Finally, we begin to discuss the key point, which is the part that motivated me to learn exterior calculus. Using exterior calculus, one can conveniently derive aspects of differential geometry and sometimes simplify calculations. The main reason is that exterior calculus itself is a formal generalization of differentiation; thus, it is not surprising that differential geometry can be described using it. More importantly, exterior calculus treats dx^{\mu} as a set of basis elements, effectively introducing two sets of bases into geometry: one is the vector basis itself (the basis for contravariant vectors in tensor language), which allows for symmetric inner products, and the other is dx^{\mu}, which allows for anti-symmetric exterior products. Therefore, when exterior calculus is introduced into geometry, differential geometry gains an “ideal arsenal” including differentiation, integration, symmetric products, and anti-symmetric products. This is the primary reason why exterior calculus can accelerate derivations in differential geometry.
Motion of Frames
We previously 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} 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} Since d\hat{\boldsymbol{e}}_{\mu} is the differential of a vector, the result is also a vector, and thus 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 anti-symmetric, having only n(n-1)/2 components. In particular, if \eta_{\mu\nu} is the identity matrix, then \omega_{\mu}^{\alpha} is anti-symmetric.
Next, we assert: d^2 \boldsymbol{r}=0 \tag{50} Note that this is not trivial. Although we know that d^2 f=0 for any function f, d\boldsymbol{r} is not truly the differential of a function, but rather an arbitrarily given differential form vector; thus 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-dimensional 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 of 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 actually 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, but with the ordinary product replaced by the exterior product.
Going Further
The motion of the orthonormal frame was discussed above, yielding: 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 anti-symmetric. 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 viewed as the result of the frame moving from \boldsymbol{x} to an infinitesimally neighboring position \boldsymbol{x}+d\boldsymbol{x}. What is the transformation formula for moving 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 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 above multiplication 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}, i.e., the matrix multiplication is commutative. If the multiplication of \omega_{\nu}^{\mu}(\boldsymbol{x}) at different positions is commutative (which always holds in two-dimensional space but not necessarily in others), then we have: \begin{aligned} &\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x}_2) \\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1) \exp\left[\sum_k\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 result of the frame’s motion also depends on the path.
Motion of Vectors
Consider a vector \boldsymbol{A}=\hat{\boldsymbol{e}}_{\mu}\hat{A}^{\mu}, where 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 it can be seen that the extra term arises from the motion of the frame.
Next, consider its exterior derivative: \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 geometric meaning of d\omega as "circling around," and recalling the definition of the Riemann curvature tensor in component language, we can guess that \mathcal{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} \mathcal{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 use the \hat{} on R to indicate that it is measured in the orthonormal frame. Recalling the geometric meaning of the exterior product of differentials discussed earlier, we find that \omega^{\beta}\land \omega^{\gamma} is exactly the projection of the area element. The left side of the above equation represents the change of a vector moving along a closed curve, and the right side represents the same change, but described in component language. Therefore, they have the same geometric meaning, and the equality is inevitable without needing to substitute specific components for verification.
If we need to switch back to the original coordinate system, we substitute \omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}, \hat{A}^{\mu}=h_{\alpha}^{\mu} A^{\alpha}, and \hat{\boldsymbol{e}}_{\mu} = \boldsymbol{e}_{\alpha}(h^{-1})_{\alpha}^{\mu} 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} yielding: 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} As seen here, I have run out of indices and had to use primes to denote summation indices. Ultimately, we see: 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} Of course, 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}\mathcal{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'} and write it in the form of a sum over dx^{\beta}\land dx^{\gamma} to read off R^{\mu}_{\nu\beta\gamma}. Specific operational examples will be seen in the next section.
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