# The Unapologetic Mathematician

## 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 $H$.

November 25, 2010 -

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