
#1
Apr2110, 02:32 AM

P: 394

Hi, Topologists/Geometers:
I am trying to show that CP^2 and CP^2bar (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
Apr2110, 02:02 PM

P: 491

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 2torsion in the cobordism group, contradiction. 



#3
Apr2210, 01:51 PM

P: 662

Zhentil:
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 CP^{2}2 is <1>, and that for CP^{2}^ (defined as CP^{2} with reversed orientation is <1>. This means that surfaces intersect positively , and at single points in CP^{2}, and surfaces intersect negatively at single points in CP^{2}^. Now assume there is a diffeomorphism f: CP^{2} >CP^{2}^ Then f is either i) orientationpreserving, or ii)orientationreversing. If we have: i) , then the image under f of both homology 2surfaces will still intersect positively. If we have ii), then the image under f of the 2surfaces will also end up intersecting positively at one point. In neither case can the 2homology surfaces intersect negatively at one point the way they do in CP^{2}^. Does that Work.? 



#4
Apr2210, 01:57 PM

P: 491

CP^2 , and CP^2 bar, are not diffeomorphic.
You have to be careful (why does an orientationreversing map reverse signs on homology? it doesn't in general), but in this case, you're translating my way #1 via Poincare duality.




#5
Apr2710, 11:41 PM

P: 662

H_{4}(CP^{2};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. 



#6
Apr2810, 06:55 AM

P: 491

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).




#7
May310, 11:40 PM

P: 662

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. 



#8
May410, 06:31 AM

P: 491

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 MayerVietoris). 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.




#9
Sep211, 10:55 AM

P: 2

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. 



#10
Sep211, 11:08 AM

P: 662

Thanks, Dolan, I'll look it up.




#11
Sep211, 11:12 AM

P: 2

Sorry, I meant page 330, not 320!



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