Sobolev spaces on manifolds

Sobolev spaces on manifolds February 22, 2011

Posted by David Speyer in Uncategorized.
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I just spent the last few days working through the proof of the Hodge theorem in Wells Differential Analysis on Complex Manifolds. There were a few things which confused me right at the start, and which could have been easily cleared up if I had just understood which constructions were canonical and which were not. The point of this blog entry is to record that data. I’ll also add a little extra background, so people other than me have a chance of understanding it. Since I am very much not an analyst, there is a high risk of errors here; I would appreciate anyone who points them out to me. Also, to keep life simple, $s$ will be an integer and $p$ will be $2$ .

Let $M$ be a smooth $n$ -dimensional Riemmannian manifold and $E$ a complex vector bundle equipped with a positive definite Hermitian form.

Things that are true, important, and canonical

For every integer $s$ , there is a topological vector space $W^s(M)$ . There are inclusions $W^{s+1}(M) \hookrightarrow W^s(M)$ . These are compact maps, meaning that if $v_k$ is a bounded sequence in $W^s$ then $v_k$ has a convergent subsequence in $W^{s+1}$ . The smooth functions, $C^{\infty}(M)$ , have compatible inclusions into all of the $W^s$ ‘s, with dense image.

For $s \geq 0$ , a sequence $f_i$ of smooth functions is Cauchy in $W^s$ if, for any differential operator $D$ , of order $\leq s$ , the sequence $D f_i$ is Cauchy in the $L^2$ topology. So $W^0 \cong L^2$ and, more generally, $W^s$ can be thought of all the completion of $C^{\infty}$ in this variant of the $L^2$ topology.

We have containments $C^{s-\lfloor n/2 \rfloor -1}(M) \supseteq W^s(M) \supseteq C^s(M)$ , compatible with the injections $W^{s+1} \hookrightarrow W^s$ and $C^{k+1} \hookrightarrow C^k$ . In other words, if we have a sequence in $C^{\infty}(M)$ which is $W^s$ -Cauchy, then it approaches a limit in the $C^{s-\lfloor n/2 \rfloor -1}$ -topology and, if we have a sequence in $C^{\infty}$ approaching a limit in the $C^s$ -topology, then it is $W^s$ -Cauchy.

There is a sesquilinear pairing $W^s \times W^{-s} \to \mathbb{C}$ with respect to which $W^{s+1} \hookrightarrow W^s$ and $W^{-s} \hookrightarrow W^{-s-1}$ are adjoint. This exhibits $W^{-s}$ as the conjugate-linear-dual of $W^s$ . (And, I think, vice versa, although Wells doesn’t say this.) When $s=0$ , this is the standard pairing on $L^2(M)$ . The fact that the dual of the smaller space is the larger space is strange. My current intuition is that, although $W^{s+1} \hookrightarrow W^{s}$ is an injection, it schrunches $W^{s+1}$ down, so that it is easier for sequences to converge in the image. Thus, many linear functions which are continuous on $W^{s+1}$ are no longer continuous as functions on $W^{s+1}$ with the topology restricted from $W^s$ . I still find this hard to think about, though.

An order $m$ differential operator $L$ will extends to a continuous map $W^s$ to $W^{s-m}$ . The above pairing will induce an adjoint map $L^*: W^{s} \to W^{s-m}$ . In the cases I care about, $L^*$ is also a differential operator, but I think this is in general not true.

$W^s$ has the structure of a normed vector space, which is canonical in the sense that the norm is well defined up to a constant distortion factor.

The following is true, but should be ignored

The topology on $W^s(M)$ arises from a Hilbert space structure. Nothing is ever adjoint with respect to this inner product. Orthogonality in this inner product is not a canonical property. Since it is a Hilbert space, $W^s$ is isomorphic to its own dual and, hence, $W^s \cong W^{-s}$ ; this isomorphism is never used. It would have been better had I not known this structure existed.

Below the fold, for the curious, I will repeat Wells’ definitions in the order I would have given them.

Sobolev Structures on $\mathbb{R}^n$

Terry Tao has a very comprehensive blog post on this, so I’ll be brief.

Let $N$ be the standard $L^2$ norm on functions on $\mathbb{R}^n$ . For $s \geq 0$ , define a norm $N_{s}$ by

$\displaystyle{ N_s(f)^2 = \sum_{a_1+\cdots + a_n \leq s} c(a_\bullet) N \left(\frac{\partial^{\sum a_i}}{(\partial x_1)^{a_1} \cdots (\partial x_n)^{a_n}} f \right)^2}$

where $c(a_{\bullet})$ are positive reals whose exact values don’t matter. (This is the first evidence that the Hilbert space structure on $W^s$ doesn’t matter.)

