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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!

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    $\begingroup$ The characteristic of the field makes a big difference here, since in prime characteristic tensor products of simple modules are seldom semisimple even if $G$ is (connected) reductive; moreover, such groups often have faithful irreducible representations but only tori are linearly reductive. On the other hand, in characteristic 0 linearly reductive = reductive. (And I guess the modules here are all finite dimensional?) Note too that the additive group has a faithful irreducible 1-dimensional module. $\endgroup$
    – Jim Humphreys
    Commented Mar 9, 2011 at 21:37
  • $\begingroup$ P.S. My final sentence is nonsense, but the rest of the comment emphasizes the need to focus the question more. $\endgroup$
    – Jim Humphreys
    Commented Mar 10, 2011 at 0:04
  • $\begingroup$ Thanks Jim. What I had in mind is characteristic 0, and representations are rational representations of algebraic groups (so finite dim). So in this case $V\otimes W$ is semi-simple? And what's the 1-dim module of $\mathbb G_a?$ I thought there was no algebraic homomorphism $\mathbb G_a\to\mathbb G_m.$ $\endgroup$
    – shenghao
    Commented Mar 10, 2011 at 0:05
  • $\begingroup$ Under your assumptions, see Chevalley's old theorem quoted in the paper by Serre to which Herzig links. For the additive group, see my P.S. above. $\endgroup$
    – Jim Humphreys
    Commented Mar 10, 2011 at 0:20
  • $\begingroup$ I'm just a little worried that if I should distinguish rep. of an abstract group from rational rep. of an algebraic one, as it appears that Chevalley's theorem is for abstract groups... $\endgroup$
    – shenghao
    Commented Mar 10, 2011 at 0:27

3 Answers 3

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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).

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  • $\begingroup$ Dear George, thank you for providing the answer (which also solves my concern in earlier comment). Can you explain a bit more about why the unipotent radical of $G$ is trivial as soon as $G$ has a faithful semisimple rep.? $\endgroup$
    – shenghao
    Commented Mar 10, 2011 at 23:43
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    $\begingroup$ Dear Shenghao, first consider the case where $G$ has a faithful simple representation $L$. The unipotent radical $R$ of $G$ must have a non-0 fixed point in $L$, and $G$ leaves invariant the $R$-fixed points $L^R$. Since $L$ is simple for $G$, conclude $L = L^R$ so that $R$ acts trivially on $L$. Since $L$ is faithful for $G$, $R=1$. In the general case, can view $G$ as a subgroup of $\prod_i GL(L_i)$ for simple $G$-modules $L_i$. Now the image of $R$ in each $GL(L_i)$ is trivial, so indeed $R$ is trivial. $\endgroup$
    – George McNinch
    Commented Mar 11, 2011 at 1:22
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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.

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  • $\begingroup$ In fact $r + s \ge p$ suffices, as Serre also points out in section 1.3 of his paper. The point is that there is a surjection $V_r \otimes V_s \to V_p$ and $V_p$ isn't semisimple. $\endgroup$
    – fherzig
    Commented Mar 10, 2011 at 0:08
  • $\begingroup$ Excuse me but, do you mean "simple" or "not semi-simple" regarded as representations of the finite group of $\mathbb F_p$-points or rational representations of $SL_2$ over $\mathbb F_p?$ $\endgroup$
    – shenghao
    Commented Mar 10, 2011 at 0:08
  • $\begingroup$ At least when $r+s = p$ and $p > 2$ it doesn't matter, it's true in both cases. (First of all the $V_k$ remain irreducible as reps. of the finite group. And $V_p$ isn't semisimple even as rep. of the finite group.) $\endgroup$
    – fherzig
    Commented Mar 10, 2011 at 0:23
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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.

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  • $\begingroup$ It's also worth recalling an earlier related question on MO: mathoverflow.net/questions/18280 $\endgroup$
    – Jim Humphreys
    Commented Mar 10, 2011 at 0:12
  • $\begingroup$ Yes, that's a good point. You listed the relevant paper of Jantzen in your answer. $\endgroup$
    – fherzig
    Commented Mar 10, 2011 at 0:19

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