective filtrations.

Lemma 28

The pair (Z, L) defines another BFV model for (P, J).

Proof

Since the bracket between elements of X and elements of Y is zero, the element L solves the master equation. The Künneth formula implies together with Lemma 27 the conditions on the cohomology. $\square$

We call Z the product of X and Y and write $Z = X \hat{\otimes } Y$. Adding the new variables in ${{\mathrm{\mathcal {N}}}}$ does not change the cohomology of the BRST complex X:

Lemma 29

The natural map $X \rightarrow Z$ defines a quasi-isomorphism of differential graded commutative algebras.

Proof

We define the maps $\iota : X_0 \rightarrow Z_0$ as the natural map and $p: Z_0 \rightarrow X_0$ as the map taking $x \otimes y$ to $x \pi (y)$, where $\pi : Y_0 \rightarrow Y_0$ is the projection onto ${{\mathrm{\mathbb {R}}}}= {{\mathrm{\text {Sym}}}}^0( {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$ along the ${{\mathrm{\text {Sym}}}}^j({{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$ with $j>0$. Both maps extend to the respective completion. We claim that they define mutual inverses on cohomology.

From Eq. 2, we infer that there exists a map t on $Y_0$ such that $d_S t + t d_S = {{\mathrm{\text {id}}}}- \pi$. In particular, $\pi d_S = d_S \pi = 0$. We have $t = \frac{1}{j} s$ on form degree $j>0$ and $t = s$ on ${{\mathrm{\mathbb {R}}}}$. Hence, t preserves the filtration up to degree shift. Hence, ${{\mathrm{\text {id}}}}\otimes t$ extends from $Z_0$ to the completion Z. By tensoring the other maps too, we obtain the identity

$$\begin{aligned} d_{R+S} ({{\mathrm{\text {id}}}}\otimes t) + ({{\mathrm{\text {id}}}}\otimes t) d_{R+S}&= ({{\mathrm{\text {id}}}}\otimes d_S)({{\mathrm{\text {id}}}}\otimes t) + ({{\mathrm{\text {id}}}}\otimes t)({{\mathrm{\text {id}}}}\otimes d_S) = {{\mathrm{\text {id}}}}\otimes ({{\mathrm{\text {id}}}}- \pi ) \\&= {{\mathrm{\text {id}}}}- \iota \circ p \end{aligned}$$

on Z. The first equality is true since t shifts degree by one. We are left to show that both maps $\iota$ and p descend to cohomology. For $\iota$ this is trivial. For p note that for homogeneous x of degree k,

$$\begin{aligned} p(d_{R+S}(x\otimes y))&= p ( (d_R x) \otimes y + (-1)^k x \otimes d_S y) \\&= \pi (y) d_R x + (-1)^k x \pi d_S y = d_R( x\pi y) = d_R( p (x\otimes y)). \end{aligned}$$

$\square$

Now, we are ready to formulate the notion of stable equivalence introduced in [8]:

Definition 30

Let (X, R) and $(X', R')$ be two simple BFV models for (P, J). We say that (X, R) and $(X',R')$ are stably equivalent if there exist trivial BFV models (Y, S) and $(Y', S')$ and a Poisson isomorphism $X \hat{\otimes } Y \longrightarrow X' \hat{\otimes } Y'$ taking $R + S$ to $R' + S'$.

5.5 Relating Tate resolutions

Now, we want to consider BFV models (R, X) and $(R', X')$ whose Tate resolutions $(X/I, d_R)$ and $(X'/I',d_{R'})$ are not equal. We have the notion of stable equivalence. Our aim is to prove that any two such BFV models are stably equivalent and that stably equivalent BFV models are quasi-isomorphic. As a tool, we need the following lifting statement:

Lemma 31

Let $P = {{\mathrm{\mathbb {R}}}}[V]$ with bracket induced by a symplectic structure of a finite-dimensional vector space V and consider $T = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ and $T'= P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}')$. Assume there is an isomorphism $\phi : T \rightarrow T'$ of graded commutative algebras which is the identity in degree zero. Let X be the completion of $X_0 = P \otimes {{\mathrm{\text {Sym}}}}( {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {M}}}}^*)$. Construct analogously the space $X'$. Then, $\phi$ lifts to a Poisson isomorphism $\Phi : X \rightarrow X'$.

Proof

Since T and $T'$ are negatively graded and isomorphic as graded algebras, we have ${{\mathrm{\mathcal {M}}}}\cong {{\mathrm{\mathcal {M}}}}'$ as graded vector spaces. Hence, we may assume ${{\mathrm{\mathcal {M}}}}= {{\mathrm{\mathcal {M}}}}'$ and thus $T = T'$ and $X = X'$.

Pick standard coordinates $\{x_1, \dots , x_n, y_1, \dots , y_n\}$ on the space V for the symplectic structure, so that ${{\mathrm{\mathbb {R}}}}[V] = {{\mathrm{\mathbb {R}}}}[x_i,y_j]$ and $\{x_i, y_j\} = \delta _{ij}$. Let $\{e_j^{(l)}\}_j$ be a basis of ${{\mathrm{\mathcal {M}}}}^{-l}$ and $\{{e_j^{(l)}}^*\}_j$ be the respective dual bases. Then there are elements $a_{j_1 \dots j_k}^{l_1 \dots l_k}(j,l)(x_i,y_i) \in {{\mathrm{\mathbb {R}}}}[x_i, y_i]$ and invertible matrices $a_{jk}^{(l)} \in {{\mathrm{\mathbb {R}}}}[x_i,y_i]$ such that

$$\begin{aligned} \phi (e_j^{(l)}) = \sum _k a_{jk}^{(l)}(x_i,y_i) e_k^{(l)} + \sum a_{j_1 \dots j_k}^{l_1 \dots l_k}(j,l)(x_i,y_i) e_{j_1}^{(l_1)} \cdots e_{j_k}^{(l_k)}, \end{aligned}$$

where the sum runs over all integers $k \ge 2$ and $(j_1, l_1), \ldots (j_k, l_k)$ with $l_1 + \cdots + l_k = l$ and is thus finite. Consider indeterminats $Y_i, {E_j^{(l)}}^* \in X_0$ of degree 0 and l,  respectively, defining

$$\begin{aligned} S(x_i, Y_i, e_j^{(l)}, {E_j^{(l)}}^*)&= \sum _i x_i Y_i + \sum _{j,k,l} a_{jk}^{(l)}(x_i,Y_i) {E_j^{(l)}}^* e_k^{(l)} \\&\quad + \sum _{(j,l)} \sum a_{j_1 \dots j_k}^{l_1 \dots l_k}(j,l)(x_i,Y_i) {E_j^{(l)}}^* e_{j_1}^{(l_1)} \cdots e_{j_k}^{(l_k)}. \end{aligned}$$

Consider the equations

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

which read

$$\begin{aligned} y_i&= Y_i + \sum _{j,k,l} \frac{\partial a_{jk}^{(l)}(x_i,Y_i)}{\partial x_i} {E_j^{(l)}}^* e_k^{(l)} + \sum _{(j,l)} \sum \frac{\partial a_{j_1 \dots j_k}^{l_1 \dots l_k}(j,l)(x_i,Y_i)}{\partial x_i} \\&\quad \times {E_j^{(l)}}^* e_{j_1}^{(l_1)} \cdots e_{j_k}^{(l_k)},\\ X_i&= x_i+ \sum _{j,k,l} \frac{\partial a_{jk}^{(l)}(x_i,Y_i)}{\partial Y_i} {E_j^{(l)}}^* e_k^{(l)} + \sum _{(j,l)} \sum \frac{ \partial a_{j_1 \dots j_k}^{l_1 \dots l_k}(j,l)(x_i,Y_i)}{\partial Y_i} \\&\quad \times {E_j^{(l)}}^* e_{j_1}^{(l_1)} \cdots e_{j_k}^{(l_k)},\\ {e_j^{(l)}}^*&= \sum _k a_{kj}^{(l)}(x_i,Y_i) {E_k^{(l)}}^* + \sum _{(j',l')} \sum a_{j_1 \dots j_k}^{l_1 \dots l_k}(j',l')(x_i,Y_i) (-1)^{l (l'+1)} {E_{j'}^{(l')}}^*\\&\quad \times \frac{\partial (e_{j_1}^{(l_1)} \cdots e_{j_k}^{(l_k)})}{\partial e_j^{(l)}},\\ E_j^{(l)}&= \sum _k a_{jk}^{(l)}(x_i,Y_i) e_k^{(l)} + \sum a_{j_1 \dots j_k}^{l_1 \dots l_k}(j,l)(x_i,Y_i) e_{j_1}^{(l_1)} \cdots e_{j_k}^{(l_k)}. \end{aligned}$$

The linear part is invertible. Hence, we can solve the equations for $(X_i, Y_i, E_j^{(l)}, {E_j^{(l)}}^*)$ in terms of $(x_i, y_i, e_j^{(l)}, {e_j^{(l)}}^*)$ (and vice versa) and hence also for $(x_i, Y_i, e_j^{(l)}, {E_j^{(l)}}^*)$ in terms of $(x_i, y_i, e_j^{(l)}, {e_j^{(l)}}^*)$ (and vice versa) in the completion X. Hence, the function S generates a Poisson automorphism $\Phi : X \rightarrow X$ by Lemma 86. Let I be the ideal generated by positive elements as defined previously. We have $\Phi (x_i) = X_i \equiv x_i = \phi (x_i) { \pmod I}$ and $\Phi (y_i) = Y_i \equiv y_i = \phi (y_i) {\pmod I}$; thus also $\Phi (e_j^{(l)}) = E_j^{(l)} \equiv \phi (e_j^{(l)}) {\pmod I}$. Hence, $\Phi$ is a lift of $\phi$. $\square$

Theorem 32

Consider $P = {{\mathrm{\mathbb {R}}}}[V]$ with bracket induced by a symplectic structure on a finite-dimensional vector space V. Any two BFV models for (P, J) are stably equivalent.

