Interpretation of vanishing of cohomology groups

Question

Let $G$ be a group and $M$ a left $G-$module. It is well know that for some conditions all the cohomology groups $H^{i}(G,M)$, $i=0,1,2,.....$ vanish. The same can be do for the Hochschild cohomology groups $HH^{i}(A,M)$ of an associative algebra $A$ and an $A-$bomodule $M$....and so on in other contexts like cohomology of Lie algebras.

Mathematicians like Hochschild-Serre, Morris Hirsch and Rolf Farnsteiner stablished such that theorems.

My question is: why is it interesting that all the cohomology groups of a given algebraic structure vanish?

Is there a geometric interpretation for $H^{i}(G,M)=0$ for all $i=0,1,2....$?

What tells us about an algebra $A$ and a given $A-$bimodule $M$ the condition $HH^{i}(A,M)=0$ for all $i=0,1,2....$?

I'm sorry for my bad english.

Answer 1

This is a broad question. Let me comment it for Lie algebras, which you have mentioned.

For Lie algebras $L$, even for the adjoint module $M=L$, the vanishing of $H^n(L,L)$ for all $n\ge 0$ has very strong consequences, both algebraically and geometrically. Concerning geometric interpretations, $H^2(L,L)=0$ alone yields that $L$ is geometrically rigid. Also, the vanishing of $H^2(L,L)$ or $H^3(L,L)$ is important for deformation theory of Lie algebras.

Furthermore, Roger Carles has proved many algebraic properties following from the vanishing, namely that the solvable radical ${\rm rad}(L)$ then is necessarily nilpotent. Of course the center $Z(L)$ is trivial because of $H^0(L,L)=0$. If we also assume that $L$ is perfect, then Pirashvili has conjectured that $L$ is semisimple.

In general, if $H^1(L,M)=0$ for all finite-dimensional $L$-modules, then $L$ is semisimple. This is a "converse" of Whitehead's theorem in characteristic zero. Also $H^2(L,M)=0$ has strong consequences, almost forcing $L$ to be semisimple.

Answer 2

As it already answered for the case of Lie algebras, you can have some similar results for associative algebras with the Hochschild cohomology.

Firstly, one of the most simple computations shows that $$HH^0(A,A)=Center(A). $$ So, the zero degree measures how commutative the algebra is (if it is equal to A, then A is commutative).

Now, for a (finite-dimensional) algebra A over a field of characteristic zero (or, more generally, a perfect field), you have that $$A \text{ is semisimple} \iff HH^1(A,M)=0\,\,\,\forall M $$ Actually, when you can also consider an algebra over a general field or commutative ring $k$, this is a characterization of a class algebras called separable (which may be stricter than being semisimple), i.e. $$ A \text{ is separable} \iff HH^1(A,M)=0\,\,\,\forall M $$ For example, if $k$ is semisimple, then this is equivalent to saying that $A\otimes A^{op}$ is semisimple.

This is somewhat mentioned in the original article of G. Hochschild, where he also comments the case of degree 2, saying that $$A \text{ is segregated} \iff HH^2(A,M)=0\,\,\,\forall M. $$

Another interesting fact, if you are interested in path algebras, is that there is a proof of Happel that shows that if $k$ is an algebraically closed field and ${Q}$ is a finite (connected) quiver without oriented cycles, then $$Q\text{ is a tree} \iff HH^1(kQ,kQ)=0. $$