Discriminant locus in knot space

Question

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?

Answer 1

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.

Answer 2

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").