By 苏剑林 | November 07, 2016
We finally begin to discuss the highlight, which is the content that prompted me to learn exterior differentiation in the first place. Using exterior differentiation, one can conveniently derive many aspects of differential geometry, and it occasionally simplifies calculations. The primary reason for this is that exterior differentiation is itself a generalization of the differential in form; thus, it is not surprising that the concepts of differential geometry can be described using exterior differentiation. Most importantly, exterior differentiation treats $dx^{\mu}$ as a set of basis elements. This effectively introduces two sets of bases into geometry: one is the inherent vector basis (the basis for contravariant vectors in tensor language), which allows for symmetric inner products, and the other is the $dx^{\mu}$ basis, which allows for antisymmetric exterior products. Consequently, when exterior differentiation is introduced into geometry, differential geometry gains a full suite of "ideal equipment" including differentiation, integration, symmetric products, and antisymmetric products. This is the main reason why exterior differentiation can accelerate derivations in differential geometry.
Previously, we obtained:
\[\begin{aligned}&\omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}\\ &d\boldsymbol{r}=\hat{\boldsymbol{e}}_{\mu} \omega^{\mu}\\ &ds^2 = \eta_{\mu\nu} \omega^{\mu}\omega^{\nu}\\ &\langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle = \eta_{\mu\nu}\end{aligned} \tag{45} \]By applying $d$ to both sides of $\langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle = \eta_{\mu\nu}$, we get:
\[d\eta_{\mu\nu} = \langle d\hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle + \langle \hat{\boldsymbol{e}}_{\mu}, d\hat{\boldsymbol{e}}_{\nu}\rangle \tag{46} \]$d\hat{\boldsymbol{e}}_{\mu}$ is the differential of a vector, and the result is also a vector. Therefore, it can be expressed as a linear combination of $\hat{\boldsymbol{e}}_{\mu}$, namely:
\[d\hat{\boldsymbol{e}}_{\mu} = \hat{\boldsymbol{e}}_{\alpha} \omega_{\mu}^{\alpha} \tag{47} \]Thus, we have:
\[\begin{aligned}d\eta_{\mu\nu} =& \langle \hat{\boldsymbol{e}}_{\alpha}\omega_{\mu}^{\alpha} , \hat{\boldsymbol{e}}_{\nu}\rangle + \langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\alpha}\omega_{\nu}^{\alpha} \rangle \\ =&\langle \hat{\boldsymbol{e}}_{\alpha}, \hat{\boldsymbol{e}}_{\nu}\rangle\omega_{\mu}^{\alpha} + \langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\alpha} \rangle\omega_{\nu}^{\alpha}\\ =&\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha} \end{aligned} \tag{48} \]In most practical cases, $\eta_{\mu\nu}$ is a constant diagonal matrix, so:
\[\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}=0 \tag{49} \]Then $\omega_{\mu\nu} = \eta_{\mu \alpha}\omega_{\nu}^{\alpha}$, viewed as a matrix, is antisymmetric and has only $n(n-1)/2$ components. Specifically, if $\eta_{\mu\nu}$ is the identity matrix, then $\omega_{\mu}^{\alpha}$ is antisymmetric.
Next, we assert:
\[d^2 \boldsymbol{r}=0 \tag{50} \]It should be noted that this is not obvious. Although we know that $d^2 f=0$ for any function $f$, $d\boldsymbol{r}$ is not strictly the differential of a function, but rather an arbitrarily given vector-valued differential form. Therefore, $d^2 \boldsymbol{r}=0$ is not an obvious conclusion. However, we can imagine that any $n$-dimensional curved space (manifold) can be embedded into a sufficiently high $m$-dimensional flat space (Euclidean space) as a subset, much like a surface in three-dimensional space. In this way, we have the parametric equations for this subspace:
\[\begin{aligned}&X^{1} = X^1(x^1,\dots,x^n)\\ &X^{2} = X^2 (x^1,\dots,x^n)\\ &\dots\\ &X^{m} = X^m(x^1,\dots,x^n) \end{aligned} \tag{51} \]Thus:
\[d^2 \boldsymbol{r} = d^2 (X^1, X^2,\dots,X^m)=(d^2 X^1, d^2 X^2,\dots,d^2 X^m)=0 \tag{52} \]This proves $d^2 \boldsymbol{r}=0$, which yields:
\[\begin{aligned}0=& d(d\boldsymbol{r})\\ =&d(\hat{\boldsymbol{e}}_{\mu} \omega^{\mu})\\ =&\hat{\boldsymbol{e}}_{\mu} d\omega^{\mu}+d\hat{\boldsymbol{e}}_{\mu} \land \omega^{\mu}\\ =&\hat{\boldsymbol{e}}_{\mu} d\omega^{\mu} + \hat{\boldsymbol{e}}_{\nu} \omega_{\mu}^{\nu} \land \omega^{\mu}\\ =&\hat{\boldsymbol{e}}_{\mu} d\omega^{\mu} + \hat{\boldsymbol{e}}_{\mu} \omega_{\nu}^{\mu} \land \omega^{\nu}\\ =&\hat{\boldsymbol{e}}_{\mu}(d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu})\\ \end{aligned} \tag{53} \]This implies:
\[d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0 \tag{54} \]It can be seen that the term $\omega_{\nu}^{\mu}\land \omega^{\nu}$ corresponds exactly to matrix multiplication, except that the ordinary product is replaced by the exterior product.
