0
$\begingroup$

Suppose $X$ is a surface, are there examples for curves on X which not 1-connected, but $h^0(C,O_C)=1$? (Here a curve is an effective divisor, not necessarily reduced or irreducible) (1-connectedness means when it splits into two effective divisors, the two parts have intesection number greater than or equal to 1)

Another question is I am not sure if the dimension of cohomology and 1-connectedness is preserved by linear equivalence ?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

Here is an example with $X$ smooth : consider a regular surface $X$ with an elliptic fibration having a double fiber $2E$ (typically, an Enriques surface). Then $2E=E+E$ is not 1-connected ($E^2=0$), but one checks easily that $H^1(X,\mathcal{O}_X(-2E))=0$, hence $h^0(\mathcal{O}_{2E})=1$.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Thanks for your answer! But how do you get the example? Also I find it hard to check the $H^1(X,O_X(-2E))=0$, is there an easy way to show that? $\endgroup$
    – user39380
    Apr 8, 2014 at 12:13
  • $\begingroup$ 1) Every Enriques surface has such a pencil, look at any book on surfaces (e.g. Barth etc.). 2) The linear system $|2E|$ is a pencil of elliptic curves; take a smooth $F\in |2E|$. Since $H^1(X,\mathcal{O}_X)=0$, the exact sequence $0\rightarrow \mathcal{O}_X(-F)\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}_F\rightarrow 0$ gives $H^1(X,\mathcal{O}_X(-F))=0$. $\endgroup$
    – abx
    Apr 8, 2014 at 12:18
  • $\begingroup$ Is that because $H^0(O_X)=H^0(O_F)=k$? Also is pencil here just mean the existence of an elliptic fibration? $\endgroup$
    – user39380
    Apr 8, 2014 at 12:31
  • $\begingroup$ Yes and yes ... $\endgroup$
    – abx
    Apr 8, 2014 at 12:53
  • $\begingroup$ Since we have $F$ is linear equivalent to $2E$, $O_F$ and $O_{2E}$ define the same invertible sheaf, can we argue $H^0(O_{2E})=k$ directly by this? $\endgroup$
    – user39380
    Apr 8, 2014 at 14:02
1
$\begingroup$

Take $X$ to be a quadratic cone and $C = L_1 + L_2$, the union of two lines on it. Then $L_1\cdot L_2 = 1/2$, so by your definition it is not 1-connected.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.