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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3$\begingroup$ Isomorphic categories would be surprising. Maybe what you want to ask is whether they have equivalent categories of representations? $\endgroup$– Mikael Vejdemo-JohanssonCommented Oct 14, 2009 at 17:58
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1$\begingroup$ 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. $\endgroup$– Sean TilsonCommented Nov 18, 2010 at 0:47
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6 Answers
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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).
answered Oct 14, 2009 at 18:44
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17$\begingroup$ "(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. $\endgroup$– Ben Webster ♦Commented Oct 14, 2009 at 18:46
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4$\begingroup$ I'm also willing to bet that the coauthors on the paper are a nonempty strict subset of Gelaki, Nikshych, and Ostrick. $\endgroup$ Commented Oct 14, 2009 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!
answered Oct 20, 2009 at 10:38
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5$\begingroup$ Could you give a pointer to the deep connections between GNS reconstruction and DR reconstruction? $\endgroup$– LSpiceCommented Mar 28, 2018 at 17:09
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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. arXiv: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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2$\begingroup$ I think it's worth pointing out that the linked Etingof-Gelaki paper doesn't just give such examples, it gives a complete characterization of when two non-isomorphic groups can have equivalent tensor categories (and, there, non-equivalent symmetric tensor categories). $\endgroup$ Commented Oct 16, 2019 at 18:25
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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.
answered Oct 14, 2009 at 18:13
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Late to this game, but the answers here seem to me to have changed the question. Would it not be most faithful to read this as asking of the categories enriched by all the qualities of a module category are equivalent?
First is to agree on what one means by:
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?
I think for me the most direct interpretation is to take this to mean the category of $\mathbb{C}G$-modules.
Take therefore the characterization of module categories by Morita (see [1,4.11]. Here one finds that to be a category of modules you need to be:
- Abelian.
- Grothendieck (some intersection conditions)
- Finitely pro-generated.
Every category of $R$-modules for a ring $R$ has those properties and furthermore, if $P$ is some pro-generator then define $S:=\operatorname{End}(P)$ and you get a ring such that all the objects in this category become $S$-modules and the morphisms become $S$-linear and in fact the $S$-module category is equivalent to your given category. This is what Morita says.
So the question thus becomes, when are two rings $R$ and $S$ going to give equivalent module categories despite perhaps not being isomorphic rings. That is, more-or-less by definition, to say that the rings $R$ and $S$ are Morita equivalent.
So now we have a computable question. When are two group algebras over $\mathbb{C}$ (or over any ring of coefficients) Morita equivalent? Once we compute:
\begin{gather*} \mathbb{C}D_8=\mathbb{C}\oplus \mathbb{C}\oplus \mathbb{C}\oplus \mathbb{C}\oplus M_2(\mathbb{C}) \\ \mathbb{C}Q_8=\mathbb{C}\oplus \mathbb{C}\oplus \mathbb{C}\oplus \mathbb{C}\oplus M_2(\mathbb{C}). \end{gather*}
These are not only Morita equivalent they are isomorphic rings. So these two categories are in fact equivalent.
I think many later distinctions are enriching the intended categories. For example, one could say treat the rings $\mathbb{C}D_8$ and $\mathbb{C}Q_8$ as rings with involutions given by the inverse on elements (Hopf algebras). Then they cease to be isomorphic and you see a difference. Likewise other posts seem to enrich the categories with other qualities. But to me those seem like exceptions to the most superficial interpretation of the question. In any case, those who find confusion may benefit from teasing out the implications of the implicit assumptions.
[1] Pareigis, B. (1970). Categories and Functors. Translated from the German. Pure and Applied Mathematics, Vol. 39. New York: Academic Press.
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$\begingroup$ Isn't the observation that two groups with the same number of irreducible representations give isomorphic module categories a long-winded version of Ben Webster's answer? The dimension of matrix algebras you get is irrelevant as they are all Morita equivalent. $\endgroup$ Commented Nov 2 at 21:14
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$\begingroup$ Ben Webster's clearly thinking along the same lines. However abelian categories need not be module categories. Module categories have much beyond that. I simply meant to point out what additional qualities are defining of module categories. $\endgroup$– AlgeboyCommented Nov 3 at 23:34
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As far as I know, If you consider the corresponding categories as Tannakian categories, they are not isomorphic, for you can rediscover the group from the Tannakian category (as its fundamental gp?)
