Abstract Nonsense

Crushing one theorem at a time

The Irreps of the Product of Finitely Many Finite Groups


Point of post: In this post we shall discuss how one can find the set of all irreps, up to equivalence, of a group of the form G_1\times\cdots\times G_n given the irreps of G_k for k\in[n].

Motivation

We’ve developed quite an extensive theory regarding how to find the irreps of a finite group G and how to relate those irreps to one another, as well as their characters.The question remains though if there is a natural way that the irreps of G relate naturally to constructions based on G. In particular, in this post we are interested in determining the relationships between the irreps (in particular the characters) of the product G\times H of two groups G and H given the knowledge of the characters of G and H. So for example, it’s easy (as we’ve shown) to construct the character table for S_3 and it’s equally easy to construct the character table for \mathbb{Z}_2. A next logical step would be to combine them and find the character table for S_3\times\mathbb{Z}_2. It would be nice if one would have to not go tromping through all the extra work to create this (larger) character table having gone through the (admittedly small amount of work) to construct the ones for S_3 and \mathbb{Z}_2. In this post we shall show that our greatest wish is true–we can’t just easily get some characters of S_3\times\mathbb{Z}_2 from those of S_3 and \mathbb{Z}_2 but we can easily get all of them.


Finding The Irreducible Characters of the Finite Product of Finitely Many Groups

Let G and H be groups, and \rho:G\to\mathcal{U}\left(\mathscr{V}\right) and \psi:H\to\mathcal{U}\left(\mathscr{W}\right) be representations on G and H respectively. We can define a new representation, called the tensor product of \rho and \psi, on G\times H by

\text{ }

\rho\boxtimes\psi:G\times H\to\mathscr{U}\left(\mathscr{V}\otimes\mathscr{W}\right)

 

by (\rho\boxtimes\psi)(g,h)=\rho(g)\otimes\psi(h) where \mathscr{V}\otimes\mathscr{W} is the tensor product of vector spaces and \rho(g)\otimes\psi(h) is the tensor product of linear transformations (compare with the tensor product of representations given one group). This is indeed a representation since the tensor product of two transformations is unitary and it’s a homomorphism since

\text{ }

\begin{aligned}\left(\rho\boxtimes\psi\right)\left((g,h)(u,v)\right) &=\left(\rho\boxtimes\psi\right)\left(gu,hv\right)\\ &=\left(\rho(gu)\right)\otimes\left(\psi(hv)\right)\\ &=\left(\rho(g)\rho(u)\right)\otimes\left(\psi(h)\psi(v)\right)\\ &=\left(\rho(g)\otimes\psi(h)\right)\left(\rho(u)\otimes\psi(v)\right)\\ &=\left(\rho\boxtimes\psi\right)(g,h)\left(\rho\boxtimes\psi\right)(u,v)\end{aligned}

 

The interesting difference this type of tensor representation and the one we previously discussed is that the tensor product of two irreps of G and H is an irrep on G\times H. The fascinating thing is that the set of all such tensor product of representations constitutes all of the irreps. Indeed

Theorem: Let G and H be finite groups and for each \alpha\in\widehat{G} and each \beta\in\widehat{H} choose a representative \rho^{(\alpha)}\in\alpha and \psi^{(\beta)}\in\beta. Then, for every such \alpha,\beta one has that \rho^{(\alpha)}\boxtimes\psi^{(\beta)} is an irreducible representation for G\times H and \rho^{(\alpha)}\boxtimes\psi^{(\beta)}\cong\rho^{(\gamma)}\boxtimes\psi^{(\gamma)} if and only if (\alpha,\beta)=(\gamma,\delta). Moreover

\text{ }

\widehat{G\times H}=\left\{\left[\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right]:\alpha\in\widehat{G}\text{ and }\beta\in\widehat{H}\right\}

\text{ }

Proof: Let \alpha\in\widehat{G} and \beta\in\widehat{H} be arbitrary. To prove that \rho^{(\alpha)}\boxtimes\psi^{(\beta)} is an irrep for G\times H we note that evidently \text{tr}\left(\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right)=\chi^{(\alpha)}\chi^{(\beta)} and so

