Local uniqueness of metric for locally maximally symmetric spaces I am currently studying maximally symmetric spaces, physics style. So I am mainly interested in purely local results.
I define a (locally) maximally symmetric space as a pseudo-Riemannian manifold which has $n(n+1)/2$ independent Killing vector fields.
I have managed to derive that this is equivalent to the curvature tensor being of the form $$ R_{\kappa\lambda\mu\nu}=K(g_{\mu\kappa}g_{\nu\lambda}-g_{\mu\lambda}g_{\nu\kappa}), $$ where $K$ is a constant.
The book Gravitation and Cosmology by Weinberg has a theorem that if $\bar g_{\mu^{\prime}\nu^\prime}(x^\prime)$ and $g_{\mu\nu}(x)$ are two metrics (since this is purely local, I am basically working in an open set of $\mathbb R^n$) that have the same signature, and both are maximally symmetric such that (Einstein summation convention assumed throughout this post) $$ \bar R_{\kappa^\prime\lambda^\prime\mu^\prime\nu^\prime}=K(\bar g_{\mu^\prime\kappa^\prime}\bar g_{\nu^\prime\lambda^\prime}-\bar g_{\mu^\prime\lambda^\prime}\bar g_{\nu^\prime\kappa^\prime}) \\ R_{\kappa\lambda\mu\nu}=K(g_{\mu\kappa}g_{\nu\lambda}-g_{\mu\lambda}g_{\nu\kappa}) $$for the same $K$ constant, then the two metrics $\bar g_{\mu^\prime\nu^\prime}$ and $g_{\mu\nu}$ differ by a coordinate transformation, eg. there exist functions $$ x^{\mu^\prime}=\Phi^{\mu^\prime}(x) $$ such that $$ g_{\mu\nu}(x)=\bar g_{\mu^\prime\nu^\prime}(\Phi(x))\frac{\partial\Phi^{\mu^\prime}}{\partial x^\mu}(x)\frac{\partial\Phi^{\nu^\prime}}{\partial x^\nu}(x). $$
Weinberg proceeds to prove this by constructing a coordinate transformation explicitly via a power series. It is ugly and long.

I figured there is probably an easier way.
Namely, if $\bar g$ and $g$ are two metrics of the same signature and $\bar\theta^{a^\prime}$ is a $\bar g$-orthonormal coframe, while $\theta^a$ is a $g$-orthonormal coframe, then the two metrics will be equivalent if and only if the two coframes differ by a generalised orthogonal transformation (Lorentz transformation for general relativity), eg. there is an $\mathrm O(n-s,s)$-valued function $\Lambda$ on the open set such that $$ \bar\theta^{a^\prime}=\Lambda^{a^\prime}_{\ a}\theta^a. $$
However, even if this is not true, there must be a $\mathrm{GL}(n,\mathbb R)$-valued function $L$ such that $$ \bar\theta^{a^\prime}=L^{a^\prime}_{\ a}\theta^a. $$
So I figured I could probably prove this statement by proving that $L$ is actually a (generalized) orthogonal transformation.

The curvature form for (locally) maximally symmetric spaces have a simple form $$ \mathbf R^{ab}=K\theta^a\wedge\theta^b \\  \bar{\mathbf R}^{a^\prime b^\prime}=K\bar\theta^{a^\prime}\wedge\bar\theta^{b^\prime}. $$
My strategy was to take the "barred" quantities in the "primed" frame and transform them (via the possibly nonorthogonal $L$) into the "unprimed" frame.
Eg. for the metric we have $\bar g_{a^\prime b^\prime}\equiv \eta_{a^\prime b^\prime}$ (where $\eta$ is the canonical symbol associated with the metric of a given signature, eg. the Minkowski symbol for general relativity) but in the unprimed frame it is $ \bar g_{ab} $ which is not necessarily "Minkowskian".
I have tried to construct the curvature form from the frame directly and compare it with the expression I have listed above in hopes that I may arrive at some relation that implies one of $$ \bar g_{ab}=\eta_{ab} \\ \bar\Gamma^{ab}=-\bar\Gamma^{ba}, $$ which would immediately mean that $L$ is actually a generalised orthogonal transformation, however I arrived at no useful result.
Question: Can I prove this statement (namely that two locally maximally symmetric spaces of the same dimension, signature and same value of $K$ will be locally isometric) using this orthonormal frame method?
If so how to do that? I am quite stuck with it.
 A: I finally managed to solve this problem, and do so close to the way I wanted to. I am reproducing my solution here without the heavy calculations.
Let $x=(x^\mu)=(x^1,...,x^n)$ and $y=(y^\alpha)=(y^1,...,y^n)$ denote two sets of variables, I will use greek indices from the middle of the alphabet for the $x$ variables and greek indices from the beginning of the alphabet for the $y$ variables. One is given two sets of metric coefficients $\bar g_{\alpha\beta}(y)$ and $g_{\mu\nu}(x)$, both are assumed to be of the same signature.
We wish to examine the integrability conditions for the existence of functions $y^\alpha=\phi^\alpha(x)$ such that $$ \bar{g}_{\alpha\beta}(\phi(x))\frac{\partial\phi^\alpha}{\partial x^\mu}\frac{\partial\phi^\beta}{\partial x^\nu}=g_{\mu\nu}(x). $$ We are equipped with the additional assumption that the curvature tensor of the metric $\bar g_{\alpha\beta}$ has the form $$ \bar R_{\gamma\delta\alpha\beta}=K(\bar g_{\gamma\alpha}\bar g_{\beta\delta}-\bar g_{\gamma\beta}\bar g_{\alpha\delta}) $$ and the curvature tensor of the metric $g_{\mu\nu}$ has the form $$ R_{\kappa\lambda\mu\nu}=K(g_{\kappa\mu}g_{\nu\lambda}-g_{\kappa\nu}g_{\mu\lambda}), $$ where the constants $K$ agree.

