TITLE: Topology of plane curve complements after blow-ups
QUESTION [9 upvotes]: Let $C$ be a (singular, reducible) curve in the complex projective plane $\mathbb{P}^2$. An old problem is to study the topology of the complement $\mathbb{P}^2-C$. A famous result in this direction was originally stated by Zariski, but not proven until 1980 by Fulton. It states that if $C$ has only nodes for singularities (that is, double points with distinct tangents), then the fundamental group $\pi_1(\mathbb{P}^2-C)$ is abelian. 
My problem is as follows: again, let $C$ be a (singular, reducible) curve in $\mathbb{P}^2$. Now, blow up the plane at one of the singular points of $C$. Denote the resulting surface $S_1$, the exceptional divisor by $E_1$, and the proper transform of $C$ by $\hat{C}_1$. Then blow up $S_1$ at a singular point of $\hat{C}_1$, and repeat this procedure to get a surface $S=S_n$, exceptional divisors $E_1,\ldots,E_n\subset S$, and the proper transform $\hat{C}\subset S$ of the original curve. I am interested in the topology of the complement $X=S-(\hat{C} \; \cup E_{i_1} \cup \ldots \cup E_{i_k})$. In other words, $X$ is obtained from $S$ by deleting $\hat{C}$ as well as some (but not all) of the exceptional divisors obtained from blowing up. In particular, I would like to prove that the fundamental group $\pi_1(X)$ is abelian.
If the $C$ has only nodes for singularities, then $\pi_1(\mathbb{P}^2-C)$ is already abelian, and it is not hard to see that $\pi_1(X)$ will also be abelian. However, I am more interested in some cases where $C$ has slightly more complicated singularities. For example, if $C$ has a triple point with distinct tangents, $\pi_1(\mathbb{P}^2-C)$ is non-abelian; but I suspect that if I blow up the plane at the triple point, then $\pi_1(S-\hat{C})$ will be abelian. Does anyone know of any techniques or references for dealing with this situation?
(Note: It is certainly not the case that $\pi_1(X)$ will always be abelian. I am interested in techniques that will allow me to prove it is abelian for certain specific curves and configurations of exceptional divisors.)

REPLY [5 votes]: The following  Theorem by Nori  (Proposition 3.27 of this paper)  is  in the spirit of what you are looking for.

Theorem. Let $D$ and $E$ be curves on
  a smooth projective surface $X$. Assume that 

the only singularities of $D$ are nodes;
$D$ and $E$ intersect transversally;
every irreducible component $C$ of $D$ satisfies $C^2 > 2 r(C)$ where
  $r(C)$ is the number of singularities of  $C$.

Then the kernel of the natural
  morphism $$ \pi_1(X-D-E)
> \longrightarrow \pi_1(X-E) $$ is
  abelian.

In particular if you take an irreducible plane curve $C$ having a triple point with distinct tangents as its only singularity then the fundamental group of the complement 
of its strict transform is abelian (apply Theorem above to $D  = \hat C$ and  $ E = \emptyset$ ). As Dmitri pointed out, if you drop irreducibility this is no longer true. 
You might want   to take a look at this survey. There you will find Nori's Theorem 
in Section 2.3.

Further comments added later: 

If $C \subset \mathbb P^2$ is an irreducible curve with only one singularity having 
smooth branches intersecting pairwise transversely then 
the complement of $C$ in $\mathbb P^2$ and well as the complement of its strict transform in the blow-up of $\mathbb P^2$ are abelian ($D = \hat C$,  $E =$ exceptional divisor in the first case; and $D = \hat C$, $E= \emptyset$ in the second).
If you have a reduced connected curve $C = E_1 + \ldots + E_k$ with rational irreducible components,  the  intersection matrix $(E_i\cdot E_j)$ is negative definite, and the dual graph is a tree  then the fundamental group of the complement of a neighborhood of $C$ has been determined by Mumford in this paper.