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?
3 Answers
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.
The simplest example of a singular algebraic variety which is a topological manifold is given by the cusp $$z_1^2z_0^3=0.$$ 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:
$$z_1^2+ z_2^2+\cdots z_{2k+1}^2z_0^3=0,\quad \text{where} \quad k\in \mathbb{N}_0.$$
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 BrieskornPham variety: $$ \Gamma(a): \quad z_1^{a_1}+\cdots z_n^{a_n}=0. $$
 Brieskorn has proved the following conjecture of Milnor:
$$\Gamma(a)\quad \text{is a topological manifold} \iff \prod_{1\leq k_l\leq a_k1}(1\epsilon_1^{k_1} \epsilon_1^{k_2}\cdots \epsilon_n^{k_n} )=1,$$ 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.
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.

$\begingroup$ Why is the hypersurface homeomorphic to R<sup>8</sup>? $\endgroup$ Aug 26, 2010 at 18:44

1$\begingroup$ Let me first give a refference for the fact that these 28 7manifolds are indeed homepomorphic to spheres  en.wikipedia.org/wiki/Exotic_sphere . Once you believe this, one can see that these hypersurfaces are R^8. Indeed, these polynomials are quasihomogenious, and so there is diagonal action of R^+ on on C^5, that preserves a polynomial (z_i > t^a z_i). This shows that each hypersurface is a cone over a sphere, i.e. it is homeoporphic to R^8. $\endgroup$ Aug 26, 2010 at 20:25