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Let $Y\subset\mathbb{P}^n$ be a smooth variety of codimension two. Consider the blow-up $X = Bl_Y\mathbb{P}^n$ of $\mathbb{P}^n$ along $Y$, and let $E$ be the exceptional divisor over $Y$. Then $E$ has a structure of $\mathbb{P}^1$-bundle over $Y$. The anticanonical divisor is $$-K_X = (n+1)H-E.$$ I would like to compute $(-K_X)^n$. For example if $Y\subset\mathbb{P}^3$ is a curve of degree $d$ and genus $g$, by Shafarevich "Algebraic Geometry V" Lemma 2.2.14, we have: $$E^3 = -deg(N_{Y/\mathbb{P}^3}) = K_{\mathbb{P}^3}\cdot C-2g+2 =-4d-2g+2.$$ and $$(-K_X)^3 = 62+2g-8d.$$ I would like to have a formula for the top self-intersection $(-K_X)^n$ when $X$ is the blow-up of a smooth subvariety $Y\subset\mathbb{P}^n$ of codimension two. Even for complete intersections of two hypersurfaces it would be good. More generally does there exists a formula for $(-K_X)^n$ when $X$ is the blow-up of $\mathbb{P}^n$ along a smooth subvariety of codimension $c$ (again, I would be happy with complete intersections)?

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up vote 6 down vote accepted

So you want to compute the intersection numbers $(H^p\cdot E^q)$, $p+q=n$.

Let me start with some notation. Let $b: X\rightarrow \mathbb{P}^n$ be the blowing up, $i:E \hookrightarrow X$ the embedding, $p:E\rightarrow Y$ the projection. Write one factor $E$ as $i_*1$ (in $CH(X)$, say) and use the projection formula twice: $$(H^p\cdot E^q)=(i^*E^{q-1}\cdot p^*H_Y^p)=(p_*i^*E^{q-1}\cdot H_Y^p)\quad \mbox{where }\ H_Y:=c_1(\mathcal{O}_{\mathbb{P}^n}(1))_{|Y}\ .$$ We have $i^*E=-h$, where $h$ is the first Chern class of the tautological line bundle $\mathcal{O}_E(1)$ on $E=\mathbb{P}(N^*_{Y/\mathbb{P}^n})$. The classes $s_m:=p_*h^{m+c-1}$ (with $c=\mathrm{codim}(Y)$) are the Segre classes of $N_{Y/\mathbb{P}^n}$; they are easily calculated from the Chern classes, see Fulton Intersection Theory, Chapter 3. So the numbers $(H^p\cdot E^q)=(-1)^{q-1}s_{q-c}\cdot H_Y^p$ can be calculated when you know the Chern classes of the normal bundle. In codimension 2 you just have two nonzero Chern classes, so the computation is relatively easy. For instance, if $n=3$ and $Y$ is a curve of degree $d$ and genus $g$, you find $(-K_X^3)= 64-8d+2g-2$.

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I guess there is something wrong in your formula. For instace if we blow-up a point in $\mathbb{P}^2$ the first two Segre classes of the conormal bundle are $s_0=1, s_1 = 2$. With your formula we get $E^2 = -s_1 = -2$ and $HE = s_0 = 1$, instead of $E^2=-1, HE = 0$. – user49214 Apr 9 '14 at 23:22
I see the point: my subscript of the Segre class was not the standard one, I edited. The (co)normal bundle of a point is of course trivial, so $s_0=1$, all the others $s_i$ are zero. This gives what you expect. – abx Apr 10 '14 at 5:13
Sorry to bother you abx but I think there is something wrong. For instance blow up a conic $Y\subset\mathbb{P}^3$. Then using your formula $E^3 = H^0\cdot E^3 = s_1$. The conormal bundle of the conic is $\mathcal{O}_Y(-1)\oplus\mathcal{O}_Y(-2)$. Therefore the first Chern class is $c_1 = -3$ and the first Segre class is $s_1 = 3$. We get $E^3 = 3$. But I guess the right number is $E^3 = -6$. – S_Z_S Apr 15 '14 at 15:39
You are right for the sign, and I am right for the number. You must take $c_1$ on $Y$: so $\mathcal{O}_Y(1)$ has degree 2, and my formula gives $E^3=6$. Now I am wrong on the sign, because I followed Fulton and I didn't realize he doesn't use Grothendieck notation for $\mathbb{P}(E)$. So $N^*$ should be replaced by $N$, and then one gets indeed $E^3=-6$. I have edited. Thanks! – abx Apr 15 '14 at 16:01

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