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# Computation of Kontsevich Weights of Connection and Curvature Graphs for
Symplectic Poisson Structures

Nima Moshayedi  Institut für Mathematik
Universität Zürich
Winterthurerstrasse 190 CH-8057 Zürich<EMAIL_ADDRESS>and  Fabio
Musio  Institut für Mathematik
Universität Zürich
Winterthurerstrasse 190 CH-8057 Zürich<EMAIL_ADDRESS>
###### Abstract.

We give an explicit computation of weights of Kontsevich graphs which arise
from connection and curvature terms within the globalization picture as in
[12] for symplectic manifolds. Moreover, we consider the case of a cotangent
bundle, which will simplify the curvature expression significantly.

###### Contents

  1. 1 Introduction
    1. 1.1 Motivation
    2. 1.2 Notions of Formal Geometry
  2. 2 Computation of Kontsevich Weights
    1. 2.1 Case 1: No Boundary Vertices
      1. 2.1.1 Step 1
      2. 2.1.2 Step 2
      3. 2.1.3 Step 3
      4. 2.1.4 Putting everything together
    2. 2.2 Case 2: One Boundary Vertex
      1. 2.2.1 Step 1
      2. 2.2.2 Step 2
      3. 2.2.3 Step 3
      4. 2.2.4 Putting everything together
    3. 2.3 Case 3: Two Boundary Vertices
  3. 3 Including the Weights
    1. 3.1 The Product $P(\mathsf{T}\phi^{*}\pi)$
    2. 3.2 The Connection 1-form $A(R,\mathsf{T}\phi^{*}\pi)$
    3. 3.3 The Curvature 2-form $F(R,R,\mathsf{T}\phi^{*}\pi)$
    4. 3.4 The lifted Curvature 2-form $F(\overline{R},\overline{R},\mathsf{T}\overline{\phi}^{*}\pi)$
  4. A Binomial Sums
    1. A.1 $B(n)$
    2. A.2 $C(n)$
    3. A.3 $A(n)$

##  1\. Introduction

###  1.1. Motivation

In [17] Kontsevich proved that the differential graded Lie algebra
(DGLA)111Endowed with the zero differential and the Schouten–Nijenhuis
bracket. of multivector fields on an open subset $M\subset\mathbb{R}^{d}$ is
$L_{\infty}$-quasi-isomorphic to the DGLA222Endowed with the Hochschild
differential and the Gerstenhaber bracket. of multidifferential operators on
functions on $M$, i.e. there exists an $L_{\infty}$-quasi-isomorphism

(1) $\mathcal{U}\colon T_{poly}(M)\to D_{poly}(M),$

such that its zeroth Taylor component $\mathcal{U}^{(0)}$ is given by the
Hochschild–Kostant–Rosenberg map. This result is known as the _formality_
theorem. If one restricts to the case of bivector fields and bidifferential
operators, one can recover deformation quantization for Poisson manifolds. The
resulting star product was also constructed in [17] by an explicit formula. In
[7, 5, 12] a globalization picture was presented for this star product on any
Poisson manifold $M$, including the construction of the local star product by
using techniques of field theory, in particular the _Poisson Sigma Model_ [16,
20, 4]. In [12] this construction was extended to manifolds with boundary as
in the _BV-BFV formalism_ [8, 9, 10] which is a perturbative quantum gauge
formalism compatible with cutting and gluing. A similar approach, as the one
presented by Fedosov in [13] for symplectic manifolds, was used, by
considering notions of formal geometry. In particular, one starts with a
_formal exponential map_ $\phi$ on the manifold $M$ and constructs a _flat_
connection $D_{G}$, called the _classical Grothendieck connection_ , on the
completed symmetric algebra of the cotangent bundle
$\widehat{\textnormal{Sym}}(T^{*}M)$. This construction can be deformed to the
_Weyl bundle_ $\widehat{\textnormal{Sym}}(T^{*}M)[[\hbar]]$ and, as it was
shown in [5, 6, 12], it induces a similar equation as the one in Fedosov’s
construction. In [12] it was shown that the different terms of this equation
are given by a certain class of graphs. We want to give an explicit
computation of the weights for these graphs. Let $G_{n,m}$ denote the set of
all _admissible graphs_ as in [17] with $n$ vertices in the bulk of the upper
half-plane $\mathbb{H}:=\\{z\in\mathbb{C}|\;\text{Im}(z)>0\\}$ and $m$
vertices on $\mathbb{R}$. Define a map

(2) $\mathcal{U}_{\Gamma}\colon\bigwedge^{n}T_{poly}(M)\to D_{poly}(M)[1-n]$

using the $L_{\infty}$-morphism $\mathcal{U}$. Let $\pi$ be a Poisson
structure on $\mathbb{R}^{d}$ and let $\xi,\zeta$ be any two vector fields on
$\mathbb{R}^{d}$. Let us define

(3) $\displaystyle P(\pi)$ $\displaystyle:=\sum\limits_{n\geq
0}\,\sum\limits_{\Gamma\in
G_{n,2}}\frac{\hbar^{n}}{n!}w_{\Gamma}\mathcal{U}_{\Gamma}(\pi\wedge\cdots\wedge\pi),$
(4) $\displaystyle A(\xi,\pi)$ $\displaystyle:=\sum\limits_{n\geq
0}\,\sum\limits_{\Gamma\in
G_{n+1,1}}\frac{\hbar^{n}}{n!}w_{\Gamma}\mathcal{U}_{\Gamma}(\xi\wedge\pi\wedge\cdots\wedge\pi),$
(5) $\displaystyle F(\xi,\zeta,\pi)$ $\displaystyle:=\sum\limits_{n\geq
0}\,\sum\limits_{\Gamma\in
G_{n+2,0}}\frac{\hbar^{n}}{n!}w_{\Gamma}\mathcal{U}_{\Gamma}(\xi\wedge\zeta\wedge\pi\wedge\cdots\wedge\pi),$

where $w_{\Gamma}\in\mathbb{R}$ denotes the Kontsevich weight of the graph
$\Gamma$. The term (3) represents Kontsevich’s star product, (4) represents
the _deformed Grothendieck connection_ $\mathcal{D}_{G}:=D_{G}+O(\hbar)$ (see
construction below) and (5) its curvature. Let us emphasize a bit more on the
formal geometry construction.

###  1.2. Notions of Formal Geometry

Let $M$ be a smooth manifold and let $\phi\colon U\to M$ be a map where
$U\subset TM$ is an open neighbourhood of the zero section. For $x\in M,y\in
T_{x}M\cap U$ we write $\phi_{x}(y):=\phi(x,y)$. We say that $\phi$ is a
_generalized exponential map_ if for all $x\in M$ we have that
$\phi_{x}(0)=x$, and ${\mathrm{d}}\phi_{x}(0)=\mathrm{id}_{T_{x}M}$. In local
coordinates we can write

(6)
$\phi_{x}^{i}(y)=x^{i}+y^{i}+\frac{1}{2}\phi_{x,jk}^{i}y^{j}y^{k}+\frac{1}{3!}\phi^{i}_{x,jk\ell}y^{j}y^{k}y^{\ell}+\dotsm$

where the $x^{i}$ are coordinates on the base and the $y^{i}$ are coordinates
on the fibers. We identify two generalized exponential maps if their jets
agree to all orders. A _formal exponential map_ is an equivalence class of
generalized exponential maps. It is completely specified by the sequence of
functions $\left(\phi^{i}_{x,i_{1}\ldots i_{k}}\right)_{k=0}^{\infty}$. By
abuse of notation, we will denote equivalence classes and their
representatives by $\phi$. From a formal exponential map $\phi$ and a function
$f\in C^{\infty}(M)$, we can produce a section
$\sigma\in\Gamma(\widehat{\textnormal{Sym}}(T^{*}M))$ by defining
$\sigma_{x}=\textsf{T}\phi_{x}^{*}f$, where T denotes the Taylor expansion in
the fiber coordinates around $y=0$ and we use any representative of $\phi$ to
define the pullback. We denote this section by $\textsf{T}\phi^{*}f$, it is
independent of the choice of representative, since it only depends on the jets
of the representative.

As it was shown [5, 3, 11], one can define a _flat_ connection $D_{G}$ on
$\widehat{\textnormal{Sym}}(T^{*}M)$ with the property that $D_{G}\sigma=0$ if
and only if $\sigma=\textsf{T}\phi^{*}f$ for some $f\in C^{\infty}(M)$. As
already mentioned before, this connection is called the _classical
Grothendieck connection_. In fact, $D_{G}={\mathrm{d}}_{x}+L_{R}$ where
$R\in\Omega^{1}(M,{\mathrm{Der}}(\widehat{\textnormal{Sym}}(T^{*}M)))$ is a
1-form with values in derivations of $\widehat{\textnormal{Sym}}(T^{*}M)$,
which we identify with $\Gamma(TM\otimes\widehat{\textnormal{Sym}}(T^{*}M))$.
We have denoted by ${\mathrm{d}}_{x}$ the de Rham differential on $M$ and by
$L$ the Lie derivative. In coordinates we have

(7) $R(\sigma)_{\ell}=-\frac{\partial\sigma}{\partial
y^{j}}\left(\left(\frac{\partial\phi}{\partial
y}\right)^{-1}\right)_{k}^{j}\frac{\partial\phi^{k}}{\partial x^{\ell}}.$

Define $R(x,y):=R_{\ell}(x,y){\mathrm{d}}x^{\ell}$,
$R_{\ell}(x,y):=R_{\ell}^{j}(x,y)\frac{\partial}{\partial y^{j}}$,
$R^{j}(x,y):=R^{j}_{\ell}(x,y){\mathrm{d}}x^{\ell}$, and

(8) $R_{\ell}^{j}=-\left(\left(\frac{\partial\phi}{\partial
y}\right)^{-1}\right)_{k}^{j}\frac{\partial\phi^{k}}{\partial
x^{\ell}}=-\delta_{\ell}^{j}+O(y).$

For $\sigma\in\Gamma(\widehat{\textnormal{Sym}}(T^{*}M))$, $L_{R}\sigma$ is
given by the Taylor expansion (in the $y$ coordinates) of

$-{\mathrm{d}}_{y}\sigma\circ({\mathrm{d}}_{y}\phi)^{-1}\circ{\mathrm{d}}_{x}\phi\colon\Gamma(TM)\to\Gamma(\widehat{\textnormal{Sym}}(T^{*}M)),$

where we denote by ${\mathrm{d}}_{y}$ the de Rham differential on the fiber.
This shows that $R$ does not depend on the choice of coordinates. One can
generalize this also for any fixed vector
$\xi=\xi^{i}(x)\frac{\partial}{\partial x^{i}}\in T_{x}M$, instead of just
considering the de Rham differential ${\mathrm{d}}_{x}$, by

(9) $D_{G}^{\xi}=\xi+\widehat{\xi},$

where

(10)
$\widehat{\xi}(x,y)=\iota_{\xi}R(x,y)=\xi^{i}(x)R_{\ell}^{j}(x,y)\frac{\partial}{\partial
y^{j}}.$

Here $\xi^{i}(x)$ would replace the 1-form part ${\mathrm{d}}x^{i}$.

