4
$\begingroup$

this is elementary question about classical knot equivalence.

I know that just isotopy which need not to be ambient is not proper to define knot equivalence because bachelor's unknotting.

but this example is unsatisfied to me, because this unknotting isotopy is not differentiable.

If the isotopy is given "smooth" condition, it seems to define knot equivalence intuitively. i.e $ F(x,t): S^1 \times [0,1] \to S^3 $ s.t $F(x,t)$ is smooth map and level restriction $F(x,t_0)$ is embedding. Can it define knot equivalence?

I think it may be false.

Any reference is grateful to me.

Improve this question
$\endgroup$
2
  • 6
    $\begingroup$ The answer to your question is the "isotopy extension theorem". Take a look at Hirsch's "Differential Topology" textbook. $\endgroup$
    – Ryan Budney
    Sep 7, 2010 at 2:55
  • 1
    $\begingroup$ Ryan, thank you for your kind response to my stupid question. Few minute ago, I found exact answering article about it. The American Mathematical Monthly Vol. 112, No. 5 (May, 2005), pp. 417-425 <Florian Deloup> The Fundamental Group of the Circle Is Trivial $\endgroup$
    – Seonhwa Kim
    Sep 7, 2010 at 8:48

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.