Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a smooth algebraic variety over an algebraically closed field $k$ of characteristic $p>\dim X$. The motivation for my question comes from the following results.

1. If $X$ is Frobenius split (the $p$-th power map $\mathcal{O}_X \to F_* \mathcal{O}_X$ admits n $\mathcal{O}_X$-linear splitting) then the Kodaira vanishing theorem holds for $X$.

The proof uses nothing but Serre vanishing and the projection formula.

2. If the complex $F_* \Omega^\bullet_X$ is quasi-isomorphic to a complex with zero differentials, then the Kodaira-Akizuki-Nakano vanishing theorem holds for $X$.

The proof uses Cartier isomorphism, hypercohomology spectral sequences, Serre vanishing and the projection formula and is similar to that of 1.

3 (Deligne-Illusie 1987). If $X$ lifts to $W_2(k)$, then the complex $F_* \Omega^\bullet_X$ is quasi-isomorphic to a complex with zero differentials.

4 (Buch-Thomsen-Lauritzen-Mehta 1995). If $X$ is strongly Frobenius split (that is, $0\to B_1\to Z_1\to \Omega^1_X\to 0$ splits, where $Z_i$ and $B_i$ are cocycles/coboundaries in $F_* \Omega^\bullet_X$), then $X$ and $F$ lift to $W_2(k)$ and the Bott vanishing theorem holds for $X$.

My (maybe incorrect) feeling is that strong Frobenius splitting and lifting of the Frobenius to $W_2(k)$ are quite uncommon, Frobenius splitting is a common behavior "on the Fano side" and that lifting of $X$ to usually $W_2(k)$ exists.

Question. Are there examples of Frobenius split varieties for which $F_* \Omega^\bullet_X$ is not quasi-isomorphic to a complex with zero differentials (for example, because the Hodge spectral sequence does not degenerate, see also this question on the Hodge spectral sequence)? If yes (that's my intuition here), does Frobenius splitting imply some weaker property of $F_* \Omega^\bullet_X$ which implies Kodaira vanishing?

Edit. Note that Frobenius splitting just states that the complex $F_* \Omega^\bullet_X$ is quasi-isomorphic to a complex whose first differential $C^0 \to C^1$ is zero.

share|improve this question
1  
I believe that even if Frobenius doesn't lift you still get that a Frobenius split variety (with nice assumptions) will lift to $W_2(k)$. This is because even though the obstruction to lifting Frobenius $\eta\in Ext^1(\Omega^1, B^1)$ is non-zero, the obstruction to merely lifting $X$ is the image of $\eta$ under the connecting homomorphism $Ext^1(\Omega^1, B^1)\to Ext^2(\Omega^1, \mathcal{O}_X)$. By the splitting assumption this map is $0$ and hence there is no obstruction to lifting $X$. See V. Srinivas Decomposition of the de Rham Complex. –  Matt May 29 '11 at 19:10
    
Great! Why don't you post this as an answer? –  Piotr Achinger May 30 '11 at 11:23
add comment

2 Answers

up vote 5 down vote accepted

Keeping the notation of the question $X$ is a smooth variety over an algebraically closed field $k$ of characteristic $p>\dim X$. Just following along from Decomposition of the de Rham Complex by V Srinivas, we see that the obstruction to lifting the pair $(X,F)$ to a pair $(X^{(2)}, F^{(2)})$ where $X^{(2)}$ is a lift of the variety to $W_2(k)$ and $F^{(2)}$ is a lift of Frobenius consistent with all diagrams is exactly the class $\zeta\in \mathrm{Ext}^1(\Omega_{X/k}^1, B_X^1)$ that corresponds to the sequence $0\to B_X^1\to Z_X^1\to \Omega_{X/k}^1\to 0$.

If we look at the sequence $0\to \mathcal{O}_X\to F_* \mathcal{O}_X\to B_X^1\to 0$ and take $\mathrm{Hom}(\Omega^1, -)$ we get a connecting homomorphism in the long exact sequence $\mathrm{Ext}^1(\Omega^1, B^1)\stackrel{\delta}{\to} \mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)$. It is well known that the obstruction to lifting lies in $\mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)\simeq H^2(X, \mathcal{T}_X)$, but what is not well-known is that the obstruction class in this case is exactly the image of $\zeta$ under $\delta$. So $\delta$ acts as sort of a forgetful map for obstruction to lifting the pair to obstruction for lifting the variety without lifting Frobenius.

While it is possible that a Frobenius split variety has non-zero obstruction class $\zeta$ (no example comes to mind right now) and hence the pair doesn't lift, this splitting assumption actually gives us lots of information when coupled with the above information.

We see that $0\to \mathcal{O}_X\to F_* \mathcal{O}_X\to B_X^1\to 0$ splitting gives us $\mathrm{Ext}^1(\Omega^1, F_*\mathcal{O}_X)\twoheadrightarrow \mathrm{Ext}^1(\Omega^1, B^1)\stackrel{\delta}{\to} \mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)$, so in fact $\delta=0$. This means that it doesn't matter whether or not we can lift the pair, all we had to know was that the obstruction to lifting $X$ was the image of $\zeta$ under $\delta$ which is $0$.

Thus any smooth Frobenius split variety lifts to $W_2(k)$ and since we assumed $p>\dim X$ we also get that the Hodge-de Rham spectral sequence degenerates at $E_1$ by work of Deligne and Illusie.

share|improve this answer
    
It is indeed strange that in the Brion-Kumar book, the main reference on the subject, they prove Kodaira vanishing but don't even remark that the full Kodaira-Akizuki-Nakano vanishing holds. –  Piotr Achinger May 31 '11 at 7:15
    
Do you know any examples of Frobenius split varieties which do not lift all the way to characteristic zero? –  mdeland May 31 '11 at 13:18
    
I would love to know such an example. This is something I've been thinking a lot about lately. If you get a lift of the pair to $W_2$ then you should be able to repeat a similar argument to get a lift all the way up, but you'd still also need algebraizability. My guess is that there must be some fairly easy examples. –  Matt May 31 '11 at 16:03
    
I had wondered that also but I could never prove it. Maybe lifting the pair to W_2 only guarantees a lifting of X to W_3? –  mdeland May 31 '11 at 17:42
    
That is quite possible (although my gut says if you can do the next lift, then there is some sort of common argument that should keep going through). The main candidate for a Frobenius split that does not lift to characteristic $0$ is something that lifts to $W_2(k)$ but doesn't lift to characteristic $0$, but these are rather hard to come by. –  Matt May 31 '11 at 21:15
add comment

I just found a nice reference for this question: K. Joshi "Exotic Torsion, Frobenius Splitting and the Slope Spectral Sequence" (Canad. Math. Bull. Vol. 50 (4), 2007), section 9.

Theorem 9.1 (unpublished work of V. B. Mehta). Let X be a smooth, projective, F-split variety over an algebraically closed field $k$ of characteristic $p>0$. Then for all $i+j < p$, the Hodge to de Rham spectral sequence degenerates at $E^{i,j}_1$. In particular, for $i+j = 1$ we have the following exact sequence $$ 0\to H^0(X, \Omega^1_X)\to H^1_{DR}(X/k)\to H^1(X, \mathcal{O}_X)\to 0. $$ Moreover, any F-split variety with $dim(X)<p$ satisfies Kodaira-Akizuki-Nakano vanishing.

Corollary 9.2. Let $X/k$ be a smooth, projective, Frobenius split variety. Then $X$ admits a flat lifting to $W_2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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