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{{\bf Proof:}\,\,}
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  \title{On Harder's $\Sl(2,\R)$\,-$\Sl (3,\R)$\,-\,identity}

\begin{document}
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 %\begin{title}{On Harder's $\Sl(2,\R)$\,-$\Sl (3,\R)$\,-\,identity}\end{title}

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\centerline{\bf  Appendix by  Don Zagier} \bigskip
  
The object of this appendix to prove the experimentally obtained formula stated in Section 2.2.4 of the paper.

\section {  Notations and statement of the identity}  Our notations differ slightly from those
of the paper.  Fix an integer $D\ge0\,$.  (This is the $d+2$ of the paper, but we
assume neither $D$ even nor $D>2\,$.)  We define coefficients
  \beq  C_{a,b}\;=\;\sum_{\nu=0}^{\min(a,b)}(-1)^{a+b-\nu}\binom a\nu\binom{2D-a}{b-\nu}
  \qquad\qquad(0\le a,\,b\le 2D)\,.   \eeq
(This is  $(-1)^a\,2^D\,C_{a,b}^{(d)}$ in the notation of the paper.)  For $n\in\Z$ and $z\in\C$ we define 
\beq \g_n(z)\;=\;\Gamma\bigl(\frac{z+|n|+1}2\bigr)^{-1}\;
 \sum_{k=0}^{|n|/2}(-1)^k\,\binom{|n|}{2k}\,\Gam\bigl(k+\h\bigr)\,\Gam\bigl(\frac{z+|n|}2-k\bigr)\,. \eeq
In the paper the number $c(z,n)=i^{-n}\g_n(z-1)$ is used instead, but  $\g_n(z)$ is 
slightly easier to work with since it is an even function of $n$ and is real for $z$ real,
and the shift of $z$ by~1 also simplifies some of the formulas.  Harder's identity says that
 \beq\begin{aligned}  &\sum_{\substack{ 0\le b\le2D\\ b\equiv m_\alpha\,\text{(mod 2)}}}
     \sum_{\substack{ 0\le c\le2D\\ c\equiv m_\beta\,\text{(mod 2)}}}
  i^{D+b+c}\,C_{2D,b}\,C_{b,c}\, C_{c,e}\,\g_{D-b}(z-D+1)\,\gamma_{D-c}(z) \\
  &\qquad =\; \frac{2^{3D}\pi}z\,\Bigl(\delta_{e,2D}\,+\,(-1)^{m_\bg}\delta_{e,0}\Bigr)
  \end{aligned}\eeq
for any $m_\ag ,\,m_\bg \in\Z/2\Z$ satisfying $m_\ag +m_\bg \equiv D$, $e\in2\Z$ satisfying $0\le e\le 2D$,
and $z\in\Z$ ($z=d-n_\ag =n_\bg +1$ in Harder's notation) satisfying $0<z<D-1$ and $z\equiv m_\ag \!\pmod 2$.
In fact this identity is true without the restriction on the parity of $m_\ag +m_\bg $ (but with the 
right-hand side replaced by~0 if $m_\ag +m_\bg \not\equiv D\pmod2$) or of $e$, and for all complex
numbers~$z$ (in the sense of meromorphic functions).  If we restrict to integral values of $z$, then
the individual terms on the left sometimes have poles and the identity does not make sense, but if 
$z\equiv m_\ag \!\pmod 2$ then all terms are finite and the identity is true whenever $z>0$, without
the assumption $z<D-1$.

