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\title{Density of Zarisky density for surface groups, II }
\author{I. Kim, P. Pansu}
\date{February 8, 2012}

\newtheorem{thm}{Theorem}
\newtheorem{lem}{Lemma}
\newtheorem{prop}[lem]{Proposition}
\newtheorem{cor}[lem]{Corollary}
\newtheorem{defi}[lem]{Definition}
\newtheorem{rem}[lem]{Remark}
\newtheorem{nota}[lem]{Notation}
\newtheorem{ques}[lem]{Question}
\newtheorem{prob}[lem]{Problem}
\newtheorem{sol}[lem]{Solution}
\newtheorem{conc}[lem]{Conclusion}
\newtheorem{ex}[lem]{Example}
\newtheorem{exs}[lem]{Examples}
\newtheorem{res}[lem]{Results}
\newtheorem{plan}[lem]{Plan of lecture}
\newtheorem{scheme}[lem]{Scheme of proof}
\newtheorem{work}[lem]{Work in progress}

\def\proof{\par\medskip\noindent {\bf Proof}}
\def\qed{~q.e.d.}

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\begin{document}

\begin{frame}
\titlepage
\end{frame}


\begin{frame}

\begin{defi}
Let $G$ be the group of real points of an algebraic group. Let $\Gamma$ be a finitely generated group. Say a homomorphism $\phi:\Gamma\to G$ is 
\begin{itemize}
  \item {\em fully flexible} if Zariski dense homomorphisms are dense in a neighborhood of $\phi$ in $Hom(\Gamma,G)$.
  \item {\em partially rigid} if there are no Zariski dense homomorphisms in a neighborhood of $\phi$ in $Hom(\Gamma,G)$.
\end{itemize}
\end{defi}

\begin{prob}
Determine which homomorphisms of surface groups to reductive real algebraic groups are fully flexible, or partially rigid, or neither.
\end{prob}\ppause


The story started with W. Goldman's flexibility result for non-maximal surface groups in $SU(2,1)$ (1985). The full flexibility of surface groups in $Sl(n,\R)$ is due to O. Guichard (2007). 


\end{frame}

\section{Results}

\subsection{No third option}

\begin{frame}

\begin{thm}[No third option]
A homomorphism from a high genus surface group $\Gamma$ to a reductive real algebraic group $G$ is either fully flexible or partially rigid.
\end{thm}

\begin{center}
\includegraphics[width=3in]{cone.pdf}

\end{center}
\begin{eqnarray*}
\left\{\begin{array}{ccc}
x^2 +y^2 -tz-t^2 &=&0, \\
x^2 +y^2+z^2 -t^2&=&0.
\end{array}\right.
\end{eqnarray*}
\end{frame}

\subsection{Balanced homomorphisms}

\begin{frame}

Let $G$ be semi-simple, $\phi:\Gamma\to G$ a reductive homomorphism. Let $\mathfrak{c}$ be the center of the centralizer of $\phi(\Gamma)$. It splits the complexified Lie algebra of $G$ into root spaces $\mathfrak{g}_{\lambda}$. When $\lambda$ is pure imaginary, the sesquilinear Killing form $\bar{X}\cdot X$ on $\mathfrak{g}_{\lambda}$ is non-degenerate, giving rise to a representation $\rho_{\lambda}$ in $U(p_{\lambda},q_{\lambda})$, and a Toledo invariant $T_{\lambda}$. \ppause

\begin{defi}
\label{defc}
Among the above roots, let $\textcolor{red}{P}$ be the subset of pure imaginary roots $\lambda$ such that $\rho_{\lambda}$ is a maximal representation with $T_{\lambda}>0$ and vanishing signature ($p_{\lambda}=q_{\lambda}$). Say $\mathfrak{c}$ is \emph{balanced} with respect to $\phi$ if 0 belongs to the interior of the sum of the convex hull of the imaginary parts of elements of $\textcolor{red}{P}$ and the linear span of the real and imaginary parts of roots not in $\pm \textcolor{red}{P}$.
\end{defi}

\begin{center}
\includegraphics[width=1.5in]{balanced.pdf}
\end{center}\ppause

\begin{thm}[Fully flexible $\Leftrightarrow$ balanced]
A homomorphism $\phi$ of a high genus surface group to a semisimple real algebraic group is fully flexible if and only if $\mathfrak{c}$ is balanced with respect to $\phi$.
\end{thm}

