If $\chi^{\lambda}$ and $\chi^{\mu}$ are the characters of two irreducible representations $V^{\lambda}$ and $V^{\mu}$ of a finite group $G$, is there a simple way of proving that :
$$ \chi^{\lambda} *\chi^{\mu} = \delta_{\lambda, \mu} \frac{|G|}{\dim V^{\lambda}} \,\chi^{\lambda}$$
where $\chi^{\lambda} *\chi^{\mu}(\sigma)=\sum_{g\in G}\chi^{\lambda}(\sigma g^{-1})\chi^{\mu}(g)$ is the product of convoluition and $\delta_{\lambda, \mu}$ equals 1 if $\lambda=\mu$ and 0 if $\lambda\neq\mu$.
I'm actually interested in representations of the symetric group $\mathcal{S}_n$, and this relation was instrumental in the definition of an isomorphism between the center of $\mathbb{C}[\mathcal{S}_n]$ and the complex functions on the conjugacy classes of $\mathcal{S}_n$.
I found a demonstration on this page
http://drexel28.wordpress.com/2011/03/02/representation-theory-using-orthogonality-relations-to-compute-convolutions-of-characters-and-matrix-entry-functions/:
\begin{aligned}\left(\chi^{(\alpha)}\ast\chi^{(\beta)}\right)(x)
&= \sum_{g\in G}\chi^{(\alpha)}\left(xg^{-1}\right)\chi^{(\beta)}(g)\\
&=\sum_{g\in G} \sum_{p=1}^{d_\alpha}\sum_{q=1}^{d_\beta}D^{(\alpha)}_{p,p}\left(xg^{-1}\right)D^{(\beta)}_{q,q}(g)\\
&=\sum_{p,s=1}^{d_\alpha}\sum_{q=1}^{d_\beta}D^{(\alpha)}_{p,s}(x)\sum_{g\in G}\overline{D^{(\alpha)}_{p,s}(g)}D^{(\beta)}_{q,q}(g)\\ &= \sum_{p,s=1}^{d_\alpha}\sum_{q=1}^{d_\beta}D^{(\alpha)}_{p,s}(x)\frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\delta_{p,q}\delta_{q,s}\\
&=\frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\sum_{p,s=1}^{d_\alpha}D^{(\alpha)}_{p,s}(x)\delta_{p,q}\delta_{p,s}^2\\
&= \frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\sum_{p=1}^{d_\alpha}D^{(\alpha)}_{p,p}(x)\\
&= \frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\chi^{(\alpha)}(x)\end{aligned}
but it is not clear for me why $\sum_{g\in G}\overline{D^{(\alpha)}_{p,s}(g)}D^{(\beta)}_{q,q}(g)$ is equal to $\frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\delta_{p,q}\delta_{q,s} $. It seems to me that it is a not-so-trivial consequence of Schur's lemma. So my question is : Is there a simpler way of proving this relation about convolution of characters ?
Best Answer
Not only do characters enjoy orthogonality relations, the entries of any two matrix representations also exhibit orthogonality. Indeed, the relations for characters follow as corollary to the latter.
First, a quick corollary of Schur's. Let $A$ be an irreducible matrix representation of $G$ over $\mathbb{C}$, and suppose $T$ is a matrix that commutes with $A(g)$ for all $g\in G$. Then the same is true for $T-cI$ for any $c\in\Bbb C$; choosing an eigenvalue $c$ of $T$ means that $T-cI$ is noninvertible, so it must be the zero matrix by Schur's, and hence $T$ is a scalar multiple of the identity matrix. (For arbitrary fields $k$ this can be adapted by tensoring $V$ with the algebraic closure $k^{\operatorname{alg}}$ and then restricting back down to $k$, as long as the representation $A$ is absolutely irreducible (i.e., stays irreducible upon tensoring with $k^{\operatorname{alg}}$). If it is not, then this is not generally true.)
Let $k$ be a field, $A:G\to M_{r\times r}(k)$ and $B:G\to M_{s\times s}(k)$ two irreducible matrix representations of $G$, and then finally $X$ an $r\times s$ matrix of unknowns. Consider the matrix
$$Y=\sum_{g\in G} A(g)XB(g^{-1}). \tag{1}$$
With $h\in G$ arbitrary, we have
$$\begin{array}{c l} A(h)Y & =\sum_{g\in G} A(hg)XB(g^{-1}) \\ & = \sum_{g\in G} A(g)XB\big(\underbrace{(h^{-1}g)^{-1}}_{g^{-1}h}\big) \\ & = \sum_{g\in G}A(g)XB(g^{-1})B(h) \\ & = YB(h). \end{array} \tag{2}$$
If $Y$ is invertible then $A\cong B$ are equivalent representations. Otherwise, if $A\not\cong B$ are inequivalent, we find $Y$ is not invertible and so by Schur's lemma is the zero matrix. Assume the latter case. Then the $(i,j)$ entry of the equation $Y=0$ is
$$\sum_{g\in G}\sum_{k,l} a_{ik}(g)x_{kl}b_{lj}(g^{-1})=\sum_{k,l}x_{kl}\left(\sum_{g\in G}a_{ik}(g)b_{lj}(g^{-1})\right)=0. \tag{3}$$
The $x_{lk}$'s are arbitrary unknowns however, yet the equality above holds regardless, therefore the coefficient of each is zero. In other words, $\langle a_{ik},b_{lj}\rangle_G=0$ when $A,B$ are inequivalent. (This is the inner product defined on class functions of $G$, i.e. functions that are constant on conjugacy classes.)
Otherwise we might as well say $A=B$ (if they're equivalent it's not going to make any difference to the character's values anyway), and we consider
$$\operatorname{tr}Y=\operatorname{tr} \frac{1}{|G|}\sum_{g\in G} A(g)XA(g^{-1})=\frac{1}{|G|}\sum_{g\in G}\operatorname{tr} X=\operatorname{tr}X. \tag{4}$$
Let $k$ be algebraically closed. Since $Y$ is a scalar multiple of the identity, we have $y_{ii}=\operatorname{tr}X/r$. Hence we write $Y=\frac{\operatorname{tr}X}{r}I$ as
$$\frac{1}{|G|}\sum_{k,l}x_{kl}\left(\sum_{g\in G}a_{ik}(g)a_{lj}(g^{-1})\right)=\begin{cases} \frac{x_{11}+\cdots+x_{rr}}{r} & i=j \\ 0 & \text{otherwise}.\end{cases} \tag{5}$$
Equating coefficients above and then putting this together with $A,B$ inequivalent, we have derived
$$\langle a_{ik},b_{lj}\rangle_G=\frac{\delta_{AB}\delta_{ij}\delta_{kl}}{r}. \tag{6}$$
Note that in $\Bbb C$, $B(g^{-1})=\overline{B(g)^T}$ because $B$ is unitary, which switches the indices in $b_{\circ\circ}$ above too.
This is still a brutish method of index juggling, and I'm not familiar with an easier way in these matters. (Of course I wouldn't though; I just came across the above a few hours ago.) Perhaps someone else has enlightenment on that count.