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Let's define a graph manifold as 3-manifold which is obtained by plumbing of a circle bundle over $\Sigma _g$ (with Euler number 0) with a circle bundle over $\Sigma_h$ (with Euler number k). Is there any way to find an open book decomposition of this manifold explicitly?(I think this is the same as asking how to switch between a spinal open book decomposition of a 3-manifold and an open book decomposition of it)

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By plumbing do you mean the boundary of the plumbing of the associated disk bundles? If not what do you mean? And are the 3-manifolds surface bundles over the circle? Otherwise I don't see what is "the monodromy of these two surface bundles". – BS. Aug 11 '13 at 17:05
BS: I edited the statement of problem. – nikita Aug 11 '13 at 17:15
up vote 1 down vote accepted

One can produce open books for graph manifolds in basically the same way that one produces an open book when cabling a fibered knot. Let's begin there.

With a fibered knot you've got a torus $T$ that separates a $D^2 \times S^1$ neighborhood of the knot from a surface bundle over the circle. Near $T$, these both look like fibrations that each meet $T$ in different slopes. An essential simple closed curve in $T$ with slope different from these two gives the binding for an open book that `interpolates' between $D^2$ fibers and surface fibers. Outside a neighborhood of $T$, the pages of this open book are comprised of $D^2$ fibers and surface fibers. Further details of such a construction are in for example. Here's a picture of this sort of thing:

This works equally well when $T$ (or a collection of tori) decomposes your manifold into surface bundles over the circle whose fibers meet in different slopes on $T$, like when you have a spinal open book. Do this cabling procedure along all your decomposing tori with a bit of care, and you'll get an open book.

For a graph manifold where $T$ decomposes the manifold into circle bundles (or more generally Seifert fibered spaces) you just need to further decompose along tori that cut off $D^2 \times S^1$ neighborhoods of enough circle fibers until the resulting pieces are all surface bundles. In this case you'll end up with the pieces all being a surface times a circle. Do this sort of cabling procedure again.

(And I reckon if the base space of one of your Seifert fibered spaces is non-orientable, you'll have to work a wee bit harder. But it's doable.)

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Thank you, that is exactly what I was vaguely thinking about. – nikita Aug 12 '13 at 1:54
You're welcome. – Ken Baker Aug 12 '13 at 2:11
Ken Baker: I think the monodromy calculations which you do in the above paper, can be carried over to the spinal open book case to calculate the monodromy of the new open book, because the the center disks (to which we attach handles of a genus h surface to get $\Sigma _h \times S^1$ instead of $D^2 \times S^1$) do not appear in the monodromy calculation (and you just have to take care of the rotation map). Is that right? – nikita Aug 18 '13 at 19:55
That sounds right. – Ken Baker Aug 23 '13 at 13:57

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