This paper is based on the master thesis [19].

### Acknowledgements

We would like to thank Alberto Cattaneo and Konstantin Wernli for useful
discussions. This research was (partly) supported by the NCCR SwissMAP, funded
by the Swiss National Science Foundation, and by the SNF grant No.
200020_172498/1.

##  2\. Computation of Kontsevich Weights

Let $(M,\omega)$ be a symplectic manifold regarded as a Poisson manifold with
Poisson structure $\pi$ induced by the symplectic form $\omega$. Moreover let
$\phi\colon TM\supset U\rightarrow M$ be a formal exponential map and denote
by $\mathsf{T}$ the Taylor expansion in fiber coordinates around zero.
Anticipating the computation of the star product $P(\mathsf{T}\phi^{*}\pi)$,
the connection $1$-form $A(R,\mathsf{T}\phi^{*}\pi)$ and its curvature
$2$-form $F(R,R,\mathsf{T}\phi^{*}\pi)$ as in [12], we will explicitly compute
the Kontsevich weights of three families of graphs in this section. Throughout
the paper we use the harmonic angle function

(11)
$\displaystyle\varphi(u,v)=\text{arg}\left(\frac{v-u}{v-\bar{u}}\right)=\frac{1}{2i}\text{log}\left(\frac{(v-u)(\bar{v}-u)}{(v-\bar{u})(\bar{v}-\bar{u})}\right)$

which measures the angle on $\mathbb{H}\cup\mathbb{R}$ as depicted in Figure
1.

$u$$v$$\varphi$ Figure 1. Illustartion of the angle function $\varphi$.

The propagator used in the computation of the Kontsevich weights is then
simply given by ${\mathrm{d}}\varphi(u,v)$ and is usually called the
_Kontsevich propagator_. Now let $\Gamma\in G_{n,m}$ be an admissible graph
with $n$ vertices of first type, $m$ vertices of second type and $2n+m-2$
edges. We use this propagator to compute the Kontsevich weight [17]
$w_{\Gamma}$ of $\Gamma$ as

(12) $\displaystyle w_{\Gamma}=\int\limits_{\bar{C}_{n,m}}\omega_{\Gamma}.$

Here $\bar{C}_{n,m}$ denotes the _Fulton–MacPherson/Axelrod–Singer (FMAS)
compactification_ [15, 1] of the configuration space $C_{n,m}$ of $n$ points
in $\mathbb{H}$ and $m$ points on $\mathbb{R}$ modulo scaling and translation.
Let us briefly recall the construction of the needed configuration spaces. We
define the open configuration space

(13)
$\mathrm{Conf}_{n,m}=\\{(x_{1},\ldots,x_{n},q_{1},\ldots,q_{m})\in\mathbb{H}^{n}\times\mathbb{R}^{m}\mid
x_{i}\not=x_{j},\,\forall i\not=j,\,q_{1}<\ldots<q_{m}\\}.$

The 2-dimensional real Lie group of orientation preserving affine
transformations of the real line

(14) $G^{(1)}=\\{z\mapsto az+b\mid a,b\in\mathbb{R},\,a>0\\}$

acts freely on $\mathrm{Conf}_{n,m}$. One can check that the quotient space
$C_{n,m}:=\mathrm{Conf}_{n,m}/G^{(1)}$ is in fact a smooth manifold of
dimension $2n+m-2$. The differential form $\omega_{\Gamma}$ on $\bar{C}_{n,m}$
is given by

(15)
$\displaystyle\omega_{\Gamma}=\frac{1}{(2\pi)^{2n+m-2}}\bigwedge\limits_{\text{edges
$e$}}{\mathrm{d}}\varphi_{e},$

where the wedge product is over all $2n+m-2$ edges $e$ of the graph $\Gamma$.
Let $n\geq 2$ and define

(16) $\mathrm{Conf}_{n}=\\{(x_{1},\ldots,x_{n})\in\mathbb{C}^{n}\mid
x_{i}\not=x_{j},\,\forall i\not=j\\}.$

We have an action on $\mathrm{Conf}_{n}$ by the 3-dimensional Lie group

(17) $G^{(2)}=\\{z\mapsto az+b\mid a\in\mathbb{R},\,b\in\mathbb{C},\,a>0\\}.$

Again, one can check that the quotient space
$C_{n}:=\mathrm{Conf}_{n}/G^{(2)}$ is a smooth manifold of dimension $2n-3$.
Also here, we will denote its FMAS compactification by $\bar{C}_{n}$. We refer
to [17] for a more detailed construction. To simplify the notation we will use
graphical language, where the figure below corresponds to a factor of
${\mathrm{d}}(\varphi(u,v)^{n})/(2\pi)^{n}$ in $w_{\Gamma}$. If there is no
$n$ above the arrow it simply means that $n=1$.

We know that the dimension of the configuration space $C_{n,m}$ is $2n+m-2$,
and since we work on a symplectic manifold $M$ (with Darboux coordinates
around each point $x\in M$), a vertex of first type is either a vertex
representing the tensor $\mathsf{T}\phi^{*}_{x}\pi$, which we will call a
$\mathsf{T}\phi^{*}_{x}\pi$-vertex, with precisely two outgoing and no
incoming edges, or a vertex representing the 1-form $R$, which we will call an
$R$-vertex, with precisely one outgoing edge and arbitrarily many incoming
edges [12, 18]. So we may write $n=p+r$, where $p$ is the number of
$\mathsf{T}\phi^{*}_{x}\pi$-vertices and $r$ is the number of $R$-vertices. We
then have that $\text{deg}(\omega_{\Gamma})=2p+r$, and in order for the
integral (12) not to vanish, we must have that $\omega_{\Gamma}$ is a top
form, i.e. that $2n+m-2=2p+r$. This then implies that

(18) $\displaystyle r+m=2$

So we have to distinguish three different cases, namely $(r,m)=(2,0)$,
$(r,m)=(1,1)$ and $(r,m)=(0,2)$, which we will treat separately in what
follows.

######  Remark 2.1.

Actually, we will see below that all the integrals over the non-compactified
configuration spaces $C_{n,m}$ of the graphs we are considering converge and
are thus finite. So it is not necessary to work with the compactifications.

###  2.1. Case 1: No Boundary Vertices

We will first treat the case $(r,m)=(2,0)$, i.e. the case where we have no
boundary vertices and exactely two $R$-vertices. In that case we get a family
of graphs $(\Gamma_{n})_{n\geq 0}$, where $\Gamma_{n}$ is the graph with $n$
wedges as in Figure 2(a) (stemming from $n$
$\mathsf{T}\phi^{*}_{x}\pi$-vertices) attached to the wheel as in Figure 2(b)
(stemming from the two $R$-vertices).

Figure 2. Graphs in the case $(r,m)=(2,0)$ consist of: (a) wedges stemming
from $\mathsf{T}\phi^{*}_{x}\pi$-vertices attached to (b) a wheel stemming
from the two $R$-vertices.

Examples of the graphs $\Gamma_{n}$ are given in Figure 3 below for $n=0,1,2$.

Figure 3. Graphs $\Gamma_{n}$ for (a) $n=0$, (b) $n=1$ and (c) $n=2$ wedges
attached to the wheel.

The Kontsevich weight of the graph $\Gamma_{n}$ for $n\geq 0$ is given by

(19) $\displaystyle
w_{\Gamma_{n}}=\frac{1}{(2\pi)^{2n+2}}\int\limits_{C_{n+2,0}}{\mathrm{d}}\varphi(x,y){\mathrm{d}}\varphi(y,x){\mathrm{d}}\varphi(z_{1},x){\mathrm{d}}\varphi(z_{1},y)\cdots{\mathrm{d}}\varphi(z_{n},x){\mathrm{d}}\varphi(z_{n},y).$

######  Remark 2.2.

We will omit the wedge product if it is clear. Moreover, for $n=0$ we simply
set
${\mathrm{d}}\varphi(z_{1},x){\mathrm{d}}\varphi(z_{1},y)\cdots{\mathrm{d}}\varphi(z_{n},x){\mathrm{d}}\varphi(z_{n},y)=1$
in the integral above.

######  Remark 2.3.

The sign of the weight $\omega_{\Gamma}$ depends on the ordering of the edges
of the graph $\Gamma$ (i.e. the ordering of the propagator $1$-forms in the
integrand), and thus the ordering must always be specified. Throughout this
whole section, we will stick to the ordering given in (19).

The goal now is to compute (19) explicitly. We will do this in several steps,
mainly using Stokes’ theorem as in [21].

####  2.1.1. Step 1

In a first step, we want to integrate out the wedges. More precisely, for a
wedge as in Figure 2(a) we want to compute the corresponding integral

(20)
$\displaystyle\frac{1}{(2\pi)^{2}}\int\limits_{z\in\mathbb{H}\backslash\\{x,y\\}}{\mathrm{d}}\varphi(z,x){\mathrm{d}}\varphi(z,y),$

i.e. we want to integrate out $z$ (with $x,y\in\mathbb{H}$ fixed). To do this
we make a branch cut such that $\varphi(z,x)\in(0,2\pi)$ and use Stokes’
theorem

(21)
$\displaystyle\int\limits_{z\in\mathbb{H}\backslash\\{x,y\\}}{\mathrm{d}}\varphi(z,x){\mathrm{d}}\varphi(z,y)=\int\limits_{\partial}\varphi(z,x){\mathrm{d}}\varphi(z,y),$

where $\partial$ is the boundary of the integration domain depicted in Figure
4 below.

Figure 4. Boundary $\partial$ of the integration domain: $C$ is the half-
circle at infinity, $B_{+}$ and $B_{-}$ are infinitesimally close together,
the circles $C_{1}$ and $C_{2}$ have infinitesimal radius and $H_{+}\cup
H_{-}$ is the real line.

