I am reading the John Lee's Introduction to Smooth manifolds, Second Edition, Proof of Proposition 16.45 and stuck at understanding some statement :
Proposition 16.45 (The Riemannian Densitiy ) Let $(M,g)$ be a Riemannian Manifold with or without boundary. There is a unique positive density $\mu_g$ on $M$, called the Riemannian density,
with the property that $$ \begin{equation} \mu_g(E_1, \dots, E_n)=1 \tag{16.20} \end{equation} $$
for any local orthonormal frame $(E_i)$.
Proof. Uniqueness is immediate, because any two densities that agree on a basis must be equal ( c.f. His book Proposition 16.35 ). Given any point $p\in M$, let $U$ be a connected smooth coordinate neighborhood of $p$. Since $U$ is diffeomorphic to an open subset of Euclidean space, it is orientable. Any choice of orientation of $U$ uniquely determines a Riemannian volume form $\omega_g$ on $U$, with the property that $\omega_g(E_1, \dots ,E_n )=1$ for any 'oriented' orthonormal frame. If we put $\mu_g := | \omega_g| $ ( c.f. p.430 ), it follows easily that $\mu_g$ is a smooth positive density on $U$ satisfying (16.20). If $U$ and $V$ are two overlapping smooth coordinate neighborhoods, the two definitions of $\mu_g$ agree where they overlap by uniqueness, so this defines $\mu_g$ globally.
I don't understand why the bold statement is true. An issue that bothers me is, $\omega_g(E_1, \dots ,E_n )=1$ is only gauranteed for 'oriented' orthonormal frame, but (16.20) requires
$\mu_g(E_1, \dots, E_n)=1 $ for 'any' local orthonormal frame, possibly not oriented.
My one guess is to use the definition of the density function on finite dimensional vector space.
Definition ( his book p.428 ) Let $V$ be an $n$-dimensional vector space. A density on $V$ is a function $$ \mu : V \times \cdots \times V \to \mathbb{R}$$
($n$-copies) satisfying the following condition : if $T : V \to V$ is any linear map, then
$$ \mu (Tv_1, \dots ,Tv_n) = |\operatorname{det}T | \mu(v_1, \dots ,v_n) $$
Assume that we are given local othogonal frame $E_1 , \dots , E_n : V \subseteq U \to TU$. We want to show
$(\mu_g)_p (E_1 |_p , \dots , E_n|_p ) =1 $ for each $p\in V$. Assume that there exists an local neighborhood $ p \in W \subseteq V $ and local oriented orthonormal frame $E'_1 , \dots, E'_n : W \to TU$. Then the base change linear map $T$ between $(E_i |_p)$ and $(E'_i|_p)$ has determinant $\pm 1$ ( since each are orthonormal bases ). So by the above definition, we can deduce $(\mu_g)_p (E_1 |_p , \dots , E_n|_p ) =1 $. And my question is, the existence of such local oriented orthonormal frame $(E'_i)$ on $W$ is true? Or is there any other easy route to show the $(\mu_g)_p (E_1 |_p , \dots , E_n|_p ) =1 $? Can anyone helps?
Best Answer
There are a few things to verify