\documentstyle[12pt]{article}
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\newtheorem{cor}{Corollary}[section]
\newtheorem{thm}{Theorem}[section]
\newtheorem{lmm}{Lemma}[section]
\newenvironment{pf}{ Proof:}{ $\triangle$ \newline}
\title{ Smooth Cocycles Rigidity for Lattice Actions.}
\author{A. Kononenko}
\date{January, 1995 }
%\address{Pennsylvania State University}
%\email{avk@@math.psu.edu}
\begin{document}
\newfont{\Blackbb}{msbm10 scaled \magstep 0}
\newfont{\frakfurt}{eufm10 scaled \magstep 0}
\maketitle
\begin{abstract}
We apply duality method to
prove a rigidity theorem for smooth cocycles of actions of
higher rank cocompact lattices on a class of homogeneous spaces.
In particular, in split case, we
explicitly find all smooth cocycles of actions of lattices on the
imaginary
boundaries of symmetric spaces.
\end{abstract}
\tableofcontents
%\section{ Introduction.}
\newpage
Cocycles of a group action are a fundamental tool for understanding the
action. They lie at the heart of many questions about rigidity,
existence
of invariant structures, and other properties of the action (see
\cite{katokhurder}, \cite{katokspatzier}, \cite{zimmer},
\cite{katokcocycles},
\cite{hurder-cocycles}.)
We developed
a new duality method for calculating
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 $F$ 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
$F$.
Then, we just need to get a hold on the kernel of $F$.
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 another
action, which dynamically may be completely different
from the original one, and might turn out to be easier to study.
Successful implementation of this program leads to different results
on
cohomological, differential and infinitesimal
rigidity of actions of lattices which will be presented in
\cite{kononenko-infinitesimal} and \cite{kononenkoyue}.
In this paper, we apply our methods to prove a
rigidity theorem
(Theorem~\ref{thm-rigidity}) for the $\Ci $ real
cocycles of the actions of cocompact
lattices in a semi-simple Lie group $G$ of higher rank, without compact
factors
(with some additional conditions), on $N \backslash G $ for a class of
subgroups $N$ in $G$.
In case when
$N$ is a minimal parabolic subgroup of $G$, and thus $ N \backslash G$
is
the maximal boundary of the symmetric space corresponding to $G$ we
obtain
a complete description of all
$\Ci $ real cocycles of the $\Gamma $ action
on it (Theorem~\ref{thm-boundary}).
To prove this result
we use our duality to transform the question about cocycles of
the $\Gamma$ action on $N \backslash G$ (which does not seem
to have any "nice" properties) into a question about cocycles of the
$N$ action on $G/\Gamma $ (which is ``nice'' in many respects: it is
partially hyperbolic, it preserves the natural measure $\mu$ on
$G/\Gamma$, etc...)
Also, along the way, we prove some facts about trivializations
of the cocycles for certain classes of actions
(Theorem~\ref{thm-isometries}
and Theorem~\ref{thm-gamma}) and some
Katok-Spatzier type constant cocycles
theorem (Theorem~\ref{thm-const}).
Some results for H\"older cocycles are mentioned in
Section~\ref{sec-holder}.
I would like to thank my advisor, Prof.A.Katok, for generously
sharing his
knowledge and time with me, for his guidance and encouragement
%which made this work possible.
during the development of this work.
\section{General Duality Theorem.}
Recall that if a group $G$ acts on a space $X$ 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 $ will be called cohomologous if
there exists a function $P : X \rightarrow H $ such that
$$ \beta (g,m) = P(gm)^{-1} \alpha (g,m) P(m). $$
By $\operatorname{H}(G,X) $ we will denote a set of equivalence
classes of cocycles.
Imposing additional regularity conditions 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.
%We put on $\operatorname{H}(G,X) $ factor topology induced from the
%set of all cocycles, endowed with the topology of uniform convergence
on
%compact subsets of $G\times X.$
Most of the time we will work with the cocycles with coefficients
in $\Bbb{R}$, but it is easy to see that all our results will
automatically
be true for cocycles with coefficients in $\Bbb{R}^l, l \in \Bbb{N}$ as
well.
Also, notice that for real cocycles we will use additive notation.
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 $X_1$.
Let us try to study $\operatorname{H}(G,X_2)$ by lifting
its elements to $X_1$.
