In introductory knot theory books, authors usually make a choice of smooth knots or piecewise-linear knots. I often find myself wanting to work in the larger setting of piecewise-smooth knots which subsumes both smooth and PL knots. To do this I would need to prove a "piecewise smooth isotopy-extension theorem" in order to show that knots which are equivalent in the piecewise-smooth sense are equivalent in the usual sense that there exists a continuous ambient isotopy or an ambient homemorphism throwing one onto the other:

Suppose $H: S^1 \times I \rightarrow S^3$ is a continuous isotopy and there exist $$0=t_0 < t_1 < t_2 < ... < t_k = 2\pi$$ such that $H|_{[t_i, t_{i+1}] \times I}$ is $C^\infty$ for all $i$. Then show that $H$ extends to a continuous isotopy of $S^3$, or at the very least there exists a homeomorphism of $S^3$ carrying $H_0$ to $H_1$. Feel free to add hypotheses as needed.

One approach to a proof would replace the piecewise smooth knots $H_0$ and $H_1$ with smooth knots using continuous ambient isotopies of $S^3$ (I can do this). Then if I knew that every piecewise smooth isotopy (defined as above) of smooth knots can be replaced by a smooth isotopy, I could use the smooth isotopy extension theorem to get a continuous ambient isotopy from $H_0$ to $H_1$.

Any thoughts?


migrated from math.stackexchange.com Aug 7 '13 at 22:55

This question came from our site for people studying math at any level and professionals in related fields.

  • 1
    $\begingroup$ This strikes me as a question that might interest folks over at MathOverflow. $\endgroup$ – Kevin Carlson Aug 5 '13 at 22:05
  • $\begingroup$ @Kevin Carlson: Thanks, I will post there if it isnt answered here $\endgroup$ – Tim kinsella Aug 5 '13 at 22:33
  • 2
    $\begingroup$ @Timkinsella: Dear Tim, If you decide to move this quetion to MO, don't cross-post; rather, flag for moderator attention and ask them about migrating your question. Regards, $\endgroup$ – Emerton Aug 6 '13 at 0:42

Everything that you wish for is true. You can approximate "ambient" (in the sense explained below) isotopies by smooth isotopies and keep the ends (the knots) fixed or not. This is spelled out in great detail in:

  • MR0674117, Bröcker, Theodor; Jänich, Klaus: Introduction to differential topology. Translated from the German by C. B. Thomas and M. J. Thomas. Cambridge University Press, Cambridge-New York, 1982. vii+160 pp.

Also the book [Hirsch: Differential topology] has many details.


Sorry, I looked up ambient isotopy in Wikipedia just now, and it means something different than what I remember. I meant an isotopy which can be extended to an isotopy of tubular neighborhoods. The Wikipedia notion was called diffeotopy, or something like that, was it not? Of course it gives a finer equivalence relation on closed curves (embedded with a tubular neighborhood).

  • $\begingroup$ Thank you, Peter. Did you mean to say "ambient isotopy" when you did? As I look through my copies of Hirsch and Janick/Brocker, I'm still not sure how to turn a piecewise smooth isotopy (as defined in my second paragraph) into any kind of ambient isotopy of $S^3$, smooth or $C^0$. $\endgroup$ – Tim kinsella Aug 8 '13 at 6:51
  • $\begingroup$ I know that it is not the case that "continuously isotopic smooth knots are smoothly isotopic" since then all knots could be pulled out to get the unknot. So it seems that the piecewise smoothness of the isotopy must be used and not just the fact that its $C^0$. $\endgroup$ – Tim kinsella Aug 8 '13 at 7:01
  • $\begingroup$ I see! Thank you for the clarification. Yes I should probably have used the term "diffeotopy." So, to be clear, do you claim that (a) a Piecewise smooth isotopy of smooth knots extends to a $C^0$ isotopy of tubular neighborhoods, and that (b) this, in turn, implies the smooth knots are smoothly isotopic? $\endgroup$ – Tim kinsella Aug 8 '13 at 11:33
  • $\begingroup$ Yes, I claim this. It is simpler, because piecewise smooth isotopies can be approximated by smooth isotopies: just smooth them by locally convolving with smoothing kernels so tightly, that they stay isotopies (embeddings are open in the Whitney $C^1$ topology.) $\endgroup$ – Peter Michor Aug 8 '13 at 11:47
  • $\begingroup$ Thanks for your help. Sounds like I need to buy your book! :) $\endgroup$ – Tim kinsella Aug 8 '13 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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