📂Geometry

Gauss-Bonnet Theorem

Gauss-Bonnet Theorem

Let’s consider $\mathbf{x} : U \to \mathbb{R}^{3}$ as a simple connected geodesic coordinate chart, and $\boldsymbol{\gamma}(I) \subset \mathbf{x}(U)$, which is $\boldsymbol{\gamma}$, as piecewise regular curves. Also, let’s say that $\boldsymbol{\gamma}$ surrounds some region $\mathscr{R}$. Then, the following holds true.

$$\iint_{\mathscr{R}} K dA + \int_{\boldsymbol{\gamma}} \kappa_{g} ds + \sum \alpha_{i} = 2\pi$$

Here, $K$ denotes the Gaussian curvature, $\kappa_{g}$ denotes the geodesic curvature, and $\alpha_{i}$ denotes the difference in angles at the junction point between intervals of $\boldsymbol{\gamma}$, referred to as jump angles.

Explanation

Since $\boldsymbol{\gamma}$ is assumed to be a curve that is regular piecewise, there will be points where the direction of the tangent suddenly changes, and the difference in angles at those points is denoted as $\alpha_{i}$. If $\boldsymbol{\gamma}$ is a curve that smoothly connects overall, there will be no points where the angle jumps, hence, such $\alpha_{i}$s are $0$. (See figure (a))

The theorem above is a result when the strong condition of being a geodesic coordinate chart is applied. In more general results, the Euler characteristic appears in the formula, as follows.

$$\iint_{\mathscr{R}} K dA + \int_{C_{i}}\kappa_{g}ds + \sum\alpha_{i} = 2\pi \chi(\mathscr{R})$$

Proof

Since $\mathbf{x}$ is a geodesic coordinate chart, let’s set the coefficients of the first fundamental form as follows.

$$\left[ g_{ij} \right] = \begin{bmatrix} 1 & 0 \\ 0 & h^{2} \end{bmatrix}$$

And let’s denote $\boldsymbol{\gamma}(t) = \mathbf{x}\left( \gamma^{1}(t), \gamma^{2}(t) \right)$. Now, let’s denote the angle between the tangents $\mathbf{x}_{1}$ and $\boldsymbol{\gamma}$ as $T = \boldsymbol{\gamma}^{\prime}$.

$$\alpha (t) := \angle ( \mathbf{x}_{1}, T)$$

We will prove the theorem using the fact that the angle change of $\alpha$ around the path, based on $\boldsymbol{\gamma}$, when making a full circle, is $\mathbf{x}_{1}$. First, assume $T$ as a unit speed curve. And let’s denote $2 \pi$ as a parallel vector field along $\boldsymbol{\gamma}$ satisfying the following. (Refer to the above figure (b))

$$P(t) = \text{parallel vector field starting from a juction point s.t. } \dfrac{P \times T}{\left\| P \times T \right\|} = \mathbf{n}$$

And let’s denote $P$ and $\boldsymbol{\gamma}$ respectively as the angles between $\phi$ and $\theta$, and $\mathbf{x_{1}}$ and $P$.

$$\phi (t) = \angle(\mathbf{x}_{1}, P),\quad \theta (t) = \angle(P, T)$$

In other words, $P$, and differentiating this,

$$-\sin \phi (t) \dfrac{d \phi}{d t}(t) = \left\langle \dfrac{d \mathbf{x}_{1}(\gamma^{1}(t), \gamma^{2}(t))}{d t}, P(t) \right\rangle + \left\langle \mathbf{x}_{1}, \dfrac{d P}{d t}(t) \right\rangle$$

