Tensor product of simple representations 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!
 A: 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).
A: Let $G=SL_2(F_p)$. Put $V_k$ the $k+1$-dimensional representation. Then $V_k$ is simple for $0\le k\le p-1$. Take $0\le r,s\le p-1$ with $r+s>p$. Then $V_r\otimes V_s$ is not semisimple.
A: If the characteristic is $p$ then, by a theorem of Serre, it's true provided $dim(V) + dim(W) < p+2$. To be safe, let me assume that the base field is algebraically closed. (One should be safe over a perfect field though.) The example given by Bruce Westbury shows that the above condition is (in some sense) best possible. In fact, Serre showed this result is even true for arbitrary groups (not even algebraic)! Serre's paper is extremely beautiful and well worth reading. It in fact reduces the general case to the case of algebraic groups, and in that situation uses some ideas of Jantzen. The paper is available here for free:
http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN356556735_0116&DMDID=dmdlog35
As Jim Humphreys wrote in the mathscinet review (MR1253203):
There are not many interesting theorems of the form: "If $G$ is any group, then $\ldots$…''. But an old theorem of Chevalley and a new theorem proved in this paper certainly qualify.
