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It is well-known that the only genus one fibered knots are the trefoil and the figure-eight. On the other hand, there exist infinitely many fibered links for any fixed higher genus.

My question is about what happens if we go to more boundary components: fixing the number of boundary components, are there infinitely many genus one fibered links?

Another related question: the monodromy of a fibered genus one link corresponds, after collapsing the boundary, to an element of $SL_n(\mathbb Z)$. For example the trefoil yields $((0,1),(-1,1))$ and the figure-eight $((2,1),(1,1))$. Which conjugacy classes of $SL_n(\mathbb Z)$ can be obtained as monodromies of many-components genus one fibered links?

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2 Answers 2

This will not answer your questions explicitly, but perhaps give a hint on how to go about thinking about them. In his paper, How to construct all fibered knots and links (Topology 21 (1982), no. 3, 263–280, MR0649758), John Harer explained how to do just that: start from the unknot, and perform a finite sequence of Hopf plumbings/deplumbings and Stallings twists. Adding a Hopf band increases the genus of the fiber by 1, and multiplies the monodromy by the corresponding Dehh twist; performing a Stallings twist along a suitable curve leaves the genus unchanged, but again composes the monodromy with a certain Dehn twist. Starting from a known fibered $n$-component link (such as the one consisting of $n$ fibers of the Hopf map), and performing an arbitrarily long sequence of Stallings twists should give you at least some candidate to answer the first question, and perhaps a starting point towards answering the second one.

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Thanks! The Stallings twist seems a good idea, I will think about it. –  Pierre Dehornoy Apr 11 '13 at 13:01
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This works for the Borromean rings, which is a positive genus 1 link, and therefore fibered. Each boundary component is unknotted, and since the linking numbers are zero, the induced framing of the Seifert surface is the longitude. So one may perform Dehn twists around a single boundary component to get an infinite family of fibered genus 1 links (in fact with the same complement). –  Ian Agol Apr 11 '13 at 20:42

Agol's comment works to give 3 component fibered links with genus 1 fiber, but we can dial it back a bit further.

The connect sum of a Hopf band and its mirror is a 3-component link of unknots with genus 0 fiber. Moreover one component is 0-framed so Dehn twists along it will give an infinite family of 3 component fibered links with genus 0 fiber. So to get an infinite family of 2 component links with genus 1 fiber, you can just plumb on another Hopf band along an arc joining the other two boundary components. This plumbing will be disjoint from the twisting.

I think the Alexander polynomials of these links will distinguish them, perhaps even the Conway polynomials.

It would be interesting to see the Hopf stabilization equivalences in these families.

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Thanks Ken. Along the same line, one can start from a trefoil and plumb a Hopf band. This can be done in several ways, depending on a joice of a segment on the punctured torus, that is, a rational number. The monodromy in this case is the product of a matrix of trace 1 and a matrix of trace 2, and we can obtain many conjugacy classes in this way (I think at least one per trace). If we start from a figure-eight, it works in the same way, except that we obtain the product of a matrix of trace 3 with a matrix of trace 2. –  Pierre Dehornoy Apr 23 '13 at 17:28

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