Timeline for Irregularity of surfaces for dominant maps
Current License: CC BY-SA 4.0
11 events
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| Nov 12 at 12:52 | comment | added | user267839 | here another say "low level" argument - but which works only in complex setting: Firstly note that for complex surfaces we have $q(S)= h^0(S, \Omega_S)$; think that's classical Hodge theory + comparison results with sheaf coho. (Note, this fails badly in positive characteristic) Then we are in situation when surjective, (generically) finite maps of smooth schemes $f: X \to Y$ induce injective maps $f^*\Omega_Y \to \Omega_X$ of differerentials. Have still take some time to check it (or find reference) but expect that this should holds always when $K(X)/K(Y)$ separable, esp. in char $0$ | |
| Nov 12 at 11:43 | comment | added | user267839 | related to this | |
| Nov 11 at 17:00 | comment | added | R. van Dobben de Bruyn | More generally, for any surjective map $f \colon Y \to X$ of smooth projective varieties, the pullback $H^i(X,\Omega_X^j) \to H^i(Y,\Omega_Y^j)$ is injective for all $i,j$, at least in characteristic $0$. It is true in any characteristic for any Weil cohomology theory (but Hodge and de Rham cohomology have torsion coefficients if $\operatorname{char} k > 0$, and there the result is false); one reference is Kleiman's paper in Dix exposés sur la cohomologie des schémas, Proposition 1.2.4. The proof is a simple consequence of Poincaré duality and the projection formula. | |
| Nov 11 at 16:52 | review | Close votes | |||
| Nov 16 at 3:08 | |||||
| Nov 11 at 16:30 | comment | added | Jason Starr | You do not need $X$ to be unbranched over $S$. Every unbranched cover of $S$ pulls back to an unbranched cover of $X$. Since $X$ is simply connected, there is a section of the pullback unbranched cover. This equals the graph of an $S$-morphism from $X$ to the original unbranched cover. In particular the degree of that unbranched cover divides the degree of $X$ over $S$. So the fundamental group of $S$ is finite. | |
| Nov 11 at 16:25 | comment | added | Jason Starr | By the Universa Coefficients Theorem, the first cohomology group is the dual of the first homology group. If the irregularity is positive, so is the first Betti number. So the first homology group is infinite. Therefore the fundamental group is infinite. | |
| Nov 11 at 16:16 | comment | added | user267839 | @JasonStarr: Why finiteness of fundametal group of $S$ implies that irregularity of $S$ is zero? (...by the way, the cover in the question was a priori branched; may I assume that in your answer you firstly passed to locus $S' \subset S$ over which $f$ is etale and then exploited that $q(S)$ is a birational invariant; so it's ok to assume $f$ etale, right?) | |
| Nov 11 at 15:59 | comment | added | Jason Starr | Since the K3 surface is simply-connected, every finite unbranched cover of $S$ is intermediate between $S$ and its finite cover $X$. Thus the fundamental group of $S$ is finite. Therefore the irregularity of $S$ equals zero. | |
| Nov 11 at 15:46 | history | edited | user267839 | CC BY-SA 4.0 |
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| Nov 11 at 15:41 | history | edited | YCor | CC BY-SA 4.0 |
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| Nov 11 at 15:28 | history | asked | user267839 | CC BY-SA 4.0 |