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Let $K$ be a field (of course, of positive characteristic, unless you want a trivial question). Let $G$ be a finite group, and $V$ and $W$ be two completely reducible (finite-dimensional) representations of $G$ over $K$. Is the (interior) tensor product $V\otimes_K W$ a completely reducible representation of $G$ ?

I know that this holds for exterior tensor products: Let $K$ be a field. Let $G$ and $H$ be two finite groups, and $V$ and $W$ be two completely reducible (finite-dimensional) representations of $G$ and $H$, respectively, over $K$. Then, the tensor product $V\otimes_K W$ is a completely reducible representation of $G\times H$. (Proof: Combine Curtis/Reiner "Methods of Representation Theory I" Theorems 7.10 and 10.38 (i).)

Note that my above question is equivalent to the Jacobson radical of the group ring $KG$ being a coideal (the coalgebra structure on $KG$ is the canonical one, of course: $\Delta g=g\otimes g$). It may be total nonsense but unfortunately I don't have any nontrivial examples of modular representations to check with.

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Not sure why you're mentioning "coideals" or "coalgebras". If $A = K[G]$ is the group algebra (or any finite-dim. assoc. $K$-algebra) and $J$ is its Jacobson radical then a left $A$-module is semisimple if and only if $J$ acts as 0 on it (see Lang's "Algebra", possibly the exercises, in the section on semisimplicity or Jacobson radical, or both). Your $V \otimes_K W$ as a left $A$-module also has $J$ acting as 0, hence it is semisimple. (Main point is that $J$ is 2-sided ideal and $A/J$ is semisimple ring.) So answer is "yes". –  BCnrd Aug 23 '10 at 13:08
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have a look at Example 4.10 of math.rwth-aachen.de/~Gerhard.Hiss/Preprints/projsum.pdf which gives a group with two simple modules whose tensor product has a non-simple projective summand. Also, shouldn't your comultiplication be g--> g\otimes g? –  Matthew Towers Aug 23 '10 at 14:24
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When I become Emperor of the Universe, one of my first decrees will be the one establishing, once and for all eternity, the term 'simple' as the unique way to refer to things without subthings. –  Mariano Suárez-Alvarez Aug 23 '10 at 17:04
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After they finally classified the finite simple groups, some emperor comes along and wants to redefine "simple" to prohibit sub-things (presumably meaning nontrivial sub-things) rather than (non-trivial) quotient things. I suppose I should rejoice that, in this empire, even I can classify the simple groups. –  Andreas Blass Aug 23 '10 at 19:53
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The non-abelian group of order 6 is a counterexample in characteristic 2. It is probably wise to have at least worked with this example. –  Jack Schmidt Aug 24 '10 at 1:07

3 Answers 3

up vote 6 down vote accepted

Jim has already given the correct answer "no".

Here is a hopefully instructive example. Let $k$ be an alg. closed field of pos. char $p$, and let $G = SL_2(k)$. Write $V = k^2$ for the "natural" 2-dimensional representation of $G$ say with basis $e_1,e_2$. Let $W = S^pV$ be the $p$-th symmetric power of $V$. Then $W$ contains a 2 dimensional submodule $A$ spanned by the $p$-th powers $e_1^p$ and $e_2^p$; the module $A$ is isom. to the "first Frobenius twist" of $V$.

It is an exercise to check that there is no $G$-stable complement to $A$ in $W$; i.e. the SES $$0 \to A \to W \to W/A \to 0$$ is not split. Thus $W$ is not completely reducible. Evidently there is a surjective mapping $V^{\otimes p} \to W$, thus also the $p$-th tensor power $V^{\otimes p}$ is not completely reducible.

But $V$ is a simple (hence completely reducible) $G$-module; thus tensor powers of a completely reducible module are not in general completely reducible. In fact, the $(p-1)$-th tensor power $V^{\otimes p-1}$ is completely reducible; arguing as before, one sees that $V \otimes (V^{\otimes p-1})$ is not completely reducible; thus in general the tensor product of two completely reducible modules is not completely reducible.

I gave some further remarks about semisimplicity of tensor products in an answer to this question.