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.
It turns out that the spaces of $\operatorname{H}^{tr}$-cocycles enjoy
the following duality property:
\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 $\operatorname{H}^{tr}$ spaces are defined
with
respect to the lift of actions to the whole $M.$
\end{thm}
Rather then proving Theorem~\ref{thm-generalduality} we will
prove its particular case, Theorem~\ref{thm-duality},
which will be sufficient for all our applications. The proof
of Theorem~\ref{thm-duality} is very similar to the proof of
Theorem~\ref{thm-generalduality} but has the advantage of being much
more transparent and much less technically involved.
The full proof of Theorem~\ref{thm-generalduality} will be presented in
\cite{kononenko-infinitesimal} where, unlike in this
article, we will need to use it in its general form.
\section{Duality Theorem for Subgroup Actions.}
\label{sec-duality}
Let $G$ be a topological group (Lie group), $P$ and $Q$ its
subgroups.
Let $P$ act on $G/Q$ from the left, and $Q$ act on $\PG$ from the
right.
Lift both actions to the whole $G.$
Then, we prove the following:
\begin{thm}
\label{thm-duality}
For real cocycles:
\begin{enumerate}
\item There is a naturally defined, linear imbedding
$$\KPQ : \HtrPGQ \rightarrow \HQPG.$$
\item $\operatorname{Im}(\KPQ)=\HtrQPG.$
\item $\KPQ\circ \KQP = \operatorname{Id}$ on $\HtrPGQ$, i.e. $\KPQ$
defines a natural duality between $\HtrPGQ$ and $\HtrQPG$.
\end{enumerate}
\end{thm}
\begin{cor}
\label{cor-duality}
If\/ $\HtrPGQ = \HPGQ$ and $\HtrQPG = \HQPG$ then there is a
natural
duality between $\HPGQ$ and $\HQPG$.
\end{cor}
Remark: Later we will see that quite often not only all lifts of
the cocycles to the whole $G$ are trivial, but that also all the
cocycles of the
action of a subgroup on the whole group are trivial
(Theorems~\ref{thm-gamma} and Remark in the end of
Section~\ref{sec-trivial}.) )
\begin{pf}
Let $\alpha : P \times G/Q \rightarrow \Bbb{R}$ be a cocycle,
$\beta $$ : P \times G \rightarrow \Bbb{R}$ be its lifting to $G$.
Assume that $\beta $ is trivial.
We will say that a function $f:G \rightarrow \Bbb{R}$ represents
$\beta $ if $ \beta (p, g)=f(g)-f(pg)$.
Let $C_P$ (resp.\ $C_Q$) be the set of functions on $\PG $
(resp.\ $G/Q$), which is
naturally identified with the set of all left $P$ (resp.\ right $Q$)
invariant functions on $G$.
\begin{lmm}
If $f_1$ and $f_2$ represent $\beta $ then $f_1-f_2$ belongs to
$C_P$.
\end{lmm}
\begin{pf}
If $$ \beta (p, g)=f_1(g)-f_1(pg)=f_2(g)-f_2(pg)$$
then $$f_1(pg)-f_2(pg)=f_1(g)-f_2(g).$$ So,
$f_1-f_2$ is $P$ invariant.
\end{pf}
Let $f$ be some function representing $\beta $. Then since $\beta $
is a lift of $\alpha $, we
have $$ \beta (p, g)= \beta (p, gq^{-1})=f(gq^{-1})-f(pgq^{-1}).$$
So, $g \rightarrow f(gq^{-1})$
also represents $\beta $. So,
$$f(gq^{-1})=f(g)+ \phi (q, g),$$
where $\phi (q, g) $ belongs to $C_P$.
\begin{lmm}
$\phi (q, g)$ defines a cocycle of the action of $Q$ on
$\PG $.
\end{lmm}
\begin{pf}
We have, $f(g(q_1q_2)^{-1})=f(g)+ \phi (q_1q_2, g)$.
On the other hand,
$$f(gq_2^{-1}q_1^{-1})=f(gq_2^{-1})+ \phi (q_1, gq_2^{-1})
=f(g)+ \phi (q_2, g)+ \phi (q_1, gq_2^{-1}).$$
So, $\phi (q_1q_2, g)= \phi (q_1, gq_2^{-1})+ \phi (q_2, g)$.
