Suppose that $C\subset\mathbb P^2$ is an irreducible projective curve over $\mathbb C$ such that the normalization morphism $\bar C\to C$ is unramified (i.e., the induced morphism $\bar C\to\mathbb P^2$ is an immersion). Could you help me with examples of curves with this property for which $\pi_1(\mathbb P^2-C)$ is not commutative?

Thank you in advance.

  • $\begingroup$ I think that quartic with 3 cusps has fundamental group of complement dihedral of order 12. Also brief sketch with van Kampen suggests to me that (...almost) any curve with as many cusps as genus of normalisation should have the group in qusetion infinite nonabelian; but there's still a problem is to find such curves (it may as well be that three-cusp quartic is the only one?) $\endgroup$
    – Denis T
    Commented Sep 24, 2022 at 6:40
  • $\begingroup$ @DenisT If a curve has a cusp, then its normalization morphism is not unramified, so such curves don't qualify as required examples. $\endgroup$ Commented Sep 24, 2022 at 6:52
  • $\begingroup$ Oh. One should properly wake up before writing anything on the Internet. (...Cusps are unibranch, so in my simple topologist mind they have unramified normalisation, because "homeomorphisms cannot be ramified"). Am I right that you are asking for an example of immersed plane curve $C: \,q(x, y, z)$ such that its Milnor fiber $F:\, q(x, y, z) = 1$ has nontrivial $\pi_1$? For irreducible curve $\pi_1(F)$ should be equal exactly to $\pi_1(P^2 \setminus C)'$, unless I'm making some obvious mistake again. (Numerous articles by Degtyarev, Shimada and Dimca should contain multifarious examples..) $\endgroup$
    – Denis T
    Commented Sep 24, 2022 at 8:18
  • $\begingroup$ @DenisT Thank you, I will look up ther papers. $\endgroup$ Commented Sep 24, 2022 at 9:09
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    $\begingroup$ @JasonStarr It is, yes, but it does not tell us much about global fundamental groups. For example, if $C$ is a quartic with one simple triple point (that is formally isomorphic to the union of three concurrent lines), then $\pi_1$ of its complement is abelian. $\endgroup$ Commented Sep 24, 2022 at 13:21


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