Two definitions of an integral on a Riemannian manifold

Let $M$ be a Riemannian manifold.

• If $M$ is orientable then we are familiar with the definition of $\int_{M}f$; where $f$ is a smooth function with a compact support, e.g. J.Lee's book.

• However, we are able to define a measure (Riemannian measure) on any Riemannian manifold that is not necessarily orientable, e.g. Grigoryan, book: Heat kernel and analysis on manifolds. Therefore we can define $\int_{M}f$ on any Riemannian manifold $M$ as long as $f$ is Borel measurable (usual measure theory).

My question is: which definition do we frequently prefer to use? Thanks!

• I don't have either book, but does the second definition not generalize the first? (That is, does orientation matter in the first definition?) – Andrew D. Hwang May 14 '17 at 2:08
• In the first definition we need the assumption of orientation to make sense it (def. by using an atlas then prove it doesn't depend on the chosen atlas, we need the orientation here). However, I don't see where we need it in the second definition! Weird? – user36548 May 14 '17 at 5:42
• A volume form on a smooth $n$-manifold $M$ is a non-vanishing $n$-form $\mu$ on $M$. A volume density (less standard term?) is, loosely, the absolute value of a volume form, $|\mu|$. On a Riemannian manifold, one often requires a compatibility: The value of a volume density on an orthonormal basis is $1$. A volume form exists if and only if $M$ is orientable; a volume density always exists. <> My question was, does your first definition refer to a volume form, but then compute the integral of $f$ using positively-oriented coordinates? If so, the second definition generalizes the first. – Andrew D. Hwang May 14 '17 at 12:39
• Yes, my first definition refers to the volume form, that is the form you mentioned, and its existence due to an available orientation. I don't know what is a volume density. But do you mean that: for any Riemannian manifold (not necessarily orientable), we can still define the integral in which it somehow gives a generalisation to the case of having an orientation? Thanks. – user36548 May 14 '17 at 16:05

If $(M, g)$ is an arbitrary Riemannian $n$-manifold (orientable or not), then in a coordinate neighborhood, there exist precisely two $n$-forms (which we might denote $\pm\mu$) such that $\mu(\Basis_{1}, \dots, \Basis_{n}) = \pm 1$ for every orthonormal frame $(\Basis_{j})_{j=1}^{n}$. Consequently, there is a unique volume density $|\mu|$, satisfying $|\mu|(\Basis_{1}, \dots, \Basis_{n}) = 1$ for every orthonormal frame.
(If you like, there is a trivial real line bundle over $M$ whose transition maps are the absolute values of the Jacobians of ordinary coordinate changes. By definition, a volume density is a section of this bundle.)
Either a volume form or a volume density suffices to integrate measurable functions on $M$, locally in the same way that the Euclidean measure $|dx_{1}\, dx_{2} \cdots dx_{n}|$/ordinary $n$-dimensional Lebesgue measure suffices, and globally using partitions of unity.
If $M$ is connected and orientable, then fixing a choice of orientation (say, picking $\mu$) at one point imposes a unique compatible orientation globally, yielding a smooth $n$-form $\mu$. The value of $\mu$ on a positively-oriented frame coincides with the value of the corresponding volume density, and for the purpose of integrating functions, it makes no difference whether we use a volume form or the associated volume density. In that sense, the second definition generalizes the first.
• That's kind of new to me because I used to think that it must be orientable when speaking of integration on manifolds. Now I 've learnt that in case of Riemannian manifolds $\int_{M} f$ makes sense unconditionally. – user36548 May 14 '17 at 18:53