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Vector Fields on S^2, 0 only at 1 point.

  1. Feb 27, 2009 #1


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    Hi, everyone:

    I am trying to produce a V.Field that is 0 only at one point of S^2.

    I have been thinking of using the homeo. between S^2-{pt.} and

    R^2 to do this. Please tell me if this works:

    We take a V.Field on R^2 that is nowhere zero, but goes to 0

    as (x,y) grows (in the sense that it "goes to oo" in the Riemann sphere), and

    then pulling it back via the stereo projection.

    We could use, e.g:

    V(x,y)=( 1/(X^2+1), 1/(Y^2+1))

    For the pullback. Does this work?

  2. jcsd
  3. Feb 28, 2009 #2
    Yes, though a priori this vector field will only be continuous. To check differentiability you need a chart containing the zero.
  4. Mar 4, 2009 #3


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    Thanks, YYat, a followup, please:

    Is it then possible to use this technique of pullbacks of vector fields (seen as sections)
    to show that orientability is a topological (maybe "diffeo-topological") property?.

    Specifically, I was thinking of using a diffeomorphism between M orientable and smooth
    and N smooth. Then we could pullback a nowhere-zero form w in M into a nowhere-zero
    form f_*(w) in N. Would this work?
    Is it true that orientability is a topological property (i.e., if X,Y are homeo. and X
    is orientable. Is Y also orientable?)

    Thanks Again.
  5. Mar 4, 2009 #4
    There are several different, but equivalent definitions of orientability (see also the http://en.wikipedia.org/wiki/Orientability#Orientability_of_manifolds" article). For a smooth n dimensional manifold the existence of a nowhere vanishing n-form can be taken as a definition, and a diffeomorphism pulls back nowhere vanishing n-forms to nowhere vanishing n-forms. So indeed, orientability is a diffeomorphism invariant.
    However, orientability can also be defined for topological manifolds (these have charts, but the chart transitions need not be differentiable), quoting wikipedia:

    "An n-dimensional manifold is called non-orientable if it is possible to take the homeomorphic image of an n-dimensional ball in the manifold and move it through the manifold and back to itself, so that at the end of the path, the ball has been reflected."

    Other, less intuitive but often easier to work with definitions can be made using homology theory. With a topological definition, it is immediate that orientability is in fact a topological (homeomorphism) invariant.
    Last edited by a moderator: Apr 24, 2017
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