The course I am teaching on representation theory is essentially the first time I’ve taught a topic which I myself never had the occasion to learn from a regular class. Because of this, some things are already coming up which I did not really know, or did not appreciate properly, before. Here is one I find amusing: one of the basic uses of representation theory, for a finite group , is to analyze in a convenient way the complex-valued functions on the group. Indeed, suitable “matrix coefficients”
of irreducible unitary representations
of turn out to form an orthonormal basis of this space of functions, for the inner product associated to the uniform probability measure on :
This fact is obtained from the inner-product relations
(the representations are the same here because, if one uses non-isomorphic ones, the inner product of their respective matrix coefficients are all zero).
Now suppose one considers the irreducible representations of over an algebraically-closed field of possibly positive characteristic, coprime to however. Then things work out quite similarly as in the complex case, and the analogue of the inner product for -valued functions is
which makes sense since one can divide by the order of in . The matrix coefficients are defined now by
for some -linear form on the space of , and some vector in that space, using the duality-bracket notation.
The inner product formula — with obvious notation — is now
The cute thing is that this makes sense because the dimensions of irreducible representations of over are also invertible in . (The proof really proceeds by showing the identity
where the dimension of arises as the trace of the identity matrix). And these dimensions are indeed invertible in because it is known that they divide the order of , and so our assumption on the field gives the result.
Now for the query: is there a simpler way to see that is non-zero in ? My quick look at the books on my shelf show how to derive this from (the easy part of) Brauer characters (which shows in particular that the (multi)set of dimensions over and over are the same) and the corresponding fact for the complex numbers (where the divisibility is proved using simple facts about algebraic integers), but it seems to me that a more straightforward argument might be possible…
[Update (18.03.2011): as many readers certainly realized, this is indeed the case: the start of the proof for the divisibility of by the dimension of irreducible representations is to show something like
for some by taking the trace of a suitable average of the . This shows that if the order of the group is invertible in the base field, so is the dimension.]