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\begin{document}
\title[Rigidity in quaternionic space]
{Local rigidity of complex hyperbolic lattices in quaternionic
hyperbolic space}
\author{Inkang Kim and Pierre Pansu}
\date{}
\maketitle

\begin{abstract}
In this note, we study infinitesimal deformations of complex hyperbolic space lattices in quaternionic hyperbolic space of the same dimension and prove that they are obstructed. A local rigidity result follows.
\end{abstract}

\footnotetext[1]{I. Kim gratefully acknowledges the partial support
of KRF Grant (0409-20060066) and a warm support of IHES during his
stay.} \footnotetext[2]{P. Pansu, Univ Paris-Sud, Laboratoire de
Math\'ematiques d'Orsay, Orsay, F-91405} \footnotetext[3]{\hskip
47pt CNRS, Orsay, F-91405.}


\section{Introduction}

A. Weil's local rigidity theorem, \cite{Weil}, states that a uniform lattice $\Gamma$ in a semisimple Lie group $G$ cannot be continuously deformed in $G$ (unless its Lie algebra $\mathfrak{g}$ equals $\mathfrak{sl}(2,\br)$). In fact, Weil proves that the space of first order infinitesimal deformations (the first cohomology of the $\Gamma$-module $\mathfrak{g}$ for the adjoint action of $\Gamma$) vanishes. 

When $G$ is embedded in a larger group $G'$, infinitesimal deformations of $\Gamma$ in $G'$ may exist. This can happen only if $G$ is simple of rank one. The case when $G=SO(n,1)$, $G'$ arbitrary, is rather well understood, see \cite{Millson}. Each time no general vanishing theorem applies to $H^{1}(\Gamma,\mathfrak{g}')$, examples of actual deformations exist.
When $G=SU(n,1)$ and $G'=SU(n+1,1)$, $n\geq 2$, W. Goldman and J. Millson \cite{GM} have shown that all infinitesimal deformations are obstructed. 

In this paper, we carry over the same computation in case $G=SU(n,1)$ and $G'=Sp(n,1)$, $SU(2n,2)$ or $SO(4n,4)$. Let $\Gamma$ be a uniform lattice in $SU(n,1)$. The second order obstruction is a quadratic map $H^{1}(\Gamma,\mathfrak{g}')\ra H^{2}(\Gamma,\mathfrak{g}')$, the {\em cup-square}. We show that the cup-square of a nontrivial element of $H^{1}(\Gamma,\mathfrak{g}')$ does not vanish. 

This is done in two steps. First, using Matsushima and Murakami's method, \cite{MM}, we show that harmonic 1-forms representing nontrivial classes in $H^{1}(\Gamma,\mathfrak{g}')$ are severely restricted : most of their components vanish, and one can interpret them as $(1,0)$-forms $\alpha$ with values in a certain complex vectorbundle. Then we show that the cup-square $[\alpha\wedge\alpha]$ paired with the K\"ahler form of complex hyperbolic space is proportional to the squared $L^2$-norm of $\alpha$.

This leads to the following theorem.
\begin{sthm}
\label{thm}
Let $\Gamma\subset SU(n,1)$ be a uniform lattice. Let $\phi:\Gamma\ra G'$ be one of the composed inclusions $\Gamma\ra SU(n,1)\ra U(n,1)\ra Sp(n,1)\ra SU(2n,2)\ra SO(4n,4)$. Then every homomorphism $\phi':\Gamma\ra G'$ which is close enough to $\phi$ has its image contained in a conjugate of $SU(n,1)Z$, where $Z$ denotes the centralizer of $SU(n,1)$ in $G'$.
\end{sthm}

Using a different method, B. Klingler \cite{Klingler} gives a more geometric proof of Theorem \ref{thm}. However, our approach gives restrictions on harmonic 1-forms, which may be useful in order to understand these objects.

\section{Infinitesimal deformations}

Let $M\in \mathfrak{gl}(n+1,\bh)$ and $M=C+jD$ where $C,D\in
\mathfrak{gl}(n+1,\bc)$. Then $M\in \mathfrak{sp}(n,1)$, if and only if $C\in \mathfrak{u}(n,1)$ and $JD$ is symmetric where $J$ is the diagonal matrix with entries $1,1,-1$. Write $E=JD$. If $A\in U(n,1)$, 
\begin{eqnarray*}
A^{-1}MA=A^{-1}(C+jJE)A=A^{-1}CA+jJA^{\bot}EA.
\end{eqnarray*}
So, under $SU(n,1)$,
$$\mathfrak{sp}(n,1)=\mathfrak{u}(n,1)\oplus S^2 V_{\bc}^{*}=\br\oplus\mathfrak{su}(n,1)\oplus S^2 V_{\bc}^{*},$$ 
where $V_{\bc}=\bc^{n+1}$ denotes the standard representation of $GL(n+1,\bc)$ restricted to $SU(n,1)$. In a similar manner, we shall see in sections \ref{su2n2} and \ref{so4n4} that
\begin{eqnarray*}
\mathfrak{su}(2n,2)&=&\br\oplus 2\mathfrak{su}(n,1)\oplus\Lambda^2 V_{\bc}^* 
\oplus S^2 V_{\bc}^* ,\\
\mathfrak{so}(4n,4)&=&2\br\oplus 2\mathfrak{su}(n,1)\oplus 2\Lambda^{2}V_{\bc}^* \oplus 2\Lambda^{2}\bar{V}_{\bc}^{*}\oplus S^2 V_{\bc}^* \oplus S^{2}\bar{V}_{\bc}^{*}.
\end{eqnarray*}

Let $\Gamma$ be a lattice in $SU(n,1)$. Let $\mathfrak{z}$ denote the centralizer of $\mathfrak{su}(n,1)$ in $\mathfrak{g}'$. Then $H^1(\Gamma,\mathfrak{z})=H^1(\Gamma,\br)\otimes\mathfrak{z}$ need not vanish. Such infinitesimal deformations integrate into homomorphisms of $\Gamma$ to the centralizer $Z$ of $SU(n,1)$ in $G'$, yielding uninteresting deformations in $SU(n,1)Z$.

Weil's vanishing theorem, \cite{Weil}, implies that $H^1(\Gamma,\mathfrak{su}(n,1))=0$. Raghunathan's vanishing theorem, \cite{Ra}, implies that $H^1(\Gamma,\Lambda^2 V_{\bc}^{*})$ and $H^1(\Gamma,\Lambda^2 \bar{V}_{\bc}^{*})$ vanish, but the other factors $H^1(\Gamma,S^2 V_{\bc}^{*})$ and $H^1(\Gamma,S^2 \bar{V}_{\bc}^{*})$ need not vanish. Indeed, the symmetric powers $S^j \bar{V}_{\bc}$ and $S^j V_{\bc}$ are the exceptions in Raghunathan's vanishing theorem, \cite{Ra}. In fact, following D. Kazhdan, \cite{Kazhdan}, A. Borel and N. Wallach give a proof that $H^1(\Gamma,S^2 V_{\bc}^{*})\not=0$ for sufficiently deep uniform congruence subgroups $\Gamma$ constructed from a degenerate Hermitian form of signature $(n,1)$ defined over a totally real number field, see \cite{BW}, corollary VIII.5.9. When $n=2$, using techniques from \cite{Rogawski}, L. Clozel has been able to show that for all $\bq$-structures on $SU(2,1)$, either isotropic or anisotropic, $H^1(\Gamma,S^2 V_{\bc}^{*})\not=0$ for sufficiently deep congruence subgroups $\Gamma$, \cite{Clozel}.

