You are currently browsing the tag archive for the ‘GL(n)’ tag.

Let G={\rm GL}(n,\mathbb{F}_q) and let N be the nilpotent radical of the standard Borel subgroup, i.e., N is the subgroup of G consisting of upper triangular matrices with 1’s on the diagonal. Given a non-trivial character \psi: \mathbb{F}_q\rightarrow \mathbb{C}^\times, we define a character \psi_N of N by

\psi_N( \begin{pmatrix}  1 & x_{12} & x_{13} & \cdots & x_{1n}\\  & 1 & x_{23} & \cdots & x_{2n}\\  & & 1 & \cdots & \vdots \\  & & & \ddots & \vdots\\  & & & & 1\\  \end{pmatrix}) = \psi(x_{12}+x_{23}+\cdots+x_{n-1, n}).

This defines a one-dimensional representation of N. One can show that the induced representation \mathcal{G}={\rm Ind}_N^G(\psi_N) is multiplicity-free, i.e., when we decompose \mathcal{G}={\rm Ind}_N^G(\psi_N) into a sum of irreducible constituents,

\mathcal{G}={\rm Ind}_N^G(\psi_N)=\oplus a_i\pi_i,

we have a_1=1 for all i. Each constituent \pi_i is called a generic irreducible representation. There are many reasons for calling them generic. One is that no matter which non-trivial character \psi we choose, we obtain the same induced representation up to equivalence. One can see this by looking at the induced character values on each conjugacy class. These values will be polynomials in q and, once simplified, do not depend on the character \psi. Moreover, this Gelfand-Graev representation \mathcal{G} contains most irreducible representations of the group.

To see this, we can compare the number of irreducible constituents of \mathcal{G} with the number of irreducible representations of G. By a standard result in the representation theory of finite groups, the number of irreducible representations of G is equal to its number of conjugacy classes.

Read the rest of this entry »

%d bloggers like this: