Showing that $\sum_{\rho}\chi_\rho(g)\overline{\chi_\rho(g)} = \frac{|G|}{|C(g)|}$ for a finite group $G$, $g \in G$ and $C(g)$ a conjugacy class

Question

Suppose that it is known that a finite group $G$ has equally many conjugacy classes as it has irreducible representations $\rho_1,\dots,\rho_k$. How can we then show that $\sum_{\rho}\chi_\rho(g)\overline{\chi_\rho(f)} = \frac{|G|}{|C(g)|}$ if $f = g$ and $\sum_{\rho}\chi_\rho(g)\overline{\chi_\rho(g)} = 0$ if $f \neq g$ for $f, g \in G$? We do also know that the irreducible characters $\chi_{\rho_1},\dots,\chi_{\rho_k}$ form an orthonormal basis for the conjugacy class functions $\mathbb{C}_{\text{class}, G}$ of $G$. Therefore, $\sum_{g \in G}\chi_{\rho_i}(g)\overline{\chi_{\rho_j}(g)} = \delta^i_j$. But then how to extend this to the equality $\sum_{\rho}\chi_\rho(g)\overline{\chi_\rho(f)} = \frac{|G|}{|C(g)|}$?

Edit: Since $G$ is finite it must have a finite number of conjugacy classes and thus a finite number of irreducible characters. Hence by orthonormality of the characters (and reordering of finite sums), we know that $\sum_{g\in G}\sum_{\rho}\overline{\chi_{\rho}(g)}\chi_\rho(g) = \sum_{\rho}\sum_{g\in G}\overline{\chi_{\rho}(g)}\chi_\rho(g) = \#\text{Conjugacy classes of } G$. We know from algebra, that the order of the conjugacy class of a group must divide the order of the group.

Currently, I'm stuck at the proof, as I don't know what is the final straw to put these pieces together to show the claimed result.

Answer 1

What you are asking is just the Schur orthogonality relations. But there are some typos in your statements. The exposition below fixes the typos. Let $\chi_{\rho_1},\cdots,\chi_{\rho_k}$ be the irreducible characters (over $\mathbb C$) and $C(g_1),\cdots,C(g_k)$ be the conjugacy classes of $G$, so $$\bigcup_{\ell=1}^k C(g_{\ell})=G,\qquad (1)$$ where $g_1,\cdots,g_k\in G$ are some chosen representatives. Writing the centralizer of $g$ in $G$ by $C_G(g),$ one then has $$|G|=|C_G(g)|\cdot |C(g)|.$$ Since $\chi_{\rho_i}$’s form an orthonormal basis of the space of class functions, one has $$\frac 1{|G|}\sum_{g\in G}\chi_{\rho_i}(g)\overline{\chi_{\rho_j}(g)}=\delta_{ij},$$ which together with (1) implies that $$\frac 1{|G|}\sum_{\ell=1}^k\sum_{g\in C(g_{\ell})}\chi_{\rho_i}(g)\overline{\chi_{\rho_j}(g)}=\sum_{\ell=1}^k\frac{|C(g_{\ell}|}{|G|}\chi_{\rho_i}(g_{\ell})\overline{\chi_{\rho_j}(g_{\ell})}=\delta_{ij}.\qquad (2)$$ The relation in (2) shows that the $k\times k$ matrix $U=(a_{ij})$ defined by $$a_{ij}=\sqrt{\frac{|C(g_j)|}{|G|}}\chi_{\rho_i}(g_j)$$ is unitary. It follows that the column vectors of $U$ are orthonormal, i.e. $$\frac{\sqrt{|C(g_{\ell})|\cdot|C(g_m)|}}{|G|}\sum_{i=1}^k \chi_{\rho_i}(g_{\ell})\overline{\chi_{\rho_i}(g_m)}=\delta_{\ell m},$$ hence for $f,g\in G,$ one has if $f\not\sim g$ (i.e. not conjugate), then $$\sum_{i=1}^k \chi_{\rho_i}(f)\overline{\chi_{\rho_i}(g)}=0,$$ and if $f\sim g,$ then $$\sum_{i=1}^k \chi_{\rho_i}(f)\overline{\chi_{\rho_i}(g)}=\frac {|G|}{|C(g)|}=|C_G(g)|,$$ as required.

Answer 2

While Pythagoras' answer is without flaw, I thought I might present another way of approaching this that might also be insightful.

For this, let $V_\rho$ denote a simple $\mathbb{C}G$-module with character $\rho$ and $V_\rho^* = Hom_{\mathbb{C}}(V_\rho,\mathbb{C})$ its dual representation (which admits character $\overline{\rho}$). We note that $$\mathbb{C}G \to \bigoplus_{\rho} End_\mathbb{C}(V_\rho), g \mapsto (v \mapsto gv)_{\rho} $$ is an isomorphism of $\mathbb{C}$-algebras by Artin-Wedderburn.

Moreover, we have a $\mathbb{C}$-linear isomorphism $$V_\rho \otimes V_{\rho}^* \to End_{\mathbb{C}}(V_\rho), v \otimes f \mapsto (w \mapsto f(w)v).$$

Note that $\mathbb{C}G$, $End_{\mathbb{C}}(V_\rho)$ and $V_\rho \otimes V_\rho^*$ are all $\mathbb{C}(G \times G)$-modules:

• For $\mathbb{C}G$, we can set $(g,h).a = gah^{-1}$,
• for $\varphi \in End_{\mathbb{C}}(V_\rho)$, we can set $((g,h).\varphi )(v) = g \varphi(h^{-1}v)$ and
• for $v \otimes f \in V_\rho \otimes V_\rho^*$ we can set $(g,h) \dot (v \otimes f)= (gv) \otimes hf) $ where $(hf)(w) = f(h^{-1}w)$ for $w \in V_\rho$.

It is easy to see that these bimodule structures are preserved by the above isomorphisms. Thus, we obtain an isomorphism $\mathbb{C}G \cong \bigoplus_{\rho} V_\rho \otimes V_{\rho}^* $ of $\mathbb{C}(G\times G)$-modules.

The $G \times G$-character of $V_\rho \otimes V_{\rho}^*$ is given by $(g,h) \mapsto \rho(g)\overline{\rho(h)}$ and so we find that the $G \times G$-character of $\mathbb{C}G$ is $\chi(g,h) = \sum_\rho \rho(g)\overline{\rho(h)}$.

At the same time, note that the $\mathbb{C}(G\times G)$-module structure of $\mathbb{C}G$ is induced by the action of $G \times G$ on $G$. In particular, $\chi(g,h)$ is just the number of fixed points of $(g,h)$ with respect to this action. Since $gxh^{-1} = x$ is equivalent to $g = x h x^{-1}$, we find that $\chi(g,h) = 0$ if $g$ and $h$ are not conjugate and $\chi(g,h) = |C_G(g)|$ otherwise.