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This is a question from a physicist, so please be kind.* Suppose that $M$ is an orientable smooth manifold without boundaries and $\omega$ a form of an appropriate degree such that it can be integrated over $M$, $$ I=\int_M \omega. $$ The objective is to compute $I$. According to the Poincaré or Dolbeault-Grothendieck lemma (for real and complex manifolds respectively), locally (in some coordinate neighbourhood, $U_i$, where $M$ looks like an open subset of $\mathbb{R}^n$ or $\mathbb{C}^n$, with $n$ a positive integer, and because i don’t want to restrict the scope i will write $\mathbb{K}$ for either of the two fields) $\omega$ is exact, $$ \omega_i=dA_i, $$ where $A_i$ is some differential form defined in $U_i$ of the appropriate degree. If $A_i$ were globally defined (in which case I suppose, $A_i=A$, for all $i$) then we could apply Stokes' theorem, $$ \int_MdA = \oint_{\partial M}A=0, $$ and learn that $I=0$ (because $\partial M$ is null). I am in the unfortunate (and I believe very common) situation where I only know $\omega$ locally (in particular I have an explicit expression for $A_i$) and I want to compute $I$. So my first question is the following:

  1. What precisely does it mean for $A_i$ to be only locally and not globally defined? How can I check whether my $\omega_i$ is globally or only locally exact given only an explicit local expression for $A_i$ and the corresponding transition functions on chart overlaps? For instance, might it be true that in order for $A_i$ to be globally defined it must transform under changes of coordinates as an antisymmetric tensor (on chart overlaps, $U_i\cap U_j$), and might this be sufficient for $A_i$ to be globally defined?

Suppose now that I am in the fortunate situation where I know the answer to this question, and I have concluded that $A_i$ is not globally defined and hence that $\omega$ is not globally exact. The next question is:

  1. How can I explicitly reconstruct $I$ given the explicit local expression $\omega_i=dA_i$ on $U_i$ and the corresponding transition functions on patch overlaps $U_i\cap U_j$?

To be slightly more precise here I am implicitly considering an atlas for $M$, i.e. a family of charts, $\{(U_i,\phi_i)\}$, with $\{U_i\}$ a family of open sets such that $\cup_i U_i=M$, and $\phi_i:U_i\rightarrow \mathbb{K }$ a homeomorphism from $U_i$ to an open subset of $\mathbb{K}$ (in particular, the maps $\phi_i$ are to be considered known and identified with a convenient set of local coordinates). The case of interest is when the transition functions $f_{ij}=\phi_i\circ \phi_j^{-1}$ from $\phi_j(U_i\cap U_j)$ to $\phi_i(U_i\cap U_j)$ are $C^{\infty}$ and also known.

For question 2 Stokes' theorem comes to the rescue, given that in a patch $U_i$, $$ \int_{U_i}dA_i = \oint_{\partial U_i}A_i, $$ but how exactly does one sum over $i$ to reconstruct the full integral $I$ making use of the transition functions which map $A_j$ to $A_i$ on patch overlaps?

*I have rewritten the question completely because following a fairly extensive discussion with @John Hughes (whom I think I annoyed quite a bit, see below) and his correspondingly good comments, it became very clear that my intended question was not clearly formulated.

Wakabaloola
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I'm not certain what you mean here. It's certainly true that if $M$ is, say, $S^1$, then there's a 1-form whose integral over $M$ is nonzero. You'll often see, for instance, $$ \int_{S^1} ~d\theta = 2\pi $$

Sadly, that's a cruel joke, because "$d\theta$" is the name given to a particular 1-form that is not, in fact, "d" of any $0$-form. Explicitly, $$ d\theta (x, y) = \frac{-y}{x^2 + y^2} dx + \frac{x}{x^2 + y^2} dy $$ on $\Bbb R^2 - \{ (0,0) \}$ and the form we call $d\theta$ on $S^1$ is the restriction of this to the subset of all points $(x, y)$ with $x^2 + y^2 = 1$.

This may not be the enlightenment you were looking for, but your first main point "There are situations where the integral of the total derivative does not vanish" just seems wrong to me (because of Stokes' theorem!). (I'm assuming here that we're talking about forms $A$ and $dA$ that are everywhere defined on a compact manifold $M$, possibly with boundary.)

