# When is $\mathbb C^d\setminus\mathcal Z$ simply connected?

Let $\Delta$ be a fan in the lattice $N\cong\mathbb Z^n$ with $d$ edges $\{\rho_1,\cdots,\rho_d\}$. Consider the co-ordinate ring $\mathbb C[x_1,\cdots,x_n]$. Let $\mathcal Z=\bigcup_C\mathcal V(x_i\ |\ \rho_i\in C)$ where the union is taken over all primitive collections $C$ of edge vectors in $\Delta$.

Definition - A subset $C\subseteq \Delta(1)$ (the set of edges in $\Delta$) is a primitive collection if :

1. $C\nsubseteq \sigma(1)$ for all $\sigma \in \Delta$

2. For every proper subset $C'$ of $C$ there is a $\sigma\in\Delta$ with $C'\subseteq \sigma(1)$

Question - Is there a way in general to check if $\mathbb C^d\setminus\mathcal Z$ is simply connected in the analytical topology?

I know that $2\le\text{codim}\mathcal Z\le d$.

Suppose further that $\Delta$ is compete and simplicial then we have either

1. $2\le \text{codim}\mathcal Z\le \lfloor \frac{n}{2}\rfloor+1$ or

2. $d=n+1$ and $\mathcal Z=\{0\}$

It is easy to see that if $\Delta$ is simplicial and 2. is true then $\mathbb C^d\setminus\mathcal Z=\mathbb C^{n+1}\setminus\{0\}$ is simply connected. What about in general? I understand that it will not always be true, but for what codimension of $\mathcal Z$ can one expect $\mathbb C^d\setminus\mathcal Z$ to be simply connected? (Assuming there is a way to relate codimension of $\mathcal Z$ to the simply connectedness of $\mathbb C^d\setminus\mathcal Z$).

Thank you.

Theorem. If $Z \subseteq \mathbb{C}^d$ is Zariski-closed of codimension $c$, then $\pi_i(\mathbb{C}^d\smallsetminus Z) = 0$ for $0<i\leq 2c-2$, and if $Z$ is nonempty, then $\pi_{2c-1}(\mathbb{C}^d\smallsetminus Z) \neq 0$.
Proposition 5.8 here gives three conditions for the inclusion map $\mathbb{C}^d\setminus \mathcal{Z}\hookrightarrow \mathbb{C}^d$ to be 2-connected. In particular when those conditions are satisfied $\mathbb{C}^d\setminus \mathcal{Z}$ is simply-connected.