\section{ Properties of the coefficients $C_{a,b}\,$}
The key property of the numbers (1) is the identity
  \beq \sum_{b=0}^{2D} C_{a,b}\,x^b\;=\;(x+1)^a\,(x-1)^{2D-a}\qquad\qquad(0\le a\le 2D)\,, \eeq
which follows straight from the binomial theorem.  This immediately implies the symmetry properties
  \beq C_{a,b}\;=\;(-1)^b\,C_{2D-a,b}\;=\;(-1)^a\,C_{a,2D-b}\qquad\qquad(0\le a,\,b\le 2D)  \eeq
(the first of which is also obvious from (1)).  A further symmetry property is
 \beq\binom{2D}a\,C_{a,b}\;=\;\binom{2D}b\,C_{b,a}\,, \eeq
which follows either from (1) and the identity
 $$ \binom{2D}a\,\binom a\nu\,\binom{2D-a}{b-\nu}\;=\; \binom{2D}b\,\binom b\nu\,\binom{2D-b}{a-\nu} $$
or else from (4) and the generating function calculation 
 $$ \sum_{a=0}^{2D}\sum_{b=0}^{2D}\binom{2D}a C_{a,b}\,x^ay^b\;=\;
 \sum_{a=0}^{2D}\binom{2D}a\,x^a(y+1)^a(y-1)^{2D-a}\;=\;(xy+x+y-1)^{2D} $$
Finally, substituting $x=\pm1$ into (4) and using (6) gives two identities which we will use below:
 \beq \sum_{c=0}^{2D}\binom{2D}cC_{c,e}=2^{2D}\delta_{e,2D},\quad
 \sum_{c=0}^{2D}(-1)^c\binom{2D}cC_{c,e}=2^{2D}\delta_{e,0}\quad\quad(0\le e\le 2D) \eeq 

Observe that (4) says simply that the $(2D+1)\times(2D+1)$ matrix $\bigl(C_{a,b}\bigr)_{0\le a,b\le2D}$ is the
image of the matrix $\bigl(\smallmatrix1&1\\1&{-1}\endsmallmatrix\bigr)$ under the $(2D)$th symmetric
power map $GL(2)\to GL(2D+1)$.  In particular, the square of this matrix is $\,2^{2D}$ times the identity. 


\section{Properties of the functions $\g_n(z)\,$} These are summarized in the following proposition and its corollary.
\begin{prop}  (i) For $n\in\Z$ and $z\in\C$ with $\Re(z)>0$ we have the integral representation
\beq \g_n(z)\;=\; \int_{-\pi/2}^{\pi/2}e^{in\theta}\,(\cos \theta)^{z-1}\,d\theta\,. 
\eeq
(ii) For all $n\in\Z$ the function $\g_n(z)$ is given by the closed formula
\beq \g_n(z)\;=\; \frac\pi{2^{z-1}}\,\frac{\Gam  (z)}{\Gam \bigl(\frac{z+1+n}2\bigr)\Gam \bigl(\frac{z+1-n}2\bigr)}\,
 \eeq
\end{prop}
\begin{cor} For $n$ and $z$ in $\,\Z\,$ the values of $\g_n(z)$ are given by
\beq \g_n(z)\;=\;\begin{cases}\qquad2^{1-z}\pi\,\binom{z-1}{\frac{|n|+z-1}2} &\text{if $z\ge1$, $z\equiv n+1\!\pmod2$, } \\
\;(-1)^{\frac{|n|-z}2}\,2^{1-z}\,z^{-1}\,/\,\binom{\frac{|n|+z-1}2}z &\text{if $z\ge1$, $z\equiv n\!\pmod2$, } \\
(-1)^{\frac{|n|+z-1}2}\,2^{-z}\,\pi\,\binom{\frac{|n|-z-1}2}{-z} &\text{if $z\le0$, $z\equiv n+1\!\pmod2$, } \\
\qquad\qquad\quad\infty &\text{if $z\le0$, $z\equiv n\!\pmod2$.} \end{cases}\eeq 
In particular, $\g_n(z)=0$ if $z\equiv n+1\!\pmod2$ and either $z<-|n|$ or $0<z<|n|$. 
 \end{cor}
\begin{bew} Since the left- and right-hand sides of both (8) and (9) are even functions of~$n$, we may
assume that $n\ge0$. Setting $m=2k$ in (2) and using the beta integral followed by the substitution $t=\sin^2\theta$, we find
\begin{align*}\g_n(z)&\;=\;\sum_{m=0}^n\,\binom nm\,\Re(i^m)\,\int_0^1t^{(m-1)/2}\,(1-t)^{(z+n-m-2)/2}\,dt\\
  &\;=\; 2\int_0^{\pi/2}\cos(n\theta)\,\cos(\theta)^{z-1}\,d\theta
\end{align*}
for $\Re(z)>0$, establishing (8).  If $z$ is a positive integer congruent to $n+1$ modulo~2, then (8) gives

