Outdated Archival Version

These pages are not updated anymore. They reflect the state of 20 August 2005. For the current production of this journal, please refer to http://www.math.psu.edu/era/.

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%\newenvironment{pf}{ Proof:}{ $\triangle$ \newline}
\title{Twisted cocycles and rigidity problems.}
\author{A. Kononenko}
\date{February 8, 1995\inRevisedForm March 8, 1995}
\address{Department of Mathematics, Pennsylvania State University,
\newline
218 McAllister Building, PA 16802.}
\email{avk@@math.psu.edu}
\communicated_by{Svetlana Katok}
\subjclass{58}



\begin{document}
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\begin{abstract}
 
We consider a class of cohomologies associated to a 
group action, outline a duality method for their calculation, and apply it to
study different questions related to the group action.  In particular,
we prove a number of results on infinitesimal and cohomological rigidity
of higher rank cocompact lattice actions on imaginary boundaries of some 
symmetric spaces (as well as results on cohomologies of some partially 
hyperbolic actions and lattice actions on a broader class of homogeneous 
spaces).  We also obtain a very transparent proof of local $C^3$ rigidity of 
projective actions of cocompact lattices in $PSL(2,\Bbb{R})$.

\end{abstract}

\maketitle

% \tableofcontents
\section{Twisted cocycles. Definitions and notations.}
\label{sec-definitions}

Measurable twisted cocycles of a group action, as described below,
 are the first cohomologies of
generalized induced representations (see \cite{kirillov}) associated
with the action. (If the  action is transitive then they are simply
the representations induced by representations of a stationary
subgroup.)
In explicit form twisted cocycles has, probably,
appeared for the first time in the work of Feldman and Moore
\cite{feldmanmoore-1} in the
context of  equivalence relations and their cohomologies.  Our main
motivation for giving rigorous definitions here is that we want to consider
continuous and smooth cohomologies, and the essentially measurable methods
and definitions of \cite{feldmanmoore-1} are not immediately applicable.

 Recall that if  a group $G$ acts on a space $M$ then the cocycle
 of this action with coefficients in a group $H$ is a
 map $\alpha : G \times X \rightarrow H $ such that
 $$ \alpha (g_1 g_2,m) = \alpha (g_1,g_2(m)) \alpha (g_2,m). $$
Two cocycles $\alpha $ and $\beta $ are called cohomologous if
there exists a function $P : M \rightarrow H $ such that
$$ \beta (g,m) = P(gm)^{-1} \alpha (g,m) P(m). $$
By $\operatorname{H}(G,H,M) $ we will denote a set of equivalence
classes of cocycles. (Most of the times we will shorten the notation to
 $\operatorname{H}(G,M) $ since it will be clear from the text which
 coefficients we work with).

Imposing additional regularity condition on the functions
$\alpha(g,\cdot )$ and $P$ ($\Ci$, measurable, continuous, etc...), in the above
definition, we define the spaces of cocycles of the corresponding regularity.


Before we define twisted cocycles, we would like to notice the
close connection between cocycles and cohomologies.
 Let $\cal{R}$ be some abelian group.
 Then we can endow  the space $F(M)$ --- space of functions on $M$ with
 values in $\cal{R}$, with a
$G$-module structure  defining
$$g\cdot f(m) = f(g^{-1}m).$$
Consider first cohomologies of $G$ with coefficients in this $G$-module.
Then we notice that
 $z:G \rightarrow F(M)$ satisfies the cohomology equation if and only if
$$\alpha_z (g,m) = z(g^{-1})(m)$$ satisfies the equation for cocycles
with coefficients in $\cal{R}.$ And, it is easy to check that
 $z \sim z_1$ if and only if $\alpha_z $ is cohomological to $\alpha_{z_1}$ as
 cocycles.  So,
 we have
 $$ \operatorname{H}^1(G,F(M)) \cong \operatorname{H}(G,\cal{R}, M).$$

