# Showing every knot has a regular projection using differential topology

Can we use differential topology to prove that every smooth knot has a regular projection?

Here is some background:

Let $\gamma : S^1 \rightarrow \mathbb{R}^3$ be a smooth unit-speed imbedding. For $v \in S^2$ we say that the projection, $\pi _v$, from $\mathbb{R}^3$ to the plane orthogonal to $v$ is a regular projection of $\gamma$ if the curve $\gamma _v := \pi _v\circ \gamma$ has the following properties:

1. $\gamma _v$ is an immersion

2. There exists a finite set $I := \{a_1, a_2, ..., a_k, b_1, b_2,... b_k\} \subset S^1$ such that $\gamma _v |_{S^1-I}$ is injective, and $\gamma _v (a_i) = \gamma _v (b_i)$ are distinct points such that for each $i$, $\gamma _v'(a_i)$ and $\gamma _v' (b_i)$ are linearlly independent.

Now fix a particular $\gamma$. To me, the task of proving the existence of a regular projection seems like a job for differential topology, and in particular, Sard's theorem. For instance it follows immediately from Sard's theorem that the first condition is satisfied for almost all $v \in S^2$ since the image of the map $S^1 \rightarrow S^2, s \mapsto \gamma '(s)$ has measure zero.

I have been trying to prove something similar about the set of $v$ which satisfy the second condition but I haven't been successful. To give an idea for the sorts of tools I was hoping to use I'll explain my attempts.

One idea I had was to consider the map $$f: S^1 \times S^1 - \Delta \rightarrow \mathbb{R}, (s,t) \mapsto (\gamma '(s) \times \gamma ' (t)) \cdot (\gamma (s) - \gamma (t))$$ where $\Delta$ is the diagonal in the torus, $\times$ is the cross product and $\cdot$ is the dot product. Then $f(s, t) = 0$ iff either {$\gamma (s) - \gamma (t)$ lies in the plane spanned by $\gamma ' (s)$ and $\gamma ' (t)$} or {$\gamma '(s)$ is parallel to $\gamma '(t)$}. In either case, this means $\gamma (s)$ and $\gamma (t)$ must not project to the same point if the projection is to be regular. This rules out the one dimensional subspace spanned by $\gamma (s) -\gamma (t)$. Now if $0$ were a regular value for $f$ then $f^{-1}(0)$ would be a submanifold of dimension $1$. We could then consider the smooth map $$f^{-1}(0) \rightarrow \mathbb{R}P^2, (s,t) \rightarrow [\gamma (s) -\gamma (t)]$$ and apply Sard's theorem to find that almost every $v \in S^2$ also satisfies condition 2. However, as far as I can tell, there is no good reason $0$ should be a regular value for $f$. Maybe we could pick an isotope of $\gamma$ with this property? Moreover this still doesn't rule out the possibility that 3 or more points on the curve project to the same point.

• Did you ever find a good proof of the regular knot projection? I'm trying to puzzle it out from the replies and comments, but I'm unable to reach any solid conclusion. Commented Aug 13 at 20:04

Consider the chord map $$\nu\colon S^1\times S^1 - \Delta \to S^2\,, \quad \nu(s,t) = \frac{\gamma(s)-\gamma(t)}{\|\gamma(s)-\gamma(t)\|}\,.$$ I'll let you compute that if $\nu(s,t) = v$, then [I think!] $$d\nu_{(s,t)}(w,z) = \frac1{\|\gamma(s)-\gamma(t)\|}\text{proj}_{T_v S^2}(w-z)\,.$$ So if $\pm v$ are regular values of $\nu$, you are guaranteed that the projections of $\gamma'(s)$ and $\gamma'(t)$ onto the orthogonal complement of $v$ span.
That there are finitely many points in the preimage of such $[v]\in\mathbb RP^2$ can be argued by making sure that $\pm v$ are never tangent to $\gamma$ and getting a compact $0$-dimensional submanifold of $C\times C - \Delta$.
• Um, this calculation applies only when the chord for $s$ and $t$ lines up in the direction we're going to project. It doesn't tell us about anything other than $(s,t)\in\nu^{-1}(v)$. Are we ok? Commented Jul 18, 2013 at 0:50
• You can make an independent argument to see that having trisecants is "rare." Take $S^1\times S^1\times S^1$, map by two chords to $S^2\times S^2$, and take the preimage of the diagonal. You then make sure $v$ misses this (at most) curve of directions. Commented Jul 18, 2013 at 1:25