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Dividing Sets in Contact Structures, and Induced Orientations

  1. Mar 14, 2014 #1


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    Hi everyone, a couple of technical questions :

    1) Definition: Anyone know the definition of the induced orientation of a submanifold S of an orientable manifold M?

    2)Dividing sets in contact manifolds: We have a contact 3-manifold (M3,ζ ). We
    define a surface S embedded in M3 to be a convex surface if there exists a contact
    vector field X that is transverse to S, i.e., X does not live on the surface; X is not in the span of any basis for TpS. Now, we define the dividing set of the surface S to be the set of points of X that live in the contact planes , i.e., p is in the dividing set if X(p) is in ζ(p) ; ζ(p) is the contact plane at the point p, and X(p) is the contact vector field at p ( a contact vector field for (M3,ζ ) is a vector field whose flow preserves ζ , i.e., LXζ=gζ , where L is the
    Lie derivative of the form ζ along the vector field X, and g is a positive smooth function.

    So, say I have the standard contact structure in R3 given by ker(cos(πr)dx+sin(πr)dθ) . I know ∂/∂z is a contact field , so that it is transverse to any disk in the xy-plane. How do I find the dividing set in this case? I need to find the points in R3 so that ∂/∂z (p)
    ( basically, the z-axis "based at p " ) lies in the contact plane at p.

    I'm kind of stuck in a loop here; any suggestions, please ?
    Last edited: Mar 14, 2014
  2. jcsd
  3. Mar 14, 2014 #2


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    Well, it seems I may have to go thru the pain of deriving a ( basis for ) the contact structure explicitly:

    We set :

    (cosrdz+ rsinrdθ)(a∂/∂z+ b∂/∂θ)=0 ; the r variable is free. Then:

    acosr+ brsinr =0 , so that acosr = -brsinr

    So that { ( 0, b, -brtanr ), (c,0,0) }; c any Real number is a basis for the contact planes, and the vector field clearly has singularities at r=k∏. But now I have to figure out how to crank out the contact planes using the basis. Not hard, but pretty tedious, since this is a curved system ( it is cylindrical-coordinates-based ). Would be great if someone had a simpler approach.
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