# Relationship between Stokes's theorem and the Gauss-Bonnet theorem

Stokes's theorem and the Gauss-Bonnet theorem are clearly very spiritually similar: they both relate the integral of a quantity $$A$$ over a region to the integral of some quantity $$B$$ over the boundary of the region, where $$A$$ can in some sense be thought of as a "curvature at one higher derivative" of $$B$$ or a closely related quantity. Is either of these theorems a special case of the other? If not, is there a more general theorem of which they are both special cases (which isn't too many levels higher up in abstraction)?

Edit: the answers to this follow-up question provide derivations of the Gauss-Bonnet theorem from Stokes's theorem in this paper, on pg. 105 of this textbook, and in Chapter 6 Section 1 of this textbook. Unfortunately, the derivations are too advanced for me to understand, as I haven't formally studied graduate-level differential geometry. I would appreciate any answer that summarizes the basic idea of the derivation.

• this is a good question! – GiantTortoise1729 Sep 4 '16 at 7:16
• Gauss Bonnet can be proved using Stokes theorem, but I cannot recall a way to prove Stokes using Gauss Bonnet. Indeed they are quite different: Stokes theorem consider only object in the smooth category while in Gauss Bonnet you consider object related to metric, curvatures... – user99914 Sep 4 '16 at 8:07
• @JohnMa Thank you, that is a very insightful distinction. Could you very briefly outline the proof of the GB theorem from Stokes' theorem (like in one or two sentences with no equations)? – tparker Sep 4 '16 at 17:08
• Good question, I'd like to know the answer. This might be related: mathoverflow.net/questions/50051/… – David Herrero Martí Oct 30 '16 at 1:08

## 1 Answer

I thought it would be nice to have a complete proof of the Gauss Bonnet formula, which is a great achivment of mathematics. Starting from the begining, I introduce the Pfaff forms and show the way they work. Then I use them to get the structure equations of the surface, the moving frame of the surface and the general curve on a surface. Combining all results I arive to $$(8)$$ which is a classical formula of differential geometry. With the help of $$(8)$$ I prove Liouville's formula and then the Gauss Bonnet formula. I make no use, or explain, the Levi-Civita theory, since it may be skiped and demands a lot of material to be taken into acount. I have also add some notes of mine and prove the fundamental theorem of Gauss and the equations of Mainardi and Godazzi.

Assume a two-dimesional surface $$\textbf{S}$$ of the Eucledean space $$E_3\cong \textbf{R}^3$$ which is of class $$C^3$$. That is, the surface is given by $$\overline{x}=\overline{x}(u,v)=\{x_1(u,v),x_2(u,v),x_3(u,v)\}\textrm{, }u,v\in D$$ and $$x_{i}(u,v)\in C^3$$, $$\overline{x}_u\times \overline{x}_v\neq \overline{0}$$, $$\overline{x}_u=\frac{\partial\overline{x}}{\partial u}$$, $$\overline{x}_v=\frac{\partial\overline{x}}{\partial v}$$. In every point $$P$$ of the surface we attach a moving frame of three orthonormal vectors (that is $$\{\overline{e}_1,\overline{e}_2,\overline{e}_3\}$$ and $$\left\langle \overline{e}_i,\overline{e}_j\right\rangle=\delta_{ij}$$), with the assumption that $$\overline{n}=\overline{e}_3$$ is orthonormal to the tangent plane of surface (in every $$P$$).Then there exist Pfaff (differentiatable) forms, $$\omega_i$$ and $$\omega_{ij}$$ such that $$d\overline{x}=\sum^{3}_{j=1}\omega_j\overline{e}_j\textrm{, }(\omega_3=0\Leftrightarrow \overline{n}=\overline{e}_3)$$ $$d\overline{e}_i=\sum^{3}_{j=1}\omega_{ij}\overline{e}_j\textrm{, }i=1,2,3$$ This can be seen as: $$d\overline{x}=\{\partial_1 x_1du+\partial_2x_1dv,\partial_1 x_2du+\partial_2x_2dv,\partial_1 x_3du+\partial_2x_3dv\}$$ and the Pfaff derivatives $$\nabla_kf$$ and $$\nabla_k\overline{F}$$ for any function $$f$$ or vector $$\overline{F}$$ (resp.) are defined as $$df=\sum^{3}_{k=1}(\nabla_kf)\omega_k=\sum^{3}_{k=1}\partial_kfdu_k.\tag 1$$ Set now $$q_1=\frac{d\omega_1}{\omega_1\wedge\omega_2}\textrm{, }q_2=\frac{d\omega_2}{\omega_1\wedge\omega_2}.$$ For to holds $$(1)$$ it must be $$\nabla_1\nabla_2f-\nabla_2\nabla_1f+q_1\nabla_1f+q_2\nabla_2f=0\textrm{, (condition)}.$$ From the relations $$d\left\langle\overline{e}_i,\overline{e}_j\right\rangle=0$$, $$d(d\overline{x})=\overline{0}$$, $$d(d\overline{n})=\overline{0}$$, we get the structure equations of the surface: $$\omega_{ij}+\omega_{ji}=0\textrm{, }i,j=1,2,3,$$ $$d\omega_j=\sum^{3}_{i=1}\omega_i\wedge\omega_{ij}\textrm{, }j=1,2,3$$ $$d\omega_{ij}=\sum^{3}_{k=1}\omega_{ik}\wedge\omega_{kj}\textrm{, }i,j=1,2,3.$$ Observe that $$\omega_3=\omega_{11}=\omega_{22}=\omega_{33}=0$$ and we can write $$d\overline{x}=\omega_1\overline{e}_1+\omega_2\overline{e}_2$$ $$d\overline{e}_1=\omega_{12}\overline{e}_2-\omega_{31}\overline{e}_3\tag 2$$ $$d\overline{e}_2=-\omega_{12}\overline{e}_1-\omega_{32}\overline{e}_3$$ $$d\overline{e}_3=\omega_{31}\overline{e}_1+\omega_{32}\overline{e}_2$$ Moreover it is (structure equations): $$d\omega_1=\omega_{12}\wedge\omega_2$$ $$d\omega_2=-\omega_{12}\wedge\omega_1$$ $$\omega_1\wedge\omega_{31}+\omega_2\wedge\omega_{32}=0$$ $$d\omega_{12}=-\omega_{31}\wedge\omega_{32}\tag 3$$ $$d\omega_{31}=\omega_{12}\wedge\omega_{32}$$ $$d\omega_{32}=-\omega_{12}\wedge\omega_{31}$$ If we write the connections (of $$\omega_{ij}$$ in terms of $$\omega_i$$): $$\omega_{12}=\xi\omega_1+\zeta \omega_2$$ $$\omega_{31}=-a\omega_1-b\omega_2$$ $$\omega_{32}=\eta\omega_1-c\omega_2$$ We easily get (from the structure equations) $$\xi=q_1$$, $$\zeta=q_2$$, $$\eta=-b$$. Hence $$\omega_{12}=q_1\omega_1+q_2 \omega_2\tag 4$$ $$\omega_{31}=-a\omega_1-b\omega_2\tag 5$$ $$\omega_{32}=-b\omega_1-c\omega_2.\tag 6$$

Assume now a ''strange'' operator $$\theta$$ such that $$\theta(A,B)=\left| \begin{array}{cc} \nabla_1\textrm{ }\nabla_2\\ A\textrm{ }B \end{array} \right|+q_1A+q_2B=\nabla_1B-\nabla_2A+q_1A+q_2B.$$ This is not so ''strange'' since if $$\omega=A\omega_1+B\omega_2$$ and $$f$$ function of $$u,v$$, then $$d(f\omega)=\theta(Af,Bf)\omega_1\wedge\omega_2=\left(\left| \begin{array}{cc} \nabla_1f\textrm{ }\nabla_2f\\ A\textrm{ }B \end{array} \right|+\theta(A,B)f\right)\omega_1\wedge\omega_2.$$ By this way we have $$d\omega_{12}=\theta(q_1,q_2)\omega_1\wedge\omega_2=-K\omega_1\wedge\omega_2\textrm{, }K=ac-b^2\textrm{, (Gauss curvature)}$$ $$d\omega_{31}=\theta(-a,-b)\omega_1\wedge\omega_2=(q_2b-q_1c)\omega_1\wedge\omega_2=\frac{q_2b-q_1c}{K}\omega_{31}\wedge\omega_{32}$$ and $$d\omega_{31}=q^{III}_1\omega_{31}\wedge\omega_{32}$$ $$d\omega_{32}=q^{III}_2\omega_{31}\wedge\omega_{32}$$ Hence the theorem of Gauss is: $$\theta(q_1,q_2)=-K=b^2-ac.$$ The Mainardi and Godazzi equations are: $$\theta(a,b)=\left| \begin{array}{cc} q_1\textrm{ }q_2\\ a\textrm{ }b \end{array} \right|$$ $$\theta(b,c)=\left| \begin{array}{cc} q_1\textrm{ }q_2\\ b\textrm{ }c \end{array} \right|.$$ The condition of Pfaff derivatrives becomes $$\theta(\nabla_1f,\nabla_2f)=0$$ and the Beltrami derivative is $$\Delta_2f=\theta(-\nabla_2f,\nabla_1f)$$ $$\theta(\overline{e}_1,\overline{e}_2)=0.$$ ... etc

Now assume a curve $$\Gamma$$ on the surface and its moving frame in $$P\in\textbf{S}$$ as follows: $$\overline{t}$$ is tangent of the curve in $$P$$, $$\overline{n}$$ is orthonormal of the surface in $$P$$ and $$\overline{n}_g$$ is orthonormal both in $$\overline{t}$$ and $$\overline{n}$$. Then we can easily see that exists $$\frac{1}{\rho_g}$$,$$\frac{1}{R}$$ and $$\frac{1}{\tau_g}$$ such that $$\frac{d\overline{t}}{ds}=\frac{\overline{n}_g}{\rho_g}+\frac{\overline{n}}{R}\tag 7$$ $$\frac{d\overline{n}_g}{ds}=-\frac{\overline{t}}{\rho_g}+\frac{\overline{n}}{\tau_g}$$ $$\frac{d\overline{n}}{ds}=-\frac{\overline{t}}{R}-\frac{\overline{n}_g}{\tau_g}.$$ Where $$s$$ being the natural parameter of $$\Gamma$$. All $$\frac{1}{\rho_g}$$,$$\frac{1}{R}$$,$$\frac{1}{\tau_g}$$ are invariants. If we consider also the Frenet frame $$\{\overline{t},\overline{h},\overline{b}\}$$, which is such $$\frac{d\overline{t}}{ds}=\frac{h}{\rho}$$ $$\frac{d\overline{h}}{ds}=-\frac{\overline{t}}{\rho}+\frac{\overline{b}}{\tau}$$ $$\frac{d\overline{b}}{ds}=-\frac{\overline{h}}{\tau}$$ and take the angle $$\psi$$ between $$\overline{h}$$ and $$\overline{n}$$, we get $$(\psi\in[0,2\pi))$$ $$\frac{1}{\rho_g}=\frac{\sin(\psi)}{\rho}$$ $$\frac{1}{R}=\frac{\cos(\psi)}{\rho}$$ $$\frac{1}{\tau_g}=\frac{1}{\tau}+\frac{d\psi}{ds}.$$

