Looking for reference: If a Riemanian manifold is foliated by max symmetric submanifolds, then coordinates can always be chosen such that ...

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In Weinberg's book on General Relativity in section 13.5 it is shown that, loosely stated, if a Riemannian manifold $(M,g)$ of dim $m$ is composed of maximally symmetric submanifolds $(N,h)$ of dim $n$ that are all isomorphic, then coordinates $(v^1,\dots, v^{m-n}, u^1,\dots,u^n)$ can always be chosen such that

$$g = g_{ab}(v)\mathbb d v^a \mathbb d v^b + f(v)h_{ij}(u)\mathbb d u^i \mathbb d u^j.$$

I was wondering if anyone knows a more mathematics oriented text that proves or gives an account of this result. (Does the result have a name?) Thanks!

EDIT: And if a reference cannot be provided, a an answer providing the precise mathematical statement, preferably with proof, would also be very much appreciated.

EDIT: I see now that the original statement in my question was quite flawed. But frankly that was the whole reason that I posted the question. In any case, I found the correct statement in this original article: B. Schmidt, “Isometry gropus with surface-orthogonal trajectories,” z naturforsch sect A 22 (1967). See also my answer.

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I found the original article that proves the result I was looking for, albeit in a somewhat different language:

B. Schmidt, “Isometry gropus with surface-orthogonal trajectories,” z naturforsch sect A 22 (1967)

The proper statement is the following:

Let $G$ be a Lie group that acts with isometries on a Riemannian manifold $M$ (of any signature) of dim $m$ such that its orbits are connected $n$-dimensional Riemannian manifolds and the the stabilizer subgroup of any point $p\in M$ leaves no vector in the tangent space $T_pN$ invariant. If $\dim G = n(1+1)/2$ (i.e. if the orbits are maximally symmetric) then suitable coordinates $(v^1,\dots, v^{m-n}, u^1,\dots,u^n)$ can always be chosen such that the metric on $M$ has the form

$$ g=g_{ab}(v)\text{d}v^a \text{d} v^b+f(v)h_{ij}\text{d}u^i\text{d}u^j$$

with $u^i$ being the coordinates on $N$.