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Transversality of a Vector Field in terms of Forms (Open Books)

  1. Dec 28, 2013 #1

    WWGD

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    Hi, All:
    Sorry for the length of the post, but I think it is necessary to set things up so that the post is understandable:

    I'm going through an argument in which we intend to show that a given vector field [ itex]R_ω [/ itex]
    (actually a Reeb field associated with a contact form ω) is positively-tangent to a
    link ( as in a space whose connected components are [itex] S^1 [/itex] -knots. This means the Reeb field
    lives in the tangent space to the Link, along the positive direction) , and [itex] R_{\omega}[/itex] is positively-
    transverse to a surface S ( so that [itex]R_{\omega}[/itex] intersects S positively at points).
    The argument is made using properties of differential forms, in the context of open book decompositions of contact 3-manifolds.
    A needed definition is this: A contact structure ζ is _supported by an open book decomposition (B, π )_ if ζ can be isotope thru contact structures so that there is a contact 1-form ω for ζ satisfying:

    1)dω is a positive area form on each page [itex]∑_{\theta}[/itex] of the open book, and:

    2)ω>0 on the binding B ; both B and the pages are oriented.

    End of setup.

    ------------------------------------------------------------------------------
    Actual Question:

    To be more specific, I'm trying to understand the following arguments purporting to show
    the equivalence between these two conditions:

    (1) The contact manifold (M,ζ ) is supported by the open book (B,π)

    (3) There is a Reeb field [itex]R_ω[/itex] for a contact structure isotopic to ω , so that [ itex]R_ω[/itex] is positively-tangent to the binding B, and positively-transverse to the pages of the open book.

    Proof:
    (3)->(1) : Since [itex] R_ω[/itex] is assumed positively -tangent to the binding B , we have ω>0 on oriented tangent vectors to B. Since the Reeb field [itex]R_ω [/itex] is positively-transverse to the pages of the OB (open book) , we have that [itex]dω=i_{R_ω}[/itex](ω /\ dω) >0 on the pages of the OB (where i is --I am? -- the interior product , or contraction of the form ω /\ dω by the vector field
    [itex] R_ω[/itex]

    Questions:
    i)How does [itex]R_{ω}[/itex] being positively-tangent to the ( knots in the ) binding imply ω >0 ?
    I know this means the vector field being positively-tangent to the binding means that [itex]R_{ω}[/itex] lies along the tangent space ; a 1-d tangent space, to each of the knots, along the chosen positive direction orientation.

    ii)Why is dω equal to the contraction of ω /\ dω ? , and how does the positive transversality imply that dω>0?

    (1)->(3): Assume (1), and let ω be the form with the given conditions, and let [itex]R_{ω}[/itex] be the Reeb field associated with ω. Then "It is clear that [itex]R_{ω}[/itex] is positively-transverse to the pages of the OB, since dω is an area form on the pages of the open book"

    I have no clue about the connection between the Reeb field being positively-transverse to the pages, and dω being an area form on the pages. I know if [itex]dw[/itex] is a positive area form on the pages, then dω (X,Y)>0 at any pair of positively-oriented tangent vectors. And I know a Reeb field associated to a contact structure is transverse to the planes in the contact structure .But I can't see how this relates to $dω being an area form for the pages of the open book.

    Thanks for any suggestions, ideas.
     
    Last edited: Dec 28, 2013
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