Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a local field, $G=GL_2(F)$, $\chi_1,\chi_2$ quasicharacters, and $B(\chi_1,\chi_2)$ be the principal series (we are assuming $\chi_1,\chi_2$ to be such that the representation is irreducible).

Is $B(\chi_1,\chi_2)$ a constituent of $L^2(Z\backslash G)$?

I'm pretty sure it never is, but can't prove it. Am familiar with Bump up to 4.5.

On the other hand, say $\chi_1\chi_2^{-1} = |\cdot |^{\pm 1}$ (so not a principal series). Same question?

This time I think it is...

share|cite|improve this question
up vote 7 down vote accepted

A principal series representation is never square integrable. Probably in the case of $GL_2$ one can compute matrix coefficients explicitly to verify this, but it also follows from the (somewhat more theoretical) computation of matrix coefficients via the Jacquet module. A good reference for this is Casselman's notes (available on his web-page), which develop the theory of the Jacquet module and its application to matrix coefficients essentially from scratch. (Casselman work's in the context of an arbitrary reductive group, and so uses some basic Lie theoretic results as input; but in the case of $GL_2$ one can just replace that input by explicit facts and computations about $GL_2$ and its Borel subgroups.)

In the case when the induced representation is trivial, the infinite dimensional constituent (which I would call special; but I don't know what Bump's terminology is) will be square integrable (up to a twist; one may need to apply a twist to ensure that the central character is unitary, if it is not already).

Again, this can be seen via a Jacquet module computation: the Jacquet module of an induced representation for $GL_2$ is two-dimensional, and so there are two exponents, one of which gives rise to square integrable matrix coefficients, but the other of which doesn't. In the reducible case, the one-dimensional constituent is contributing the "bad" exponent, and so when we remove that constituent, the infinite-dimensional piece that's left has just a one-dimensional Jacquet module, whose exponent is good enough to ensure square integrability.

My memory is that Casselman discusses this example specifically in his notes.

share|cite|improve this answer

If the principal series would be square integrable, they would be discrete points in the Fell topology. It can be seen that the $B(\chi_1, \chi_2)$ varies continuously with the pair $(\chi_1, \chi_2)$ with the usual topology of local characters as long as it is irreducible.

Now, the point $\chi_1 \chi_2^{-1}$ being the norm or its inverse, the representation contains the Steinberg representation (times a one-dim'l twist) as a submodule or -quotient, being not irreducible anymore. Otherwise, it is irreducible.

The Steinberg (times a one-dim'l twist) and the supercuspidal representations are square-intgerable, if the central character is unitary.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.