[A Gentle Introduction to Exterior Differentiation] 7. The Power of the Method Computationally

【外微分浅谈】7. 有力的计算 苏剑林 · 2016-11-11 · Mathematics computation riemannian geometry exterior differentiation

Here we will show just how powerful the method from the previous section is when it comes to computing the Riemann curvature tensor! Let's list once more all the formulas we've derived. First, the conceptual one

$$\begin{aligned}&\omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}\\ &d\boldsymbol{r}=\hat{\boldsymbol{e}}_{\mu} \omega^{\mu}\\ &ds^2 = \eta_{\mu\nu} \omega^{\mu}\omega^{\nu}\\ &\langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle = \eta_{\mu\nu}\end{aligned} \tag{65} $$

then

$$\begin{aligned}&d\eta_{\mu\nu}=\omega_{\nu\mu}+\omega_{\mu\nu}=\eta_{\nu\alpha}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}\\ &d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0\end{aligned} \tag{66} $$

these two help us pin down $\omega_{\nu}^{\mu}$; next comes

$$\mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha} \tag{67} $$

Finally, if you want $\hat{R}^{\mu}_{\nu\beta\gamma}$ in an orthonormal frame, you need to write

$$\mathscr{R}_{\nu}^{\mu}=\sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma} \tag{68} $$

If you want $R^{\mu}_{\nu\beta\gamma}$ in the original frame, you need to write

$$(h^{-1})_{\mu'}^{\mu}\mathscr{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'} = \sum_{\beta < \gamma} R^{\mu}_{\nu\beta\gamma}dx^{\beta}\land dx^{\gamma} \tag{69} $$

and then read off $R^{\mu}_{\nu\beta\gamma}$ one by one, just like filling in a table. more

A Two-Dimensional Example: The Sphere

Let's warm up with a two-dimensional example and compute the Riemann curvature tensor of the sphere $ds^2 = d\theta^2 + \sin^2 \theta d\phi^2$.

We take

$$\omega^1 = d\theta, \quad \omega^2 = \sin\theta d\phi \tag{70} $$

that is

$$\boldsymbol{h}=\begin{pmatrix}1&0\\0&\sin\theta\end{pmatrix},\quad \boldsymbol{\eta}=\begin{pmatrix}1&0\\0&1\end{pmatrix} \tag{71} $$

Since $\boldsymbol{\eta}$ is the identity matrix, $d\eta_{\mu\nu}=\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}$ tells us that $\omega_{\nu}^{\mu}$ is an antisymmetric matrix. Writing $d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0$ in matrix form,

$$\begin{pmatrix} 0 & \omega_2^1 \\ -\omega_2^1 & 0 \end{pmatrix}\land \begin{pmatrix} d\theta \\ \sin\theta d\phi \end{pmatrix}=-d\begin{pmatrix} d\theta \\ \sin\theta d\phi \end{pmatrix}=-\begin{pmatrix} 0 \\ \cos\theta d\theta\land d\phi \end{pmatrix} \tag{72} $$

Due to the antisymmetry, $\omega_{\nu}^{\mu}$ has only one independent component, and it's not hard to see that $\omega_2^1=-\cos\theta d\phi$. This solving process can be done by guessing and trying things out. Next, we compute $\mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha}$, i.e.

$$\begin{aligned}\mathscr{R}_{\nu}^{\mu} = &d\begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix}\\ &+\begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix}\land \begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix} \end{aligned} \tag{73} $$

The final matrix-multiplication term is clearly zero — in fact one can show that in a 2-dimensional space $\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha}$ is always identically zero — so

$$\mathscr{R}_{\nu}^{\mu} = d\begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix}=\begin{pmatrix} 0 & \sin\theta d\theta\land d\phi \\ -\sin\theta d\theta\land d\phi & 0 \end{pmatrix} \tag{74} $$

And since

$$\mathscr{R}_{\nu}^{\mu}=\sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma}=\hat{R}^{\mu}_{\nu 1 2 }\sin\theta d\theta \land d\phi \tag{75} $$

comparing the two we get

$$\hat{R}^{\mu}_{\nu 1 2 } = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \tag{76} $$

