Abstract Nonsense

Crushing one theorem at a time

Another Way of Viewing the Dual Group

Point of post: In this post we discuss another way to isomorphically view the linear group of a group G.



In our last post we discussed how to naturally turn a subset of \widehat{G} into a group, namely the subset \widehat{G_\mathfrak{L}} of linear characters. That said, the definition we gave is not very conducive to proving things about \widehat{G_\mathfrak{L}}. In this post we prove that \widehat{G_\mathfrak{L}} is isomorphic to \text{Hom}\left(G,\mathbb{T}\right) where \mathbb{T} is the circle group and the group operation is pointwise multiplication. This shall prove to be just the sort of concrete object we need to do some serious analysis concerning \widehat{G_\mathfrak{L}}.

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Different Way of Viewing the Linear Group of a Group G

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The first interesting result that we have is that \widehat{G_\mathfrak{L}} is isomorphic to something very simple (or at least simpler to say). Namely, let \mathbb{T} denote the circle group then what we’ll now show is that \widehat{G_\mathfrak{L}}\cong\text{Hom}\left(G,\mathbb{T}\right), the set of all homomorphisms G\to\mathbb{T} with point wise multiplication. Indeed:

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Theorem: Let G be a finite group then \widehat{G_\mathfrak{L}}\cong\text{Hom}\left(G,\mathbb{T}\right).

Proof: Define F:\widehat{G_\mathfrak{L}}\to\text{Hom}\left(G,\mathbb{T}\right) by \alpha\mapsto \chi^{(\alpha)}. Firstly this map is well-defined since for linear \alpha\in\widehat{G} and each \lambda\in\alpha, one has that \chi_\lambda=\chi^{(\alpha)}\in\text{Hom}\left(G,\mathbb{T}\right). Indeed, it’s a homomorphism because if \lambda:G\to\mathcal{U}\left(\mathscr{V}\right) then we know that \dim_\mathbb{C}\mathscr{V}=1 so that for every g\in G, \lambda(g)=z_g\mathbf{1} for some z_g\in\mathbb{C}. Thus, for every g,h\in G,

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\begin{aligned}\chi_\lambda(gh) &=\text{tr}\left(\lambda(gh)\right)\\ &=\text{tr}\left(\lambda(g)\lambda(h)\right)\\ &=\text{tr}\left(z_g\mathbf{1}z_h\mathbf{1}\right)\\ &=\text{tr}\left(z_gz_h\mathbf{1}\right)\\ &=z_gz_h=\chi_\lambda(g)\chi_\lambda(h)\end{aligned}

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and as has been said many times before each \chi_\lambda(g) is a |G|-root of unity and thus trivially in \mathbb{T}.

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So, to see that F is a homomorphism we merely note that for each g\in G and \alpha,\beta\in\widehat{G_\mathfrak{L}}

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so that F\left(\alpha\otimes\beta\right)=F(\alpha)F(\beta). To see that F is injective we merely note that if F(\alpha)=F(\beta) then the characters of \alpha and \beta are the same and thus as has been said before we know that \alpha=\beta. Lastly, to show that F is surjective we merely note that if \chi\in\text{Hom}\left(G,\mathbb{T}\right) then by definition \chi:G\to\mathcal{U}\left(\mathbb{C}\right) where \mathbb{C} is given the usual inner product and thus \chi is a linear representation. Moreover, if \chi\in\alpha then by definition \chi^{(\alpha)}=\chi so that F(\alpha)=\chi and so F is surjective. The conclusion follows. \blacksquare




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


April 12, 2011 - Posted by | Algebra, Representation Theory | , , , ,


  1. […] it’s difficult to deal directly with the dual group even with our alternate characterization of it. In this post we show that the lay of the land becomes much nicer. Namely, we shall show that […]

    Pingback by Representation Theory: Dual Group of a Cyclic Group « Abstract Nonsense | April 13, 2011 | Reply

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