A Mind for Madness

Musings on art, philosophy, mathematics, and physics

A-W Consequences

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I said I’d do the uniqueness part of Artin-Wedderburn, but I’ve decided not to prove it. Here is the statement: Every left semisimple ring R is a direct product R\cong M_{n_1}(\Delta_1)\times\cdots \times M_{n_m}(\Delta_m) where \Delta_i are division rings (so far the same as before), and the numbers m, n_i, and the division rings \Delta_i are uniquely determined by R.

The statement here is important since if we can figure one of those pieces of information out by some means, then we’ve completely figured it out, but I think the proof is rather unenlightening since it is just fiddling with simple components.

Let’s use this to write down the structure of kG where G is finite and k algebraically closed with characteristic not dividing |G|. This is due to Molien: then kG\cong M_{n_1}(k)\times\cdots \times M_{n_m}(k).

By Maschke we know that kG is semisimple, and by Artin-Wedderburn, we get then that kG\cong\prod M_{n_i}(\Delta_i). In fact, the proof of Artin-Wedderburn even tells us that \Delta_i=End_{kG}(L_i) where L_i is a minimal left ideal of kG. Thus, given some minimal left ideal L, it suffices to show that \Delta=End_{kG}(L)\cong k.

Note that \Delta is a subspace of kG as a vector space over k. Thus it is finite dimensional. Now we have both L and \Delta as finite dimensional vector spaces (over k). Let a\in k, then this element acts on L by u\mapsto au. But au=ua, so k\subset Z(\Delta). Choose d\in \Delta, then adjoin it to k: k(d). Since this is commutative, and a subdivision ring, it is a field. i.e. k(d)/k as a field extension is finite, and hence algebraic, so d is algebraic over k. But we assumed k algebraically closed, so d\in k. Thus \Delta=k, and we are done.

As a Corollary to this, we get that under the same hypotheses, |G|=n_1^2+\cdots + n_m^2. This is just counting dimensions under the isomorphism above, since dim_k(kG)=|G| and dim_k(M_{n_i})=n_i^2. Note also that we can always take one of the n_i to be 1, since we always have the trivial representation.

Let’s end today with an example to see how nice this is. Without needing to peek inside or know anything about representations of S_3, we know that \mathbb{C}S_3\cong \mathbb{C}\times\mathbb{C}\times M_2(\mathbb{C}), since the only way to write 6 as the sum of squares is 1+1+1+1+1+1, or 1+1+4, and the first one gives \mathbb{C}^6 which is abelian which can’t happen since S_3 is non-abelian. Thus it must be the second one.

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

One thought on “A-W Consequences

  1. Artin-Wedderburn is definitely a cleaner and more natural approach to some of these results in group representation theory than characters, but I think you do need characters in some cases and applications.

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