# When is induction to Siegel parabolic of cuspial representation reducible

Hello,

I'm currently looking at the symplectic group $Sp_4(\mathbb{F}_q)$ ($q$ odd), trying to understand some of its comlpex representation theory. I note that it has been completely calculated by Srinivasan (1968). A partial description appears in "Seminar on Algebraic Groups and Related Finite Groups", Springer et al (1970). My question regards what is written in the second to last section of the description: my calculations turn out a tiny bit different from theirs.

There are two parabolics, $P_1$, $P_2$, with Levi decompositions $P_i=M_iU_i$, and $$M_1 \simeq SL_2\times GL_1\,\ M_2 \simeq GL_2$$

Let $\tau$ be an irreducible cuspidal representation of $GL_2$, and $\hat\tau$ be its lift to $P_2$ (by way of $M_2$). Let $\theta$ be the corresponding character of the anisotropic torus $T$ in $GL_2$ (well, one of them, doesn't matter which). Let $\rho$ be the representation $Ind_P^G \hat{\tau}$

Enough notation:

If I did my first exercise correctly, a simple application of Mackey theory shows that $\rho$ is reducible if and only if the central character of $\tau$ is trivial.

If I did my second exercise correctly, the central character of $\tau$ is trivial if and only if $\theta^{q+1}\equiv 1$ (the center being exactly the $q+1$ powers), if and only if $\theta^{-1}=\theta^q$.

This is not what it is written in "Seminar on". Instead, the condition $\theta=\theta^q$ is claimed. In fact, if I did a third exercise correctly, if $\theta=\theta^q$ then the virtual character $R_T^\theta$ is reducible and $\theta$ doesn't actually give rise to an irreducible cuspidal representation of $GL_2$.

What is the correct condition on $\theta$ for $\rho$ to be reducible?

-
In any case, in case it helps, this situation is the same as a theta-correspondence (over the finite field) sending functions on an anisotropic O(2) to Sp(4-by-4), by acting on functions on $\mathbb F_q^{2\times 2}$. This does group the characters of the $SO(2)$ usefully. –  paul garrett Aug 1 '11 at 22:30