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CP^2 , and CP^2 -bar, are not diffeomorphic.

  1. Apr 21, 2010 #1


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    Hi, Topologists/Geometers:

    I am trying to show that CP^2 and CP^2-bar (meaning CP^2 with reverse choice of
    orientation, are not diffeomorphic.

    I am kind of rusty; all I can think of using is the intersection form, but I do not
    remember the precise result that we can use to show this. I have checked Rohklin's
    theorem and other results in intersection forms, without success.

    I would appreciate your suggestions.
  2. jcsd
  3. Apr 21, 2010 #2
    Way #1: Remember H^2(CP^2,Z) = Z, call the positive generator a, and the volume form on CP^2 is a positive multiple of a ^ a. Then f*(a)=ca, which implies f*(a^a)=c^2 a^a. Then c^2 can't possibly equal negative one.

    Way #2: CP^2 freely generates the oriented cobordism group in dimension 4. If CP^2 admits an orientation reversing diffeomorphism, it's 2-torsion in the cobordism group, contradiction.
  4. Apr 22, 2010 #3

    Thanks Again. I wonder if this would also work: ( I will try to be more succint this time.)

    We know that the intersection form for CP22 is <1>, and that for CP2^ (defined as CP2 with reversed orientation is <-1>.

    This means that surfaces intersect positively , and at single points in CP2, and
    surfaces intersect negatively at single points in CP2^.

    Now assume there is a diffeomorphism f: CP2 -->CP2^

    Then f is either i) orientation-preserving, or ii)orientation-reversing.

    If we have:

    i) , then the image under f of both homology 2-surfaces will still intersect

    positively. If we have

    ii), then the image under f of the 2-surfaces will also end up intersecting

    positively at one point.

    In neither case can the 2-homology surfaces intersect negatively at one point

    the way they do in CP2^.

    Does that Work.?
  5. Apr 22, 2010 #4
    You have to be careful (why does an orientation-reversing map reverse signs on homology? it doesn't in general), but in this case, you're translating my way #1 via Poincare duality.
  6. Apr 27, 2010 #5
    Zhentil: I am a little confused here: are you using some sort of deRham duality to construct a volume form by using the fundamental class.?. Otherwise, how do we go from a fundamental class in H_2(CP2,Z) , to a top form , i.e., a form in
    H4(CP2;Z).?. The fundamental class is , AFAIK, a purely
    topological object, while a volume form assumes a differentiable structure.

    I hope this is not too far off.
  7. Apr 28, 2010 #6
    The general answer is because CP^2 is Kahler. The more specific reason (and the reason this question is simple to answer) is because the cohomology ring of CP^2 is a truncated polynomial ring on one generator (i.e. a in H^2, a^3 = 0).
  8. May 3, 2010 #7
    Wow, you blew me away there, Zhentil. Are you using Chern classes or something.?
    My pithy one semester of algebraic topology cannot deal with that.

    If I am too far off, just ignore it.
  9. May 4, 2010 #8
    Based on one semester of algebraic topology, you can prove that the cohomology of CP^2 is Z in the even dimensions and 0 otherwise (by using the cell structure or Mayer-Vietoris). The only other bit of information comes from the fact that the generator of the top cohomology is actually the square of the generator of the middle cohomology. This is where the Kahler comment comes in: if you have a Kahler form, its top power actually gives you a volume form.
  10. Sep 2, 2011 #9
    I came across this thread the other day and thought you might be
    interested in the fact that exactly this point is discussed in the review by Eguchi, Gilkey and Hanson, Physics Reports, vol 66 (1980) 213. On page 320 they discuss the Dolbeault complex and show that the index is only integral if one chooses the correct oriantation.
  11. Sep 2, 2011 #10
    Thanks, Dolan, I'll look it up.
  12. Sep 2, 2011 #11
    Sorry, I meant page 330, not 320!
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