$x\in\mathcal{M}^{\prime}$. Of course we also require that the integrals are
well-defined and that the denominator of (5.3.1) does not vanish.

Let $\mathcal{F}\subset C^{\infty}(\mathcal{M}^{\prime},\mathfrak{g})$ be the
space of allowed gauge-fixing functions. As in this case $\langle
g\rangle_{F}$ is well-defined, it makes sense to consider it at more general
instances. Observe also that if the Lie group is not compact, an invariant
function is not a test function. It is then understood that it is replaced by
a test function that in a neighborhood of the zeros of $F$ coincides with the
given function. To avoid cumbersome notation, we will never explicitly change
the function, so we will write e.g.
$\int_{\mathbb{R}}\delta_{0}(x){\mathrm{d}}x=1$, without mentioning that the
constant function 1 is replaced by a test function which is one in a
neighborhood of zero. However, since the gauge-fixing function $F$ is
arbitrarily chosen, we want the conditions on $\langle g\rangle_{F}$ to be
locally independent of $F$.

######  Definition 5.3.3 (Gauge-fixing independence).

A locally constant function on $\mathcal{F}$ is called _gauge-fixing
independent_.

######  Theorem 5.3.4.

Let $X\colon\mathfrak{g}\to\mathcal{M}$ be an infinitesimal action of the Lie
algebra $\mathfrak{g}$ on the manifold $\mathcal{M}$. If $S$ and $g$ are
invariant functions and

(5.3.3)
$\operatorname{\textnormal{div}}X_{\gamma}+\operatorname{Tr}\mathrm{ad}_{\gamma}=0,\qquad\forall\gamma\in\mathfrak{g},$

then $\langle g\rangle_{F}$ is gauge-fixing independent.

The condition in Theorem 5.3.4 basically tells us that the divergence of $X$
is a constant function. In particular, for the case when the Lie algebra is
_unimodular_ (i.e. $\operatorname{Tr}\mathrm{ad}_{\gamma}=0$ for all
$\gamma$), the condition says that the infinitesimal action must be measure-
preserving. The discussions before are covered by Theorem 5.3.4 since the Lie
algebra of a compact Lie group is indeed unimodular.

##### The BRST operator and the proof of Theorem 5.3.4

The infinitesimal action of $\mathfrak{g}$ on $\mathcal{M}$ gives
$C^{\infty}(\mathcal{M})$ the structure of a $\mathfrak{g}$-module. Therefore,
we can consider the Lie algebra complex $\bigwedge\mathfrak{g}^{*}\otimes
C^{\infty}(\mathcal{M})$. Note that the Lie algebra differential $\delta$ is
in particular a derivation on $\bigwedge\mathfrak{g}^{*}\otimes
C^{\infty}(\mathcal{M})$ and moreover defines a vector field on
$\mathcal{M}\times\Pi\mathfrak{g}$. Recall that a vector field on the
superspace $\Pi\mathfrak{g}$ is an element of
$\bigwedge\mathfrak{g}\otimes\mathfrak{g}$. If $\mathfrak{g}$ is a Lie
algebra, the commutator can be regarded as an element of
$\bigwedge^{2}\mathfrak{g}^{*}\otimes\mathfrak{g}$ and as such it is a vector
field on $\Pi\mathfrak{g}$. Vector fields on
$\mathcal{M}\times\Pi\mathfrak{g}$ can then be identified with elements of
$\bigwedge\mathfrak{g}^{*}\otimes\mathfrak{g}\otimes
C^{\infty}(\mathcal{M})\oplus\bigwedge\mathfrak{g}^{*}\otimes\mathfrak{X}(\mathcal{M})$.
The commutator tensor the constant function 1 is an element of
$\bigwedge^{2}\mathfrak{g}^{*}\otimes\mathfrak{g}\otimes
C^{\infty}(\mathcal{M})$ while the infinitesimal action of $\mathfrak{g}$ on
$\mathcal{M}$ is an element of
$\mathfrak{g}^{*}\otimes\mathfrak{X}(\mathcal{M})$. As such they define vector
fields on $\mathcal{M}\times\Pi\mathfrak{g}$ and $\delta$ is their sum. If we
choose a basis $\\{e_{i}\\}$ of $\mathfrak{g}$ and denote by
$\\{f^{k}_{ij}\\}$ the corresponding structure constants, the algebra of
functions on $\Pi\mathfrak{g}$ can be identified with the graded commutative
algebra with odd generators $c^{i}$ (ghost variables) and

$\delta c^{k}=-\frac{1}{2}f^{k}_{ij}c^{i}c^{j}.$

On functions $f\in C^{\infty}(\mathcal{M})$, we get that $\delta$ acts by

$\delta f=c^{i}L_{X_{e_{i}}}f.$

The vector field $\delta$ can be extended to $\widetilde{\mathcal{M}}$ by
adding it to the vector field on $\mathfrak{g}^{*}\otimes\Pi\mathfrak{g}^{*}$.
Moreover, using the dual basis $\\{e^{i}\\}$ of $\mathfrak{g}^{*}$, the
algebra of functions can be identified with the graded commutative algebra
with odd generators $\bar{c}_{i}$ and even generators $\lambda_{i}$. There we
define

(5.3.4) $\delta\bar{c}_{i}=\lambda_{i},\qquad\delta\lambda_{i}=0,\quad\forall
i.$

More abstractly, observe that a polynomial vector field on
$\mathfrak{g}^{*}\times\Pi\mathfrak{g}^{*}$ is an element of
$\bigwedge\mathfrak{g}\otimes\textnormal{Sym}(\mathfrak{g})\otimes(\mathfrak{g}^{*}\oplus\mathfrak{g}^{*})$.
The identity operator can then be regarded as an element of
$\mathfrak{g}\otimes\mathfrak{g}^{*}$. Using the inclusion map

$\displaystyle i\colon\mathfrak{g}\otimes\mathfrak{g}^{*}$
$\displaystyle\to\bigwedge\mathfrak{g}\otimes\textnormal{Sym}(\mathfrak{g})\otimes(\mathfrak{g}^{*}\oplus\mathfrak{g}^{*}),$
$\displaystyle a\otimes b$ $\displaystyle\mapsto 1\otimes a\otimes(b\oplus
0),$

we can regard it as a vector field on
$\mathfrak{g}^{*}\times\Pi\mathfrak{g}^{*}$. Moreover, note that this vector
field corresponds to the de Rham differential on $\Pi\mathfrak{g}^{*}$.

######  Lemma 5.3.5.

The map $\delta$ is a differential (i.e. an odd derivation that squares to
zero) on $\bigwedge\mathfrak{g}^{*}\otimes
C^{\infty}(\mathcal{M})\otimes\bigwedge\mathfrak{g}\otimes\textnormal{Sym}(\mathfrak{g})$.
It has degree one with respect to the grading

$\deg(\alpha\otimes f\otimes\beta\otimes\gamma):=\deg(\alpha)-\deg(\beta).$

In fact, we can rewrite Lemma 5.3.5 by saying that $\delta$ is a cohomological
vector field on $\widetilde{\mathcal{M}}$. In the physics literature, $\delta$
is called the _BRST operator_. A function on $\mathcal{M}$ can also be
considered as a function on $\widetilde{\mathcal{M}}$. By definition of
$\delta$, a function $f$ is invariant if and only $\delta f=0$.

######  Exercise 5.3.6.

Show that

$\operatorname{\textnormal{div}}\delta=\operatorname{\textnormal{div}}X_{c}+\operatorname{Tr}\mathrm{ad}_{c}.$

Note that gauge-fixing function $F\colon\mathcal{M}\to\mathfrak{g}$ can be
regarded as an element of $C^{\infty}(\mathcal{M})]\otimes\mathfrak{g}$. Using
the inclusion

$C^{\infty}(\mathcal{M})\otimes\mathfrak{g}\hookrightarrow
C^{\infty}(\mathcal{M})\otimes\bigwedge\mathfrak{g}\hookrightarrow\bigwedge\mathfrak{g}^{*}\otimes
C^{\infty}(\mathcal{M})\otimes\bigwedge\mathfrak{g}\otimes\textnormal{Sym}(\mathfrak{g}),$

we can associate to $F$ a function $\Psi_{F}$ on $\widetilde{\mathcal{M}}$.
With the same notations as above, we have

$\Psi_{F}=\bar{c}_{i}F^{i}.$

The odd function $\Psi_{F}$ is called _gauge-fixing fermion_.

