$\displaystyle=\frac{1}{2}\int_{M}\partial_{M}\mathsf{A}^{l,l}_{I\,\mathrm{res}}\bar{\partial}_{M}\mathsf{A}^{l,l}_{I\,\mathrm{res}}+\int_{M}\mathsf{A}^{k,-}_{\mathrm{in}}d_{\mathfrak{g}}^{+}\mathsf{A}^{k-1}_{I\,\mathrm{res}}$
$\displaystyle\qquad-G(\mathsf{A}^{k,-}_{\mathrm{in}}+d_{\mathfrak{g}}^{-}\mathsf{A}^{k-1}_{I\,\mathrm{res}},\mathsf{A}^{Q}_{\mathrm{out}}),$
(109) $\displaystyle
S_{\mathrm{gh}}[\mathsf{A}^{[>0]}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{out}},\mathsf{A}_{I\,\mathrm{res}},\mathsf{A}_{\mathrm{res}}]$
$\displaystyle=\int_{M}(\mathsf{A}^{[>0]}_{\mathrm{out}}-\mathsf{A}^{[>0]}_{\mathrm{in}})\mathsf{A}^{[<0]}_{\mathrm{res}}+\int_{M}\mathsf{A}_{\mathrm{res}}^{[<0]}d_{M}\mathsf{A}_{I\,\mathrm{res}}^{[>0]}.$

See also the toy model in considered [16, Section 12].

####  6.3.2. Hitchin polarization on 6-dimensional manifolds and effective
action

In 7-dimensional Chern–Simons theory, there is an interesting — nonlinear —
polarization on the boundary phase space, coming from the special geometry of
three-forms in six dimensions first described by Hitchin in [29]. The idea is
as follows. A complex 3-form $A$ on a six-dimensional manifold with a complex
structure decomposes as

(110) $A=A^{+,\mathrm{nl}}+A^{-,\mathrm{nl}}$

where $A^{+,\mathrm{nl}}$ and $A^{-,\mathrm{nl}}$ are decomposable complex
3-forms, i.e., triple wedge products:

$A^{\pm,\mathrm{nl}}=\theta_{1}^{\pm}\wedge\theta_{2}^{\pm}\wedge\theta_{3}^{\pm},\qquad\theta_{i}^{\pm}\in\Omega^{1}(M,\mathbb{C}).$

The $3$-form $A$ is called _nondegenerate_ if $A^{+,\mathrm{nl}}\wedge
A^{-,\mathrm{nl}}$ is everywhere nonvanishing (which is equivalent to the fact
that the form $A$ is not decomposable). In this case $A^{+,\mathrm{nl}}$ and
$A^{-,\mathrm{nl}}$ are uniquely determined by $A$ and define a polarization
of $\Omega^{3}(M)_{\mathrm{nd}}$, the subset of nondegenerate forms. For more
details on this polarization, we refer to [26]. The effective action on the
cylinder with $\mathcal{P}^{[<0],-,\mathrm{l}}$-polarization on the in-
boundary595959For consistency with [26] and [9], we switch here to the
polarization with base parametrized by $\mathsf{A}^{3,+,\mathrm{l}}$, as
opposed to the rest of the paper. and
$\mathcal{P}^{[<0],+,\mathrm{nl}}$-polarization on the out-boundary thus reads
$S_{\mathrm{eff}}=S_{\mathrm{ph}}+S_{\mathrm{gh}}$ with $S_{\mathrm{gh}}$
given by (109) and

(111)
$S_{\mathrm{ph}}[\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}},\mathsf{A}^{-,\mathrm{nl}}_{\mathrm{out}},\mathsf{A}^{2,0}_{I\,\mathrm{res}},\mathsf{A}^{1,1}_{I\,\mathrm{res}},\mathsf{A}^{0,2}_{I\,\mathrm{res}}]\\\
=\frac{1}{2}\int_{M}\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}+\int_{M}\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}}d\mathsf{A}^{0,2}_{I\,\mathrm{res}}+\mathsf{A}^{2,1}_{\mathrm{in}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}-G(\mathsf{A}^{+,\l}_{\mathrm{in}}+d\mathsf{A}^{2,0}_{I\,\mathrm{res}}+\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}};\mathsf{A}^{-,\mathrm{nl}}_{\mathrm{out}}).$

######  Proposition 6.1.

The partition function is given by

$Z=\exp\frac{i}{\hbar}\left(S_{\mathrm{ph}}+S_{\mathrm{gh}}\right),$

with $S_{\mathrm{ph}}$ given by (111) and $S_{\mathrm{gh}}$ given by (109). It
satisfies the modified quantum master equation

$(\Omega_{\mathrm{in}}+\Omega_{\mathrm{out}}-\hbar^{2}\Delta_{\mathrm{res}})Z=0,$

with $\Omega_{\mathrm{in}}$, $\Omega_{\mathrm{out}}$ given by the standard
quantization of the BFV boundary action
$\frac{1}{2}\int_{M}\mathsf{A}d\mathsf{A}$:

$\displaystyle\Omega_{\mathrm{in}}$
$\displaystyle=\int_{M}d\mathsf{A}_{\mathrm{in}}^{2}\left(\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}}-i\hbar\frac{\delta}{\delta\mathsf{A}_{\mathrm{in}}^{+,\mathrm{l}}}\right)-i\hbar\int_{M}d\mathsf{A}_{\mathrm{in}}^{1}\frac{\delta}{\delta\mathsf{A}_{\mathrm{in}}^{4}}-i\hbar\int_{M}d\mathsf{A}_{\mathrm{in}}^{0}\frac{\delta}{\delta\mathsf{A}_{\mathrm{in}}^{5}},$
$\displaystyle\Omega_{\mathrm{out}}$
$\displaystyle=\int_{M}d\mathsf{A}_{\mathrm{out}}^{2}\left(\mathsf{A}^{-,\mathrm{nl}}_{\mathrm{out}}-i\hbar\frac{\delta}{\delta\mathsf{A}^{-,\mathrm{nl}}_{\mathrm{out}}}\right)-i\hbar\int_{M}d\mathsf{A}_{\mathrm{out}}^{1}\frac{\delta}{\delta\mathsf{A}_{\mathrm{out}}^{4}}-i\hbar\int_{M}d\mathsf{A}_{\mathrm{out}}^{0}\frac{\delta}{\delta\mathsf{A}_{\mathrm{out}}^{5}}.$

###### Proof.

We are in the situation of Remark 5.8 here because of the fact that the
boundary action is linear in the canonical variables defining the Hitchin
polarization, see equation (110). It follows that the mQME is satisfied for
the standard quantization of the boundary action. ∎

######  Remark 6.2.

We want to stress that in this gauge, the semiclassical approximation to this
partition function that was used as an ansatz in [26] (the integral kernel of
the generalized Segal–Bargmann transform606060See Appendix A for a motivation
for the comparison of BV-BFV partition function with standard Segal–Bargmann
transform. from the linear to the nonlinear polarization) is found to be
_exact_ : there are no quantum corrections. Adapting the proof of [16,
Appendix B] to the infinite-dimensional setting one can show that changing the
gauge fixing by changing the propagator on the interval results in a change of
the partition function by a $(\hbar^{2}\Delta_{\mathrm{res}}-\Omega)$-cocycle.

######  Remark 6.3.

To be precise in comparison with [26], one should identify the residual fields
in (111) with the Lagrange multipliers enforcing the constraint
$d_{M}\mathsf{A}=0$ in [26]. This, of course, is precisely their role when
interpreting (111) as the Hamilton–Jacobi action for 7D Chern–Simons theory in
the chosen polarizations.

