On a Hirzebruch surface

Question

I am trying to solve exercise in Huybrechts's book 'Complex geometry'

While solving problems, one problem kept me from going forward.

That is,

The surface $\Sigma_n=\mathbb{P}$ $(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))$ is n-th Hirzebruch surface.

Show that $\Sigma_n$ is isomorphic to the hypersurface

{${([x_0 , x_1 ],[y_0,y_1,y_2]):x_0^n y_1 - x_1^n y_2 =0}$}$\subset \mathbb{P}^1 \times \mathbb{P}^2$.

How to attack this?

I hope someone shed me a light!

Answer 1

The first thing I would like to point out is that the variety we care about is actually

{$([x_0 , x_1 ],[y_0,y_1,y_2]):{x_0}^n y_1 - {x_1}^n y_2 =0$}$ =V \subset \mathbb{P}^1 \times \mathbb{P}^2$

(the defining equations you gave are not usually homogenous, and this is what Huybrechts gives in his book: http://books.google.com/books?id=eZPCfJlHkXMC&q=hirzebruch).

We want to start out by thinking about $\mathbb{P}^1 \times \mathbb{P}^2$ as a $\mathbb{P}^2$ bundle over $\mathbb{P}^1$. In particular, it comes with the distinguished line bundle $\pi_2^*\mathscr{O}_{\mathbb{P}^2}(1)$ ($\pi_2$ is the projection onto $\mathbb{P}^2$). Now we examine the short exact sequence of sheaves of modules which defines V:

$0\rightarrow \pi_1^*\mathscr{O}_{\mathbb{P}^1}(-n) \otimes \pi_2^*\mathscr{O}_{\mathbb{P}^2}(-1)\rightarrow \mathscr{O} _{\mathbb{P}^2\times\mathbb{P}^1} \rightarrow \mathscr{O}_V\rightarrow0$.

We would like to apply ${\pi_1}_*$ on this sequence after twisting by $\pi_2^*\mathscr{O}_{\mathbb{P}^2}(1)$. The point of this is that if you believe that $V$ is a $\mathbb{P}^1$ bundle over $\mathbb{P}^1$ (which you should because you can check it on charts) then ${\pi_1}_*(\mathscr{O}_V \otimes \pi_2^*\mathscr{O}_{\mathbb{P}^2}(1))$ should be the corresponding locally free sheaf of rank 2. So we can then check to see if it is isomorphic to $\mathscr{O} _{\mathbb{P}^1}\oplus\mathscr{O} _{\mathbb{P}^1}(n)$.

After twisting and applying ${\pi_1}_*$, we get the sequence:

$0\rightarrow \mathscr{O}_{\mathbb{P}^1}(-n)\rightarrow \mathscr{O} _{\mathbb{P}^1}^{\oplus 3} \rightarrow {\pi_1}_*(\mathscr{O}_V \otimes \pi_2^*\mathscr{O}_{\mathbb{P}^2}(1))\rightarrow0$.

(In general one does not get exactness on the right, but in a situation like this we do)

And by figuring out what our maps are doing we should be able to argue that the sheaf on the right is some line bundle twist of $\mathscr{O}_ {\mathbb{P}^1}\oplus \mathscr{O}_{\mathbb{P}^1} (n)$, which implies $V \cong \mathbb{P}(\mathscr{O} _{\mathbb{P}^1}\oplus\mathscr{O} _{\mathbb{P}^1}(n))$.

Answer 2

The holomorphic section of $\mathcal O_\mathbb{P^1}(n)$ is canonically isomorphic to $\mathbb C[z_0,z_1]_n$ of all homogeneous polynomial of degree n, which is an n+1 dimension vector space. Let $s_0$, $s_1$ be two holomorphic section corresponds to the polynomial $z_{0}^n$ and $z_{1}^n$ respectively.

