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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.

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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 $C\times C - \Delta$.

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  • $\begingroup$ Thank you, Ted. I am probably just confused but wouldn't the first part imply that there exists a projection so that no two distinct points on the curve have parallel derivatives? $\endgroup$ Jul 18, 2013 at 0:33
  • $\begingroup$ 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? $\endgroup$ Jul 18, 2013 at 0:50
  • $\begingroup$ Ohh ok got it! And the only other tiny detail is that the fibers of intersection points might contain more than two points. But maybe there is a simple remedy $\endgroup$ Jul 18, 2013 at 1:01
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    $\begingroup$ 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. $\endgroup$ Jul 18, 2013 at 1:25
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    $\begingroup$ Yeah, I wish he were a bit more detailed though. He also doesn't usually work in the smooth setting or use much differential topology. $\endgroup$ Jul 18, 2013 at 2:08

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