####  6.3.3. Comparison with Kodaira–Spencer gravity

Following [26], we want to compare the Chern–Simons effective action (111)
with the Kodaira–Spencer action functional (145). Let us fix a reference
holomorphic 3-form $\omega_{0}\in\Omega^{3,0}(M)$. Its conjugate is an
antiholomorphic 3-form $\overline{\omega}_{0}\in\Omega^{0,3}(M)$. We can then
parametrize $A^{+,\mathrm{nl}}$ and $A^{-,\mathrm{nl}}$ as

$\displaystyle A^{+,\mathrm{nl}}$ $\displaystyle=\rho e^{\mu}\omega_{0},$
$\displaystyle A^{-,\mathrm{nl}}$
$\displaystyle=\overline{\rho}e^{\overline{\mu}}\overline{\omega}_{0},$

where $\rho,\overline{\rho}\in\Omega^{0}_{\mathbb{C}}(M)$,
$\mu\in\Omega^{-1,1}(M)$, $\overline{\mu}\in\Omega^{1,-1}(M)$ and

(112) $\rho
e^{\mu}\omega_{0}=\rho\left(\omega_{0}+\mu\omega_{0}+\frac{\mu^{2}}{2}\omega_{0}+\frac{\mu^{3}}{6}\omega_{0}\right),$

where $\mu\omega_{0}$ should be interpreted as extension of contraction to
forms with values in vector fields.

Of course, a complex 3-form still has a decomposition
$A=A^{+,\mathrm{l}}+A^{-,\mathrm{l}}$, with
$A^{+,\mathrm{l}}\in\Omega^{3,0}(M)\oplus\Omega^{2,1}(M)$, and
$A^{-,\mathrm{l}}\in\Omega^{1,2}(M)\oplus\Omega^{0,3}(M)$.

The following expression for $G$ is given in [26]:

(113) $G(A^{3,0},A^{2,1},\overline{\rho},\overline{\mu})=\\\
=\int_{M}\overline{\rho}(A^{3,0}\overline{\omega}_{0}+A^{2,1}\overline{\mu}\,\overline{\omega}_{0})+\overline{\rho}^{2}\langle\overline{\mu}^{3}\rangle\omega_{0}\overline{\omega}_{0}-\frac{\left\langle\left((A^{2,1}-\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}_{0})^{\vee}\right)^{3}\right\rangle}{(A^{3,0})^{\vee}-\overline{\rho}\langle\overline{\mu}^{3}\rangle}\omega_{0}\overline{\omega}_{0}.$

For completeness, we present a derivation of this formula in Appendix B.2.
Here for $\overline{\mu}\in\Omega^{1,-1}(M)$ the function
$\langle\mu^{3}\rangle$ is defined in (127) and for any $A\in\Omega^{p,q}$ we
define $A^{\vee}\in\Omega^{p-3,q}$ by $A^{\vee}\omega_{0}=A$.

Consider now the state

(114)
$\psi\left(\overline{\rho}_{\mathrm{out}},\overline{\mu}_{\mathrm{out}},\overline{\mathsf{A}^{[>0]}_{\mathrm{out}}}\right)=\delta(\overline{\mu}_{\mathrm{out}})\delta\left(\mathsf{A}^{[>0]}_{\mathrm{out}}\right)\exp\frac{i}{\hbar}\int_{M}\overline{\rho}_{\mathrm{out}}\omega_{0}\overline{\omega}_{0}.$

This is an extension of the physical state616161This state
$\psi_{\mathrm{ph}}$ was proposed in [26] as a way of “fixing the string
coupling constant.” It is a quantization of the lagrangian given by
$\overline{\mu}=0$, $p_{\overline{\rho}}=1$, where $p_{\overline{\rho}}$
denotes the canonical momentum of $\overline{\rho}$.

$\psi_{\mathrm{ph}}(\overline{\mu},\overline{\rho})=\delta(\overline{\mu}_{\mathrm{out}})\exp\frac{i}{\hbar}\int_{M}\overline{\rho}_{\mathrm{out}}\omega_{0}\overline{\omega}_{0}$

to the ghost sector. Note that this state trivially satisfies the mQME because
$\Omega_{\mathrm{out}}\psi_{\mathrm{ph}}$ is linear in the ghost fields, but
we are multiplying with the ghost delta function.626262This is just the
identity $x\delta(x)=0$.  Now we compute (formally) the vector
$Z\left|\psi\right\rangle$ (put differently, we are specifying a boundary
condition on the $\mathrm{out}$-boundary for the quantum theory). This means
computing the following functional integral:

$\displaystyle Z\left|\psi\right\rangle$
$\displaystyle=\int\mathcal{D}\overline{\rho}_{\mathrm{out}}\mathcal{D}\overline{\mu}_{\mathrm{out}}\mathcal{D}\mathsf{A}_{\mathrm{out}}^{[>0]}Z\cdot\psi$
$\displaystyle=\int\mathcal{D}\overline{\rho}_{\mathrm{out}}Z[\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{in}},\overline{\rho}_{\mathrm{out}},0,0,\mathsf{A}^{\geq
4}_{\mathrm{res}},\mathsf{A}^{\leq
2}_{I\,\mathrm{res}}]\exp\frac{i}{\hbar}\int_{M}\overline{\rho}_{\mathrm{out}}\omega_{0}\overline{\omega}_{0}$
$\displaystyle=\mathrel{\mathop{\mathchar
58\relax}}Z^{\prime}[\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{in}},\mathsf{A}^{\geq
4}_{\mathrm{res}},\mathsf{A}^{\leq 2}_{I\,\mathrm{res}}].$

The partition function $Z$ depends on $\overline{\rho}_{\mathrm{out}}$ only
through $G$ and we have

(115)
$\int\mathcal{D}\overline{\rho}\;e^{-\frac{i}{\hbar}G(A^{3,0},A^{2,1},\overline{\rho},0)+\frac{i}{\hbar}\int_{M}\overline{\rho}\omega_{0}\overline{\omega}_{0}}=\\\
=\delta(A^{3,0}-\omega_{0})\,\exp\left(-\frac{i}{\hbar}\int_{M}\frac{1}{6}\langle
A^{2,1},A^{2,1},A^{2,1}\rangle\right).$

Here for $A\in\Omega^{2,1}(M)$ we have $\frac{1}{6}\langle
A,A,A\rangle\mathrel{\mathop{\mathchar
58\relax}}=\langle(A^{\vee})^{3}\rangle\omega_{0}\overline{\omega}_{0}$. Thus,
$Z^{\prime}$ has the following expression:

$\displaystyle Z^{\prime}$
$\displaystyle=Z^{\prime}_{\mathrm{ph}}Z^{\prime}_{\mathrm{gh}},\quad\mbox{where}$
(116) $\displaystyle Z^{\prime}_{\mathrm{ph}}$
$\displaystyle=\delta(\mathsf{A}^{3,0}_{\mathrm{in}}+\partial\mathsf{A}^{2,0}_{I\,\mathrm{res}}-\omega_{0})\,\exp\frac{i}{\hbar}\bigg{(}\frac{1}{2}\int_{M}\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}+\int_{M}\mathsf{A}^{3,0}_{\mathrm{in}}\bar{\partial}\mathsf{A}^{0,2}_{I\,\mathrm{res}}$
$\displaystyle+\int_{M}\mathsf{A}^{2,1}_{\mathrm{in}}\partial\mathsf{A}^{0,2}_{I\,\mathrm{res}}+\mathsf{A}^{2,1}_{\mathrm{in}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}-\langle((\mathsf{A}_{\mathrm{in}}^{2,1}+\bar{\partial}\mathsf{A}^{2,0}_{I\,\mathrm{res}}+\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}})^{\vee})^{3}\rangle\omega_{0}\overline{\omega}_{0}\bigg{)},$
(117) $\displaystyle Z^{\prime}_{\mathrm{gh}}$
$\displaystyle=\exp\frac{i}{\hbar}\left(\int_{M}-\mathsf{A}^{\leq
2}_{\mathrm{in}}\mathsf{A}^{\geq 4}_{\mathrm{res}}+\int_{M}\mathsf{A}^{\geq
4}_{\mathrm{res}}d_{M}\mathsf{A}^{\leq 1}_{I\,\mathrm{res}}\right).$

We stress that $Z^{\prime}$ was obtained from $Z$ through a formal functional
integral. However, we have the following result.

