Showing that Sobolev norms on manifolds are equivalent

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:

1. A finite cover of $$M$$ by sets $$U_i$$.

2. charts $$h_i:U_i \cong \Bbb R^n$$.

3. Trivilizations $$\phi_i$$ of $$E|_{U_i}$$

4. $$\mu_i$$ 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.

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.

• Thanks, yes! linearity part is quite crucial. – CL. Dec 25 '18 at 0:35
• Thanks ktoi, I wonder if you have time commenting on my recent post on elliptic regularity. – CL. Dec 25 '18 at 7:39

If you are familiar with pseudodifferential operators, you may want to look at Proposition 7.3 in Shubin's book Pseudodifferential Operators and Spectral Theory.

Shubin defines two different Sobolev norms $$\Vert \cdot \Vert_s$$ and $$\Vert \cdot \Vert_s'$$, where the first is the same as yours (depending on the choice of charts, frames and partitions of unity) and the second depends on the choice of certain pseudodifferential operators.

He then proves for each norm independently that it is complete on the Sobolev space $$H^s$$ and uses the open mapping theorem to conclude that they are equivalent. In particular the induced topology is the same in each case and thus it does not depend on either choice.

• Thanks, is there an introductory text to pseudodifferential operator? – CL. Dec 25 '18 at 0:36
• Yes, he introduces pseudodifferential operators. But now that I think about it, Proposition 7.3 gives you much more than you need. If you are really just interested in the first kind of norm and want to prove independence of choice, the open mapping theorem can be used right away (and you can forget about pseudodifferential operators): – Jan Bohr Dec 25 '18 at 8:45
• The crucial point is, that one can first establish $H^s$ as a vector space (that is what Shubin does) and a little work goes into showing that this definition is independent of all kinds of choices. Then, afterwards, one defines $\Vert\cdot\Vert_s$ (depending on charts, frames and p.o.u.) and shows that it defines a complete norm on $H^s$. Now all complete norms on a vector space have to be equivalent (open mapping), hence a different choice (of charts, frames and p.o.u.) yields an equivalent norm. – Jan Bohr Dec 25 '18 at 8:48
• Ok thanks, I will definitely check this book out. If you have time, I would be happy (as I commented on the other post too) that you have a look on my recent post on elliptic regularity. – CL. Dec 25 '18 at 9:51