orientation ${\boldsymbol{q}}\prec{\boldsymbol{q}}^{\dagger}$ expressed by (9)
is built into the last term of (22) even if it is written as follows,

$(F)\overleftarrow{\delta_{{\boldsymbol{q}}}}\cdot\overrightarrow{\delta_{{\boldsymbol{q}}^{\dagger}}}(G)+(F)\overleftarrow{\delta_{{\boldsymbol{q}}^{\dagger}}}\cdot\overrightarrow{\delta_{{\boldsymbol{q}}}}(G).$

Should this be the notation for input, one then usually proclaims that
“differential 1-forms anticommute” so that
$\langle\delta{\boldsymbol{q}}^{\dagger}\wedge\delta{\boldsymbol{q}}\rangle=-\langle\delta{\boldsymbol{q}}\wedge\delta{\boldsymbol{q}}^{\dagger}\rangle=-1$
in $\lshad F,G\rshad=\langle\overrightarrow{\delta
F}\wedge\overleftarrow{\delta G}\rangle$.

We now are almost in a position to (re)configure the couplings in the four
terms of (21). The first term will of course become $\Delta F\cdot G$, and the
last will provide $(-)^{\operatorname{gh}(F)}F\cdot\Delta G$; one is here
allowed to integrate by parts (as explained in section 1.3) in order to shake
the derivatives off $\delta s^{i_{1}}$ and $\delta s^{\dagger}_{i_{2}}$ prior
to evaluation of couplings in the resulting object’s numeric value at its
argument $s$. Yet there remains one more logical step to be done with (22):
let us reverse back
$\overleftarrow{\delta}\\!\\!s\mapsto\overrightarrow{\delta}\\!\\!s$ and
$\overleftarrow{\delta}\\!\\!s^{\dagger}\mapsto\overrightarrow{\delta}\\!\\!s^{\dagger}$
so that on one hand, the vertical differentials fall on $F$ but on the other
hand, the normalization of the basis which stands near $\delta
s^{i}({\boldsymbol{y}}_{1})$ and $\delta
s^{\dagger}_{i}({\boldsymbol{y}}_{2})$ is the _first_ not second column in
(13). This yields the following integrand of
$(-)^{\operatorname{gh}(F)}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{1}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{2}\int_{M}\operatorname{dvol}({\boldsymbol{x}}_{1})\int_{M}\operatorname{dvol}({\boldsymbol{x}}_{2})$,
with a summation over $i_{1},i_{2},j_{1},j_{2}$, and $|\sigma_{1}|\geq 0,\
|\sigma_{2}|\geq 0$,

$\displaystyle\left.\left(f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{j_{1}}_{\sigma_{1}}}\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{1}}(s)}\langle\vec{e}^{{}\,\dagger
j_{1}}({\boldsymbol{x}}_{1}),+\vec{e}_{i_{1}}({\boldsymbol{y}}_{1})\rangle\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\sigma_{1}}(\delta
s^{i_{1}})({\boldsymbol{y}}_{1})\cdot$ (23)
$\displaystyle{}\qquad{}\cdot(\delta
s^{\dagger}_{i_{2}})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\sigma_{2}}({\boldsymbol{y}}_{2})\langle-\vec{e}^{{}\,\dagger
i_{2}}({\boldsymbol{y}}_{2}),\vec{e}_{j_{2}}({\boldsymbol{x}}_{2})\rangle\left.\left(\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{j_{2},\sigma_{2}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{2}}(s)}+{}$
$\displaystyle{}+$
$\displaystyle\left.\left(f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{j_{1},\sigma_{1}}}\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{1}}(s)}\cdot{}$
$\displaystyle\Biggl{\\{}\begin{matrix}&(\delta
s^{i_{1}})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\sigma_{2}}({\boldsymbol{y}}_{1})\langle\vec{e}_{i_{1}}({\boldsymbol{y}}_{1})|&&|\vec{e}^{{}\,\dagger
j_{2}}({\boldsymbol{x}}_{2})\rangle\\\
\langle\vec{e}_{j_{1}}({\boldsymbol{x}}_{1})|&&|-\vec{e}^{{}\,\dagger
i_{2}}({\boldsymbol{y}}_{2})\rangle\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\sigma_{1}}(\delta
s_{i_{2}}^{\dagger})({\boldsymbol{y}}_{2})\rangle&\end{matrix}\Biggr{\\}}\cdot$
$\displaystyle\mbox{\hbox to275.99173pt{{ }\hfil{
}}}\left.\left(\frac{\overrightarrow{\partial}}{\partial
q^{j_{2}}_{\sigma_{2}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{2}}(s)}.$

The integrations by parts are performed and couplings are reconfigured at the
end of the day in exactly same manner as it has been done in Definition 1; let
us recall that we now define the BV-Laplacian on a larger space. Namely, the
variations couple with the dual variations whereas the differentials of
functionals’ densities attach to each other.

######  Definition 2.

The _variational Schouten bracket_ of two integral functionals

${F=\int
f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]\cdot\operatorname{dvol}({\boldsymbol{x}}_{1})}\qquad\text{and}\qquad{G=\int
g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]\cdot\operatorname{dvol}({\boldsymbol{x}}_{2})}$

is the on-the-diagonal couplings surgery which, by using a normalized test
shift $\delta{\boldsymbol{s}}=(\delta s;0)+(0;\delta
s^{\dagger})\in\Gamma(T\pi_{{\text{{BV}}}})$, yields the functional from
$\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$ whose
construction at a BV-section $s\in\Gamma(\pi_{{\text{{BV}}}})$ is232323Note
that the directions of $\partial/\partial{\boldsymbol{y}}_{i}$ are reversed so
that the minus signs appear. We emphasize that, prior to the evaluation of
reconfigured couplings, the (co)vectors at ${\boldsymbol{x}}_{j}$ channel the
partial derivatives to $f$ or $g$ according to the couplings’ old arrangement.

$\displaystyle\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{1}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{2}\int_{M}{\mathrm{d}}{\boldsymbol{x}}_{1}\int_{M}\operatorname{dvol}({\boldsymbol{x}}_{2})\Biggl{[}\left.\left(f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{j_{1}}_{\sigma_{1}}}\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{1}}(s)}$
$\displaystyle\Biggl{\\{}\begin{matrix}\left(-\frac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\sigma_{1}}\delta
s^{i_{1}}({\boldsymbol{y}}_{1})\overbrace{\langle\vec{e}_{i_{1}}({\boldsymbol{y}}_{1}),-\vec{e}^{{}\,\dagger
i_{2}}({\boldsymbol{y}}_{2})\rangle}^{-1}\,\delta
s^{\dagger}_{i_{2}}({\boldsymbol{y}}_{2})\left(-\frac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\sigma_{2}}\\\
\underbrace{\langle\vec{e}^{{}\,\dagger
j_{1}}({\boldsymbol{x}}_{1})|\qquad\qquad\qquad\mathstrut,\mathstrut\qquad\qquad\qquad|\vec{e}_{j_{2}}({\boldsymbol{x}}_{2})\rangle}_{-1}\end{matrix}\Biggr{\\}}\left.\left(\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{j_{2},\sigma_{2}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{2}}(s)}$
${}+\left.\left(f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{j_{1},\sigma_{1}}}\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{1}}(s)}\\\
{}{}\qquad\quad\Biggl{\\{}\begin{matrix}\delta
s^{i_{1}}({\boldsymbol{y}}_{1})\left(-\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\sigma_{2}}\overbrace{\langle\vec{e}_{i_{1}}({\boldsymbol{y}}_{1}),(-\vec{e}^{{}\,\dagger
i_{2}})({\boldsymbol{y}}_{2})\rangle}^{-1}\cdot\left(-\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\sigma_{1}}\delta
s^{\dagger}_{i_{2}}({\boldsymbol{y}}_{2})\\\
\underbrace{\langle\vec{e}_{j_{1}}({\boldsymbol{x}}_{1})|\qquad\qquad\qquad\qquad,\mathstrut\qquad\qquad\qquad\qquad|\vec{e}^{{}\,\dagger
j_{2}}({\boldsymbol{x}}_{2})\rangle}_{+1}\end{matrix}\Biggr{\\}}\\\
{}\mbox{\hbox to275.99173pt{{ }\hfil{
}}}\left.\left(\frac{\overrightarrow{\partial}}{\partial
q^{j_{2}}_{\sigma_{2}}}g({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{2}}(s)}\Biggr{]}.$

Note that the inner couplings between variations provide a restriction to the
diagonal ${\boldsymbol{y}}_{1}={\boldsymbol{y}}_{2}$ and yield the singular
integral operators which then act to the right via multiplication by $-1$ only
if ${\boldsymbol{y}}_{2}={\boldsymbol{x}}_{2}$ and
${\boldsymbol{y}}_{1}={\boldsymbol{x}}_{1}$, respectively. The outer coupling
then furnishes the main diagonal
${\boldsymbol{x}}_{1}={\boldsymbol{y}}_{1}={\boldsymbol{y}}_{2}={\boldsymbol{x}}_{2}$,
restricting the objects further to the same BV-fibre point in the total space
of the BV-bundle. This reveals why over each point of the base $M^{n}$ the
(derivatives of the) densities $f$ and $g$ are restricted to the infinite jet
of the same section $s$; this also means that, since the moment when the
couplings are reconfigured, the volume element
$\operatorname{dvol}({\boldsymbol{x}}_{1})$ is discarded because appears a new
singular linear integral operator with a standard sign
$\int{\mathrm{d}}{\boldsymbol{x}}_{1}$.

We conclude the reasoning and sum up the definitions and notations in the
following theorem.

######  Theorem 3.

The BV-Laplacian $\Delta$ is the linear operator

$\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})\times\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})\to\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$

which acts on products of functionals $F$ and
$G\in\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$ by
the rule

$\Delta(F\cdot G)=\Delta(F)\cdot G+(-)^{\operatorname{gh}(F)}\lshad
F,G\rshad+(-)^{\operatorname{gh}(F)}F\cdot\Delta G.$ (24)

The variational Schouten bracket $\lshad\,,\,\rshad$ measures the deviation
for the BV-Laplacian $\Delta$ from being a derivation.
$\bullet$ After integration by parts, Definition 2 implies the renouned
coordinate formula

$\lshad
F,G\rshad=\int\operatorname{dvol}({\boldsymbol{x}})\Biggl{(}\frac{\overrightarrow{\delta}\\!f({\boldsymbol{x}},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])}{\delta{\boldsymbol{q}}}\cdot\frac{\overleftarrow{\delta}\\!g({\boldsymbol{x}},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])}{\delta{\boldsymbol{q}}^{\dagger}}-\\\
-\frac{\overrightarrow{\delta}\\!f({\boldsymbol{x}},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])}{\delta{\boldsymbol{q}}^{\dagger}}\cdot\frac{\overleftarrow{\delta}\\!g({\boldsymbol{x}},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])}{\delta{\boldsymbol{q}}}\Biggr{)}.$
(25)

######  Remark 2.11.

Let us recall from Remark 1.5 that the building blocks of local functionals
are encoded by equivalence classes of their densities, whereas the underlying
integration manifold $M^{n}$ is endowed with the field-dependent volume
element $\operatorname{dvol}({\boldsymbol{x}},\phi)$. The variational Schouten
bracket transforms two given integral functionals $F$ and $G$ into $\lshad
F,G\rshad$. For every configuration of physical fields $\phi\in\Gamma(\pi)$,
the integration measure is the same in $F$, $G$, and $\lshad F,G\rshad$. This
is because the couplings are local over points
$\bigl{(}{\boldsymbol{x}},\phi({\boldsymbol{x}})\bigr{)}$ in the total space
of the bundle $\pi$ of physical fields, see Remark 2.2 on p. 2.2 ; the
equality of local sections $\phi$ at which all (derivatives of) functionals’
densities are evaluated ensures the equality of metric tensor elements
$g_{\mu\nu}$ in all functionals by virtue of Einstein’s general relativity
equations.

The operational definition of the antibracket $\lshad\,,\,\rshad$ determines
the way how this structure acts on the square
$\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})\times\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$
of entire space
$\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$
containing formal products of functionals.

######  Theorem 4.

Let $F$, $G$, and
$H\in\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$ be
ghost parity-homogeneous functionals. The variational Schouten bracket
$\lshad\,,\,\rshad\colon\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})\times\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})\to\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$
has the following properties:

  * (i)

The value of $\lshad\,,\,\rshad$ at two arguments $F$ and $G\cdot H$ is

$\lshad F,G\cdot H\rshad=\lshad F,G\rshad\cdot
H+(-)^{(\operatorname{gh}(F)-1)\operatorname{gh}(G)}G\cdot\lshad F,H\rshad.$
(26)

This formula recursively extends to products of arbitrary finite number of
factors in the second argument.

  * (ii)

The bracket $\lshad\,,\,\rshad$ is shifted-graded skew-symmetric:

$\lshad
F,G\rshad=-(-)^{(\operatorname{gh}(F)-1)\cdot(\operatorname{gh}(G)-1)}\lshad
G,F\rshad,$ (27)

which extends $\lshad\,,\,\rshad$ to products of arbitrary finite number of
factors taken as its first argument in (26).

  * (iii)

The bracket $\lshad\,,\,\rshad$ satisfies the shifted-graded Jacobi identity

$(-)^{(\operatorname{gh}(F)-1)(\operatorname{gh}(H)-1)}\lshad F,\lshad
G,H\rshad\rshad+(-)^{(\operatorname{gh}(F)-1)(\operatorname{gh}(G)-1)}\lshad
G,\lshad H,F\rshad\rshad+{}\\\
{}+(-)^{(\operatorname{gh}(G)-1)(\operatorname{gh}(H)-1)}\lshad H,\lshad
F,G\rshad\rshad=0,$ (28)

which stems from graded Leibniz rule (36) for evolutionary vector fields
${\boldsymbol{Q}}^{F}$ defined by the rule
${\boldsymbol{Q}}^{F}(\cdot)\cong\lshad F,\,\cdot\,\rshad$ (here the
equivalence up to integration by parts is denoted by $\cong$ ).

Finally, the variational Schouten bracket extends by linearity to formal sums
of elements from
$\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})$.

###### Proof.

The bilinearity of $\lshad\,,\,\rshad$ is obvious. It is also clear that the
terms in $\lshad F,G\cdot H\rshad$ are grouped in two parts: those in which
the ghost-parity graded derivations
$\overrightarrow{\partial}/\partial{\boldsymbol{q}}^{\dagger}$ act on $G$ and
those for $H$; the former do not contribute with any extra sign factors
whereas the latter do — in a way which depends on the parity
$\operatorname{gh}(G)$. This means that $\lshad F,G\cdot H\rshad=\lshad
F,G\rshad\cdot H+\ldots$; to grasp the sign in front of the term which has
been omitted, let us swap the graded multiples $G$ and $H$. We have that
$G\cdot H=(-)^{\operatorname{gh}(G)\operatorname{gh}(H)}H\cdot G$, whence
$\lshad F,G\cdot H\rshad=(-)^{\operatorname{gh}(G)\operatorname{gh}(H)}\lshad
F,H\rshad\cdot G+\cdots$. By recalling that $\operatorname{gh}(\lshad
F,H\rshad)=\operatorname{gh}(F)+\operatorname{gh}(H)-1$, we conclude that

$\lshad F,G\cdot H\rshad=\lshad F,G\rshad\cdot
H+(-)^{\operatorname{gh}(G)\operatorname{gh}(H)}(-)^{(\operatorname{gh}(F)+\operatorname{gh}(H)-1)\cdot\operatorname{gh}(G)}G\cdot\lshad
F,H\rshad,$

which yields formula (26).

