
#1
Feb709, 10:49 PM

P: 393

:
I hope this is not too dumb. I am kind of unclear on some issues: i) I understand that every bilinear map over a fin. dim. V space /F has a matrix representation (depending on the choice of basis ). Still, the intersection form ( in an orientable 4mfld M.) is a map: f: H^2(M;Z)xH^2(M;Z) >Z with: f(a,b) =(a\/b)[M] where M is a fundamental (orientation ) class , i.e., M is a generator of H^4(M;Z) ~ Z (since M is assumed orientable). Now: It is not clear that H^2(M;Z) is a vector space. It may be a ring, and therefore may not have a basis ( I think we consider rings as modules over themselves, and see if they are free or not. ). If H^2(M;Z) is not a ring: How do we represent the form above as a quadratic form?. ii) If we do have a representation: how do we use the relation between cap and cup (together with Poincare duality), to show that the cupping of H_2(M;Z)\/H_2(M;Z) is equivalent to the intersection number a.b , where a,b are the Poincare duals to the H^2(M;Z)'s ( I am working under the result that every class H^2(M;Z) can be represented by an embedded submanifold, and that the submanifolds can be disturbed if they do not intersect transversally ) Thanks For any Help. 



#2
Feb909, 08:11 AM

P: 707

However since the cohomology is Abelian you can still define the pairing in terms of a set of generators of the group. I assume that you mean only to use the nontorsion part because in part 2 of your question you use Poincare duality. Poincare duality over Z doesn't work unless the manifold is orientable over Z in which case cup product with torsion classes in the bilinear pairing will always give zero and is thus uninteresting. For part 2 you need to show that the Poincare dual to an orientable embedded submanifold of an orientable manifold is the Thom class of its normal bundle.You then show that the cup product of two Thom classes is the Poincare dual to the intersection manifold where you intersect after choosing representative manifolds of the two homology classes that intersect transversally. 



#3
Feb1109, 12:04 AM

P: 393

Thanks, Wofsy, that was helpful. A followup, please:
I have been trying to follow your advice. Something I am unclear on still, tho, is on what a torsion cycle/cocycle is. Since I have not done heavy homology/ cohomology in a while, I am kind of rusty; all I can remember is that in some cases, like when constructing a K(M,G) (an EilenbergMcLane space) , if we wanted G=Z/pZ , we would introduce torsion by introducing a relation to Z , like declaring that twisting a cycle p times would give us a zero. Something else I am a bit confused in what I have been looking for is the relation between simpleconnectedness and torsion : I see that if H_1(X,Z) is zero, then H_2(X,Z) has no torsion, but I am not too clear (not being clear on what a torsion (co)cycle is) on what is going on here. I wonder if we are using Hurewicz (since simpleconnectedness is assumed as including pathconnectedness. ) Hope the wheels will be betteroiled soon. 



#4
Feb1109, 12:39 AM

P: 393

Intersection Form in 4D. Followup.
I am back: just wanted to clarify that the algebraic torsion seems clear; depending
on the generators of the (co)homology. It is the geometric perspective that is not clear. Thanks Again. 



#5
Feb1109, 12:41 AM

P: 393

Sorry, I hit the 'Reply' button involuntarily: if you could please give me a source, ref.
for the concept. 



#6
Feb1109, 06:05 AM

P: 707

All homology and cohomology groups regardless of coefficients are abelian groups. Abelian groups have a simple structure. They are the direct product of a free Abelian group (direct product of copies of Z) and a torsion part which is a direct product of finite cyclic groups. Torsion just means that some integer multiple of the group element is zero. So in Z/2 twice any element is zero. In Z/3, three times any element is zero. The structure of Abelian groups has nothing to do with cohomology per se. It is a purely algebraic fact. The most elegant and wonderful way to learn this is to learn the structure theorem for modules of finite type over a principal ideal domain. Finitely generated Abelian groups are modules over Z. So a typical Abelian group will look like ZxZ...xZ xZ/n1x...xZ/n2. For instance ZxZ/2xZ/3. Or Z/6xZ/10xZ/2000. If a finite dimensional manifold is orientable over Z then its top dimension homology group is isomorphic to Z. Any top dimensional coholomology class is a Zlinear map of the top homology group into Z. So if this cohomology class is a torsion class it must be zero. For instance suppose this cohomology class is 2 torsion and call it S and call the generator of the top homology class x. Then 0 = (2S)(x) = 2S(x) (by Z linearity) But S(x) is an integer and the only integer whose double is zero is zero itself. Thus S is zero. I do not know why a simply connected manifold can not have torsion Zcohomology in dimension 2. In fact I don't believe it. Can you give me a proof? Regards wofsy 



#7
Feb1109, 06:56 AM

P: 707

In manifolds this can happen in any dimension. In dimension one the simplest example is the projection of half of a great circle onto the projective plane via the antipodal mapping ( the 2 fold cover of P2 by S2). This semigreat circle projects to a cycle in P2 because its end points are antipodal and are thus identified in P2. The entire great circle projects to twice this circle in P2. But the entire great circle is a boundary as is easily seen intuitively. Thus its projection, which is twice the original cycle in P2, is also a boundary. Further, P2 is a nonorientable manifold. This means that any attempt to form a Zcycle in dimension 2 fails. Its top homology over Z is zero. It is instructive to figure this out with diagrams. It all becomes clear. wofsy 



#8
Feb2809, 12:06 AM

P: 393

If a finite dimensional manifold is orientable over Z then its top dimension homology group is isomorphic to Z. Any top dimensional coholomology class is a Zlinear map of the top homology group into Z. So if this cohomology class is a torsion class it must be zero. For instance suppose this cohomology class is 2 torsion and call it S and call the generator of the top homology class x.
Then 0 = (2S)(x) = 2S(x) (by Z linearity) But S(x) is an integer and the only integer whose double is zero is zero itself. Thus S is zero. Wofsy: I was going over this post again, and I was a bit confused about something: I understand that, by duality of cohomology w.resp. to homology, every cohomology class can be represented by a Zlinear map from a cycle into Z. Still: what do you mean by a Zlinear map into Z in cohomology being torsion. Doesn't this imply (C is a cycle class rep.): S: [M]>Z ; M is the fundamental class. S(C)=/0 , but 2S(C)=0 doesn't this imply that there are no torsion elements in here, i.e., the only way 2S(C)=0 , is if S(C)==0 . But then S must be the zero map. I hope this makes sense. I am reviewing my Hatcher, just in case. I do not know why a simply connected manifold can not have torsion Zcohomology in dimension 2. In fact I don't believe it. Can you give me a proof? Regards wofsy[/QUOTE] 



#9
Mar109, 01:43 PM

P: 707

You are right  a torsion class in the top dimension must be zero. My point was that when taking cup products of two cohomology classes, one of them may be a torsion class in a lower dimension. When cupped up to the top dimension you still have a torsion class so it in fact must be zero. 



#10
Mar209, 12:17 AM

P: 393

Just to say thanks again, Wofsy. Unfortunately, as you see , at this point, my
algebraic topology is still relatively weak, so I am not able to answer your questions at this point. I am stronger at this point in the pointset , and analysisside; hopefully if you need help in these, I may be able to answer a question. 



#11
Mar309, 02:34 PM

P: 707

It was a pleasure to help you. Anytime.




#12
Mar1609, 12:48 AM

P: 393

I do not know why a simply connected manifold can not have torsion Zcohomology in dimension 2.
In fact I don't believe it. Can you give me a proof? Regards wofsy[/QUOTE] I think this is false: Take an EilenbergMclane K(G,2), with G a torsion group. Then ,Pi_1 =0 , and by Hurewicz (Hip, Hip, Hurewicz! :) ) Pi_2=G =H_2 has torsion. 



#13
Mar1609, 12:49 AM

P: 393

I meant to say that I think my claim was false.




#14
Mar1609, 04:44 PM

P: 707

Then ,Pi_1 =0 , and by Hurewicz (Hip, Hip, Hurewicz! :) ) Pi_2=G =H_2 has torsion.[/QUOTE] is there one of these that is a manifold? 



#15
Mar1609, 05:46 PM

P: 322

By the universal coefficient theorem in cohomology:
[tex]0\to\text{Ext}(H_1(M),\mathbb{Z})\to H^2(M;\mathbb{Z})\to \text{Hom}(H_2(M),\mathbb{Z})\to 0[/tex] is exact. Furthermore if [tex]H_2(M)[/tex] is finitely generated, then [tex]\text{Hom}(H_2(M),\mathbb{Z})[/tex] is isomorphic to the free part of [tex]H_2(M)[/tex]. Thus, in the case [tex]H_1(M)=0[/tex], [tex]H^2(M)[/tex] is torsion free. Also, if M is a closed connected orientable 3manifold, then the torsion subgroup of [tex]H_{2}(M)[/tex] is trivial (see Hatcher p. 238). 



#16
Mar1709, 09:29 AM

P: 707





#17
Mar2009, 05:42 PM

P: 393

[QUOTE=wofsy;2068664]The cup product pairing over Z is usually restricted to the nontorsion part of the cohomology groups. In this case, the cohomology is a free abelian group and so has a basis.
However since the cohomology is Abelian you can still define the pairing in terms of a set of generators of the group. I assume that you mean only to use the nontorsion part because in part 2 of your question you use Poincare duality. Poincare duality over Z doesn't work unless the manifold is orientable over Z in which case cup product with torsion classes in the bilinear pairing will always give zero and is thus uninteresting. Wofsy: I just thought of another argument for the quotient of the Abelian group A by torsion has a basis: A torsionfree module over a PID is free . In our case, Z is the PID, ( since Z is a Euclidean ring, it is a PID) . 



#18
Mar2009, 06:11 PM

P: 707

[QUOTE=WWGD;2125664]



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