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?
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4$\begingroup$ There are problems with the question. ( 1) Affine varieties are already holomorphically convex. (2) Affine varieties do not contain any compact Riemann surfaces whatsoever. $\endgroup$– Donu ArapuraCommented Mar 13, 2012 at 15:38
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1$\begingroup$ @Donu: I believe you have answered the question... $\endgroup$– Igor RivinCommented Mar 13, 2012 at 16:32
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1 Answer
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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.
answered Mar 14, 2012 at 2:32