Proving (27) amounts to a count of signs whenever the bracket $\lshad
F,G\rshad$ of an ordered pair of ghost parity-graded objects is virtually
transformed into $\lshad G,F\rshad$. By using the rule of signs for odd-parity
coordinates, $q^{\dagger}_{\alpha,\sigma}\cdot
q^{\dagger}_{\beta,\tau}=-q^{\dagger}_{\beta,\tau}\cdot
q^{\dagger}_{\alpha,\sigma}$, we first note that

$\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{j_{2},\sigma_{2}}}g\bigl{(}{\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]\bigr{)}=(-)^{\operatorname{gh}(G)-1}\left(g\bigl{(}{\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]\bigr{)}\right)\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{j_{2},\sigma_{2}}},$

with a similar formula for the left- and right-acting graded derivative of
$f$. By swapping the (variational) derivatives of the densities $f$ and $g$,
we gain the signs $(-)^{\operatorname{gh}(F)\cdot(\operatorname{gh}(G)-1)}$
and $(-)^{(\operatorname{gh}(F)-1)\cdot\operatorname{gh}(G)}$ for the
respective terms in (23) on p. 23. Combined together, the two steps accumulate
equal factors
$(-)^{(\operatorname{gh}(F)+1)\cdot(\operatorname{gh}(G)-1)}=(-)^{(\operatorname{gh}(F)-1)\cdot(\operatorname{gh}(G)+1)}=(-)^{(\operatorname{gh}(F)-1)\cdot(\operatorname{gh}(G)-1)}$.
Thirdly, by comparing
$(-)^{(\operatorname{gh}(F)-1)\cdot(\operatorname{gh}(G)-1)}\,\lshad
F,G\rshad$ – in which the derivatives of $f$ and $g$ are interchanged and the
derivations’ directions are reversed – with $\lshad G,F\rshad$, we conclude
that the reconfiguration of couplings in the second term in (23) for the
former expression yields _minus_ the first term in $\lshad G,F\rshad$.
Likewise, the couplings reattachment in the first term of such (23) produces
minus the second term in $\lshad G,F\rshad$. This is because the (co)vectors
in the differentials of densities remain unswapped, now going in the ‘wrong’
order.

We now refer to [32, Proposition 3] for a proof of property (iii) in a wider,
non-commutative setup of cyclic words (cf. [29, 36, 46]). It is remarkable
that the reasoning persists within a naïve theory of variations, not referring
to our main idea that each test shift brings its own copy of the base $M^{n}$
into the picture. A key point in the proof is that the rule
${\boldsymbol{Q}}^{F}(\cdot)\cong\lshad F,\,\cdot\,\rshad$ naturally
associates with functionals $F$ the evolutionary fields ${\boldsymbol{Q}}^{F}$
on the infinite jet superbundles at hand, and with _evolutionary_ vector
fields it does not matter under ‘whose” total derivatives such fields dive,
obeying their defining property
$[{\boldsymbol{Q}}^{F},\overrightarrow{{\mathrm{d}}}/{\mathrm{d}}{\boldsymbol{x}}]=0$
(i. e., any integrations by parts, which transform the derivatives
$\overrightarrow{\partial}/\partial{\boldsymbol{y}}_{i}$ falling on test
shifts into total derivatives
$\overrightarrow{{\mathrm{d}}}/{\mathrm{d}}{\boldsymbol{x}}$ falling on the
functionals’ densities, do not mar the outcome even if one attempts to perform
such integrations ahead of time). ∎

###  2.5 Main result : the proof of properties (1c–1d)

We are ready to _prove_ the main interrelations between the BV-Laplacian
$\Delta$ and variational Schouten bracket $\lshad\,,\,\rshad$. Let us recall
that either a validity of these properties was postulated (see [21]) or an ad
hoc regularization technique was formally employed in the literature in order
to mask the seemingly present divergencies (which are actually not there), cf.
[22, §15].

Let us fix the terms. In what follows we refer to building blocks from
$\overline{H}^{n}(\pi_{{\text{{BV}}}})$ and their descendants – containing
reconfigured variations – from
$\overline{H}^{n(1+k)}(\pi_{{\text{{BV}}}}\times
T\pi_{{\text{{BV}}}}\times\ldots\times T\pi_{{\text{{BV}}}})$ as integral
functionals. Such objects will be used for bases of inductive proofs of Lemmas
5 and 7. We then extend the properties (1c) and $\Delta^{2}=0$ to the space
$\overline{\mathfrak{N}}^{n}(\pi_{{\text{{BV}}}},T\pi_{{\text{{BV}}}})\supseteq\overline{\mathfrak{M}}^{n}(\pi_{{\text{{BV}}}})$
of local functionals, that is, of formal sums of products of (varied
descendants of) building blocks.

######  Lemma 5.

Let $F\in\overline{H}^{n(1+k)}\bigl{(}\pi_{{\text{{BV}}}}\times
T\pi_{{\text{{BV}}}}\times\ldots\times T\pi_{{\text{{BV}}}}\bigr{)}$ and
$G\in\overline{H}^{n(1+\ell)}\bigl{(}\pi_{{\text{{BV}}}}\times
T\pi_{{\text{{BV}}}}\times\ldots\times T\pi_{{\text{{BV}}}}\bigr{)}$ be two
integral functionals; here $k,\ell\geqslant 0$. Then

$\Delta\bigl{(}\lshad{F,G}\rshad\bigr{)}=\lshad{\Delta
F,G}\rshad+(-)^{\operatorname{gh}(F)-1}\lshad{F,\Delta G}\rshad.$ (29)

###### Proof.

The key idea is that the structures $\Delta$ and $\lshad\,,\,\rshad$ yield
equivalence classes of integral functionals which, after an integration by
parts at the end of the day, are _independent_ of a choice of the built-in
test shifts normalized by (16). Consequently, the composite structure
$\Delta(\lshad{\cdot},{\cdot}\rshad)$ does not change under swapping $\delta
s_{1}^{\alpha}\rightleftarrows\delta s_{2}^{\beta}$, $\delta
s_{1,\alpha}^{\dagger}\rightleftarrows\delta s_{2,\beta}^{\dagger}$ of the
respective variations $\delta{\boldsymbol{s}}_{1}$ and
$\delta{\boldsymbol{s}}_{2}$ in $\Delta$ and $\lshad\,,\,\rshad$. Hence the
terms which are skew-symmetric under such exchange necessarily vanish.

For the sake of clarity, let us assume that $F=\int
f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\,\operatorname{dvol}({\boldsymbol{x}}_{1})$
and $G=\int
g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\,\operatorname{dvol}({\boldsymbol{x}}_{2})$
are just building blocks from the cohomology group
$\overline{H}^{n}(\pi_{{\text{{BV}}}})$; this simplification is legitimate
because new variations which come from $\Delta$ and $\lshad\,,\,\rshad$ do not
interfere with any other test shifts if those are already absorbed by the
densities $f$ and $g$. Suppose that $\delta{\boldsymbol{s}}_{1}$ and
$\delta{\boldsymbol{s}}_{2}$ are two normalized variations of a section
$s\in\Gamma(\pi_{{\text{{BV}}}})$. By definition, we have that242424To keep
track of their origin, we let the directed derivatives
$\partial/\partial{\boldsymbol{y}}_{i}$ or
$\partial/\partial{\boldsymbol{z}}_{j}$ remain falling on the respective
coefficients in $\delta{\boldsymbol{s}}_{1}$ and $\delta{\boldsymbol{s}}_{2}$;
the integration by parts is performed in a standard way prior to the
reconfigurations which are shown in the formula.

$\displaystyle\Delta\left(\lshad
F,G\rshad\right)(s)=\int_{M}{\mathrm{d}}{\boldsymbol{z}}_{1}\int_{M}{\mathrm{d}}{\boldsymbol{z}}_{2}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{1}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{2}\int_{M}{\mathrm{d}}{\boldsymbol{x}}_{1}\int_{M}\operatorname{dvol}({\boldsymbol{x}}_{2})\
\cdot$ $\displaystyle\Biggl{\\{}(\delta
s_{1}^{\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{1}}\right)^{\sigma_{1}}({\boldsymbol{z}}_{1})\,\left\langle\vec{e}_{\alpha}({\boldsymbol{z}}_{1}),-\vec{e}^{{}\,\dagger\alpha}({\boldsymbol{z}}_{2})\right\rangle\,(\delta
s^{\dagger}_{1,\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{2}}\right)^{\sigma_{2}}({\boldsymbol{z}}_{2})\langle\vec{e}^{{}\,\dagger\alpha}(\cdot),\vec{e}_{\alpha}(\cdot)\rangle\frac{\overrightarrow{\partial}}{\partial
q^{\alpha}_{\sigma_{1}}}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}$
$\displaystyle\quad\smash{\Biggl{[}}f({\boldsymbol{x}}_{1}.[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\smash{\frac{\overleftarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}}\,\underline{\langle\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}_{1})|}\,\Bigl{\langle}\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}(\delta
s_{2}^{\beta})({\boldsymbol{y}}_{1})\,\vec{e}_{\beta}({\boldsymbol{y}}_{1}),$
$\displaystyle\mbox{\hbox to71.13188pt{{ }\hfil{
}}}{-}\vec{e}^{{}\,\dagger.\beta}({\boldsymbol{y}}_{2})\,(\delta
s^{\dagger}_{2,\beta})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}({\boldsymbol{y}}_{2})\Bigr{\rangle}\,\underline{|\vec{e}_{\beta}({\boldsymbol{x}}_{2})\rangle}\,\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])+{}$
$\displaystyle{}\quad{}+f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\,\underline{\langle\vec{e}_{\beta}({\boldsymbol{x}}_{1})|}\,\Bigl{\langle}(\delta
s_{2}^{\beta})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}({\boldsymbol{y}}_{1})\,\vec{e}_{\beta}({\boldsymbol{y}}_{1}),$
$\displaystyle\mbox{\hbox to71.13188pt{{ }\hfil{
}}}{-}\vec{e}^{{}\,\dagger\beta}({\boldsymbol{y}}_{2})\,\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}(\delta
s^{\dagger}_{2,\beta})({\boldsymbol{y}}_{2})\Bigr{\rangle}\,\underline{|\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}_{2})\rangle}\,\frac{\overrightarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\smash{\Biggr{]}\left.\Biggr{\\}}\right|_{\begin{subarray}{c}j^{\infty}(s)\\\
{\boldsymbol{x}}_{i}={\boldsymbol{y}}_{j}={\boldsymbol{z}}_{k}\end{subarray}}}.$

The partial derivatives $\overrightarrow{\partial}/\partial
q^{\alpha}_{\sigma_{1}}\circ\overrightarrow{\partial}/\partial
q^{\dagger}_{\alpha,\sigma_{2}}$ are distributed between the arguments $f$ and
$g$ by the graded Leibniz rule. Whenever _none_ of the two operators overtakes
the density of $F$, the reconfiguration yields $\lshad\Delta
F,G\rshad({\boldsymbol{s}})$. Likewise, if _both_ derivatives indexed by
$\alpha$ overtake $F$ and an old derivative that fell on $g$, then we obtain
$(-)^{\operatorname{gh}(F)-1}\lshad F,\Delta G\rshad({\boldsymbol{s}})$, which
is the second term in the right-hand side of (29). We claim that the remaining
four terms cancel out by virtue of independence of $\Delta$ and
$\lshad\,,\,\rshad$ from a choice of normalized variations. To prove this
claim, we consecutively inspect the behaviour of those four terms under a swap
$\delta{\boldsymbol{s}}_{1}\rightleftarrows\delta{\boldsymbol{s}}_{2}$ of
coefficients in the normalized test shifts.

The first and second terms sum up to the difference

$\left\langle(\delta
s^{\alpha}_{1})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{1}}\right)^{\sigma_{1}}({\boldsymbol{z}}_{1})\,\overbrace{\vec{e}_{\alpha}({\boldsymbol{z}}_{1}),(-\vec{e}^{{}\,\dagger\alpha})({\boldsymbol{z}}_{2})}^{-1}\,(\delta
s^{\dagger}_{1,\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{2}}\right)^{\sigma_{2}}({\boldsymbol{z}}_{2})\right\rangle\underbrace{\langle\vec{e}^{{}\,\dagger\alpha}({\boldsymbol{x}}_{2}),\vec{e}_{\alpha}({\boldsymbol{x}}_{1})\rangle}_{-1}\,\cdot{}\\\
\cdot\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}\,\underline{\bigl{\langle}\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}_{1})\bigr{|}}\Biggl{\langle}\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}(\delta
s_{2}^{\beta})({\boldsymbol{y}}_{1})\,\overbrace{\vec{e}_{\beta}({\boldsymbol{y}}_{1}),(-\vec{e}^{{}\,\dagger\beta})({\boldsymbol{y}}_{2})}^{-1}\\\
(\delta
s^{\dagger}_{2,\beta})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}({\boldsymbol{y}}_{2})\Biggr{\rangle}\underbrace{\bigl{|}\vec{e}_{\beta}({\boldsymbol{x}}_{2})\bigr{\rangle}}_{-1}\frac{\overrightarrow{\partial}}{\partial
q^{\alpha}_{\sigma_{1}}}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])+{}\\\
{}+\left\langle(\delta
s^{\alpha}_{1})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{1}}\right)^{\sigma_{1}}({\boldsymbol{z}}_{1})\,\overbrace{\vec{e}_{\alpha}({\boldsymbol{z}}_{1}),(-\vec{e}^{{}\,\dagger\alpha})({\boldsymbol{z}}_{2})}^{-1}\,(\delta
s^{\dagger}_{1,\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{2}}\right)^{\sigma_{2}}({\boldsymbol{z}}_{2})\right\rangle\underbrace{\langle\vec{e}^{{}\,\dagger\alpha}({\boldsymbol{x}}_{1}),\vec{e}_{\alpha}({\boldsymbol{x}}_{2})\rangle}_{-1}\,\cdot\\\
\cdot(-)^{\operatorname{gh}(F)-1}\frac{\overrightarrow{\partial}}{\partial
q^{\alpha}_{\sigma_{1}}}f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\underline{\bigl{\langle}\vec{e}_{\beta}({\boldsymbol{x}}_{1})\bigr{|}}\Biggl{\langle}(\delta
s_{2}^{\beta})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}({\boldsymbol{y}}_{1})\,\overbrace{\vec{e}_{\beta}({\boldsymbol{y}}_{1}),(-\vec{e}^{{}\,\dagger\beta})({\boldsymbol{y}}_{2})}^{-1}\\\
\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}(\delta
s^{\dagger}_{2,\beta})({\boldsymbol{y}}_{2})\Biggr{\rangle}\underbrace{\bigl{|}\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}_{2})\bigr{\rangle}}_{+1}\,\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\frac{\overrightarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]).$
(30)

Recalling that

$f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}=(-)^{\operatorname{gh}(F)-1}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]),$

let us swap the derivations which fall on $f$ from the left and right; this
eliminates the sign $(-)^{\operatorname{gh}(F)-1}$. We proceed likewise for
$g$ and then transport the variations $\delta{\boldsymbol{s}}_{1}$ and
$\delta{\boldsymbol{s}}_{2}$, exchanging their places (and their rôles with
respect to $\Delta$ and $\lshad\,,\,\rshad$). The second term in formula (30)
becomes

