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Let $V$ be a subvariety of $\mathbb C^n$ with irreducible components of dimension >$0$. Is $H_{2n-1}(\mathbb C^n\setminus V)=0$?

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2 Answers 2

up vote 6 down vote accepted

It's certainly true, though I don't know the best way of proving it.

One way to see it is that if you take the one-point compactifications $V^+\subset(\mathbb{C}^n)^+=S^{2n}$, you get a connected space, since all components contain the point at infinity. The result then follows by Alexander duality in $S^{2n}$.

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Thanks you, I guess you are right. –  aglearner May 11 '11 at 16:00
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The sharp bound is this: For any closed algebraic set $V$ of codimension $d$ in ${\Bbb C}^n$, with $U={\Bbb C}^n \setminus V$, one has $\pi_i(U) = 0$ for $0 < i\leq 2d-2$ and $\pi_{2d-1}(U) \neq 0$. Using the Hurewicz isomorphism, you get the same vanishing and non-vanishing for homology.

A simple proof is in the appendix to my notes (with Fulton) on equivariant cohomology, http://www.math.washington.edu/~dandersn/eilenberg/ . A slick reason for vanishing was pointed out by David Speyer: given a (nice) map of an $i$-sphere into $U$, the (real) lines between the points in image of the sphere and points in $V$ sweep out a space of dimension at most $(2n-2d)+i+1$. When $i<2d-1$, you can pick a point in $U$ not lying on any such line, and contract your sphere down to that point. The non-vanishing happens because all algebraic sets have nontrivial fundamental classes in Borel-Moore homology. (Vanishing can also be proved using B-M homology.)

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Oops -- now that my browser is processing math, I realized I completely misread the question. Sorry, this doesn't answer it! (Of course, you already have James Cranch's fine answer.) –  Dave Anderson May 11 '11 at 17:43
    
Dave, nevertheless, thanks for writing this, I even understand what question you were answering :) ! –  aglearner May 11 '11 at 17:52
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