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After a good look for anything in the way of an explicit example, and harassing the algebraic geometers in the department, I still am unable to find an answer to my question, so any light will be very much appreciated.

Suppose we have blown up $\mathbb{P}^2$ at some singular points (of a mapping), and we find that in this new space we have 3 curves which have self-intersection -1, which we would like to blow down. In my particular case I have two degree 1 curves (which pass through 2 base points in $\mathbb{P}^2$) and a degree 2 curve which passes through 5 base points in $\mathbb{P}^2$. Given that I have these curves explicitly, is it possible to find the blow down of each curve, and if so how?

Also, what happens to the Picard group in this blown down space? If one of my curves is $H-E_1-E_2$ in my blown up space, where $H$ is the representative of a generic degree 1 curve in $\mathbb{P}^2$ and $E_1$ and $E_2$ are the total transforms of two base points, then what is the Picard group in this space where we have blown down the line $H-E_1-E_2$?

My experience has been that all these things are theoretically calculable but nobody seems to apply the theory in specific cases.

Many thanks!

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It would be better if you explain on what surface you have these curves (I suspect it is $P^2$ blown up in 5 points) and what are the curves−−−one is $H-E_1-E_2$, what are the other two? –  Sasha Sep 6 '12 at 6:43
    
Yes, $\mathbb{P}^2$ blown up in 5 points, the other curves are $H-E_1-E_3$ and $2H-E_1-E_2-E_3-E_4-E_5$. –  philiph Sep 6 '12 at 7:04

2 Answers 2

Linear systems $|2H-E_1-E_2-E_3-E_4|$ and $|2H-E_1-E_2-E_3-E_5|$ define two maps from your surface to $P^1$. Using both you get a map to $P^1\times P^1$. It blows down all your curves and additionally the curve $H-E_2-E_3$. If you want to keep the last curve, you can blow up its image in $P^1\times P^1$.

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This works because these linear systems have intersection number =0 with the (-1)-curves that have to be blown down. In general what you need to find is a (complete) linear system that does not intersect the cuves to blow down, and (to avoid blowing up other things, down anything else, that cuts every other (-1) curve positively. The blown down surface can be identified it as the image of the map to projective space given by the linear system. To get the ideal of the surface you need elimination of variables: this can be done by computer algebra programs. –  quim Sep 6 '12 at 9:29
    
Yes this seems to work. As for the Picard group, I know when we blow up we just add another element, the line which is the blowup of a point. But in the case of blow down, we must remove an element, and in this case, what are we removing? Some combination of the lines blown down? –  philiph Sep 9 '12 at 0:19
    
Each line which is blown down should be removed. –  Sasha Sep 9 '12 at 7:24

In general, it is not easy to understand directly the contraction of a curve in a blow-up of points, but it is possible to do it in the following way:

if your surface is rational you can contract other curves disjoint from the ones you have chosen, in order to go to $\mathbb{P}^2$ or to an Hirzebruch surface (any projective smooth rational surface with no $(-1)$-curve is $\mathbb{P}^2$ or an Hirzebruch surface $\mathbb{F}_n$ with $n\not=1$), and you will find that your surface is the blow-up of points in $\mathbb{P}^2$ or a Hirzebruch surface.

Let us compute your specific example. You have chosen $p_1,p_2,p_3,p_4,p_5\in \mathbb{P}^2$ and have blown-up them, and obtain $X\to \mathbb{P}^2$. You want to contract $H-E_1-E_2$, $H-E_1-E_3$ and $2H-E_1-E_2-E_3-E_4-E_5$, and get $X\to Y$.

The Picard group of the surface $X$ has rank $6$ (generated by $H$ and the $E_i$) so $Y$ has rank $3$.

We can also contract $H-E_1-E_4$, $H-E_1-E_5$ and obtain a smooth surface of Picard rank $1$, which is therefore $\mathbb{P}^2$.

The surface $Y$ is then the blow-up of two points in $\mathbb{P}^2$ (the unique del Pezzo surface of degree $7$) and has exactly three $(-1)$-curves.

Everything can also be done explicitly in charts and coordinates, but takes a bit more time...

PS: I assumed that in your post your points where all in $\mathbb{P}^2$ (no infinitely near point) and hence general (no $3$ collinear) because you said that a conic passed through the five points. In a case where the points are in a more special configuration everything also works, but your curves contracted in $Y$ are maybe $(-2)$-curves of $X$ or $(-3)$-curves,...

PS2: The construction of Sasha also does what I suggested, in your special example. The morphism $X\to \mathbb{P}^1\times \mathbb{P}^1$ contracts $H-E_1-E_2$, $H-E_1-E_3$, $H-E_2-E_3$ and $2H-E_1-E_2-E_3-E_4-E_5$. So $Y$ is also the blow-up of one point in $\mathbb{P}^1\times \mathbb{P}^1$.

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