3-connection on nontrivial topological 3-manifold

I'm studying Chern-Simons theory on topological nontrivial 3-manifold (I come from a physics background, so I'm new to some mathematical concepts). If the first homology group $H_1(M)$ is nontrivial one needs to consider a good cover of the manifold and a polyhedral decomposition. Then, we can define a U(1) gauge field as follows:



1. Why if the homology group is trivial i can take $d\lambda_{ab}$ so that $v_a=v_b$, without speaking of bundles?

On the other hand, in the language of bundles, if the manifold is globally trivializable (so admits a global section) the connection could be described by 1-form globally defined.

2. Why if the first homology group is trivial then the manifold is globally trivializable?

it might help if we could read the rest of the sentence explaining the implications of A.7. Vanishing 1st de Rham cohomology means every closed global 1-form is exact, i.e. equals df for some function f.

well, if both the 1st and second cech cohomologies vanish, then the cech 2-cocycle represented by the nabc, would be a cech coboundary, of form nab+nbc-nac, with the nab defined on the Uab. Then one could modify the lambdas, by subtracting these nab and get new lambdas, Lab, which would now themselves be cech 1-cocycles. hence if the 1st cech cohomology also vanishes, these would be cech 1 -coboundaries of form Lab = Lb-La, where the La are defined on the Ua. Then finally subtracting the dLa from the 1-forms va, would give a family of 1-forms that agree on the overlaps Uab, hence would give a global 1-form v.

this requires vanishing of the integral 2nd cech cohomology, and the 1st real cech cohomology, (since the nabc have integral coefficients, and the Lab have real coefficients). but it may be what they mean, since at least the computations work.

well reading more carefully, they are using constant as coefficients for the 2 - cech cohomology, functions as coefficients for the 1st cohomology, and 1-forms as coefficients for the zeroth cohomology. i.e. they say; "as the degree of the deRham complex goes down, the degree of the cech complex goes up."

Thank you very much!

mathwonk said:

it might help if we could read the rest of the sentence explaining the implications of A.7. Vanishing 1st de Rham cohomology means every closed global 1-form is exact, i.e. equals df for some function f.

here it is:


and this is the paper: https://arxiv.org/abs/1402.3140

mathwonk said:

well, if both the 1st and second cech cohomologies vanish, then the cech 2-cocycle represented by the nabc, would be a cech coboundary, of form nab+nbc-nac, with the nab defined on the Uab. Then one could modify the lambdas, by subtracting these nab and get new lambdas, Lab, which would now themselves be cech 1-cocycles. hence if the 1st cech cohomology also vanishes, these would be cech 1 -coboundaries of form Lab = Lb-La, where the La are defined on the Ua. Then finally subtracting the dLa from the 1-forms va, would give a family of 1-forms that agree on the overlaps Uab, hence would give a global 1-form v.

I have some questions because I'm new to many of these concepts. In polyhedral decomposition I do a triangulation of the manifold and associate a cycle to every open set, and I can obtain the boundaries; could I think of the cocycle as the duals of cycle, so something I can integrate on the cycles, such as the forms? what are the differences between cocycles and forms? Are the boundary relationship the same for cycle and cocycle? I also read in an article that the components of this triple connection are p-form valued cochain. What does it mean?

Then, if you can let me know if I got what you are saying: if H_1 is trivial every cycle with relative cocycle n_abc should be a boundary of something, with cocycle L_ab, which is itself a boundary of something, with cocycle L_a: I can take this coboundary dL_ab such that it exactly cancel d\lambda_ab, so that v_b=v_a. Is it right? If so, since every intersection set is contractible, I expect that a form is closed iff exact (i.e. every cycle is the border of something): so why this procedure is not possible if H_1 is non trivial?

the n_abc are cech 2-cocycles. i.e. a cech n-cochain with coefficients in p-forms, is a family of p-forms, each defined on an intersection of n+1 open sets of the cover. here the tech 2-cochains are constants (which are apparently (-1)-forms, whatever that means), defined on triple intersections. the cech 1-cochains are functions (0-forms), each defined on double intersections of opens, and the cech 0-cochains have values in the 1-forms, va, each defined on one open set of the cover.

but then they take the deRham coboundary, i.e. the exterior derivative d, of the 1-forms, and show that A.7 implies this gives a family of 2-forms, each defined on a single open, and which agree on overlaps., Hence they get a closed global 2-form, i.e. a 0-cech cocycle with coefficients in 2-forms, namely the curvature.

