By 苏剑林 | November 05, 2016
As is well known, mastering Riemannian geometry requires a strong sense of geometric intuition. But beyond that, Riemannian geometry described in the language of components also requires good analytical skills to untangle, because there are numerous indices representing components and summations, appearing everywhere at first glance. This cumbersome component language is not always pleasing; in many places, it is even notorious.
In the language of components, we can essentially establish a coordinate system of any form locally, i.e., using an arbitrary set of bases $\{\boldsymbol{e}_{\mu}\}$, or natural frames. However, it is undeniable that under an orthogonal frame (orthonormal basis), many equations become much simpler. Given our familiarity with Euclidean space, research conducted under an orthogonal frame may feel more intuitive. Therefore, if conditions permit, we should use orthogonal frames $\{\hat{\boldsymbol{e}}_{\mu}\}$, even if they are moving frames. Here, we use the $\hat{}$ notation to mark the orthogonal frame.
For example, we have the differential element
\[d\boldsymbol{r} = \boldsymbol{e}_{\mu}dx^{\mu} \tag{12}\]
which is measured in a general frame. Then the Riemannian metric can be obtained as
\[ds^2 = \langle d\boldsymbol{r}, d\boldsymbol{r}\rangle= g_{\mu\nu}dx^{\mu} dx^{\nu} \tag{13}\]
where
\[g_{\mu\nu} = \langle \boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\rangle \tag{14}\]
is a matrix that might contain complex functions. We write the Riemannian metric in matrix form:
\[g_{\mu\nu}dx^{\mu} dx^{\nu}=d\boldsymbol{x}^T \boldsymbol{g}d\boldsymbol{x} \tag{15}\]
Then we attempt to perform such a decomposition:
\[\boldsymbol{g}=\boldsymbol{h}^T \boldsymbol{\eta}\boldsymbol{h} \tag{16}\]
where $\boldsymbol{h}$ and $\boldsymbol{\eta}$ are matrices of the same shape as $\boldsymbol{g}$, so
\[ds^2 = (\boldsymbol{h}d\boldsymbol{x})^T\boldsymbol{\eta}(\boldsymbol{h}d\boldsymbol{x}) \tag{17}\]
In component language, this is written as
\[ds^2 = \eta_{\mu\nu}(h_{\alpha}^{\mu} dx^{\alpha} )(h_{\beta}^{\nu} dx^{\beta}) \tag{18}\]
We denote
\[\omega^{\mu} = h_{\alpha}^{\mu} dx^{\alpha} \tag{19}\]
In fact, $h_{\alpha}^{\mu}$ is a transformation matrix that transforms the original arbitrary frame $\{\boldsymbol{e}_{\mu}\}$ into a (moving) orthogonal frame $\{\hat{\boldsymbol{e}}_{\mu}\}$, i.e.,
\[\hat{\boldsymbol{e}}_{\mu} = \boldsymbol{e}_{\alpha}(h^{-1})^{\alpha}_{\mu} , \quad \boldsymbol{e}_{\mu} = \hat{\boldsymbol{e}}_{\alpha} h^{\alpha}_{\mu} \tag{20}\]
At this point, we have
\[d\boldsymbol{r} = \boldsymbol{e}_{\mu} dx^{\mu} =\hat{\boldsymbol{e}}_{\mu} \omega^{\mu} \tag{21}\]
This indicates that $\omega^{\mu}$ in an orthogonal frame is equivalent to $dx^{\mu}$ in a general frame, and
\[ds^2 = \eta_{\mu\nu} \omega^{\mu} \omega^{\nu} \tag{22}\]
The above equation shows that orthogonal frames help simplify the Riemannian metric; now the metric tensor is the simpler $\eta_{\mu\nu}$. It should be pointed out that the ideal decomposition is for $\boldsymbol{\eta}$ to be the identity matrix. However, if we consider general Riemannian metrics (especially those in general relativity) and restrict ourselves to the real number domain, this ideal might not always be achievable. For example, it cannot be done for a simple $\boldsymbol{g}=\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$. Therefore, we only hope for $\boldsymbol{\eta}$ to be as simple as possible, such as a constant diagonal matrix, without requiring it to be the identity matrix. Such a decomposition is always possible, especially in many practical cases where $\boldsymbol{g}$ is a diagonal matrix, making it quite easy to implement. Therefore, we assume here that $\eta_{\mu\nu}$ is a diagonal matrix whose diagonal elements are 1 or -1.
Next, we write
\[ds^2 = \langle d\boldsymbol{r},d\boldsymbol{r}\rangle=\langle \hat{\boldsymbol{e}}_{\mu},\hat{\boldsymbol{e}}_{\nu}\rangle \omega^{\mu}\omega^{\nu} \tag{23}\]
which is
\[\eta_{\mu\nu} = \langle \hat{\boldsymbol{e}}_{\mu},\hat{\boldsymbol{e}}_{\nu}\rangle \tag{24}\]
Here, $\hat{\boldsymbol{e}}$ is the "orthogonal frame" in the sense of the above equation.