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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?

flag	asked Mar 13 2012 at 15:28 Antoine 1,020●1●11

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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 2012 at 15:38

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@Donu: I believe you have answered the question... – Igor Rivin Mar 13 2012 at 16:32

Donu Arapura · Accepted Answer · 2012-03-14 02:32:25Z UTC

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.

Compact Riemann surfaces as holomorphically convex subsets of affine algebraic varieties

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