Algebraic varieties which are topological manifolds 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?
 A: *

*The simplest example of a singular
algebraic variety which is a
topological manifold is given by the
cusp $$z_1^2-z_0^3=0.$$   The cusp  is   a topological manifold
homeomorphic to a real plane
$\mathbb{R}^2$ as can    be seen by
the parametrization $t\mapsto (z_1,z_0)= (t^2,t^3)$  where $t$ is a complex variable.

*Mumford has proven that a two 
dimensional normal complex space
which is a topological manifold is
always nonsingular. 

*Mumford's result does not generalize
to (odd) dimensions higher than 2 as 
proven  by Brieskorn using the
following counter examples which generalizes the case of the cusp:
$$z_1^2+ z_2^2+\cdots
   z_{2k+1}^2-z_0^3=0,\quad \text{where} \quad k\in
   \mathbb{N}_0.$$ 

*More generally, given  $a=(a_1,
   \cdots, a_n)\in \mathbb{N}^n_0$ with $a_j>1$ for all $j$, one
can define the following  variety
$\Gamma(a)$ known as a 
Brieskorn-Pham variety: $$ \Gamma(a): \quad z_1^{a_1}+\cdots
   z_n^{a_n}=0. $$ 

*Brieskorn has proved the following conjecture of Milnor:
$$\Gamma(a)\quad \text{is a topological manifold}
   \iff \prod_{1\leq k_l\leq   
   a_k-1}(1-\epsilon_1^{k_1}   
   \epsilon_1^{k_2}\cdots   
   \epsilon_n^{k_n}  )=1,$$ where
$\epsilon_k=\mathrm{exp}\Big({\frac{2\pi
   }{a_k}\mathrm{i} }\Big)$ for $k=1,\cdots, n$.


References.
Mumford, D., "The topology of normal singularities of an algebraic surface and a criterion for simplicity," Publ. Math. de l'Institut des Hautes Etudes Scientifiques (Paris: 1961), no. 9.
Brieskorn, Egbert V. (1966), "Examples of singular normal complex spaces which are topological manifolds", Proceedings of the National Academy of Sciences, 55 (6): 1395–1397.
A: Another good example are Brieskorn singularities 
$z_1^2+z_2^2+z_3^2+z_4^3+z_5^{6k-1}=0$, $1\le k\le 28$, if you take a little sphere in 
$C^5$ centered at zero, then its intersection with the hypersurface is $S^7$ with a non-standard smooth structue. So the hypersurface is homeomerphic to $R^8$ but does not have a smooth structure.
A: 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.