That is, in the orthonormal frame we have $\hat{R}^{1}_{2 1 2 } = 1, \hat{R}^{2}_{1 1 2 } = -1$. Then, using $(h^{-1})_{\mu'}^{\mu}\mathscr{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'} = \sum_{\beta < \gamma} R^{\mu}_{\nu\beta\gamma}dx^{\beta}\land dx^{\gamma}$ to compute $R^{\mu}_{\nu\beta\gamma}$, we have

$$\begin{aligned}&\begin{pmatrix}1&0\\0&\sin\theta\end{pmatrix}^{-1}\begin{pmatrix} 0 & \sin\theta d\theta\land d\phi \\ -\sin\theta d\theta\land d\phi & 0 \end{pmatrix}\begin{pmatrix}1&0\\0&\sin\theta\end{pmatrix}\\ =&R^{\mu}_{\nu 12}d\theta\land d\phi\end{aligned} \tag{77} $$

that is

$$R^{\mu}_{\nu 1 2 } = \begin{pmatrix} 0 & \sin^2\theta \\ -1 & 0 \end{pmatrix} \tag{78} $$

which is to say $R^{1}_{2 1 2 } = \sin\theta, R^{2}_{1 1 2 } = -1$. This whole process only involves matrix multiplication, something we're all familiar with — a huge saving of mental effort compared to the tedious multi-index summations. Comparing the forms of $\hat{R}^{\mu}_{\nu\beta\gamma}$ and $R^{\mu}_{\nu\beta\gamma}$, we can indeed see that the orthonormal frame plays a genuine simplifying role.

A Four-Dimensional Example: The Schwarzschild Metric

The first exact solution of Einstein's field equations is the Schwarzschild metric, obtained by solving a metric of the form

$$ds^2= -e^{2\Phi}dt^2 + e^{2\Lambda} dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2 \tag{79} $$

Its starting point is to consider an isotropic metric, so $\Phi, \Lambda$ is assumed to be a function of $r$ alone. Let's compute the Riemann curvature tensor in this case (moderately-to-quite difficult).

Quite naturally, we take

$$\omega^1 = e^{\Phi}dt, \quad \omega^2 = e^{\Lambda }dr, \quad \omega^3 = rd\theta, \quad \omega^4 = r\sin\theta d\phi \tag{80} $$

Then

$$\boldsymbol{h}=\begin{pmatrix}e^{\Phi}&0&0&0\\0&e^{\Lambda }&0&0\\0&0&r&0\\0&0&0&r\sin\theta\end{pmatrix},\quad \boldsymbol{\eta}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} \tag{81} $$

From $d\eta_{\mu\nu}=\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}$ we know that $\omega_{\mu\nu}=\omega_{\mu}^{\alpha}\eta_{\alpha \nu}$ is antisymmetric, and combining this with a bit of analysis of the form of $\boldsymbol{\eta}$, we can conclude that $\omega_{\nu}^{\mu}$ takes the following form

$$\begin{pmatrix}0&\omega_2^1&\omega_3^1&\omega_4^1\\ \omega_2^1&0&\omega_3^2&\omega_4^2\\ \omega_3^1&-\omega_3^2&0&\omega_4^3\\ \omega_4^1&-\omega_4^2&-\omega_4^3&0\end{pmatrix} \tag{82} $$

Its distinguishing feature is that, viewed as a block matrix

$$\left(\begin{array}{c:ccc}0&\omega_2^1&\omega_3^1&\omega_4^1\\ \hdashline \omega_2^1&0&\omega_3^2&\omega_4^2\\ \omega_3^1&-\omega_3^2&0&\omega_4^3\\ \omega_4^1&-\omega_4^2&-\omega_4^3&0\end{array}\right) =\left(\begin{array}{c:c}E & F\\ \hdashline G&H\end{array}\right) \tag{83} $$

it is symmetric, but the diagonal blocks $E,H$ are each antisymmetric. Exactly how we partition it into blocks depends on how we split $\boldsymbol{\eta}$ into $\left(\begin{array}{c:c}-I & 0\\ \hdashline 0&I\end{array}\right)$, where $I$ denotes the identity matrix.

Once we have the concrete form of $\omega_{\nu}^{\mu}$, we can write down $d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0$:

$$\begin{aligned}&\begin{pmatrix}0&\omega_2^1&\omega_3^1&\omega_4^1\\ \omega_2^1&0&\omega_3^2&\omega_4^2\\ \omega_3^1&-\omega_3^2&0&\omega_4^3\\ \omega_4^1&-\omega_4^2&-\omega_4^3&0\end{pmatrix}\land \begin{pmatrix}e^{\Phi}dt\\ e^{\Lambda }dr\\rd\theta\\r\sin\theta d\phi\end{pmatrix}\\ =&-d\begin{pmatrix}e^{\Phi}dt\\ e^{\Lambda }dr\\rd\theta\\r\sin\theta d\phi\end{pmatrix}=-\begin{pmatrix}e^{\Phi} \dot{\Phi} dr\land dt\\ 0 \\dr\land d\theta\\ \sin\theta dr\land d\phi+ r\cos\theta d\theta\land d\phi \end{pmatrix}\end{aligned} \tag{84} $$