######  Lemma 6.4.

The function
$Z^{\prime}[\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{in}},\mathsf{A}^{\geq
4}_{\mathrm{res}},\mathsf{A}^{\leq 2}_{I\,\mathrm{res}}]$ satisfies the
modified quantum master equation, i.e., is an
$(\Omega_{\mathrm{in}}-\hbar^{2}\Delta_{\mathrm{res}})$-cocycle, where
$\Omega_{\mathrm{in}}=\Omega^{(0)}_{\mathrm{in}}+\Omega^{(1)}_{\mathrm{in}}$
is the standard quantization of
$-\frac{1}{2}\int_{M}\mathsf{A}d_{M}\mathsf{A}$ in the
$\mathcal{P}^{[<0],-}$-polarization, explicitly given by

$\displaystyle\Omega^{(0)}_{\mathrm{in}}$
$\displaystyle=-\int_{M}\mathsf{A}^{2,1}_{\mathrm{in}}\bar{\partial}\mathsf{A}^{1,1}_{\mathrm{in}}+\mathsf{A}^{2,1}_{\mathrm{in}}\partial\mathsf{A}_{\mathrm{in}}^{0,2}+\mathsf{A}_{\mathrm{in}}^{3,0}\bar{\partial}\mathsf{A}_{\mathrm{in}}^{0,2},$
$\displaystyle\frac{i}{\hbar}\Omega^{(1)}_{\mathrm{in}}$
$\displaystyle=-\int_{M}\frac{\delta}{\delta\mathsf{A}^{3,0}_{\mathrm{in}}}\partial\mathsf{A}^{2,0}_{\mathrm{in}}+\frac{\delta}{\delta\mathsf{A}^{2,1}_{\mathrm{in}}}\bar{\partial}\mathsf{A}^{2,0}_{\mathrm{in}}+\frac{\delta}{\delta\mathsf{A}^{2,1}_{\mathrm{in}}}\partial\mathsf{A}^{1,1}_{\mathrm{in}}+\frac{\delta}{\delta\mathsf{A}^{\leq
2}_{\mathrm{in}}}d\mathsf{A}^{\leq 1}_{\mathrm{in}}.$

This is expected because $\psi$ satisfies $\Omega_{\mathrm{out}}\psi=0$ and
$Z$ satisfies the mQME. See also the discussions of gluing in [16, Sections
11.4,12.2]. The interpretation of this Lemma is that $Z^{\prime}$ is a valid
state in the linear polarization.

###### Proof.

We will only prove the claim in ghost number 0, since in positive ghost number
the effective action is the same as for linear polarizations. To begin, we
note that any function of
$\mathsf{A}^{3,0}_{\mathrm{in}}+\partial\mathsf{A}^{2,0}_{I\,\mathrm{res}}$ or
$\mathsf{A}_{\mathrm{in}}^{2,1}+\bar{\partial}\mathsf{A}^{2,0}_{I\,\mathrm{res}}+\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}}$
is
$(i\hbar\\{S_{\mathrm{gh}},\bullet\\}_{\mathrm{res}}-\Omega^{(1)}_{\mathrm{in}})$-closed
since
$\\{S_{\mathrm{gh}},\bullet\\}_{\mathrm{res}}\big{|}_{\mathrm{gh}=0}=\int_{M}\mathsf{A}^{i,j}_{\mathrm{in}}\frac{\delta}{\delta\mathsf{A}^{i,j}_{I\,\mathrm{res}}}$.
This implies that the delta function in (116) and the last term in (117) are
$\hbar^{2}\Delta_{\mathrm{res}}-\Omega^{(1)}$ closed. It is a straightforward
check that the remaining exponential terms are
$(\hbar^{2}\Delta_{\mathrm{res}}-\Omega^{(0)}-\Omega^{(1)})$-closed, which
concludes the proof. ∎

We will now argue that formally integrating out the residual fields, in ghost
number 0 we obtain the Kodaira–Spencer action. Let us restrict to the gauge-
fixing lagrangian $\mathcal{L}$ defined similarly to (108), but given in ghost
number 0 by $\partial^{*}$-exact 2-forms. We will denote

$Z^{\prime\prime}[\mathsf{A}^{3,0}_{\mathrm{in}},\mathsf{A}^{2,1}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{in}}]=\int_{\mathcal{L}}Z^{\prime}[\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{in}},\mathsf{A}^{\geq
4}_{\mathrm{res}},\mathsf{A}^{\leq 2}_{I\,\mathrm{res}}].$

The modified quantum master equation implies that for a $(2,0)$-form $\chi$
one has

$Z^{\prime}[\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}}+d\chi,\mathsf{A}^{[>0]}_{\mathrm{in}},\mathsf{A}^{\geq
4}_{\mathrm{res}},\mathsf{A}^{\leq
2}_{I\,\mathrm{res}}]=Z^{\prime}[\mathsf{A}^{+,\mathrm{l}}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{in}},\mathsf{A}^{\geq
4}_{\mathrm{res}},\mathsf{A}^{\leq 2}_{I\,\mathrm{res}}-\chi].$

By a change of variables, this implies

$Z^{\prime\prime}[\mathsf{A}^{3,0}_{\mathrm{in}}+\partial\chi,\mathsf{A}^{2,1}_{\mathrm{in}}+\bar{\partial}\chi,\mathsf{A}^{[>0]}_{\mathrm{in}}]=Z^{\prime\prime}[\mathsf{A}^{3,0}_{\mathrm{in}},\mathsf{A}^{2,1}_{\mathrm{in}},\mathsf{A}^{[>0]}_{\mathrm{in}}].$

We can use this property to reduce the computation of $Z^{\prime\prime}$ to
the case $\mathsf{A}^{3,0}=\rho_{0}\omega_{0}$, where $\rho_{0}$ is a
constant. The $\delta$ function in $Z^{\prime}_{\mathrm{ph}}$ then factorizes
as $\delta(\rho_{0}-1)\delta(\partial A^{2,0}_{I\,\mathrm{res}})$. Since
$\partial$ is an isomorphism on the gauge-fixing lagrangian, the integral over
$\mathsf{A}^{2,0}_{I\,\mathrm{res}}$ gives

$Z^{\prime\prime}_{\mathrm{ph}}=\int\mathcal{D}A^{0,2}_{I\,\mathrm{res}}A^{1,1}_{I\,\mathrm{res}}\delta(\rho_{0}-1)\exp\frac{i}{\hbar}\bigg{(}-\frac{1}{2}\int_{M}\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}+\\\
+\int_{M}\mathsf{A}^{2,1}_{\mathrm{in}}\partial\mathsf{A}^{0,2}_{I\,\mathrm{res}}+\mathsf{A}^{2,1}_{\mathrm{in}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}-\langle((\mathsf{A}_{\mathrm{in}}^{2,1}+\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}})^{\vee})^{3}\rangle\omega_{0}\overline{\omega}_{0}\bigg{)}$

and the integral over $A^{0,2}_{I\,\mathrm{res}}$ then gives

$Z^{\prime\prime}_{\mathrm{ph}}=\int\mathcal{D}\mathsf{A}^{1,1}_{I\,\mathrm{res}}\delta(\rho_{0}-1)\delta(\partial\mathsf{A}_{\mathrm{in}}^{2,1})\cdot\\\
\cdot\exp\frac{i}{\hbar}\bigg{(}\int_{M}\frac{1}{2}\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}+\mathsf{A}^{2,1}_{\mathrm{in}}\bar{\partial}\mathsf{A}^{1,1}_{I\,\mathrm{res}}-\langle((\mathsf{A}_{\mathrm{in}}^{2,1}+\partial\mathsf{A}^{1,1}_{I\,\mathrm{res}})^{\vee})^{3}\rangle\omega_{0}\overline{\omega}_{0}\bigg{)}.$

Finally, writing $\mathsf{A}^{2,1}_{\mathrm{in}}$ in the Hodge decomposition
$\mathsf{A}^{2,1}_{\mathrm{in}}=x+\partial\lambda+\partial^{*}\tau$, we obtain
by another change of variables636363We can choose $\lambda$ such that
$\partial^{*}\lambda=0$, so that $b\in\mathcal{L}\cap\Omega^{1,1}(M)$.
$b=\mathsf{A}^{1,1}_{I\,\mathrm{res}}+\lambda$ the expression

(118)
$Z^{\prime\prime}_{\mathrm{ph}}[\rho_{0},x,\lambda,\tau]=\delta(\rho_{0}-1)\delta(\partial\partial^{*}\tau)\cdot\\\
\cdot\int_{\mathcal{L}\cap\Omega^{1,1}(M)}\mathcal{D}b\;\exp\frac{i}{\hbar}\int_{M}\left(-\frac{1}{2}\partial\lambda\bar{\partial}\lambda+\frac{1}{2}\partial
b\bar{\partial}b+\bar{\partial}b\partial\lambda+\frac{1}{6}\langle(x+\partial
b),(x+\partial b),(x+\partial b)\rangle\right),$

which coincides with eq. (2.50) in [26]. Thus, we see that the Chern–Simons
partition function on a cylinder, paired with the state (114), coincides with
the partition function of Kodaira–Spencer theory with background $x$ and
action functional given in (145) for $\lambda=0$. The latter integral can be
evaluated perturbatively in terms of Feynman graphs and rules. It would be
interesting to compare our results to other constructions of the BCOV theory,
such as in [17].