\end{pf}
We have constructed $\phi$ using a particular $f$ representing
$\beta$. Now we
will show that, in fact, the equivalence class of $\phi$ is
independent
from the choice of $f$.
\begin{lmm}
If $f_1$ and $f_2$ represent $\beta $, they generate equivalent
$\phi _1$ and
$\phi _2$.
\end{lmm}
\begin{pf}
If $f_2=f_1+ \eta$, and $\eta$ is in $C_P$, then
$$f_2(gq^{-1})=f_1(gq^{-1})+ \eta (gq^{-1})=f_1(g)+ \phi _1(q, g)+
\eta (gq^{-1})= $$
$$=f_2(g)+ \phi _1(q, g)+ \eta (gq^{-1})- \eta (g).$$
So, $\phi _2(q, g)=\phi _1(q, g)- \eta (gq^{-1})+ \eta (g)$.
\end{pf}
So, in fact we have defined a map $\KPQ$ from $\HtrPGQ $ to $\HQPG $.
From it definition by the formula
$f(gq^{-1})=f(g)+ \phi (q, g),$
we see that
$\phi $ is in $\HtrQPG$ and is
represented by the function $ -f$. Applying our construction we have
$\beta (p, g)=f(g)-f(pg)= \KQP ( \phi )$, which immediately implies
that $\KPQ\circ \KQP = \operatorname{Id}$ on $\HtrPGQ$,
which proves the Theorem~\ref{thm-duality}.
\end{pf}
\section{Trivializations of Cocycles on the Whole Group.}
\label{sec-trivial}
In this section we prove some results about trivializations of
cocycles
of the action of a subgroup on a group. We will consider only the
left action,
although all the results are, of course, true for the right action
as well. For brevity, we only deal with the $\Ci $ cocycles,
although all the results have natural analogs in other cases.
The statements and proofs are
completely analogous to the $\Ci$ case, except for some minor
technical differences. So, we will not go into it.
\begin{thm}
\label{thm-isometries}
Let $\Gamma $ be a discrete group, acting freely and totally
discontinuously on a Riemanian manifold $M$.
Assume that
\begin{enumerate}
\item It acts by isometries.
\item There exists $\delta > 0 $ such that $ \forall m \in M $,
$\forall \gamma \in \Gamma $, and $ \forall \epsilon \leq
\delta $,
$\gamma B(m, \epsilon )$ are disjoint, where $B(m, \epsilon
)$ is
an open ball of radius $\epsilon $ around $m$.
\end{enumerate}
Then $\operatorname{H}( \Gamma, M)=0$.
\end{thm}
First we prove two technical lemmas.
Call a set $U = \bigcup_{\gamma \in \Gamma } \gamma B$,
where $B$ is an open ball in $M$
such that all $\gamma B$ are pairwise disjoint a $\Gamma$-ball.
Call a radius of $B$ a radius of $U$.
Call a cover of $M$ by $\Gamma$-balls a $\Gamma$-cover.
\begin{lmm}
\label{lmm-cover}
Under the assumptions of the Theorem~\ref{thm-isometries}
for every $ 0 < \epsilon \leq \delta $ there exists a locally finite
$\Gamma$-cover $U_n$,
$ n \in \Bbb{N} $, such that the radii of all $U_n$ are
equal to $\epsilon$.
\end{lmm}
\begin{pf}
Let $K_1 \subset K_2 \subset K_3 \subset \cdots \subset K_n \subset
\cdots M $
be such that $K_n $ are all compact and $\bigcup_{n=1}^{\infty} K_n =
M$.
Then we will construct the $\Gamma$-cover $U_n$ in the following way.
Step 1 : take any point $m_1 \in K_1$ and put $U_1 = \bigcup_{\gamma
\in \Gamma } \gamma B(m_1, \epsilon ) $.
Step k : Let $A=\bigcup_{n=1}^{k-1} U_n $. Suppose $A \bigcap K_i =K_i,
i0$. Then take any point
$m_k \in K_j \backslash (A \bigcap K_j)$ and put
$U_k = \bigcup_{\gamma \in \Gamma} \gamma B(m_k, \epsilon)$.
Now, notice that the set $S=\{\gamma m_n, n \in \Bbb{N}, \gamma \in
\Gamma \}$ is
$\epsilon$-separated set in M. Therefore, $S \bigcap K $ is finite for
any
compact set $K \subset M $.
Thus, $K_i$ is covered by a finite number of $U_n$, and
so $K_i \subset \bigcup_{n=1}^{\infty} U_n$.
Therefore, $U_n$ covers $M$.
Let us prove that $U_n$ is locally finite. Suppose it is not. Let $x
\in M$
be covered by infinitely many $U_n$. Then $x$ is less then at
$\epsilon$
distance from infinitely many points in $S$, or in other words
$S \bigcap \overline{B(x, \epsilon)}$ is infinite set which is
impossible.
The lemma is proved by contradiction.
\end{pf}
``Blow up'' the cover $U_n$ a little. Namely, fix $\epsilon_1$ such
that
$\epsilon < \epsilon_1 < \delta $ and consider the new cover
$U_{n}'= \bigcup_{\gamma \in \Gamma}\gamma B(m_n, \epsilon_1).$ Then,
from the proof
of the Lemma~\ref{lmm-cover}, we easily see that $U_{n}'$ is also a
locally
finite $\Gamma$-cover.
\begin{lmm}
There exists a $\Ci $ partition of unity $e_n$,
such that
\begin{itemize}
\item $e_n(x) =0$ for $x \overline{\in} U_{n}'$,
\item $e_n(x) >0$ for $x \in U_n $ and
\item all functions $e_n$ are $\Gamma $ invariant.
\end{itemize}
\end{lmm}
\begin{pf} Take ``bump'' functions $b_n$ such that
\begin{itemize}
\item $b_n$ is non-negative.
\item $b_n(x) > 0$ for $x \in B(m_n, \epsilon) $,
\item $b_n(x) =0 $ for $x \overline{\in} B(m_n, \epsilon_1) $.
\end{itemize}
Then, ``spread'' $b_n$. Namely, define
\begin{itemize}
\item $e_{n}'(x)=b_n(\gamma^{-1} x)$ for $x \in \gamma B(m_n,
\epsilon_1)$,
\item $e_{n}'(x) =0$ otherwise.
\end{itemize}
Then $e_{n}'$ are $\Gamma$ invariant $\Ci $ functions. Since $U_{n}'$ is
locally finite $a(x) = \sum e_{n}'(x) $ is a well defined $\Ci $
function.
Since $U_n$ is also a cover, $a(x) > 0 $ for all $x \in M $.
Then set $e_n = e_{n}'/a(x) $ to prove the lemma.
\end{pf}
Now we are ready to prove Theorem~\ref{thm-isometries}.
\begin{pf}
Let $\alpha ( \gamma , m)$ be a cocycle, $\gamma \in \Gamma, m \in M
$.
Construct functions $f_n$ in the following way: First define,
\begin{itemize}
\item
$h_n(m)= \alpha ( \gamma , \gamma ^{-1} (m))$,
if $m \in \gamma B(m_n, \epsilon_1) $,
\item
$h_n(m)=0$, otherwise.
\end{itemize}
Now, put $f_n =e_nh_n$. Then obviously $f_n$ are $\Ci $ functions.
We have $$f_n( \gamma m)-f_n(m)=e_n(m) \alpha ( \gamma , m).$$
Put $f= \sum_{n=1}^{ \infty } f_n$. Then since $U_{n}'$ is locally
finite,
$f$ is a well defined $\Ci $ function on $M$.
Furthermore,
$$f( \gamma m)-f(m)= \sum_{n=1}^{ \infty }(f_n( \gamma m)-f(m))
= \sum_{n=1}^{\infty}e_n(m) \alpha ( \gamma , m)= \alpha (
\gamma , m).$$
Therefore, $-f$ trivializes $\alpha $.
\end{pf}
Remarks: It is easy to see that the assumptions 1) and 2) of the
Theorem~\ref{thm-isometries}
are not really essential for it to be true. All we need, is
the existence of the partition of unity by $\Gamma$ invariant functions
subordinate to the locally finite cover $U_n = \bigcup_{\gamma \in
\Gamma}
\gamma B_n$, where $B_n$ are open sets such that $\gamma B_n$,
$\gamma \in \Gamma $ are disjoined.
