# The Unapologetic Mathematician

## Characters of Induced Representations

We know how to restrict and induce representations. Now we want to see what this looks like on the level of characters.

For restricted representations, this is easy. Let $X$ be a matrix representation of a group $G$, and let $H\subseteq G$ be a subgroup. Then $X\!\!\downarrow^G_H(h)=X(h)$ for any $h\in H$. We just consider an element of $H$ as an element in $G$ and construct the matrix as usual. Therefore we can see that

\displaystyle\begin{aligned}\chi\!\!\downarrow^G_H(h)&=\mathrm{Tr}\left(X\!\!\downarrow^G_H(h)\right)\\&=\mathrm{Tr}\left(X(h)\right)\\&=\chi(h)\end{aligned}

That is, we get the restricted character by restricting the original character.

As for the induced character, we use the matrix of the induced representation that we calculated last time. If $X$ is a matrix representation of a group $H$, which is a subgroup $H\subseteq G$, then we pick a transversal of $H$ in $G$. Using our formula for the induced matrix, we find

\displaystyle\begin{aligned}\chi\!\!\uparrow_H^G(g)&=\mathrm{Tr}\left(X\!\!\uparrow_H^G(g)\right)\\&=\mathrm{Tr}\left(\begin{array}{cccc}X(t_1^{-1}gt_1)&X(t_1^{-1}gt_2)&\cdots&X(t_1^{-1}gt_n)\\X(t_2^{-1}gt_1)&X(t_2^{-1}gt_2)&\cdots&X(t_2^{-1}gt_n)\\\vdots&\vdots&\ddots&\vdots\\X(t_n^{-1}gt_1)&X(t_n^{-1}gt_2)&\cdots&X(t_n^{-1}gt_n)\end{array}\right)\\&=\sum\limits_{i=1}^n\mathrm{Tr}\left(X(t_i^{-1}gt_i)\right)\\&=\sum\limits_{i=1}^n\chi(t_i^{-1}gt_i)\end{aligned}

Where we define $\chi(g)=0$ if $g\notin H$. Now, since $\chi$ is a class function on $H$, conjugation by any element $h\in H$ leaves it the same. That is,

$\displaystyle\chi(h^{-1}gh)=\chi(g)$

for all $g\in G$ and $h\in H$. So let’s do exactly this for each element of $H$, add all the results together, and then divide by the number of elements of $H$. That is, we write the above function out in $\lvert H\rvert$ different ways, add them all together, and divide by $\lvert H\rvert$ to get exactly what we started with:

\displaystyle\begin{aligned}\chi\!\!\uparrow_H^G(g)&=\frac{1}{\lvert H\rvert}\sum\limits_{h\in H}\sum\limits_{i=1}^n\chi(h^{-1}t_i^{-1}gt_ih)\\&=\frac{1}{\lvert H\rvert}\sum\limits_{h\in H}\sum\limits_{i=1}^n\chi\left((t_ih)^{-1}g(t_ih)\right)\end{aligned}

But now as $t_i$ varies over the transversal, and as $h$ varies over $H$, their product $t_ih$ varies exactly once over $G$. That is, every $x\in G$ can be written in exactly one way in the form $t_ih$ for some transversal element $t_i$ and subgroup element $h$. Thus we find:

$\displaystyle\chi\!\!\uparrow_H^G(g)=\frac{1}{\lvert H\rvert}\sum\limits_{x\in G}\chi(x^{-1}gx)$