\section{A classical vanishing theorem}

\subsection{Matsushima and Murakami's vanishing theorem}
\label{MM} 

Let $X=G/K$ and $M=\Gamma\setminus X$ be a compact locally symmetric manifold. Let $\rho:G\ra End(F)$ be a representation. The cohomology $H^p (M,F_\rho)$ is isomorphic to the cohomology of $F$-valued forms $\eta$ on $X$ such that
\begin{eqnarray}
\label{eqninv}
(L_\gamma)^* \eta=\rho(\gamma)\eta,\quad \gamma\in \Gamma,
\end{eqnarray}
where $L_\gamma$ denotes the left action of $\Gamma$ on $X$. Let $\pi:G\to X$ be the orbital map. Following p. 376 of \cite{MM}, define an $F$-valued differential form $\eta^0$ on $G$ as follows.
\begin{eqnarray}
\label{eqnadj}
\eta^0_s=\rho(s^{-1})\pi^* \eta_s, \quad s\in G.
\end{eqnarray}
Then the equivariance condition (\ref{eqninv}) becomes a simpler invariance condition
\begin{eqnarray*}
(L_\gamma)^* \eta^0=\eta^0,\quad \gamma\in \Gamma,
\end{eqnarray*}
where $L_\gamma$ is a left translation on $G$.

Fix a Killing-orthonormal basis $X_1,\ldots,X_N$ of $\mathfrak{p}$. Pick an inner product $(,)_F$ on $F$ which is $\rho(K)$-invariant and for which elements of $\rho(\mathfrak{p})$ are symmetric. The induced inner product on $Hom(\mathfrak{p},F)$ is
$$(\eta,\zeta)=\sum_{h=1}^N (\eta(X_h),\zeta(X_h))_F .$$ 
Define a symmetric operator on $Hom(\mathfrak{p},F)$ as follows.
\begin{eqnarray*}
T\eta(Y)=\sum_{k=1}^N \frac{1}{p}\rho(X_k)^2 \eta(Y)+\rho([Y,X_k])\eta(X_k).
\end{eqnarray*}

\begin{thm}[Matsushima-Murakami]
If $\eta$ is a harmonic $p$-form on $M=\Gamma\setminus G/K$, then
\begin{eqnarray}
\label{harmonic}
\int_{\Gamma\setminus G}(T\eta^0 ,\eta^0 )\leq 0.
\end{eqnarray}
In particular, if $T$ is positive definite, then the cohomology
group $H^p (\Gamma,F_\rho)$ vanishes.
\end{thm}

In our setting, $G=SU(2,1)$, $p=1$ and $F=\mathfrak{w}$ equipped with the obvious $U(3)$-invariant inner product. It turns out that $H$ is positive indefinite, as we shall see.

\subsection{Case of 1-forms}

\begin{Prop}
\label{calculT}
Let $\eta\in Hom(\mathfrak{p},F)$. Let $\beta$ denote the $F$-valued bilinear form on $\mathfrak{p}$ defined by
\begin{eqnarray*}
\beta(X,Y)=\rho(X)(\eta(Y)).
\end{eqnarray*}
Split $\beta=\sigma+\alpha$ into its symmetric and skew-symmetric parts. Then
\begin{eqnarray*}
(T\eta,\eta)=2|\alpha|^2 +|\mathrm{Trace}(\beta)|^2 .
\end{eqnarray*}
\end{Prop}

\begin{pf}
The first term in $(T\eta,\eta)$ is
\begin{eqnarray*}
(T_1 \eta,\eta)
&:=&\sum_{k,\,\ell=1}^N (\rho(X_k)^2 \eta(X_\ell),\eta(X_\ell))_F\\
&=&\sum_{k,\,\ell=1}^N (\rho(X_k) \eta(X_\ell),\rho(X_k)\eta(X_\ell))_F\\
&=&\sum_{k,\,\ell=1}^N |\beta(X_k,X_\ell)|^2_F\\
&=&|\beta|^2 .
\end{eqnarray*}
The second term in $(T\eta,\eta)$ is
\begin{eqnarray*}
(T_2 \eta,\eta)
&:=&\sum_{k,\,\ell=1}^N (\rho([X_\ell ,X_k])\eta(X_k),\eta(X_\ell))_F\\
&=&(T_3 \eta,\eta)-(T_4 \eta,\eta),
\end{eqnarray*}
where
\begin{eqnarray*}
(T_3 \eta,\eta)
&:=&
\sum_{k,\,\ell=1}^N (\rho(X_\ell)\circ \rho(X_k)\eta(X_k),\eta(X_\ell))_F\\
&=&\sum_{k,\,\ell=1}^N (\rho(X_k)\eta(X_k),\rho(X_\ell)\eta(X_\ell))_F\\
&=&\sum_{k,\,\ell=1}^N (\beta(X_k ,X_k),\beta(X_\ell ,X_\ell))_F\\
&=&|\sum_{k=1}^N \beta(X_k ,X_k)|^2_F\\
&=&|\mathrm{Trace}(\beta)|^2 ,
\end{eqnarray*}
and
\begin{eqnarray*}
(T_4 \eta,\eta)
&:=&
\sum_{k,\,\ell=1}^N (\rho(X_k)\circ \rho(X_\ell)\eta(X_k),\eta(X_\ell))_F\\
&=&\sum_{k,\,\ell=1}^N (\rho(X_\ell)\eta(X_k),\rho(X_k)\eta(X_\ell))_F\\
&=&(\beta,\beta\circ\phi).
\end{eqnarray*}
Here, $\phi(X,Y)=(Y,X)$. Note that $\phi$ merely permutes vectors in the basis of $\mathfrak{p}\otimes\mathfrak{p}$. Therefore
\begin{eqnarray*}
(\sigma,\alpha)=(\sigma\circ\phi,\alpha\circ\phi)=(\sigma,-\alpha)=-(\sigma,\alpha),
\end{eqnarray*}
Thus $(\sigma,\alpha)=0$. Hence
\begin{eqnarray*}
|\beta|^2 =|\sigma|^2 +|\alpha|^2 ,
\end{eqnarray*}
\begin{eqnarray*}
(\beta,\beta\circ\phi)&=&
(\sigma+\alpha,\sigma-\alpha)\\
&=&|\sigma|^2-|\alpha|^2\\
&=&|\beta|^2-2|\alpha|^2 .
\end{eqnarray*}
Summing up,
\begin{eqnarray*}
(T\eta,\eta)=|\beta|^2 +|\mathrm{Trace}(\beta)|^2 -(|\beta|^2-2|\alpha|^2)=2|\alpha|^2 +|\mathrm{Trace}(\beta)|^2 .
\end{eqnarray*}
\end{pf}

\section{Consequences of Matsushima-Murakami's method}

\subsection{Subspaces in $\mathfrak{sp}(n,1)$}

From now on, $G=SU(n,1)$, $F=\mathfrak{w}$ is the space of complex quadratic forms on $\bc^{n+1}$, with the usual action of $GL(n+1,\bc)$, $(X,Q)\mapsto X^{\top}QX$, restricted to $SU(n,1)$. The inner product on $F$ is the usual $U(n+1)$-invariant hermitian form. 

