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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 ?

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  • $\begingroup$ Do you need the full strength of the formula, or are you just in need of the orthogonality relations enjoyed by characters? Either way, we have already developed a purely algebraic approach to this relation, through the use of structure theorems and double centralizer theorems for non-commutative algebras. But, after that development, the proof is quite simple. In fact, that approach $\endgroup$
    – awllower
    Jun 1, 2012 at 7:33
  • $\begingroup$ even said more about characters: they can express the primitive central idempotent elements of the module A in terms of characters. And this of course leads to general relations between representation matrices, for those elements are orthonormal. $\endgroup$
    – awllower
    Jun 1, 2012 at 7:37

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much for the clarification of the orthogonality of matrix coefficients as a consequence of Schur's lemma. Thus the proof is a little bit longer than what i expected, but we cannot do better using coefficients. $\endgroup$
    – saposcat
    Jun 1, 2012 at 13:40
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Actually Awllower replied to my question.

The theorem 2.13 page 19 in "Character Theory of finite groups" by I. Martin Isaacs, is exactly the relation I wanted, and the author prove it in a pure algebraic way, though not as elementary as I expected.

The relation on the convolution product of irreducible characters appears here to be a consequence of a decomposition of the primitive central idempotent elements (noted $e_i$ in Isaac's book) of $\mathbb{C}[G]$ in terms of irreducible characters, as awllower said.

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