# Abstract Nonsense

## Representation Theory of Semidirect Products: The Preliminaries (Pt. III)

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

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The interesting thing about this ‘dualization’ of $\varphi$ is its relation ship between the original $\rho$ and the aforementioned $\mathscr{V}_\chi$. Namely:

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Theorem: For any $k\in K$ and $\chi\in\text{irr}(A)$  one has that $\rho_{\widetilde{k}}\left(\mathscr{V}_{\chi}\right)\subseteq \mathscr{V}_{\widehat{\varphi}_k\left(\chi\right)}$.

Proof: Let $x\in\mathscr{V}_{\chi}$ be arbitrary, then for any $\widetilde{n}\in \widetilde{N}$ one has

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\begin{aligned}\rho_{\widetilde{n}}\rho_{\widetilde{k}}(x) &= \rho_{\widetilde{k}}\rho_{\widetilde{k}^{-1}\widetilde{n}\widetilde{k}}(x)\\ &= \rho_{\widetilde{k}}\rho_{\widetilde{\varphi_k^{-1}(n)}}(x)\\ &=\rho_{\widetilde{k}}\chi\left(\varphi_k^{-1}(n)\right)(x)\\ &= \rho_{\widetilde{k}}\left(\widehat{\varphi}_k(\chi)\right)(n)x\\ &= \left(\widehat{\varphi}_k(\chi)\right)(n)\rho_{\widetilde{k}}(x)\end{aligned}

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And, since $\widetilde{n}\in\widetilde{N}$ was arbitrary it follows that $\rho_{\widetilde{k}}(x)$ is in $\mathscr{V}_{\widehat{\varphi}_k(\chi)}$ as claimed. Since $x\in\mathscr{V}_{\chi}$ was arbitrary the conclusion follows. $\blacksquare$

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From this we have the following corollary:

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Corollary: Let $\chi_0\in\text{irr}(A)$ and $\mathcal{O}_{\chi_0}$ the associated orbit under the action of $K$ on $\text{irr}(A)$ induced by $\widehat{\varphi}:K\to\text{Aut}(\text{irr}(A))$. Then, $\displaystyle \bigoplus_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}$ is $\rho$-invariant.

Proof:  We first show that the space in question is invariant under $\rho_{\widetilde{k}}$ for every $k\in K$. Let $v$ be an element of $\displaystyle \bigoplus_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}$. Then, there exists $v_{\chi}\in\mathscr{V}_{\chi}$ for every $\chi\in\mathcal{O}_{\chi_0}$ such that $\displaystyle v=\sum_{\chi\in\mathcal{O}_{\chi_0}}v_\chi$. So, by our last theorem

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$\displaystyle \rho_{\widetilde{k}}(x)=\sum_{\chi\in\mathcal{O}_{\chi_0}}\rho_{\widetilde{k}}(v_{\chi})\in \sum_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\widehat{\varphi}_{k}(\chi)}=\bigoplus_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}$

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Since $k\in K$ was arbitrary the conclusion follows.

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We next show that the space in question is invariant under $\rho_{\widetilde{n}}$ for every $n\in N$. Indeed, if $x$ is as above then

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$\displaystyle \rho_{\widetilde{n}}(x)=\sum_{\chi\in\mathcal{O}_{\chi_0}}\rho_{\widetilde{n}}(v_\chi)=\sum_{\chi\in\mathcal{O}_{\chi_0}}\chi(n)v_\chi\in\sum_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}=\bigoplus_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}$

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Since $x$ and $n$ were arbitrary this part of the proof follows.

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To finish the proof we merely note that for any $g\in G=A\rtimes_\varphi K$ one has $g=\widetilde{n}\widetilde{k}$ for some $n\in N$ and $k\in K$ and so for any $x$ in the direct sum in question one has

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$\displaystyle \rho_{g}=\rho_{\widetilde{n}\widetilde{k}}(g)=\rho_{\widetilde{n}}\left(\rho_{\widetilde{k}}(x)\right)\in \rho_{\widetilde{n}}\left(\bigoplus_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}\right)\subseteq\bigoplus_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}$

$\blacksquare$

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From this we conclude as a corollary:

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Corollary: Let $\rho:A\rtimes_\varphi K\to\mathcal{U}\left(\mathscr{V}\right)$ be an irrep. Then, there exists precisely one orbit $\mathcal{O}$ of the action $\widehat{\varphi}:K\to\text{Aut}\left(\text{irr}(A)\right)$ for which $\mathscr{V}_\chi\ne\{\bold{0}\}$ if and only if $\chi\in\mathcal{O}$.

Proof: We know since

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$\displaystyle \bigoplus_{\chi\in\text{irr}(A)}\mathscr{V}_{\chi}=\mathscr{V}$

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that there exists some $\chi_0\in\text{irr}(A)$ such that $\mathscr{V}_{\chi}\ne\bold{0}$. We claim that $\mathcal{O}_{\chi_0}$ is the orbit we seek. Indeed, to see that $\mathscr{V}_{\lambda}=\{\bold{0}\}$ for $\lambda\notin\mathcal{O}_{\chi_0}$ we note that since $\displaystyle \bigoplus_{\chi\in\mathcal{O}_{\chi_0}}\mathscr{V}_{\chi}$ is non-zero and $\rho$-invariant we may conclude from the irreducibility of $\rho$ that is the full space $\mathscr{V}$. Thus, one has that $\displaystyle \bigoplus_{\chi\in\mathscr{O}_{\lambda}}\mathscr{V}_{\chi}=\{\bold{0}\}$ and so of course $\mathscr{V}_{\lambda}=\{\bold{0}\}$. Now, if $\chi\in\mathcal{O}_{\chi_0}$ then $\chi=\widehat{\varphi}_{k}(\chi_0)$ for some $k\in K$. Thus, by an earlier theorem we have that $\rho_{\widehat{k}}\left(\mathscr{V}_{\chi_0}\right)\subseteq\mathscr{V}_{\chi}$ and so $\dim\mathscr{V}_{\chi}\geqslant\dim\mathscr{V}_{\chi_0}$ and so $\mathscr{V}_{\chi}\ne\bold{0}$. $\blacksquare$

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From this it’s easy to see that

Corollary: If $\mathcal{O}$ is the non-zero orbit described in the above proof then $\dim\mathscr{V}_{\chi}=\dim\mathscr{V}_{\lambda}$ for every $\chi,\lambda\in\mathcal{O}$. In particular, $\dim\mathscr{V}=\#\left(\mathcal{O}\right)\dim\mathscr{V}_{\chi}$.

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References:

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