Sobolev space on a compact Riemannian manifold does not depend of the metric I am reading Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities by Emmanuel Hebey and he stated on page $22$ this result and then he stated

This leads to the following:


Proposition $2.2$ If $M$ is compact, $H_k^p(M)$ does not depend on the metric.

The author defines Sobolev spaces as follows

Let $(M,g)$ be a smooth Riemannian manifold. For $k$ integer, and $u: M \longrightarrow \mathbb{R}$ smooth, we denote by $\nabla^k u$ the $k^{\text{th}}$ covariant derivative of $u$, and $|\nabla^k u|$ the norm of $\nabla^k u$ defined in a local chart by


$$|\nabla^k u| = g^{i_1j_1} \cdots g^{i_kj_k} (\nabla^k u)_{i_1 \cdots i_k} (\nabla^k u)_{j_1 \cdots j_k}$$


Recall that $(\nabla u)_i = \partial_i u$, while


$$(\nabla^2 u)_{ij} = \partial_{ij} u - \Gamma_{ij}^k \partial_k u$$


Given $k$ an integer, and $p \geq 1$ real, set


$$\mathscr{C}_k^p(M) = \left\{ u \in C^{\infty}(M) / \forall j = 0, \cdots, k, \int_M |\nabla^j u|^p dv(g) < +\infty \right\}$$


When $M$ is compact, one clearly has that $\mathscr{C}_k^p(M) = C^{\infty}(M)$ for any $k$ integer, and $p \geq 1$. For $u \in \mathscr{C}_k^p(M)$, set also


$$||u||_{H_k^p} = \sum_{j=0}^k \left( \int_M |\nabla^j u|^p dv(g) \right)^{\frac{1}{p}}$$


We define the Sobolev space $H_k^p(M)$ as follows:


Definition $2.1$ Given $(M,g)$ a smooth Riemannian manifold, $k$ an integer, and $p \geq 1$ real, the Sobolev space $H_k^p(M)$ is the completion of $\mathscr{C}_k^p(M)$ with respect to $||\cdot||_{H_k^p}$.

According this definition and the auxiliary result, I think it is sufficient show that there exist $C_1, C_2 > 0$ such that $C_1 |\nabla^k u|_{\tilde{g}} \leq |\nabla^k u|_g \leq C_2 |\nabla^k u|_{\tilde{g}}$ for arbitrary Riemannian metrics $g$ and $\tilde{g}$ defined on $M$, but I do not have idea how to show it. I would like a hint how to prove it.
Thanks in advance!
 A: In your first equation, it seems the left side should be $|\nabla^k u|^2$. I'll assume this is the case.
This might not be the most efficient path, but here's a way to establish a bound bound $\sum_{m=0}^k|\nabla^mu|^2_g\le a\sum_{m=0}^k|\widetilde{\nabla}^mu|^2_{\widetilde{g}}$, which, with some additional manipulations, should imply the equivalence $\|u\|_{H^k_p}\le c\|u\|_{\widetilde{H}^k_p}$.
First, we can define some auxiliary vector bundles which will be convenient later.
Let $T_{\le k}M$ denote the whitney sum over $M$ of all covariant tensor bundles of rank $\le k$, i.e.
$$
T_{\le k}M=T_0M\oplus T_1M\oplus\dots\oplus T_kM
$$
Where $T_0M\cong M\times\mathbb{R}$. Let $\nabla^{\le k}$ denote an operator which takes values in sections of this bundle, given by
$$
(\nabla^{\le k}u)(p)=u(p)\oplus(\nabla u)(p)\oplus\dots\oplus(\nabla^ku)(p)
$$
Let $S\subset T_{\le k}M$ be the image of $\nabla^{\le k}$, i.e.
$$
S=\left\{(\nabla^{\le k}u)(p):u\in C^\infty M,\ p\in M\right\}
$$
Characterizing $S$ by working locally
Let $x^1,\dots,x^n:U\to\mathbb{R}^n$ be a local coordinate chart. Within $U$, we can define an operator $D^{\le k}_\varphi$ which gives all partial derivatives of $u$ up to order $k$ as a vector in $\mathbb{R}^N$, i.e.
$$
\left(D^{\le k}_\varphi u\right)_{i_1\dots i_m}=\frac{\partial^m u}{\partial x^{i_1}\dots\partial x^{i_m}}\ \ \ \ \ \ \ 0\le m\le k,\ \ \ 1\le i_1\le\dots\le i_m<n
$$
Within this coordinate chart, $\nabla^{\le k} u$ is a linear function of $D^{\le k}_\varphi u$ (seen via unravelling the expression for $\nabla$ in terms of $\partial$ and Christoffel symbols), and by reversing the role of $\partial_i$ and $\nabla_i$, we see the same is true in reverse, that is $D^{\le k}_\varphi u$ is a linear function of $\nabla^{\le k}u$. Since we have essentially constructed a smooth local trivialization, $S$ is a smooth vector subbundle of $T_{\le k}M$ of dimension $N=\sum_{m=0}^k\binom{n+m-1}{m}$.
Construct inner products on $U\times\mathbb{R}^N$.
$g$ induces a positive definite inner product on $T_{\le k}M$ (the orthogonal direct sum of the inner products on $T_mM$), and this inner product is inherited by $S$. We can of course perform exactly the same construction for the other metric and obtain another subbundle $\widetilde{S}$ with a metric induced by $\widetilde{g}$. Both of these inner products induce inner products on $U\times\mathbb{R}^N$, given by
$$
g\left(D^{\le k}u,D^{\le k}u\right)=g\left(\nabla^{\le k}u,\nabla^{\le k}u\right)\ \ \ \ \ \widetilde{g}\left(D^{\le k}u,D^{\le k}u\right)=\widetilde{g}\left(\widetilde{\nabla}^{\le k}u,\widetilde{\nabla}^{\le k}u\right)
$$
Local inequalities
Let $K\subset U$ be compact. define two constants $a,b$ by
$$
a=\max_{\substack{(p,v)\in K\times\mathbb{R}^N \\ v\neq 0}}\frac{g(v,v)}{\widetilde{g}(v,v)}\ \ \ \ \ b=\max_{\substack{(p,v)\in K\times\mathbb{R}^N \\ v\neq 0}}\frac{\widetilde{g}(v,v)}{g(v,v)}
$$
Note that we can restrict both maxima to the compact set $K\times S^{N-1}$ without change, so both are well defined. This gives us the bounds $\|\nabla^{\le k}u\|^2_{g}\le a\|\widetilde{\nabla}^{\le k}u\|^2_{\widetilde{g}}$ and $\|\widetilde{\nabla}^{\le k}u\|^2_{\widetilde{g}}\le b\|\nabla^{\le k}u\|^2_{g}$ on $K$.
Global comparisons
Choosing a finite atlas $\phi_\alpha:U_\alpha\to\mathbb{R}^n$ and a subbordinate covering by compact sets $K_\alpha\subset U_\alpha$. We may repeat the above procedure to obtain $a_\alpha,b_\alpha$ for each chart. Let $a=\max a_\alpha$, $b=\max b_\alpha$, and the inequalities stated above hold for all of $M$.