oh yes, there are two coboundary operators here, cech coboundary, and exterior differentiation. so you can differentiate a p form valued n-cochain, and get a (p+1_form valued n-cochain. you can also take the cech coboundary of a p-form valued cech n-cochain, and get a p-form valued cech (n+1) cochain. in all cases, the number of subscripts is one more than the cech degree of the cochain.

since there are two different coboundary operators, there are two different definitions of cohomology here, deRham cohomology, and cech cohomology, and there is a theorem relating them. there is also a third topological version of homology and cohomology, namely singular (co)homology, which is also related to these.

The deRham cohomology in degree n, (with real coefficients), is just the quotient space of all closed global n-forms, modulo all global exact n-forms. the cech cohomology in degree n, with coefficients in the p-forms, is the quotient of all cech n-cocycles, mod all cech n-coboundaries, with those coefficients.

aha! in the book by Bott-Tu, they discuss a Cech-deRham "double complex". They show that the cohomology of the double complex is always the same as the deRham cohomology (with real coefficients), and is also the same as the Cech cohomology (with real coefficients) if the finite intersections of the cover are contractible (a good cover).

now I believe these cohomologies are also the same ( for manifolds) as singular cohomology, and singular real cohomology is also dual to singular real homology, so if your book means that, when they say that H_1(M) = 0, that the 1st real singular homology vanishes, then that implies vanishing of the 1st real deRham and real cech cohomologies.

however, as I said, to get the conclusion that the 1-forms va can be chosen to agree globally, seems to require that the cech 2-cocycle n_abc is a cech coboundary, as well as that the resulting cech 1-cocycle L_ab, is also a cech coboundary. The second statement does follow from the vanishing of H_1(M), since it is a statement about vanishing of 1st cohomology with real coefficients, but I don't know why it would imply the cech 2-cocycle n_abc would be a cech 2-coboundary.

there is a lot of topology being assumed here. you might (or might not) enjoy chapters (0,1 and) 2 of the book by Bott-Tu, differential forms in algebraic topology.

anything that assigns a real number to a chain is a cochain with real coefficients. since singular 1-chains are just paths, differential 1-forms can be viewed as 1-cochains with real coefficients. a 1-chain is a cycle if it is a loop, i.e. a path whose endpoints agree. Since an exact 1-form df is evaluated on a path by subtracting the values of the function f at the two endpoints, an exact 1-form assigns zero to a 1-cycle. By stokes theorem, a closed 1-form assigns zero to a 1-boundary, i.e. to the boundary of a piece of surface. so there is a pairing of the deRham cohomology with the singular cohomology, and the theorem is that it is a perfect pairing in good cases, so that deRham and singular cohomology agree on (compact) manifolds, (without boundary?).

Recall from above: "Then one could modify the lambdas, by subtracting these nab and get new lambdas, Lab, which would now themselves be cech 1-cocycles. hence if the 1st cech cohomology also vanishes, these would be cech 1 -coboundaries of form Lab = Lb-La, where the La are defined on the Ua. Then finally subtracting the dLa from the 1-forms va, would give a family of 1-forms that agree on the overlaps Uab, hence would give a global 1-form v."

in this process, finding the L_a, such that L_ab = L_b - L_a, requires the vanishing of 1st cech cohomology, a global statement. This is where H_1(M)=0 is used. The fact that on contractible opens, a closed form is exact does not help. That just says the deRham cohomology of the contractible open set vanishes.

We are not saying that the cech 1 -cocycle L_ab is a deRham coboundary, of form dL_a, rather we are saying it is a cech coboundary of form ∂(L_*)_ab = L_b - L_a.

well I looked at the paper. one observation is that they mention the especially nice case of a "homology 3-sphere", for which in fact both H_1 and H_2 are zero, (with integer and real coefficients. In particular we would have all the vanishing we want, to do the calculation above, for a homology sphere.

the second observation is that they make clear that the group H_1 they are talking about is (say simplicial or singular) homology with integer coefficients, since they decompose it as a sum of components, one "free" of form Z^n for some n, and one "torsion" of form Z/n1 + Z/n2+...+Z/nk. and they give examples of computations, depending on which component the homology cycle gamma is chosen from. In particular vanishing of their H_1 gives also vanishing of both H^1 deRham and H^1 cech, with real coefficients.

