The Unapologetic Mathematician

Mathematics for the interested outsider

Induced Matrix Representations

Sorry I missed posting this back in the morning…

We want to work out the matrices of induced representations. Explicitly, if V is a left H-module of degree d, where H is a subgroup of G, then V\!\!\uparrow_H^G is a left G-module. If we pick a basis of V, we get a matrix representation X:H\to\mathrm{Mat}_d(\mathbb{C}). We want to describe a matrix representation corresponding to V\!\!\uparrow_H^G. In the process, we’ll see that we were way off with our first stabs at the dimensions of tensor products over H.

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

To see this, notice that the coset t_iH is a subset of G. Thus it describes a subspace of \mathbb{C}[G] — that spanned by the elements of the coset, considered as basis vectors in the group algebra. The action of H on \mathbb{C}[G] 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 \mathbb{C}[H], considered as a right H-module.

Okay, so when we consider the tensor product \mathbb{C}[G]\otimes_HV, we can pull any action by H across to the right and onto V. What remains on the left? A vector space spanned by the transversal elements \{t_i\}, which essentially index the left cosets of H in G. We have one copy of V for each of these cosets, and so the dimension of the induced module V\!\!\uparrow_H^G is d\lvert G\rvert/\lvert H\rvert.

How should we think about this equation, heuristically? The tensor product multiplies the dimensions of vector spaces, which gives d\lvert G\rvert. Then the action of H on the tensor product divides by a factor of \lvert H\rvert — at least in principle. In practice, this only works because in our example the action by H is free. That is, no element in the bare tensor product \mathbb{C}[G]\otimes V is left fixed by any non-identity element of H.

So how does this give us a matrix representation of G? Well, g acts on \mathbb{C}[G] by shuffling around the subspaces that correspond to the cosets of H. In fact, this is exactly the coset representation of G corresponding to H! If we write g=t_ih for some i, then this uses up the transversal element t_i. The h is left to “pass through” and act on V.

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

\displaystyle X\!\!\uparrow_H^G(g)=\begin{pmatrix}X(t_i^{-1}gt_j)\end{pmatrix}=\left(\begin{array}{cccc}X(t_1^{-1}gt_1)&X(t_1^{-1}gt_2)&\cdots&X(t_1^{-1}gt_n)\\X(t_2^{-1}gt_1)&X(t_2^{-1}gt_2)&\cdots&X(t_2^{-1}gt_n)\\\vdots&\vdots&\ddots&\vdots\\X(t_n^{-1}gt_1)&X(t_n^{-1}gt_2)&\cdots&X(t_n^{-1}gt_n)\end{array}\right)

where n is the number of cosets, and we simply define X(t_i^{-1}gt_j) to be a zero block if t_i^{-1}gt_j does not actually fall into H.

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November 25, 2010 - Posted by | Algebra, Group theory, Representation Theory

5 Comments »

  1. [...] 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 , [...]

    Pingback by Characters of Induced Representations « The Unapologetic Mathematician | November 29, 2010 | Reply

  2. [...] we must check that induction is additive. Here, the induced matrices will come in handy. If and are matrix representations of , then the direct sum is the matrix [...]

    Pingback by Induction and Restriction are Additive Functors « The Unapologetic Mathematician | December 1, 2010 | Reply

  3. [...] 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 [...]

    Pingback by (Real) Frobenius Reciprocity « The Unapologetic Mathematician | December 3, 2010 | Reply

  4. [...] we have a matrix representation, so we look at the induced matrix representation. We have to pick a transversal for the subgroup in . Then we have the induced matrix in block [...]

    Pingback by Inducing the Trivial Representation « The Unapologetic Mathematician | December 6, 2010 | Reply

  5. [...] its dimension is a little easier if we recall what induction looks like for matrix representations: the direct sum of a bunch of copies of , one [...]

    Pingback by The Branching Rule, Part 4 « The Unapologetic Mathematician | February 1, 2011 | Reply


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