Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the fundamental group of the following space in $\mathbf C^n$?

This is the topological space given by $$\{(x_1,\ldots,x_n)\in \mathbf C^n-\{0\} : \vert x_1\vert < 1, \ldots, \vert x_n\vert <1\} - \{x_1\cdot \ldots \cdot x_n =0\}.$$

For $n=1$, I know it's $\mathbf Z$. I'm guessing it should be $\mathbf Z^n$ in general. Is this true? And why?

share|cite|improve this question
Are you sure it's $\mathbb{Z}$ for $n=1$? The way you defined the space it's contractible, so the fundamental group should vanish. – Piotr Pstrągowski Mar 17 '13 at 14:08
Whoops. I was thinking about the punctured polydisc! My apologies. – Tom Mar 17 '13 at 14:09
Can you describe the unit disc (not punctured!) up to homeomorphism? This is a reasonable first step. – Piotr Pstrągowski Mar 17 '13 at 14:11
Ow I think I see what you're getting at. When $n=1$, this is going to be like $\mathbf C$ minus the origin topologically, right? When $n>1$, this is probably a simply connected space, no? It might even be contractable because of all the extra space you get to move around. – Tom Mar 17 '13 at 14:15
@PiotrPstragowski I changed the question again. This should be a less trivial question. – Tom Mar 17 '13 at 14:23
up vote 1 down vote accepted

Your space deformation retracts to the $n$-torus $\prod_i\{\vert x_i\vert=\frac {1}{2} \}$, so has the same fundamental group : $$\mathbb Z^n $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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