$\smash{\Biggl{\langle}(\delta
s_{2}^{\beta})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}({\boldsymbol{y}}_{1})\,\overbrace{\vec{e}_{\beta}({\boldsymbol{y}}_{1}),(-\vec{e}^{{}\,\dagger\beta})({\boldsymbol{y}}_{2})}^{-1}\,\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}(\delta
s^{\dagger}_{2,\beta})({\boldsymbol{y}}_{2})\Biggr{\rangle}}\underbrace{\bigl{\langle}\vec{e}_{\beta}({\boldsymbol{x}}_{1}),\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}_{2})\bigr{\rangle}}_{+1}\\\
\cdot\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\alpha}_{\sigma_{1}}}\underline{\bigl{\langle}\vec{e}^{{}\,\dagger\alpha}({\boldsymbol{x}}_{1})\bigr{|}}\Biggl{\langle}\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{z}}_{1}}\right)^{\sigma_{1}}(\delta
s^{\alpha}_{1})({\boldsymbol{z}}_{1})\overbrace{\vec{e}_{\alpha}({\boldsymbol{z}}_{1}),(-\vec{e}^{{}\,\dagger\alpha})({\boldsymbol{z}}_{2})}^{-1}(\delta
s^{\dagger}_{1,\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{2}}\right)({\boldsymbol{z}}_{2})\Biggr{\rangle}\\\
\underbrace{\bigl{|}\vec{e}_{\alpha}({\boldsymbol{x}}_{2})\bigr{\rangle}}_{-1}\,\frac{\overrightarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]).$

It is now readily seen that the first term in (30) and this equivalent
expression of its second term are opposite to each other. Indeed, relabel the
summation indexes $\alpha\rightleftarrows\beta$, $\sigma\rightleftarrows\tau$
so that $\delta s^{\alpha}_{1}\rightleftarrows\delta s^{\beta}_{2}$, $\delta
s^{\dagger}_{1,\alpha}\rightleftarrows\delta s^{\dagger}_{2,\beta}$, and swap
the copies of base manifold $M^{n}$ by
${\boldsymbol{y}}\rightleftarrows{\boldsymbol{z}}$. Due to the second factors
in the products $(-1)\cdot(-1)\cdot(-1)\cdot(-1)=+1$ versus
$(-1)\cdot(+1)\cdot(-1)\cdot(-1)=-1$, the two terms in (30) cancel out after
the integration by parts and evaluation of the couplings in view of (16).

Next, the integrand of $\Delta\bigl{(}\lshad F,G\rshad\bigr{)}(s)$ contains a
restriction to the infinite jet $j^{\infty}(s)$ of the third term, which is

$\displaystyle\Biggl{\langle}(\delta
s_{1}^{\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{1}}\right)^{\sigma_{1}}({\boldsymbol{z}}_{1})\,\overbrace{\vec{e}_{\alpha}({\boldsymbol{z}}_{1}),(-\vec{e}^{{}\,\dagger\alpha})({\boldsymbol{z}}_{2})}^{-1}\,(\delta
s^{\dagger}_{1,\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{2}}\right)^{\sigma_{2}}({\boldsymbol{z}}_{2})\Biggr{\rangle}\underbrace{\bigl{\langle}\vec{e}^{{}\,\dagger\alpha}({\boldsymbol{x}}_{2}),\vec{e}_{\alpha}({\boldsymbol{x}}_{1})\bigr{\rangle}}_{-1}$
$\displaystyle\ {}\cdot\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\left(f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\right)$
$\displaystyle\quad\underline{\bigl{\langle}\vec{e}_{\beta}({\boldsymbol{x}}_{1})\bigr{|}}\Biggl{\langle}(\delta
s^{\beta}_{2})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}({\boldsymbol{y}}_{1})\,\overbrace{\vec{e}_{\beta}({\boldsymbol{y}}_{1}),(-\vec{e}^{{}\,\dagger\beta})({\boldsymbol{y}}_{2})}^{-1}\,\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}(\delta
s^{\dagger}_{2,\beta})({\boldsymbol{y}}_{2})\Biggr{\rangle}\underbrace{\bigl{|}\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}_{2})\bigr{\rangle}}_{+1}$
$\displaystyle\mbox{\hbox to304.44447pt{{ }\hfil{
}}}\frac{\overrightarrow{\partial}}{\partial
q^{\alpha}_{\sigma_{1}}}\frac{\overrightarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]).$

Let the summation indexes be relabelled as above:
$\alpha\rightleftarrows\beta$, $\sigma\rightleftarrows\tau$, and
$\smash{\delta s^{\alpha}_{1}\rightleftarrows\delta s^{\beta}_{2}}$, $\delta
s^{\dagger}_{1,\alpha}\rightleftarrows\delta s^{\dagger}_{2,\beta}$ on top of
${\boldsymbol{y}}\rightleftarrows{\boldsymbol{z}}$. The transformation of
graded derivations falling from the left and right on $f$ is then

$\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\left(f\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\right)\longmapsto\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\left(f\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\right)=\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\left((-)^{\operatorname{gh}(F)-1}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}f\right)=\\\
=(-)^{\operatorname{gh}(F)-2}\cdot(-)^{\operatorname{gh}(F)-1}\left(\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}f\right)\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}=-\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\left(f\frac{\overleftarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\right).$

This minus sign shows that the third term as it was written initially, and the
newly produced one in which the variations $\delta{\boldsymbol{s}}_{1}$ and
$\delta{\boldsymbol{s}}_{2}$ are interchanged have opposite signs. At the same
time, these integral functionals must be equal to each other due to
independence of $\Delta$ and $\lshad\,,\,\rshad$ of a choice of the test
shifts. Therefore, each of those expressions vanishes.

The fourth term is processed analogously; its integrand is

$\displaystyle(-)^{\operatorname{gh}(F)}\Biggl{\langle}(\delta
s_{1}^{\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{1}}\right)^{\sigma_{1}}({\boldsymbol{z}}_{1})\,\overbrace{\vec{e}_{\alpha}({\boldsymbol{z}}_{1}),(-\vec{e}^{{}\,\dagger\alpha})({\boldsymbol{z}}_{2})}^{-1}\,(\delta
s^{\dagger}_{1,\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{2}}\right)^{\sigma_{2}}({\boldsymbol{z}}_{2})\Biggr{\rangle}\cdot\underbrace{\bigl{\langle}\vec{e}^{{}\,\dagger\alpha}({\boldsymbol{x}}_{1}),\vec{e}_{\alpha}({\boldsymbol{x}}_{2})\bigr{\rangle}}_{-1}$
$\displaystyle\ {}\cdot\frac{\overrightarrow{\partial}}{\partial
q^{\alpha}_{\sigma_{1}}}f({\boldsymbol{x}}_{1},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\frac{\overleftarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}$
$\displaystyle\qquad\underline{\bigl{\langle}\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}_{1})\bigr{|}}\Biggl{\langle}\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}(\delta
s^{\beta}_{2})({\boldsymbol{y}}_{1})\,\overbrace{\vec{e}_{\beta}({\boldsymbol{y}}_{1}),(-\vec{e}^{{}\,\dagger\beta})({\boldsymbol{y}}_{2})}^{-1}\,(\delta
s^{\dagger}_{2,\beta})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}({\boldsymbol{y}}_{2})\Biggr{\rangle}\underbrace{\bigl{|}\vec{e}_{\beta}({\boldsymbol{x}}_{2})\bigr{\rangle}}_{-1}$
$\displaystyle\mbox{\hbox to304.44447pt{{ }\hfil{
}}}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}g({\boldsymbol{x}}_{2},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]).$

The very same procedure of two variations interchange and relabelling restores
an almost identical expression in which, however, the parity-odd derivations
go in the inverse order $\overrightarrow{\partial}/\partial
q^{\dagger}_{\beta,\tau_{2}}\circ\overrightarrow{\partial}/\partial
q^{\dagger}_{\alpha,\sigma_{2}}$. Equal to minus itself, the fourth term
vanishes. This concludes the proof. ∎

The following example illustrates the assertion of Lemma 5 (but not a
technique of its proof which itself accompanies Lemma 1). We use the
convention from Remark 2.7, denoting by
${\mathrm{d}}/{\mathrm{d}}{\boldsymbol{y}}_{i}$ or
${\mathrm{d}}/{\mathrm{d}}{\boldsymbol{z}}_{j}$ the total derivatives which
act on the functionals’ densities at points ${\boldsymbol{x}}_{k}$; this keeps
track of those derivatives origin and lets us indicate the couplings’ values
as they appear after the integrations by parts, contributing only with sign
factors $\pm 1$. For the sake of brevity we do not write the (co)vectors
$\vec{e}_{i}$ and $\vec{e}^{{}\,\dagger\,i}$ in the formulas below, referring
to the proofs in preceding sections. Likewise, we do not indicate the base
point congruences that occur due to the absolute locality of couplings.

An overall comment to Example 2.4 below is that, fully aware of the goal which
is to calculate $\Delta\left(\lshad F,G\rshad\right)$ or, respectively,
$\lshad\Delta F,G\rshad$ and $\lshad F,\Delta G\rshad$, we do not interrupt
the logic of our reasoning by attempting to view the intermediate objects
$\lshad F,G\rshad$ or $\Delta F$ and $\Delta G$ as mappings
$\Gamma(\pi_{{\text{{BV}}}})\to\Bbbk$, cf. Corollary 2 on p. 2. Such mappings
would not be elements of the structures which stand in the left- and right-
hand sides of the identity under examination. The slogan is that a step-by-
step evaluation is illegal; derivations of the end-product from input data
must not be interrupted at half-way.

We also emphasize that the example below is a prototype reasoning which is
equally well applicable to any other arguments $F$ and $G$ in (29); a choice
of the functionals is here not specific to any model. The point is that
equality (29) holds and does not require any manual regularization.

######  Example 2.4.

Consider the integral functionals

$F=\int q^{\dagger}qq_{x_{1}x_{1}}\,{\mathrm{d}}x_{1}\quad\text{and}\quad
G=\int q^{\dagger}_{x_{2}x_{2}}\cos q\,{\mathrm{d}}x_{2}.$

Let us show that equality (29) is satisfied for $F$ and $G$, that is,

$\Delta\left(\lshad F,G\rshad\right)=\lshad\Delta F,G\rshad+\lshad F,\Delta
G\rshad,\qquad\operatorname{gh}(F)=1,$ (31)

in the frames of product-bundle geometry of variations and operational
definitions of the BV-Laplacian $\Delta$ and variational Schouten bracket
$\lshad\,,\,\rshad$.

We have

$\lshad
F,G\rshad=\iiiint{\mathrm{d}}x_{1}{\mathrm{d}}x_{2}{\mathrm{d}}y_{1}{\mathrm{d}}y_{2}\Bigl{\langle}\Bigl{(}\underbrace{q^{\dagger}q_{xx}+\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{1}^{2}}(q^{\dagger}q)}_{x_{1}}\Bigr{)}\cdot\underbrace{\langle\delta
s_{2}(y_{1}),\delta
s_{2}^{\dagger}(y_{2})\rangle}_{+1}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{2}^{2}}\bigl{(}\underbrace{\cos
q}_{x_{2}}\bigr{)}\Bigr{\rangle}\\\
{}+\iiiint{\mathrm{d}}x_{1}{\mathrm{d}}x_{2}{\mathrm{d}}y_{1}{\mathrm{d}}y_{2}\Bigl{\langle}\bigl{(}\underbrace{qq_{xx}}_{x_{1}}\bigr{)}\cdot\underbrace{\langle\delta
s_{2}^{\dagger}(y_{2}),\delta
s_{2}(y_{1})\rangle}_{-1}\cdot\bigl{(}\underbrace{-q^{\dagger}_{xx}\,\sin
q}_{x_{2}}\bigr{)}\Bigr{\rangle}.$

Therefore, one side of the expected equality is

$\Delta\bigl{(}\lshad
F,G\rshad\bigr{)}=\int\\!\\!{\mathrm{d}}z_{1}\int\\!\\!{\mathrm{d}}z_{2}\int\\!\\!{\mathrm{d}}x_{1}\int\\!\\!{\mathrm{d}}x_{2}\int\\!\\!{\mathrm{d}}y_{1}\int\\!\\!{\mathrm{d}}y_{2}\underbrace{\langle\delta
s_{1}(z_{1}),\delta
s_{1}^{\dagger}(z_{2})\rangle}_{+1}\cdot\underbrace{\langle\delta
s_{2}(y_{1}),\delta s_{2}^{\dagger}(y_{2})\rangle}_{+1}\cdot{}\\\
{}\cdot\Bigl{\langle}\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{1}^{2}}\underbrace{(1)}_{x_{1}}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{2}^{2}}\bigl{(}\underbrace{\cos
q}_{x_{2}}\bigr{)}+\underline{\underbrace{q_{xx}}_{x_{1}}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{2}^{2}}\bigl{(}\underbrace{-\sin
q}_{x_{2}}\bigr{)}}+\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{1}^{2}}\underbrace{(1)}_{x_{1}}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{2}^{2}}\bigl{(}\underbrace{\cos
q}_{x_{2}}\bigr{)}+\underline{\underline{\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{1}^{2}}\underbrace{(q)}_{x_{1}}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{2}^{2}}\bigl{(}\underbrace{-\sin
q}_{x_{2}}\bigr{)}}}\Bigr{\rangle}\\\
+\int\\!\\!{\mathrm{d}}z_{1}\int\\!\\!{\mathrm{d}}z_{2}\int\\!\\!{\mathrm{d}}x_{1}\int\\!\\!{\mathrm{d}}x_{2}\int\\!\\!{\mathrm{d}}y_{1}\int\\!\\!{\mathrm{d}}y_{2}\underbrace{\langle\delta
s_{1}(z_{1}),\delta
s_{1}^{\dagger}(z_{2})\rangle}_{+1}\cdot\underbrace{\langle\delta
s_{2}^{\dagger}(y_{2}),\delta s_{2}(y_{1})\rangle}_{-1}\cdot{}\\\
\cdot\Bigl{\langle}\underline{\underbrace{q_{xx}}_{x_{1}}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{2}^{2}}\bigl{(}\underbrace{-\sin
q}_{x_{2}}\bigr{)}}+\underline{\underline{\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{1}^{2}}\underbrace{(q)}_{x_{1}}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{2}^{2}}\bigl{(}\underbrace{-\sin
q}_{x_{2}}\bigr{)}}}+\bigl{(}\underbrace{qq_{xx}}_{x_{1}}\bigr{)}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{2}^{2}}\bigl{(}\underbrace{-\cos
q}_{x_{2}}\bigr{)}\Bigr{\rangle}\ .$

The respective pairs of underlined terms cancel out and there remains only

${}=\int{\cdots}\int{\mathrm{d}}z_{1}\,{\mathrm{d}}z_{2}\,{\mathrm{d}}x_{1}\,{\mathrm{d}}x_{2}\,{\mathrm{d}}y_{1}\,{\mathrm{d}}y_{2}\underbrace{\langle\delta
s_{1}(z_{1}),\delta
s_{1}^{\dagger}(z_{2})\rangle}_{+1}\cdot\bigl{\langle}\bigl{(}\underbrace{qq_{xx}}_{x_{1}}\bigr{)}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{2}^{2}}\bigl{(}\underbrace{-\cos
q}_{x_{2}}\bigr{)}\bigr{\rangle}\cdot\\\ \underbrace{\langle\delta
s_{2}^{\dagger}(y_{2}),\delta s_{2}(y_{1})\rangle}_{-1}.$ (32)

On the other hand, we obtain that

$\Delta
F=\iiint{\mathrm{d}}z_{1}{\mathrm{d}}z_{2}{\mathrm{d}}x_{1}\underbrace{\langle\delta
s_{1}(z_{1}),\delta
s_{1}^{\dagger}(z_{2})\rangle}_{+1}\cdot\bigl{\langle}\underbrace{q_{xx}}_{x_{1}}+\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{1}^{2}}\underbrace{(q)}_{x_{1}}\bigr{\rangle},$

which yields

$\lshad\Delta
F,G\rshad=\int\\!\\!{\mathrm{d}}z_{1}\int\\!\\!{\mathrm{d}}z_{2}\int\\!\\!{\mathrm{d}}x_{1}\int\\!\\!{\mathrm{d}}x_{2}\int\\!\\!{\mathrm{d}}y_{1}\int\\!\\!{\mathrm{d}}y_{2}\,\underbrace{\langle\delta
s_{1}(z_{1}),\delta s_{1}^{\dagger}(z_{2})\rangle}_{+1}\cdot{}\\\
\Bigl{\langle}\Bigl{(}\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{1}^{2}}\underbrace{(1)}_{x_{1}}+\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{1}^{2}}\underbrace{(1)}_{x_{1}}\Bigr{)}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}y_{2}^{2}}\bigl{(}\underbrace{\cos
q}_{x_{2}}\bigr{)}\Bigr{\rangle}\cdot\underbrace{\langle\delta
s_{2}(y_{1}),\delta s_{2}^{\dagger}(y_{2})\rangle}_{+1}=0.$

From the fact that the other BV-Laplacian,

$\Delta
G=\iiint\,{\mathrm{d}}z_{1}\,{\mathrm{d}}z_{2}\,{\mathrm{d}}x_{2}\underbrace{\langle\delta
s_{1}(z_{1}),\delta
s_{1}^{\dagger}(z_{2})\rangle}_{+1}\cdot\bigl{\langle}\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{2}^{2}}\bigl{(}\underbrace{-\sin
q}_{x_{2}}\bigr{)}\bigr{\rangle},$

does not contain $q^{\dagger}$ so that the first half of the Schouten bracket
$\lshad F,\Delta G\rshad$ drops out, we deduce that

$\lshad F,\Delta
G\rshad=\int{\cdots}\int\,{\mathrm{d}}z_{1}\,{\mathrm{d}}z_{2}\,{\mathrm{d}}x_{1}\,{\mathrm{d}}x_{2}\,{\mathrm{d}}y_{1}\,{\mathrm{d}}y_{2}\underbrace{\langle\delta
s_{1}(z_{1}),\delta s_{1}^{\dagger}(z_{2})\rangle}_{+1}\cdot{}\\\
\bigl{\langle}\bigl{(}\underbrace{qq_{xx}}_{x_{1}}\bigr{)}\cdot\tfrac{{\mathrm{d}}^{2}}{{\mathrm{d}}z_{2}^{2}}\bigl{(}\underbrace{-\cos
q}_{x_{2}}\bigr{)}\bigr{\rangle}\cdot\underbrace{\langle\delta
s_{2}^{\dagger}(y_{2}),\delta s_{2}(y_{1})\rangle}_{-1}.$ (33)

Consequently, the two sides of (31), namely, $\Delta\bigl{(}\lshad
F,G\rshad\bigr{)}$ expressed by (32) and $\lshad\Delta F,G\rshad+\lshad
F,\Delta G\rshad$ accumulated in (33), match perfectly for the functionals $F$
and $G$ at hand.

######  Theorem 6.

Let $F$,
$G\in\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{{\text{{BV}}}})$ be
two functionals. The Batalin–Vilkovisky Laplacian $\Delta$ satisfies the
relation

$\Delta\bigl{(}\lshad{F,G}\rshad\bigr{)}=\lshad{\Delta
F,G}\rshad+(-)^{\operatorname{gh}(F)-1}\lshad{F,\Delta G}\rshad.$ (29)

In other words, the operator $\Delta$ is a graded derivation of the
variational Schouten bracket $\lshad{\,,\,}\rshad$.

###### Proof.

We prove this by induction over the number of building blocks in each argument
of the Schouten bracket in the left hand side of (29). If $F$ and $G$ both
belong to $\overline{H}^{*}(\pi_{\text{{BV}}}\times
T\pi_{\text{{BV}}}\times\ldots\times T\pi_{\text{{BV}}})$, then Lemma 5 states
the assertion, which is the base of induction. To make an inductive step,
without loss of generality let us assume that the second argument of
$\lshad{\,,\,}\rshad$ in (29) is a product of two elements from
$\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{{\text{{BV}}}})$, each of
them containing less multiples from $\overline{H}^{*}(\pi_{\text{{BV}}}\times
T\pi_{\text{{BV}}}\times\ldots T\pi_{\text{{BV}}})$ than the product. Denote
such factors by $G$ and $H$ and recall that by Theorem 4,

$\lshad{F,G\cdot H}\rshad=\lshad{F,G}\rshad\cdot
H+(-)^{(\operatorname{gh}(F)-1)\cdot\operatorname{gh}(G)}G\cdot\lshad{F,H}\rshad.$

Therefore, using Theorem 3 we have that

$\displaystyle\Delta(\lshad$ $\displaystyle F,G\cdot H\rshad)$
$\displaystyle={}$ $\displaystyle\Delta(\lshad{F,G}\rshad)\cdot
H+(-)^{\operatorname{gh}(F)+\operatorname{gh}(G)-1}[\\![\lshad{F,G}\rshad,H]\\!]+(-)^{\operatorname{gh}(F)+\operatorname{gh}(G)-1}\lshad{F,G}\rshad\cdot\Delta
H$
$\displaystyle+(-)^{(\operatorname{gh}(F)-1)\operatorname{gh}(G)}\left(\Delta
G\cdot\lshad{F,H}\rshad+(-)^{\operatorname{gh}(G)}[\\![G,\lshad{F,H}\rshad]\\!]+(-)^{\operatorname{gh}(G)}G\cdot\Delta(\lshad{F,H}\rshad)\right).$
Using the inductive hypothesis in the first and last terms of the right-hand
side in the above formula, we continue the equality and obtain
$\displaystyle={}$ $\displaystyle\lshad{\Delta F,G}\rshad\cdot
H+(-)^{\operatorname{gh}(F)-1}\lshad{F,\Delta G}\rshad\cdot
H+(-)^{\operatorname{gh}(F)+\operatorname{gh}(G)-1}[\\![\lshad{F,G}\rshad,H]\\!]$
$\displaystyle+(-)^{\operatorname{gh}(F)\operatorname{gh}(G)}[\\![G,\lshad{F,H}\rshad]\\!]+(-)^{\operatorname{gh}(F)+\operatorname{gh}(G)-1}\lshad{F,G}\rshad\cdot\Delta
H$ $\displaystyle+(-)^{(\operatorname{gh}(F)-1)\operatorname{gh}(G)}\Delta
G\cdot\lshad{F,H}\rshad+(-)^{\operatorname{gh}(F)\operatorname{gh}(G)}G\cdot\lshad{\Delta
F,H}\rshad$
$\displaystyle+(-)^{\operatorname{gh}(F)\operatorname{gh}(G)+\operatorname{gh}(F)-1}G\cdot\lshad{F,\Delta
H}\rshad.$ (34)

On the other hand, let us expand the formula

$\lshad{\Delta F,G\cdot
H}\rshad+(-)^{\operatorname{gh}(F)-1}\lshad{F,\Delta(G\cdot H)}\rshad,$

which is the right hand side of (29) in the inductive claim. We obtain

$\displaystyle={}$ $\displaystyle\lshad{\Delta F,G}\rshad\cdot
H+(-)^{(\operatorname{gh}(\Delta F)-1)\operatorname{gh}(G)}G\cdot\lshad{\Delta
F,H}\rshad$ $\displaystyle+(-)^{\operatorname{gh}(F)-1}[\\![F,\ \Delta G\cdot
H+(-)^{\operatorname{gh}(G)}\lshad{G,H}\rshad+(-)^{\operatorname{gh}(G)}G\cdot\Delta
H\ ]\\!]$ $\displaystyle={}$ $\displaystyle\lshad{\Delta F,G}\rshad\cdot
H+(-)^{\operatorname{gh}(F)\operatorname{gh}(G)}G\cdot\lshad{\Delta
F,H}\rshad+(-)^{\operatorname{gh}(F)-1}\lshad{F,\Delta G}\rshad\cdot H$ (35)
$\displaystyle+(-)^{\operatorname{gh}(F)-1}(-)^{(\operatorname{gh}(F)-1)(\operatorname{gh}(G)-1)}\Delta
G\cdot\lshad{F,H}\rshad+(-)^{\operatorname{gh}(F)-1}(-)^{\operatorname{gh}(G)}[\\![F,\lshad{G,H}\rshad]\\!]$
$\displaystyle+(-)^{\operatorname{gh}(F)-1}(-)^{\operatorname{gh}(G)}\lshad{F,G}\rshad\cdot\Delta
H+(-)^{\operatorname{gh}(F)-1}(-)^{\operatorname{gh}(G)}(-)^{(\operatorname{gh}(F)-1)\operatorname{gh}(G)}G\cdot\lshad{F,\Delta
H}\rshad.$

Comparing (35) with (34), which was derived from the inductive hypothesis, we
see that all terms match except for

$(-)^{\operatorname{gh}(F)+\operatorname{gh}(G)-1}[\\![\lshad{F,G}\rshad,H]\\!]+(-)^{\operatorname{gh}(F)\operatorname{gh}(G)}[\\![G,\lshad{F,H}\rshad]\\!]$

from (34) versus

$(-)^{\operatorname{gh}(F)+\operatorname{gh}(G)-1}[\\![F,\lshad{G,H}\rshad]\\!]$

from (35). However, these three terms constitute Jacobi’s identity (28) for
the variational Schouten bracket. Namely, we have that (cf. [32])

$[\\![F,\lshad{G,H}\rshad]\\!]=[\\![\lshad{F,G}\rshad,H]\\!]+(-)^{(\operatorname{gh}(F)-1)(\operatorname{gh}(G)-1)}[\\![G,\lshad{F,H}\rshad]\\!],$
(36)

so that by multiplying both sides of the identity by
$(-)^{\operatorname{gh}(F)+\operatorname{gh}(G)-1}$, we fully balance (34) and
(35). This completes the inductive step and concludes the proof. ∎

######  Lemma 7.

The linear operator

$\Delta\colon\overline{H}^{n(1+k)}\bigl{(}\pi_{\text{{BV}}}\times
T\pi_{\text{{BV}}}\times\ldots\times
T\pi_{\text{{BV}}}\bigr{)}\longrightarrow\overline{H}^{n(2+k)}\bigl{(}\pi_{\text{{BV}}}\times
T\pi_{\text{{BV}}}\times\ldots\times T\pi_{\text{{BV}}}\bigr{)}$

is a differential for every $k\geqslant 0$.

The proof of Lemma 7 is conceptually close to the second and third steps in
the proof of Lemma 5. Namely, two normalized variations are swapped in an
integral functional within the image of $\Delta^{2}$, which yields an
indistinguishable result of opposite sign.

###### Proof.

Let $\delta{\boldsymbol{s}}_{1}$ and $\delta{\boldsymbol{s}}_{2}$ be
normalized test shifts of a section $s\in\Gamma(\pi_{{\text{{BV}}}})$, and let
$H=\int
h({\boldsymbol{x}},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\cdot\operatorname{dvol}({\boldsymbol{x}})$
be an integral functional. (It suffices to consider a simplified picture
$H\in\overline{H}^{n}(\pi_{{\text{{BV}}}})$, not taking into account any
built-in variations in the construction of $H$.) By definition, we have that

$\Delta(\Delta
H)(s)=\int_{M}{\mathrm{d}}{\boldsymbol{z}}_{1}\int_{M}{\mathrm{d}}{\boldsymbol{z}}_{2}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{1}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{2}\int_{M}\operatorname{dvol}({\boldsymbol{x}})\cdot\\\
\cdot\Biggl{\\{}\left\langle(\delta
s^{\alpha}_{1})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{1}}\right)^{\sigma_{1}}({\boldsymbol{z}}_{1})\,\overbrace{\vec{e}_{\alpha}({\boldsymbol{z}}_{1}),(-\vec{e}^{{}\,\dagger\alpha})({\boldsymbol{z}}_{2})}^{-1}\,(\delta
s^{\dagger}_{1,\alpha})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{z}}_{2}}\right)^{\sigma_{2}}({\boldsymbol{z}}_{2})\right\rangle\underbrace{\left\langle\vec{e}^{{}\,\dagger\alpha}({\boldsymbol{x}}),\vec{e}_{\alpha}({\boldsymbol{x}})\right\rangle}_{-1}\\\
{}\quad\left\langle(\delta
s^{\beta}_{2})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\tau_{1}}({\boldsymbol{y}}_{1})\,\overbrace{\vec{e}_{\beta}({\boldsymbol{y}}_{1}),(-\vec{e}^{{}\,\dagger\beta})({\boldsymbol{y}}_{2})}^{-1}\,(\delta
s^{\dagger}_{2,\beta})\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\tau_{2}}({\boldsymbol{y}}_{2})\right\rangle\underbrace{\left\langle\vec{e}^{{}\,\dagger\beta}({\boldsymbol{x}}),\vec{e}_{\beta}({\boldsymbol{x}})\right\rangle}_{-1}\\\
\frac{\overrightarrow{\partial}}{\partial
q^{\alpha}_{\sigma_{1}}}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\frac{\overrightarrow{\partial}}{\partial
q^{\beta}_{\tau_{1}}}\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}h({\boldsymbol{x}},[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\left.\Biggr{\\}}\right|_{j^{\infty}_{{\boldsymbol{x}}}(s)}.$

By exchanging the integrand’s upper two lines and then relabelling
$\alpha\rightleftarrows\beta$,

$\sigma\rightleftarrows\tau$ so that $\delta
s_{1}^{\alpha}\rightleftarrows\delta s_{2}^{\beta}$ and $\delta
s^{\dagger}_{1,\alpha}\rightleftarrows\delta s^{\dagger}_{2,\beta}$, and by
swapping the reference ${\boldsymbol{y}}\rightleftarrows{\boldsymbol{z}}$ to
copies of the base manifold $M^{n}$, we almost recover the initial expression
(which should be the case), yet the order in which the parity-odd partial
derivatives follow is inverse,

$\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\circ\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\longmapsto\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}\circ\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}=-\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\alpha,\sigma_{2}}}\circ\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{\beta,\tau_{2}}}.$

Therefore the integrand of functional $\Delta^{2}H$ vanishes, which proves the
assertion. ∎

######  Theorem 8.

The Batalin–Vilkovisky Laplacian $\Delta$ is a differential: for all
$H\in\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}}$, $T\pi_{{\text{{BV}}}})$
we have

