[Understanding Riemannian Geometry] 7. The Gauss–Bonnet Formula

【理解黎曼几何】7. 高斯-博内公式 苏剑林 · 2016-10-21 · Mathematics geometry riemannian geometry

What's exciting is that the path we took to derive Riemannian curvature also gives us a glimpse of the Gauss–Bonnet formula, letting us truly experience the flavor of studying intrinsic geometry.

The Gauss–Bonnet formula is a classic result in global differential geometry, one that connects the local and global properties of a space. Starting from a geometric approach, and combining it with some matrix transformations and mathematical analysis, we gradually derived geodesics, covariant derivatives, and the curvature tensor — and now we can also arrive at the classical Gauss–Bonnet formula. This shows that we've come quite far along this path. Although the process isn't perfectly rigorous, it hasn't strayed from the core theme of this series: geometric intuition. The purpose of this post is precisely to share an intuitive way of thinking about Riemannian geometry — and since it's about intuition, the focus is on exchanging ideas rather than on rigorous proof. So, readers should treat this series as supplementary material for Riemannian geometry, at best.

Rewriting the Formula

First, let's rewrite equation $(48)$ into a more geometrically meaningful form. Starting from

$$\Delta A^{\mu} =-R^{\mu}_{\alpha\beta\gamma} A^{\alpha} dx^{\beta}\delta x^{\gamma}=-g^{\mu\nu}R_{\nu\alpha\beta\gamma} A^{\alpha} dx^{\beta}\delta x^{\gamma} \tag{54} $$

and swapping the positions of $\beta, \gamma$, we obtain

$$\Delta A^{\mu} =-g^{\mu\nu}R_{\nu\alpha\gamma\beta} A^{\alpha} dx^{\gamma}\delta x^{\beta} \tag{55} $$

Using $R_{\nu\alpha\beta\gamma}=-R_{\nu\alpha\gamma\beta}$, and adding the two equations together, we get

$$\begin{aligned}\Delta A^{\mu} =&-\frac{1}{2}g^{\mu\nu}R_{\nu\alpha\beta\gamma} A^{\alpha} (dx^{\beta}\delta x^{\gamma}-dx^{\gamma}\delta x^{\beta})\\ =&-\frac{1}{2}g^{\mu\nu}R_{\nu\alpha\beta\gamma} A^{\alpha} \det\begin{pmatrix} dx^{\beta} & \delta x^{\beta}\\ dx^{\gamma} & \delta x^{\gamma}\end{pmatrix} \end{aligned} \tag{56} $$

The general geometric meaning of this expression requires tools like exterior differentiation and surface integrals to interpret, which we won't get into here. But we can consider the special case of two-dimensional space (i.e., a two-dimensional surface embedded in three-dimensional Euclidean space), where the geometric meaning becomes much clearer. When $n=2$, each summation index really only has two terms in its sum. That is,

$$\Delta A^{\mu} = -\frac{1}{2}\sum_{\nu=1}^2 \sum_{\alpha=1}^2\sum_{\beta=1}^2\sum_{\gamma=1}^2 g^{\mu\nu}R_{\nu\alpha\beta\gamma} A^{\alpha} \det\begin{pmatrix} dx^{\beta} & \delta x^{\beta}\\ dx^{\gamma} & \delta x^{\gamma}\end{pmatrix} \tag{57} $$

We can first work out the sum over $\beta,\gamma$: because of the determinant structure, this is only nonzero when $\beta \neq \gamma$, and then using $R_{\nu\alpha\beta\gamma}=-R_{\nu\alpha\gamma\beta}$, we get

$$\Delta A^{\mu} = -\sum_{\nu=1}^2 \sum_{\alpha=1}^2 g^{\mu\nu}R_{\nu\alpha 12} A^{\alpha} \det\begin{pmatrix} dx^{1} & \delta x^{1}\\ dx^{2} & \delta x^{2}\end{pmatrix} \tag{58} $$

Next consider the sum over $\nu,\alpha$, which similarly is only meaningful when $\nu\neq\alpha$, and again there's antisymmetry $R_{\nu\alpha\beta\gamma}=-R_{\alpha\nu\beta\gamma}$, so we obtain

$$\Delta A^{\mu} = - (g^{\mu 1} A^{2}-g^{\mu 2} A^{1}) R_{12 12}\det\begin{pmatrix} dx^{1} & \delta x^{1}\\ dx^{2} & \delta x^{2}\end{pmatrix} \tag{59} $$

which can be rewritten as

$$\Delta A^{\mu} = - \sqrt{g}(g^{\mu 1} A^{2}-g^{\mu 2} A^{1}) \frac{R_{12 12}}{g}\sqrt{g}\det\begin{pmatrix} dx^{1} & \delta x^{1}\\ dx^{2} & \delta x^{2}\end{pmatrix} \tag{60} $$

Notice that in two-dimensional space, $\sqrt{g}\det\begin{pmatrix} dx^{1} & \delta x^{1}\\ dx^{2} & \delta x^{2}\end{pmatrix}$ has a clear geometric meaning: it is precisely the area of the quadrilateral spanned by the vectors $(dx^1, dx^2)$ and $(\delta x^1, \delta x^2)$ (recall the result from equation $(15)$), which we'll denote simply as $\Delta S$. And $\frac{R_{12 12}}{g}$ is exactly the Gaussian curvature $K$ as defined in differential geometry, so this can be written as

$$\Delta A^{\mu} = - \sqrt{g}(g^{\mu 1} A^{2}-g^{\mu 2} A^{1}) K\Delta S \tag{61} $$