Here $\dot{}$ means differentiation with respect to $r$. With a bit of brainstorming, one can quickly pin down the answer. For instance, looking at row 2 being identically zero, we can conclude that $\omega_2^1,\omega_3^2,\omega_4^2$ depend only on $dt,d\theta,d\phi$ respectively; combining this with row 1, we can determine $\omega_2^1 = e^{\Phi-\Lambda}\dot{\Phi} dt$, and further deduce that $\omega_3^1,\omega_4^1$ depend only on $d\theta,d\phi$ respectively. Then, combining with row 3, we can determine $\omega_3^1=0$ as well as $\omega_3^2=-e^{-\Lambda}d\theta$, and deduce that $\omega_4^3$ depends only on $d\phi$; finally, looking at row 4, we quickly determine $\omega_4^1=0, \omega_4^2 = -\sin\theta e^{-\Lambda} d\phi, \omega_4^3 = -\cos\theta d\phi$, arriving at

$$\omega_{\nu}^{\mu}=\begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi&\cos\theta d\phi&0\end{pmatrix} \tag{85} $$

Now we can compute $\mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha}$, namely

$$\begin{aligned}\mathscr{R}_{\nu}^{\mu} =&d\begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi&\cos\theta d\phi&0\end{pmatrix}\\ &+\begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi&\cos\theta d\phi&0\end{pmatrix}\\ &\land \begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi&\cos\theta d\phi&0\end{pmatrix} \end{aligned} \tag{86} $$

which gives

$$\begin{aligned}&\mathscr{R}_1^1=\mathscr{R}_2^2=\mathscr{R}_3^3=\mathscr{R}_4^4=0\\ &\mathscr{R}^1_2=\mathscr{R}^2_1=-e^{\Phi-\Lambda}(\ddot{\Phi}+\dot{\Phi}^2-\dot{\Phi}\dot{\Lambda}) dt\land dr\\ &\mathscr{R}^1_3=\mathscr{R}^3_1=-e^{\Phi-2\Lambda} \dot{\Phi} dt\land d\theta\\ &\mathscr{R}^1_4=\mathscr{R}^4_1=-e^{\Phi-2\Lambda} \dot{\Phi} \sin\theta dt\land d\phi\\ &\mathscr{R}^2_3=-\mathscr{R}^3_2=e^{-\Lambda}\dot{\Lambda} dr\land d\theta\\ &\mathscr{R}^2_4=-\mathscr{R}^4_2=e^{-\Lambda}\dot{\Lambda}\sin\theta dr\land d\phi\\ &\mathscr{R}^3_4=-\mathscr{R}^4_3=(1-e^{2\Lambda})\sin\theta d\theta\land d\phi\end{aligned} \tag{87} $$

and from this we can read off $\hat{R}^{\mu}_{\nu\beta\gamma}$ one at a time, for example

$$\hat{R}^{1}_{212}=-e^{-2\Lambda}(\ddot{\Phi}+\dot{\Phi}^2-\dot{\Phi}\dot{\Lambda}), \quad\hat{R}^{1}_{312}=-\frac{1}{r}e^{-2\Lambda}\dot{\Phi} \tag{88} $$

and so on. If you like, you can go on to compute $R^{\mu}_{\nu\beta\gamma}$ as well, since $\boldsymbol{h}$ is a diagonal matrix, so this doesn't add much extra work.

If readers actually carry out this computation themselves, they may still complain that it takes a lot of time, and perhaps feel that it doesn't simplify things all that much. But following the steps above one by one, it's a calculation that can be done even by hand. This is, at the very least, a method that a human can feasibly carry out — the computation above was in fact done by hand by the author, without resorting to software like Mathematica. I doubt anyone has ever computed the Riemann curvature tensor in more than three dimensions starting directly from the raw expression $R^{\mu}_{\nu\beta\gamma}=\frac{\partial \Gamma^{\mu}_{\nu\gamma}}{\partial x^{\beta}}-\frac{\partial \Gamma^{\mu}_{\nu\beta}}{\partial x^{\gamma}}+\Gamma^{\mu}_{\alpha\beta}\Gamma^{\alpha}_{\nu\gamma}-\Gamma^{\mu}_{\alpha\gamma}\Gamma^{\alpha}_{\nu\beta}$. Forget about actually computing it — even keeping track of the summation indices is no easy task. By comparison, the techniques of exterior differentiation are far more effective. Of course, no matter which method you use, it always takes some time to train yourself to become proficient with it; and even once you're proficient, it still takes some thought to work through a problem — it's not something you can see at a glance, unless you're a computer.