Recently I reviewed some knowledge about topology in the book The Geometry of Physics-An Introduction by Theodore Frankel, contemplating a lot and having some understandings or speculations. What about exact forms and closed forms perplexed me. The definition of them seems to be involved with the topological strecture of the manifold they are defined. And de Rham’s theorem intrigued me much. After this review, I understand them deeper and find some new problems at the same time. I will note them in the following paragraphs.
Corollary (5.4) says If α is a (continuously differentiable) 1-form whose integral over all closed curves on Mn vanishes, then ß is exact. I deem this principle can be generalized to a p-form, that is if ß is a continuously differentiable p-form whose integral over all p-cycles on Mn vanishes, then ß is exact. Furthermore I deem this generalization can be separated into two procedures, closedness then exactness: First, if ß is a continuously differentiable p-form whose integral over the p-cycles which bounds on Mn vanishes, then ß is closed. Second, if this closed form ß whose integral over the remaining p-cycles on Mn vanishes, then ß is exact. If the p-th Betti number of Mn doesn’t vanish, then the remaining p-cycles are the linear combinations of the basis p-cycles of the p-th homology group. If the p-th Betti number of Mn vanishes, then there are not remaining p-cycles; in this case, if ß is closed, it must be exact.
Among the analysis in the former paragraph, several principles are involved. In the first place, in Section 5.2 Closed Forms and Exact Forms, it describes two observations: First, The integral of an exact form over an orientable closed manifold (i.e., compact without boundary) is 0; Second, The integral of a closed form over the boundary of an oriented compact manifold is 0. I deem their converses (the generalized version—consider all cycles on a manifold M) are If the integral of a continuously differentiable p-form ß over all p-cycles on M vanishes, then ß is exact (This is the generalized version of Corollary (5.4))[The argument underlying it is: the boundary of any p-cycle is 0, then if there is a p-1 form κ such that ß=dκ, Stokes’s theorem dictates ∫zpß=∫∂zpκ=0, which is satisfied only when ∫Mß=0.]; and If the integral of a continuously differentiable p-form ß over all bounding p-cycles on M vanishes, then ß is closed[The argument underlying it is: if dß=0, then Stokes’s theorem dictates ∫zpß=∫Cp+1dß=0, which is satisfied only when ∫zpß=0, where zp is the boundary of cp+1.]. Then there is a tricky place for the case p=n. If M is closed, then M must not bound because the manifold it bounds must be of dimension n+1 but n has been the largest dimension. Then if the integral of a continuously differentiable n-form β over M vanishes, then β is exact because M is the only n-cycle on M. And an exact form must be closed. There are no bounding n-cycles on M to test if β is closed but we still know it is closed because dß=0 for dß is an n+1 form, which must be 0 on a n-manifold. So in this case though there is no bounding cycle, we can still use other way to judge if a form is closed. Then let’s turn to the case when M is not closed. In this case, we know ß is closed by the same argument as the previous case. Is ß exact? We want to know if there is a n-1 form κ such that ß=dκ. But there is no n-cycle on M, so we can’t use the vanishing integral value to ensure the existence of κ. So we may need other way to check if κ exists. In the second place, Theorem (5.3) says If M is a manifold with first Betti number 0, then every closed 1-form α on M is exact. I think it can be generalized to Every closed p-form on M whose p-th Betti number is 0 is exact. This is involved in the final sentence of the last paragraph.
Then let us come to de Rham’s Theorem, which is involved in the sentence before the final one of the second paragraph. RP, the de Rham vector space, is the collection of equivalence classes of closed p-forms on M. HP, the p-th homology group, is the collection of equivalence classes of p-cycles on M. Associated with a representative p-form β from a certain de Rham class, we can define a linear functional I: RP→HP(M,R) by I(ß)(z)=∫z ß, where z is a representative p-cycle from a certain homology class and HP is the dual vector space of HP and is called the p-th cohomology vector space. The integral value is inependent of the representatives from their respective classes because the integral of a closed p-form over a bounding p-cycle vanishes and the integral of an exact p-form over a p-cycle vanishes. When M is compact, HP being finite-dimensional, one can choose bp p-cycles on M, none of whose linear combinations is bounding, as the basis for HP, bp being the p-th Betti number, the dimension of HP, the number of classes in HP. I is an isomorphism: first, I is onto, that is if zp(1),…, zp(bp), is a p-cycle basis of HP, and π(i) are arbitrary bp real numbers, then there is a closed form ß in RP such that ∫zp(i) ß=π(i). Second, I is 1:1; that means if I(ß)(z)=∫z ß=0 for all cycles zp on M, then ß is exact. Because a finite-dimensional vector space has the same dimension as its dual space, dim HP= dim HP, and because I is an isomorphism, dim RP=dim HP=dim HP=bp. Thus one can choose bp closed p-forms, none of their linear combinations being exact, as the basis for RP. In the above, I deem one can view any bounding p-cycle and any excat p-form as some kind of 0-vector on HP and RP, respectively. Then the integral of any ß over a bounding p-cycle must vanish; when input with a 0-vector, one should get the value no matter what functional ß is employed. And the integral of an exact p-form over any p-cycle should vanish (the converse of it is the explanation following 1:1; if all p-cycles in HP are mapped to 0, then it can only be the case the functional defined by ß is a 0-vector; that is β is exact.); 0-functional must get the value 0 no matter what p-cycle is input. Furthermore, that no linear combination of the bp basis p-cycles (p-forms) is bounding means no one p-cycle (p-form) among HP (RP) is equivalent to the linear combination of the other bp-1 p-cycles (p-forms). Also, the statement before the final one of the last paragraph can also be explained this way: Because the p-th Betti number of M vanishes, there are only bounding p-cycles on M, and because dim RP=bp=0, there must be only exact p-form in RP. Another explanation is in the end of the second paragraph: the norm for an (closed) exact p-form is its integral for all (bounding) p-cycles vanishes but for the case bp=0, there are only bounding p-cycles, thus closedness is also exactness.
The functional I resembles the vectors v and 1-forms on M, the spaces of them being dual to each other; given a specific 1-form α, input with a vector gives a value; the functional α(v) is an isomorphism. On a manifold of n-dimension M, given n basis vectors e1,…,en, if τi are arbitrary n numbers, then there is a 1-form α on M such that α(ei)=τi , i=1,…,n; there are n equations for n variables, n components of α, so they must be able to be determined uniquely. That means α(v) is onto. If α(v)=0 for any v, then α=0 (all components of α vanish). This means α(v) is 1:1. Let’s advance to the p-form ß on M. The functional β(v(1),…,v(p)) defines an isomorphism, too, among which v(1),…,v(p) are p linear independent vectors on M. Given cnp arbitrary numbers λ(i) , i=1,…,cnp, there is a p-form ß such that ß(e(p))=λ(i), where e(p) denotes certain p basis vectors among e1,…,en. There are cnp sets of e(p). Again, there are cnp equations for cnp variables, the cnp components of ß, so they can be decided uniquely. That means β(v(1),…,v(p)) is onto. And if β(v(1),…,v(p)) =0 for all sets of v(1),…,v(p), then β=0(all components of β vanish). This means β(v(1),…,v(p)) is 1:1. All these reasoning can be applied to de Rham’s theorem. Let us summary as follows. A general closed p-form β in RP must be able to be expanded in terms of bp basis closed p-forms in RP, thus β having bp components. Thus the bp equations ∫zp(i) ß=π(i), i=1,…,bp can determine bp components of β uniquely. What differ between the resemblance are 1. For de Rham’s theorem, we are restricted to closed p-forms and unbounding p-cycles, which leads the need to redefine the “o-vectors” for them, ie. taking exact p-forms and bounding p-cycles as “0-vectors”; in the case of its counterpart, we just take the usual 0-vector (its all components vanish) and the usual 0 form (all of its covariant components vanish); thus saying two cycles (closed forms) are equivalent to each other in de Rham’s theorem is the analogue of saying two vectors (two forms) are linear dependent; 2. The dimension for RP and HP are both bp while the dimension of the vector space for vectors on M and its dual 1-form space are both n and the dimension of the vector space for p-forms are cnp and there are cnp sets of p vectors. Therefore for the basis p-cycles on HP, we can define its dual basis closed p-forms on RP such that I (μ(i))(zp(j))=∫zp(j)μ(i)=δ ij , where μ(i), i=1,…,bp is the basis for RP and zp(j), j=1,…,bp is the basis for HP.