Now, let $\widehat{f}$ be the Fourier transform of $f$ and let $\omega_i$ be the frequency variable dual to $x_i$ . Then we have the convenient formula

$\displaystyle{ N_s(f)^2 = \sum_{a_1+\cdots + a_n \leq s} c(a_\bullet) N\left( \prod \omega_i^{a_i} \widehat{f} \right)^2}$

$\displaystyle{ =\sum_{a_1+\cdots + a_n \leq s} c(a_\bullet) \int_{\omega} \prod |\omega_i|^{2 a_i} |\widehat{f}|^2}.$

Up to a constant distortion, this is the same as

$\displaystyle{ \int_{\omega} (1+|\omega|)^{2s} |\widehat{f}|^2}$

We use this last formula to define the $W^s$ topology for negative $s$ .

There are now two ways we can think of $W^s$ , at least for $s \geq 0$ .

From the inside: $W^s$ is the completion of $C_c^{\infty}$ in the above norms. (The subscript means compact support.) One can easily see that $C_c^s$ is in $W^s$ , so one can also think of it as the corresponding completion of $C_c^s$ . The injection $W^{s+1} \hookrightarrow W^s$ is by noticing that any $W^{s+1}$ Cauchy sequence is also $W^s$ Cauchy.

From the outside: $W^s$ is the subspace of $L^2(\mathbb{R}^d)$ where the integral $\int |\widehat{f}(\omega)|^2 (1+|\omega|)^{2s}$ converges. By a nontrivial lemma, this is actually a subspace of $C^{s-\lfloor n/2 \rfloor -1}$ so, for $s$ large enough, we can think of it as a subspace of these somewhat smoother functions. The inclusion is given by noticing that this condition on the integral becomes weaker as $s$ becomes more negative.

The pairing between $W^{s}$ and $W^{-s}$ is extended from the standard $L^2$ pairing on $C^{\infty}_c$ : $\langle f, g \rangle = \int f \overline{g}$ . That this extends continuously to $W^s \times W^{-s}$ follows from Cauchy-Schwartz.

Sobolev spaces on compact manifolds

The above definition exhibited $W^s$ as a Hilbert space: our norm obeys the parallelogram law. I’m not sure whether or not there is any benefit to knowing this. But, when we get to manifolds, there definitely isn’t.

Let $M$ be a compact Riemmannian manifold and $E$ a Hermitian vector bundle. For $s\geq 0$ , define the $W^s$ topology on $C^{\infty}(M)$ by saying that $f_i$ is Cauchy if, for every differential operator $D$ of order $\leq s$ , the sequence $D f_i$ is $L^2$ Cauchy.

Choose a finite open cover $U_i$ , so that $U_i$ embeds into $\mathbb{R}^n$ and such that $E$ is trivial on $U_i$ ; choose a trivialization of $E$ on $U_i$ , and choose a partition of unity $\rho_i$ subordinate to $U_i$ . Then any section $f \in C^{\infty}(E)$ gives a collection of $C^{\infty}$ functions on $\mathbb{R}^n$ ; the components of $\rho_i f$ .

Lemma: A sequence $f_i$ is $W^s$ -Cauchy if and only if all the sequences $(\rho_j f_i)_k$ , for all $j$ and $k$ , are $W^s$ -Cauchy.

This lemma is not completely trivial: I goofed up the first two times I tried to write a proof. Hint: There is some $\epsilon$ such that $M$ is covered by the sets $\{ x : \rho_i(x) > \epsilon \}$ .

This lets us define $W^s$ -Cauchy sequences for negative $s$ . We can then define $W^s(M)$ as the completion of $C^{\infty}(E)$ in this topology. The fact that this topology can be given by a norm is somewhat important; the fact that this norm can be chosen to obey the paralellogram law is completely unimportant.

When $s\geq 0$ , we can think of $W^s(M)$ as a subspace of $L^2(M)$ , given by those functions whose finite smooth cutoffs have sufficiently slow growing Fourier transforms. The fact that the precise details of the smooth cutoff do not effect the high-frequency behavior of the Fourier transform is really cool and nonobvious, and is perhaps most easily seen from the above lemma.

And, finally, the pairing between $W^{s}$ and $W^{-s}$ is still given by the continuous extension of $\int_M f \overline{g}$ . One shows that the integral converges by, once again, passing to local coordinates and using Cauchy-Schwartz.

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Comments»

1. Emmanuel Kowalski - February 22, 2011: >The fact that the dual of the smaller space is the larger space is strange.

The algebraic way I think about this is that it is (somewhat) analogue to the fact that the dual of a direct sum (“small space”) is a direct product (“large space”).
2. Mark Meckes - February 23, 2011: “The fact that the dual of the smaller space is the larger space is strange.”

