Abstract Nonsense

Crushing one theorem at a time

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


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

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What we next do is ‘dualize’  the action \varphi:K\to\text{Aut}\left(A\right) to an action \widehat{\varphi}:K\to\text{Aut}\left(\text{irr}(A)\right) (where we, for notational convenience, have denoted \text{Hom}(A,\mathbb{T}) by \text{irr}(A). Since A is abelian we know the two coincide) by \left(\widehat{\varphi}_k(\chi)\right)(a)=\chi\left(\varphi_k^{-1}(a)\right). It is not at all clear that 1) \widehat{\varphi}_k(\chi) is really a homomorphism, 2) \widehat{\varphi}_k is an automorphism, and 3) \widehat{\varphi} is a homomorphism. Indeed:

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Theorem: Let \widehat{\varphi}:K\to\text{Aut}\left(\text{irr}(A)\right) be defined as above. Then:

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\begin{aligned}&\mathbf{(1)}\quad \widehat{\varphi}_k(\chi)\textit{ is a homomorphism }G\to\mathbb{T}\textit{ for all }k\in K,\chi\in\text{irr}(A)\\ &\mathbf{(2)}\quad \widehat{\varphi}_k:\text{irr}(A)\to\text{irr}(A)\textit{ is an automorphism}\\ &\mathbf{(3)}\quad \widehat{\varphi}\textit{ is a homomorphism }K\to\text{Aut}\left(A\right)\end{aligned}

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Proof:

1) We note that for any k\in K, a,a'\in A, and \chi\in\text{irr}(A) one has that

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\displaystyle \begin{aligned}\left(\widetilde{\varphi}_k(\chi)\right)(aa') &= \chi\left(\varphi_k^{-1}\left(aa'\right)\right)\\ &= \chi\left(\varphi_k^{-1}(a)\varphi_k^{-1}(a')\right)\\ &= \chi\left(\varphi_k^{-1}(a)\right)\chi\left(\varphi_k^{-1}(a')\right)\\ &= \left(\widehat{\varphi}_k(\chi)\right)(a)\left(\widehat{\varphi}_k(\chi)\right)(a')\end{aligned}

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(note the use of the abelianess of A).

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2) We note that for any k\in K, a\in A, and any \chi,\chi'\in\text{irr}(A) one has

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\displaystyle \begin{aligned}\left(\widehat{\varphi}_k\left(\chi\chi'\right)\right)(a) &= \left(\chi\chi'\right)\left(\varphi_k^{-1}(a)\right)\\ &= \chi\left(\varphi_k^{-1}(a)\right)\chi'\left(\varphi_k^{-1}(a)\right)\\ &= \left(\widehat{\varphi}_k(\chi)\right)(a)\left(\widehat{\varphi}_k\left(\chi'\right)\right)(a)\end{aligned}

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so that \widehat{\varphi}_k is an endomorphism. To prove it is an automorphism it suffices to prove (since A is finite) to prove it is injective. To do this we note that if \chi\in\ker\widehat{\varphi}_k and a\in A then \varphi_k(a)\in A and so

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1=\left(\widehat{\varphi}_k(\chi)\right)(\varphi_k(a))=\chi\left(\varphi_k^{-1}\left(\varphi_k(a)\right)\right)=\chi(a)

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Thus, \chi(a)=1 for every a\in A, and thus \chi=\chi^{\text{triv}}.

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3) We note that for any \chi\in\text{irr}(G) and k,k'\in K one has

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\begin{aligned}\left(\widehat{\varphi}_{kk'}(\chi)\right)(a) &= \chi\left(\varphi_{kk'}^{-1}(a)\right)\\ &= \chi\left(\varphi_{k'}^{-1}\left(\varphi_k^{-1}(a)\right)\right)\\ &= \left(\widehat{\varphi}_{k'}(\chi)\right)\left(\varphi_k^{-1}(a)\right)\\ &= \left(\widehat{\varphi_k}\left(\widehat{\varphi}_{k'}(\chi)\right)\right)(a)\end{aligned}

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and so \widehat{\varphi}_{kk'}=\widehat{\varphi}_k\circ\widehat{\varphi}_{k'}. \blacksquare

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References:

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

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May 8, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,

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