Let $S$ be a complex K3 surface, and $P\subset S$ a finite set of points in $S$. It is known that $$ H^i(S,\mathbb{Z})\cong H^i(S\setminus P,\mathbb{Z}) $$ for $0\le i \le 2$. Then the Euler characteristic computation implies that $b_3(S\setminus P)=|P|$. I want to confirm that $$ H^3(S\setminus P,\mathbb{Z}) \cong \mathbb{Z}^{|P|}, $$ that is, there is no torsion.
$\begingroup$
$\endgroup$
4
-
$\begingroup$ What do you get from the Universal Coefficient Theorem? $\endgroup$– S. Carnahan ♦Sep 8, 2013 at 12:41
-
$\begingroup$ I don't think $P$ yields any torsion in $H_2(S\setminus P,\mathbb{Z})$, so the UCT implies that $H^3(S\setminus P,\mathbb{Z})\cong Hom(H_3(S\setminus P,\mathbb{Z}),\mathbb{Z})$, but does this help? $\endgroup$– SohrabSep 8, 2013 at 12:53
-
$\begingroup$ I think it does, since $\mathrm{Hom}(T,\mathbb Z)$ vanishes if $T$ is torsion... $\endgroup$– Dan PetersenSep 8, 2013 at 14:22
-
$\begingroup$ Sohrab, I think your Euler characteristic is off by one. Using Poincaré-Lefschetz duality will help give you a clean answer. $\endgroup$– Tim PerutzSep 8, 2013 at 14:46
Add a comment
|