Inspired by [this thread](https://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifolds), 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?



Algebraic varieties which are topological manifolds
The answer from Dmitri motivates this partial answer from the topological side of the question.


It is [a theorem of Mark Goresky](http://www.math.ias.edu/~goresky/pdf/triangulations.pdf) 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](http://www.ams.org/journals/proc/1999-127-10/S0002-9939-99-04860-1/S0002-9939-99-04860-1.pdf) (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.




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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](http://www.scichina.com:8081/sciAe/fileup/PDF/06ya1441.pdf) 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.