$\Delta^{2}(H)=0.$

###### Proof.

We prove Theorem 8 by induction over the number of building blocks from
$\overline{H}^{*}\bigl{(}\pi_{\text{{BV}}}\times
T\pi_{\text{{BV}}}\times\ldots\times T\pi_{\text{{BV}}}\bigr{)}$ in the
argument
$H\in\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{{\text{{BV}}}})$ of
$\Delta^{2}$. If $H\in\overline{H}^{*}\bigl{(}\pi_{\text{{BV}}}\times
T\pi_{\text{{BV}}}\times\ldots\times T\pi_{\text{{BV}}}\bigr{)}$ itself is an
integral functional, then by Lemma 7 there remains nothing to prove. Suppose
now that $H=F\cdot G$ for some
$F,G\in\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{{\text{{BV}}}})$.
Then Theorem 3 yields that

$\displaystyle\Delta^{2}$ $\displaystyle(F\cdot G)=\Delta\left(\Delta F\cdot
G+(-)^{\operatorname{gh}(F)}\lshad{F,G}\rshad+(-)^{\operatorname{gh}(F)}F\cdot\Delta
G\right).$ Using Theorem 3 again and also Theorem 6, we continue the equality:
$\displaystyle={}$ $\displaystyle\Delta^{2}F\cdot
G+(-)^{\operatorname{gh}(\Delta F)}\lshad{\Delta
F,G}\rshad+(-)^{\operatorname{gh}(\Delta F)}\Delta F\cdot\Delta G$
$\displaystyle{}+(-)^{\operatorname{gh}(F)}\lshad{\Delta
F,G}\rshad+(-)^{\operatorname{gh}(F)}(-)^{\operatorname{gh}(F)-1}\lshad{F,\Delta
G}\rshad$ $\displaystyle{}+(-)^{\operatorname{gh}(F)}\Delta F\cdot\Delta
G+(-)^{\operatorname{gh}(F)}(-)^{\operatorname{gh}(F)}\lshad{F,\Delta
G}\rshad+(-)^{\operatorname{gh}(F)}(-)^{\operatorname{gh}(F)}F\cdot\Delta^{2}G.$

By the inductive hypothesis, the first and last terms in the above formula
vanish; taking into account that $\operatorname{gh}(\Delta
F)=\operatorname{gh}(F)-1$ in $\mathbb{Z}_{2}$, the terms with $\Delta
F\cdot\Delta G$ cancel against each other, as do the terms containing
$\lshad{\Delta F,G}\rshad$ and $\lshad{F,\Delta G}\rshad$. The proof is
complete. ∎

##  3 The quantum master-equation

###  3.1 The Laplace equation

In this section we inspect the conditions upon functionals
$F\in\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$ under
which the Feynman path integrals
$\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds]\,F([s],[s^{\dagger}])$ are
(infinitesimally) independent of the unphysical anti-objects
$s^{\dagger}\in\Gamma(\boldsymbol{\zeta}^{1})$. The derivation of such a
condition (see equation (39) below) relies on an extra assumption of the
translation invariance of a measure in the path integral. It must be noted,
however, that we do not define Feynman’s integral here and do not introduce
that measure which essentially depends on the agreement about the classes of
‘admissible’ sections $\Gamma(\pi)$ or $\Gamma(\boldsymbol{\zeta}^{(0|1)})$.
Consequently, our reasoning is to some extent heuristic.

The basics of path integration, which we recall here for consistency, are
standard: they illustrate how the geometry of the BV-Laplacian works in
practice. We draw the experts’ attention only to the fact that in our notation
$\Psi$ is not the gauge fixing fermion $\boldsymbol{\Psi}$ such that the odd-
component’s section $s^{\dagger}\in\Gamma(\boldsymbol{\zeta}^{1})$ is the
restriction of $\delta\boldsymbol{\Psi}/\delta q$ to the jet of a section for
$\boldsymbol{\zeta}^{0}$ ; instead, we let $\Psi$ determine the infinitesimal
shift $\dot{q}^{\dagger}=\delta\Psi/\delta q$ of coordinates along the fibre’s
parity-odd half. We also note that the preservation of parity is not mandatory
here and thus an even-parity
$\Psi\in\overline{H}^{n}(\boldsymbol{\zeta}^{0})\hookrightarrow\overline{H}^{n}(\pi_{\text{{BV}}})$
is a legitimate choice.

Let $F=\int
f({\boldsymbol{x}},{\boldsymbol{q}},{\boldsymbol{q}}^{\dagger})\,\operatorname{dvol}({\boldsymbol{x}})\in\overline{H}^{n}(\pi_{\text{{BV}}})$
be a functional; here and in what follows we proceed over the building blocks
of elements from
$\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$ by the
graded Leibniz rule. Let
$\Psi=\int\psi({\boldsymbol{y}},{\boldsymbol{q}})\,\operatorname{dvol}({\boldsymbol{y}})\in\overline{H}^{n}(\boldsymbol{\zeta}^{0})\hookrightarrow\overline{H}^{n}(\pi_{\text{{BV}}})$
be an integral functional which, by assumption, is constant along ghost
parity-odd variables:
$\Psi(s^{\alpha},s^{\dagger}_{\beta})=\Psi(s^{\alpha},t^{\dagger}_{\beta})$
for any sections $\\{s^{\alpha}\\}\in\Gamma({\boldsymbol{\zeta}}^{0})$ and
$\\{s^{\dagger}_{\beta}\\},\\{t^{\dagger}_{\beta}\\}\in\Gamma({\boldsymbol{\zeta}}^{1})$.
We investigate under which conditions the path integral
$\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\,F(s^{\alpha},s^{\dagger}_{\beta})\colon\Gamma({\boldsymbol{\zeta}}^{1})\to\Bbbk$
is infinitesimally independent of a choice of the anti-objects:

$\displaystyle\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon^{\dagger}}\right|_{\varepsilon^{\dagger}=0}\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\,F\Bigl{(}s^{\alpha},s^{\dagger}_{\beta}+\varepsilon^{\dagger}\,\frac{\vec{\delta}\psi}{\delta
q^{\beta}}\bigg{|}_{s^{\alpha}}\Bigr{)}=0\quad\text{for all
$s^{\dagger}\in\Gamma({\boldsymbol{\zeta}}^{1})$.}$ (37)

Note that this formula makes sense because the bundles
${\boldsymbol{\zeta}}^{0}$ and ${\boldsymbol{\zeta}}^{1}$ are dual so that a
variational covector in the geometry of ${\boldsymbol{\zeta}}^{0}$ acts as a
shift vector in the geometry of ${\boldsymbol{\zeta}}^{1}$. The left-hand side
of (37) equals

$\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\int_{M}\operatorname{dvol}({\boldsymbol{x}})\,\frac{\overrightarrow{\delta}\\!\psi}{\delta
q^{\beta}}({\boldsymbol{x}},{\boldsymbol{q}})\bigr{|}_{j^{\infty}_{\boldsymbol{x}}(s^{\alpha})}\cdot\frac{\overleftarrow{\delta}\\!f}{\delta
q^{\dagger}_{\beta}}({\boldsymbol{x}},{\boldsymbol{q}},{\boldsymbol{q}}^{\dagger})\bigr{|}_{j^{\infty}_{\boldsymbol{x}}(s^{\alpha},s^{\dagger}_{\gamma})},\qquad
s^{\dagger}\in\Gamma({\boldsymbol{\zeta}}^{1}).$

Take any auxiliary section $\delta{\boldsymbol{s}}=(\delta s^{\alpha},\delta
s^{\dagger}_{\beta})\in\Gamma\bigl{(}T{\boldsymbol{\zeta}}^{(0|1)})\bigr{)}$
normalized by $\delta s^{\alpha}(x)\cdot\delta s^{\dagger}_{\alpha}(x)\equiv
1$ at every ${\boldsymbol{x}}\in M^{n}$ for each
$\alpha=1,\dots,m+m_{1}+\cdots+m_{\lambda}=N$ and blow up the scalar integrand
to a pointwise contraction of dual object taking their values in the fibres
$T_{({\boldsymbol{x}},\phi({\boldsymbol{x}}),s({\boldsymbol{x}}))}V_{\boldsymbol{x}}$
and
$T_{({\boldsymbol{x}},\phi({\boldsymbol{x}}),s^{\dagger}({\boldsymbol{x}}))}\Pi
V^{\dagger}_{{\boldsymbol{x}}}$ of $T(\pi_{\text{{BV}}})$ over $\phi(x)$: for
$s=(s^{\alpha},s^{\dagger}_{\beta})$ we have

$\int_{M}\operatorname{dvol}({\boldsymbol{x}})\,\left.\left(\frac{\overrightarrow{\delta}\\!\psi}{\delta
q^{\alpha}}\cdot\frac{\overleftarrow{\delta}\\!f}{\delta
q^{\dagger}_{\alpha}}\right)\right|_{j^{\infty}_{\boldsymbol{x}}(s)}\\\
{}=\int_{M}\operatorname{dvol}({\boldsymbol{x}}_{1})\int_{M}\operatorname{dvol}({\boldsymbol{x}}_{2})\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{1}\int_{M}{\mathrm{d}}{\boldsymbol{y}}_{2}\left.\left(\psi({\boldsymbol{x}}_{1},{\boldsymbol{q}})\frac{\overleftarrow{\partial}}{\partial
q^{j_{1}}_{\sigma_{1}}}\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{1}}(s)}\mbox{\hbox
to56.9055pt{{ }\hfil{ }}}\\\ {}\cdot\underline{\langle\vec{e}^{\,\dagger
j_{1}}({\boldsymbol{x}}_{1})|}\,\bigl{\langle}\left(\tfrac{\overleftarrow{\partial}}{\partial{\boldsymbol{y}}_{1}}\right)^{\sigma_{1}}\delta
s^{i_{1}}({\boldsymbol{y}}_{1})\,\vec{e}_{i_{1}}({\boldsymbol{y}}_{1}),-\vec{e}^{\,\dagger
i_{2}}({\boldsymbol{y}}_{2})\,\delta
s^{\dagger}_{i_{2}}({\boldsymbol{y}}_{2})\left(\tfrac{\overrightarrow{\partial}}{\partial{\boldsymbol{y}}_{2}}\right)^{\sigma_{2}}\bigr{\rangle}\,\underline{|\vec{e}_{j_{2}}({\boldsymbol{x}}_{2})\rangle}\\\
{}\cdot\left.\left(\frac{\overrightarrow{\partial}}{\partial
q^{\dagger}_{j_{2},\sigma_{2}}}f({\boldsymbol{x}}_{2},{\boldsymbol{q}},{\boldsymbol{q}}^{\dagger})\right)\right|_{j^{\infty}_{{\boldsymbol{x}}_{2}}(s)}.$

In fact, the integrand refers to a definition of the evolutionary vector field
${\boldsymbol{Q}}^{\Psi}$ such that
${\boldsymbol{Q}}^{\Psi}(F)\cong\lshad{\Psi,F}\rshad$ modulo integration by
parts in the building blocks of $F$, cf. [32]. Due to a special choice of the
dependence of $\Psi$ on $s$ only, this is indeed the Schouten bracket
$\lshad{\Psi,F}\rshad$.

To rephrase the indifference of the path integral to a choice of $\Psi$ in
terms of an equation upon the functional $F$ alone, we perform integration by
parts in Feynman’s integral. For this we employ the translation invariance
$[Ds]=[D(s-\mu\cdot\delta s)]$ of the functional measure.

######  Lemma 9.

Let $H=\int
h({\boldsymbol{x}},{\boldsymbol{q}},{\boldsymbol{q}}^{\dagger})\,\operatorname{dvol}({\boldsymbol{x}})\in\overline{H}^{n}(\pi_{\text{{BV}}})\subset\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$
be an integral functional and $\delta
s\in\Gamma(T{\boldsymbol{\zeta}}^{0})\hookrightarrow\Gamma\bigl{(}T{\boldsymbol{\zeta}}^{(0|1)}\bigr{)}$
be a shift. Then we have that

$\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\int_{M}\operatorname{dvol}({\boldsymbol{x}})\,\delta
s^{\nu}({\boldsymbol{x}})\cdot\frac{\overleftarrow{\delta}\\!h}{\delta
q^{\nu}}\bigg{|}_{j^{\infty}_{\boldsymbol{x}}(s^{\alpha},s^{\dagger}_{\beta})}=0,$

where the section $s^{\dagger}\in\Gamma({\boldsymbol{\zeta}}^{1})$ is a
parameter.

###### Proof.

Indeed,

$\displaystyle 0$
$\displaystyle=\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\mu}\right|_{\mu=0}\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\,H(s^{\alpha},s^{\dagger}_{\beta}),$
because the integral contains no parameter $\mu\in\Bbbk$. We continue the
equality:
$\displaystyle=\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\mu}\right|_{\mu=0}\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[D(s^{\alpha}-\mu\,\delta
s^{\alpha})]\,H(s^{\alpha},s^{\dagger}_{\beta})$
$\displaystyle=\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\mu}\right|_{\mu=0}\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\,H(s^{\alpha}+\mu\,\delta
s^{\alpha},s^{\dagger}_{\beta})=\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\mu}\right|_{\mu=0}H(s^{\alpha}+\mu\,\delta
s^{\alpha},s^{\dagger}_{\beta}),$

which yields the helpful formula in the lemma’s assertion. ∎

Returning to the functionals $\Psi$ and $F$ and denoting
$G(s):=\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\ell}\right|_{\ell=0}F(s+\ell\cdot\overleftarrow{\delta
s}^{\dagger})$, we use the Leibniz rule for the derivative of $H=\Psi\cdot G$:

$\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\mu}\right|_{\mu=0}(\Psi\cdot
G)(s+\mu\cdot\overleftarrow{\delta
s})=\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\mu}\right|_{\mu=0}(\Psi)(s+\mu\cdot\overleftarrow{\delta
s})\cdot
G(s)+\Psi(s)\cdot\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\mu}\right|_{\mu=0}(G)(s+\mu\cdot\overleftarrow{\delta
s}).$

Because the path integral over $[Ds^{\alpha}]$ of the entire expression
vanishes by Lemma 9 in which we were ready to proceed by the Leibniz rule over
building blocks, we infer that the path integrals of the two terms are
opposite. Now take the traces over indexes in both variations. The integral of
the first term equals the initial expression for the path integral containing
$F$, i. e., the left-hand side of equation (37). Consequently, if

$\displaystyle\int_{\Gamma({\boldsymbol{\zeta}}^{0})}[Ds^{\alpha}]\,\Psi(s^{\alpha})\cdot\Delta
F(s^{\alpha},s^{\dagger}_{\beta})=0$ (38)

for $\\{s^{\dagger}_{\beta}\\}\in\Gamma({\boldsymbol{\zeta}}^{1})$ and for all
$\Psi\in\overline{H}^{n}({\boldsymbol{\zeta}}^{0})\hookrightarrow\overline{H}^{n}(\pi_{\text{{BV}}})$,
then the path integral over $F$ is infinitesimally independent of a section
$\\{s^{\dagger}_{\beta}\\}\in\Gamma({\boldsymbol{\zeta}}^{1})$.

The condition

$\Delta F=0$ (39)

is sufficient for equation (38), and therefore equation (37), to hold. By
specifying a class $\Gamma(\pi_{\text{{BV}}})$ of admissible sections of the
BV-bundle for a concrete field model, and endowing that space of sections with
a suitable metric, one could reinstate a path integral analogue of the main
lemma in the calculus of variations and then argue that the condition $\Delta
F=0$ is also necessary.

