Ring homomorphism from the representation ring into group algebra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:42:08Z http://mathoverflow.net/feeds/question/26874 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26874/ring-homomorphism-from-the-representation-ring-into-group-algebra Ring homomorphism from the representation ring into group algebra? Jamie Vicary 2010-06-02T22:05:07Z 2010-06-03T10:27:03Z <p>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 <code>$\mathrm{Tr}_\rho: g \mapsto \mathrm{Tr}(\rho(g))$</code>. 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.</p> <p>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?</p> http://mathoverflow.net/questions/26874/ring-homomorphism-from-the-representation-ring-into-group-algebra/26882#26882 Answer by Yemon Choi for Ring homomorphism from the representation ring into group algebra? Yemon Choi 2010-06-02T23:11:17Z 2010-06-02T23:11:17Z <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/26874/ring-homomorphism-from-the-representation-ring-into-group-algebra/26908#26908 Answer by Bugs Bunny for Ring homomorphism from the representation ring into group algebra? Bugs Bunny 2010-06-03T10:27:03Z 2010-06-03T10:27:03Z <p>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.</p>