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

A complex analytic space is a topological space (say, Hausdorff and second countable) such that each point has an open neighborhood homeomorphic to some zero set $V(f_1,\ldots,f_k)$ of finitely many holomorphic functions in $\mathbb{C}^n$, in a way such that the transition maps (restricted to their appropriate domains) are biholomorphic functions.

I have an embarrassing basic question. It seems weird to me that we are isomorphing the open neighborhoods to the closed sets $V(\ldots)$. I realize that we can think of $V(\ldots)$ as being open in its own induced topology, but I can't really picture what such homeomorphism would look like. Is there possibly a mistake in the definition I read? Should the open neighborhoods be homeomorphic to open subsets of the $V(\ldots)$?

share|cite|improve this question
Well, the punctured complex plane $\mathbb{C} \setminus \{ 0 \}$ is homeomorphic and even biholomorphic to $V(x y - 1) \subset \mathbb{C}^2$. So it's certainly possible. This definition appears to be imitating the definition of a variety as being locally isomorphic to affine varieties (i.e. the zero locus of some set of multivariate polynomials). – Zhen Lin Oct 19 '11 at 6:36
Okay thanks, that's helpful. – Ben K Oct 19 '11 at 22:51

Your Answer


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

Browse other questions tagged or ask your own question.