\documentstyle[12pt]{article}
\def\Bbb#1{\mbox{\Blackbb #1}}
\def\frak#1{\mbox{\frakfurt #1}}
\def\operatorname#1{\mbox{\rm #1}}
\newcommand{\Ho}{\operatorname{H}}
\newcommand{\Ci}{C^\infty}
\newtheorem{prop}{Proposition}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{df}{Definition}[section]
\newtheorem{thm}{Theorem}[section]
\newtheorem{lmm}{Lemma}[section]
\newenvironment{pf}{ Proof:}{ $\triangle$ \newline}
\title{ Infinitesimal tensor rigidity of higher rank lattice actions.}
\author{A. Kononenko}
\date{ }
%\address{University of Pennsylvania}
%\email{avk@@math.upenn.edu}
\begin{document}
\newfont{\Blackbb}{msbm10 scaled \magstep 0}
\newfont{\frakfurt}{eufm10 scaled \magstep 0}
\maketitle
\section{Introduction}
Any $\Ci$ action of a group $G$ on a manifold $X$ induces a
representation of
$G$ in the space $(\otimes_1^p S(T^* X))\otimes( \otimes_1^q S(TX))$
of smooth
$(p,q)$-tensor fields on $X.$ Here $S(T^*X)$ and $S(TX)$ denote
the spaces of smooth
sections of, correspondingly,
the cotangent and tangent bundles, i.e., covector and vector fields.
We will denote the first cohomologies of that representation by
$$\operatorname{H}^1 (G, X, (\otimes_1^p S(T^* X))\otimes ( \otimes_1^q
S(TX)))$$
or, to shorten the notation, by
$\operatorname{H}^1_{(p,q)} (G, X).$
\begin{df}
\label{df-rigidity}
A smooth action of a group $G$ on a manifold $X$ is called $\Ci$
infinitesimal $(p,q)$-tensor rigid if the
$\Ci$ first cohomologies
$\operatorname{H}^1_{(p,q)} (G, X)$ are
trivial.
\end{df}
Note that for $p=0,$ $q=1,$ this is a standard definition of
infinitesimal
rigidity \cite{zimmer-infinitesimal}.
Clearly, for any $C^1$ action,
similar notions and definitions may be introduced in
$L^2,$ continuous or Hoelder categories.
For the most part we will work with the $\Ci$ actions and $\Ci$
cohomologies. We will mention some $L^2,$ continuous and Hoelder
results in Section~\ref{sec-continuous}.
The main result of this paper is the following rigidity
\begin{thm}
\label{thm-main}
Let $G$ be a semi-simple connected Lie group of
$\Bbb{R}$-rank $\geq 3$, with finite center and without either
compact factors or factors of rank one.
Let $\Gamma$ be an irreducible cocompact lattice in $G.$
Let $N$ be a subgroup of $G$ that contains the maximal connected split
Cartan subgroup $A$ of $G.$ Let $N_0$ be the connected component of
the unit in $N.$
$(\times)$ Suppose that $N_0 \backslash G$ is smoothly parallelizable.
Then every element of
$\operatorname{H}^1_{(p,q)} (\Gamma, N \backslash G)$
uniquely extends to an element of
$\operatorname{H}^1_{(p,q)} (G, N \backslash G).$
\end{thm}
Remark 1: In particular, Theorem~\ref{thm-main} applies to the actions
of $\Gamma$
on the boundarues of the symmatric space associated with $G$ ($N$ is a
parabolic
subgroup).
Remark 2: Moreover, it will follow from our proof that
$$\operatorname{H}^1_{(p,q)} (\Gamma, N \backslash G) =
\operatorname{H}^1_{(p,q)} (G, N \backslash G) \cong$$
$$\cong (\operatorname{H}^1)^{tr} (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N)) ,$$
where $Ad_N$ is the factor
of the restriction of the adjoined representation of $G$ to $N$ by the
subrepresentation in the tangent space $\cal{N}$ to $N,$ $Ad_N^*$ is
its
conjugate, and $(\operatorname{H}^1)^{tr} (N, (\otimes_1^p
Ad_N^*)\otimes
( \otimes_1^q Ad_N))$ is a certain subspace of
$\operatorname{H}^1 (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N))$ (see Section~\ref{sec-proof} for the details).