John Hughes
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  • The statement "There are situations where the integral of the total derivative does not vanish" is correct: consider the integral $I = \int_{M_g}d(\bar{\omega}\int^z\omega)$, where $\omega$ (or $\bar{\omega}$) is one of the $g$ globally defined (anti-)holomorphic differential 1-forms (whose existence is guaranteed by a famous index theorem) on a compact genus $g$ Riemann surface $M_g$. So $I$ is an integral of a total derivative. One can show that $I = -2i{\rm Im}\Omega$, and so is not zero. This is an example, but my question and interest above is much broader than this. – Wakabaloola Mar 13 '18 at 12:46
  • Perhaps you could explicitly write out what those 1-forms are for $g = 1$, where we're talking about a torus, which we can parameterize via $\theta$ and $\phi$. I'm wondering whether it'll turn out that your $\omega$ is just $d\theta$ in disguise. Frankly, I don't understand your notation, so I can't say whether what you've written is right or wrong. I encourage you to get explicit and write down what you're talking about in coordinates in a nice simple case like the torus, and possibly enlighten us both. – John Hughes Mar 13 '18 at 12:51
  • Certainly: for $g=1$, $\omega=dz$, $\bar{\omega}=d\bar{z}$, $d=dz\partial_z+d\bar{z}\partial_{\bar{z}}$, where $z,\bar{z}$ are global (anti-)holomorphic coordinates with identifications $z\sim z+1$ and $z\sim z+\tau$, where $\tau\in \mathbb{C}$ (with ${\rm Im} \tau>0$) characterises the complex structure of the torus. Then, $I=\int_{T^2}d(zd\bar{z})$ and $zd\bar{z}$ is a 1-form. So this trivial example reduces to $I=\int_{T^2}dz\wedge d\bar{z}=-2i{\rm Im}\tau$. The genus-$g$ generalisation follows from the Riemann bilinear identity. – Wakabaloola Mar 13 '18 at 13:09
  • Excellent. This shows that there's a 2-form $\omega$ whose integral is nonzero. [In real coordinates $z = x + iy$, it appears that $dz \wedge d\bar{z}$ is something like $C~ dx \wedge dy$ for some constant $C$.] Can you tell me a 1-form $\eta$ such that $\omega = d\eta$? Because what you claimed was that there was a 2-form that was a total derivative, and whose integral does not vanish. (It'd be great if you could tell me in real coords, because I'm much weaker on complex stuff (grad school was a long time ago!), but ...) – John Hughes Mar 13 '18 at 13:25
  • Yes, in real coordinates, $\eta = (x+iy)dx+(y-ix)dy$, $d=dx\partial_x+dy\partial_y$, and so $\omega = d\eta=-2i dx\wedge dy$. – Wakabaloola Mar 13 '18 at 13:42
  • Incidentally, by the Poincare or Dolbeault-Grothendieck lemma (for real and complex manifolds respectively) ``everything'' is locally a total derivative, and the torus is a little special because the manifold can be covered by a single chart, which is why I gave the genus-$g$ case first in my counter-example to your claim. My interest is in the general case where a single chart that covers the whole manifold doesn't exist, so that one needs to construct transition functions and establish relations between the $A$'s on different charts and then glue everything together to compute $\int_MdA$. – Wakabaloola Mar 13 '18 at 14:13
  • So at the point $(x, y) = (0,0)$, I get that $\eta = 0dx + 0 dy$; and at $(x, y) = (1, 0)$, I get that $\eta = dx - idy$. Now $(0,0)$ and $(1, 0)$ are the same point on the torus, so this seems inconsistent. In particular, just because $\eta$ is defined on the universal cover $\Bbb R^2$ doesn't mean it's defined on the quotient manifold $T^2$. So: I'm still not seeing a well-defined 1-form on the torus whose exterior derivative is your $\omega$. – John Hughes Mar 13 '18 at 14:18
  • Very good, the quantity $\eta$ is multiple-valued because of the identification, but nevertheless the integral $\int d\eta$ is perfectly well-defined. (I never required that $\eta$ be well-defined globally.) So I return to (a simpler version of) my question: what properties must the quantity $A$ satisfy in order for $\int_MdA =0$ when $M$ has no boundary? As stated in the question, I suspect that it should be globally well-defined, but what precisely does that mean (in terms of transition functions and charts, etc.)? – Wakabaloola Mar 13 '18 at 14:40
  • In case I'm not already crystal clear: I'm looking for the necessary and sufficient conditions on $A$ which ensure $\int_MdA=0$ when the smooth manifold $M$ (of any dimension $d$) is connected, orientable and has no boundary. – Wakabaloola Mar 13 '18 at 14:54
  • May we also assume that $M$ is compact? If so, then I believe the condition on $A$ is "it should be defined and differentiable on all of $M$, for then Stokes theorem tells us that the integral you've written is zero." – John Hughes Mar 14 '18 at 01:15
