By 苏剑林 | October 21, 2016
Excitingly, our approach to deriving Riemann curvature also allows us to catch a glimpse of the beauty of the Gauss–Bonnet formula, offering a true experience of researching intrinsic geometry. The Gauss–Bonnet formula is a classic formula in global differential geometry that establishes a link between the local and global properties of a space. Starting from a geometric path and combining elements of matrix transformations and mathematical analysis, we have step-by-step derived geodesics, covariant derivatives, and the curvature tensor. Now, we can even arrive at the classical Gauss–Bonnet formula, demonstrating how far we have come on this journey. Although the process is not perfect, it remains true to the core of this series: geometric intuition. The purpose of this article is to share an intuitive approach to Riemannian geometry; as it is a way of thinking, the focus is on the exchange of ideas rather than rigorous proof. Therefore, for all readers, please treat this series as supplementary material for Riemannian geometry.
Formal Rewrite
First, we can rewrite Equation $(48)$ into a form with more geometric significance. 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 exchanging the positions of $\beta$ and $\gamma$, we obtain
$$ \Delta A^{\mu} = -g^{\mu\nu}R_{\nu\alpha\gamma\beta} A^{\alpha} dx^{\gamma}\delta x^{\beta} \tag{55} $$
Using the property $R_{\nu\alpha\beta\gamma} = -R_{\nu\alpha\gamma\beta}$ and adding the two equations together, we derive
$$ \begin{aligned}\Delta A^{\mu} =& -\frac{1}{2}g^{\mu\nu}R_{\nu\alpha\beta\gamma} A^{\alpha} (dx^{\beta}\delta x^{\gamma}-dx^{\gamma}\delta x^{\beta})\\
=& -\frac{1}{2}g^{\mu\nu}R_{\nu\alpha\beta\gamma} A^{\alpha} \det\begin{pmatrix} dx^{\beta} & \delta x^{\beta}\\ dx^{\gamma} & \delta x^{\gamma}\end{pmatrix}
\end{aligned} \tag{56} $$
The general geometric meaning of this formula requires interpretation using exterior derivatives, surface integrals, and other advanced topics, which we will not discuss here. However, we can consider the special case of a two-dimensional space (i.e., a two-dimensional surface in three-dimensional Euclidean space), where the geometric meaning becomes clear. When $n=2$, each summation index actually covers only two terms. 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 calculate the sum over $\beta$ and $\gamma$. Due to the nature of the determinant, it is only meaningful when $\beta \neq \gamma$. Then, utilizing $R_{\nu\alpha\beta\gamma} = -R_{\nu\alpha\gamma\beta}$, we obtain
$$ \Delta A^{\mu} = -\sum_{\nu=1}^2 \sum_{\alpha=1}^2 g^{\mu\nu}R_{\nu\alpha 12} A^{\alpha} \det\begin{pmatrix} dx^{1} & \delta x^{1}\\ dx^{2} & \delta x^{2}\end{pmatrix} \tag{58} $$
Next, considering the sum over $\nu$ and $\alpha$, similarly, it is only meaningful when $\nu \neq \alpha$. Furthermore, because of the antisymmetry $R_{\nu\alpha\beta\gamma} = -R_{\alpha\nu\beta\gamma}$, we can obtain
$$ \Delta A^{\mu} = - (g^{\mu 1} A^{2}-g^{\mu 2} A^{1}) R_{12 12}\det\begin{pmatrix} dx^{1} & \delta x^{1}\\ dx^{2} & \delta x^{2}\end{pmatrix} \tag{59} $$
Rewriting this as
$$ \Delta A^{\mu} = - \sqrt{g}(g^{\mu 1} A^{2}-g^{\mu 2} A^{1}) \frac{R_{12 12}}{g}\sqrt{g}\det\begin{pmatrix} dx^{1} & \delta x^{1}\\ dx^{2} & \delta x^{2}\end{pmatrix} \tag{60} $$
Note 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 the area of the parallelogram spanned by the vectors $(dx^1, dx^2)$ and $(\delta x^1, \delta x^2)$ (referring back to the result of Equation $(15)$), which we denote simply as $\Delta S$. Since $\frac{R_{12 12}}{g}$ is exactly the Gaussian curvature $K$ as defined in differential geometry, we can write
$$ \Delta A^{\mu} = - \sqrt{g}(g^{\mu 1} A^{2}-g^{\mu 2} A^{1}) K\Delta S \tag{61} $$
Change in Angular Difference
Now, we can analyze the angle between the vector $A^{\mu}$ and its changed state $A^{\mu} + \Delta A^{\mu}$. Assuming $A^{\mu}$ is a unit vector, we first calculate the inner product:
