$S$ is the boundary of a genus $n$ handlebody in $S^3$. $\{m_1, m_2,..., m_n\}$ is the collection of the meridian circles of $S$; $\{l_1,l_2,...,l_n\}$ is the collection of the longitude circles on $S$. They are both unlinks.

Suppose there are 2 orientation preserving self-homeomorphisms of $S$, $f$ and $g$, such that $\{f(m_1),f(m_2),\dots,f(m_n)\}$ is isotopic to $\{g(m_1),g(m_2),\dots g(m_n)\}$ as links (that is, there is an isotopy $I_t:\sqcup_{i=1}^n S^1\to S^3$ such that $I_0(S^1_i)=f(m_i)$, $I_1(S^1_i)=g(m_i))$ and they have the same framings ($S$ induces a framing on each $f(m_i), g(m_j)$ such that the framing of $f(m_i)$ is taken by the isotopy to the framing of $g(m_i)$). Similarly for $f(l_i)$ and $g(l_i)$. What can I say about $f$ and $g$?

When $n=1$, I think $f$ has to be isotopic to $g$. Have no idea about $n>1$. In the simple case when $\{f(m_i)\}$ and $\{f(l_i)\}$ are unlinks with framing $0$, what is $f$? (I think uniqueness of Heegaard splitting implies $f$ is identity modulo some simple self-homeomorphisms, right?)


  • $\begingroup$ When $n=1$ your thought is correct. But if $n > 1$ then $f$ and $g$ can differ by a Dehn twist around a circle in $S$ that separates $\{m_1,l_1,...,m_k,l_k\}$ from $\{m_{k+1},l_{k+1},...,m_n,l_n\}$, where $1 \le k < n$. $\endgroup$
    – Lee Mosher
    Jul 26, 2018 at 14:28
  • $\begingroup$ Yes, Can we say f is isotopic to g modulo these homeomorphisms? When {f(m_i)} and{f(l_i)} are unlinks with framing 0, f could also be the homeomorphism that send {m_1,m_2, ..., m_n},{l_1,l_2,...,l_n} to {m_1, m_2, ...,m_n},{l_1+l_2,l_2,...l_n}. Are these the only possibilities? I am trying to see if it is possible to exist other "complicated" homeomorphisms. $\endgroup$
    – guest123
    Jul 26, 2018 at 14:52

1 Answer 1


Just an observation: consider the subgroup $\mathcal{G}_S\leq Mod(S)$ of the mapping class group of $S$ represented by a homeomorphisms $h:S\to S$ which extend to an orientation-preserving homeomorphisms $H:S^3\to S^3$, such that $H_{|S}=h$. Then $\mathcal{G}_S$ is called the Goeritz group of the Heegaard splitting $S$. For genus $2$, a presentation of this group is known, and generators in genus $3$, but in general we have only a conjectural set of generators.

In any case, given a homeomorphism $f:S\to S$, we can obtain a $g$ with the properties you describe by taking $g=h\circ f$, where $[h]\in \mathcal{G}_S$. This follows because any orientation-preserving homeomorphism $H:S^3 \to S^3$ is isotopic to the identity by an isotopy $H_t$. Then $H_t$ gives the required isotopies for both the meridian and longitudinal links.

So I suppose one can refine your question to ask whether such $f$ and $g$ exist that don't differ by an element of $\mathcal{G}_S$?


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.