Now using (11) we can discuss the different boundary components:

  * •

On $H_{+}\cup H_{-}$: $z\in\mathbb{R}$ and hence
${\mathrm{d}}\varphi(z,y)={\mathrm{d}}\text{arg}(1)=0$

  * •

On $B_{+}$: $\varphi(z,x)=2\pi$

  * •

On $B_{-}$: $\varphi(z,x)=0$

  * •

On $C_{1}$: $z=x+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ $\implies$
${\mathrm{d}}\varphi(z,y)={\mathrm{d}}\text{arg}\left(\frac{y-x}{y-\bar{x}}\right)=0$

  * •

On $C_{2}$: $z=y+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ and $\theta\in[0,2\pi)$ $\implies$ $\varphi(z,x)\rightarrow\varphi(y,x)$,
${\mathrm{d}}\varphi(z,y)=-{\mathrm{d}}\theta$

  * •

On $C$: $z=R\mathrm{e}^{i\theta}$ for $R\rightarrow\infty$ and
$\theta\in[0,\pi]$ $\implies$ $\varphi(z,x)=\varphi(z,y)=2\theta$,
${\mathrm{d}}\varphi(z,y)=2{\mathrm{d}}\theta$

We then finally get

(22)
$\displaystyle\begin{split}\int\limits_{z\in\mathbb{H}\backslash\\{x,y\\}}{\mathrm{d}}\varphi(z,x){\mathrm{d}}\varphi(z,y)&=\int\limits_{\partial}\varphi(z,x){\mathrm{d}}\varphi(z,y)=2\pi\int\limits_{B_{+}}{\mathrm{d}}\varphi(z,y)+\int\limits_{0}^{\pi}4\theta{\mathrm{d}}\theta-\varphi(y,x)\int\limits_{0}^{2\pi}{\mathrm{d}}\theta\\\
&=2\pi\big{(}\varphi(x,y)-\varphi(y,x)+[x;y]\pi\big{)},\end{split}$

where

(23) $\displaystyle[x;y]=\begin{cases}+1,&\text{if
}\mathrm{Re}(x)>\mathrm{Re}(y)\\\ -1,&\text{if
}\mathrm{Re}(x)<\mathrm{Re}(y)\end{cases}$

Dividing the result (22) by $(2\pi)^{2}$ (as in (20)), we get

(24)
$\displaystyle\frac{1}{2\pi}\big{(}\varphi(x,y)-\varphi(y,x)\big{)}\pm\frac{1}{2},$

which agrees with the result given in Lemma 3.3 of [2].

Finally, consider the limit $(x,y)\rightarrow(p,q)$ for $p,q\in\mathbb{R}$
with $p<q$. Using (24) and the fact that $\varphi(p,q)=2\pi$ and
$\varphi(q,p)=0$, we can compute the Kontsevich weight of the graph below.

We get that $\int\limits_{C_{1,2}}\Big{(}\mathord{\leavevmode\hbox to0pt{\vbox
to0pt{\pgfpicture\makeatletter\hbox{\hskip 0.0pt\lower 0.0pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}} \feynman \vertex[dot,minimum
size=1mm] (c){}; \vertex[below left=0.5cm and 0.4cm of c,dot,minimum
size=1mm](f1){}; \vertex[right=0.8cm of f1,dot, minimum size=1mm] (f3){};
\vertex[left=0.2cm of f1] (r); \vertex[right=0.2cm of f3] (s);
\par\par\par\diagram* { (c) -- [fermion2] (f1) (c) -- [fermion2] (f3) (r) --
(s) }; \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}\Big{)}=\frac{1}{2}$,
which agrees with the result in Section 6.4.3 of [17].

####  2.1.2. Step 2

In a second step we want to compute the weight of the graph

where $n,m\geq 1$ and we use the notation introduced above, i.e. we want to
explicitly compute the integral

(25)
$\displaystyle\frac{1}{(2\pi)^{n+m}}\int\limits_{C_{2,0}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}.$

As before we make the branch cut such that $\varphi(x,y)\in(0,2\pi)$ and use
Stokes’ theorem

(26)
$\displaystyle\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}=\int\limits_{\partial}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}$

with boundary $\partial$ of the integration domain depicted in Figure 5 below.

Figure 5. Boundary $\partial$ of the integration domain: $C_{-}\cup C_{+}$ is
the half-circle at infinity, $B_{+}$ and $B_{-}$ are infinitesimally close
together, the circle $C_{1}$ has infinitesimal radius and $H$ is the real
line.

Again, we discuss the different boundary components:

  * •

On $H$: $y\in\mathbb{R}$ $\implies$
${\mathrm{d}}\varphi(y,x)={\mathrm{d}}\text{arg}(1)=0$

  * •

On $B_{-}\cup B_{+}$: ${\mathrm{d}}\varphi(y,x)=0$

  * •

On $C_{-}$: $y=R\mathrm{e}^{i\theta}$ for $R\rightarrow\infty$ $\implies$
$\varphi(x,y)=2\pi$

  * •

On $C_{+}$: $y=R\mathrm{e}^{i\theta}$ for $R\rightarrow\infty$ $\implies$
$\varphi(x,y)=0$

  * •

On $C_{1}$: $y=x+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ and $\theta\in(-\frac{\pi}{2},\frac{3\pi}{2})$ $\implies$
$\varphi(x,y)=\frac{3\pi}{2}-\theta$ and $\\\
\varphi(y,x)=\begin{cases}\frac{\pi}{2}-\theta,&\text{for
}\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\\\ \frac{5\pi}{2}-\theta,&\text{for
}\theta\in(\frac{\pi}{2},\frac{3\pi}{2})\end{cases}$

With this, we compute the integral

(27)
$\displaystyle\begin{split}&\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}=\int\limits_{\partial}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}=(2\pi)^{m}\int\limits_{C_{-}}{\mathrm{d}}\varphi(y,x)^{n}+\int\limits_{C_{1}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}\\\
&=(2\pi)^{m}\pi^{n}-n\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{3\pi}{2}-\theta\right)^{m}\left(\frac{\pi}{2}-\theta\right)^{n-1}{\mathrm{d}}\theta-n\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\left(\frac{3\pi}{2}-\theta\right)^{m}\left(\frac{5\pi}{2}-\theta\right)^{n-1}{\mathrm{d}}\theta.\end{split}$

Now we use the substitution $a=\frac{\pi}{2}-\theta$ to compute

(28)
$\displaystyle\begin{split}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{3\pi}{2}-\theta\right)^{m}\left(\frac{\pi}{2}-\theta\right)^{n-1}{\mathrm{d}}\theta&=\int\limits_{0}^{\pi}(\pi+a)^{m}a^{n-1}{\mathrm{d}}a=\sum\limits_{k=0}^{m}{m\choose
k}\pi^{k}\int\limits_{0}^{\pi}a^{m+n-k-1}{\mathrm{d}}a\\\
&=\sum\limits_{k=0}^{m}{m\choose k}\frac{\pi^{m+n}}{m+n-k}.\end{split}$

Similarly, we use the substitution $a=\frac{3\pi}{2}-\theta$ to compute

(29)
$\displaystyle\begin{split}\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\left(\frac{3\pi}{2}-\theta\right)^{m}\left(\frac{5\pi}{2}-\theta\right)^{n-1}{\mathrm{d}}\theta&=\int\limits_{0}^{\pi}a^{m}(\pi+a)^{n-1}{\mathrm{d}}a=\sum\limits_{k=0}^{n-1}{n-1\choose
k}\pi^{k}\int\limits_{0}^{\pi}a^{m+n-k-1}{\mathrm{d}}a\\\
&=\sum\limits_{k=0}^{n-1}{n-1\choose k}\frac{\pi^{m+n}}{m+n-k}.\end{split}$

Putting everything together we get

(30)
$\displaystyle\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}=\left(2^{m}-\sum\limits_{k=0}^{m}{m\choose
k}\frac{n}{m+n-k}-\sum\limits_{l=0}^{n-1}{n-1\choose
l}\frac{n}{m+n-l}\right)\pi^{m+n}.$

It is not hard to see that for $n=1$ the above formula simplifies to

(31)
$\displaystyle\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)=2^{m}\left(1-\frac{2}{m+1}\right)\pi^{m+1},$

which agrees with the result in Section 4 of [21].

####  2.1.3. Step 3

In a third step we want to compute an integral similar to (25), but with an
additional factor $[x;y]$ as defined in (23). So we want to compute the
integral

(32)
$\displaystyle\frac{1}{(2\pi)^{n+m}}\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}$

As usual, we use Stokes’ theorem

(33)
$\displaystyle\begin{split}\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}&=\int\limits_{\begin{subarray}{c}y\in\mathbb{H}\backslash\\{x\\}\\\
\mathrm{Re}(y)<\mathrm{Re}(x)\end{subarray}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}-\int\limits_{\begin{subarray}{c}y\in\mathbb{H}\backslash\\{x\\}\\\
\mathrm{Re}(y)>\mathrm{Re}(x)\end{subarray}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}\\\
&=\int\limits_{\partial_{+}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}-\int\limits_{\partial_{-}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}\end{split}$

with boundaries $\partial_{+}$ and $\partial_{-}$ of the integration domain
depicted in Figure 6 below.

Figure 6. Boundaries $\partial_{+}$ (on the left) and $\partial_{-}$ (on the
right) of the integration domain: $C_{-}\cup C_{+}$ is the half-circle at
infinity, $B_{+}$ and $B_{-}$ as well as $L_{+}$ and $L_{-}$ are
infinitesimally close together, the circle $D_{+}\cup D_{-}$ has infinitesimal
radius and $H_{+}\cup H_{-}$ is the real line.

As before, we discuss the different boundary components:

  * •

On $H_{+}\cup H_{-}$: ${\mathrm{d}}\varphi(y,x)=0$

  * •

On $B_{\pm}\cup L_{\pm}$: ${\mathrm{d}}\varphi(y,x)=0$

  * •

On $C_{-}$: $\varphi(x,y)=2\pi$

  * •

On $C_{+}$: $\varphi(x,y)=0$

  * •

On $D_{-}$: $y=x+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ and $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$ $\implies$
$\varphi(x,y)=\frac{3\pi}{2}-\theta$, $\varphi(y,x)=\frac{\pi}{2}-\theta$

  * •

On $D_{+}$: $y=x+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ and $\theta\in(\frac{\pi}{2},\frac{3\pi}{2})$ $\implies$
$\varphi(x,y)=\frac{3\pi}{2}-\theta$, $\varphi(y,x)=\frac{5\pi}{2}-\theta$

With this we compute the integral

(34)
$\displaystyle\begin{split}&\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}=\int\limits_{\partial_{+}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}-\int\limits_{\partial_{-}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}\\\
&=-n\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\left(\frac{3\pi}{2}-\theta\right)^{m}\left(\frac{5\pi}{2}-\theta\right)^{n-1}d\theta-(2\pi)^{m}\pi^{n}+n\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{3\pi}{2}-\theta\right)^{m}\left(\frac{\pi}{2}-\theta\right)^{n-1}d\theta\\\
&=\left(-2^{m}+\sum\limits_{k=0}^{m}{m\choose
k}\frac{n}{m+n-k}-\sum\limits_{l=0}^{n-1}{n-1\choose
l}\frac{n}{m+n-l}\right)\pi^{m+n},\end{split}$

where we used (28) and (29) in the last step.

