TITLE: Showing that Sobolev norms on manifolds are equivalent
QUESTION [7 upvotes]: Let me first define a "Sobolev space on manifold". Let $M$ be a closed $n$-dimensional manifold, $E \rightarrow M$ a complex vector bundle. 

Let us pick: 

A finite cover of $M$ by sets $U_i$. 
charts $h_i:U_i \cong \Bbb R^n$.
Trivilizations $\phi_i$ of $E|_{U_i}$ 
$\mu_i$ particion of unity of subordinate to $\{U_i\}$. 

Define the Sobolev norm of a section $u \in \Gamma(M,E)$ by 
$$ ||u||_k^2 := \sum_i ||(\mu_i \circ h_i^{-1}) (\phi_i \circ u \circ h_i^{-1} ) ||_k^2$$ 
this is well defined, the RHS being a fintie sum of Sobolev $k$-norm of compactly supported functions on $\Bbb R^n$. 

So I want to show 

The equivalence class of $|| \cdot ||_k$ is independent of the choices made. 


What I know: 

Result 1: Let $a \in C^\infty_c$. Then $f \mapsto af$ extends to a bounded operator $M_a:W^s \rightarrow W^s$ for each $s \in \Bbb Z$. $$||au||_s \le C(a)||u||_s$$
Result 2:    Let $\phi:U' \rightarrow V'$ be a diffeomoprhism of open subsets of $\Bbb R^n$ with $U \subseteq U'$ and $V= \phi(U) \subseteq V'$ be relatively compact. Then $u \mapsto u\circ \phi$ extends to a bounded map  for all $s \in \Bbb Z$. 
      $$W^s (V) \rightarrow W^s(U) $$ 


Thoughts so far:
Edit: I believe we start by proving 4. 
Let us see first vary the partition of unity, with $\tau:= \{ \tau_i \}$.  So that 
$$ \tau _j  = \sum_i \tau_j \mu_i $$ 
\begin{align*}
 || (\tau_j \circ h_j^{-1}) (\phi_j \circ u \circ h_j^{-1}) ||_k^2 & \le \sum _i || (\tau_j \circ h_j^{-1}) (\mu_i \circ h_j^{-1})  (\phi_j \circ u \circ h_j^{-1}) ||_k^2  \\ 
 & \le C(\tau) ||u||_{k}^2
\end{align*}
constant $C(\tau)$ dependent on partition. 
This uses Result 1.  Then if we take a another cover $\{V_j\}$. From independence of 4, we may choose a partition wrt $U_i \cap V_j$. Using Result 2, we take care of 2. 

Now I am stuck at addressing 3. 

This post should be pretty much self contained, but for those who might find it helpful in consulting original source, I am concered with Lemmea 3.6.2, pg 47.

REPLY [2 votes]: The key observation is to recall that trivialisations of vector bundles are fibrewise linear, and so you can prove an analogous result to result 2 for certain mappings of the form $A \circ u.$
Given the above setup, suppose $\psi_j : E|_{U_i} \rightarrow U_i \times \mathbb R^s$ is another choice of trivialisation for each $i.$ Then the composition,
$$ \varphi_i \circ \psi_i^{-1} : U_i \times \mathbb R^s \longrightarrow U_i \times \mathbb R^s $$
maps $(x,v) \mapsto (x,A_i(x)v),$ where $A_i : U_i \rightarrow \mathrm{GL}(s,\mathbb R).$ By shrinking $U_i$ slightly if necessary and identifying $\mathrm{GL}(s,\mathbb R) \subset \mathbb R^{s^2},$ we can assume that $A_i \circ h_i^{-1}, A_i^{-1} \circ h_i^{-1}$ are bounded in $C^k(\mathbb R^n, \mathbb R^{s^2}).$
We want to show there is $C>0$ such that,
$$ \lVert (\mu_j \circ h_j^{-1})(\varphi_j \circ u \circ h_j^{-1})\rVert_k^2 \leq C \lVert (\mu_j \circ h_j^{-1})(\psi_j \circ u \circ h_j^{-1})\rVert_k^2. $$
For this we write,
\begin{align*}
  \lVert (\mu_j \circ h_j^{-1})(\varphi_j \circ u \circ h_j^{-1})\rVert_k^2 &= \lVert (\mu_j \circ h_j^{-1})(\varphi_j \circ \psi_j^{-1} \circ \psi_j \circ u \circ h_j^{-1}) \rVert_k^2 \\
  &= \lVert(\mu_j \circ h_j^{-1}) (\mathrm{id},A_j \circ h_j^{-1})(\psi_j \circ u \circ h_j^{-1}) \rVert_k^2 \\
  &\leq C \left(1+\lVert A_j \circ h_j^{-1} \rVert_{C^k(\mathbb R^n,\mathbb R^{s^2})}\right) \lVert(\mu_j \circ h_j^{-1})(\psi_j \circ u \circ h_j^{-1}) \rVert_k^2
\end{align*}
The last line requires checking; the idea is to expand out each $\nabla^m\left( (\mathrm{id},A_j \circ h_j^{-1}) \psi \circ u \circ h_j^{-1}\right)$ and prove the estimate pointwise. Interchanging $\varphi_j$ and $\psi_j$ establishes the equivalence.
From here you've done most of the work. The rest is mostly a matter of notation and putting everything together.