# Is there a DeRham _Homology_?

1. Sep 8, 2011

### 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?

2. Sep 9, 2011

### 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.

3. Sep 9, 2011

### Bacle

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

4. Sep 9, 2011

### 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.

5. Sep 13, 2011

### lavinia

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.

6. Sep 13, 2011

### 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.

7. Jun 26, 2012

### 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)

8. Jun 26, 2012

### lavinia

Smooth simplexes

9. Jun 26, 2012

### 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.

10. Jun 26, 2012

### lavinia

It should be a smooth triangulation of the manifold.

11. Jun 26, 2012

### 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.