######  Exercise 5.3.7.

Show that

$S_{F}=S+\delta\Psi_{F}.$

Assume now that the action $S$ is invariant, i.e. $\delta S=0$.

######  Lemma 5.3.8.

Let $g$ be a function on $\widetilde{\mathcal{M}}$. If $\delta g=0$ and
$\delta$ is divergence-free, then

$I_{F}:=\int_{\widetilde{\mathcal{M}}}\exp(\mathrm{i}S_{F}/\hbar)g\widetilde{\mu},$

is gauge-fixing independent.

###### Proof.

Let $(F_{t})$ be a family of gauge-fixing functions. Then

$\frac{{\mathrm{d}}}{{\mathrm{d}}t}I_{F_{t}}=\frac{\mathrm{i}}{\hbar}\int_{\widetilde{\mathcal{M}}}\delta\left(\frac{{\mathrm{d}}}{{\mathrm{d}}t}\Psi_{F_{t}}\right)\exp(\mathrm{i}S_{F}/\hbar)g\widetilde{\mu}\\\
=\frac{\mathrm{i}}{\hbar}\int_{\widetilde{\mathcal{M}}}\delta\left(\frac{{\mathrm{d}}}{{\mathrm{d}}t}\Psi_{F_{t}}\exp(\mathrm{i}S_{F}/\hbar)g\right)\widetilde{\mu}=\frac{\mathrm{i}}{\hbar}\int_{\widetilde{\mathcal{M}}}\operatorname{\textnormal{div}}\delta\exp(\mathrm{i}S_{F}/\hbar)g\widetilde{\mu}=0.$

∎

###### Proof of Theorem 5.3.4.

Using Lemma 5.3.8 together with Exercise 5.3.6 we immediately get the proof. ∎

An particular case of a $\delta$-closed function is a $\delta$-exact function.
These functions are irrelevant as for computing expectation values. In fact,

$\int_{\widetilde{\mathcal{M}}}\exp(\mathrm{i}S_{F}/\hbar)\delta
h\widetilde{\mu}=\int_{\widetilde{\mathcal{M}}}\delta\left(\exp(\mathrm{i}S_{F}/\hbar)h\right)\widetilde{\mu}=0$

if $S$ is invariant and $\delta$ is divergence-free. We can now extend Theorem
5.3.4 to the following theorem.

######  Theorem 5.3.9.

Let $X\colon\mathfrak{g}\to\mathcal{M}$ be an infinitesimal action of the Lie
algebra $\mathfrak{g}$ on the manifold $\mathcal{M}$. If the action $S$ is
invariant and the BRST operator $\delta$ is divergence-free (i.e. (5.3.3)
holds), then

  1. (1)

$\langle g\rangle_{F}$ is gauge-fixing independent for all $g\in\ker\delta$,

  2. (2)

$\langle g\rangle_{F}$ for all $g\in\mathrm{im}\,\delta$.

Hence, the expectation value defines a linear function on the
$\delta$-cohomology.

######  Remark 5.3.10 (Ward identities).

Note that point (2) produces identities relating expectation values of
different quantities. Such identities as called _Ward identities_ and usually
have nontrivial content.

######  Example 5.3.11 (Translations).

Let $\mathcal{M}=\mathfrak{g}=\mathbb{R}$. Moreover, let $\mathfrak{g}$ act by
infinitesimal translations. Denoting by $x$ the coordinate on $\mathcal{M}$,
we have $\delta x=c$ and $\delta c=0$. Assume $S$ and $g$ are constant. Then

$\langle
g\rangle_{F}=\frac{\exp(\mathrm{i}S/\hbar)\displaystyle{\int}\delta_{0}(F(x))F^{\prime}(x)}{\exp(\mathrm{i}S/\hbar)\displaystyle{\int}\delta_{0}(F(x))F^{\prime}(x)}=g$

if the denominator does not vanish. Clearly, $\langle g\rangle_{F}$ is gauge-
fixing independent. Similarly, one can treat rotation-invariant functions on
$\mathcal{M}=S^{1}$. A section here is just a point. We take
$\mathcal{M}^{\prime}$ to be a neighborhood of this point. After identifying
$\mathcal{M}^{\prime}$ with $\mathbb{R}$, we can proceed as above with $F$
being any function with a single nondegenerate zero corresponding to the image
of the section.

######  Example 5.3.12 (Plane rotations).

Let $\mathcal{M}=\mathbb{R}^{2}\setminus\\{0\\}$ and
$\mathfrak{g}=\mathfrak{so}(2)$ acting by infinitesimal rotations. Denoting by
$x$ and $y$ the coordinates on $\mathbb{R}^{2}$, we have

$\delta x=yc,\quad\delta y=-xc,\quad\delta c=0.$

Let $\mathcal{M}^{\prime}=\\{x>0\\}$. A possible choice for $F$ is the
function $F(x,y)=y$. Then

$S_{F}(x,y,c,\lambda,\bar{c})=S(x,y)+\lambda y+\bar{c}xc,$

where $S$ is the given rotation-invariant action. Then we get

$\langle
g\rangle_{F}=\frac{\displaystyle{\int}_{0}^{+\infty}\exp(\mathrm{i}S(x,0)/\hbar)g(x,0)x{\mathrm{d}}x}{\displaystyle{\int}_{0}^{+\infty}\exp(\mathrm{i}S(x,0)/\hbar)x{\mathrm{d}}x},$

which is equal to the expected expression

$\frac{\displaystyle{\int}_{\mathcal{M}}\exp(\mathrm{i}S(x,y)/\hbar)g(x,y){\mathrm{d}}x{\mathrm{d}}y}{\displaystyle{\int}_{\mathcal{M}}\exp(\mathrm{i}S(x,y)/\hbar){\mathrm{d}}x{\mathrm{d}}y}$

for a rotation-invariant function $g$.

####  5.3.3. Infinite dimensions

In the infinite-dimensional setting, we want to consider (5.3.2) whenever it
makes sense as a perturbative expansion. In this case $\mathcal{M}$ is an
infinite-dimensional manifold, $\mathfrak{g}$ an infinite-dimensional Lie
algebra acting freely on $\mathcal{M}$ and $S$ some local action function on
$\mathcal{M}$. The space $\widetilde{\mathcal{M}}$ and the BRST operator
$\delta$ will be exactly defined as above and $\delta$ will still be a
cohomological vector field on $\widetilde{\mathcal{M}}$. A gauge-fixing
function $F$ will be allowed whenever the corresponding $A$ is nondegenerate
and the critical point of the action at a zero of $F$ is also nondegenerate.
Then, for suitable functions $g$, we will be able to define $\langle
g\rangle_{F}$ as a perturbative expansion using Feynman diagrams. Note that an
observable in this setting is a $\delta$-cohomology class $g$ for which
$\langle g\rangle_{F}$ is well-defined. In fact, Theorem 5.3.9 will hold
whenever $\operatorname{\textnormal{div}}\delta=0$. Note that, however, the
divergence of $\delta$ is not defined a priori and has to be understood in
terms of expectation values. The usual way to proceed is to assume Theorem
5.3.9 to hold and to derive from it properties of the expectation values. Once
they are properly defined in terms of Feynman diagrams, one can check whether
the identities hold and call any deviation an _anomaly_.

Note also that if our aims are of mathematical nature, Theorem 5.3.9 provides
a source for a lot of interesting conjectures (which, fortunately, in most
cases turn out to be true).

