text:
\section{Introduction and statement of the main result}
\label{ref-1-0}
This is a companion note to \cite{vdbcalaque}. We refer to \cite[\S 8]{vdbcalaque} and \cite{Ko3} for unexplained notations and conventions.

Below we will compute the Kontsevich weight $w_n$
of the ``opposite'' wheel
\[
\psfrag{m}[][]{}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\includegraphics[width=4cm]{wheel_n.eps}
\]
with edge ordering $(z_1,z_2)<\cdots<(z_n,z_1)<(w,z_1)<\cdots<(w,z_n)$.

Thus\footnote{This is the traditional Kontsevich weight. In
  \cite[\S8]{vdbcalaque} we use $W_n=
  (-1)^{2n(2n-1)/2}w_n=(-1)^nw_n$. Since $w_n=0$ for $n$ odd we actually
have $W_n=w_n$.}
\[
w_n=\frac{1}{(2\pi)^{2n}}
\int_{C_{n+1,0}} d\varphi(z_1,z_2)\cdots d\varphi(z_1,z_n) d\varphi(w,z_1)\cdots d\varphi(w,z_n)
\]
where $\varphi(u,v)$ is computed as in the following diagram.
\begin{center}
\psfrag{h}[l][l]{$\varphi(u,v)$}
\psfrag{p}[][]{$u$}
\psfrag{q}[][]{$v$}
\psfrag{e}[][]{$$}
\includegraphics[width=7cm]{kontsevich.eps}
\end{center}
and where $C_{n+1,0}$ is the configuration space of $n+1$ points in
the complex upper half plane ${\cal H}$.  If we normalize things by
putting $w=i$ then the integration domain is equal to the complement of
the big diagonal
in $({\cal H}-\{i\})^n$.

Recall (e.g.\
\cite{Wolfram}) that the modified Bernoulli numbers $\ss_n$
are defined by
\begin{align*}
\sum_{n} \ss_n x^n&=\frac{1}{2}\log \frac{e^{x/2}-e^{-x/2}}{x}\\
&=\frac{x^2}{48}
-\frac{x^4}{5760}+\frac{x^6}{ 362880}-\frac{x^8}{19353600}+
\frac{x^{10}}{958003200} - \frac{691 x^{12}}{31384184832000}
+
\cdots
\end{align*}
The main result of this note is a relatively elementary proof of the following
(see \S\ref{ref-9.6-62})
\begin{proposition} \cite[Theorem 18]{Willwacher}
\begin{equation}
\label{ref-1.1-1}
w_n=-(-1)^{n(n-1)/2}n\ss_n
\end{equation}
\end{proposition}
We needed \eqref{ref-1.1-1} for the proof of \cite[Thm
1.3]{vdbcalaque}. At that time we didn't have a reference so we did
the computation ourselves. After this note was posted on the arXiv
Torossian sent us an argument which shows that \eqref{ref-1.1-1}
follows from the work of Cattaneo and Felder \cite{CF2,CF3} on the
quantization of coisotropic manifolds combined with Shoikhet's
vanishing results for wheels with spokes pointing inwards
\cite{Shoikhet} and some facts related to the
Campbell-Baker-Hausdorff formula. In this way one obtains a shorter and more
conceptual proof of \eqref{ref-1.1-1}.
Torossian also pointed out Willwacher's
recent work \cite{Willwacher} which uses the same ingredients.

While more conceptual it seems this approach is substantially less
elementary than ours so we think that this note is still useful.

Let us finally also mention that a result similar to
\eqref{ref-1.1-1} was announced by Shoikhet
in \cite[\S2.3.1]{Shoikhet}.  It can presumably be obtained from the
methods in \cite{Shoikhet1}.

\medskip

The structure of this note is as follows: in \S\ref{ref-3-3} we represent
our computations graphically. This makes it easy to see
which boundary components in Stokes theorem yield non-trivial
contributions. We ignore signs as these are not easy to handle graphically.

In \S\ref{ref-4-14}-\S\ref{ref-8-47} we compute the contributions of the
relevant boundary components with precise signs. Thanks to some
``lucky'' cancellations we obtain a system of recursions for computing
$w_n$.

Finally in \S\ref{ref-9-53} we solve the recursions.

The reader will notice that  some formulas
and figures appear more than once in this paper.
This has been done to make the three logical parts
\S\ref{ref-3-3},\S\ref{ref-4-14}-\S\ref{ref-8-47},\S\ref{ref-9-53} more or less
independently readable. This should make our computations easier to follow.

\medskip

As indicated above we use almost nothing but Stokes theorem in our
computations. For a general introduction to the use of Stokes theorem in the
computation of Kontsevich weights see Torossian's article in \cite{CKTB}

\section{Acknowledgment}
\label{ref-2-2}
The symbolic computations were done using the SAGE \cite{SAGE}
interface to the computer algebra package Maxima \cite{Maxima}. Many
identities and integrals were tested numerically
using the Monte Carlo integration library which is part of GSL
\cite{GSL}.

We thank Charles Torossian for pointing how \eqref{ref-1.1-1} follows
from the work of Cattaneo and Felder.

Finally I would like to specially thank the anonymous referee for his very careful
reading of the manuscript. This has allowed me to improve the
exposition and to correct numerous ``misprints''.
\section{Graphical representation of  the computation}
\label{ref-3-3}
In this section we give a graphical overview of our computations. We will
ignore signs.  Hence we don't care about orientations or the ordering
of the factors in a product of forms.

\subsection{Graphical language}
\label{ref-3.1-4}
In the Kontsevich integral we associate to a graph $\Gamma$ a form $\omega_\Gamma$
on a suitable configuration space $C^+_{p,q}$. The form $\omega_\Gamma$ is the product of $d\varphi(u,v)$, where $(u,v)$ runs over the edges of $\Gamma$, divided by $(2\pi)^e$ where $e$ is the number of edges.

Besides the usual Kontsevich graphs we introduce graphs with some new
kind of arrows.

An arrow of the form
\[
\psfrag{n}[b][b]{$n$}
\psfrag{e}[t][t]{$$}
\psfrag{p}[t][t]{$u$}
\psfrag{q}[t][t]{$v$}
\includegraphics[width=2cm]{arrow_with_multiplicity.eps}
\]
corresponds to a factor $d(\varphi(u,v)^n)/(2\pi)^n$ in $\omega_\Gamma$.
For such an arrow
we assume that we have made a branch cut such that $\varphi(u,v)\in ]0,2\pi[$.
I.e.\ if $\operatorname{Re} u=\operatorname{Re} v$ then $\operatorname{Im} v<\operatorname{Im} u$. Note that we have to specify the branch cut
as the value of the integral depends on it.

A bold arrow of the form
\[
\psfrag{n}[b][b]{$n$}
\psfrag{e}[t][t]{$$}
\psfrag{p}[t][t]{$u$}
\psfrag{q}[t][t]{$v$}
\includegraphics[width=2cm]{bold_arrow.eps}
\]
corresponds to a factor $\varphi(u,v)^n/(2\pi)^n$  in $\omega_\Gamma$. We make the same
branch cut as above.  If $n$ is not indicated then we assume $n=1$.

We introduce a third kind of special arrow. A dashed arrow of the form
\[
\psfrag{e}[t][t]{$$}
\psfrag{p}[t][t]{$u$}
\psfrag{q}[t][t]{$v$}
\includegraphics[width=2cm]{dotted_arrow.eps}
\]
indicates a restriction of the integration domain to $\operatorname{Re} v=\operatorname{Re} u+\epsilon$,
$\operatorname{Im} v>\operatorname{Im} u$ with $\epsilon$ a positive infinitesimal.

Here is the basic computation rule for computing $\int \omega_\Gamma$ for one of the
generalized graphs (Rule 1).
{\psfrag{n}[b][b]{$n$}
\psfrag{e}[t][t]{$$}
\psfrag{p}[t][t]{$u$}
\psfrag{q}[t][t]{$v$}
\begin{align*}
\int \includegraphics[width=2cm]{arrow_with_multiplicity.eps}&=
\pm\int_\partial \includegraphics[width=2cm]{bold_arrow.eps}
\end{align*}
\begin{align*}
\int_\partial \includegraphics[width=2cm]{bold_arrow.eps}
&=\left(\int_{\partial_{\text{classical}}} \includegraphics[width=2cm]{bold_arrow.eps}\right)\pm \left(\int \includegraphics[width=2cm]{dotted_arrow.eps}\right)
\end{align*}
}
Here $\partial$ denotes the boundary of the integration domain (including the
branch cut). $\partial_{\text{classical}}$ denotes the boundary without
the branch cut. $\partial_{\text{classical}}$ can be determined combinatorially
by contracting vertices in $\Gamma$ and moving vertices to the real line.

The difference between $\partial$ and $\partial_{\text{classical}}$ is given by the
two sides of the branch cut. The difference in value of $\varphi(u,v)$ on
opposite sides of the branch cut is $2\pi$. This gives the contribution with the
dashed arrow.

Here is another obvious computation rule (Rule 2)
\[
\psfrag{n}[b][b]{$n$}
\psfrag{n1}[b][b]{$n+1$}
\psfrag{e}[t][t]{$e$}
\psfrag{p}[t][t]{$p$}
\psfrag{q}[t][t]{$q$}
\raisebox{-2.1mm}{\includegraphics[width=2cm]{double_arrow.eps}}\quad=\quad
\frac{1}{n+1}\times
\raisebox{-1.5mm}{\includegraphics[width=2cm]{double_arrow_simplified.eps}}
\]
Below we will also consider some integrals over the configuration
spaces $C_p$ consisting of $p$ points in the complex plane. These are it terms of $\theta(u,v)$ where $\theta(u,v)$
is as in the following diagram
\begin{center}
\psfrag{h}[l][l]{$\theta(u,v)$}
\psfrag{p}[][]{$u$}
\psfrag{q}[][]{$v$}
\psfrag{e}[][]{$$}
\includegraphics[width=2cm]{theta_def.eps}
\end{center}
To indicate that we are integrating over some $C_p$ we will put the graph
in a box. Otherwise we use the same conventions as above (with
$\theta(u,v)$ replacing $\varphi(u,v)$). Since $d\theta(u,v)=d\theta(v,u)$
some arrow indicators are superfluous. We will omit those.

We will repeatedly use the following vanishing result
\begin{equation}
\label{lem66}
\int \raisebox{-0.4cm}{\includegraphics[width=1cm]{simple_box.eps}}=0
\end{equation}
provided the graph in the box contains only ordinary edges and
the number of edges is $\ge 3$. An integral as in \eqref{lem66} often arises
as a factor of an integral over a boundary component of $C_{n+1,0}$.
\subsection{Applying Stokes theorem to $w_n$}
\label{ref-3.2-5}
We will now apply this graphical language to computations with wheels.
We assume first $n\ge 2$

The first step is (using Rule 1)
\begin{equation}
\label{ref-3.1-6}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$$}
\int_{C_{n+1,0}} \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_n.eps}}=\pm
\int_{\partial_{\text{classical}}}\raisebox{-2.5cm}{ \includegraphics[width=4cm]{wheel_n_bold.eps}}
\pm \int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed.eps}}
\end{equation}
We now simplify the integral over  $\partial_{\text{classical}}$ by looking
at the different components of $\partial_{\text{classical}}$.
\begin{enumerate}
\item A group of vertices moves to the real line. Since every vertex
  has a least one normal outgoing arrow this means that to obtain a
  non-zero result all vertices should move to the real line which is
  excluded. So components of this type contribute nothing.
\item A group of vertices $S$ comes together. Assume first $w\in S$.
  If $z_i\in S$, $i<n$ then to avoid double arrows after contraction
  we should also have $z_{i+1}\in S$.  If $z_1\in S$ this means we
  have to contract all vertices. I.e.\ the integral over this
  component becomes an integral over $C_{n+1}$
\[
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[][]{$$}
\psfrag{w}[][]{$w$}
\int_{C_{n+1}}\raisebox{-2cm}{\includegraphics[width=4cm]{wheel_n_bold_euclidean.eps}}
\]
\item
Assume $z_1\not\in S$. If $z_i\in S$ for $i<n$ then we contract more than one edge and the resulting
integral is zero by \eqref{lem66}. Thus $S=\{w,z_n\}$. The resulting
integral is (using Rule 2)
\begin{equation}
\label{ref-3.2-7}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$2$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
\frac{1}{2}\int_{C_{n,0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_n_bold_contracted.eps}}
\end{equation}
\item Now assume $w\not \in S$. This means that we contract an edge
  $(z_i,z_{i+1})$ or $(z_n,z_1)$ (if we contract more edges then the
  integral is zero by \eqref{lem66}). The only contractions which do not
  create double edges are $(z_1,z_2)$ and~$(z_n,z_1)$. In
  that case the resulting integrals are (using Rule 2)
\begin{equation}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[r][r]{$2$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
\frac{1}{2}\int_{C_{n.0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_bold_contract_variant.eps}}
\end{equation}
and
\begin{equation}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_{n-1}$}
\psfrag{z2}[][]{$z_1$}
\psfrag{m}[r][r]{$2$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
\frac{1}{2}\int_{C_{n,0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_bold_contract_variant.eps}}
\end{equation}
\end{enumerate}

\subsection{The boundary component associated to the branch cut}
\label{ref-3.3-8}
We need to compute the integral associated to the graph with the dashed
arrow in \eqref{ref-3.1-6}. In order to derive a recursion relation
we work more generally, starting with the following identity.
\begin{equation}
\label{ref-3.5-9}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{z3}[r][r]{$z_3$}
\psfrag{m}[l][l]{$m$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
 \int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed_m.eps}}
=\pm\int_{\partial_{\text{classical}}} \raisebox{-2.5cm}{%
\includegraphics[width=4cm]{wheel_dashed_bold.eps}}
\pm \int \raisebox{-2.5cm}{%
\includegraphics[width=4cm]{wheel_dashed_dashed.eps}}
\end{equation}
However the integral containing the two dashed arrows is zero since  if
$z_1$, $z_2$ move on the same vertical line then $d\varphi(z_1,z_2)=0$.

