I am looking for a reference for the following assertion that I believe to be true. All representations are assumed to be finite-dimensional.

Let $G_1$ and $G_2$ be affine algebraic group schemes over a field $k$ such that every simple $G_2$-representation $V_2$ has endomorphism ring $\mathrm{End}_{G_2}(V_2) = k$. Then for any two simple representations $V_1$ and $V_2$ of $G_1$ and $G_2$, respectivey, $$V_1\otimes V_2$$ is a simple $G_1\times G_2$-representation. Moreover, any simple $G_1\times G_2$-representation is of this form.

The corresponding statements regarding complex representations of compact Lie groups are given as Lemmas 3.66 and 3.67 in Adams’ Lectures on Lie Groups, and may probably also be found in any other relevant textbook.

For the given statement concerning affine algebraic groups, I believe to have a proof based on the following lemma in Jantzen's Representations of Algebraic Groups (see 6.15 (2)): Let $V$ and $W$ be representations of an affine algebraic group scheme $G$ over a field $k$. Suppose $V$ restricts to a simple representation of a normal subgroup scheme $N\triangleleft G$ such that $\mathrm{End}_{N}(V) = k$. Then $(\mathrm{soc}_N W)_V$, the $V$-isotypical part of the $N$-socle of $W$, is a $G$-sub-representation of $W$, and we have an isomorphism of $G$-modules $\mathrm{Hom}_N(V,W)\otimes V \cong (\mathrm{soc}_N W)_V$.

However, I have not been able to locate the above statement regarding products as such in the literature, which I still find a little puzzling.

  • $\begingroup$ See also mathoverflow.net/questions/80558/… $\endgroup$ – Dietrich Burde Jul 27 '13 at 12:23
  • $\begingroup$ Thanks. I had actually seen that post. The discussion there is related to the fact that $V_1\otimes V_2$ indeed is a $G_1\times G_2$-representation, but there is no assertion of simplicity etc. $\endgroup$ – Communicative Algebra Jul 27 '13 at 14:15
  • $\begingroup$ For the time being, I've included my own argument as Proposition 4.1 in arXiv:1308.0796. $\endgroup$ – Communicative Algebra Aug 7 '13 at 7:39
  • $\begingroup$ @CA: I'm not sure offhand what is written down explicitly in books or such, but I guess the argument for simplicity is the same as for abstract groups. But people don't always need the general case, so for instance Serre's Part I on finite group representations over $\mathbb{C}$ just gives an ad hoc proof using the group algebra and complete reducibility. (P.S. The term "Communicative Algebra" actually appears prominently by mistake in a French chapter of the Bourbaki series on commutative algebra.) $\endgroup$ – Jim Humphreys Aug 8 '13 at 22:44
  • $\begingroup$ @JimHumphreys: Thanks for helping me trace my roots! I don't like being discredited as a “mistake”, though. Eisenbud has dedicated a whole monograph to me, available exclusively from the University Library of Wuppertal (see this larger version of my profile picture.) And in any case, I mathoverflow, so I am. $\endgroup$ – Communicative Algebra Aug 11 '13 at 6:51

I may have a more "direct" argument for the first part of the statement. Assume $V_1$ and $V_2$ are simple. Let $W$ be a sub-$G_1\times G_2$-representation of $V_1\times V_2$. Considering it as a $G_2$-representation and using the assumption on endomorphisms of $V_2$, one can obtain that $W=M\otimes V_2$ for some vector subspace $M$ of $V_1$. Then, one can show that $M$ is a sub-$G_1$-representation of $V_1$. To do that, one may choose for instance a nonzero linear form $\phi$ on $V_2$, which induces a map of $G_1$-representations $id\otimes\phi\colon V_1\otimes V_2\to V_1$. Then, the composition $M\otimes V_2\to V_1\otimes V_2\to V_1$ is a map of $G_1$-representations whose image is $M$. If $M$ is nonzero and $V_1$ simple, one obtains that $M=V_1$ and $W=V_1\otimes V_2$. Thus, $V_1\otimes V_2$ is simple.

  • $\begingroup$ This argument is indeed more direct—and, I'd say, more morally correct. (My own argument, which I've linked to in the comments to the question, deduces the first part of the statement from the second.) $\endgroup$ – Communicative Algebra Aug 11 '13 at 6:28

Out of curiosity I consulted a Usally Reliable Source, who pointed to an explicit older argument in Steinberg's 1967-68 Yale lecture notes here: see pages 205-206 at the beginning of $\S12$. Though Steinberg assumes an algebraically closed field, in order to invoke Schur's Lemma, Jantzen's Lemma I.6.15 which you mention incorporates your alternate assumption on End spaces. (As noted, the result you want can be deduced from Jantzen's formulation but isn't as straightforward as Steinberg's.)

When working with affine algebraic groups (or group schemes) the emphasis naturally shifts to algebras of regular functions and their comodules, etc. But the general principle here has certainly been around for a long time in various parts of representation theory. Presumably it started over $\mathbb{C}$ for finite groups and Lie groups but then got more general as modular representation theory developed. (Curtis and Reiner may have written down a suitable version in one of their books, but I don't immediately see where.)

Anyway, it's perfectly reasonable in a research paper to sketch a proof in your own language "for the convenience of the reader" without claiming that the result itself is new.

  • $\begingroup$ Thank you for following up on this! Steinberg's lecture notes certainly seem a reference worth mentioning. On the other hand, I always find it irritating when a claim is proved by reference to a source that relies on stronger assumptions—it always leaves me wondering whether the author has really thought about the question or is simply being sloppy. I guess the last paragraph still outlines the best solution. $\endgroup$ – Communicative Algebra Aug 11 '13 at 6:36

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