Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $X$ is a smooth threefold, and $C \subset X$ a smooth curve. Let $Y$ be the blowup of $X$ along $C$, with exceptional divisor $E$. What is the intersection number $E^3$ on $Y$? (in terms of the genus and normal bundle of $C$, etc)

I assume that I could extract the answer from Theorem 6.7 of Fulton's book on intersection theory, were I better familiar with the contents of chapters one through five -- I'd be happy to hear either a direct method or a pointer about how to get it from Fulton!

share|improve this question

2 Answers 2

up vote 3 down vote accepted

The intersection number $E^3$ equals $-\deg N_{C|X}$ the negative of the degree of the normal bundle of $C$. Here, as usual, $\deg N_{C|X}=2g-2-K_X.C$. This statement and the proof can be found in Griffiths-Harris and in Iskovskikh-Prokhorov Algebraic Geometry V III. $\S$ 2.3.

share|improve this answer

Let $Y\subset\mathbb{P}^n$ be a smooth variety, and let $\epsilon:X = Bl_Y\mathbb{P}^n\rightarrow\mathbb{P}^n$ be the blow-up of $\mathbb{P}^n$ along $Y$.

Let $\widetilde{H}$ be the pull-back of the hyperplane section $H$ of $\mathbb{P}^n$, and $E$ be the exceptional divisor. If $H_Y =H\cdot Y$ we have $$\widetilde{H}^{h-i}E^i = p^*H_Y^{n-i}\cdot i^*E^{i-1} = H_Y^{n-i}\cdot p_*i^*E^{i-1}.$$ Recall that $E = \mathbb{P}(N_{Y/\mathbb{P}^n})$, and $i^*E = -e$, where $e = c_1(\mathcal{O}_E(1))$. Let use denote by $s_j$ the Segre classes of $N_{Y/\mathbb{P}^n}$, and let $c = codim_{\mathbb{P}^n}(Y)$. We have the following intersection numbers:

  • $\widetilde{H}^n = 1$;
  • $\widetilde{H}^{n-i}\cdot E^i = 0$ for $i < c$;
  • $\widetilde{H}^{n-i}\cdot E^i = (-1)^{i-1}s_{i-c}H_Y^{n-i}$ for $i\geq c$.

Let $C\subset\mathbb{P}^3$ be a smooth curve of degree $d$ and genus $g$. By the exact sequence $$0\mapsto T_{C}\mapsto T_{\mathbb{P}^3|C}\rightarrow N_{C/\mathbb{P}^3}\mapsto 0$$ we get $s_1(N_{C/\mathbb{P}^3}) = -c_{1}(N_{C/\mathbb{P}^3}) = -4d-2g+2$. Then $\widetilde{H}^3 = 1$, $\widetilde{H}^2\cdot E = 0$, $\widetilde{H}\cdot E^2 = -s_0H_Y = -d$, and $E^3 = s_1 = 2-2g-4d$. For instance we can compute the cube of the anti-canonical divisor: $$(-K_{Bl_C\mathbb{P}^3})^3 = (4\widetilde{H}-E)^3 = 64-12d+4d+2g-2 = 62-8d+2g.$$

share|improve this answer

Your Answer

 
discard

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.