## Induced Matrix Representations

Sorry I missed posting this back in the morning…

We want to work out the matrices of induced representations. Explicitly, if is a left -module of degree , where is a subgroup of , then is a left -module. If we pick a basis of , we get a matrix representation . We want to describe a matrix representation corresponding to . In the process, we’ll see that we were *way* off with our first stabs at the dimensions of tensor products over .

The key point is to realize that is a free right module over . That is, we can find some collection of vectors in so that any other one can be written as a linear collection of these with coefficients (on the right) in . Indeed, we can break up into the left cosets of . Picking one representative of each coset — we call this a “transversal” for — we have essentially chopped up into chunks, each of which looks exactly like .

To see this, notice that the coset is a subset of . Thus it describes a subspace of — that spanned by the elements of the coset, considered as basis vectors in the group algebra. The action of on shuffles the basis vectors in this coset around amongst each other, and so this subspace is invariant. It should be clear that it is isomorphic to , considered as a right -module.

Okay, so when we consider the tensor product , we can pull any action by across to the right and onto . What remains on the left? A vector space spanned by the transversal elements , which essentially index the left cosets of in . We have one copy of for each of these cosets, and so the dimension of the induced module is .

How should we think about this equation, heuristically? The tensor product multiplies the dimensions of vector spaces, which gives . Then the action of on the tensor product divides by a factor of — at least in principle. In practice, this only works because in our example the action by is free. That is, no element in the bare tensor product is left fixed by any non-identity element of .

So how does this give us a matrix representation of ? Well, acts on by shuffling around the subspaces that correspond to the cosets of . In fact, this is exactly the coset representation of corresponding to ! If we write for some , then this uses up the transversal element . The is left to “pass through” and act on .

To write this all out explicitly, we get the following block matrix:

where is the number of cosets, and we simply define to be a zero block if does not actually fall into .

[...] for the induced character, we use the matrix of the induced representation that we calculated last time. If is a matrix representation of a group , which is a subgroup , [...]

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[...] we’ve defined . But if we choose a transversal for — like we did when we set up the induced matrices — then we can break down as the direct sum of a bunch of copies of [...]

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