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?

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    $\begingroup$ 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". $\endgroup$ Oct 22, 2023 at 14:37
  • $\begingroup$ @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. $\endgroup$ Oct 22, 2023 at 16:12
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    $\begingroup$ 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. $\endgroup$ Oct 22, 2023 at 18:06
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    $\begingroup$ 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? $\endgroup$ Oct 22, 2023 at 21:52
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    $\begingroup$ 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. $\endgroup$ Oct 23, 2023 at 11:46

2 Answers 2


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.

  • $\begingroup$ 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 $U$? $\endgroup$ Oct 23, 2023 at 13:58
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    $\begingroup$ @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. $\endgroup$ Oct 23, 2023 at 14:35
  • $\begingroup$ @David E Speyer Of course. Thanks. $\endgroup$ 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.)


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


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