This observation leads to many analogs of the
Theorem~\ref{thm-isometries} for
different classes of totally discontinuous actions.
For example, the assumptions 1) and 2) can be substituted by an
assumption
that the factor $M/ \Gamma $ is compact.
Also, notice that if $M$ has transitive group of isometries, then the
assumption
2) follows from the other assumptions of the
Theorem~\ref{thm-isometries}.
We will not discuss the other versions of the
Theorem~\ref{thm-isometries},
since we only use it to get the following:
\begin{thm}
\label{thm-gamma}
Let $G$ be a Lie group, $\Gamma $ a discrete subgroup,
acting on $G$
from the left. Then $\operatorname{H}( \Gamma, G)=0$.
\end{thm}
\begin{pf}
Put a left invariant Riemanian metric on $G$ and apply
Theorem~\ref{thm-isometries}.
\end{pf}
Remark: Another useful observation about trivializations of cocycles of
subgroup actions on groups is the following:
assume $G$ is a Lie group, $P$ -- Lie subgroup and the orbit
foliation for
the left
action of $P$ on $G$ admits a global transverse section $T$,
which is a submanifold of $G$ and
every $g$ in $G$ decomposes as
$p(g)t(g)$, where $p(g) \in P$, and $t(g) \in T$, and $p(g),
t(g):
G \rightarrow G$ are $\Ci.$
Let $\beta $ be a cocycle, such that $\beta: P\times G \rightarrow
\Bbb{R}$
is $\Ci$ (which, of course, is much more then is required in the
definition
of $\Ci$ cocycle), then $f(g)= -\beta (p(g), t(g)) $
trivializes $\beta $. Indeed,
$$f(g)-f(pg)= \beta (pp(g), t(g))- \beta (p(g), t(g))=
\beta(p,p(g)t(g)) = \beta (p, g).$$
Notice, that all elements of $H^{tr}$-subspaces have representatives
that
satisfy the above assumption.
\section{Constant Cocycles Theorem.}
\label{sec-constant}
For the rest of the paper $G$ is a
semi-simple connected Lie group of
$\Bbb{R}$-rank $n\geq 2$, with finite center and
without compact factors. Let $\algG $ be its
Lie algebra, $A$ a connected component of the maximal split Cartan
subgroup, $\algA $ the corresponding commutative subalgebra, $N$ a
closed subgroup containing $A$, $\algN $ the corresponding
subalgebra,
$\lambda _i$
the roots with respect to $\algA $, $\Vlami $ the corresponding root
spaces.
Let
$$\Wlami = \bigoplus_{c \in \Bbb{R}} V_{c\lambda_i} . $$
The elements of $\Wlami $ we will denote by
$\glami $. Let $\Gamma $
be an irreducible cocompact lattice in $G$.
%For the rest of the first part of this paper, we will deal with the
$\Ci$ case.
We will prove the following:
\begin{thm}
\label{thm-const}
Let $G$ be a semi-simple connected Lie group of
$\Bbb{R}$-rank $n\geq 2$, with finite center and without compact
factors,
$A$ a connected component of the maximal split Cartan
subgroup, $N$ a closed subgroup containing $A$. Let $\Gamma $
be an irreducible cocompact lattice in $G$.
(*) Also we will assume that $\cal G $ has no factors isomorphic to
$\frak {so} (m, 1) $ or $\frak {su} (m, 1) $.
Then all $\Ci $ cocycles of the action of $N$ on $G/ \Gamma $ are
$\Ci $ cohomologous to constant cocycles.
\end{thm}
Remark: The (*) condition is needed due to a non-uniformity in
certain estimates on the decay of correlation coefficients for unitary
representations of groups $\frak {so} (m, 1) $ and $\frak {su} (m, 1).$
It is quite possible that for the particular representations considered
in
the proof of Theorem~\ref{thm-katok} sufficient estimates still can be
obtained and then
the (*) condition will disappear from all our results.
First we will prove two lemmas which allow us to reduce the proof
of Theorem~\ref{thm-const} to the case when $N$ is connected. Let
$N_0$ be the connected component of identity.
\begin{lmm}
\label{lmm-triv}
If a cocycle $\alpha $ is trivial on $N_0 $, it is constant
on the whole $N $.
