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March 10

Basis for the centre of a complex group algebra

Hello everyone,

A few friends and I are teaching ourselves some simple representation theory at the moment, and I was hoping you could clear a disagreement up. Supposing you're working with group $G\leq GL(n,p)$ (the general linear group), and you're looking at the center of the group algebra over the field $K=\mathbb {C}$ (as in Group Ring) - your wikipedia article says the set of irreducible characters forms a basis for $Z(K[G])$ with respect to the given inner product. My friend claims that if $B_{1},\,\ldots ,\,B_{r}$ form the conjugacy classes of G, then the sum of the elements $s_{j}=\sum _{m\in B_{j}}m$ in each conjugacy class are a basis for $Z(K[G])$ . However, when I tried to use MAGMA to calculate an example, I got a number of the $s_{i}$ to be equal to one another. So, my question is - is it my computing skills which are at fault or is my friend wrong about the conjugacy class sums being a basis? The evidence seems to point strongly to my own calculations being at fault, I simply want to confirm so I can go and track down my mistake! Thankyou in advance :-) Spalton232 (talk) 16:26, 10 March 2011 (UTC)[reply]

The center of K[G] consists exactly of those functions G→K that are constant on each conjugacy class, so your friend is right. Beware that the sum

\textstyle \sum _{m\in B_{j}}m

must happen in the group ring -- that is,

\textstyle \sum _{m\in B_{j}}e_{m}

-- not in the matrix ring M(n,p) that GL(n,p) embeds in. They have quite different additive structures. –Henning Makholm (talk) 17:31, 10 March 2011 (UTC)[reply]

When you say 'happens in the group ring', what precisely do you mean, could you clarify? I don't quite follow, sorry. Say I have my matrices M₁, ..., M_k making up some conjugacy class, how do I obtain my element of the basis? Just 'adding up the matrices' doesn't seem to be working, I presume I'm doing something wrong. Spalton232 (talk) 19:15, 10 March 2011 (UTC)[reply]

The group ring comes with an addition operation and the multiplication operation. Its multiplication operation happens (by design) to coincide with the multiplication in GL(n,p) in the cases where there is a risk of confusion, but the addition is totally different from matrix addition, and the sum you want here means a repeated application of the group ring's addition. As regards addition, the group ring is simply the vector space C^|G|, and the elements you're adding are its canonical basis vectors. Another way to express the result of the sum is the vector (0,0,...,0,1,1,...,1,0,...,0,0), where the positions that have ones in them are those that correspond to members of the conjugacy class. –Henning Makholm (talk) 19:28, 10 March 2011 (UTC)[reply]

I see - so if I were to take two of these sums and 'multiply' them together, s_is_j, we would take each of the matrices in each conjugacy class i and j, multiply them together and get some sum of various multiples of other s_k? I am trying to calculate the structure coefficients for the

(s_{t})_{t=1}^{r}

(or rather, find a general method for calculating them) but I think I am making a mess of it, as evidenced by the above. I imagine it is a triviality once you understand it! Would I calculate it as

c_{rst}=|\{x,y\}\in B_{r}\times B_{s}:xy=z\}|,\,{\text{some }}z\in B_{t}

? Thank you for the help so far, Spalton232 (talk) 20:43, 10 March 2011 (UTC)[reply]

Yes, that's right. –Henning Makholm (talk) 21:46, 10 March 2011 (UTC)[reply]