## Hom Space Duals

Again, sorry for the delay but I was eager to polish something up for my real job this morning.

There’s something interesting to notice in our formulæ for the dimensions of spaces of intertwinors: they’re symmetric between the two representations involved. Indeed, let’s take two -modules:

where the are pairwise-inequivalent irreducible -modules with degrees . We calculate the dimensions of the -spaces going each way:

but these are equal! So does this mean these spaces are isomorphic?

Well, yes. Any two vector spaces having the same dimension are isomorphic, but they’re not “naturally” isomorphic. Roughly, there’s no universal method of giving an explicit isomorphism, and so it’s regarded as sort of coincidental. But there’s something else around that’s *not* coincidental.

It turns out that these spaces *are* naturally isomorphic to each other’s dual spaces. That is, for any -modules and we have an isomorphism

Luckily, we already know that their dimensions are equal, so the rank-nullity theorem tells us all we need is to find an injective linear map from one to the other.

So, let’s take an intertwinor and use it to build a linear functional on . For any intertwinor we define

Where is the trace of an endomorphism. Given a matrix, it’s the sum of the diagonal entries. Since the composition of linear maps is linear in each variable, and the trace is a linear function, this is a linear functional as desired. It should also be clear that the construction is itself a linear map.

Now, we must show that this map is injective. That is, for no to we find . This will follow if we can find for every nonzero at least one so that . To do so, we pick a basis for each irreducible representation that shows up in either or so we can replace and with matrix representations. Now we can write

where is an complex matrix. To construct our , we simply take the conjugate transpose of each of these matrices:

where now is an complex matrix, as desired. We multiply the two matrices:

and find that each is a square matrix. Thus the trace of this composition is the sum of their traces.

We’ve already seen that the composition of a linear transformation and its adjoint is self-adjoint and positive-definite. In terms of complex matrices, this tells us that the product of a matrix and its conjugate transpose is conjugate-symmetric and positive-definite. This means that it’s diagonalizable with all nonnegative real eigenvalues down the diagonal. And thus its trace is a nonnegative real number, and it can only be zero if the original matrix was zero.

The upshot, if you didn’t follow that, is that if we have an so that . And thus the map is injective, as we asserted. Proving naturality is similarly easy to proving it for additivity of -spaces, and you can work it out if you’re interested.

[...] product? We get the complex conjugate: . What happens when we swap the arguments to the functor? We get the dual space: . Complex conjugation corresponds to passing to the dual [...]

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[...] we can use the duality on hom spaces and apply it to yesterday’s Frobenius [...]

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