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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.

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  • $\begingroup$ 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

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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.

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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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