(Both here and elsewhere in the paper we shall denote
the first cohomologies of a representation $T$ of a group $G $
by $\operatorname{H}^1(G,T).$)
Therefore, the infinite dimensional problem of the description of
$\operatorname{H}^1_{(p,q)} (\Gamma, N \backslash G) $ reduces to a
problem of calculating the first cohomologies of some finite
dimensional
representation of $N.$
In particular, one can see that $H^1_{(p,q)}(\Gamma, N\backslash G)$
is finite dimensional and has
dimension no greater then $(dim(G)-dim(N))^{(p+q)dim(N)}.$
In general the properties of
$(\operatorname{H}^1)^{tr} (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N))$ depend on the combinatorical relations between the
roots of $\cal{G}$ -- the Lie algebra of $G$ -- and $\cal{N}.$
The different types of phenomena that play a crucial role in the study
of
$(\operatorname{H}^1)^{tr} (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N))$ are discussed in Section~\ref{sec-projective}, for
the particular case when $G=SL(n,\Bbb{R})$ and $N$ is a parabolic
subgroup.
Remark 3: The $(\times)$ condition looks very unnatural. It seems
reasonable
to conjecture that Theorem~\ref{thm-main} should be true without it, and
that
it is just a reflection of the limitations of our methods rather than of
some
real phenomena. Unfortunately, for the methods we use,
it is essential for us to be able to work
with twisted cocycles instead of the cohomologies in the tensor bundles,
and
for that we need the parallelization. One possible way to get rid of
this
condition is to use the $L^2$ analog of Theorem~\ref{thm-main}
(for the $L^2$ case $(\times)$ is, obviously, not a restriction at all)
and then
try to prove that the $L^2$ extensions of $\Ci$ or continuous elements
of
$\operatorname{H}^1_{(p,q)} (\Gamma, N \backslash G)$ are,
correspondingly,
$\Ci$ or continuous themselves. Thus far we have not been able to
verify
this, but we plan to pursue this direction in future work.
\section{Some definitions, notations an preliminaries}
\label{sec-preliminaries}
In this section we recall standard definitions of cocycles of group
actions
and state some results from ~\cite{kononenko-infinitesimal},
\cite{kononenko-cocyclesrigidity} (also announced in
\cite{kononenko-era})
that we will use in this article.
\begin{df}
If a group $G$ acts on a space $X$ then a 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 : X \rightarrow H $ such that
$$ \beta (g,m) = P(gm)^{-1} \alpha (g,m) P(m). $$
\end{df}
We will denote a set of equivalence
classes of cocycles by $\operatorname{H}(G,X,H). $
\begin{df}
Let $\cal{R}$ be an $H$-module. Let $ \alpha $ be a cocycle with
coefficients
in H. A map $\beta(g,m) :G \times X \rightarrow \cal{R}$
will be called a twisted cocycle if it satisfies
$$\beta(g_1g_2,m)= \beta(g_2,m) +\alpha(g_2,m)^{-1}\beta(g_1, g_2m),
\forall
g_1,g_2 \in G, m \in X.$$
Two cocycles $\beta$ and $\beta_1$ will be called cohomologous (or
equivalent) if
and only if there exists a function $f: X \rightarrow \cal{R}$ such
that
$$\beta(g,m)-\beta_1(g,m) = f(m) - \alpha(g,m)^{-1}f(gm), \forall g
\in G,
m \in X.$$
\end{df}
We will denote the set of equivalence classes by
$\operatorname{H}(G,X,\cal{R},\alpha).$
\begin{df}
We will call a cocycle or twisted cocycle constant if it does not
depend on $m \in X.$
\end{df}
Remark 1: One can easily see that, for cohomologous cocycles
$\alpha_1$ and
$\alpha_2,$ $\operatorname{H}(G,X,\cal{R}, \alpha_1)$ and
$\operatorname{H}(G,X,\cal{R}, \alpha_2)$ are isomorphic.
This justifies the notation
$\operatorname{H}(G,X,\cal{R},\alpha).$
Remark 2: We define cocycles of different regularity, for example
$\Ci,$ $L^2,$ continuous, Hoelder, imposing the corresponding
regularity condition
on the functions $\alpha(g, \cdot)$ and $\beta(g,\cdot ).$
The following definitions were introduced in
\cite{kononenko-infinitesimal}
and \cite{kononenko-cocyclesrigidity}.
\begin{df}
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 $X_1$.
