Intersection numbers

Shadow · **Posted:** Thu, 29 Mar 2012 22:47:42 UTC

I'm supposed to show that there is no neighborhood of $\mathbb{C}P^1$ in $\mathbb{C}P^2$ on which my copy of $\mathbb{C}P^1$ is globally cut out by independent functions, and I think this should just be a byproduct of $I_2(\mathbb{C}P^1,\mathbb{C}P^1)=1$ , but I cannot see how to show this, as far as I can see I should be able to separate $\mathbb{C}P^1$ from itself since $\mathbb{C}P^2$ is just $\mathbb{C}P^1$ with a copy of $\mathbb{R}^4$ glued in, so why not just move the sphere away from itself into this copy or Euclidean space. Any help appreciated.

outermeasure · **Posted:** Fri, 30 Mar 2012 03:50:00 UTC

Shadow wrote:

I'm supposed to show that there is no neighborhood of $\mathbb{C}P^1$ in $\mathbb{C}P^2$ on which my copy of $\mathbb{C}P^1$ is globally cut out by independent functions, and I think this should just be a byproduct of $I_2(\mathbb{C}P^1,\mathbb{C}P^1)=1$ , but I cannot see how to show this, as far as I can see I should be able to separate $\mathbb{C}P^1$ from itself since $\mathbb{C}P^2$ is just $\mathbb{C}P^1$ with a copy of $\mathbb{R}^4$ glued in, so why not just move the sphere away from itself into this copy or Euclidean space. Any help appreciated.

To be able to move the sphere away from itself into the copy of $\mathbb{R}^4$ means you have a global section of the principal S^1

-bundle $S^1\to S^3\to S^2$ and hence you would get $S^3\cong S^2\times S^1$ , which is nonsense.

Another way to see why $[\mathbb{C}P^1]\cdot[\mathbb{C}P^1]=1$ is using the cohomology ring of $\mathbb{C}P^2$ : $H^*(\mathbb{C}P^n)=\mathbb{Z}[\omega]/(\omega^{n+1})$ where $\omega$ has degree 2 and represented by the dual of any hyperplane $\mathbb{C}P^{n-1}$ .

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Intersection numbers

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