It remains to determine the integral over $\partial_{\text{classical}}$. We consider
the various boundary components. We now assume $n\ge 3$.
\begin{enumerate}
\item A group of vertices moves to the real line. Since every vertex
  has at least one normal outgoing arrow (using $n\ge 2$) this means
  that to obtain a non-zero result all vertices should move to the
  real line which is excluded. So no contribution from these components.
\item A group of vertices $S$ comes together. Assume first $w\in S$.
  If $z_i\in S$, $i\neq 1,n$ then to avoid double arrows after
  contraction we should also have $z_{i+1}\in S$. However if $z_2\in
  S$ or $z_n\in S$ then $z_1\in S$ for otherwise after contraction a
  solid and a dashed arrow coincide and thus the result is zero. Thus
  we either contract all vertices or we contract
  $w,z_i,\ldots,z_n,z_1$ for $i\ge 3$ or we contract $(w,z_1)$. If we contract all vertices
  then the integral becomes
\[
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{z3}[r][r]{$z_3$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[][]{$w$}
\int \raisebox{-2cm}{\includegraphics[width=4cm]{wheel_dashed_bold_boxed.eps}}
\]
\item Assume we contract $w,z_i,\ldots,z_n,z_1$ for $3\le i\le n$. Then the resulting
integral is \emph{a multiple of}
\begin{equation}
\label{ref-3.6-10}
\psfrag{z1}[][]{$z_1$}
\psfrag{zi}[][]{$z_i$}
\psfrag{zi1}[l][l]{$z_{i+1}$}
\psfrag{zn}[r][r]{$z_n$}
\psfrag{m}[][]{$m$}
\psfrag{w}[][]{$w$}
\int \raisebox{-2cm}{\includegraphics[width=4cm]{something_zero.eps}}
\end{equation}
\item It remains to consider the case $S=\{z_1,w\}$. In that case we get using
Rule 2:
\[
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_2$}
\psfrag{z2}[][]{$z_3$}
\psfrag{m}[l][l]{$m+1$}
\psfrag{zn-1}[l][l]{$z_{n}$}
\frac{1}{m+1}\int_{C_{n,0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_n_bold_contracted.eps}}
\]
\item Now assume $w\not \in S$. This means that we contract an edge
  $(z_i,z_{i+1})$ or $(z_n,z_1)$ (if we contract more edges then the
  integral is zero by \eqref{lem66}). The only two possibilities
  which do not give zero immediately are $(z_1,z_2)$ and $(z_2,z_3)$. The
  resulting integrals are
\[
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_2$}
\psfrag{z2}[][]{$z_3$}
\psfrag{m}[l][l]{}
\frac{1}{2} \int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed.eps}}
\]
and (using Rule 2)
\[
\psfrag{w}[][]{$w$}
\psfrag{t}[r][r]{$m+1$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_2$}
\psfrag{z2}[][]{$z_3$}
\psfrag{m}[l][l]{}
\frac{1}{m+1} \int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_contracted_variant.eps}}
\]
The factor $\frac{1}{2}$ in the first integral comes from the fact
that  in this boundary component we have $\varphi(w,z_2)=0$ if $z_2$
is to the left of $z_1$.
\end{enumerate}
As will be explained below we need a second method for attacking the term with
the dashed arrow in \eqref{ref-3.1-6}. Therefore we use
\[
\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed_m_mirror.eps}}
=\pm\int_{\partial_{\text{classical}}} \raisebox{-2.5cm}{%
\includegraphics[width=4cm]{wheel_dashed_bold_m_mirror.eps}}
\pm \int \raisebox{-2.5cm}{%
\includegraphics[width=4cm]{wheel_dashed_dashed_m_mirror.eps}}
\]
Again the integral containing the two dashed arrows is zero. Likewise is the
integral over the boundary components where some vertices move to the real
line. We discuss the other boundary components of $\partial_{\text{classical}}$
assuming $n\ge 3$.
\begin{enumerate}
\item A group of vertices $S$ comes together. Assume first $w\in S$.
  If $z_i\in S$, $1\le i<n-1$  then to avoid double arrows after
  contraction we should also have $z_{i+1}\in S$. However if $z_n\in
  S$ of $z_2\in S$ then $z_1\in S$ for otherwise after contraction
a solid and a dashed arrow coincide and thus the result is zero.

It follows that the possibilities for contraction are $\{z_1,\ldots,z_n,w\}$,
$\{z_1,\ldots,z_{n-1},w\}$ or
$\{z_i,\ldots,z_{n-1},w\}$ for $i\ge 3$. In  the last case we must have
$i=n-1$ by \eqref{lem66}.

If we contract all vertices then the contribution to the integral is
\[
\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-1.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_m_mirror_boxed.eps}}
\]
\item If we contract
$w,z_1,\ldots,z_{n-1}$ then the resulting integral is \emph{a multiple of}
\begin{equation}
\label{ref-3.7-11}
\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{zn2}[l][l]{$z_{n-2}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-1.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_contracted_euclidean.eps}}
\end{equation}
\item It remains to consider the case
$S=\{z_{n-1},w\}$. In that case we obtain (using Rule 2)
\[
\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{zn2}[l][l]{$z_{n-2}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m+1$}
\psfrag{w}[l][l]{$w$}
\frac{1}{m+1}\int \raisebox{-1.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_n_1_contraction.eps}}
\]
\item
Now assume $w\not \in S$. This means that we contract an edge
$(z_n,z_1)$ or $(z_{n-1},z_{n})$ (otherwise the result is zero by  \eqref{lem66} or by the fact that we create double arrows). The
resulting integrals are
\begin{equation}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_{n-1}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[r][r]{$m+1$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
\frac{1}{2}\int_{C_{n.0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_dashed.eps}}
\end{equation}
and (using Rule 2)
\[
\psfrag{w}[][]{$w$}
\psfrag{m}[l][l]{$m+1$}
\psfrag{zn1}[][]{$z_{n-1}$}
\psfrag{z1}[][]{$z_2$}
\psfrag{z2}[][]{$z_3$}
\frac{1}{m+1} \int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_m_mirror_contracted.eps}}
\]
The factor $1/2$ in the first integral comes from the fact
that due to our branch cut  $\varphi(w,z_n)=0$ if $z_2$
is to the left of $z_1$.
\end{enumerate}

\subsection{Further comments}
\label{ref-3.4-12}
We now indicate how the relations we have derived will be used.
We still do not specify precise signs.
\begin{enumerate}
\item We first discuss the partial evaluation of $w_n$ using
  \eqref{ref-3.1-6}. One ingredient that enters is \eqref{ref-3.2-7}. We
  define
\[
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{zn-1}[l][l]{$z_{n}$}
\alpha_{n,m}=\int_{C_{n+1,0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_n_bold_contracted.eps}}
\]
We will evaluate $\alpha_{n,m}$ by subsequently integrating over
$z_1$, $z_2$, etc\dots. See \S\ref{ref-4-14} below.
\item Another ingredient that enters is \eqref{ref-8.2-51}.
\[
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[][]{$$}
\psfrag{w}[][]{$w$}
\int_{C_{n+1}}\raisebox{-2cm}{\includegraphics[width=4cm]{wheel_n_bold_euclidean.eps}}
\]
We do not know how to evaluate this integral in general but we will show
by a symmetry argument that it is zero if $n$ is even (see \S\ref{ref-8.2-50}).
\item The final ingredient is given by the two identically looking integrals
\[
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[r][r]{$2$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
\int_{C_{n.0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_bold_contract_variant.eps}}
\]
and
\[
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_{n-1}$}
\psfrag{z2}[][]{$z_1$}
\psfrag{m}[r][r]{$2$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
\int_{C_{n,0}}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_bold_contract_variant.eps}}
\]
We will show that the signs are such that these two integrals cancel (see \S\ref{ref-8.3-52}).
\item So ultimately we get an expression for $w_n$ in terms
  of $\alpha_{n-1,m+1}$ (which we know how to compute) and
\[
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed.eps}}
\]
which is a special case of both the following integrals
\[
\beta_{n,m}\overset{\text{def}}{=}
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{z3}[r][r]{$z_3$}
\psfrag{m}[l][l]{$m$}
\psfrag{zn-1}[l][l]{$z_{n-1}$}
\int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed_m.eps}}
\]
and
\[
\bar{\beta}_{n,m}\overset{\text{def}}{=}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed_m_mirror.eps}}
\]
\item We now discuss $\beta_{n,m}$ using \eqref{ref-3.5-9}. We will show
  below (see \S\ref{ref-5.2-32}) that
\[
\psfrag{z1}[][]{$z_1$}
\psfrag{zi}[][]{$z_i$}
\psfrag{zi1}[l][l]{$z_{i+1}$}
\psfrag{zn}[r][r]{$z_n$}
\psfrag{m}[][]{$m$}
\psfrag{w}[][]{$w$}
\int \raisebox{-2cm}{\includegraphics[width=4cm]{something_zero.eps}}
\]
is zero. Hence we obtain an expression for $\beta_{n,m}$ in terms of
$\alpha_{n-1,m+1}$, $\beta_{n-1,1}$, $\beta_{n+1,m-1}$ and
\[
\epsilon_{n,m} \overset{\text{def}}{=}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{z3}[r][r]{$z_3$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[][]{$w$}
\int \raisebox{-2cm}{\includegraphics[width=4cm]{wheel_dashed_bold_boxed.eps}}
\]
\item In a similar way we discuss  $\bar{\beta}_{n,m}$. As already pointed out
we will show that the integral
\[
\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{zn2}[l][l]{$z_{n-2}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-1.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_contracted_euclidean.eps}}
\]
is zero (see \S\ref{ref-5.2-32} below).
We will also show that the integral
\[
\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{zn2}[l][l]{$z_{n-2}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m+1$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-1.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_n_1_contraction.eps}}
\]
is zero (see \S\ref{ref-6.1-40}).

