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It is well-known that there exist pseudo-Anosov automorphisms of surfaces that act trivially on the homology: they form the Torelli group. Similarly there exists pseudo-Anosov automorphisms that act periodically.

On the other hand, given a fibered knot with pseudo-Anosov monodromy, the Alexander polynomial is the characteristic polynomial of the homological monodromy, and, at the same time, its degree is twice the genus of the fiber. This implies that no monodromy of a fibered knot lies in the Torelli group.

My question is: can the homological monodromy of a knot be periodic and the geometric monodromy pseudo-Anosov? My guess is no, and this is probably known, but I could not find a reference nor a proof.


Now I think that the answer is yes, see comment below!

An improved question could then be: can the homological monodromy of a knot be periodic, the geometric monodromy be pseudo-Anosov, and the fiber surface not contain an unlinked zero-framed link?

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up vote 4 down vote accepted

There exists an infinite family of quasipositive fibre surfaces of genus 3 with the same Alexander polynomial as the torus knot T(2,7). This answers Pierre's improved question, since quasipositive surfaces do not contain any essential unlinked annuli.

Let us first recall that the fibre surface S' of T(2,7) is obtained from the fibre surface S of the torus link T(2,6) by a single positive Hopf plumbing operation. The plumbing operation is determined by a relatively embedded arc J in S. Given that the genus of S is two, there exists an infinite family of pairwise non-isotopic embedded intervals in the relative homology class of J. Plumbing positive Hopf bands along those instead of J gives rise to an infinite family of quasipositive fibred knots with the same Seifert matrix. All of them have pseudo-Anosov monodromy, since there is only one fibred knot of genus 3 with periodic monodromy of order 14: the torus knot T(2,7).

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The simplest fibered knot with periodic homology monodromy is probably K = 11n74. Robert Riley dubbed this his "favorite knot" (see "Parabolic Representations of Knot Groups, I"). It is doubly slice, which means that K is the cross-section of an unknotted 2-sphere in 4-space.

The knot is a member of a family of knots obtained from the square knot by twisting along an unknot that is a push-off of a simple closed null-homologous curve C on the fiber surface. Since C is null-homologous, the Seifert linking form is unchanged.

Richard Hitt and I wrote about these knots in an paper, "Ribbon Knot Families via Stallings' Twists."

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My previous answer was flawed: the example I came up with was not unlinked.

The fibered knots with periodic monodromy are the torus knots. If one has a separating multicurve on the Seifert surface of a torus knot which is an unlink in $S^3$ and has zero framing, then Dehn twisting about it will yield fibered knots with periodic homological monodromy.

However, on a torus knot Seifert surface, such a multicurve cannot exist. The easiest case is for a separating curve on the Seifert surface. This cannot be unknotted, since this would give a lower 4-ball genus for the torus knot (in fact, the knot is quasi-positive).

If one has a separating multicurve which has framing zero, then for a non-separating component which is non-separating, one can surger the Seifert surface in the 4-ball to get a smaller genus surface, again a contradiction.

So this gives some positive evidence for your conjecture. One could attempt to analyze Giroux's classification of fibered links by Murasugi summing with Hopf bands, and see how the monodromy changes.

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You should have kept the older answer! – Mariano Suárez-Alvarez Apr 23 '13 at 23:27
The old answer claimed to have found such a link on the Seifert surface of a torus knot, which is clearly impossible by the new answer. I think you can look at old versions if you like. – Ian Agol Apr 24 '13 at 0:10
Thanks for your answer. Now I realize that Morton's construction (H. R. MORTON, 'Fibred knots with a given Alexander polynomial', Enseign. Math. 31 (1983), 205-222) actually gives counter-examples: Morton constructs infinitely many distinct fibered knots with prescribed Alexander polynomial (actually he starts from Burde's knots and twists). If we take his construction for a cyclotomic Alexander polynomial, then the homological monodromy has to be periodic (since diagonalizable), but infinitely many of Morton's knots will have pseudo-Anosov monodromy. – Pierre Dehornoy Apr 24 '13 at 0:30
Only idiots fail to contradict themselves three times a day, quoth Nietzsche. – Mariano Suárez-Alvarez Apr 24 '13 at 1:04

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