Let $G = \operatorname{GL}_n(F)$ with the usual Borel subgroup $P = TU$. Let $\chi = \chi_1 \otimes \cdots \otimes \chi_n$ be an unramified character of $T$. Suppose that $\chi$ is regular, which is to say that $\chi_i \neq \chi_j$ for $i \neq j$. Equivalently, $\chi \neq w.\chi$ for all $w \in W(T,G)$.

Then $I(\chi) = \operatorname{Ind}_{TU}^G \chi \delta^{1/2}$ is irreducible if and only if $\chi_i \neq \chi_j | \cdot |_F$ for all $i \neq j$. Is this true without the regularity assumption on $\chi$? This is claimed in Example 4.2 of Prasad and Raghuram's notes on representation theory for $\operatorname{GL}_n$, but I wasn't sure if they were making an underlying assumption of regularity.


1 Answer 1


Suppose $F$ is a non-archimedean local field. This regularity assumption is not needed, nor is the assumption that $\chi$ is unramified. The sufficient part for irreducibility is part of Theorem 4.2 of Bernstein and Zelevinsky - Induced representations of reductive $\mathfrak p$-adic groups. I and the other part is part of Theorem 6.1 in Zelevinsky - Induced representations of reductive $\mathfrak p$-adic groups. II. On irreducible representations of $\mathrm{GL}(n)$.

Edit: To respond to Vincent's comment, let me say what I know (very limited): Note that when $F$ is a finite field, $\delta$ and $|\cdot|_F$ are trivial and the assertion follows from Mackey theory that $$\mathrm{Hom}_G(I(\chi),I(\chi))=\bigoplus_{w\in N_G(T)/T}\mathrm{Hom}_T(\chi,{}^w\chi)$$ When one tries to adapt this to (any) local field, the normalization factor $\delta$ and $|\cdot|_F$ should naturally arise. In general $I(\chi)$ is no longer semisimple and one needs some tricks to know a bit more than the $\mathrm{Hom}$. Beyond this I know nothing about how the case $F$ archimedean has in common with the $p$-adic case.

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    $\begingroup$ This is for $p$-adic groups right? Do you know if there is an analogous statement for $F = \mathbb{R}$? It sounds like everything should work the same there, but for different reasons, but maybe there is some subtlety that I am overlooking? $\endgroup$
    – Vincent
    Jul 13, 2018 at 7:48

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