# Abstract Nonsense

## 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 $G$.

Motivation

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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$\left(F\left(\alpha\otimes\beta\right)\right)(g)=\chi^{(\alpha\otimes\beta)}(g)=\chi^{(\alpha)}(g)\chi^{(\beta)}(g)=\left(F(\alpha)\right)(g)\left(F(\beta)\right)(g)$

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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. $\blacksquare$

References:

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