Proof

Let (X, R) and $(X',R')$ be BFV models with associated Tate resolutions $T := X/I \cong P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ and $T' := X'/I' \cong P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}')$. By [8, Theorem A.2], there exist negatively graded vector spaces ${{\mathrm{\mathcal {N}}}}$ and ${{\mathrm{\mathcal {N}}}}'$ with finite-dimensional homogeneous components, differentials $\delta _{{{\mathrm{\mathcal {N}}}}} : {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}) \rightarrow {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}), \delta _{{{\mathrm{\mathcal {N}}}}'} : {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}') \rightarrow {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}')$ with cohomology ${{\mathrm{\mathbb {R}}}}$, and an isomorphism $\phi$ of differential graded commutative algebras

$$\begin{aligned} P\otimes {{\mathrm{\text {Sym}}}}( {{\mathrm{\mathcal {M}}}}\oplus {{\mathrm{\mathcal {N}}}}) \rightarrow&P\otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}}' \oplus {{\mathrm{\mathcal {N}}}}') \end{aligned}$$

restricting to ${{\mathrm{\text {id}}}}_P : P \rightarrow P$ in degree 0. Let Y and $Y'$ be the trivial BFV models corresponding to ${{\mathrm{\mathcal {N}}}}$ and ${{\mathrm{\mathcal {N}}}}'$ with BRST charges S and $S'$, respectively. Consider the spaces $Z = X \hat{\otimes } Y$ and $Z' = X' \hat{\otimes } Y'$. Together with the operators $L = R + S$ and $L' = R' + S',$ they form BFV models (Z, L) and $(Z', L')$ for (P, J) by Lemma 28.

We now construct a Poisson isomorphism $\Phi : X \hat{\otimes } Y \rightarrow X' \hat{\otimes } Y'$ sending $R + S$ to $R' + S'$. By Lemma 31, the map $\phi$ lifts to a Poisson isomorphism $\Psi : X \hat{\otimes } Y \rightarrow X' \hat{\otimes } Y'$. Now, $L''= \Psi (L)$ solves $\{-,-\}=0$ in $X' \hat{\otimes } Y'$. Moreover, $\{L'', -\}$ induces $\delta '$ on $P \otimes {{\mathrm{\text {Sym}}}}( {{\mathrm{\mathcal {M}}}}' \oplus {{\mathrm{\mathcal {N}}}}')$. By Theorem 24, there exists a Poisson isomorphism $\chi$ of $X' \hat{\otimes } Y'$ with $L' = \chi ( L '')$. Set $\Phi = \chi \circ \Psi$.

We are now in the situation

where the vertical arrows represent natural maps which are quasi-isomorphisms by Lemma 29. $\square$

Lemma 33

The complexes of two stably equivalent BFV models are quasi-isomorphic. In particular, they have cohomologies which are isomorphic as graded commutative algebras.

Proof

Let (X, R) and $(X',R')$ be two stably equivalent BFV models. Hence, we are in the situation

where the downward arrows are quasi-isomorphisms of differential graded commutative algebras by Lemma 29 and the bottom arrow is a Poisson isomorphism $X \hat{\otimes } Y \rightarrow X' \hat{\otimes } Y'$ sending $R + S$ to $R' + S'$. $\square$

From Theorem 32 and Lemma 33, we obtain results analogously to the treatment of the Lagrangian case in [8]

Corollary 34

Let $P={{\mathrm{\mathbb {R}}}}[V]$ with bracket induced by a symplectic structure on a finite-dimensional vector space V. Any two BRST complexes arising from BFV models for the same coisotropic ideal $J\subset P$ are quasi-isomorphic. Hence, the BRST cohomology is uniquely determined by $(P=R[x_i,y_i],J)$ up to an isomorphism of graded commutative algebras.

6 Cohomology

Let P be a unital, Noetherian Poisson algebra and J a coisotropic ideal. Let (X, R) be a BFV model for $J \subset P$. In this section, we analyse the cohomology of the complex $(X,d_R)$. We follow the strategy from [8].

6.1 Cohomology and filtration

The associated graded of X is defined by ${{\mathrm{\text {gr}}}}^p X = {{\mathrm{\mathcal {F}}}}^pX / {{\mathrm{\mathcal {F}}}}^{p+1} X$. The differential $d_R$ induces a map $\delta$ on $X/I = T = P \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {M}}}})$ and the results from Sect. 4 apply.

Lemma 35

$H^j( \mathrm{gr}^p~X, d_R )\cong B^p \otimes _P P/J$ for $j=p$ and $H^j( \mathrm{gr}^p~X, d_R ) \cong 0$ for $j \ne p$.

Proof

Fix p. $B^p$ is a free P-module. By Lemma 16, we have

$$\begin{aligned} H^j( {{\mathrm{\mathcal {F}}}}^p X / {{\mathrm{\mathcal {F}}}}^{p+1} X, d_R )&\cong H^j( B^p \otimes _P T^{\bullet - p} , 1 \otimes \delta ) \\&\cong B^p \otimes _P H^{j-p}(T, \delta ) \cong B^p \otimes _P H^{j-p}(X/I, d_R). \end{aligned}$$

$\square$

Next, we want to prove that, to compute the cohomology in a fixed degree, one may disregard elements of high filtration degree.

Lemma 36

Let $j < p$ be integers with $p \geqslant 0$. Then, $H^j( {{\mathrm{\mathcal {F}}}}^p X, d_R ) = 0$.

Proof

Let $x \in {{\mathrm{\mathcal {F}}}}^p X^j$ be a cocycle representing a cohomology class in $H^j({{\mathrm{\mathcal {F}}}}^p X, d_R)$. Then, $x + {{\mathrm{\mathcal {F}}}}^{p+1} X^j$ defines a cocycle in $H^j({{\mathrm{\mathcal {F}}}}^p X/ {{\mathrm{\mathcal {F}}}}^{p+1} X, d_R)$. By Lemma 35, there is $y_0 \in {{\mathrm{\mathcal {F}}}}^p X^{j-1}$ with $x - d_Ry_0 \in {{\mathrm{\mathcal {F}}}}^{p+1} X^j$. Hence, this element defines a cocycle in $H^j({{\mathrm{\mathcal {F}}}}^{p+1} X / {{\mathrm{\mathcal {F}}}}^{p+2} X, d_R) = 0$. Hence, there is $y_1 \in {{\mathrm{\mathcal {F}}}}^{p+1} X^{j-1}$ with $x - d_Ry_0 - d_Ry_1 \in {{\mathrm{\mathcal {F}}}}^{p+2}X^j$. Iterating this procedure, we find a sequence $y_0, y_1, \dots$ of elements $y_j \in {{\mathrm{\mathcal {F}}}}^{p+j} X^{j-1}$ with $x - d_R(y_0 + \cdots + y_j) \in {{\mathrm{\mathcal {F}}}}^{j+1} X^j$. By Lemma 77, the element $y := y_0 + \cdots \in X^{j-1}$ is well defined and $y_0 + \cdots + y_j \rightarrow y$. Since all $y_j$ are in ${{\mathrm{\mathcal {F}}}}^p X^{j-1}$ and this set is closed by Lemma 75, we have $y \in {{\mathrm{\mathcal {F}}}}^p X^{j-1}$. Finally, for n fixed, and all j,

$$\begin{aligned} d_R y_0 + \cdots + d_R y_n + \cdots + d_R y_{n+j} - x \in {{\mathrm{\mathcal {F}}}}^{n+1} X^j. \end{aligned}$$

Since $d_R = \{R, -\}$ is continuous (Lemma 74), we have $d_R y - x \in {{\mathrm{\mathcal {F}}}}^{n+1} X^j$. Since n was arbitrary, $d_R y = x$. $\square$

Corollary 37

The cohomology of $(X, d_R)$ is concentrated in a non-negative degree.

Corollary 38

The natural map $H^j( X, d_R ) \rightarrow H^j( X / {{\mathrm{\mathcal {F}}}}^{p+1} X, d_R)$ is an isomorphism for $j < p$ and injective for $j=p$.

Proof

The short exact sequence $0 \rightarrow {{\mathrm{\mathcal {F}}}}^{p+1} X \rightarrow X \rightarrow X / {{\mathrm{\mathcal {F}}}}^{p+1} X \rightarrow 0$ defines the long exact sequence

$$\begin{aligned}&\cdots \rightarrow H^j( {{\mathrm{\mathcal {F}}}}^{p+1}X, d_R ) \rightarrow H^j( X, d_R ) \rightarrow H^j ( X / {{\mathrm{\mathcal {F}}}}^{p+1} X, d_R )\\&\quad \rightarrow H^{j+1}( {{\mathrm{\mathcal {F}}}}^{p+1} X, d_R ) \rightarrow \cdots . \end{aligned}$$

For $j \leqslant p,$ the first term is zero and for $j < p,$ and both the first and the last terms are zero by Lemma 36. $\square$

6.2 Spectral sequences

Lemma 39

Let $E^{p,q}_r$ be the spectral sequence corresponding to the filtered complex ${{\mathrm{\mathcal {F}}}}^p X^{p+q}$ with differential $d_R$. We have $H^\bullet ( X, d_R ) \cong E^{\bullet ,0}_2$ as graded commutative algebras.

