Let $X$ be a connected smooth complex algebraic variety and $Z=\bigcup_{i=1}^r Z_i$ be a union of smooth connected hypersurfaces, satisfying that each two intersect transversally. Assume for simplicity that $Z$ is connected and choose a point $x\in X\setminus Z$. As $Z$ has real codimension $2$, it is known that the map $\Phi:\pi_1(X\setminus Z,x)\rightarrow \pi_1(X,x)$ is a surjection. I am looking for a description of the kernel of this map.

I believe that this kernel should be the normalizer in $\pi_1(X\setminus Z,x)$ of the subgroup generated by homotopy classes of loops $\tau_i$ such that each $\tau_i$ circles once around $Z_i$. This is, for example, the case when $X=\mathbb{C}^n$ and $Z$ is the union of complex hyperplanes. See, for instance, The fundamental group of the complement of a union of complex hyperplanes.

I have made some attempts at this, but haven't been successful thus far. My latest attempt goes as follows: Assume that $Z$ is a smooth connected hypersurface and take a tubular neighborhood $Z\subset U\subset X$ (which we assume to contain $x$). Applying Seifert-van Kampen to the cover $X\setminus Z$ and $U$, we have an isomorphism:

$$(\pi_1(X\setminus Z,x)* \pi_1(U,x))/N(\pi_1(U\setminus Z,x)\rightarrow \pi_1(X,x)$$.

As before, the map $\pi_1(U\setminus Z,x)\rightarrow \pi_1(U,x)$ is surjective. Hence, all elements coming from $\pi_1(U,x))$ are trivial inside the amalgamated product. My feeling is that $U\setminus Z$ should deformation retract to a spherical bundle. For example, if $U$ is trivial then $U\setminus Z$ is homotopy equivalent to $Z\times \mathbb{S}^1$, and the loop corresponding to $(id, 1)\in \pi_1(Z\times \mathbb{S}^1,x))\simeq \pi_1(Z,x)\times \mathbb{Z}$ generates the kernel of $\Phi$. I would very much appreciate any help regarding this matter.

  • $\begingroup$ Are you willing to assume that $Z$ is a divisor is a normal crossings? If not, you might need some weaker transversality condition to get things to work. I agree it's an $S^1$-bundle away from the the singularities of $Z$, but more subtle close to these points. $\endgroup$ Aug 4, 2023 at 20:26
  • $\begingroup$ Yes, the $Z_i$ intersect transversally. I will add this to the main question. Thanks! $\endgroup$ Aug 4, 2023 at 20:44
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    $\begingroup$ There should be a handle decomposition of $U$ with handles of indices at most $2m := 2\dim Z = 2\dim X - 2$, which has one $m$-handle for each component of $Z$. Attaching $U$ to $\overline{X \setminus U}$ means attaching only handles of indices at least 2, and the 2-handles you have to attach kill exactly the meridians of the components of $Z$. $\endgroup$ Aug 5, 2023 at 9:30

2 Answers 2


I found an affirmative answer to this question on Complex reflection groups, braid groups, and Hecke algebras. The proof is in Appendix A1, specifically on page 181.


This is not an answer, just an idea too long for a comment. Maybe this helps: The group you are after is the fundamental group of the preimage of $X\smallsetminus Z$ in the universal covering of $X$. This basically means that you can replace $X$ with its universal cover and assume it's simply connected from start. Then, looking at $U=X\smallsetminus Z$ and its universal covering, you may want to look at the deck transformation group, pick a fundamental domain for the action, say a Dirichlet domain for a Riemannian metric. Then the Fundamental group is generated by all translations mapping the fundamental domain to its "neighbours". Maybe these transformations are what you want.


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