# Hochschild (co)homology and representation theory

Dear members of Mathoverflow,

I just discovered the notion of Hochschild (co)homology. I understand well the formalism however I am wondering about the meaning of this (co)homology for representation theory.

I consider an algebra $\mathcal{A}$, such as $\mathcal{U}(su(n))$, a finite dimensional representation space $V$ for $\mathcal{A}$ (they are well known in the case of $\mathcal{U}(su(n))$) and the $\mathcal{A}$-bimodule $\mathcal{M}=\mathrm{End}(V)$.

In this example, what would be the interpretation of Hochschild (co)homology from the point of view of representation theory ? To what is it an obstruction ?

I can compute things in simple cases such as $\mathcal{U}(su(2))$ but I do not the global interpretation emerge in this case...

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Not a proper answer, but just a comment: the view of 1st and 2nd degree Hochschild cohomology as obstructions to the splitting of certain extensions ought to be explained in several sources: Weibel's book goes into some detail on this. –  Captain Oates Oct 4 '12 at 8:44
In the case of enveloping algebras of semisimple Lie algebras: my guess (but I am not sure this is true in detail) is that the vanishing of higher cohomology groups corresponds to the Lie algebra being semisimple, so all submodules of a given module split off as module summands. Presumably if one replaces su(n) by something solvable then not every indecomposable module is irreducible, and my instinct is that this should correspond to some non-trivial $H^1$ –  Captain Oates Oct 4 '12 at 8:48
@Damien, I give you a +1 not for the question but for having discovered Hochschild (co)homology, which I think, according to my own experience, is a matter of celebration in one's life. –  Fernando Muro Oct 4 '12 at 9:48
About Yemon's comment: If $\mathfrak{g}$ is a finite-dimensional complex semisimple Lie algebra, then the higher (Hochschild) cohomology groups for the universal enveloping algebra of $\mathfrak{g}$, taken with trivial coefficients, are nonzero in degree $\dim \mathfrak{g}$, and sometimes in lower nonzero degrees as well, depending on the Lie type of $\mathfrak{g}$. But in this case, $H^1$ taken with an arbitrary finite-dimensional coefficient module is always zero, which does correspond to the semisimplicity of finite-dimensional representations. –  Christopher Drupieski Oct 4 '12 at 15:28
Look at D. Happel, Hochschild cohomology of finite-dimensional algebras, in Seminaire d'Algebre Paul Dubreil et Marie-Paul Malliavin, 39eme Annee (Paris, 1987/1988), Lecture Notes in Mathematics 1404 (1989), 108-126. –  Benjamin Steinberg Oct 4 '12 at 15:34
If $A$ is a $k$-algebra and $V$, $W$ are $A$-modules, then $Hom_k(V,W)$ is an $A$-bimodule in a natural way (acting contravariantly in $V$ and covariantly in $W$). It is well known (can be found in Cartan-Eilenberg) that $n^{th}$-Hochschild cohomology of $Hom_k(V,W)$ is $Ext_A^n(V,W)$. Your case of $End(V)$ would then give $Ext_A(V,V)$.