\end{frame}

\subsection{Classification}

\begin{frame}

\begin{thm}[Hernandez, Bradlow-Garc\'{\i}a-Prada-Gothen, Burger-Iozzi-Wienhard]
Maximal representations in simple real algebraic groups whose symmetric space is Hermitian and not of tube type are partially rigid: they factor through tube type Hermitian subgroups.
\end{thm}\ppause

\begin{thm}[Classification]
Converse holds for classical simple real algebraic groups (i.e. real forms of $Sl(n,\C)$, $O(n,\C)$ or $Sp(n,\C)$). 
\end{thm}
In other words, flexibility is the rule unless $G=SU(p,q)$, $q>p$ and
$\phi(\Gamma)$ is contained in and maximal in a conjugate of $S(U(p,p)\times U(q-p))\subset SU(p,q)$, or $G=SO^* (2n)$, $n$ odd, and
$\phi(\Gamma)$ is contained in and maximal in a conjugate of $SO^* (2n-2)\times SO(2)\subset SO^* (2n)$.

\ppause

\begin{rem}
1. The restriction on genus is probably irrelevant.

2. Non constructive : deformations are not given by explicit formulae nor geometric constructions.
\end{rem}


\end{frame}

\section{Proof of balancedness criterion}
\subsection{Tools}

\begin{frame}

\begin{thm}
{\em (W. Goldman, 1985)}. If $\Gamma$ is a surface group and $\rho$ is reductive, then, in a neighborhood of the conjugacy class of $\rho$, $Hom(\Gamma,G)$ is analytically equivalent to 
\begin{eqnarray*}
\{u\in H^1 (\Gamma,\mathfrak{g}_{ad\circ\rho})\,|\,[u\smile u]=0\}.
\end{eqnarray*}
\end{thm}
Here, smiling bracket denotes cup-product : $H^1 (\Gamma,\mathfrak{g}_{ad\circ\rho})\to H^2 (\Gamma,\mathfrak{g}_{ad\circ\rho})$.\ppause

\begin{rem}
This can prove flexibility without providing explicit deformations.
\end{rem}\ppause

\bigskip

The dimension of $H^1 (\Gamma,\mathfrak{g}_{ad\circ\rho})$ can be computed via Euler characteristic and Poincar\'e duality: $H^2 (\Gamma,\mathfrak{g})=(H^0 (\Gamma,\mathfrak{g}^*))^*$. \ppause

\medskip

Cup-products can be computed thanks to

\begin{thm}
{\em (W. Meyer, 1972).} Let $(E,\Omega)$ be a flat symplectic vector bundle over $\Sigma$. The quadratic form
$Q(a)=\int_{\Sigma}\Omega(a\smile a)$ on $H^{1}(\Sigma,E)$ is
nondegenerate of signature $4c_1 (E,\Omega)$.
\end{thm}

\end{frame}

\subsection{Why high genus ?}

\begin{frame}

\begin{prop}
\begin{itemize}
  \item The dimension of $Hom(\Gamma,G)$ at points with trivial centralizers is $(|\chi(\Sigma)|+1)\dim(G)$.\ppause
  \item If the genus of $\Sigma$ is large enough, non Zariski dense homomorphisms form a subset of $Hom(\Gamma,G)$ of dimension less than $(|\chi(\Sigma)|+1)\dim(G)$.\ppause
\item Therefore it is sufficient to prove density of smooth points in neighborhoods of homomorphisms with nontrivial centralizers.\ppause
\end{itemize}
\end{prop}

\begin{prop}
Levi factors of centralizers are treated by explicitly deforming given cohomology classes $u$ such that $[u\smile u]=0$.
\end{prop}
Such deformations $v$, satisfying $[v\smile u]=0$, $[v\smile v]=0$, are obtained as zeroes of sections of bundles on the Grassmannian of symplectic subspaces of $H^1 (\Gamma,\R)$. They exist due to nonvanishing of powers the top Chern class of the universal bundle. This requires a high dimensional Grassmannian and thus high genus.


