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Given two $C^1$ immersed curves $f, g: S^1 \to {\mathbb R}^3$ with disjoint image, I would like a simple proof, working only in the smooth category, that there exists a unit direction $y \in {\mathbb R}^3$ such that projection orthogonal to $y$ is regular: that is, (i) the projected curves intersect transversely in the plane and (ii) no two such intersections pile up (in other words, the fibers of the projection of the link into the plane have at most two points).

Part (i) is easily obtained by choosing a regular value of the link map $h:S^1 \times S^1 \to S^2$, $h(s,t) = (f(s)-g(t))/|f(s)-g(t)|$.

As for part (ii), the proofs I've seen are for polygonal knots (for example Proposition I.3.1 of Crowell, Fox, Introduction to Knot Theory) and can be directly adapted to polygonal links. The main ideia to exclude fibers of the projection with more than three points is to consider triples of skewed edges and exclude the directions of the lines that connect them. This idea does not seem to apply directly to smooth links. To obtain (ii) one could of course approximate the smooth links by appropriate polygonal links. My question is: does there exist a nice way to prove (ii) directly for smooth links without using polygonal approximations?

PS: One could slightly move one of the links to get (ii) but, since I would like the proof to be as pedestrian as possible, I would like to keep the links fixed and just change the direction.

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    $\begingroup$ Here you can find hints for a proof: dropbox.com/s/xnud8u03uiwkiil/math132ps5.pdf?dl=0 You can also find some discussion of this in the answers and comments here: math.stackexchange.com/questions/446091/… $\endgroup$ – Ian Agol Jun 4 '17 at 21:59
  • $\begingroup$ Thank you very much @IanAgol! I was hoping precisely for a transversality argument that excludes the trisecants and it seems very likely that the nice two chord arguments you showed me can be adapted to links. When I have checked the details I will comment here again. $\endgroup$ – Lucas Seco Jun 5 '17 at 10:02
  • $\begingroup$ @IanAgol, I could not follows the arguments of the class notes you linked to me. I provided an answer below following the ideas in your second link which I find more geometrical. Perhaps the last argument can be simplified further. $\endgroup$ – Lucas Seco Jun 14 '17 at 16:42
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I could find the following satifactory answer to my question with the ideas provided in the comments above.

Consider the natural projection onto the projective plane, of a non-null vector into the line it spans

${\mathbb R}^3 - 0 \to {\mathbb P}^2 \qquad v \mapsto [v]$

and identify the tangent space $T{\mathbb P}^2_{[v]}$ with the corresponding tangent space of the sphere $S^2_v = v^\perp$.

For part (i), consider the chord map between $f$ and $g$, given by

$c = c(f,g): S^1 \times S^1 \to {\mathbb P}^2 \qquad c(x,y) = [f(x) - g(y)]$

defined for $(x,y)$ such that $f(x) \neq g(y)$. It can be shown that intersecting the regular values of $c(f,g), c(f,f), c(g,g)$ and removing the tangent directions $[f'], [g']$ we get a full set of directions $[v] \in R$ such that under orthogonal projection to $v$, the projected curves are regular, intersect transversely and self-intersect transversely.

For part (ii), consider the secant map between $f$ and $g$, given by

$s = s(f,g): S^1 \times S^1 \times S^1 \to {\mathbb P}^2 \times {\mathbb P}^2 \qquad s(x,y,z) = (\,[f(x) - g(y)],\, [g(y) - f(z)]\,)$

defined for $(x,y,z)$ such that $f(x) \neq g(y) \neq f(z)$. To avoid triple points we must avoid directions $v$ such that $s(x,y,z) = ([v],[v]) = \Delta([v])$, where $\Delta: {\mathbb P}^2 \to {\mathbb P}^2 \times {\mathbb P}^2$ is the diagonal embedding. Note that we want to avoid directions whose image under $\Delta$ are in $$ {\rm Im}\,s \cap \Delta({\mathbb P}^2) = s( s^{-1}( \Delta({\mathbb P}^2) ) ) $$ and that the image $\Delta({\mathbb P}^2)$ is an embedded submanifold of ${\mathbb P}^2 \times {\mathbb P}^2$. If the secant map was transversal to it, then $s^{-1}( \Delta({\mathbb P}^2) )$ would be a curve, so that the above intersection would be a differentiable image of a curve into a surface, hence a null set of ${\mathbb P}^2$.

Unfortunately it can only be shown that the secant map is transversal to $\Delta(R)$. Nevertheless, the above argument is local and still follows. Indeed, since the secant is transversal to the point $\Delta(w)$ for $w \in R$ (meaning that $s(x,y,z) = \Delta(w)$ implies that the image of $s'(x,y,z)$ complements the image of $\Delta'(w)$ in $T({\mathbb P}^2\times{\mathbb P}^2)_{\Delta(w)})$, by the same proof involving the local form of immersions there exists a neighbourhood $U$ of $\Delta(w)$ in $\Delta( {\mathbb P}^2 )$ such that $$ {\rm Im}\,s \cap U = s( s^{-1}( U ) ) $$ is a differentiable image of a curve into a surface, hence a null set of $\Delta({\mathbb P}^2)$. These open sets give an open cover of $\Delta(R)$. Since ${\mathbb P}^2$ is a manifold and thus its subsets are Lindelöf, we get an enumerable cover $U_i$ of $\Delta(R)$. Removing from $\Delta(R)$ the null sets ${\rm Im}\,s \cap U_i$, we get another full set $R'$ contained in $R$ such that $$ \Delta(R') \cap {\rm Im}\, s = \emptyset $$

Note that for links we must consider the construction above with the secant map $s(f,g)$ and also $s(f,f)$ and $s(g,g)$. This is since each $f$ and $g$ can have self-intersections in ${\mathbb R}^3$ which we do not want to get projected to new self-intersections of $f$ and $g$ in the plane.

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