# The Unapologetic Mathematician

## Tensor Products over Group Algebras

So far, we’ve been taking all our tensor products over the complex numbers, since everything in sight has been a vector space over that field. But remember that a representation of $G$ is a module over the group algebra $\mathbb{C}[G]$, and we can take tensor products over this algebra as well.

More specifically, if $V_G$ is a right $G$-module, and ${}_GW$ is a left $G$-module, then we have a plain vector space $V\otimes_GW$. We build it just like the regular tensor product $V\otimes W$, but we add new relations of the form

$\displaystyle(vg)\otimes w=v\otimes(gw)$

That is, in the tensor product over $\mathbb{C}[G]$, we can pull actions by $g$ from one side to the other.

If $V$ or $W$ have extra group actions, they pass to the tensor product. For instance, if ${}_HV_G$ is a left $H$-module as well as a right $G$-module, then we can define

$\displaystyle h(v\otimes w)=(hv)\otimes w$

Similarly, if $V_{GH}$ has an additional right action by $H$, then so does $V\otimes_GW$, and the same goes for extra actions on $W$. Similar to the way that hom spaces over $G$ “eat up” an action of $G$ on each argument, the tensor product $\otimes_G$ “eats up” a right action by $G$ on its left argument and a left action by $G$ on its right argument.

We can try to work out the dimension of this space. Let’s say that we have decompositions

\displaystyle\begin{aligned}V_G&=V^{(1)}_G\oplus\dots\oplus V^{(m)}_G\\{}_GW&={}_GW^{(1)}\oplus\dots\oplus{}_GW^{(n)}\end{aligned}

into irreducible representations (possibly with repetitions). As usual for tensor products, the operation is additive, just like we saw for $\hom$ spaces. That is

$\displaystyle V\otimes_GW\cong\bigoplus\limits_{i=1}^m\bigoplus\limits_{j=1}^nV^{(i)}\otimes_GW^{(j)}$

So we really just need to understand the dimension of one of these summands. Let’s say $V$ is irreducible with dimension $d_V$ and $W$ is irreducible with dimension $d_W$.

Now, we can pick any vector $v\in V$ and hit it with every group element $g\in G$. These vectors must span $V$; they span some subspace, which (since $V$ is irreducible) is either trivial or all of $V$. But it can’t be trivial, since it contains $v$ itself, and so it must be $V$. That is, given any vector $v'\in V$ we can find some element of the group algebra $g'\in\mathbb{C}[G]$ so that $v'=vg'$. But then for any $w\in W$ we have

$\displaystyle v'\otimes w=(vg')\otimes w=v\otimes(g'w)$

That is, every tensor can be written with $v$ as the first tensorand. Does this mean that $\dim(V\otimes_GW)=\dim(W)$? Not quite, since this expression might not be unique. For every element of the group algebra that sends $v$ back to itself, we have a different expression.

So how many of these are there? Well, we have a linear map $\mathbb{C}[G]\to V$ that sends $g\in\mathbb{C}[G]$ to $vg\in V$. We know that this is onto, so the dimension of the image is $d_V$. The dimension of the source is $\dim(\mathbb{C}[G])=\lvert G\rvert$, and so the rank-nullity theorem tells us that the dimension of the kernel — the dimension of the space that sends $v$ back to itself — is $\lvert G\rvert-d_V$.

So we should be able to subtract this off from the dimension of the tensor product, due to redundancies. Assuming that this works as expected, we get $\dim(V\otimes_GW)=d_V+d_W-\lvert G\rvert$, which at least is symmetric between $V$ and $W$ as expected. But it still feels sort of like we’re getting away with something here. We’ll come back to find a more satisfying proof soon.

November 9, 2010