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Let $M$ be a Riemannian manifold, and define the Sobolev spaces $H^k(M)$ to be the set of distributions $f$ on $M$ such that $Pf \in L^2(M)$ for all differential operators $P$ on $M$ of order less than or equal to $k$ with smooth coefficients.

If $F : M\rightarrow N$ is a diffeomorphism, must it hold that $H^k(M)$ and $H^k(N)$ are linearly isomorphic? If $F$ and all its derivatives are bounded, then composition with $F$ gives the desired isomorphism. Is there still a way to produce an isomorphism if this is not the case?

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2 Answers

up vote 2 down vote accepted

Of course, there is always an (abstract) isomorphism between two separable Hilbert spaces. For the question to be interesting, we should ask for a concrete isomorphism in terms of $F$.

I am skeptical about the idea of "fixing" the composition operator by putting multiplication on top of it. This sort of thing can work for Lebesgue spaces, but getting the desired Sobolev norm with multiplication is unlikely.

I suspect that if the composition operator does not work, there is no natural isomorphism. As you might know already, composition does work without boundedness of derivatives in one exceptional case: for $k=1$ in two dimensions. Any conformal or even quasiconformal map between surfaces of finite area induces an isomorphism of $H^1$ by composition. (The finiteness of area is needed only for the zero-order term.)

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For a compact Riemannian manifold $(M, g)$, $H^k(M; g)$ only depends on the diffeomorphism type of $M$. Your construction should work in the compact case, but you should also make sure that $H^k(M; g)$ doesn't depend on the choice of Riemannian metric $g$ on $M$.

In the non-compact case, the metric matters. Let $M$ be non-compact. Then we can find metrics $g_1$, $g_2$ on $M$ such that $\mathrm{vol}_{g_1}(M) < \infty$ and $\mathrm{vol}_{g_2}(M) = \infty$. Then, for example, all constant functions $u \equiv C$ for $C \in \mathbb{R}$ satisfy $u \in H^k(M; g_1)$ but $u \notin H^k(M; g_2)$ for all $k$. So for non-compact manifolds you should at least require $F: (M, g_M) \longrightarrow (N, g_N)$ to be an isometry; but it's very likely that you also require some decay conditions on the derivatives of $F$ in the infinite volume case.

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I know that the construction $f\mapsto f\circ F$ is not always an isomorphism of Sobolev spaces. But, I am wondering if there is a different way to produce such an isomorphism, for example $f \mapsto \phi f \circ F$ for some function $\phi:M\rightarrow \mathbb R$ which will cancel out the effect of $F$ on the derivatives of $f$. –  user15464 Jul 27 '12 at 1:33
    
@user15464 I think these notes settle the issue for compact manifolds: staff.science.uu.nl/~ban00101/anman2009/lecture3.pdf. See especially sections 3.3 and 3.8 as well as Theorem 3.7.4. –  Henry T. Horton Jul 27 '12 at 1:55
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For the non-compact case your construction appears to work for $H^k_{\text{loc}}(M)$ as well. I will need to think about if/how you could find a map $\phi: M \to \Bbb R$ so that $(\phi f) \circ F \in H^k(M)$ in the non-compact case, however. –  Henry T. Horton Jul 27 '12 at 2:02
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