As a result we obtain an expression for $\bar{\beta}_{n,m}$ in terms
of  $\bar{\beta}_{n-1,m+1}$, $\beta_{n-1,1}$ and
\[
\bar{\epsilon}_{n,m}\overset{\text{def}}{=}\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-1.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_m_mirror_boxed.eps}}
\]
\item We will show below by a symmetry argument that $\epsilon_{n,m}=-\bar{\epsilon}_{n,m}$ (see \S\ref{ref-7.1-44}).
\item Unfortunately we do not know how to compute $\epsilon_{n,m}$.
  However we will show below that by considering a suitable linear
  combination of $\beta_{n,m}$ and $\bar{\beta}_{n,m}$ the unknown
  quantity $\epsilon_{n,m}$ cancels out in the computation. This will ultimately
give us a recursive procedure for computing $w_n$.
\end{enumerate}
\section{A recursion relation for  $\alpha_{n,m}$}
\label{ref-4-14}
 By our definition in \S\ref{ref-3.4-12}, $\alpha_{n,m}$ is the
integral associated to the enhanced graph
\[
\psfrag{w}[][]{$w$}
\psfrag{zn}[][]{$z_n$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{zn-1}[l][l]{$z_{n}$}
\raisebox{-2.5cm}{\includegraphics[width=4.5cm]{wheel_n_bold_contracted.eps}}
\]
Thus with our standard ordering of edges we have
\[
\alpha_{n,m}=\frac{1}{(2\pi)^{2n+m-1}}\int_{C_{n+1,0}} d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n) d\varphi(z_n,w) d\varphi(w,z_1)^m d\varphi(w,z_2)\cdots d\varphi(w,z_{n})
\]
In this section we prove
\begin{equation}
\label{ref-4.1-15}
\alpha_{1,m}=-\left(\frac{1}{2}-\frac{1}{m+1}\right)
\end{equation}
and for $n\ge 2$.
\begin{equation}
\label{ref-4.2-16}
\alpha_{n,m}=(-1)^n \left( \frac{1}{2}\alpha_{n-1,2}-\frac{1}{m+1} \alpha_{n-1,m+1}\right)
\end{equation}
We put $\alpha_{0,m}=1$ such that \eqref{ref-4.2-16}
holds for $n\ge 1$.
\subsection{Step 1} We will prove
\label{ref-4.1-17}
\begin{equation}
\label{ref-4.3-18}
\int_{z\in {\cal H}\setminus\{z_1,z_2\}}
d\varphi(z_1,z)^m d\varphi(z,z_2)=(2\pi)^m \varphi(z_1,z_2)-2\pi \varphi(z_1,z_2)^m
\end{equation}
where we have made a branch cut such that $\varphi(z_1,z)\in ]0,2\pi[$.

In Figure \ref{ref-1-13}
\begin{figure}
\psfrag{C1}[][]{$C_1$}
\psfrag{C2}[][]{$C_2$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{Lp}[r][r]{$B_+$}
\psfrag{Lm}[l][l]{$B_-$}
\psfrag{Cp}[][]{$C_+$}
\psfrag{Cm}[][]{$C_-$}
\psfrag{H}[][]{$H$}
\includegraphics[width=10cm]{branch_cut_th_case.eps}
\caption{}
\label{ref-1-13}
\end{figure}
we have indicate the integration domain and its boundary. Here $C_+\cup C_-$ is the
circle at infinity, $B_+$, $B_-$ are infinitesimally close together and $C_1$, $C_2$
have infinitesimal radius. Finally $H$ is the real line. Thus we have
\[
\int_{z\in \CC\setminus \{z_1,z_2\}} d\varphi(z_1,z)^m d\varphi(z,z_2)=
\int_{\text{boundary}} \varphi(z_1,z)^m d\varphi(z,z_2)
\]
We first compute the integral over $C_{\pm}\cup B_{\pm}\cup C_1\cup H$.
On $B_-\cup C_-$
we have that $\varphi(z_1,z)$ is constant and equal to $2\pi$. On $B_+\cup C_+$ we
have that $\varphi(z_1,z)=0$.  On $H\cup C_1$ we have that $d\varphi(z,z_2)=0$.

Thus we find
\[
\int_{B_{\pm}\cup C_{\pm}\cup C_1\cup H}\varphi(z_1,z)^md\varphi(z,z_2)=
(2\pi)^m \int_{B_{-}\cup C_{-}}d\varphi(z,z_2)=(2\pi)^m \varphi(z_1,z_2)
\]
The integral over $C_2$ is equal to $-2\pi \varphi(z_1,z_2)^m$. This
finishes the proof of \eqref{ref-4.3-18}.
\subsection{Step 2}  We will prove
\label{ref-4.2-20}
\begin{equation}
\label{ref-4.4-21}
\int_{z\in {\cal H}\setminus \{w\}} d\varphi(w,z)^m d\varphi(z,w)=
(2\pi)^{m+1} \left(\frac{1}{2}-\frac{1}{m+1}\right)
\end{equation}
 In Figure \ref{ref-2-19} we have indicated the integration domain and its boundary.
\begin{figure}
\psfrag{z1}[][]{$w$}
\psfrag{C1}[][]{$C_1$}
\psfrag{Lp}[r][r]{$B_+$}
\psfrag{Lm}[l][l]{$B_-$}
\psfrag{Cp}[][]{$C_+$}
\psfrag{Cm}[][]{$C_-$}
\psfrag{H}[][]{$H$}
\includegraphics[width=10cm]{branch_cut_once_more}
\caption{}
\label{ref-2-19}
\end{figure}
 Thus we
have
\[
\int_{z\in \CC\setminus \{w\}} d\varphi(w,z)^m d\varphi(z,w)=
\int_{\text{boundary}} \varphi(w,z)^m d\varphi(z,w)
\]
On $B_-\cup C_-$
we have that $\varphi(w,z)$ is constant and equal to $2\pi$. On $B_+\cup C_+$ we
have that $\varphi(w,z)=0$.  On $H$ we have that $d\varphi(z,w)=0$.
Thus
\[
\int_{C_{\pm}\cup B_{\pm}\cup H} \varphi(w,z)^m d\varphi(z,w)=(2\pi)^m\pi
\]
It remains to compute
\[
\int_{C_1} \varphi(w,z)^m d\varphi(z,w)
\]
We put $z=w+re^{-i\theta}$ for $\theta\in ]-\pi/2,3\pi/2[$. We have
$\varphi(w,z)=3\pi/2-\theta$ and hence $d\varphi(z,w)=-d\theta$. Thus
we get
\[
\int_{C_1} \varphi(w,z)^m d\varphi(z,w)=-\int_{-\pi/2}^{3\pi/2}  (3\pi/2-\theta)^md\theta
=-\int_0^{2\pi}\alpha^m d\alpha=-\frac{(2\pi)^{m+1}}{m+1}
\]
This finishes the proof of \eqref{ref-4.4-21}.
\subsection{Step 3} Now we prove \eqref{ref-4.1-15}.
\label{ref-4.3-22}
\[
\alpha_{1,m}=\frac{1}{(2\pi)^{m+1}}\int d\varphi(z_1,w) d\varphi(w,z_1)^m=
-\frac{1}{(2\pi)^{m+1}}\int  d\varphi(w,z_1)^m d\varphi(z_1,w)=
-\left(\frac{1}{2}-\frac{1}{m+1}\right)
\]
by \eqref{ref-4.4-21}.
\subsection{Step 4}
\label{ref-4.4-23}
Now we prove \eqref{ref-4.2-16}.
\[
\alpha_{n,m}=(-1)^{n}\frac{1}{(2\pi)^{2n+m-1}}\int d\varphi(w,z_1)^m d\varphi(z_1,z_2) \cdots d\varphi(z_{n-1},z_n) d\varphi(z_n,w) d\varphi(w,z_2)\cdots d\varphi(w,z_{n})
\]
Integrating over $z_1$ and using \eqref{ref-4.3-18} first we find
\begin{multline*}
\alpha_{n,m}=(-1)^{n}\frac{1}{(2\pi)^{2n+m-1}}\int ((2\pi)^m \varphi(w,z_2)-
2\pi \varphi(w,z_2)^m)\times\\
d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n) d\varphi(z_n,w) d\varphi(w,z_2)\cdots d\varphi(w,z_{n})
\end{multline*}
which is equal to
\begin{multline*}
(-1)^{n}\frac{1}{2(2\pi)^{2n-1}}\int  d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n) d\varphi(z_n,w)  d\varphi(w,z_2)^2   \cdots d\varphi(w,z_{n-1})\\
-
(-1)^{n}\frac{1}{(m+1)(2\pi)^{2n+m-2}} \int d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n) d\varphi(z_n,w) d\varphi(w,z_2)^{m+1}\cdots d\varphi(w,z_{n})
\end{multline*}
So that we indeed find
\[
\alpha_{n,m}=(-1)^n \left( \frac{1}{2}\alpha_{n-1,2}-\frac{1}{m+1} \alpha_{n-1,m+1}\right)
\]
\section{A recursion relation for $\beta_{n,m}$}
\label{ref-5-24}
 By our definition in \S\ref{ref-3.4-12}, $\beta_{n,m}$ is the
integral associated to the enhanced graph
\[
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[][]{$w$}
\includegraphics[width=4cm]{wheel_dashed_m.eps}
\]
Thus with our standard edge ordering
\[
\beta_{n,m}=
\frac{1}{(2\pi)^{2n+m-2} }
\int_{B_-\setminus D}
 d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
 d\varphi(w,z_2)^m\cdots d\varphi(w,z_{n})
\]
where
\[
B_-=\{\operatorname{Re} z_1=\operatorname{Re} w+, \operatorname{Im} {z_1}>\operatorname{Im} w\}\subset C_{n+1,0}
\]
\[
D=\{\operatorname{Re} z_2=\operatorname{Re} w, \operatorname{Im} {z_2}>\operatorname{Im} w\}\subset C_{n+1,0}
\]
($D$ represents the branch cut associated to the arrow
$w\xrightarrow{m} z_2$, see \S\ref{ref-3.1-4}).  We need to be careful
about specifying the orientation of the integration domain. We do this
next. We first normalize things by putting $w=i$.  If $z_j=x_j+iy_j$
then $C_{n+1,0}$ is oriented by \cite{AMM1}
\[
dx_1dy_1dx_2dy_2\cdots dx_ndy_n
\]
In the neighborhood of $B_-$ we have coordinates
$(x_1,y_1,\ldots,x_n,y_n)$ with $x_1\ge 0$. Therefore if $B_-$ is
oriented with its outgoing normal (as is necessary for the application
of Stokes theorem) then it is oriented by
\begin{equation}
\label{ref-5.1-25}
-dy_1dx_2dy_2\cdots dx_ndy_n
\end{equation}
Having determined the orientation of $B_-$ we let $w$ be free again. In this section we will prove
\begin{equation}
\label{ref-5.2-26}
\beta_{2,m}=-\frac{1}{8}+\frac{1}{m+1}\left( \frac{1}{2}-\frac{1}{m+2}\right)
\end{equation}
and for $n\ge 3$
\begin{equation}
\label{ref-5.3-27}
\beta_{n,m}=-\frac{1}{m+1}\alpha_{n-1,m+1}
+(-1)^{n+1} \frac{1}{2}\beta_{n-1,1}+(-1)^{n} \frac{1}{m+1} \beta_{n-1,m+1}
+(-1)^{n}\epsilon_{n,m}
\end{equation}
where $\epsilon_{n,m}$ is the integral associated to the graph
\[
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{z3}[r][r]{$z_3$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[][]{$w$}\raisebox{-2cm}{\includegraphics[width=4cm]{wheel_dashed_bold_boxed.eps}}
\]
(see \S\ref{ref-3.4-12}). Thus with our standard orientation on edges
\[
\epsilon_{n,m}=
\frac{1}{(2\pi)^{2n+m-2}}\int \theta(w,z_2)^m d\theta(z_1,z_2)d\theta(z_2,z_3)
\cdots d\theta(z_n,z_1) d\theta(w,z_3)\cdots d\theta(w,z_n)
\]
Here we put $w=0$, $z_1=i$ and the integral is over the complement
of the diagonal in $(\CC-\{0,i\})^{n-1}$.
\subsection{Step 1} We first consider the case $n=2$. Thus
\label{ref-5.1-28}
\[
\beta_{2,m}=
\frac{1}{(2\pi)^{2+m}}
\int_{B_-\setminus D}
 d\varphi(z_1,z_2)d\varphi(z_{2},z_1)
 d\varphi(w,z_2)^m
\]
We first integrate over $w$. We get
\[
\int_{B_-\setminus D}  d\varphi(w,z_2)^m=
\varphi(z_1,z_2)^m-(2\pi)^m [z_1,z_2]
\]
where
\[
[z_1,z_2]=
\begin{cases}
1&\text{$z_1$ is to the left of $z_2$}\\
0&\text{otherwise}
\end{cases}
\]
Thus we must compute
\begin{equation}
\label{ref-5.4-29}
\frac{1}{(2\pi)^{2+m}}
\int
\varphi(z_1,z_2)^m d\varphi(z_1,z_2)d\varphi(z_{2},z_1)
\end{equation}
and
\begin{equation}
\label{ref-5.5-31}
\frac{1}{(2\pi)^{2}}
\int [z_1,z_2] d\varphi(z_1,z_2)d\varphi(z_{2},z_1)
\end{equation}
Using Stokes theorem we rewrite \eqref{ref-5.5-31} as a path integral as
in Figure \ref{ref-3-30}
\begin{figure}
\psfrag{L1}[r][r]{$L_1$}
\psfrag{L2}[r][r]{$L_2$}
\psfrag{C1}[l][l]{$C_1$}
\psfrag{C}[l][l]{$C$}
\psfrag{z1}[r][r]{$z_1$}
\psfrag{H}[r][r]{$H$}
\includegraphics[width=8cm]{yet_another_one.eps}
\caption{}
\label{ref-3-30}
\end{figure}
\[
\frac{1}{(2\pi)^{2}}
\int_{L_1\cup C_1\cup L_2\cup H\cup C} \varphi(z_1,z)d\varphi(z,z_1)
\]
If $z\in L_{1,2}\cup H$ then $d\varphi(z,z_1)=0$. If $z\in C$ then
$\varphi(z_1,z)=2\pi$.

