1

let $S$ be a compact surface with non-empty boundary, let $\alpha : [0,1] \rightarrow S$ be a Properly-embedded arc (meaning both endpoints of the arc are in $\partial S$) and let $f$ be an element in $MCG(S)$, the mapping class group of $S$. Let $\alpha(0)=x$. I am trying to understand the geometric motivation behind the property of "right-veeringness" of $f$. We say that $f$ is right-veering if the orientation given at the point $x$ by the pair $(f'(\alpha(0)) ,\alpha'(0))$ agrees with the orientation of the surface $S$ at the point $x$. Any ideas ?

RikOsuave
  • 337
  • 1
  • 9
  • This first showed up in Honda-Kazez-Matic relating to tight contact structures. The basic idea is that $f$ not right-veering allows you to find an overtwisted disk in the contact manifold associated to the open book with monodromy $f$. Very roughly, this is because the way the contact structure associated to the open book works involves "continuing to twist further" in the right-veering way as you go closer to the binding. So non-right-veering means there'd have to be "extra twisting" and an overtwisted disk, meaning the contact structure is not tight. – aes Dec 21 '14 at 07:28
  • Thanks; that is helpful. HKM is where I read it, actually. Problem is I don't have much geometric intuition for the contact structure associated to, or supporting, an open book. I know the formal definition, but I don't have a good geometric feel for it yet, but I will look into it more carefully. – RikOsuave Dec 21 '14 at 07:36
  • Also, there's something missing in your definition. Read the definition on page 2 of the paper I linked (and the preceding discussion). You need to either work in the universal cover and homotop lifts of $\alpha$ and $f(\alpha)$ so they don't intersect before checking the condition or work on the surface and homotop them so they intersect minimally. Also, $f$ is right veering if for all $\alpha$ this is true. – aes Dec 21 '14 at 07:37
  • But why do we need to isotope $\alpha, f(\alpha)$ into not intersecting, how would that change $(f'(\alpha(0)), \alpha'(0))$. Besides, can we always isotope $\alpha$ into being differentiable at $0$? – RikOsuave Dec 21 '14 at 07:41
  • The condition is that the arc to $f(\alpha)$ starts out to the right of $\alpha$ at the boundary (i.e. at $0$). This isn't robust under homotopy, so the correct condition is to first homotop $f(\alpha)$ to intersect $\alpha$ minimally, and then see if it starts out heading rightward. (It would have to come across the arc an extra time if you change which direction is starts, and this is a well-defined concept.) Maybe this was actually your question. I'll paste these all into an answer in case we need to edit/discuss further. – aes Dec 21 '14 at 07:45

1 Answers1

1

For the overall definition: You need to either work in the universal cover and homotop lifts of $\alpha$ and $f(\alpha)$ so they don't intersect before checking the condition or work on the surface and homotop them so they intersect minimally. Then, $f$ is right veering if for all $\alpha$ the condition is true.

The condition: The condition is that the arc to $f(\alpha)$ starts out to the right of $\alpha$ at the boundary (i.e. at $0$). (That is, the initial tangent directions $(f \circ \alpha)'(0)$ and $\alpha'(0)$ form an oriented basis.) But this isn't robust under homotopy (or rather, isotopy of $f$ fixing the boundary), so the correct condition is to first homotop $f(\alpha)$ to intersect $\alpha$ minimally, and then see if it starts out heading rightward. (It would have to come across the arc an extra time if you change which direction it starts, so this is a well-defined concept.)

This first showed up in Honda-Kazez-Matic relating to tight contact structures. The basic idea is that $f$ not right-veering allows you to find an overtwisted disk in the contact manifold associated to the open book with monodromy $f$. Very roughly, this is because the way the contact structure associated to the open book works involves "continuing to twist further" in the right-veering way as you go closer to the binding. So non-right-veering means there'd have to be "extra twisting" and an overtwisted disk, meaning the contact structure is not tight.

aes
  • 7,803
  • Thanks, that was helpful. I will look into it more carefully. – RikOsuave Dec 21 '14 at 08:07
  • Thanks, I have a third one you may be able to help me with :http://math.stackexchange.com/questions/1076373/universal-cover-of-a-surface-with-boundary-what-does-cantor-set-on-boundary-cor – RikOsuave Dec 21 '14 at 08:09