Proof

Begin with $E_0^{p,q} := {{\mathrm{\mathcal {F}}}}^p X^{p+q} / {{\mathrm{\mathcal {F}}}}^{p+1} X^{p+q}$. It is concentrated in degree $p \geqslant 0, q \leqslant 0$. By Lemma 35, we have the following isomorphism of differential bi-graded algebras:

$$\begin{aligned} E^{p,q}_1&= H^q ( E_0^{p, \bullet } , d_R ) = H^q( {{\mathrm{\mathcal {F}}}}^p X^{p+ \bullet } / {{\mathrm{\mathcal {F}}}}^{p+1} X^{p + \bullet }, d_R ) \\&= H^{p+q}({{\mathrm{\mathcal {F}}}}^p X / {{\mathrm{\mathcal {F}}}}^{p+1} X, d_R)\cong {\left\{ \begin{array}{ll} B^p \otimes _P P/J, &{} \text {if } q=0 \\ \nonumber 0, &{} \text {if } q\ne 0. \end{array}\right. } \end{aligned}$$

Hence, $E^{p,q}_1$ is concentrated in degree $p \geqslant 0$ and $q = 0$. Moreover, $d_1^{p,q}$ maps $E_1^{p,q}$ to $E_1^{p+1, q}$. Hence, also $E_2^{p,q}$ is concentrated in $p \geqslant 0, q = 0$. Since $d_2$ maps $E^{p,0}_2$ to $E_2^{p+2,-1},$ it is zero for degree reasons and hence the spectral sequence degenerates at $E_2$.

We are left to prove that the spectral sequence converges to the cohomology. By [6, chapter XV, proposition 4.1], this follows from Lemma 36. $\square$

We could use this lemma to prove $H^0(X, d_R) \cong (P/J)^J$ as algebras. However, we want to consider an additional structure on the latter space.

6.3 The Poisson algebra structure on $(P/J)^J$

The following two remarks are well known and easily checked.

Remark 40

The Poisson algebra structure on P induces a Poisson algebra structure on $(P/J)^J$.

Remark 41

The graded Poisson algebra structure on X induces a Poisson algebra structure on the cohomology $H^0(X, d_R)$ in degree zero.

Those two structures are in fact isomorphic. We will explicitly construct a Poisson isomorphism. By Corollary 38, we have $H^0(X, d_R) \cong H^0(X/{{\mathrm{\mathcal {F}}}}^2 X, d_R)$ as vector spaces.

Lemma 42

Representatives in $X^0$ of cocycles in $X^0/{{\mathrm{\mathcal {F}}}}^2 X^0$ defining elements in $H^0(X/{{\mathrm{\mathcal {F}}}}^2 X , d_R)$ may be taken of the form

$$\begin{aligned} x = x_0 + \sum _{i,j \in L} a_{ij} e^*_i e_j, \end{aligned}$$

where $L = \{ n \in {{\mathrm{\mathbb {N}}}}: \deg (e^*_j) = 1 \}$, $x_0 \in P$, the $\{e_j\}$ are a homogeneous basis of ${{\mathrm{\mathcal {M}}}}$, and the $a_{ij} \in P$ are chosen such that

$$\begin{aligned} \{\delta (e_j), x_0\} = \sum _{i \in L} a_{ji} \delta (e_i). \end{aligned}$$

Conversely, every such element defines a cohomology class.

Proof

We have

$$\begin{aligned} X^0 / {{\mathrm{\mathcal {F}}}}^2 X^0&= P \oplus ( P \otimes ({{\mathrm{\mathcal {M}}}}^*)^{-1} \otimes {{\mathrm{\mathcal {M}}}}^{-1} ) \\ X^{-1} / {{\mathrm{\mathcal {F}}}}^2 X^{-1}&= (P \otimes {{\mathrm{\mathcal {M}}}}^{-1}) \oplus ( P \otimes ( {{\mathrm{\mathcal {M}}}}^*)^1 \otimes {{\mathrm{\mathcal {M}}}}^{-2})\\&\quad \oplus (P \otimes ({{\mathrm{\mathcal {M}}}}^*)^1 \otimes ({{\mathrm{\mathcal {M}}}}^{-1} \wedge {{\mathrm{\mathcal {M}}}}^{-1})). \end{aligned}$$

Hence, an arbitrary cochain may be taken to be of the form

$$\begin{aligned} x = x_0 + \sum _{i,j \in L} a_{ij} e^*_i e_j \end{aligned}$$

for some $x_0, a_{ij} \in P$. We compute with the help of Lemma 18,

$$\begin{aligned} d_R x&= \{R, x_0\} + \sum _{i,j \in L} ( \{R, a_{ij} \}e^*_i e_j + \{R, e^*_i \} e_j a_{ij} - \{R, e_j\} e^*_i a_{ij}) \\&\equiv \{R, x_0\} - \sum _{i,j \in L} \{R, e_i\} e_j^* a_{ji}\\&\equiv \sum _{j \in L} \bigg ( (-1)^{1+d_j}\{\delta (e_j), x_0\} - \sum _{i \in L} a_{ji} \delta (e_i) \bigg ) e^*_j \pmod { {{\mathrm{\mathcal {F}}}}^2 X^1}. \end{aligned}$$

$\square$

Theorem 43

$H^0(X, d_R) \cong (P/J)^J$ are Poisson algebras.

Proof

Let $\pi : X \rightarrow P = X/(I + I_-)$ denote the projection onto all monomials which contain no factors of nonzero degree. Here, $I_- \subset X$ denotes the ideal generated by all elements of negative degree. Define the map $\Phi : H^0(X, d_R) \rightarrow P/J$ by $\Phi ([x]) := \pi (x) + J$. This map is well defined: Let $x = d_R y$ be exact. Consider again the differential $\delta$ on $T=X/I$ that is induced by $d_R$ and its representation as an element $Q_0 \in X^1$. Also, pick a homogeneous basis $e_j$ of ${{\mathrm{\mathcal {M}}}}$ as done before. By Lemma 18, we obtain

$$\begin{aligned} d_R(y) \equiv \{Q_0, y\} \equiv \sum _i \{ (-1)^{1+d_i} \delta (e_i) e_i^*, y\} \equiv \sum _{i : \deg e_i \!= -1} \delta (e_i) \{ e_i^*, y\}\!\!\!\! \pmod {I {+} I_-}. \end{aligned}$$

The last sum is finite. Hence,

$$\begin{aligned} \pi (x) = \pi (d_R y) = \sum _{i : \deg e_i = -1} \delta (e_i) \pi \{e_i^*, y\} \in J. \end{aligned}$$

By Lemma 38, we have $H^0(X, d_R) \cong H^0(X/{{\mathrm{\mathcal {F}}}}^2 X, d_R)$ as vector spaces. Hence, we have a corresponding linear map $X/{{\mathrm{\mathcal {F}}}}^2 X \rightarrow P/J$.

The image of either of those maps is J-invariant: Let $[x] \in H^0( X / {{\mathrm{\mathcal {F}}}}^2 X, d_R)$. According to Lemma 42, we may pick a representative $x_0 = \pi (x_0) + \sum _{i,j \in L} a_{ij} e^*_i e_j$ of x,  where $a_{ij} \in P$ satisfy $\{\delta (e_j), \pi (x_0) \} = \sum _{i\in L} a_{ji} \delta (e_i)$. In particular, $\{\delta (e_j), \pi (x_0)\} \in J$. Fix $b \in J$. Then there exist $b_j \in P$ with $b = \sum _{i \in L} b_i \delta (e_i)$ and thus $\{b, \pi (x_0)\} = \sum _{i \in L} \big (b_i \{\delta (e_i), \pi (x_0)\} + \delta (e_i) \{b_i, \pi (x)\} \big ) \in J$.

Hence, we have two linear maps:

$$\begin{aligned} \phi : H^0(X/{{\mathrm{\mathcal {F}}}}^2 X, d_R)&\rightarrow (P/J)^J ,\\ \Phi : H^0(X, d_R)&\rightarrow (P/J)^J, \end{aligned}$$

given by projection onto the P component followed by modding out J, which correspond to each other under the isomorphism $H^0(X, d_R) \cong H^0(X/{{\mathrm{\mathcal {F}}}}^2 X, d_R)$.

The map $\phi$ is surjective: Let $p \in P$ with $\{J,p\}\subset J$. By Lemma 42, the element $x = p + \sum _{ij \in L} a_{ij} e^*_i e_j$ is a cocycle if $\{\delta (e_j), p\} = \sum _{i\in L} a_{ji} \delta (e_i)$. But those $a_{ij} \in P$ exist since $\{\delta (e_j)\}_{ j \in L}$ generate J. Hence, also the map $\Phi$ is surjective.