\end{lmm}
\begin{pf}
Let $p \in N $ and $ \overline{p} $ be the corresponding coset in
$N_0 \backslash N $. Consider the right action of $A$ on $ N_0
\backslash N $.
Since $ N_0 \backslash N $ is not more then countable, the stabilizer
of
$ \overline{p} $ is not empty, i.\ e.\ there exists $a \in A $ such
that
$\overline{pa} = \overline{p} $, or, in other words, there exists
$p_1 \in N_0 $
such that $p_1p=pa $. Then we have $$\alpha (pa, m) = \alpha (p, am) +
\alpha (a, m) = \alpha (p_1p, m) = \alpha(p_1, pm) + \alpha (p, m) , m
\in G/\Gamma .$$
So, $\alpha (p, am) = \alpha(p, m) $.
Since $a$ has an infinite order, by Moore`s ergodicity theorem
(see \cite{zimmer})
it acts ergodicly
on $G/ \Gamma$. Thus, it acts topologically transitively (since every
open set
has positive measure), i.e. it has a dense orbit.
Therefore $\alpha (p, \cdot )$
is constant on $G/ \Gamma$.
\end{pf}
\begin{lmm}
\label{lmm-const}
If $\alpha (p, \cdot ) $ is constant for every $p \in N_0 $, then
it is
constant for all $p \in N $.
\end{lmm}
\begin{pf}
Let $\mu $ be the normalized Haar measure on $ G/\Gamma $.
Let $$\beta (p) =\gral \alpha (p, m) d \mu .$$ Then, due to the
invariance
of $\mu $, we get
$$ \beta (p_1p_2) = \gral \alpha (p_1, p_2m) d \mu +\gral \alpha (p_2,
m) d \mu =
\gral \alpha (p_1, m) d \mu +$$
$$+ \gral \alpha (p_2, m) d \mu = \beta (p_1) +
\beta (p_2) .$$ So, $\beta (p)$ is a one-dimensional representation of
$N$, i.e.
a constant cocycle.
Let us prove that $\beta (p) = \alpha (p, m) $. Consider $\theta (p, m)
=
\alpha (p, m) - \beta (p) $. Then $\theta $ is a cocycle, and for $p
\in N_0 $,
$\theta (p, m) = 0 $. So, by the Lemma~\ref{lmm-triv}, $\theta (p, m)
$
is constant for all $p$. But from its definition as
$\alpha(p,m)-\beta(p)$,
we see that it is orthogonal to the constants. Thus, $\theta =0$,
and $\alpha(p, m) = \beta (p) $.
\end{pf}
To prove the Theorem~\ref{thm-const} it will be enough to prove it for
$N$
connected.
Indeed, then we will know that every cocycle is cohomologous to a
cocycle
constant
on $N_0 $, and thus constant on the whole $N$ by the
Lemma~\ref{lmm-const}.
We start by proving the following simple structural lemma:
\begin{lmm}
\label{lmm-structure}
If $\algN $ is a subalgebra containing $\algA $ then it is generated as
a vector space (and thus as an algebra) by a set of $\glami \in \Wlami
$.
\end{lmm}
\begin{pf}
It will be enough to prove that if $v \in \algN $ and
$$v= \frak g_0 + \sum_{i=1}^{k} \glami , $$
with $\lambda_i \neq 0$, and
$\lambda_i $ and $\lambda_j $ not proportional for $i \neq j$, then
every
$\glami \in \algN $.
Let us prove that $\glamk \in \algN $. Pick up $a_m \in
\Ker(\lambda_m )$
and $a_m \overline{\in } \Ker(\lambda_i ), i \neq m $.
Then $$ [a_{k-1}, [a_{k-2}, [ \ldots [a_1, v] \ldots ]]] = \lambda_k
(a_1)
\lambda_k (a_2) \cdots \lambda_k (a_{k-1} ) \glamk \in \algN .$$ So,
$\glamk \in \algN $.
\end{pf}
To prove Theorem~\ref{thm-const} when $N$ is connected we will need
the following
Theorem due Katok and Spatzier (\cite{katokspatzier-mathreslet}):
\begin{thm}
\label{thm-katok}
For $A$, $G$ and $\Gamma $ as in Theorem~\ref{thm-const} every $\Ci$
cocycle
$\alpha : A \times G/ \Gamma \rightarrow \Bbb{R} $ is $\Ci$
cohomologous
to a constant cocycle.
