I'm currently studying maximally symmetrical spaces, physics style. So I am mainly interested in purely local results.

I define a (locally) maximally symmetric space as a pseudo-Riemannian manifold that has $ n (n + 1) / 2 $ independent killing vector fields.

I have found that this is equivalent to the curvature tensor that is of the shape $$ R _ { kappa lambda mu nu} = K (g _ { mu kappa} g _ { nu lambda} -g _ { mu lambda} g _ { nu kappa} ), $$ from where $ K $ is a constant.

The book *Gravity and cosmology* von Weinberg has a sentence that if $ bar g _ { mu ^ { prime} nu ^ prime} (x ^ prime) $ and $ g _ { mu nu} (x) $ are two metrics (since this is purely local, I basically work in an open set of $ mathbb R ^ n $) which have the same signature and are both maximally symmetric, so that (assuming Einstein summation convention in 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 *equal* $ K $ constant, then the two metrics $ bar g _ { mu ^ prime nu ^ prime} $ and $ g _ { mu nu} $ differ by a coordinate transformation, z. There are functions $$ x ^ { mu ^ prime} = Phi ^ { mu ^ prime} (x) $$ so 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 proves this by explicitly constructing a coordinate transformation over a power series. It is ugly and long.

I thought 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 $while $ theta ^ a $ is a $ g $-orthonormal coframe, then the two metrics are the same if and only if the two coframes differ by a generalized orthogonal transformation (Lorentz transformation for general relativity), z. There is a $ mathrm O (n-s, s) $-valued function $ Lambda $ on the open set so that $$ bar theta ^ {a ^ prime} = Lambda ^ {a ^ prime} _ { a} theta ^ a. $$

But even if that is not true, there is *must be* on $ mathrm {GL} (n, mathbb R) $-valued function $ L $ so that $$ bar theta ^ {a ^ prime} = L ^ {a ^ prime} _ { a} theta ^ a. $$

So I thought I could probably prove that statement by proving that $ L $ is actually a (generalized) orthogonal transformation.

The curvature shape for (locally) maximally symmetrical spaces has a simple form $$ mathbf R ^ {ab} = K theta ^ a wedge theta ^ b \ mathbf R ^ {a ^ prime b ^ prime} = K bar theta ^ {a ^ prime} wedge bar theta ^ {b ^ prime}. $$

My strategy was to take the "locked" sizes in the "prepared" frame and transform them (possibly through non-orthogonal ones) $ L $) in the "unprimed" framework.

For example, for the metric we have $ bar g_ {a ^ prime b ^ prime} equiv eta_ {a ^ prime b ^ prime} $ (from where $ eta $ is the canonical symbol associated with the metric of a given signature, e.g. the Minkowski symbol for general theory of relativity), but in the unpainted frame it is $ bar g_ {ab} $ This is not necessarily "Minkowskian".

I tried to construct the curvature shape directly out of the frame and compare it with the expression that I listed above in the hope that I come to a relationship that implies one of $$ bar g_ {ab} = eta_ {ab} \ bar gamma ^ {ab} = – bar gamma ^ {ba}, $$ that would mean that immediately $ L $ is actually a generalized orthogonal transformation, but I have come to no useful conclusion.

**Question:** Can I prove this statement (namely, that two locally maximal symmetric spaces of the same dimension, signature, and the same value of $ K $ will be locally isometric) using this orthonormal framework method?

If yes, how does it work? I'm pretty stuck with it.