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Given a finite group G, its complexified ring of finite-dimensional complex representations is isomorphic to its algebra of class functions, by the trace map $\mathrm{Tr}_\rho: g \mapsto \mathrm{Tr}(\rho(g))$. These class functions can in turn can be shown to correspond to the elements in the center of the group algebra, by sending a class function $c:G \to \mathbb{C}$ to the element $\sum_{g \in G} c(g) \cdot g$ in the group algebra.

This map from class functions to the group algebra isn't a ring homomorphism. In particular, it doesn't preserve the unit. However, I'm pretty sure that the representation ring and the center of the group algebra are nevertheless isomorphic as algebras. Am I right? Even better, is there a canonical such homomorphism with some group-theoretical importance?

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They should be canonically dual, if anything. Both are Hopf algebras and the multiplication on one corresponds to the comultiplication on the other. –  Qiaochu Yuan Jun 2 '10 at 23:11
    
See perhaps section 6.3 of Serre. –  Steve D Jun 2 '10 at 23:47
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The two algebras are finite dimensional semisimple (check that they do not have nilpotents, for example) and commutative of the same dimension, so they are isomorphic. –  Mariano Suárez-Alvarez Jun 3 '10 at 2:09
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(Serre is, in all likelyhood, Serre's book on representations of finite groups in Steve's comment---as far as I know Serre himself is not divided in sections!) –  Mariano Suárez-Alvarez Jun 3 '10 at 2:10
    
@Mariano: LOL, yes that's what I meant. The isomorphism is constructed there, but it isn't canonical. –  Steve D Jun 3 '10 at 4:07
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2 Answers 2

up vote 5 down vote accepted

Off the top of my head: the representation ring is really indexed by the set of irreps (well, OK, a set of representatives from equivalence classes thereof) and is indeed commutative and semisimple (at least over the complex numbers) so isomorphic to an appropriate direct product of copies of the ground field.

The centre of the group algebra is the (sub)algebra of class functions, so is most naturally indexed by the set of conjugacy classes. Again, it is commutative and semisimple so is isomorphic to a direct product of the appropriate number of copies of the ground field.

Thus, while I agree that the two algebras are isomorphic when you have a finite group, the isomorphisms seems quite heinously uncanonical -- if I am correct in recalling the maxim that there is no canonical bijection between the set of conjugacy classes and a representative set of irreps.

I also have the feeling that if there were some "canonical" homomorphism from the representative ring to the group ring in the case of finite groups, then there should be some analogue for compact groups or abelian groups. But this seems not to be the case: consider, for example, the group of integers with its usual additive structure.

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Thanks for that comment, Yemon. I have a vague feeling that the Fourier transform should provide just such a canonical homomorphism, but the details are slipping through my fingers at present. –  Jamie Vicary Jun 2 '10 at 23:46
    
Jamie: different multiplication in general, I think. (I am more familiar with all this in the context of Fourier algebras and L^1-convolution algebras, but when the group is finite this word "Banach" can be dropped largely without qualm.) Convolution of class functions (on the group algebra side) would be taken by the Fourier transform to the pointwise product -- and we know that isn't how the representation ring works, because the "product" of two irreps in that ring is the formal decomposition of their tensor product as a combination of irreps, not the "Kronecker delta" product. –  Yemon Choi Jun 3 '10 at 0:10
    
The Fourier transform should provide a homomorphism between L^2(one thing) and L^2(the other), roughly speaking, not between the two things themselves. –  Qiaochu Yuan Jun 3 '10 at 3:25
    
But since everything is finite, $L^2(L^2(X)) \simeq X$ --- so what's the distinction? –  Jamie Vicary Jun 3 '10 at 10:44
    
Jamie, that is not true of most $X$s! Consider. for example, the case where $X=\{p\}$ is a one elemnt set. –  Mariano Suárez-Alvarez Jun 3 '10 at 13:53
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The centre of the group algebra has also a basis indexed by irreducible representations, the basis of central irreducible idempotents. But the corresponding natural linear map is not an isomorphism (irreducibles are not usually idempotents) unless you the group is an elementary abelian 2-group. So the natural map is some sort of Fourier transformation as pointed out by Qiaochu.

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