\end{thm}
Now, we are ready to finish the proof of Theorem~\ref{thm-const}.
\begin{pf}
Assume that $N$ is connected.
By Theorem~\ref{thm-katok} $\alpha $ is $\Ci $ cohomologous to a
cocycle
constant on $A$. Let us prove that it will then be constant on the
whole $N$.
First of all, subtracting $\alpha$'s integral, like in the proof of the
Lemma~\ref{lmm-const}, we can assume that $\alpha(a,m)=0, \forall a \in
A,
m \in G/\Gamma.$
Suppose that $p \in N$ commutes with some element $a \in A$. Then
we have:
$$\alpha (ap, m)= \alpha (a, pm) + \alpha (p, m) = \alpha (p, am) +
\alpha (a, m).$$
Thus, $\alpha (p, m) = \alpha (p, am) , m \in G/ \Gamma$ i.\ e.\
$\alpha (p, \cdot )$ is constant on orbits of $a$ on $G/ \Gamma$ .
By the same ergodicity argument as in the Lemma~\ref{lmm-triv} we
conclude that
$\alpha (p, \cdot )$ is constant on $G/ \Gamma$.
Let $p \in \Exp( \Wlami \bigcap \algN)$. Take $a \in \Exp(\Ker
\lambda_{i})$. Then the
previous argument applies and thus $\alpha (p, \cdot )$ is constant.
If $\alpha (p_1, \cdot )$ and $\alpha (p_2, \cdot )$ are constant,
then $\alpha (p_1p_2, \cdot )$ is also constant. Indeed,
$$\alpha (p_1p_2, m) = \alpha (p_1, p_2 m) + \alpha (p_2, m) =
\alpha (p_1, m) + \alpha (p_2, m).$$
Now, to conclude the proof we just notice that
due to the Lemma~\ref{lmm-structure} every $p$ in a small enough
neighborhood of the unit
in $N$ can be represented as a finite product of elements from
$\Exp( \Wlami \bigcap \algN)$.
And every element of $N$ can be represented
as a finite product of elements from that neighborhood.
\end{pf}
\section{Rigidity Theorem.}
\label{sec-rigidity}
We are now ready to prove the following rigidity theorem for $\Ci $
cocycles
of $\Gamma $ action on $ N \backslash G $:
\begin{thm}
\label{thm-rigidity}
Let $G$ be a semi-simple connected Lie group of
$\Bbb{R}$-rank $n\geq 2$, with finite center and without compact
factors,
$A$ a connected component of the maximal split Cartan
subgroup, $N$ a closed subgroup containing $A$. Let $\Gamma $
be an irreducible cocompact lattice in $G$.
(*) Also we will assume that $\cal G $ has no factors isomorphic to
$\frak {so} (m, 1) $ or $\frak {su} (m, 1) $.
Then every $\Ci $ cocycle of the $\Gamma $
action on $N \backslash G$ uniquely extends to a cocycle of the $G$
action.
\end{thm}
\begin{pf}
By the Theorem~\ref{thm-gamma}, $\operatorname{H} (\Gamma , G )=0
$, and so
$$\HGaNG = \HtrGaNG \cong \HtrNGGa .$$
On the other hand,
$\operatorname{H} (G, G) =0 $ (every cocycle $\alpha(g_1,g_2)$ is
trivialized
be a function $\alpha(g, e),$ where $e$ is the unit element in $G$),
and thus,
$$\operatorname{H} (G, N \backslash G) = \operatorname{H}^{tr} (G, N
\backslash
G) \cong \operatorname{H}^{tr} (N, G/G) .$$
But by Theorem~\ref{thm-const}, $\HNGGa = \widehat{N}$, where
$\widehat{N} $ is the space of 1-dimensional real representations of
$N$. Also, obviously,
$\operatorname{H} (N, G/G) = \widehat{N} $.
So, both $\operatorname{H}^{tr} (N, G/G) $ and $\HtrNGGa $ are
isomorphic to
the space $ \widehat{N}_{G}^{tr} $ of the 1-dimensional representations
of N which
give rise to trivial cocycles of the action of $N$ on $G$.
Also, it is easy to see that given an element
$ \pi \in \widehat{N}_{G}^{tr} $
the duality produces such cocycles $\alpha \in \HGaNG $ and
$\beta \in \HGNG $ that $\beta $ is an extension of $\alpha $.
\end{pf}
>From the proof we easily get:
\begin{cor}
\label{cor-rigidity}
$$\HGaNG = \HGNG =\widehat{N}_G^{tr}. $$
\end{cor}
\section{Application to the $\Gamma $ Action on the Maximal Boundary.}
\label{sec-application}
Assume $G$ is split.
If $P$ is a minimal parabolic subgroup, then $P \backslash G $ is the
Furstenberg's maximal boundary of the symmetric space
corresponding to $G$ (see \cite{mostow}).
We can, easily find explicit formulas for
all the cocycles of the $\Gamma $ action on it.
By the Iwasawa decomposition $G$ splits into a topological direct
product
$G=KP$, where $K$ is the connected subgroup corresponding to the
maximal
compactly imbedded subalgebra in $G$. And, we can use $T=K$ as the
cross
section described in Remark in the end of Section~\ref{sec-trivial}.
And since the Remark is obviously applicable to constant cocycles
we get $\widehat{P} = \widehat{P}_G^{tr} $.
So, by the Corollary~\ref{cor-rigidity}, we have:
$$ \operatorname{H} ( \Gamma , P \backslash G ) \cong
\widehat{P}_G^{tr} = \widehat{P} .$$
Notice that $\widehat{P} $ is naturally identified with the
$\widehat{A}$.
Moreover, using our duality construction we can find the
explicit formulas for all $\Gamma$-cocycles on $P \backslash G$:
\begin{thm}
\label{thm-boundary}
Let $G$ be a semi-simple connected Lie group of
$\Bbb{R}$-rank $n\geq 2$, with finite center and without compact
factors.
Assume $G$ is split. Let $P$ be a minimal parabolic subgroup.
Let $\Gamma $ be an irreducible cocompact lattice in $G$.
(*) Also we will assume that $\cal G $ has no factors isomorphic to
$\frak {so} (m, 1) $ or $\frak {su} (m, 1) $.
Then, for $\Ci$ cocycles, we have:
\begin{enumerate}
\item $ \operatorname{H} ( \Gamma , P \backslash G ) $ is $n$
dimensional.
\item By the Iwasawa decomposition $G$ splits into a
topological direct product
$G=KP$, where $K$ is the connected subgroup corresponding to the
maximal
compactly imbedded subalgebra in $G$. So, every $g \in G$ is
uniquely represented as $g=p(g)t(g)$, where $p(g) \in P$ and $t(g) \in
K$.
Then $ \operatorname{H} ( \Gamma , P \backslash G )$ is
isomorphic to
$\widehat{P} $, and the cocycle corresponding to
the representation $\pi \in \widehat{P} $ is
$$ \alpha (\gamma , g) = \pi (p(g \gamma^{-1}))- \pi(p(g)). $$
\end{enumerate}
\end{thm}
Remark: we are abusing the notation writing $\alpha(\gamma , g) $ for
a cocycle on $\PG $, but it is easy to check that since $\pi $ is a
representation of $P$, $\alpha (\gamma , g )$ depends only on the
coset of $g$ in $\PG $, and thus, really defines a cocycle on $\PG $.
\section{H\"older cocycles.}
\label{sec-holder}
If $G$ is split the $A$ action on $G/\Gamma$ is not just partially
hyperbolic but Anosov. This allows
Katok and Spatzier to get the H\"older analog of the
Theorem~\ref{thm-katok}
(see \cite{katokspatzier}).
Then, for split $G$, we can get the H\"older analogs of
Theorems~\ref{thm-const},
~\ref{thm-rigidity} and ~\ref{thm-boundary}.
In particular, we get the following regularity result:
\begin{thm}
If $G$ is split, then
every H\"older cocycle of the $\Gamma $ action on
$ N \backslash G $ is H\"older cohomologous to a $\Ci $ cocycle.
\end{thm}
The proofs are completely parallel to the proofs of the corresponding
$\Ci$
results, except for the fact that in the appropriate moment we should
use
the H\"older analog of the Theorem~\ref{thm-katok}.
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Department of Mathematics, Pennsylvania State University, 218 McAllister
Building, PA 16802.
E-mail:avk@math.psu.edu
\end{document}