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Let $X$ be an (as nice as you prefer) alg. variety (or alg. stack) defined over $\mathbb{F}_{q}$ and let $\mathcal{F}$ be an l-adic sheaf on $X_n = X {\times_{\mathbb{F}q}} \mathbb{F}_{q^n}$. Fix an isomorphism between $\mathbb{C}$ and $\overline{\mathbb{Q}}_l$ (otw replace below $\mathbb{C}$ by the $l$-adic numbers). Suppose we know that the sheaf $\mathcal{F}$ comes (by pullback) from a sheaf, say $\mathcal{G}$, defined on $X$.

I would like to know if it is possible to recover the trace function of $\mathcal{G}$, say $g = f_\mathcal{G}:X(\mathbb{F}_q) \to \mathbb{C}$ defined by

$f_\mathcal{G}(x) = tr(Frob_x,\mathcal{G}_x)$ from the (analogously defined) trace function of $\mathcal{F}$, say

$f:=f_\mathcal{F}:X(\mathbb{F}_{q^n})\to\mathbb{C}$

I see that already for $X=\mathbb{F}_q$ you can only recover the trace function $g$ up to an $n$-th root of unity. So the following questions might look a bit silly:

  1. What's the best thing that we can do in general? i.e. if I have $f$ can I find $g$ up to some constant ($n$-th root of unity; need properness?).
  2. If I have $f$ what additional structure/information would I need to recover (uniquely) the function $g$?

Also, if you know any reference in which similar problems are treated please let me know.

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  • $\begingroup$ Since the Frobenius for $\mathbb{F}_n$ is the nth power of the Frobenius for $\mathbb{F}$, you seem to be asking: what can you tell about (the trace of) a matrix from knowing (the trace of) its nth power? $\endgroup$
    – anon
    Commented Dec 17, 2012 at 20:08

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As anon points out, the answer in general is that one cannot recover it up to trace. Indeed, a sheaf on $\mathbb F_q$ is just a matrix up to conjugation. The corresponding sheaf on $\mathbb F_{q^n}$ is just that matrix to the $n$th power. There is no relation between the traces.

However, if you have the characteristic polynomial of the matrix, you're in luck. The eigenvalues of the sheaf on $\mathbb F_q$ must of course be $n$th roots of the eigenvalues of teh sheaf on $\mathbb F_q^n$, which gives you at most $n^k$ possibilities for a sheaf of rank $k$.

Similarly, if we want to get down to a list of possibilities for a sheaf on a larger space, we need to look at the whole sheaf. So a different version of your question would be this:

Given a sheaf $\mathcal G$ on $X_n$, how many descents does it have to $X$, and what are they?

Then we compute the trace on each descent. A descent is just going to be a map $\mathcal G \to \left(Frob_q\right)^*\mathcal G$ such that the composition of the map with itself $n$ times is the identity.

One case in which you are alright is if $\mathcal G$ is lisse and irreducible, and the base is connected, so the automorphism group of $\mathcal G$ is just $\mathbb G_m$. Then any two descents must differ by an $n$th root of unity, and their traces differ by an $n$th root of unity, and you're off.

If $\mathcal G$ is lisse and semisimple, then you can write it as a sum of irreducible pieces of this form. You can throw away the pieces that are not fixed by $Frob_q$, because the trace on those parts will always be zero. Then, on a piece that is fixed, the automorphism group will be $\mathbb G_m$, so the trace will be defined by the sheaf up to a $n$th root of unity.

If you are not lisse or not semisimple, things get a bit messier.

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  • $\begingroup$ @Will: thank you for the answer. My sheaves are actually perverse sheaves. I'll see if I can develop what you said. $\endgroup$
    – Dragos Fratila
    Commented Dec 18, 2012 at 13:09
  • $\begingroup$ Well perverse sheaves are an abelian category, so you can decompose into irreducible components. If your irreducible components are lisse sheaves on smooth subvarieties you're in good shape, because you can use the same method as for a lisse semisimple sheaf, since semisimplification is functorial. However this gives you an overestimate. It could also be helpful to use the adjoint functor approach. $\endgroup$
    – Will Sawin
    Commented Dec 18, 2012 at 16:11
  • $\begingroup$ what is the adjoint functor approach? $\endgroup$
    – Dragos Fratila
    Commented Dec 18, 2012 at 16:42
  • $\begingroup$ Let $\pi: X_n \to X$ be the obvious map. Then you're trying to find $\mathcal F$ such that $\pi^* \mathcal F \cong \mathcal G$, so you're trying to find pairs of a sheaf $\mathcal F$ and a special element in $Hom( \pi^* \mathcal F,\mathcal G)$. But this is just the same as $Hom(\mathcal F, \pi_* \mathcal G)$. Moreover this preserves injections, so we are tasked with finding certain subobjects of $\pi_* \mathcal G$. If we can develop a good understanding of $\pi_* \mathcal G$ this may be the correct approach. $\endgroup$
    – Will Sawin
    Commented Dec 18, 2012 at 22:55
  • $\begingroup$ @Will Ok, I see. A thought: we know $\mathcal{G}$ and supposedly the eigenvalues of the Frobenius on its stalks so I would like to take $n$-th roots of these eigenvalues. From what you said, if the sheaf is irreducible (be it lisse or perverse) there should be only one way to do this up to an $n$-th root of unity. Is this correct? $\endgroup$
    – Dragos Fratila
    Commented Dec 19, 2012 at 17:31

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