Flatness in Algebraic Geometry vs. Fibration in Topology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:36:38Z http://mathoverflow.net/feeds/question/20184 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20184/flatness-in-algebraic-geometry-vs-fibration-in-topology Flatness in Algebraic Geometry vs. Fibration in Topology Daniel Loughran 2010-04-02T19:47:22Z 2010-04-03T01:41:17Z <p>I am currently trying to get my head around flatness in algebraic geometry. In particular, I'm trying to relate the notion of flatness in algebraic geometry to the notion of fibration in algebraic topology, because they do formally seem quite similar. I'm guessing that the answers to my questions are "well-known", but I am struggling to find anything decent in the literature. Any help/references will be most useful.</p> <p>The set up is this: Let $E,B$ be smooth projective complex algebraic varieties, and let $\pi:E \to B$ be a surjective flat map such that the fibres $E_b:=\pi^{-1}(b), b \in B$ are smooth projective complex algebraic varieties.</p> <p>I am aware that each fibre has the same Hilbert polynomial, so cohomologically they are quite similar. But each fibre can certainly be non-isomorphic as algebraic varieties (e.g. moduli spaces). However:</p> <ol> <li><p>Using GAGA type methods, we can consider $E$ and $B$ as complex manifolds. Is it true that $(\pi,E,B)$ is a fibration? That is, satisfies the homotopy lifting property with respect to any topological space?</p></li> <li><p>Again considering $E,B$ and each fibre $E_b$ as a complex manifold, is it true that each fibre $E_b$ is homotopy equivalent to each other? What about homeomorphic/diffeomorphic?</p></li> </ol> <p>Thanks,</p> <p>Dan</p> http://mathoverflow.net/questions/20184/flatness-in-algebraic-geometry-vs-fibration-in-topology/20187#20187 Answer by Emerton for Flatness in Algebraic Geometry vs. Fibration in Topology Emerton 2010-04-02T20:19:22Z 2010-04-03T01:41:17Z <p>A surjective flat (equals faithfully flat) map with smooth fibres is in fact a smooth morphism, and hence induces a submersion on the underlying manifolds obtained by passing to complex points. Since the fibres are projective, it is furthermore proper (in the sense of algebraic geometry) [see the note added at the end; this is not a logical deduction from the given condition on the fibres, but nevertheless seems to be a reasonable reinterpretation of that condition], and hence proper (in the sense of topology). A theorem of Ehresmann states that any proper submersion of smooth manifolds is a fibre bundle. In particular, it is a fibration in the sense of homotopy theory, and the fibres are diffeomorphic (thus also homeomorphic, homotopic, ... ). </p> <p>Note: Your specific question is really about smooth morphisms (these are flat morphisms with smooth fibres, although there are other definitions too, which are equivalent under mild hypotheses on the schemes involved, and in particular, are equivalent for maps of varieties over a field). One point about the notion of flat map is that it allows one to consider cases in which the fibres over certain points degenerate, but still vary continuously (in some sense). It may well be a special feature of algebraic geometry (and closely related theories such as complex analytic geometry) that one can have such a reasonable notion, a feature related to the fact that one can work in a reasonable manner with singular spaces in algebraic geometry, because the singularities are so mild compared to what can occur in (say) differential topology.</p> <p>[Added: I should add that I took a slight liberty with the question, in that I interpreted the condition that the fibres are projective stronger than is literally justified, in so far as I replaced it with the condition that the map is proper. As is implicit in Chris Schommer-Pries's comment below, we can find non-proper smooth surjections whose fibres are projective varieties: e.g. if, as in his example, we consider the covering of $\mathbb P^1$ by two copies of $\mathbb A^1$ in the usual way, then the fibres consists of either one or two points (one point for $0$ and the point at $\infty$, two points for all the others), and any finite set of points is certainly a projective variety.</p> <p>Nevertheless, my interpretation of the question seems to have been helpful; hopefully, with the addition of this remark, it is not too misleading.]</p>