Derivation of metric flatness locally

Question

In Einstein Gravity by Zee chapter I.6, he discussed the local flatness of the metric. There are two steps in what he did. First, he showed that the metric at a point can always be written as a flat metric. Second, he showed that in general, a metric can be expanded around the flat metric and that the first order expansion vanishes.

The first part, consider a point $P$ in coordinates $x^\mu$ with metric $g_{\mu\nu}(x)$, if we go to a new coordinate $x'^\mu$, the new metric $g'_{\mu\nu}(x')$ transforms as,

\begin{equation} g'_{\lambda\sigma}(x') = g_{\mu\nu}(x) \frac{\partial x^\mu}{\partial x'^\lambda} \frac{\partial x^\nu}{\partial x'^\sigma} \end{equation}

We can always shift our coordinates so that $P$ is labeled by $x = 0$. Expand the metric around $P$ out to second order,

\begin{equation} g_{\mu\nu}(x) = g_{\mu\nu}(0) + A_{\mu\nu,\rho} x^\rho + B_{\mu\nu,\rho \gamma} x^\rho x^\gamma + \ldots \end{equation}

where the commas in the subscripts carried by $A$ and $B$ are purely for notational clarity to separate two sets of indices. Now, we change coordinates $$x^\mu = K^\mu_{\; \nu} x'^\nu + L^\mu_{\; \nu \lambda} x'^\nu x'^\lambda + \ldots. $$ At $P$, the new metric up to linear in $x'^\nu$ is then,

\begin{equation} g'_{\lambda\sigma}(0) = g_{\mu\nu}(0) K^\mu_{\; \lambda} K^\nu_{\; \sigma} \quad \rightarrow \quad g' = K^T g K \end{equation}

Since $g_{\mu\nu}(0)$ is symmetric and real there always exists a $K$ that diagonalizes it. After $g_{\mu\nu}(0)$ becomes diagonal we could scale each coordinate by an appropriate factor, $x^\mu \rightarrow x^\mu/\sqrt{g_{\mu\mu}(0)}$ (no sum over repeated indices), so that the diagonal elements become 1. So, $g_{\mu\nu}(0) = \delta_{\mu \nu}$. Since $g_{\mu\nu}(0)$ had $\frac{1}{2} D(D + 1)$ arbitrary elements to begin with, we had to use this many elements in $K^\mu_{\; \nu}$ to adjust these to $\delta_{\mu \nu}$. Hence, $K^\mu_{\; \nu}$ has $D^2 − \frac{1}{2}D(D + 1) = \frac{1}{2} D(D − 1)$ elements left over.

I don't understand the last statement "we had to use this many elements in $K^\mu_{\; \nu}$ to adjust these to $\delta_{\mu \nu}$", what elements in $K^\mu_{\; \nu}$?

As I understood, suppose $g_{\mu\nu}(0) \rightarrow (g_{xx},g_{yy},g_{xy},g_{yx})$ is a non-diagonal metric. There should be a $K^\mu_{\; \lambda}$ say, $(K^x_{\; \theta},K^y_{\; \theta}, \ldots)$ for some coordinate transformation $x^\mu (x') = (x(\theta,\phi),y(\theta,\phi))$, which makes the metric diagonal. Suppose the diagonal metric has the form $$ds^2 = g_{\theta \theta} d\theta^2 + g_{\phi \phi} d\phi^2.$$ Then, we can rescale $\theta \rightarrow \theta/\sqrt{g_{\theta \theta}}$, $\phi \rightarrow \phi/\sqrt{g_{\phi \phi}}$, which gives $ds^2 = d\theta^2 + d\phi^2$. This is just the flat metric which we can rewrite as $ds^2 = dX^2 + dY^2$.

The second part, the metric for some arbitrary point can be expanded such that $g_{\mu\nu} (x) = \delta_{\mu\nu} + A_{\mu\nu,\rho} x^\rho + \ldots$. The linear term $A_{\mu\nu,\rho} x^\rho$ can be removed by a suitable choice of $L^\mu_{\nu \lambda}$ in $$x^\mu = x'^\mu + L^\mu_{\nu \lambda} x'^\nu x'^\lambda + \ldots.$$ Evidently, $A_{\mu\nu,\rho}$ and $L^\mu_{\nu \lambda}$ have been modified already by what we have done thus far, but we do not want to introduce more letters.

What happened to $K^\mu_{\; \nu}$ in $$x^\mu = K^\mu_{\; \nu} x'^\nu + L^\mu_{\; \nu \lambda} x'^\nu x'^\lambda + \ldots~?$$ It seems like $K^\mu_{\; \nu}$ was set to $\delta^\mu_{\; \nu}$.