$$ \begin{aligned}&\frac{g_{\mu\nu}A^{\mu}(A^{\nu}+\Delta A^{\nu})}{\sqrt{g_{\mu\nu}A^{\mu}A^{\nu}}\sqrt{g_{\mu\nu}(A^{\mu}+\Delta A^{\mu})(A^{\nu}+\Delta A^{\nu})}}\\
=& \frac{1+g_{\mu\nu}A^{\mu}\Delta A^{\nu}}{\sqrt{1+2g_{\mu\nu}A^{\mu}\Delta A^{\nu}+g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu}}} \end{aligned}\tag{62} $$
Because $\cos \Delta \theta = 1-\frac{\Delta\theta^2}{2} + \dots$, we need to calculate up to the second-order terms, namely the $\Delta A^{\mu} \Delta A^{\nu}$ term. Approximating to the second order, the result is
$$ 1-\frac{1}{2}\left[g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu}-(g_{\mu\nu}A^{\mu}\Delta A^{\nu})^2 \right] \tag{63} $$
Therefore,
$$ \begin{aligned}\Delta \theta =& \sqrt{g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu}-(g_{\mu\nu}A^{\mu}\Delta A^{\nu})^2}\\
=& \sqrt{(g_{\mu\nu} A^{\mu} A^{\nu})(g_{\mu\nu}\Delta A^{\mu}\Delta A^{\nu})-(g_{\mu\nu}A^{\mu}\Delta A^{\nu})^2} \end{aligned}\tag{64} $$
This is actually the area of the parallelogram spanned by $A^{\mu}$ and $\Delta A^{\mu}$. When $n=2$, this 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 the expressions for $\Delta A^{1}$ and $\Delta A^{2}$ gives
$$ \Delta \theta = g\left[g^{22}(A^1)^2-g^{12}A^2 A^1 - g^{21}A^1 A^2+g^{11}(A^2)^2\right]K\Delta S \tag{65} $$
For a second-order matrix, there is an inverse formula
$$ \begin{aligned}\begin{pmatrix} g_{11} & g_{12} \\ g_{21} & g_{22}\end{pmatrix}^{-1}=& \frac{1}{g_{11}g_{22}-g_{12}g_{21}}\begin{pmatrix} g_{22} & -g_{12} \\ -g_{21} & g_{11} \end{pmatrix}\\
=& \frac{1}{g}\begin{pmatrix} g_{22} & -g_{12} \\ -g_{21} & g_{11} \end{pmatrix} \end{aligned}\tag{66} $$
Thus
$$ g^{11}=\frac{g_{22}}{g}, \, g^{12}=-\frac{g_{12}}{g}, \, g^{21}=-\frac{g_{21}}{g}, \, g^{22}=\frac{g_{11}}{g} \tag{67} $$
Substituting into Equation $(65)$ reveals
$$ \Delta \theta = [g_{11} (A^1)^2 + g_{12} A^1 A^2 + g_{21} A^2 A^1 + g_{22} (A^2)^2 ] K\Delta S = K\Delta S \tag{68} $$
The final equality holds because we initially assumed $A^{\mu}$ is a unit vector. Thus, the angular difference generated after a vector returns along a small closed curve is equal to the product of the Gaussian curvature $K$ and the area element $\Delta S$. Consequently, we can infer that if a vector is parallel transported along a large-scale closed curve $\mathbb{C}$ and returns, the amount of change is represented by the surface integral
$$ \Delta \theta = \int_{\mathbb{C}} K d S \tag{69} $$
This is the primary content of the Gauss–Bonnet formula in differential geometry: that the angular difference is equal to the surface integral of the Gaussian curvature. Properties such as the sum of interior angles of a spherical triangle are related to it. It is one of the founding works of global differential geometry.
Commentary
It is worth noting that our discussion above is entirely intrinsic, meaning we did not introduce the concept of a surface embedded in three-dimensional space, which is very appealing. The great mathematician Shiing-Shen Chern once said that the best work of his life was the intrinsic proof of the generalized Gauss–Bonnet formula (previous proofs were extrinsic). It can be seen that purely intrinsic work is the pursuit of Riemannian geometry research. Of course, ours is at most an illustrative guide and does not constitute a full proof. But for this series, this level is quite sufficient.
One point that might confuse readers is that, as a volume (or area), it should be non-negative; however, if written in the form of a matrix determinant, it can be positive or negative, which seems contradictory. This is indeed difficult to clarify without introducing exterior derivatives. Within the scope of elementary analysis, the only solution is to take the absolute value if a negative volume appears.