The Unapologetic Mathematician

Mathematics for the interested outsider

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 Posted by | Algebra, Group theory, Representation Theory | 6 Comments

   

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