######  Remark 6.5 (On gauge invariance of $Z^{\prime\prime}$).

If one uses formally the properties of the BV integral, it is immediate that
the $Z^{\prime\prime}$ gives a class in $\Omega_{\mathrm{in}}$-cohomology
independent of the gauge-fixing lagrangian $\mathcal{L}$.646464To give a
rigorous proof of this statement would require to give a strict interpretation
of the functional integral in (118), which is beyond the scope of the present
paper. It should be noted that the restriction of $b$ to the subspace
$\partial^{*}b=0$ is a gauge-fixing condition for the KS theory, so one should
consider also the the gauge independence of $Z_{KS}$. We see here that this
cohomology class has a representative given in terms of the KS partition
function. The partition function $Z^{\prime\prime}$ can be also interpreted as
the BV-BFV partition function on the cylinder paired with the state $\psi$ at
the $\mathrm{out}$-boundary, with _all_ fields integrated out using an axial-
type gauge (the components of the gauge field involving $dt$ are set to zero).
Another open question is how $Z^{\prime\prime}$ behaves when we deform away
from this type of gauge to a general gauge fixing on the cylinder (say, one
given by a Riemannian metric). This is a subject of ongoing research.

##  Appendix A Segal–Bargmann transform via BV-BFV

Recall (see [28] for details) that the Segal–Bargmann space
$\mathcal{H}^{\mathrm{SB}}$ is the Hilbert space of holomorphic functions
$\psi(z)$ on $\mathbb{C}$ satisfying

$\int_{\mathbb{C}}\frac{i}{4\pi\hbar}dz\,d\bar{z}\;e^{-\frac{|z|^{2}}{2\hbar}}\;\overline{\psi(z)}\psi(z)<\infty$

(here we assume that $\hbar$ is a fixed positive number), equipped with inner
product

(119)
$\langle\psi_{1},\psi_{2}\rangle=\int_{\mathbb{C}}\frac{i}{4\pi\hbar}dz\,d\bar{z}\;e^{-\frac{|z|^{2}}{2\hbar}}\;\overline{\psi_{1}(z)}\psi_{2}(z).$

The Segal–Bargmann space is isomorphic to the Hilbert space
$L^{2}(\mathbb{R})$ of square-integrable functions on $\mathbb{R}$, with the
unitary isomorphism $L^{2}(\mathbb{R})\rightarrow\mathcal{H}^{\mathrm{SB}}$
given by the Segal–Bargmann transform:

(120)
$\chi(x)\mapsto\psi(z)=(\pi\hbar)^{-\frac{1}{4}}\int_{\mathbb{R}}dx\;e^{-\frac{1}{\hbar}\left(\frac{z^{2}}{4}-zx+\frac{x^{2}}{2}\right)}\;\chi(x).$

Now we would like to show how the transformation (120) can be seen as the
partition function for topological quantum mechanics on an interval with
appropriate boundary polarizations.

Consider topological quantum mechanics on the interval $I$ parametrized by
$0\leq t\leq 1$ — the theory with $0$-form fields $x,p\in\Omega^{0}(I)$ and
action

(121) $S=\int_{I}p\,dx.$

In the BV-BFV formalism, we adjoin the anti-fields
$x^{*},p^{*}\in\Omega^{1}(I)$ — $1$-form fields carrying ghost number $-1$
(while $x,p$ carry ghost number $0$), so that the odd symplectic form on BV
fields is: $\int_{I}\delta x\wedge\delta x^{*}+\delta p\wedge\delta p^{*}$.
The BFV phase space assigned to a point $\mathrm{pt}^{\pm}$ (where $\pm$ is
the orientation) is: $\mathcal{F}^{\partial}=\mathbb{R}^{2}\;\ni(x,p)$ with
the Noether $1$-form $\alpha_{\mathrm{pt}^{\pm}}=\pm p\,\delta x$ and
vanishing BFV action $S_{\mathrm{pt}}=0$.

Alongside the real coordinates $x,p$ on the phase space, we will consider the
complex coordinates $z=x-ip$, $\bar{z}=x+ip$. The symplectic structure on the
phase space is
$\omega_{\mathrm{pt}^{\pm}}=\delta\alpha_{\mathrm{pt}^{\pm}}=\pm\delta
p\wedge\delta x$. Written in complex coordinates it has the form
$\mp\frac{i}{2}\delta\bar{z}\wedge\delta z$.

Consider the polarization $\mathrm{Span}\\{\frac{\partial}{\partial p}\\}$
(i.e., $x$ fixed) at $t=0$ and the polarization
$\mathrm{Span}\\{\frac{\partial}{\partial\bar{z}}\\}$ (i.e., $z$ fixed) at
$t=1$. The corresponding modification of the action (121) by a boundary term
is:

(122)
$S^{f}=S\underbrace{-\big{(}\frac{i}{4}x^{2}+\frac{1}{2}xp+\frac{i}{4}p^{2}\big{)}}_{f}\Big{|}_{t=1}$

— this boundary term is chosen so that one has $-\frac{i}{2}\bar{z}\delta
z=p\delta x+\delta f$. Thus, the corresponding boundary Noether $1$-form is:

(123) $\alpha_{\partial I}^{f}=-\frac{i}{2}\bar{z}\,\delta
z\big{|}_{t=1}-p\,\delta x\big{|}_{t=0}$

— it vanishes along the chosen polarization, as desired.

Next consider the following splitting of the (complexified) phase space

$\mathcal{F}^{\partial}_{\mathbb{C}}=\mathfrak{g}^{+}\oplus\mathfrak{g}^{-},$

where $\mathfrak{g}^{+}$ is parametrized by $z$ and $\mathfrak{g}^{-}$ is
parametrized by $x$ (we are borrowing the notations from (5) here). The space
of fields
$\mathcal{F}=\Omega^{\bullet}(I,\mathfrak{g}^{+})\oplus\Omega^{\bullet}(I,\mathfrak{g}^{-})$
is fibered over $\mathcal{B}\ni(z_{\mathrm{out}},x_{\mathrm{in}})$ with the
fiber

$\mathcal{Y}=\Omega^{\bullet}(I,\\{1\\};\mathfrak{g}^{+})\oplus\Omega^{\bullet}(I,\\{0\\};\mathfrak{g}^{-})=\\\
=\underbrace{\big{(}\Omega^{0}(I,\\{1\\};\mathfrak{g}^{+})\oplus\Omega^{0}(I,\\{0\\};\mathfrak{g}^{-})\big{)}}_{\mathcal{Y}^{\prime}_{K-ex}}\bigoplus\underbrace{\big{(}\Omega^{1}(I;\mathfrak{g}^{+}[-1])\oplus\Omega^{1}(I;\mathfrak{g}^{-}[-1])\big{)}}_{\mathcal{Y}^{\prime}_{d-ex}}.$

This is an acyclic complex, and thus we can choose the space of residual
fields to be zero (cf. Remark 3.1). The corresponding propagator — the
integral kernel of the chain contraction $K$ — is:

$\eta(t,t^{\prime})=-\pi^{+}\otimes\theta(t^{\prime}-t)+\pi^{-}\otimes\theta(t-t^{\prime}).$

