The Unapologetic Mathematician

Mathematics for the interested outsider

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 - Posted by | Algebra, Group theory, Representation Theory

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 […]

    Pingback by The Road Forward « The Unapologetic Mathematician | December 7, 2010 | Reply


Leave a comment

« Previous | Next »