Is There Such a Thing as DeRham Homology?

[WWGD]

Hi, I hope this is not too ignorant, but my Algebraic Topology is rusty:

Is there such a thing as DeRham _homology_? I always hear and read about

DeRham cohomology, but I have never heard of DeRham homology. Is there

such a thing?

 

[hunt_mat]

DeRahm cohomology is all to do with the cohomology of differential forms, you set up your Meyer-Veotoris sequence via the exterior derivative. There is alway a pairing between homology and cohomology, so I guess that there is that but there is no formal theory called DeRahm homology as far as I am aware.

 

[Bacle]

Why stop there with the (good) question? How about Cech homology?

 

[quasar987]

Cech homology exists, but it is not a homology theory in the sense of the Eilenberg-Steenrod axioms. The little book by Hocking and Young talks a bit about it.

 

[lavinia]

DeRahm cohomology is all to do with the cohomology of differential forms, you set up your Meyer-Veotoris sequence via the exterior derivative. There is alway a pairing between homology and cohomology, so I guess that there is that but there is no formal theory called DeRahm homology as far as I am aware.

The connection of the De Rham cochain complex to cohomology with real coefficiants is Stokes theorem. So the duality is through the isomorphism with real cohomology.

 

[Jamma]

Hmm, this is an interesting question and is applicable to something I have been thinking about. Apparently there is a notion of Cech homology- but it doesn't satisfy all of the properties you'd like (mainly, it doesn't do well with giving you sequences which should be exact). However, it can be altered, to a thing called (I think) strong homology, which does give a well behaved homology theory and for which cech cohomology is the dual theory.

 

[Sina]

We have the homology of simplixes (was it called simplical homology?) whose cohomology is precisely isomorphic to deRham cohomology I believe (you can show that each linear functional on the free group of simplices is precisely integration by some form of appropriate degree)

 

[lavinia]

We have the homology of simplixes (was it called simplical homology?) whose cohomology is precisely isomorphic to deRham cohomology I believe (you can show that each linear functional on the free group of simplices is precisely integration by some form of appropriate degree)

Smooth simplexes

 

[Sina]

erm yes smooth it should be to relate it to some triangulation of the manifold. I was a bit unclear sorry, I think it should be a triangulation of the smooth manifold whose deRham cohomology you want to dualize. I never proved this so I am a bit unsure about the precise conditions.

 

[lavinia]

erm yes smooth it should be to relate it to some triangulation of the manifold. I was a bit unclear sorry, I think it should be a triangulation of the smooth manifold whose deRham cohomology you want to dualize. I never proved this so I am a bit unsure about the precise conditions.

It should be a smooth triangulation of the manifold.

 

[lavinia]

I think there is a theorem which says that any real (maybe Z/2Z) homology class of a smooth manifold can be represented by an embedded submanifold. I will check it. If this is true, then embedded submanifolds might give you a de Rham homology. This is just an impression. Let's think it through.