Is the square of a curve minus its diagonal affine? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:03:34Z http://mathoverflow.net/feeds/question/78563 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78563/is-the-square-of-a-curve-minus-its-diagonal-affine Is the square of a curve minus its diagonal affine? Victor Rotger 2011-10-19T10:33:29Z 2012-09-28T17:09:45Z <p>Let $X$ be a smooth irreducible projective algebraic curve of genus $g\geq 1$ and $S=X^2$ the surface one obtains as the cartesian product of $X$ with itself. Let $\Delta$ be the diagonal in $S$, that is to say, a copy of $X$ embedded diagonally in $X^2$. </p> <p>Is the quasi-projective surface $U=S\setminus \Delta$ affine? </p> <p>One criterion for showing affinness is showing that $\Delta$ is an ample divisor in $S$. But the self-intersection of $\Delta$ is $\Delta^2=2-2g&lt;0$ and therefore $\Delta$ can not be ample. Does this already imply that $U$ is not affine? I guess not.</p> <p>Besides, Serre's criterion provides a necessary and sufficient condition for $U$ to be affine: this is the case if and only if $H^i(U,\mathcal F)=0$ for all $i>0$ and all coherent sheaves $\mathcal F$ on $U$. But I don't know how to check this in this example. </p> http://mathoverflow.net/questions/78563/is-the-square-of-a-curve-minus-its-diagonal-affine/78564#78564 Answer by rita for Is the square of a curve minus its diagonal affine? rita 2011-10-19T10:51:35Z 2011-10-19T13:03:17Z <p>The answer is no.</p> <p>Assume for contradiction that $U$ is affine. Then regular functions on $U$ separate points. On the other hand a regular function on $U$ extends to a rational function on $S$ with poles at most on $\Delta$, namely to a section of $H^0({\mathcal O}_S(n\Delta))$. If $g>1$, $h^0({\mathcal O}_X(n\Delta))=1$ for every $n\ge 0$, and we have a contradiction.</p> <p>If $g=1$, then $\Delta$ is a fiber of the morphism $X\times X\to X$ defined by $(x,y)\mapsto x-y$, hence $U$ is fibered in smooth elliptic curves and therefore it is not affine. </p> http://mathoverflow.net/questions/78563/is-the-square-of-a-curve-minus-its-diagonal-affine/108361#108361 Answer by Sándor Kovács for Is the square of a curve minus its diagonal affine? Sándor Kovács 2012-09-28T17:09:45Z 2012-09-28T17:09:45Z <p>This answer is inspired by Rita's second paragraph.</p> <p>It turns out that this $U$ contains projective curves for all $g\geq 1$ and hence cannot be affine:<br> Embed $X$ into its Jacobian $X\hookrightarrow J$ and let $Z\subseteq J$ be the image of $X\times X$ via the map $J\times J\to J$, $(x,y)\mapsto x-y$. This contracts the diagonal of $X\times X$. If $g>1$, then $Z$ is singular, but still projective. Let $H\subset Z$ be a general hyperplane cut of $Z$ inside $J$ and $C\subset X\times X$ the preimage of $H$. By choice, $H$ avoids the image of $\Delta$ and hence the projective curve $C$ is contained in $U$.</p>