The Change in Angle

Now let's analyze: after vector $A^{\mu}$ becomes $A^{\mu}+\Delta A^{\mu}$, what is the angle between these two vectors? Assuming $A^{\mu}$ is a unit vector, we first need to compute the inner product

$$\begin{aligned}&\frac{g_{\mu\nu}A^{\mu}(A^{\nu}+\Delta A^{\nu})}{\sqrt{g_{\mu\nu}A^{\mu}A^{\nu}}\sqrt{g_{\mu\nu}(A^{\mu}+\Delta A^{\mu})(A^{\nu}+\Delta A^{\nu})}}\\ =&\frac{1+g_{\mu\nu}A^{\mu}\Delta A^{\nu}}{\sqrt{1+2g_{\mu\nu}A^{\mu}\Delta A^{\nu}+g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu}}} \end{aligned}\tag{62} $$

Since $\cos \Delta \theta = 1-\frac{\Delta\theta^2}{2}+\dots$, we need to expand up to the second-order term, i.e., $\Delta A^{\mu} \Delta A^{\nu}$; approximating to second order, the result is

$$1-\frac{1}{2}\left[g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu}-(g_{\mu\nu}A^{\mu}\Delta A^{\nu})^2 \right] \tag{63} $$

so

$$\begin{aligned}\Delta \theta =& \sqrt{g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu}-(g_{\mu\nu}A^{\mu}\Delta A^{\nu})^2}\\ =&\sqrt{(g_{\mu\nu} A^{\mu} A^{\nu})(g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu})-(g_{\mu\nu}A^{\mu}\Delta A^{\nu})^2} \end{aligned}\tag{64} $$

This is in fact exactly the area of the parallelogram spanned by $A^{\mu}$ and $\Delta A^{\mu}$. When $n=2$, that is, $\sqrt{g} \det\begin{pmatrix}A_1 & \Delta A_1\\ A_2 & \Delta A_2\end{pmatrix}=\sqrt{g} (A^{1}\Delta A^{2}-A^{2}\Delta A^{1})$, substituting into the expression for $\Delta A^{1}, \Delta A^{2}$, we get

$$\Delta \theta = g\left[g^{22}(A^1)^2-g^{12}A^2 A^1 - g^{21}A^1 A^2+g^{11}(A^2)^2\right]K\Delta S \tag{65} $$

For a 2×2 matrix, there's an inversion formula

$$\begin{aligned}\begin{pmatrix} g_{11} & g_{12} \\ g_{21} & g_{22}\end{pmatrix}^{-1}=&\frac{1}{g_{11}g_{22}-g_{12}g_{21}}\begin{pmatrix} g_{22} & -g_{12} \\ -g_{21} & g_{11} \end{pmatrix}\\ =&\frac{1}{g}\begin{pmatrix} g_{22} & -g_{12} \\ -g_{21} & g_{11} \end{pmatrix} \end{aligned}\tag{66} $$

therefore

$$g^{11}=\frac{g_{22}}{g},\,g^{12}=-\frac{g_{12}}{g},\,g^{21}=-\frac{g_{21}}{g},\,g^{22}=\frac{g_{11}}{g} \tag{67} $$

Substituting into equation $(65)$ gives

$$\Delta \theta = [g_{11} (A^1)^2 + g_{12} A^1 A^2 + g_{21} A^2 A^1 + g_{22} (A^2)^2 ] K\Delta S = K\Delta S \tag{68} $$

The last equality holds because we assumed from the start that $A^{\mu}$ is a unit vector. In this way, the angular deviation produced when a vector is parallel-transported around a small closed curve and returns to its starting point equals the product of the Gaussian curvature $K$ and the area element $\Delta S$. From this, we can deduce that if a vector is parallel-transported around a large-scale closed curve $\mathbb{C}$ and brought back, the total deviation is given by the surface integral

$$\Delta \theta = \int_{\mathbb{C}} K d S \tag{69} $$

This is the main content of the Gauss–Bonnet formula in differential geometry: the angular deviation equals the surface integral of the Gaussian curvature. Results such as the sum of interior angles of a spherical triangle are all connected to this formula. It stands as one of the founding works of global differential geometry.

A Few Remarks

It's worth noting that our discussion above is entirely intrinsic — that is, we never introduced the notion of a surface embedded in three-dimensional space. This is quite appealing. The great mathematician Shiing-Shen Chern himself said that the best work of his life was the intrinsic proof of the higher-dimensional Gauss–Bonnet formula (proofs before his were extrinsic). As we can see, purely intrinsic work is the ideal pursuit in the study of Riemannian geometry. Of course, what we've done here is at most a suggestive guide, not a complete proof. But for the purposes of this series, this level of treatment is sufficient.

One point that may puzzle readers: since it represents a volume, it should be non-negative, but when written as a matrix determinant, it can be positive or negative — which seems contradictory. Without introducing exterior differentiation, this is indeed difficult to fully clarify. Within the scope of elementary analysis, the only workaround is: whenever a negative volume shows up, simply take its absolute value.