The above discussed the motion of the orthonormal frame, obtaining:
\[d\hat{\boldsymbol{e}}_{\mu}=\hat{\boldsymbol{e}}_{\nu}\omega_{\mu}^{\nu} \tag{55} \]Assuming $\eta_{\mu\alpha}$ is a constant matrix, then $\omega_{\mu\nu}=\eta_{\mu\alpha}\omega_{\nu}^{\alpha}$ is antisymmetric. The above equation can actually be written as:
\[\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x}+d\boldsymbol{x}) =\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x}) + d\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x})=\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x})[\delta_{\nu}^{\mu}+\omega_{\mu}^{\nu}(\boldsymbol{x})] \tag{56} \]This can be seen as the result of the frame moving from $\boldsymbol{x}$ to an infinitesimally neighboring position $\boldsymbol{x}+d\boldsymbol{x}$. What then is the transformation formula from an arbitrary point $\boldsymbol{x}_1$ to another point $\boldsymbol{x}_2$? We can divide the path from $\boldsymbol{x}_1$ to $\boldsymbol{x}_2$ into several small segments, moving one small segment at a time, approximating each segment with the above formula, then superimposing them and taking the limit:
\[\begin{aligned}&\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x}_2) \\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1)\prod_{k} [\delta_{\nu}^{\mu}+\omega_{\mu}^{\nu}(\boldsymbol{x}_1+kd\boldsymbol{x})]\\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1)\prod_{k} \exp[\omega_{\mu}^{\nu}(\boldsymbol{x}_1+kd\boldsymbol{x})] \end{aligned} \tag{57} \]Note that $\omega_{\mu}^{\nu}$ is a matrix, and the multiplication above is matrix multiplication. For matrices $\boldsymbol{A}$ and $\boldsymbol{B}$, $\exp(\boldsymbol{A})\exp(\boldsymbol{B})=\exp(\boldsymbol{A}+\boldsymbol{B})$ if and only if $\boldsymbol{A}\boldsymbol{B}=\boldsymbol{B}\boldsymbol{A}$, meaning matrix multiplication is commutative. If the multiplication of $\omega_{\nu}^{\mu}(\boldsymbol{x})$ at different positions is commutative (this is always true in two-dimensional space, but not always in other spaces), then we have:
\[\begin{aligned}&\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x}_2) \\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1) \exp\left[\sum_i\omega_{\mu}^{\nu}(\boldsymbol{x}_1+kd\boldsymbol{x})\right]\\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1) \exp\left(\int_{\boldsymbol{x}_1}^{\boldsymbol{x}_2}\omega_{\mu}^{\nu}\right) \end{aligned} \tag{58} \]The integral here is performed along a certain path from $\boldsymbol{x}_1$ to $\boldsymbol{x}_2$. It is evident that the result of the integral depends on the path; therefore, the resulting motion of the frame also depends on the path.
Consider a vector $\boldsymbol{A}=\hat{\boldsymbol{e}}_{\mu}\hat{A}^{\mu}$. We have added a $\hat{}$ to $A$ to indicate that its components are measured in the orthonormal frame. Now consider its differential:
\[\begin{aligned}d\boldsymbol{A}=&\hat{\boldsymbol{e}}_{\mu}d\hat{A}^{\mu}+d\hat{\boldsymbol{e}}_{\mu} \hat{A}^{\mu}\\ =&\hat{\boldsymbol{e}}_{\mu}d\hat{A}^{\mu}+\hat{\boldsymbol{e}}_{\nu}\omega_{\mu}^{\nu}\hat{A}^{\mu}\\ =&\hat{\boldsymbol{e}}_{\mu}(d\hat{A}^{\mu}+\omega_{\nu}^{\mu}\hat{A}^{\nu})\end{aligned} \tag{59} \]This is equivalent to the covariant derivative of a vector, and we can see that the extra term arises due to the motion of the frame.
