How to show that $\mathbb{P}^2$ blowing up at a point is not isomorphic to $\mathbb{P}^1*\mathbb{P}^1$?

Question

I know that $\mathbb{P}^2$ blowing up at two points is isomorphic to $\mathbb{P}^1*\mathbb{P}^1$ blowing up at one point, so why $\mathbb{P}^2$ blowing up at a point is not isomorphic to $\mathbb{P}^1*\mathbb{P}^1$? and is $\mathbb{P}^2$ blowing up at one point not isomorphic to $\mathbb{P}^2$ blowing up at another point? I think the answer is no but I don't know why.

Answer 1

$\newcommand{\Proj}{\mathbf{P}}$I assume you're working over the complex numbers.

• Because the projective plane $\Proj^{2}$ is homogeneous, the blow-up of $\Proj^{2}$ at a point $p$ does not depend, up to isomorphism, on the choice of $p$.

• The blow-up of $\Proj^{2}$ at one point has an exceptional divisor with self-intersection $-1$. No curve in $\Proj^{1} \times \Proj^{1}$ has self-intersection $-1$.

• A rational map $\Proj^{2} \to \Proj^{1} \times \Proj^{1}$ can be defined by picking two points $p \neq q$, blowing up at $p$ and $q$, then blowing down the proper transform of the line $\overline{pq}$.

The diagrams below show the respective surfaces with embedded rational curves and their self-intersection numbers; dots indicate points being blown up. (You can think of these pictures as metaphorical, or as the literal images of moment maps for real $2$-torus actions. In that case, if you want to nitpick, the diagonal edge in the upper-left diagram should have slope $-1$.)

The standard algebro-geometric rational map from $\Proj^{1} \times \Proj^{1} \to \Proj^{2}$ should be mentioned: Consider a smooth quadric surface $Q \simeq \Proj^{1} \times \Proj^{1}$ in $\Proj^{3}$. Pick a point $q$ of $Q$ and a plane $H \simeq \Proj^{2}$ not containing $q$. Projection away from $q$ defines a map $Q \setminus\{q\} \to H$ that collapses two lines, the rulings of $Q$ through $q$, and is otherwise bijective (since every line through $q$ that is not a ruling hits $Q$ in exactly one other point). This map is well-defined on the blow-up of $Q$ at $q$. That is, blowing up the quadric at one point then blowing down two points gives $H$.

Answer 2

Over the complex numbers, you can compute their Euler characteristics as manifolds. The euler characteristic of $Bl_p P^2$ is 4 (I will explain below). The euler characteristic of $P^1 \times P^1$ is 5 - this comes from computing the cohomology grups via Kunneth. Thus they cannot be even homeomorphic (in the classical topology). Note that an isomorphism of varieties will induce a homeomorphism in the classical topology.

To see that the Euler characteristic of $Bl_p P^2$ is 4, think about the blow up as removing a point and replacing it with a $S^2$. This total change in Euler characteristic is 1, so $\chi(P^2) + 1 = 3 + 1 = 4$. I'm not completely sure how to formalize this (other than "integrating against the euler charactersitic", which you can see here for: A.G. Khovanskii and A.V. Pukhlikov, Integral transforms based on Euler characteristic and their applications ). You can also tediously show that the plane blown up at a point is a Hirzebruch surface, and then refer to here: http://www.map.mpim-bonn.mpg.de/Hirzebruch_surfaces

Double check my computations?

Maybe a similar argument works in positive characteristic?

Edit: As Mariano points out, one can do this computation with Mayer-Vietoris. That is, we want to know $\chi(Bl_0(\mathbb{C}^2))$, and there is a natural deformation retraction between the blow up of the affine plane at a point and the $\mathbb{CP}^1$ of lines through $0$. A little more explicitly, we fix the exceptional divisor and for $t \in (0,1]$ send $v \in C^2 \setminus p$ to $tv$, and in the limit send it to the line spanned by $v$.

From here we can use inclusion exclusion for the Euler characteristic (which is what Mayer Vietoris implies for Euler characteristic): Taking $U$ to an $A^2$ neigborhood of $p$ and $V = P^2 \setminus p$, $Blp P^2 = (Bl_p U) \cup V$, $Bl_pU \cap V = U \cap V$. So $\chi(P^2) = \chi(U) + \chi(V) - \chi(U \cap V) = \chi(Bl_p U) - 1 + \chi(V) - \chi(U \cap V) = \chi(Bl_p P^2) - 1$. (This works in general for blowing up a smooth variety at a point.)