Timeline for Is the square of a curve minus its diagonal affine?

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Nov 3, 2011 at 21:20	comment	added	rita		Added: $D$ and $D_1$ above are meant to be EFFECTIVE.
Nov 3, 2011 at 20:46	comment	added	rita		One has $\Delta^2=2-2g<0$ and $\Delta$ irreducible, so there cannot be a curve linearly equivalent to $\Delta$ and distinct from $\Delta$. This is the statement for $n=1$. Then one proceeds inductively: let $D$ be linearly equivalent to $(n+1)\Delta$, then $D\Delta<0$, hence $D=\Delta+D_1$ with $D_1$ linearly equivalent to $n\Delta$.
Nov 3, 2011 at 18:36	comment	added	vic		I guess you meant that for $g>1$ one has $h^0(\mathcal O_S(n\Delta))=1$ for all $n\geq 0$. How do you prove that, say for $n=1$, for notational simplicity? It does not follow automatically from Riemann-Roch, which only tells you that $h^0(\mathcal O_S(\Delta))=h^1(\mathcal O_S(\Delta))+1-g^2-2g+3-3g$, as the self-intersection of the diagonal is $-K_X$ and the canonical divisor of $X^2$ is $k_X\times X + X\times K_X$. This means that you know how to compute $h^1(\mathcal O_S(\Delta))$, or rather take a different approach?
Oct 19, 2011 at 13:03	history	edited	rita	CC BY-SA 3.0	added 1 characters in body
Oct 19, 2011 at 10:56	vote	accept	Victor Rotger
Oct 8, 2012 at 12:43
Oct 19, 2011 at 10:51	history	answered	rita	CC BY-SA 3.0