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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.

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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

Yes, see Prop. 3.1 and Remark 3.2 here.

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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

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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

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