  Now, assume that $\cal{R}$ is endowed with a structure of $H$-module,
  for some group $H.$
  Then, aside from $g \cdot f(m) = f(g^{-1}m)$, we have other ways to turn
  $F(M)$ into $G$-module.  Namely, in addition to the shift by $g$ we can
  also multiply (``twist'') $f\in F(M)$ by a function $x(g,m):M \rightarrow H.$
  Then, $$g \cdot f(m) = x(g,m) f(g^{-1}m)$$ defines the structure of $G$-module
  on $F(M)$ if and only if $x(g,m)$ satisfies the following equation
$$x(g_1g_2,m) =x(g_1,m)x(g_2, g_1^{-1}m),$$
  which is equivalent to $\alpha(g,m)= x(g^{-1},m)^{-1}$ being a cocycle
  with coefficients in $H.$  Denote the module  constructed using cocycle
  $\alpha $ as $F_{\alpha}(M).$

  Now, it is easy to check that  $F_{\alpha_1}(M)$ and $F_{\alpha_2}(M)$
   are isomorphic as $G$-modules if $\alpha_1$ and $\alpha_2$ are
   equivalent cocycles.  Thus, we have constructed $G$-modules
   $F_{\alpha}(M)$ for each class of cocycles $\alpha
   \in \Ho(G,H,M).$  We will call this modules twisted modules.
   The trivial module $F(M)$ corresponds to the
   class of trivial cocycles: $\alpha(g,m)= e$, where $e$ is the
   unit in $H$.


   Consider cohomologies of $G$ with coefficients in $F_{\alpha}(M).$
 That naturally leads us to the following definition:
   map $\beta(g,m) :G \times M \rightarrow \cal{R}$ is called twisted cocycle
 iff $$\{z : g \rightarrow \beta(g^{-1},m) \in F(M))\} \in
 \operatorname{H}^1(G,F_{\alpha}(M)).$$  And two twisted cocycles
 are called cohomologous iff they generate
 the same element of $\operatorname{H}^1(G,F_{\alpha}(M))$.

 Denote the space of equivalence classes of twisted cocycles constructed
 using $\alpha \in \operatorname{H}(G,H,M)$ by
 $\operatorname{H}(G,\cal{R},M,\alpha)$ (again, most of the times we will omit
 reference to $\cal{R}$ in this notation, since it should be clear from the
 text which coefficients we work with).  Then, it is well defined since by
  the definition
 $$\operatorname{H}(G,M,\alpha) \cong
 \operatorname{H}^1(G,F_{\alpha}(M))$$
 and, thus, only depends on the equivalence class of $\alpha$, since the
 cohomologies with coefficients in isomorphic modules coincide.

We define twisted cocycles of different regularity by restricting $F(M)$
to be the space of functions of corresponding regularity.

 From this definition of twisted cocycles we can easily derive the following
 equivalent definition, which is more convenient in some
 calculations.


 Fix $ \alpha \in \operatorname{H}(G,H,M)$.
 A map $\beta(g,m) :G \times M \rightarrow \cal{R}$
 will be called twisted cocycle if it satisfies
$$\beta(g_1g_2,m)= \beta(g_2,m) +\alpha(g_2,m)^{-1}\beta(g_1, g_2m), \forall
 g_1,g_2 \in G, m \in M.$$
  Two cocycles $\beta$ and $\beta_1$ will be called cohomologous (or
  equivalent) if
  and only if there exists $f \in F(M)$ such that
  $$\beta(g,m)-\beta_1(g,m) = f(m) - \alpha(g,m)^{-1}f(gm), \forall g \in G,
  m \in M.$$

 Then, the set of equivalence classes can be identified with
 $\operatorname{H}(G,M,\alpha).$



\section{Duality method.}

In this section we outline  duality method for calculating cocycles
and twisted cocycles of group actions.

 The main idea is to try to lift the action to a well understood action on a
 bigger space.  This lifting induces a map $L$ on cocycle spaces.  Then, naturally,
 the image and the kernel of this map contain most of the information about
 the cocycles of the original action.  And since, presumably, we have a good
 understanding of the lifted action, we may hope to be able to easily obtain
 significant information about its cocycles, and, thus, the image of $L$.
 Then, we just need to get a hold on the kernel of $L$.
It turns out, that the kernel subspaces enjoy
 some  very useful duality properties,
which make it possible, in some cases, to substitute a question about
  cocycles  of an action by a  question about cocycles of a
  different action, which dynamically may be completely different
  from the original one, and might turn out to be easier to study.


