Inspired by this thread, which concludes that a nonsingular 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?

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 $(n1)$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 odddimensional 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 BrieskornPham 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 $S^{2n1}$ to $\mathbb{C}P^{n1}$. In the famous Brieskorn examples, the link is a topological sphere with a circle action, but the circle action yields a nontrivial Seifert fibration over an orbifoldtype complex variety. On the other hand, I don't think that this circle action always exists. 


Another good example are Brieskorn singularities $z_1^2+z_2^2+z_3^2+z_4^3+z_5^{6k1}=0$, $1\le k\le 28$, if you take a little sphere in $C^5$ centered at zero, then its intersection with the hypersurface is $S^7$ with a nonstandard smooth structue. So the hypersurface is homeomerphic to $R^8$ but does not have a smooth structure. 




