# Orthogonality of matrix coefficients in positive characteristic

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 $G$, is to analyze in a convenient way the complex-valued functions on the group. Indeed, suitable “matrix coefficients”
$f(g)=\langle \rho(g)v,w\rangle$
of irreducible unitary representations
$\rho\,:\, G\rightarrow \mathrm{U}(d_{\rho},\mathbf{C}),$
of $G$ turn out to form an orthonormal basis of this space of functions, for the inner product associated to the uniform probability measure on $G$:
$\langle f_1,f_2\rangle=\frac{1}{|G|}\sum_{g\in G}{f_1(g)\overline{f_2(g)}}.$
This fact is obtained from the inner-product relations
$\langle f_{v,w},f_{s,t}\rangle =\frac{\langle v,s\rangle\overline{\langle w,t\rangle}}{\dim(\rho)}$
where
$f_{v,w}=\langle \rho(g)v,w\rangle,\quad\quad f_{s,t}=\langle \rho(g)s,t\rangle,$
(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 $G$ over an algebraically-closed field $k$ of possibly positive characteristic, coprime to $|G|$ however. Then things work out quite similarly as in the complex case, and the analogue of the inner product for $k$-valued functions is
$\langle f_1,f_2\rangle=\frac{1}{|G|}\sum_{g\in G}{f_1(g)f_2(g^{-1})},$
which makes sense since one can divide by the order of $G$ in $k$. The matrix coefficients are defined now by
$f(g)=\langle \lambda,\rho(g)v\rangle,$
for some $k$-linear form $\lambda$ on the space of $\rho$, and some vector $v$ in that space, using the duality-bracket notation.

The inner product formula — with obvious notation — is now
$\langle f_{v,\lambda},f_{w,\mu}\rangle =\frac{\langle \lambda,w\rangle\langle \mu,v\rangle}{\dim(\rho)}.$

The cute thing is that this makes sense because the dimensions $\dim\rho$ of irreducible representations of $G$ over $k$ are also invertible in $k$. (The proof really proceeds by showing the identity
$\dim(\rho)\langle f_{v,\lambda},f_{w,\mu}\rangle =\langle \lambda,w\rangle\langle \mu,v\rangle,$
where the dimension of $\rho$ arises as the trace of the identity matrix). And these dimensions are indeed invertible in $k$ because it is known that they divide the order of $G$, and so our assumption on the field gives the result.

Now for the query: is there a simpler way to see that $\dim(\rho)$ is non-zero in $k$? 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 $k$ and over $\mathbf{C}$ 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 $|G|$ by the dimension of irreducible representations is to show something like
$\alpha \dim(\rho)=|G|,$
for some $\alpha\in k$ by taking the trace of a suitable average of the $\rho(g)$. This shows that if the order of the group is invertible in the base field, so is the dimension.]