The differential equation we wish to solve is equivalent to the equation $$ \phi^\ast\bar\vartheta^a(x)=\Lambda^a_i(x)\vartheta^i(x), $$ where $\Lambda^a_i(x)$ is a point-dependent generalized orthogonal transformation, and $\bar\vartheta^a(y)$ is an orthonormal coframe for $\bar g_{\alpha\beta}$ and $\vartheta^i(x)$ is an orthonormal coframe for $g_{\mu\nu}$. Here and from now on the indices $a,b,...$ denote "orthonormal frame indices" for the $y$-space and $i,j,...$ denote orthonormal frame indices for the $x$-space.
The equation is for the functions $\phi^\alpha(x)$ and the orthogonal transformations $\Lambda^a_i(x)$.
In order to get an equation for the orthogonal transformation as well, we use the relationship between the connection 1-forms to get $$ d\Lambda^a_j=\Lambda^a_i\omega^i_{\ j}-\phi^\ast\bar\omega^a_{\ b}\Lambda^b_j, $$ where $\omega^i_{\ j}$ are the connection forms of the $\vartheta^i$ and $\bar\omega^a_{\ b}$ are the connection forms of the $\bar\vartheta^a$.

Now introduce a coordinate space with coordinates $(x^\mu,y^\alpha,z^a_i)$ where the $z^a_i$ are matrix elements of an orthogonal transformation, so these coordinates are "overdetermined", and introduce the $1$-forms $$ X^a(x,y,z)=\bar\vartheta^a(y)-z^a_i\vartheta^i(x) \\ Y^a_j(x,y,z)=\mathrm dz^a_j+\bar\omega^a_{\ b}(y)z^b_j-z^a_i\omega^i_{\ j}(x). $$ A "function" (more like a section) $(\phi,\Lambda):(x^\mu)\mapsto(x^\mu,y^\alpha,z^a_i)=(x^\mu,\phi^\alpha(x),\Lambda^a_i(x))$ solves the differential equations if and only if $(\phi,\Lambda)^\ast X^a=(\phi,\Lambda)^\ast Y^a_j=0$, so the necessary and sufficient condition for the local existence of solutions is the local existence of integral submanifolds for the differential system $X^a,Y^a_j$, and thus by Frobenius' theorem, solutions for generic "initial conditions" exist if the differential system is closed i.e. $$dX^a=\xi^a_b\wedge X^b+\xi^{aj}_b\wedge Y^b_j \\  dY^a_j=\zeta^{ak}_{bj}\wedge Y^b_k+\zeta^a_{jb}\wedge X^b$$ for some 1-forms $\xi$ and $\zeta$.
Differentiation of the $X^a$ gives $$ dX^a=\vartheta^i\wedge Y^a_i-\bar\omega^a_{\ b}\wedge X^b, $$ and the differentiation of the $Y^a_j$ gives $$ dY^a_j=z^b_j\bar\Omega^a_{\ b}-z^a_i\Omega^i_{\ j}+Y^b_j\wedge\bar\omega^a_{\ b}-Y^a_i\wedge\omega^i_{\ j}, $$ where the $\bar\Omega^a_{\ b}$ are the curvature 2-forms of $\bar\vartheta^a$ and the $\Omega^i_{\ j}$ are the curvature forms of the $\vartheta^i$.
In the orthonormal frames, the assumption of constant curvature gives $$ \Omega^{ij}=K\vartheta^i\wedge\vartheta^j,\quad\bar\Omega^{ab}=K\bar\vartheta^a\wedge\bar\vartheta^b. $$ Inserting this into the equation for $dY^a_j$ and using the fact that $z^a_i$ is an orthogonal transformation (i.e. it has an inverse and $z^a_iz^b_j\eta_{ab}=\eta_{ij}$, where $\eta$ is the canonical symbol of the metric of given signature) gives $$ dY^a_j=Kz^b_j\eta_{bc}(\bar\vartheta^a\wedge X^c+X^a\wedge\bar\vartheta^c-X^a\wedge X^c)+Y^b_j\wedge\bar\omega^a_{\ b}-Y^a_i\wedge\omega^i_{\ j}, $$ which shows that under the assumption on the forms of the curvature 2-forms, the differential system generated by the $X^a,Y^a_j$ closes, and the equations are integrable, thus the two metrics are locally isometric.
