# A Is there a natural paring between homology and cohomology?

1. Feb 12, 2017

### lichen1983312

I am looking at the definition of the characteristic numbers from the wikipedia
https://en.wikipedia.org/wiki/Characteristic_class#Characteristic_numbers
"one can pair a product of characteristic classes of total degree n with the fundamental class"
I am not sure how is this paring defined here? I know that for the de Pham cohomology the pairing could be defined by integrating differential forms over the manifold. But here, the definition does not need de Pham cohomology, right?

2. Feb 12, 2017

### lichen1983312

Is it defined this way? If the cochain is defined as $C_n^ * = Hom({C_n},R)$, and let $\alpha \in Hom({C_n},R)$ and $\beta \in {C_n}$, so I can define a product
$< \alpha ,\beta > = \alpha (\beta ) \in R$
is this right?

3. Feb 13, 2017

### lavinia

yes

De Rham's theorem says that de Rham cohomology is isomorphic to singular cohomology with real(or complex) coefficients. The isomorphism is to view a differential form as a homomorphism of a triangulation of a smooth manifold (every smooth manifold has a smooth triangulation) into the real numbers by integrating over the smooth simplices in the smooth chain.

Last edited: Feb 14, 2017
4. Feb 13, 2017

### lichen1983312

THanks very much!

5. Apr 8, 2017

### marjudo

There is an exemple like that:
Let M be a differential manifold, let us describe the differential algebra (in) (Ω * (M), d) of its differential forms. For any natural integer p:

Ωp (M) is the space of the differential forms of degree p on M.
Dp: Ωp (M) → Ωp + 1 (M) is the operator of external differentiation on the differential forms of degree p.
We denote by dω the external derivative of ω when we do not want to specify its degree; Then we must assume dpω where p is the degree of ω.

6. Apr 8, 2017

### WWGD

Just to add that this is just a generalization of the pairing between a finite-dimensional V.Space and its dual , since in most cases both homology and cohomology will be finite-dimensional (I am no expert but I cannot think of a case of either being infinite-dimensional, let alone a case a non-specialist would run into of the infinite-dimensional case.)

EDIT: I think this may be true up to the universal coefficient theorem, since we may be working over rings and so not have a vector space structure.

7. Apr 8, 2017

### Staff: Mentor

There is no need to restrict the dimension. Formally $V^*=\operatorname{Hom}(V,\mathbb{F})$, so it works for every vector space. Duality is a well-defined functor of categories.

8. Apr 8, 2017

### WWGD

But what do you do when you do not have a vector space structure, e.g., you work with coefficients in a ring, so you may have a module?

9. Apr 8, 2017

### Staff: Mentor

The formal definition goes:

Let $\mathcal{C}$ be a category with objects $\mathcal{Ob(\mathcal{ C})}$ and morphisms $\mathcal{M}_\mathcal{C}(.,.)$.
Then the dual category $\mathcal{C}^*$ is given by
1. $\mathcal{Ob(\mathcal{C}^*)}=\mathcal{Ob(\mathcal{C})}$
2. $A,B \in \mathcal{Ob(\mathcal{C}^*)} \Longrightarrow \mathcal{M}_{\mathcal{C}^*}(A,B)=\mathcal{M}_{\mathcal{C}}(B,A)$
3. For $A,B,C \in \mathcal{Ob(\mathcal{C}^*)}$ we have a composition $$\circ \, : \,\mathcal{M}_{\mathcal{C}^*}(A,B) \times \mathcal{M}_{\mathcal{C}^*}(B,C) \longrightarrow \mathcal{M}_{\mathcal{C}^*}(A,C)$$ by the composition $\circ (f,g) = f \circ g$ in $\mathcal{M}_\mathcal{C}$.
This definition works as well for modules, sets, functions or whatever. $^*$ is a functor $\mathcal{C} \rightarrow \mathcal{C}^*$.

The duality principle says: A statement that is true for all categories is also true for all dual categories.

10. Apr 8, 2017

### lavinia

The pairing referred to here is between the top homology of an oriented $n$-dimensional manifold without boundary and a cup product of characteristic classes that has cohomology dimension $n$. For Chern classes and Pontryagin classes the coefficients are integers and the $n$'th homology of the manifold $H_{n}(M;Z)$ is isomorphic to $Z$. The fundamental cycle of the manifold is a generator of $H_{n}(M;Z)$ that is determined by the orientation. The pairing is just the value of the product of characteristic classes on the fundamental cycle. If the manifold is not orientable, it still has a fundamental class with $Z_2$ coefficients. In this case, the $n'th$ homology group is isomorphic to $Z_2$. An $n$ dimensional product of Stiefel-Whitney classes can be evaluated on this fundamental class. The value is either $0$ or $1$ and is called a Stiefel-Whitney number. If $Z$ coefficients are used for a non-orientable manifold then there is no pairing because $H_{n}(M;Z) = 0$ even though $H^{n}(M;Z)=Z_2$

11. Apr 8, 2017

### WWGD

Yes, good point, I was just wondering if the pairing of a vector space with its dual extended to rings and other algebraic objects. I think the pairing you describe is a generalization of this pairing between a f.d vector space and its dual.

12. Apr 8, 2017

### lavinia

An abelian group is a module over $Z$. If the group is torsion then all homomorphisms into $Z$ are zero. Not sure if that is an example of what you were asking about.

13. Apr 8, 2017

### lavinia

A technical point about De Rham cohomology and Chern and Pontryagin classes:

A De Rham cohomology class is an element of $H^{j}(M;R)$. A Chern or Pontryagin class is an integer cohomology class. That is: it is an element of $H^{j}(M;Z)$. A famous theorem known as the "Weil Homomorphism" shows that there are De Rham cohomology classes that are the images of the Chern classes under the homomorphism $H^{j}(M;Z) →H^{j}(M:R)$ induced by the inclusion of coefficient groups $Z→R$.

14. Apr 8, 2017

### WWGD

Yes, thanks, that is what I was curious about. When considering the case of the dual of a finite Abelian group.

15. Apr 8, 2017

### WWGD

I am not sure I understand what you meant. The dual of a finite Abelian group G is 0 , so there is no isomorphism G to G*.

16. Apr 8, 2017

### Staff: Mentor

No, it was my fault. I thought you talked about duality in general.

17. Apr 8, 2017

### WWGD

Sorry, Herr/Prof Frisch, my algebra and Category theory are not as Frisch as they should be :(.

18. Apr 9, 2017

### lavinia

The group does not need to be finite. An infinite direct product of finite cyclic groups works.

19. Apr 9, 2017

### WWGD

Just curious, is the dual space always defined as the space of linear maps in the given category or is it given by the morphisms in the given category, e.g., if we deal with Abelian groups, do we consider as its dual the space of homomorphisms into the integers, or is it always linear maps? EDIT: I know for R-modules M, we use homomorphisms into R, but I wonder if the dual can be defined more generally than that.

Last edited: Apr 9, 2017
20. Apr 9, 2017

### lavinia

For Abelian groups one views them as $Z$ modules. But the module of linear homomorphisms into $Z$ may not be isomorphic to the group (viewed as a Z-module). So you do not have duality. The abelian group would have to be torsion free or what is the same, it would have to be a free abelian group.

I do not know anything about a general theory of duality.

For manifolds, duality theorems are linked to orientablility of the manifold. The key theorem is Poincare Duality which says that if the manfold is orientable and without boundary then $H_{i}(M;Z) ≈H^{n-i}(M;Z)$ The cohomology classes in the complementary dimension define homomorphisms of the homology classes through intersections. So this is a true duality.

Last edited: Apr 9, 2017