Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a field of characteristic zero, and let $G \subseteq \mathrm{GL}_V$ be an algebraic group over $K$, acting faithfully on a finite dimensional vector space $V$. Let $H \subseteq \mathrm{GL}_V$ be the largest algebraic subgroup with the following propertes:

(1) If a subspace $V_1 \subseteq V$ is invariant under $G$, then it is also invariant under $H$.

(2) Given $G$-invariant subspaces $V_1$ and $V_2$ of $V$, and integers $a,b\geq 0$, the equality $$\mathrm{Hom}_G(V_1^{\otimes a}, (V/V_2)^{\otimes b}) = \mathrm{Hom}_H(V_1^{\otimes a}, (V/V_2)^{\otimes b})$$ holds.

The second condition means that $G$ and $H$ have the same fixed points in any tensor space that can be formed out of subquotients of $V$. The inclusion $G\subseteq H$ is tautological, and my question is:

do we have $G=H$?

If $G$ is reductive, then the answer iy yes, because in that case $V$ and all its tensor powers are semisimple, but the equality $G=H$ also holds for example if $G$ is the group of upper triangular matrices.

share|improve this question
Probably the characteristic 0 assumption is needed here to make the question interesting (and it's required especially in the reductive case). But in situations like this I always tend to wonder what can be salvaged in prime characteristic, where much of the set-up still makes sense. –  Jim Humphreys Apr 24 '12 at 0:40
add comment

2 Answers

Yes, see Prop. 3.1 and Remark 3.2 here.

share|improve this answer
add comment

Yes, it is the basic premise on which Tannakian theory rests! see corollary 2.9 (page 20 of the texed version) of Deligne-Milne's article Tannakian categories

share|improve this answer
I do not think that this answers the question. In order to use loc.cit., one should be able to deduce from the conditions (1) and (2) that that the inclusion $G\subseteq H$ induces an equivalence of categories $\mathrm{Rep}(H)\to \mathrm{Rep}(G)$. –  Xandi Tuni Apr 11 '12 at 11:47
Hi, I agree. Sorry about that. Paul's reference is right on target. –  SGP Apr 14 '12 at 19:28
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.