In light of my previous question, I am interested in the following scenario: Let $\tilde Y$ be the blow-up of $Y=\mathbb{P}^n$ along a linear subvariety $X\subseteq Y$ of codimension $d$, i.e. $X\cong\mathbb{P}^{n-d}$. Let $E$ denote the exceptional divisor. Let $H$ be a hyperplane in $Y$ and denote by $P$ the strict transform. I am now interested in the degrees of $P^a E^b$ for $a+b=n$. The only thing related I could find was this post, but I am in a more general setting.

PS: I am working over $\mathbb{C}$, so you may assume that, or just any algebraically closed field, or less of course.


Assuming $H$ contains $X$, I think $P^{n-b} \cdot E^b = (-1)^{b-1+\dim X} {b-1 \choose \dim X}$, where $\dim X = n-d$. One can, if one wishes, reduce to the case $a=0$ by letting $Y'$ be the intersection of $a$ generic hyperplanes through $X$, for then $P^a\cdot E^b = (E|_{Y'})^b$. Or, without making this reduction, let $T=P+E$ be the total transform of $H$ in $\tilde{Y}$, and note that (i) $E$ is a $\mathbb{P}^{d-1}$-bundle over $X$, (ii) $P|_E$ is a $\mathbb{P}^{d-2}$-subbundle over $X$, meeting every fiber of $E\to X$ in a hyperplane, (iii) $T|_E$ is the pullback to $E$ of a hyperplane section of $X$. It follows geometrically that, if $a+b=n$, then $P^{a}\cdot T^{b-1} \cdot E = (P|_E)^{a}\cdot (T|_E)^{b-1}$ equals 1 if $(a,b)=(d-1,n-d+1)$ and 0 otherwise. Then we can expand $P^a \cdot E^b = P^a \cdot (T-P)^{b-1} \cdot E$ using multilinearity of the intersection form.

  • $\begingroup$ This is great! However, why exactly is $P|_E^a\cdot T|_E^{b-1} = 1$ iff $(a,b)=(d-1,n-d+1)$? You say it follows geometrically, but I cannot quite see it. Thanks a lot already and also in advance. $\endgroup$ – Jesko Hüttenhain Mar 9 '12 at 13:53
  • $\begingroup$ If I'm not mistaken, (ii) says $P|_E^a$ is represented by a $\mathbb{P}^{d-a-1}$-subbundle over $X$, which is empty if $a>d-1$, and (iii) says $T|_E^{b-1}$ is represented by the preimage in $E$ of a linear $\mathbb{P}^{n-d+1-b}$ in $X$, which is empty if $b>n-d+1$. If $(a,b)=(d-1,n-d+1)$, then $P|_E^a$ is a section of $E\to X$ and $T|_E^{b-1}$ is a fiber, and these things meet transversally in one point. $\endgroup$ – Tiankai Mar 9 '12 at 16:21

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