\end{frame}

\subsection{Centers of centralizers}

\begin{frame}

Let $\mathfrak{c}$ denote the center of the centralizer of $\phi(\Gamma)$. $\mathfrak{g}\otimes\C$ splits under $\mathfrak{c}$ into root spaces $\mathfrak{g}_{\lambda}$. $H^1 (\Gamma,\mathfrak{g})\otimes\C$ splits accordingly.

\begin{lem}
$[\cdot\smile \cdot]$ vanishes on each $H^1 (\Gamma,\mathfrak{g}_{\lambda})$.

$H^1 (\Gamma,\mathfrak{g}_{\lambda})$ and $H^1 (\Gamma,\mathfrak{g}_{\mu})$ are orthogonal with respect to $[\cdot\smile\cdot]$ unless $\lambda+\mu=0$.
\end{lem}\ppause

\smallskip

On each $\mathfrak{g}_{\lambda,\R}=\mathfrak{g}\cap(\mathfrak{g}_{\lambda}\oplus\mathfrak{g}_{-\lambda})$, all $ad_Z$, $Z\in\mathfrak{c}$ are proportional. Therefore the corresponding alternating forms $(X,Y)\to Z\cdot[X,Y]$ are proportional to the symplectic form $\Omega_{\lambda}=\Im m(s_{\lambda})$. On $H^1 (\Gamma,\mathfrak{g}_{\lambda,\R})$, all $Z\cdot[\cdot\smile\cdot]$ are proportional to the quadratic form $Q_{\lambda}(u,u)=\int_{\Sigma}\Omega_{\lambda}(u\smile u)$. \ppause 
Let $\rho_{\lambda}:\Gamma\to Sp(\mathfrak{g}_{\lambda,\R},\Omega_{\lambda})$ denote the composed symplectic linear representation, and $T_{\lambda}$ the corresponding Toledo invariant. Meyer's formula yields

\begin{lem}
If $\lambda\not=0$, $Q_{\lambda}$ is nondegenerate and its index is equal to $4T_{\lambda}$. Therefore
\begin{eqnarray*}
4|T_{\lambda}|\leq \dim(H^1 (\Gamma,\mathfrak{g}_{\lambda,\R}))=-\chi(\Sigma)\mathrm{dim}(\mathfrak{g}_{\lambda,\R}).
\end{eqnarray*}
In particular, $Q_{\lambda}$ is definite if and only if $\rho_{\lambda}$ is a maximal representation.
\end{lem}


\end{frame}

\subsection{End of proof}

\begin{frame}

On $H^1 (\Gamma,\mathfrak{g})$, $[\cdot\smile\cdot]=\sum_{\lambda}Q_{\lambda}\lambda$. Smooth points of $\chi(\Gamma,G)$ correspond to classes $u=\sum_{\lambda}u_{\lambda}$, $u_{\lambda}\in H^1 (\Gamma,\mathfrak{g}_{\lambda,\R})$, such that $[u\smile u]=0$ and $u_{\lambda}\not=0$ for a spanning set of $\lambda$'s. If $u$ is not a smooth point, too many $u_{\lambda}$'s vanish, perturb them to nonzero $v_{\lambda}$'s. Indefinite quadratic forms $Q_{\lambda}$ allow arbitrary moves in the direction of $\lambda$, but positive definite ones allow one to reach only points of a convex hull. If 0 belongs to it, nonzero $v_{\lambda}$'s can be found such that $[v\smile v]=0$.\ppause

\medskip

Real or $\C$-linear roots have vanishing Toledo invariant. Let $\lambda$ be pure imaginary with $Q_{\lambda}$ definite. Then Toledo invariant equals $$T_{\lambda}=\frac{|\chi(\Gamma)|}{2}(p_{\lambda}+q_{\lambda}).$$ Milnor-Wood inequality implies that $$|T_{\lambda}|\leq |\chi(\Gamma)|\mathrm{rank}(U(p_{\lambda},q_{\lambda}))=|\chi(\Gamma)|\min\{p_{\lambda},q_{\lambda}\}).$$ This implies $p_{\lambda}=q_{\lambda}$. Whence the definition of $\textcolor{red}{P}$.

