Manifolds with volume forms on every submanifold

If we equip a manifold with an inner product (i.e. we have a Riemannian Manifold) then we get a canonical volume form on that manifold (please mentally insert the prefix "pseudo" into my question whenever you feel it necessary).

Whenever we have a submanifold we can pull back the metric along the inclusion map to get an inner product on the submanifold. We then get a volume form on the submanifold.

I'm wondering if there are any converse results to this. I'm looking for some result of the form:

Suppose $\mathcal{M}$ is a manifold with a volume form assigned to every submanifold, such that these volume forms satisfy (some consistency condition). Then $\mathcal{M}$ is Riemannian, and the volume forms arise as above.

My only progress so far is to notice that you can define some kind of "norm" on such an $\mathcal{M}$ by, for any vector $v$, picking a curve $\gamma:(-1,1)\rightarrow\mathcal{M}$ with $\gamma'(0)=v$, letting $\mu$ be the measure on the submanifold which is the image of this curve and then defining $$\lVert v \rVert=\lim_{\varepsilon\rightarrow0}\frac{\mu(\gamma((0,\varepsilon))}{\varepsilon}.$$ But I don't see what properties I need to put on the volume forms in order to get the triangle inequality or the parallelogram law.

Solution 1:

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.