Statement 1: Knots of opposite chirality have ambient isotopy, but not regular isotopy.

Statement 2: We can then define two such knots to be equivalent if they are ambient isotopic, meaning that there exists an (orientation-preserving) homeomorphism $\mathbb{R}^3\to\mathbb{R}^3$ carrying one to the other.

Am wondering if these two statements contradict each other: How could an orientation-preserving homeomorphism $\mathbb{R}^3\to\mathbb{R}^3$ always carry one knot to its mirror image? Is it because ambient isotopy of knots are defined differently in the two statements? Which definition is "standard", or more commonly accepted?

  • 1
    $\begingroup$ I have no idea what Mathworld is talking about there. The equivalence relation defined by smooth isotopy of embeddings is the same thing as the equivalence relation defined by orientation-preserving homeomorphisms of $\mathbb R^3$. See my answer here. $\endgroup$
    – user98602
    Jul 17 '15 at 18:16

The key reference on regular isotopy seems to be the following paper:

Kauffman, Louis H. “An invariant of regular isotopy.” Transactions of the American Mathematical Society 318, no. 2 (1990): 417–471.

Note that regular isotopy is an equivalence relation on knots diagrams, not the knots themselves.

The statements on MathWorld don't make a whole lot of sense, so I'm not quite sure how to respond to the rest of your question. It is possible for a knot to be ambient isotopic to its mirror image, e.g. the figure-eight knot-- see chiral knot on Wikipedia. There is only one commonly used definition of ambient isotopy of knots.


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.