The BV-BFV partition function is then given by the following path integral

(124)
$Z(z_{\mathrm{out}},x_{\mathrm{in}})=\int\mathcal{D}z_{\mathrm{fl}}\;\mathcal{D}x_{\mathrm{fl}}\;e^{\frac{i}{\hbar}S^{f}(\widetilde{z_{\mathrm{out}}}+z_{\mathrm{fl}},\widetilde{x_{\mathrm{in}}}+x_{\mathrm{fl}})}\\\
=\int\mathcal{D}z_{\mathrm{fl}}\;\mathcal{D}x_{\mathrm{fl}}\;e^{\frac{i}{\hbar}\Big{(}i\int_{I}z_{\mathrm{fl}}\,dx_{\mathrm{fl}}+\big{(}-ix_{\mathrm{fl}}(1)z_{\mathrm{out}}+\frac{i}{4}z^{2}_{\mathrm{out}}\big{)}+\big{(}-iz_{\mathrm{fl}}(0)\,x_{\mathrm{in}}+\frac{i}{2}x_{\mathrm{in}}^{2}\big{)}\Big{)}}\\\
=e^{-\frac{1}{\hbar}\big{(}\frac{z^{2}_{\mathrm{out}}}{4}-z_{\mathrm{out}}x_{\mathrm{in}}+\frac{x^{2}_{\mathrm{in}}}{2}\big{)}}.$

Here we have a contribution from the Feynman diagram with single propagator
connecting $z_{\mathrm{out}}$ and $x_{\mathrm{in}}$. In (124) we recognize the
integral kernel of the Segal–Bargmann transform (120). This is, of course, to
be expected: the partition function for a cylinder (in this case, an
interval), with polarization $\mathcal{P}_{1}$ on the in-boundary and
polarization $\mathcal{P}_{2}$ on the out-boundary, maps
$\mathcal{P}_{1}$-states to $\mathcal{P}_{2}$-states.

######  Remark A.1.

Note that the Hamilton–Jacobi action for the theory (121) on the interval with
our choice of in/out polarizations is:

$S_{\text{HJ}}=S^{f}\big{(}x(t)=x_{\mathrm{in}},z(t)=z_{\mathrm{out}}\big{)}\quad=\quad\frac{i}{4}z_{\mathrm{out}}^{2}-iz_{\mathrm{out}}x_{\mathrm{in}}+\frac{i}{2}x_{\mathrm{in}}^{2}.$

Here we again recognize the expression in the exponent in (120).

Finally, the measure $e^{-\frac{|z|^{2}}{2\hbar}}$ in (119) from the BV-BFV
standpoint originates from the gluing of intervals (more precisely, from
gluing the out-end of one interval with $z$-fixed polarization to the in-end
of another interval with $\bar{z}$-fixed polarization). Indeed, consider the
theory (121) on the interval $I=[t_{0},t_{1}]$ with some polarization
$\mathcal{P}$ at $t_{0}$ and $z$-fixed polarization at $t_{1}$, and also the
same theory on the interval $I^{\prime}=[t_{1},t_{2}]$, with some polarization
$\mathcal{P}^{\prime}$ at $t_{2}$ and with $\bar{z}$-fixed polarization at
$t_{1}$.

The respective actions including the boundary terms adjusting for the
polarization are

$S^{f}_{I}=-\frac{i}{2}\int_{I}\bar{z}dz+f^{\mathcal{P}}\big{|}_{t_{0}}\quad,\quad
S^{f}_{I^{\prime}}=-\frac{i}{2}\int_{I^{\prime}}\bar{z}dz-\frac{i}{2}z\bar{z}\big{|}_{t_{1}}+f^{\mathcal{P^{\prime}}}\big{|}_{t_{2}},$

with $f^{\mathcal{P}},f^{\mathcal{P}^{\prime}}$ the appropriate656565I.e.,
chosen in such a way that $\frac{i}{2}\bar{z}\delta z+\delta f^{\mathcal{P}}$
vanishes along $\mathcal{P}$ and $-\frac{i}{2}\bar{z}\delta z+\delta
f^{\mathcal{P}^{\prime}}$ vanishes along $\mathcal{P}^{\prime}$. boundary
terms at $t_{0}$, $t_{2}$. It follows that the partition function for the
glued interval $I\cup I^{\prime}=[t_{0},t_{2}]$ is:

(125) $Z_{I\cup
I^{\prime}}(v_{\mathrm{out}},u_{\mathrm{in}})=\int\mathcal{D}z(t)\,\mathcal{D}\bar{z}(t)\;e^{\frac{i}{\hbar}(S^{f}_{I}+\frac{i}{2}z\bar{z}\big{|}_{t_{1}}+S^{f}_{I^{\prime}})}=\\\
=\int_{\mathbb{C}}\frac{i\,dz\,d\bar{z}}{4\pi\hbar}\;Z_{I^{\prime}}(v_{\mathrm{out}},\bar{z})\;e^{-\frac{z\bar{z}}{2\hbar}}\;Z_{I}(z,u_{\mathrm{in}}).$

In the first integral, the term $\frac{i}{2}z\bar{z}\big{|}_{t_{1}}$
compensates the boundary term of $S^{f}_{I^{\prime}}$ at $t_{1}$. The final
integral is over the values of $z,\bar{z}$ at $t=t_{1}$. Also, we denoted
$u_{\mathrm{in}}$ a coordinate parametrizing the space of leaves of the
polarization $\mathcal{P}$ and similarly $v_{\mathrm{out}}$ a coordinate
paramterizing the space of leaves of $\mathcal{P}^{\prime}$. In (125), we see
the Segal–Bargmann measure
$\frac{i\,dz\,d\bar{z}}{4\pi\hbar}\,e^{-\frac{|z|^{2}}{2\hbar}}$ , cf. (119),
appearing. The normalization factor $\frac{i}{4\pi\hbar}$ is chosen in such a
way that, choosing
$\mathcal{P}^{\prime}=\mathrm{Span}\\{\frac{\partial}{\partial\bar{z}}\\}$, we
have

$Z_{I\cup
I^{\prime}}(z_{\mathrm{out}},u_{\mathrm{in}})=Z_{I}(z_{\mathrm{out}},u_{\mathrm{in}}),$

which in turn follows from the identity

$\int\frac{i}{4\pi\hbar}dz\,d\bar{z}\;\underbrace{e^{\frac{z_{\mathrm{out}}\bar{z}}{2\hbar}}}_{Z_{I^{\prime}}(z_{\mathrm{out}},\bar{z})}e^{-\frac{|z|^{2}}{2\hbar}}\Psi(z)=\Psi(z_{\mathrm{out}})$

true for any holomorphic function $\Psi(z)$ (for which the l.h.s. converges),
applied to $\Psi(z)=Z_{I}(z,u_{\mathrm{in}})$.

###  A.1. Aside: contour integration in the complexified space of fields, a
lattice toy model

Throughout the paper we are dealing with complexified phase spaces (so that we
can impose the convenient holomorphic/antiholomorphic polarizations) and
complexified spaces of fields where the path integral should be understood as
an integral over a real contour.

A toy model is provided by topological quantum mechanics $S=\int_{I}pdx$, as
above. The phase space for a point is $\Phi=\mathbb{R}^{2}$. We consider the
model with boundary polarization
$\mathrm{Span}\left\\{\frac{\partial}{\partial z}\right\\}$ (i.e., $\bar{z}$
fixed) at $t=0$ and
$\mathrm{Span}\left\\{\frac{\partial}{\partial\bar{z}}\right\\}$ (i.e., $z$
fixed) at $t=1$; these polarizations are defined on the complexified phace
space $\Phi_{\mathbb{C}}=\mathbb{C}^{2}$. The action modified by the
appropriate boundary term is
$S^{f}=\int_{I}(-\frac{i}{2}\bar{z}dz)-\frac{i}{2}z\bar{z}\big{|}_{t=0}$.

