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I am trying to calculate some intersection numbers and would appreciate help on the following problem:

Consider two divisors $D_1$ and $D_2$. Blowing up their intersection yields $\varphi^{*}(D_i) = \widetilde{D_i} + m_i E$, $\varphi$ is the blowup morphism and E is the exceptional Divisor. I hope there are no mistakes so far.

Now I am interested in calculating intersection numbers that would look like $\varphi^*(D_i) . \alpha$, where $\alpha$ is a cycle which is contained in the exceptional locus with the property that the dimension of $\varphi_{*}\alpha$ is lower than that of $\alpha$. My first intuition was to apply the projection formula, but on second thought I am not sure if this is the right way to go because of the difference in the dimensions. Any thoughts on a good way to calculate an intersection number like this?

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    $\begingroup$ Why would $\varphi _*(\alpha )$ have lower dimension that $\alpha $? This depends very much on the situation. If it is, it just means that the product is $0$. $\endgroup$
    – abx
    Apr 25, 2021 at 18:53
  • $\begingroup$ You are right, I was trying to make the question general, even though I am interested in the case where $dim(\varphi_* \alpha)< dim( \alpha)$. I will change the question. Can you explain though why the product would be zero in this case? It was my guess but I can't really prove it $\endgroup$
    – Galathea
    Apr 25, 2021 at 19:40
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    $\begingroup$ Because already $\varphi _*\alpha =0$, by the very definition of $\varphi _*$. $\endgroup$
    – abx
    Apr 25, 2021 at 19:52
  • $\begingroup$ I'm sorry, I dont understand yet. How can I use $\varphi_* \alpha = 0$ to calculate $\varphi^* D . \alpha$? I'm aware it must be simple but I do not see it.. $\endgroup$
    – Galathea
    Apr 25, 2021 at 19:57
  • $\begingroup$ If it's an intersection number, and not a class that you're after, you can use the projection formula. $\endgroup$
    – Will Sawin
    Apr 25, 2021 at 21:37

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