One place to look for the consequences of assuming their H_1 = 0, seems to be equation (16), an exact sequence that computes the Deligne-Beilinson cohomology in that case, as just the quotient of the space of all 1-forms, modulo the subspace of closed 1-forms with integral periods. They mention there that DB cohomology parametrizes the configuration space, whose connected components are parametrized by H_1. In particular if H_1 = 0, then this configuration space is connected.

They also do some examples later in the section following equation (97) that may be of interest, since they may show how the choice of a topologically non trivial torsion cycle gamma in H_1 affects matters.

Those references sure include an awful lot of heavy duty mathematics! My hat is off to anyone who has read even a few of those tomes. To me this is really a differential geometry paper using physics terminology, which of course is why your question is posted here.

Lavinia is someone who could possibly be interested in this topic, and helpful. I myself have little else to say, i.e. I am about at the limit of my understanding.

well here are a couple more elementary definitions;

A chain complex is an indexed sequence of abelian groups An with homomorphisms ∂n going from An to An-1, such that the composition of any 2 consecutive maps is zero. the nth homology of the complex is the quotient of these two subgroups of An: H_n = kernel(∂n)/image (∂n+1) = cycles/boundaries.

A cochain complex is an indexed sequence of abelian groups Bn with homomorphisms dn going from Bn to Bn+1, whose compositions are again zero. The nth cohomology H^n of the complex is the quotient of the following two subgroups of Bn: H^n = kernel(dn)/image(dn-1) = cocycles/coboundaries.

Given any chain complex (An,∂n), we can define a dual cochain complex by setting Bn = Hom(An,Z) = group homomorphisms from An to the integers Z. We can define a differential dn on Bn, sending f in Hom(An,Z) to df in Bn+1 = Hom(An+1,Z), as follows. If h is in An+1, then ∂h is in An, and we let the integer df(h) be defined as f(∂h), i.e. df = fo∂:An+1—>An—>Z.

We can also take R-vector spaces as our chain groups, and R-linear functions as cochains.

If we have a triangulation of some space, we can define the group An as all Z-linear (or R-linear) combinations of (oriented or ordered) n dimensional simplexes, and the homomorphism ∂ as the boundary operator sending an n simplex to a linear combination of its n-1 diml faces. The resulting complexes compute the simplicial homology and cohomology of the space. Instead of simplicial n diml faces, we can also use continuous or smooth pieces of n diml surfaces. In good cases all give the same resulting homologies and cohomologies.

In case of smooth paths, surfaces, etc… we can define the n-cochains to be not all homomorphisms on the n-chains, but just as those homomorphisms given by integration of n-forms. In this way we can look at (integration over) n-forms as n-cochains.

Miraculously, i.e. by Stokes theorem, the coboundary operator here agrees with the exterior derivative, i.e. the integral of dw over S equals the integral of w over ∂S. Then, thought of as a cochain, an n-form is a cocycle if and only if it is closed, i.e. locally exact, and an n-form is a coboundary if and only if it is (globally) exact. Thus the nth cohomology of this “deRham” complex is just (closed n-forms)/ (exact n- forms). The theorem is that this complex has the same cohomology as the simplicial or singular complexes (with real coefficients).

[Actually this resemblance of differential calculus to cohomology is not a miracle, since, according to Hermann Weyl's Concept of a Riemann surface, the definition of cohomology arose by generalizing the behavior of integrals of (closed) differential 1-forms. In the case of a Riemann surface, the basic properties of an "integral function" (1-cocycle), which assigns a number to every closed curve, are to be linear over formal sums of curves, and to assign zero to a closed curve which is wholly contained in a contractible set such as a disc. Two such functions are cohomologous if they assign the same number to every closed curve. It is then of interest to determine just how nice a representative one can choose for a given cohomology class. For a (compact) Riemann surface, the deRham theorem says one can choose a smooth closed 1-form as representative of any class, unique up to addition mod an exact 1-form; and the Hodge theorem says one can even choose a unique harmonic representative, (unique since the only harmonic functions on a compact Riemann surface are constants). These theorems are apparently also true for higher degree forms on (compact) smooth and Riemannian manifolds, respectively. ]