Then, for any group $H,$
every element of $\operatorname{H}(G,X_2,H) $ can be uniquely lifted
to an element of $\operatorname{H}(G,X_1,H) .$
By $\operatorname{H}^{tr}(G,X_2,H)$ we will denote a
set of those elements of $\operatorname{H}(G,X_2,H) $ that lift
to a trivial element in $\operatorname{H}(G,X_1,H).$
We define
$\operatorname{H}^{tr}(G,X_2,\cal{R},\alpha),$
$\alpha \in \operatorname{H}(G,X_2,H)$
accordingly, where, as before, $\cal{R}$ is an $H$-module.
\end{df}
Remark: Sometimes, when it is not likely to cause any confusion, we
will omit
$H$ and $\cal{R}$ in the notations introduced above.
Also, assume that there is a certain structure on $X_1$ and $X_2$ which
is
preserved by the factor map.
This could 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, measurable,
or smooth cocycles). The following duality theorems are true for all
classes of cocycles which make sense for the spaces and factor maps
involved.
(For the proofs see \cite{kononenko-infinitesimal}.)
\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 $\operatorname{H}^{tr}(G_1,
M/G_2).$
Here $M/G_2 $ and $G_1 \backslash M$
are the factor spaces on which the other group
acts in the obvious way; the $\Ho^{tr}$-spaces are defined with respect
to the
lifts of the actions to the whole $M.$
\end{thm}
The duality for twisted cocycles is ``built over'' the duality for
non-twisted ones:
\begin{thm}
\label{thm-twistedduality}
Let $G_1,$ $G_2 ,$ $H$ and $M$ be as above.
Let $\cal{R}$ be any $H$-module. Let
$\alpha \in \operatorname{H}^{tr}(G_1, M/G_2,H).$
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
(the spaces of twisted cocycles are naturally
equipped with the 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
$K_1(G_1,G_2).$
The following result is proved in \cite{kononenko-infinitesimal}.
\begin{thm}
\label{thm-twistsmooth}
Let $G$ and $\Gamma$ be as in Theorem~\ref{thm-main}.
Let $A$ be a maximal connected split Cartan subgroup of $G.$
Let $\alpha$ be a constant real multiplicative cocycle, i.e.
any representation of $A$ in $\Bbb{R}^+.$
Then for the $\Ci$ cocycles, every element of
$\Ho(A,G/\Gamma,\Bbb{R},\alpha)$ is $\Ci$
cohomologous to a constant twisted cocycle. (We think of $\Bbb{R}$ as
being naturally equipped with the $\Bbb{R}^+$-module structure.)
Moreover, if $\alpha$ is non-trivial, i.e., not identically equal to
$1,$ then
$$\Ho(A,G/\Gamma,\Bbb{R},\alpha)=0.$$ That is, its every element
of $\Ho(A,G/\Gamma,\Bbb{R},\alpha)$
is $\Ci$ cohomologous to a zero cocycle.
\end{thm}
A slight modification of the arguments used to prove Theorem $4.1$ in
\cite{kononenko-cocyclesrigidity} yields the following
\begin{thm}
\label{thm-gamma}
If $\cal{R},$ in addtition to being an $H$-module, also has a
structure of
an $\Bbb{R}^+$ module, and the actions
of $H$ and $\Bbb{R}^+$ on $\cal{R}$ commute, then, for any
$\alpha \in \Ho(\Gamma, G,H),$
$$ \Ho(\Gamma ,G ,\cal{R}, \alpha)=0,$$
where $\Gamma$ acts on $G$ by multiplication.
\end{thm}
Remark 1: In fact, similarly to Theorem $4.1$ in
\cite{kononenko-cocyclesrigidity},
the conclusion of Theorem~\ref{thm-gamma} holds for a broad
variety of totally discontinuous actions. For the details,
see \cite{kononenko-cocyclesrigidity}.
Remark 2:
Another useful observation about trivializations of cocycles of
subgroup actions on groups is the following:
Assume $G$ is a Lie group, $P$ is its 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 or a twisted cocycle such that
$\beta: P\times G \rightarrow H$
($\beta: P\times G \rightarrow \cal{R}$)
is $\Ci$ (which, of course, is much more than is required in the
definition
of a $\Ci$ cocycle). Then $\beta$ is $\Ci$ cohomologous to a trivial
cocycle or, correspondingly, trivial twisted cocycle.