##### The trivial Poisson sigma model on the plane

Consider a 2-dimensional generalization of Section 5.2 which is also the basis
for the study of the Poisson sigma model. Let $\xi$ and $\eta$ be 0-form and
1-form on the plane respectively. We assume that they vanish at infinity
sufficiently fast (e.g., as Schwarz functions) so that we may define the
action

(5.3.5) $S:=\int_{\Sigma}\eta\,{\mathrm{d}}\xi,$

with $\Sigma=\mathbb{R}^{2}$. Note that we have dropped the wedge product to
avoid cumbersome notation and we will stick to this convention for the rest of
the discussion. The space of fields is given by
$\mathcal{M}=\Omega^{0}_{0}(\mathbb{R}^{2})\oplus\Omega^{1}_{0}(\mathbb{R}^{2})$.
On $\mathcal{M}$ we have an action of the abelian Lie algebra
$\mathfrak{g}=\Omega^{0}_{0}(\mathbb{R}^{2})$, given by the monomorphism
$i\circ{\mathrm{d}}$, where

$\mathfrak{g}\xhookrightarrow{{\mathrm{d}}}\Omega^{1}_{0}(\mathbb{R}^{2})\xhookrightarrow{i}\mathcal{M}.$

The action function $S$ is clearly invariant. The BRST differential $\delta$
is given on coordinates by

$\delta\xi=0,\qquad\delta\eta={\mathrm{d}}c.$

In order to define a gauge-fixing function, we want to choose a Riemannian
metric on $\mathbb{R}^{2}$. Denote by $*$ the _Hodge star operator_ induced by
the Riemannian metric. Then, using the Hodge star, we can define the pairing

(5.3.6)
$\langle\alpha,\beta\rangle_{*}:=\int_{\mathbb{R}^{2}}(*\alpha)\beta,\quad\alpha\in\Omega^{j}(\mathbb{R}^{2}),\beta\in\Omega^{k}(\mathbb{R}^{2}),\quad
j,k=0,1,2.$

Let ${\mathrm{d}}^{*}:=*{\mathrm{d}}*$ be the formal adjoint of the de Rham
differential and choose the gauge-fixing function
$F(\xi,\eta)={\mathrm{d}}^{*}\eta$. Note that different metrics will give
different gauge-fixing functions. one can show that the corresponding operator
$A$ is then given by the Laplacian on 0-forms which is invertible for the
given conditions at infinity. There is also a unique critical point, i.e. a
solution to ${\mathrm{d}}\xi={\mathrm{d}}\eta=0$, satisfying the gauge-fixing
condition ${\mathrm{d}}^{*}\eta=0$, in particular, $\xi=\eta=0$. Hence, this
gauge-fixing is indeed allowed.

Using integration, we can identify $\mathfrak{g}^{*}$ with
$\Omega^{2}(\mathbb{R}^{2})$. The corresponding gauge-fixing fermion is

$\Psi_{F}=\int_{\mathbb{R}^{2}}\bar{c}\,{\mathrm{d}}^{*}\eta.$

This gives the gauge-fixed action

(5.3.7)
$S_{F}=\int_{\mathbb{R}^{2}}\Big{(}\eta\,{\mathrm{d}}\xi+\lambda\,{\mathrm{d}}^{*}\eta-\bar{c}\,{\mathrm{d}}^{*}{\mathrm{d}}c\Big{)}.$

Using the pairing as in (5.3.6), we can rewrite the gauge-fixed action as

$S_{F}=\frac{1}{2}\langle\phi,\boldsymbol{{\mathrm{d}}}\phi\rangle_{*}-\langle*\bar{c},\Delta
c\rangle_{*},$

where $\Delta:={\mathrm{d}}^{*}{\mathrm{d}}+{\mathrm{d}}{\mathrm{d}}^{*}$
denotes the _Hodge Laplacian_ ,

$\phi:=\begin{pmatrix}\xi\\\ \eta\\\
\lambda\end{pmatrix}\in\Omega^{0}_{0}(\mathbb{R}^{2})\oplus\Omega^{1}_{0}(\mathbb{R}^{2})\otimes\Omega^{2}_{0}(\mathbb{R}^{2})$

and

$\boldsymbol{{\mathrm{d}}}:=\begin{pmatrix}0&*{\mathrm{d}}&0\\\
*{\mathrm{d}}&0&{\mathrm{d}}*\\\ 0&{\mathrm{d}}*&0\end{pmatrix}.$

The propagators between $\phi$ and $c$ or $\bar{c}$ are clearly zero. Hence,
we get

$\langle*\bar{c}(z)c(w)\rangle_{0}=-\mathrm{i}\hbar G_{0}(w,z),$

where $G_{0}$ denotes the Green function of the Hodge Laplacian acting on
functions. Now in order to get the propagator between two fields $\phi$, we
need to invert the symmetric operator $\boldsymbol{{\mathrm{d}}}$. For this,
we first compute its square

$\boldsymbol{{\mathrm{d}}}^{2}=\begin{pmatrix}\Delta&0&0\\\ 0&\Delta&0\\\
0&0&\Delta\end{pmatrix}.$

Then, we can observe that
$\boldsymbol{{\mathrm{d}}}^{-1}=\boldsymbol{{\mathrm{d}}}\boldsymbol{{\mathrm{d}}}^{-2}$
and hence

$\boldsymbol{{\mathrm{d}}}^{-1}=\begin{pmatrix}0&*{\mathrm{d}}\Delta^{-1}&0\\\
*{\mathrm{d}}\Delta^{-1}&0&{\mathrm{d}}*\Delta^{-1}\\\
0&{\mathrm{d}}*\Delta^{-1}&0\end{pmatrix}.$

In particular, we have

$\langle\xi(z)\eta(w)\rangle_{0}=\mathrm{i}\hbar*_{z}{\mathrm{d}}_{z}G_{1}(z,w)=\mathrm{i}\hbar*_{w}{\mathrm{d}}_{w}G_{0}(w,z),$

where $G_{1}$ denotes the Green function of the Hodge Laplacian acting on
1-forms. Let us introduce the _superfields_

(5.3.8)
$\displaystyle\begin{split}\boldsymbol{\xi}&:=\xi-{\mathrm{d}}^{*}\bar{c},\\\
\boldsymbol{\eta}&:=c+\eta.\end{split}$

Then we can define a _superpropagator_

(5.3.9)
$\mathrm{i}\hbar\theta(z,w):=\left\langle\boldsymbol{\xi}(z)\boldsymbol{\eta}\right\rangle_{0}=\langle\xi(z)\eta(w)\rangle_{0}-\langle{\mathrm{d}}^{*}\bar{c}(z)c(w)\rangle_{0}\\\
=\mathrm{i}\hbar(*_{z}{\mathrm{d}}_{z}+*_{w}{\mathrm{d}}_{w})G_{0}(w,z)\in\Omega^{1}(C_{2}(\mathbb{R}^{2})),$

where $C_{2}(\mathbb{R}^{2})$ denotes the configuration space of two points in
$\mathbb{R}^{2}$ as in Remark 4.3.7.

######  Lemma 5.3.13.

If we choose the Euclidean metric, then

$\theta=\frac{{\mathrm{d}}\phi_{E}}{2\pi},$

where ${\mathrm{d}}$ denotes the differential on $C_{2}(\mathbb{R}^{2})$ and
$\phi_{E}(z,w)$ the Euclidean angle between a fixed reference line and the
line passing through $z$ and $w$.