Gauss consider first the geodesic curvature $$\frac{1}{\rho_g}$$ of a curve in a surface. From (7) we have $$\frac{1}{\rho_g}=\left\langle\frac{d\overline{t}}{ds},\overline{n}_g\right\rangle.$$ Also $$\overline{t}=\frac{d\overline{x}}{ds}\textrm{, }\frac{d\overline{t}}{ds}=\frac{d^2\overline{x}}{ds^2}\textrm{, }n_g=\overline{n}\times \overline{t}$$ and $$\frac{1}{\rho_g}=\left(\frac{d\overline{x}}{ds},\frac{d^2\overline{x}}{ds^2},\overline{n}\right)=\textrm{det}\left(\frac{d\overline{x}}{ds},\frac{d^2\overline{x}}{ds^2},\overline{n}\right)\tag 8$$ Assuume now the surface curve coresponding to $$\omega_2=0$$ and we ask about its geodesic curvature. We have $$d\overline{x}=\omega_1\overline{e}_1+\omega_2\overline{e}_2\Rightarrow \left(\frac{d\overline{x}}{ds}\right)_{\omega_2=0}=\frac{\omega_1}{ds}\overline{e}_1.$$ From (2) we get $$\left(\frac{d^2\overline{x}}{ds^2}\right)_{\omega_2=0}=\frac{d}{ds}\left(\frac{\omega_1}{ds}\right)\overline{e}_1+\frac{\omega_1\omega_{12}}{ds^2}\overline{e}_2-\frac{\omega_1\omega_{31}}{ds^2}\overline{e}_3$$ From (2) and (8) we find $$\left(\frac{1}{\rho_g}\right)_{\omega_2=0}=q_1.$$ In the same manner for the curve $$\omega_1=0$$: $$\left(\frac{1}{\rho_g}\right)_{\omega_1=0}=q_2$$ Assume now a curve $$\Gamma$$ on a surface. Let $$\overline{t}$$ be its tangent vector and $$\phi$$ is the angle between $$\overline{t}$$ and $$\overline{e}_1$$ (the tangent $$\overline{t}$$ is on the $$\{\overline{e}_1,\overline{e}_2\}$$ plane).

We have $$\left(\frac{d\overline{x}}{ds}\right)_{\Gamma}=\overline{t}=\cos(\phi)\overline{e}_1+\sin(\phi)\overline{e}_2$$ and $$\left(\frac{d^2\overline{x}}{ds^2}\right)_{\Gamma}=\frac{\omega_{12}+d\phi}{ds}(-\sin(\phi)\overline{e}_1+\cos(\phi)\overline{e}_2)-\frac{\omega_{31}\cos(\phi)+\omega_{32}\sin(\phi)}{ds}\overline{e}_3$$ From relation (8) we get $$\frac{1}{\rho_g}=\frac{d\phi}{ds}+\frac{\omega_{12}}{ds}=\frac{d\phi}{ds}+q_1\frac{\omega_1}{ds}+q_2\frac{\omega_2}{ds}.$$ But $$\cos(\phi)=\frac{\omega_1}{ds}$$, $$\sin(\phi)=\frac{\omega_2}{ds}$$. Hence we find $$\frac{1}{\rho_g}=\frac{d\phi}{ds}+q_1\cos(\phi)+q_2\sin(\phi)\textrm{, Liouville formula}.$$ or in $$\theta$$ notation $$\frac{1}{\rho_g}=\theta(\cos(\phi),\sin(\phi)).$$

Proof of the Gauss Bonnet Formula

From the formulas $$\frac{1}{\rho_g}=\frac{\omega_{12}}{ds}+\frac{d\phi}{ds},$$ $$d\omega_{12}=-\omega_{31}\wedge\omega_{32}=-K\omega_1\wedge\omega_2$$ and Stokes theorem we get: $$\int_{\partial D}\frac{ds}{\rho_g}=\int_{\partial D}\omega_{12}+\int_{\partial D}\frac{d\phi}{ds}ds=\int\int_{D}d(\omega_{12})+\int_{\partial D}d\phi=$$ $$=-\int\int_{D}\frac{\omega_{31}\wedge\omega_{32}}{\omega_1\wedge\omega_2}\omega_1\wedge\omega_2+\int_{\partial D}d\phi=-\int\int_{D}K\omega_1\wedge\omega_2+2\pi,$$ since $$\int_{\partial_D}d\phi=2\pi.$$ Hence we get the Gauss Bonnet formula $$\int_{\partial D}\frac{ds}{\rho_g}+\int\int_{D}K\omega_1\wedge\omega_2=2\pi.$$