18
$\begingroup$

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?

$\endgroup$
1
  • 4
    $\begingroup$ unibranch with smooth normalization :) $\endgroup$ Oct 8, 2010 at 6:12

3 Answers 3

14
$\begingroup$

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.

$\endgroup$
25
$\begingroup$
  • 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.

$\endgroup$
2
  • $\begingroup$ May I ask you for a reference for the result of Mumford? I did not manage to find it by myself. $\endgroup$
    –  V. Rogov
    May 30, 2019 at 9:00
  • $\begingroup$ @V.Rogov Sure, I added references to the answer. $\endgroup$
    – JME
    Jun 1, 2019 at 10:31
19
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 7-manifolds 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.