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Let $S$ be a connected scheme. We say that $S$ is simply connected if every smooth and proper morphism $X \to S$ of relative dimension $0$ has a section. This is equivalent to the standard definition, since finite etale morphisms are smooth and proper morphisms of relative dimension $0$, and, after reducing to the connected case, havinga section is equivalent to being trivial.

Now say that $S$ is $d$-trivial if every smooth and proper morphism $X \to S$ of relative dimension $d$ has a section. So being $0$-trivial is the same as being simply connected.

We might extend the definition to call $\infty$-trivial schemes that are $d$-trivial for all $d$.

For what numbers $d>0$ can we say something about which schemes are $d$-trivial?

This question was originally inspired by this excellent question in which it is proven that $\operatorname {Spec} \mathbb Z$ is $1$-trivial but not $6$-trivial. The apparent difficulty of further progress on $\operatorname{Spec} \mathbb Z$ led me to wonder if other cases, such as varieties over $\mathbb C$, might be easier. In particular:

For what $d$ is $\mathbb P^1_\mathbb C$ $d$-trivial? $\mathbb P^n_\mathbb C$?

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It is easy to see that $\mathbb{P}^1_{\mathbb{C}}$ is $1$-trivial: if the fibres have genus $0$ it follows from Tsen's theorem that there must be a section and if the fibres have genus $>0$ it follows from the fact that $\mathbb{P}^1$ is simply connected and that Teichmuller space is a bounded domain that any such morphism must be a product. –  ulrich Mar 13 '13 at 5:43
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For $g=1$ a simpler argument works: $M_1$ itself is affine! –  Will Sawin Mar 13 '13 at 6:59
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What about the analogous question for fibrations of topological spaces, or at least fiber bundles? –  David Corwin Mar 13 '13 at 7:21
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Characteristic 0? I think the counterexamples I gave you before, using Fulton-Deligne, might show that $\mathbb{P}^1$ is not $2$-trivial in positive characteristic (there are also Moret-Bailly pencils). –  Jason Starr Mar 13 '13 at 10:54
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@Dan Petersen: Any family of elliptic curves is a torsor of order $n$, for some $n$, over the family of Jacobians. If the base is simply connected the $n$-torsion local system of the family of Jacobians must be trivial and then its $H^1$ is also trivial. So the torsor is trivial, hence the original family has a section. –  ulrich Mar 13 '13 at 11:17

1 Answer 1

Already $\mathbb{P}^2_{\mathbb{C}}$ is not $1$-trivial, e.g., the relative Proj of the symmetric algebra of the sheaf of relative differentials (the projectivized tangent bundle) cannot have any section.

$\textbf{Edit}.$ Also you can use Serre's construction to construct similar locally free sheaves of rank $2$ over $\mathbb{P}^1\times \mathbb{P}^1$. This raises a red flag in considering "$d$-simple" as a generalization of "simply connected". For every generalization of simply connected that I know of, a product of two "simply connected" varieties is again "simply connected". Yet that fails for "$1$-simple".

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Could it be that the correct generalization is that a product of a $d$-simple and a $d'$-simple variety is $(d+d')$-simple? –  Dan Petersen Mar 13 '13 at 15:56
    
@Jason: Are there any higher-dimensional simply-connected varieties for which Serre's construction fails? I don't think simply-connectedness in characteristic $p$ has that property. @Dan: $d+d'$-simple implies $d$-simple, because you can always take the fiber product of a smooth proper cover with $\mathbb P^{d'}$. –  Will Sawin Mar 13 '13 at 17:52
    
@Will: You can put a double structure on $\mathbb{P}^n$ as in Hartshorne, Exercise III.5.9 in such a way that the resulting scheme has no nontrivial locally free sheaves of rank $2$ coming from Serre's construction. In general, I believe it is an open question whether every proper scheme over a field has "enough locally free sheaves". Burt Totaro has a nice survey about this. –  Jason Starr Mar 13 '13 at 18:23

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