Product of character values

Question

From Isaac's character theory book;

$3.12$ Let $x\in Irr(G)$ and $g,h\in G$. Show that

$$\chi(g)\chi(h)=\dfrac{\chi(1)}{|G|}\sum_{z\in G}\chi(gh^z)$$

I had thought that it was related to $3.9)$;

$$K_iK_j=\sum_{v}a_{ijv}K_v$$ where $K_i$ is the sum of the $i$ th conjugacy classes and

$$a_{ijv}=\dfrac{|\mathfrak{K_i}||\mathfrak{K_j}|}{|G|}\sum_{\chi\in Irr(G)}\dfrac{\chi(g_i)\chi(g_j)\overline{\chi(g_v)}}{\chi(1)}$$

But by using above equalites, I found that $$\chi(g)\chi(h)=\sum_{\chi\in Irr(G)}\dfrac{\chi(g)\chi(h)}{\chi(1)}$$.

I could not proceed more. Any help would be appreciated.

Answer 1

Let $\rho:G\rightarrow GL(V)$ be your irreducible representation.

1-Prove that $f=\frac{\chi(1)}{|G|}\cdot\sum_{z\in G}\rho(z)\rho(h)\rho(z^{-1})$ commutes with every $\rho(t)$ and therefore by Schur's lemma it is a homothety of ratio $\frac{Tr(f)}{\chi(1)}=\chi(h)$.

2-It follows that $\chi(gf)=\operatorname{Tr}(\rho(g)(\chi(h)\cdot1_V))=\chi(h). \chi(g)$

Answer 2

Theorem Let $\chi \in Irr(G)$, and $x,y \in G$, then $$\chi(x)\chi(y)=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi(xy^z)$$

Before proving this theorem we need to set notation and an observation. Wrtie $\mathfrak{X}$ for any representation affording the character $\chi$. Let $x \in G$, then the $\color{blue}{conjugacy \ class}$ of $x$ in $G$ is denoted by $\color{blue}{K_x}$ and the $\color{darkgreen}{sum \ of \ its \ elements}$ as central element of the group algebra $\mathbb{C}[G]$ is denoted by $\color{darkgreen}{\hat{K}_x}$. As is well-known, the Schur Lemma implies the formula $$\mathfrak{X}(\hat{K}_x)=\omega_{\chi}(\hat{K}_x)I$$ where $I$ is the identity matrix of dimension $\chi(1)$, and taking traces yields $$\omega_{\chi}(\hat{K}_x)=\frac{\chi(x)|K_x|}{\chi(1)}$$ Observe that if we let $y$ run over $G$, and look at $x^y$, each element of $K_x$ appears $|C_G(x)|$ times. Hence we have the useful equality in $\mathbb{C}[G]$:

$$|C_G(x)|\hat{K}_x=\sum_{y \in G} x^y$$

Now let us proceed proving the theorem. Working in $\mathbb{C}[G]$ and applying the previous formula, we have $$\sum_{z \in G}xy^z=\sum_{z \in G}xz^{-1}yz=x\sum_{z \in G}y^z=x|C_G(y)|\hat{K}_y$$ This gives $$\mathfrak{X}(\sum_{z \in G}xy^z)=\sum_{z \in G}\mathfrak{X}(xy^z)=\mathfrak{X}(x)|C_G(y)|\omega_{\chi}(\hat{K}_y)I$$ Taking traces at both sides in the formula above gives $$\sum_{z \in G}\chi(xy^z)=\chi(x)|C_G(y)|\omega_{\chi}(\hat{K}_y)=\chi(x)|C_G(y)|\frac{\chi(y)|\hat{K}_y|}{\chi(1)}=\chi(x)\chi(y)\frac{|G|}{\chi(1)}$$ which proves the theorem.$\square$

Corollary Let $\chi \in Irr(G)$. Then

$$\sum_{x,y \in G}\chi([x,y])=\frac{|G|^2}{\chi(1)}.$$

Proof In the Theorem, put $y=x^{-1}$. Then the formula yields $$|\chi(x)|^2=\chi(x)\chi(x^{-1})=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi([x^{-1},z]).$$ Since $\chi$ is irreducible we have $\sum_{x^{-1} \in G}|\chi(x^{-1})|^2=\sum_{x \in G}|\chi(x)|^2=|G|$, so summing over all $x \in G$ in the formula above gives the desired result.$\square$

Corollary Let $\chi \in Irr(G)$, and $x,y \in G$, then $$\chi(x)\chi(y)=\frac{\chi(1)}{|G|^2}\sum_{g,h \in G}\chi(x^gy^h)$$

Proof The theorem tells us that $$\sum_{g,h \in G}\chi(x^gy^h)=\sum_{g\in G}(\sum_{h \in G}\chi(x^gy^h))=\sum_{g\in G}\frac{|G|}{\chi(1)}\chi(x^g)\chi(y)=\sum_{g\in G}\frac{|G|}{\chi(1)}\chi(x)\chi(y)=\frac{|G|^2}{\chi(1)}\chi(x)\chi(y).$$