# Proof of Existence of the Riemannian Density ( John Lee's Smooth manifodls)

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 $$$$\mu_g(E_1, \dots, E_n)=1 \tag{16.20}$$$$
for any local orthonormal frame $$(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?

If we put $$\mu_g:=|\omega_g|$$ it follows easily that $$\mu_g$$ is a smooth positive density on $$U$$ satisfying (16.20).

There are a few things to verify

• $$\mu_g$$ is a density on $$U$$: this follows because $$\omega_g$$ is an $$n$$-form on $$U$$, and $$\mu_g:=|\omega_g|$$.
• $$\mu_g$$ is smooth on $$U$$: we have $$\omega_g$$ is smooth and nowhere-vanishing, so $$\mu_g$$, being the ‘absolute value’, is smooth as well (recall that an equivalent way of characterizing smoothness is to say that for all local frames of smooth vector fields $$\{\xi_1,\dots,\xi_n\}$$, the function $$\mu_g(\xi_1,\dots, \xi_n)$$ must be smooth; now by virtue of $$\omega_g$$ being smooth, we have the function $$\omega_g(\xi_1,\dots, \xi_n)$$ is a smooth function, and since $$\omega_g$$ is nowhere-vanishing, this evaluated function is nowhere-vanishing, so taking its absolute value still gives a smooth function).
• $$\mu_g$$ satisfies (16.20): fix any point $$p$$, and an orthonormal basis $$\{e_1,\dots, e_n\}$$ of $$T_pM$$. This is of course linearly independent, and so $$\omega_g$$ being nowhere vanishing tells us the number $$(\omega_g)_p(e_1,\dots, e_n)$$ is either positive or negative. In the first case, this by definition means the basis is positively-oriented with respect to the orientation determined by $$(\omega_g)_p$$ on $$T_pM$$. So, by the defining condition of the Riemannian volume form, this number has to be $$+1$$, so $$(\mu_g)_p(e_1,\dots, e_n)=|1|=1$$. In the second case, the number is negative, and so (by multilinearity) $$(\omega_g)_p(-e_1,\dots, e_n)>0$$, so $$\{-e_1,e_2,\dots, e_n\}$$ is positively-oriented and of course still orthonormal, and so $$(\omega_g)_p(-e_1,\dots, e_n)=1$$, and thus $$(\omega_g)_p(e_1,\dots, e_n)=-1$$. Therefore, $$(\mu_g)_p(e_1,\dots, e_n)=|-1|=1$$. So, we’ve shown (16.20) holds pointwise, and thus it obviously holds at the local level as well (and global level as well, if global orthonormal frames exist).
• $$\mu_g$$ is a positive density: this should be obvious based on the previous bullet point and the transformation law for densities.
• this is all of course obvious :) Commented May 14, 2023 at 15:12
• Thank you. And I am suck at some point. You wrote, " In the first case, this by definition means the basis is positively-oriented with respect to the orientation determined by $(\omega_{g})_p$ on $T_pM$. So, by the defining condition of the Riemannian volume form, this number has to be $+1$". An issue that confusing me is, the definition of the Riemannian volume form only requires that $\omega_g(E_1, \dots , E_n) =1$ for every local oriented orthonormal 'frame' $(E_i)$ for $M$. Perhaps, positively oriented basis on tangent vector space extends to local oriented orthonormal frame ? Commented May 15, 2023 at 0:35
• @Plantation for every orthonormal basis at a point, you can find a local orthonormal frame. This is something you should prove for yourself. Anyway, I could have phrased everything in my answer at the level of local sections as well. You should make the necessary changes yourself. Commented May 15, 2023 at 2:00
• Yes what you wrote is also fine. Commented May 15, 2023 at 2:07
• That’s why I emphasized the function is nowhere vanishing, so the absolute value is smooth (absolute value is smooth away from the origin) Commented May 15, 2023 at 2:37