Summarizing, whenever equation (39) holds, one can assign arbitrary admissible
values to the odd-parity coordinates; for example, one can let
$s^{\dagger}_{\beta}({\boldsymbol{x}})=\delta\boldsymbol{\psi}/\delta
q^{\beta}\bigr{|}_{j^{\infty}_{\boldsymbol{x}}(s^{\alpha})}$ for a gauge-
fixing integral
$\boldsymbol{\Psi}=\int\boldsymbol{\psi}({\boldsymbol{x}},{\boldsymbol{q}})\,\operatorname{dvol}({\boldsymbol{x}})\in\overline{H}^{n}({\boldsymbol{\zeta}}^{0})$.
This choice is reminiscent of the substitution principle, see [32] and [45].

Laplace’s equation (39) ensures the infinitesimal independence from non-
physical anti-objects for path integrals of functionals over physical fields –
not only in the classical BV-geometry of the bundle $\pi_{\text{{BV}}}$, but
also in the quantum setup, whenever all objects are tensored with formal power
series $\Bbbk[[\hbar,\hbar^{-1}]]$ in the Planck constant $\hbar$. It is
accepted that each quantum field $s^{\hbar}$ contributes to the expectation
value of a functional $\mathcal{O}^{\hbar}$ with the factor
$\exp({{\boldsymbol{i}}}S^{\hbar}(s^{\hbar})/{\hbar})$, where $S^{\hbar}$ is
the quantum BV-action of the model. Solutions $\mathcal{O}^{\hbar}$ of the
equation
$\Delta\bigl{(}\mathcal{O}^{\hbar}\cdot\exp({{\boldsymbol{i}}}S^{\hbar}/{\hbar})\bigr{)}=0$
are the observables. In particular, the postulate that the unit $1\colon
s^{\hbar}\mapsto 1\in\Bbbk$ is averaged to unit by the Feynman integral of
$1\cdot\exp({{\boldsymbol{i}}}S^{\hbar}(s^{\hbar})/{\hbar})$ over the space of
quantum fields $s^{\hbar}$ normalizes the integration measure and constrains
the quantum BV-action by the quantum master-equation (see, e.g., [7, 8, 20,
22, 54]).

######  Proposition 10.

Let $S^{\hbar}$ be the even quantum BV-action (i. e., let it have a density
that has an even number of ghost parity-odd coordinates in each of its terms).
If the identity

$\Delta\left(\exp\bigl{(}\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\bigr{)}\right)=0$

holds, then $S^{\hbar}$ satisfies the quantum master-equation:

${\boldsymbol{i}}\hbar\,\Delta
S^{\hbar}=\tfrac{1}{2}\lshad{S^{\hbar},S^{\hbar}}\rshad.$ (40)

######  Proposition 11.

If an even functional $\mathcal{O}$ and the quantum BV-action $S^{\hbar}$ are
such that
$\Delta\bigl{(}\mathcal{O}\exp({\boldsymbol{i}}S^{\hbar}/\hbar)\bigr{)}=0$ and
$\Delta\bigl{(}\exp({\boldsymbol{i}}S^{\hbar}/\hbar)\bigr{)}=0$ hold,
respectively, then $\mathcal{O}$ satisfies

$\Omega^{\hbar}(\mathcal{O})\mathrel{{:}{=}}-{\boldsymbol{i}}\hbar\,\Delta\mathcal{O}+\lshad{S^{\hbar},\mathcal{O}}\rshad=0.$
(41)

We quote the standard proofs of Propositions 10 and 11 from [35] in A — yet
now we gain a deeper insight on a construction of the quantum BV-differential
$\Omega^{\hbar}$.

######  Remark 3.1.

A practical way to fix the signs which arise in the BV-Laplacian and Schouten
bracket from the ghost parity and a grading in the case of a superbundle
$\pi\colon E^{(m_{0}+n_{0}|m_{1}+n_{1})}\to M^{(n_{0}|n_{1})}$ of superfields
is by a re-derivation of the Laplace equation
$\Delta(\mathcal{O}\exp(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}))=0$ upon an
observable $\mathcal{O}$ starting from the Schwinger–Dyson condition,

$\vec{\partial}^{\,({\boldsymbol{q}}^{\dagger})}_{\vec{\delta}\Psi/\delta{\boldsymbol{q}}}\left(\int[D{\boldsymbol{q}}]\,\mathcal{O}([{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\bigl{(}[{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}]\bigr{)}\right)\right)=0,$
(42)

which postulates the Feynman path integral’s independence of the non-physical
BV-coordinates ${\boldsymbol{q}}^{\dagger}$ with odd ghost parity. Note that
the measure in the path integral involves only ghost parity-even objects
(whatever be their $\mathbb{Z}_{2}$-grading).

######  Theorem 12.

Let
$\mathcal{O}\in\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$
be a functional and let the even functional
$S^{\hbar}\in\overline{H}^{n}(\pi_{\text{{BV}}})$ satisfy quantum master-
equation (40). Then the operator $\Omega^{\hbar}$, defined in (41), squares to
zero:

${(\Omega^{\hbar})}^{2}(\mathcal{O})=0.$

###### Proof.

We calculate, using Theorem 6,

$\displaystyle{(\Omega^{\hbar})}^{2}(\mathcal{O})$
$\displaystyle=\lshad{S^{\hbar},\lshad{S^{\hbar},\mathcal{O}}\rshad-{\boldsymbol{i}}\hbar\,\Delta\mathcal{O}}\rshad-{\boldsymbol{i}}\hbar\,\Delta\big{(}\lshad{S^{\hbar},\mathcal{O}}\rshad-{\boldsymbol{i}}\hbar\,\Delta\mathcal{O}\big{)}$
$\displaystyle=\lshad{S^{\hbar},\lshad{S^{\hbar},\mathcal{O}}\rshad}\rshad-{\boldsymbol{i}}\hbar\,\lshad{S^{\hbar},\Delta\mathcal{O}}\rshad-{\boldsymbol{i}}\hbar\,\lshad{\Delta
S^{\hbar},\mathcal{O}}\rshad+{\boldsymbol{i}}\hbar\,\lshad{S^{\hbar},\Delta\mathcal{O}}\rshad+({\boldsymbol{i}}\hbar)^{2}\Delta^{2}\mathcal{O}.$
The last term vanishes identically by Theorem 8, while the second term cancels
against the fourth term. Using Jacobi’s identity (28) for the Schouten bracket
on the first term, we obtain:
$\displaystyle{(\Omega^{\hbar})}^{2}(\mathcal{O})$
$\displaystyle=-{\boldsymbol{i}}\hbar\,\lshad{\Delta
S^{\hbar},\mathcal{O}}\rshad+{\textstyle\frac{1}{2}}\lshad{\lshad{S^{\hbar},S^{\hbar}}\rshad,\mathcal{O}}\rshad=\lshad{-{\boldsymbol{i}}\hbar\,\Delta
S^{\hbar}+{\textstyle\frac{1}{2}}\lshad{S^{\hbar},S^{\hbar}}\rshad,\mathcal{O}}\rshad.$
Now is the crucial moment in the entire proof. By the logic of our reasoning’s
objective, the theorem’s claim is that the operator $(\Omega^{\hbar})^{2}$
yields zero whenever acting on a functional $\mathcal{O}$. We accordingly
transform the variational Schouten bracket of two terms to the operator
realization,
$\displaystyle\cong\vec{{\boldsymbol{Q}}}^{-{\boldsymbol{i}}\hbar\,\Delta
S^{\hbar}+\frac{1}{2}\lshad S^{\hbar},S^{\hbar}\rshad}(\mathcal{O}),$

with the evolutionary derivation now acting on the argument. Let us emphasize
that a transition from the variational Schouten bracket – which increases the
number of bases $M\times\ldots\times M$ by construction – to the evolutionary
vector field chops a multiplication of geometries by uniquely fixing the
field’s generating section.252525It might happen otherwise that a co-multiple
of $\mathcal{O}$ under $\lshad\,,\,\rshad$ looks like zero as a map of the
space $\Gamma(\pi_{\text{{BV}}})$ yet the bracket with it could still be
nonzero, see, e. g., $\Delta G$ on p. 33 in Example 2.4. But by our initial
assumption, this generating section is zero by virtue of (40). Therefore the
image of $\mathcal{O}$ under such map vanishes, which proves the assertion. ∎

###  3.2 Gauge automorphisms of quantum BV-cohomology groups

By using the quantum BV-differential $\Omega^{\hbar}$, let us construct a
closed algebra of infinitesimal gauge symmetries for the quantum master-
equation (40).

######  Proposition 13.

Let $F\in\overline{\mathfrak{N}^{n}}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$ be
an arbitrary odd-parity functional and $S^{\hbar}$ the quantum master-action
satisfying (40). Then the infinitesimal shift of the functional $S^{\hbar}$,

$\dot{S}^{\hbar}=\Omega^{\hbar}(F)\quad\Longleftrightarrow\quad
S^{\hbar}\mapsto
S^{\hbar}(\varepsilon)=S^{\hbar}+\varepsilon\cdot\Omega^{\hbar}(F)+\overline{o}(\varepsilon),\
\varepsilon\in\mathbb{R},$ (43)

is a symmetry of (40) so that
$\Delta\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}(\varepsilon)\right)\right)=\overline{o}(\varepsilon)$
in Peano’s notation.

$\bullet$ The algebra of infinitesimal gauge symmetries (43) of the quantum
master-equation is closed,

$\left.\left(\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon_{1}}\circ\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon_{2}}-\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon_{2}}\circ\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon_{1}}\right)\right|_{\varepsilon_{i}=0}(S^{\hbar})=\Omega^{\hbar}\bigl{(}\lshad
F_{1},F_{2}\rshad\bigr{)},$ (44)

i.e., the commutator of two even-parity symmetries with respective generators
$F_{1}$ and $F_{2}$ is the infinitesimal gauge symmetry whose generator is the
odd Poisson bracket of $F_{1}$ and $F_{2}$.

######  Remark 3.2.

The odd-parity generators
$F_{i}\in\overline{\mathfrak{N}^{n}}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$
never evolve in the course of a transformation which is induced by any
generator $F_{j}$ on the quantum BV-action functional $S^{\hbar}$.

###### Proof.

Assuming a smooth dependence of $S^{\hbar}(\varepsilon)$ on $\varepsilon$, we
obtain that262626This proof is standard: it originates from the cohomological
deformation theory for solutions of the Maurer–Cartan equation (e. g., of
(40)), see [37].

$\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon}\Delta\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}(\varepsilon)\right)\right)=\tfrac{{\boldsymbol{i}}}{\hbar}\dot{S}^{\hbar}\cdot\Delta\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}(\varepsilon)\right)\right)+\Omega^{\hbar}(\dot{S}^{\hbar})\cdot\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}(\varepsilon)\right).$

Because $(\Omega^{\hbar})^{2}=0$ by Theorem 12, for $\dot{S}^{\hbar}$ to be an
infinitesimal symmetry of the equation
$\Delta\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\right)=0$
it is sufficient that $S^{\hbar}=\Omega^{\hbar}(F)$ for some odd-parity
functional $F$.

Second, let

$\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon_{i}}(S^{\hbar})=-{\boldsymbol{i}}\hbar\,\Delta
F_{i}+\lshad S^{\hbar},F_{i}\rshad\qquad\text{for
}i=1,2,\qquad\varepsilon_{i}\in\mathbb{R},$

and postulate that
$\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon_{i}}(F_{j})\equiv 0$ for all $i$
and $j$. Then commutator (44) of even-parity infinitesimal transformations
(43) generated by the functionals $F_{1}$ and $F_{2}$ is

$\lshad-{\boldsymbol{i}}\hbar\,\Delta F_{1}+\lshad
S^{\hbar},F_{1}\rshad,F_{2}\rshad-\lshad-{\boldsymbol{i}}\hbar\,\Delta
F_{2}+\lshad S^{\hbar},F_{2}\rshad,F_{1}\rshad\\\
{}=-{\boldsymbol{i}}\hbar\,\left(\lshad\Delta F_{1},F_{2}\rshad-\lshad\Delta
F_{2},F_{1}\rshad\right)+\left(\lshad\lshad
S^{\hbar},F_{1}\rshad,F_{2}\rshad-\lshad\lshad
S^{\hbar},F_{2}\rshad,F_{1}\rshad\right).$

Because $F_{1}$ has odd parity, we swap the factors in $-\lshad\Delta
F_{2},F_{1}\rshad=\lshad F_{1},\Delta F_{2}\rshad$; likewise, $+\lshad
F_{1},\lshad S^{\hbar},F_{2}\rshad\rshad$ is the last term in the above
expression. From our main Theorem 6 and by Jacobi identity (28) we conclude
that the commutator is equal to

$-{\boldsymbol{i}}\hbar\,\Delta\bigl{(}\lshad F_{1},F_{2}\rshad\bigr{)}+\lshad
S^{\hbar},\lshad F_{1},F_{2}\rshad\rshad=\Omega^{\hbar}\bigl{(}\lshad
F_{1},F_{2}\rshad\bigr{)},$

that is, the Schouten bracket of $F_{1}$ and $F_{2}$ is the new gauge symmetry
generator. ∎

######  Remark 3.3.

(cf. [51, §5]). The transformation
$\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\mapsto\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}(\varepsilon)\right)$
for a finite $\varepsilon\in\mathbb{R}$ is determined by the operator
$\exp(\varepsilon[\Delta,F])$, where $[\ ,\ ]$ is the anticommutator of two
odd-parity objects. Indeed, by Theorem 3 we have that

$\displaystyle\Delta\bigl{(}$ $\displaystyle
F\cdot\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\bigr{)}+F\cdot\Delta\bigl{(}\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\bigr{)}$
$\displaystyle=\Delta
F\cdot\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)-\lshad
F,\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\rshad-F\cdot\Delta\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\right)+F\cdot\Delta\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\right)$
$\displaystyle=\tfrac{{\boldsymbol{i}}}{\hbar}(-{\boldsymbol{i}}\hbar\,\Delta
F+\lshad
S^{\hbar},F\rshad)\cdot\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)=\tfrac{{\boldsymbol{i}}}{\hbar}\dot{S}^{\hbar}\cdot\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)=\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon}\right|_{\varepsilon=0}\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)\right).$

Note that the Schouten bracket acts on
$\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right)$ by the Leibniz
rule (see Theorem 4) and we then use the equality $-\lshad
F,\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\rshad=\tfrac{{\boldsymbol{i}}}{\hbar}\lshad
S^{\hbar},F\rshad$ which holds by Theorem 4 again.

Let us now regard the full quantum BV-action as the generating functional for
ghost parity-even observables $\mathcal{O}$, see [54].

######  Lemma 14.

There are no observables $\mathcal{O}$, other than the identically zero
functional, which would be ghost parity-odd.

