Defining the Cup Product in Cubular Homology Theory

[wofsy]

I would like to know how to define the cup product in cubular homology theory.

 

[jim mcnamara]

Do you mean the Massey product? Please define cubular.

 

[wofsy]

Do you mean the Massey product? Please define cubular.

Cubular homology is where you use cubes instead of simplexes.

 

[morphism]

How do you define the cup product in simplicial homology? Did you mean cohomology? If so, then the cup product is defined pretty much the same way. Perhaps you can show us your definitions and explain what you're having trouble adapting.

 

[wofsy]

How do you define the cup product in simplicial homology? Did you mean cohomology? If so, then the cup product is defined pretty much the same way. Perhaps you can show us your definitions and explain what you're having trouble adapting.

I see why I've been confusing. I meant cohomology. Sorry.

For simplices two cochains cup by multiplying their values on front and back faces of the simplex. a cup b (simplex) = a(front face).b(back face) where the face splitting is according to the degree of a and b.

For cubes I imagine it would be similar but am not sure. A cube can be thought of a having front and back faces but that may be just an analogy.

Generally speaking, you want a "diagonal approximation" from C(x) -> C(x) tensor C(x) . For simplices this is called the Alexander diagonal approximation.

 

[morphism]

You still have that in cubical homology. In fact, you have the Eilenberg-Zilber theorem, which tells you that for any topological spaces X and Y there is a chain homotopy equivalence C_\ast(X\times Y) \to C_\ast(X)\otimes C_\ast(Y).

 

[wofsy]

You still have that in cubical homology. In fact, you have the Eilenberg-Zilber theorem, which tells you that for any topological spaces X and Y there is a chain homotopy equivalence C_\ast(X\times Y) \to C_\ast(X)\otimes C_\ast(Y).

right. But what is the diagonal approximation?

 

[morphism]

I don't really understand your question. The diagonal approximation is the same.

The diagonal map \Delta : X \to X \times X induces a map C_\ast(X) \to C(X \times X) which we can then compose with the Eilenberg-Zilber map given above to get a map C_\ast(X) \to C_\ast(X) \otimes C_\ast(X).

If you want all the details, then check out chapter XIII in Massey's book "A Basic Course in Algebraic Topology" (Springer GTM127), where the cup product is defined. (Incidentally, Massey's book is the only place I've seen the cubical approach to singular (co)homology. Is Massey the guy who invented this?)

 

[wofsy]

I don't really understand your question. The diagonal approximation is the same.

The diagonal map \Delta : X \to X \times X induces a map C_\ast(X) \to C(X \times X) which we can then compose with the Eilenberg-Zilber map given above to get a map C_\ast(X) \to C_\ast(X) \otimes C_\ast(X).

If you want all the details, then check out chapter XIII in Massey's book "A Basic Course in Algebraic Topology" (Springer GTM127), where the cup product is defined. (Incidentally, Massey's book is the only place I've seen the cubical approach to singular (co)homology. Is Massey the guy who invented this?)

so take the example of a single 5 dimensional cube. What is the diagonal approximation? Can you write it out? It can not be the same as the simplicial approximation.

 

[morphism]

Ah, I see what you're asking. You basically want the explicit Eilenberg-Zilber map for cubical homology. This isn't the most straightforward thing to write down, as you need a few intermediaries. But like I said, all the details are found in Massey. If you really need this stuff and can't get your hands on a copy, I can probably write it all out for you later.

 

[wofsy]

Ah, I see what you're asking. You basically want the explicit Eilenberg-Zilber map for cubical homology. This isn't the most straightforward thing to write down, as you need a few intermediaries. But like I said, all the details are found in Massey. If you really need this stuff and can't get your hands on a copy, I can probably write it all out for you later.

You are right. I should just try to derive it myself rather than asking. Will do.