We have seen that, under $U(n,1)$, $\mathfrak{sp}(n,1)=\mathfrak{u}(n,1)\oplus \mathfrak{w}$.

Under $U(n)$, $\mathfrak{sp}(n)$ splits as
\begin{eqnarray*}
\mathfrak{sp}(n)=\mathfrak{u}(n)\oplus \mathfrak{w}',
\end{eqnarray*}
where $\mathfrak{sp}(n)$ consists of skew-hermitian $n\times n$
quaternionic matrices, $\mathfrak{u}(n)$ of skew-hermitian complex
matrices and $\mathfrak{w}'$ of matrices of the form $jD$ where $D$ is a
symmetric complex matrix. In other words, 
\begin{eqnarray*}
\mathfrak{w}'=\mathfrak{w}\cap \mathfrak{sp}(n).
\end{eqnarray*}

As representations of $U(n)$, $\mathfrak{p}$ and $\mathfrak{w}'$ are $\bc$-linear. $\mathfrak{p}$ identifies with the standard $n$-dimensional complex representation of $U(n)$ and $\mathfrak{w}'$ with $S^2 \mathfrak{p}^{*}$. Then
\begin{eqnarray*}
Hom_{\br}(\mathfrak{p},\mathfrak{w}')=Hom_{\bc}(\mathfrak{p},S^2 \mathfrak{p}^{*})\oplus Hom_{\bar{\bc}}(\mathfrak{p},S^2 \mathfrak{p}^{*})
\end{eqnarray*}

The space $Hom_{\bc}(\mathfrak{p},S^2 \mathfrak{p}^{*})$ contains a copy $U$ of $S^3 \mathfrak{p}^{*}$. The embedding $S^3 \mathfrak{p}^{*}\to Hom_{\bc}(\mathfrak{p},S^2 \mathfrak{p}^{*})$ associates to a symmetric trilinear form $\tau$ the linear map $w\mapsto \iota_w \tau :(z_1 ,z_2)\mapsto \tau(w,z_1 ,z_2)$.

\begin{Prop}
\label{pi} 
Let $\Gamma$ be a cocompact lattice in $G=SU(n,1)$. Let $\mathfrak{w}\simeq S^2 V_{\bc}^*$ denote the space of complex quadratic forms on $\bc^{n+1}$, and $\mathfrak{w}'\simeq S^2 \mathfrak{p}^*$ the subspace of complex quadratic forms on $\bc^{n}$. Let $\alpha$ be a $\Gamma$-equivariant harmonic $\mathfrak{w}$-valued 1-form on $H^n_\bc$. Then, for all $Y\in T_{x_0}H^n_\bc=\mathfrak{p}$, $\alpha_{x_0}(Y)\in \mathfrak{w}'$. Furthermore, $\alpha_{x_0}\in Hom_{\br}(\mathfrak{p},\mathfrak{w}')$ is $\bc$-linear and belongs to the summand $U\simeq S^3 \mathfrak{p}^{*}$. In particular, $\alpha$ is a $S^2 \mathfrak{p}^{*}$-valued $(1,0)$-form.
\end{Prop}

Beware that $\mathfrak{w}'\simeq S^2 \mathfrak{p}^*$ defines a non-parallel complex vector subbundle in the flat bundle $\mathfrak{w}\simeq S^2 V_{\bc}^*$.

\begin{rem}
\label{pibar}
Since $S^2 \bar{V}_{\bc}^*$ is the conjugate vector space of $S^2 V_{\bc}^*$, Proposition \ref{pi} implies that $\Gamma$-equivariant harmonic $S^2 \bar{V}_{\bc}^*$-valued 1-forms on $H^n_\bc$ are in fact $S^2 \bar{\mathfrak{p}}^{*}$-valued $(0,1)$-forms.
\end{rem}


\subsection{Proof of Proposition \ref{pi}}

Let $\mathfrak{su}(n,1)=\mathfrak{su}(n)\oplus \mathfrak{p}$ be the Cartan
decomposition of $\mathfrak{su}(n,1)$.
Here, $\mathfrak{su}(n)=s(\mathfrak{u}(n)\oplus \mathfrak{u}(1))$
consists of traceless block-diagonal skew-hermitian complex $(n+1)\times
(n+1)$ matrices, and $\mathfrak{p}$ consists of complex matrices of the form
$\begin{pmatrix}
0 & x   \\
x^{*}&0
\end{pmatrix}$, $x\in\bc^n$. A straightforward calculation yields

\begin{lemma}
\label{adjoint}
Let $X=\begin{pmatrix}
0 & x   \\
x^{*} & 0
\end{pmatrix}$, $x\in\bc^n$, be a vector of $\mathfrak{p}$. Let $Q=\begin{pmatrix}
A & B   \\
B^{\bot} & d
\end{pmatrix}$ be a $(n+1)\times (n+1)$ symmetric complex matrix. Then
\begin{eqnarray*}
\rho(X)(Q)=X^{\top}Q+QX=\begin{pmatrix}
(Bx^{*})^{\top}+Bx^{*} & Ax+d\bar{x}\\
(Ax+d\bar{x})^{\top} &  2x^{\top}B
\end{pmatrix}.
\end{eqnarray*}
\end{lemma}

Let $\eta\in Hom(\mathfrak{p},F)$ be represented by a matrix $Q=Y\mapsto\begin{pmatrix}
A(Y) & B(Y)   \\
B(Y)^{\bot} & d(Y)
\end{pmatrix}$ of $\br$-linear forms on $\mathfrak{p}$. Then the bilinear form $\beta=\begin{pmatrix}
\beta_{1}&\beta_{2}\\
\beta_{2}^{\top} & \beta_{3}
\end{pmatrix}$
becomes a triple of matrix valued bilinear forms on $\bc^n$,
\begin{eqnarray*}
\beta_{1}(x,y)&=&(B(y)x^{*})^{\top}+B(y)x^{*},\\
\beta_{2}(x,y)&=&A(y)x+d(y)\bar{x},\\
\beta_{3}(x,y)&=&2x^{\top}B(y).
\end{eqnarray*}

