I am looking for a proof of the result that any open book $(B, \pi)$ ; $B$ a fibered link , $\pi$ a map of $M^3-B \rightarrow S^1$ on a 3-manifold $M^3$ , so that $\pi$: $(M^3-B)$ fibers over $S^1$ , where the fibers $\pi^{-1}(\theta)$ are surfaces $\Sigma_{\theta}$ with $\partial \Sigma_{\theta}=B$ , can be obtained as an abstract open book (AOB). An AOB is a pair $(\Sigma, \phi)$ , for $\Sigma$ a compact surface with non-empty boundary , so that $\phi|_{\partial \Sigma}=Id$; $Id(x)=x$ , for $\Sigma$ and $\phi: \Sigma \rightarrow \Sigma$, a diffeomorphism called the monodromy of the AOB (i.e., $\phi$ is an element of the MCG of $\phi$).

The "actual" book is then obtained from the AOB by first doing the mapping torus $\Sigma_{\phi}$, and then filling-in a solid torus $D^2 \times S^1$ for each boundary component. The binding B then lives in the solid tori ( which are tubular neighborhoods of the respective components in the binding.) I would also appreciate a proof of the fact that the binding $B$ alone does not determine the open book, i.e., $M^3 -B$ may be fibered over $S^1$ in non-isotopic ways.

I'm specially interested on how to determine the monodromy once an "actual" , i.e., non-abstract, open book is given, i.e., given the data $(B, \pi)$ for $M^3$ , how do I obtain the needed $\phi$ ?

Thanks for any help, refs, etc.


The "proof" that you are looking for is essentially just understanding definitions. The fibration of the $M^3-B$ by pages describes $M^3-B$ as mapping torus of the monodromy on any chosen page. The meridians of $B$ nail down the action of the monodromy near the boundary of the page. (Changing the monodromy by a Dehn twists along a curves isotopic to boundary components of the page changes the curves in the boundary of the mapping torus that are to become meridians. This however typically changes the ambient manifold as it is topologically tantamount to doing Dehn surgeries on the binding components.)

With an orientation, the binding $B$ does determine the open book, almost. Require that the orientation on the binding agrees with the boundary orientation from the pages. Non-isotopic fibrations of $M^3-B$ with this condition occur when there are essential tori in $M^3-B$, and these alternative fibrations come from "adding" copies of these tori to the fibration.

With different orientations on the binding you can get different open books. For example, reverse the orientation on one binding component of the positive Hopf band and you get the binding for the negative Hopf band. Also, if you consider rational open books, then you'll have infinitely many fibrations for the same binding (well, if you have at least two binding components). Look into the Thurston norm to understand the ways that a manifold with toral boundary may fiber over the circle.

Finally, teasing out the monodromy from an embedded open book is not always so easy to do. There are a few basic strategies.

  • Murasugi desum (generalized Hopf deplumbing -- see Gabai's work) as much as possible, breaking it smaller open books.

  • Add Dehn twists to transform the open book (and manifold) into a more manageable one. This can be useful if the the open book admits a Stallings twist.

Ideally you can do the above two to produce open books with monodromies you recognize. Failing that, you can still try:

  • Take a set of arcs on the page whose complement is a disk, and push them through the mapping torus until they land back on the page you started with.

This gets you a set of arcs and their images under the monodromy. Since the complement of these arcs is a disk, this actually is a description of the monodromy. However people often want the monodromy written as a product of Dehn twists. Passing from the arcs to their images to a product of Dehn twists is somewhat of a challenge. Though I reckon one should be able to devise an algorithm that does this.

Perhaps people have some other approaches for this.


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.