Thus
\[
\frac{1}{(2\pi)^{2}} \int_{L_1\cup L_2\cup H\cup C}
\varphi(z_1,z)d\varphi(z,z_1)=\frac{1}{(2\pi)}(\pi-0)=\frac{1}{2}
\]
Now we compute
\[
\frac{1}{(2\pi)^{2}} \int_{C_1}
\varphi(z_1,z)d\varphi(z,z_1)
\]
We put $z=z_1+re^{-i\theta}$ for $\theta\in ]-\frac{\pi}{2},\frac{\pi}{2}[$.
Then $\varphi(z_1,z)=\frac{3}{2}\pi-\theta$  and hence $d\varphi(z,z_1)=-d\theta$.

Thus
\begin{align*}
\frac{1}{(2\pi)^{2}} \int_{C_1}
\varphi(z_1,z)d\varphi(z,z_1)
&=
-\frac{1}{(2\pi)^{2}}\int_{-\pi/2}^{\pi/2} (\frac{3}{2}\pi-\theta)d\theta\\
&=\frac{1}{(2\pi)^{2}}\int_{-\pi/2}^{\pi/2} (\theta-\frac{3}{2}\pi)d\theta\\
&=\frac{1}{(2\pi)^{2}}\int_{-2\pi}^{-\pi} \theta d\theta\\
&=-\frac{3}{8}
\end{align*}
and hence
 \[
\frac{1}{(2\pi)^{2}}
\int [z_1,z_2] d\varphi(z_1,z_2)d\varphi(z_{2},z_1)=\frac{1}{2}-\frac{3}{8}=\frac{1}{8}
\]
One the other hand by \eqref{ref-4.3-18}
\begin{align*}
\frac{1}{(2\pi)^{2+m}}
\int
\varphi(z_1,z_2)^m d\varphi(z_1,z_2)d\varphi(z_{2},z_1)&=
\frac{1}{(m+1)(2\pi)^{2+m}}
\int
\ d\varphi(z_1,z_2)^{m+1}d\varphi(z_{2},z_1)\\ &=
\frac{1}{m+1}\left( \frac{1}{2}-\frac{1}{m+2}\right)
\end{align*}
so that our final formula is
\[
\beta_{2,m}=-\frac{1}{8}+\frac{1}{m+1}\left( \frac{1}{2}-\frac{1}{m+2}\right)
\]
\subsection{Step 2} We claim
\label{ref-5.2-32}
\begin{equation}
\label{ref-5.6-33}
\int_{z\in \CC\setminus \{z_1,z_2\}} d\theta(z_1,z)d\theta(z,z_2)=0
\end{equation}
Since $d\theta(z,z_2)=d\theta(z_2,z)$ we may as well prove
\[
\int_{z\in \CC\setminus \{z_1,z_2\}} d\theta(z_1,z)d\theta(z_2,z)=0
\]
We may assume $z_1\neq z_2$ since otherwise the claim is trivial.
 Consider the following figure
\[
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{z}[][]{$z$}
\psfrag{zp}[][]{$z'$}
\includegraphics[width=4cm]{point_reflection.eps}
\]
Then we have $d\theta(z_1,z)d\theta(z_2,z)=-d\theta(z_1,z')d\theta(z_2,z')$.
Since the map $z\mapsto z'$ is orientation preserving this proves what we want.
\subsection{Step 3} \label{ref-5.3-34} Now we assume $n\ge 3$. First we observe
\label{ref-5.3-35}
\begin{align*}
\beta_{n,m}&=
(-1)^n\frac{1}{(2\pi)^{2n+m-2} }
\int_{B_-\setminus D}
  d\varphi(w,z_2)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n}) \\
&=(-1)^n\frac{1}{(2\pi)^{2n+m-2} }
\int_{\partial(B_-\setminus D)}
 \varphi(w,z_2)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})
\end{align*}
The components of $\partial(B_-\setminus D)$ which yield non-zero
contributions to the integral have been listed in \S\ref{ref-3.3-8}. Note
that the contribution of \eqref{ref-3.6-10} is zero by \eqref{ref-5.6-33} (integrate
first over $z_i$).
We now list the contributing integrals with the correct signs.
\begin{enumerate}
\item \label{ref-1-36} Consider the boundary component $\partial_1$ given by contracting
  all edges.  The coordinates in the neighborhood of this component
  are given by $r,x'_2,y'_2,\ldots,x'_n,y'_n$
\begin{align*}
y_1&=1+r\\
x_i&=rx'_i\\
y_i&=1+ry'_i
\end{align*}
with $r\ge 0$. In these coordinate the orientation form on $B_-$ is (up to
a positive multiple) given by
\[
-drdx'_2dy'_2\cdots dx'_ndy'_n
\]
and hence $\partial_1$ is oriented by $dx'_2dy'_2\cdots dx'_ndy'_n$. Whence
the contribution to the integral is equal to $(-1)^n\epsilon_{n,m}$.
\item Now consider the boundary component $\partial_2$ given by
  contracting $\{z_1,w\}$. We claim $\partial_2\cong C_{n,0}$,
  taking orientations into account.  Coordinates in a neighborhood of
  $\partial_2$ are given by $r,x_2,y_2,\ldots,x_n,y_n$ where $y_1=1+r$
  and $r\ge 0$. Thus the orientation form is $-dr dx_2 dy_2\cdots
  x_n,y_n$ and hence $\partial_2$ is oriented by $dx_2 dy_2\cdots
  x_n,y_n$ which is the normal orientation on $C_{n,0}$ \cite{AMM1}.
  We obtain
\begin{align*}
&(-1)^n\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_2 }
  \varphi(w,z_2)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&= (-1)^n\frac{1}{(2\pi)^{2n+m-2} }\int_{C_{n,0}} \varphi(w,z_2)^md\varphi(w,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_1)
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&=(-1)^n\frac{1}{(m+1)(2\pi)^{2n+m-2} }
\int_{C_{n,0}} d\varphi(w,z_2)^{m+1}d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_1)
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&=-\frac{1}{(m+1)(2\pi)^{2n+m-2} }\int_{C_{n,0}} d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_1)d\varphi(w,z_2)^{m+1}
d\varphi(w,z_3)\cdots d\varphi(w,z
_{n})\\
&=-\frac{1}{m+1}\alpha_{n-1,m+1}
\end{align*}
\item
\label{ref-3-37}
Let $\partial_3$ be obtained by contracting $(z_1,z_2)$.  We
  claim that $\partial_3=-C'_2\times B'_-$ where $B'_-$ is the
  analogue of $B_-$ inside $C_{n,0}$ and where $C'_2$ is the part of
  $C_2$ where the two points are not on the same vertical line. Thus
  $C'_2$ and hence $\partial_3$ has two connected components.

  Coordinates in a neighborhood of $\partial_3$ are given by
  $r,\theta,y_1,x_3,y_3,\ldots, x_n,y_n$ where
\begin{align*}
x_2=&r\cos\theta\\
y_2=&y_1+r\sin\theta
\end{align*}
and $r\ge 0$ and $\theta\not\in \{\pi/2,-\pi/2\}$ (the latter because
of the branch cut involving $\varphi(w,z_2)$). Hence we find that the
orientation form on $B_-$ is up to a positive factor given by
\[
-dy_1drd\theta dx_3dy_3\cdots dx_ndy_n
\]
Therefore the orientation form on $\partial_3$ (using the outward normal) is
given by
\[
-dy_1d\theta dx_3dy_2\cdots dx_ndy_n=(-d\theta)\times (-dy_1 dx_3dy_3\cdots dx_ndy_n)
\]
which does indeed represent the orientation on $-C'_2\times B'_-$.

