Let $G$ be a semisimple algebraic group. (I'm already interested in the case $G=SL_2$.)

Let $\mathcal C$ be a semisimple rigid monoidal abelian category endowed with pair of exact tensor functors $Rep_G \to \mathcal C \to Vect$ that whose composition is the forgetful functor $Rep_g \to Vect$. Assume every object of $\mathcal C$ is a summand of an object in the image of this functor $Rep_G \to \mathcal C$.

To check that this functor $Rep_G \to \mathcal C$ is an equivalence, it is sufficient to check that it is full. (Faithfulness follows from the composition being the forgetful functor, and essential surjectivity follows from fullness and the summand condition.)

Is it sufficient to check the fullness condition on finitely many objects of $Rep_G$?

I know the answer is "yes" for $G=SL_n, Sp_{2n},$ or $SO_n$ if $\mathcal C$ is symmetric, at least under the mild assumption that both functors are symmetric tensor functors. Then $\mathcal C$ corresponds by the Tannakian correspondence to a subgroup $H$ of $G$. Then as long as the space of $H$-homomorphisms from the fourth tensor power of the standard representation of $G$ to itself has the correct dimension then $H$ is all of $G$. This is "Larsen's alternative", and you can see why this might be true by considering that as long as the group of $H$-homomorphisms from the adjoint representation of $G$ to itself is one-dimensional, the adjoint representation of $G$ is $H$-irreducible, so the Lie algebra of $H$ is either all the Lie algebra of $G$ or none of it. The second case can be eliminated by considering a slightly more complicated representation of $G$.

Is it still true without using the Tannakian correspondence?


1 Answer 1


In the situation you describe here, the category $\mathcal{C}$ will automatically be symmetric, and the functors will automatically be symmetric functors. The reason is the following: If you have two objects $X$ and $Y$ of $\mathcal{C}$ which are direct summands of $F(A)$ and $F(B)$ respectively (where $F:Rep-G\rightarrow \mathcal{C}$), then we have the following composition of morphisms in $\mathcal{C}$: $X\otimes Y\rightarrow F(A)\otimes F(B)\rightarrow F(A\otimes B)\rightarrow F(B\otimes A)\rightarrow F(B)\otimes F(A)$. By composing with the map $F(B)\otimes F(A)\rightarrow F(B)/Y\otimes F(A)\oplus F(B)\otimes F(A)/X$ and apply the second functor $F':\mathcal{C}\rightarrow Vect$ which is faithful, we see that the first map must split via $X\otimes Y\mapsto Y\otimes X$. One can then show that we get in this way a well defined symmetric structure on $\mathcal{C}$, for which both functors $F$ and $F'$ are symmetric. So your argument about assuming that $\mathcal{C}$ is symmetric and using Tannaka reconstruction is always valid.

About the general question: I only know to say that semisimplicity of $G$ is necessary here: if for example $G=\mathbb{G}_m$ then the collection of subgroups $H_n = n$th roots of unity in $G$ (for different $n$'s) gives a counterexample.

  • $\begingroup$ Thanks, that makes sense! I know how to do the group theory part, that's fine. Do you know what happens if I don't have a functor to Vect? Is there some counterexample you can construct? $\endgroup$
    – Will Sawin
    Feb 5, 2016 at 17:04
  • $\begingroup$ The thing is, that if you have an abelian symmetric monoidal category which has some nice finiteness properties, then there is a theorem of Deligne which tells you how to construct a fiber functor to $Vect$ (or, if you assume some weaker finiteness properties, a fiber functor to $sVect$). See the categories $Rep-GL(t)$ of Deligne for a counterexample, where these finite properties do not hold, and objects have non-intergral dimensions. $\endgroup$
    – Ehud Meir
    Feb 5, 2016 at 17:29
  • $\begingroup$ Yes, but that functor does not necessarily factor through some arbitrary other functor. For instance, I claim every monoidal abelian category containing a rigid object admits a functor from $Rep_{SL_2}$ sending the standard representation to that object. See my other question: mathoverflow.net/questions/188090/non-abelian-freeness-of-su-2 $\endgroup$
    – Will Sawin
    Feb 5, 2016 at 19:13
  • $\begingroup$ By "functor" you mean here additive functor, right? Because then there is a counterexample by taking the second category to be $Vec_K$ and the object to be the unit. $\endgroup$
    – Ehud Meir
    Feb 5, 2016 at 21:51
  • $\begingroup$ Good point. But I think the conditions for it to work are pretty mild. And anyways it's still true that we haven't proved the map factors through the fiber functor. $\endgroup$
    – Will Sawin
    Feb 6, 2016 at 2:31

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