According to Proposition \ref{calculT}, $(T\eta,\eta)=0$ if and only if the following 6 equations hold.
\begin{eqnarray*}
\beta_{1},~\beta_{2}~\mathrm{and}~\beta_{3}~\mathrm{are~symmetric},\mathrm{Trace}(\beta_{1})=0,~\mathrm{Trace}(\beta_{2})=0,~\mathrm{Trace}(\beta_{3})=0.
\end{eqnarray*}

\begin{lemma}
\label{beta3}
If $\beta_{3}$ and $\beta_1$ are symmetric, then $B=0$.
\end{lemma}

\begin{pf}
Let $B_{\bc}$ and $B_{\bar{\bc}}$ denote the $\bc$-linear (resp. anti $\bc$-linear) components of the $\br$-linear map $B$. Matrixwise, each of $B_{\bc}$ and $B_{\bar{\bc}}$ is given by a $n\times n$ complex matrix $\mathcal{B}_{\bc}$ (resp. $\mathcal{B}_{\bar{\bc}}$), and $B(y)=\mathcal{B}_{\bc}y+\mathcal{B}_{\bar{\bc}}\bar{y}$. Thus
\begin{eqnarray*}
\beta_3 (x,y)=x^{\top}\mathcal{B}_{\bc}y+x^{\top}\mathcal{B}_{\bar{\bc}}\bar{y}
\end{eqnarray*}
is the sum of a $\bc$-bilinear and a sesquilinear form. If $\beta_3$ is symmetric, the sesquilinear part vanishes (i.e. $\mathcal{B}_{\bar{\bc}}=0$), and $\mathcal{B}_{\bc}$ is symmetric. 

Next,
\begin{eqnarray*}
\beta_1 (x,y)=(\mathcal{B}_{\bc}yx^{*})^{\top}+\mathcal{B}_{\bc}yx^{*}
\end{eqnarray*}
is sesquilinear. If $\beta_1$ is symmetric, it is identically zero. Since rank one matrices of the form $yx^{*}$ span all $n\times n$ complex matrices, $(\mathcal{B}_{\bc}M)^{\top}+\mathcal{B}_{\bc}M=0$ for all $n\times n$ complex matrices $M$. Take $M=\mathcal{B}_{\bc}^{*}$ and take the trace to conclude that $\mathcal{B}_{\bc}=0$.
\end{pf}

\begin{lemma}
\label{beta2}
If $\beta_{2}$ is symmetric and $\mathrm{Trace}(\beta_{2})=0$, then $d=0$ and $A(y)$ depends $\bc$-linearly on $y$. Furthermore, identifying $\bc^n$-valued bilinear maps with trilinear forms, $(x,y)\mapsto A(y)x$ is fully symmetric.
\end{lemma}

\begin{pf}
Let $A_{\bc}$ and $A_{\bar{\bc}}$ denote the $\bc$-linear (resp. anti $\bc$-linear) components of the $\br$-linear map $A:\bc^n \ra S^{2}(\bc^n)$. Similarly, let $d_{\bc}$ and $d_{\bar{\bc}}$ denote the $\bc$-linear (resp. anti $\bc$-linear) components of the $\br$-linear form $d$. If $\beta_2 : (x,y)\mapsto A(y)x+d(y)\bar{x}$ is symmetric, then
\begin{eqnarray*}
\begin{cases}
\forall x,\,y\in\bc^n ,~A_{\bar{\bc}}(y)x &=d_{\bc}(x)\bar{y},\\
(x,y)\mapsto d_{\bar{\bc}}(y)\bar{x}  & \text{is symmetric},\\
(x,y)\mapsto A_{\bc}(y)x  & \text{is symmetric}.
\end{cases}
\end{eqnarray*}

The trace of the restriction of $\beta$ to a complex line $\bc e$, $|e|=1$, depends only on its sesquilinear part 
\begin{eqnarray*}
\beta_2^{sq}(x,y)=A_{\bar{\bc}}(y)x+d_{\bc}(y)\bar{x}=d_{\bc}(x)\bar{y}+d_{\bc}(y)\bar{x}.
\end{eqnarray*}
and is equal to $2\beta_2^{sq}(e,e)=4d_{\bc}(e)\bar{e}$. Let $e_1 ,\ldots,e_n$ be a Hermitian basis of $\bc^n$. Then 
\begin{eqnarray*}
\mathrm{Trace}(\beta_2)=\mathrm{Trace}(\beta_2^{sq})=4\sum_{k=1}^{n}d_{\bc}(e_k)\bar{e}_k .
\end{eqnarray*}
Since $\mathrm{Trace}(\beta_{2})=0$, we get $d_{\bc}=0$. This implies that $A_{\bar{\bc}}(y)x=0$ for all $x$ and $y$, i.e. $A_{\bar{\bc}}=0$. 

Next, pick a nonzero vector $y\in\mathrm{ker}(d_{\bar{\bc}})$. Since $(x,y)\mapsto d_{\bar{\bc}}(y)\bar{x}$ is symmetric, for all $x\in\bc^n$, $d_{\bar{\bc}}(x)\bar{y}=0$, thus $d_{\bar{\bc}}=0$.

Finally, view the components of $A_{\bc}(y)x$ in some Hermitian basis of $\bc^n$ as bilinear forms on $\bc^n$, with respective matrices $\mathcal{A}^{1},\ldots,\mathcal{A}^{n}$. Since $(x,y)\mapsto A_{\bc}(y)x$ is symmetric, these matrices are symmetric, $\mathcal{A}^{\ell}_{jk}=\mathcal{A}^{\ell}_{kj}$. But for every $y=(y_{1},\ldots,y_{n})\in\bc^n$, $A(y)_{k\ell}=\sum_{j}y_{j}\mathcal{A}^{\ell}_{jk}$ is symmetric, i.e. $\mathcal{A}^{\ell}_{jk}=\mathcal{A}^{k}_{j\ell}$. This implies that $\mathcal{A}^{\ell}_{jk}$ is fully symmetric.
\end{pf}

\section{Second order obstruction}

\subsection{Cup-product}

\begin{Def}
\label{deflambda}
On $\mathfrak{sp}(n,1)$, there is an $SU(n,1)$-invariant linear form
$\lambda$. It is the Killing inner product with the $SU(n,1)$-invariant vector $iI_{n+1}$, which generates the centralizer of $SU(n,1)$ in $Sp(n,1)$. 
\end{Def}

We shall need an expression of the restriction of $\lambda$ to $\mathfrak{u}(n)$. The restriction of the Killing form of $Sp(n,1)$ to $U(n)$ is proportional to the Killing form of $U(n)$, which is proportional to $\Re e(\mathrm{Trace}_{\bc}(A^{*}A))$. Therefore, for $A\in \mathfrak{u}(n)\subset \mathfrak{sp}(n,1)$,
\begin{eqnarray*}
\lambda(A)=A\cdot iI_{n+1}=\Re e(i\,\mathrm{Trace}_{\bc}(A))
\end{eqnarray*}
up to a nonzero multiplicative constant.

