reference for simplicial homology and cohomology

mathwonk

a more serious criticism of Whitney's proof is that he apparently did not even prove the isomorphism is functorial. This is a serious drawback to the simplicial approach.

see:

http://www.jstor.org/discover/10.230...21100789689321

lavinia

Not sure yet. Thanks for that link. It opens up a whole new idea for me, the idea of PL forms.

BTW: While this paper I am reading does not discuss the cup product it does give what the authors call the simplicial analogue of the Hodge star operator.

mathwonk

homology and cohomology seem quite intersting. i.e. a simpliciual complex has a well defined homology and cohomology but it is not clear how to associate one simplicial complkex to a topological space like a manifold. I.e. every triangulation dioes so but there arte many of them. the cech process also does so but there arte a lot of open sets.
'
'the deRham process associates a complex but the space are infinite dimensional. the same holds for the singular procedure.

so the processes that associate finite dimensional complexes do not seem canonical, and the canonical ones seem to give infinite dimensional spaces. so one has to relate all this.

lavinia

I just noticed that the forms

r![itex]\Sigma[/itex][itex]^{r}_{i=0}[/itex]x[itex]_{i}[/itex]dx[itex]_{0}[/itex]^...^dx[itex]_{i-1}[/itex]^dx[itex]_{i+1}[/itex]^ ...dx[itex]_{r}[/itex]

where the x[itex]_{i}[/itex]'s are barycentric coordinates

are not smooth at the boundaries of r simplices so they are not in the de Rham complex. I think this is why Singer and Thorpe construct smooth partitions of unity around the simplicial complex.

But I wonder if the de Rham complex is extended to continuous forms that are smooth except possibly on a set of measure zero where they are identically zero whether it still gives the real cohomology and whether the barycentric coordinate based forms give an isomorphism of cohomology,

mathwonk

I don't know quite how you would take d of a form that is not smooth.


to repeat my philosophical remarks above, it seems the job of computing (co)homology is to approximate an infinite dimensional complex by a finite dimensional one having the same (co)homology groups.

e.g. the deRham complex is infinite dimensional and intrinsic, while the simplicial one is finite dimensional, and depends on choices, as is the cech complex associated to a finite cover.

if they give isomorphic cohomology then you deduce that the one depending on choices was independent of the choices, and also that the intrinsic one had finite dimensional cohomology. double bonus. obvious perhaps, but cool.

Jamma

In terms of your non-smooth forms, it feels like you may want some smooth structure on your simplicial complex - I've certainly heard of piecewise linear triangulations. It seems though that the most obvious way to prove things is:

Simplicial [itex]\cong[/itex] singular [itex]\cong[/itex] De Rham

Mathwonk is correct about proving that the De Rham complex is dual to the singular homology group (integrate along the chain) - I believe that this induces a pairing between the singular and De Rham cohomology groups which gives the desired isomorphism.

[Edit]:

Actually, I should probably put:

Simplicial [itex]\cong[/itex] singular [itex]\cong[/itex] smooth singular [itex]\cong[/itex] De Rham

since there is a pairing not for arbitrary chains, but for smooth chains. I presume it isn't too difficult to show that the chain complex of smooth chains is quasi-isomorphic to that of singular chains (chain homotopy will involve "smoothing out" chains).

mathwonk

no wonder eilenberg and maclane decided they needed to axiomatize this topic.

in addition to those 4 theories above, there are cech, alexander, cellular, derived functor sheaf definitions,...etc.....etc ....

the book by eilenberg maclane is also quite illuminating.