Given a countable discrete group $G$ with [Kazhdan's property (T)](http://en.wikipedia.org/wiki/Kazhdan%27s_property_%28T%29), consider $\mathbb{C}G$ or $l^2(G)$ as a left $G$-module, then we can consider the [group cohomology](http://en.wikipedia.org/wiki/Group_cohomology#H2),


Is it known that $H^n(G, l^2(G))=0, H^n(G, \mathbb{C}(G))=0,~~ \forall n>1$?


I am mainly interested in the $n=2$ case.



vanishing higher cohomology group for property T group?
If you are interested in cohomology with coefficients in the group ring $\mathbb CG$, then you should know the Bieri-Eckmann theory of duality for a group of finite cohomological dimension. The group ring is a bimodule, with commuting left and right actions of the group. Taking cohomology uses up one of these actions, but the other action makes the cohomology a $G$-module (or complex thereof) and there is an isomorphism $H^i(G;M)\cong H\_{n-i}(G; M\otimes D)$, where $D$ is the $n$ is the homological dimension and $D$ is the dualizing complex, the cohomology with coefficients in the group ring.


In particular, if the group has a classifying space that is a oriented closed manifold of dimension $n$ (a case that probably predates Bieri-Eckmann), cohomology satisfies Poincaré duality, so the dualizing module is $\mathbb Z$ concentrated in a single degree; that is, $H^n(G;\mathbb ZG)\cong \mathbb Z$ and the cohomology vanishes in other degrees. So if $G$ is a cocompact torsion-free lattice in $SL\_3(\mathbb R)$, it has property T and $H^5(G;\mathbb ZG)\cong \mathbb Z$. 


More general than a PD group is a (Bieri-Eckmann) duality group where $H^\*(G;\mathbb ZG)$ is concentrated in the top degree, though not cyclic. And Bieri-Eckmann prove for general groups subject to finiteness conditions that the top cohomology is non-trivial. So any lattice in a Lie group with property T gives an example of a property T group with nontrivial cohomology. In particular, I believe that $SL\_3(\mathbb Z)$ has virtual dimension 3, so that gives an example with $H^3(G; \mathbb ZG)$ nontrivial. I do not believe that there are any lattices of dimension 2 in Lie groups with T. I suspect that lattices are all duality groups, so the cohomology of the group ring vanishes below the top dimension.


YCor mentions 2-dimensional groups with property T. The Bierri-Eckmann theory shows that these have nontrivial $H^2(G;\mathbb ZG)$.




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If you want something useful to come out of $H^\*(G; \ell^2G)$, you should not use ordinary cohomology, a tool for discrete coefficients, but you should change your definitions to take into account the topology. Study the theory of $L^2$ cohomology for suggestions. It may have a good definition already, but I think it usually restricts to coefficients with trivial action.




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A question came up in the comments of an example of a property T group with nontrivial second cohomology with trivial coefficients. An easy source of second cohomology classes are Kähler forms. The natural metric on a Hermitian symmetric space is Kähler. For example, $Sp\_{2g}(\mathbb R)/U(g)$ is a Hermitian symmetric space. Thus $A\_g=Sp\_{2g}(\mathbb Z)\backslash Sp\_{2g}(\mathbb R)/U(g)$, the moduli of principally polarized abelian varieties is Kähler, and the Kähler class is nontrivial for $g>1$. If instead we take a cocompact lattice, the quotient is closed and the Kähler class is nontrivial for all $g$; moreover, it is easy to see that it is nontrivial, since a power of the Kähler form is the volume form, which is now cohomologically nontrivial. But we must take $g>1$ to get property T.


