\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