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Can't comb hair on a 2-sphere. On a Calabi Yau manifold?

  1. Nov 12, 2014 #1
    Does it make sense to ask if we can "comb the hair" on a Calabi yau manifold?

    Hope this is the right place for this question.

    Thanks for any help!
     
  2. jcsd
  3. Nov 12, 2014 #2
    The proper way to formulate the question for a general manifold, and hence in a way that makes sense for Calabi Yau manifolds in particular, is to ask if there exists a nowhere vanishing section of the tangent bundle (ie. a a nowhere vanishing vector field.) This is always true for a Calabi Yau manifold though since one way to define a Calabi Yau manifold is as a Kahler manifold on which there exists a nowhere vanishing holomorphic one form. Using a Hermitian metric, this implies the existence of a nowhere vanishing holomorphic vector field so the hair on Calabi Yau manifolds can always be combed.
     
  4. Nov 13, 2014 #3

    Ben Niehoff

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    Not quite. A Calabi-Yau manifold is a complex manifold with trivial canonical bundle, which is equivalent to saying it has a nowhere-vanishing holomorphic top form. That is, if the CY manifold has ##n## complex dimensions (or ##2n## real dimensions), then there is a nowhere-vanishing holomorphic ##n##-form. Yet another equivalent statement is that it has a covariantly-constant spinor; the holomorphic top-form is then constructed as a bilinear of this spinor (and I am pretty sure it is the only non-vanishing bilinear, in general).

    So the Calabi-Yau condition does not imply the existence of a nonvanishing vector field.

    What *does* imply the existence of a nonvanishing vector field is having a nonzero first Betti number ##b_1##. The simplest non-trivial Calabi-Yaus are the K3 surfaces. Looking at the Hodge diamond of a K3 surface on the Wiki page:

    http://en.wikipedia.org/wiki/K3_surface

    I see that the Hodge numbers ##h^{1,0} = h^{0,1} = 0##, and hence ##b_1 = h^{1,0} + h^{0,1} = 0##. Thus one cannot comb the "hair" of a K3 surface.
     
  5. Nov 13, 2014 #4

    Ben Niehoff

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    A further note: in 1 complex dimension (i.e. 2 real dimensions), a Calabi-Yau is just a torus. The holomorphic top form is a 1-form, and as Terandol points out, it implies the existence of a nowhere-vanishing vector field. In fact, there are two nowhere-vanishing vector fields, because you have two nowhere-vanishing 1-forms: the holomorphic top-form and its complex conjugate. As expected, the first Betti number of the torus is 2, and it should be obvious how to visualize the two linearly-independent nonvanishing vector fields.
     
  6. Nov 13, 2014 #5
    Much for the brain to chew on and digest! Thanks to you both!
     
  7. Nov 13, 2014 #6

    lavinia

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    There is a general theorem that says that an orientable smooth compact manifold without boundary has a vector field without zeros if only if its Euler characteristic is zero.

    The Euler characteristic can be computed by integrating the Euler class over the manifold. If the tangent bundle has a complex structure as in Calabi-Yau manifolds,then the Euler class is the top dimensional Chern class.

    According to the Wikipedia article,for K3 surfaces, the second Chern class- which is the top dimensional Chern class for these manifolds- integrates to to 24 so no K3 surface can be combed.
     
    Last edited: Nov 13, 2014
  8. Nov 13, 2014 #7
    Yeah my first post was nonsense.

    I'm not sure I understand this. Do you mean this is true only in the special case of a complex manifold (or maybe even more specifically a Calabi-Yau manifold)? It seems to me for an arbitrary real manifold the vanishing of the first betti class does not control the existence of a nonvanishing vector field either way. For example, a compact genus [itex] g>1[/itex] surface has [itex] b_1=2g \neq 0 [/itex] but it can't have a nonvanishing vector field since it's Euler characteristic [itex] 2-2g\neq 0 [/itex] but by the Hopf index theorem a nonvanishing vector field implies the euler characteristic is zero

    Similarly any odd-dimensional (>1) sphere shows that [itex] b_1=0 [/itex] does not imply a nonzero vector field can't exist.

    Edit: Looks like lavinia beat me to it and more or less answered this already.
     
  9. Nov 13, 2014 #8

    lavinia

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    I know nothing about Calabi-Yau manifolds and was asking the same question. Let me add on to your points.


    - Every surface of genus 2 or more is a complex manifold( Riemann surface ) and has non-zero first betti number. Yet all have non-zero Euler characteristic. All have non-zero holomorphic 1 forms.

    So it seems that it is special to Calabi-Yau manifolds. But maybe there is a less restrictive condition.
     
  10. Nov 13, 2014 #9

    Ben Niehoff

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    Ack, you're right. The first Betti number counts the number of linearly-independent 1-cycles. Somehow I got it in my head that having 1-cycles had something to do with being comb-able.

    The Euler number is the correct topological invariant to use (and it is equal to the alternating sum of the Betti numbers).

    I don't think all Riemann surfaces have non-zero holomorphic 1-forms, though. In fact, they must not, unless their Euler characteristic vanishes. So that leaves only the torus. Since another equivalent Calabi-Yau condition is Kahler + Ricci-flatness, this had better be the case.
     
  11. Nov 13, 2014 #10

    lavinia

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    Right. I meant holomorphic 1 forms that are not everywhere zero.
     
  12. Nov 18, 2014 #11
    Putting pieces together from two sources it appears a Calibi-Yau manifold can be combed?

    From, http://universe-review.ca/R15-26-CalabiYau01.htm#Kahler

    "The first Chern class is vanishing (or zero) if all the tangent vectors on a manifold can be oriented to the same direction. "

    And from, http://books.google.com/books?id=b3R31u-HbcIC&pg=PA61&dq=kahler manifolds differential geometry&hl=en&sa=X&ei=FflrVKOPA8uxsATc9oHoAw&ved=0CEgQ6AEwBw#v=onepage&q=kahler manifolds differential geometry&f=false

    "A Calibi Yau manifold is defined as a kahler manifold with a vanishing first Chern class."

    Thanks for help!
     
  13. Nov 19, 2014 #12

    lavinia

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    As has been explained above, a manifold can be combed if and only if its Euler class is zero. This is not the same as the first Chern class equal to zero unless the complex manifold is a Riemann surface.

    A simply connected Ricci flat Kahler n- manifold has an everywhere non-zero holomorphic volume form. I would guess that this is the orientability property referred to in your quoted sentence.

    Orientability and combability are not the same. Orientability implies that there is an Euler class for the vector bundle. More generally a sphere bundle that is orientable has an Euler class. But combability means that the sphere bundle has a section and for this the Euler class must be zero.
     
    Last edited: Nov 20, 2014
  14. Nov 20, 2014 #13
    Thank you for taking time to point out my error!
     
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