There are two ways to stabilize a Legendrian knot $k$ in standard contact sphere $(S^3,\xi_{st})$ i.e. adding right cusps or left cusps, let's call these two stabilized Legendrian knots $k_R$ and $k_{L}$ respectively. If the knot $k$ is embedded on the page of an open book $(\Sigma,\phi)$ of $(S^3,\xi_{st})$, then $k_R$ and $k_L$ can also be embedded on page of an open book obtained by positive stabilization of $(\Sigma,\phi)$. I want to know what are the monodromies of the two new open books relative to $(\Sigma,\phi)$.

I know they cannot be equivalent, since $(-1)$-contact surgery on each of the stabilized knots gives two non-homotopic contact structures (their first chern class is different), so the contact structures are not isotopic and therefore the open books cannot be equivalent.

  • $\begingroup$ What kind of relationship are you looking for? For example, the stabilized page and monodromies can be written down in terms of the original one. $\endgroup$ – Kyle Hayden Feb 1 '16 at 22:03
  • $\begingroup$ I edited the problem, that's what I want to know, how to write down the two monodrmies. $\endgroup$ – nikita Feb 1 '16 at 22:08
  • $\begingroup$ I believe this is done in Section 3 of Ozbagci's "A Note on Contact Surgery Diagrams" (home.ku.edu.tr/~bozbagci/diagram.pdf). I know I've seen it elsewhere, though. $\endgroup$ – Kyle Hayden Feb 1 '16 at 22:46

As you mentioned, $k_L$ and $k_R$ both live in a stabilisation of the open book $(\Sigma,\phi)$.

Namely, suppose you have the triple $(\Sigma,\phi,k)$, where now I think of $k$ as an embedded, nonseparating curve in $\Sigma$. Recall that $k$ is oriented (in order to make sense of "left" and "right" stabilisations), therefore it makes sense to consider a path $\gamma$ that goes from $k$ to $\partial \Sigma$ to the left of $k$ or to the right of $k$, to a given point $p\in\partial\Sigma$.

Once you've chosen such a path, stabilise $(\Sigma,\phi)$ near $p$, choosing a stabilisation arc that is boundary-parallel. There is a natural way to make $k$ go through the new handle, running along $\gamma$ and than back along $\gamma^{-1}$.

The choice of a left/right path $\gamma$ determines whether the stabilisation is a left or a right stabilisation (although the standard terminology is positive vs negative).

I'm not 100% sure that a left path corresponds to a left stabilisation in your language, but there's a 50% chance that I'm right. In any event, I have learnt this from Lisca-Ozsváth-Stipsicz-Szabó (the definition of the LOSS invariant in Heegaard Floer homology), and they credit this to John Etnyre, in his (extremely valuable) Lectures on open book decompositions and contact structures.

  • $\begingroup$ Note that if you do two stabilisations, one "on the left of $k$" and one "on the right of $k$", you arrive at a single open book on which each of $k_L$ and $k_R$ can be Legendrian realised. $\endgroup$ – magicker72 Feb 3 '16 at 18:05
  • $\begingroup$ That is trickier than you make it sound, though; what I understand from your wording is that you have a double stabilisation in the page. $\endgroup$ – Marco Golla Feb 3 '16 at 21:34
  • 1
    $\begingroup$ Indeed, you can Legendrian realise a double stabilisation on the page. However, if you push $k$ over only one of the two handles, you can realise a single stabilisation (and which one depends on which handle you push over). $\endgroup$ – magicker72 Feb 3 '16 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.