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I am wondering whether there is a notion of restriction of a divisor to itself? More specifically, let $X=\mathbb{P}^3$ be the projective space and let $x, y$ be two points in $X$, let $l$ be the line passing through $x,y$. Let $Y$ be the blowup of $X$ along $x, y$ with exceptional divisors $E_x', E_y'$. Let $\pi: Z\to Y$ be the blowup of $Y$ along the proper transforms of the line $l$. Let $E_x=\pi^{-1}(E_x'), E_y=\pi^{-1}(E_y')$.

We have $E_x$ is the blowup of $E_x'\cong\mathbb{P}^2$ at one point corresponding to the line $l$. Then how to calculate $E_x|_{E_x}$? Is it equal to $-H$, where $H$ is the hyperplane class on $E_x$?

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$E_x$ is the preimage or the proper transform of $E'_x$ ? – user18119 Jan 13 '13 at 14:49
@QiL: Sorry I made it wrong, $E_x$ is the preimage of $E_x'$. – minimax Jan 15 '13 at 2:25
in fact I realized afterward that they are the same because the preimage of $x$ in $Z$ is a line and the preimage of $x$ in the blowup of $E'_x$ along $x$ is also a line, so they concide. – user18119 Jan 15 '13 at 12:05
@QiL: Thank you, but why is the preimage of $x$ in $Z$ a line? – minimax Jan 16 '13 at 8:33
because $Y$ has dimension $3$ and you blow-up a line $\ell$, so the exceptional divisor is a familly of projective lines parametrized by $\ell$ (it is the projective bundle of relative dimension $1$ over $\ell$. – user18119 Jan 16 '13 at 22:35

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