# Abstract Nonsense

## Different Formula For the Character of an Induced Representation (Pt. II)

Point of Post: This post is a continuation of this one.

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Now, note that if $t_i\ne t_p$ then $t_i^{-1}g\in H$ and $t_p^{-1}g\in H$ can never happen. Indeed, if $t_i^{-1}g,t_p^{-1}g\in H$ then $(t_i^{-1}g)\left(t_p^{-1}g\right)^{-1}=t_i^{-1}t_p\in H$ which is impossible since $t_iH\ne t_p H$. Thus, if $t_i\ne t_p$ then by definition either $t_i^{-1}g\notin H$ or $t_p^{-1}g\notin H$ for every $g\in G$ and thus by definition for every $g\in G$ either $f_j\left(t_i^{-1}g\right)=\bold{0}$ or $f_q\left(t_p^{-1}g\right)=\bold{0}$ and so evidently $\left\langle f_j\left(t_i^{-1}g\right),f_q\left(t_p^{-1}g\right)\right\rangle=0$ for every $g\in G$. Moreover, along the same lines of reasoning we can consider the sum only over $t_iH$ since the term for associated to a $g\notin t_i H$ is zero since $t_i^{-1}g\notin H$. Thus by $\mathbf{(1)}$ we know that

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\displaystyle \begin{aligned}\left\langle f_{i,j},f_{p,q}\right\rangle &= \frac{1}{|G|}\sum_{g\in G}\left\langle f_j\left(t_i^{-1}g\right),f_q\left(t_p^{-1}g\right)\right\rangle_\mathscr{V}\\ &= \frac{\delta_{i,p}}{|G|}\sum_{g\in G}\left\langle f_j\left(t_i^{-1}g\right),f_q\left(t_i^{-1} g\right)\right\rangle_\mathscr{V}\\ &= \frac{\delta_{i,p}}{|G|}\sum_{g\in t_iH}\left\langle f_j\left(t_i^{-1}g\right),f_q\left(t_i^{-1}h\right)\right\rangle_\mathscr{V}\\ &= \frac{\delta_{i,p}}{|G|}\sum_{h\in H}\left\langle f_j\left(h\right),f_q(h)\right\rangle_\mathscr{V}\\ &= \frac{\delta_{i,p}}{|H|}\sum_{h\in H}\left\langle \rho_{h}^{-1}(v_j),\rho_h^{-1}(v_q)\right\rangle_\mathscr{V}\\ &= \frac{\delta_{i,p}}{|H|}\sum_{h\in H}\left\langle v_j,v_q\right\rangle\\ &= \delta_{i,p}\delta_{j,q}\end{aligned}

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from where the fact that $\mathcal{B}$ is an orthonormal basis follows. Note though that evidently $\displaystyle \varrho\left(\sum_{x\in C}\delta_x\right)$ commutes with $\text{Ind}^G_H(\rho)_{t_i}$. Indeed, since $\varrho$ is a $\ast$-representation and $\text{Ind}^G_H(\rho)_{t_i}=\varrho(\delta_{t_i})$ one has that

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\displaystyle \begin{aligned}\text{Ind}^G_H(\rho)_{t_i}^{-1}\varrho\left(\sum_{x\in C}\delta_x\right)\text{Ind}^G_H(\rho)_{t_i} &=\varrho\left(\delta_{t_i^{-1}}\right)\varrho\left(\sum_{x\in C}\delta_x\right)\varrho\left(\delta_{t_i}\right)\\ &= \varrho\left(\delta_{t_i^{-1}}\ast\sum_{x\in C}\delta_x\ast\delta_{t_i}\right)\\ &= \varrho\left(\sum_{x\in C}\delta_{t_i^{-1}xt_i}\right)\\ &= \varrho\left(\sum_{x\in C}\delta_x\right)\end{aligned}

But, since $\mathcal{B}$ is orthonormal it’s a common fact the matrix entry on the diagonal associated to $f_{i,j}$ is $\displaystyle \left\langle \varrho\left(\sum_{x\in C}\delta_x\right)f_{i,j},f_{i,j}\right\rangle$ but we may rewrite this as

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\displaystyle \begin{aligned}\left\langle \varrho\left(\sum_{x\in C}\delta_g\right)\text{Ind}^G_H(\rho)_{t_i}(f_j),\text{Ind}^G_H(\rho)_{t_i}(f_j)\right\rangle &=\left\langle \text{Ind}^G_H(\rho)_{t_i}^{-1}\varrho\left(\sum_{x\in C}\delta_g\right)\text{Ind}^G_H(\rho)_{t_i}(f_j),f_j\right\rangle\\ &=\left\langle \varrho\left(\sum_{x\in C}\delta_g\right)(f_j),f_j\right\rangle\end{aligned}

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Thus,

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$\displaystyle \text{tr}\left(\varrho\left(\sum_{x\in C}\delta_g\right)\right)=\sum_{i=1}^{r}\sum_{j=1}^{n}\left\langle \varrho\left(\sum_{x\in C}\delta_x\right)(f_j),f_j\right\rangle=\left(G:H\right)\sum_{j=1}^{m}\left\langle \varrho\left(\sum_{x\in C}\delta_x\right)\left(f_j\right),f_j\right\rangle$

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But, note that if $x\notin H$ one has that it is impossible to simultaneously have $g\in H$ and $x^{-1}g\in H$ and so $f_j(g)$ or $f_j(x^{-1}g)$ is $\bold{0}$ for every $g\in G$ and so evidently

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\displaystyle \begin{aligned}\left\langle \varrho(\delta_x)(f_j),f_j\right\rangle &=\frac{1}{|G|}\sum_{g\in G}\left\langle \text{Ind}^G_H(\rho)_x(f_j)(g),f_j(g)\right\rangle_\mathscr{V}\\ &=\frac{1}{|G|}\sum_{g\in G}\left\langle f_j\left(x^{-1}g\right),f_j(g)\right\rangle_\mathscr{V}\\ &=0\end{aligned}

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.