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Let $W\subset \mathbb P^4$ be a smooth surface and let $X$ be the blow-up of $\mathbb P^4$ along $W$ with exceptional divisor $E$. By Griffiths-Harris, we have $$A^2(X)=A^2(\mathbb P^4)\oplus A^1(W),$$ so if $H$ denotes the pullback of $O(1)$ on $\mathbb P^n$, there is a basis of $A^2(X)$ given by $H^2$ and line bundles on $W$.

However I can also form the 2-cycles $H\cdot E$ and $E^2$. How can these be expressed in terms of the above basis?

Edit: It seems that $H\cdot E$ should correspond to the line bundle $O_{\mathbb P^4}(1)$ pulled back to $E=\mathbb P N_W^*$. What about $E^2$?

In other words, if $i:E\to X$ is the inclusion and $\pi:E=\mathbb P N_W^*\to W$ is the bundle, then what is $i_*O_E(-1)$ in terms of $H$ and $A^1(W)$?)

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    $\begingroup$ I'm guessing $H \cdot E$ corresponds to the line bundle $\mathcal O(1)$ on $W$. $\endgroup$
    – Will Sawin
    Commented Sep 14, 2014 at 15:41
  • $\begingroup$ Yeah, I think you are right. I'm more curious about $E^2$ though, this should correspond to $O_{E}(-1)$ on the exceptional divisor $E=\mathbb P(N_W^*)$. But I don't see how to relate this to a line bundle down-stairs. $\endgroup$ Commented Sep 14, 2014 at 19:17

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Some notation : $i:E\hookrightarrow X$ the inclusion, $b: X\rightarrow \mathbb{P}^n$ the blowing-up, $p:E\rightarrow W$ the projection, $N$ the normal bundle of $W$ in $\mathbb{P}^n$. The map $A^1(W)\rightarrow A^2(X)$ that you mention is $i_*p^*$. Now $$H\cdot E=H\cdot i_*1=i_*i^*H= i_*p^*\mathscr{O}_{\mathbb{P}^n}(1)_{|W}\ .$$ The computation of $E^2$ is trickier. We have $E^2=i_*i ^*E$; by construction of the blow up $i^*E =-h$, where $h$ is the class of the tautological line bundle $\mathscr{O}_E(1)$. The so-called "key formula" gives $i_*(h+p^*c_1(N))=b^*[W]=\deg(W)\,H^2$, so $$E^2= i_*p^*c_1(N)-\deg(W)\,H^2\ .$$ For the "key formula", see the classical paper of Lascu-Mumford-Scott.

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  • $\begingroup$ Thanks for the answer and the reference. This is indeed very helpful. $\endgroup$ Commented Sep 14, 2014 at 21:59

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