###### Proof.

The Green function for the Euclidean Laplacian in two dimensions is given by

$G_{0}(z,w)=\frac{1}{2\pi}\log(|z-w|),$

where $|\enspace|$ denotes the Euclidean norm. In complex coordinates we have

$G_{0}(z,w)=\frac{1}{4\pi}\log((z-w)(\bar{z}-\bar{w})).$

Then

${\mathrm{d}}_{z}G_{0}(z,w)=\frac{1}{4\pi}\left(\frac{{\mathrm{d}}z}{z-w}+\frac{{\mathrm{d}}\bar{z}}{\bar{z}-\bar{w}}\right).$

The Euclidean Hodge star operator in complex coordinates gives
$*{\mathrm{d}}z=\mathrm{i}{\mathrm{d}}z$ and
$*{\mathrm{d}}\bar{z}=\mathrm{i}{\mathrm{d}}\bar{z}$. Hence, we have

$*_{z}{\mathrm{d}}_{z}G_{0}(z,w)=\frac{1}{4\pi\mathrm{i}}\left(\frac{{\mathrm{d}}w}{w-z}-\frac{{\mathrm{d}}\bar{w}}{\bar{w}-\bar{z}}\right).$

Summing everything up, we get

$\theta=\frac{1}{4\pi\mathrm{i}}\left(\frac{{\mathrm{d}}z-{\mathrm{d}}w}{z-w}-\frac{{\mathrm{d}}\bar{z}-{\mathrm{d}}\bar{w}}{\bar{z}-\bar{w}}\right)=\frac{1}{4\pi\mathrm{i}}{\mathrm{d}}\log\left(\frac{z-w}{\bar{z}-\bar{w}}\right).$

One the other hand, we have $z-w=|w-z|\exp(\mathrm{i}\phi)$, which gives

$\phi=\frac{1}{2\mathrm{i}}\log\left(\frac{z-w}{\bar{z}-\bar{w}}\right),$

and hence we get the claim. ∎

######  Remark 5.3.14.

The cohomology class of $\theta$ is in fact the generator of
$H^{1}(C_{2}(\mathbb{R}^{2}),\mathbb{Z})$. It is not difficult to see that
other choices of metric will still give the same cohomolgy class. One can
easily note that $*_{w}\theta(z,w)$ is the Green function of the operator
$P:=*{\mathrm{d}}\Delta^{-1}*$ which is a _parametrix_ for the de Rham
differential on forms that vanish at infinity, in particular,

${\mathrm{d}}P+P{\mathrm{d}}=\mathrm{id}.$

The convolution relating $P$ to $\theta$ can be written as

$P\alpha=-\pi_{2}(\theta\pi_{1}^{*}\alpha),\quad\alpha\in\Omega^{j}_{0}(\mathbb{R}^{2}),\quad
j=0,1,2,$

with $\pi_{1}$ and $\pi_{2}$ being the projections to $\mathbb{R}^{2}$. Then

${\mathrm{d}}P\alpha-P{\mathrm{d}}\alpha=-(\pi_{2})_{*}({\mathrm{d}}\theta\pi_{1}^{*}\alpha)+\pi_{*}^{\partial}(\theta)\alpha,$

where $\pi_{*}^{\partial}(\theta)(w)$ denotes the integral of $\theta$ along a
limiting small circle around $w$. Since $P$ is a parametrix and $\alpha$ is
arbitrary, we can see that in general $\theta$ is closed and integrates to 1
along the generators of $H_{1}(C_{2}(\mathbb{R}^{2}),\mathbb{Z})$.

##### Expectation values

Note that any function of $\xi$ is BRST-invariant, i.e. we can consider the
evaluation of $\xi$ at some point $u$. A function of $\int_{\gamma}\eta$ is
also invariant for any closed curve $\gamma$. Hence, the expectation value

$\left\langle\xi(u)\int_{\gamma}\eta\right\rangle_{0}=:\mathrm{i}\hbar
W_{\gamma}(u),\quad u\not\in\mathrm{im}\,\gamma,$

is independent of the gauge-fixing. Since we also have

$\mathrm{i}\hbar
W_{\gamma}(u)=\left\langle\boldsymbol{\xi}(u)\int_{\gamma}\boldsymbol{\eta}\right\rangle_{0},$

we can immediately see that $W_{\gamma}$ is in fact the winding number of
$\gamma$ around $u$. This number is in fact invariant under deformations of
$\gamma$ or displacements of $u$, which indicates that the theory is
topological. For example, let us deform $\gamma$ to $\gamma^{\prime}$.
Denoting by $\sigma$ a 2-chain whose boundary is $\gamma-\gamma^{\prime}$, we
get

$W_{\gamma}(u)-W_{\gamma^{\prime}}(u)=\left\langle\xi(u)\int_{\sigma}{\mathrm{d}}\eta\right\rangle_{0}.$

Moreover, introduce the sequence of divergence-free vector fields
$X_{r}(\xi,\eta)=\lambda_{r}\oplus 0$, where $(\lambda_{r})$ is a sequence of
functions that converges almost everywhere to the characteristic function of
the image of $\sigma$. Then we get

$W_{\gamma}(u)-W_{\gamma^{\prime}}(u)=\lim_{r\to\infty}\langle\xi(u)X_{r}(S)\rangle_{0}=\mathrm{i}\hbar\lim_{r\to\infty}\langle
X_{r}(\xi(u))\rangle_{0}=0,$

under the assumption that $u$ does not belong to $\sigma$.

##### The trivial Poisson sigma model on the upper half-plane

Consider now the action (5.3.5) where $\Sigma=\mathbb{H}^{2}$. As a boundary
condition, we impose that the 1-form $\eta$ vanishes when restricted to the
boundary $\partial\mathbb{H}^{2}=\mathbb{R}\times\\{0\\}$. The Lie algebra
acting on the space of fields is given by 0-forms on $\mathbb{H}^{2}$
vanishing on $\partial\mathbb{H}^{2}$. The BRST complex can then be defined
exactly as before and we can choose the same gauge-fixing function. We define
the superpropagator as in (5.3.9) with the difference that we will denote it
by $\vartheta$ instead of $\theta$ in order to avoid any confusion.

######  Lemma 5.3.15.

If we choose the Euclidean metric, then

$\vartheta=\frac{{\mathrm{d}}\phi_{h}}{2\pi},$

where ${\mathrm{d}}$ denotes the differential on $C_{2}(\mathbb{H}^{2})$ and
$\phi_{h}(z,w)$ denotes the angle between the vertical line through $w$ and
the geodesic joining $w$ to $z$ in the hyperbolic Poincaré metric (recall
Figure 4.3.4).

###### Proof.

The Green function $G^{\mathbb{H}^{2}}_{0}$ of the Laplacian on
$\mathbb{H}^{2}$ is the restriction to $\mathbb{H}^{2}$ of the Green function
of the Laplacian on $\mathbb{R}^{2}$ plus a harmonic function such that the
sum satisfies the boundary conditions. In complex coordinates, we need
$G^{\mathbb{H}^{2}}_{0}(w,z)=0$ whenever $w$ is real. This can be done by
setting

$G^{\mathbb{H}^{2}}_{0}(w,z)=G_{0}(w,z)-G_{0}(\bar{w},z).$

Then we get

$\vartheta(z,w)=\theta(z,w)-\theta(z,\bar{w}).$

Since the hyperbolic angle is given by

$\phi_{h}(z,w)=\frac{1}{2\mathrm{i}}\log\left(\frac{(z-w)(\bar{z}-w)}{(\bar{z}-\bar{w})(z-\bar{w})}\right),$

we conclude the claim. ∎

######  Remark 5.3.16.

Note that $\vartheta$ is the generator of
$H^{1}(C_{2}(\mathbb{H}^{2}),\mathbb{H}^{2}\times\partial\mathbb{H}^{2};\mathbb{Z})$.

##### Generalizations

We can also consider a collection of $n$ 0-forms $\xi^{i}$ and $n$ 1-forms
$\eta_{i}$ with $i=1,\ldots,n$. Then we want to look at the action

$\int_{\Sigma}\eta_{i}\,{\mathrm{d}}\xi^{i}.$

We can also think of $\xi$ and $\eta$ as forms taking values in
$\mathbb{R}^{n}$. The Lie algebra $\mathfrak{g}$ of symmetries will then
consist of the direct sum of $n$ copies of the previous one; in other words it
will be the abelian Lie algebra of $\mathbb{R}^{n}$-valued 0-forms. Let
$c_{i}$ for $i=1,\ldots,n$ denote the generators of the algebra of functions
of $\Pi\mathfrak{g}$. Then we can define the BRST operator through
$\delta\xi^{i}$ and $\delta\eta_{i}=c_{i}$. If we choose the gauge-fixing
function to be $F_{i}(\xi,\eta)={\mathrm{d}}^{*}\eta_{i}$, everything remains
the same as before. In particular, we can again introduce superfields
$\boldsymbol{\xi}^{i}:=\xi^{i}-{\mathrm{d}}^{*}\bar{c}^{i}$ and
$\boldsymbol{\eta}_{i}:=c_{i}+\eta_{i}$ and compute the superpropagator