The map $\Phi$ is injective: Let $x \in X^0$ represent $[x] \in H^0(X, d_R)$ with $\pi (x) \in J$. We claim that there exist $y_j \in {{\mathrm{\mathcal {F}}}}^j X^{-1}$ with $x - d_R( y_0 + \cdots + y_n ) \in {{\mathrm{\mathcal {F}}}}^{n+1} X^0$. By Lemma 35, we know that $H^j( {{\mathrm{\mathcal {F}}}}^p X / {{\mathrm{\mathcal {F}}}}^{p+1} X, d_R )$ is concentrated in degree zero with $H^0(X / {{\mathrm{\mathcal {F}}}}^{1} X, d_R ) \cong P/J$ via the natural map. Now, $x + {{\mathrm{\mathcal {F}}}}^1 X^0$ defines the zero cohomology class in $H^0( X / {{\mathrm{\mathcal {F}}}}^1 X, d_R),$ since $\pi (x) \in J$. Hence there exists $y_0 \in {{\mathrm{\mathcal {F}}}}^0 X^{-1}$ with $x - d_R y_0 \in {{\mathrm{\mathcal {F}}}}^1 X^0$. Again, $x - d_R y_0 + {{\mathrm{\mathcal {F}}}}^2 X^0$ defines the zero cohomology class in $H^0( {{\mathrm{\mathcal {F}}}}^1 X /{{\mathrm{\mathcal {F}}}}^2 X, d_R) = 0$. Hence, there exists $y_1 \in {{\mathrm{\mathcal {F}}}}^1 X^{-1}$ with $x - d_R(y_0 + y_1) \in {{\mathrm{\mathcal {F}}}}^2 X^0$ and so on. Hence, $y_j$ exist and their sum converges to an element $y \in X^{-1}$ by Lemma 77, which satisfies $x - d_R y = 0$ by Lemma 75.

Hence, the map $\Phi$ is an isomorphism of vector spaces. This map also respects the product structure

$$\begin{aligned} \Phi ( [x] [y] )&= \Phi ( [xy] ) = \pi (x y) + J = \pi (x) \pi (y) + J = (\pi (x) + J)(\pi (y) + J)\\&= \Phi ([x])\Phi ([y]) \end{aligned}$$

and is hence an isomorphism of algebras. Finally, map $\Phi$ respects the bracket:

$$\begin{aligned} \Phi ( \{ [x], [y] \})&= \Phi ( [ \{x,y\}]) = \pi (\{x,y\}) + J = \pi (\{\pi (x), y\} ) + J \\&= \pi ( \{\pi (x), \pi (y)\}) + J = \{\pi (x), \pi (Y)\} + J = \{\pi (x) + J, \pi (y) + J\}\\&= \{\Phi (x), \Phi (y)\}, \end{aligned}$$

since $\{ \pi (x) - x , X^0\} \subset \{ (I + I_0) \cap X^0, X^0\} \subset \ker \pi ,$ where $\ker \pi = I + I_0 \subset X$ is the ideal generated by all elements of nonzero degree. The last inclusion holds by the Leibnitz rule since all summands of elements in $I + I_0$ that are of degree zero contain at least two factors of nonzero degree. $\square$

7 Examples

We present two well-known examples. More interesting examples can be found in [13].

7.1 Rotations of the Plane

Here, we present an example, where the cohomology in degree zero has a nontrivial bracket and the cohomology in degree 1 does not vanish. It is obtained by considering the symplectic lift of the rotations of the plane to the cotangent bundle of the plane.

Consider $P = {{\mathrm{\mathbb {R}}}}[x_1, x_2, y_1, y_2]$ with $\{x_i, y_j\} = \delta _{ij}$. The ideal $J \subset P$ generated by $\mu = x_1 y_2 - x_2 y_1$ is coisotropic. A Tate resolution of J is given by

$$\begin{aligned} 0 \rightarrow P \cdot e \rightarrow P \rightarrow P/J \rightarrow 0 \end{aligned}$$

where the differential $\delta$ is the P-linear derivation defined by $\delta (e) = \mu$. Indeed, this complex is a Koszul complex which is exact, since $\mu \ne 0$ defines a regular sequence. Hence, $X = \big ( P \cdot e \big ) \oplus \big ( P \oplus P \cdot e^* e \big ) \oplus \big ( P \cdot e^* \big )$. We now apply the construction from Sect. 3. We obtain $Q_0 = e^* \mu$ and $R = Q_0,$ since $\{Q_0, Q_0\} = 0$. One easily calculates

$$\begin{aligned} H^0(X, d_R) = \frac{ \{ a + b e^*e : \{\mu , a\} = \mu b , a,b \in P\} }{ \{ \mu c + \{\mu ,c\} e^* e : c \in P \} }. \end{aligned}$$

Notice that the isomorphism $H^0(X, d_R) \rightarrow (P/J)^J$ given by projection onto P is evident here. Moreover, the bracket on this space does not vanish: $x_1^2 + x_2^2$ and $y_1^2 + y_2^2$ define cohomology classes, for which $\{x_1^2 + x_2^2, y_1^2 + y_2^2\} = 4 (x_1 y_1 + x_2 y_2)$ is not in J. Furthermore,

$$\begin{aligned} H^1(X, d_R) = \frac{\{ a e^* : a \in P\}}{\{ d_R( a + b e^*e ) : a, b \in P\}} \cong \frac{ P }{ \{\{\mu , a\} + \mu b : a, b \in P \}} \end{aligned}$$

does not vanish, since $\deg _0\{\mu , a\} \ge 1$ and $\deg _0(\mu b) \ge 2$. Here, $\deg _0$ denotes the degree in $P={{\mathrm{\mathbb {R}}}}[x_i, y_i]$.

7.2 Rotations of space

Let $X = {{\mathrm{\mathbb {R}}}}^3$ and $M = T^*X \cong X \oplus X^*$. Consider the group $G = SO(3)$ acting on X via the standard representation $\rho _0 : G \rightarrow {{\mathrm{\text {End}}}}{X}$. The symplectic lift is given by $\rho : G \rightarrow {{\mathrm{\text {End}}}}{M}$, $\rho (A)(x,p) = (Ax, p \circ A^{-1})$. Mapping the standard basis of $X = {{\mathrm{\mathbb {R}}}}^3$ to its dual basis, we obtain an isomorphism $\iota : X \rightarrow X^*$. A possible moment map is the angular momentum mapping $\mu : M \rightarrow {{\mathrm{\mathbb {R}}}}^3$, $\mu (x,p) = x \times \iota {p}$. Here, $\times$ refers to the vector product, and we identified ${{\mathrm{\mathfrak {g}}}}\cong {{\mathrm{\mathbb {R}}}}^3$ using the basis

$$\begin{aligned} e_1&= \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad -1 \\ 0 &{}\quad 1 &{}\quad 0 \end{pmatrix},&\quad e_2&\quad = \begin{pmatrix} 0 &{}\quad 0 &{}\quad -1 \\ 0 &{}\quad 0 &{}\quad 0 \\ 1 &{}\quad 0 &{}\quad 0 \end{pmatrix},&e_3&= \begin{pmatrix} 0 &{}\quad -1 &{}\quad 0 \\ 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}. \end{aligned}$$

We define $M_0 = \mu ^{-1}(0) = \{ (x,p) \in X \oplus X^* : \iota p \parallel x\}$. This is not a manifold. If it was one, it had dimension 4. However, all $(x,0), x \in {{\mathrm{\mathbb {R}}}}^3$ and all $(0,p), p \in ({{\mathrm{\mathbb {R}}}}^3)^*$ belong to $M_0$. Hence, they also belong to the tangent space at the origin, provided $M_0$ was a manifold. Since the tangent space at the origin is linear, it would have dimension $6 > 4$. Since the constraint surface $M_0$ is not a manifold, results from [9, 14] do not apply.

We take the Tate resolution T of the vanishing ideal J of $M_0$ in $P = {{\mathrm{\mathbb {R}}}}[ M ] = {{\mathrm{\mathbb {R}}}}[x_1,x_2,x_3,p_1,p_2,p_3]$ with $\{x_i,p_j\} = \delta _{ij}$. It exists since P is Noetherian. We obtain the existence and uniqueness of a BRST charge R as described in the previous sections.

8 Quantization

In this section, we discuss quantization. In Sect. 8.1, we define a quantum algebra quantizing the Poisson algebra X from the previous part of this note. We rigorously define multiplication via normal ordering in the presence of infinitely many ghost variables.

In Sect. 8.2, we construct a solution of the quantum master equation associated with a given solution of the classical master equation. This means we construct an element of the quantum algebra that agrees with the quantization of the classical solution up to an error of order $\hbar$ and squares to zero.

In Sect. 8.3, we discuss the uniqueness of such solutions of the quantum master equation. We parallel our discussion in the classical case. In Sect. 8.3.1, we define the notion of a quantum gauge equivalence. In Sect. 8.3.2, we prove that two solutions of the quantum master equation that agree up to an error of order $\hbar$ are related via an automorphism of associative algebras. In Sect. 8.3.3, we show that the solutions of the quantum master equation associated with two BRST models associated with the same Tate resolution are also related by an automorphism of associative algebras. In Sect. 5.5, we have shown that any two BRST models associated with the same coisotropic ideal J are stably equivalent. We would like to find a quantum analogue of this theorem. We were able to prove in Sect. 8.3.5 that the process of adding extra variables yields a quasi-isomorphism of differential graded algebras on the quantum level. However, as discussed in Sect. 8.3.6, we were unable to quantize the general Poisson isomorphism of Lemma 31.

8.1 Quantum algebra

Assumption 44

Assume that $P = {{\mathrm{\mathbb {R}}}}[V]$ where V is a finite-dimensional real symplectic vector space and the bracket on P is induced by the symplectic structure.

