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\title{Flexibility of surface groups in semi-simple Lie groups}
\author{I. Kim, P. Pansu}
\date{\today}

\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}
\frametitle{Plan}
\section{Plan}

\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 {\em flexible} if Zariski dense homomorphisms are dense in a neighborhood of $\phi$ in $Hom(\Gamma,G)$.
\end{defi}

\begin{prob}
Determine which homomorphisms of surface groups to almost simple Lie groups are flexible.
\end{prob}\ppause

\begin{plan}
We survey (global) rigidity results concerning surface groups in semisimple Lie groups.

\medskip

We complement them with new flexibility results.
\end{plan}

\end{frame}

\begin{frame}
\frametitle{Toledo invariants}
\section{Toledo invariants}
Let $X$ be a Hermitian symmetric space, with K\"ahler form $\Omega$ (the metric is normalized so that the minimal sectional curvature equals $-1$). Let $\Sigma$ be a closed surface of negative Euler characteristic, let $\Gamma=\pi_1 (\Sigma)$ act isometricly on $X$. Pick a smooth equivariant map $\tilde{f}:\tilde{\Sigma}\to X$.

\begin{defi}
Define the Toledo invariant of the action $\rho:\Gamma\to Isom(X)$ by
\begin{eqnarray*}
T_{\rho}=\frac{1}{2\pi}\int_{\Sigma}\tilde{f}^{*}\Omega.
\end{eqnarray*}
\end{defi}

Then 
\begin{enumerate}
  \item $T_{\rho}$ depends continuously on $\rho$.
  \item There exists $\ell_X \in\Q$ such that $T_{\rho}\in\ell_X \Z$.
  \item $|T_{\rho}|\leq|\chi(\Sigma)||\mathrm{rank}(X)|$.
\end{enumerate}


\end{frame}

\begin{frame}
\frametitle{Rigidity in rank 1}
\section{Rigidity in rank 1}
\begin{ex}
When $X=H^1_{\C}$ is a disk, inequality $|T_{\rho}|\leq|\chi(\Sigma)|$ is due to J. Milnor (1958). Furthermore $\ell_X =1$, $T$ takes all integer values between $-|\chi(\Sigma)|$ and $|\chi(\Sigma)|$. 
\end{ex}\ppause

\begin{thm}
{\em (W. Goldman, 1980).} Let $X=H^1_{\C}$. The level sets of $T$ coincide with the connected components of the character variety $\chi(\Gamma,PU(1,1))$. Furthermore $|T_{\rho}|=|\chi(\Sigma)|$ if and only if $\rho(\Gamma)$ is discrete and cocompact in $PU(1,1)=Isom(H^1_{\C})$.
\end{thm}

Note that all components of $\chi(\Gamma,PU(1,1))$ have the same dimension $3|\chi(\Sigma)|$.\ppause

\begin{thm}
{\em (D. Toledo, 1979, 1989).} Let $X=H^n_{\C}$ have rank 1. Then $|T_{\rho}|\leq|\chi(\Sigma)|$. Furthermore, $|T_{\rho}|=|\chi(\Sigma)|$ if and only if $\rho(\Gamma)$ stabilizes a complex geodesic $H^1_{\C}$ in $X$ and acts cocompactly on it.
\end{thm}

It follows that, for $n\geq 2$, different components of $\chi(\Gamma,PU(n,1))$ can have different dimensions.

\end{frame}


\begin{frame}
\frametitle{Higher rank}
\section{Higher rank}

\begin{defi}
Actions $\rho$ such that $|T_{\rho}|=|\chi(\Sigma)|\mathrm{rank}(X)$ are called {\em maximal representations}.
\end{defi}

\begin{exs}
Pick cocompact actions $\rho_1 ,\ldots,\rho_r$ of $\Gamma$ on $H^1_{\C}$. Then the direct sum representation on the polydisk $(H^1_{\C})^r$ is maximal. When the polydisk is embedded in a larger symmetric space of rank $r$, it remains maximal. It follows that all Hermitian symmetric spaces admit maximal representations.
\end{exs}\ppause

\begin{prop}
{\em (Toledo, 1987).} In case $X$ is Siegel's upper half space (i.e. $Isom(X)=Sp(n,\R)$), such actions can be deformed to become Zariski dense.
\end{prop}\ppause

But this may fail for other Hermitian symmetric spaces.

\begin{thm}
{\em (L. Hern\`andez Lamoneda, 1991, S. Bradlow, O. Garc\'{\i}a-Prada, P. Gothen, 2003).} Maximal reductive representations of $\Gamma$ to $PU(p,q)$, $p\leq q$, can be conjugated into $P(U(p,p)\times U(q-p))$.
\end{thm}
\end{frame}

\begin{frame}
\frametitle{Tube type}
\section{Tube type}

Gromov, Thurston, Domic-Toledo : Surfaces admit ideal triangulations. $T_{\rho}$ is the sum of Kähler areas of ideal triangles in an ideal triangulation.\ppause

Observation : in $H_{\C}^{n}$, the Kähler area of ideal triangles takes a full interval of values for $n>1$, whereas only one value for $n=1$. In a polydisc, finitely many values.\ppause

\begin{thm}
{\em (J.-L. Clerc, B. \O rsted 2003).} The Kähler area of an ideal triangle either takes a full interval of values (non tube type case) or finitely many values (tube type case), in both cases bounded by $\frac{1}{2}\mathrm{rank}(X)$. Equality holds iff the triangle is contained in a subsymmetric space of tube type.
\end{thm}

\begin{defi}
Say a Hermitian symmetric space is of {\em tube type} if it can be realized as a domain in $\C^n$ of the form $\R^n +iC$ where $C\subset\R^n$ is a proper open cone.
\end{defi}\ppause

\begin{exs}
Siegel's upper half spaces and Grassmannians with isometry groups $PO(2,q)$ are of tube type. The Grassmannian $\mathcal{D}_{p,q}$, $p\leq q$, with isometry group $PU(p,q)$ is of tube type iff $p=q$. Products of tube type spaces are of tube type, so polydisks are of tube type.
\end{exs}