Suppose that $G$ acts on $X_1$ and $X_2$ in such a way that
the action on $X_2$ is a factor of the action on $X_1$.
By $\operatorname{H}^{tr}(G,X_2)$ we will denote a
set of those elements of $\operatorname{H}(G,X_2) $ that lift
to a trivial element in $\operatorname{H}(G,X_1)$.

 Also, assume that there is a certain structure on $X_1$ and $X_2$
 preserved by the factor map.
  (It can be a structure of
a topological space,  measurable space, smooth manifold etc).
 And suppose that we consider some class of cocycles with respect to this
structure (for example,  continuous cocycles,  measurable,
smooth etc). If the reference to which class of
cocycles we are working with is absent
 it means that we work with any of the above
described classes, which are applicable to the factor maps involved.
 In that case cohomologous means cohomologous in the
corresponding class.

Definitions for $\Ho^{tr}$-spaces of twisted cocycles are absolutely
analogous.

In \cite{kononenko-infinitesimal} we proved:
\begin{thm}
\label{thm-generalduality}
Let $G_1$ and $G_2 $ be two groups acting on a space $M$. Consider cocycles
with values in an arbitrary  group $H$.

If those actions commute, then:

 $$\operatorname{H}^{tr}(G_1, M/G_2) \cong \operatorname{H}^{tr}(G_2,
 G_1 \backslash M). $$


 To be more precise, there is a canonically defined  map
 $$K(G_1, G_2):\operatorname{H}^{tr}(G_1, M/G_2) \rightarrow
 \operatorname{H}^{tr}(G_2,G_1 \backslash M) $$
  such that
$K(G_2, G_1)\circ K(G_1, G_2)=Id$ on $\operatorname{H}^{tr}(G_1, M/G_2).$


 Where $M/G_2 $ and $G_1 \backslash M$
 are the factor spaces on which the other group
 acts in the obvious way, and $\Ho^{tr}$-spaces defined with respect to the
 lifts of actions to the whole $M.$
\end{thm}


 And the duality for twisted cocycles is ``built over'' the duality for
 non-twisted ones:



\begin{thm}
\label{thm-twistedduality}
Let $G_1$ and $G_2 $ be two groups acting on a space $M$.
Let $H$ be any group, and
 $\cal{R}$ be any $H$-module.  Let $\alpha$ be a
cocycle with values in $H$.  Assume that
$\alpha \in  \operatorname{H}^{tr}(G_1,H, M/G_2).$


If the $G_1$ and $G_2$ actions commute, then:

 $$\operatorname{H}^{tr}(G_1, M/G_2, \alpha) \cong
 \operatorname{H}^{tr}(G_2,G_1 \backslash  M, K(G_1, G_2)(\alpha) ). $$


 To be more precise, there is a canonically defined group  isomorphism
 (obviously, the  spaces of twisted cocycles are naturally
 equipped with abelian group structure induced from $\cal{R}$ ):

 $$K_1(G_1, G_2):\operatorname{H}^{tr}(G_1, M/G_2, \alpha) \rightarrow
 \operatorname{H}^{tr}(G_2,G_1 \backslash M, K(G_1, G_2)(\alpha) )$$
such that
$K_1(G_2, G_1)\circ K_1(G_1, G_2)=Id$ on $\operatorname{H}^{tr}(G_1, M/G_2,
 \alpha).$


\end{thm}

Remark: Whenever it is not likely to cause a misunderstanding we will denote
the duality maps simply by $K,$ instead of $K(G_1,G_2)$ and $K_1(G_1,G_2).$