The path integral for this model can be presented by a lattice approximation
(which happens to be exact):

(126)
$Z(z_{\mathrm{out}}=z_{N},\bar{z}_{\mathrm{in}}=\bar{z}_{0})=\int_{C\subset
Y_{\mathbb{C}}}\prod_{k=1}^{N-1}\frac{i\,dz_{k}d\bar{z}_{k}}{4\pi\hbar}\;\;e^{\frac{1}{2\hbar}\big{(}\sum_{k=1}^{N}(z_{k}-z_{k-1})\bar{z}_{k-1}+z_{0}\bar{z}_{0}\big{)}}\\\
=\int_{C\subset
Y_{\mathbb{C}}}\prod_{k=1}^{N-1}\frac{i\,dz_{k}d\bar{z}_{k}}{4\pi\hbar}\;\;e^{\frac{1}{2\hbar}\big{(}z_{1}\bar{z}_{0}-z_{1}\bar{z}_{1}+z_{2}\bar{z}_{1}-z_{2}\bar{z}_{2}+\cdots-
z_{N-1}\bar{z}_{N-1}+z_{N}\bar{z}_{N-1}\big{)}}.$

Here:

  * •

We understand that the interval $I=[0,1]$ is partitioned into $N\geq 1$
smaller intervals $[t_{0}=0,t_{1}]$, $[t_{1},t_{2}]$, …, $[t_{N-1},t_{N}=1]$.

  * •

The space $Y_{\mathbb{C}}=(\mathbb{C}^{2})^{N-1}$ — the fiber of the
complexified space of fields over boundary conditions — is the product of
complexified phase spaces corresponding to $t_{1},\ldots,t_{N-1}$. In
particular, we understand that $z_{k}$ and $\bar{z}_{k}$ are independent
complex variables: they do not have to be complex conjugates of each other.

  * •

The integration is over a “contour” $C\subset Y_{\mathbb{C}}$ — a real
$2(N-1)$-dimensional subspace. In particular, the integrand in (126) is a
holomorphic $2(N-1)$-form on $Y_{\mathbb{C}}$ pulled back to $C$ by the
inclusion.

For the contour $C$, we can consider the following two examples:

  1. (i)

Contour $C_{1}$ given by $\bar{z}_{k}=z_{k}^{*}$, $k=1,\ldots,N-1$, where $*$
is the complex conjugation.

  2. (ii)

Contour $C_{2}$ given by reality conditions
$z_{k}\in\mathbb{R}\subset\mathbb{C}$, $\bar{z}_{k}\in
i\mathbb{R}\subset\mathbb{C}$.

For $C=C_{1}$, (126) is an absolutely convergent Gaussian integral (for
arbitrary boundary conditions) and yields

$Z(z_{\mathrm{out}},\bar{z}_{\mathrm{in}})=e^{\frac{1}{2\hbar}\bar{z}_{\mathrm{in}}z_{\mathrm{out}}}.$

For $C=C_{2}$, (126) is an oscillatory Fresnel integral which is only
conditionally convergent and even that only under special assumptions on the
boundary conditions ($\bar{z}_{\mathrm{in}}\in
i\mathbb{R},z_{\mathrm{out}}\in\mathbb{R}$).666666 Absolute/conditional
convergence property is particularly clear in the simplest case $N=2$: here
(126) becomes
$\int_{C\subset\mathbb{C}^{2}}\frac{i\,dz_{1}d\bar{z}_{1}}{4\pi\hbar}\;e^{\frac{1}{2\hbar}(z_{1}\bar{z}_{0}-z_{1}\bar{z}_{1}+z_{2}\bar{z}_{1})}$.
When the integral over $C_{2}$ converges, its value coincides with the result
of integration over $C_{1}$ (which is clear, e.g., from a contour deformation
argument).

In summary, we have the complexified space of fields of the lattice theory
$F^{I}_{\mathbb{C}}\ni(\bar{z}_{0},z_{1},\bar{z}_{1},\ldots,z_{N-1},\bar{z}_{N-1},z_{N})$
fibered over the complex space of boundary conditions $\mathcal{B}^{\partial
I}_{\mathbb{C}}\ni\\{\bar{z}_{\mathrm{in}},z_{\mathrm{out}}\\}$ with complex
fiber $Y_{\mathbb{C}}$ (lattice fields with zero boundary conditions), and the
integration in the lattice path integral (126) is over a contour $C\subset
Y_{\mathbb{C}}$ — a half-dimensional real submanifold.

##  Appendix B Kodaira–Spencer theory

We briefly review the definition of the Kodaira–Spencer action functional that
was introduced in [9, Section 5], where it was used to analyze the target
space physics of the $B$-model. See also [26, Section 2.1].

###  B.1. Some operations on complex forms

Let $M$ be a 6-dimensional Calabi–Yau manifold with a reference holomorphic
3-form $\omega_{0}$ (sometimes the pair $(M,\omega_{0})$ is called a gauged
Calabi–Yau manifold). We denote by $\Omega^{p,q}(M)$ complex forms of Hodge
type $(p,q)$ — sections of the bundle
$\wedge^{p}(T_{\mathbb{C}}^{*}M)^{1,0}\otimes\wedge^{q}(T^{*}_{\mathbb{C}}M)^{0,1}$
and by $\Omega^{-p,q}(M)$ sections of the bundle
$\wedge^{p}(T_{\mathbb{C}}M)^{1,0}\otimes\wedge^{q}(T^{*}_{\mathbb{C}}M)^{0,1}$,
i.e., $(0,q)$-forms with values in $(p,0)$-vector fields. Contraction with the
reference holomorphic 3-form provides a map

$\displaystyle\Omega^{-p,q}(M)$ $\displaystyle\to\Omega^{3-p,q}(M)$
$\displaystyle A$ $\displaystyle\mapsto A^{\vee}=A\omega_{0}$

(we omit the symbols for wedge products and contractions). For a $(p,q)$-form
$A$ with $p\geq 0$, we set $A^{\vee}=A\omega_{0}^{-1}\in\Omega^{p-3,q}$, in
particular we have $(A^{\vee})^{\vee}=A$. For $A,B,C\in\Omega^{-1,1}(M)$, we
further define the operations

$\displaystyle A^{\vee}\circ B^{\vee}$
$\displaystyle=(AB)^{\vee}=(AB)\omega_{0}$ $\displaystyle\in\Omega^{1,2}(M),$
$\displaystyle\langle A^{\vee},B^{\vee},C^{\vee}\rangle$
$\displaystyle=A^{\vee}(B^{\vee}\circ C^{\vee})=A^{\vee}(BC)\omega_{0}$
$\displaystyle\in\Omega^{3,3}(M),$ (127) $\displaystyle\langle A^{3}\rangle$
$\displaystyle=-\frac{1}{6}\frac{\langle
A^{\vee},A^{\vee},A^{\vee}\rangle}{\omega_{0}\overline{\omega}_{0}}=\frac{1}{6}(A^{3}\omega_{0})(\overline{\omega}_{0})^{-1}$
$\displaystyle\in\Omega^{0,0}(M),$

and the same operations make sense for
$\bar{A},\bar{B},\bar{C}\in\Omega^{1,-1}(M)$ if we replace $\omega_{0}$ by
$\bar{\omega}_{0}$. The minus sign in (127) ensures that $\langle
A^{3}\rangle\overline{\omega}_{0}=\frac{1}{6}A^{3}\omega_{0}$. Also, in the
last equation we made use of the fact that one can divide by sections of a
line bundle. By a lemma of Tian [40], if $\partial A=\partial B=0$, we have

(128) $[A,B]^{\vee}=\partial(A^{\vee}\circ B^{\vee}).$

###  B.2. The generating function for Hitchin polarization

For completeness, we include here a derivation of the generating function
(113) for the transformation from the linear polarization to the nonlinear
polarization. A complex 3-form $A$ has decompositions
$A=A^{+,\mathrm{l}}+A^{-,\mathrm{l}}$, with
$A^{+,\mathrm{l}}\in\Omega^{3,0}(M)\oplus\Omega^{2,1}(M)$, and
$A^{-,\mathrm{l}}\in\Omega^{1,2}(M)\oplus\Omega^{0,3}(M)$, and

$A=A^{+,\mathrm{nl}}+A^{-,\mathrm{nl}},$

where $A^{+,\mathrm{nl}}$ and $A^{-,\mathrm{nl}}$ are decomposable complex
3-forms. The $3$-form $A$ is called _nondegenerate_ if
$A^{+,\mathrm{nl}}\wedge A^{-,\mathrm{nl}}$ is everywhere nonvanishing. We
parametrize

$\displaystyle A^{+,\mathrm{nl}}$ $\displaystyle=\rho e^{\mu}\omega_{0},$
$\displaystyle A^{-,\mathrm{nl}}$
$\displaystyle=\overline{\rho}e^{\overline{\mu}}\overline{\omega}_{0},$

where $\rho,\overline{\rho}\in\Omega^{0}_{\mathbb{C}}(M)$,
$\mu\in\Omega^{-1,1}(M)$, $\overline{\mu}\in\Omega^{1,-1}(M)$ and

