Timeline for Can proper-smooth base change be used to show that varieties cannot be lifted to characteristic zero?
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| Feb 19, 2020 at 22:53 | comment | added | R. van Dobben de Bruyn | @shenghao: Serre's original example of the failure of Hodge symmetry is a $\mathbf Z/p$-quotient of a complete intersection. This can easily be made to lift to $\mathbf Z_p$ by starting with a $\mathbf Z/p$-representation that lifts and lifting the equations of the complete intersection. See for example Prop 1.3 in this preprint of mine. This was already known to Raynaud if the residue field is infinite (apply Prop 4.2.3 to Serre's construction). | |
| Jul 15, 2011 at 20:42 | comment | added | Matt | Just for completeness I'll point out that Matt Emerton has sketched a construction of such an example here: math.stackexchange.com/questions/51146/… | |
| Jul 12, 2011 at 19:08 | comment | added | Matt | I'm pretty sure this second statement of shenghao is false. The Hodge numbers don't have to be constant in this family, they are only upper semicontinuous which means they could jump at the special fiber. Take any $Y\to \mathrm{Spec}(R)$ where $R$ a complete DVR of mixed characteristic with res field $k$ in which a Hodge number $h^{st}$ with $s+t=n$ jumps on the special fiber $Y_0$. By construction $Y_0/k$ is liftable and by the previous argument $h^n(X, \mathbb{Q}_\ell)\neq \sum h^{pq}$. This is a counterexample, but I don't know of a concrete realization of this off the top of my head. | |
| Jul 5, 2011 at 18:48 | comment | added | Matt | Ah. (Almost) figured it out. $Y\to S$ a lifting of $X$ as in question, then by the Theorem in the question $h^n(X, \mathbb{Q}_\ell)=h^n(\overline{Y}_\eta, \mathbb{Q}_\ell)$. But now we have a smooth variety over an algebraically closed field of char $0$, so $\ell$-adic is the same as de Rham and Hodge-de Rham SS degenerates to give $h^n(\overline{Y}_\eta, \mathbb{Q}_\ell)=h^n_{dR}(\overline{Y}_\eta)=\sum_{p+q=n} h^{pq}$. Actually, to finish we need a reason $\sum h^{pq}$ stays constant in the family as well... | |
| Jul 3, 2011 at 23:37 | comment | added | Matt | @shenghao Can you expand on that second point at all? That isn't obvious to me. In fact, several counterintuitive examples have been constructed with respect to liftings. Essentially what you've claimed is that any liftable variety has equal Hodge and $\ell$-adic Betti numbers. William Lang constructed an example of a liftable surface for which the de Rham and Hodge Betti numbers are not equal, and I've thought a bit about this comparison. I haven't actually put thought into the $\ell$-adic vs Hodge comparison which is why this surprises me a little. | |
| Jun 11, 2011 at 16:44 | vote | accept | Akhil Mathew | ||
| Jun 11, 2011 at 15:46 | comment | added | shenghao | Here are two instances: if Hodge symmetry fails, then the proper smooth variety cannot be lifted (e.g. Serre's example of a proj. smooth surface); if $h^n(X,\mathbb Q_l)\ne\sum_{p+q=n}h^{pq},$ then it cannot be lifted. | |
| Jun 11, 2011 at 13:05 | history | edited | Akhil Mathew | CC BY-SA 3.0 |
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| Jun 11, 2011 at 6:24 | answer | added | naf | timeline score: 12 | |
| Jun 11, 2011 at 6:20 | comment | added | naf | What you state is not quite correct: for example, if $\mathcal{F}$ is the constant sheaf $\mathbb{Z}/l$ then the residue characteristic should be prime to $l$. To see this is necessary consider a family of elliptic curves and $R^1\pi_*$. | |
| Jun 11, 2011 at 3:01 | history | asked | Akhil Mathew | CC BY-SA 3.0 |