  • "Very good, the quantity η is multiple-valued because of the identification, but nevertheless the integral ∫dη is perfectly well-defined. (I never required that η be well-defined globally.)" If $\eta$ isn't defined, how are you doing to define $d\eta$? You say "I'm looking for necessary and sufficient conditions on $A$ (which in the example I asked for, you called $\eta$) for ..."; I asked for an explicit $A$; you gave me something that was not in fact a form on $M$ and was, in fact, exactly analogous to the $d\theta$ example in my answer. At this point, you're on your own. – John Hughes Mar 14 '18 at 01:18
  • I've checked; for general $M$, the condition is that $A$ be compactly supported (i.e., it's zero everywhere outside some compact set). For compact $M$, that condition is met by every form, hence the answer is "the integral over $M$ of $dA$ is $0$ for every smooth $(n-1)$ form on a compact manifold of dimension $n$". In short, in the compact case, no conditions at all are necessary. – John Hughes Mar 14 '18 at 01:46
  • Many thanks for your further comments. The cases of interest to me are when $A$ is not globally defined, but is well-defined in local patches and has well-defined transition functions from one patch to another. (This kind of situation arises all the time in gauge theories and string theory for example, and I'm trying to understand how it works.) So the cases you are explicitly referring to are in a sense trivial from this viewpoint, because you have only explicitly discussed cases where $A$ is well-defined throughout $M$. See e.g. the Appendix of https://inspirehep.net/record/20725?ln=en – Wakabaloola Mar 14 '18 at 12:04
  • Sigh. You asked a perfectly well-formulated mathematical question, using pretty standard notation...only it turns out you had a different question in mind. Your informal description of $A$ in this comment (well defined in local patches, has well-defined transition functions from one patch to another) is almost a perfect informal description of a differential form (i.e., the exact thing I've been asking you to exhibit, and the exact thing to which Stokes' Theorem applies). I still have no idea what the heck you're talking about. But I think you should read about diff'l forms, and do homework... – John Hughes Mar 14 '18 at 12:29
  • ...problems on them until you're really pretty solid on exactly what they are. Then you can really understand the proof of Stokes' theorem (which is, after all, a theorem), and perhaps determine whether the objects you care about bear any relation to those that are well-known and well-studied in mathematics. Even if that studying turns out not to be directly relevant to what you want to study in physics, it might help you in other ways (and save the time of folks like me). BTW, "The cases of interest to me are when A is not globally defined" directly contradicts your question's title. :( – John Hughes Mar 14 '18 at 12:32
  • Thanks for taking the time and for the advice, i appreciate it greatly and don't take it for granted. My intended question is indeed not clearly formulated, because I have written $\int_M dA$ when actually what I mean is $\int_M \omega$, where $\omega$ is a well-defined form of the appropriate dimension which locally is of the form $dA$, not globally (somewhat akin to the Kahler form for example), so that $A$ is also not defined globally. I will amend the question appropriately. – Wakabaloola Mar 14 '18 at 14:52
  • Furthermore, question (1) as formulated remains unanswered, and only when this is answered will I be able to fully understand the answer to question (2). (BTW, the title doesn't contradict anything: if I understand what 'globally defined' means I will also understand what characterises non-globally defined quantities.) – Wakabaloola Mar 14 '18 at 15:01
  • And thanks! your comments have been very helpful. – Wakabaloola Mar 14 '18 at 15:02
  • Question 1 is the essence of the definition of a differential form on a manifold; it's covered in virtually every book on differential topology and most of those on differential geometry. There's a very nice reading list at https://math.stackexchange.com/questions/13575/teaching-myself-differential-topology-and-differential-geometry which might have a book that points you in the right direction. – John Hughes Mar 14 '18 at 17:32
  • Great, thanks for the link. By the way, I have completely rewritten the main post, and hopefully it is clearer now what exactly I intended to ask in the previous version! – Wakabaloola Mar 14 '18 at 17:45
  • The differential topology book I have is Bott and Tu (and also Nakahara discusses this stuff), and the answers (I am sure) are in there, but I have just not been able to extract a simple answer to the correspondingly simple question 1 (and when it comes to question 2 all I have been able to find is general answers whereas I want to see the details, possibly in a specific example) -- hence my post here – Wakabaloola Mar 14 '18 at 18:02