Abstract Nonsense

Crushing one theorem at a time

The Dual Group


Point of post: In this post we discuss how one can turn a subset of \widehat{G} naturally into a group.

Motivation

We’ve done extensive work with the set \widehat{G} of all equivalency classes of irreps on a finite group G. A natural question that we have put off until now is if there is a natural way to make \widehat{G} into a group itself. In this post we provide a partial affirmative answer to this question by showing that the set \widehat{G_\mathfrak{L}} of all linear representations (representations of degree one) forms a group under a natural product.

Linear Representations and Their Tensor Products

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Let G be a finite group an irrep \rho of G is said to be linear if \deg\rho=1. We call an equivalency class \alpha\in\widehat{G} linear if d_\alpha=1 (or equivalently if every one of its members is a linear irrep). If \lambda is a linear irrep then \chi_\lambda is called a linear character of G. We denote the set of all linear \alpha\in\widehat{G} by \widehat{G_\mathfrak{L}}.

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What we first show is that if \rho is an irrep and \lambda a linear irrep then the tensor product \lambda\otimes\rho is an irrep. Indeed:

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Theorem: Let G be a finite group. Then, if \lambda is a linear irrep of G and \rho an irrep then \lambda\otimes\rho is an irrep of G with \deg(\lambda\otimes\rho)=\deg\rho.

Proof: To see that \lambda\otimes\rho is irreducible we note that

\displaystyle \begin{aligned}\left\langle\chi_{ \lambda\otimes\rho},\chi_{\lambda\otimes\rho}\right\rangle &= \frac{1}{|G|}\sum_{g\in G}\left|\chi_\lambda(g)\chi_\rho(g)\right|^2\\ &=\frac{1}{|G|}\sum_{g\in G}\left|\chi_\lambda(g)\right|^2\left|\chi_\rho(g)\right|^2\\ &= \frac{1}{|G|}\sum_{g\in G}\left|\chi_\rho(g)\right|^2\\ &=\left\langle\chi_\rho,\chi_\rho\right\rangle\\ &= 1\end{aligned}

 

(where we’ve used the simple fact that for any linear representation \lambda one has that \left|\chi_\lambda(g)\right|=1 for every g\in G) from where the conclusion follows from our previous characterization of irreducibility. To see that \deg(\lambda\otimes\rho)=\deg\rho we merely note that

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\deg(\lambda\otimes\rho)=\chi_{\lambda\otimes\rho}(e)=\chi_\lambda(e)\chi_\rho(e)=\chi_\rho(e)=\deg\rho.

\blacksquare

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Now, it’s clear from this that for any two linear irreps \lambda,\mu their tensor product \lambda\otimes\mu is also a linear irrep, and since it’s evident that if \lambda\cong\lambda' and \mu\cong\mu' that \lambda\otimes\mu\cong\lambda'\otimes\mu' we may unambiguously consider the product of \alpha,\beta\in\widehat{G_\mathfrak{L}}, denoted \alpha\otimes \beta to be the equivalency class [\lambda\otimes\mu] for any \lambda\in\alpha and \mu\in\beta.

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It’s evident that for any three representations \rho,\psi,\nu that \rho\otimes(\psi\otimes \nu)\cong(\rho\otimes \psi)\otimes \nu so that if \alpha,\beta,\gamma are the equivalency classes of \rho,\psi, and \nu respectively then \alpha\otimes(\beta\otimes\gamma)=(\alpha\otimes\beta)\otimes\gamma so that the map \otimes:\widehat{G_\mathfrak{L}}\times\widehat{G_\mathfrak{L}}\to\widehat{G_\mathfrak{L}} is associative. Moreover, it’s clear that if \alpha_{\text{triv}} denotes the equivalence class of the trivial character \tau:G\to\mathbb{C}^\times:g\mapsto 1 then \alpha_\text{triv} is a two-sided identity for \otimes. Indeed, recall that two irreps are equivalent if and only if the admit the same character and for any \alpha\in\widehat{G_\mathfrak{L}} and \lambda\in\alpha one has that

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\chi_{\tau\otimes\lambda}(g)=\chi_\tau(g)\chi_\lambda(g)=\chi_\lambda(g)=\chi_\lambda(g)\chi_\tau(g)=\chi_{\lambda\otimes\tau}(g)

 

for every g\in G so that \tau\otimes\lambda\cong\lambda\cong\lambda\otimes\tau or, said equivalently, \alpha_{\text{triv}}\otimes\alpha=\alpha=\alpha\otimes\alpha_{\text{triv}}. Lastly, we claim that if \alpha\otimes\beta=\alpha\otimes\gamma then \beta=\gamma. Indeed,

\displaystyle \begin{aligned}\delta_{\beta,\gamma} &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\beta)}(g)\overline{\chi^{(\gamma)}(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\beta)}(g)\overline{\chi^{(\gamma)}(g)}\left|\chi^{(\alpha)}(g)\right|^2\\ &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha)}(g)\chi^{(\beta)}(g)\overline{\chi^{(\alpha)}(g)\chi^{(\gamma)}(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha\otimes\beta)}(g)\overline{\chi^{(\beta\otimes\gamma)}(g)}\\ &= \left\langle \chi^{(\alpha\otimes\beta)},\chi^{(\alpha\otimes\gamma)}\right\rangle\\ &=1\end{aligned}

 

Thus, \widehat{G_\mathfrak{L}} is a finite set which has the associative binary operation \otimes with the two-sided identity \alpha_\text{triv} and which satisfies the left and right cancellative property. There’s a name for such a structure: a group. Moreover, it’s clear that \widehat{G_\mathfrak{L}} is commutative since for any any two representations \rho and \psi it’s clearly true that \rho\otimes\psi\cong\psi\otimes \rho. Also, since for any \alpha\in\widehat{G_\mathfrak{L}} and any \lambda\in\alpha we have that 1=\left|\chi_\lambda(g)\right|^2=\chi_\lambda (g)\overline{\chi_\lambda(g)} it’s clear that \chi_{\lambda\otimes\text{Conj}_\lambda^J}=\chi^{\text{triv}} for any complex conjugate J on \lambda's representation space, and so \alpha^{-1}=\overline{\alpha}. Summing this all up we have:

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Theorem: Let G be a finite group, then \left(\widehat{G_\mathfrak{L}},\otimes\right) is an abelian group with identity \alpha_\text{triv} and \alpha^{-1}=\overline{\alpha}.

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We call the group \widehat{G_\mathfrak{L}} the dual group of G.

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As a last note, we know that \left|\widehat{G_\mathfrak{L}}\right|=\left|G^\text{ab}\right|=\#\left(G/[G,G]\right) where G^\text{ab} is the abelianization of G and [G,G] is the commutator subgroup.

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References:

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

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

4 Comments »

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