An answer to this question would also answer Isotopy of periodic homeomorphisms of a surface along periodic homeomorphisms

Let $M$ be a topological manifold and let $f,g$ be two orientation preserving homeomorphisms of $M$. Suppose that $f$ and $g$ are isotopic and that they are conjugate in $Homeo^+(M)$.

Are they conjugate by a homeomorphism isotopic to the identity?


1 Answer 1


No: Let $M$ be the 2-dim. surface consisting of tori welded to each other so that they form a string (sorry, I do not know how to draw here). Let $f$ be the the time 1 flow of a vector field supported in a small part of the $i$-torus, and let $g$ be the time 1 flow of the same vector field, but now supported in the $i+1$-torus. They are isotopic, because each one isotopic to the identity. They are conjugated by the shift of the tori by 1. But this shift can never be isotoped to the identity. If you want $M$ to be compact, you can arrange the tori in a (large) circle.

Edit: More details.

Suppose that the vector field rotates a circle (which is not involved in the welding) by a maximal angle which characterizes this circle. Then any conjugation has to map this circle to the corresponding one in the next torus. But these two circles are different generators of $\pi_1(M)$. So the conjugating homeomorphism cannot be isotopic to the identity.

  • $\begingroup$ Thanks for the edit! I didn’t understand it completely. I will think about the question in the other link or which conditions to impose on f and g for this to be true (if it is true at all in some setting). $\endgroup$
    – Paul
    Aug 26, 2018 at 1:55

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