Then we have a map $\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}\longrightarrow\mathcal{O}_{\mathbb{P}^1}(n) $ by sending $(f,g)$ to $fs_0+gs_1.$ This map will induces an isomorphism from $\Sigma_n$ to $\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))$.

If you want to write the isomorphism explicitly. Here we define the embbeding $\phi:\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))\longrightarrow \mathbb P^1\times \mathbb P^2$

On $\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))\mid_{U_0}$, this map is give by

$U_{0}\times \mathbb P^{1}\longrightarrow \mathbb P^1\times \mathbb P^2\\ [x_0,x_1]\times[a,b]\mapsto[x_0,x_1]\times[a,b\frac{{x_1}^n}{{x_0}^n},b]$

On $\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))\mid_{U_1}$, this map is give by

$U_{1}\times \mathbb P^{1}\longrightarrow \mathbb P^1\times \mathbb P^2\\ [x_0,x_1]\times[a,b]\mapsto[x_0,x_1]\times[a,b,b\frac{{x_0}^n}{{x_1}^n}]$

With the definition above $\phi$ is well defined and is the embedding we are looking for.

Answer 3

Sections of $\mathcal{O}_{\mathbb{P}^1}(n)$ are homogeneous degree $n$ polynomials in $x_0,x_1$. You can consider the two sections $x_0^n,x_1^n,$ and construct the morphism $$g:\mathbb{P}^1\rightarrow\mathbb{P}^1\times\mathbb{P}^2,\:[x_0,x_1]\mapsto ([x_0,x_1][0,x_0^n,x_1^n]).$$ You have also the morphism $$e:\mathbb{P}^1\rightarrow\mathbb{P}^1\times\mathbb{P}^2,\:[x_0,x_1]\mapsto ([x_0,x_1][1,0,0]).$$ Note that the images of these two morphism are disjoint. The image of $e$ is going to be the unique curve with self-intersection $-n$. Let $[y_0,y_1,y_2]$ be the homogeneous coordinates on $\mathbb{P}^2$. The fiber of $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$ over the point $[x_0,x_1]\in\mathbb{P}^{1}$ is exactly the line $$F_{x_0,x_1} = \left\langle [0,x_0^n,x_1^n],[1,0,0]\right\rangle = \{x_1^ny_1-x_0^ny_2 = 0\}.$$ Then $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$ is the surface $\mathbb{F}_n$ in $\mathbb{P}^{1}\times\mathbb{P}^2$ defined by $x_1^ny_1-x_0^ny_2 = 0$.

Another method: let $E= e(\mathbb{P}^1)$ and $F$ a fiber. Then $E+F$ represents the sections of type $$p:\mathbb{P}^1\rightarrow\mathbb{P}^1\times\mathbb{P}^2,\:[x_0,x_1]\mapsto ([x_0,x_1][p_n(x_0,x_1),x_0^n,x_1^n])$$ where $p_n(x_0,x_1)$ is a homogeneous polynomial of degree $n$. The curve $E$ is given by $y_1 = y_2 = 0$ in $\mathbb{F}_n$, therefore $E^{2} = -n$. The linear system $|E+nF|$ has degree zero on $E$, one on $F$ and $n$ on $E+F$. This means that it maps $\mathcal{F}_n$ to a surface $C$ in $\mathbb{P}^{n+1}$ whith a unique singular point (corresponding to the contraction of $E$) and whose hyperplane section is a rational normal curve of degree $n$. Then $C$ is a cone over a rational normal curve of degree $n$ and the vertex is a singularity of type $\frac{1}{n}(1,1)$. The surface $\mathbb{F}_n$ is the blow up of the vertex of the cone which is isomorphic to the scroll $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$.

Finally, in a less direct way, one could observe that both $\mathbb{F}_n$ and $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$ are geometrically ruled surfaces over $\mathbb{P}^1$, and that both contain a unique curve with negative self-intersection $-n$. This is enough to conclude that $\mathbb{F}_n\cong \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$.