Le: Tensor of Cochain Complexes, Isomorphism of Complexes

[Bacle]

Hi, everyone:
A couple of things, please:
1) I am going over the Leray-Hirsch theorem in Hatcher's AT , which gives the conditions
under which we can obtain the cohomology of the top space of the bundle
from the tensor product of the cohomology of the fiber, and that of the base
( a sort of relative to Kunneth's theorem), and I see the statement, that
(paraphrase) the isomorphism:

H* (E;R)=H*(B;R)(x)H*(F;R)

where R is a ring, and (x) is the tensor product "is not always a ring homomorphism"

question: is this then an isomorphism of cochain complexes.?. If so, does
anyone know the def. of iso. of cochain complexes.?.

2)How do we tensor cochains.?. How do we tensor Cochain complexes

The isomorphism above is described explicitly, and uses the tensor product of chains.

Anyone know how to define this.?

How about the tensor product of cochain complexes C,C'.?. My naive guess would be:

H_n( C(x)C') = (+)(H_i(C;R)(x)H_(n-i)(C';R)) as a set

but I don't see how to define the coboundary. I tried to imitate the construction
of the tensor of chain complexes, but I am just going in circles.

Any Ideas.?

Thanks.

 

[lavinia]

I think the Leray-Hirsch Theorem says this.

First, it requires certain cohomology classes to exist in the total space of the bundle. These classes restrict to a basis for the finite dimensional cohomology of the fiber.

Given these cohomology classes, the isomorphism is an isomorphism of H*(B:R) - modules where the module structure is derived directly from the tensor product on one case and derives from the cup product in the other.

 

[lavinia]

the tensor product of two cochain complexes is just the direct sum of all pairwise tensor products of modules, the first module coming from the first cochain complex, the second module coming for the second cochain complex. usually the module of degree m in the tensor product is the sum of the tensor products of pairs of modules the sum of whose degrees equals m. The coboundary operator is easily defined and is described in your textbook.

 

[Bacle]

O.K, thanks, yes, it is a module, guess graded module isomorphism. But I don't
see anywhere in Hatcher's AT where tensor of cochain complexes is defined;
I tried defining the coboundary in a similar way in which the boundary of the
tensor product of chain complexes is defined, but it did not seem to work. Anyway,
I will expand on what failed a bit later.

Also: do you know how we define the tensor product of cochains.?. This
tensor product is part of the statement of the actual isomorphism.

 

[lavinia]

I do not believe that the tensor product of cochains is part of the statement of the isomorphism. I believe that it is the tensor product of cohomology classes.

 

[Bacle]

You're right, Lavinia, but isn't it ultimately a product on cochains, since homology classes
are represented by cochains.?. Sorry if this is dumb; I am out of my element (tho
trying to learn). I understand that the basis of the tensor product is the sum of the
tensors of the respective bases, but I am still kind of confused. But the elements on
the right-hand side of the isomorphism:

http://www.math.cornell.edu/~hatcher/AT/AT.pdf (p.432)


i.e., p*(b)\/c , (with \/ cupping) is a cochain, so the expression on the left should also
be a cochain, albeit a representing cochain.




I think I figured out
the def. of the coboundary, but there is a small problem with a (-1) I need to take
care of , to show that the square of the coboundary as I defined it (same as with the tensor of chain complexes) is zero.

Thanks for Any Suggestions.

 

[Bacle]

Actually, I may be stuck, since it is too late tomorrow I will look at it again.

 

[lavinia]

You are right that the cup product is defined on cochains but it induces a product on cohomology classes. Cohomology classes are equivalence classes of cochains - not cochains.

In the Leray Hirsch theorem the map that takes

$$H^*(B,R) \otimes H^*(F,R) -> H^*(E,R)$$

is defined on the level of cohomology classes.

Perhaps you are thinking that the tensor product of the base cohomology with the fiber cohomology derives from a tensor product of the cochain complexes. But I do not think so. I think it is just a tensor product of R-modules. The reason that you get a mapping into the cohomology of the total space is that H*(B) pulls back into H*(E) via the bundle projection map and H*(F) maps into H*(E) by assumption the the total space contains cohomology classes that restrict to a basis for the cohomology of each fiber. One just takes these basis elements in the fiber and maps them back into their corresponding classes in the fiber.

It may be though that actually demonstrating the isomorphism requires an argument on the level of cochains. Perhaps you could outline how the proof actually works. I do not have Hatcher's book.

 

[lavinia]

in the case of a product bundle you get a map on cochains.

The two projections

$$E -> B$$

and

$$E -> F$$

induce maps on cochains. The cup product of these cochains gives you the map,

$$C^*(B,R) \otimes C^*(F,R) -> C^*(E,R)$$

In the case where B is a manifold, I think the Leray-Hirsh theorem should follow from the case of a product bundle and the usual Meyer-Vietoris argument. A Meyer-Vietoris argument can be reduced to the level of cochains so this may be the way to do the whole thing with cochains - at least in the case of a manifold.

