Abstract Nonsense

Crushing one theorem at a time

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

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

\text{ }

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

\text{ }

\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}

\text{ }

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

\text{ }

\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

\text{ }

\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}

\text{ }


\text{ }

\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

\text{ }

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

\text{ }

\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}

\text{ }

\text{ }


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


April 26, 2011 - Posted by | Algebra, Representation Theory | , , , ,

1 Comment »

  1. […] Point of Post: This post is a continuation of this one. […]

    Pingback by Different Formula For the Character of an Induced Representation (Pt. III) « Abstract Nonsense | April 26, 2011 | Reply

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