\begin{lemma}
\label{square} 
Let $\alpha$ be an $\mathfrak{sp}(n)$-valued $(1,0)$-form
on $T_{x_{0}}H_{\bc}^{n}=\mathfrak{p}$. Assume that $\alpha$ belongs to $Hom_{\bc}(\mathfrak{p},\mathfrak{w}')$. Thanks to the Lie bracket of $\mathfrak{sp}(n)$, $[\alpha\wedge\alpha]$ becomes an $\mathfrak{sp}(n)$-valued 2-form on $\mathfrak{p}$. Let $\omega$ denote the K\"ahler form on $\mathfrak{p}$. There is a nonzero constant $c$ such that
\begin{eqnarray*}
\lambda\circ[\alpha\wedge\alpha]\wedge\omega^{n-1}=c|\alpha|^2
\omega^{n}.
\end{eqnarray*}
\end{lemma}

\begin{pf}
Recall that the embedding of $\mathfrak{w}'$ to $\mathfrak{sp}(n)$ is defined by $A\mapsto jA$. Write $\alpha=j\delta$ where $\delta$ is a symmetric complex matrix of $(1,0)$-forms. Then, for all $Y$, $Y'\in\mathfrak{p}$,
\begin{eqnarray*}
\alpha\wedge\alpha(Y,Y')=\alpha(Y)\otimes\alpha(Y')-\alpha(Y')\otimes\alpha(Y)\in\mathfrak{sp}(n)\otimes\mathfrak{sp}(n),
\end{eqnarray*}
\begin{eqnarray*}
[\alpha\wedge\alpha](Y,Y')=[\alpha(Y),\alpha(Y')]-[\alpha(Y'),\alpha(Y)]=2[\alpha(Y),\alpha(Y')]\in\mathfrak{sp}(n).
\end{eqnarray*}
Let $A$, $B$ be two symmetric complex matrices. The Lie bracket of their images in $\mathfrak{sp}(n)$ is
\begin{eqnarray*}
[jA,jB]=jAjB-jBjA=-\bar{A}B+\bar{B}A,
\end{eqnarray*}
(note it belongs to $\mathfrak{u}(n)$), thus
\begin{eqnarray*}
[\alpha\wedge\alpha](Y,Y')=-2(\overline{\delta(Y)}\delta(Y')-\overline{\delta(Y')}\delta(Y)),
\end{eqnarray*}
showing that $[\alpha\wedge\alpha]$ is a matrix of $(1,1)$-forms. Up to a nonzero constant,
\begin{eqnarray*}
\lambda\circ[\alpha\wedge\alpha](Y,Y')=\Im m(\mathrm{Trace}_{\bc}(\delta(Y)^{*}\delta(Y'))).
\end{eqnarray*}
Note that $\lambda\circ[\alpha\wedge\alpha](Y,iY)=\mathrm{Trace}(\delta(Y)^{*}\delta(Y))=|\delta(Y)|^2 >0$.

If $\phi$ is a $(1,1)$-form on $\bc^n$, then
\begin{eqnarray*}
\frac{\phi\wedge\omega^{n-1}}{\omega^{n}}=\frac{2}{n}\sum_{k=1}^{n}\phi(E_k ,iE_k),
\end{eqnarray*}
where $E_1 ,\ldots,E_n$ is a unitary basis of $\bc^n$ (i.e. $(E_1 , iE_1 ,\ldots,E_n ,iE_n)$ is an orthonormal basis of the underlying real Euclidean vectorspace). Therefore
\begin{eqnarray*}
\frac{\lambda\circ[\alpha\wedge\alpha]\wedge\omega^{n-1}}{\omega^{n}}=\frac{2}{n}\sum_{k=1}^{n}|\delta(E_k)|^2
\end{eqnarray*}
is a nonzero multiple of $|\alpha|^2$.
\end{pf}

\begin{Prop}
\label{cup} 
Let $\alpha$ be a nonzero $\Gamma$-equivariant harmonic $\mathfrak{sp}(n,1)$-valued 1-form on $H^n_\bc$. Assume that the component of $\alpha$ on the centralizer of $\mathfrak{su}(n,1)$ in $\mathfrak{sp}(n,1)$ vanishes. Then $[\alpha]\smile[\alpha]\not=0$ in $H^2 (\Gamma,\mathfrak{sp}(n,1))$. In particular, $\alpha$ does not integrate into a nontrivial deformation of the conjugacy class of the embedding $\Gamma\ra Sp(n,1)$.
\end{Prop}

\begin{pf}
By contradiction. According to Weil's vanishing theorem, the $\mathfrak{su}(n,1)$-component of $\alpha$ vanishes, thus $\alpha$ is $\mathfrak{w}$-valued. According to Proposition \ref{pi}, $\alpha$ can be viewed as section of the homogeneous bundle over $H^2_\bc$ whose fiber is the subspace $U$ of $Hom_{\bar{\bc}}(\mathfrak{p},\mathfrak{w}')$. In particular, $\alpha$ can be viewed pointwise as a $\mathfrak{sp}(n)$-valued $(1,0)$-form. Assume that $[\alpha]\smile[\alpha]=0$, i.e., that there exists a $\Gamma$-equivariant $\mathfrak{sp}(n,1)$-valued 1-form $\eta$ on $H^n_\bc$
such that $d\eta=\alpha\wedge\alpha$. View these forms as defined on $M=\Gamma\setminus H^n_\bc$. Then, with Lemma \ref{square},
\begin{eqnarray*}
2c\parallel \alpha \parallel_{L^{2}(M)}^{2}
&=&\int_{M}c |\alpha|^{2}\omega^{n}\\
&=&\int_{M}\lambda\circ[\alpha\wedge\alpha]\wedge\omega^{n-1}\\
&=&\int_{M}\lambda\circ(d\eta)\wedge\omega^{n-1}\\
&=&\int_{M}d(\lambda\circ\eta\wedge\omega^{n-1})\\
&=&0,
\end{eqnarray*}
thus $\alpha=0$, contradiction.
\end{pf}

