Abstract Nonsense

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.

April 11, 2011 -

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