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I would like generalize the following theorem from Calc III: "If $U$ is an open, connected subset of $\mathbb{R}^2$ or $\mathbb{R}^3$ and $\omega$ is a 1-form on $U$ with $d\omega = 0$ on $U$, then TFAE: 1) $\omega = df$ on $U$ 2) $\omega$ is path-independent 3) For any closed loop $L$ lying entirely inside $U$, $\oint_{L} \omega = 0$"

The generalization I would like prove is "If $U$ is an open, connected subset of $\mathbb{R}^3$ with $\pi_1(U)$ finitely generated and $H_{*,dR}(U)$ finitely generated and $\omega$ is a 2-form on $U$ with $d\omega = 0$ on $U$, then TFAE: 1) $\omega = d\theta$ for some 1-form on U 2) For any 2 compact, oriented surfaces with boundary $S_1$ and $S_2$ bounding the same loop $L$ in $U$ and inducing the same orientation on $L$ , $\iint_{S_1} \omega = \iint_{S_2} \omega$ 3) For any closed, oriented surface $S$ lying entirely inside $U$, $\unicode{x222F}_{S}\ \omega = 0$"

The implications other than 3) $\Rightarrow$ 1) [or 2) $\Rightarrow$ 1)] follow from Stokes's Theorem and some standard TFAE round-robin/cycle proof nonsense.

Thanks much in advance. Any assistance is appreciated.

  • If $U$ is a simply connected open subset of $\Bbb{R}^2$ (otherwise $U = {0<|x+iy|<2}, \omega = \frac{dx+idy}{x+iy}, d\omega= (d\frac{1}{x+iy}) \wedge (dx+idy)=\frac{-1}{(x+iy)^2}(dx+idy) \wedge (dx+idy)=0$, $\int_{|x+iy|=1} \omega =\int_{|x+iy|=1} d\log(x+iy)= 2i\pi$ – reuns Apr 24 '19 at 13:19
  • I'm not sure that your reply is on point, but log is not single-valued; there is no 1-form $\omega$ with a single-valued 0-form anti-derivative $f$ with $\oint_L \omega \ne 0$. Thanks for replying anyways! – Jeffrey Rolland Apr 24 '19 at 13:39
  • $ \frac{dz}{z}$ is well-defined on $U : |z| > 0\subset \Bbb{C}$, on simply connected open subsets $V \subset U$ it is $d \log z$ for some branch of $\log$ so it integrates to $0$ on closed curves $\subset V$ and $d( \frac{dz}{z})=0$, and it integrates to $2i\pi$ on $|z|=1$. Ie. you need to ask $U$ being simply connected. To integrate a 2-form I think you need to find a change of variable so that you are integrating $f(u,v,w)du \wedge dv$ on $(u+t,v,w), t \in [0,T]$ – reuns Apr 24 '19 at 13:45
  • Yes, in integrates to 0 on closed curves which a subset of the simply-connected $V$ and has an anti-derivative there; it down not integrate to zero on |z| = 1, which not a subset of any simply-connected subset of |z| > 0, and so has no anti-derivative on any subset W with {|z| = 1} $\subset$ W $\subset$ {|z| > 0}. You don't need simply-connected for the theorem; simply-connected is sufficient, but not necessary. – Jeffrey Rolland Apr 24 '19 at 13:48
  • Also, $\frac{dz}{z}$ is not path-independent: its integral over the first quarter circle, oriented CCW, is $\frac{\pi}{2}i$, whereas its integral over the last 3/4, oriented CW, is $-\frac{3\pi}{2}i$. – Jeffrey Rolland Apr 24 '19 at 14:00
  • I think this may help. The (real) differential form $\omega = \frac{2x}{x^2+y^2}dx + \frac{2y}{x^2+y^2}dy$ is defined on the non-simply connected connected region |z| > 0, but its integral over |z| = 1 is 0; it has anti-derivative$ f = \ln(x^2+y^2)$ on U (where f is single-valued) and is also path-independent on U. – Jeffrey Rolland Apr 24 '19 at 14:18
  • If $U$ is not simply connected then $d \omega = 0$ isn't enough for $\omega$ having an anti-derivative. That $d \omega = 0$ means $\int_\gamma \omega = 0$ for each curve $\omega$ which is $U$-homotopic to a point. $\omega$ has an anti-derivative on $U$ iff $d \omega = 0$ and $\int_\beta\omega = 0$ for each generator $\beta$ of the homotopy group $\pi_1(U)$. – reuns Apr 24 '19 at 15:42
  • Your last statement is another theorem, derivable from the one I have at the top of my question. I generally give the proof that your theorem follows from mine (as well as a proof of mine) in my Calc III course. But my actual question is about the anaolg for 2-forms. – Jeffrey Rolland Apr 24 '19 at 18:40

1 Answers1

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(Many special thanks to Craig Guilbault) I have a proof for 3) $\Rightarrow$ 1). It is orders and orders of magnitude harder/more complicated than the proof of 2) $\Rightarrow$ 1) for the first theorem as outlined in Anton. It involves "Munkres-deRham cohomology" (as outlined in Analysis on Manifolds by Munkres - his definition of $H^0(U)$ is different from standard deRham cohomology but is otherwise the same) and the standard Mayer-Vietoris sequence.