 We will mostly use Theorems~\ref{thm-generalduality} and
 ~\ref{thm-twistedduality} in case of two closed subgroups $P$ and $Q$ acting
 on Lie group $G$, $P$ from the left, $Q$ from the right.  Then for
 smooth, continuous and H\"older cocycles, for
 $\alpha \in \Ho^{tr}(P, G/Q)$ we have:
 $$ \Ho^{tr}(P,G/Q, \alpha) \cong \Ho^{tr} (Q, P \backslash G, K(\alpha)).$$


\section{Cohomological rigidity}
\label{sec-cohomological}

In this section we work with cocycles with coefficients in $\Bbb{R}^l$,
$l \in \Bbb{N}.$
Let $G$ be a
 semi-simple connected Lie group  with finite center and
    without compact factors. Let $\algG $ be its
  Lie algebra,  $A$ --- connected component of the maximal split Cartan
  subgroup,  $\algA $ the corresponding commutative subalgebra,  $N$ ---
closed  subgroup containing $A$. Let $\Gamma $
  be an irreducible cocompact lattice in $G$.

 In \cite{kononenko-cocyclesrigidity} we proved:
 \begin{thm}
  \label{thm-gamma}

   Let $G$ be a Lie group,  $\Gamma $ a discrete subgroup,
   acting on $G$
        from the left. Then,  for $\Ci$, continuous or H\"older
        cocycles we have:
        $\operatorname{H}( \Gamma, G)=0.$

\end{thm}
 Then, for $\alpha \in \Ho (\Gamma, N \backslash G),$ we have:
 $$\Ho(\Gamma, N \backslash G, \alpha)=
 \Ho^{tr} (\Gamma, N \backslash G, \alpha) \cong
 \Ho^{tr} (N, G/\Gamma, K(\alpha)).$$

So, our duality method allows us to transform the question
about cocycles of lattice action (which is ``not nice in any way'' --- no
obvious hyperbolicity properties, no invariant measures, we do not
have a very good understanding of the structure of $\Gamma$) into
a question about cocycles of $N$-action (which is ``much better action'' ---
partially hyperbolic,
has finite smooth invariant measure, $N$ is relatively ``simple'' group).
\subsection{Smooth cocycles.}
For partially hyperbolic actions we develop special analytic
technique (\cite{kononenko-infinitesimal})
similar to the technique used by Katok and Spatzier in their study
of the cocycle rigidity of standard
 abelian actions (\cite{katokspatzier-mathreslet}).

The following  result is proved in
\cite{kononenko-cocyclesrigidity}:
 \begin{thm}
 \label{thm-const}
 Let $G$ be   of
  $\Bbb{R}$-rank $n\geq 2$.


   (*) Assume that $\cal G $ has no factors isomorphic to
$\frak {so} (m, 1) $ or $\frak {su} (m, 1) $.

Then  all $\Ci $ cocycles of the action of $N$ on $G/ \Gamma $ are
$\Ci $ cohomologous to constant cocycles.

\end{thm}

Remark: The (*) condition is needed due to a non-uniformity in
certain estimates on the decay of correlation coefficients for unitary
representations of  groups $\frak {so} (m, 1) $ and $\frak {su} (m, 1).$
It is quite likely that for the particular representations considered in
the proofs of our results sufficient estimates still can be
obtained and then
    the (*) condition will disappear from all our results (for details see
    \cite{katokspatzier-mathreslet} or \cite{kononenko-infinitesimal}).


Due to Theorem~\ref{thm-const},
studying  twisted cocycles of $N$-action we may restrict our
attention to the case of constant twisting.  Then the following
result proved in   \cite{kononenko-infinitesimal}
 gives an ``almost  complete'' description of all twisted cocycles
of $N$-action:
  \begin{thm}
  \label{thm-nsmooth}
 Let $G$ be   of
  $\Bbb{R}$-rank $n\geq 3$.

   (*) Assume that $\cal G $ has no factors isomorphic to
$\frak {so} (m, 1) $ or $\frak {su} (m, 1) $.


 Let $\pi$ be a non-zero  linear form on $\algA$,
  extendable to a representation of $N$ and not proportional to
  a root of rank one factor of $\cal{G}.$
   Then every element of $\HN$ is $\Ci$ cohomologous to a constant
   twisted cocycle.