The resulting integral is now
\begin{align*}
&(-1)^n\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_3}
  \varphi(w,z_2)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&=(-1)^{n+1}\frac{1}{2}\frac{1}{(2\pi)^{2n-1} }\int_{B'_-}
  d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{2})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&=(-1)^{n+1} \frac{1}{2}\beta_{n-1,1}
\end{align*}
where the factor $1/2$ comes from the fact that $\varphi(w,z_2)=0$ if $z_2$
is to the left of $z_1$.
\item Let $\partial_4$ be obtained by contracting $(z_2,z_3)$. A
  similar computation as above shows $\partial_4\cong -C_2\times B'_-$.
The resulting integral is
\begin{align*}
&(-1)^n\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_4}
  \varphi(w,z_2)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&=(-1)^{n+1}\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_4}
  \varphi(w,z_2)^md\varphi(z_2,z_3)d\varphi(z_1,z_2)d\varphi(z_3,z_4)\cdots d\varphi(z_{n-1},z_n)\times \\& \qquad \qquad\qquad d\varphi(z_n,z_{1})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&=(-1)^{n}\frac{1}{(2\pi)^{2n+m-3} }\int_{B'_-}
  \varphi(w,z_3)^md\varphi(z_1,z_3)d\varphi(z_3,z_4)\cdots d\varphi(z_{n-1},z_n)\times \\& \qquad \qquad\qquad d\varphi(z_n,z_{1})
d\varphi(w,z_3)\cdots d\varphi(w,z_{n})\\
&=(-1)^{n}\frac{1}{(m+1)(2\pi)^{2n+m-3} }\int_{B'_-}
  d\varphi(z_1,z_3)d\varphi(z_3,z_4)\cdots d\varphi(z_{n-1},z_n)\times \\& \qquad \qquad\qquad d\varphi(z_n,z_{1})
d\varphi(w,z_3)^{m+1}\cdots d\varphi(w,z_{n})\\
&=(-1)^{n} \frac{1}{m+1} \beta_{n-1,m+1}
\end{align*}
\end{enumerate}
Combining all contributions we get \eqref{ref-5.3-27}.
\section{A recursion relation for $\bar{\beta}_{n,m}$}
\label{ref-6-38}
 By our definition in \S\ref{ref-3.4-12}, $\bar{\beta}_{n,m}$ is the
integral associated to the enhanced graph
\[
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
 \raisebox{-2.5cm}{\includegraphics[width=4cm]{wheel_dashed_m_mirror.eps}}
\]
Thus with our standard edge ordering
\[
\bar{\beta}_{n,m}=\frac{1}{(2\pi)^{2n+m-2} }
\int_{B_-\setminus D}
 d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
 d\varphi(w,z_2)\cdots d\varphi(w,z_{n})^m
\]
where
\[
B_-=\{\operatorname{Re} z_1=\operatorname{Re} w+, \operatorname{Im} {z_1}>\operatorname{Im} w\}\subset C_{n+1,0}
\]
\[
D=\{\operatorname{Re} z_n=\operatorname{Re} w, \operatorname{Im} {z_2}>\operatorname{Im} w\}\subset C_{n+1,0}
\]
and where $B_-$ is oriented with the outgoing normal (see \eqref{ref-5.1-25}).
Clearly we have
\[
\bar{\beta}_{2,m}=\beta_{2,m}
\]
\[
\bar{\beta}_{n,1}=\beta_{n,1}
\]
In this section we will prove for $n\ge 3$
\begin{equation}
\label{ref-6.1-39}
\bar{\beta}_{n,m}=
(-1)^{n} \frac{1}{2}\beta_{n-1,1}+(-1)^{n+1} \frac{1}{m+1} \bar{\beta}_{n-1,m+1}
+\bar{\epsilon}_{n,m}
\end{equation}
where $\bar{\epsilon}_{n,m}$ is the integral associated to the graph
\[
\psfrag{zn}[][]{$z_n$}
\psfrag{zn1}[l][l]{$z_{n-1}$}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{m}[l][l]{$m$}
\psfrag{w}[l][l]{$w$}
\int \raisebox{-1.5cm}{\includegraphics[width=4cm]{wheel_dashed_bold_m_mirror_boxed.eps}}
\]
(see \S\ref{ref-3.4-12}). Thus with our standard orientation on edges
\[
\bar{\epsilon}_{n,m}=
\frac{1}{(2\pi)^{2n+m-2}}\int \theta(w,z_n)^m d\theta(z_1,z_2)d\theta(z_2,z_3)
\cdots d\theta(z_n,z_1) d\theta(w,z_2)\cdots d\theta(w,z_{n-1})
\]
Here we put $w=0$, $z_1=i$ and the integral is over the complement
of the diagonal in $(\CC-\{0,i\})^{n-1}$.
\subsection{Proof}
\label{ref-6.1-40}
We assume  $n\ge 3$. We first observe
\begin{align*}
\bar{\beta}_{n,m}&=
\frac{1}{(2\pi)^{2n+m-2} }
\int_{B_-\setminus D}
  d\varphi(w,z_n)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1}) \\
&=\frac{1}{(2\pi)^{2n+m-2} }
\int_{\partial(B_-\setminus D)}
 \varphi(w,z_n)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})
\end{align*}
The components of $\partial(B_-\setminus D)$ which yield non-zero
contributions to the integral have been listed in \S\ref{ref-3.3-8}. Note
that the contribution of \eqref{ref-3.7-11} is zero by \eqref{ref-5.6-33} (integrate
first over $z_{n-1}$).
We now list the contributing integrals with the correct signs.
\begin{enumerate}
\item Consider the boundary component $\partial_1$ given by contracting
  all edges. Then a computation similar to \S\ref{ref-5.3-34}(\ref{ref-1-36})
yields that the contribution is $\bar{\epsilon}_{n,m}$.
\item Now consider the boundary component $\partial_2$ given by
contracting $(z_{n-1},w)$.  It this case we get an integral which contains
as a subintegral (up to a scalar factor)
\[
\int_{z_n\in {\cal H}\setminus\{w,z_1\}} d\varphi(w,z_n)^{m+1} d\varphi(z_n,z_1)
\]
Using \eqref{ref-4.3-18} this is equal to
\[
(2\pi)^{m+1}\varphi(w,z_1)-2\pi\varphi(w,z_1)^{m+1}=(2\pi)^{m+1} \times 2\pi-
2\pi\times (2\pi)^{m+1}=0
\]
So we are lucky that there is no contribution in this case.
\item Let $\partial_3$ be obtained by contracting $(z_1,z_n)$. A
  similar computation as in \S\ref{ref-5.3-34}\ref{ref-3-37}) shows
  $\partial_3=-C'_2\times B'_-$. The resulting integral is
\begin{align*}
&\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_3}
  \varphi(w,z_n)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})\\
&=(-1)^{n-1}\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_3}
 \varphi(w,z_n)^m d\varphi(z_n,z_{1})d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})\\
&=(-1)^{n}\frac{1}{2}\frac{1}{(2\pi)^{2n-2} }\int_{B'_-}
d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_1)
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})\\
&=(-1)^{n} \frac{1}{2}\bar{\beta}_{n-1,1}=(-1)^{n}\frac{1}{2}\beta_{n-1,1}
\end{align*}
where the factor $1/2$ comes from the fact that $\varphi(w,z_n)=0$ if $z_2$
is to the left of $z_1$.
\item Let $\partial_4$ be obtained by contracting $(z_{n-1},z_n)$.  A
  similar computation as in \S\ref{ref-5.3-34}\ref{ref-3-37}) shows
  $\partial_4\cong -C_2\times B'_-$. The resulting integral is
\begin{align*}
&\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_4}
  \varphi(w,z_n)^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1})
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})\\
&=(-1)^{n}\frac{1}{(2\pi)^{2n+m-2} }\int_{\partial_4}
  \varphi(w,z_n)^md\varphi(z_{n-1},z_n)d\varphi(z_1,z_2)\cdots d\varphi(z_{n-2},z_{n-1})\times \\& \qquad \qquad\qquad d\varphi(z_n,z_{1})
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})\\
&=(-1)^{n+1}\frac{1}{(2\pi)^{2n+m-3} }\int_{B'_-}
  \varphi(w,z_{n-1})^md\varphi(z_1,z_2)\cdots d\varphi(z_{n-2},z_{n-1})\times \\& \qquad \qquad\qquad d\varphi(z_{n-1},z_{1})
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})\\
&=(-1)^{n+1}\frac{1}{(m+1)(2\pi)^{2n+m-3} }\int_{B'_-}
  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-2},z_{n-1})\times \\& \qquad \qquad\qquad d\varphi(z_{n-1},z_{1})
d\varphi(w,z_2)\cdots d\varphi(w,z_{n-1})^{m+1}\\
&=(-1)^{n+1} \frac{1}{m+1} \bar{\beta}_{n-1,m+1}
\end{align*}
Combining all contributions we get \eqref{ref-6.1-39}.
\end{enumerate}
\section{A recursion not involving unknown quantities.}
\label{ref-7-41}
Put
\[
\hat{\beta}_{n,m}=\frac{1}{2}(\beta_{n,m}+(-1)^{n}\bar{\beta}_{n,m})
\]
We claim
\begin{equation}
\label{ref-7.1-42}
\hat{\beta}_{n,m}=-\frac{1}{2(m+1)}\alpha_{n-1,m+1}
+(-1)^{n+1} \frac{1}{2}\hat{\beta}_{n-1,1}+(-1)^n\frac{1}{m+1} \hat{\beta}_{n-1,m+1}
\end{equation}
with initial condition
\begin{equation}
\label{ref-7.2-43}
\hat{\beta}_{2,m}=-\frac{1}{8}+\frac{1}{m+1}\left( \frac{1}{2}-\frac{1}{m+2}\right)
\end{equation}
\subsection{Step 1}
\label{ref-7.1-44}
We claim first that
\begin{equation}
\label{ref-7.3-45}
\epsilon_{n,m}=-\bar{\epsilon}_{n,m}
\end{equation}
We have
\[
\epsilon_{n,m}=
\frac{1}{(2\pi)^{2n+m-2}}\int \theta(w,z_2)^m d\theta(z_1,z_2)d\theta(z_2,z_3)
\cdots d\theta(z_n,z_1) d\theta(w,z_3)\cdots d\theta(w,z_n)
\]
Here we put $w=0$, $z_1=i$ and the integral is over a suitable open
subset of $\CC^{2n-1}$.

We apply the   permutation $(2n)(3\,
n-1)\cdots $ to $z_1,\ldots,z_n$. This does not change the orientation. We find
\begin{align*}
\epsilon_{n,m}&=
\frac{1}{(2\pi)^{2n+m-2}}\int \theta(w,z_n)^m d\theta(z_1,z_{n})d\theta(z_n,z_{n-1})
\cdots d\theta(z_2,z_1) d\theta(w,z_{n-1})\cdots d\theta(w,z_2)\\
&=\frac{1}{(2\pi)^{2n+m-2}}\int \theta(w,z_n)^m d\theta(z_n,z_{1})d\theta(z_{n-1},z_{n})
\cdots d\theta(z_1,z_2) d\theta(w,z_{n-1})\cdots d\theta(w,z_2)\\
&=(-1)^s \bar{\epsilon}_{n,m}
\end{align*}
The sign $(-1)^s$ comes from the fact that we have invert the order of the
1-forms in $d\theta(z_n,z_1)d\theta(z_{n-1} z_{n})\cdots d\theta(z_1,z_2)$ and
in $d\theta(w,z_{n-1})\cdots d\theta(w,z_2)$.
Thus
\[
s=\frac{n(n-1)}{2}+\frac{(n-2)(n-3)}{2} =n^2-3n+3
\]
which is always an odd number. This proves \eqref{ref-7.3-45}.
\subsection{Step 2}
\label{ref-7.2-46}
Formula \eqref{ref-7.2-43} is obvious from \eqref{ref-5.2-26}
and the fact that $\hat{\beta}_{2,m}=\beta_{2,m}$. Formula \eqref{ref-7.1-42} follows from \eqref{ref-5.3-27} and \eqref{ref-6.1-39} together
with the following computation
\begin{align*}
\hat{\beta}_{n,m}&=\frac{1}{2}\left(-\frac{1}{m+1}\alpha_{n-1,m+1}
+(-1)^{n+1} \frac{1}{2}\beta_{n-1,1}+\frac{1}{2}\beta_{n-1,1}
+(-1)^{n} \frac{1}{m+1} \beta_{n-1,m+1}-\frac{1}{m+1} \bar{\beta}_{n-1,m+1}\right)\\
&=-\frac{1}{2(m+1)}\alpha_{n-1,m+1}
+(-1)^{n+1} \frac{1}{2}\hat{\beta}_{n-1,1}+(-1)^n\frac{1}{m+1} \hat{\beta}_{n-1,m+1}
\end{align*}
\section{A formula for $w_n$}
\label{ref-8-47}
In this section we prove the following fact
\begin{equation}
\label{ref-8.1-48}
w_n=
\begin{cases}
\hat{\beta}_{n,1}-\frac{1}{2}\alpha_{n-1,2}&\text{if $n$ is even}\\
0&\text{if $n$ is odd}
\end{cases}
\end{equation}
\subsection{Step 1} We assume $n$ odd. Put $w=i$. We identify
\label{ref-8.1-49}
$C_{n,0}$ with an open subset of ${\cal H}^{n}$. The orientation is
derived from the standard orientation on ${\cal H}^n$ (according to
\cite{AMM1}).

We consider the map $\alpha:{\cal H}\rightarrow {\cal H}: x+iy\mapsto -x+iy$ and we extend
$\alpha$ diagonally to a map ${\cal H}^n\rightarrow {\cal H}^n$ also denoted by $\alpha$.
$\alpha$ multiplies the orientation by $(-1)^n$.

Hence we have
\[
w_n=(-1)^n\frac{1}{(2\pi)^{2n}} \int_{C_{n,0}}
d\varphi(\alpha(z_1),\alpha(z_2))\cdots d\varphi(\alpha(z_1),\alpha(z_n))
d\varphi(w,\alpha(z_1))\cdots d\varphi(w,\alpha(z_n))
\]
(since $\alpha(w)=w$).

It is now easy to see that $\varphi(\alpha(u),\alpha(v))=2\pi-\varphi(u,v)$ and
hence $d\varphi(\alpha(u),\alpha(v))=-d\varphi(u,v)$. Thus
\[
w_n=(-1)^n(-1)^{2n}w_n=(-1)^nw_n
\]
Since $n$ is odd this implies $w_n=0$.
\subsection{Step 2} Consider the following enhanced graph
\label{ref-8.2-50}
\begin{equation}
\label{ref-8.2-51}
\psfrag{z1}[][]{$z_1$}
\psfrag{z2}[][]{$z_2$}
\psfrag{zn}[][]{$z_n$}
\psfrag{m}[][]{$$}
\psfrag{w}[][]{$w$}
\raisebox{-2cm}{\includegraphics[width=4cm]{wheel_n_bold_euclidean.eps}}
\end{equation}
We claim that its corresponding integral
\[
\delta=\frac{1}{(2\pi)^{2n}} \int_{C_{n+1}} \theta(w,z_1) d\theta(z_1,z_2)\cdots
d\theta(z_n,z_1) d\theta(w,z_2)\cdots d\theta(w,z_n)
\]
is zero when $n$ is even.