Pick a decomposition $V = L \oplus L^*$ where L is a Lagrangian subspace of V and $\{v, \lambda \} = \lambda (v)$ for $v \in L$ and $\lambda \in L^*$. Set ${{\mathrm{\mathcal {N}}}}:= {{\mathrm{\mathcal {M}}}}\oplus L$ so that we may write $X_0 = {{\mathrm{\text {Sym}}}}( {{\mathrm{\mathcal {N}}}}\oplus {{\mathrm{\mathcal {N}}}}^*)$. We define $G_0 = G_- \otimes G_+$ where $G_- = {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}})$ and $G_+ = {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}})$ as vector spaces. We introduce a formal parameter $\hbar$ and want to define a product on $G_0[\hbar ]$ quantizing $X_0[\hbar ]$. Intuitively, we want to define the product $(x^1_{-} \otimes x^1_+)(x^2_{-} \otimes x^2_+)$ of monomials $x^{i}_- \otimes x^{i}_+ \in G_0 \subset G_0[\hbar ]$ by commuting $x^1_+$ to the right of $x^2_{-}$ using the canonical commutation relations we obtain from the split $V = L \oplus L^*$. A rigorous definition works as follows:

Let $(W[\hbar ], \star )$ be the deformation quantization [15] of $X_0$ defined by the (graded, see e.g. [7]) Wick-type, or normal ordered-type [20] Moyal product $\star$, defined by the split $V=L \oplus L^*$. Since $X_0$ contains only polynomials, the Moyal product of two elements of $W[\hbar ]$ is a polynomial in $\hbar$. Since naturally $G_0[\hbar ] \cong X_0[\hbar ]$ as ${{\mathrm{\mathbb {R}}}}[\hbar ]$-modules and, by construction of the deformation quantization, $X_0[\hbar ] \cong W[\hbar ]$ as ${{\mathrm{\mathbb {R}}}}[\hbar ]$-modules, the Moyal product defines a product on $G_0[\hbar ]$ turning $G_0[\hbar ]$ into a unital associative algebra.

We do the construction involving $G_0[\hbar ]$ instead of $X_0[\hbar ],$ since we want to avoid convergence issues. For example, the relation $\sum _j e^*_j e_j = \sum _j ( e_j e_j^* + \hbar )$ is problematic in the completion of $X_0[\hbar ]$. Using a suitable completion of $G_0[\hbar ]$, we avoid expressions involving infinite sums of terms that are not normally ordered.

Next, we define the completion and extend the product to it. We introduce the filtration on $G_0$ defined by the subspaces ${{\mathrm{\mathcal {F}}}}^p G_0 = \bigoplus _{q \ge p} G_- \otimes G_+^p$. Set ${{\mathrm{\mathcal {F}}}}^p G_0^n = {{\mathrm{\mathcal {F}}}}^p G_0 \cap G_0^n = \bigoplus _{q \ge p} G_-^{n-p} \otimes G_+^p$. We complete $G_0$ to the graded vector space $G = \bigoplus _n G^n$ where

$$\begin{aligned} G^n = \lim _{\leftarrow p} \frac{G_0^n}{{{\mathrm{\mathcal {F}}}}^p G_0^n}. \end{aligned}$$

This graded vector space is again filtered by the subspaces ${{\mathrm{\mathcal {F}}}}^p G^n = \lim _{\leftarrow q} \frac{{{\mathrm{\mathcal {F}}}}^p G_0^n}{{{\mathrm{\mathcal {F}}}}^{p+q} G_0^n}$.

Define the graded vector space $G_\hbar$ by its homogeneous components $H_\hbar ^n = G^n[[\hbar ]]$. We have a family of projections $p_j : G_\hbar \rightarrow G$ mapping $\sum _{k \ge 0} x_k \hbar ^k \mapsto x_j$. The space $G_0[\hbar ]$ can be considered a graded subspace of $G_\hbar$. We want to extend the algebra structure on $G_0[\hbar ]$ to $G_\hbar$. For this task, we need to analyse the compatibility of the product on $G_0[\hbar ]$ with the filtration on $G_0$.

Lemma 45

We have for all $j, p \ge 0$ and $n,m \in {{\mathrm{\mathbb {Z}}}},$

• (1) $p_j( G_0^n \cdot {{\mathrm{\mathcal {F}}}}^p G_0^m ) \subset {{\mathrm{\mathcal {F}}}}^p G_0^{n+m},$

• (2) $p_j( {{\mathrm{\mathcal {F}}}}^p G_0^n \cdot G_0^m ) \subset {{\mathrm{\mathcal {F}}}}^{p+m} G_0^{n+m}.$

Proof

Consider $x = a\otimes u \cdot b\otimes v \in G_0^n \cdot G_0^m$ where $a \in G_-^{n-l}, u \in G_+^l, b \in G_-^{m-k}$ and $v \in G_+^k$ for some $l, k \ge 0$. Then,

$$\begin{aligned} p_j(a\otimes u \cdot b\otimes v) = \sum _{b' u'} \pm ab' \otimes u'v, \end{aligned}$$

where $u'$ and $b'$ arise from u and b by deleting j matching pairs $(e,e^*),$ in which $e \in {{\mathrm{\mathcal {N}}}}$ is a factor in b and $e^* \in {{\mathrm{\mathcal {N}}}}^*$ is a factor in u of opposite degree. The sum is finite.

To prove the first statement, it suffices to note that $\deg u' \ge 0$ and $k \ge p$. For the second statement, we note that $\deg u' + \deg b' = \deg u + \deg b$ and hence we can estimate $\deg u' + \deg v \ge \deg u' + \deg b' + \deg v = \deg u + \deg b + \deg v = l + m-k + k \ge p + m$. $\square$

Lemma 46

The product (and hence also the commutator) extend to the completion turning $(G_\hbar , \cdot )$ into a graded algebra.

Proof

First, we consider $x = (x_p + {{\mathrm{\mathcal {F}}}}^p G_0^n)_p \in G^n$ and $y = (y_p + {{\mathrm{\mathcal {F}}}}^p G_0^m)_p \in G^m$. By Lemma 45, the limit $\lim _{p \rightarrow \infty } x_p \cdot y_p \in H^{n+m}$ is well defined and the definition $x \cdot p = \lim _{p\rightarrow \infty } x_p \cdot y_p$ does not depend on the choice of representatives $x_p, y_p$. The multiplication extends to $G_\hbar$ by bi-linearity in ${{\mathrm{\mathbb {R}}}}[ [\hbar ]]$. $\square$

Lemma 47

The product on $G_\hbar ^{n} \times G_\hbar ^m$ is continuous in each entry.

Proof

Let $x_r, x \in G_\hbar ^n$ with $x_r \rightarrow x$ and $y \in G_\hbar ^m$. Write $p_j(x_r) = (x_{r,p}^{(j)} + {{\mathrm{\mathcal {F}}}}^p G_0^n)_p$, $p_j(x) = (x_p^{(j)} + {{\mathrm{\mathcal {F}}}}^p G_0^n)_p$, and $p_j(y)= (y_p^{(j)} + {{\mathrm{\mathcal {F}}}}^p G_0^m)_p$. Fix $j, p \in {{\mathrm{\mathbb {N}}}}_0$. Take $r_0$ such that for all $0 \le k \le j$ and all $r \ge r_0,$ we have $x_{r, p-m}^{(k)} \equiv x_{p-m}^{(k)} \pmod { {{\mathrm{\mathcal {F}}}}^{p-m} G_0^n}$. Such $r_0$ exists since $x_r \rightarrow x$. For $r \ge r_0,$ we have by Lemma 45

$$\begin{aligned} p_j(x_r \cdot y)-p_j(x\cdot y)&= \bigg ( \sum _{l=0}^j p_l\Big ( \sum _{k=0}^{j-l} (x_{r,p-m}^{(k)} - x_{p-m}^{(k)}) \cdot y_p^{(j-l-k)} \Big ) + {{\mathrm{\mathcal {F}}}}^p G_0^{n+m}\bigg )_p\\&= 0. \end{aligned}$$

Continuity in the other entry follows analogously. $\square$

Now that we have set up the algebra, we define the quantization mapping. We have the canonical graded vector space isomorphism $q_0 : {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}\oplus {{\mathrm{\mathcal {N}}}}^*) \rightarrow {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}) \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}^*)$. Since this map respects the respective filtrations, we can extend it to

$$\begin{aligned} q : X \rightarrow G \subset G_\hbar . \end{aligned}$$

Since the inverse of $q_0$ also respects the filtration and thus extends, the map $q: X \rightarrow G$ is an isomorphism of graded vector spaces.

To relate the multiplicative structure on X to the one on $G_\hbar$, we set $A = G_\hbar /(\hbar )$ where $(\hbar )\subset G_\hbar$ is the two-sided ideal generated by $\hbar$. Hence, $A \cong G$ act as graded vector spaces, but not as graded algebras since G is not closed under multiplication.

Remark 48

We have $[G_\hbar , G_\hbar ] \subset (\hbar )$. Hence, $(\hbar )$ is the two-sided ideal generated by the commutator. Hence, the map $\frac{1}{\hbar }[ - , - ] : G_\hbar \otimes G_\hbar \rightarrow G_\hbar$ is well defined and turns $G_\hbar$ into a graded noncommutative Poisson algebra. This map descends to A,  turning it into a graded commutative Poisson algebra.

