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Let $G$ be a linear algebraic group over field $k$ of characteristic zero. It is well known that the category of finite dimensional $k$--linear representations of $G$ is abelian, and that it is semisimple if and only if $G$ is reductive.

What can be said about the group $G$, if $\mathrm{Ext}^2(V,W)=0$ for all finite dimensional $k$--linear representations $V$ and $W$ of $G$?

One can of course ask the same question with $\mathrm{Ext}^n(V,W)=0$ for any $n\geq 0$. For the moment, I would already be happy to see some examples where a nonzero $\mathrm{Ext}^2$ occurs.

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  • $\begingroup$ If $G$ is $k^n$ with its additive group structure and $n\geq2$, then you can easily find $V$ and $W$ such that $\operatorname{Ext}^2(U,V)$ is not zero. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 28, 2013 at 17:52
  • $\begingroup$ Since the usual homological proof of the semisimplicity of semisimple algebras is based on the second Whitehead lemma, one could guess that vanishing of $H^2$ implies semisimplicity. At the level of Lie algebras, this was proved by our fellow MOer Pasha Zusmanovich here, arxiv.org/abs/0704.3864 $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 28, 2013 at 18:00
  • $\begingroup$ If $H^2(L,M)=0$ for all finite-dimenional $L$ modules then $L$ is either a $1$-dimensional algebra, or a semisimple algebra, or a direct sum of both (in characteristic zero). $\endgroup$
    – Dietrich Burde
    Commented May 28, 2013 at 18:11
  • $\begingroup$ Dietrich, that is precisely the result of Zusmanovich I linked to :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 28, 2013 at 18:12
  • $\begingroup$ Yes, II know. I wanted to point this out again, because $L$ need not be semisimple then, but almost. $\endgroup$
    – Dietrich Burde
    Commented May 28, 2013 at 18:13

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