Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have come across the following exercise in Kosinski's 'Differential Manifolds':

Exercise: Consider an imbedding $\mathbb R\to \mathbb R^3$ where the image is "the line with a knot": enter image description here

Show that this imbedding is isotopic to the standard imbedding $\mathbb R\subset \mathbb R^3$.

But I cannot see what such an isotopy could possibly do to unknot this knot? The only thing I can think of is pulling the knot at both ends to tie it completely tight at the origin. But then I would necessarily cause the second derivative to explode, wouldn't I? What else is there to do? Or am I possibly wrong in thinking that the pulling gives me trouble with higher derivatives?

Thanks for your help!

share|cite|improve this question
up vote 2 down vote accepted

Isotopy requires in my understanding only homeomorphisms, not diffeomorphisms, hence your concern about exploding derivatives can be put aside.

Thus asume that the knot embedding $f\colon\mathbb R\to\mathbb R^3$ is identical to the standard embedding for $x\notin(-1,1)$. Letting $$F(t,x)=\begin{cases}(x,0,0)&\text{if }|x|\ge t\\tf(\frac xt)&\text{if }|x|<t\end{cases}$$ this gives an isotopy with $F(1,x)=f(x)$, $F(0,x)=(x,0,0)$, as one can check. (And this is exactly what your idea was).

For a smooth $F$, as seems to be required by Kosinsky, one can just push the knot out to infinity, i.e. (again assumin $f(x)=(x,0,0)$ for $x\notin(-1,1)$) let $$F(t,x)=\begin{cases}(x,0,0)&\text{if }t=0\\f(x+\frac1t-1)+1-\frac1t&\text{if }0<t\le 1\end{cases}$$ This is obviously smooth at $(t,x)$ with $t\ne0$ and is just $(t,x)\mapsto(x,0,0)$ in a neighbourhood of points at the $t=0$ boundary.

share|cite|improve this answer
Thank you for your answer. I have to disagree with your first observation, however. Isotopies are by definition smooth in his book. (Def. (4.1) on page 33 if you have the book) – Sam Oct 17 '12 at 20:07
OK, that's what I would have called a smooth isotopy. So maybe one has to play with $e^{-\frac1t}$ or the like here? – Hagen von Eitzen Oct 18 '12 at 6:01
Thank you for your answer! – Sam Oct 24 '12 at 9:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.