Theorem 49

The graded (commutative) Poisson algebras X and A are isomorphic via the map $\phi = \pi \circ \iota \circ q : X \rightarrow G \rightarrow G_\hbar \rightarrow A$, where $\pi : G_\hbar \rightarrow A=G_\hbar /(\hbar )$ is the canonical projection and $\iota : G \rightarrow G_\hbar$ is the inclusion. In particular, for all $x,y \in X$

$$\begin{aligned} \frac{1}{\hbar }[q(x),q(y)] \equiv q(\{x,y\}) \quad \pmod { (\hbar )}. \end{aligned}$$

(3)

Proof

We already know that $\phi$ is an isomorphism of graded vector spaces. We have to prove the compatibility with the product and bracket structures. By density of $X_0 \subset X$ and continuity of all maps involved, it suffices to consider $X_0$. Without the completion the statement is standard; see e.g. [20]. $\square$

Corollary 50

If $R \in X^1$ solves the classical master equation, then $\frac{1}{\hbar } [q(R),q(R)] \equiv 0 \pmod { (\hbar ) }$. Conversely, if $r = q(R) + \hbar (\cdots )$ solves the quantum master equation $[r,r]=0$, then R solves the classical master equation.

8.2 Solving the quantum master equation

We want to construct a solution $r \in G_\hbar ^1$ of the quantum master equation $[r,r] = 0$. We seek a solution of the form

$$\begin{aligned} r = q(R) + \hbar q(R_1) + \hbar ^2 q(R_2) + \cdots \end{aligned}$$

for some $R_j \in X^1$ where $R \in X^1$ is a given solution of the classical master equation.

Assumption 51

We assume $H^2(X, d_R) = 0$.

8.2.1 A differential on the quantum algebra

Define $D = \frac{1}{\hbar }[ q(R), -]$. This defines a map $G_\hbar \rightarrow G_\hbar$ by Remark 48. It preserves the ideal $(\hbar )$ and hence descends to a derivation $D_0$ on $A = G_\hbar /(\hbar )$. We calculate

$$\begin{aligned} D^2(x) = \frac{1}{\hbar ^2}[ q(R), [q(R), x]] = \frac{1}{\hbar ^2} [ [q(R), q(R)], x] - \frac{1}{\hbar ^2} [q(R) , [q(R), x]]. \end{aligned}$$

Hence by Corollary 50 and Remark 48,

$$\begin{aligned} D^2(x)&= \frac{1}{2\hbar } \left[ \frac{1}{\hbar }[q(R), q(R)], x \right] \equiv 0 \quad \pmod { \hbar }, \end{aligned}$$

(4)

so $D_0$ is a differential on A.

Theorem 52

We have $D_0 \circ \phi = \phi \circ d_R$. In particular, $\phi : X \rightarrow A$ is an isomorphism of differential graded commutative algebras and $H^\bullet (X, d_R) \cong H^\bullet (A, D_0)$.

Proof

Let $x \in X$. By Theorem 49,

$$\begin{aligned} D_0(\phi (x))&= D_0( \pi ( q(x) ) )= \pi ( D(q(x))) = \pi \left( \frac{1}{\hbar } [q(R), q(x)] \right) \\&= \pi ( q( \{R,x\} )) = \pi (q(d_R(x))) = \phi (d_R(x)). \end{aligned}$$

$\square$

Corollary 53

Under Assumption 51, $H^2(A, D_0) = 0$.

8.2.2 Construction of a solution of the quantum master equation

Theorem 54

Let $n \ge 0$ be an integer. For $n \ge 1$, assume we have constructed $R_1, R_2, \dots , R_n \in X^1$ such that $r_n := q(R) + \sum _{l=1}^n \hbar ^l q(R_l)$ satisfies $\frac{1}{\hbar }[r_n, r_n] \equiv 0 \pmod { (\hbar ^{n+1}) }$. For $n=0,$ set $r_n = q(R)$ which also satisfies this assumption by Corollary 50. We claim that there exists $R_{n+1} \in X^1$ such that

$$\begin{aligned} \frac{1}{\hbar } [r_n + \hbar ^{n+1} q(R_{n+1}), r_n + \hbar ^{n+1} q(R_{n+1}) ] \equiv 0 \quad \pmod { (\hbar ^{n+2}) }. \end{aligned}$$

Proof

We compute for any $R_{n+1} \in X^1$, using Corollary 50,

$$\begin{aligned}&\frac{1}{\hbar } [r_n + \hbar ^{n+1} q(R_{n+1}), r_n + \hbar ^{n+1} q(R_{n+1}) ] \\&\quad = \frac{1}{\hbar } [r_n, r_n] + 2 \hbar ^{n+1} \frac{1}{\hbar } [r_n, q(R_{n+1}) ] + \hbar ^{2n+2} \frac{1}{\hbar } [q(R_{n+1}), q(R_{n+1})] \\&\quad \equiv \frac{1}{\hbar } [r_n, r_n] + 2 \hbar ^{n+1} \frac{1}{\hbar } [q(R), q(R_{n+1}) ] \quad \pmod { (\hbar ^{n+2})}. \end{aligned}$$

By the induction assumption, we can write the right hand side as $\hbar ^{n+1}$ times

$$\begin{aligned} \frac{1}{\hbar ^{n+1}} \frac{1}{\hbar }[r_n,r_n] + 2 D(q(R_{n+1})) \in H \end{aligned}$$

and we want this to be a multiple of $\hbar$. This means we need to show that we can pick $R_{n+1} \in X^1,$ such that $\pi ( \frac{1}{\hbar ^{n+1}} \frac{1}{\hbar }[r_n,r_n] ) + 2 D_0 (\phi (R_{n+1}))$ vanishes in A. By Corollary 53 and the fact that $\phi$ is surjective, it suffices to prove that $\pi ( \frac{1}{\hbar ^{n+1}} \frac{1}{\hbar }[r_n,r_n] )$ is $D_0$-closed. By the Jacobi identity,

$$\begin{aligned} 0 = \frac{1}{\hbar ^2} [r_n, [r_n, r_n]] = D\left( \frac{1}{\hbar } [r_n, r_n]\right) + \sum _{l=1}^{n} \hbar ^l \frac{1}{\hbar } [q(R_l), \frac{1}{\hbar }[r_n, r_n]]. \end{aligned}$$

By the induction assumption, we may divide this equation by $\hbar ^{n+1}$ to arrive at

$$\begin{aligned} D\left( \frac{1}{\hbar ^{n+1}} \frac{1}{\hbar } [r_n, r_n] \right) = - \sum _{l=1}^n \hbar ^l \frac{1}{\hbar } \left[ q(R_l), \frac{1}{\hbar ^{n+1}} \frac{1}{\hbar }[r_n, r_n]\right] \equiv 0 \pmod \hbar . \end{aligned}$$

$\square$

Hence, we have constructed $r = q(R) + \hbar q(R_1) + \cdots$ with $[r,r]=0$.

8.3 Uniqueness of the solution

In this paragraph, we consider questions of uniqueness of solutions of the quantum master equation that arise from quantization of a solution of the classical master equation.

8.3.1 Quantum gauge equivalences

We define the subspace $K = \{ x \in G_\hbar ^0 : p_0(x) \in q(I^{(2)}) \}$.

Lemma 55

K is closed under the commutator.

Proof

Let $x,y \in K$. Theorem 49 allows us to calculate

$$\begin{aligned} \frac{1}{\hbar } [x,y] \equiv \frac{1}{\hbar } [p_0(x),p_0(y)] \equiv q( \{ q^{-1}(p_0(x)), q^{-1}(p_0(x)) \} ) \pmod {\hbar }. \end{aligned}$$

By Lemma 76, the claim follows. $\square$

We call elements of K generators of quantum gauge equivalences. Typical elements of K are quantizations of generators of classical gauge equivalences or any degree zero multiple of $\hbar$. To exponentiate the Lie algebra K to a group acting on $G_\hbar$ by isomorphisms of associative algebras, we show that in each degree in $\hbar$ the Lie algebra L acts pro-nilpotent with respect to the filtration ${{\mathrm{\mathcal {F}}}}^p G^n$.

Lemma 56

We define ${{\mathrm{\text {ad}}}}_a b = \frac{1}{\hbar }[a,b]$ for $a \in K$ and $b \in G_\hbar$. The Lie algebra ${{\mathrm{\text {ad}}}}K \subset {{\mathrm{\text {End}}}}(G_\hbar )$ acts pro-nilpotently in each degree of $\hbar$. In particular, it exponentiates to a group of automorphisms of associative algebras.

