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Say we have $C = Z(x_0^2-x_1^2-x_2^2)$ in $\mathbb{C}P^2$ and we define $X = \mathbb{C}P^2 \setminus C$. Is this space simply connected?

RUBÉN IZQUIERDO LÓPEZ
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  • Can you draw some pictures please? What are these? – Bob Dobbs Nov 02 '22 at 22:18
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    $\mathbb{C}P^2$ is the complex proyective plane obtained from $\mathbb{C}^3\setminus 0$ from the equivalence $u \sim v$ iff they are propotional. And $Z(x_0^2 - x_1^2-x_2^2)$ is the set of those equivalence classes $[x:y:z]$ where $x^2-y^2-z^2= 0$. – RUBÉN IZQUIERDO LÓPEZ Nov 03 '22 at 06:34
  • It is cone with 45 degrees. – Bob Dobbs Nov 03 '22 at 06:50
  • Could you elaborate? C is not singular since naming $F = x_0^2 - x_1^2 - x_2^2$ we have that $\frac{\partial F}{\partial x_i} = 0$ iff $x_i=0$ which is not possible. – RUBÉN IZQUIERDO LÓPEZ Nov 03 '22 at 06:59
  • https://math.stackexchange.com/questions/486668/x2-y2-z2-is-irreducible-in-mathbb-c-x-y-z?noredirect=1 Related? – Bob Dobbs Nov 03 '22 at 07:03
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    I cannot find a reference right now, but I'm fairly certain $H_1(\mathbb CP^2 \setminus C, \mathbb Z) \cong \mathbb Z/d\mathbb Z$, where $C$ is a smooth plane curve of degree $d$. So here the homology would be $\mathbb Z/2\mathbb Z$, hence the fundamental group would be some extension thereof. – Tabes Bridges Nov 03 '22 at 18:53
  • Potential key words: "Zariski-Van Kampen theorem". – KReiser Nov 03 '22 at 19:01

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