Munkres shows that for $U = \mathbb{R}^3 \backslash \{(0,0,0)\}$, $$H^k_{M-dR}(U) \cong \begin{cases} \mathbb{R} & k=2 \\ 0 & \text{ow} \end{cases}$$ and shows this implies 3) $\Rightarrow$ 1) for this $U$. Note $U$ strong deformation retracts onto the standard 2-sphere $S^2$.

For $U = \mathbb{R}^3 \backslash ([S^1 \times \{0\}] \cup \{(0,0,z)\ |\ z \in \mathbb{R}\})$, set $X = \mathbb{R}^3 \backslash [\{(x,y,z)\ |\ x^2+y^2=1, z \ge 0\} \cup \{0,0,z)\ |\ z \in \mathbb{R}\}]$, $Y = \mathbb{R}^3 \backslash [\{(x,y,z)\ |\ x^2+y^2=1, z \le 0\} \cup \{0,0,z)\ |\ z \in \mathbb{R}\}]$, and $A = X \cap Y$. Then U = $X \cup Y$. Note $U$ strong deformation retracts onto the genus-1 torus $T^2$.

I will outline later why $$H^k_{M-dR}(A) \cong \begin{cases} \mathbb{R} & k = 0 \\ \mathbb{R} \oplus \mathbb{R} & k=1 \\ 0 & \text{ow} \end{cases}$$ and $$H^k_{M-dR}(X) \cong H^k_{M-dR}(Y) \cong\begin{cases} \mathbb{R} & k=1 \\ 0 & \text{ow} \end{cases}$$ I will also outline later why this leads to the Mayer-Vietoris sequence

\begin{equation} \scriptstyle 0 \to H^0(U) \to^{j^0} H^0(X)\oplus H^0(Y)\to^{i^0} H^0(A) \to^{d} H^1(U) \to^{j^1} H^1(X)\oplus H^1(Y) \to^{i^1} H^1(A) \to^{d} H^2(U) \to^{j^2} H^2(X)\oplus H^2(Y) \end{equation}

which leads to the three exact sequences

$$0 \to H^0(U) \to^{j^0} H^0(X) \oplus H^0(Y)\to^{i^0} \text{im}(i^0)$$ $$0 \to H^0(U) \to 0 \to 0$$

then

$$H^0(X)\oplus H^0(Y) \to^{i^0} H^0(A) \to^{d} H^1(U) \to^{j^1} \text{im}(j^1) \to 0$$ $$0 \to \mathbb{R} \to H^1(U) \to \mathbb{R} \to 0$$

and

$$0 \to \text{coker}(i^1) \to^{d} H^2(U) \to^{j^2} H^2(X)\oplus H^2(Y)$$ $$0 \to \mathbb{R} \to H^2(U) \to 0$$

and so leads to $H^k_{M-dR}(U) \cong \begin{cases} 0 & k = 0 \\ \mathbb{R} \oplus \mathbb{R} & k=1 \\ \mathbb{R} & k = 2 \\ 0 & \text{ow} \end{cases}$

I will then outline later how to prove 3) $\Rightarrow$ 1), actually computing the antiderivative 1-form for a closed 2-form $\omega$ on $U$ with $\unicode{x222F}\ _{T^2}\ \omega = 0$ using a "co-chain homotopy operator" ($\mathcal{P}$ in Munkres, $K$ in Bott and Tu, and $L$ in this post) - basically integrating the 2-form with respect to $x_3 = z$ to get a 1-form on the projection of $A$ into the plane and then taking the pull-back along this projection - from this cohomology calculation for the $U$ in question.

Finally, I will show later for the various other $U$'s that have points and/or wedges of circles and/or vertical lines deleted how to find subsets $U_i$ with an associated closed, oriented surface $S_i$ onto which they strong deformation retract and associated $A_i$'s, $X_i$'s, and $Y_i$'s, compute $H^*_{M-dR}(U_i)$ for each $i$, use Mayer-Vietoris with this to compute $H^*_{M-dR}(U)$, and, for a closed 2-form $\omega$ with $\unicode{x222F}_{S_1}\ \omega = 0$ for each $i$, compute the antiderivative 1-form from the co-chain homotopy operators for the various $A_i$ with identifications via the overlaps of the various $U_i$'s from Mayer-Vietoris. The final answer will likely not be suitable for a traditional Calc III student (boo-hiss!).

"If you can't explain it simply, you don't understand it well enough." - Albert Einstein

  • Please avoid excessive edits, as those bump the question up to the front page. You should prepare the answer in advance, especially when self-answering a question. You can also use https://math.meta.stackexchange.com/questions/4666/sandbox-for-drafts-of-long-complex-posts if you want to do this in an environment compatible with the site. – Asaf Karagila Jul 04 '19 at 13:16