(129) $\rho
e^{\mu}\omega_{0}=\rho\left(\omega_{0}+\mu\omega_{0}+\frac{\mu^{2}}{2}\omega_{0}+\frac{\mu^{3}}{6}\omega_{0}\right).$

To write the generating function from the linear to the nonlinear
polarization, we use

(130)
$G(A^{+,\mathrm{l}},A^{-,\mathrm{nl}})=\frac{1}{2}\int_{M}A^{+,\mathrm{l}}A^{-,\mathrm{l}}-A^{-,\mathrm{nl}}A^{+,\mathrm{nl}}.$

######  Lemma B.1.

In the variables $A^{3,0},A^{2,1},\overline{\rho},\overline{\mu}$, the
generating function is given by

(131) $G(A^{3,0},A^{2,1},\overline{\rho},\overline{\mu})=\\\
=\int_{M}\overline{\rho}(A^{3,0}\overline{\omega}_{0}+A^{2,1}\overline{\mu}\,\overline{\omega}_{0})+\overline{\rho}^{2}\langle\overline{\mu}^{3}\rangle\omega_{0}\overline{\omega}_{0}-\frac{\left\langle\left((A^{2,1}-\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}_{0})^{\vee}\right)^{3}\right\rangle}{(A^{3,0})^{\vee}-\overline{\rho}\langle\overline{\mu}^{3}\rangle}\omega_{0}\overline{\omega}_{0}.$

###### Proof.

This is a tedious but straightforward computation. One way to do it is to
express $G$ in terms of $\rho,\mu,\overline{\rho},\overline{\mu}$ first. To
this end, notice that the decomposition in (129) is a decomposition into forms
of definite Hodge type. Thus, we can write

(132) $\displaystyle A^{3,0}$
$\displaystyle={\rho}\,{\omega}_{0}+\overline{\rho}\frac{\overline{\mu}^{3}}{6}\overline{\omega}_{0}=\left(\rho+\overline{\rho}\langle\overline{\mu}^{3}\rangle\right)\omega_{0},$
(133) $\displaystyle A^{2,1}$
$\displaystyle={\rho}\,{\mu}\,{\omega}_{0}+\overline{\rho}\frac{\overline{\mu}^{2}}{2}\overline{\omega}_{0},$

and similarly for $A^{0,3}$ and $A^{1,2}$. We thus obtain

(134) $\displaystyle A^{3,0}A^{0,3}$
$\displaystyle=\rho\overline{\rho}(1+\langle\mu^{3}\rangle\langle\overline{\mu}^{3}\rangle)\omega_{0}\overline{\omega}_{0}+\left(\rho^{2}\langle\mu^{3}\rangle\
+\overline{\rho}^{2}\langle\overline{\mu}^{3}\rangle\right)\omega_{0}\overline{\omega}_{0},$
(135) $\displaystyle A^{2,1}A^{1,2}$
$\displaystyle=\rho\overline{\rho}\left(\mu\omega_{0}\overline{\mu}\,\overline{\omega}_{0}+\frac{1}{4}\overline{\mu}^{2}\overline{\omega}_{0}\mu^{2}\omega_{0}\right)+\frac{1}{2}\left(\rho^{2}\mu\omega_{0}{\mu}^{2}\omega_{0}-\overline{\rho}^{2}\overline{\mu}\,\overline{\omega}_{0}\overline{\mu}^{2}\overline{\omega}_{0}\right).$

On the other hand, we have

(136)
$A^{-,\mathrm{nl}}A^{+,\mathrm{nl}}=\rho\overline{\rho}\left(\overline{\omega}_{0}\omega_{0}+\overline{\mu}\,\overline{\omega}_{0}\mu\omega_{0}+\frac{1}{4}\overline{\mu}^{2}\overline{\omega}_{0}\mu^{2}\omega_{0}+\langle{\overline{\mu}^{3}}\rangle\langle\mu^{3}\rangle\omega_{0}\overline{\omega}_{0}\right).$

Summing (134) and (135) and subtracting (136), the last two terms in (136)
cancel and we obtain

(137)
$A^{+,\mathrm{l}}A^{-,\mathrm{l}}-A^{-,\mathrm{nl}}A^{+,\mathrm{nl}}=2\rho\overline{\rho}\left(\omega_{0}\overline{\omega}_{0}-\overline{\mu}\,\overline{\omega}_{0}\mu\omega_{0}\right)\\\
+\rho^{2}\left(\langle\mu^{3}\rangle\omega_{0}\overline{\omega}_{0}+\frac{1}{2}\mu\omega_{0}\mu^{2}\omega_{0}\right)+\overline{\rho}^{2}\left(\langle\overline{\mu}^{3}\rangle\omega_{0}\overline{\omega}_{0}-\frac{1}{2}\overline{\mu}\,\overline{\omega}_{0}\overline{\mu}^{2}\overline{\omega}_{0}\right).$

Recall that
$\mu\omega_{0}\mu^{2}\omega_{0}=(\mu^{3}\omega_{0})\omega_{0}=6\langle\mu^{3}\rangle\overline{\omega}_{0}\omega_{0}$,
hence we can simplify this expression to

$A^{+,\mathrm{l}}A^{-,\mathrm{l}}-A^{-,\mathrm{nl}}A^{+,\mathrm{nl}}=2\rho\overline{\rho}\left(\omega_{0}\overline{\omega}_{0}-\overline{\mu}\,\overline{\omega}_{0}\mu\omega_{0}\right)-2\left(\rho^{2}\langle\mu^{3}\rangle+\overline{\rho}^{2}\langle\overline{\mu}^{3}\rangle\right)\omega_{0}\overline{\omega}_{0}.$

From equations (132), (133) we get

$\displaystyle\rho\omega$
$\displaystyle=A^{3,0}-\overline{\rho}\langle\overline{\mu}^{3}\rangle\omega_{0},$
$\displaystyle\rho\mu\omega_{0}$
$\displaystyle=A^{2,1}-\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}_{0},$

which we use to rewrite the first term as

$2\rho\overline{\rho}\left(\omega_{0}\overline{\omega}_{0}-\overline{\mu}\,\overline{\omega}_{0}\mu\omega_{0}\right)=2\overline{\rho}A^{3,0}\overline{\omega}_{0}-2\overline{\rho}^{2}\langle\overline{\mu}^{3}\rangle\omega_{0}+2\overline{\rho}A^{2,1}\mu\omega_{0}+6\overline{\rho}^{2}\langle\overline{\mu}^{3}\rangle\omega_{0}\overline{\omega}_{0}.$

In total, (130) evaluates to

(138)
$G=\int_{M}\overline{\rho}A^{3,0}\overline{\omega}_{0}+\overline{\rho}A^{2,1}\overline{\mu}\,\overline{\omega}_{0}+\overline{\rho}^{2}\langle\overline{\mu}^{3}\rangle\omega_{0}\overline{\omega}_{0}-\frac{\langle(\rho\mu)^{3}\rangle}{\rho}\omega_{0}\overline{\omega}_{0}.$

Formula (131) may now be obtained by solving equations (132),(133) for $\rho$
and $\rho\mu$, which gives

(139) $\displaystyle\rho$
$\displaystyle=\frac{A^{3,0}-\frac{1}{6}\overline{\rho}\,\overline{\mu}^{3}\overline{\omega}_{0}}{\omega_{0}}=(A^{3,0})^{\vee}-\overline{\rho}\langle\overline{\mu}^{3}\rangle,$
(140) $\displaystyle\rho\mu$
$\displaystyle=\frac{A^{2,1}-\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}_{0}}{\omega_{0}}=\left(A^{2,1}-\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}_{0}\right)^{\vee}.$

Plugging (139),(140) into (138) we obtain (131). ∎

The defining property of $G$ is the following.

######  Lemma B.2.

We have $\delta G=\theta^{\mathrm{l}}-\theta^{\mathrm{nl}}$ where
$\theta^{\mathrm{l}}=A^{-,\mathrm{l}}\delta A^{+,\mathrm{l}}$ and
$\theta^{\mathrm{nl}}=A^{+,\mathrm{nl}}\delta A^{-,\mathrm{nl}}$.

