On $\mathbb{P}^2$ we have the notion of general positions: no 3 points on a line, no 6 on a conic, etc. In particular, blowing up points (up to 8) in general positions give ample anti-canonical class, hence del-Pezzo surfaces. Do we have a characterization of this general position on $\mathbb{P}^1\times\mathbb{P}^1$? I am considering from the perspective that blowing up $\mathbb{P}^1\times\mathbb{P}^1$ at a point is isomorphic to $\mathbb{P}^2$ blown up two points. So for example the second point we blow up on $\mathbb{P}^1\times\mathbb{P}^1$ cannot be on the two rulings of the first point. What about, say, 5 points on $\mathbb{P}^1\times\mathbb{P}^1$?
1 Answer
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5
Five points are in a general position if
pairs of points do not lie on the same ruling;
quadruples of points do not lie on a conic.
For six points you get an extra condition
- six points do not lie on a curve of bidegree $(1,2)$ or $(2,1)$.
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1$\begingroup$ I guess the idea is that the linear system of curves of bidegree $(a,b)$ has (projective) dimension $(a+1)(b+1)-1$. So a set of $(a+1)(b+1)$ points imposes independent conditions on curves of bidegree $(a,b)$ if and only if the points don't lie on a curve of bidegree $(a,b)$. And so the "general position" here is really something like "every subset imposes independent conditions on linear systems". $\endgroup$ Commented Feb 21 at 7:19
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$\begingroup$ You don't need to consider all $(a,b)$; only small ones. $\endgroup$– SashaCommented Feb 21 at 7:41
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1$\begingroup$ It'd be great if you could give a brief explanation on why this is the case, or at least a reference. $\endgroup$ Commented Feb 21 at 8:21
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1$\begingroup$ For instance, you can project to the plane from one of these points (say $P_0$) and write down the genericity assumptions on the images of the other four points (say, $P_i$, $1 \le i \le 4$) and the images (say $Q_1$ and $Q_2$) of the two rulings through $P_0$. Say, if the images of $P_1$, $P_2$, and $P_3$ are colinear, then $P_0$, $P_1$, $P_2$, and $P_3$ are coplanar, hence lie on a conic. If $Q_1$ and the images of $P_1$ and $P_2$ are colinear then $P_0$, $P_1$, and $P_2$ lie on a conic that contains a ruling, hence $P_1$ and $P_2$ lie on the other ruling. $\endgroup$– SashaCommented Feb 21 at 8:55
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$\begingroup$ @Sasha Thanks for the answer and comments. Do you happen to know any literature or paper that either explains or applies this? I just want to study this in more details $\endgroup$– fp1Commented Feb 25 at 0:45