Notice, that all elements of $H^{tr}$-subspaces have representatives
that
satisfy the above assumption.
Also, it is easy to see that all cocycles and twisted cocycles of $G$
acting on $G$ are trivial, regardless of the above mentioned regularity
condition.
\section{Proof of the main theorem.}
\label{sec-proof}
Clearly it is enough to prove Theorem~\ref{thm-main} for $N=N_0.$
Since $M=N \backslash P$ is parallelizable, $S(TM)$ can be
identified with $F(M)$ which denotes the space of smooth
functions on $P \backslash G$ with values in
$\Bbb{R}^k$, where $k= dim M.$
Then the derivative of this action can be written as a cocycle
$D(\gamma, m)$ with values in $GL(k,\Bbb{R})$: $D(\gamma, m)$ is
equal to the derivative of the transformation
$m \rightarrow m \gamma^{-1}$, evaluated at point $m \in P \backslash
G$
in the coordinate system
used to trivialize the tangent bundle over $P \backslash G.$
Then the action of $\Gamma $ on $S(TM)$ translates into the
following action on $F(M)$:
$$ \gamma \cdot f(m) = D(\gamma, m \gamma) f(m \gamma)=
D(\gamma^{-1}, m)^{-1}f(m \gamma),
f(m) \in F(M).$$
Similarly the action of $\Gamma $ on $S(T^* M)$ translates
into the following action on $F( M)$:
$$ \gamma \cdot f(m) = D(\gamma, m \gamma)^* f(m \gamma)=
(D(\gamma^{-1}, m)^{-1})^* f(m \gamma),
f(m) \in F( M),$$
where $D^*$ denotes the inverse transpose of the matrix $D.$
We pass from the cohomologies to twisted cocycles by
considering $$\beta(\gamma,m) =\chi(\gamma^{-1},m)$$ for
$\chi(\gamma,m) \in \operatorname{H}^1_{(p,q)} (\Gamma, M).$
Then $$\beta(\gamma,m) \in
\operatorname{H}(\Gamma, N\backslash G, \Bbb{R}^{k^{(p+q)}},
(\otimes_1^p D^*)\otimes ( \otimes_1^q D)),$$ and it is clear that
$\operatorname{H}^1_{(p,q)} (\Gamma, M)$ is isomorphic to
$$\operatorname{H}(\Gamma, N\backslash G, \Bbb{R}^{k^{(p+q)}},
(\otimes_1^p D^*)\otimes ( \otimes_1^q D)).$$
Lifting the action and the cocycle $(\otimes_1^p D^*)\otimes (
\otimes_1^q D)$
to the whole $G,$ and using Theorems ~\ref{thm-gamma},
~\ref{thm-generalduality}
and ~\ref{thm-twistedduality}
(applied to the right action of $\Gamma $ and the left action of
$N$ on $G$) we have
$$\operatorname{H}(\Gamma, N\backslash G, \Bbb{R}^{k^{(p+q)}},
(\otimes_1^p D^*)
\otimes (\otimes_1^q D))=
\operatorname{H}^{tr}(\Gamma, N\backslash G,
\Bbb{R}^{k^{(p+q)}},(\otimes_1^p D^*)
\otimes (\otimes_1^q D)) \cong$$
$$ \cong \operatorname{H}^{tr}(N, G/\Gamma , \Bbb{R}^{k^{(p+q)}},
K((\otimes_1^p D^*)\otimes ( \otimes_1^q D))).$$
It is clear from the construction of the duality map $K$ (see
\cite{kononenko-infinitesimal} or \cite{kononenko-cocyclesrigidity})
that
$$K((\otimes_1^p D^*) \otimes ( \otimes_1^q D))=
(\otimes_1^p K(D)^*) \otimes ( \otimes_1^q K(D)).$$
It is proved in \cite{kononenko-infinitesimal} that $K(D)$ is
cohomologous to
a constant cocycle given by the representation $Ad_N$ of $N$ which is
the factor
of the restriction of the adjoined representation of $G$ to $N$ by the
subrepresentation in the tangent space $\cal{N}$ to $N.$
The exact matrix form of the cocycle $K(D)=Ad_N$ depends on the way we
parallelize $ M.$ But different parallizations lead to equivalent
representations, i.e., to cohomologous cocycles. Therefore, since the
spaces of twised cocycles depend only on the cohomology class of the
twisting, in order to study
$$\operatorname{H}(N, G/\Gamma , \Bbb{R}^{k^{(p+q)}},
(\otimes_1^p K(D)^*)\otimes ( \otimes_1^q K(D)))$$
we may use whatever matrix representation of $Ad_N$ we want.
