open book decompositions of $\Sigma\times S^1$

Question

Let $\Sigma$ be a closed orientable surface. Is there a standard open book decomposition on the $3$-manifold $M=\Sigma\times S^1$?

If the binding is allowed to be empty in the definition of an open book decomposition, then this is obvious, since $M$ is the mapping torus of the identity of $\Sigma$. The literature is not clear on this, however.

If the binding must be non-empty, then since it is contained in every page, it seems that the obvious fibering $M\to S^1$ is never the fibering of an open book.

Answer 1

I'm not sure about "preferred", but here is the simplest open book decomposition of $\Sigma_g \times S^1$ that I know:

Take the open book $(\Sigma_{g,1},\rm{id})$ with page the genus-$g$ surface with one boundary component and trivial monodromy. The result of page-framed surgery (ie $0$-surgery) on the binding component will give us $\Sigma_g \times S^1$. There are several ways to implement that surgery on the open book, but the simplest (AFAIK) is to do a positive stabilisation (along a boundary-parallel arc) to get $(\Sigma_{g,2},T_{\partial_1})$ and then add a negative Dehn twist to $\partial_2$ (which is the same knot as we started with, but now with page framing $-1$).

So $(\Sigma_{g,2},T_{\partial_1}T^{-1}_{\partial_2})$ will give you an open book for $\Sigma_g \times S^1$. Notice that when $g=0$, this recovers the standard open book $(S^1 \times [0,1],\rm{id})$ for $S^2 \times S^1$.

Answer 2

Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

Edited

So, $M = \Sigma \times S^1$ has the "obvious" open book decomposition (if we allow empty bindings). By the above $M$ also has (infinitely many) open book decompositions with non-empty bindings. I very much doubt that there is a preferred one among the latter.

Answer 3

That every (closed oriented) $3$-manifold has an open book decomposition is attributed to Alexander

Alexander, J. W. Note on Riemann spaces. American M. S. Bull. 26, 370-372 (1920).

and there are several explicit ways to construct open book decompositions on $3$-manifolds of which none will yield a simple open book in general.

I think most authors do not allow the binding to be empty in the definition of an open book.

For $S^1\times S^2$ there exists a preferred open book, namely the open book with trivial monodromy and annulus page.

For the other surfaces, one can see that the monodromy cannot be the identity and thus the open books have to be more complicated.

Pattrick Massot has beautiful animations of a simple open book on $T^3$ with $4$ binding components and a page of genus $1$:

https://www.math.u-psud.fr/~pmassot/en/exposition/giroux_correspondance.html

If you look at open books from another viewpoint there may be simpler open books. For example, it is known that on any $3$-manifold there exist open books with genus-$0$-pages or open books with only $1$ binding component.

I could imagine that one can construct open books on general $S^1\times \Sigma$ similar to that one above. Note, that in the open book of $S^1\times T^2$ we get one binding component for every handle of the surface $T^2$ in $S^1\times T^2$. If one starts with a genus-$g$ surface presented as a $2g$-gon with edges identified, one can probably draw a similar open book.