8
$\begingroup$

Let $X$ be an affine variety of dimension $n$ over $\mathbb{C}$.

Does the analytic space associated with $X$ have the homotopy type of a $n$-dimensional CW complex?

$\endgroup$
1
11
$\begingroup$

If $X$ is smooth, and if you ask to have the same homotopy type of a CW complex of real dimension at most $n$, this is precisely the statement of the Andreotti-Frankel theorem.

It is true, more generally, for a Stein manifold of complex dimension $n$.

If $X$ is arbitrarily singular, the same theorem holds, provided $X$ is irreducible. This was shown by Karčjauskas for algebraic varieties and by Hamm for Stein spaces.

$\endgroup$
2
  • 5
    $\begingroup$ When $X$ is any algebraic variety (in particular, any affine variety ), the aforementioned CW-compex is also finite. This implies that the fundamental group $\pi_1(X)$ and the homology and cohomology groups $H_i(X, \, \mathbb{Z})$, $H^i(X, \, \mathbb{Z})$ are all finitely generated. This is not true anymore fror an arbitrary smooth Stein space, for which the CW-complex can be infinite. Take for instance $$X=\{x \in \mathbb{C}\, | \, \sin x \neq 0 \}.$$ $\endgroup$ Jan 20 '17 at 16:01
  • $\begingroup$ Well, $\mathbb{C}$ has the homotopy type of a point, right? Compactness is not preserved under homotopy equivalence. $\endgroup$ Jan 20 '17 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy