2
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.
$\endgroup$
10
  • $\begingroup$ this is all of course obvious :) $\endgroup$
    – peek-a-boo
    Commented May 14, 2023 at 15:12
  • $\begingroup$ 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 ? $\endgroup$
    – Plantation
    Commented May 15, 2023 at 0:35
  • 1
    $\begingroup$ @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. $\endgroup$
    – peek-a-boo
    Commented May 15, 2023 at 2:00
  • 1
    $\begingroup$ Yes what you wrote is also fine. $\endgroup$
    – peek-a-boo
    Commented May 15, 2023 at 2:07
  • 1
    $\begingroup$ That’s why I emphasized the function is nowhere vanishing, so the absolute value is smooth (absolute value is smooth away from the origin) $\endgroup$
    – peek-a-boo
    Commented May 15, 2023 at 2:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .