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As a complex affine variety or projective variety, is it possible it is a manifold with boundary?

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up vote 6 down vote accepted

No, this is not possible (unless you allow the boundary to be empty). If $X$ is a complex algebraic variety (affine or projective, this doesn't matter), there are two possibilities. If $X$ is smooth (as an algebraic variety), then $X$ is a smooth manifold with empty boundary. Otherwise let $Y$ be the singular locus of $X$. Then $X\setminus Y$ is a smooth variety and hence a smooth manifold with empty boundary. On the other hand, the \emph{real} codimension of $Y$ in $X$ is at least two (because the complex codimension is at least one). However, if $M$ is a (real) manifold with nonempty boundary, it is not possible to find a subset $Y$ of $M$ that has real codimension two such that $M\setminus Y$ has empty boundary. (The boundary $\partial M$ of $M$ is either empty or has real codimension one in $M$.) This implies that $X$ is not a manifold with boundary.

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Thanks! but how about the variety is defined over $R$? – HYYY Jul 5 '10 at 0:15

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