6
$\begingroup$

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

$\endgroup$
0

2 Answers 2

6
$\begingroup$

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 $S^{2n}$.

$\endgroup$
0
4
$\begingroup$

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.)

$\endgroup$
2
  • $\begingroup$ 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.) $\endgroup$ May 11, 2011 at 17:43
  • $\begingroup$ Dave, nevertheless, thanks for writing this, I even understand what question you were answering :) ! $\endgroup$
    – aglearner
    May 11, 2011 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.