   Moreover, if $\pi$ is such that it is not proportional
  to any root of  $\cal{G}$,
then $$\HN=0.$$




  \end{thm}


Since we now have a good understanding of cocycles of $N$-action,
using duality, we
easily get the following rigidity results for $\Gamma$-action
(\cite{kononenko-cocyclesrigidity} and \cite{kononenko-infinitesimal}):


\begin{thm}
\label{thm-rigidity}

 Let $G$ be  of
  $\Bbb{R}$-rank $n\geq 2$.


   (*) Assume that $\cal G $ has no factors isomorphic to
$\frak {so} (m, 1) $ or $\frak {su} (m, 1) $.

Then  every $\Ci $ cocycle of the $\Gamma $
action on $N \backslash G$ uniquely extends to a cocycle of the $G$ action.


\end{thm}


\begin{thm}
  \label{thm-twistsmooth}
 Let $G$ be of
  $\Bbb{R}$-rank $n\geq 3$.

      (*) Assume that $\cal G $ has no factors isomorphic to
$\frak {so} (m, 1) $ or $\frak {su} (m, 1) $.

Let $\alpha$ be any element of $\HGaNG$. Then for the $\Ci$ cocycles we have


  1) if $\alpha$'s dual is not proportional to a root
 of $\cal{G}$, then
                              $$\HGaNGa =0.$$

  2) if $\alpha$'s dual is  proportional to a root
 of $\cal{G}$, but not to a root of rank one factor of $\cal{G}$,
 then $\HGaNGa$ may be non-trivial, but
 its every element is extendable to a cocycle of the whole $G$-action twisted
 by the extension of $\alpha$ guaranteed by Theorem~\ref{thm-rigidity}.

 \end{thm}

Notice that in the case when $\cal{G}$ does not have rank one factors
Theorems~\ref{thm-rigidity} and ~\ref{thm-twistsmooth} give a complete
description of all twisted and non-twisted cocycles of the $\Gamma$ action
on $N \backslash G.$
\subsection{Continuous cocycles.}
The usual non-twisted cocycles,  are known to show some
rigidity on measurable (\cite{zimmer}), $\Ci$ and H\"older
(\cite{katokspatzier}, \cite{katokspatzier-mathreslet})
levels and to be highly unstable on continuous level.  In sharp
contrast to it, the non-trivially twisted cocycles show strong
regularity already on the continuous level.  This phenomena proves to
be extremely useful in our studies of differential rigidity
(\cite{kononenkoyue}).

The following results are proved in \cite{kononenko-infinitesimal}:
\begin{thm}
\label{thm-ncont}

 Let $G$ be of $\Bbb{R}$-rank $n\geq 2$.
   Then, for continuous cocycles, there is an open set $U$ in
   $\operatorname{H}(N,G/\Gamma)$ such that
 for any  $\alpha \in U,$
 $$\operatorname{H}(N,G/\Gamma,\alpha)=0.$$


    Moreover, $U$ contains all constant cocycles with  the
    exception, may be, of some constant cocycles, which are
    proportional to the roots of $\cal{G}$.

   Also, if $\alpha$ is constant and proportional to a root of $\cal{G}$,
   $\operatorname{H}(N,G/\Gamma, \alpha)$ is not necessarily trivial, but
   may consist of only constant cocycles.
\end{thm}


 \begin{thm}
 \label{thm-twistcont}

 Let $G$ be  of
  $\Bbb{R}$-rank $n\geq 2$.
Then there exists an open set $V \in \HGaNG$, such that for any
 $\alpha \in V$, $\HGaNGa =0.$

 Moreover
  $V$ contains all elements of $\HGaNG$,with constant duals, with the
 exception, may be, of those whose dual is proportional to a
 root of $\cal{G}.$

 \end{thm}

 In particular, notice that $V$ contains ``almost all'' $\Ci$ cocycles.

\subsection{H\"older cocycles.}
For H\"older cocycles one can obtain results analogous to  our $\Ci$
results, in the non-twisted case, and our continuous results in the twisted
one.  However, for the non-twisted H\"older results we have to impose
additional condition that $G$ is split.



