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In Singular points of complex hypersurfaces, John Milnor proves the following theorem:

Let $x \in V$ be a point on a variety $V$ in $\mathbb{R}^n$ or $\mathbb{C}^n$. Assume $x$ is either a smooth point or an isolated singularity. Let $D_{\epsilon}$ be the closed $\epsilon$-ball about $x$, $S_{\epsilon}$ its boundary (the sphere about $x$ of radius $\epsilon$), and $K = V \cap S_{\epsilon}$. Then for $\epsilon$ sufficiently small, the pair $(D_{\epsilon}, V \cap D_{\epsilon})$ is homeomorphic to the pair $(CS_{\epsilon}, CK)$, where $C$ denotes taking the cone. (Theorem 2.10)

In Remark 2.11, Milnor observes that this theorem "likely" holds even if $x$ is a non-isolated singularity; in particular, it is known even in this case that "a suitably chosen neighborhood of any point is homeomorphic to the cone over something."

This book was written in 1968. What is the current status of this problem?

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  • $\begingroup$ Someone mentioned, in the comments to my recent question, that complex varieties are known to be triangulable. This implies that kind of "locally conical". $\endgroup$
    – Tom Goodwillie
    Jul 24, 2010 at 14:36
  • $\begingroup$ This was the basis for Milnor's statement that a suitably chosen neighborhood is homeomorphic to a cone over something. The question is whether that "something" is $K$. $\endgroup$
    – Charles Staats
    Jul 24, 2010 at 15:07
  • $\begingroup$ Sorry. I did not read the question carefully. $\endgroup$
    – Tom Goodwillie
    Jul 24, 2010 at 15:59
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    $\begingroup$ I think what Greg Kuperberg is saying is that if the result exists in the literature it's likely to be proved as an intermediate step to something else. But anyway, take a look at the book "Stratified Morse theory" by Goresky and MacPherson. That seems to be the most likely place to find something like this. $\endgroup$
    – Donu Arapura
    Jul 24, 2010 at 20:49
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    $\begingroup$ What I mean is some combination of these possibilities. In particular, that it's the sort of result that people care about, but only enough to wish that someone else has written it up. Because it's a folk semi-result: People can guess that it's true and suggest approaches to prove it, but it's still messy and maybe no one has done the work. Theorems of this type often don't have a status, and are often proven as intermediate steps. $\endgroup$
    – Greg Kuperberg
    Jul 24, 2010 at 22:28

2 Answers 2

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There is a good paper of Goresky, "Triangulation of Stratified Objects", that I think reasonably quickly implies Milnor's result and its generalization to non-isolated singularities. The result is that any Whitney-stratified set, and in particular any algebraic variety in $\mathbb{C}^n$, is supported on a smooth triangulation. I think that you just need that and the inverse function theorem.

As I meant to explain in the comments, this theorem has sometimes been regarded as a "chore" theorem. You can look at what Goresky says: "Triangulation theorems for stratified objects have been obtained independently by Hendricks (unpublished), Johnson (unpublished), and Kato (in Japanese)". When Goresky wrote his paper, it was a messy question that did not have a well-defined status. Now the situation is a bit better and I think that this generalization of Milnor's result can be called settled. Sometimes a good author not only proves a chore theorem, but also cleans it up an elevates it to non-chore status. But a lot of chore theorems are never proven in a clean form or are never proven at all.

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  • $\begingroup$ The irony here is that Milnor goes through a quite careful proof of his version of this "chore" theorem. He does, however, feel the need to justify this by mentioning that "similar methods will be important later." $\endgroup$
    – Charles Staats
    Jul 25, 2010 at 0:21
  • $\begingroup$ As far as I know, Milnor has always been extremely careful in all of his mathematical writings. But, as you say, his justification seems to suggest that he also considered a chore theorem. Arguably Goresky turned it into something better. $\endgroup$
    – Greg Kuperberg
    Jul 25, 2010 at 6:19
  • $\begingroup$ Looks like this also answers my question mathoverflow.net/questions/32438/… $\endgroup$
    – Igor Belegradek
    Jul 25, 2010 at 13:45
  • $\begingroup$ @Igor: Goresky's result can cetairnly be applied to your other question, but you also show that the $r$-neighborhoods intersect in a standard way with each other. I tried to sketch a solution to that part of the problem. $\endgroup$
    – Greg Kuperberg
    Jul 25, 2010 at 14:41
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Indeed the following theorem to me seems exactly you were looking for (see J. Bochnak, M. Coste, M-F. Roy, "Real algebraic geometry", Theorem 9.3.6 [Local conic structure]):

Let $E$ be a semialgebraic susbet of $\mathbb{R}^n$ and $x$ be a nonisolated point of $E.$ Let also $D_\epsilon$ be the closed $\epsilon$-ball around $x$ and $S_\epsilon$ its boundary. Set $K=S_\epsilon \cap E$. Then there for $\epsilon>0$ small enough the pair $(D_\epsilon,E∩D_\epsilon)$ is semialgebraically homeomorphic to the pair $(CS_\epsilon,CK)$, where $C$ denotes taking the cone. Moreover the semialgebraic homeomorphism can be chosen as to preserve the distance from $x.$

Two words of remarks on the previous statement:

  1. Every real or complex algebraic set in $\mathbb{R}^n$ or in $\mathbb{C}^n\simeq \mathbb{R}^{2n}$ is a semialgebraic set.
  2. The point $x$ is any nonisolated point of $E$ (no matter singular - in whatever meaning this word has for a general semialgebraic set - or regular).
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