Excuse me if this question is trivial or trivially false, or not at this sites level. Lets work over an algebraically closed field of characteristic $p>0$, say $k$. The action of the Frobenius on $H^r_{crys}(X/W(k))$ gives us a sequence of positive rational numbers $a_i$ by taking the order of the eigenvalues.

Now if $X$ has a good lift to $W(k)$, $Y$, say with torsion free $H^p(Y, \Omega_Y/W)$, I believe we can decompose the Frobenius morphism using Hodge decomposition to obtain an inequality $a_i\le r$.

My question is whether this is true in general. This may be trivial, but I am a bit new to Crystalline cohomology so I can't figure it out.

  • 1
    $\begingroup$ What do you mean by "order of the eigenvalues"? $\endgroup$ – user25309 Sep 4 '16 at 21:07
  • 1
    $\begingroup$ @user25309 $p$-adic order of the eigenvalues. $\endgroup$ – Pax Kivimae Sep 4 '16 at 21:24

I assume that you mean $H^r_{crys}(X/W(k))$ because $H^r_{crys}(X/k)$ is just the de Rham cohomology over $k$.

Next, over arbitrary base field this definition of $a_i$'s(they are called slopes of Frobenius) does not make sense because Frobenius is only $p$-linear and its eignevalues a priori depend on the choice of a linearization(see Example 8.1.3 here). Over a finite field one can use that some power of Frobenius is linear, so eigenvalues $\beta_i$ of $F^n$ are well-defined and we put $a_i=\mathrm{ord}_p (\beta_i)/n$. Over arbitrary $k$ one should use Dieudonne-Manin classification(Thm 8.1.4 in the linked notes).

Now, let us see what restrictions we have on these slopes. I know to proofs of the fact that on $H^r$ slopes are in the interval $[0,r]$ for $r\leq n$ and are in the interval $[r-n,n]$ for $r\geq n$.

First one works in the case $X$ is projective. In what follows, let us denote by $H^r$ the cohomology $H^r_{crys}(X/W(k))/torsion$. First, there is a perfect pairing given by Poincare duality $$H^r\otimes H^{2n-r}\to W(k)(n)$$ where $W(k)(n)$ means $1$-dimensonal module with action of $F$ via $p^n$. This means that if $a_1,\dots,a_k$ are slopes of $F$ on $H^r$ then $n-a_1,\dots,n-a_k$ are slopes of $F$ on $H^{2n-r}$. So, we already see that slopes are in the interval $[0,n]$. For a proof of Poincare duality see, for example, Berthelot's book "Cohomologie Cristalline des Schemas de Characteristique p>0".

Next, for porjective $X$ we have Hard Lefschetz theorem(it holds only rationally):$$L^{n-r}:H^r[1/p]\cong H^{2n-r}[1/p]$$ where $L$ is multiplication by hyperplane section, the Lefschetz operator. For $x\in H^r[1/p]$ we have $F(L^{n-r} x)=F(L^{n-r})F(x)=p^{n-r}L^{n-r}F(x)$ because Frobenius on projective space multiplies hyperplane section by $p$. This means that slopes on $H^{2n-r}$ are obtained fromf slopes on $H^{r}$ by adding $n-r$, so $\{a_1+n-r,\dots,a_k+n-r\}=\{n-a_1,\dots,n-a_k\}$. In other words, for every $a_i$ there exists $a_j$ such that $a_i+n-r=n-a_j$ so $a_i+a_j=r$ and thus all slopes $a_i$ are less or equal than $r$. The proof of Hard Lefschetz can be found in Katz and Messing "Some Consequences of the Riemann Hypothesis for Varieties over Finite Fields"

Another proof which works, more generally, for arbitrary proper $X$ can be obtained using de Rham-Witt complex. A great account for this theory is given in the original paper by Illusie "Complexe de de Rham-Witt et cohomologie cristalline". In short, crystalline cohomology can be computed as hypercohomology of certain complex $W\Omega^{\bullet}$ of sheaves of abelian groups on $X$. It turns out that, modulo torsion, the corresponding spectral sequence degenrates(as in Hodge theory in characteristic zero) giving, in particular, that the subspace of $H^r[1/p]$ with slopes in the interval $[i,i+i[$ is equal to $H^{r-i}(W\Omega^i_X)[1/p]$ which gives that there are no slopes on $H^r$ less than $r-n$ which is exactly what we wanted combined with Poincare duality.

  • $\begingroup$ This answer is exactly what I was looking for! Glad to have an intuitive proof on hand before braving Illusie (especially since my french is horrendous). The equality $H^{r-i}(W\Omega^i)[1/p]=H^r[1/p]$ is very interesting. I believe we have that $H^r(W\Omega^0)=H^r(X, W)=H^r(X, D\mathbb{G}_m)=D(H^r(X, \mathbb{G}_m))$, for the last object being the formal group associated to $Ker(H^r(X\times Spec(A))\to H^r(X))$ (we might need some smoothness conditions here). Are there similar interpretations of the higher $H^{r-i}(W\Omega^i)[1/p]$ in terms of formal groups? I believe these are the formal $\endgroup$ – Pax Kivimae Sep 4 '16 at 23:09
  • $\begingroup$ groups obtained from the shift operator and duality, $D^{-1}(C[-i][i])$ as remarked by Groethendieck (I forget where though). $\endgroup$ – Pax Kivimae Sep 4 '16 at 23:11
  • $\begingroup$ @PaxKivimae No, I do not claim that $H^{r-i}(W\Omega^i)[1/p]=H^r[1/p]$. There is a direct sum decomposition(exactly as in Hodge theory)$H^r[1/p]=\bigoplus_i H^{r-i}(W\Omega^i)[1/p]$ compatible with Frobenius action and summand $H^{r-i}(W\Omega^i)[1/p]$ has slopes in the interval $[i,i+1[$. $\endgroup$ – SashaP Sep 5 '16 at 9:48
  • $\begingroup$ Yes, that is what I meant, I seem to be getting worse at writing Tex. But since $H^{r-i}(W\Omega^i)[1/p]$ has sloops in $[i, i+1[$, there is an isogenous $(H^{r-i}(W\Omega^i )[1/p][-i])[i]$, with slopes in $[0, 1[$, and so is the $H^{r-i}(W\Omega^i )[1/p][-i]$ is a Dieudonne module, and can be written as $D(G_i)$. I was just wondering if the $G_i$ had interpretations, since I think that $G_0$ is just the Artin-Mazur formal group. $\endgroup$ – Pax Kivimae Sep 5 '16 at 13:53
  • $\begingroup$ @PaxKivimae I do not now the answer from the top of my head. Probably it is worth asking a separate question. $\endgroup$ – SashaP Sep 9 '16 at 13:15

Here is an alternative way to prove that for $X$ projective of dimension $n$, the slopes on $H^r$ are in $[0,r]$, using instead of the Hard Lefshetz theorem (used in Sasha's answer), the Lefschetz hyperplane theorem, which is more elementary.

By Poincaré duality, we know that the slopes are in $[0,n]$, so it is enough to consider $r \leq n$.

We proceed by induction on the dimension $n$ of $X$. The result is obvious for $n=0$. For $X$ of dimension $n$, the result is true for $r=n$ by Poincaré duality. Fix a projective embedding of $X$ and let $Y$ be a generic hyperplane section. By Lefschetz hyperplane theorem, the natural maps $H^r(X) \rightarrow H^r(Y)$ are isomorphisms for $r<n-1$ and injective for $r=n-1$, and so the result follows from the result for $Y$ which is known by induction.


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.