MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Take two intersecting lines $L_1,L_2$ in $\mathbb{P}^3$. Blow-up $L_1$, and then blow up the strict transform $L_2'$ of $L_2$ and call $E_2$ the exceptional divisor of this second blow-up. $E_2$ is a $\mathbb{P}^1$-bundle over $L_2'\cong\mathbb{P}^1$ and so is some Hirzebruch surface $F_n$.

By doing local computations, I think that $E_2$ is $F_1$. Is there a good way (I mean, without coordinates, or at least without having to use affine charts) to figure out which $F_n$ it is?

share|cite|improve this question

Here is another approach, not even referring to exact sequences. Let's do this torically - using Fulton's fan language. The fan of ${\mathbb P}^3$ is given by the cones spanned by the vectors $e_1,e_2,e_3,-e_1-e_2-e_3$ in the vector space spanned by them. The first blowup corresponds to subdividing the fan by inserting the vector $e_1+e_2$, whereas the second blowup corresponds to subdividing the fan further by inserting the vector $e_1+e_3$ - please draw a picture! Staring at the picture will tell you that this last vector will give an edge of the four cones $(e_1, e_1+e_2, e_1+e_3)$, $(e_1+e_2, e_3, e_1+e_3)$, $(e_3, -e_1-e_2-e_3, e_1+e_3)$, $(-e_1-e_2-e_3, e_1, e_1+e_3)$. The geometry of the divisor corresponding to the vector $v=e_1+e_3$, the exceptional divisor $E_2$ you are looking for, is given by projecting these four cones onto the quotient of the vector space by the span of $v$; you can arrange this simply by setting $e_1+e_3=0$ in the above cones, getting the cones $(e_1, e_1+e_2)$, $(e_1+e_2, -e_1)$, $(-e_1, -e_2)$ and $(-e_2, e_1)$ in two dimensions. This is the standard fan of ${\mathbb F}_1$.

share|cite|improve this answer
Wow, this is great. – Enrique Jul 20 '11 at 16:02

Here is an alternative approach without referring to deformations.

Let $P=\mathbb P^3$, $B$ denote the first blow up, $E$ the exceptional divisor of the blow up $\pi:B\to P$, and $L$ the proper transform of $L_2$. Also let $k$ denote the base field.

Then $E\simeq \mathbb{P^1\times P^1}$ and $\Omega_{B/P}$ is supported on $E$ where it is isomorphic to a line bundle (actually $\mathscr O_{E/L_1}(-2)$, but this does not matter). So, we have that $$\Omega_{B/P}\otimes \mathscr O_L\simeq k.$$

On $B$ there is a short exact sequence $$ 0\to \pi^*\Omega_P \to \Omega_B \to \Omega_{B/P} \to 0$$ which remains exact after restricting to $L$: $$ 0\to \pi^*\Omega_P\otimes \mathscr O_L \to \Omega_B\otimes \mathscr O_L \to \Omega_{B/P}\otimes \mathscr O_L \to 0$$

Then there is the conormal sequence of $L$ in $B$: $$ 0\to \mathscr N_{L/B}^{\ \ \vee} \to \Omega_B\otimes \mathscr O_L \to \Omega_L \to 0, $$ and there is a similar short exact sequence on $L_2$, which we pull back to $L$: $$ 0\to \pi^*\mathscr N_{L_2/P}^{\ \ \vee} \to \pi^*\Omega_P\otimes \mathscr O_{L} \to \pi^*\Omega_{L_2} \to 0. $$

Notice that the second short exact sequence maps to the first and that $\pi$ is an isomorphism between $L$ and $L_2$, so using the previous short exact sequence and the Snake Lemma, we obtain that there is a short exact sequence:

$$ 0\to \pi^*\mathscr N_{L_2/P}^{\ \ \vee}\to \mathscr N_{L/B}^{\ \ \vee} \to k\to 0. $$

Finally, $\pi^*\mathscr N_{L_2/P}^{\ \ \vee}\simeq \mathscr O_L(-1)\oplus\mathscr O_L(-1)$, so it follows that $\mathscr N_{L/B}^{\ \ \vee}\simeq \mathscr O_L(a)\oplus\mathscr O_L(b)$ such that $a,b\geq -1$ and $a+b=-1$. This can only happen if $$\mathscr N_{L/B}\simeq \mathscr O_L\oplus\mathscr O_L(1),$$

and then blowing up $L$ the exceptional divisor is the Hirzebruch surface $\mathbb F_1$.

share|cite|improve this answer
Thanks for witting this. It is nice to see as many approaches as possible. I'll also need to sit down an understand all the details in this one! – Enrique Jul 19 '11 at 21:35

Let $\pi \colon X \to \mathbb{P}^3$ be the first blow-up.

The normal bundle of $L_2$ in $\mathbb{P^3}$ is $$N_{L_2 / P^3}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}(1);$$ in fact $L_2$ moves into a family of dimension $4=h^0(N_{L_2/ P^3})$, namely the projective Grassmannian $\mathbb{G}(1,3)$.

The strict transform $L_2'$ of $L_2$, instead, moves into a family of smaller dimension, namely the strict transform of the family $\mathfrak{X}$ of lines intersecting $L_1$. Since $\mathfrak{X}$ is a divisor in $\mathbb{G}(1,3)$, we have $h^0(N_{L_2'/ X})=3$. In fact, one proves that $$N_{L_2' / X}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1},$$ see the edit at the endo of the post.

It follows that the exceptional divisor of the blow-up of $X$ along $L_2'$ is the projective bundle $$\mathbb{P}(\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}),$$ which is in turn isomorphic to the Hirzebruch surface $\mathbb{F}_1$.

EDIT. Let me prove in full details that $N_{L_2' / X}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}.$

Let us consider the family $\mathcal{S}$, given by the strict transforms in $X$ of the planes containing $L_2$, and let $S$ be a general element of $\mathcal{S}$. Then $S$ is isomorphic to $\mathbb{F}_1$, since it is just a plane blown-up at the point $L_1 \cap L_2$. Moreover $L_2' \subset S$ is a fibre of the ruling. Finally, since two elements of $\mathcal{S}$ intersect in a reducible curve made by a fibre and the $(-1)$-curve, it follows $$N_{S/X}=\mathcal{O}_S(S) \cong \mathcal{O}_{\mathbb{F}_1}(C_0+f),$$
where $C_0$ is the $(-1)$-curve and $f$ is a fibre of the ruling (of course, these $\mathbb{F}_1$ have nothing to do with the exceptional divisor of the second blow-up...)

Now, by using the normal bundle sequence associated with $L_2' \subset S \subset X$ we obtain $$0 \to N_{L_2' / S} \to N_{L_2' / X} \to N_{S / X} \otimes \mathcal{O}_{L_2'} \to 0,$$ that is $$ 0 \to \mathcal{O}_{P^1} \to N_{L_2' / X} \to \mathcal{O}_{P^1}(1) \to 0$$ (the last line bundle on the right comes from the fact that $(C_0+f)f=1$).

Since $\operatorname{Ext}^1(\mathcal{O}_{P^1}(1), \mathcal{O}_{P^1})=H^1(\mathcal{O}_{P^1}(-1))=0,$ the last exact sequence splits and the claim follows.

share|cite|improve this answer
Dear Francesco: Thanks! This seems much nicer, but it will take me a while to understand it completely! Would you mind explaining in more detail how the argument that $h^0(N_{L_2'/ X})=3$ implies the equality $N_{L_2' / X}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}$? – Enrique Jul 19 '11 at 17:21
You are right, some details were needed. I've added them. – Francesco Polizzi Jul 19 '11 at 18:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.