# Self-intersection of exceptional divisor

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!

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.

• I think this is Algebraic Geometry V. §§ 2.2.14 (just before §§2.3). Unfortunately there is no proof there, just a vague reference to Griffiths-Harris. I came across this post when looking how to generalise the formula to higher dimensions. – Jesus Martinez Garcia Aug 12 '16 at 14:57

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: $$(-K_{Bl_C\mathbb{P}^3})^3 = (4\widetilde{H}-E)^3 = 64-12d+4d+2g-2 = 62-8d+2g.$$

• I know it is a long shot because it's been more than 2 years, but what are the morphism $i$ and $p$? I thought they would be the inclusion of $i$ in the blow-up and the blow-up morphism respectively, but I am not sure due to the first display formula. – Jesus Martinez Garcia Aug 12 '16 at 15:47