I am trying to have a better understanding of how one goes , "travels" between the different formats/layouts of open books for a fixed given 3-manifold M; between the abstract type and the "actual" type. I am new to this, and I have not been able to find clear explanations, or whether there are canonical choices that help make things more clear. Please critique/suggest :

The abstract type is given by ( $\Sigma, f$) , where $\Sigma$ is a compact, oriented surface with non-empty boundary, and $f$ is an element of $MCG( \Sigma$), the Mapping Class Group of $\Sigma$ , and the "actual" type ($ B, \pi$ ), where $B$ is a fibered link, and $\pi$ is a map , so that $\pi: (M-B) \rightarrow S^1$ is a locally-trivial bundle with fiber the surface $\Sigma$, so that $ \partial \Sigma =B$ , and $\pi|_N(B)$, (where N is a tubular neighborhood of B --a fibered link has a trivial normal bundle) is the argument map . It seems reasonable to assume here that $\Sigma$ is a Seifert surface for the fibered link $B$.

So, let me lay out what I think about the two transitions:

1) Between $(B, \pi)$ and $(\Sigma_\pi, f_\pi$ ).

The ambiguity I see here is that , assuming $\Sigma$ is a Seifert surface for $B$, does not narrow $\Sigma$ down even up to homeomorphism, let alone isotopy, because there are many possible Seifert surfaces for any $B$ unless maybe we fix the genus of $\Sigma$ ( So that $\Sigma$ is given up to boundary components by $B$ and up to genus , so we have $\Sigma_{n,g}$ up to homeomorphism ). Is there a canonical way of choosing $\Sigma$ here?. We could , I guess just define $ \Sigma_\pi:= \pi^{-1} (\theta)$ , for any $\theta \in S^1 $ . But, how does one determine $f_\pi$? I know we could calculate the total monodromy of $\pi$ , by choosing a Riemann metric g, computing the dual vector field to $d\pi$, and then flowing it. Is this how this is usually done?

Now, for the opposite direction,

2)Between $( \Sigma, f )$ and $( B, \pi )$

The mapping torus $M_f$ clearly gives us the map $\pi$ onto $S^1$, by following along the copy of $S^1$ in the mapping torus of $f$ along $\partial (S^1 \times D^2)$. And, if $\Sigma$ must be a Seifert surface for B, then we would just let $B:= \partial \Sigma$. Is this how things are usually done?

An additional question, please:

3) Why do we define equivalence AOBs $( \Sigma_1, f_1)$ and $(\Sigma_2, f_2)$ to be equivalent if there exists a homeo. $h: \Sigma_1 \rightarrow \Sigma_2$ satisfying :

$h \circ f_1 = f_2 \circ h$ ?

This last shows that $f_1$ and $f_2$ are conjugate elements in $MCG(\Sigma)$. Does this last condition guarantee that $f_1, f_2$ are isotopic $(in MCG(f)$? ; since $f_1, f_2$ are isomorphic, set $f =f_1= f_2)$?


You are asking definitional questions about surface homeomorphisms, so I suggest you read one of the standard references. The book by Casson and Bleiler is short and, as I recall, very clear.

1) Remember that $f_\pi$ is a mapping class, not an actual surface homeomorphism.

2) Yes, this is how things are done. There is a important point here - we assume that the representative of $f$ is a surface homeomorphism that fixes the boundary of $\Sigma$ pointwise.

3) Conjugate elements in the mapping class group can be distinct - the group is not Abelian. When this happens no two representatives are isotopic, by definition. As a particular example, fix $\alpha$ and $\beta$ a pair of non-separating curves that are not isotopic. Then the left Dehn twists $T_\alpha$ and $T_\beta$ are conjugate, but not equal, in the mapping class group. (That they are conjugate uses the classification theorem for surfaces. That they are not equal is an exercise in the definitions.) In particular, no representatives of $T_\alpha$ and $T_\beta$ respectively are isotopic.

What is going on, instead, is that conjugate mapping classes give homeomorphic mapping tori. Note also that the conjugating map $h$ fixes $\partial \Sigma$ pointwise.

  • $\begingroup$ Could you please elaborate on 1)? I understand $f_\pi$ is given up to isotopy, but how can I compute it up to isotopy by only knowing only the pair $(B, \pi)$? Thanks for the Casson-Bleiler ref. $\endgroup$ – user56470 Jul 26 '14 at 21:17
  • $\begingroup$ If you are given the bundle and the projection map, then cut the bundle along a fiber. The (re-)gluing (to recover the bundle from $\Sigma \times I$) is the desired homeomorphism. $\endgroup$ – Sam Nead Jul 26 '14 at 21:24
  • $\begingroup$ Thanks. And, re 1), is it correct to say that $\Sigma_{\pi}$ is just a fiber $\pi^{-1} (\theta)$ for $\theta \in S^1$? $\endgroup$ – user56470 Jul 26 '14 at 21:33
  • $\begingroup$ Yes - that is right. However you should probably write $\Sigma_\theta$, not $\Sigma_\pi$. $\endgroup$ – Sam Nead Jul 26 '14 at 23:16
  • $\begingroup$ Thank you. Please give me a few days to read more carefully. $\endgroup$ – user56470 Jul 26 '14 at 23:56

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