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.)


2 Answers 2


Corrected according to the comment of Scott.

Let me show that in the case you blow up $\mathbb P^2$ in one point an throw away preimages of three lines trough the point you get something with non-abelian fundamental group. The blow up of $\mathbb P^2$ at one point is a $\mathbb P^1$ bundle over $\mathbb P^1$, so when we throw away $3$ fibers we get a fibration $\mathbb P^1\to S\to \ C$, where $C$ is $\mathbb P^1$ punctured in $3$ points. The long exact sequence of homothopy groups reads as follows:

$...\to \pi_1(\mathbb P^1)\to \pi_1(S)\to \pi_1(C)\to \pi_0 \mathbb (\mathbb P^1)\to ...$

$...\to 0 \to \pi_1(S)\to \pi_1(C)\to 0 $

So $\pi_1(S)=F_2$, where $F_2$ is a free group on two generators.

I can not say anything about the general situation...

  • $\begingroup$ If I'm not mistaken, the long exact sequence of homotopy groups should decrease in degree (but I think this makes your proof even easier). $\endgroup$
    – S. Carnahan
    Mar 3, 2011 at 12:05
  • $\begingroup$ Scott, huge thanks, I was a bit sleepy while writing this :) $\endgroup$ Mar 3, 2011 at 12:47
  • $\begingroup$ Thanks for the counterexample--obviously the situation is even more complicated than I thought. $\endgroup$ Mar 4, 2011 at 4:22

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

  1. the only singularities of $D$ are nodes;
  2. $D$ and $E$ intersect transversally;
  3. 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:

  1. 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).
  2. 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.
  • $\begingroup$ Thanks for the references. Obviously this is a very relevant theorem to my situation. Unfortunately it does not completely solve my problem since I am blowing up reducible plane curves, and typically all components will have negative self-intersection. Any ideas of what I can do if $C^2<0$? $\endgroup$ Mar 4, 2011 at 4:25
  • $\begingroup$ You are welcome. I am far from a specialist on the subject, but I would take a look at the paper by Mumford I linked above. $\endgroup$ Mar 4, 2011 at 12:21

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