It's easy to see that permuting the points does not change the
orientation on $C_{n+1}$.  We apply the permutation $(2n)(3\,
n-1)\cdots $ for $z_1,z_2,\ldots,z_n$. Hence we also have
\begin{align*}
\delta&=\frac{1}{(2\pi)^{2n}} \int_{C_{n+1}}
\theta(w,z_1) d\theta(z_1,z_n)d\theta(z_n z_{n-1})\cdots d\theta(z_2,z_1)
d\theta(w,z_{n})\cdots d\theta(w,z_2)\\
&=\frac{1}{(2\pi)^{2n}} \int_{C_{n+1}}
\theta(w,z_1) d\theta(z_n,z_1)d\theta(z_{n-1} z_{n})\cdots d\theta(z_1,z_2)
d\theta(w,z_{n})\cdots d\theta(w,z_2)
\end{align*}
To put this back in standard form we have to invert the order of the
1-forms in $d\theta(z_n,z_1)d\theta(z_{n-1} z_{n})\cdots d\theta(z_1,z_2)$ and
in $d\theta(w,z_{n})\cdots d\theta(w,z_2)$. This introduces a cumulative sign
of $(-1)^s$ where
\[
s=\frac{n(n-1)}{2}+\frac{(n-1)(n-2)}{2} =(n-1)^2
\]
Thus
\[
\delta=-(-1)^n \delta
\]
and hence if $n$ is even then $\delta=0$.
\subsection{Step 3} Now we assume $n$ even.
\label{ref-8.3-52}
As a pedagogical device we keep signs of the form $(-1)^n$ below.
 We first observe
\begin{align*}
w_n&=(-1)^n \frac{1}{(2\pi)^{2n}}
\int_{C_{n+1,0}}
 d\varphi(w,z_1)  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_n)\\
&=(-1)^n \frac{1}{(2\pi)^{2n}}\int_{\text{boundary}}
 \varphi(w,z_1)  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_n)
\end{align*}
The boundary consists of two parts. One part consists of both sides
of the branch cut $B_{\pm}$.
\[
B_-=\{\operatorname{Re} z_1=\operatorname{Re} w+, \operatorname{Im} {z_1}>\operatorname{Im} w\}\subset C_{n+1,0}
\]
\[
B_+=\{\operatorname{Re} z_1=\operatorname{Re} w-, \operatorname{Im} {z_1}>\operatorname{Im} w\}\subset C_{n+1,0}
\]
The values of $\varphi(w,-)$ on both sides of the branch cut differ by
$2\pi$. Hence the contribution of this  part of the boundary is equal to
\[
(-1)^n \frac{1}{(2\pi)^{2n-1}}\int_{B_-}
 \varphi(w,z_1)  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_n)
\]
which is nothing but
\[
(-1)^n \beta_{n,1}
\]
In \S\ref{ref-3.2-5} we have listed the ``classical'' parts of the boundary which
contribute to the integral. Here we give the contributions with the precise
signs. Note that the contribution of \eqref{ref-8.2-51} is zero
by Step 2.
\begin{enumerate}
\item Let $\partial_1$ be the boundary component given by contracting
  $(w,z_n)$. According to \cite{AMM1} we have $\partial_1\cong-C_2\times C_{n,0}$.
The contribution to the integral is
\begin{multline*}
(-1)^n \frac{1}{(2\pi)^{2n}}\int_{\partial_1}
 \varphi(w,z_1)  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_{n-1})d\varphi(w,z_n)\\
=(-1)^n \frac{1}{(2\pi)^{2n}}\int_{\partial_1}
 \varphi(w,z_1)  d\varphi(w,z_n) d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_{n-1})\\
=(-1)^{n+1} \frac{1}{(2\pi)^{2n-1}}\int_{C_{n,0}}
 \varphi(w,z_1)  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},w)d\varphi(w,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_{n-1})\\
=(-1)^{n+1} \frac{1}{2(2\pi)^{2n-1}}
\int_{C_{n,0}}
 d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},w)d\varphi(w,z_{1})^{2} d\varphi(w,z_2)\cdots
d\varphi(w,z_{n-1})\\
=(-1)^{n+1} \frac{1}{2}\alpha_{n-1,2}
\end{multline*}
\item Let $\partial_2\cong -C_2\times C_{n,0}$ be obtained by
  contracting $(z_1,z_2)$. We obtain
\begin{multline*}
(-1)^n \frac{1}{(2\pi)^{2n}}\int_{\partial_2}
 \varphi(w,z_1)  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_n)\\
=(-1)^{n+1} \frac{1}{(2\pi)^{2n-1}}\int_{C_{n,0}}
 \varphi(w,z_2)  d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{2}) d\varphi(w,z_2)\cdots
d\varphi(w,z_n)\\
=(-1)^{n+1} \frac{1}{(2)(2\pi)^{2n-1}}
\int_{C_{n,0}}
d\varphi(z_2,z_3)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{2}) d\varphi(w,z_2)^{2}\cdots
d\varphi(w,z_n)
\end{multline*}
\item Let $\partial_3\cong -C_2\times C_{n,0}$ be obtained by
  contracting $(z_n,z_1)$. We obtain
\begin{multline*}
(-1)^n \frac{1}{(2\pi)^{2n}}\int_{\partial_3}
 \varphi(w,z_1)  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_n)d\varphi(z_n,z_{1}) d\varphi(w,z_2)\cdots
d\varphi(w,z_n)\\
=- \frac{1}{(2\pi)^{2n}}\int_{\partial_3}
 \varphi(w,z_1) d\varphi(z_n,z_{1})  d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_{n})  d\varphi(w,z_2)\cdots
d\varphi(w,z_n)\\
=\frac{1}{(2\pi)^{2n-1}}\int_{C_{n,0}}
 \varphi(w,z_1)   d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_{1})  d\varphi(w,z_2)\cdots
d\varphi(w,z_{n-1}) d\varphi(w,z_1)\\
=(-1)^{n-2}\frac{1}{(2\pi)^{2n-1}}\int_{C_{n,0}}
 \varphi(w,z_1)   d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_{1})  d\varphi(w,z_1) d\varphi(w,z_2)\cdots
d\varphi(w,z_{n-1})\\
=(-1)^{n-2}\frac{1}{(2)(2\pi)^{2n-1}}\int_{C_{n,0}}
   d\varphi(z_1,z_2)\cdots d\varphi(z_{n-1},z_{1})  d\varphi(w,z_1)^{2} d\varphi(w,z_2)\cdots
d\varphi(w,z_{n-1}
\end{multline*}
Hence the contributions of $\partial_2$ and $\partial_3$ cancel. The
contributions of $\partial_1$ and $B_-$ yield \eqref{ref-8.1-48} in case $n$
is even.
\end{enumerate}
\section{Solving the recursion}
\label{ref-9-53}
\subsection{The problem}
\label{ref-9.1-54}
For the benefit of the reader we restate the recursion relations we have
derived. See \eqref{ref-4.1-15}\eqref{ref-4.2-16}\eqref{ref-7.1-42}\eqref{ref-7.2-43}\eqref{ref-8.1-48}.
\begin{align*}
\alpha_{0,m}&= 1\\
\alpha_{n,m}&=(-1)^n \left( \frac{1}{2}\alpha_{n-1,2}-\frac{1}{m+1} \alpha_{n-1,m+1}\right) \qquad \text{(for $n\ge 1$)}\\
\hat{\beta}_{2,m}&=-\frac{1}{8}+\frac{1}{m+1}\left( \frac{1}{2}-\frac{1}{m+2}\right)\\
\hat{\beta}_{n,m}&=-\frac{1}{2(m+1)}\alpha_{n-1,m+1} +(-1)^{n+1}
\frac{1}{2}\hat{\beta}_{n-1,1}+(-1)^n\frac{1}{m+1}
\hat{\beta}_{n-1,m+1 }
\qquad \text{(for $n\ge 3$)}
\\
w_n&= \hat{\beta}_{n,1}-\frac{1}{2}\alpha_{n-1,2}\qquad\text{(for $n\ge 2$ even)}
\end{align*}
\subsection{Eliminating some signs and fractions}
\label{ref-9.2-55}
We simplify  the equations by putting
\[
\tilde{\alpha}_{n,m}=(-1)^{\frac{n(n+1)}{2}}\frac{1}{m!} \alpha_{n,m}
\]
Then we have
\[
\tilde{\alpha}_{0,m}=\frac{1}{m!}
\]
and
\begin{equation}
\label{ref-9.1-56}
\tilde{\alpha}_{n,m}=
\frac{1}{m!}\tilde{\alpha}_{n-1,2}- \tilde{\alpha}_{n-1,m+1}
\end{equation}
Similarly we put
\[
\tilde{\beta}_{n,m}=(-1)^{\frac{n(n+1)}{2}} \frac{\hat{\beta}_{n,m}}{m!}
\]
so that we get.
\[
\tilde{\beta}_{2,m}=\frac{1}{8m!}-\frac{1}{2(m+1)!}+\frac{1}{(m+2)!}
\]
and
\begin{equation}
\label{ref-9.2-57}
\tilde{\beta}_{n,m}=-(-1)^{n}\frac{1}{2}\tilde{\alpha}_{n-1,m+1}-\frac{1}{2m!}\tilde{\beta}_{n-1,1}+\tilde{\beta}_{n-1,m+1}
\end{equation}
\subsection{Computing $\tilde{\alpha}$}
\label{ref-9.3-58}
 Iterating \eqref{ref-9.1-56} we find
\begin{align*}
\tilde{\alpha}_{n,2}&=  \frac{1}{2!}\tilde{\alpha}_{n-1,2}
- \tilde{\alpha}_{n-1,3}\\
&=\frac{1}{2!}\tilde{\alpha}_{n-1,2}
-\frac{1}{3!} \tilde{\alpha}_{n-2,2}+\tilde{\alpha}_{n-2,4}\\
&=\left(\frac{1}{2!}\tilde{\alpha}_{n-1,2}
-\frac{1}{3!} \tilde{\alpha}_{n-2,2}+\cdots + (-1)^{n+1}\frac{1}{(n+1)!} \tilde{\alpha}_{0,2}\right)+(-1)^n\tilde{\alpha}_{0,n+2}
\end{align*}
Thus if we put
\[
A_2=\sum_{n\ge 0} \tilde{\alpha}_{n,2} x^n
\]
we get
\[
A_2=\left(\frac{x}{2!}-\frac{x^2}{3!}+\cdots\right)A_2+\sum_{n\ge 0}(-1)^n \frac{x^n}{(n+2)!}
\]
Here
\[
\frac{x}{2!}-\frac{x^2}{3!}+\cdots=\frac{e^{-x}-1+x}{x}
\]
and
\[
\sum_{n\ge 0}(-1)^n \frac{x^n}{(n+2)!}=\frac{e^{-x}-1+x}{x^2}
\]
from which we deduce
\[
A_2=-\frac{1}{x}+\frac{1}{1-e^{-x}}
\]
For use below we record
\[
A'_2\overset{\text{def}}{=}A_2-A_2(0)=-\frac{1}{x}+\frac{1+e^{-x}}{2(1-e^{-x})}
\]

Knowing $\tilde{\alpha}_{n,2}$ we can compute arbitrary $\tilde{\alpha}_{n,m}$ via
\begin{equation}
\label{ref-9.3-59}
\begin{split}
\tilde{\alpha}_{n,m}&=\frac{1}{m!}\tilde{\alpha}_{n-1,2}- \tilde{\alpha}_{n-1,m+1}\\
&=\frac{1}{m!}\tilde{\alpha}_{n-1,2}-\frac{1}{(m+1)!}\tilde{\alpha}_{n-2,2}
+\cdots+(-1)^{n-1}\frac{1}{(n+m-1)!}\tilde{\alpha}_{0,2}
+
(-1)^n\frac{1}{(m+n)!}
\end{split}
\end{equation}
We will not write down a closed expression for $\tilde{\alpha}_{n,m}$
as we won't need it.

