# blowing curves down

If I blow up $\mathbb{P}^2$ in $2$ points I get $3$ exceptional divisors - two are disjoint and one intersects both of them. What object do I get if I blow down this intersecting curve? It's a minimal surface with Picard number $2$ and no $(-1)$-curves, what is that? $\mathbb{P}^1\times\mathbb{P}^1$?

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Yes, this is exactly what you get. – Dmitri May 7 '11 at 9:43
And that's why? – math_donk May 7 '11 at 9:49
How did you came to your guess in the first place? – Dmitri May 7 '11 at 9:58
Because it's only surface with Picard number 2 with no (-1) curves I know :) – math_donk May 7 '11 at 10:01
You can actually check this with affine charts, I remember doing that once. However, a very slick way to see'' this is to use the toric description of $\mathbb{P}^2$, as well as the toric description of blow-up and blow-down. – Karl Schwede May 7 '11 at 12:23

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Did you mean to make all the text invisible? – user5117 May 23 '15 at 10:16

A nice geometric way to see this is the following.

Take a smooth quadric $Q \cong \mathbb{P}^1 \times \mathbb{P}^1 \subset \mathbb{P}^3$, a point $p \in Q$ and consider the projection

$\pi_p \colon X \dashrightarrow \Pi$

where $\Pi \cong \mathbb{P}^2$ is a general plane.

The map $\pi_p$ is birational and contracts the two lines $\ell_1, \ell_2 \subset Q$ passing through $p$. Therefore the indeterminacy locus of $\pi$ consists of the point $p$ alone, whereas the indeterminacy locus of $\pi^{-1} \colon \Pi \dashrightarrow Q$ consists of the two points

$q_1 := \ell_1 \cap \Pi, \quad q_2:= \ell_2 \cap \Pi$.

This shows that the blow-up $\textrm{Bl}_pQ$ of $Q$ at $p$ is isomorphic to the blow-up of $\Pi$ at $q_1$ and $q_2$.

Moreover the directions in $T_pQ$, i.e. the points of the exceptional divisor $E \subset \textrm{Bl}_pQ$, are in natural correspondence with the points of the line $\ell_3 \subset \Pi$ joining $q_1$ and $q_2$.

Hence $Q$ can be obtained by blowing up the plane $\Pi$ at the two points $q_1$ and $q_2$ and then blowing down the strict transform of the line $\ell_3$ joining them.

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Here is another way to look at it; it is a more explicit variant of Sandor's answer: If you look at the family of (proper transforms of) lines passing through each of the two points you blew up, you can see directly that you get two rulings, the members of one crossing the members of the other transversally in exactly one point. This will let you construct an explicit isomorphism with $\mathbb P^1\times \mathbb P^1$.

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Let me give a somewhat less direct answer than Francesco's, which might shed some light as well.

Over an algebraically closed field, the class of minimal rational surfaces consist of $\mathbb{P}^2$ along with the rational ruled surfaces $\mathbb{F}_n = \mathbb{P}(\mathcal{O}\oplus \mathcal{O}(n))$, $n=0,1,2\ldots$. (Standard references, at least over $\mathbb{C}$, are Barth-Peters-Van de Ven, Beauville,...)

The ruled surfaces are exactly the rational surfaces with Picard number $2$. All of these with the exception of $\mathbb{F}_0= \mathbb{P}^1\times \mathbb{P}^1$ posses curves with negative self intersection. And that should be enough to determine your surface.

In general, there is a procedure called an elementary transformation, for moving from once ruled surface to an adjacent one. It consists of blowing up once, and then down in a "different direction". Your example, is really an elementary transformation from $\mathbb{F}_1$ (which is $\mathbb{P}^2$ blown up once) to $\mathbb{F}_0$.

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Here is yet another way to see this:

Blowing up a point on $\mathbb P^2$ makes it a rationally ruled surface mapping to the exceptional curve where the fibers are the proper transforms of the lines going through the blown up point.

Blowing up a point on a ruled surface and then blowing down the proper transform of the fiber it was sitting on produces another ruled surface still mapping to the exceptional curve of the first blow up.

However, the two blow ups are symmetric so the same surface is a ruled surface in a different direction, mapping onto the other exceptional curve. The only ruled surface that has two rulings is $\mathbb P^1\times \mathbb P^1$.

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Up to automorphisms, the map $\mathbb{P}^2\to \mathbb{P}^1\times \mathbb{P}^1$ is in fact given by $(x:y:z)\mapsto ((x:y)(x:z))$, one easily sees that it blows-up $(0:0:1)$ and $(0:1:0)$ and that it contracts the line of equation $x=0$ onto the point $((0:1),(0:1))$. The inverse is $((a:b),(c:d))\to (ac:bc:ad)$, and it blows-up $((0:1),(0:1))$ and contracts the two lines $a=0$ and $c=0$ (passing through the point $((0:1),(0:1))$) onto $(0:0:1)$ and $(0:1:0)$.

One funny other comment is that if you take the real sphere $S$ of dimension $2$ and take the stereographic projection, it gives a birational map $S\to \mathbb{R}^2$ which is not defined at the north pole (the point of the projection). Seeing $S$ in $\mathbb{P}^3$ as a a smooth quadric and putting $\mathbb{R}^2$ into the projective plane, we obtain a birational map $S\to \mathbb{P}^2$ which blows-up the point $q$ (north pole) and contract the two imaginary lines being the intersection of the quadric with the tangent plane at $q$.

Over the reals, $S$ is not isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ (look at the topology of the real part) but extending to $\mathbb{C}$ it is, and the stereographic projection is exactly the inverse of the map you wanted (up to change of coordinates).

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