 
\begin{equation*}\g_n(z) \;=\;\frac12\,\int_{-\pi}^\pi e^{in\theta}\,\bigl(\frac{e^{i\theta}+e^{-i\theta}}2\bigr)^{z-1}\,d\theta
 \;=\; \frac\pi{2^{z-1}}\,\binom{z-1}{\frac{z-1-n}2}\,, 
\end{equation*}
proving (9) for these values of $z$. But this is sufficient, since the quotient of each term in~(2)
by the right-hand side of~(9) is a rational function of~$z$ and two rational function which agree at infinitely
many arguments are equal.  We can also give a purely combinatorial proof of (9), e.g., if $n=2m\ge0$
(the case of odd $n$ is similar), then using the duplication formula of the gamma function we find
$$\frac{\g_n(z)}{\text{RHS of (9)}}\;=\;\binom{m-\frac{z+1}2}m^{-1}\,\sum_{k=0}^m\binom{m-\h}k\binom{-\frac z2}{m-k}\;=\;1\,.$$
The corollary follows immediately from (9) by computing the values or limiting values of the right-hand side
as~$z$ tends to an integer.  The second and third lines of (10) are   identities given in Section~2.2.2 of the
paper.
 \end{bew}

\section{Proof of the identity (3)}  For $0\le c\le2D$ we define a meromorphic function $\wg_c(z)$ by
  \beq  \wg_c(z)\;=\;\sum_{b=0}^{2D}i^{D+b+c}\,C_{c,b}\,\g_{D-b}(z-D+1)\,.   \eeq
Note that this is a real function (in the sense that $\overline{\wg_c(\overline z)}=\wg_c(z)$) because of
the symmetry properties $C_{c,2D-b}=(-1)^cC_{c,b}$ and $\g_n(z)=\g_{-n}(z)$.  The key fact is the identity 
  \beq  \wg_c(z) \;=\; \frac{2^{D+1}\,\pi}{z\,\g_{D-c}(z)}\,.  \eeq
To prove this, we may assume that $\Re(z)>D-1$ since both sides of~(12) are meromorphic functions of~$z$.  Then~(8) and (4) 
together with the substitution $t=\cos\bigl(\frac\pi4-\frac\theta2\bigr)^2$ and the beta integral give
  \begin{align*} \wg_c(z) &\;=\; i^{D+c}\,\int_{-\pi/2}^{\pi/2}(\cos\theta)^{z-D}\,e^{iD\theta}\,(ie^{-i\theta}+1)^c\,(ie^{-i\theta}-1)^{2D-c}\,d\theta \\
  &\;=\; 2^{z+D}\, \int_{-\pi/2}^{\pi/2}\cos\bigl(\frac\pi4-\frac\theta2\bigr)^{z+c-D}\,\sin\bigl(\frac\pi4-\frac\theta2\bigr)^{z+D-c}\,d\theta \\
  &\;=\; 2^{z+D}\, \int_0^1 t^{(z+c-D-1)/2}\,(1-t)^{(z+D-c-1)/2}\,dt  \\
 & \;=\; 2^{z+D}\,\Gam \bigl(\frac{z+c-D+1}2\bigr)\,\Gam \bigl(\frac{z-c+D+1}2\bigr)\,/\,\Gam (z+1)\,,   
   \end{align*}  
and together with (9) this proves the claim.
 
Combining equations~(12) and~(7), we obtain
\begin{align*} & \sum_{b=0}^{2D}\sum_{c=0}^{2D} i^{D+b\pm c}\,\,C_{2D,b}\,C_{b,c}\,C_{c,e}\,\g_{D-b}(z-D+1)\,\g_{D-c}(z) \\
  &\quad\;=\;\sum_{c=0}^{2D}\,(\pm1)^c\,{ \binom{2D}c}\,\wg_c(z)\,\g_{D-c}(z)\,C_{c,e}
         \qquad\bigl(\text{by (6), since $  C_{2D,b}=\binom{2D}b\,$}\bigr)\\
  &\quad\;=\; \frac{2^{D+1}\pi}z\,\sum_{c=0}^{2D} \,(\pm1)^c\,{ \binom{2D}c}\,C_{c,e}\qquad\text{(by (12))}\\
  &\quad\;=\;  \frac{2^{3D+1}\pi}z\,\delta_{e,D\pm D}\qquad\text{(by (7))}\,, 
   \end{align*}
and (3) follows by taking the real or imaginary part of the sum or difference of these two identities.
\end{document} 
