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This is a refinement of one of my earlier questions (I failed to put into words what I really wanted to ask). First of all, I'm not sure "singularity" is the correct word to use hence the quotes. Consider the following wild knot:

enter image description here

Then what exactly happens where the curls get infinitely small? Is the knot still differentiable there? I'm asking because I'm trying to understand why requiring a knot to be differentiable is not enough to prevent knots from being wild. On the other hand, smoothness is enough.

Thanks for help!

(If anyone knows the parametric equation of this curve it might make it easier to see what happens.)

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May I ask how you draw that beautiful picture? What software did you use? How did you insert it here? –  Hui Yu Oct 24 '12 at 11:20
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@HuiYu I downloaded the picture from the Wikipedia article about wild knots. Then I uploaded it using the upload image button (you can see the button if you edit one of your posts). –  Rudy the Reindeer Oct 24 '12 at 11:23

2 Answers 2

The image given is only an embedding of manifolds, and doesn't look like an embedding of differential manifolds, and even less of an embedding of smooth manifolds.

To be an embedding of differential manifolds, you need your map to induce an injective map on tangent vectors. Here, at the end of the infinitely shrinking knot, there is no tangent line to your embedding : on one side you want to send the tangent vector to a horizontal vector, but on the other side, the slope keeps turning in circle as you get closer to the singularity.

However, all is not lost, you can squeeze the chain : instead of using a map of the kind $t \mapsto (tx(\log t), ty(\log t), tz(\log t))$ for some periodic functions $x,y,z$, use $t \mapsto (tx(\log t), t^2x(\log t)y(\log t), t^2x(\log t)z(\log t))$ instead. It is still not differentiable at $t=0$ because $x(\log t)$ isn't convergent, but at least there is a tangent line. I think you may find a differentiable wild knot with enough tinkering.

But you can't have a map of $C^1$ manifolds (nor of smooth manifolds) that looks like that. If the map was $C^1$, the tangent line has to vary continuously when you move along the knot, so there has to be a whole neighboorhood around the singularity where all the tangent lines don't differ from the horizontal line by, say, a $\pi/4$ angle. But you can't make any wild knotting with only using tangent lines in that cone. In particular, you can locally thicken your knot by attaching small disks all orthogonal to the original horizontal tangent vector. Then since the circle is compact, you should be able to get a global thickening from a finite number of those local ones.

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That last statement is commonly known as the tubular neighborhood theorem. –  Alexander Thumm Oct 24 '12 at 17:06

This is not really an answer so much as me writing out some confusion, so I'm making it community wiki. Consider the curve $$ f(x) = \begin{cases} x^2\sin(1/x) & x < 0 \\ 0 & x \geq 0. \end{cases} $$

Then $f$ exhibits wild-knot like behavior at the origin and is differentiable but not twice differentiable there; you could probably make your wild knot parametrically by bending the graph of $f$ into three-space. But on the other hand

$$ f(x) = \begin{cases} \exp\left(\frac{-1}{x^2}\right)\sin(1/x) & x < 0 \\ 0 & x \geq 0. \end{cases} $$

exhibits infinite oscillation and is smooth, so it seems like I ought to be able to make a smooth wild knot too, if my argument in the earlier part is right.

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