Let $G$ be a finite group and $g\in G$. Let $\{\rho_{i}\}$ be the set of irreducible representations of $G$, where $\rho_{i}: G \rightarrow GL(V_{i})$ and $\chi_{i}$ be the character of $\rho_{i}$. Why is it that

$$\delta_{i,j} = \frac{1}{|G|} \sum \operatorname{dim}(V_{i})\chi_{i}(g_{i}^{-1}g_{j})$$

where $\delta_{i,j}$ is the Kronecker Delta? If I just choose a group and try it out, I am convinced it works, but I can't see why. Even in the (easy?) case when $i=j$ and $\chi_{i}(1) = \dim(V_{i})$, I have no intuition.

The only meaningful approach I think that might work is to look at $\rho_{1}\otimes \rho_{2}\otimes \rho_{3}\otimes\dots$ and try to prove properties of its trace, but I don't know what I would even want to prove or if this would help.

  • $\begingroup$ The elements $g_i$ and $g_j$ are fixed and so the index of the summation should denoted by a different index letter, say $k$. Better yet, just denote the group elements $g$ and $h$, as there's simply no need to enumerate the elements of the group. $\endgroup$ – whacka Apr 20 '15 at 17:25

One can write the formula more simply as so:

$$\sum_{V~{\rm irr}} (\dim V)\chi_V(g)=\begin{cases}|G| & g=e \\ 0 & \rm otherwise \end{cases} $$

One can replace $g$ with $h^{-1}g$ and $g=e$ with $g=h$ in order to get the original version. This equality comes from computing the trace of the linear map $x\mapsto gx$ on the regular representation $\Bbb C[G]$ in two different ways.

For the way on the right, the elements of $G$ form a basis and left multiplication by elements of $G$ simply permute these basis elements, so these left multiplication maps are represented by permutation matrices. The trace of a permutation matrix equals the number of $1$s in the diagonal, which is the number of fixed points of the permutation. The maps $x\mapsto gx$ have no fixed points except if $g=e$ in which case all $|G|$ elements of $G$ are fixed points. Thus $\chi_{\Bbb C[G]}(g)=|G|\delta_{g,e}$.

For the way on the left, we need the Peter-Weyl decomposition of the regular representation:

$$\Bbb C[G]\cong \bigoplus_{V~{\rm irr}}V^{\oplus \dim V}.$$

Note that if $G$ acts on $V\oplus W$ then $\chi_{V\oplus W}(g)=\chi_V(g)+\chi_W(g)$ for all $g$. So the above tells us

$$\chi_{\Bbb C[G]}(g)=\sum_{V~{\rm irr}}(\dim V)\chi_V(g).$$

Equating the two formulas for $\chi_{\Bbb C[G]}(g)$ finishes the proof.

Background: I've furnished what I think is a slick proof of the PW decomposition for finite groups, which goes on to use the lemma discussed here to derive the Schur orthogonality relations.

  • $\begingroup$ Thanks! I substantially underestimated how much was necessary to justify that formula. I've never seen any of the ideas in this solution and expected it to be a simpler exercise in using basic definitions. $\endgroup$ – JessicaK Apr 21 '15 at 3:43

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