Suppose I take it to be correct, then expanding the metric transformation to linear in $x'$,

\begin{align} g'_{\lambda\sigma}(x') & = (\delta_{\mu\nu} + A_{\mu\nu,\rho} x'^\rho) (\delta^\mu_\lambda + L^\mu_{\; \alpha \lambda} x'^\alpha) (\delta^\nu_\sigma + L^\nu_{\; \beta \sigma} x'^\beta)\\ & = \delta_{\lambda\sigma} + \delta_{\mu \sigma} L^\mu_{\; \alpha \lambda} x'^\alpha + \delta_{\lambda \nu} L^\nu_{\; \beta \sigma} x'^\beta + A_{\lambda\sigma,\rho} x'^\rho \end{align}

I think what he meant is that $\delta_{\mu \sigma} L^\mu_{\; \alpha \lambda} x'^\alpha + \delta_{\lambda \nu} L^\nu_{\; \beta \sigma} x'^\beta + A_{\lambda\sigma,\rho} x'^\rho = 0$ after some relabelling of indices.

Any guidance is greatly appreciated.

Answer 1

For the first question, because $K$ is a general matrix with $D^2$ elements but $g$ has only $\frac{1}{2}D(D+1)$ independent elements, in order to diagonalize $g$ you only need $\frac{1}{2}D(D+1)$ elements from $K$.

In other words $K^{T}gK = \delta$ is a system of $\frac{1}{2}D(D+1)$ equations for $D^2$ variables. So after the diagonalization you are left with $D^2 - \frac{1}{2}D(D+1) = \frac{1}{2}D(D-1)$ independent elements in $K$.

For the second part, suppose to keep $K^{\mu}_{\,\,\nu}$ obtaining \begin{align} g'_{\lambda \sigma}(x') &= (\delta_{\mu\nu} + A_{\mu\nu,\rho}x'^{\rho})(K^{\mu}_{\,\,\lambda} + L^{\mu}_{\,\,\alpha\lambda}x'^{\alpha})(K^{\nu}_{\,\,\sigma} + L^{\nu}_{\,\,\beta\sigma}x'^{\beta})\\ &= (\delta_{\mu\nu} + A_{\mu\nu,\rho}x'^{\rho})\left(K^{\mu}_{\,\,\lambda} K^{\nu}_{\,\,\sigma}+K^{\mu}_{\,\,\lambda}L^{\nu}_{\,\,\beta\sigma}x'^{\beta} +K^{\nu}_{\,\,\sigma}L^{\mu}_{\,\,\alpha\lambda}x'^{\alpha} \right)\\ &= \delta_{\lambda\sigma} + \tilde{L}_{\lambda\beta\sigma}x'^{\beta} +\tilde{L}_{\sigma\alpha\lambda}x'^{\alpha} + K^{\mu}_{\,\,\lambda}K^{\nu}_{\,\,\sigma}A_{\mu\nu,\rho}x'^{\rho}\\ &= \delta_{\lambda\sigma} + (\tilde{L}_{\lambda\alpha\sigma} +\tilde{L}_{\sigma\alpha\lambda} + K^{\mu}_{\,\,\lambda}K^{\nu}_{\,\,\sigma}A_{\mu\nu,\alpha})\,x'^{\alpha}\\ &= \delta_{\lambda\sigma} + (\tilde{L}_{\lambda\alpha\sigma} +\tilde{L}_{\sigma\alpha\lambda} + \tilde{A}_{\lambda\sigma,\alpha})\,x'^{\alpha} \\ &=\delta_{\lambda\sigma} + (\tilde{L}_{\lambda\sigma\alpha} +\tilde{L}_{\sigma\lambda\alpha} + \tilde{A}_{\lambda\sigma,\alpha})\,x'^{\alpha} \\ &\equiv \delta_{\lambda\sigma} + (L_{\lambda\sigma\alpha} +L_{\sigma\lambda\alpha} + A_{\lambda\sigma,\alpha})\,x'^{\alpha} \end{align} where only linear terms in $x'$ have been kept. In the third row the $L$ objects has been changed by $K$ and renamed $\tilde{L}$. In the fourth row the muted indices has been renamed to factor out $x'^{\alpha}$. In the fifth $A$ has been changed by $K$ and renamed $\tilde{A}$. In the sixth one notices that $L$ should be symmetric in the last two indices, indeed $L^{\mu}_{\,\, \nu \lambda}x'^{\nu}x'^{\lambda} =L^{\mu}_{\,\, \lambda \nu}x'^{\lambda}x'^{\nu} = L^{\mu}_{\,\, \lambda \nu}x'^{\nu}x'^{\lambda} $. Finally in the last row one simply forget about the tilde, as stated on the book: "Evidently, $A_{\mu\nu,\rho}$ and $L^{\mu}_{\,\, \nu \lambda}$ have been modified already by what we have done thus far".

So, your final comment is correct, we can choose the components of $L$ such that \begin{equation} L_{\lambda\sigma\alpha} +L_{\sigma\lambda\alpha} + A_{\lambda\sigma,\alpha} = 0. \end{equation}