Proof

Fix integers $j \ge 0$ and $k \ge 1$. For $i = 1, \dots , k$, take $u_i = u_{i0} + \hbar u_{i1} + \cdots$ where $u_{i0} \in q(I^{(2)})$ and $u_{ij} \in G^0$. Let $x \in G^n$. Fix $l = l_1 + l_k$ with integers $l_i \ge 0$. Then,

$$\begin{aligned}&p_j({{\mathrm{\text {ad}}}}_{u_1}^{l_1} \cdots {{\mathrm{\text {ad}}}}_{u_k}^{l_k} x) \\&\quad = p_j\left( \sum _{\begin{array}{c} j_i: \{1, \dots , l_i\} \rightarrow {{\mathrm{\mathbb {N}}}}_0 \\ i = 1, \ldots , k \end{array}} \hbar ^{\sum _{i=1}^k \sum _{s=1}^{l_i} j_i(s)} {{\mathrm{\text {ad}}}}_{u_{1j_1(1)}} \circ \cdots \circ {{\mathrm{\text {ad}}}}_{u_{1j_1(l_1)}} \circ \cdots \circ {{\mathrm{\text {ad}}}}_{u_{kj_k(1)}} \circ \cdots \circ {{\mathrm{\text {ad}}}}_{u_{kj_k(l_k)}} (x) \!\right) \\&\quad = \sum _{\begin{array}{c} j_i: \{1, \dots , l_i\} \rightarrow {{\mathrm{\mathbb {N}}}}_0 \\ i = 1, \dots , k \\ n = \sum _{i=1}^k \sum _{s=1}^{l_i} j_i(s) \le j \end{array}} p_{j-n} ({{\mathrm{\text {ad}}}}_{u_{1j_1(1)}} \circ \cdots \circ {{\mathrm{\text {ad}}}}_{u_{1j_1(l_1)}} \circ \cdots \circ {{\mathrm{\text {ad}}}}_{u_{kj_k(1)}} \circ \cdots \circ {{\mathrm{\text {ad}}}}_{u_{kj_k(l_k)}} (x) ). \end{aligned}$$

This is a finite sum. We now write out each argument of $p_{j-n}$ as a sum of products of the $u_{pq} \in G$. Each such product satisfies the following conditions. It contains $(l+1)$ factors. The factor x appears once. The number of factors $u_{pq}$ with $q \ge 1$ is bounded above by n. Hence, the number of factors $u_{i0} \in q(I^{(2)})$ is bounded below by $(l-n)$. Thus, the number of positive factors before normal ordering is bounded from below by $2(l-n)$. After normal ordering and applying $p_{j-n},$ the number of positive factors that still remain are bounded from below by $2(l-n) - (j-n) \ge 2(l-j)$. Hence, $p_j({{\mathrm{\text {ad}}}}_{u_1}^{l_1} \cdots {{\mathrm{\text {ad}}}}_{u_k}^{l_k} x)$ is a finite sum of elements in $G^n$, which contain at least $2(l-j)$ factors of positive degree. This bound is independent of x.

Finally, fix $p,j \ge 0$ and let $x = \sum _j x_j \hbar ^h \in G_\hbar ^n$. We have

$$\begin{aligned} p_j( {{\mathrm{\text {ad}}}}_{u_1}^{l_1} \cdots {{\mathrm{\text {ad}}}}_{u_k}^{l_k} x) = \sum _{k=0}^j p_{j-k} ( {{\mathrm{\text {ad}}}}_{u_1}^{l_1} \cdots {{\mathrm{\text {ad}}}}_{u_k}^{l_k} x_k ). \end{aligned}$$

Pick $l_0$ such that for all $m = 0, \dots , j$, for all $r \ge 0$ and $l=l_1 + \cdots + l_k \ge l_0$ we have $p_m( {{\mathrm{\text {ad}}}}_{u_1}^{l_1} \cdots {{\mathrm{\text {ad}}}}_{u_k}^{l_k} x_r ) \in {{\mathrm{\mathcal {F}}}}^p G^n$. Then for all $l = l_1 + \cdots + l_k \ge l_0,$ we have $p_j ( {{\mathrm{\text {ad}}}}_{u_1}^{l_1} \cdots {{\mathrm{\text {ad}}}}_{u_k}^{l_k} x) \in {{\mathrm{\mathcal {F}}}}^p G^n$. Hence, ${{\mathrm{\text {ad}}}}K$ acts pro-nilpotently with respect to this filtration and thus ${{\mathrm{\text {ad}}}}K$ exponentiates to a group of vector space automorphisms $\{\exp {{\mathrm{\text {ad}}}}_u : G_\hbar \rightarrow G_\hbar , u \in K\}$. These maps preserve the multiplicative structure since ${{\mathrm{\text {ad}}}}_u$ is a derivation for the product. $\square$

8.3.2 Ambiguity for a given solution of the classical master equation

Let $R \in X^1$ be a solution of the classical master equation. Throughout this paragraph, we assume

Assumption 57

We have $H^1(X,d_R) = 0$. Thus, $H^1(A, D_0) = 0$.

Let

$$\begin{aligned} r&= q(R) + \hbar q(R_1) + \cdots ,\\ r'&= q(R) + \hbar q(R_1') + \cdots \end{aligned}$$

be two solutions to the quantum master equation, so that $r \equiv r' \pmod {\hbar }$.

Lemma 58

Let $n \in {{\mathrm{\mathbb {N}}}}_0$. Assume that for $l = 1, \dots , n,$ we have $R_l = R_l'$. Then there exists a generator $c \in (\hbar ^{n+1}) \subset K$ of a quantum gauge equivalence such that $\exp {{\mathrm{\text {ad}}}}_c r \equiv r' \pmod { (\hbar ^{n+2}) }$.

Proof

Let $v = q(R_{n+1})-q(R_{n+1}') \in G^1$. Then, $0 = [r + r', r-r']$ since $r,r'$ solve the quantum master equation. Moreover, $r-r' \equiv \hbar ^{n+1} v \pmod {\hbar ^{n+2}}$. Hence,

$$\begin{aligned} 0 = \frac{1}{\hbar } [r+r', v + \hbar \cdots ] \equiv \frac{1}{\hbar }[ 2 q(R), v] \equiv 2 Dv \pmod { \hbar }. \end{aligned}$$

Thus, $D_0 \pi v = 0$. Hence by Assumption 57, $\pi v = D_0 \pi u$ for some $u \in G^0_\hbar$, so $v \equiv D u \pmod {\hbar }$. Since $v \in G^1$ is constant in $\hbar$, we may also assume that $u \in G^0$. Set $c = \hbar ^{n+1} u \in K$. We check that

$$\begin{aligned} \exp {{\mathrm{\text {ad}}}}_c r - r'&= r-r' + \frac{1}{\hbar }[c,r] + \sum _{l=2}^{+\infty } \frac{1}{l!} {{\mathrm{\text {ad}}}}_c^l r \\&\equiv \hbar ^{n+1} \left( v - \frac{1}{\hbar }[r, u]\right) \equiv \hbar ^{n+1} ( v - D u) \equiv 0 \pmod { (\hbar ^{n+2})}. \end{aligned}$$

$\square$

Theorem 59

Under Assumption 57, there is a quantum gauge equivalence mapping r to $r'$.

Proof

By Lemma 58, there exists a sequence of generators $c_j \in (\hbar ^{j+1}) \subset K$ of quantum gauge equivalences $\exp {{\mathrm{\text {ad}}}}_{c_j}$ which define a sequence $r_{(j)}$ of solutions of the quantum master equation via $r_{(0)} = r$ and $r_{(j+1)} = \exp {{\mathrm{\text {ad}}}}_{c_j} r_{(j)}$, so that $r_{(j+1)} \equiv r' \pmod { \hbar ^{j+2}}$. We have

$$\begin{aligned} r_{(j+1)} = \exp {{\mathrm{\text {ad}}}}_{c_j} \exp {{\mathrm{\text {ad}}}}_{c_{j-1}} \cdots \exp {{\mathrm{\text {ad}}}}_{c_0} r = \exp {{\mathrm{\text {ad}}}}_{\gamma _j} r, \end{aligned}$$

for some $\gamma _j \in K$. We are left to show that $\gamma _j \rightarrow \gamma \in K$ and $\exp {{\mathrm{\text {ad}}}}_\gamma r = r'$. By the Campbell–Baker–Hausdorff formula, we have $\gamma _0 = c_0$ and $\gamma _{j+1} = \gamma _j + c_{j+1} + \cdots$, where the terms we have dropped involve sums of nested commutators $\frac{1}{\hbar }[-,-]$, each of which contain at least one $c_{j+1} \in (\hbar ^{j+2})$. Hence, $\gamma _{j+1} \equiv \gamma _{j} \pmod {\hbar ^{j+2}}$ and thus $\lim \gamma _j = \gamma \in K$ exists.

Finally, fix $k \in {{\mathrm{\mathbb {N}}}}_0$. We will prove that $p_k(\exp {{\mathrm{\text {ad}}}}_\gamma r - r') = 0$. We already know that $p_k( \exp {{\mathrm{\text {ad}}}}_{\gamma _{k-1}}r - r') = 0,$ since $\exp {{\mathrm{\text {ad}}}}_{\gamma _{k-1}} r = r_{(k)}$. Hence, it suffices to prove that $p_k( \exp {{\mathrm{\text {ad}}}}_{\gamma _{k-1}}r - \exp {{\mathrm{\text {ad}}}}_\gamma r) = 0$. We show by induction in $l \in {{\mathrm{\mathbb {N}}}}_0$ that ${{\mathrm{\text {ad}}}}_{\gamma _{k-1}}^l r \equiv {{\mathrm{\text {ad}}}}_\gamma ^l r \pmod {\hbar ^{k+1}}$. The case $l=0$ is trivial. Now, suppose the statement holds for some $l \in {{\mathrm{\mathbb {N}}}}_0$. Then,

$$\begin{aligned} {{\mathrm{\text {ad}}}}_{\gamma _{k-1}}^{l+1} r - {{\mathrm{\text {ad}}}}_\gamma ^{l+1} r&= {{\mathrm{\text {ad}}}}_{\gamma _{k-1}} ( {{\mathrm{\text {ad}}}}_{\gamma _{k-1}}^{l} r - {{\mathrm{\text {ad}}}}_\gamma ^{l} r + {{\mathrm{\text {ad}}}}_{\gamma }^l r) - {{\mathrm{\text {ad}}}}_\gamma ^{l+1} r \\&\equiv {{\mathrm{\text {ad}}}}_{\gamma _{k-1}} {{\mathrm{\text {ad}}}}_\gamma ^l r - {{\mathrm{\text {ad}}}}_\gamma ^{l+1} r = {{\mathrm{\text {ad}}}}_{\gamma _{k-1} - \gamma } {{\mathrm{\text {ad}}}}_\gamma ^l r \equiv 0 \pmod {\hbar ^{k+1}}, \end{aligned}$$

since $\gamma \equiv \gamma _{k-1} \pmod {\hbar ^{k+1}}$. $\square$

8.3.3 Ambiguity for two classical solutions corresponding to the same Tate resolution

Let $R, R'$ be two solutions of the classical master equation associated with the same Tate resolution. Let

$$\begin{aligned} r&= q(R) + \hbar q(R_1) + \cdots , \\ r'&= q(R') + \hbar q(R_1') + \cdots \end{aligned}$$

be two solutions of the quantum master equation.