In fact there is another interesting way to define a simplicial complex from a manifold using an open cover {Ua} as a “triangulation”. I.e. the open sets Ua are considered the vertices; the pairwise intersections are considered the edges, the triple intersections are the faces, and so on…

This is in fact reminiscent of an actual triangulation. I.e. given a triangulated space, associate to each vertex, the open set which is (the interior of) the union of all top dimensional simplices sharing that vertex. So to each vertex we have associated an open set, and these open sets give us an open cover of our space. Then notice the intersection of two of these open sets, associated to two vertices p,q, will equal (the interior of) the union of all top dimensional simplices which share the edge joining p and q. So the abstract “edge” joining the two open sets Up and Uq, i.e. their pairwise intersection, corresponds to the actual edge joining p and q. I believe these open sets are called "stars", so the correspondence between the usual simplices and the Cech simplices, associates to each simplex, the star of that simplex.

This gives us a “Cech” complex with boundary operator. Thus any method of assigning numbers, or functions to these open sets and their intersections will give us an abstract cochain complex.

E.g. if we assign a 1-form va to each open set Ua, we get a Cech 0-cochain with values in the 1-forms. Its Cech coboundary is the Cech 1-cochain that assigns to the “edge” Uab, just the difference vb-va, a Cech 1-cochain with values again in the 1-forms.

If we turn to the other differentiation, or coboundary, operator, the exterior derivative, we apply it to this Cech 0-cochain, and get (dva) a family of (locally) exact 2-forms associated to each open set, i.e. another Cech 0-cochain, but this time with values in the exact 2-forms. Indeed, as shown in your article, under the assumptions that the 1-forms va have a Cech 1-coboundary vb-va which is exact (= dLab), in particular that Cech 1-coboundary has values in the closed 1-forms, then the Cech 0-cochain (dva) with values in 2-forms, has Cech 1-coboundary equal to zero, hence (dva) defines a global 2-form which is locally exact, hence a global closed 2-form.

cheez wizz, is this as complicated to you as to me? sorry if it is unhelpful.

but one thing to try to keep straight is that what is meant by the words "cocycle", "coboundary", etc, depend on which coboundary operator is referred to, Cech or DeRham. I.e. a deRham p-cocycle is a closed p-form; and a Cech 0-cocycle with values in p-forms, is a globally well - defined p-form.

@polology:

some direct answers:

"could I think of the cocycle as the duals of cycle, so something I can integrate on the cycles, such as the forms?": qualified yes, see next answer (i.e. forms are cochains, closed forms are cocycles.)

" what are the differences between cocycles and forms?"
a real p-cochain assigns a real number to a p-chain. hence (integration over) a p-form, is a special type of p-cochain. a cocycle is a cochain with coboundary zero. hence a p-form is a cocycle if it has exterior derivative zero, i.e. a closed p-form is a particular example of a p-cocycle.

" Are the boundary relationship the same for cycle and cocycle?"
I am not sure what this is asking. if ∂ is the boundary operator on a p-chain, and if f is a (p-1) cochain, then the coboundary operator d acts on f as follows: df = f∂, i.e. since chains and cochairs are dual, the coboundary is dual to, or "adjoint" to the boundary. Since you are asking specifically about cycles and cocycles, then in a sense yes, a chain c is a cycle iff ∂c = 0, and a cochain f is a cocycle iff df = 0.

" I also read in an article that the components of this triple connection are p-form valued cochain. What does it mean?"
these are p-form valued Cech cochains. these are dual to Cech chains. previously we have been talking about n-chains as if they were pieces of n-dimensional surfaces, i.e. simplicial or singular chains. here we are talking about Cech n-chains, which are intersections of (n+1) tuples of open sets. so a Cech n-cochain assigns a number, (or p-form) to an intersection of (n+1) open sets. this assignment is not done by integrating, but just by telling you what form goes with what intersection, e.g. va goes with Ua, and Lab goes with Uab.

there are two coboundary operators here, the Cech coboundary and the exterior derivative. hence these guys can be a cocycle in two different senses. i.e. either the Cech coboundary or the deRham exterior derivative can be zero. (These triples are made up of elements with two different indexings, the indexing by the number (n+1) of open sets, and the indexing by the degree p of the form.) the deRham coboundary is zero if the form is closed, and the Cech coboundary tells you something about compatibility of the forms on overlaps of their open sets.

e.g. a Cech 0-cochain with values in 0- forms, is a family of (smooth) functions fa, one on each open set Ua of the cover. It is a deRham cocycle iff the functions are all (locally) constant dfa = 0. It is a Cech cocycle iff the functions agree on overlaps, i.e. if the fa together define a global smooth function.

these properties are related, since if a family of functions fa are such that they differ on overlaps by constants, i.e. if the fa are not constant but the differences fb-fa are constant, then the Cech 1- coboundary (fb-fa) on Uab is a deRham 0-cocycle, d(fb-fa)=0. It follows that the family of derivatives (dfa) defines a Cech 0-cocycle with values in 1-forms, i.e. that this family is a globally defined 1-form.

I guess the point here is that the two coboundaries apparently commute, i.e. (I am out of different letters, but if I write ∂ now for the Cech coboundary) then d∂ = 0, implies also ∂d = 0. i.e. if the Cech coboundary is deRham closed, then the deRham coboundary is well defined. this sort of argument was used in the paper to explain how to get a well defined connection 2-form (dva), from the family (va) of 1-forms. I.e. vb-va =dLab being deRham closed, implied (dva) was globally well defined.

[one way to prove the deRham theorem that topological cohomology can be computed by differential forms, is to look at such things as elements in a big double complex, and compare three different cohomologies, the deRham cohomology using d on one index and keeping the other index zero, or doing the opposite using ∂, or combining them somehow into a cohomology of the full double complex using ∂+d, then showing they are all the same. something like showing the vertical, horizontal, and diagonal cohomologies all agree. I am not expert here and have forgotten the details. this is explained in Bott-Tu.]

Hi @mathwonk thank you a lot for those answers! It's been a while since I was searching for someone who could help me understand those concepts. I can't answer you in the conversation you opened because for unknown reasons the site detects some spam-like content in my reply. Sorry if I haven't responded yet, I just need some time to grasp them so that I can ask meaningful questions if I still have any :)

Okay, here I am. Sincerely it was hard for me to have a deep understanding of some things of the previous answers due to my unfamiliarity with these concepts (i think after the exam i'll come back here to have a better understanding), but the last answer was extremely helpful. I'll try to give a (hope not over)simplified picture of what I understood about the intial question, if you want let me know if it's correct, even if not so formal.

In order to study the gauge theory on a nontrivial manifold we have to consider a good cover (every open set and intersections are contractible) and a triangulation (associate vertex to open sets, edges to intersection and so on), obtaining a simplicial complex. Combination of n-simplexes with boundary operator gives us a "Cech" complex: dual to this is a cochain complex obtained assigning p-forms to each sets and intersection. So we have two boundary operator: the exterior derivative and the Cech coboundary (illuminating quote: "the deRham coboundary is zero if the form is closed, and the Cech coboundary tells you something about compatibility of the forms on overlaps of their open sets").

Now: we can start assigning a 0-cochain 1-form valued to every open set. If the Cech coboundary is trivial, this means v_a=v_b in the intersection and so it's a global 1-form: this is the case of the homology sphere, indeed for the argument of the double complex in this case cech, de rham and singular cohomologies (which is dual to real singular homology) are the same, and then vanishes. But if the cohomology is not trivial, the coboundary is v_b-v_a=l_{ab} (where l_{ab} it's a 1-cochain 0-form valued not yet specified).

Here it comes the crucial part. You said:

"under the assumptions that the 1-forms va have a Cech 1-coboundary vb-va which is exact (= dLab), in particular that Cech 1-coboundary has values in the closed 1-forms, then the Cech 0-cochain (dva) with values in 2-forms, has Cech 1-coboundary equal to zero, hence (dva) defines a global 2-form which is locally exact, hence a global closed 2-form."

In the beginning I was trying to understand why v_b-v_a should be an exact form, but it seems now that it is an assumption made to obtain a global closed 2-form, as Chern Simons theory wants. Indeed, if we simply want dv_a=dv_b => d(v_b-v_a)=0, so l_{ab} should be a closed form.

If the last statement it's correct, I have one last question: the reason why l_{ab} it's not only closed but also exact, is due to the fact that it is defined on a contractible open set (the intersection), so every closed form is exact? (I know, you said that this doesn't help, but I can't think to other reasons)

yes, on a contractible set, closed and exact forms are the same. i.e. closed is the same as locally exact.

Thank you again for your answers and patience, the exam went well :)

thank you for the feedback. congratulations!