# Intersection Form in 4-D. Followup.

1. Feb 7, 2009

### WWGD

:

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 4-mfld 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. Feb 9, 2009

### wofsy

The cup product pairing over Z is usually restricted to the non-torsion 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 non-torsion 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 sub-manifold
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. Feb 11, 2009

### WWGD

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 Eilenberg-McLane 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 simple-connectedness 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
simple-connectedness is assumed as including path-connectedness. )

Hope the wheels will be better-oiled soon.

4. Feb 11, 2009

### WWGD

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. Feb 11, 2009

### WWGD

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

6. Feb 11, 2009

### wofsy

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 Z-linear 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 Z-cohomology in dimension 2.

In fact I don't believe it. Can you give me a proof?

Regards

wofsy

Last edited: Feb 11, 2009
7. Feb 11, 2009

### wofsy

In homology, torsion arises when finite multiples of cycles are boundaries.
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 semi-great 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 non-orientable manifold. This means that any attempt to form a Z-cycle 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. Feb 28, 2009

### WWGD

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 Z-linear 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 Z-linear map from a cycle into Z. Still: what do you mean by a Z-linear
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 Z-cohomology in dimension 2.

In fact I don't believe it. Can you give me a proof?

Regards

wofsy[/QUOTE]

9. Mar 1, 2009

### wofsy

[/QUOTE]

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. Mar 2, 2009

### WWGD

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 point-set , and analysis-side; hopefully

if you need help in these, I may be able to answer a question.

11. Mar 3, 2009

### wofsy

12. Mar 16, 2009

### WWGD

I do not know why a simply connected manifold can not have torsion Z-cohomology 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 Eilenberg-Mclane 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. Mar 16, 2009

### WWGD

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

14. Mar 16, 2009

### wofsy

I think this is false: Take an Eilenberg-Mclane 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.[/QUOTE]

is there one of these that is a manifold?

15. Mar 16, 2009

### yyat

By the universal coefficient theorem in cohomology:

$$0\to\text{Ext}(H_1(M),\mathbb{Z})\to H^2(M;\mathbb{Z})\to \text{Hom}(H_2(M),\mathbb{Z})\to 0$$

is exact. Furthermore if $$H_2(M)$$ is finitely generated, then $$\text{Hom}(H_2(M),\mathbb{Z})$$ is isomorphic to the free part of $$H_2(M)$$. Thus, in the case $$H_1(M)=0$$, $$H^2(M)$$ is torsion free.

Also, if M is a closed connected orientable 3-manifold, then the torsion subgroup of $$H_{2}(M)$$ is trivial (see Hatcher p. 238).

Last edited: Mar 16, 2009
16. Mar 17, 2009

### wofsy

very cool. thanks

17. Mar 20, 2009

### WWGD

18. Mar 20, 2009

### wofsy

19. Mar 25, 2009

### WWGD

Hi again:

I assume that you mean only to use the non-torsion 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 sub-manifold
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.[/QUOTE]

I have followed up on this. Now I am trying to show that if M is compact and boundaryless, then the intersection form is unimodular ( i.e., has determinant +/- 1
which implies that it is invertible over Z ). The results I am trying (without much
success ) are:

i)
"
If g is an element in a free Abelian group g , and Q is a quadratic form over
(i.e., defined on) G , then the map:

L: G-->G* , with G* the dual to G ( i.e., all linear maps from G into Z)

with L: g--> Q(g,s) ( s in G)

is an isomorphism iff Q is unimodular . "

ii) Just Poincare duality, since M is closed.

So I need to take an element x in H^2(X;Z) ( maybe working in H_2(X;Z) may

make more sense) and I am trying to show that the map : x--> Q( x,y) , is an

isomorphism. I am pretty sure Poincare duality would help here, but I am not too clear

on how.

Thanks for any ideas.

20. Mar 27, 2009

### wofsy

Hi WW I have tried to do this proof and have failed so far. But you can see how Poincare duality is involved. If the dual map is an isomorphism from the second homology to the second cohomology then it is easy to see in simple cases that the intersection product must be unimodular. Take the case where a Z -basis of the homology lattice is also a set of eigen-vectors of the intersection matrix. Then all of the eigen-values must be non-zero integers. If one of these eigen-values has absolute value greater than 1 then the linear map that maps its eigen vector to 1 and all others to zero can not be represented with the quadratic form.

Similarly if you had a quadratic form whose matrix interchanged basis vectors with a scale factor like (1,0) -> (0,2) (0,1) - > (2,0) then the linear map that sends (1,0) to 1 and (0,1) to zero can not be represented with the quadratic form.

But in general I'm not sure. Let's think about it some more.