\section{$SU(n,1)$-lattices in $SU(2n,2)$}
\label{su2n2}

\subsection{Splitting of the Lie algebra}

Let $q=a+jb$ be a quaternion, with $a$, $b\in\bc$. The matrix of left multiplication by $q$ is $\begin{pmatrix}a&-\bar{b}\\ b&\bar{a}\end{pmatrix}$. Therefore, if $A\subset Gl(n+1,\bc)$, its image under the embeddings $Gl(n+1,\bc)\to Gl(n+1,\bh)\to Gl(2n+2,\bc)$ is $\begin{pmatrix}A&0\\ 0&\bar{A}\end{pmatrix}$. If $\mathfrak{u}(n,1)\subset\mathfrak{gl}(n+1,\bc)$ is the subspace of matrices $A$ such that $A^* Q+QA=0$ with $Q=\begin{pmatrix}I_n&0\\ 0&-1\end{pmatrix}$, then $\mathfrak{u}(n,1)$ is mapped to $\mathfrak{u}(2n,2)$ defined as the subspace of matrices $M\in\mathfrak{gl}(2n+2)$ such that $M^* Q'+Q'M=0$ with $Q'=\begin{pmatrix}Q&0\\ 0&Q\end{pmatrix}$. Thus, under the adjoint action of $U(n,1)$,
\begin{eqnarray*}
\mathfrak{u}(2n,2)=\mathfrak{u}(n,1)\oplus \mathfrak{u}(n,1)\oplus Hom_{\bc}(\bc^{n+1},\bc^{n+1}),
\end{eqnarray*}
where $U(n,1)$ acts on a square matrix $N\in Hom_{\bc}(\bc^{n+1},\bc^{n+1})$ as follows,
\begin{eqnarray*}
(A,N)\mapsto A^{-1}N\bar{A}.
\end{eqnarray*}
Putting $B=NQ$ conjugates this action to
\begin{eqnarray*}
(A,B)\mapsto A^{-1}B(A^{-1})^{\top},
\end{eqnarray*}
i.e. $Hom_{\bc}(\bc^{n+1},\bc^{n+1})=V_{\bc}^* \otimes_{\bc} V_{\bc}^*$, where $V_{\bc}$ denotes the standard representation of $U(n,1)$. Then
\begin{eqnarray*}
\mathfrak{u}(2n,2)=2\mathfrak{u}(n,1)\oplus\Lambda^2 V_{\bc}^* \oplus S^2 V_{\bc}^* ,
\end{eqnarray*}
where $S^2 V_{\bc}^* $ corresponds to matrices of the form $\begin{pmatrix}0&BQ\\ -B^* Q&0\end{pmatrix}$ in $\mathfrak{u}(2n,2)$ with $B$ symmetric. In particular,
\begin{eqnarray*}
\mathfrak{su}(2n,2)=\mathfrak{z}\oplus2\mathfrak{su}(n,1)\oplus\Lambda^2 V_{\bc}^* \oplus S^2 V_{\bc}^* ,
\end{eqnarray*}
where $\mathfrak{z}=\br$ is generated by $J'=\begin{pmatrix}iI_{n+1}&0\\ 0&-iI_{n+1}\end{pmatrix}$. 

\subsection{Cup-product}

\begin{Def}
\label{deflambdap}
On $\mathfrak{su}(2n,2)$, there is an $SU(n,1)$-invariant linear form
$\lambda'$. It is the Killing inner product with the $SU(n,1)$-invariant diagonal matrix $J'$. 
\end{Def}

Let $Z=\begin{pmatrix}0&BQ \\ -B^* Q&0\end{pmatrix}$ and $Z'=\begin{pmatrix}0&B'Q \\ -B'^* Q&0\end{pmatrix}$ belong to $S^2 V_{\bc}^* \subset \mathfrak{u}(2n,2)$. Then
\begin{eqnarray*}
[Z,Z']=\begin{pmatrix}-BQB'^* Q+B'QB^*cQ&0\\ 0&-B^* QB'Q+B'^* QBQ\end{pmatrix}
\end{eqnarray*}
belongs to $\mathfrak{u}(n,1)\oplus\mathfrak{u}(n,1)$,
\begin{eqnarray*}
\lambda'([Z,Z'])&=&\Re e(\mathrm{Trace}_{\bc}(i(-BQB'^* Q+B'QB^* Q)\\
&&-i(-B^* QB'Q+B'^* QBQ)))\\
&=&-4\Im m(\mathrm{Trace}_{\bc}(B'QB^* Q)).
\end{eqnarray*}
If $Z$, $Z'\in S^2 \mathfrak{p}^* \subset S^{2}V_{\bc}^{*}$, then $BQ=QB$, $B'Q=QB'$, thus the formula simplifies to $-4\Im m(\mathrm{Trace}_{\bc}(B'B^*))$. 

Let $\eta$ be an equivariant harmonic $\mathfrak{su}(2n,2)$-valued 1-form on $H^{n}_{\bc}$. Then $\eta=\tau+\alpha Q$ where $\tau$ is $\br$-valued and, according to Proposition \ref{pi}, $\alpha$ is $\bc$-linear and $S^2 \mathfrak{p}^*$-valued. Thus $\alpha Q=Q\alpha$ is $\mathfrak{u}(2n)$-valued. Then, for $Y$, $Y'\in\mathfrak{p}$,
\begin{eqnarray*}
\lambda'\circ([\alpha\wedge\alpha])(Y,Y')=-8\Im m(\mathrm{Trace}_{\bc}(\alpha(Y')\alpha(Y)^*))
\end{eqnarray*}
is a negative multiple of the expression $\Im m(\mathrm{Trace}_{\bc}(\alpha(Y')\alpha(Y)^*))$ already encountered in the proof of Lemma \ref{square}. Therefore $\lambda'\circ([\alpha\wedge\alpha])\wedge\omega^{n-1}$ is a negative multiple of $|\alpha|^2 \omega^n$. Thus vanishing of the cup-product cohomology class $[\eta]\smile[\eta]\in H^2 (\Gamma\setminus H^{2}_{\bc},\mathfrak{su}(2n,2))$ implies vanishing of $\alpha$, i.e. of the class of $\eta$ in $H^{1}(\Gamma,\mathfrak{su}(2n,2))/H^{1}(\Gamma,\mathfrak{z})$. 


\section{$SU(n,1)$-lattices in $SO(4n,4)$}
\label{so4n4}

\subsection{Splitting of the Lie algebra}

We have seen that the embedding $Gl(n+1,\bc)\to Gl(n+1,\bh)\to Gl(2n+2,\bc)$ lands into the block diagonal subgroup $Gl(n+1,\bc)\times Gl(n+1,\bc)\subset Gl(4n+4,\br)$. In particular, $U(n,1)$ lands into $U(n,1)\times U(n,1)\subset O(2n,2)\times O(2n,2)$. Under $O(2n,2)\times O(2n,2)$, 
\begin{eqnarray*}
\mathfrak{so}(4n,4)=\mathfrak{so}(2n,2)\oplus \mathfrak{so}(2n,2)\oplus End_{\br}(\br^{2n+2}).
\end{eqnarray*}
Since $U(n,1)$ preserves a complex structure, $\br^{2n+2}=\bc^{n+1}$, every $\br$-linear map $L$ is the sum of a $\bc$-linear and an anti-$\bc$-linear one, $L=L_{\bc}+L_{\bar{\bc}}$, and the action of $A\in U(n,1)$ on $L$ is $A^{-1}L_{\bc}\bar{A}+\bar{A}^{-1}L_{\bar{\bc}}A$. Thus $End_{\br}(\bc^{n+1})$ equals the sum of $End_{\bc}(\bc^{n+1})=V_{\bc}^* \otimes_{\bc} V_{\bc}^*$ and its conjugate $\bar{V}_{\bc}^{*}\otimes_{\bc}\bar{V}_{\bc}^{*}$.

