MathOverflow is a question and answer site for professional mathematicians. 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

Is there a simple argument (or a counterexample) to show that a holomorphically convex subset of an affine algebraic variety is a subvariety which is a compact Riemann surface?

share|cite|improve this question
There are problems with the question. ( 1) Affine varieties are already holomorphically convex. (2) Affine varieties do not contain any compact Riemann surfaces whatsoever. – Donu Arapura Mar 13 '12 at 15:38
@Donu: I believe you have answered the question... – Igor Rivin Mar 13 '12 at 16:32
up vote 7 down vote accepted

Perhaps I should turn my comment into an "answer". Affine algebraic varieties over $\mathbb{C}$ are Stein spaces. That is, they are already holomorphically convex, and points can be separated by global holomorphic functions. The latter property implies that affine varieties can never contain compact Riemann surfaces.

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.