# Presentation of the fundamental group of a complex variety

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.

• 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. Aug 4, 2023 at 20:26
• Yes, the $Z_i$ intersect transversally. I will add this to the main question. Thanks! Aug 4, 2023 at 20:44
• 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 $Z$. Aug 5, 2023 at 9:30

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.