Let $X$ be the union of $3$ hypersurfaces in $\mathbb{C}P^2$, then how to compute the $\pi_1(\mathbb{C}P^2\setminus X)$?

What I know is the complement of a hypersurface in $\mathbb{C}P^2$ is $\mathbb{C}^2$. I can't go further.

  • 1
    $\begingroup$ It seems that your "hypersurface" complex projective line. If so, put one of these lines at infinity, and make two other lines coordinate axes. Then think about fibering the rest over some Riemann surface. $\endgroup$ – Moishe Kohan Apr 21 at 2:07
  • $\begingroup$ @Moishe can you explain more. Your words always confused me. $\endgroup$ – 6666 Apr 21 at 3:54
  • $\begingroup$ what do you mean by disjoint in this case ? $\endgroup$ – Ignorant Mathematician Apr 21 at 9:01
  • $\begingroup$ It depends on the hypersurface! $\endgroup$ – Andres Mejia Apr 21 at 19:48

In the lines of @MoisheKohan after a change of co-ordinates you can assume one of your hypersurfaces is $H_3=\{[z_1:z_2:z_3]\in \mathbb C\mathbb P^2 | \ z_3= 0 \}$

$ \mathbb C \mathbb P^2-H_3\cong\mathbb C^2$

Thus you are reduced to computing $\pi_1(\mathbb C^2-l_1\cup l_2)$ where $l_1, \ l_2$ are two complex straight lines. Again after a change of co-ordinates you can assume $l_1=\{(z_1,z_2)\in \mathbb C^2 :z_1=0\}$ and $l_2=\{(z_1,z_2)\in \mathbb C^2 :z_2=0\}$

Then you have $\mathbb C^2 -l_1\cup l_2=\mathbb C^*\times \mathbb C^*$

So you get $$\pi_1(\mathbb C^2-l_1\cup l_2)=\pi_1(\mathbb C^*)\times \pi_1(\mathbb C^*)=\mathbb Z^2$$

Thus $$\pi_1(\mathbb C\mathbb P^2-H_1\cup H_2 \cup H_3)=\mathbb Z^2$$

  • $\begingroup$ But your $l_1\cap l_2=(0,0)$ while i require $l_1\cap l_2\cap l_3=\emptyset$ $\endgroup$ – 6666 Apr 21 at 18:02
  • $\begingroup$ This is in $\mathbb C^2$ not in $ \mathbb C \mathbb P^2$ $\endgroup$ – Ignorant Mathematician Apr 21 at 18:22
  • $\begingroup$ The homeomorphism from $CP^2-H_3$ to $C^2$ makes them intersect? $\endgroup$ – 6666 Apr 21 at 19:28
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    $\begingroup$ If not they will be parallel and since both of them pass through $(0,0)$ they are the same line. They pass through the origin since they are subspaces after all . $\endgroup$ – Ignorant Mathematician Apr 21 at 19:50

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