This is a problem concerning a lemma in Oka's paper "On the fundamental group of the complement of a reduced curve in $\mathbb{P}^2$". Let $C$ be a curve in $\mathbb{P}^2$ and $L$ be a general line to $C$. The lemma says that $\pi_1(\mathbb{P}^2-C\cup L)$ is abelian if and only $\pi_1(\mathbb{P}^2-C)$ is abelian. Now consider the arrangement of a conic $C$ and three lines $L_1$, $L_2$ and $L_3$ in $\mathbb{P}^2$. Assume that the conic passes through the three double points $L_1\cap L_2$, $L_1\cap L_3$ and $L_2\cap L_3$. We can show that $\pi_1(\mathbb{P}^2-C\cup L_1\cup L_2\cup L_3)$ is abelian. Let $L_\infty$ be a general line to the arrangement. According to Oka's lemma, the fundamental group $\pi_1(\mathbb{C}^2-C\cup L_1\cup L_2\cup L_3)=\pi_1(\mathbb{P}^2-C\cup L_1\cup L_2\cup L_3\cup L_\infty)$ should be abelian. However, without the projective relation, it is impossible to prove that the group is abelian. I don't know where I made mistakes. The only mistake that I suspect is that $\mathbb{P}^2-C\cup L_1\cup L_2\cup L_3\cup L_\infty$ does not equal $\mathbb{C}^2-C\cup L_1\cup L_2\cup L_3$. But why they do not equal.

$\textbf{Added:}$ Here are the fundamental groups. The computation uses braid monodromy method. The fundamental group of the affine complement is $A=<1, 2, 3, 4 \mid 431=314=143, 432=324=243, 132=321=213>$. The fundamental group of the projective complement has an extra relation. The group is $G=<1, 2, 3, 4 \mid 431=314=143, 432=324=243, 132=321=213, 43^221=e>$, where $e$ is the group identity.

  • $\begingroup$ If you just need to prove that $A$ and $G$ are not isomorphic: the abelianisation of $A$ is $\mathbb{Z}^4$, the abelianisation of $G$ is $\mathbb{Z}^3$. $\endgroup$ – Marco Golla Dec 19 '11 at 10:14
  • $\begingroup$ @Marco: Actually, I want to know if $A$ is abelian. If $A$ is not abelian, where is the mistake in the argument. $\endgroup$ – Fei YE Dec 20 '11 at 14:05
  • $\begingroup$ The mistake is that Oka requires his curve to be the union of irreducible curves that don't intersect in triples: in your case, $C\cap L_1 \cap L_2$ is nonempty. $\endgroup$ – Marco Golla Dec 20 '11 at 22:51

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