###### Proof.

Indeed, Eq. (42) implies that the path integral

$I=\int_{\Gamma(\zeta^{0})}[D{\boldsymbol{q}}]\,\mathcal{O}([{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}([{\boldsymbol{q}}],[{\boldsymbol{q}}^{\dagger}])\right)$

over the space of ghost parity-even BV-section components is effectively
independent of the ghost parity-odd BV-variables ${\boldsymbol{q}}^{\dagger}$.
Notice further that the ghost parity $\operatorname{gh}(I)$ of this constant
function $I([{\boldsymbol{q}}^{\dagger}])$ is equal to that of $\mathcal{O}$;
the quantum master-action $S^{\hbar}$ is parity-even. Under a (speculative)
assumption that an observable $\mathcal{O}$ could be ghost parity-odd, we
obtain an odd parity constant. Unless a possibility of their existence is
postulated by brute force, this odd-parity constant must be equal to zero,
whence the ghost parity-odd functional
$\mathcal{O}\in\overline{H^{n}}(\pi_{\text{{BV}}})\subseteq\overline{\mathfrak{N}^{n}}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$
itself is zero. ∎

In what follows we accept for transparency that there is no grading in the
initial geometry of physical fields, i.e., for sections of the bundle
$\pi\colon E^{n+m}\to M^{n}$. Let us focus on the standard cohomological
approach to quantum BV-models and to their gauge symmetries (cf. [37]).

######  Lemma 15.

Suppose that an infinitesimal shift $S^{\hbar}\mapsto
S^{\hbar}+\lambda\cdot\mathcal{O}+\overline{o}(\lambda)$ of the quantum BV-
action by using an even-parity functional
$\mathcal{O}\in\overline{H^{n}}(\pi_{\text{{BV}}})\subseteq\overline{\mathfrak{N}^{n}}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$
does not destroy the quantum master-equation,

$\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\lambda}\right|_{\lambda=0}\Delta\left(\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}(S^{\hbar}+\lambda\cdot\mathcal{O})\right)\right)=0.$

Then the observable $\mathcal{O}$ is $\Omega^{\hbar}$-closed:
$-{\boldsymbol{i}}\hbar\,\Delta\mathcal{O}+\lshad
S^{\hbar},\mathcal{O}\rshad=0.$

###### Proof.

The proof literally repeats that of Proposition 13. ∎

For a given odd-parity functional $F\in\overline{H^{n}}(\pi_{\text{{BV}}})$,
we organize the infinitesimal shift (43) of the master-functional $S^{\hbar}$
as follows:

$\displaystyle\dot{S}^{\hbar}$
$\displaystyle=-{\boldsymbol{i}}\hbar\,\Delta(F)+\lshad S^{\hbar},F\rshad,$
$\displaystyle\dot{\mathcal{O}}$ $\displaystyle=\lshad\mathcal{O},F\rshad.$

Note that, unless one has that $\Delta F=0$ incidentally, the transformation
of the integral _functional_ $S^{\hbar}$ is not induced by any infinitesimal
transformation of the BV-_variables_ , that is, by an evolutionary vector
field on the horizontal infinite jet space at hand. No earlier than the
transformation law $S^{\hbar}\mapsto S^{\hbar}(\varepsilon)$ is postulated, it
becomes an act of will to think that the functional $F$ is the generator of
parity-preserving evolutionary vector field
$\overleftarrow{Q}^{F}=\overrightarrow{Q}^{F}$ acting on the BV-variables so
that $\dot{\mathcal{O}}\cong\overrightarrow{Q}^{F}(\mathcal{O})$ for all
observables $\mathcal{O}$.

Furthermore, let us extend the deformation
$\mathcal{O}\mapsto\mathcal{O}(\varepsilon)$ of even-parity cocycles
$\mathcal{O}\in\ker\Omega^{\hbar}$ to the space of odd-parity functionals
$\xi\in\overline{H^{n}}(\pi_{\text{{BV}}})\subseteq\overline{\mathfrak{N}^{n}}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$
which produce the coboundaries $\Omega^{\hbar}(\xi)$. Namely, we postulate
that

$\dot{\xi}=\lshad\xi,F\rshad$

for all such functionals $\xi$; here we denote by the dot over $\xi$ its
velocity in the course of the transformation generated by a given $F$. Let us
remember however that the law for evolution of the odd-parity functionals
$\xi$ which produce the $\Omega^{\hbar}$-coboundaries is different from our
earlier postulate (see Proposition 13) that the odd-parity generators $F_{i}$
of gauge symmetries do not evolve: $dF_{i}/d\varepsilon_{j}\equiv 0$ or, in
shorthand notation,

$\dot{F}\equiv 0.$ (45)

We claim that under these hypotheses, the structure of quantum BV-cohomology
group remains intact in the course of gauge symmetry transformations of the
quantum master-action, $S^{\hbar}\mapsto S^{\hbar}(\varepsilon)$, even though
the quantum BV-differential is modified,
$\Omega^{\hbar}\mapsto\Omega^{\hbar}(\varepsilon)$, and the cocycles and
coboundaries are also deformed.

######  Theorem 16.

An infinitesimal shift of the quantum BV-cohomology classes induced by (43),
(45), and

$\displaystyle\dot{\mathcal{O}}$ $\displaystyle=\lshad\mathcal{O},F\rshad,$
$\displaystyle\mathcal{O}$ $\displaystyle\in\ker\Omega^{\hbar},$
$\displaystyle\dot{\xi}$ $\displaystyle=\lshad\xi,F\rshad,$ $\displaystyle\xi$
$\displaystyle\in\overline{H^{n}}(\pi_{\text{{BV}}})\subseteq\overline{\mathfrak{N}^{n}}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}}),\
\xi\text{ odd},$

yields an isomorphism of the $\Omega^{\hbar}$-cohomology group: under such
mapping, every $\Omega^{\hbar}$-closed, even-parity $\Omega^{\hbar}$-cocycle
$\mathcal{O}$ becomes $\Omega^{\hbar}(\varepsilon)$-closed, whereas the
transformation of an even-parity coboundary $\Omega^{\hbar}(\xi)$ produces an
$\Omega^{\hbar}(\varepsilon)$-coboundary:
$(\Omega^{\hbar}(\xi))(\varepsilon)=\Omega^{\hbar}(\varepsilon)\bigl{(}\xi(\varepsilon)\bigr{)}$.

###### Proof.

Let $\mathcal{O}\in\ker\Omega^{\hbar}$ be an even-parity observable and $F$ an
odd-parity generator of gauge transformation. Consider the equation
$\Omega^{\hbar}(\varepsilon)(\mathcal{O}(\varepsilon))=0$ which states that
the transformed functional $\mathcal{O}(\varepsilon)$ remains a coboundary.
The term which is proportional to $\varepsilon$ in this equation’s left-hand
side is equal to

$\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon}\right|_{\varepsilon=0}\left(-{\boldsymbol{i}}\hbar\,\Delta\mathcal{O}(\varepsilon)+\lshad
S^{\hbar}(\varepsilon),\mathcal{O}(\varepsilon)\rshad\right)=\Omega^{\hbar}(\dot{\mathcal{O}})+\lshad\dot{S}^{\hbar},\mathcal{O}\rshad=\Omega^{\hbar}(\lshad\mathcal{O},F\rshad)+\lshad\Omega^{\hbar}(F),\mathcal{O}\rshad;$

recalling once again that
$\Omega^{\hbar}=-{\boldsymbol{i}}\hbar\,\Delta+\lshad
S^{\hbar},\,\cdot\,\rshad$, we continue the equality

$=-{\boldsymbol{i}}\hbar\,\Delta(\lshad\mathcal{O},F\rshad)+\lshad
S^{\hbar},\lshad\mathcal{O},F\rshad\rshad+\lshad-{\boldsymbol{i}}\hbar\,\Delta
F+\lshad S^{\hbar},F\rshad,\mathcal{O}\rshad.\mbox{\hbox to119.50157pt{{
}\hfil{ }}}$

Now by Theorem 6 we obtain that, the observable $\mathcal{O}$ being parity-
even,

$=-{\boldsymbol{i}}\hbar\,\lshad\Delta\mathcal{O},F\rshad+{\boldsymbol{i}}\hbar\,\lshad\mathcal{O},\Delta
F\rshad+\lshad
S^{\hbar},\lshad\mathcal{O},F\rshad\rshad-{\boldsymbol{i}}\hbar\,\lshad\Delta
F,\mathcal{O}\rshad+\lshad\mathcal{O},\lshad S^{\hbar},F\rshad\rshad=\\\
=\lshad\Omega^{\hbar}(\mathcal{O}),F\rshad\cong-\vec{{\boldsymbol{Q}}}^{F}\bigl{(}\Omega^{\hbar}(\mathcal{O})\bigr{)}=0,$

because $\lshad S^{\hbar},\lshad\mathcal{O},F\rshad\rshad=\lshad\lshad
S^{\hbar},\mathcal{O}\rshad,F\rshad-\lshad\mathcal{O},\lshad
S^{\hbar},F\rshad\rshad$ by Jacobi identity (28), because we are inspecting
the $\varepsilon$-linear term in the operator
$\Omega^{\hbar}(\varepsilon)\circ\bigl{(}\varepsilon=0\longmapsto\varepsilon\neq
0\bigr{)}$ applied to $\mathcal{O}$, and $\mathcal{O}$ is an
$\Omega^{\hbar}$-cocycle. Therefore, the zero initial condition
$\Omega^{\hbar}(\mathcal{O})=0$ evolves at zero velocity to the
$\Omega^{\hbar}(\varepsilon)$-cocycle equation
$\Omega^{\hbar}(\varepsilon)\bigl{(}\mathcal{O}(\varepsilon)\bigr{)}=0$ upon
$\mathcal{O}(\varepsilon)$.

Likewise, let $\Omega^{\hbar}(\xi)$ be a coboundary for some odd-parity
functional $\xi$ which evolves by $\dot{\xi}=\lshad\xi,F\rshad$. Then the
even-parity observable $\Omega^{\hbar}(\xi)\in\ker\Omega^{\hbar}$ evolves as
fast as $\lshad\Omega^{\hbar}(\xi),F\rshad$ but simultaneously we have that
the mapping $\Omega^{\hbar}$ and its argument $\xi$ change. We claim that the
two evolutions match so that $(\Omega^{\hbar}(\xi))(\varepsilon)$ is
$\Omega^{\hbar}(\varepsilon)$-exact. Indeed, we have that

$\left.\frac{{\mathrm{d}}}{{\mathrm{d}}\varepsilon}\right|_{\varepsilon=0}\bigl{(}\Omega^{\hbar}(\varepsilon)(\xi(\varepsilon))\bigr{)}=\Omega^{\hbar}\bigl{(}\lshad\xi,F\rshad\bigr{)}+\lshad\Omega^{\hbar}(F),\xi\rshad\\\
{}=-{\boldsymbol{i}}\hbar\,\lshad\Delta\xi,F\rshad\underline{{}-{\boldsymbol{i}}\hbar\,\lshad\xi,\Delta
F\rshad}+\lshad
S^{\hbar},\lshad\xi,F\rshad\rshad+\lshad\underline{-{\boldsymbol{i}}\hbar\,\Delta
F}+\lshad S^{\hbar},F\rshad,\underline{\xi}\rshad;$

by cancelling out the underlined Schouten brackets and then using the Jacobi
identity we obtain

$=\lshad-{\boldsymbol{i}}\hbar\,\Delta\xi,F\rshad+\lshad\lshad
S^{\hbar},\xi\rshad,F\rshad+\lshad\xi,\lshad
S^{\hbar},F\rshad\rshad-\lshad\xi,\lshad
S^{\hbar},F\rshad\rshad=\lshad\Omega^{\hbar}(\xi),F\rshad,$

which proves our claim.

Summarizing, we see that gauge symmetries of the quantum master-equation
induce automorphisms of the $\Omega^{\hbar}$-cohomology group. ∎

We conclude that it would be conceptually incorrect to say that the
infinitesimal gauge transformations of all functionals in a quantum BV-model
are induced by a canonical transformation, determined by the evolutionary
vector field $\overrightarrow{Q}^{F}$ acting on the BV-variables. Let us
remember that the even-parity quantum master-action
$S^{\hbar}\in\overline{H^{n}}(\pi_{\text{{BV}}})$ and its descendants, the
observables $\mathcal{O}$ evolve by

$\displaystyle\dot{S}^{\hbar}$ $\displaystyle=-{\boldsymbol{i}}\hbar\,\Delta
F+\lshad S^{\hbar},F\rshad=\Omega^{\hbar}(F),\qquad
F\in\overline{H^{n}}(\pi_{\text{{BV}}})\subseteq\overline{\mathfrak{N}^{n}}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}}),\quad
F\text{ odd},$ and $\displaystyle\dot{\mathcal{O}}$
$\displaystyle=\lshad\mathcal{O},F\rshad.$ We note that the evolution of the
generating functional $S^{\hbar}_{{\text{{BV}}}}$ is not determined by a
vector field on the space of BV-variables. Likewise, we recall that the odd-
parity arguments $\xi$ of $\Omega^{\hbar}$ for the coboundaries
$\Omega^{\hbar}(\xi)\sim 0$ do evolve, $\displaystyle\dot{\xi}$
$\displaystyle=\lshad\xi,F\rshad,$ whereas the generators $F$ of gauge
symmetries for (40) never change: symbolically, $\displaystyle\dot{F}$
$\displaystyle=0$

(see Eq. (45) above). In fact, one may think that each $F$ determines a
parity-preserving evolutionary vector field $\overrightarrow{Q}^{F}$ on the
space of BV-variables, but it is not the objects $\overrightarrow{Q}^{F}$ but
the full systems of four distinct evolution equations which encode the
deformation of respective functionals. Neither the functionals’ attribution to
the space of building blocks $\overline{H^{n}}(\pi_{\text{{BV}}})\ni
S^{\hbar}$, $\mathcal{O}$, $F$ nor a functional’s parity,
$\operatorname{gh}(S^{\hbar})=\operatorname{gh}(\mathcal{O})$ and
$\operatorname{gh}(F)=\operatorname{gh}(\xi)$, completely determines their
individual transformation laws.

######  Remark 3.4.

The supports of test shifts $\delta{\boldsymbol{s}}$ can be arbitrarily
small272727We recall that the smoothness class of variations
$\delta{\boldsymbol{s}}$ is determined by smoothness of the frame fields
$\vec{e}_{i}({\boldsymbol{x}}),\ \vec{e}^{{}\,\dagger i}({\boldsymbol{x}})$
and coefficient functions $\delta s^{i}({\boldsymbol{x}}),\ \delta
s_{i}^{\dagger}({\boldsymbol{x}})$.  and they can be chosen in such a way that
all boundary terms vanish in the course of integration by parts within
equivalence classes from the horizontal cohomology groups
$\overline{H}^{n(1+k)}(\pi_{{\text{{BV}}}}\times
T\pi_{{\text{{BV}}}}\times\ldots\times T\pi_{{\text{{BV}}}})$. Let us note
also that these integrations by parts (see section 1.3) transport the
derivatives from one copy of the base manifold $M^{n}$ to another copy; this
reasoning stays local with respect to base points ${\boldsymbol{x}}$ and local
volume elements $\operatorname{dvol}({\boldsymbol{x}})$ because the geometric
mechanism of locality yields the diagonal in powers of the base manifold.
However, an integration by parts in functionals from
$\overline{H}^{n}(\pi_{{\text{{BV}}}})$ is a different issue. In fact, it
refers to the topology of $M^{n}$ or to a choice of the class
$\Gamma(\pi_{{\text{{BV}}}})$ of admissible sections (so that there appear no
boundary terms as well). Let us recall that the only place where such global,
de Rham cohomology aspect is explicitly used is the proof of Jacobi’s identity
for the variational Schouten bracket (see [32]). In turn, Theorems 8 and 12
relate these properties of the bracket $\lshad\,,\,\rshad$ to cohomology
generators $\Delta^{2}=0$ and $(\Omega^{\hbar})^{2}=0$. (The converse is also
true: Jacobi’s identity for $\lshad\,,\,\rshad$ stems from $\Delta^{2}=0$.)
This motivates why the de Rham and quantum BV-cohomologies are interrelated
(cf. [5]).

