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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?

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    $\begingroup$ This is true for the $\mathrm{GL}_2(F)$ principal series containing the special representation as a subrepresentation. $\endgroup$
    – Peter Humphries
    Commented Feb 17, 2020 at 13:57
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    $\begingroup$ 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
    Commented Feb 17, 2020 at 14:22
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    $\begingroup$ Look up the Steinberg representation $\endgroup$
    – Peter Humphries
    Commented Feb 17, 2020 at 14:33
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    $\begingroup$ @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
    Commented Feb 17, 2020 at 14:42
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    $\begingroup$ @Monty no, it's not necessarily trivial---you can also get unramified twists. $\endgroup$
    – Kimball
    Commented Feb 17, 2020 at 16:16

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

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  • $\begingroup$ 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
    Commented Feb 17, 2020 at 14:40
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    $\begingroup$ Look at the contragriedient representation, it will be unramified principal series and ti will have trivial rep as a quotient. $\endgroup$
    – user152491
    Commented Feb 17, 2020 at 15:05
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    $\begingroup$ @user152491, perhaps it would be worthwhile to enlarge your answer to include the point about taking contragredients... $\endgroup$
    – paul garrett
    Commented 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
    Commented Mar 18, 2020 at 16:46

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