8
$\begingroup$

My question is about a local computation in the paper of Gelbart and Jacquet, "A relation between automorphic representations of GL2 and GL3", from 1978.

Let $\sigma$ be an irreducible smooth complex representation of $GL_2(F)$, where $F$ is a non-archimedean local field (characteristic 0 if it helps), and $\chi$ a smooth character of $F^\times$.

Jacquet has defined a Rankin--Selberg $L$-factor $L(\sigma \otimes \sigma^\vee \otimes \chi, s)$, which is always of the form 1/(polynomial in $q^{-s}$) where $q$ is the size of the residue field. Gelbart and Jacquet define the adjoint L-factor by $$ L(Ad(\sigma), \chi, s) := \frac{L(\sigma \otimes \sigma^\vee \otimes \chi, s)}{L(\chi, s)}.$$

In the Gelbart--Jacquet paper, they write down an integral $I(s, f, \Phi, \Psi, W)$ on the metaplectic group $Mp_2(F)$, which depends on $s$ and various choices of auxiliary data ($W$ is a vector in the Whittaker model of $\sigma$, $\Phi$ and $\Psi$ are Schwartz functions on $F$, etc).

If either:

  • $\sigma$ is unramified (or a twist of an unramified representation);
  • or $\sigma$ is ramified, but $\chi$ is much more ramified than $\sigma$ is, so both $L(\sigma \otimes \sigma^\vee \otimes \chi, s)$ and $L(\chi,s )$ are identically 1,

then they show that the auxiliary data can be chosen in such a way that $I(s, \dots) = L(\operatorname{Ad}(\sigma), \chi, s)$.

Does this hold more generally? Is it true, for arbitrary $\sigma$ and $\chi$, that we can choose a finite collection of quadruples $(f_i, \Phi_i, \Psi_i, W_i)$, $i=1 \dots r$, such that $$\sum_{i=1}^r I(s, f_i, \Phi_i, \Psi_i, W_i) = L(\operatorname{Ad} \sigma, \chi, s)?$$

(EDIT: I didn't make it plain originally that I was willing to allow a finite collection of test data, rather than just one. This form of the statement is true, essentially by definition, for the integral representations of the standard L-function on $GL_n$, and the Rankin--Selberg L-function on $GL_m \times GL_n$, so I'm looking for a generalisation of that to the adjoint L-function.)

$\endgroup$
2
  • $\begingroup$ In general, it is quite challenging to choose test data (Whittaker functions and possibly Schwartz functions) for Eulerian integrals that give $L$-functions. For example, this is already very difficult for Ranking-Selberg integrals for $\mathrm{GL}_2 \times \mathrm{GL}_2$ at nonarchimedean places with both representations ramified. $\endgroup$ May 6, 2016 at 20:44
  • $\begingroup$ On reflection, my question was badly worded -- I was thinking of test data as living in the tensor product (Schwartz functions) x (Whittaker functions) x etc, so "linear combinations" are meaningful. I didn't want to necessarily ask for a pure tensor in this space. I've clarified the question. $\endgroup$ May 7, 2016 at 16:01

1 Answer 1

3
$\begingroup$

I'm not familiar with Gelbart and Jacquet's integral representation you mention (so this is more of a long comment than an answer). One can see some techniques from the local Rankin--Selberg GLn\times GLm convolutions setting in:

  • the 2010 dissertation of Kyung-Mi Kim (called Test Vectors of Rankin--Selberg convolutions of general linear groups and available online);

  • and in my work with Nadir Matringe last year (arXiv.1501.07587v3) which treats the cuspidal case and gives the existence of data such that a single Rankin-Selberg integral realises the L-factor (we give explicit data using Bushnell--Kutzko type theory when the L-factor is non-trivial (hence in particular n=m); that there exists such data when the L-factor is trivial is known from the work of Jacquet--Piatetski-Shapiro--Shalika as we mention in our introduction.)

You can find more details in the introductions to both of the above. In the Rankin--Selberg case, we expect that an amalgamation of these techniques would answer the general question (giving a finite number of explicit vectors such that the sum of the corresponding integrals realises the L-factor), however determining under which conditions there is a single integral I'm not sure.

$\endgroup$
1
  • 1
    $\begingroup$ This is a great answer to the question I asked, but in retrospect I actually wanted to ask a slightly different question (one which is true by definition for the Rankin--Selberg L-factor). I've edited the question to clarify. $\endgroup$ May 7, 2016 at 16:03

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.