Initially, we need consider only the tangent space $T_p\mathcal{M}$ at each point $p$ of the manifold $\mathcal{M}$ independently (for this, credit to Oscar Cunningham for this comment on 31 March 2015); every submanifold $\mathcal{S}$ of $\mathcal{M}$ that contains $p$ has a tangent space $T_p\mathcal{S}$ that is a subspace of $T_p\mathcal{M}$. Further, by your hypothesis, there is a quadratic form (or, equivalently, a symmetric bilinear form) on the tangent space. (I assume characteristic $0$ for simplicity.)

For every restriction of the tangent space $T_p\mathcal{M}$ to a $1$-dimensional subspace $\mathcal{S}$, the quadratic form determines (up to a sign) a $1$-form (the volume form for $T_p\mathcal{S}$), **and vice versa** (again for $T_p\mathcal{S}$). The entire quadratic form on $T_p\mathcal{M}$ is then determined up to a single overall sign (this should be evident through continuity across $1$-dimensional subspaces, but does not need continuity to prove). Through continuity on $\mathcal{M}$, we can extend this to a single selection of sign of the quadratic form for each connected component of the manifold.

Thus, it is clear that the volume forms on the one-dimensional submanifolds are sufficient to determine the quadratic form up to a sign, and hence also the pseudo-Riemannian metric, with the sign being free to choose separately on each connected component of the manifold. Intuitively, a suitable consistency condition (aside from the usual continuity requirements) is that whenever two submanifolds are contained in a third submanifold and intersect at only one point, the exterior product relates their volume forms at that point.

One way to derive this result is by applying a more general result which allows one to integrate along fibers of a submersion:

**Oriented Submersions**

Let $M^m$, and $N^n$ be oriented smooth manifolds with $m>n$, and $\pi:M^m\to N^n$ be a smooth submersion. We can define *adapted oriented coordinates* as a set of oriented coordinate charts such that $\varphi$ has local representative equal to the projection onto the first $n$ coordinates $\pi(x^1,\cdots,x^m)=(x^1\cdots,x^n)$. There is a unique induced orientation on each fiber $\pi^{-1}(p)$ such that $(x^{n+1},\cdots,x^m)$ is an oriented chart on the fiber for any adapted oriented coordinates $x^i$.

Let $\alpha\in\Omega^nN$ and $\beta\in\Omega^{m-n}M$ be compactly supported differential forms. We can separate the following integral:
$$
\int_M\pi^*(\alpha)\wedge\beta=\int_Nf_\beta\alpha \\
\text{where}\ \ \ f_\beta(p):=\int_{\pi^{-1}(p)}\beta|_{\pi^{-1}(p)}
$$
To show this, it suffices to first show it holds in the case that $\beta$ is supported in a single set of adapted oriented coordinates with domain $(0,1)^m\mapsto(0,1)^n$, where it is a straightforward computation. The general case can then be obtained using a partition of unity.

**Riemannian Submersions**

Let $\pi:M^m\to N^n$ be a Riemannian submersion with $m>n$ ($M$ and $N$ now have metrics, but need not be oriented). Locally, we can find orthonormal coframes $E^1,\cdots,E^m$ on $U\subseteq M$ and $e^1,\cdots,e^n$ on $V(U)\subseteq N$ such that $\pi^*(e^i)=E^i$ for all $p\in U$ and $i\in[1,n]$. Note that $e^1\wedge\cdots\wedge e^n$ is a Riemann volume form on $N$, $E^1\wedge\cdots\wedge E^m$ is a Riemann volume form on $M$, and $E^{n+1}\wedge\cdots\wedge E^m$ restricts a Riemann volume form on each fiber $\pi^{-1}(q)\cap U$, Choosing a smooth $f:M\to \mathbb{R}$ supported in $U$, we can choose local orientations induced by these volume forms, we can apply the above result to the integral of $f$, yielding the following:
$$
\int_Mf\ dV=\int_N\hat{f}dV \\
\text{where}\ \ \ \hat{f}(q)=\int_{\pi^{-1}(q)}f\ dV
$$
Where $dV$ denotes the Riemannian measure on each manifold. Once again this result holds globally, provided $f$ is compactly supported, by a partition of unity argument.

**Homogeneous Spaces**

In your case, you have defined the metrics so that the quotient map $\pi:G\to X$ is a Riemannian submersion, and each fiber $\pi^{-1}(x)$ is isometric to $H$. Thus we have
$$
\operatorname{vol}(G)=\int_G1dV=\int_X fdV \\
\text{where}\ \ \ f(x):=\int_{\pi^{-1}(x)}1dV=\int_H1dV=\operatorname{vol}(H) \\
\implies \operatorname{vol}(G)=\int_X \operatorname{vol}(H)dV=\operatorname{vol}(X)\operatorname{vol}(H) \\
$$

## Best Answer

As Ted Shifrin kindly pointed out, this turns out to have been researched thoroughly, starting with Weyl. The proposed formula should be true possibly with some additional conditions.

In the beginning of this paper, we see the Weyl's formula: if $M$ is a $d$-dimensional compact, connected submanifold of $\mathbb{R}^D$ with boundary and if $M^\perp _r = \{ x + y | x \in M, y \in T_x^\perp M, \|x\| \le r \}$ denotes the normal thickening of $M$, then there exist constants $k_l(M)$ such that $$ \text{Vol}(M_r^\perp) = \omega_{D-d} r^{D-d} \sum_{\substack{0 \le l \le D \\ l \text{ even} }} \frac{k_{l}(M) r^{l} }{(D-d+2) \cdots (D-d+l) } $$ where $\text{Vol}(M_r)$ is the Euclidean volume (which I believe is the Lebesgue measure). In particular, $k_0(M)$ is the Riemannian volume of $M$. Therefore, if $M_r = \{ x+y | x \in M, \|y\| \le r \}$ is the

full thickeningof $M$, then we can sandwich $\text{Vol}(M_r)$ as follows: $$\text{Vol}(M_r^\perp) \le \text{Vol}(M_r) \le \text{Vol}(M_r^\perp) + \text{Vol}((\partial M)_r^\perp)$$ Here, to get the second inequality, we assume that for each point $z \in \mathbb{R}^D$ there is a point $z' \in M$ such that either (1) $z' \in M$ and $z - z' \in T_{z'}^\perp M$ or (2) $z' \in \partial M$ and $z - z' \in T_{z'}^\perp (\partial M)$.Seeing that the leading term of $\text{Vol}((\partial M)_r^\perp)$ in the Weyl's formula is of order $r^{D-d+1}$, we then see by a sandwiching argument that: $$\lim_{r \rightarrow 0} \frac{\text{Vol}(M_r)}{\omega_{D-d} r^{D-d}} = k_0(M)$$