###### Proof.

This follows from Equation (130). But one can also check it through direct
computation: we have

$\frac{\delta G}{\delta
A^{3,0}}=\overline{\rho}\,\overline{\omega}_{0}+\frac{\left\langle\left(\left(A^{2,1}-\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}\right)^{\vee}\right)^{3}\right\rangle}{(A^{3,0})^{\vee}-\overline{\rho}\langle\overline{\mu}^{3}\rangle^{2}}(\omega_{0})^{-1}\omega_{0}\overline{\omega}_{0}=\overline{\rho}\,\overline{\omega}_{0}+\frac{\langle(\rho\mu)^{3}\rangle}{\rho^{2}}\overline{\omega}_{0}=A^{3,0}.$

Notice that we have

$\frac{\delta}{\delta\mu}\langle\mu^{3}\rangle=\frac{1}{2}(\mu^{2}\omega)(\overline{\omega})^{-1}.$

It follows that

$\frac{\delta}{\delta
A^{2,1}}\langle((A^{2,1})^{\vee})^{3}\rangle=\frac{1}{2}(((A^{2,1})^{\vee})^{2}\omega_{0})\overline{\omega}_{0}^{-1}\omega_{0}^{-1}=-\frac{\frac{1}{2}(((A^{2,1})^{\vee})^{2}\omega_{0})}{\omega_{0}\overline{\omega}_{0}^{-1}}$

(note the sign) and therefore

$\frac{\delta G}{\delta
A^{2,1}}=\overline{\rho}\,\overline{\mu}\,\overline{\omega}+\frac{1}{2}\rho\mu^{2}\omega_{0}=A^{1,2}.$

This proves that $\delta G/\delta A^{+,\mathrm{l}}=A^{-,\mathrm{l}}$.
Computing $\delta G/\delta\overline{\rho}$ gives

$\displaystyle\frac{\delta G}{\delta\bar{\rho}}\delta\overline{\rho}$
$\displaystyle=\delta\overline{\rho}\left(A^{3,0}\overline{\omega}_{0}+A^{2,1}\overline{\mu}\,\overline{\omega}_{0}+2\overline{\rho}\langle\overline{\mu}^{3}\rangle\omega_{0}\overline{\omega}_{0}+\frac{1}{2}\rho\mu^{2}\omega_{0}(\frac{1}{2}\overline{\mu}^{2}\overline{\omega}_{0})+\rho\langle\mu^{3}\rangle\omega_{0}\langle\overline{\mu}^{3}\rangle\overline{\omega}_{0}\right)$
$\displaystyle=-\rho
e^{\mu}\omega_{0}e^{\overline{\mu}}\overline{\omega}_{0}\delta\overline{\rho}.$

Finally, computing $\delta G/\delta\overline{\mu}$ gives

$\displaystyle\frac{\delta G}{\delta\overline{\mu}}\delta\overline{\mu}$
$\displaystyle=-\overline{\rho}A^{2,1}(\delta\overline{\mu}\,\overline{\omega}_{0})+\frac{1}{2}\overline{\rho}^{2}(\overline{\mu}^{2}\overline{\omega}_{0})(\delta\overline{\mu}\,\overline{\omega}_{0})+\frac{1}{2}\rho\mu^{2}\omega_{0}(\overline{\rho}\,\overline{\mu}\,\delta\overline{\mu}\,\overline{\omega}_{0})+\rho\langle\mu^{3}\rangle(\delta\overline{\mu}\,\overline{\omega}_{0})\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}_{0}$
$\displaystyle=-\overline{\rho}\rho\,\mu\,\omega_{0}(\delta\overline{\mu}\,\overline{\omega}_{0})+\frac{1}{2}\rho\mu^{2}\omega_{0}(\overline{\rho}\,\overline{\mu}\,\delta\overline{\mu}\,\overline{\omega}_{0})+\rho\langle\mu^{3}\rangle(\delta\overline{\mu}\,\overline{\omega}_{0})\frac{1}{2}\overline{\rho}\,\overline{\mu}^{2}\overline{\omega}_{0}$
$\displaystyle=-\rho\overline{\rho}(e^{\mu}\omega_{0})\delta\overline{\mu}(\overline{\omega}_{0}+\overline{\mu}\omega_{0}+\frac{1}{2}\overline{\mu}^{2}\overline{\omega}_{0}).$

Using

$\delta
A^{-,\mathrm{nl}}=\delta\overline{\rho}e^{\overline{\mu}}\overline{\omega}_{0}=e^{\overline{\mu}}\overline{\omega}_{0}\delta\overline{\rho}+\overline{\rho}\delta\overline{\mu}(\overline{\omega}_{0}+\overline{\mu}\omega_{0}+\frac{1}{2}\overline{\mu}^{2}\overline{\omega}_{0}),$

we obtain

$\delta G=A^{+,\mathrm{l}}\delta A^{-,\mathrm{l}}-A^{+,\mathrm{nl}}\delta
A^{-,\mathrm{nl}}.$

∎

###  B.3. Deformations of complex structures

Let $M$ be a compact Calabi–Yau manifold supplied with a reference holomorphic
3-form $\omega_{0}$. A deformation of the complex structure is equivalent to a
deformation of the $\bar{\partial}$ operator
$\bar{\partial}\to\bar{\partial}_{\bar{A}}=\bar{\partial}+\bar{A}$, where
$\bar{A}\in\Omega^{-1,1}(M)=\Gamma(T^{1,0}M\otimes(T^{*})^{0,1}M)$. The
integrability condition $\bar{\partial}_{A}^{2}=0$ is equivalent to ([31])

(141) $\bar{\partial}\bar{A}+\frac{1}{2}[\bar{A},\bar{A}]=0.$

The moduli space of complex structures is thus given by solutions of (141)
modulo gauge transformations

(142) $\delta\bar{A}=\bar{\partial}\varepsilon+[\bar{A},\varepsilon],$

with $\varepsilon\in\Omega^{-1,0}(M)$. The tangent space to the moduli space
of complex structures is given by the linearization of (141), i.e., it is the
quotient of $\\{\alpha\colon\bar{\partial}\bar{\alpha}=0\\}$ by linearized
gauge transformations $\delta\alpha=\bar{\partial}\varepsilon$.
After Tian ([40]), this problem can be reformulated using $\bar{A}^{\vee}$ as
follows. Imposing the constraint $\partial\bar{A}^{\vee}=0$ and using (128),
we can rewrite (141) as

(143)
$\bar{\partial}\bar{A}^{\vee}+\partial(\bar{A}^{\vee}\circ\bar{A}^{\vee})=0.$

###  B.4. Kodaira–Spencer action

The Kodaira–Spencer action functional as introduced in [9] is

(144)
$S_{KS}[\bar{A}^{\vee}]=\int_{M}\frac{1}{2}\bar{A}^{\vee}\partial^{-1}\bar{\partial}\bar{A}^{\vee}+\frac{1}{6}\langle\bar{A}^{\vee},\bar{A}^{\vee},\bar{A}^{\vee}\rangle.$

Here the first term is well-defined due to $\partial\bar{\partial}$-lemma. The
equation of motion of (144) is (143). One can resolve the nonlocality by
writing $\bar{A}^{\vee}=x+\partial b$, where $x$ is a $\partial$-harmonic
$(2,1)$-form. The action functional then becomes

(145) $S_{KS}(x;b)=\int_{M}\frac{1}{2}\partial
b\bar{\partial}b+\frac{1}{6}\langle(x+\partial b),(x+\partial b),(x+\partial
b)\rangle.$

This action functional has the following remarkable property. From eq. (143),
it follows that any harmonic $(2,1)$-form $x=\bar{A}^{\vee}_{1}$ can be
interpreted as a first order deformation of the complex structure. The tree
level diagrams of (145) then generate forms $\bar{A}^{\vee}_{n}$ with the
property that $\bar{A}^{\vee}=\sum\varepsilon^{n}\bar{A}^{\vee}_{n}$ is a
solution of the Kodaira–Spencer equation (143). We refer to [9, Section 5.2]
for details.

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