5
$\begingroup$

If $f: X \to S$ is a smooth, proper morphism of nonsingular complex algebraic varieties with $S$ connected, then the dimensions of the (algebraic) de Rham cohomology groups of the fibers of $f$ are constant. One way to see this is to use that, complex analytically, $f$ is a locally trivial fibration (Ehresmann's theorem), hence all fibers are homeomorphic, and to conclude by using the topological invariance of analytic de Rham cohomology and the comparison between algebraic and analytic de Rham cohomology.

Is there a 'purely algebraic' proof of this result, i.e. one which does not appeal to GAGA and Ehresmann's theorem?

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ By smoothness, the de Rham cohomology groups carry canonical connections called the Gauß–Manin connection. It follows that they are locally free, 2.5.2.1 of Yves André, Différentielles non commutatives et théorie de Galois différentielle ou aux différences. $\endgroup$
    – Z. M
    Commented Dec 20, 2022 at 6:36
  • $\begingroup$ You can use Deligne-Illusie to prove the degeneration of the Hodge-de Rham spectral sequence. $\endgroup$
    – Jason Starr
    Commented Dec 20, 2022 at 11:29
  • $\begingroup$ . . . By vanishing theorems (also proved algebraically by Deligne-Illusie), the coherent sheaf cohomology groups arising in that spectral sequence are locally constant. Thus, also the de Rham cohomology groups are locally constant. $\endgroup$
    – Jason Starr
    Commented Dec 20, 2022 at 14:41
  • $\begingroup$ @Z.M thanks. What is the argument for the existence of the Gauss-Manin connection in characteristic 0? I thought this involved something like Ehresmann's theorem. $\endgroup$
    – Legendre
    Commented Dec 21, 2022 at 19:57
  • $\begingroup$ @JasonStarr thank you. Could you elaborate on the relation between vanishing theorems and local constancy of the de Rham cohomology of the fibers? $\endgroup$
    – Legendre
    Commented Dec 21, 2022 at 19:58

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .