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

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 $$\mathbb R^3$$.

• In general, the answer is no. Consider a region in $\Bbb R^3$ that is homotopy equivalent to a torus, then the second homotopy group is trivial by considering its universal covering space, but the second de Rham cohomology group is non-trivial. The implication is valid if you also require $\Omega$ to be simply connected, since Hurewicz theorem then gives you an isomorphism between 2nd homotopy group and integral homology. Commented Apr 18, 2023 at 1:39
• @Kevin.S If we assume that any domain enclosed by a closed surface $\Sigma\subset\Omega$ is contained within $\Omega$，Can this condition guarantee the theorem holds?
– HGF
Commented Apr 18, 2023 at 2:30
• @Kevin.S Let $\Omega$ be the region the annulus $1\leq (x-3)^2+z^2\leq 4$ rotates around the $z$-axis, we know that $\pi_2(\Omega)=0$ and $H^2(\Omega)=\mathbb R$. How to construct a closed form which is not exact and how to find two suface which have the same boundary but the suface integral is not equal?
– HGF
Commented Apr 18, 2023 at 5:23
• The condition you proposed seems to be an intuitive way of saying "the second homology vanishes" because you're essentially killing non-trivial classes in $H_2$ by making them boundaries of three dimensional chains, so it sort of makes the result about cohomology trivially true. Commented Apr 18, 2023 at 7:52
• You can transfer all the calculations on the rotating annulus to a torus and then pull everything back since they are homotopy equivalent. You can use the fact that a generator of $H^2(T^2)$ (which is a 2-form) is given by the wedge product of two generators of $H^1$ to do some kind of explicit calculation, but it's probably still messy I think. Commented Apr 18, 2023 at 8:05

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: $$\omega = \frac{-z\,dx + (x-3)\,dz}{(x-3)^2+z^2}\wedge\frac{-y\,dx+x\,dy}{x^2+y^2}.$$

• Here we assume that if $\Sigma_1$ and $\Sigma_2$ have the same boundary, then the boundary curve have the same orientation. If we assume that any domain enclosed by a closed surface $\Sigma\subset\Omega$ is contained within $\Omega$，Can we deduce that the second de Rham cohomology group $H^2(\Omega)$ is trivial？ In your example, we can find a surface the domain enclosed by the surface does not contain in $\Omega$.
– HGF
Commented Apr 20, 2023 at 10:10
• Is it right that any domain enclosed by a closed surface $\Sigma\subset\Omega$ is contained within $\Omega$ is equivalent to $H^2(\Omega)=0$.
– HGF
Commented Apr 20, 2023 at 10:14
• No, when I cut $\Sigma$ in half, the boundary orientations of the two pieces are opposite. (Just try it with a sphere and two hemispheres.) If every closed surface in $\Omega$ bounds, this will tell us that every $2$-cycle is a boundary, and hence that the homology $H_2(\Omega,\Bbb Z)=0$, and from this it follows with a bit of algebraic topology that $H^2_{\text{deRham}}(\Omega)=0$. Commented Apr 20, 2023 at 17:53

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 $$\iint\limits_{\Sigma_1}P dydz+Q dzdx+R dxdy =\oint_{\Gamma}f_1dx+f_2dy+f_3dz=\iint\limits_{\Sigma_2}P dydz+Q dzdx+R dxdy.$$

• This is not an answer but a new question, which, to make things worse, is meaningless as stated. Commented Apr 20, 2023 at 13:04