Let $E / F$ be a quadratic extension of nonarchimedean local fields (characteristic 0 if it matters), and $\pi$ an irreducible infinite-dimensional smooth representation of $GL_2(E)$. Let $B$ be the upper-triangular Borel of $GL_2$. I'd like to know: do we always have $$\operatorname{dim} Hom_{B(F)}(\pi, \mathbf{C}) = 1 ?$$ I know how to construct a non-zero element in this space, so my concern is to prove its dimension is $\le 1$, i.e. to show that $(GL_2(E), B(F))$ is a Gelfand pair.

Here's why I think it should be true. Firstly, the analogous result is true if we replace the field extension $E$ with $F \oplus F$; this is the main result of the Harris--Scholl paper cited below. Secondly, it's true if $\pi$ is discrete-series: by Frobenius reciprocity we have $Hom_{B(F)}(\pi, \mathbf{C}) = Hom_{GL_2(F)}(\pi, Ind_{B(F)}^{GL_2(F)}(\mathbf{C}))$. The representation $Ind_{B(F)}^{GL_2(F)}(\mathbf{C})$ has a trivial subrepresentation and the quotient is the Steinberg representation $\mathrm{St}$, so we have a left-exact seq $$0 \to Hom_{GL_2(F)}(\pi, \mathbf{C}) \to Hom_{B(F)}(\pi, \mathbf{C}) \to Hom_{GL_2(F)}(\pi, \mathrm{St}).$$ From the results in Prasad's 1992 paper cited below, exactly one of $Hom_{GL_2(F)}(\pi, \mathbf{C})$ and $Hom_{GL_2(F)}(\pi, \mathrm{St})$ is 1-dimensional and the other is zero, so we are done.

However, if $\pi$ is a principal series, we have the same exact sequence but it can happen that $Hom_{GL_2(F)}(\pi, \mathrm{St})$ and $Hom_{GL_2(F)}(\pi, \mathbf{C})$ are both non-zero (this occurs iff $\pi$ is the normalised induction of a pair of characters of $E^\times$ which are distinct, and both trivial on $F^\times$). I tried to bash out this case via Mackey theory, but I couldn't get it to work.

Harris, M.; Scholl, A. J., A note on trilinear forms for reducible representations and Beilinson’s conjectures., J. Eur. Math. Soc. (JEMS) 3, No. 1, 93-104 (2001). Preprint version Published version

Prasad, Dipendra, Invariant forms for representations of $\text{GL}_2$ over a local field, Am. J. Math. 114, No. 6, 1317-1363 (1992).

  • $\begingroup$ If $E/F$ is a finite field extension (not characteristic 2) and $\pi$ is a generic, principal, non-Steinberg representation, that is not the normalised induction of a pair of characters of $E^\times$ which are distinct, and both trivial on $F^\times$, then I think that $(\pi, 1)_{B(F)}=0$. Are you sure that over local fields the hom-space is at least one dimensional? And if $\pi$ is the normalised induction ... then $(\pi, 1)_{B(F)}=2$, so I would guess that over local fields too $dim\ Hom=2$. $\endgroup$ Dec 18, 2020 at 11:07
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    $\begingroup$ Yes, I'm sure it's non-zero, because one can construct a canonical element of this space using the Asai zeta integral. Over a finite field the reps are all semisimple, so the exact seq is obviously exact on the right; however smooth reps over local fields are not semisimple in general, so the sequence continues into a long exact sequence of Ext groups, and the content of my question is whether or not the map $Hom_{GL_2(F)}(\pi, \mathrm{St}) \to Ext^1_{GL_2(F)}(\pi, \mathbb{C})$ is non-zero. $\endgroup$ Dec 18, 2020 at 12:52
  • $\begingroup$ Did you ask Nadir Matringe ? (Nadir.Matringe@math.univ-poitiers.fr) $\endgroup$ Dec 22, 2020 at 10:47

1 Answer 1


You can approach the problem via the mirabolic subgroup $P_2(F)\subset B_2(F)$. First we can restrict to $\pi$ with central character trivial on $F^\times$. Then you want to know if in this situation $Hom_{P_2(F)}(\pi,1)$ is of dimension at most $1$. This is indeed the case for unitary representations for example by [Matringe, Pacific Journal 2014, Proposition 2.5]. In particular this seems to take care of the remaining case you had of representations of the form $\chi_1\times \chi_2$ with $(\chi_i)_{|F^\times}=1$, which are clearly unitary as both $\chi_i$'s have to be unitary in this case.

  • $\begingroup$ I guess you mean this paper: msp.org/pjm/2014/271-2/pjm-v271-n2-p06-s.pdf. It doesn't seem to have a Proposition 2.3 -- do you mean Proposition 2.5? $\endgroup$ Dec 22, 2020 at 14:59
  • $\begingroup$ Yes I meant Proposition 2.5 sorry $\endgroup$
    – Nadir
    Dec 22, 2020 at 15:22
  • $\begingroup$ Brilliant, thanks Nadir! $\endgroup$ Dec 22, 2020 at 15:45

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