####  2.1.4. Putting everything together

Finally, we are able to compute the Kontsevich weight (19) of the graphs
$\Gamma_{n}$ described at the beginning of Section 2.1. Integrating over
$z_{i}$ for $i=1,\ldots,n$, and applying the result for (22) obtained in the
first step we get

(35)
$\displaystyle\begin{split}w_{\Gamma_{n}}&=\frac{1}{(2\pi)^{2n+2}}\int\limits_{C_{n+2,0}}{\mathrm{d}}\varphi(x,y){\mathrm{d}}\varphi(y,x){\mathrm{d}}\varphi(z_{1},x){\mathrm{d}}\varphi(z_{1},y)\cdots{\mathrm{d}}\varphi(z_{n},x){\mathrm{d}}\varphi(z_{n},y)\\\
&=\frac{1}{(2\pi)^{n+2}}\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}\big{(}\varphi(x,y)-\varphi(y,x)+[x;y]\pi\big{)}^{n}{\mathrm{d}}\varphi(x,y){\mathrm{d}}\varphi(y,x)\\\
&=\frac{1}{(2\pi)^{n+2}}\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}{n\choose
k}{n-k\choose
l}(-1)^{l}\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}\varphi(x,y)^{n-k-l}\varphi(y,x)^{l}([x;y]\pi)^{k}{\mathrm{d}}\varphi(x,y){\mathrm{d}}\varphi(y,x)\\\
&=\frac{1}{2^{n+2}}\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}{n\choose
k}{n-k\choose
l}\frac{(-1)^{l}}{\pi^{n-k+2}(n-k-l+1)(l+1)}\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]^{k}{\mathrm{d}}\varphi(x,y)^{n-k-l+1}{\mathrm{d}}\varphi(y,x)^{l+1}.\end{split}$

Note that for even $k$ we have

(36)
$\displaystyle\begin{split}&\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]^{k}{\mathrm{d}}\varphi(x,y)^{n-k-l+1}{\mathrm{d}}\varphi(y,x)^{l+1}=\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{n-k-l+1}{\mathrm{d}}\varphi(y,x)^{l+1}\\\
&=\left(2^{n-k-l+1}-\sum\limits_{r=0}^{n-k-l+1}{n-k-l+1\choose
r}\frac{l+1}{n-k-r+2}-\sum\limits_{s=0}^{l}{l\choose
s}\frac{l+1}{n-k-s+2}\right)\pi^{n-k+2},\end{split}$

where we have used (30). Similarly, for odd $k$ we get

(37)
$\displaystyle\begin{split}&\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]^{k}{\mathrm{d}}\varphi(x,y)^{n-k-l+1}{\mathrm{d}}\varphi(y,x)^{l+1}=\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]{\mathrm{d}}\varphi(x,y)^{n-k-l+1}{\mathrm{d}}\varphi(y,x)^{l+1}\\\
&=\left(-2^{n-k-l+1}+\sum\limits_{r=0}^{n-k-l+1}{n-k-l+1\choose
r}\frac{l+1}{n-k-r+2}-\sum\limits_{s=0}^{l}{l\choose
s}\frac{l+1}{n-k-s+2}\right)\pi^{n-k+2},\end{split}$

where we have used (34).

We will now simplify the expressions we got. We start by observing a few
things:

First of all, we clearly have that

(38)
$\displaystyle{\mathrm{d}}\varphi(x,y)^{n-k-l+1}{\mathrm{d}}\varphi(y,x)^{l+1}=-{\mathrm{d}}\varphi(y,x)^{l+1}{\mathrm{d}}\varphi(x,y)^{n-k-l+1}.$

Similarly, we also have that

(39) $\displaystyle[x;y]=-[y;x].$

Furthermore, we can obviously swap $x$ and $y$ in the integral and get the
same result, i.e.

(40)
$\displaystyle\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]^{m}{\mathrm{d}}\varphi(x,y)^{n-k-l+1}{\mathrm{d}}\varphi(y,x)^{l+1}=\int\limits_{x\in\mathbb{H}\backslash\\{y\\}}[y;x]^{m}{\mathrm{d}}\varphi(y,x)^{n-k-l+1}{\mathrm{d}}\varphi(x,y)^{l+1}.$

Now assume that $n$ is even. Applying (38), (39) and (40) to the last line of
(35), it follows that

(41) $\displaystyle
w_{\Gamma_{n}}=\frac{1}{2^{n+2}}\sum\limits_{\begin{subarray}{c}k=0\\\ k\text{
even}\end{subarray}}^{n}{n\choose
k}{n-k\choose\frac{n-k}{2}}\frac{(-1)^{\frac{n-k}{2}}}{\pi^{n-k+2}\left(\frac{n-k}{2}+1\right)^{2}}\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{\frac{n-k}{2}+1}{\mathrm{d}}\varphi(y,x)^{\frac{n-k}{2}+1}.$

So most of the terms cancel for $n$ even.
Now using (30) we observe that

(42)
$\displaystyle\begin{split}&\frac{1}{\pi^{2m}}\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{m}=2^{m}-\sum\limits_{k=0}^{m}{m\choose
k}\frac{m}{2m-k}-\sum\limits_{l=0}^{m-1}{m-1\choose l}\frac{m}{2m-l}\\\
&=2^{m}-\sum\limits_{k=0}^{m-1}\left({m\choose k}+{m-1\choose
k}\right)\frac{m}{2m-k}-1=2^{m}-\sum\limits_{k=0}^{m-1}{m\choose
k}-1=2^{m}-\sum\limits_{k=0}^{m}{m\choose k}=0.\end{split}$

Plugging this result into (41) with $m=\frac{n-k}{2}$ we finally get that

(43) $\displaystyle w_{\Gamma_{n}}=0\qquad\text{for even }n\geq 0.$

For $n$ odd the different terms in the last line of (35) do not cancel
anymore. Instead, we will try to write (30) and (34) more compactly. To do
this, let us introduce the so-called _hypergeometric function_
${}_{2}F_{1}(a,b;c;z)$. It is defined by the series

(44)
${}_{2}F_{1}(a,b;z;c):=\sum\limits_{k=0}^{\infty}\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{z^{k}}{k!}$

for $z\in\mathbb{C}$ with $|z|<1$, where $(a)_{k}$ is the _Pochhammer symbol_
given by

(45) $\displaystyle(a)_{k}=\begin{cases}1,&\text{if }k=0\\\
a(a+1)\cdots(a+k-1),&\text{if }k>0\end{cases}$

It is not hard to see that the series terminates if either $a$ or $b$ is a
non-positive integer. In that case the hypergeometric function reduces to a
polynomial and can therefore also be defined for $|z|\geq 1$.
We can now use the hypergeometric function to write

(46) $\displaystyle\sum\limits_{k=0}^{m}{m\choose
k}\frac{1}{m+n-k}=\frac{{}_{2}F_{1}(-m,-m-n;1-m-n;-1)}{m+n},$

and

(47) $\displaystyle\sum\limits_{k=0}^{n-1}{n-1\choose
k}\frac{1}{m+n-k}=\frac{{}_{2}F_{1}(1-n,-m-n;1-m-n;-1)}{m+n}.$

This allows us to write (30) as

(48)
$\displaystyle\begin{split}&\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}\\\
&=\Big{(}2^{m}-\frac{n}{m+n}\big{(}{{}_{2}F_{1}}(-m,-m-n;1-m-n;-1)+{{}_{2}F_{1}}(1-n,-m-n;1-m-n;-1)\big{)}\Big{)}\pi^{m+n},\end{split}$

and (34) as

(49)
$\displaystyle\begin{split}&\int\limits_{y\in\mathbb{H}\backslash\\{x\\}}[x;y]{\mathrm{d}}\varphi(x,y)^{m}{\mathrm{d}}\varphi(y,x)^{n}\\\
&=\left(-2^{m}+\frac{n}{m+n}\big{(}{{}_{2}F_{1}}(-m,-m-n;1-m-n;-1)-{{}_{2}F_{1}}(1-n,-m-n;1-m-n;-1)\big{)}\right)\pi^{m+n}.\end{split}$

Plugging those results into (35) we finally get for all $n\geq 0$

(50)
$\displaystyle\begin{split}w_{\Gamma_{n}}&=\frac{1}{2^{n+2}}\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}{n\choose
k}{n-k\choose l}\frac{(-1)^{l}}{(n-k-l+1)(l+1)}\Big{(}(-1)^{k}2^{n-k-l+1}\\\
&-\frac{l+1}{n-k+2}\big{(}{{}_{2}F_{1}}(-l,-n+k-2;-n+k-1;-1)\\\
&+(-1)^{k}{{}_{2}F_{1}}(-n+k+l-1,-n+k-2;-n+k-1;-1)\big{)}\Big{)}.\end{split}$

The Kontsevich weights of the first few graphs are given in Table 1 below.

$n$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$
---|---|---|---|---|---|---|---|---|---|---
$w_{\Gamma_{n}}$ | $0$ | $\frac{1}{24}$ | $0$ | $\frac{1}{320}$ | $0$ | $\frac{1}{2688}$ | $0$ | $\frac{1}{18432}$ | $0$ | $\frac{1}{112640}$
Table 1. Kontsevich weights of the graphs $\Gamma_{n}$ for $n=0,1,\ldots,9$

As a sanity check we have the following: For $n=0$ the graph $\Gamma_{0}$ is
just a wheel with two vertices (see Figure 3(a)) and its weight is zero
according to Lemma 7.3 of [17]. For $n=1$ the graph $\Gamma_{1}$ is just a
wheel with two spokes pointing outward (see Figure 3(b)) and its weight is
$\frac{1}{24}$ according to Proposition 1.1 of [21]. So at least for $n=0,1$
our formula (50) for the Kontsevich weights $w_{\Gamma_{n}}$ produces the
correct values.

###  2.2. Case 2: One Boundary Vertex

We will now treat the case $(r,m)=(1,1)$, i.e. the case where we have one
boundary vertex and one $R$-vertex. In that case we get a family of graphs
$(\Upsilon_{n})_{n\geq 0}$, where $\Upsilon_{n}$ is the graph with $n$ wedges
as in Figure 7(a) (stemming from $n$ $\mathsf{T}\phi_{x}^{*}\pi$-vertices)
attached to the graph containing a single edge from the $R$-vertex to the
boundary vertex as in Figure 7(b) below.

