6
$\begingroup$

Let $\pi$ be a generic cuspidal automorphic representation on $GSp(4)$, with level $\Gamma_0(q)$ (the group of symplectic matrices with lower left block divisible by $q$), i.e. $$\Gamma_0(q) = \left\{ \begin{pmatrix} A & B \\ C & D \end{pmatrix} \in Sp_4(\mathbb{Z}) \ : \ C \equiv 0 \mod q \right\}$$

I would like to understand how this notion of level $q$ interacts with the epsilon factor in the functional equations of the attached L-functions, i.e. the classical notion of conductor.

There are two natural representations of the Langlands dual group of $(G)Sp_4$ into a general linear group, viz. the standard $\rho_1 : GSp_4(\mathbb{C}) \to GL_4(\mathbb{C})$ and the spinor $\rho_2 : SO_5(\mathbb{C}) \to GL_5(\mathbb{C})$. These two representations conjecturally induce a compatible map between associated automorphic representations, and allow to define associated L-functions by pulling back the corresponding notions already defined in the $GL_n$ setting. We therefore set $L(s, \pi, \rho_1)$ (resp. $L(s,\pi, \rho_2)$) to be the corresponding L-functions.

These L-functions have been studied precisely in various works (e.g. Piatetskii-Shapiro or Takloo-Bighash), and have been related to local newforms of Roberts and Schmidt, providing explicit formulas in some cases. The global L-function, completed by suitable Archimedean factors, has a functional equation of the form $$\Lambda(1-s, \tilde\pi) = \varepsilon_\pi N_\pi^{s-\tfrac12} \Lambda(s, \pi)$$ for a certain integer $N_\pi$. We call it the arithmetic conductor of $\pi$.

My question is: do we know $N_\pi$ in terms of the level $q$? (I emphasize level since $N_\pi$ is in fact known, by Roberts-Schmidt local newform theory, in terms of the paramodular level; but here I am using a kind of different level, mimicking $GL_n$ congruence subgroups).

$\endgroup$
2
  • 1
    $\begingroup$ As it stands, I'm not sure that this question is well-posed. What do you mean by the level of a cuspidal automorphic representation $\pi$ of $\mathrm{GSp}_4$? Do you mean that there exists a cuspidal automorphic form that is right-invariant by the adelisation of $\Gamma_0(q)$? Is $q$ the minimal such level among all such automorphic forms in $\pi$? Also when you write $\Lambda(s,\pi)$, do you mean $\Lambda(s,\pi,\rho_1)$ or $\Lambda(s,\pi,\rho_2)$? $\endgroup$ Commented Apr 5, 2023 at 14:29
  • $\begingroup$ @PeterHumphries Yes I do mean right-invariance, and $q$ is minimal there. And I would like to understand how these two notions of "level" relate in the case of both $\rho_1$ and $\rho_2$ $\endgroup$ Commented Apr 6, 2023 at 6:41

1 Answer 1

4
$\begingroup$

To make the question well-posed, I'm going to suppose that we fix $\pi$ and take $q$ to be the smallest integer such that $\pi$ has nonzero invariants under $\Gamma_0(q)$. Then $q$ gives you some information about $\pi$, but I don't think you can reliably expect it to nail down the arithmetic conductor exactly; it is rather a miracle that the paramodular invariants do encode the conductor in all cases, you can't just expect a random family of subgroups always to do this.

The question is, of course, a local one: for $\pi$ a representation of $\operatorname{GSp}_4(\mathbb{Q}_\ell)$ having invariants under the Siegel congruence subgroup $Si(\ell^r) \subseteq \operatorname{GSp}_4(\mathbb{Z}_\ell)$, but not under $Si(\ell^{r-1})$, what can the conductor of $\pi$ be?

For $r = 1$ this can be completely answered using the results of Schmidt's paper "Iwahori-spherical representations" (see the tables in the Roberts–Schmidt book for a handy summary). In particular, if we suppose $\pi$ is generic, then $\pi$ has invariants under $Si(\ell)$, but not $Si(1) = \operatorname{GSp}_4(\mathbb{Z}_\ell)$, if and only if $\pi$ is a representation of type IIa, IIIa or VIa with unramified inducing data; and its conductor is $\ell$ if the type is IIa, but $\ell^2$ if the type is IIIa or VIa. This shows that the minimal $r$ at which $\pi$ has nontrivial $Si(\ell^r)$-invariants is not enough to encode the conductor.

(PS. Your question seems to implicitly suggest that the "arithmetic conductors" $N_\pi$ appearing in the functional equation of the degree 4 $L$-factor and the degree 5 $L$-factor are the same. Are you sure about this? I am not sure it is true.)

$\endgroup$
5
  • 3
    $\begingroup$ Your suspicion is correct - It is easy to verify that the arithmetic conductors appearing in the two different $L$-functions are not the same already for Eisenstein series. $\endgroup$
    – Will Sawin
    Commented Apr 5, 2023 at 18:55
  • $\begingroup$ Thanks for clarifying my phrasing, you are absolutely right about the meaning I wanted to give to the level. I tried to make the question uniform in $\rho_1$ and $\rho_2$, but without doubt it has to be understood as two separate questions, one for $\rho_1$ and one for $\rho_2$, and in particular $N_\pi$ also does depend on $\rho_i$. One of the reasons I am bothered about this relation, without hoping much since indeed the two subgroups are pretty different, are the recent works on GSp(4) Kuznetsov, that use the notion of level with respect to these congruence subgroups, not the paramodular $\endgroup$ Commented Apr 6, 2023 at 6:45
  • $\begingroup$ I think the "moral" has to be that (once you go beyond $GL_2$) there are lots of ways in which a representation can be ramified, and you can't expect any single number to measure the amount of ramification perfectly. $\endgroup$ Commented Apr 6, 2023 at 14:25
  • $\begingroup$ @DesideriusSeverus Adding to David Loeffler's comment above, you already see this for $\mathrm{GL}_n$ with $n \geq 3$; one can derive a Kuznetsov formula for the principal congruence subgroup $\Gamma(q)$ instead of $\Gamma(q)$ (e.g. Assing-Blomer), which has important applications regarding density theorems, but measures ramification of automorphic representations of $\mathrm{GL}_n$ in a way that is unrelated to the arithmetic conductor. $\endgroup$ Commented Apr 6, 2023 at 14:43
  • $\begingroup$ Whoops, I meant "$\Gamma(q)$ instead of $\Gamma_0(q)$" $\endgroup$ Commented Apr 6, 2023 at 17:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .