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Reference for simplicial homology and cohomology

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lavinia
#1
May4-12, 01:23 PM
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I would like to read more about this version of simplicial homology and cohomology. Any reference or explanation is welcome.

I hope this description is correct.

Triangulate a smooth manifold and assign an orientation to each simplex in the triangulation.

Define the boundary operator with respect to this orientation.
This defines the simplicial chain complex.

for the cochain complex use exactly the same simplices but define the coboundary operator as the transpose of the boundary operator. It maps k chains to K + 1 chains while the boundary operator maps k chains to k-1 chains.

Thanks in advance.
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mathwonk
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May4-12, 02:10 PM
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Spanier is a standard reference for simplicial homology. Other classics are Hilton and Wylie, and Eilenberg and Steenrod.

An actual textbook,decently short, and containing a proof of the deRham theorem is Singer and Thorpe.

http://www.abebooks.com/servlet/Sear...ts=t&x=63&y=11

It seems that once one has a few basic computational theorems, especially Maier Vietoris, it does not matter what theory one is using. Theorems on what happens when you "add cells" are also basic.


I am not sure I understood the subtlety of your question. Did it involve the use of oriented as opposed to ordered simplices, or the application to triangulated manifolds, or was it a general question about simplicial homology?

a triangulation seems just to be a homeomorphism from a simplicial com-lex so the question seems to reduce to understanding the homology of a simplicial complex. At any rate Hilton and Wylie seem to do the oriented version as do Singer and Thorpe, while perhaps Eilenberg Steenrod do the ordered version. Singer Thorope apply it to a triangulkated manifold.

The only place I have seen it proved that a smooth manifold has a triangulation is in a book by Munkres that I used to have. It took up the whole book.

However one can give a short description of a proof by saying one embeds the manifold in R^n and restricts, then tweaks slightly, a fine triangulation of R^n.
Jamma
#3
May6-12, 02:55 PM
P: 429
Doesn't every singular simplex canonically inherit an orientation (since it is defined as a homeomorphism from the standard singular simplex, which has an orientation, into the space)? Of course, these orientations may not define an orientation of the space - obviously simplicial complexes (triangulated spaces) are more general than oriented manifolds (actually, not all manifolds are triangulable, so that's not quite true).

I too don't understand the question - you don't need manifolds to define simplicial (co)homology. It is relatively easy to show that the simplicial (co)homology of any triangulated space is isomorphic to its singular (co)homology.

lavinia
#4
May6-12, 09:00 PM
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Reference for simplicial homology and cohomology

You are right that the homology question does not require a manifold. Sorry for being confusing.


it is just that the problem I am trying to understand requires a 4k dimensional manifold.

I think the authors think of each geometric simplex as the image of a single mapping of the standard simplex. The other mappings, those that map the standard simplex with a different ordering of vertexes are not included in the chain complex.

You can choose an orientation of a simplex by choosing an ordering of its vertices. Equivalent orderings are considered the same as the chosen one so I guess there is an equivalence relation where they are all identified.

To me this all seems a little unusual. For instance,the boundary operator must be determined by a function that assigns a sign to each face of a simplex and this sign depends on the simplex and the chosen orderings.

Also the cochains are the same as the chains. there are no homomorphisms of chains into the integers as in singular cohomology. There is a single graded group of ordered simplices made into a chain complex in two ways - one with the boundary operator, the other with its transpose.

I will look at Spanier to make sure there isn't some subtlety to this way of doing things.
Jamma
#5
May7-12, 06:03 AM
P: 429
Arghh, it lost my long post - let's write it all up again...

I think I see what you are getting at. The definition of a simplicial complex, as it stands on wiki, seems to me to be very strict. We require that the faces of the simplices not only "match up" but "are the same" whenever they overlap.

Now, in the definition of the singular chain complex, there is nothing to say that for any given σ, the chain -σ is equal to σ with the opposite orientation. For example, if we take the chain given by σ:[0,1]->[0,1], σ(t)=t, there is nothing in the definition of the singular chain complex which says that -σ = σ' where σ' is the map σ':[0,1]->[0,1], σ'(t)=1-t.

However, the following probably can be said: [(-1)^n]σ' is homologous to σ where σ' is given by a reordering of vertices with sign n.

Hence, it seems to me that we arrive at this point: we can either think of σ with a negative orientation and -σ to be different in the chain complex, which means we will require a much stricter version of a simplicial complex (and probably require a finer triangulation), or we can acknowledge that they will end up being homologous and work in the chain complex where they are identified (in fact, we may like to identify more than just these to make things even easier on ourselves).

You should just view the simplicial chain complex as being a subcomplex of the singular chain complex, where this inclusion is a quasi-isormorphism. Of course, the correct statement of this depends on the above - we may want to work with the chain complex where -σ and σ with the negative orientation are already identified. But in the end, we should still get quasi-isomorphisms and the same homology and everything should follow smoothly.

Also the cochains are the same as the chains. there are no homomorphisms of chains into the integers as in singular cohomology. There is a single graded group of ordered simplices made into a chain complex in two ways - one with the boundary operator, the other with its transpose.
I didn't follow this - the cochains of the simplicial cochain complex are still homomorphisms of chains into the integers (or other coefficient group) - this is how they are defined! The only difference now is that, in most cases, we now are working with a finitely generated module where the dual will be isomorphic to the original with the obvious identification. Taking the transpose maps gives precisely the cochain complex.

We know this is the right cochain complex - for example, the UCT says that the homology determines the cohomology. If we accept that the simplicial homology is isomorphic to the singular homology, then their cohomology must also be the same, so indeed taking the transpose maps and then taking homology gives us precisely the singular cohomology.
lavinia
#6
May7-12, 10:03 AM
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Thanks Jamma and Mathwonk.

From what you are saying it seems that I can view a simplicial complex just as a collection of ordered finite subsets of a set with the condition that if a subset is ordered then each of its subsets must also be ordered.

The boundary operator is then determined by the orderings.
Jamma
#7
May8-12, 04:32 AM
P: 429
From what you are saying it seems that I can view a simplicial complex just as a collection of ordered finite subsets of a set with the condition that if a subset is ordered then each of its subsets must also be ordered.

The boundary operator is then determined by the orderings.
I'm not sure if it is defined in this way - but I think it can (and for all intents and purposes, we act as if it is).

For example, we "know" that we can triangulate a Mobius band, we can even draw a picture to show that we can. However, the technical definition of a simplicial chain complex would require the boundaries of the little triangles to be exactly the same maps. Can we be sure that we can find the correct maps so this all works out?

I think the resolution probably goes something like this. We can triangulate our space, and don't really need to worry when the maps at the faces aren't exactly the same, we only need to know that they are the same up to a boundary. From there, one can probably make statements like you have in terms of the subsets of vertices, and their orderings and so on. If one can triangulate a space using simplexes in a way such that the overlapping faces are the same subsets, then one can probably be sure it all works out up to homology (or that one could potentially find the exact maps, if required). So in the end, you are right.

Incidentally, it seems you've independently stumbled upon the idea of a "simplicial set". These can basically be thought of as purely algebraic models of topological spaces (actually, I think that they could be more general than just simplicial complexes - I know there is a geometric realisation functor). One can study the homotopy theory of these objects and so on. The idea behind them is basically right inline with what you were saying about collections of ordered finite subsets - you have your vertices, 1-simplices which are ordered pairs of vertices and so on, and you have "face maps" and "degeneracy maps".

I'd advise you to check them out!
lavinia
#8
May9-12, 07:31 AM
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Quote Quote by Jamma View Post
I think the resolution probably goes something like this. We can triangulate our space, and don't really need to worry when the maps at the faces aren't exactly the same, we only need to know that they are the same up to a boundary. From there, one can probably make statements like you have in terms of the subsets of vertices, and their orderings and so on. If one can triangulate a space using simplexes in a way such that the overlapping faces are the same subsets, then one can probably be sure it all works out up to homology (or that one could potentially find the exact maps, if required). So in the end, you are right.
Interesting post Jamma.

One thought: A triangulation determines a topological space that is homeomorphic to the original. It is just the subset of the standard simplex in Euclidean space (dimension = number of vertices in the triangulation minus 1) spanned by the simplices in the triangulation (simplices identified by their vertices). This homeomorphism makes the maps on overlapping faces the same.