Figure 7. Graphs in the case $(r,m)=(1,1)$ consist of: (a) wedges stemming
from $\mathsf{T}\phi^{*}_{x}\pi$-vertices attached to (b) a single edge from
the $R$-vertex to the boundary vertex

Examples of the graphs $\Upsilon_{n}$ are given in Figure 8 below for
$n=0,1,2$.

Figure 8. Graphs $\Upsilon_{n}$ for (a) $n=0$, (b) $n=1$ and (c) $n=2$ wedges
attached to the single edge from the $R$-vertex to the boundary vertex

The Kontsevich weight of the graph $\Upsilon_{n}$ for $n\geq 0$ is given by

(51) $\displaystyle
w_{\Upsilon_{n}}=\frac{1}{(2\pi)^{2n+1}}\int\limits_{C_{n+1,1}}{\mathrm{d}}\varphi(x,q){\mathrm{d}}\varphi(z_{1},x){\mathrm{d}}\varphi(z_{1},q)\cdots{\mathrm{d}}\varphi(z_{n},x){\mathrm{d}}\varphi(z_{n},q).$

######  Remark 2.4.

As before, the ordering of the edges of the graph $\Upsilon_{n}$ specified in
(51) above determines the sign of $w_{\Upsilon_{n}}$. Throughout this whole
section we will stick to this ordering.

Again, the goal is to compute (51) explicitly. As before, we will do this in
several steps.

####  2.2.1. Step 1

For a wedge as in Figure 7(a) we want to compute the corresponding integral

(52)
$\displaystyle\frac{1}{(2\pi)^{2}}\int\limits_{z\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(z,x){\mathrm{d}}\varphi(z,q),$

i.e. we want to integrate out $z$ (with $x,q\in\overline{\mathbb{H}}$ fixed).
The computation is almost the same as the one we have already done in Section
2.1.1 above: Again we make a branch cut such that $\varphi(z,x)\in(0,2\pi)$
and use Stokes’ theorem

(53)
$\displaystyle\int\limits_{z\in\mathbb{H}\backslash\\{x\\}}{\mathrm{d}}\varphi(z,x){\mathrm{d}}\varphi(z,q)=\int\limits_{\partial}\varphi(z,x){\mathrm{d}}\varphi(z,q),$

where $\partial$ is the boundary of the integration domain depicted in Figure
9 below.

Figure 9. Boundary $\partial$ of the integration domain: $C$ is the half-
circle at infinity, $B_{+}$ and $B_{-}$ are infinitesimally close together,
the (half-)circles $C_{1}$ and $C_{2}$ have infinitesimal radius and
$H_{+,1}\cup H_{+,2}\cup H_{-}$ is the real line.

Let us have a look at the different boundary components:

  * •

On $H_{+,1}\cup H_{+,2}\cup H_{-}$: $z\in\mathbb{R}$ and hence
${\mathrm{d}}\varphi(z,q)=0$

  * •

On $B_{+}$: $\varphi(z,x)=2\pi$

  * •

On $B_{-}$: $\varphi(z,x)=0$

  * •

On $C_{1}$: $z=x+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ $\implies$
${\mathrm{d}}\varphi(z,q)={\mathrm{d}}\text{arg}\left(\frac{q-x}{q-\bar{x}}\right)=0$

  * •

On $C_{2}$: $z=q+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ and $\theta\in[-\pi,0]$ $\implies$ $\varphi(z,x)\rightarrow\varphi(q,x)$,
$\varphi(z,q)=-2\theta$

  * •

On $C$: $z=R\mathrm{e}^{i\theta}$ for $R\rightarrow\infty$ and
$\theta\in[0,\pi]$ $\implies$ $\varphi(z,x)=\varphi(z,q)=2\theta$

We can then compute the integral

(54)
$\displaystyle\begin{split}\int\limits_{z\in\mathbb{H}\backslash\\{x,q\\}}{\mathrm{d}}\varphi(z,x){\mathrm{d}}\varphi(z,q)&=\int\limits_{\partial}\varphi(z,x){\mathrm{d}}\varphi(z,q)=2\pi\int\limits_{B_{+}}{\mathrm{d}}\varphi(z,q)+\int\limits_{0}^{\pi}4\theta{\mathrm{d}}\theta-2\varphi(q,x)\int\limits_{-\pi}^{0}{\mathrm{d}}\theta\\\
&=2\pi\big{(}\varphi(x,q)-\varphi(q,x)+[x;q]\pi\big{)}\end{split}$

where

(55) $\displaystyle[x;q]=\begin{cases}+1,&\text{if }\mathrm{Re}(x)>q\\\
-1,&\text{if }\mathrm{Re}(x)<q\end{cases}$

Dividing the result (54) by $(2\pi)^{2}$ we get

(56)
$\displaystyle\frac{1}{2\pi}\big{(}\varphi(x,q)-\varphi(q,x)\big{)}\pm\frac{1}{2},$

which agrees with the result in Lemma 5.3 of [21].
Finally, observe that one obtains (54) by simply taking the limit
$y\rightarrow q\in\mathbb{R}$ in (22).

####  2.2.2. Step 2

In a second step we want to compute the integral

(57)
$\displaystyle\frac{1}{(2\pi)^{m+n}}\int\limits_{C_{1,1}}\varphi(q,x)^{m}{\mathrm{d}}\varphi(x,q)^{n}$

for $n\geq 1$ and $m\geq 0$.
First note that $C_{1,1}$, shown in Figure 10 below, is a smooth manifold of
dimension $1$ which is homeomorphic to an open interval and $\bar{C}_{1,1}$ is
homeomorphic to a closed interval.

######  Remark 2.5.

We work with the standard orientation on $C_{1,1}$, which is induced by the
standard orientation on the plane $\mathbb{R}^{2}$.

Figure 10. The manifold $C_{1,1}$ is the product of a (fixed) single point $q$
on the real line and an open half circle

It is not hard to see that the boundary $\partial C_{1,1}$ is just a two-
element set. More precisely $\partial C_{1,1}=\\{(q,s),(q,t)\\}$ with $s<q$
and $t>q$ (for a more detailed treatment, see [17, 12, 10]).
But now we have to make a branch cut such that $\varphi(q,x)\in(0,2\pi)$. Then
the boundary $\partial$ of the integration domain, depicted in Figure 11
below, contains four points, namely

(58) $\displaystyle\partial=\\{(q,s),(q,t),(q,y_{+}),(q,y_{-})\\},$

where $y$ is the point on the half circle directly above $q$, i.e. with
Re$(y)=q$, and $y_{+}$ and $y_{-}$ are the limits $x\rightarrow y$ on the half
circle from the left (i.e. from the region Re$(x)<q$ of the half-circle) and
from the right (i.e. from the region Re$(x)>q$) respectively.

Figure 11. $C_{1,1}$ with branch cut and its boundary $\partial$ consisting of
four points

Finally, using Stokes’ theorem and the fact that ${\mathrm{d}}\varphi(q,x)=0$
for $q\in\mathbb{R}$, we get

(59)
$\displaystyle\begin{split}&\int\limits_{C_{1,1}}\varphi(q,x)^{m}{\mathrm{d}}\varphi(x,q)^{n}=\int\limits_{\partial}\varphi(q,x)^{m}\varphi(x,q)^{n}\\\
&=\varphi(q,s)^{m}\varphi(s,q)^{n}-\varphi(q,y_{+})^{m}\varphi(y_{+},q)^{n}+\varphi(q,y_{-})^{m}\varphi(y_{-},q)^{n}-\varphi(q,t)^{m}\varphi(t,q)^{n}\\\
&=\begin{cases}(2\pi)^{n},&\text{if }m=0\\\ 2^{m}\pi^{m+n},&\text{if
}m>0\end{cases}\end{split}$

####  2.2.3. Step 3

Let us start with writing (54) differently as

(60) $\displaystyle
2\pi\big{(}\varphi(x,q)-\varphi(q,x)+\pi[x;q]\big{)}=2\pi\big{(}\varphi(x,q)-\varphi(q,x)-\pi+2\pi(x;q)\big{)}$

where

(61) $\displaystyle(x;q)=\begin{cases}+1,&\text{if }\mathrm{Re}(x)>q\\\
0,&\text{if }\mathrm{Re}(x)<q\end{cases}$

In this step we then want to compute an integral similar to (57), but with an
additional factor $(x;y)$ as defined above. So we want to compute

(62)
$\displaystyle\frac{1}{(2\pi)^{m+n}}\int\limits_{C_{1,1}}(x;q)\varphi(q,x)^{m}{\mathrm{d}}\varphi(x,q)^{n}$

for $n\geq 1$ and $m\geq 0$.
As before we use Stokes’ theorem and find that

(63)
$\displaystyle\begin{split}&\int\limits_{C_{1,1}}(x;q)\varphi(q,x)^{m}{\mathrm{d}}\varphi(x,q)^{n}=\int\limits_{\begin{subarray}{c}C_{1,1}\\\
\mathrm{Re}(x)>q\end{subarray}}\varphi(q,x)^{m}{\mathrm{d}}\varphi(x,q)^{n}\\\
&=\varphi(q,y_{-})^{m}\varphi(y_{-},q)^{n}-\varphi(q,t)^{m}\varphi(t,q)^{n}=2^{m}\pi^{m+n}\end{split}$

for all $m\geq 0$ and all $n\geq 1$.