The map $\mathfrak{so}(2n,2)\to \Lambda^2 (\br^{2n+2})^*$, $C\mapsto QC$ conjugates the adjoint $SO(2n,2)$ action with its action on real alternating 2-forms. In presence of the $U(n,1)$-invariant complex structure $J$, alternating 2-forms split into two subspaces $\Lambda_+$ and $\Lambda_-$. Indeed, $\Lambda^2 J$ is an involution. The inverse map $B\mapsto QB$ maps $\Lambda_+$ to $\mathfrak{u}(n,1)\subset\mathfrak{so}(2n,2)$. $J$ also acts as a derivation on alternating 2-forms, yielding a complex structure on $\Lambda_-$. Since
\begin{eqnarray*}
\mathfrak{so}(2n,2)\otimes\bc =\Lambda^{2,0}(\br^{2n+2})^* \oplus\Lambda^{1,1}(\br^{2n+2})^* \oplus\Lambda^{0,2}(\br^{2n+2})^* ,
\end{eqnarray*}
$\Lambda^{1,1}(\br^{2n+2})^* =\Lambda_+ \otimes\bc$, $\Lambda^{2,0}(\br^{2n+2})^* \oplus\Lambda^{0,2}(\br^{2n+2})^* =\Lambda_-\otimes\bc$, thus, as a complex representation of $U(n,1)$, the $\Lambda_-$ factor in the first diagonal block is isomorphic to $\Lambda^{2}V_{\bc}^*$, and the $\Lambda_-$ factor in the second diagonal block is isomorphic to $\Lambda^{2}\bar{V}_{\bc}^{*}$. 

We conclude that
\begin{eqnarray*}
\mathfrak{so}(4n,4)=\mathfrak{z}\oplus 2\mathfrak{su}(n,1)\oplus 2\Lambda^{2}V_{\bc}^* \oplus 2\Lambda^{2}\bar{V}_{\bc}^{*}\oplus S^2 V_{\bc}^* \oplus S^{2}\bar{V}_{\bc}^{*},
\end{eqnarray*}
where $\mathfrak{z}=\br^2$ is the sum of the centers of the 2 copies of $\mathfrak{u}(n,1)$, generated respectively by $\begin{pmatrix}J&0\\ 0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\ 0&J\end{pmatrix}$.

Choose $J=\begin{pmatrix}
0 & -I_{n+1}\\
I_{n+1} & 0
\end{pmatrix}$ as a complex structure on $\br^{2n+2}$. Then $S^{2}V_{\bc}^{*}$ (resp. $S^{2}\bar{V}_{\bc}^{*}$) corresponds to matrices of the form $\begin{pmatrix}0&BQ'\\ -B^{\top}Q'&0\end{pmatrix}$ where $B$ is symmetric and $BJ=JB$ (resp. $BJ=-JB$). Symmetric and $\bc$-linear matrices can be written $B=\begin{pmatrix}C&-D\\ D&C\end{pmatrix}$ with $C$ symmetric and $D$ skew-symmetric. Symmetric and anti-$\bc$-linear matrices can be written $B=\begin{pmatrix}C&D\\ D&-C\end{pmatrix}$ with $C$ and $D$ symmetric.

\subsection{Cup-product}

\begin{Def}
\label{deflambdapp}
On $\mathfrak{so}(4n,4)$, there is a $SU(n,1)$-invariant linear form
$\lambda'$, the Killing inner product with the $SU(n,1)$-invariant vector $J'=\begin{pmatrix}J&0\\ 0&J\end{pmatrix}$. 
\end{Def}

Let $Z=\begin{pmatrix}0&BQ'\\ -B^*Q'&0\end{pmatrix}$, $Z'=\begin{pmatrix}0&B'Q'\\ -B'^* Q'&0\end{pmatrix}\in
S^2 V_{\bc}^* \oplus S^2 \bar{V}_{\bc}^*$. View $Z$ and $Z'$ as vectors in $\mathfrak{so}(4n,4)$. Then
\begin{eqnarray*}
\lambda'([Z,Z'])=4\mathrm{Trace}(JB'Q'BQ').
\end{eqnarray*}

If $Z\in S^2 V_{\bc}^*$ and $Z'\in S^2 \bar{V}_{\bc}^*$, this vanishes. Indeed, since $BJ=JB$ and $B'J=-JB'$, $JB'Q'BQ'=-B'Q'BQ'J$,
$$\mathrm{Trace}(JB'Q'BQ')=-\mathrm{Trace}(B'Q'BQ'J)=-\mathrm{Trace}(JB'Q'BQ').$$

If $Z$ and $Z'\in S^2 \mathfrak{p}^* \subset S^2 V_{\bc}^*$, write $B=\begin{pmatrix}C&-D\\ D&C\end{pmatrix}$ and $B'=\begin{pmatrix}C'&-D'\\ D'&C'\end{pmatrix}$ where $C$, $C'$ are symmetric and $D$, $D'$ are skew-symmetric. Since $B$ and $B'$ commute with $Q'$, 
\begin{eqnarray*}
\lambda'([Z,Z'])=4\mathrm{Trace}(JB'B)=8\mathrm{Trace}(DC'+CD').
\end{eqnarray*}
The complex structure on $S^2 V_{\bc}^*$ is $B\mapsto \mathcal{J}(B)=JB$, i.e. $(C,D)\mapsto (-D,C)$. Thus
\begin{eqnarray*}
\lambda'([Z,\mathcal{J}(Z)])&=&8\mathrm{Trace}(-D^{2}+C^{2})\\
&=&8\mathrm{Trace}(D^{\top}D+C^{\top}C)=4|B|^2 =2|Z|^2 .
\end{eqnarray*}

If $Z$ and $Z'\in S^2 \bar{\mathfrak{p}}^* \subset S^2 \bar{V}_{\bc}^*$, write $B=\begin{pmatrix}C&D\\ D&-C\end{pmatrix}$ and $B'=\begin{pmatrix}C'&D'\\ D'&-C'\end{pmatrix}$ where $C$, $C'$, $D$ and $D'$ are symmetric. Then 
\begin{eqnarray*}
\lambda'([Z,Z'])=8\mathrm{Trace}(DC'-CD').
\end{eqnarray*}
The complex structure on $S^2 \bar{V}_{\bc}^*$ is $B\mapsto \mathcal{J}(B)=-JB$, i.e. $(C,D)\mapsto (D,-C)$. Thus
\begin{eqnarray*}
\lambda'([Z,\mathcal{J}(Z)])&=&8\mathrm{Trace}(-D^{2}-C^{2})\\
&=&-8\mathrm{Trace}(D^{\top}D+C^{\top}C)=-4|B|^2 =-2|Z|^2 .
\end{eqnarray*}

