Locally Flat Understanding I wanted to make sure that I was definitely understanding the proof of locally flat correctly.  I can't see to find a similar proof to the one in the book, so I'm not super sure if my understanding/interpretation is correct.  The book that I'm using is "Einstein Gravity in a Nutshell" by Anthony Zee.
Starting with the coordinate transform for a metric,
$$g'_{\lambda\sigma}(x')=g_{\mu\nu}(x)\frac{\partial x^\mu}{\partial x'^\lambda}\frac{\partial x^\nu}{\partial x'^\sigma}$$
if we expand the metric in unprimed coordinates up to the second order
$$g_{\mu\nu}\approx g_{\mu\nu}(0)+A_{\mu\nu,\lambda}x^\lambda+B_{\mu\nu,\lambda\sigma}x^\lambda x^\sigma + ...$$
where we have centered the metric on $x=0$.
We can likewise expand the coordinate transformation as
$$x^\mu\approx K^\mu_\lambda x'^\lambda+L^\mu_{\lambda\nu}x'^\lambda x'^\nu+M^\mu_{\lambda\nu\sigma}x'^\lambda x'^\nu x'^\sigma+...$$
If we take the partial derivative,
$$\frac{\partial x^\mu}{\partial x'^\lambda}\approx K^\mu_\lambda+L^\mu_{\lambda\nu}x'^\nu+M^\mu_{\lambda\nu\sigma} x'^\nu x'^\sigma+...$$
Since we are looking for the local result, we drop all other terms s.t.
$$\frac{\partial x^\mu}{\partial x'^\lambda}\approx K^\mu_\lambda$$
so at a local level the metric transform is
$$g'_{\lambda\sigma}(x')\approx g_{\mu\nu}(0)K^\mu_\lambda K^\nu_\sigma$$
Since $K^\mu_\lambda$ is summing over the row index, in matrix representation we can express this as
$$g'\approx K^TgK$$
since the metric is by definition real and symmetric, g can always be diagonalised.
Is anything in the above wrong?  The power series is from the book, and I think it kind of makes sense?  Something about it feels a bit off, but I can't quite put my finger on what, so I'd love some confirmation on whether I've understood the proof correctly.  Thanks!
 A: Almost correct, save for two things.
One is more about pedantry, but I really, really dislike when people say "expand" or "drop", it makes it sound like this is some nebulous infinitesimal-y, approximation-y thing. It isn't.
Let's say we want to construct a "locally flat" (which is itself a kinda bad term, let's use "Riemann normal coordinates" instead) coordinate system at $x_0$.
As a first step, we can make a translational coordinate transformation such that $x_0^i=0$ (I am gonna use latin indices in this answer).
Now, we are going to perform a coordinate transformation where we do not "expand" anything, we simply want to make the coordinate transformation a linear homogenous polynomial, so we write $$ x^i=A^i_{\ j}x^{\prime j}, $$ then $$ \frac{\partial x^i}{\partial x^{\prime j}}=A^i_{\ j}. $$
A thing to note is that adding higher order terms to the coordinate transformation won't change a thing, since if $$ x^i=A^i_{\ j}x^{\prime j}+B^i_{\ jk}x^{\prime j}x^{\prime k}, $$ then $$ \frac{\partial x^i}{\partial x^{\prime j}}=A^i_{\ j}+B^i_{\ jk}x^{\prime k}, $$ but the point $x_0$ has coordinates $0$, so if we evaluate the derivative at $x^i=0$ we get $$ \frac{\partial x^i}{\partial x^{\prime j}}(0)=A^i_{\ j}. $$ We did not have to "drop" anything at all.
The transformation of the metric components at $0$ is $$ g^\prime_{ij}(0)=\frac{\partial x^k}{\partial x^{\prime i}}(0)\frac{\partial x^l}{\partial x^{\prime j}}(0)g_{kl}(0)=A^k_{\ i}A^l_{\ j}g_{kl}(0), $$ so we may choose the coefficients $A^i_{\ j}$ such that $g^\prime_{ij}=\eta_{ij}$, since now we are working at a single point and all things here are constant matrices.
However what makes a coordinate system a Riemannian normal system (or a "locally flat" system if one'd like) about $x_0$ is not that the metric is Minkowskian at the point $x_0$, but that in addition to that, the Christoffel symbol vanishes at that point. The transformation of the Christoffel symbol is $$ \Gamma^{\prime n}_{\ lm}=\Gamma^k_{\ ij}\frac{\partial x^i}{\partial x^{\prime l}}\frac{\partial x^j}{\partial x^{\prime m}}\frac{\partial x^{\prime n}}{\partial x^k}+\frac{\partial x^{\prime n}}{\partial x^i}\frac{\partial^2 x^i}{\partial x^{\prime l}\partial x^{\prime m}}, $$ if the LHS is zero, then this simplifies to $$ -\frac{\partial^2 x^k}{\partial x^{\prime l}\partial x^{\prime m}}=\Gamma^k_{\ ij}\frac{\partial x^i}{\partial x^{\prime l}}\frac{\partial x^j}{\partial x^{\prime m}}, $$ and if we evaluate at $0$ and insert the above-written quadratic transformation, we get $$ -B^k_{\ lm}=\Gamma^k_{\ ij}(0)A^i_{\ l}A^j_{\ m}, $$ thus if $A^i_{\ j}$ is any matrix that transforms the metric at $0$ into Minkowskian form, then the coordinate transformation $$ x^i=A^i_{\ j}x^{\prime j}-\Gamma^i_{\ jk}(0)A^j_{\ l}A^k_{\ m}x^{\prime l}x^{\prime m} $$ has the effect that at $x_0=0$ we have $$ g_{ij}(0)=\eta_{ij} \\ \Gamma^i_{\ jk}(0)=0 \\ (\partial_i g_{jk})(0)=0. $$
