# The Unapologetic Mathematician

## Inducing the Trivial Representation

We really should see an example of inducing a representation. One example we’ll find extremely useful is when we start with the trivial representation.

So, let $G$ be a group and $H$ be a subgroup. Since this will be coming up a bunch, let’s just start writing $1$ for the trivial representation that sends each element of $H$ to the $1\times1$ matrix $\begin{pmatrix}1\end{pmatrix}$. We want to consider the induced representation $1\!\!\uparrow_H^G$.

Well, we have a matrix representation, so we look at the induced matrix representation. We have to pick a transversal $\{t_i\}$ for the subgroup $H$ in $G$. Then we have the induced matrix in block form: $\displaystyle1\!\!\uparrow_H^G(g)=\begin{pmatrix}1(t_1^{-1}gt_1)&1(t_1^{-1}gt_2)&\cdots&1(t_1^{-1}gt_n)\\1(t_2^{-1}gt_1)&1(t_2^{-1}gt_2)&\cdots&1(t_2^{-1}gt_n)\\\vdots&\vdots&\ddots&\vdots\\1(t_n^{-1}gt_1)&1(t_n^{-1}gt_2)&\cdots&1(t_n^{-1}gt_n)\end{pmatrix}$

In this case, each “block” is just a number, and it’s either $1$ or $0$, depending on whether $t_i^{-1}gt_j$ is in $H$ or not. But if $t_i^{-1}gt_j\in H$, then $t_i^{-1}gt_jH=H, and$latex g(t_jH)=(t_iH)\$. That is, this is exactly the coset representation of $G$ corresponding to $H$. And so all of these coset representations arise as induced representations.

December 6, 2010 -

## 1 Comment »

1. […] now we can define the -module by inducing the trivial representation from the subgroup to all of . Now, the are not all irreducible, but we will see how to identify a […]

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