It is easy to prove (see \cite{kononenko-infinitesimal} or
\cite{kononenko-cocyclesrigidity}) that $\cal{N}$ has a basis (as a
linear
space) that cosists
of root vectors of $G.$ Therefore, we may find a subspace $\cal{M}$
which is a complement of $\cal{N}$ in $\cal{G}$ and also has a basis
$v_i,$ $i=1,\ldots,k,$ consisting of root vectors. Let $\lambda_i$ be
the corresponding roots. Now, if $\frak{a} \in \cal{A}$ and
$a=exp(\frak{a}),$
$Ad_N (a)v_i =exp(\lambda_i(\frak{a}))v_i.$ We see that the matrix
presentation
of $Ad_N (a)$ is diagonal if $a\in A.$ Therefore, the matrix
$(\otimes_1^p K(D)^*(a))\otimes ( \otimes_1^q K(D)(a)) $ is also
diagonal, if $a \in A.$
Let $\beta(p,m) \in
\operatorname{H}(N, G/\Gamma , \Bbb{R}^{k^{(p+q)}}, \alpha(p)),$
$p \in N,$ $m \in G/\Gamma,$ where,
both here and elsewhere, we denote the representation
$(\otimes_1^p K(D)^*)\otimes ( \otimes_1^q K(D)))$ of $N$ by $\alpha.$
Consider the restriction of $\beta$ to $A.$ Then, the vector valued
cocycle
$\beta $ splits into $k^{(p+q)}$ real cocycles. According to
Theorem~\ref{thm-twistsmooth} each of the components is $\Ci$
cohomologous
to a constant cocycle. Thus, the whole $\beta(a,m)$ is cohomologous to
a
constant cocycle. Therefore, from now on we may assume that
$\beta(a,\cdot)$ is constant if $a \in A$.
Let $\theta(p)=\int_{M}\beta(p,m) d\mu,$ where $\mu$ is the normalized
$G$ invariant smooth measure on $G/\Gamma.$ Since $\alpha$ is a
constant
cocycle, and, thus, the integration commutes with multiplication by
$\alpha,$ we see that $\theta$ is a constant cocycle belonging to
$\operatorname{H}(N, G/\Gamma, \alpha).$ We will prove that
$\beta(p,m) =\theta(p).$
Indeed, let $\Delta(p,m)=\beta(p,m) -\theta(p).$ Then, $\Delta(a,m)=0,$
i.e., $\Delta(a,m) $ is a zero vector, if $a \in A.$
Let $p_{\lambda}$ be such an element of $N$ that
$p_{\lambda}=exp(\frak{p}_{\lambda}),$ where
$\lambda$ is a root of $G,$ and $\frak{p}_{\lambda}$ is a vector from
the
corresponding root space. Let $a \in Ker(\lambda).$ Then, since
$a$ and $p_{\lambda}$ commute we have
$$\Delta(a p_{\lambda},m)=\Delta(p_{\lambda},m)
+\alpha(p_{\lambda})^{-1}
\Delta(a,p_{\lambda}m)=\Delta(p_{\lambda},m)=$$
$$\Delta(p_{\lambda} a,m)=\Delta(a,m)+\alpha(a)^{-1}\Delta(p_{\lambda},
am).$$
Therefore,
$$\Delta(p_{\lambda},m)=\alpha(a)^{-1}\Delta(p_{\lambda}, am).$$
Then for the $j$-th component $\Delta(p_{\lambda},m)_j$ of
$\Delta(p_{\lambda},m),$
$j=1,\ldots, k^{(p+q)},$ we have:
$$\Delta(p_{\lambda},m)_j=\alpha(a)_j^{-1}\Delta(p_{\lambda}, am)_j,$$
where $\alpha(a)_j$ denotes the $j$-th component of $\alpha.$
But for a non-zero function $\Delta(p_{\lambda},m)_j$ such an equality
is
possible only if $\alpha(a)_j=1.$ And even then, due to Moore's
ergodicity theorem, $\Delta(p_{\lambda},m)_j$ has to be a constant
function.