I strongly recommend that you read the section on De Rham theory in Bott and Tu's book, Differential Forms in Algebraic Topology. This is a wonderfully clear book and the use of calculus on manifolds simplifies many of the arguments.

 

[lavinia]

I was thinking about how this theorem applies to circle bundles over the 2-sphere.

In the case of $$S^2 \times S^1$$

you just have the Kunneth formula and the generator of the cohomology of the circle gives you the class in total space that restricts to a basis in the cohomology of the fiber.

But for the Hopf fibration there can not be such a class that restricts to a basis of the cohomology of the fiber circles because $$S^3$$ is simply connected. The same thing applies to the tangent circle bundle using real coefficients.

What about the cohomology of the tangent circle bundle using Z/2 coefficients? I am not sure here.

 

[zhentil]

You should jump ahead to the Serre spectral sequence for a fibration. Then it all becomes either crystal clear or not, depending on your perspective :)

 

[zhentil]

What about the cohomology of the tangent circle bundle using Z/2 coefficients? I am not sure here.

What's the Z/2 cohomology of RP^3?

 

[Bacle]

But the spaces I am working with at the moment are simple-enough that Spectral
Sequences are not necessary, and Leray-Hirsch, or even simpler techniques are enough. I am off spectral sequences after having done some work on the Vasiliev.

 

[lavinia]

What's the Z/2 cohomology of RP^3?

Z/2 in every dimension.

So LeRay-Hirsch might be right - but I don't know about the fiber orientation classes.
I think that in the case of the tangent circle bundle the cochain that restricts to the generator of the cohomology of the fiber circles is not closed. In fact now that I think of it just consider any connection 1-form wrsp to a Riemannian metric. The exterior derivative of the connection 1-form is minus the Gauss curvature times the pull back of the volume element. But the 2 sphere can not have identically zero Gauss curvature so the form is not closed.

Another look at this just observes that the real cohomology of RP^3 is zero so there can be no orientation class over the reals.

Over Z the same is true - I think - because isn't the integer cohomology of RP^3 is Z/2?

Yes, this seems right. Then there can be no orientation class over Z/2 as is seen form the commutative diagram,

$$H^1(RP^3,Z/2) \leftarrow H^1(RP^3,Z)$$

$$\mbox{ \ \ \ \ \ \ \ \ \ \ i* } \downarrow \mbox{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i* } \downarrow$$

$$H^1(S^1,Z/2) \mbox{\ \ \ \} \longleftarrow \mbox{\ \ \ \ \ \ 0 }$$

Where $$S^1}$$ is an arbitrary fiber and i is the inclusion map.

 

[lavinia]

How about the cohomology of the Klein bottle considered as a circle bundle over the circle?

Here you do have an orientation class - if you like you can prove this using the Hochschild-Serre spectral sequence - but the cup product structure is not the same as in a product of two circles.

Does this mean that the Leray-Hirsch isomorphism does not derive from a mapping of the cochain complexes?

 

[lavinia]

You should jump ahead to the Serre spectral sequence for a fibration. Then it all becomes either crystal clear or not, depending on your perspective :)

can you explain this a little?

 

[zhentil]

can you explain this a little?

The E2 page of the Serre spectral sequence is what one would obtain by Leray-Hirsch. So in a sense, the Serre spectral sequence can be seen as measuring the obstruction to finding cohomology classes that restrict to generators of the fiber.

 

[zhentil]

Does this mean that the Leray-Hirsch isomorphism does not derive from a mapping of the cochain complexes?

I'm not sure quite sure what this means. In the case of a Kunneth theorem, you would have this, but in general there's no map from a fiber bundle to the fiber. This may be misunderstanding your question, but if you mean a Kunneth-style argument where you pull back cohomology classes based on two projections, that can't work in a nontrivial fiber bundle. Think of the proof of Thom isomorphism in de Rham cohomology: you don't get the fiber cohomology class for free, i.e. it's not pulled back from anything.

 

[lavinia]

I'm not sure quite sure what this means. In the case of a Kunneth theorem, you would have this, but in general there's no map from a fiber bundle to the fiber. This may be misunderstanding your question, but if you mean a Kunneth-style argument where you pull back cohomology classes based on two projections, that can't work in a nontrivial fiber bundle. Think of the proof of Thom isomorphism in de Rham cohomology: you don't get the fiber cohomology class for free, i.e. it's not pulled back from anything.

no but since there is a mapping that respects the product structure on might think that there is a mapping of chain complexes. that was the original question I think. No one was expecting a Kunneth argument. Still there definitely are non-trivial bundles where the fiber cohomology is pulled back from classes in the total space of the bundle.