1
$\begingroup$

Let $F$ be a local field of characteristic 0.

I am wondering whether an unramified principal series representation of $\operatorname{GL}_n(F)$ can have 1-dimensional quotient when $n>1$.

In some paper, the author claims that it can’t.

Do you know the reason?

Improve this question
$\endgroup$
9
  • 3
    $\begingroup$ This is true for the $\mathrm{GL}_2(F)$ principal series containing the special representation as a subrepresentation. $\endgroup$
    – Peter Humphries
    Feb 17, 2020 at 13:57
  • 1
    $\begingroup$ I just answered the case $n=2$. I don’t know the answer off the top of my head for $n>2$. $\endgroup$
    – Peter Humphries
    Feb 17, 2020 at 14:22
  • 1
    $\begingroup$ Look up the Steinberg representation $\endgroup$
    – Peter Humphries
    Feb 17, 2020 at 14:33
  • 3
    $\begingroup$ @PeterHumphries It's the same thing for general $n$---for the analogous induction of powers of absolute values from the Borel, one irreducible subquotient is Steinberg and one is trivial. The trivial one is of course the unique unramified subquotient of the corresponding unramified reducible principal series. $\endgroup$
    – Kimball
    Feb 17, 2020 at 14:42
  • 1
    $\begingroup$ @Monty no, it's not necessarily trivial---you can also get unramified twists. $\endgroup$
    – Kimball
    Feb 17, 2020 at 16:16

1 Answer 1

Reset to default
3
$\begingroup$

Think of $Ind_B^G 1$ as smooth functions on $G/B$ and look at the subspace of constant functions.

Improve this answer
$\endgroup$
4
  • $\begingroup$ The constant function forms a 1-dimensional subrepresentation of $Ind_B^G 1$. But how about quotients? Does it can have 1-dimensional quotients? $\endgroup$
    – Monty
    Feb 17, 2020 at 14:40
  • 2
    $\begingroup$ Look at the contragriedient representation, it will be unramified principal series and ti will have trivial rep as a quotient. $\endgroup$
    – user152491
    Feb 17, 2020 at 15:05
  • 1
    $\begingroup$ @user152491, perhaps it would be worthwhile to enlarge your answer to include the point about taking contragredients... $\endgroup$
    – paul garrett
    Mar 18, 2020 at 15:36
  • $\begingroup$ @Monty, your question currently asks about subquotients, so it's probably best to edit it if you only want to discuss quotients. $\endgroup$
    – LSpice
    Mar 18, 2020 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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