The Unapologetic Mathematician

Inner Tensor Products

Let’s say we have two left $G$-modules — ${}_GV$ and ${}_GW$ — and form their outer tensor product ${}_{GG}V\otimes W$. Note that this carries two distinct actions of the same group $G$, and these two actions commute with each other. That is, $V\otimes W$ carries a representation of the product group $G\times G$. This representation is a homomorphism $\rho\otimes\sigma:G\times G\to\mathrm{End}(V\otimes W)$.

It turns out that we actually have another homomorphism $\Delta:G\to G\times G$, given by $\Delta(g)=(g,g)$. Indeed, we check that

\displaystyle\begin{aligned}\Delta(gh)&=(gh,gh)\\&=(g,g)(h,h)\\&=\Delta(g)\Delta(h)\end{aligned}

If we compose this with the representing homomorphism, we get a homomorphism $(\rho\otimes\sigma)\circ\Delta:G\to\mathrm{End}(V\otimes W)$. That is, $V\otimes W$ actually carries a representation of $G$ itself!

We’ve actually seen this before, a long while back. The group algebra $\mathbb{C}[G]$ is an example of a bialgebra, with the map $\Delta$ serving as a “comultiplication”. If this seems complicated, don’t worry about it. The upshot is that this “inner” tensor product $V\otimes W$, considered as a representation of $G$, behaves as a monoidal product in the category of $G$-modules. Sometimes we will write $V\hat{\otimes}W$ (and similar notations) when we need to distinguish the inner tensor product from the outer one, but often we can just let context handle it.

Luckily, characters are just as well-behaved for inner products. Indeed, we can check that

\displaystyle\begin{aligned}\chi\hat{\otimes}\psi(g)&=\mathrm{Tr}(\rho\otimes\sigma(\Delta(g)))\\&=\mathrm{Tr}(\rho\otimes\sigma(g,g))\\&=\chi\otimes\psi(g,g)\\&=\chi(g)\psi(g)\end{aligned}

However, unlike for outer tensor products the inner tensor product of two irreducible representations is not, in general, itself irreducible. Indeed, we can look at the character table for $S_3$ and consider the inner tensor product of two copies of $V^\perp$.

What we just proved above tells us that the character of $V^\perp\hat{\otimes}V^\perp$ is $\left(\chi^\perp\right)^2$. This takes the values

\displaystyle\begin{aligned}\chi^\perp(e)^2&=4\\\chi^\perp\left((1\,2)\right)^2&=0\\\chi^\perp\left((1\,2\,3)\right)^2&=1\end{aligned}

which does not occur as a line in the character table, and thus cannot be an irreducible character. Indeed, calculating inner products, we find

\displaystyle\begin{aligned}\langle\chi^\mathrm{triv},\left(\chi^\perp\right)^2\rangle&=1\\\langle\mathrm{sgn},\left(\chi^\perp\right)^2\rangle&=1\\\langle\mathrm{sgn},\left(\chi^\perp\right)^2\rangle&=1\end{aligned}

And so we find that

$\displaystyle\left(\chi^\perp\right)^2=\chi^\mathrm{triv}+\mathrm{sgn}+\chi^\perp$

which means that

$\displaystyle V^\perp\hat{\otimes}V^\perp\cong V^\mathrm{triv}\oplus V^\mathrm{sgn}\oplus V^\perp$

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November 5, 2010 -

4 Comments »

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