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In Demailly's icm2006 p21, there is a statement:

If $X$ is the surface obtained by blowing-up $\mathbb P^2$ in one point, then the exceptional divisor $E ≃ \mathbb P^1$ has a cohomology class {$\alpha$} such that $\int_E\alpha = E^2 = −1$.

Can someone provide more details about what happens in $\int_E\alpha$? If I'm ritht, $E$ can be treated as a Riemann sphere, and the cohomology class $\alpha$ can be represented by a (1,1) form, so $\int_E\alpha$ can be seen as an integral of a (1,1) form over a sphere, but why it should be $-1$? And is there a canonical way to construct $\alpha$? Any comments or references are welcome!

Arctic Char
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Tom
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  • Does this answer your question? Geometric motivation for negative self-intersection – Arctic Char Aug 22 '20 at 08:58
  • @Arctic Char, I don't think it answered my question, what I expect is a simple and clear answer like the one you gave several days ago, but the answers to the question you post seen too long and complicated which only makes me more puzzled and frustrated, actually it never explain why $\int_E\alpha = E^2$, so I'm still waiting for a better answer. – Tom Aug 22 '20 at 11:38
  • Concerning $\int_E\alpha = E\cdot E$, what is your definition of $E\cdot E$? – Arctic Char Aug 22 '20 at 11:48
  • @ Arctic Char, so you mean once the divisor is determined, the corresponding cohomology class $\alpha$ is uniquely determined? and $E\cdot E$ is by definition equals to the integral above? – Tom Aug 22 '20 at 11:54
  • Yes, in general the intersection no $E_1\cdot E_2$ is defined by $\int_{E_1} \alpha_2$, where $\alpha_2$ is the Poincare dual to $E_2$. – Arctic Char Aug 22 '20 at 12:10
  • @Arctic Char, ok, thanks, but is the Poincare dual to $E_2$ is uniquely defined or differs by a positive or negative signature? – Tom Aug 22 '20 at 12:16
  • @Arctic Char, I'm looking forward to your answer. – Tom Aug 22 '20 at 13:15
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    To add another comment (because I don't want to write a complete answer right now): The crucial thing is that the normal bundle of $E$ in $X$ is the tautological line bundle on $\Bbb P^1$, i.e., $\mathscr O_E(-1)$. – Ted Shifrin Aug 22 '20 at 20:55
  • Dear Ted, then why it should be $\mathcal O(-1)$ other than $\mathcal O(1)$? – Tom Aug 23 '20 at 14:20
  • The tautological line bundle is the dual of the hyperplane section bundle. It has no global holomorphic sections. If you write down the local description (blowing up the origin in $\Bbb C^n$ but you can do just $n=2$ if you prefer), you'll see this immediately when you thinking about the projection down to $\Bbb P^{n-1}$. – Ted Shifrin Aug 23 '20 at 23:11
  • Dear Ted, for $\mathbb C^2=(z_0,z_1)$, the local description for the blow up is $z_0\cdot\omega_1=z_1\cdot\omega_0$, $(\omega_0,\omega_1)$ is the homogeneous coordinate of $\mathbb P^1$, then is it enough to conclude that $\mathbb C^2$ is the tautological bundle of $\mathbb P^1$? – Tom Aug 24 '20 at 04:03

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