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Actually, the symmetric powers were the thing I was interested in first. But your group is not finite ;) –  darij grinberg Aug 23 '10 at 19:54
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The story shouldn't change upon replacing alg closed $k$ by the finite field $F=\mathbf{F}_p$ (or $F = \mathbf{F}_q$...). The $F$-points $V(F)$ of $V$ form an abs irred module (over $F$) for the finite group $G(F)$, and the $F$-points of $W$ given by $W(F) = S^p V(F)$ are not completely reducible as $G(F)$-module, at least if #$F$ >> 0. (Well, $W$ is indecomposable as $G$-module, hence for #$F$ large, $W(F)$ is indec as $G(F)$-module. Since I haven't thought this through recently, I fret a bit about the indecomposability of $W(F)$ as $G(F)$-module for tiny $F$.). –  George McNinch Aug 23 '10 at 20:25

The answer to your question is usually no (which is fortunate because the lack of complete reducibility gives modular representation theorists something to do), starting for example with the tensor product of two irreducible representations of $G$ over an algebraically closed field whose prime characteristic divides the group order. Examples for finite groups of Lie type are legion and come up naturally when you tensor the Steinberg representation with an arbitrary one: then you get a projective module whose indecomposable direct summands are rarely irreducible. Textbooks like those by Jon Alperin, Curtis-Reiner, Serre, or me on modular representations illustrate such outcomes of tensoring.

ADDED: Concerning failure of complete reducibility in general, see also the related MO question 18280. For references to some older literature on tensoring with the Steinberg representation, see the third section of my 1987 AMS Bulletin survey here.

http://www.ams.org/journals/bull/1987-16-02/S0273-0979-1987-15512-1/S0273-0979-1987-15512-1.pdf">here.

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Mea culpa for giving a bogus argument for the wrong answer above. Jim, apart from the rationale I gave in terms of poor behavior of Jacobson radical with respect to the comultiplication $K[G] \rightarrow K[G] \otimes K[G]$ to explain how it can fail, is there a better "ring-theoretic" explanation for this ubiquitous phenomenon? –  BCnrd Aug 23 '10 at 17:43
    
@BCnrd: I haven't seen a ring-theoretic approach to this, which may get complicated: some fairly nontrivial tensor products do turn out to be completely reducible. (In other words, "ubiquitous" is tricky here.) Historically, the fact that tensor products of modules for group algebras arise from Hopf algebra structure wasn't so explicit. By now there are other interesting classes of Hopf algebras for which complete reducibility is also an issue. –  Jim Humphreys Aug 23 '10 at 19:07

I claim that semisimple KG-modules are closed under tensor product in the modular setting iff G has a unique p-Sylow subgroup where p is the characteristic.

Pf. Let P be the p-radical of G. That is P is the largest normal p-subgroup of G. It is well known that P is the intersection of the kernels of all irreps of G over K. So we have $$KG\to K[G/P]\to KG/Rad(KG).$$

If P is a p-Sylow then $K[G/P]$ is semisimple by Maschke and so the last map is an isomorphism. Thus Rad(KG) is a Hopf ideal and so the completely reducible reps are closed under tensor product.

On the other hand if the completely reducible reps are closed under tensor, the radical is a Hopf ideal. Since the Hopf algebra quotients of a group algebra are the algebras of quotient groups it follows the last map is an iso (since g-1 is in the radical iff g is in P). But then by Maschke p does not divide the order of G/P so P is a p-Sylow.

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Oh, so what you call the $p$-radical is the $p$-core, as far as I understand. I fear I need some more proofs or references here. I've got a reference for the fact that completely reducible reps are closed under tensor products if and only if the Jacobson radical is a Hopf ideal (Satz 5.3 in Theresia Nolte's diploma thesis math.rwth-aachen.de/~Gerhard.Hiss/Students/… ). But I'm missing a proof that $P$ is the intersection of the kernels of all irreps of $G$ over $K$. (This generalizes the fact that all irreps of a $p$-group over $K$ are trivial, but ... –  darij grinberg Dec 28 '12 at 2:51
    
... the proof of this fact that I know doesn't carry over.) –  darij grinberg Dec 28 '12 at 2:51
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There are 2 proofs on mathoverflow.net/questions/69039/… that every normal p-subgroup is contained in the kernel of each irrep. Now consider the regular rep of G. It can be written in block triangular form with the diagonal blocks irreducible reps. The kernel of the projection to the diagonal is precisely the intersection of the kernels of the irreps. But the kernel is unitriangular hence a p-group. –  Benjamin Steinberg Dec 28 '12 at 5:17
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The last map follows because g-1 is in the radical iff g is in the kernel of each irrep. By above this occurs iff g is in P. The kernel of $KG\rightarrow K[G/P]$ is generated by the elements g-1 with g in P. –  Benjamin Steinberg Dec 28 '12 at 5:20
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math.wisc.edu/~passman/balgebra.pdf is a good reference for the Hopf ideal result and the fact that each Hopf ideal in a group algebra is generated by the elements g-1 ranging over some g in some normal subgroup N. The moral is that the largest Hopf ideal contained on the radical is generated by g-1 with g in the p-radical. –  Benjamin Steinberg Dec 28 '12 at 5:24

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