Now consider its exterior differential:
\[\begin{aligned}d^2\boldsymbol{A}=&d[\hat{\boldsymbol{e}}_{\mu}(d\hat{A}^{\mu}+\omega_{\nu}^{\mu}\hat{A}^{\nu})]\\ =&\hat{\boldsymbol{e}}_{\mu}d(d\hat{A}^{\mu}+\omega_{\nu}^{\mu}\hat{A}^{\nu})+d\hat{\boldsymbol{e}}_{\mu}\land (d\hat{A}^{\mu}+\omega_{\nu}^{\mu}\hat{A}^{\nu}) \\ =&\hat{\boldsymbol{e}}_{\mu}d(\omega_{\nu}^{\mu}\hat{A}^{\nu})+\hat{\boldsymbol{e}}_{\alpha}\omega_{\mu}^{\alpha}\land (d\hat{A}^{\mu}+\omega_{\nu}^{\mu}\hat{A}^{\nu})\\ =&-\hat{\boldsymbol{e}}_{\mu}\omega_{\nu}^{\mu}\land d\hat{A}^{\nu} + \hat{\boldsymbol{e}}_{\mu}d\omega_{\nu}^{\mu}\hat{A}^{\nu}+\hat{\boldsymbol{e}}_{\alpha}\omega_{\mu}^{\alpha}\land (d\hat{A}^{\mu}+\omega_{\nu}^{\mu}\hat{A}^{\nu})\\ =&\hat{\boldsymbol{e}}_{\mu}(d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha})\hat{A}^{\nu}\end{aligned} \tag{60} \]Based on our previous discussion of the "winding" meaning of $d\omega$, and recalling the definition of the Riemann curvature tensor in component language, we can guess that $\mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha}$ must be related to the Riemann curvature tensor. In fact, we have:
\[\begin{aligned}\mathscr{R}_{\nu}^{\mu}=&\frac{1}{2}R^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma}\\ =&\sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma}\end{aligned} \tag{61} \]Again, we added a $\hat{}$ to $R$ to show it is measured in the orthonormal frame. Recalling the geometric meaning of the exterior product of differentials discussed earlier, $\omega^{\beta}\land \omega^{\gamma}$ is exactly the projection of the area element. The left side of the equation represents the change in a vector moved along a closed curve, and the right side also represents the change moved along a closed curve, except the right side is expressed in component language. Thus, since they have the same geometric meaning, the equality is inevitable and does not require substituting specific components to verify.
If we need to convert back to the original coordinate system, we substitute $\omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}$ into:
\[d^2\boldsymbol{A} = \hat{\boldsymbol{e}}_{\mu} \left( \sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma} \right)\hat{A}^{\nu} \tag{62} \]along with $\hat{A}^{\mu}=h_{\alpha}^{\mu} A^{\alpha}$ and $\hat{\boldsymbol{e}}_{\mu} = \boldsymbol{e}_{\alpha}(h^{-1})_{\alpha}^{\mu}$, finally obtaining:
\[d^2\boldsymbol{A} = \boldsymbol{e}_{\mu}\left( \sum_{\beta < \gamma} (h^{-1})_{\mu'}^{\mu}\hat{R}^{\mu'}_{\nu' \beta' \gamma'} h_{\nu}^{\nu'}h_{\beta}^{\beta'}h_{\gamma}^{\gamma'} dx^{\beta}\land dx^{\gamma} \right)A^{\nu} \tag{63} \]I have run out of indices and had to use primes to represent summation indices. Ultimately, it can be seen that:
\[R^{\mu}_{\nu\beta\gamma}=(h^{-1})_{\mu'}^{\mu}\hat{R}^{\mu'}_{\nu' \beta' \gamma'} h_{\nu}^{\nu'}h_{\beta}^{\beta'}h_{\gamma}^{\gamma'} \tag{64} \]In actual calculations, we do not need to calculate $\hat{R}^{\mu}_{\nu\beta\gamma}$ first and then $R^{\mu}_{\nu\beta\gamma}$. Instead, we can directly calculate $(h^{-1})_{\mu'}^{\mu}\mathscr{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'}$ and write it in the summation form of $dx^{\beta}\land dx^{\gamma}$ to read off $R^{\mu}_{\nu\beta\gamma}$. Specific operational examples will be shown in the next section.