# 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

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

November 5, 2010 -

1. [...] Character Table of S4 Let’s use our inner tensor products to fill in the character table of . We start by listing out the conjugacy classes along with their [...]

Pingback by The Character Table of S4 « The Unapologetic Mathematician | November 8, 2010 | Reply

2. [...] just like we found with inner tensor products, we can combine these two actions of into one. Now we have one left action of on linear maps by [...]

Pingback by Group Invariants « The Unapologetic Mathematician | November 12, 2010 | Reply

3. [...] actions by , we can take the outer tensor product, which carries an action by . Then we pass to the inner tensor product, acting on each tensorand by the same group element. To be more [...]

Pingback by Tensors Over the Group Algebra are Invariants « The Unapologetic Mathematician | November 15, 2010 | Reply

4. [...] the character of . If — as a left -module — has character and has character , then we know that the inner tensor product has [...]

Pingback by The Dimension of the Space of Tensors Over the Group Algebra « The Unapologetic Mathematician | November 16, 2010 | Reply