####  2.2.4. Putting everything together

Now we can use the results from steps $1$-$3$ to compute the Kontsevich weight
(51) of the graphs $\Upsilon_{n}$, $n\geq 0$, described at the beginning of
Section 2.2. Integrating over $z_{i}$ for $i=1,\ldots,n$, and applying (60),
we get

(64)
$\displaystyle\begin{split}w_{\Upsilon_{n}}&=\frac{1}{(2\pi)^{2n+1}}\int\limits_{C_{n+1,1}}{\mathrm{d}}\varphi(x,q){\mathrm{d}}\varphi(z_{1},x){\mathrm{d}}\varphi(z_{1},q)\cdots{\mathrm{d}}\varphi(z_{n},x){\mathrm{d}}\varphi(z_{n},q)\\\
&=\frac{1}{(2\pi)^{n+1}}\int\limits_{C_{1,1}}\big{(}\varphi(x,q)-\varphi(q,x)-\pi+2\pi(x;q)\big{)}^{n}{\mathrm{d}}\varphi(x,q)\\\
&=\frac{1}{(2\pi)^{n+1}}\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}\sum\limits_{s=0}^{n-k-l}{n\choose
k}{n-k\choose l}{n-k-l\choose s}(-1)^{l+s}\\\
&\times\int\limits_{C_{1,1}}\varphi(x,q)^{n-k-
l-s}\varphi(q,x)^{s}\pi^{l}(2\pi(x;q))^{k}{\mathrm{d}}\varphi(x,q)\\\
&=\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}\sum\limits_{s=0}^{n-k-l}{n\choose
k}{n-k\choose l}{n-k-l\choose s}\frac{(-1)^{l+s}}{2^{n-k+1}\pi^{n-k-l+1}(n-k-
l-s+1)}\\\
&\times\int\limits_{C_{1,1}}(x;q)^{k}\varphi(q,x)^{s}{\mathrm{d}}\varphi(x,q)^{n-k-
l-s+1}.\end{split}$

We note that for $k=0$ we have

(65)
$\displaystyle\int\limits_{C_{1,1}}\varphi(q,x)^{s}{\mathrm{d}}\varphi(x,q)^{n-l-s+1}=\begin{cases}(2\pi)^{n-l+1},&\text{if
}s=0\\\ 2^{s}\pi^{n-l+1},&\text{if }s>0\end{cases}$

where we have used (59). Similarly, for $k\geq 1$ we have

(66)
$\displaystyle\int\limits_{C_{1,1}}(x;q)^{k}\varphi(q,x)^{s}{\mathrm{d}}\varphi(x,q)^{n-k-
l-s+1}=\int\limits_{C_{1,1}}(x;q)\varphi(q,x)^{s}{\mathrm{d}}\varphi(x,q)^{n-k-
l-s+1}=2^{s}\pi^{n-k-l+1},$

where we have used (63).
Plugging the above two results into the last line of (64) we get

(67)
$\displaystyle\begin{split}w_{\Upsilon_{n}}&=\underbrace{\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}\sum\limits_{s=0}^{n-k-l}{n\choose
k}{n-k\choose l}{n-k-l\choose s}\frac{(-1)^{l+s}}{2^{n-k-s+1}(n-k-
l-s+1)}}_{=:A(n)}\\\ &-\underbrace{\sum\limits_{l=0}^{n}{n\choose
l}\frac{(-1)^{l}}{2^{n+1}(n-l+1)}}_{=:B(n)}+\underbrace{\sum\limits_{l=0}^{n}{n\choose
l}\frac{(-1)^{l}}{2^{l}(n-l+1)}}_{=:C(n)}.\end{split}$

As shown in Appendix A, we have that

(68) $\displaystyle\begin{split}A(n)&=\frac{(-1)^{n}}{2^{n+1}(n+1)},\\\
B(n)&=\frac{(-1)^{n}}{2^{n+1}(n+1)},\\\
C(n)&=\frac{1+(-1)^{n}}{2^{n+1}(n+1)}.\end{split}$

Hence, we finally have

(69) $\displaystyle w_{\Upsilon_{n}}=\frac{1+(-1)^{n}}{2^{n+1}(n+1)},\qquad
n\geq 0.$

In particular, we see that

(70) $\displaystyle w_{\Upsilon_{n}}=0\qquad\text{for odd }n\geq 1.\;$

The Kontsevich weights of the first few graphs are given in Table 2 below.

$n$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$
---|---|---|---|---|---|---|---|---|---|---
$w_{\Upsilon_{n}}$ | $1$ | $0$ | $\frac{1}{12}$ | $0$ | $\frac{1}{80}$ | $0$ | $\frac{1}{448}$ | $0$ | $\frac{1}{2304}$ | $0$
Table 2. Kontsevich weights of the graphs $\Upsilon_{n}$ for $n=0,1,\ldots,9$

As a sanity check we have the following: For $n=0$ the graph $\Upsilon_{0}$ is
just a single edge as in Figure 7(b) and its weight is zero according to
Section 6.4.3 of [17]. For $n=1$ the graph $\Upsilon_{1}$ is just a single
edge with one wedge attached as in Figure 8(b) and its weight is $0$ according
to Appendix B of [14]. For $n=2$ the graph $\Upsilon_{2}$ is a single edge
with two wedges attached as in Figure 8(c) and its weight is $\frac{1}{12}$
according to Appendix A of [21]. So at least for $n=0,1,2$ our formula (69)
for the Kontsevich weights $w_{\Gamma_{n}}$ produces the correct values.

###  2.3. Case 3: Two Boundary Vertices

Finally, we will treat the case $(r,m)=(0,2)$, i.e. the case where we have two
boundary vertices and no $R$-vertex. In that case we get a family of graphs
$(\Lambda_{n})_{n\geq 0}$, where $\Lambda_{n}$ is the graph with $n$ wedges as
in Figure 12 (stemming from $n$ $\mathsf{T}\phi^{*}_{x}\pi$-vertices) attached
to the two boundary vertices.

Figure 12. Graphs in the case $(r,m)=(0,2)$ consist of wedges attached to the
two boundary vertices

Examples of the graphs $\Lambda_{n}$ are given in Figure 13 below for
$n=0,1,2$.

Figure 13. Graphs $\Lambda_{n}$ for (a) $n=0$, (b) $n=1$ and (c) $n=2$ wedges
attached to the two boundary vertices

The Kontsevich weight of the graph $\Lambda_{n}$ for $n\geq 0$ is given by

(71) $\displaystyle
w_{\Lambda_{n}}=\frac{1}{(2\pi)^{2n}}\int\limits_{C_{n,2}}{\mathrm{d}}\varphi(z_{1},p){\mathrm{d}}\varphi(z_{1},q)\cdots{\mathrm{d}}\varphi(z_{n},p){\mathrm{d}}\varphi(z_{n},q).$

######  Remark 2.6.

For $n=0$ we simply set
${\mathrm{d}}\varphi(z_{1},p){\mathrm{d}}\varphi(z_{1},q)\cdots{\mathrm{d}}\varphi(z_{n},p){\mathrm{d}}\varphi(z_{n},q)=1$
in the integral above.

######  Remark 2.7.

As before, the ordering of the edges of the graph $\Lambda_{n}$ specified in
(71) above determines the sign of $w_{\Lambda_{n}}$. Throughout this whole
section we will stick to this ordering.

######  Remark 2.8.

Since we work with the configuration space $\mathrm{Conf}_{n,m}$ as in (13),
and in particular with the quotient $C_{n,m}=\mathrm{Conf}_{n,m}/G^{(1)}$, it
follows that $C_{0,2}$ is a single point (and not a two-element set).

Finally, our goal is to compute (51) explicitly for the given family of
graphs. However, this time the computation is much easier and shorter than
before.
For the boundary vertices $p,q,\in\mathbb{R}$ with $p<q$ we have already
computed the Kontsevich weight of a wedge as in Figure 12 at the end of
Section 2.1.1. Our result was

(72)
$\displaystyle\frac{1}{(2\pi)^{2}}\int\limits_{C_{1,2}}{\mathrm{d}}\varphi(z,p){\mathrm{d}}\varphi(z,q)=\frac{1}{2}.$

For the sake of completeness and to make sure that we get the same result, let
us nonetheless do a direct computation. For a wedge as in Figure 12, we
compute the corresponding integral

(73)
$\displaystyle\frac{1}{(2\pi)^{2}}\int\limits_{z\in\mathbb{H}}{\mathrm{d}}\varphi(z,p){\mathrm{d}}\varphi(z,q),$

with $p,q\in\mathbb{R}$, $p<q$ fixed. As before, we use Stokes’ theorem

(74)
$\displaystyle\int\limits_{z\in\mathbb{H}}{\mathrm{d}}\varphi(z,p){\mathrm{d}}\varphi(z,q)=\int\limits_{\partial}\varphi(z,p){\mathrm{d}}\varphi(z,q),$

where $\partial$ is the boundary of the integration domain depicted in Figure
14 below.

Figure 14. Boundary $\partial$ of the integration domain: $C$ is the half-
circle at infinity, the half circles $C_{1}$ and $C_{2}$ have infinitesimal
radius and $H_{1}\cup H_{2}\cup H_{3}$ is the real line.

As usual, let us have a look at the different boundary components:

  * •

On $H_{1}\cup H_{2}\cup H_{3}$: $z\in\mathbb{R}$ and hence
${\mathrm{d}}\varphi(z,q)=0$

  * •

On $C_{1}$: $z=p+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ $\implies$ ${\mathrm{d}}\varphi(z,q)={\mathrm{d}}\text{arg}(1)=0$

  * •

On $C_{2}$: $z=q+\varepsilon\mathrm{e}^{-i\theta}$ for $\varepsilon\rightarrow
0$ $\implies$ $\varphi(z,p)\rightarrow\varphi(q,p)=0$

  * •

On $C$: $z=R\mathrm{e}^{i\theta}$ for $R\rightarrow\infty$ and
$\theta\in[0,\pi]$ $\implies$ $\varphi(z,p)=\varphi(z,q)=2\theta$

We can then compute the integral

(75)
$\displaystyle\int\limits_{z\in\mathbb{H}}{\mathrm{d}}\varphi(z,p){\mathrm{d}}\varphi(z,q)=\int\limits_{\partial}\varphi(z,p){\mathrm{d}}\varphi(z,q)=\int\limits_{0}^{\pi}4\theta
d\theta=2\pi^{2},$

which indeed agrees with (72) after dividing by $(2\pi)^{2}$.
With this result at hand, it is now easy to compute the Kontsevich weight of
the graph $\Lambda_{n}$ for $n\geq 1$. We get

(76) $\displaystyle
w_{\Lambda_{n}}=\frac{1}{(2\pi)^{2n}}\int\limits_{C_{n,2}}{\mathrm{d}}\varphi(z_{1},p){\mathrm{d}}\varphi(z_{1},q)\cdots{\mathrm{d}}\varphi(z_{n},p){\mathrm{d}}\varphi(z_{n},q)=\frac{1}{(2\pi)^{2n}}(2\pi^{2})^{n}=\frac{1}{2^{n}}.$

For $n=0$ it is not hard to see that

(77) $\displaystyle w_{\Lambda_{0}}=\int\limits_{C_{0,2}}1=1,$

since $C_{0,2}$ is a single point (cf Remark 2.8). So all in all we finally
have

(78) $\displaystyle w_{\Lambda_{n}}=\frac{1}{2^{n}},\qquad n\geq 0.$

##  3\. Including the Weights

###  3.1. The Product $P(\mathsf{T}\phi^{*}\pi)$

Let $\sigma,\tau\in\Gamma(\widehat{\textnormal{Sym}}(T^{*}M)[[\hbar]])$ and
let $x\in M$. Using the Kontsevich weights computed above, we get

(79)
$P(\mathsf{T}\phi^{*}_{x}\pi)(\sigma_{x}\otimes\tau_{x})=\sum\limits_{n=0}^{\infty}\frac{\hbar^{n}}{2^{2n}n!}(\mathsf{T}\phi^{*}_{x}\pi)^{i_{1}j_{1}}\cdots(\mathsf{T}\phi^{*}_{x}\pi)^{i_{n}j_{n}}(\sigma_{x})_{\,,i_{1}\cdots
i_{n}}(\tau_{x})_{\,,j_{1}\cdots j_{n}}$

where we sum over all the indices $i_{1},\ldots,i_{n},j_{1},\ldots,j_{n}$.
Moreover, we use the notation, where the indices on the right of the comma
denote derivatives with respect to the corresponding variable, e.g.