\section{Differential rigidity.}
\label{sec-differential}

It turns out that in the study of differentiable perturbations
 of the action often one can single out the cohomology
 class of twisted cocycles which is responsible for the existence or
 non-existence of smooth conjugacy.  That is, the perturbed action is
 smoothly conjugate to the original action (or one of its
``canonical'' perturbations)
 iff the corresponding cocycle is cohomologically trivial. This
 takes place for projective actions on $S^1$, actions by rotations,
 and some other classes of actions.


 In \cite{kononenkoyue} we show that for $C^3$ actions on $S^1$ the
 cohomology class of the Schwarzian twisted
 cocycle  is the obstacle to the existence of $C^3$ conjugacy to a projective
 action.  This, together with our duality technique
  yields a very transparent proof of $C^3$
 local differential rigidity for
 projective actions of cocompact lattices in $PSL(2, \Bbb{R})$:
 \begin{thm}
 \label{thm-main}

 Let $\Gamma \subset PSL(2,\Bbb{R})$ be a cocompact lattice.  Let
 $\tilde{\Gamma}$ be a representation of $\Gamma$ into $Diff^3(S^1)$
 which is sufficiently $C^1$ close to $\Gamma$ among a set of finitely many
 generators.  Then there exists another cocompact lattice $\Gamma_1
 \subset PSL(2,\Bbb{R})$ and a $C^3$ diffeomorphism $h \in Diff^3(S^1)$
 such that $\tilde{\Gamma}=h^{-1} \Gamma_1 h.$
 \end{thm}


 E.Ghys, in \cite{ghys},  proved the $C^{\infty}$ version of this
result.
And we learned that recently Ghys (\cite{ghys-c3}) has extended his result to a
 global one and proved the $C^3$ version as well.
 The main motivation for giving our proof is to exhibit the
 intricate geometric structure associated to these actions by a simple cocycle.


{\bf Outline of a proof of Theorem~\ref{thm-main}.}
 Recall that for a diffeomorphism $f \in Diff^3(S^1)$, its Schwarzian
  derivative $S(f)$ is defined by
$$ S(f)(x)=\frac{f'''(x)}{f'(x)} -\frac{3}{2} \left(\frac{f''(x)}{f'(x)}
\right)^2=\left(\frac{f''(x)}{f'(x)}\right)'-\frac{1}{2}
\left(\frac{f''(x)}{f'(x)}\right)^2.$$
 Its most important properties are:
  \begin{enumerate}  \item It is projectively
  invariant; in fact, $S(f)=S(g)$ if and only if
  $$g(x)=\frac{af(x)+b}{cf(x)+d}$$ for some numbers $a,b,c,d$ such that
  $ad-bc=1.$

  \item $S(f)=0$ if and only if $f \in SL(2,\Bbb{R}).$

  \item Cocycle property: $S(f\circ g)(x)=S(f)(g(x))[g'(x)]^2+S(g)(x).$

  \end{enumerate}

  We observe the following fact.

  \begin{prop}
  \label{prop-h}
  A diffeomorphism $f \in Diff^3(S^1)$ is conjugate to a projective
  transformation $g \in SL(2,\Bbb{R})$ via a $C^3$ diffeomorphism $h$
  if and only if its Schwarzian $S(f)$ is cohomologous to $0$, i.e.
  there exists a continuous function $\sigma(x)$ on $S^1$ such that
$$     S(f)(x)=\sigma(f(x))[f'(x)]^2 - \sigma(x).$$

\end{prop}

  Thus, if group $G$ acts on $S^1$ then the question whether the action is
  $C^3$ conjugate to some projective action is equivalent to the question
  whether its Schwarzian twisted cocycle
  $S \in \Ho(G, \Bbb{R}, S^1, D^{-2})$, where $D$ is derivative cocycle, is
  cohomologically trivial as continuous twisted cocycle.