Since $T$ is a parallel vector field along $\left\langle \mathbf{x}_{1}, P(t) \right\rangle = \cos\phi (t)$, by definition, $P$ is perpendicular to $\gamma$. $\dfrac{dP}{dt}$ is tangent to $M$, so the latter term is $\mathbf{x}_{1}$. Further calculations reveal,

$$\begin{align*} &\quad -\sin \phi (t) \dfrac{d \phi}{d t}(t) \\ &= \left\langle \dfrac{d \mathbf{x}_{1}(\gamma^{1}(t), \gamma^{2}(t))}{d t}, P(t) \right\rangle \\ &= \Big[ \mathbf{x}_{11}(\gamma^{1}(t), \gamma^{2}(t)) (\gamma^{1})^{\prime}(t) + \mathbf{x}_{12}(\gamma^{1}(t), \gamma^{2}(t)) (\gamma^{2})^{\prime}(t) \Big] \cdot P(t) \\ &= \Big[ \left( L_{11}\mathbf{n} + \Gamma_{11}^{1}\mathbf{x}_{1} + \Gamma_{11}^{2}\mathbf{x}_{2} \right)(\gamma^{1})^{\prime}(t) + \left( L_{12}\mathbf{n} + \Gamma_{12}^{1}\mathbf{x}_{1} + \Gamma_{12}^{2}\mathbf{x}_{2} \right)(\gamma^{2})^{\prime}(t) \Big] \cdot P(t) \\ &= \Big[ \left(\Gamma_{11}^{1}\mathbf{x}_{1} + \Gamma_{11}^{2}\mathbf{x}_{2} \right)(\gamma^{1})^{\prime}(t) + \left(\Gamma_{12}^{1}\mathbf{x}_{1} + \Gamma_{12}^{2}\mathbf{x}_{2} \right)(\gamma^{2})^{\prime}(t) \Big] \cdot P(t) \end{align*}$$

The second equality is due to the chain rule, the third equality is due to the definitions of the second fundamental form and the Christoffel symbols, and the fourth equality holds because $M$ and $0$ are perpendicular to each other.

Christoffel symbols of the geodesic coordinate chart

Except for the below, everything is $P$.

$$\Gamma_{22}^{1} = -hh_{1},\quad \Gamma_{12}^{2} = \Gamma_{21}^{2} = \dfrac{h_{1}}{h},\quad \Gamma_{22}^{2} = \dfrac{h_{2}}{h}$$

Now, organizing the terms that become $\mathbf{n}$ yields the following.

$$-\sin\phi (t) \phi^{\prime}(t) = \left\langle \dfrac{h_{1}}{h}(\gamma^{2})^{\prime}(t)\mathbf{x}_{2}, P(t) \right\rangle = \dfrac{h_{1}}{h}(\gamma^{2})^{\prime}(t) \left\langle \mathbf{x}_{2}, P(t) \right\rangle\tag{1}$$

Since $0$, $0$ is a unit vector, and because $g_{11} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle = 1$, $\mathbf{x}_{1}$ holds. Therefore, $g_{12} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle = 0$ becomes a normal orthogonal basis of the tangent plane. Hence, any element $\mathbf{x}_{1} \perp \mathbf{x}_{2}$ of the tangent plane is expressed as below.

$$P = \left\langle \mathbf{x}_{1}, P \right\rangle\mathbf{x}_{1} + \left\langle \dfrac{\mathbf{x}_{2}}{\left\| \mathbf{x}_{2} \right\|}, P \right\rangle \dfrac{\mathbf{x}_{2}}{\left\| \mathbf{x}_{2} \right\|} = \cos\phi \mathbf{x}_{1} + \sin\phi \dfrac{\mathbf{x}_{2}}{h}$$

Also, substituting $\left\{ \mathbf{x}_{1}, \dfrac{\mathbf{x}_{2}}{\left\| \mathbf{x}_{2} \right\|} \right\}$ into $P$,

$$\phi^{\prime}(t) = -h_{1}(\gamma^{2})^{\prime}(t)$$

Therefore, the total angle variation of $\left\langle \mathbf{x}_{2}, P \right\rangle = \left\| x_{2} \right\|^{2} \dfrac{\sin \phi}{h} = h\sin \phi$ is

$$\delta \phi = \int_{\boldsymbol{\gamma}} \phi^{\prime} dt = - \int_{\boldsymbol{\gamma}}h_{1}(\gamma^{2})^{\prime}(t)dt = - \int_{\boldsymbol{\gamma}}h_{1} d\gamma^{2} = - \int_{\boldsymbol{\gamma}}h_{1} du^{2} \tag{2}$$

