measure on non-oriented Riemannian manifold

Question

Let $M$ be a non-oriented Riemannian manifold of dimension $m$. Nash embedding theorem implies that there exists an isometric embedding $\phi: M\longrightarrow \mathbb{R}^n$ for $n$ sufficiently large. The Lebesgue measure of $\mathbb{R}^n$ induces a measure on $\phi(M)$ (we induce the measure on $m$-submanifold $\phi(M)$ locally, similar as in Fubini Theorem). Hence we obtain a measure on $M$. If $\phi$, $\psi$ are two such isometric embeddings of $M$ into $\mathbb{R}^n$, $\mathbb{R}^N$ resp., then since they are isometries, the induced measure from $R^n$ is the same with the induced measure from $\mathbb{R}^N$. Thus we get a well-defined measure on the non-oriented manifold $M$.

Is my argument valid or not? I am very confused...

Answer 1

The argument is valid, but you don't need Nash's theorem to come up with the canonical measure on a nonorientable Riemannian manifold $M$. There are other ways:

1. $M$ is a metric space, and thus carries the $m$-dimensional Hausdorff measure. This measure (up to normalization) agrees with what you get from the volume form when $M$ is orientable. See Riemannian measure and Hausdorff measure in a general Riemannian Manifold.
2. $M$ has an orientable double cover. You can push the measure from the cover to $M$ by the covering map, which is a local isometry. Divide by $2$ for proper normalization.
3. Use the concept of $m$-dimensional density.
4. Use a partition of unity $(\varphi_i)$ subordinate to coordinate patches. Define the measure on each patch using its volume form, multiply by $\varphi_i$, sum over $i$.