  Now, let $\Gamma \subset PSL(2,\Bbb{R})$ be a cocompact lattice.
  And let $\tilde{\Gamma} \subset Diff^3(S^1).$  If $\tilde{\Gamma}$
  is sufficiently $C^1$ close to $\Gamma$ among a set of finitely many
  generators, then by \cite{sullivan} there exists a homeomorphism
  $h \in Homeo(S^1)$ such that $\tilde{\Gamma}=h^{-1}\Gamma h.$  Via the
  conjugacy $h$ the Schwarzian cocycle $S$ of $\tilde{\Gamma}$ can be
  pulled back to a twisted cocycle $T$ of the $\Gamma$ action:
  $\Gamma \times S^1 \rightarrow \Bbb{R}$,

  $$T(\gamma,x)=S(h^{-1}\gamma h, h^{-1}(x)),$$
  which satisfies
  $$T(\gamma_1 \gamma_2,x)=T(\gamma_2,x)+T(\gamma_1, \gamma_2(x))
  [\tilde{\gamma'_2}(h^{-1}(x))]^2=$$
  $$=T(\gamma_2,x)+T(\gamma_1, \gamma_2(x))
  [\gamma'_2(x)]^2 \delta(\gamma_2,x),$$
  where $\delta (\gamma,x) = [\tilde{\gamma'}(h^{-1}(x))/\gamma'(x)]^2.$
   Moreover, if the perturbation
  $\tilde{\Gamma}$ is $C^1$ close to $\Gamma$
  among its finitely many generators, then  $\delta$
  is close to $1$ among those generators.
To finish the proof of Theorem~\ref{thm-main} all we have to do now
  is to prove that $T$ is a continuous coboundary.

Let $G=PSL(2,\Bbb{R})$, $P$ be the subgroup consisting of upper triangular
matrices, and $\Gamma$, as before, a cocompact Fuchsian subgroup.  Then the
action by projective transformations is isomorphic to the right action of
$\Gamma$ on $P\backslash G.$  Let $\Delta_1=(\gamma'(x))^{-2}$ be the inverse of
the square
of the derivative cocycle for the $\Gamma$ action.
  Let, $\Delta=\delta(\gamma,x)^{-1}\Delta_1$ ---
  be its pullback.
Then $T$  is an element of
$\operatorname{H}(\Gamma,P\backslash G, \Delta).$
Then we have:
$$\operatorname{H}(\Gamma,P\backslash G, \Delta)=
\operatorname{H}^{tr}(\Gamma,P\backslash G, \Delta) \cong
\operatorname{H}^{tr}(P,G/\Gamma,K( \Delta)).$$

 Direct calculation shows that $K(\Delta_1)$ is constant cocycle equal
 to the forth power of the bigger eigenvalue of $p \in P.$

 So, now instead of dealing with cocycles of $\Gamma$-action twisted
 by cocycle close to inverse of square of derivative cocycle
 (``complicated'' group and ``complicated'' twisting) we have a
 much simpler object to work with --- cocycles of $P$ action
 twisted by a cocycle close to a known constant cocycle
 (``simple'' group and ``simple'' twisting).
And, indeed, we fairly easily
 prove that if $K(\Delta) $ is a continuous cocycle  close
 enough to $K(\Delta_1)$  then
 $\operatorname{H}(P,G/\Gamma, K(\Delta))=0$, and thus
 $\operatorname{H}(\Gamma, P \backslash G, \Delta)=0.$

\section{Infinitesimal rigidity.}

Recall that an action of a group $G$ on a differentiable manifold $X$
is called infinitesimally rigid if $\Ho^1 (G, Vect(X))=0$, where
$Vect(X)$ is the space of $\Ci$ vector fields on $X$ (which is
endowed with a structure of $G$-module in obvious way).  The definition
is, of course, motivated by Weil's local rigidity theorems
(\cite{weil-cohomology}) and the fact that $Vect(X)$ is the Lie
algebra of the infinite dimensional Lie group of diffeomorphisms
of $X$ --- $Diff(X),$ and the natural action of $G$ on $X$ is the
composition of the action of $G$ in $Diff(X)$ with the adjoint
representation of $Diff(X)$ in its Lie algebra.