(80) $R_{i,i_{1}\dotsm i_{k}}:=\partial_{i_{1}}\dotsm\partial_{i_{k}}R_{i}.$

###  3.2. The Connection 1-form $A(R,\mathsf{T}\phi^{*}\pi)$

In Section 2.2 we have obtained the Kontsevich weights

(81) $\displaystyle w_{\Upsilon_{n}}=\frac{1+(-1)^{n}}{2^{n+1}(n+1)},\qquad
n\geq 0$

of the family of graphs $(\Upsilon_{n})_{n\geq 0}$. Let
$\sigma\in\Gamma(\widehat{\textnormal{Sym}}(T^{*}M)[[\hbar]])$ and fix $x\in
M$. For $R$ as in Section 1.2 we set $R_{x}(y):=R(x,y)$ and
$(R_{x})^{k}_{i}(y):=R^{k}_{i}(x,y)$. Using the Kontsevich weights above, we
get

(82)
$\displaystyle\begin{split}A(R_{x},\textsf{T}\phi^{*}_{x}\pi)(\sigma_{x})&={\mathrm{d}}x^{i}A\left((R_{x})^{k}_{i}\frac{\partial}{\partial
y^{k}},\textsf{T}\phi^{*}_{x}\pi\right)(\sigma_{x})\\\
&={\mathrm{d}}x^{i}\sum\limits_{n=0}^{\infty}\frac{\hbar^{n}}{2^{n}n!}\frac{1+(-1)^{n}}{2^{n+1}(n+1)}(\textsf{T}\phi^{*}_{x}\pi)^{i_{1}j_{1}}\cdots(\textsf{T}\phi^{*}_{x}\pi)^{i_{n}j_{n}}(R_{x})^{k}_{i,i_{1}\cdots
i_{n}}(\sigma_{x})_{\,,kj_{1}\cdots j_{n}}\end{split}$

where we again sum over all indices
$i,k,i_{1},\ldots,i_{n},j_{1},\ldots,j_{n}$.
This allows us to write down an explicit expression for the deformed
Grothendieck connection, namely

(83)
$\displaystyle\begin{split}(\mathcal{D}_{G})_{x}&={\mathrm{d}}_{x}+A(R_{x},\textsf{T}\phi^{*}_{x}\pi)=\left(\frac{\partial}{\partial
x^{i}}+A\big{(}(R_{x})_{i},\textsf{T}\phi^{*}_{x}\pi\big{)}\right){\mathrm{d}}x^{i}\\\
&=\left(\frac{\partial}{\partial
x^{i}}+\sum\limits_{n=0}^{\infty}\frac{\hbar^{n}}{2^{n}n!}\frac{1+(-1)^{n}}{2^{n+1}(n+1)}(\textsf{T}\phi^{*}_{x}\pi)^{i_{1}j_{1}}\cdots(\textsf{T}\phi^{*}_{x}\pi)^{i_{n}j_{n}}(R_{x})^{k}_{i,i_{1}\cdots
i_{n}}\frac{\partial^{n+1}}{\partial y^{j_{n}}\cdots\partial y^{j_{1}}\partial
y^{k}}\right){\mathrm{d}}x^{i}\end{split}$

###  3.3. The Curvature 2-form $F(R,R,\mathsf{T}\phi^{*}\pi)$

In Section 2.1 we have obtained the Kontsevich weights

(84)
$\displaystyle\begin{split}w_{\Gamma_{n}}&=\frac{1}{2^{n+2}}\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}{n\choose
k}{n-k\choose l}\frac{(-1)^{l}}{(n-k-l+1)(l+1)}\Big{(}(-1)^{k}2^{n-k-l+1}\\\
&-\frac{l+1}{n-k+2}\big{(}{{}_{2}F_{1}}(-l,-n+k-2;-n+k-1;-1)\\\
&+(-1)^{k}{{}_{2}F_{1}}(-n+k+l-1,-n+k-2;-n+k-1;-1)\big{)}\Big{)}\end{split}$

for the family of graphs $(\Gamma_{n})_{n\geq 0}$. Using these weights above
we then get for $x\in M$

(85)
$\displaystyle\begin{split}F(R_{x},R_{x},\textsf{T}\phi^{*}_{x}\pi)&={\mathrm{d}}x^{i}\wedge{\mathrm{d}}x^{j}F\big{(}(R_{x})_{i},(R_{x})_{j},\textsf{T}\phi^{*}_{x}\pi\big{)}\\\
&={\mathrm{d}}x^{i}\wedge{\mathrm{d}}x^{j}\sum\limits_{n=0}^{\infty}\frac{\hbar^{n}}{2^{n}n!}w_{\Gamma_{n}}(\textsf{T}\phi^{*}_{x}\pi)^{i_{1}j_{1}}\cdots(\textsf{T}\phi^{*}_{x}\pi)^{i_{n}j_{n}}(R_{x})^{k}_{i,li_{1}\cdots
i_{n}}(R_{x})^{l}_{j,kj_{1}\cdots j_{n}}\end{split}$

where, as usual, we sum over the indices
$i,j,k,l,i_{1},\ldots,i_{n},j_{1},\ldots,j_{n}$.
We can then write down the _modified deformed Grothendieck connection_ as
defined in [5, 12], namely

(86)
$\displaystyle\overline{\mathcal{D}}_{G}=\mathcal{D}_{G}+[\gamma,\enspace]_{\star},$

where the deformed Grothendieck connection $\mathcal{D}_{G}$ is explicitly
given by (83), the star product is explicitly given by (79),
$[\enspace,\enspace]_{\star}$ denotes the star commutator and
$\gamma\in\Omega^{1}(M,\widehat{\textnormal{Sym}}(T^{*}M)[[\hbar]])$ is such
that

(87) $\displaystyle F^{M}+\mathcal{D}_{G}\gamma+\gamma\star\gamma=0,$

with Weyl curvature $F^{M}=F(R,R,\textsf{T}\phi^{*}\pi)$ explicitly given by
(85). The existence of such a $\gamma$ was given in [5, 6]. Let us emphasize a
bit more on this existence result. Since $\gamma$ takes values in
$\widehat{\textnormal{Sym}}(T^{*}M)[[\hbar]]$ we may write

(88)
$\displaystyle\gamma=\gamma_{0}+\hbar\gamma_{1}+\hbar^{2}\gamma_{2}+\ldots$

Similarly, for the deformed Grothendieck connection we may write

(89)
$\displaystyle\mathcal{D}_{G}=D_{G}+\hbar^{2}\mathcal{D}_{2}+\hbar^{4}\mathcal{D}_{4}+\ldots$

where $D_{G}={\mathrm{d}}+L_{R}$ is the classical Grothendieck connection and
where we have used that the Kontsevich weights (81) satisfy
$w_{\Upsilon_{0}}=1$ and $w_{\Upsilon_{n}}=0$ for all odd $n\geq 1$.
Finally, for the curvature we can write

(90) $\displaystyle F^{M}=\hbar F_{1}+\hbar^{3}F_{3}+\hbar^{5}F_{5}+\ldots$

where we have used that that the Kontsevich weights (84) satisfy
$w_{\Gamma_{n}}=0$ for all even $n\geq 0$.
This now allows us to decompose Equation (87) into a system of equations
depending on the order of $\hbar$.
In order $\hbar^{0}$ we get the equation

(91) $\displaystyle D_{G}\gamma_{0}=0,$

which, according to Section 1.2, can be solved by
$\gamma_{0}=\textsf{T}\phi^{*}f$ for some smooth function $f\in
C^{\infty}(M)$, since the cohomology of the classical Grothendieck connection
$H^{\bullet}_{D_{G}}(\Gamma(\widehat{\textnormal{Sym}}(T^{*}M)))$ is
concentrated in degree zero (by the Poincaré Lemma) and

(92)
$H^{0}_{D_{G}}(\Gamma(\widehat{\textnormal{Sym}}(T^{*}M)))\cong\textsf{T}\phi^{*}C^{\infty}(M)\cong
C^{\infty}(M).$

In order $\hbar^{1}$ we get the equation

(93) $\displaystyle F_{1}+D_{G}\gamma_{1}+(\gamma_{0}\star\gamma_{0})_{1}=0.$

Using the Bianchi identity we see that $D_{G}F_{1}=0$ and by Equation (91) it
also immidiately follows that $D_{G}(\gamma_{0}\star\gamma_{0})_{1}=0$. So we
get that $D_{G}\gamma_{1}$ is equal to a $D_{G}$-closed form, but the
corresponding cohomology group is trivial. Hence it follows that
$D_{G}\gamma_{1}$ is equal to a $D_{G}$-exact form, and thus it is possible to
find a $\gamma_{1}$ that solves Equation (93).
By induction, one can show that in each order $\hbar^{k}$ for $k\geq 1$,
$D_{G}\gamma_{k}$ is equal to a $D_{G}$-closed and hence $D_{G}$-exact form
depending on the lower order coefficients of $F^{M}$ and $\gamma$. In
particular it follows that there exists a $\gamma_{k}$ solving the equation
for the corresponding order.

