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I am very much a beginner in the theory of automorphic forms and I might (will?) make mistakes in what follows. Please correct me.

A loose interpretation of the Langland's philosophy is that to any variety $X/\mathbb Z$, we should be able to find an automorphic form $f$ so that $\zeta(X) = \zeta(f)$ where $\zeta(f)$ is the Hasse-Weil zeta function and $\zeta(f)$ is the L-function associated to an automorphic form.

Question 1: What automorphic objects do the zeta functions of varieties over finite fields correspond to?

Tentative answer: It seems to me that these should correspond to euler factors at a prime of an automorphic form. (Lift the variety to characteristic 0, find the zeta function of that and take the corresponding automorphic form).

Is this on the right track?

Question 2: If so, what property of the automorphic forms does the Riemann hypothesis for the Weil conjectures over finite fields (proved by Deligne) correspond to? More generally, what do the Weil Conjectures correspond to on the automorphic side.

I can see two possibilities:

1) They correspond to some conjectural property of automorphic forms and the proof of the Weil conjectures actually tells us something about (a subclass of) automorphic forms.

2) They correspond to some known (easy to prove?) property of automorphic forms and the "reason" Deligne/Grothendieck had to work so hard is because we don't know how to associate automorphic forms to motivic L functions.

Question 3: Which one is right?

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  • $\begingroup$ There should be admissible representations of $GL_n(K_v)$ somewhere with $K_v/\mathbf{Q}_p$ and its residue field $k_v$ being your finite field ? (automorphic representations assemble admissible representations of $GL_n(K_v)$ with the restriction of being trivial on the diagonal embedding $GL_n(K) \to \prod_v GL_n(K_v)$) $\endgroup$
    – reuns
    Commented Oct 24, 2018 at 21:01

1 Answer 1

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This is a brief answer; possibly others have different opinions about this.

Question 1: The Langlands conjectures gives a correspondence between Galois representations and automorphic forms. So a very naive way to view the zeta function of a variety over a finite fields from this philosophy is to look for a Galois representation which it comes from. However, this is exactly the point of the Weil conjectures; the zeta function has a description in terms of the action of the absolute Galois group of the finite field on the etale cohomology of the variety. These indeed give the Euler factors.

Question 2: The Riemann hypothesis for the Weil conjectures over finite fields corresponds to the Ramanjuan conjecture for automorphic forms. In fact, this was one of Deligne's original applications of the Weil conjectures, to proving the Ramanjuan conjecture for the Ramanjuan tau function.

Question 3: The answer is that it is a mixture of 1) and 2). For 1), as I said above, the Weil conjectures give you results towards the Ramanjuan conjecture for automorphic forms. For 2), it is a standard result that each Euler factor of an automorphic $L$-function is a rational function in $p^{s}$, but proving the rationality of the zeta functions of varieties over finite fields was one of the first difficult steps in the proof of the Weil conjectures.

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    $\begingroup$ Thanks, this is along the lines of what I was thinking. However, doesn't the ramanujan conjecture seem weaker than the statement that the roots have a prescribed absolute value bound? As far as I know, the ramanujan conjecture only seems to bound the trace (I only know the modular form case...) $\endgroup$
    – Asvin
    Commented Oct 24, 2018 at 20:36
  • $\begingroup$ Also i edited in a third question (which i think you partially answer in your answer to the second question). $\endgroup$
    – Asvin
    Commented Oct 24, 2018 at 21:03
  • $\begingroup$ I'm not sure about in general, but at least in the case of curves, the Hasse-Weil bound $|C(\mathbb{F}_p) - p +1| \leq 2g \sqrt{p}$ is equivalent to the Riemann hypothesis for the zeta function (see en.wikipedia.org/wiki/Hasse%27s_theorem_on_elliptic_curves). $\endgroup$
    – Daniel Loughran
    Commented Oct 24, 2018 at 21:19
  • $\begingroup$ Right, but that is because we already know the discriminant. Maybe the two are equivalent on general but I don't know... $\endgroup$
    – Asvin
    Commented Oct 24, 2018 at 21:19
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    $\begingroup$ @Asvin For newforms we know the product of eigenvalues of Frobenius, so the bound on the trace is equivalent to the statement on the absolute values of the eigenvalues. For general automorphic forms the generalized Ramanujan conjecture is about the absolute values of the eigenvalues. $\endgroup$
    – François Brunault
    Commented Oct 25, 2018 at 18:54

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