Let $G$ be a linear algebraic group over some field, and let $V$ and $W$ be two simple rational representations of $G.$ Is $V\otimes W$ semi-simple?


I was trying to convince myself that if $G$ has a faithful semi-simple representation, then $G$ is linearly reductive, and was reduced to the question above. The problem I have in mind is over characteristic 0, but answers addressing char. $p$ is equally appreciated too!



Tensor product of simple representations
If $G$ is a(ny) group, if $k$ is a field of characteristic 0, and if $V$ and $W$ are semisimple finite dimensional $kG$ modules, then $V \otimes\_k W$ is indeed semisimple as a $kG$-module. This is due to Chevalley, and (I think I'm not off-base in saying this) inspired the characteristic $p>0$ result of Serre mentioned in other answers/comments.


The argument goes as follows: it is enough to prove the result after replacing $k$ by an algebraic closure. Now replace $G$ by the Zariski closure of its image in $GL(V) \times GL(W)$ -- this Zariski closure leaves invariant the same subspaces of $V \otimes\_k W$ as does $G$, so we may suppose $G$ to be a linear algebraic group over $k$.


Since representations of finite groups in char. 0 are semisimple, a $G$-representation
is semisimple just in case that is true upon restriction to the connected component $G^0$. Thus we may and will suppose $G$ to be connected.


Finally, note that $G$ has a faithful semisimple representation, namely $V \oplus W$. Thus
the unipotent radical of $G$ is trivial so that $G$ is a connected and reductive group over $k$. Now the semisimplicity of $V \otimes W$ follows (*every* finite dimensional rational representation of $G$ is semisimple).