## Conclusion

Mathematical models are designed for description of phenomena of Nature ; a
construction of the models’ objects is not the same as their evaluation at
given configurations of the models, which would associate $\Bbbk$-numbers to
physical fields $\phi\in\Gamma(\pi)$ in terms of such objects. Namely,
consider an Euler–Lagrange model whose primary element is the action
functional $S\colon\Gamma(\pi)\to\Bbbk$. By definition, derivative objects are
obtained from $S$ by using natural operations such as $\smash{\vec{\delta}}$
or $\lshad\,,\,\rshad$ and $\Delta$. The derivative objects’ geometric
complexity is greater than that of $S$ because they absorb the domains of
definition for test shifts $\delta s_{1},\,\ldots\,,\delta s_{k}$ of field
configurations. We emphasize that such composite structure objects do not yet
become maps $\Gamma(\pi)\to\Bbbk$ which would suit well for their evaluation
at sections ${\boldsymbol{s}}\in\Gamma(\pi)$ yielding $\Bbbk$-numbers. The
intermediate objects can rather be used as arguments of $\lshad\,,\,\rshad$ or
$\Delta$ in a construction of larger, logically and geometrically more complex
objects ; we illustrate by Fig. 9

$S\colon\Gamma(\pi)\to\Bbbk$; obsevables $\mathcal{O}_{\mu}$ in
$S+\lambda_{\mu}\mathcal{O}_{\mu}$$\text{Object}\in\overline{H}^{n(1+k)}(\pi\times\underbrace{T\pi\times\ldots\times
T\pi}_{k})$$\text{Map}\colon\Gamma(\pi)\to\Bbbk$$\overleftarrow{\delta}\\!\\!,\
\lshad\,,\,\rshad,\ \Delta$by parts,surgery of $\langle\,,\,\rangle$ Figure 9:
The action $S$ as a generator of observables, building blocks of derivative
objects as horizontal cohomology classes in products of bundles over $M\times
M\times\ldots\times M$, and resulting mappings as the objects’ contractions
over Whitney’s sum of bundles.

the expansion of analytic structures and their shrinking in the course of
integration by parts and multiplication of normalized test shifts in
reconfigured couplings. Indeed, the derivative objects become multi-linear
maps with respect to $k$-tuples of the variations $\delta s_{1}$, $\ldots$,
$\delta s_{k}\in\Gamma(T\pi)$ only when the integrations by parts carry all
derivatives away from the test shifts, channelling the derivations to
densities of the object’s constituent blocks such as the Lagrangian in the
action functional. A surgery of couplings then contracts the values of
normalized test shifts by virtue of (16) at every point of the base manifold.
This is how maps $\Gamma(\pi)\to\Bbbk$ are obtained.

We conclude that a calculation of composite-structure object may not be
interrupted ahead of time. Otherwise speaking, every calculation stretches
from its input data to the end value at ${\boldsymbol{s}}$ ; independently
existing values at ${\boldsymbol{s}}$ for the resulting object’s constituent
elements not always contribute to the sought-for value of the large structure
(e. g., consider (1c) on p. 1c and Example 2.4 on p. 2.4 and try to calculate
consecutively the objects $\Delta F$, $\Delta G$, and their Schouten brackets
with $G$ and $F$, respectively, for that example’s functionals $F$ and $G$).
Summarizing, it is illegal to construct composite objects step by step,
redundantly inspecting the elements’ values at field configurations. One must
not deviate from a way towards the appointed end of logical reasoning.

In fact, it is us but not Nature who calculates (e. g., the left-hand sides of
equations of motion): Nature neither calculates nor evaluates ; for there is
no built-in mechanism for doing that.282828The probabilistic approach to
evolution of Nature suggests that maxima of transition (and correlation)
functions concentrate near the zero loci of such deterministic equations’
left-hand sides. At the same time, Noether symmetries of the action $S$ are
abundant in the models. Not referring to any actual transformation of a
system’s components, such symmetries reflect the model’s geometry. The
analytic machinery of self-regularizing structures yields the invariants – e.
g., cohomology classes as in section 3.2 – which constrain the probabilistic
laws of evolution. This implies that there is no ever-growing logical
complexity in a description of the Universe ; the flow of local, observer-
dependent time does not require any perpetual increase of the number
$k\geqslant 0$ of factors in the product-bundle location of objects over $k+1$
copies of the space-time. Conversely, there always remains a unique copy of
the space-time for all local functionals.

The space-time geometry of information transfer is very restrictive: its
pointwise locality of events of couplings between dual objects is an absolute
principle ; by weakening this hypothesis one could create a source of
difficulties through causality violation. Consequently, a count of space-time
points where the couplings with a given (co)vector occur makes the formalism
of singular linear integral operators truly adequate in mathematical models of
physical phenomena.292929We recall from Remark 1.5 on p. 1.5 that the volume
elements
$\operatorname{dvol}\bigl{(}{\boldsymbol{x}},\phi({\boldsymbol{x}})\bigr{)}=\sqrt{|\det(g_{\mu\nu})|}\,{\mathrm{d}}{\boldsymbol{x}}$
are present in the building blocks of composite-structure objects.Let us note
further that an association of the weight factors
$\operatorname{dvol}({\boldsymbol{x}})$ with point ${\boldsymbol{x}}\in M^{n}$
is intrinsically related to the structure of space-time $M^{n}$ as topological
manifold (cf. [31]). It is readily seen that a discrete tiling of space-time
converts the integrations over a measure on it to weighted sums over the
points which mark the quantum domains. This links the concept with loop
quantum gravity (see e. g. [17, 47, 49]).

We finally remark that the product-base approach of bundles $\pi\times
T\pi\times\ldots\times T\pi$ over $M\times M\times\ldots\times M$ to the
geometry of variations highlights the concept of physical field as infinite-
dimensional system with degrees of freedom which are attached at every point
of space-time. The locality principle for (co)vector interaction is the
mechanism which distinguishes between space-time points with respect to its
(non)Hausdorff topology.

### Discussion

Let us finally address two logical aspects of the geometry of variations.

#### Linear vector space structures

Nature is essentially nonlinear ; for there is no mechanism which would
realize – under a uniform time bound – an arbitrarily large number of
replications of an object. This is tautological for those physical fields
$\phi$ which take values in spaces without any linear structure. Moreover,
even if there is a brute force labelling of Euler–Lagrange equations by using
the fields $\phi$, a linear vector space pattern of the equations of motion is
not utilized (the same is true for the equations’ descendants such as the
antifields $\phi^{\dagger}$ or (anti)ghosts). Indeed, it is only their the
_tangent_ spaces whose linear structure is used, in particular, in order to
split the variations in ghost parity-homogeneous components. Objects are
linearized only in the course of variations under infinitesimal test shifts.
For example, this determines the distinction between finite offsets
$\Delta{\boldsymbol{x}}$ so that
$({\boldsymbol{x}},{\boldsymbol{x}}+\Delta{\boldsymbol{x}})\in M\times M$ and
infinitesimal test shifts
$\left.\mathstrut\delta{\boldsymbol{x}}\right|_{{\boldsymbol{x}}}\in
T_{{\boldsymbol{x}}}M$ which are mapped to the number field $\Bbbk$ by
covectors
$\left.\mathstrut{\mathrm{d}}{\boldsymbol{x}}\right|_{{\boldsymbol{x}}}\in
T^{*}_{{\boldsymbol{x}}}M$.

#### Annual reproduction rate for interspecimen breeding of cats and whales

An immediate comment on the title of this paragraph is as follows. One could
proclaim that the annual reproduction rate for interspecimen breeding of –
without loss of generality – cats and whales is equal to zero for a given
year. Alternatively, one should understand that such events never happen (not
that a given year brought no brood).

This grotesque illustration works equally well for the (co)tangent spaces to
fibres of the BV-zoo or, in broad terms, for a definition of Kronecker’s
symbol $\boldsymbol{\delta}_{i}^{j}$ by zero whenever the indices $i\neq j$ do
not match so that the couplings in (11) do not eventuate. We argue that, on
top of the absolute pointwise locality for couplings (9), a superficial
definition of $\langle\,,\,\rangle$ by zero for mismatching elements
$\vec{e}_{i}$ and $\vec{e}^{{}\,\dagger j}$ of dual bases is a mere act of
will ; in reality those evaluations do not occur. Consequently, the geometry
dictates that

$\log\left\langle\vec{e}_{i}({\boldsymbol{x}}),{}^{\dagger}(\vec{e}_{j})({\boldsymbol{x}})\right\rangle=\log
1=0\quad\text{and}\quad\log\left\langle\vec{e}^{{}\,\dagger
j}({\boldsymbol{x}}),{}^{\dagger}(\vec{e}^{{}\,\dagger
i})({\boldsymbol{x}})\right\rangle=\log 1=0.$

Combined with the geometric locality principle (4) realized by singular linear
integral operators (12), this argument finally resolves the paradoxical, ad
hoc conventions ${\boldsymbol{\delta}(0)=0}$ and
${\log\boldsymbol{\delta}(0)=0}$ for Dirac’s distribution.

The author thanks the Organizing committee of XXI International conference
‘Integrable systems & quantum symmetries’ (June 11 – 16, 2013; CVUT Prague,
Czech Republic) for cooperation and warm atmosphere during the meeting. These
notes follow the lecture course which was read by the author in October 2013
at the Taras Shevchenko National University and Bogolyubov Institute for
Theoretical Physics in Kiev, Ukraine; the author is grateful to BITP for
hospitality. The author thanks M. A. Vasiliev and A. G. Nikitin for helpful
discussions and constructive criticisms.

This research was supported in part by JBI RUG project 103511 (Groningen). A
part of this research was done while the author was visiting at the IHÉS
(Bures-sur-Yvette); the financial support and hospitality of this institution
are gratefully acknowledged.

##  Appendix A Proof of Propositions 10 and 11

We need the following two lemmas.

######  Lemma 17.

Let $F\in\overline{H}^{n}(\pi_{\text{{BV}}})$ be an even integral functional,
let $G\in\overline{\mathfrak{N}}^{n}(\pi_{\text{{BV}}},T\pi_{\text{{BV}}})$ be
another functional, and let $n\in\mathbb{N}_{\geq 1}$. Then

$\lshad{G,F^{n}}\rshad=n\lshad{G,F}\rshad F^{n-1}.$

###### Proof.

We use induction on Theorem 4. Note that all signs vanish since $F$ is even,
meaning that whenever $F$ is multiplied with any other integral functional,
the factors may be freely swapped without this resulting in minus signs. For
$n=1$ the statement is trivial. Suppose the formula holds for some
$n\in\mathbb{N}_{>1}$, then $\lshad{G,F^{n+1}}\rshad={}$

$\lshad{G,F\cdot F^{n}}\rshad=\lshad{G,F}\rshad\cdot
F^{n}+F\cdot\lshad{G,F^{n}}\rshad=\lshad{G,F}\rshad\cdot
F^{n}+nF\cdot\lshad{G,F}\rshad F^{n-1}=(n+1)\lshad{G,F}\rshad\cdot F^{n},$

so that the statement also holds for $n+1$. ∎

######  Lemma 18.

Let $F\in\overline{H}^{n}(\pi_{\text{{BV}}})$ be an even integral functional,
and let $n\in\mathbb{N}_{\geq 2}$. Then

$\Delta(F^{n})=n(\Delta F)\cdot
F^{n-1}+\tfrac{1}{2}n(n-1)\lshad{F,F}\rshad\cdot F^{n-2}.$

###### Proof.

We use induction and the previous lemma. For $n=2$ the formula clearly holds
by Theorem 3. Suppose that it holds for some $n\in\mathbb{N}_{>2}$, then

$\displaystyle\Delta(F^{n+1})$ $\displaystyle=\Delta(F\cdot F^{n})=(\Delta
F)\cdot F^{n}+\lshad{F,F^{n}}\rshad+F\cdot\Delta(F^{n})$
$\displaystyle=(\Delta F)\cdot F^{n}+n\lshad{F,F}\rshad\cdot F^{n-1}+F\cdot
n(\Delta F)F^{n-1}+\tfrac{1}{2}n(n-1)F\cdot\lshad{F,F}\rshad F^{n-2}$
$\displaystyle=(n+1)(\Delta F)\cdot
F^{n}+\tfrac{1}{2}(n+1)n\,\lshad{F,F}\rshad\cdot F^{n-1},$

so that the statement also holds for $n+1$. ∎

###### Proof of Proposition 10.

For convenience, we denote $F=\frac{{\boldsymbol{i}}}{\hbar}S^{\hbar}$. Then

$\displaystyle 0$ $\displaystyle=\Delta(\exp
F)=\Delta\left(\sum_{n=0}^{\infty}\frac{1}{n!}F^{n}\right)=\sum_{n=0}^{\infty}\frac{1}{n!}\Delta(F^{n})$
$\displaystyle=\sum_{n=1}^{\infty}\frac{n}{n!}(\Delta F)\cdot
F^{n-1}+\sum_{n=2}^{\infty}\frac{1}{2n!}n(n-1)\lshad{F,F}\rshad\cdot F^{n-2}$
$\displaystyle=(\Delta
F)\cdot\sum_{n=1}^{\infty}\frac{1}{(n-1)!}F^{n-1}+\frac{1}{2}\lshad{F,F}\rshad\cdot\sum_{n=2}^{\infty}\frac{1}{(n-2)!}F^{n-2}$
$\displaystyle=\left(\Delta F+\tfrac{1}{2}\lshad{F,F}\rshad\right)\cdot\exp
F=\left(\frac{{\boldsymbol{i}}}{\hbar}\Delta
S^{\hbar}-\frac{1}{2\hbar^{2}}\lshad{S^{\hbar},S^{\hbar}}\rshad\right)\cdot\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right),$

from which the result follows. ∎

###### Proof of Proposition 11 (cf. Proposition 13 on p. 13).

Again, let us set $F=\frac{{\boldsymbol{i}}}{\hbar}S^{\hbar}$. We first
calculate, using Lemma 17,

$\lshad{\mathcal{O},\exp
F}\rshad=\sum_{n=0}^{\infty}\frac{1}{n!}\lshad{\mathcal{O},F^{n}}\rshad=\sum_{n=1}^{\infty}\frac{n}{n!}\lshad{\mathcal{O},F}\rshad
F^{n-1}=\lshad{\mathcal{O},F}\rshad\exp F.$

Then

$\displaystyle 0$ $\displaystyle=\Delta(\mathcal{O}\exp
F)=(\Delta\mathcal{O})\exp F+\lshad{\mathcal{O},\exp
F}\rshad+\mathcal{O}\cdot\Delta(\exp F)$
$\displaystyle=\big{(}\Delta\mathcal{O}+\lshad{\mathcal{O},F}\rshad\big{)}\exp
F=\left(\Delta\mathcal{O}+\tfrac{{\boldsymbol{i}}}{\hbar}\lshad{\mathcal{O},S^{\hbar}}\rshad\right)\exp\left(\tfrac{{\boldsymbol{i}}}{\hbar}S^{\hbar}\right),$

from which the assertion follows. ∎

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