## Is there a table of (fibred knot) monodromies?

Background/motivation

I'm working on contact topology (in dimension three): a fundamental theorem of Giroux gives us a bijection between contact structures (up to isotopy) and open books (up to negative stabilisation).

Moreover, a stabilisation is a "hands on" operation both at the abstract level (i.e. surface with boundary and monodromy) and at the concrete level (i.e. fibred link in a three-manifold).

Whenever we have a fibred knot $K$ (say) in $S^3$, the corresponding fibration is an open book for $S^3$, which in turn supports a contact structure $\xi$ on $S^3$, where $K$ sits as a transverse knot.

Now that we have a good theoretical framework, we'd like to put our hands on some examples, and we can start stabilising the open books coming from the Hopf bands and obtain many fibred knots/links and their monodromies, and these examples are not too hard to identify as knots/links in $S^3$.

My question is purely topological: what about the inverse approach? I want to recover the monodromy on the fibre, and I have an "algorithm" to find it, once I have compressing discs for the complement of a fibre (that is a Seifert surface of minimal genus), but to me it's rather crafty. I figure that someone must have done it at some point, so:

Is there a table associating to each fibred knot the monodromy on its fibre?

I could find only one source giving concrete examples, namely Burde and Zieschang's Knots, where monodromies for the trefoil and the figure-eight knot are computed. Also, I'm pretty sure that there's much material available for algebraic knots, and that's as far as I've got.

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 The monodromies for torus knots are computed in several sources and is perhaps easier to do yourself than to look up. I suggest treating the fibre as a horizontal incompressible surface for the Seifert fibring. Then look at how such horizontal incompressible surfaces are constructed (look in Hatcher or Jaco's notes on 3-manifolds) and deduce the monodromies from that. I believe the action is free of order $pq$ on all but two orbits, which have isotropy groups of order $p$ and $q$ respectively. – Ryan Budney Mar 20 2011 at 17:45 I'm less certain of any existing tables of monodromy for hyperbolic knots that fibre over $S^1$ -- the primary problem being, how would you index such knots? – Ryan Budney Mar 20 2011 at 18:04 I was (naively?) hoping for something that could make a good entry in Livingston's KnotInfo or in Bar-Natan's Knot Atlas, using the standard enumeration of knots. – Marco Golla Mar 20 2011 at 19:16 In the absence of any reasonable algorithm for the conjugacy problem in the mapping class group, it's unlikely to be reasonable to tabulate this, as Sam Nead points out. One thing that would help in many cases is to note that the Murasugi band-sum operation changes the monodromy in a predictable way. – Dylan Thurston Apr 2 2011 at 19:52 After Sam's and Dylan's remarks, I realise I should've stated the question as "Is there a table associating to each fibred knot a monodromy on its fibre?". I still think that such a table would save some time, in some instances. @Dylan: The Murasugi band-sum is a possibility, but I feel like it's rather ad hoc and requires a lot of training/ingenuity. – Marco Golla Apr 5 2011 at 19:48

I've produced a table of monodromies for about 40% of the hyperbolic, fibred knots listed on knotinfo. This is available at: http://surfacebundles.wordpress.com/knot-complements/

This was done by producing a triangulation of every possible surface bundle over the circle for the surfaces $S_{1,1}, \ldots, S_{5,1}$ made from a composition of $2, \ldots, 12$ or fewer Dehn twists about generators. Non-hyperbolic and non-knot complement manifolds were discarded and for each pair of isometric triangulations the short-lex later one was also discarded. Finally, for each hyperbolic, fibred knot complement listed on knotinfo, SnapPy was used to find a bundle on this list isometric to it if it existed.

As Sam points out, there is no canonical choice of generating set for $\mathop{Mod}(S_{g,1})$ so I used the Humphries generating set in each case. However, the monodromies obtained are the short lex earliest for each knot with respect to this generating set and the ordering of the generators shown at the bottom of the page. This ordering was chosen to minimise the running time; a different ordering can run several orders of magnitude slower.

I should point out that these results don't show the millions of knot complements that were also found but that don't (yet) appear on the knot tables. This simply comes from the fact that my tables are ordered by monodromy length whereas knotinfo's is ordered by crossing number.

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 Thanks! That's exactly what I was looking for! Just a question: why did you discard non-hyperbolic knots? Also, I guess that $A$ represents $a^{-1}$ in your table. – Marco Golla Apr 19 2011 at 18:55 Yes, a capital letter represent the inverse twist. Non-hyperbolic ones had to be discarded as SnapPy really struggles to determine if two non-hyperbolic triangulations are isometric or not. I don't think that it is incapable of doing it, but I've always had it return "RuntimeError: SnapPea failed to determine whether the manifolds are isometric." - even for the trefoil knot. – qwerty1793 Apr 19 2011 at 19:53

I don't know of such a table. One serious problem in making one is choosing generating sets for the mapping class groups of the surfaces $S_{g,1}$. There is no canonical choice and I suspect no good choice.

Following Lickorish, one can cook up an algorithm that turns any monodromy into a product of Dehn twists, then into a product of twists in the Lickorish generating set, and then a product of twists the Humphries generating set. But remember that the monodromy only matters up to conjugacy -- thus finding the "best" representative, if such a thing, is somehow related to the conjugacy problem. It all sounds a bit dicey.

In practice, I would go to KnotInfo to find out if a knot in the tables was fibered. If there was a particular monodromy I wanted to compute I would look at Stallings theorem, compute the action of the monodromy on the fiber subgroup (which is a free group), and then transform that into a product of Dehn twists. If that didn't work, I would consult the experts, probably starting with Nathan Dunfield or Dylan Thurston. See also their paper "A random tunnel number one 3â€“manifold does not fiber over the circle" and its many references.

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