# Kiran Kedlaya: Crystalline companions as an anabelian phenomenon

:::{.remark}
Setup: $k = \FF_q, q=p^n, X\in \Sch\slice k$ smooth geometrically connected, $\ell\neq p$ arbitrary.
Recall
\[
1\to \pi_1 X_{\kbar} \to \pi_1 X \to G_k = \ZZhat \to 1
.\]
Anabelian philosophy: everything you'd want to know about $X$ is contained in $\pi_1 X$.
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:::{.theorem title="Tamagawa"}
If $X$ is an affine curve, then $\pi_1^{\tame} X$ determines $X$.
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:::{.remark}
Problem: if $X$ is an affine genus $g$ curve with $m$ punctures, $\pi_1^{\primetop} X_{\kbar}$ is the prime-to-$p$ completion of $\Free(2g+m-1)$, independently of $X$.
Since sections induce $G_k\to \Out(\pi_1 X_{\kbar})$, we have lots of tame continuous $\bar{\QQladic}$ reps of $\pi_1 X_{\kbar}$, but very few are fixed by Frobenius.
:::

:::{.conjecture title="Deligne"}
Such representations only have "geometric origins", i.e. if $\mce$ is a lisse $\QQladic\dash$sheaf, i.e. a lisse $F\dash$sheaf with $[F:\QQladic] < \infty$, which is irreducible with determinant of finite order, then it appears on relative etale cohomology of $\pi:Y\to X$ for some $Y$.
:::

:::{.remark}
This is known in $\dim X = 1$, due to Deligne, around the same time Drinfeld proved Langlands for $\GL_2(k(X))$ for $k(X)$ a function field (or really the adeles).
So all arithmetic reps of $\pi_1$ *come from geometry*.
:::

:::{.remark}
Note that $Y$ will eventually not even be a scheme.
The determinant condition rules out transcendental twists.
Galois side: lisse sheaves on $X$; automorphic side: reps values in $\GL_n(\AA_K)$ for $K$ a field.
The proof above involves exhibiting the Galois objects as coming from relative etale cohomology in moduli of shtukas.
A priori one only knows Frobenius traces, but this turns out to be enough to uniquely characterize things in this situation.
:::

:::{.conjecture}
Later Lafforgue did this for $\GL_n$, but the corresponding statement about arithmetic reps is wide open.
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:::{.remark}
If $\mce$ as above, the Frobenius traces at all closed points $x\in \abs{X}$ are algebraic over $\QQ$.
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:::{.definition title="Companion sheaves"}
Fix an algebraic closure $\QQbar$ and two primes $\ell, \ell'\neq p$ and fix embeddings $\QQbar\injects \QQbar_\ell$ and $\QQbar \injects \QQbar_{\ell'}$.
Let $\mce,\mce'$ be lisse $\QQbar_\ell, \QQbar_{\ell'}$ sheaves (resp.) on $X$.
Say $\mce,\mce'$ are **companions** iff for every $x\in\abs{X}$ the Frobenius traces at $x$ are equal in $\QQbar$.
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:::{.remark}
Note that $\mce$ determines $\mce'$ up to semisimplification, using $L\dash$function techniques.
Moreover properties like being irreducible or having finite determinant hold simultaneously for them.
:::

:::{.theorem title="Drinfeld, L. Lafforgue, Deligne"}
With this setup, a lisse $\QQbar_{\ell}$ sheaf $\mce_\ell$ admits a compantion $\mce'$ which is a $\QQbar_{\ell'}$ for chosen embeddings of $\QQ$ for which all traces are in $\QQbar$, i.e. irreducible and finite determinant.
This is true in arbitrary dimension.
:::

:::{.remark}
What about when $\ell=p$?
There are somehow too many and too few lisse $\QQbar_p$ sheaves!
E.g. the Lefschetz trace formula doesn't hold.
Instead use the Riemann-Hilbert correspondence -- the \(p\dash \)adic analogues of lisse sheaves are certain \(p\dash \)adic integrable connections.
How to construct: start with $X$ affine and glue.
Choose a smooth affine formal scheme over $W(k)$ with special fiber $P_k \cong X$.
Let $K = \ff W(k)$ and $P_K$ be the Raynaud generic fiber.
See Tate model, Berkovich model, etc for rigid analytic geometry.
Let $\AA_K^n \leadsto \hat{\AA^n}_{W(k)}$, a closed unit disc over $K$.
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:::{.definition title="Convergent isocrystal"}
Some definitions:

- A **convergent isocrystal** is a vector bundle with an integrable connection on $P_K$.
- A **convergent $F\dash$isocrystal** is this and a compatible action of (a lift of) Frobenius.
- An **overconvergent $F\dash$isocrystal** is this on some structural enlargment of $P_K$ which takes the closed unit disc to the disc of radius $1+\eps$.
:::

:::{.remark}
These are similar to $\pi_1(X_{\kbar})\dash$representations.
Issue: \(p\dash \)adic antidifferentiation is hard; the integral of a formal power series converging on the closed disc only converges on the open disc.
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:::{.remark}
If $X$ is smooth this recovers Berthelot's rigid cohomology, which is a refinement of crystalline cohomology if $X$ is proper.
This yields a 6 functor formalism, and has the same moving parts as etale cohomology.
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:::{.theorem title="Abe"}
The Langlands correspondence extends to these when $\dim X = 1$.

The content here: Lafforgue's original proof can now be run with \(p\dash \)adic coefficients instead of just \(\ell\dash \)adic coefficients.
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:::{.theorem title="Abe-?, Drinfeld-Abe-Esmult, K"}
The companion theorem extends to both $\ell=p$ and $\ell'=p$.
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:::{.remark}
Deligne posited the existence of a *petit camarade crystalline*, a little crystalline friend. ❇🙂
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## Applications

:::{.remark}
A partial result toward a conjecture of Simpson.
Let $X\slice \CC$ be a smooth cohomologically rigid local system (so no nontrivial deformations) which is irreducible with finite determinant.
Are these of geometric origin?
In particular, is there a $\ZVHS$?
Esnault-Groecherig show that the monodromy representation factors through $\GL_n(\OO_K) \to \GL_n(\CC)$ for some $K\in \NF$.

Idea: start from complex geometry, go to \(p\dash \)adic geometry, yields an overconvergent $F\dash$crystal.
This yields integrality at $p$; use companions to go back to a lisse \(\ell\dash \)adic sheaf, then back to $\CC$ to get integrality at $\ell$.
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:::{.remark}
Going the other way, $\ell\to p$: one can prove "of geometric origin" results when $\rank \mce = 2$ (Krishnmorthy-Pal).
Idea: go from $\ell\to p$, make a candidate for the crystalline Dieudonne module for some family of AVs.
One will have a bound on the motivic weight, which is at most $\rank \mce - 1$.

A word on the proof: define a moduli stack $M_n$ of mod $p^n$ $F\dash$crystals, which is a horrendous algebraic stack.
These are roughly coherent sheaves with extra data.
Study some finite-type pieces using slops, and is universally closed since one can take flat limits along curves.
Take the Zariski closure of companion points, then take stable images to get some $M_n''$.

Show that every point in each component of it is a companion point using horizontal companions (as opposed to vertical in the fiber direction).
Then show each component maps isomorphically to $S$, which is a pointwise condition on $S$.
This only uses the companion on the fiber, which is easier to study.
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