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

[WWGD]

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.

 

[Terandol]

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 \mathbb{C}P^4 and \mathbb{C}P^2. I believe this results in \chi(\mathbb{CP^4})=5 but \chi(\mathbb{CP^2}\times \mathbb{CP^2})=\chi(\mathbb{CP^2})\chi(\mathbb{C}P^2)=9 so the homology groups of these two spaces must differ.

 

[WWGD]

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 ?

 

[mathwonk]

without any computation, it seems naively that the homology of C^4 has one generator in degree 2, whereas the product space has two.

 

[Terandol]

Still, I would like to review cup products; would you mind checking my work ?

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, H^{\bullet}(\mathbb{C}P^4,\mathbb{Z}) =\mathbb{Z[\alpha]}/(\alpha^5) and by the Kunneth formula 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) (using the standard grading on the tensor product and where \alpha has degree 2 in both cases) so you can tell from this what all the cup products have to be.