\end{frame}


\begin{frame}
\frametitle{Tube type rigidity}
\section{Tube type rigidity}
\begin{thm}
{\em (Burger, Iozzi, Wienhard, 2003).} Let $\Gamma$ be a closed surface group and $X$ a Hermitian symmetric space. Every maximal representation $\Gamma\to Isom(X)$ stabilizes a tube type subsymmetric space $Y$ and is Zariski dense in $Isom(Y)$.
\end{thm}

In particular, maximal representations of surface groups in non tube type Hermitian symmetric spaces are globally rigid. \ppause

\begin{exs}
In case $X$ is the $n$-ball $\mathcal{D}_{1,n}$ (resp. $\mathcal{D}_{p,q}$), one recovers Toledo's (resp. Barlow et al.) results.
\end{exs}

\end{frame}

\begin{frame}
\frametitle{Flexibility problem}

\begin{prob}
Characterize flexible actions of closed surface groups on (non necessarily Hermitian) symmetric spaces.
\end{prob}\ppause

\bigskip

All previously known examples of flexibility can be obtained by bending representations.

\begin{exs}
{\em (Burger, Iozzi, Wienhard).} In a tube type Hermitian symmetric space,  surface groups stabilizing a maximal polydisk and acting diagonally on it (Toledo's 1987 examples) are flexible.
\end{exs}

\bigskip

We shall give a somewhat computable flexibility criterion. When flexibility is proven, deformations do not look like bending.

\end{frame}


\begin{frame}
\frametitle{Flexibility theorem}
\section{Flexibility theorem}

Let $G$ be a semisimple Lie group. Let $\Gamma$ be the fundamental group of a closed surface of high enough genus. Let $\phi:\Gamma\to G$ be a homomorphism with reductive Zariski closure. Let $\mathfrak{c}$ be the center of the centralizer of $\phi(\Gamma)$. It splits the Lie algebra of $G$ into real root spaces $\mathfrak{g}_{\lambda,\R}$ which carry natural symplectic structures. \ppause
In particular, to each pure imaginary root $\lambda$, there corresponds a symplectic representation $\rho_{\lambda}$ 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 $T_{\lambda}>0$. 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=2in]{balanced.pdf}
\end{center}\ppause

\begin{thm}
$\phi$ is flexible if and only if $\mathfrak{c}$ is balanced with respect to $\phi$.
\end{thm}

\end{frame}


\begin{frame}
\frametitle{Consequences}
\section{Consequences}

\begin{cor}
If $\phi$ is not flexible, then one of the noncompact irreducible factors of the symmetric space of a Levi factor of the Zariski closure of $\phi(\Gamma)$ is Hermitian of tube type, and the action of $\Gamma$ on this factor is a maximal representation. 
\end{cor}
In case $G$ is of Hermitian type, this is close to a converse to the result by Burger et al. (up to the restriction on genus).\ppause

\begin{cor}
If $G$ has real rank one, then $\phi$ is flexible, unless $G$ is $PU(m,1)$ and $\phi(\Gamma)$ is discrete, cocompact in a conjugate of $P(U(1,1)\times U(m-1))\subset PU(m,1)$.
\end{cor}
This is an exact converse to Toledo's theorem (up to the restriction on genus).\ppause

\begin{work}
Classify all non flexible homomorphisms. Based on classification of root systems. 
\end{work}
Easy cases : complex Lie groups, $PSL(n,\R)$, $PSL(n,\H)$, in which cases $\textcolor{red}{P}=\emptyset$.\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}

\begin{frame}
\frametitle{Tools}
\section{Tools}

\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 $\rho$, $Hom(\Gamma,G)/G$ is analytically equivalent to 
\begin{eqnarray*}
\{u\in H^1 (\Gamma,\mathfrak{g}_{ad\circ\rho})\,|\,[u\smile u]=0\}/Z_{G}(\rho(\Gamma)).
\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}

\begin{frame}
\frametitle{Levi factors of centralizers}
\section{Levi factors of centralizers}

\begin{nota}
$\chi(\Gamma,G)=Hom(\Gamma,G)/G$.
\end{nota}

\begin{prop}
\begin{itemize}
  \item The dimension of $\chi(\Gamma,G)$ at points with trivial centralizers is $|\chi(\Sigma)|\dim(G)$.\ppause
  \item If the genus of $\Sigma$ is large enough, non Zariski dense homomorphisms form a subset of $\chi(\Gamma,G)$ of dimension less than $|\chi(\Sigma)|\dim(G)$.\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 complex Grassmannian. They exist due to nonvanishing of powers the top Chern class of the universal bundle (requires high genus).


\end{frame}

\begin{frame}
\frametitle{Centers of centralizers}
\section{Centers of centralizers}
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 a single symplectic form $\Omega_{\lambda}$. 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}


\begin{frame}
\frametitle{End of proof}
\section{End of proof}

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

\bigskip

The fact that the Zariski closure of $\phi(\Gamma)$ has to be Hermitian of tube type follows from the following theorem. 

\begin{thm}
{\em (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

\bigskip

Calculation of examples (e.g., rank one) relies on further obstructions for $\rho_{\lambda}$ to be definite : $\rho_{\lambda}$ is not only symplectic, it must be unitary with respect to a Hermitian form of vanishing signature.
\end{frame}

\end{document}


