[A Gentle Introduction to Exterior Calculus] 6. Differential Geometry

【外微分浅谈】6. 微分几何 苏剑林 · 2016-11-07 · Mathematics geometry riemannian geometry exterior differentiation

Finally we get to the main point—it's precisely this part of the material that motivated me to learn exterior calculus in the first place. Exterior calculus makes it convenient to derive various results in differential geometry, and sometimes it even simplifies computation. The main reason for this is as follows: exterior differentiation is, in form, itself a generalization of differentiation, so it isn't surprising that objects in differential geometry can be described using exterior calculus. But the most important reason is that exterior calculus treats $dx^{\mu}$ as a set of basis elements, which effectively introduces two sets of bases into geometry at once—one is the ordinary vector basis (in tensor language, the basis of contravariant vectors), which supports a symmetric inner product; the other basis is $dx^{\mu}$, which supports an antisymmetric exterior product. So once exterior calculus is brought into geometry, differential geometry gains access to a whole "ideal toolkit": differentiation, integration, symmetric products, antisymmetric products, and so on. This is the main reason exterior calculus can speed up derivations in differential geometry.

The motion of a frame

We already obtained earlier

$$\begin{aligned}&\omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}\\ &d\boldsymbol{r}=\hat{\boldsymbol{e}}_{\mu} \omega^{\mu}\\ &ds^2 = \eta_{\mu\nu} \omega^{\mu}\omega^{\nu}\\ &\langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle = \eta_{\mu\nu}\end{aligned} \tag{45} $$more

Applying $d$ to both sides of $\langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle = \eta_{\mu\nu}$ gives

$$d\eta_{\mu\nu} = \langle d\hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle + \langle \hat{\boldsymbol{e}}_{\mu}, d\hat{\boldsymbol{e}}_{\nu}\rangle \tag{46} $$

$d\hat{\boldsymbol{e}}_{\mu}$ is the differential of a vector, and the result is also a vector, so it can be expressed as a linear combination of $\hat{\boldsymbol{e}}_{\mu}$, i.e.

$$d\hat{\boldsymbol{e}}_{\mu} = \hat{\boldsymbol{e}}_{\alpha} \omega_{\mu}^{\alpha} \tag{47} $$

Hence

$$\begin{aligned}d\eta_{\mu\nu} =& \langle \hat{\boldsymbol{e}}_{\alpha}\omega_{\mu}^{\alpha} , \hat{\boldsymbol{e}}_{\nu}\rangle + \langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\alpha}\omega_{\nu}^{\alpha} \rangle \\ =&\langle \hat{\boldsymbol{e}}_{\alpha}, \hat{\boldsymbol{e}}_{\nu}\rangle\omega_{\mu}^{\alpha} + \langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\alpha} \rangle\omega_{\nu}^{\alpha}\\ =&\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha} \end{aligned} \tag{48} $$

In most practical situations, $\eta_{\mu\nu}$ is a constant diagonal matrix, so

$$\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}=0 \tag{49} $$

Then, viewed as a matrix, $\omega_{\mu\nu}=\eta_{\mu \alpha}\omega_{\nu}^{\alpha}$ is antisymmetric, having only $n(n-1)/2$ independent components. In particular, if $\eta_{\mu\nu}$ is the identity matrix, then $\omega_{\mu}^{\alpha}$ is antisymmetric.

Next, we claim that

$$d^2 \boldsymbol{r}=0 \tag{50} $$

Note that this is not obvious. Although we know that for any function $f$ we have $d^2 f=0$, $d\boldsymbol{r}$ is not actually derived as the differential of some function—it is an arbitrarily given differential-form-valued vector, so $d^2 \boldsymbol{r}=0$ is not an obviously true statement. However, we can imagine that any curved $n$-dimensional space (manifold) can be embedded, as a subset, into a sufficiently high-dimensional flat $m$-dimensional space (Euclidean space), much like a surface embedded in three-dimensional space. In that case, we have a parametric equation for this subspace,

$$\begin{aligned}&X^{1} = X^1(x^1,\dots,x^n)\\ &X^{2} = X^2 (x^1,\dots,x^n)\\ &\dots\\ &X^{m} = X^m(x^1,\dots,x^n) \end{aligned} \tag{51} $$

so that

$$d^2 \boldsymbol{r} = d^2 (X^1, X^2,\dots,X^m)=(d^2 X^1, d^2 X^2,\dots,d^2 X^m)=0 \tag{52} $$

This proves $d^2 \boldsymbol{r}=0$, which in fact gives us:

$$\begin{aligned}0=& d(d\boldsymbol{r})\\ =&d(\hat{\boldsymbol{e}}_{\mu} \omega^{\mu})\\ =&\hat{\boldsymbol{e}}_{\mu} d\omega^{\mu}+d\hat{\boldsymbol{e}}_{\mu} \land \omega^{\mu}\\ =&\hat{\boldsymbol{e}}_{\mu} d\omega^{\mu} + \hat{\boldsymbol{e}}_{\nu} \omega_{\mu}^{\nu} \land \omega^{\mu}\\ =&\hat{\boldsymbol{e}}_{\mu} d\omega^{\mu} + \hat{\boldsymbol{e}}_{\mu} \omega_{\nu}^{\mu} \land \omega^{\nu}\\ =&\hat{\boldsymbol{e}}_{\mu}(d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu})\\ \end{aligned} \tag{53} $$

and this in turn shows that

$$d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0 \tag{54} $$

Notice 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.

