TITLE: Algebraic varieties which are topological manifolds
QUESTION [18 upvotes]: Inspired by this thread, which concludes that a non-singular variety over the complex numbers is naturally a smooth manifold, does anyone know conditions that imply that the topological space underlying a complex variety is a topological manifold without necessarily implying it is smooth?

REPLY [14 votes]: The answer from Dmitri motivates this partial answer from the topological side of the question.
It is a theorem of Mark Goresky and others that every stratified space, and in particular every complex variety $V$, has a smooth triangulation.  Moreover, I would bet (although I don't know that Goresky's paper has it) that the associated piecewise linear structure is unique.  This means that the PL homeomorphism type of the link of a singular point $p$ of $V$ is a local invariant.  I don't know how to compute this local invariant in general, but there must be some way to do it from the local ring at $p$.  There can't be a simple calculation of this invariant that is fully general.  As a special case, $V$ can be the cone of a projective variety $X$.  If so, then the link at the cone point $p$ is the total space of the tautological bundle on $X$.  $X$ and therefore the link can be all sorts of things.  If $p$ is an isolated singularity, then the type of this link is obtained by "intersecting with a small sphere", as Dmitri says. 
The variety $V$ is a PL manifold if and only if the link of every vertex is a PL sphere.  This is the case for the Brieskorn examples.
On the other hand, a theorem of Edwards (or maybe Cannon and Edwards) says that a polyhedron is a topological $n$-manifold (for $n \ge 3$) if and only if the link of every vertex is simply connected and the link of every point is a homology $(n-1)$-sphere.  In particular, the link of a simplex which is not a point does not have to be simply connected!  For example, if $\Gamma \subseteq \text{SU}(2)$ is the binary icosahedral group, then $\mathbb{C}^2/\Gamma$ is not a manifold, because the link of the singular point is the Poincaré homology sphere.  But $(\mathbb{C}^2 / \Gamma) \times \mathbb{C}$ is a topological manifold, even though it is not a PL manifold.
So for the question as stated, you would want to combine Goresky's theorem with Edwards' theorem, and with a method to compute the topology of the link of a singular point.  On the other hand, whether a variety $V$ is a PL manifold could be a more natural question than whether it is a topological manifold.

At least in the case of isolated singularities, the possible topology of the link of a singular point has been studied in the language of complex analytic geometry rather than complex algebraic geometry.  I found this paper by Xiaojun Huang on this topic.  The link of the singular point is in general a strictly pseudoconvex CR manifold.  This is a certain kind of odd-dimensional analogue of a complex manifold and you could study it with algebraic geometry tools.  (I think that strict pseudoconvexity also makes it a contact manifold?)  But the analytic style seems to be more popular, maybe because a CR manifold is not a scheme.
Sometimes, for instance in the case of a Brieskorn-Pham variety, such a CR manifold has a circle action whose quotient is a complex algebraic variety.  At a smooth point, this quotient is just the usual Hopf fibration from $S^{2n-1}$ to $\mathbb{C}P^{n-1}$.  In the famous Brieskorn examples, the link is a topological sphere with a circle action, but the circle action yields a non-trivial Seifert fibration over an orbifold-type complex variety.  On the other hand, I don't think that this circle action always exists.