But since $\Delta(p,m)=\beta(p,m)-\theta (p),$ all components of
$\Delta$ must
be orthogonal to constants, for all $p \in N.$
Therefore, we see that $\Delta(p_{\lambda},m) =0.$
Since $A$ and the exponents of the root vectors from $\cal{N}$
generate
the whole $N$ we conclude that $\Delta(p,m)=0$ for all $p \in N$ and
$m \in M.$ Therefore we have that $\beta(p,m)$ is cohomologous to
a constant cocycle $\theta(p).$
We denote by
$$(\operatorname{H}^1)^{tr} (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N)) $$ the subspace in
$$\operatorname{H}^1 (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N))$$ that consists of those elements of
$\operatorname{H}^1 (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N))$ that correspond to trivial elements of
$\operatorname{H} (N, G, \Bbb{R}^{k^{(p+q)}}, \alpha).$
Then $$\operatorname{H}^1_{(p,q)} (N, M) \cong
(\operatorname{H}^1)^{tr} (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N)).$$
On the other hand,
$$\operatorname{H}^1_{(p,q)} (G, M)
=\operatorname{H}(G, N\backslash G, \Bbb{R}^{k^{(p+q)}},
(\otimes_1^p \overline{D}^*)\otimes ( \otimes_1^q \overline{D}))=$$
$$=\operatorname{H}^{tr}(G, N\backslash G, \Bbb{R}^{k^{(p+q)}},
(\otimes_1^p \overline{D}^*)\otimes ( \otimes_1^q \overline{D})) \cong
$$
$$ \cong \operatorname{H}^{tr}(N, G/G, \Bbb{R}^{k^{(p+q)}},
K(G,N)((\otimes_1^p \overline{D}^*)\otimes ( \otimes_1^q
\overline{D})))=$$
$$=\operatorname{H}^{tr}(N, G/G, \Bbb{R}^{k^{(p+q)}},
(\otimes_1^p K(G,N)(\overline{D})^*)\otimes ( \otimes_1^q K(G,N)
(\overline{D}))),$$
where $\overline{D}$ is the derivative cocycle for the action of $G$ on
$G \backslash N.$
It follows from the construction of the duality maps
\cite{kononenko-infinitesimal}, \cite{kononenko-cocyclesrigidity}
and the fact that $\overline{D}$ is an
extension of $D$
that $K(G,N)(\overline{D}) =K(\Gamma,N)(D).$
Therefore, we have (since $G/G$ consists of a single point)
$$\operatorname{H}^1_{(p,q)} (G, M) =
(\operatorname{H}^1)^{tr} (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N)).$$
And, moreover, it follows from the constructions in
\cite{kononenko-infinitesimal} that for any element $\xi \in
(\operatorname{H}^1)^{tr} (N, (\otimes_1^p Ad_N^*)\otimes
( \otimes_1^q Ad_N))$ the element
$$K_1(N,G)(\xi) \in \operatorname{H}^1_{(p,q)} (G, M)$$ is an extension
of
the element $$K_1(N,\Gamma)(\xi) \in \operatorname{H}^1_{(p,q)} (\Gamma,
M).$$
Theorem~\ref{thm-main} is proved.
\section{Projective actions.}
\label{sec-projective}
In this section we consider the case $G=SL(n,\Bbb{R}),$ $n \geq 4.$ Let
$P$ be
the minimal parabolic subgroup of $G$ consisting of the upper
triangular martices with positive diagonal entries and the determinant
equal
to $1.$
Then the following is true:
\begin{prop}
\label{prop-minimalP}
The action of $\Gamma$ on $P\backslash G$
\begin{enumerate}
\item is not $\Ci$ infinitesimal $(p,q)$-tensor rigid if \newline $p
\leq (n-1)q$ and
$q \leq (n-1)p;$
\item is $\Ci$ infinitesimal $(p,q)$-tensor rigid if \newline $p>
(n-1)(q+1),$
or if $q > (n-1)(p+1);$
\item may or may not be $\Ci$ infinitesimal $(p,q)$-tensor rigid if
\newline
$ (n-1)q