\subsection{Computing a sum}
\label{ref-9.4-60}
For use below we compute the sum for $n\ge 2$
\[
s_n=(-1)^{n-1}\tilde{\alpha}_{n-1,2}+(-1)^{n-2}\tilde{\alpha}_{n-2,3}
+\cdots +\tilde{\alpha}_{2,n-1}
\]
where an empty sum is interpreted as $0$. Using \eqref{ref-9.3-59} we get
\[
\tilde{\alpha}_{2,n-1}=\frac{1}{(n-1)!}\tilde{\alpha}_{1,2}-\frac{1}{n!}\tilde{\alpha}_{0,2}+\frac{1}{(n+1)!}
\]
\[
\tilde{\alpha}_{3,n-2}=
\frac{1}{(n-2)!} \tilde{\alpha}_{2,2}-
\frac{1}{(n-1)!} \tilde{\alpha}_{1,2}+
\frac{1}{n!} \tilde{\alpha}_{0,2}-
\frac{1}{(n+1)!}
\]
and
\[
\tilde{\alpha}_{n-1,2}=\frac{1}{2!} \tilde{\alpha}_{n-2,2}-\frac{1}{3!}
\tilde{\alpha}_{n-3,2}+\cdots + (-1)^{n-2} \frac{1}{n!}\tilde{\alpha}_{0,2}
+(-1)^{n-1} \frac{1}{(n+1)!}
\]
so that we get
\[
s_n=
(-1)^{n-1}\frac{1}{2!} \tilde{\alpha}_{n-2,2}+
(-1)^{n-2} \frac{2}{3!} \tilde{\alpha}_{n-3,2}+\cdots+
(n-2)\frac{1}{(n-1)!}\tilde{\alpha}_{1,2}-(n-2)\frac{1}{n!}\tilde{\alpha}_{0,2}+ (n-2)\frac{1}{(n+1)!}
\]
Put
\[
T=\frac{x^2}{2!}-\frac{2x^3}{3!}+\cdots=-xe^{-x}-e^{-x}+1
\]
Then we get
\[
s_n=(-1)^{n-1}(TA_2)[x^n]+\frac{1}{n!}\tilde{\alpha}_{0,2}+\frac{n-2}{(n+1)!}
\]
where $(-)[x^n]$ denotes the coefficient of $x^n$.
 We will regard this formula  as a definition if $n=0,1$.  We
obtain
\[
\sum_{n\ge 0} (-1)^n s_n x^n=-TA_2+ \sum_{n\ge 0} (-1)^n \tilde{\alpha}_{0,2}\frac{x^n}{n!}+\sum_{n\ge 0} (-1)^n\frac{n-2}{(n+1)!}x^n
\]
We have
\[
\sum_{n\ge 0} (-1)^n \tilde{\alpha}_{0,2}\frac{x^n}{n!}=\frac{1}{2}e^{-x}
\]
and
\begin{align*}
\sum_{n\ge 0} (-1)^n\frac{n-2}{(n+1)!}x^n&=\sum_{n\ge 0}(-1)^n \frac{n+1}{(n+1)!}x^n-3
\sum_{n\ge 0}(-1)^n \frac{1}{(n+1)!}x^n\\
&=e^{-x}+3\frac{e^{-x}-1}{x}
\end{align*}
A computation with a computer algebra package now yields
\[
\sum_{n\ge 0} (-1)^ns_n x^n=
\frac{-(x+4)e^{-2x}+(2x^2+3x+8)e^{-x}-(2x+4)}{2(1-e^{-x})x}
\]
Hence
\[
S\overset{\text{def}}{=}\sum_{n\ge 0} s_n x^n=
\frac{(x-4)e^{2x}+(2x^2-3x+8)e^{x}+(2x-4)}{2(e^{x}-1)x}
\]
We have $S=-3/2+O(x^4)$. For use below we record
\[
S'\overset{\text{def}}{=}\sum_{n\ge 2} s_n x^n=S+\frac{3}{2}=
\frac{(x-4)e^{2x}+(2x^2+8)e^x-x-4}{2(e^x-1)x}
\]
\subsection{Computing $\tilde{\beta}_{n,1}$}
\label{ref-9.5-61}
We will only compute $\tilde{\beta}_{n,1}$ since this the only thing we need.
Iterating \eqref{ref-9.2-57} we get for $n\ge 3$
\begin{align*}
  \tilde{\beta}_{n,1}&=(-1)^{n-1}\frac{1}{2}\tilde{\alpha}_{n-1,2}-\frac{1}{2\cdot 1!}
\tilde{\beta}_{n-1,1}+\tilde{\beta}_{n-1,2}\\
&=\frac{1}{2}\biggl((-1)^{n-1}\tilde{\alpha}_{n-1,2}+(-1)^{n-2}\tilde{\alpha}_{n-2,3}
+\cdots +\tilde{\alpha}_{2,n-1}\biggr)-\\
&\qquad\qquad\frac{1}{2}\left(\frac{1}{1!}\tilde{\beta}_{n-1,1}
+\frac{1}{2!}\tilde{\beta}_{n-2,1}+\cdots+\frac{1}{(n-2)!}\tilde{\beta}_{2,1}\right)+\tilde{\beta}_{2,n-1} \\
&=\frac{s_n}{2} -\frac{1}{2}\left(\frac{1}{1!}\tilde{\beta}_{n-1,1}
+\frac{1}{2!}\tilde{\beta}_{n-2,1}+\cdots+\frac{1}{(n-2)!}\tilde{\beta}_{2,1}\right)+\tilde{\beta}_{2,n-1}
\end{align*}
Since $s_2=0$ this identity holds for $n=2$ if we interpret an empty
sum as zero.

If we put
\[
B_1=\sum_{n\ge 2} \tilde{\beta}_{n,1}x^n
\]
then we have
\[
B_1=\frac{1}{2} S' -\frac{1}{2}\left( \frac{x}{1!}+\frac{x^2}{2!}+
\cdots\right)B_1
+
\sum_{n\ge 2}x^n\left(\frac{1}{8(n-1)!}-\frac{1}{2n!}+\frac{1}{(n+1)!}\right)
\]
We have
\[
 \frac{x}{1!}+\frac{x^2}{2!}+
\cdots=e^x-1
\]
\[
\sum_{n\ge 2} \frac{x^n}{(n-1)!}=x(e^x-1)
\]
\[
\sum_{n\ge 2} \frac{x^n}{n!}=e^x-1-x
\]
\[
\sum_{n\ge 2} \frac{x^n}{(n+1)!}=\frac{1}{x}\left(e^x-1-x-\frac{x^2}{2} \right)
\]
and hence
\begin{align*}
\sum_{n\ge
  2}x^n\left(\frac{1}{8(n-1)!}-\frac{1}{2n!}+\frac{1}{(n+1)!}\right)& =
\frac{1}{8} x(e^x-1)-\frac{1}{2}
(e^x-1-x)+\frac{1}{x}\left(e^x-1-x-\frac{x^2}{2}\right)\\
&=\left(\frac{x}{8}-\frac{1}{2}+\frac{1}{x}\right)e^x-\frac{1}{x}
-\frac{1}{2}-\frac{x}{8}
\end{align*}
Thus we get
\[
\left(1+\frac{1}{2}(e^x-1)\right)B_1=\frac{1}{2}S'+\left(\frac{x}{8}-\frac{1}{2}+\frac{1}{x}\right)e^x-\frac{1}{x}
-\frac{1}{2}-\frac{x}{8}
\]
Invoking once again a computer algebra package we find
\[
B_1=\frac{(x-2)e^x+x+2}{4(e^x-1)}
\]
\subsection{Computing $w_n$ }
\label{ref-9.6-62}
If $n$ is even we have
\[
(-1)^{\frac{n(n-1)}{2}}w_n=\tilde{\beta}_{n,1}-\tilde{\alpha}_{n-1,2}
\]
Since $A'_2$, $B_1$ are respectively an odd and an even function of
$x$, $B_1(0)=0$  and $w_n=0$ for $n$ odd we obtain
\[
\sum_{n\ge 2}(-1)^{\frac{n(n-1)}{2}}w_nx^n=B_1-xA'_2
\]
and using a computer algebra package we find
\begin{equation}
\label{ref-9.4-63}
\sum_{n\ge 2}(-1)^{\frac{n(n-1)}{2}}w_nx^n=\frac{(x+2)e^{-x}+x-2}{4(e^{-x}-1)}
\end{equation}
The derivative of
\[
\frac{1}{2}\log \frac{e^{x/2}-e^{-x/2}}{x}
\]
is equal to
\[
-\frac{(x+2)e^{-x}+x-2}{4(e^{-x}-1)x}
\]
Dividing \eqref{ref-9.4-63} by $x$ and integrating finishes the proof
of \eqref{ref-1.1-1}.

\def$'$} \def\cprime{$'$} \def\cprime{$'${$'$} \def$'$} \def\cprime{$'$} \def\cprime{$'${$'$} \def$'$} \def\cprime{$'$} \def\cprime{$'${$'$}
\ifx\undefined\leavevmode\hbox to3em{\hrulefill}\,
\newcommand{\leavevmode\hbox to3em{\hrulefill}\,}{\leavevmode\hbox to3em{\hrulefill}\,}
\fi