Let $\eta$ be an equivariant harmonic $\mathfrak{so}(4n,4)$-valued 1-form. According to \cite{Ra} and Proposition \ref{pi}, $\eta=\tau+\alpha+\alpha'$ where $\tau$ is $\mathfrak{z}$-valued, $\alpha$ is a $S^2 \mathfrak{p}^*$-valued $(1,0)$-form and $\alpha'$ a $S^2 \bar{\mathfrak{p}}^*$-valued $(0,1)$-form. The $\mathfrak{z}$-component $\tau$ disappears from bracket-wedge products. We have seen that $\lambda'\circ([\alpha\wedge\alpha'])=0$. Therefore 
$$
\lambda'\circ([\eta\wedge\eta])=\lambda'\circ([\alpha\wedge\alpha])+\lambda'\circ([\alpha'\wedge\alpha']).
$$
If $Y\in\mathfrak{p}$, since $\alpha$ has type $(1,0)$, $\alpha(iY)=\mathcal{J}(\alpha(Y))$, 
\begin{eqnarray*}
\lambda'\circ([\alpha\wedge\alpha])(Y,iY)=2|\alpha(Y)|^2 .
\end{eqnarray*}

Since $\alpha'$ has type $(0,1)$, $\alpha'(iY)=-\mathcal{J}(\alpha'(Y))$,
\begin{eqnarray*}
\lambda'\circ([\alpha'\wedge\alpha'])(Y,iY)=2|\alpha'(Y)|^2 .
\end{eqnarray*}

It follows that
\begin{eqnarray*}
\frac{\lambda\circ[(\alpha+\alpha')\wedge(\alpha+\alpha')]\wedge\omega^{n-1}}{\omega^{n}}=\frac{2}{n}(|\alpha|^2 +|\alpha'|^2).
\end{eqnarray*}
Again, if the cohomology class of $[\eta\wedge\eta]$ vanishes, then the $L^2$ norm of $\alpha$ and $\alpha'$ vanishes. This shows that the quadratic map induced by bracket-cup product on $H^{1}(\Gamma,\mathfrak{so}(4n,4))/H^{1}(\Gamma,\mathfrak{z})$ is definite.

\section{Proof of Theorem \ref{thm}}

The fact that vanishing of the cup-square is a necessary condition for integrability of an infinitesimal deformation goes back to A. Nijenhuis and R. Richardson, \cite{NR}. Here, we can even use a theorem by W. Goldman and J. Millson, \cite{GM_kahler}, completed by C. Simpson, \cite{Simpson}, providing a complete description of the character variety $Hom(\Gamma,G')/G'$ in a neighborhood of the (reductive) embedding $\phi$. It is analytically equivalent to the quotient
\begin{eqnarray*}
\mathcal{Q}_{\phi}=\{u\in H^1(\Gamma, \mathfrak{g}')\,|\,[u\smile u]=0\}/Z_{G'}(\phi(\Gamma)).
\end{eqnarray*}
The cup-square $u\mapsto [u\smile u]$ has a null-space, the subspace $H^1 (\Gamma,\mathfrak{z})$ where $\mathfrak{z}$ denotes the centralizer of $\mathfrak{su}(n,1)$ in $\mathfrak{g}'$. Therefore, it descends to the quotient
$H^1 (\Gamma,\mathfrak{g}')/H^1 (\Gamma,\mathfrak{z})$. According to Proposition \ref{cup}, the quadratic map induced on the quotient space is definite. It follows that $\mathcal{Q}_{\phi}=H^1 (\Gamma,\mathfrak{z})$. Let $Z$ denote the centralizer of $SU(n,1)$ in $G'$. In case $G'=Sp(n,1)$, $Z=U(1)$ and $SU(n,1)Z=U(n,1)$. In case $G'=SU(2n,2)$, $Z=U(1)$ and $SU(n,1)Z$ is a subgroup isomorphic to $U(n,1)$. In case $G'=SO(4n,4)$, $Z=U(1)\times U(1)$. In each case, $SU(n,1)\times Z$ is a central extension of $SU(n,1)Z$. Every homomorphism $\Gamma\to SU(n,1)Z$ which is close enough to $\phi$ lifts to $SU(n,1)\times Z$. This shows that the space $Hom(\Gamma,SU(n,1)Z)/SU(n,1)Z$ is a smooth manifold with tangent space $H^1 (\Gamma,\mathfrak{z})$. Thus, at the level of quadratic models, $Hom(\Gamma,SU(n,1)Z)/SU(n,1)Z$ and $Hom(\Gamma,G')/G'$ coincide. We conclude that homomorphisms close to $\phi$ can be conjugated into $SU(n,1)Z$.

\begin{thebibliography}{99}
{\small

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(1987), 495--520.

\bibitem{GM_kahler} William Goldman and John Millson, The deformation theory of representations of fundamental groups of compact K\" ahler manifolds. Publ. Math. I.H.\'E.S. {\bf 67}, (1988), 43--96.

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\bibitem{Klingler} Bruno Klingler, Non abelian Hodge theory and quaternionic rigidity for complex hyperbolic lattices. Manuscript (2008).

\bibitem{MM} Yozo Matsushima and Shingo Murakami, On vector bundle valued harmonic forms and automorphic forms on symmetric Riemannian
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    \end{thebibliography}

1991 {\sl{Mathematics Subject Classification.}}51M10, 57S25.

{\sl{Key words and phrases.}} Quaternionic hyperbolic space, complex hyperbolic space, harmonic form, group cohomology.

\vskip1cm

\noindent     Inkang Kim\\
     School of Mathematics\\
     KIAS, Hoegiro 87, Dongdaemun-gu\\
     Seoul, 130-722, Korea\\
     \texttt{inkang\char`\@ kias.re.kr}

\smallskip

     \noindent  Pierre Pansu\\ Laboratoire de Math{\'e}matiques
     d'Orsay\\
     UMR 8628 du CNRS\\
 Universit{\'e} Paris-Sud\\
 91405 Orsay C\'edex, France\\
  \texttt{pierre.pansu\char`\@ math.u-psud.fr}

     \end{document}

Encore, le page 8.

\lambda\circ[\alpha\wedge\alpha](Y,Y')&=&\Im m(\mathrm{Trace}_{\bc}(\delta(Y)^{*}\delta(Y')))\\
&=&\delta(Y)\cdot\delta(Y')

j\delta(Y), j\delta(Y') ils sont dans   m' < u(n)+m'=sp(n), pas dans  u(n). Alors il faut enlever "Killing inner product on u(n)".
Leur produit de killing doit etre   Re(trace \delta(Y)^{*}\delta(Y'))), n'est ce pas?
Nous avons  Im (trace \delta(Y)^{*}\delta(Y'))) dans la formule.
