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Say I have two finite groups G and H which aren't isomorphic but have the same character table (for example, the quaternion group and the symmetries of the square). Does this mean that the corresponding categories of finite dimensional complex representations are isomorphic (ignoring the forgetful functor to vector spaces), or just that the corresponding representation rings are?

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Isomorphic categories would be surprising. Maybe what you want to ask is whether they have equivalent categories of representations? –  Mikael Vejdemo-Johansson Oct 14 '09 at 17:58
    
in the particular case of $D_8$ and $Q_8$ if you look at the values of the Adams $\psi^k$ operations on the characters you get different results. so as mentioned below, the character table is a shadow of what is going on. Note that those operations do tell you something about the category of finite dimensional representations, which is obvious from their definition. –  Sean Tilson Nov 18 '10 at 0:47
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In the particular case of the non-abelian groups of order 8, their categories of modules are not equivalent as monoidal categories. That they're not equivalent as pivotal categories can be proved by looking at the Frobenius-Schur indicator (I learned this from a paper of Susan Montgomery). That they're not equivalent even as monoidal categories can be proved by counting the fiber functors to vector spaces and seeing that one has more in one case (I can't remember which paper I saw this in, but almost surely Pavel Etingof was one of the coauthors).

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"(I can't remember which paper I saw this in, but almost surely Pavel Etingof was one of the coauthors)." Right, because that narrow it down a lot. –  Ben Webster Oct 14 '09 at 18:46
    
I'm also willing to bet that the coauthors on the paper are a nonempty strict subset of Gelaki, Nikshych, and Ostrick. –  Noah Snyder Oct 14 '09 at 19:41
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This is a great question, and the answer leads to one of the best arguments for why category theory should be studied at all!

Every undergraduate mathematician should discover for themselves that character tables alone don't determine finite groups --- and then, just as their faith in the beauty of mathematics is about to shatter, they should be reassured that character tables are just a 'shadow' of the group's compact monoidal category of representations, and that DOES determine the group (or in general, groupoid).

The procedure for reconstructing a groupoid, up to equivalence, from its category of unitary complex representations, is stunningly beautiful: if G is our groupoid and Rep(G) is its representation category, then construct the groupoid which has objects given by symmetric monoidal functors Rep(G)-->Rep(1), and morphisms given by monoidal natural transformations between them. Here, Rep(1) is just the category representations of the trivial group --- in other words, just the category of finite-dimensional Hilbert spaces, with monoidal structure given by tensor product. This is known as "Doplicher-Roberts style" reconstruction, and the best reference is Muger's appendix to this paper. It's more elegant than "Tannakian" reconstruction, as there's no need to start with a given fiber functor (i.e., a specified functor Rep(G)-->Rep(1)).

This should remind you strongly of the way you recover a compact topological space from the commutative C*-algebra of functions from that space into the complex numbers ... and there are indeed deep connections!

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They're not necessarily equivalent as tensor categories.

However, there are examples of finite groups (smallest of order 64) with representation categories which are equivalent as tensor categories but not as symmetric tensor categories (see e.g. http://arxiv.org/abs/math/0007196). In other words, in some cases the same abstract tensor category might be endowed with inequivalent symmetric structures (you can think of these as the pullback of the standard symmetry of the category of vector spaces through inequivalent embedding functors).

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What structure do you want to remember on the categories? If you just remember they're abelian categories, any groups with the same number of conjugacy classes will have equivalent categories.

On the other hand, if you remember the forgetful functor to vector spaces, you can get the group back: it's the automorphism group of the forgetful functor itself.

By the way, you get more than just that the representation rings are isomorphic (if you tensor with Q, the isomorphism type of the representation ring also only depends on the number of conjugacy classes), but with the same basis, which is much stronger.

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As far as I know, If you consider the corresponding categories as Tannakian categories, they are not isomophic, for you can rediscover the group from the Tannakian category (as its fundamental gp?)

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