\end{frame}

\section{Classical groups}
\subsection{Centers of centralizers}

\begin{frame}

When $G$ is classical, one can determine centers of centralizers of reductive subgroups $H$, and the root space decomposition, to a large extent. Indeed, $G\subset Gl(V)$ is the full automorphism group of a bilinear (resp. sesquilinear) form on the standard representation $V$. If $$V=\bigoplus I_\ell$$ is the root space decomposition of the standard representation, the decomposition of the adjoint representation $$\mathfrak{g}\subset End(V)=\bigoplus Hom(I_{\ell},I_{\ell'})$$ under $H$ follows, with roots $\ell-\ell'$.


\end{frame}

\subsection{Role of tightness}

\begin{frame}

The following theorem plays a crucial role.

\begin{thm}[Burger-Iozzi-Wienhard, 2007]
Let $S$ be a semisimple Lie group whose symmetric space is Hermitian. Let $\rho:\Gamma\to S$ be a maximal representation of a surface group. Then $\rho$ is tight. Its Zariski closure $H$ is reductive of Hermitian type. The embedding $H\hookrightarrow S$ is tight. If $S$ is of tube type, so is $H$.
\end{thm}\ppause

Example: $G=SU(p,q)$.

Let $H$ be the Zariski closure of $\phi(\Gamma)$. A root $\ell-\ell'$ belongs to $\textcolor{red}{P}$ iff the sesquilinear $\Gamma$ action on $Hom(I_{\ell'},I_{\ell})$ is maximal and positive. Since this action factors through $U(I_{\ell'})\times U(I_{\ell})$, the morphism $U(I_{\ell'})\times U(I_{\ell})\to U(Hom(I_{\ell'},I_{\ell}))$ is tight, this implies that either $U(I_{\ell'})$ or $U(I_{\ell})$ is compact, the $\Gamma$ action on the other factor is maximal. The noncompact factor must be of tube type and therefore its signature vanishes.\ppause

\medskip

Assume that $\mathfrak{c}$ is not balanced with respect to $\phi$. Then one shows that there exists one root $\ell'$ such that all $\ell-\ell'$ belong to $\textcolor{red}{P}$. This implies that $\phi$ falls into $S(U(p,p)\times U(q-p))$. Also, Toledo invariants sum up with correct signs, showing that $\phi$ is maximal in $S(U(p,p)\times U(q-p))$.

\end{frame}

\end{document}


Let us discuss the easiest case, $G=SU(p,q)$. 

\bigskip

Let $H\subset G$ be reductive. $\C^{p+q}$ splits into isotypical components $I_\pi$ under $H$. $I_{\pi'}$ and $I_\pi$ are orthogonal unless $\pi'=\bar{\pi}^*$. Whence a coarser, orthogonal, decomposition into bi-isotypical components $I_\pi +I_{\bar{\pi}^*}$. The centralizer $Z(H)$ splits correspondingly as a product. 

\bigskip

Bi-isotypical components such that $\pi=\bar{\pi}^*$ contribute a unitary group to $Z(H)$, a pure imaginary line to the center $\mathfrak{c}=ZZ(H)$ and a pure imaginary root $\ell\in I$.

\bigskip

Bi-isotypical components such that $\pi\not=\bar{\pi}^*$ contribute a general linear group to $Z(H)$, a complex line to $\mathfrak{c}$ and a $\C$-linear root $\ell\in N$.

\bigskip

Roots of the adjoint action of $\mathfrak{c}$ on $\mathfrak{su}(p,q)$ are differences $\ell-\ell'$, $\ell$, $\ell'\in I \cup N \cup (-\bar{N})$, with root space $Hom(I_{\ell'},I_{\ell})$.