\text{ }

\displaystyle \begin{aligned}\left\langle \text{tr}\left(\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right),\text{tr}\left(\rho^{(\alpha)}\boxtimes \psi^{(\beta)}\right)\right\rangle &= \frac{1}{|G\times H|}\sum_{(g,h)\in G\times H}\left|\text{tr}\left(\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right)\right|^2\\ &= \frac{1}{|G\times H|}\sum_{(g,h)\in G\times H}\left|\chi^{(\alpha)}(g)\right|^2\left|\chi^{(\beta)}(h)\right|^2\\ &= \frac{1}{|G||H|}\sum_{g\in G}\sum_{h\in H}\left|\chi^{(\alpha)}(g)\right|^2\left|\chi^{(\beta)}(h)\right|^2\\ &= \left(\frac{1}{|G|}\sum_{g\in G}\left|\chi^{(\alpha)}(g)\right|^2\right)\left(\frac{1}{|H|}\sum_{h\in H}\left|\chi^{(\beta)}(h)\right|^2\right)\\ &= 1\cdot 1\\ &=1\end{aligned}

 

and thus by our alternate characterization of irreducibility we may conclude that \rho^{(\alpha)}\boxtimes\psi^{(\beta)} is an irrep of G\times H.

To prove our second we actually prove the stronger claim that

\text{ }

\left\langle\text{tr}\left(\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right),\text{tr}\left(\rho^{(\gamma)}\boxtimes\psi^{(\delta)}\right)\right\rangle=\delta_{\alpha,\gamma}\delta_{\beta,\delta}

 

Indeed,

\displaystyle \begin{aligned}\left\langle \text{tr}\left(\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right),\text{tr}\left(\rho^{(\gamma)}\boxtimes\psi^{(\delta)}\right)\right\rangle &= \frac{1}{|G\times H|}\sum_{(g,h)\in G\times H}\text{tr}\left(\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right)\overline{\text{tr}\left(\rho^{(\gamma)}\boxtimes\psi^{(\delta)}\right)}\\ &= \frac{1}{|G\times H|}\sum_{(g,h)\in G\times H}\chi^{(\alpha)}(g)\chi^{(\beta)}(h)\overline{\chi^{(\gamma)}(g)}\overline{\chi^{(\delta)}(h)}\\ &= \frac{1}{|G||H|}\sum_{g\in G}\sum_{h\in H}\chi^{(\alpha)}(g)\overline{\chi^{(\gamma)}(g)}\chi^{(\beta)}(h)\overline{\chi^{(\delta)}(h)}\\ &= \left(\frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha)}(g)\overline{\chi^{(\gamma)}(g)}\right)\left(\frac{1}{|H|}\sum_{h\in H}\chi^{(\beta)}(h)\overline{\chi^{(\delta)}(h)}\right)\\ &= \delta_{\alpha,\gamma}\delta_{\beta,\delta}\end{aligned}

 

 

To prove the last claim we note that we’ve defined a map f:\widehat{G}\times\widehat{H}\to\widehat{G\times H} by (\alpha,\beta)\mapsto \left[\rho^{(\alpha)}\boxtimes\psi^{(\beta)}\right] which by the second part of our theorem is an injection. But, by a previous theorem we have that \#\left(\widehat{G}\times\widehat{H}\right)=\#\left(\widehat{G\times H}\right) and thus by a simple set theoretic fact we may conclude that f is also a surjection from where the conclusion follows. \blacksquare

References:

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

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April 11, 2011 - Posted by | Algebra, Representation Theory | , , , ,

6 Comments »

  1. […] Point of post: This is a continuation of this post. […]

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  4. […] finds such a deomposition everything becomes easy. Namely, we know from our theorems regarding the irreps of the products of groups that every irrep of , up to equivalence, looks like where is an irrep of . But, since is […]

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  5. […] a character. Moreover, let be the trivial character for . Consider then the character , which we know is a character for . That said, it’s clear by definition that and that from where it […]

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  6. […] constructions based on and . Probably the most important that we’ve so far discussed is the relationship between irreps and and the irreps of their direct product . We continue in this vein and discuss […]

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