5
$\begingroup$

Let's take $\phi:\mathcal{X}\rightarrow B$ a family of deformations of complex surfaces above the one dimensional disk $B=D_r(0)\subset \mathbb{C}$. Suppose that $X:=\phi^{-1}(0)$ has one ordinary double point $p$, while all other fibers are smooth. So we can take $x,y,z$ complex coordinates around $p$ such that $\phi(x,y,z)=x^2+y^2+z^2$.

Now take the function $m:\widetilde{B}:=D_{\frac{1}{2}}(0)\rightarrow B$, $m(t)=t^2$ and consider the deformation $\widetilde{\mathcal{X}}=\mathcal{X}\times_B \widetilde{B}$. If $\widetilde{p}\in \widetilde{\mathcal{X}}$ is the only point mapping to $p$ then in local coordinates the germ of $\widetilde{\mathcal{X}}$ in $\widetilde{p}$ is isomorphic to the germ in the ordigin of $V(x^2+y^2+z^2-t^2)\subset \mathbb{C}^4$.

I consider the blow up of $\widetilde{\mathcal{X}}$ in $\widetilde{p}$: its exceptional divisor $E$ is a quadric in $\mathbb{P}^3$ and so is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ by Segre embedding. Finally it shoul come out that $N_{E/\widetilde{\mathcal{X}}}\simeq\mathcal{O}_{\mathbb{P}^1}(-1)\boxtimes\mathcal{O}_{\mathbb{P}^1}(-1)$ and $\mathcal{O}_{\mathbb{P}^1}(-1)\boxtimes\mathcal{O}_{\mathbb{P}^1}(-1)=p_1^*\mathcal{O}_{\mathbb{P}^1}(-1)\otimes p_2^*\mathcal{O}_{\mathbb{P}^1}(-1)$ where $p_1$ and $p_2$ are the two projections from $\mathbb{P}^1\times\mathbb{P}^1$)

My questions are:

1) I have read that for example for the blow up of the origin in $\mathbb{A}^2$, the normal bundle of the exceptional divisor $\mathbb{P}^1$ is $\mathcal{O}_{\mathbb{P}^1}(-1)$, but why is that?

2) In the specific case of my construction in which way can i prove that $N_{E/\widetilde{\mathcal{X}}}$ is actually the box product?

$\endgroup$

2 Answers 2

6
$\begingroup$

For the first question, and what Anton mentions, that the normal bundle of the exceptional divisor of the blow up of a smooth subvariety in a smooth variety is $\mathscr O(-1)$ see Thm 8.24 in [Hartshorne]. This takes care of your first question.

However, you have to be careful, because

  1. this fails if the ambient scheme is singular, as in your case, where $\widetilde{\mathcal{X}}$ is singular, and
  2. you have to keep in mind that the $(-1)$ here means a specific embedding, so it might not translate to "the natural" $(-1)$ of any given scheme, say $\mathbb P^n$ (this actually does not happen in your case, but only by luck).

To see that the normal bundle you are looking for (I assume, because what you wrote makes no sense) is the box product do this:

First let's give names to our players: Let $A$ denote the blow up of $\mathbb A^4$ at the origin, $P\simeq \mathbb P^3$ the exceptional divisor, $B\subset A$ the blow up of $\widetilde{\mathcal{X}}$ at the same point, and $E$ the exceptional divisor of that. Presumably you want $N_{E/B}$.

Observe that $A$ is smooth and hence all divisors are Cartier. Then consider the restrictions of $\mathscr O_A(P)$ to both $P$ and $B$ and then the restrictions of these to $E$:

$$ \mathscr O_A(P)|_P \simeq N_{P/A}\simeq \mathscr O_{\mathbb P^3}(-1) $$

$$ \left(\mathscr O_A(P)|_B\right)|_E \simeq \mathscr O_B(E)|_E \simeq N_{E/B} $$

and hence

$$ N_{E/B}\simeq \mathscr O_A(P)|_E \simeq \mathscr O_{\mathbb P^3}(-1)|_E\simeq \mathscr O_{\mathbb P^1}(-1)\boxtimes \mathscr O_{\mathbb P^1}(-1). $$

$\endgroup$
0
$\begingroup$

As for the first question, this is written in many places. Searching MO, by the way, is a good idea: blowing up, -1 curves, effective and ample divisors

Generally, if $X$ is a smooth subscheme in $Y$ with normal bundle $N$, then $N_{\tilde{X}}\tilde{Y}=\mathcal{O}_N(-1)$.

As for the second one, this is obvious after you combine the comment above with functoriality of the blow-up.

$\endgroup$
2
  • $\begingroup$ thank you. but what exactly do you mean by "functoriality of the blow up"? $\endgroup$ Jun 7, 2013 at 11:26
  • $\begingroup$ By functoriality I mean that the blow up of $X$ is the strict transform of $X$ in the blow up of the ambient $\mathbb{A}^4$. See Sándor's answer for details. $\endgroup$ Jun 7, 2013 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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