In this paper I am trying to read, a map between Euclidean simplices with the adjoint boundary operator and the de Rham cochain complex is given and the author says that it is an isomorphism between simplicial cohomology and de Rham cohomology. The forms are

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.


Another half thought: For a smooth triangulation the integrals of forms are the same on overlapping faces since the maps are reparameterizations of each other - I think. So the de Rham cohomology only cares about the underlying set in some sense.
mathwonk
#9
May9-12, 01:14 PM
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what coefficients is he using? deRham usually uses reals and simplicial usually uses integers.
Jamma
#10
May9-12, 03:01 PM
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I'm assuming we are looking at simplicial homology with real coefficients.
lavinia
#11
May9-12, 03:51 PM
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Quote Quote by mathwonk View Post
what coefficients is he using? deRham usually uses reals and simplicial usually uses integers.
real coefficients. The pupose of the paper is to find a combinatorial formula for the signature of a 4K manifold. Torsion cohomology classes don't count because they all die under the cup product.
mathwonk
#12
May9-12, 08:08 PM
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ok, i believe you know more algebraic topology than me, but I still comment. Over a coefficient field, cohomology is just dual to homology so to show the deRham cohomology spaces are isomorphic to the simplicial ones, one only has to show they are dual to simplicial homology.

Of course it might be easier to shown they are dual to singular homology.

And of course cohomology has a cup product making it a ring, and so one also want this to correspond to the wedge product in deRham cohomology,...blah, blah,...
lavinia
#13
May9-12, 08:45 PM
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Quote Quote by mathwonk View Post
ok, i believe you know more algebraic topology than me, but I still comment. Over a coefficient field, cohomology is just dual to homology so to show the deRham cohomology spaces are isomorphic to the simplicial ones, one only has to show they are dual to simplicial homology.

Of course it might be easier to shown they are dual to singular homology.

And of course cohomology has a cup product making it a ring, and so one also want this to correspond to the wedge product in deRham cohomology,...blah, blah,...
Mathwonk I have not tried to verify this claim and don't know how the proof would go. Let's see if we can figure it out. The authors refer to Whitney's book, Geometric Integration Theory.
mathwonk
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May9-12, 10:05 PM
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which claim? the duality is the "universal coefficients" theorem for cohomology.


more likely you meant the fact that differential forms exist dual to all homology classes.
mathwonk
#15
May9-12, 10:24 PM
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In section 6.2, Singer and Thorpe prove the isomorphism between deRham and simplicial cohomology for a smoothly triangulated manifold.

In Bott -Tu, the first result in chapter 2 proves the isomorphism between deRham cohomology and Cech cohomology. On p.191 they prove the isomorphism between singular and Cech cohomology.

In Spivak's differential GEOMETRY THE LAST EXERCISE OF CHAPTER (11) of volume I, proves the isomorphism between deRham and singular cohomology.

the last result of chapter II of Ronnie Well's "differential analysis on complex manifolds",
is the sheaf theoretic proof of the deRham theorem, probably an isomorphism of Cech and deRham cohomology.

I recommend Singer Thorpe as most elementary, and Spivak or Bott -Tu as most clear.

Wells' proof is the modern slick formulation of the idea that the theorem is proved if it is only proved locally, an idea apparently due to Weil.

So the best place to look would be the proof due to A. Weil.

here is an alternate link:

http://www.nd.edu/~lnicolae/Fanoethesis.pdf
lavinia
#16
May10-12, 05:09 AM
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Quote Quote by mathwonk View Post
In section 6.2, Singer and Thorpe prove the isomorphism between deRham and simplicial cohomology for a smoothly triangulated manifold.
Thanks Mathwonk. I just glanced at Singer and Thorpe and they use the same differential forms. I guess they are reproducing Whitney's proof. Before reading it though I would like to try proving it for myself.
mathwonk
#17
May10-12, 09:37 AM
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yes, they say in the introduction their proof is taken from Whitney's book. I just notioced that. I don't have a copy of Whitney. I don't see any attempt to prove that the ring structure given by wedge product agree nwith the one by cup product though. maybe i missed it.
mathwonk
#18
May10-12, 11:19 AM
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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


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