5
$\begingroup$

There is a well-known analogy between holonomic $\mathcal{D}$-modules on complex algebraic varieties and $\ell$-adic perverse sheaves on varieties over finite fields. Many theorems in one setting have natural analogues in the other. (Even if the proofs are sometimes completely different.)

A major breakthrough on the de Rham side is the Kedlaya–Mochizuki theorem ([Ked, Thm. 8.1.3; Moc, Thm. 1.3.3]), which morally states that any holonomic $\mathcal{D}$-module can, locally, be decomposed into a direct sum of regular holonomic pieces tensored with exponential twists. This result was instrumental in the proof of the irregular Riemann–Hilbert correspondence.

Question: What would be the appropriate $\ell$-adic analogue of the Kedlaya–Mochizuki theorem for $\ell$-adic perverse sheaves, and is such a result known to be true?

References:

  • [Ked] K. Kedlaya - Good formal structures for flat meromorphic connections, II: Excellent schemes
  • [Moc] T. Mochizuki - Wild Harmonic Bundles and Wild Pure Twistor $\mathcal{D}$-modules
$\endgroup$
4
  • 2
    $\begingroup$ I'm not an expert on the de Rham side (D-modules), but isn't the analogy between perverse sheaves and regular holonomic D-modules? So from that point of view, the irregular ones would sort of look like "junk", at least in the sense that they have no $\ell$-adic (or Betti) analogue. $\endgroup$ Commented 7 hours ago
  • 1
    $\begingroup$ Some online searching gives papers explaining exactly that: to even state an irregular Riemann–Hilbert correspondence, one first needs to find an appropriate enlargement of constructible sheaves that is engineered to upgrade the correspondence. So I don't think this translates to naturally arising questions on perverse sheaves. $\endgroup$ Commented 7 hours ago
  • 2
    $\begingroup$ Hi Remy: In positive char, irregular singularities morally correspond to wild ramifications. For example, on A1, Artin-Schreier serves as an analogue of (O, d + dt). $\endgroup$
    – Function
    Commented 7 hours ago
  • $\begingroup$ Dear @R.vanDobbendeBruyn, the Riemann-Hilbert correspondence gives an equivalence of categories between regular holonomic D-modules on a complex algebraic variety and perverse sheaves on its analytification. What I mean is truly different. For example, let X be a finite type scheme over the integers then base change it to a scheme X_0 over C and X_p over F_p. I mean that holonomic D-modules on X_0 "behave similarly" to l-adic perverse sheaves on X_p. Contrarily to RH, this is not a theorem. Only a large collection of analogies. $\endgroup$
    – Gabriel
    Commented 4 hours ago

1 Answer 1

6
$\begingroup$

In one dimension we have the following result:

Let $X$ be a curve over a finite field $k$ of characteristic $p$} and $x\in X$ a point. Let $K$ be a perverse sheaf on $X$ of generic rank less than $p$. Then the pullback of $X$ to the étale local ring of the point $x$ can be decomposed as a sum of pushforwards from tamely ramified covers of tamely ramified perverse sheaves tensored with Artin-Schreier-Witt sheaves.

However, the result is manifestly untrue if the generic rank is greater than $p$. For example, the rank $p$ Kloosterman sheaf is already a counterexample.

I do not know whether there is an analogue for low-rank sheaves in higher dimensions.

Proof sketch: We can first decompose the perverse sheaf into the direct sum of a part with unipotent monodromy and a part with no monodromy invariants, whose stalk at the point $x$ vanishes. The unipotent part is tame, allowing us to restrict attention to sheaves whose stalk at $x$ vanishes, which are equivalent to representations of the Galois group of the local field at $x$. Thus we have a representation of the Galois group of rank at least $p$. Restricted to the wild inertia group, which is a pro-$p$ group, it splits into representations of rank a power of $p$, hence splits into representations of rank at least $1$. The Galois group acts on the isomorphism classes of these representations by finitely many orbits, and the representation splits as a direct sum over orbits, so we may restrict attention to a single orbit. The stabilizer of the orbit defines a cover, necessarily tamely ramified, and we can write a representation whose characters all lie in that orbit as an induced representation from the stabilizer, i.e. a pushforward from the associated cover, of a representation where the wild inertia group acts by scalars. The order of the group of scalars that arise this way is necessarily a power of $p$. Since the dimension of the representation is still less than $p$, taking the determinant and raising to the inverse of the dimension modulo this power of $p$ produces a one-dimensional representation on which the wild inertia group has the same action. The one-dimensional representation can be decomposed into an Artin-Schreier-Witt character and a tame character, and tensoring with the inverse of the Artin-Schreier-Witt character produces a representation on which wild inertia acts trivially, i.e. a tamely ramified representation, as desired.

$\endgroup$
2
  • $\begingroup$ Hi Will! Instead of generalizing this result to higher dimensions, do you have any idea on how to generalize it to higher-rank sheaves on curves? Very naively, I imagine that one should generalize the Artin-Schreier-Witt sheaves to something akin to the exponential connection of a meromorphic function. $\endgroup$
    – Gabriel
    Commented 4 hours ago
  • $\begingroup$ @Gabriel I don't see any difference between the exponential connection of a rational function and the exponential section of a meromorphic function in this context. (We're working locally, a meromorphic function is locally a rational function plus a holomorphic function, and the exponential connection of a holomorphic function is locally trivial). The issue is that $p$-dimensional irreducible representations of the wild inertia group are not like any kind of exponential connection, since they are $p$-dimensional and exponential connections are $1$-dimensional. $\endgroup$
    – Will Sawin
    Commented 4 hours ago

You must log in to answer this question.

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