1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.