# Are singularities of complex varieties captured by topology?

Let $$X \subseteq \mathbb{C}^n$$ be an affine complex algebraic variety, with a singularity at some point $$x.$$ Let $$U \subseteq \mathbb{C}^n$$ be an open set containing $$x$$.

Can we determine if $$x$$ is a singularity just from the homeomorphism type of the triple $$(U, U \cap X, \{x\})$$?

I believe if we remember the $$C^1$$-diffeomorphism type, the answer is yes, since essentially the definition of singularity is rigged to detect points at which the variety isn't a $$C^1$$ submanifold of affine space.

Over $$\mathbb{R},$$ the topological type is too weak: $$y^2 = x^3$$ is a topological manifold, and so topology alone cannot tell that $$(0, 0)$$ is a singularity.

Over $$\mathbb{C},$$ though, can topology detect singularities?

• To see topologically the singularity, you have to intersect $X$ with a small sphere around the singular point. It is well explained in the book by Milnor "singular points of hypersurfaces". Oct 22, 2023 at 14:37
• @NicolasHemelsoet I'm not sure this is the point I'm trying to get at it. Really, I want to see if this local topology remembers the singularity; I'm not so much interested in Milnor links. Oct 22, 2023 at 16:12
• But isn't intersecting with a small sphere homotopy equivalent to intersecting with a small punctured open ball? So if you know the pointed space $(U \cap X,x)$, you also know the link. Oct 22, 2023 at 18:06
• Isn't $y^2 = x^3$ over $\mathbb C$ also homeomorphic to $\mathbb P^1$ (just geometrically "pinched" at the cusp), so the answer is no? Oct 22, 2023 at 21:52
• I think what the commenters are trying to say is that you should read the following (famous) article of David Mumford: dam.brown.edu/people/mumford/alg_geom/papers/… In particular, see the theorem at the bottom of page 1 of Mumford's article. Oct 23, 2023 at 11:46

Summarizing what has been said before and adding a bit:

1. If the normalization map $$\tilde X\to X$$ is bijective then it is a homeomorphism in the Hausdorff topology. So if $$\tilde X$$ is smooth and $$X$$ not then $$X$$ will be a topological manifold without being smooth. This happens for example for the cusp $$y^2=x^3$$.

2. If $$X$$ is normal then the problem is ony interesting for $$\dim X>1$$. For $$\dim X=2$$, the main result is the affirmative answer by Mumford, linked to by Jason Starr, that a point is non-singular if and only if it has arbitrary small simply connected punctured neighborhoods.

3. In higher dimension there is Brieskorn's paper where he shows that the variety given by $$z_1^3+z_2^2+\ldots+z_n^2=0$$ in $$\mathbb C^n$$ with $$n\ge4$$ even is normal, singular and a topological manifold. Later on he shows in a famous paper that the Milnor link of the hypersurface $$z_1^2+z_2^2+z_3^2+z_4^3+z_5^{6k-1}=0,$$ runs for $$k=1,\ldots,28$$ through all $$28$$ smooth structures of $$S^7$$. In particular, also these hypersurfaces provide counterexamples.

• In 3., why are these a counterexample? It's true that $X\cap U$ is topologically a ball, but why would it be unknotted in $U$? Oct 23, 2023 at 13:58
• @MarcoGolla If nothing else, we can reembed in a higher dimensional space. Brieksorn's examples have $X \subset \mathbb{C}^n$; if we instead considered $X \times \{ \text{pt} \} \subset \mathbb{C}^{2n}$, then we would be above the Whitney bound, so it would be an unknotted example. Oct 23, 2023 at 14:35
• @David E Speyer Of course. Thanks. Oct 23, 2023 at 16:55

Let me turn my comment into a partial answer. My reading of the question is about sufficiently small open balls $$U \subseteq \mathbf C^n$$, because the question should be of a local nature.

Lemma. If $$X$$ is smooth of dimension $$d$$ at $$x$$, then $$(U,X \cap U,x)$$ is homeomorphic to $$(\mathbf C^n,\mathbf C^d,0)$$ for $$U$$ sufficiently small, where $$\mathbf C^d \hookrightarrow \mathbf C^n$$ is a linear embedding.

Proof. Since $$X$$ is smooth of dimension $$d$$ at $$x$$, it is a complete intersection in a Zariski open neighbourhood $$V$$ of $$x$$. That is, there exists a polynomial map $$f \colon V \to \mathbf C^{n-d}$$ such that $$f^{-1}(0) = X \cap V$$. Since $$\mathrm df$$ has maximal (real) rank $$2(n-d)$$ at $$x$$, the same holds in a small neighbourhood $$U$$ of $$x$$, so $$U \to \mathbf C^{n-d}$$ is a submersion and the result follows from the rank theorem. $$\square$$

So the question is whether this is a sufficient condition. The cuspidal curve already shows that merely remembering $$(X \cap U,x)$$ is not always enough, but I suspect that the full triple might be enough. Here is a positive result (ruling out the counterexamples so far):

Lemma. Let $$X$$ be a hypersurface such that $$U \setminus X$$ is homotopy equivalent to a circle (for $$U$$ sufficiently small). Then $$X$$ is smooth at $$x$$.

The hypothesis is in particular satisfied if $$(U, X \cap U, x) \cong (\mathbf C^n,\mathbf C^{n-1},x)$$ for a linear embedding $$\mathbf C^{n-1} \hookrightarrow \mathbf C^n$$.

Proof. Suppose $$X = f^{-1}(0)$$ for some $$f \colon \mathbf C^n \to \mathbf C$$. If $$U$$ is a small ball around $$x$$, then $$E = U \setminus Z$$ has a locally trivial Milnor fibration $$f \colon E \to \mathbf C^\times$$ with Milnor fibre $$F$$. We get a long exact sequence of homotopy groups $$\ldots \to 0 \to \pi_1(F) \to \pi_1(E) \to \pi_1(\mathbf C^\times) \to \pi_0(F) \to \pi_0(E) \to \pi_0(\mathbf C^\times)$$ and isomorphisms $$\pi_i(F) \stackrel\sim\to \pi_i(E)$$ for $$i \geq 2$$. By assumption, $$E$$ is a $$K(\pi,1)$$, so $$E \to \mathbf C^\times$$ is a weak homotopy equivalence since $$\pi_0(F) = *$$. We conclude that $$F$$ is contractible, which implies that $$x$$ is a smooth point by a result of A'Campo [ACa73, Thm. 3]. $$\square$$

The proof uses nearby cycles, which for maps to higher-dimensional bases is a little more complicated (e.g. if $$X$$ is cut out by some map $$f \colon \mathbf C^n \to \mathbf C^{n-d}$$). Modern technology does allow us to consider that situation, but there might also be a more direct argument.

Example. For the cuspidal curve, we get $$(U,X \cap U,x) \cong (\mathbf C^2,\mathbf C,*)$$, but the embedding $$\mathbf C\setminus * \hookrightarrow \mathbf C^2\setminus *$$ is a (thickened) trefoil knot. So it is not isomorphic to the triple above.

(It is more traditional to consider $$\partial U$$ and $$X \cap \partial U$$, but then you can only talk about the two punctured strata and not all three strata. For hypersurfaces, we only needed the deleted link $$U \setminus X$$; I don't know if this is enough in general.)

References.

[ACa73] N. A’Campo, Le nombre de Lefschetz d’une monodromie. Indag. Math. (N.S.) 76.2 (1973), p. 113-118. ZBL0276.14004.