redpajama_set_name: RedPajamaArXiv


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You design the products that drive your company's growth. Your designs may even be advancing medical fields or other noble causes. That's why you got into the industry to begin with; you wanted to make a difference. You never wanted to be treated like a cost center, constantly asked to cut costs, to do more, faster, with less. How much time do you spend on TPS reports?
It's Friday when the mechanical team shows you the latest changes to the enclosure for a board that must go to fab before the next week. The changes are going to require changing the shape of the board and some of the circuitry. It's time for a series of ECOs. How do you keep informed of changes on the MCAD side of the equation? How do you handle changes to the schematic in the PCB? Can you get some help redesigning the board without creating even more communication breakdowns? How do you avoid getting in this whole situation in the first place?
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Kedụ, Budget Magic lovers, it's that time once again! Normally when we play budget decks in Modern, they come in close to our $100 max budget, but every once in a while, a super-cheap, ultra-budget deck manages to find its way onto the series. This is one of those weeks, as we head to Modern to play a $50-ish dollars version of Goblin Storm! The main plan of the deck is simple: resolve a Skirk Prospector along with another Goblin or two, resolve the newly reprinted Fecundity, and proceed to draw our entire deck and win the game! Is it possible for a $50 combo deck to compete in Modern? Let's get to the videos and find out; then, we'll talk more about the deck!
Goblin Storm, as its name suggests, is sort of a weird hybrid of Goblin tribal and Storm combo. While most of the creatures in our deck are Goblins, rather than using our Goblins as creatures to attack and beat our opponent down, we're actually using most of our creatures to produce mana and draw cards, which eventually allows us to get a storm-style kill with the help of Grapeshot or Voracious Dragon. To really understand the deck, the easiest thing to do is simply walk through the combo, step by step!
Goblin Storm has two main combo pieces, and to win the game, we need one copy of each on the battlefield: Skirk Prospector and Fecundity. As such, the main goal of our deck is to find these two cards and resolve them, which usually allows us to win the game on the spot. The combo itself is pretty simple—in conjunction with some other Goblins that we'll talk about in a minute, Skirk Prospector allows us to keep sacrificing Goblins to make a ton of mana, while also drawing us a card each time we sacrifice a creature thanks to Fecundity. Eventually, the drawing and mana production spiral to the point where we have more mana than we'll need and will have drawn essentially (or even literally) our entire deck. Then, we use one of our finishers to close out the game!
Faithless Looting and Commune with the Gods are in our deck for one reason: they help us find Skirk Prospector and Fecundity. One of the downsides of Goblin Storm is that we only have four copies of our two main combo pieces, and we can't really win the game without one of each, so hoping we just naturally draw into them is a risky plan. Faithless Looting allows us to churn through our deck to find our pieces and improve our hand, while Commune with the Gods digs five cards deep and can hit either half of our combo. Together, these cards help to make sure that we find both Skirk Prospector and Commune with the Gods as quickly as possible, as often as possible.
Once we've assembled Skirk Prospector and Fecundity, our next most important card is Mogg War Marshal, which is essentially a hybrid Desperate Ritual and Ancestral Recall. Since [[Mogg War Marshall] makes a Goblin token when it enters the battlefield or dies, we can cast it for two mana, immediately sacrifice it and both Goblin tokens to Skirk Prospector to make three mana (so we net one mana in the process), and also draw three cards with Fecundity, which keeps us churning through our deck to find more Mogg War Marshals and other Goblins to sacrifice.
Goblin Instigator and Hordeling Outburst aren't quite as good as Mogg War Marshall because instead of generating mana with Skirk Prospector, they just break even (with Goblin Instigator costing two mana and giving us back two mana when we sacrifice both Goblins to Skirk Prospector and Hordeling Outburst costing three but making back three once we sacrifice all of the tokens). However, both cards work extremely well with Fecundity. As we are comboing, Goblin Instigator draws us two cards for zero mana, while Hordeling Outburst draws us three! As such, even though these cards don't act like rituals like Mogg War Marshal, they are still very important to our combo, since they keep us drawing through our deck to find more copies of Mogg War Marshal and our other payoffs.
Oddly, Empty the Warrens isn't really our win condition, but it is the card that makes sure our combo can't fizzle. While in theory we could cast a bunch of cards, make a bunch of Goblin tokens with Empty the Warrens, pass the turn, and try to attack our opponent to death on our next turn, that's not our primary plan. Instead, our goal for Empty the Warrens is to use it like a super-ritual/card-draw spell. Let's say we assemble our combo and cast four copies of our random cheap Goblins as we sacrifice and draw our way through our deck with Skirk Prospector and Fecundity. We then cast an Empty the Warrens. The end result is that we get ten 1/1 Goblin tokens for just four mana. When you consider that we can sacrifice all of these Goblins to Skirk Prospector to make mana and also draw a card for each with Fecundity, in this situation, Empty the Warrens ends up being a four-mana ritual that makes 10 mana and also happens to draw us 10 cards for free! Since we have the full four copies of Empty the Warrens, odds are that in the 10 cards we draw, we'll find a bunch more Mogg War Marshals, Goblin Instigators, and the like, and hopefully another Empty the Warrens. Then, we simply cast all of our cheap stuff and cast an even bigger Empty the Warrens, which generates even more mana and draws us even more cards! The end result is that we have more mana than we could ever use and that we can also draw our entire deck!
After we draw our entire deck, it's finally time to win the game. For this, we have two options (both one-ofs) in Voracious Dragon and Grapeshot. If the plan is to win with Voracious Dragon, we simply cast our third and forth copies of Empty the Warrens, make an ungodly number of 1/1 Goblins, sacrifice five of them to Skirk Prospector to make enough mana to cast Voracious Dragon, and then devour all of our random 1/1 Goblins when Voracious Dragon enters the battlefield to deal two times as much damage to our opponent's face, which should give us 50 or 100 damage for just five mana! Just be careful to stack the triggers correctly, since when we sacrifice all of our Goblins to devour, we'll also be triggering Fecundity a whole bunch of times. If we let the Fecundity trigger resolve first, we'll accidentally kill ourselves by dying on an empty library, so make sure to manage the stack in a way that Voracious Dragon's damage resolves first, so that our opponent is dead before the Fecundity triggers resolve.
While not as much fun as killing with Voracious Dragon, our fallback plan is to Grapeshot our opponent with a bunch of copies thanks to storm. Since our combo turns naturally involve us casting 20 or more spells, it's pretty easy to get the storm count high enough to kill our opponent.
It's also worth mentioning that if all else fails (for example, our opponent has a Leyline of Sanctity and we can't target them with Voracious Dragon or Grapeshot), we can also try to win with Empty the Warrens tokens, which is a pretty effective plan in most matchups but can get blown out by sweepers like Anger of the Gods or Supreme Verdict, since our opponent will get a turn to untap and cast sorceries.
Infernal Plunge and Battle Hymn give us some extra mana production, which can help mid-combo or allow us to ramp into our combo pieces a turn or two earlier. Infernal Plunge also has the upside of allowing us to sacrifice a creature, which gives us some extra value if we happen to have our Fecundity but haven't drawn into Skirk Prospector yet.
Wild Cantor is basically our ultra-budget version of Manamorphose. Our combo is really good at making red mana, but sometimes we get pinched on green mana if we need to cast a second Fecundity, a Commune with Gods, or a sideboard card. Wild Cantor gives us a way to filter a red mana into a green mana, while also drawing us a card along the way with Fecundity. It's also a solid way to ramp into our combo in the early game if we can play it on Turn 1 and have it sit on the battlefield like an expensive, creature-based Lotus Petal.
Postmortum Lunge is just a one-of, but it's a solid addition to the deck. Its main purpose is to get Skirk Prospector back from the graveyard if our opponent manages to kill it before we are able to combo off, but it's also great when it's getting back something like Mogg War Marshal mid-combo to give us another ritual and more card draw. Since we're planning on sacrificing the creature that we reanimate anyway (or just winning the game), the downside of having to exile the creature at the end of our turn isn't really a problem, making Postmortem Lunge a very efficient option.
Lightning Bolt gives us just a touch of removal to keep our opponent's creatures under control. While we can win the game as early as Turn 3 with a solid draw, in other games, it takes us a few turns of digging with Faithless Looting and Commune with the Gods to find our combo pieces. In these slower games, having a bit of removal can be essential in staying alive long enough to find Skirk Prospector and Fecundity to combo off for the win.
All in all, we finished with a 2-3 with Goblin Storm, which isn't the best record we've ever posted, although every one of our losses was incredibly close. Against Titan Shift, our opponent had an early Chalice of the Void each game (which is really, really bad for our deck), and we still managed to win a game even through the Chalice, while also losing a super-close match to Belcher and getting brutally mana screwed against Jeskai Ascendancy in game three. As such, even though the record wasn't great, the deck was a lot of fun, and we were incredibly close to winning another match or even two.
As for the deck, it does sometimes struggle with consistency. Sometimes, we have games where we simply don't find one of our combo pieces and our deck doesn't do much of anything. On the other hand, it's incredibly explosive and fun. For a $50-ish combo deck in Modern, it's hard to complain. The deck can do some super crazy things and win as early as Turn 3, making it fast enough to keep up with the best decks in the format when everything comes together.
As far as improvements to the deck, the main challenge is that most of the potential upgrades add a lot to the budget. The mana base in our ultra-budget build is functional but pretty clunky by Modern standards, but upgrading to fetches and shocks isn't cheap. Battle Hymn probably isn't necessary; it might be worth playing an additional copy of Hordeling Outburst or even something like Goblin Bushwhacker. Speaking of Goblin Bushwhacker, playing a hybrid Goblin Storm / 8 Whack build might have some potential. We have some games where we flood the board with 1/1 Goblins but can't find our combo. In these games, a Bushwhacker or two could give us a legitimate path to winning without the combo.
In sum, Goblin Storm is a blast. It's probably in the middle in terms of level of competitiveness—it's good enough to beat just about anything in the format but does struggle with consistency (and also Chalice of the Void, which isn't that popular in Modern as a whole, but it was in our matches). If you like crazy combos, drawing your entire deck, and maybe winning on Turn 3 with Voracious Dragon of all things, give it a shot! I don't think you'll win a Grand Prix with the deck, but you can certain steal some wins at FNM and do it with style!
No ultra-budget list this week. The build we played in the videos was already in the ultra-budget range!
The non-budget build of Goblin Storm is looking to do one thing: increase consistency. The combo from the build in the videos is already strong. The main challenge is finding our combo pieces. To increase consistency, we do two things: first, we upgrade the mana with fetch lands, shock lands, and fast lands, which should ensure that we don't run into any issues with color screw. Second, we add in eight free cantrips in Street Wraith and Manamorphose. While neither of these cards do much with our combo directly, the idea is that these free redraws will essentially decrease the size of our deck from 60 cards to 52 cards, making it easier to consistently find our Skirk Prospector and Fecundity. We also add in a playset of Goblin Bushwhacker to support our backup plan of beating down with random Goblins—in the worst case, we can always cycle Goblin Bushwhacker with the help of our combo. To make room for these additions, we cut back on Wild Cantor (which is less necessary with Manamorphose and an upgraded mana base) and the underperforming rituals. All in all, this build should help fix some of the consistency issues with the budget build. While we'll still have some games where we don't draw our combo pieces, these games should happen less often.


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About: Outline of Kansas
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The following outline is provided as an overview of and topical guide to the U.S. state of Kansas: Kansas – U.S. state located in the Midwestern United States. It is named after the Kansas River which flows through it, which in turn was named after the Kansa Native American tribe, which inhabited the area. The tribe's name (natively kką:ze) is often said to mean "people of the wind" or "people of the south wind," although this was probably not the term's original meaning.
The following outline is provided as an overview of and topical guide to the U.S. state of Kansas: Kansas – U.S. state located in the Midwestern United States. It is named after the Kansas River which flows through it, which in turn was named after the Kansa Native American tribe, which inhabited the area. The tribe's name (natively kką:ze) is often said to mean "people of the wind" or "people of the south wind," although this was probably not the term's original meaning. (en)
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PLEASE NOTE: DO NOT FOLLOW GPS DIRECTIONS TO THE ADDRESS FOR Robert Frost Mountain Cabins. Those often lead to a dead end road (Seldon Mill Rd) or to a road that is not suitable for all vehicles (Green Rd). Instead, please follow the directions shown below.
Our cabins are located "off the beaten path" on a gravel road that's well maintained throughout the year. Our driveway is approximately 0.4 miles long ….. just long enough to leave all of your cares behind.
Follow Route 7 South to Route 125. Turn left onto E. Main St / Route 125 East. Continue on Route 125 for 4.3 miles. Turn left onto Lincoln Rd. Follow Lincoln Rd for 3.69 miles. Turn left onto North Branch Rd. Follow North Branch Rd for 1 mile. Robert Frost Mountain Cabins is on the left at 2430 North Branch Rd.
Start out going South on Airport Drive. At .4 miles, cross Williston Rd., at which point Airport Drive becomes Kennedy Drive. Continue on Kennedy Drive for .8 miles. Turn left onto Route 116 / Hinesburg Road. Continue on Route 116 for 27.7 miles. Turn left to stay on Route 116. Continue on Route 116 for 11.5 miles. Turn left onto E. Main St / Route 125 East. Continue on Route 125 for 3.7 miles. Turn left onto Lincoln Rd. Follow Lincoln Rd for 3.69 miles. Turn left onto North Branch Rd. Follow North Branch Rd for 1.04 miles. Robert Frost Mountain Cabins is on the left at 2430 North Branch Rd.
Start out going East on Rt. 10 to Rt. 133 to Interstate 89 in Vermont. Continue on Interstate 89 south to Burlington. Merge onto I-189 S via EXIT 13. Continue on I-189 for .2 miles. Take the US-7 exit. Turn left onto US-7/Shelburne Rd/Ethan Allen Hwy. Continue to follow US-7 S for 36 miles. Turn left onto E Main St/VT-125. Continue on Route 125 for 4.3 miles. Turn left onto Lincoln Rd. Follow Lincoln Rd for 3.69 miles. Turn left onto North Branch Rd. Follow North Branch Rd for 1 mile. Robert Frost Mountain Cabins is on the left at 2430 North Branch Rd.
Start out going north on I-87. Take the RT-149 exit, EXIT 20, toward Fort Ann/Whitehall. Follow RT-149 for 11.7 miles. Turn left onto US-4. Continue to follow US-4 for 35.9 miles to Rutland, Vermont. Take Route 7 north for approximately 30 miles. Turn right onto E Main St/VT-125. Continue on Route 125 for 4.3 miles. Turn left onto Lincoln Rd. Follow Lincoln Rd for 3.69 miles. Turn left onto North Branch Rd. Follow North Branch Rd for 1 mile. Robert Frost Mountain Cabins is on the left at 2430 North Branch Rd.
Start out going North on I-93 to I-89 North. Merge onto I-89 N. Take the VT-107 exit, EXIT 3, toward VT-14/Bethel/Royalton. Continue on Route 107 West for approximately 13 miles. Turn right onto VT-100. Continue on Route 100 North for 11.8 miles. Turn left onto VT-125 and follow for 11.8 miles. Turn right onto Lincoln Rd. Follow Lincoln Rd for 3.69 miles. Turn left onto North Branch Rd. Follow North Branch Rd for 1 mile. Robert Frost Mountain Cabins is on the left at 2430 North Branch Rd.
Start out going North on I-91. Merge onto I-89 N via EXIT 10N. Take the VT-107 exit, EXIT 3, toward VT-14/Bethel/Royalton. Continue on Route 107 West for approximately 13 miles. Turn right onto VT-100. Continue on Route 100 North for 11.8 miles. Turn left onto VT-125 and follow for 11.8 miles. Turn right onto Lincoln Rd. Follow Lincoln Rd for 3.69 miles. Turn left onto North Branch Rd. Follow North Branch Rd for 1 mile. Robert Frost Mountain Cabins is on the left at 2430 North Branch Rd.
Addison County Chamber of Commerce announces opening of Robert Frost Mountain Cabins on VermontBiz.com : VermontBiz.com published an article entitled New lodging property, bordering Green Mountain National Forest, opens in Ripton . We're grateful for this and all of the other community support and encouragement that we've been receiving.


The 2014 Business Smart Series will focus on telling your business or organizational story. Join us over the course of the year as our guest speakers discuss the methods, media, and successes of connecting with your customers through your unique story--from Once Upon a Time to Happily Ever After.
Register in advance for the entire series and receive a 25% discount! H-R Chamber Members: $90, Non-Members $135. Attend at least 4 sessions and come to the final session in Jan 2015 for free.
The Business Smarts Breakfast sessions are held on the 2nd Friday of every month at American National University (formerly National College) with networking opportunities and guest presentations on a variety of topics.
The series is sponsored by your Shenandoah Valley Small Business Development Center, the Harrisonburg-Rockingham Chamber of Commerce, and the Service Corps of Retired Executives (SCORE). Breakfast is provided by The HoneyBaked Ham Company.
Visit our Calendar or click each chapter title for details and register here.
Have a topic or speaker you'd like to hear at a future workshop? Let us know!