Going a bit further

Above we discussed the motion of an orthonormal frame and obtained

$$d\hat{\boldsymbol{e}}_{\mu}=\hat{\boldsymbol{e}}_{\nu}\omega_{\mu}^{\nu} \tag{55} $$

Assuming that $\eta_{\mu\alpha}$ is a constant matrix, $\omega_{\mu\nu}=\eta_{\mu\alpha}\omega_{\nu}^{\alpha}$ is antisymmetric. The equation above 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 moving the frame from $\boldsymbol{x}$ to an infinitesimally nearby position $\boldsymbol{x}+d\boldsymbol{x}$. So 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 a number of small segments, move a small step at a time, approximate each small step using the formula above, and then compose them and take the limit, i.e.

$$\begin{aligned}&\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x}_2) \\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1)\prod_{k} [\delta_{\nu}^{\mu}+\omega_{\mu}^{\nu}(\boldsymbol{x}_1+kd\boldsymbol{x})]\\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1)\prod_{k} \exp[\omega_{\mu}^{\nu}(\boldsymbol{x}_1+kd\boldsymbol{x})] \end{aligned} \tag{57} $$

Note that $\omega_{\mu}^{\nu}$ is a matrix, and the multiplication above is matrix multiplication; for matrices $\boldsymbol{A}$ and $\boldsymbol{B}$, $\exp(\boldsymbol{A})\exp(\boldsymbol{B})=\exp(\boldsymbol{A}+\boldsymbol{B})$ holds if and only if $\boldsymbol{A}\boldsymbol{B}=\boldsymbol{B}\boldsymbol{A}$, i.e. the matrix multiplication commutes. If the multiplication of $\omega_{\nu}^{\mu}(\boldsymbol{x})$ at different positions commutes (this always holds in two-dimensional space, but not necessarily in other spaces), then we have:

$$\begin{aligned}&\hat{\boldsymbol{e}}_{\mu}(\boldsymbol{x}_2) \\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1) \exp\left[\sum_i\omega_{\mu}^{\nu}(\boldsymbol{x}_1+kd\boldsymbol{x})\right]\\ =&\hat{\boldsymbol{e}}_{\nu}(\boldsymbol{x}_1) \exp\left(\int_{\boldsymbol{x}_1}^{\boldsymbol{x}_2}\omega_{\mu}^{\nu}\right) \end{aligned} \tag{58} $$

Here the integral is taken along some path from $\boldsymbol{x}_1$ to $\boldsymbol{x}_2$, and it's clear that the result of the integral depends on the path—so the outcome of the frame's motion also depends on the path.

The motion of a vector

Consider a vector $\boldsymbol{A}=\hat{\boldsymbol{e}}_{\mu}\hat{A}^{\mu}$; we have already attached $\hat{}$ 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 in fact exactly the covariant derivative of the vector; we can see that the extra term arises precisely because of 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 earlier discussion of the "loop" meaning of $d\omega$, together with recalling the component-language definition of the Riemann curvature tensor, we might 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. Indeed, we have

$$\begin{aligned}\mathscr{R}_{\nu}^{\mu}=&\frac{1}{2}R^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma}\\ =&\sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma}\end{aligned} \tag{61} $$

Once again, we attach $\hat{}$ to $R$ to indicate that this is measured in the orthonormal frame. Recalling the geometric meaning of the exterior product of differentials discussed earlier, we see that $\omega^{\beta}\land \omega^{\gamma}$ is exactly the projection of the area element. The left-hand side of the equation above represents the change of the vector as it moves around a closed curve, and the right-hand side represents the same thing—just expressed in component language. Since they carry the same geometric meaning, the equality must hold; there's no need to substitute the components explicitly and verify the formula by brute force.

If we need to switch back to the original coordinate system, then into

$$d^2\boldsymbol{A} = \hat{\boldsymbol{e}}_{\mu} \left( \sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma} \right)\hat{A}^{\nu} \tag{62} $$

we substitute $\omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}$, along with $\hat{A}^{\mu}=h_{\alpha}^{\mu} A^{\alpha}$ and $\hat{\boldsymbol{e}}_{\mu} = \boldsymbol{e}_{\alpha}(h^{-1})_{\alpha}^{\mu} $, and finally obtain

$$d^2\boldsymbol{A} = \boldsymbol{e}_{\mu}\left( \sum_{\beta < \gamma} (h^{-1})_{\mu'}^{\mu}\hat{R}^{\mu'}_{\nu' \beta' \gamma'} h_{\nu}^{\nu'}h_{\beta}^{\beta'}h_{\gamma}^{\gamma'} dx^{\beta}\land dx^{\gamma} \right)A^{\nu} \tag{63} $$

As you can see, I've run out of indices to use, and had to resort to primed indices to denote the summation indices. In the end we get

$$R^{\mu}_{\nu\beta\gamma}=(h^{-1})_{\mu'}^{\mu}\hat{R}^{\mu'}_{\nu' \beta' \gamma'} h_{\nu}^{\nu'}h_{\beta}^{\beta'}h_{\gamma}^{\gamma'} \tag{64} $$

Of course, in actual computation we don't need to first compute $\hat{R}^{\mu}_{\nu\beta\gamma}$ and then compute $R^{\mu}_{\nu\beta\gamma}$ from it—we can directly compute $(h^{-1})_{\mu'}^{\mu}\mathscr{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'}$, then write it as a sum over $dx^{\beta}\land dx^{\gamma}$, and read off $R^{\mu}_{\nu\beta\gamma}$ from that. For a concrete worked example, see the next section.