Frobenius splitting and derived Cartier isomorphism - MathOverflow

Frobenius splitting and derived Cartier isomorphism

2011-03-21T15:05:49Z

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.

Answer by Matt for Frobenius splitting and derived Cartier isomorphism

2011-05-31T02:51:23Z

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.

Answer by Piotr Achinger for Frobenius splitting and derived Cartier isomorphism

2011-09-29T01:13:13Z

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)

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