(5.3.10)
$\left\langle\boldsymbol{\xi}^{i}(z)\boldsymbol{\eta}_{j}(w)\right\rangle_{0}:=\begin{cases}\mathrm{i}\hbar\theta(z,w)\delta^{i}_{j},&\text{on
$\mathbb{R}^{2}$}\\\ \mathrm{i}\hbar\vartheta(z,w)\delta^{i}_{j},&\text{on
$\mathbb{H}^{2}$}\end{cases}$

Another generalization can be done by dropping the assumption that the 0-form
field vanishes at infinity. More precisely, we denote by $X^{i}$ a collection
of maps to $\mathbb{R}^{n}$ with no conditions on the boundary or at infinity
and consider the action

$S=\int_{\Sigma}\eta_{i}\,{\mathrm{d}}X^{i},$

where $\Sigma$ is either $\mathbb{R}^{2}$ or $\mathbb{H}^{2}$ and the
$\eta_{i}$s are 1-forms vanishing on the boundary and at infinity. Critical
points are then pairs of constant maps together with closed 1-forms. They will
be also degenerate modulo the action of the abelian Lie algebra of 0-forms.
However, the degeneracy will be of a very simple type as it is parametrized by
the finite-dimensional manifold $\mathbb{R}^{n}$. In fact, it is enough to
choose a measure on $\mathbb{R}^{n}$ and impose Fubini’s theorem. We will
choose a delta-measure peaked at a point $x\in\mathbb{R}^{n}$ and require $X$
to map the point $\infty$ to $x$. Note that if we write $X^{i}=x^{i}+\xi^{i}$,
the $\xi^{i}$ vanish at infinity and everything is reduced to the previous
case.

A final generalization is to replace $\mathbb{R}^{n}$ by a manifold $M$. We
think of $X^{i}$ as a local coordinate expression of a map $X\colon\Sigma\to
M$. For the action to be covariant, we need to assume that $\eta_{i}(u)$ for
$u\in\Sigma$ is the local coordinate expression of a 1-form on $\Sigma$ taking
values in the cotangent space of $M$ at $X(u)$. In particular, we assume
$\eta\in\Gamma(\Sigma,T^{*}\Sigma\otimes X^{*}T^{*}M)$. The space of fields
$\mathcal{M}$ can then be identified with the space of vector bundle maps
$T\Sigma\to T^{*}M$ and the action can be invariantly written as

$S=\int_{\Sigma}\langle\eta,{\mathrm{d}}X\rangle,$

where $\langle\enspace,\enspace\rangle$ denotes the canonical pairing between
the tangent and cotangent bundle of $M$ and ${\mathrm{d}}X$ denotes the
differential of the map $X$ regarded as a section of $T^{*}\Sigma\otimes
X^{*}TM$. if we also require $X$ to map the point $\infty$ to a given point
$x\in M$, we can expand around critical points by setting $X=x+\xi$ with
$\xi\colon\Sigma\to T_{x}M$ and by regarding $\eta$ as a 1-form taking values
in $T_{x}^{*}M$. Choosing local coordinates, we can identify $T_{x}M$ with
$\mathbb{R}^{n}$, where $n=\dim M$, and reduce everything to the previous
case.

We also allow $\Sigma$ to be any 2-manifold. The previous discussion will
change drastically if $\Sigma$ is not simply connected, as the space of
solutions modulo symmetries, with $X=x$, will now be parametrized by
$H^{1}(\Sigma,T^{*}_{x}M)$ and one has to choose a measure on this vector
space as well.

###  5.4. The Poisson sigma model

Next we want to look at deformations of the trivial Poisson sigma model as
discussed before. Here, we will describe how the action of the Poisson sigma
model is expressed in order to derive Kontsevich’s star product out of it.

####  5.4.1. Formulation of the model

We want to formulate a deformation of the action functional without
introducing extra structure on $\Sigma$ (which is either $\mathbb{R}^{2}$ or
$\mathbb{H}^{2}$). Therefore, the terms we are allowed to add must be 2-forms
on $\Sigma$ given in terms of the fields $X^{i}$ and $\eta_{i}$. In
particular, they need to be linear combinations of terms
$\alpha^{ij}(X)\eta_{i}\eta_{j}$,
$\beta^{i}_{j}(X)\eta_{i}\,{\mathrm{d}}x^{i}$ and
$\gamma_{ij}(X){\mathrm{d}}X^{i}{\mathrm{d}}X^{j}$. Note that we are not
considering a term of the form $\phi^{i}(X){\mathrm{d}}\eta_{i}$, since
integration by parts reduces it to a term of the second type. The second and
third terms can be absorbed by a redefinition of $\eta$ adding to it terms
linear in $\eta$ and ${\mathrm{d}}X$. Hence, modulo field redefinitions, the
most general deformation of the action has the form

$S=\int_{\Sigma}\left(\eta_{i}\,{\mathrm{d}}X^{i}+\frac{1}{2}\epsilon\alpha^{ij}(X)\eta_{i}\eta_{j}\right)+O(\epsilon^{2}),$

where $\epsilon$ is the deformation parameter (here typically
$\epsilon=\frac{\mathrm{i}\hbar}{2}$) and $\alpha^{ij}$ is assumed to be skew-
symmetric.

######  Remark 5.4.1.

We want to show that it makes sense to only consider those deformations in
which the $\alpha^{ij}$ are the components of a Poisson bivector field and
that the BRST formalism is only available if the Poisson structure is affine.

Recall that the BRST operator before acted by $\delta X^{i}=0$,
$\delta\eta_{i}={\mathrm{d}}c_{i}$ and $\delta c_{i}=0$, with
$c\in\Pi\mathfrak{g}$ and
$\mathfrak{g}=\Omega^{0}_{0}(\Sigma,\mathbb{R}^{n})$. We would like to deform
the trivial $\delta$ such that $\delta S=O(\epsilon^{2})$ for the new $S$.
Note that we will only consider the restriction of $\delta$ to
$\mathcal{M}\times\Pi\mathfrak{g}$ as its restriction to
$\mathfrak{g}^{*}\times\Pi\mathfrak{g}$ needs no deformation.

######  Lemma 5.4.2.

Modulo field redefinitions, there is a unique BRST operator deforming the
trivial one such that $\delta S=O(\epsilon^{2})$ and
$\delta^{2}=O(\epsilon^{2})+R$ with $R$ vanishing at critical points. It acts
by

$\displaystyle\delta X^{i}$
$\displaystyle=-\epsilon\alpha^{ij}(X)c_{j}+O(\epsilon^{2}),$
$\displaystyle\delta\eta_{i}$
$\displaystyle={\mathrm{d}}c_{i}+\epsilon\partial_{i}\alpha^{jk}(X)\eta_{j}c_{k}+O(\epsilon^{2}),$
$\displaystyle\delta c_{i}$
$\displaystyle=-\frac{1}{2}\epsilon\partial_{i}\alpha^{jk}(X)c_{j}c_{k}+O(\epsilon^{2}).$

Moreover, $R$ vanishes on the whole $\mathcal{M}\times\Pi\mathfrak{g}$ if
$\alpha$ is at most linear.

