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.


There is a simple general fact about k-connectedness of the complement of an affine algebraic set, which seems not to be so well known:

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 $\pi_{2c-1}(\mathbb{C}^d\smallsetminus Z) \neq 0$.

This is stated as Proposition 4.1 in these notes (PDF); the proof is not hard. In particular, I think the sets you're asking about are always simply connected.


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.

I haven't thought about your specific situation, but I would guess that large codimension is probably enough to imply the first two conditions in the referenced proposition above very generally. The third condition is where you might need to do some work in your specific situation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.