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Could you help me to understand what does the calculation in the bottom of the image mean? From where does $\tau^*(\zeta)$ appear?

It is page 48 from preprint 3264 & All That Intersection Theory in Algebraic Geometry.

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The crucial component of this calculation is this: If you have a $k$-dimensional subvariety of $\Bbb P^n$, to find its degree you must intersect $k$ times with a generic hyperplane (this amounts to intersecting with a generic $\Bbb P^{n-k}$). Since $\tau$ maps generically one-to-one to $\Gamma$, this computation can be done by pulling back a generic hyperplane and intersecting that pullback $k$ times back in the domain. By the comments the author makes, $\Gamma$ is $7$-dimensional in $\Bbb P^9$, and $\tau^*\zeta = \alpha+\beta$, where $\alpha$ is a hyperplane (line) in $\Bbb P^2$ and $\beta$ is a hyperplane in $\Bbb P^5$. (This simple sum comes, as the author indicates, from the fact that $\tau$ is simply bilinear.)

Then we can expand $(\alpha+\beta)^7$ by the usual binomial theorem, since these classes in $H^{2k}(\Bbb P^2\times\Bbb P^5)$ all commute, being of even degree. Thus, since $\alpha^2=\beta^5=1$ and $\alpha^j=\beta^k=0$ for $j>2$ and $k>5$, $$(\alpha+\beta)^7= \binom 72 \alpha^2\beta^5 = 21,$$ as desired.

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  • $\begingroup$ +1, but the statement that "$\alpha$ and $\beta$ do not meet" isn't quite right... $\endgroup$ – user64687 Jun 3 '14 at 13:07
  • $\begingroup$ @AsalBeagDubh: Slip of the finger, thanks! $\endgroup$ – Ted Shifrin Jun 3 '14 at 13:49

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