MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the space $K$ of all immersions of $S^1$ into $\mathbb R^3$. The set of knots with self-intersection is a discriminant in $K$ and divide it into "chambers".

Let $f$ be a knot with $n$ double points. Everybody know that neighbourhood of $f$ in $K$ looks like $\mathbb R^n$, the origin be f, and hyperplanes $x_i=0$ be a knots with self-intersection. (In other words $f$ can be considered as intersection of n planes modelling dicriminant, there are $2^n$ chambers adjacent to $f$, looks like octants in $\mathbb R^n$).

My question: what is it mean? Does there exist a continuous map of neighboorhod $f\in U$ into $\mathbb R^n$ with above prescribed properties?

We can define "tangent" vector for $f$ as piecewise-smooth vector field along $f$. So, does there exist a map from "tangent" vector space of $f$ to $\mathbb R^n$ with above properties? Can we define subspace which is "tangent" to discriminant?

share|cite|improve this question
Where did you get the $\mathbb R^n$ from? A neighbourhood of $f$ in $K$ looks like an infinite-dimensional Hilbert/Frechet space (the space of maps $S^1 \to \mathbb R^3$), depending on how differentiable your space of immersions is. – Ryan Budney Oct 25 '10 at 20:49
Oh, it sounds like you're looking at a slice of the neighbourhood, not the neighbourhood of $f$ itself. The parameters of the $\mathbb R^n$ space indicate how one chooses to resolve the $n$ double points. This isn't a neighbourhood of $f$, not at all. What you can do though is view it as part of a particular coordinate system on a neighbourhood of $f$ -- you're ignoring all but finitely-many coordinates because the other coordinates aren't very interesting. This is because the $n$ double point subspace has co-dimension $n$. – Ryan Budney Oct 25 '10 at 20:53
The actual neighbourhood of $f$ consists of $f+\epsilon$ where $\epsilon : S^1 \to \mathbb R^3$ is smooth and "small". If you were using $C^k$-immersions, that would mean $\epsilon$ through $\epsilon^{(k)}$ are all uniformly small. If you were demanding $C^\infty$-smooth maps, then you would demand arbitrarily many but finitely many derivatives of $\epsilon$ to be uniformly small. Your $\mathbb R^n$ "slice" corresponds to decomposong $\epsilon$ into the sum of some bump-functions about the double points, together with complementary functions (which are zero at the double points). – Ryan Budney Oct 25 '10 at 21:18
>Ryan Yes, I see. So, my question: what is co-dimension mean in this case? Could you give me definition? I get tired with "naive" description of this situation, but in th oter hand I can't define codimension strictly. Therefore the question may be reformulated as "Does codimension in this context be present only as intuitive conception". What does "slice of the neighbourhood" mean? Does it be the embedding $\mathbb R^n$ image? – Nikita Kalinin Oct 25 '10 at 21:23
>Ryan Does your bump-function formed a "discriminant tangent space" basis in some sense? – Nikita Kalinin Oct 25 '10 at 21:26
up vote 2 down vote accepted

If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$, the space of smooth maps $f : N \to X$ ( $Map(N,X)$) has as an open and dense subspace maps which are disjoint from $A$.

Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".

edit, continued from the comments above: The notion of co-dimension used by Vassiliev is very much in the spirit of the above comments. When you have a map $D^n \to X$ such that its restriction to $S^{n-1}$ can be perturbed to be disjoint from $A$, yet no small perturbation of the original map can be made to be disjoint from $A$, that's when you are in the co-dimension $n$ setting. A map $D^n \to X$ can't be a neighbourhood of a point in its image since $X$ isn't finite-dimensional, but you can consider such maps to be a "slice" of a neighbourhood of a point in the image. By that I mean you have a map $f : V \to X$ whose image is open in $X$, and you decompose $V$ into an $n$-dimensional subspace and a complementary subspace. The the image of the $n$-dimensional subspace under $f$ will be your "resolutions" of your knot-with-multiple (double/triple, etc) points. The complementary subspace will consist of perturbations of the knot complementary to the resolution bump functions.

share|cite|improve this answer
There should be some stronger statement saying that a neighborhood of this particular intersection looks like a transverse intersection divided into chambers, like R^n with its coordinate axes crossed with an infinite-dimensional manifold. Can this be phrased in the language above? – Dylan Thurston Oct 26 '10 at 1:54
In the comments above (before my "answer") I was looking for a specific question to answer. My response is specific to Kalinin's query about what the concept of co-dimension means, in the context of infinite-dimensional spaces. The space of knots sits in a big mapping space which has a natural stratification, but the definition of the stratification is by co-dimension, so the above is an attempt to clarify what co-dimension means in this situation. – Ryan Budney Oct 26 '10 at 2:21

This is more of a comment than an answer.
Your question looks more interesting to me in the context of holonomic knots. A holonomic knot, as defined by Vassiliev, is a knot parameterized as $(-f(t),f^\prime(t),-f^{\prime\prime}(t))$, where $f$ is a $C^\infty$ function with period $2\pi$. Vassiliev showed that any knot class has a representative which is a holonomic knot, and Birman and Wrinkle showed that any two holonomic knots representing the same knot are ambient isotopic via holonomic knots.
The whole Vassiliev finite-type theory setup which you're describing works more nicely for holonomic knots. The key property of holonomic knots which makes this so is that they have Fourier expansions. Truncating the Fourier expansions gives a sequence of successive approximations to the holonomic knots, through which the Vassiliev spectral sequence arises naturally. The space of all Fourier expansions truncated to $k$ places really is $\mathbb{R}^k$. The singular locus really is codimension 1 for each successive approximation.
Now you can ask about differential topology of the space of holonomic knots! This space a-priori has a much nicer smooth structure than I would expect classical "knot space" to have. I would also love to know the answers (I would be shocked if they were not "yes").

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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