I am reading the paper "$p$-adic cohomology: from theory to practice" by K. S. Kedlaya. I have several naive questions about section 2: Frobenius action on de Rham cohomology. As a physicist, I lack proper backgrounds needed for this paper, and I am very grateful if someone could explain my questions carefully.

Question 1: In the remark 2.1.3 (page 16), (in the notation of this paper), a morphism $\overline{f}:(\overline{X},\overline{Z}) \rightarrow (\overline{X}',\overline{Z}')$ functorially induces a homomorphism \begin{equation} H^i_{dR}(X',Z') \rightarrow H^i_{dR}(X,Z) \end{equation} why do we want $\overline{f}$ induces this morphism which is at a higher level (since there is a morphism we already easily get $H^i_{dR}(\overline{X}',\overline{Z}') \rightarrow H^i_{dR}(\overline{X},\overline{Z})$)? Is this because Theorem 2.1.2, i.e. the canonical comparison isomorphism is between $H^i_{dR}(X,Z)$ and $H^i_{crys}(\overline{X},\overline{Z})$?

Question 2: I guess the $q$-th power Frobenius map $F_{q}$ is induced by the $\mathbb{F}_q$-algebra homomorphism $A \rightarrow A$ which sends $x$ to $x^q$? The map $F_q$ induces an endomorphism $F_q^*$ of $H^i_{dR}(X,Z)$, under the comparison isomorphism, it induces an endomorphism of $H^i_{crys}(\overline{X},\overline{Z})$. Is $F_q^*$ sent to ation of geometric Frobenius (the inverse of Frobenius element of $\text{Gal}(\overline{\mathbb{F}}_q/\mathbb{F}_q)$)?


For question 1, you are correct that the map $$\bar f^* \colon H^i_{\text{dR}}(X',Z') \to H^i_{\text{dR}}(X,Z)$$ is the one obtained from $$\bar f^* \colon H^i_{\text{crys}}(\bar X',\bar Z') \to H^i_{\text{crys}}(\bar X,\bar Z)$$ using the comparison isomorphism. This is exactly the point of Remark 2.1.3: it is remarkable that the map $\bar f^*$ on $H^i_{\text{dR}}$ exists, because a priori de Rham cohomology is only functorial for morphisms $f \colon (X,Z) \to (X',Z')$ (think: a lift of $\bar f$).

For question 2, note that the geometric Frobenius is not an element of $\operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q)$. The composition of geometric Frobenius with arithmetic Frobenius is the absolute Frobenius. What happens is that on étale cohomology, the absolute Frobenius acts trivially, hence geometric Frobenius acts as the inverse of arithmetic Frobenius. See for example this MO post.

In the case of crystalline cohomology, one usually only studies the action of the absolute Frobenius. In the case of a variety over $\mathbb F_q$, the $q$-power Frobenius acts trivially on $\mathbb F_q$, hence geometric $q$-power Frobenius equals absolute $q$-power Frobenius, and the arithmetic $q$-power Frobenius is trivial. I guess one could study the arithmetic $p$-power Frobenius viewed as an element of $\operatorname{Gal}(\mathbb F_q/\mathbb F_p)$, but this is not usually considered.


I would like to add the following points to the answer of R. van Dobben de Bruyn, in order to answer the "why" part of the question "why do we want $\bar f$ to induce this morphism which is at a higher level"?

First, Kedlaya's paper is about $p$-adic cohomology of a variety in characteristic $p$. There is a well-behaved $\ell$-adic cohomology theory coming from étale cohomology when $\ell\neq p$, but some of their good properties are lost when you try to put $\ell=p$. This sort of affairs motivated Grothendieck to introduce crystalline cohomology, later developed by Berthelot, as a better substitute for $p$-adic cohomology.

So, what we are really interested in, is $H_{crys}$.

It turns out, nontrivially, that crystalline cohomology is essentially the de Rham cohomology of a lift to characteristic zero (and it doesn't depend on the lift).

So yes, we are interested in the functorial map in cohomology of the lift because this cohomology of the lift is a posteriori the thing we were looking for.

(Moreover, in my opinion, this remark is extremely interesting in itself. Namely, the fact that a map between the reductions mod p, even without coming from an actual map between the lifts char 0, it maps cohomology classes on the lifts char 0 in a well-defined manner)

Last, not only the $p$-adic étale cohomology, but also the de Rham cohomology mod p is worse-behaved than the crystalline cohomology. I have read here that the de Rham cohomology mod p just takes $p$ torsion from the crystalline cohomology.

My personal, and incomplete, explanation for why a lift char 0 is preferrable (or more "perfect", if we like to play on words) to the space mod p itself, when you consider $H_{dR}$, is as follows. First, $H_{dR}$ is used to understand the underlying space by considering differentials on it. But when you are in positive characteristic, things mess up and you might lose some important information. As a prototype for this, think of differentiating $x\mapsto x^q$, you get exactly zero. Or viceversa, you cannot find a primitive to $x\mapsto x^{p-1}$ because you lack the ability to divide by $p$.

  • $\begingroup$ I would not call these gadgets "a lift to char 0" - they lives over $\operatorname{Spec}\mathbb Z_p$, which is usually called of mixed char. After all, crystalline cohomology is an integral cohomology theory. $\endgroup$ – Yai0Phah Nov 3 '19 at 8:26
  • $\begingroup$ I would guess that the motivation to lift $\mathbb F_p$ to $\mathbb Z_p$ is that, the objects that we are really interested are $\mathbb Z$-schemes. In fact, the derived $\mathbb Z$-de Rham cohomology $\operatorname{dR}_{R/\mathbb Z}$ is not that bad (I was also misled. The $\mathbb F_p$-de Rham cohomology that you have mentioned is $\operatorname{dR}_{R/\mathbb F_p}$). On the other hand, a precise comparison of $\mathbb F_p$-de Rham and crystalline cohomologies is: $R\Gamma_{\mathrm{cris}}(X/\mathbb Z_p)\otimes_{\mathbb Z_p}^{\mathbb L}\mathbb F_p\simeq R\Gamma_{\mathrm dR}(X/\mathbb F_p)$. $\endgroup$ – Yai0Phah Nov 3 '19 at 8:40
  • $\begingroup$ (continued) where $X$ is a smooth $\mathbb F_p$-scheme. I don't know a good reference. Bhatt's lecture notes about prismatic cohomology covered this in Corollary 1.8 of lecture 6. $\endgroup$ – Yai0Phah Nov 3 '19 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.