###### Proof.

Recall that $\delta$ applied to $X$ or $\eta$ must be linear in the ghost
variables $c$. Hence, the most general deformation of $\delta$ (without adding
any extra structure on $\Sigma$) is of the form

$\displaystyle\delta X^{i}$ $\displaystyle=\epsilon
v^{ij}(X)c_{j}+O(\epsilon^{2}),$ $\displaystyle\delta\eta_{i}$
$\displaystyle={\mathrm{d}}c_{i}+\epsilon\Big{(}a^{jk}_{i}(X)\eta_{j}c_{k}+b^{j}_{i}(X){\mathrm{d}}c_{j}+d^{k}_{ij}(X){\mathrm{d}}X^{j}c_{k}\Big{)}+O(\epsilon^{2}),$

for some functions $v^{ij},a^{jk}_{i},b^{j}_{i}$ and $d^{k}_{ij}$ on
$\mathbb{R}^{n}$. Thus, we get

$\delta
S=\epsilon\int_{\Sigma}\Big{(}(a^{jk}_{i}(X)\eta_{j}c_{k}+b^{j}_{i}(X){\mathrm{d}}c_{j}+d^{k}_{ij}(X){\mathrm{d}}X^{j}c_{k}){\mathrm{d}}X^{i}\\\
+\eta_{i}({\mathrm{d}}X^{r}\partial_{r}v^{ij}(X)c_{j}+v^{ij}(X){\mathrm{d}}c_{j})+\alpha^{ij}(X){\mathrm{d}}c_{i}\eta_{j}\Big{)}+O(\epsilon^{2}).$

As the identity $\delta S=O(\epsilon^{2})$ must hold for any $\eta$, we get
the following two equations

(5.4.1) $\displaystyle
a^{jk}_{i}(X)c_{k}{\mathrm{d}}X^{i}+{\mathrm{d}}X^{r}\partial_{r}v^{jk}(X)c_{k}+v^{jk}(X){\mathrm{d}}c_{k}+\alpha^{kj}(X){\mathrm{d}}c_{k}$
$\displaystyle=0,$ (5.4.2)
$\displaystyle\int_{\Sigma}\Big{(}b_{i}^{j}(X){\mathrm{d}}c_{j}+d^{k}_{ij}(X){\mathrm{d}}X^{j}c_{k}\Big{)}{\mathrm{d}}X^{i}$
$\displaystyle=0.$

In particular, if we choose $X$ to be the constant map (with value $x$), we
deduce from (5.4.1) that

$v^{jk}(x){\mathrm{d}}c_{k}+\alpha^{kj}(x){\mathrm{d}}c_{k}=0,$

and since this has to hold for any $c$, we have

$\alpha^{jk}=v^{jk}.$

Plugging into (5.4.2), we get

$a^{jk}_{i}(X)c_{k}{\mathrm{d}}X^{i}-{\mathrm{d}}X^{i}\partial_{i}\alpha^{jk}(X)c_{k}=0,$

and since this has to hold for all $c$ and $X$, we get

$a^{jk}_{i}=\partial_{i}\alpha^{jk}.$

Using integration by parts, (5.4.2) gives

$\int_{\Sigma}\Big{(}-{\mathrm{d}}X^{r}\partial_{r}b^{j}_{i}(X)c_{j}+d^{k}_{ij}(X){\mathrm{d}}X^{j}c_{k}\Big{)}{\mathrm{d}}X^{i}=0,$

and since this has to hold for all $X$ and $c$, we finally get

$d^{k}_{ij}=\partial_{j}b^{k}_{i}.$

Hence, we have shown that

$\displaystyle\delta X^{i}$
$\displaystyle=-\epsilon\alpha^{ij}(X)c_{j}+O(\epsilon^{2}),$
$\displaystyle\delta\eta_{i}$
$\displaystyle={\mathrm{d}}c_{i}+\epsilon\Big{(}\partial_{i}\alpha^{jk}(X)\eta_{j}c_{k}+{\mathrm{d}}(b^{j}_{i}(X)c_{j})\Big{)}+O(\epsilon^{2}),$

which, after redefinition $c_{i}\mapsto c_{i}-\epsilon
b^{j}_{i}(X)c_{j}+O(\epsilon^{2})$, gives the first two equations in Lemma
5.4.2. For the last equation, we recall that the BRST operator on $c$ must be
quadratic in $c$, so its general form is

$\delta c_{i}=\frac{1}{2}\epsilon f^{jk}_{i}(X)c_{j}c_{k}+O(\epsilon^{2}).$

To determine the _structure_ functions $f^{jk}_{i}$, we can compute
$\delta^{2}$. Note that $\delta^{2}X^{i}=\delta^{2}c_{i}=O(\epsilon^{2})$. On
the other hand,

$\delta^{2}\eta_{i}=\epsilon\Bigg{(}\frac{1}{2}{\mathrm{d}}\big{(}f^{jk}_{i}(X)c_{j}c_{i}\big{)}+\partial_{i}\alpha^{jk}(X){\mathrm{d}}c_{j}c_{k}\Bigg{)}+O(\epsilon^{2})\\\
=\epsilon\Bigg{(}\big{(}f^{jk}_{i}(X)+\partial_{i}\alpha^{jk}(X)\big{)}{\mathrm{d}}c_{j}c_{k}+\frac{1}{2}{\mathrm{d}}X^{r}\partial_{r}f^{jk}_{i}(X)c_{j}c_{k}\Bigg{)}+O(\epsilon^{2}).$

At a critical point (where ${\mathrm{d}}X^{i}=O(\epsilon^{2})$) the third
summand of the last equation vanishes. Thus, $\delta^{2}=O(\epsilon^{2})$ at
critical points implies that

$f_{i}^{jk}=-\partial_{i}\alpha^{jk},$

which proves the last equation in Lemma 5.4.2, Note also that

$\delta^{2}\eta_{i}=-\frac{1}{2}\epsilon{\mathrm{d}}x^{r}\partial_{r}\partial_{i}\alpha^{jk}(X)c_{j}c_{k},$

which is zero (not only at critical points) whenever $\alpha$ is at most
linear. ∎

Next, we want to extend deformations beyond the first order in $\epsilon$.
Even without knowing the following terms, we can already state the following
Lemma.

######  Lemma 5.4.3.

$\delta^{2}=O(\epsilon^{3})$ at critical points only if $\alpha$ is Poisson.

###### Proof.

Note that we have

$\delta^{2}X^{i}=-\epsilon^{2}\Bigg{(}\alpha^{rk}(X)c_{k}\partial_{r}\alpha^{ij}(X)c_{j}+\frac{1}{2}\alpha^{ij}(X)\partial_{i}\alpha^{rj}(X)c_{r}c_{j}\Bigg{)}+O(\epsilon^{3}).$

Since this has to hold for all $c$, we get the Jacobi identity for $\alpha$. ∎

######  Remark 5.4.4.

It is in fact possible to prove, under the assumption that $\alpha$ is
Poisson, that this deformation is not only infinitesimal. In particular, we
have the following theorem.

######  Theorem 5.4.5.

Given a Poisson bivector field $\alpha$, the odd vector field

$\displaystyle\delta X^{i}$ $\displaystyle=-\epsilon\alpha^{ij}(X)c_{j},$
$\displaystyle\delta\eta_{i}$
$\displaystyle={\mathrm{d}}c_{i}+\epsilon\partial_{i}\alpha^{jk}(X)\eta_{j}c_{k},$
$\displaystyle\delta c_{i}$
$\displaystyle=-\frac{1}{2}\epsilon\partial_{i}\alpha^{jk}(X)c_{j}c_{k},$

is cohomological for $\alpha$ at most linear or at critical points for all
$\epsilon$. Moreover,

$S:=\int_{\Sigma}\Bigg{(}\eta_{i}\,{\mathrm{d}}X^{i}+\frac{1}{2}\epsilon\alpha^{ij}(X)\eta_{i}\eta_{j}\Bigg{)}$

is $\delta$-closed for all $\epsilon$.

######  Exercise 5.4.6.

Prove Theorem 5.4.5.

######  Remark 5.4.7.

The geometrical meaning of Theorem 5.4.5 is that there is a distribution of
vector fields on $\mathcal{M}$ under which the action is invariant. In
general, this distribution is involutive only on the submanifold of critical
points of $S$. It is involutive on the whole of $\mathcal{M}$ whenever
$\alpha$ is at most linear and in this case it can be regarded as the free,
infinitesimal action of a Lie algebra. The action $S$ can be generalized to
the case when one wants to consider a Poisson manifold $(M,\alpha)$ instead of
$\mathbb{R}^{n}$. For this, one regards $X$ as a map $\Sigma\to M$ and, for a
given map $X$, $\eta$ is taken to be a section of $T^{*}\Sigma\otimes
X^{*}T^{*}M$. If $\langle\enspace,\enspace\rangle$ denotes the canonical
pairing between the tangent and cotangent bundle of $M$ and by
$\alpha^{\sharp}$ the bundle map $T^{*}M\to TM$ induced by the Poisson
bivector field $\alpha$ (see also Section 3.2.1), we can write

$S=\int_{\Sigma}\Bigg{(}\langle\eta,{\mathrm{d}}X\rangle+\frac{1}{2}\epsilon\langle\eta,\alpha^{\sharp}\eta\rangle\Bigg{)}.$

####  5.4.2. Observables

For the case when $\Sigma=\mathbb{H}^{2}$, $c$ has to vanish on the boundary.
This implies that $\delta X(u)=0$ for $u\in\partial\Sigma$. Hence, we get

$\mathcal{O}_{f_{1},\ldots,f_{k};u_{1},\ldots,u_{k}}:=f_{1}(X(u_{1}))\dotsm
f_{k}(X(u_{k})),\\\ f_{1},\ldots,f_{k}\in C^{\infty}(\mathbb{R}^{n}),\quad
u_{1},\ldots,u_{k}\in\partial\Sigma\cong\mathbb{R},\quad u_{1}<\dotsm<u_{k},$

are observables, i.e. $\delta$-closed functions. In [CF00] it was shown that
with the gauge-fixing ${\mathrm{d}}^{*}\eta=0$ for the Euclidean metric on
$\Sigma$, one has

$\langle\mathcal{O}_{f_{1},\ldots,f_{k};u_{1},\ldots,u_{k}}\rangle(x)=f_{1}\star\dotsm\star
f_{k}(x),$

where $\langle\enspace\rangle(x)$ denotes the expectation value for
$X(\infty)=x$ (and expanding only around the trivial critical solution $X=x$,
$\eta=0$) while $\star$ denotes Kontsevich’s star product for the given
Poisson structure. In the next section will derive this result for the case
when $\alpha$ is at most linear so that the BRST formalism is available.

###  5.5. Deformation quantization for affine Poisson structures

An affine Poisson structure on $\mathbb{R}^{n}$ is given by a Poisson bivector
field $\alpha$ which is at most linear. The linear part $\alpha$ gives the
dual of $\mathbb{R}^{n}$ a Lie algebra structure while the constant part is a
2-cocycle in the Lie algebra cohomology with trivial coefficients. Let us
denote this Lie algebra by $\mathfrak{h}$. The fields of the Poisson sigma
model for an affine Poisson structure are then a map
$X\colon\Sigma\to\mathfrak{h}^{*}$ and a 1-form $\eta$ on $\Sigma$ with values
in $\mathfrak{h}$. Since $\mathfrak{h}$ is a Lie algebra, we can regard $\eta$
as a connection 1-form on the trivial principal bundle $P$ over $\Sigma$ (with
gauge group any Lie group whose Lie algebra is $\mathfrak{h}$). For
definiteness, we will fix $\Sigma$ to be $\mathbb{H}^{2}$ and we will require
$\eta$ to vanish at infinity and on the boundary. The action is then given by

$S=\int_{\Sigma}\Bigg{(}\eta_{i}\,{\mathrm{d}}X^{i}+\frac{1}{2}\alpha^{ij}(X)\eta_{i}\eta_{j}\Bigg{)},$

where

$\alpha^{ij}(x)=\chi^{ij}+x^{k}f^{ij}_{k}$

is the given affine Poisson structure on $\mathfrak{h}^{*}$. Using integration
by parts, we can also rewrite it as

$S=\int_{\Sigma}\Bigg{(}\langle
X,F_{\eta}\rangle+\frac{1}{2}\chi(\eta,\eta)\Bigg{)},$

where $\langle\enspace,\enspace\rangle$ denotes the canonical pairing between
$\mathfrak{h}$ and $\mathfrak{h}^{*}$ while

$(F_{\eta})_{i}={\mathrm{d}}\eta_{i}+\frac{1}{2}f_{i}^{jk}\eta_{j}\eta_{k}$

is the curvature 2-form of the connection 1-form $\eta$. Note that there is a
Lie algebra $\mathfrak{g}$, which as a vector space consists of functions
$\Sigma\to\mathfrak{h}$ vanishing at infinity and on the boundary, that acts
on the space of fields $\mathcal{M}$ and leaves the action invariant. The BRST
operator on $\mathcal{M}\times\Pi\mathfrak{g}$ has the form as in Theorem
5.4.5 with $\epsilon=1$. Geometrically, we can regard $\mathfrak{g}$ as the
Lie algebra of infinitesimal gauge transformations of the principal bundle
$P$; the field $\eta$ actually transforms as a connection 1-form, while $X$
transforms as a section of the coadjoint bundle in case $\chi=0$. For
$\chi\not=0$, we can regard $X\oplus 1$ as a section of the coadjoint bundle
for the Lie algebra $\widehat{\mathfrak{h}}\cong\mathfrak{h}\oplus\mathbb{R}$
obtained by central extension of $\mathfrak{h}$ through $\chi$. The BRST
operator on $\Pi\mathfrak{g}^{*}\times\mathfrak{g}^{*}$ has the usual form
(5.3.4).

####  5.5.1. Gauge-fixing and Feynman diagrams

Choose a metric on $\Sigma$ and define the gauge-fixing function
$F(X,\eta)={\mathrm{d}}^{*}\eta$. The gauge-fixing fermion is given by

$\Psi_{F}=\int_{\Sigma}\langle\bar{c},{\mathrm{d}}^{*}\eta\rangle$

and the gauge-fixed action is given by

$S_{F}=\int_{\Sigma}\Bigg{(}\eta_{i}\,{\mathrm{d}}X^{i}+\frac{1}{2}\alpha^{ij}(X)\eta_{i}\eta_{j}+\lambda^{i}{\mathrm{d}}^{*}\eta_{i}-\bar{c}^{k}{\mathrm{d}}^{*}\Big{(}{\mathrm{d}}c_{k}+\partial_{k}\alpha^{ij}(X)\eta_{i}c_{j}\Big{)}\Bigg{)}.$

Fix the value of $X$ at infinity to be given by $x$. We write $X=x+\xi$, where
the field $\xi$ has to vanish at infinity. We can observe that $S_{F}$ has the
same form as (5.2.7) (with $y$ the collection of $\xi,\bar{c},\lambda$ and $z$
the collection of $\eta,c$). Hence, we can write $S_{F}=S_{0}+S_{1}$ with

$\displaystyle S_{0}$
$\displaystyle=\int_{\Sigma}\Bigg{(}\eta_{i}\,{\mathrm{d}}X^{i}+\lambda^{i}{\mathrm{d}}^{*}\eta_{i}-\bar{c}^{k}{\mathrm{d}}^{*}{\mathrm{d}}c_{k}\Bigg{)},$
$\displaystyle S_{1}$
$\displaystyle=\int_{\Sigma}\Bigg{(}\frac{1}{2}\alpha^{ij}(x+\xi)\eta_{i}\eta_{j}-\bar{c}^{k}{\mathrm{d}}^{*}\Big{(}\partial_{k}\alpha^{ij}(x+\xi)\eta_{i}c_{j}\Big{)}\Bigg{)},$

and regard $S_{1}$ as a perturbation of $S_{0}$. if we consider superfields
$\boldsymbol{\xi}$ and $\boldsymbol{\eta}$ as in (5.3.8), we can write

$S_{1}=\int_{\Sigma}\frac{1}{2}\alpha^{ij}(x+\boldsymbol{\xi})\boldsymbol{\eta}_{i}\boldsymbol{\eta}_{j},$

where integration on $\Sigma$ is understood to select the 2-form component.

######  Remark 5.5.1.

This shows that as long as the considered observables can be written as
functions of the superfields and expectation values are computed in terms of
the superpropagators (5.3.10). If we denote the superpropagator graphically as
an arrow from $\boldsymbol{\eta}$ to $\boldsymbol{\xi}$, the perturbation
$S_{1}$ is represented by the two vertices as in Figure 5.5.1 with the
bivalent vertex corresponding to $\alpha^{ij}(x)=\chi^{ij}+x^{k}f^{ij}_{k}$
and the trivial vertex corresponding to the structure constants.

Figure 5.5.1. The two vertices

Let us now consider the observable
$\mathcal{O}_{f_{1},\ldots,f_{k};u_{1},\ldots,u_{k}}$ as in Section (5.4.2).
As the evaluation point, i.e., integration along a 0-cycle, is understood to
select the 0-form component of a differential form, we can write

$\mathcal{O}_{f_{1},\ldots,f_{k};u_{1},\ldots,u_{k}}:=f_{1}(x+\boldsymbol{\xi}(u_{1}))\dotsm
f_{k}(x+\boldsymbol{\xi}(u_{k})),\\\ f_{1},\ldots,f_{k}\in
C^{\infty}(\mathbb{R}^{n}),\quad
u_{1},\ldots,u_{k}\in\partial\Sigma\cong\mathbb{R},\quad u_{1}<\dotsm<u_{k}.$

Computing then the expectation value
$\langle\mathcal{O}_{f_{1},\ldots,f_{k};u_{1},\ldots,u_{k}}\rangle(x)$, we
only need the superpropagator. The Feynman diagrams then have three kind of
vertices:

  1. (1)

bivalent vertices in the upper half-plane corresponding to $\alpha^{ij}(x)$,

  2. (2)

trivalent vertices in the upper half-plane corresponding to $f_{k}^{ij}$,

  3. (3)

$\ell$-valent vertices with $\ell\geq 0$ with only incoming arrows at one of
the boundary points $u_{i}$ corresponding to the $\ell$-th derivative of
$f_{i}$.

Recall that the normal ordering excludes all graphs containing a tadpole (i.e.
an edge starting and ending at the same vertex). The combinatorics prevents
automatically vacuum subgraphs (see Figure 5.5.2 for examples).

Figure 5.5.2. Example of an allowed graph and a non-allowed graph.

For the case when $k=2$ and considering the gauge-fixing
${\mathrm{d}}^{*}\eta=0$ with respect to the Euclidean metric on
$\mathbb{H}^{2}$, i.e. with the superpropagator determined by the 1-form
$\vartheta$ as in Lemma 5.3.15, we get

(5.5.1) $\langle\mathcal{O}_{f,g;0,1}\rangle(x)=f\star g(x),$

where $\star$ denotes Kontsevich’s star product for the given affine Poisson
structure.

######  Exercise 5.5.2.

Prove (5.5.1).

####  5.5.2. Independence of the evaluation point

We want to give a formal proof of the independence of the expectation values
of $\mathcal{O}_{f_{1},\ldots,f_{k};u_{1},\ldots,u_{k}}$ from the points
$u_{1},\ldots,u_{k}$. Note first that

$f(X(v))-f(X(u))=\int_{u}^{v}{\mathrm{d}}X^{i}\partial_{i}f(X)=\int_{u}^{v}\Big{(}{\mathrm{d}}X^{i}+{\mathrm{d}}^{*}\lambda^{i}\Big{)}\partial_{i}f(X)-\delta\Phi,$

where

$\Phi:=\int_{u}^{v}{\mathrm{d}}^{*}\bar{c}\partial_{i}f(X).$

Let $(\omega_{r})$ be a sequence of 1-forms on $\Sigma$ vanishing on the
boundary and at infinity that converges to the measure concentrated on the
interval $(u,v)\in\Sigma$. Denoting by $(a,b)$, with $b\geq 0$, the
coordinates on $\Sigma=\mathbb{H}^{2}$, a possible choice for this sequence is

$\omega_{r}(a,b)=rb\exp(-rb^{2}/2)\chi_{r}(a){\mathrm{d}}a,$

where $(\chi_{r})$ is a sequence of smooth, compactly supported functions
converging almost everywhere to the characteristic function of the interval
$(u,v)$. Let $Y_{f,r}$ be the local vector field on $\widetilde{\mathcal{M}}$
corresponding to the infinitesimal displacement of $\eta_{i}$ by
$\omega_{r}\partial_{i}f(X)$. Then

$f(X(v))-f(X(u))=\lim_{r\to\infty}Y_{f,r}(S_{F})-\delta\Phi.$

If $\mathcal{O}$ is a BRST-observable depending on the fields outside the
closed interval $[u,v]$, we get

$\langle(f(X(v))-f(X(u)))\mathcal{O}\rangle=\mathrm{i}\hbar\lim_{r\to\infty}\langle
Y_{f,r}(\mathcal{O})\rangle-\langle\delta(\Phi\mathcal{O})\rangle=0.$

####  5.5.3. Associativity

The independence of the evaluation points gives us

$\lim_{v\to u^{+}}\langle\mathcal{O}_{f,g,h;u,v,w}\rangle_{0}(x)=\lim_{v\to
w^{-}}\langle\mathcal{O}_{f,g,h;u,v,w}\rangle_{0}(x).$

The left hand side corresponds intuitively to evaluating first the expectation
value of $\mathcal{O}_{f,g;u,v}$, then replacing the result at $u$ and finally
computing the expectation value of
$\mathcal{O}_{\langle\mathcal{O}_{f,g,;u,v}\rangle_{0},h;w}$. The result is
then $(f\star g)\star h$. Repeating the computation on the right hand side, we
get $f\star(g\star h)$. This rather formal argument explains why one should
expect the star product defined by the Poisson sigma model to be associative.

###  5.6. The general construction

We want to also state the theorem for general Poisson structures on
$\mathbb{R}^{d}$. However, we will not provide a proof of the general
construction here since one needs the more general gauge formalism provided by
Batalin and Vilkovisky.

####  5.6.1. The theorem of Cattaneo–Felder

######  Theorem 5.6.1 (Cattaneo–Felder[CF00]).

Consider the Poisson manifold $(\mathbb{R}^{d},\pi)$ with Poisson structure
$\pi$ together with the corresponding Poisson sigma model on a disk $D$. The
fields are then given by a map $X\colon D\to\mathbb{R}^{d}$ and a 1-form
$\eta\in\Gamma(D,T^{*}D\otimes X^{*}T^{*}\mathbb{R}^{d})$ with boundary
condition $\eta\big{|}_{S^{1}}=0$. Moreover, let $0,1$ and $\infty$ be
cyclically ordered points on $\partial D=S^{1}$, i.e. if we start at $0$ and
move counterclockwise on $S^{1}$ we will first meet $1$ and then $\infty$ (see
Figure 5.6.1), and impose the condition $\eta(\infty)=0$. Then Kontsevich’s
star product between two functions $f,g\in C^{\infty}(\mathbb{R}^{d})$
evaluated at the point $x=X(\infty)\in\mathbb{R}^{d}$ is given by

$f\star
g(x)=\int_{X(\infty)=x}f(X(1))g(X(0))\exp(\mathrm{i}S(X,\eta)/\hbar)\mathscr{D}[X]\mathscr{D}[\eta],$

where the right hand side should be understood by perturbative expansion in
terms of Feynman diagrams.

######  Remark 5.6.2.

The global field theoretic approach was given in [CMW20] using cutting and
gluing methods for manifolds with boundary by introducing a formal global
action as more generally developed in [CMW19]. Moreover, also the case for
manifolds with corners was covered. The main gauge formalism that was used is
known as the _BV-BFV formalism_ [CMR14, CMR17, CM20], which can bee seen as a
extension of the Batalin–Vilkovisky formalism for manifolds with boundary. A
first approach to a global formulation of the Poisson sigma model on closed
manifolds using the Batalin–Vilkovisky formalism was given in [BCM12].

$0$$1$$\infty$$D$ Figure 5.6.1. Cyclically ordered points on $\partial
D=S^{1}$

####  5.6.2. Other similar constructions

The way of using quantum field theory to obtain certain mathematical
constructions has been proven to be a very interesting and deep method
relating seemingly purely mathematical constructions to physics. Other
examples include Witten’s famous construction to obtain the 4-manifold
invariants described by Donaldson [Don83] using a certain deformation of the
supersymmetric _Yang–Mills_ action and a special type of observables [Wit88].
Computing the expectation value of this observable with respect to the theory
formulated by this special action, one can recover the Donaldson polynomials
of the given 4-manifold. Another important and surprising result of Witten
[Wit89] was that the expectation value of a _Wilson loop_ observable
representing a given knot with respect to the _Chern–Simons_ action [CS74,
AS91, AS94] will give the _Jones polynomial_ [Jon85] of the knot, which is an
important knot invariant. These results have lead to many mathematical
conjectures, deeper insights in physics and created a whole new perspective
towards the interplay between mathematics and quantum field theory.

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