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Non-homeomorphis between CP^4 and CP^2 x CP^2

  1. Nov 12, 2014 #1

    WWGD

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    Hi, I am trying to show that ## \mathbb CP^4 ## and ## \mathbb CP^2 \times \mathbb CP^2 ## are not homeomorphic. None of the standard methods --comparing the fundamental group of the product with the product
    of fundamental groups, nor (co)homology seem to work. So I am trying to work with the cup products and
    show these are different. Please let me know if this is correct: we need to compute the cohomology groups on
    each side, and then we just compute all possible cup products using Kunneth's formula. Is this correct?
    Thanks.
     
  2. jcsd
  3. Nov 12, 2014 #2
    Yeah computing all cup products by hand will certainly yield a proof but I don't think you need to do anything more complicated than compute the homology groups. The simplest idea that comes to mind is just to compute the Euler characteristics which we know are multiplicative so you only need to know the homology of [itex] \mathbb{C}P^4[/itex] and [itex] \mathbb{C}P^2 [/itex]. I believe this results in [itex] \chi(\mathbb{CP^4})=5 [/itex] but [itex] \chi(\mathbb{CP^2}\times \mathbb{CP^2})=\chi(\mathbb{CP^2})\chi(\mathbb{C}P^2)=9[/itex] so the homology groups of these two spaces must differ.
     
  4. Nov 13, 2014 #3

    WWGD

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    Thanks, Terandol, you're right, that is a nice, simple solution. Still, I would like to review cup products; would you mind checking my work ?
     
  5. Nov 13, 2014 #4

    mathwonk

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    without any computation, it seems naively that the homology of C^4 has one generator in degree 2, whereas the product space has two.
     
    Last edited: Nov 13, 2014
  6. Nov 13, 2014 #5
    I can take a look at your work if you would like but I'm probably the wrong person to ask about doing actual hands on computations...I usually completely mess it up the first few times I try.

    It may be helpful to check your work to observe that as graded rings, [itex] H^{\bullet}(\mathbb{C}P^4,\mathbb{Z}) =\mathbb{Z[\alpha]}/(\alpha^5) [/itex] and by the Kunneth formula [itex] H^{\bullet}(\mathbb{C}P^2\times\mathbb{C}P^2,\mathbb{Z})=\mathbb{Z}[\alpha]/(\alpha^3)\otimes_{\mathbb{Z}} \mathbb{Z}/(\alpha^3) [/itex] (using the standard grading on the tensor product and where [itex]\alpha[/itex] has degree 2 in both cases) so you can tell from this what all the cup products have to be.
     
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