For different results on infinitesimal rigidity  of lattice actions
see \cite{zimmer-infinitesimal} and \cite{lewis-infinitesimal}.

In \cite{kononenko-infinitesimal} we prove the following:


 \begin{thm}
 \label{thm-infinitesimalrigidity}
 Let  $G$ be a semi-simple connected Lie group of
  $\Bbb{R}$-rank $n\geq 3$, with finite center, without compact factors
  and factors of rank one.   Assume that $G$ is split.


  Let $M$ be the symmetric space corresponding to $G$. Let
  $\partial M$ be its maximal boundary.
  Let $\Gamma $ be an irreducible cocompact lattice in $G$.

  Then, the natural action of $\Gamma$ on $\partial M$ is infinitesimally
  rigid.

  \end{thm}


  {\bf Outline of the proof of Theorem~\ref{thm-infinitesimalrigidity}.}
  It is well known that $\partial M$ can be identified with $P \backslash G$
  (\cite{mostow}), and the action of $\Gamma $ on $\partial M$ becomes an
  action by right multiplications on $P \backslash G,$ where $P$ is minimal
  parabolic subgroup in $G.$

  From the Iwasawa decomposition we see that $P \backslash G$ is diffeomorphic
  to $K$ --- the maximal compact subgroup of $G$.  In particular,
  $P \backslash G$ is paralelizable.  Thus, the $Vect( \partial M)$ can be
  identified with $F(\partial M)$ --- smooth
  functions on $P \backslash G$ with values in
  $\Bbb{R}^k$, where $k= dim K.$

  Derivative of this action then can be written as a cocycle
  $D(\gamma, m)$ with values in $GL(m,\Bbb{R})$:  $D(\gamma, m)$ is
  equal to the derivative of the transformation
  $m \rightarrow m \gamma^{-1}$, evaluated at point $m \in P \backslash G$
  in the same coordinate system we
  used to trivialize the tangent bundle over $P \backslash G.$

  The action of $\Gamma $ on $Vect(\partial M)$ translates then into the
  following action  on $F(\partial M)$:
$$ \gamma \cdot f(m) = D(\gamma^{-1},m \gamma^{-1}) f(m \gamma^{-1})=
  D(\gamma, m)^{-1}f(m \gamma^{-1}),
  f(m) \in F(\partial M).$$

  So, we see that this action gives $F(\partial M)$ the structure of
  twisted $\Gamma$-module $F_D(\partial M),$ and that the infinitesimal rigidity of
  $\Gamma$ action on $\partial M$ is equivalent to $\Gamma $ having
  zero first cohomologies with coefficients in $F_D(\partial M).$ Or, passing,
  as usually, from
  cohomologies to twisted cocycles wee see that the infinitesimal rigidity is
  equivalent to $\Ho (\Gamma, \Bbb{R}^k, P \backslash G, D)=0.$


  First we  prove that
  $D \in \Ho^{tr} (\Gamma, GL(k, \Bbb{R}), P \backslash G),$
  and, thus, by the Theorem~\ref{thm-gamma}
$$\Ho (\Gamma,  G, \overline{D})=\Ho (\Gamma,  G)=0,$$
where $\overline{D}$ is lift of $D$ to $G.$
So, we have:
$$\Ho(\Gamma, P \backslash G, D) =
\Ho^{tr}(\Gamma, P \backslash G, D) \cong
\Ho^{tr}(P, G/\Gamma, K(D)).$$

  It turns out that $K(D)$ is a constant cocycle, i.e.
 a representation of $P.$  So, now, like in the proof of Theorem~\ref{thm-main},
 we have to deal with reasonably uncomplicated group $P$ and constant twisting.
 And, indeed, using smooth results on cohomological rigidity from
 Section~\ref{sec-cohomological} we manage to prove that every element of
 $\Ho(P, G/\Gamma, K(D))$ with the additional property of being smooth
 in group variable $p \in P$ (which is more then required by the definition
 of twisted cocycles) is a smooth coboundary.

 Now, to finish the proof all we have to do is to notice that, obviously,
 every element of $\Ho^{tr}$-spaces possesses this additional property.

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