# framed n component link on a genus n surface determines the self-homeomorphism?

$S$ 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?)

Thanks.

• 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 $1 \le k < n$. – Lee Mosher Jul 26 '18 at 14:28
• 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. – guest123 Jul 26 '18 at 14:52

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 $\mathcal{G}_S$?