######  Remark 3.1.

Note that in order to globalize Kontsevich’s star product one may be tempted
to define a bullet product

(94) $(f\bullet
g)(x):=\big{(}P(\textsf{T}\phi^{*}\pi)(\textsf{T}\phi^{*}f\otimes\textsf{T}\phi^{*}g)\big{)}(x;0).$

This is indeed a well-defined global product on $C^{\infty}(M)[[\hbar]]$, but
it is in general not associative. To make this product associative one has to
introduce a quantization map (see e.g. [5])

(95) $\rho\colon H^{0}_{D_{G}}(\Gamma(\widehat{\textnormal{Sym}}(T^{*}M)))\to
H^{0}_{\overline{\mathcal{D}}_{G}}(\Gamma(\widehat{\textnormal{Sym}}(T^{*}M)[[\hbar]]))$

which then again leads to the global star product

(96)
$f\star_{{}_{M}}g:=\left(\rho^{-1}\big{(}\rho(\textsf{T}\phi^{*}f)\star\rho(\textsf{T}\phi^{*}g)\big{)}\right)\Big{|}_{y=0}.$

Here $\star$ denotes Kontsevich’s star product and $\star_{M}$ its global
version on $M$. Using the weights, we can also get an explicit expression for
the bullet product (94) by

(97)
$\big{(}P(\mathsf{T}\phi^{*}\pi)(\mathsf{T}\phi^{*}f\otimes\mathsf{T}\phi^{*}g)\big{)}(x;0)\\\
=\left(\sum\limits_{n=0}^{\infty}\frac{\hbar^{n}}{2^{2n}n!}(\mathsf{T}\phi^{*}_{x}\pi)^{i_{1}j_{1}}\cdots(\mathsf{T}\phi^{*}_{x}\pi)^{i_{n}j_{n}}(\mathsf{T}\phi^{*}_{x}f)_{\,,i_{1}\cdots
i_{n}}(\mathsf{T}\phi^{*}_{x}g)_{\,,j_{1}\cdots j_{n}}\right)(0).$

###  3.4. The lifted Curvature 2-form
$F(\overline{R},\overline{R},\mathsf{T}\overline{\phi}^{*}\pi)$

Let $M$ be a smooth manifold and let $\phi\colon TM\rightarrow M$ be a formal
exponential map and consider the lift $\overline{\phi}\colon TN\rightarrow N$
to the cotangent bundle $N=T^{*}M$. We set $x=(q,p)\in N$ and
$y=(\bar{q},\bar{p})\in T_{x}N$. Note that this is a particular case of a
canonical symplectic manifold.
We will consider the lifted vector fields $\overline{R}$ to the cotangent
case, which induce lifted interaction vertices within the Feynman graphs which
appear in the computation of the connection $1$-form and its curvature
$2$-form and see how these terms simplify. First we note that
$A(\overline{R},\textsf{T}\overline{\phi}^{*}\pi)$ is still given by

(98) $\displaystyle
A(\overline{R}_{x},\textsf{T}\overline{\phi}^{*}_{x}\pi)(\sigma_{x})={\mathrm{d}}x^{i}\sum\limits_{n=0}^{\infty}\frac{\hbar^{n}}{2^{n}n!}\frac{1+(-1)^{n}}{2^{n+1}(n+1)}(\textsf{T}\overline{\phi}^{*}_{x}\pi)^{i_{1}j_{1}}\cdots(\textsf{T}\overline{\phi}^{*}_{x}\pi)^{i_{n}j_{n}}(\overline{R}_{x})^{k}_{i,i_{1}\cdots
i_{n}}(\sigma_{x})_{\,,kj_{1}\cdots j_{n}}.$

The simplification in this case is a small one: All summands containing a term
$(\overline{R}_{x})^{k}_{i,i_{1}\cdots i_{n}}$ with more than one derivative
with respect to $\bar{p}$ will vanish [18].
For the case of the curvature $2$-form $F^{N}$ the simplification is more
interesting. Since for each non-vanishing coefficient
$(\textsf{T}\overline{\phi}^{*}_{x}\pi)^{ij}$ one of the two outgoing edges is
always representing a $\bar{q}$-derivative and the other corresponding edge
representing a $\bar{p}$-derivative (since we work with Darboux coordinates
around $x\in N$), we see that the sum in (85) terminates at $n=2$. Or put
differently, we only have to consider the graphs $\Gamma_{n}$ up to $n=2$,
i.e. with at most two wedges attached to the wheel consisting of two
$\overline{R}$-vertices (cf. Figure 2 in Section 2.1). Moreover, since the
Kontsevich weights (84) are, up to $n=2$, given by $w_{\Gamma_{0}}=0$,
$w_{\Gamma_{1}}=\frac{1}{24}$ and $w_{\Gamma_{2}}=0$, we get

(99) $\displaystyle
F^{N}_{x}=F(\overline{R}_{x},\overline{R}_{x},\textsf{T}\overline{\phi}_{x}^{*}\pi)=\frac{\hbar}{48}(\textsf{T}\overline{\phi}^{*}_{x}\pi)^{rs}(\overline{R}_{x})^{k}_{i,lr}(\overline{R}_{x})^{l}_{j,ks}{\mathrm{d}}x^{i}\wedge{\mathrm{d}}x^{j},$

where we sum over the indices $i,j,r,s,k,l$ and where again summands
containing a term $(\overline{R}_{x})^{k}_{i,lr}$ with more than one
derivative with respect to $\bar{p}$ vanish. So in the case of a cotangent
bundle we get a much simpler expression for the Weyl curvature $F^{N}$.

##  Appendix A Binomial Sums

Here we will treat the binomial sums appearing in the expression (67) and show
that we indeed get the results stated in (68).

###  A.1. $B(n)$

Let us start with

(100) $\displaystyle B(n)=\sum\limits_{l=0}^{n}{n\choose
l}\frac{(-1)^{l}}{2^{n+1}(n-l+1)}.$

Using the well known identity

(101) $\displaystyle\sum\limits_{l=0}^{n}(-1)^{l}{n+1\choose
l}=\sum\limits_{l=0}^{n}(-1)^{l}{n\choose l}\frac{n+1}{n-l+1}=(-1)^{n}$

it immidiately follows that

(102) $\displaystyle B(n)=\frac{(-1)^{n}}{2^{n+1}(n+1)}.$

###  A.2. $C(n)$

Let us continue with

(103) $\displaystyle C(n)=\sum\limits_{l=0}^{n}{n\choose
l}\frac{(-1)^{l}}{2^{l}(n-l+1)}.$

Using the identity (101), we can write

(104) $\displaystyle(n+1)C(n)=\sum\limits_{l=0}^{n}{n+1\choose
l}\left(-\frac{1}{2}\right)^{l}.$

Using the Binomial theorem we then find

(105) $\displaystyle\sum\limits_{l=0}^{n}{n+1\choose
l}\left(-\frac{1}{2}\right)^{l}=\left(\frac{1}{2}\right)^{n+1}-\left(-\frac{1}{2}\right)^{n+1},$

and hence

(106) $\displaystyle C(n)=\frac{1+(-1)^{n}}{2^{n+1}(n+1)}.$

###  A.3. $A(n)$

Finally, let us treat the case

(107) $\displaystyle
A(n)=\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}\sum\limits_{s=0}^{n-k-l}{n\choose
k}{n-k\choose l}{n-k-l\choose s}\frac{(-1)^{l+s}}{2^{n-k-s+1}(n-k-l-s+1)}.$

Write

(108) $\displaystyle
A(n)=\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}{n\choose k}{n-k\choose
l}\frac{(-1)^{l}}{2^{n-k+1}}\sum\limits_{s=0}^{n-k-l}{n-k-l\choose
s}\frac{(-2)^{s}}{(n-k-l-s+1)}.$

We first treat the innermost sum: Set $m=n-k-l$. Then

(109) $\displaystyle\sum\limits_{s=0}^{n-k-l}{n-k-l\choose
s}\frac{(-2)^{s}}{(n-k-l-s+1)}=\sum\limits_{s=0}^{m}{m\choose
s}\frac{(-2)^{s}}{(m-s+1)}.$

Using ${m+1\choose s}=\frac{m+1}{m-s+1}{m\choose s}$, we find that

(110) $\displaystyle\sum\limits_{s=0}^{m}{m\choose
s}\frac{(-2)^{s}}{(m-s+1)}=\frac{1}{m+1}\sum\limits_{s=0}^{m}{m+1\choose
s}(-2)^{s}.$

Applying the Binomial theorem we get that

(111) $\displaystyle\sum\limits_{s=0}^{m}{m+1\choose
s}(-2)^{s}=(-1)^{m+1}\big{(}1-2^{m+1}\big{)}.$

Plugging all of this into (108), we find

(112)
$\displaystyle\begin{split}A(n)&=\sum\limits_{k=0}^{n}\sum\limits_{l=0}^{n-k}{n\choose
k}{n-k\choose
l}\frac{(-1)^{l}}{2^{n-k+1}}\frac{(-1)^{n-k-l+1}}{n-k-l+1}\big{(}1-2^{n-k-l+1}\big{)}\\\
&=\sum\limits_{k=0}^{n}{n\choose
k}\frac{(-1)^{n-k+1}}{2^{n-k+1}}\sum\limits_{l=0}^{n-k}{n-k\choose
l}\frac{1}{n-k-l+1}\big{(}1-2^{n-k-l+1}\big{)}.\end{split}$

Moreover, we have

(113) $\displaystyle\sum\limits_{l=0}^{n-k}{n-k\choose
l}\frac{1}{n-k-l+1}=\frac{1}{n-k+1}\sum\limits_{l=0}^{n-k}{n-k+1\choose
l}=\frac{1}{n-k+1}\big{(}2^{n-k+1}-1\big{)},$

and

(114) $\displaystyle\sum\limits_{l=0}^{n-k}{n-k\choose
l}\frac{2^{n-k-l+1}}{n-k-l+1}=\frac{1}{n-k+1}\sum\limits_{l=0}^{n-k}{n-k+1\choose
l}2^{n-k-l+1}=\frac{1}{n-k+1}\big{(}3^{n-k+1}-1\big{)}.$

Hence

(115) $\displaystyle\begin{split}A(n)&=\sum\limits_{k=0}^{n}{n\choose
k}\frac{(-1)^{n-k+1}}{2^{n-k+1}(n-k+1)}\big{(}2^{n-k+1}-3^{n-k+1}\big{)}\\\
&=(-1)^{n+1}\sum\limits_{k=0}^{n}{n\choose
k}\frac{(-1)^{k}}{n-k+1}\left(1-\left(\frac{3}{2}\right)^{n-k+1}\right).\end{split}$

Now

(116) $\displaystyle\sum\limits_{k=0}^{n}{n\choose
k}\frac{(-1)^{k}}{n-k+1}=\frac{1}{n+1}\sum\limits_{k=0}^{n}{n+1\choose
k}(-1)^{k}=\frac{(-1)^{n}}{n+1},$

and

(117) $\displaystyle\begin{split}\sum\limits_{k=0}^{n}{n\choose
k}\frac{(-1)^{k}}{n-k+1}\Big{(}\frac{3}{2}\Big{)}^{n-k+1}&=\frac{(-1)^{n+1}}{n+1}\sum\limits_{k=0}^{n}{n+1\choose
k}\left(-\frac{3}{2}\right)^{n-k+1}=\frac{(-1)^{n}}{n+1}\\\
&=\frac{1}{2^{n+1}(n+1)}+\frac{(-1)^{n}}{n+1}.\end{split}$

Finally, we get

(118) $\displaystyle
A(n)=(-1)^{n+1}\left(\frac{(-1)^{n}}{n+1}-\frac{1}{2^{n+1}(n+1)}-\frac{(-1)^{n}}{n+1}\right)=\frac{(-1)^{n}}{2^{n+1}(n+1)}.$

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