# Exchanging the components of a two-component link

Given a 2-component link in $$S^3$$ whose components are trivial knots, is it always possible to find a homeomorphism of $$S^3$$ that exchanges the components?

I guess the answer is "no" (but I could not find a counterexample), so here is a second question:

Which link-invariant could prevent the existence of such an exchanging homeomorphism?

## 1 Answer

Indeed, as you suspect, the answer is no. For instance, take the link $$L$$ obtained from the Hopf link by doing a $$(2,1)$$-cable of one component and a Whitehead double of the other.

A way of telling them apart is to look at the JSJ decomposition of the complement: each piece contains one component, and one component is Seifert fibred, while the other is hyperbolic.

A reflection of this fact is that if you twist each of the components with respect to the other, you obtain different knots. (By "twist" here I essentially mean "do $$\pm1$$-surgery along one component.)

I also think that the 2-variable Alexander polynomial (or branched covers, coloured signatures...) should do the trick.

• Danny Ruberman's answer to mathoverflow.net/questions/344196/… provides an explicit example of using the two-variable Alexander polynomial to provide a counterexample to the original question. – dvitek Dec 13 '20 at 23:42
• Thank dvitek, indeed the questions was already answered one year ago. However I like Marco's answer. Thanks to both of you. – Pierre Dehornoy Dec 14 '20 at 22:26