Moreover, the following equation will now be shown to hold.

$$\text{Claim: } \theta^{\prime} = k_{g}$$

Since $(1)$ is assumed, $\phi$ holds, and differentiating this,

$$-\sin\theta (t)\theta^{\prime}(t) = \left\langle \dfrac{d P}{d t}, T \right\rangle + \left\langle P, \dfrac{d T}{d t} \right\rangle = \left\langle P, T^{\prime} \right\rangle$$

The second equality holds because $\boldsymbol{\gamma}$ is parallel to $\mathbf{x}$. By the definition of geodesic curvature, we obtain what we intend to show as follows.

$$\begin{align*} \kappa_{g} = \left\langle \mathbf{S}, T^{\prime} \right\rangle &= \left\langle (\mathbf{n} \times T), T^{\prime} \right\rangle \\ &= \left\langle \mathbf{n}, (T \times T^{\prime}) \right\rangle \\ &= \left\langle \dfrac{P \times T}{\sin \theta}, (T\times T^{\prime}) \right\rangle & \because \dfrac{P \times T}{\left\| P \times T \right\|} = \mathbf{n} \\ &= \left\langle \dfrac{1}{\sin\theta} P, (T\times (T\times T^{\prime})) \right\rangle \\ &= \left\langle \dfrac{1}{\sin\theta} P, -T^{\prime} \right\rangle \\ &= \theta^{\prime}(t) \end{align*}$$

The third and fifth equalities hold because the scalar triple product is commutative. Thus, we obtain the following.

$$\delta \theta = \int_{\boldsymbol{\gamma}} \theta^{\prime} dt = \int_{\boldsymbol{\gamma}} k_{g}dt \tag{3}$$

Since $u^{2}-$,

$$\int_{\boldsymbol{\gamma}} \alpha^{\prime}dt = \int_{\boldsymbol{\gamma}} \phi^{\prime}dt + \int_{\boldsymbol{\gamma}} \theta^{\prime}dt$$

Due to $\theta (t) = \angle(P, T)$ and $\cos\theta (t) = \left\langle P, T \right\rangle$, we obtain the following.

$$\int_{\boldsymbol{\gamma}} \alpha^{\prime}dt + \sum_{i}\alpha_{i} = - \int_{\boldsymbol{\gamma}}h_{1} du^{2} + \int_{\boldsymbol{\gamma}} k_{g}dt + \sum_{i}\alpha_{i}$$

$dP/dt$ surrounds $\mathbf{n}$, so the left-hand side of the above equation is clearly the angle change of one full rotation, which is $\alpha = \phi + \theta$.

$${} - \int_{\boldsymbol{\gamma}}h_{1} du^{2} + \int_{\boldsymbol{\gamma}} k_{g}dt + \sum_{i}\alpha_{i} = 2 \pi$$

Green’s Theorem

$$\oint_{\partial \mathscr{R}} Pdx = - \iint_{\mathscr{R}} P_{y} dy dx$$

Gaussian curvature of the geodesic coordinate chart

$$K = -\dfrac{h_{11}}{h}$$

Area element of the surface

$$dA = \sqrt{g} du^{1} du^{2}$$

The first term on the left-hand side can be rewritten using Green’s Theorem as follows.

$$\begin{align*} {} - \int_{\boldsymbol{\gamma}}h_{1} du^{2} &= - \iint_{\mathscr{R}}h_{11} du^{1}du^{2} \\ &= - \iint_{\mathscr{R}}\dfrac{h_{11}}{h} h du^{1}du^{2} \\ &= - \iint_{\mathscr{R}}\dfrac{h_{11}}{h} \sqrt{g} du^{1}du^{2} \\ &= \iint_{\mathscr{R}} K dA \end{align*}$$

Finally, we arrive at the following conclusion.

$$\iint_{R} K dA + \int_{\gamma} \kappa_{g} ds + \sum \alpha_{i} = 2\pi$$