If the second homotopy group is trivial, surface integral is independent with surface

Question

Let $\Omega$ be a domain in $\mathbb R^3$ and let $\pi_2(\Omega)$ be the second homotopy group of $\Omega$. Let $\mathbf{F}=(P,Q,R)$ be a $C^1$ vector field. If $\pi_2(\Omega)=0$, can we deduce that the surface integral $$\iint\limits_{\Sigma}Pdydz+Qdzdx+Rdxdy$$ has nothing to do with the surface $\Sigma$, but only with its boundary curve of $\Sigma$ if and only the divergence of $\mathbf{F}$ is zero, that is $$\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0.$$

Here has nothing to do with the surface $\Sigma$ means that if $\widetilde{\Sigma}$ is another surface and they has the same boundary curve $$\partial\Sigma=\partial\widetilde{\Sigma},$$ then $$\iint\limits_{\Sigma}Pdydz+Qdzdx+Rdxdy=\iint\limits_{\widetilde{\Sigma}}Pdydz+Qdzdx+Rdxdy.$$

In other words, can $\pi_2(\Omega)=0$ implies the second de Rham cohomology group is zero? Here $\Omega$ is a domain in $\mathbb R^3$.

Answer 1

Since @Kevin.S has already established that you can have a region $\Omega$ with $\pi_2(\Omega) = 0$ and $H^2(\Omega)=\Bbb R$ and you've raised a good question in the comments, let me address that question.

Let $\Omega = \{(x,y,z): 1\le \left(\sqrt{x^2+y^2}-3\right)^2+z^2 \le 4\}$. Let $\Sigma = \Sigma_+\cup\Sigma_-$ be the torus $\Sigma = \{(x,y,z): \left(\sqrt{x^2+y^2}-3\right)^2+z^2 = 2\}$, and let $\Sigma_+ = \Sigma \cap \{y\ge 0\}$ and $\Sigma_- = \Sigma\cap\{y\le 0\}$ with their induced orientations. Note that $\partial\Sigma_+ = -\partial\Sigma_-$.

If we let $\omega$ be the standard area $2$-form of $\Sigma$. Then $\int_{\Sigma_+} \omega = \int_{\Sigma_-} \omega = 2\pi^2$. As we already pointed out, $\partial\Sigma_+ = -\partial\Sigma_-$, but $$\int_{\Sigma_+} \omega = 2\pi^2 = -\int_{-\Sigma_-} \omega,$$ so there's an example you wanted.

If you think of the torus as $S^1\times S^1$, you get naturally get $\omega = d\theta_1\wedge d\theta_2$. If you want Cartesian coordinates, I offer you this: $$\omega = \frac{-z\,dx + (x-3)\,dz}{(x-3)^2+z^2}\wedge\frac{-y\,dx+x\,dy}{x^2+y^2}.$$

Answer 2

Question: Is it right that any domain enclosed by a closed surface $\Sigma\subset\Omega$ is contained within $\Omega$ is equivalent to $H^2(\Omega)=0$. If so we have the following theorem for this domain. Here we convention that if $\Sigma_1$ and $\Sigma_2$ have the same boundary, then the orientation of the boundary is also the same.

Let $\Omega\subset\in\mathbb R^3$ be a domain, and $P,Q,R\in C^1(G)$. If the de Rham cohomology $H^2(G)=0$. Then the following three statements are equivalence.

$(1)$ Surface integer $\displaystyle{\iint\limits_{\Sigma}P dydz+Q dzdx+R dxdy}$ only depends on the boundary of $\Omega$.

$(2)$ For every closed surface $\Sigma\in\Omega$, we have that $\displaystyle{\iint\limits_{\Sigma}P dydz+Q dzdx+R dxdy=0}$.

$(3)$ $\displaystyle{\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0}$.

Here $(1)\Longrightarrow (2)\Longrightarrow(3)$ is easy. We only prove $(3)\Longrightarrow(1)$. Since $H^2(G)=0$, every closed form of $G$ is exact. Since $\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0$, there exists vector field $\mathbf{F}=(f_1,f_2,f_3)$ such that $$\mathrm{rot}(f_1,f_2,f_3)=(P,Q,R),$$ that is $$P=\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z},\quad Q=\frac{\partial f_1}{\partial z}-\frac{\partial f_3}{\partial x},\quad R=\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}.$$ Hence for every oriented surface $\Sigma\subset G$, by Stokes's theorem we have that \begin{eqnarray*} % \nonumber to remove numbering (before each equation) && \iint\limits_{\Sigma}P dydz+Q dzdx+R dxdy\\ &=& \iint\limits_{\Sigma}\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right) dydz+ \left(\frac{\partial f_1}{\partial z}-\frac{\partial f_3}{\partial x}\right) dzdx+\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right) dxdy \\ &=& \oint_{\partial\Sigma}f_1dx+f_2dy+f_3dz \end{eqnarray*} Now for every two surfaces $\Sigma_1,\Sigma_2$, if $$\partial\Sigma_1=\Gamma=\partial\Sigma_2,$$ then $$\iint\limits_{\Sigma_1}P dydz+Q dzdx+R dxdy =\oint_{\Gamma}f_1dx+f_2dy+f_3dz=\iint\limits_{\Sigma_2}P dydz+Q dzdx+R dxdy.$$