Theorem 60

If either of the two solutions $R,R'$ of the classical master equation satisfy Assumption 57, then there is a quantum gauge equivalence mapping r to $r'$.

Proof

By Theorem 24, there exists a classical gauge equivalence $g = \exp {{\mathrm{\text {ad}}}}_u$ mapping R to $R'$. In particular, Assumption 57 is satisfied for both solutions. We have $c=q(u) \in K$. Set $r'' = \exp {{\mathrm{\text {ad}}}}_c r$. It is a solution of the quantum master equation. We first prove that $r'' \equiv r' \pmod {\hbar }$. We have

$$\begin{aligned} r'' - r' = \exp {{\mathrm{\text {ad}}}}_c r - r' \equiv \exp {{\mathrm{\text {ad}}}}_{q(u)} q(R) - q(R') \pmod \hbar . \end{aligned}$$

Now, we prove by induction in $l \in {{\mathrm{\mathbb {N}}}}_0$ that ${{\mathrm{\text {ad}}}}_{q(u)}^l q(R) \equiv q({{\mathrm{\text {ad}}}}_u^l R) \pmod {\hbar }$. For $l=0,$ this is obvious. Suppose it holds for some $l \in {{\mathrm{\mathbb {N}}}}_0$. Then, by Eq. 3,

$$\begin{aligned} {{\mathrm{\text {ad}}}}^{l+1}_{q(u)} q(R) = {{\mathrm{\text {ad}}}}_{q(u)} {{\mathrm{\text {ad}}}}_{q(u)}^l q(R) \equiv {{\mathrm{\text {ad}}}}_{q(u)} q( {{\mathrm{\text {ad}}}}_u^l R) \equiv q({{\mathrm{\text {ad}}}}^{l+1}_u R) \pmod \hbar . \end{aligned}$$

We now have

$$\begin{aligned} \sum _{l=0}^L \frac{1}{l!} {{\mathrm{\text {ad}}}}_{q(u)}^l q(R) -q(R') \equiv q\bigg ( \sum _{l=0}^L \frac{1}{l!} {{\mathrm{\text {ad}}}}_{c_0}^l R - R'\bigg ) \pmod {\hbar }. \end{aligned}$$

For $L \rightarrow + \infty ,$ the left hand side converges to $\exp {{\mathrm{\text {ad}}}}_{q(u)} q(R) - q(R')$, and the argument of q on the right hand side converges to zero. By continuity of q and $(\hbar )$ being closed, we conclude that $r'' \equiv r' \pmod {\hbar }$.

We are now in the situation

$$\begin{aligned} r&= q(R) + \hbar q(R_1) + \cdots ,\\ r'&= q(R') + \hbar q(R_1') + \cdots ,\\ r''&= \exp {{\mathrm{\text {ad}}}}_c r = q(R') + \hbar q(R_1'') + \cdots . \end{aligned}$$

By Theorem 59, there exists a quantum gauge equivalence $\exp {{\mathrm{\text {ad}}}}_v$ with $\exp {{\mathrm{\text {ad}}}}_v r'' = r'$. In particular, $r' = \exp {{\mathrm{\text {ad}}}}_v \exp {{\mathrm{\text {ad}}}}_c r$. $\square$

8.3.4 Quantization of trivial BRST models

Let (Y, S) be a trivial BRST model, so $Y = {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}\oplus {{\mathrm{\mathcal {N}}}}^*)$ for some negatively graded vector space ${{\mathrm{\mathcal {N}}}}$ with finite-dimensional homogeneous components and $S = \sum _j e_j^* \delta (e_j)$ with $\delta (e_j) = e_k$ for some k depending on j. Let $q:Y \rightarrow G$ denote the quantization map. Then, $s = q(S)$ solves the quantum master equation. Moreover, $D_s = \frac{1}{\hbar }[ s, -]$ maps $G_0$ to $G_0$ and G to G, as can be seen using the Leibnitz rule. Hence, both $q_0 : Y_0 \rightarrow G_0$ and $Y \rightarrow G$ are isomorphisms of differential graded vector spaces, in particular, $H^j(G_0, D_s) = 0$ for $j \ne 0$ and $H^0(G_0, D_s) = {{\mathrm{\mathbb {R}}}}$ by Lemma 27.

8.3.5 Quantization of products with trivial BRST models

Let (X, R) be a BRST model and (Y, S) a trivial BRST model. Consider quantizations $q : X \rightarrow F \subset F_\hbar$ and $q: Y \rightarrow G \subset G_\hbar$ with associated solutions of the quantum master equation $s = q(S)$ and $r = q(R) + \hbar \cdots$, respectively. Write $F_0 = {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}})\otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}*)$ for some non-positively graded vector space ${{\mathrm{\mathcal {N}}}}$ and $G_0 = {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {U}}}}) \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {U}}}}^*)$ for some negatively graded vector space ${{\mathrm{\mathcal {U}}}}$. Let $Z = X \hat{\otimes }Y$ and $q: Z \rightarrow H \subset H_\hbar$ be the quantization obtained from the splitting $H_0 = {{\mathrm{\text {Sym}}}}( {{\mathrm{\mathcal {N}}}}\oplus {{\mathrm{\mathcal {U}}}}) \otimes {{\mathrm{\text {Sym}}}}({{\mathrm{\mathcal {N}}}}^* \oplus {{\mathrm{\mathcal {U}}}}^*)$.

Lemma 61

The natural map $F_\hbar \rightarrow H_\hbar$ is a quasi-isomorphism of graded associative algebras.

Proof

The natural map is a morphism of graded associative algebras, since adding new variables from ${{\mathrm{\mathcal {U}}}}$ and ${{\mathrm{\mathcal {U}}}}^*$ does not change the rules defining normal ordering in $F_\hbar$.

Consider the isomorphism $\phi _0 : F_0 \otimes G_0 \rightarrow H_0$ of graded vector spaces. It extends $\hbar$-linearly to an isomorphism

$$\begin{aligned} \phi : F_0[\hbar ] \otimes _{{{\mathrm{\mathbb {R}}}}[\hbar ]} G_0[\hbar ] \rightarrow H_0[\hbar ] \end{aligned}$$

of graded associative algebras. On the left hand side, we may take the tensor product of algebras, since elements of $F_0$ and $G_0$ commute.

Using this isomorphism, we construct the $\hbar$-linear maps

$$\begin{aligned} \iota :&F_0[\hbar ] \rightarrow F_0[\hbar ] \otimes _{{{\mathrm{\mathbb {R}}}}[\hbar ]} G_0[\hbar ] \rightarrow H_0[\hbar ] ,\\ p :&H_0[\hbar ] \rightarrow F_0[\hbar ] \otimes _{{{\mathrm{\mathbb {R}}}}[\hbar ]} G_0[\hbar ] \rightarrow F_0[\hbar ], \end{aligned}$$

where the last arrow takes $f\otimes g \in F_0 \otimes G_0$ to $f \pi (g)$. Here, $\pi : G_0 \rightarrow G_0$ is the projection onto ${{\mathrm{\mathbb {R}}}}$ along elements of nonzero form degree.

The quantization map $q : Y_0 \rightarrow G_0$ is compatible with the decompositions $Y_0 = {{\mathrm{\mathbb {R}}}}\oplus \bigoplus _{j>0} {{\mathrm{\text {Sym}}}}^j({{\mathrm{\mathcal {U}}}}\oplus {{\mathrm{\mathcal {U}}}}^*)$ and $G_0 = {{\mathrm{\mathbb {R}}}}\oplus \bigoplus _{p+q > 0} {{\mathrm{\text {Sym}}}}^p({{\mathrm{\mathcal {U}}}}) \otimes {{\mathrm{\text {Sym}}}}^q({{\mathrm{\mathcal {U}}}})$. Moreover, it intertwines the differentials $d_S$ on $Y_0$ and $D_s$ on $G_0$. Hence, the situation of the proof of Lemma 29 is established in the quantum version as well. We conclude that $\iota$ and p extend $\hbar$-linearly to the respective completions, descend to cohomology, and induce mutual inverses on cohomology. $\square$

8.3.6 Relating arbitrary BRST models

To relate the quantizations of two BRST charges defining BRST models for the same ideal, we need to quantize general automorphisms of the space X that are the identity on $P={{\mathrm{\mathbb {R}}}}[V]$ modulo I (see Lemma 31). Since the quantization procedure is not functorial, we do not directly obtain an isomorphism of differential graded algebras on the quantum level. We were unable to rigorously define a quantum analog to such an isomorphism.

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}$$

$\square$