Not at all. In fact this is a familiar phenomenon from the first nontrivial examples almost everyone learns of duality: $\ell^p$ (for which $\ell^p \subseteq \ell^q$ when $p \le q$ ) and $L^p = L^p[0,1]$ (for which $L^q \subseteq L^p$ when $p \le q$ ).

Here’s the geometric way I think of it. If you have two norms on the same space (say finite-dimensional for simplicity) and one is everywhere greater than the other, then the induced norms on the dual space satisfy the reversed inequality. In terms of the unit balls, polarity reverses inclusion. Now in infinite dimensions (when you have to start thinking about completions), a smaller norm corresponds to a larger space — there are more things for which the smaller norm is actually finite.

I think the reason the size-reversal of duality seems strange is that we get used to thinking of the dimension of a vector space as the way to measure its “size”, but that’s just not the appropriate measure of size in this context.
3. Yemon Choi - February 23, 2011: “The fact that the dual of the smaller space is the larger space is strange.”

How about “the dual of a subobject is a quotient object”? (not to be taken too seriously of course, but in some sense a manifestation of the same idea. Galois connections, etc etc)
4. Yemon Choi - February 23, 2011: Sorry, just realized my previous comment doesn’t really make sense. Please ignore, or preferably delete. (Mark Meckes’ comment is much more to the point, I think)
5. David Speyer - February 23, 2011: If you don’t mind, I’d like to leave it, because it is a perfect summary of my wrong intuition.
6. Yemon Choi - February 23, 2011: Fair enough, David.
7. S. H. Banihashemi - February 25, 2011: How is cohomology for Sobolev spaces on complex manifolds different from cohomology for ohter operator spaces as vector bundles. Does cyclic cohomnology of noncommutative geometry relate or suffice? Or am I off the tracks?
8. John Baez - February 25, 2011: David wrote:

The topology on $W^s(M)$ arises from a Hilbert space structure. Nothing is ever adjoint with respect to this inner product. Orthogonality in this inner product is not a canonical property.

I agree that it’s important to keep track of which structures are ‘canonical’ and which aren’t, but it’s also important to note that what’s canonical depends on what category you’re working with. There are a number of important categories lurking around here. One is the category of compact smooth manifolds and smooth maps. Another is the category of compact smooth Riemannian manifolds and isometries (maps that preserve the Riemannian metric).

The Hilbert space structure on $W^s(M)$ is not canonical with respect to the first category: that is, it’s not invariant under diffeomorphisms of the compact smooth manifold $M$ . But it’s canonical with respect to the second: that is, it’s invariant under diffeomorphisms that preserve the metric on the compact smooth Riemannian manifold $M$ .

A lot of important operators defined with the help of a Riemannian metric on $M$ can be thought of as unbounded self-adjoint operators on $W^s(M)$ : for example, the Laplacian! And this is very important in studying the geometry of manifolds: for example, it lets you bring in the heat equation.
9. John Baez - February 25, 2011: By the way, it’s the ‘Cauchy-Schwarz inequality’, not ‘Cauchy-Schwartz’. Laurent Schwartz invented distributions, Karl Herman Amandus Schwarz proved the inequality. And I have to stick up for Schwarz, because my advisor was Irving Segal, and his advisor was Einar Hille, and his advisor was Marcel Riesz, and his advisor was Lipot Fejer, and his advisor was Karl Herman Amandus Schwarz.
10. S. H. Banihashemi - February 26, 2011: Please allow me to thank John for the help although I still have a difficult time to connect it to ‘canonical’ pseudodifferential operator in the google book fragments. Also, it has always been a big question for me whether there is a kind of supersymmetric quantum field theory where the correspondence between the classical Riemannian manifold M and its quantized space such as $W^s(M)$ gets lost. For example, does this happen in quantum gravity, or if M is taken to be the earth, as we want to keep track of this correspondence. Is this an issue in noncommutative geometry too? Always fuzzy!
11. Max - February 26, 2011: “An order m differential operator L will extends to a continuous map W^s to W^{s-m}. The above pairing will induce an adjoint map L^*: W^{s} \to W^{s-m}. In the cases I care about, L^* is also a differential operator, but I think this is in general not true.”

This sounds a bit confusing to me. I think the following is the case.

The pairing induces the adjoint W^{m-s} to W^{-s}, which is indeed a differential operator of degree m, known as formal adjoint. It has explicit formula obtained from integration by parts (after all, the pairing is integration). As a differential operator it acts on all W^{q} to W^{q-m}, in particular for $q=s$ as well. (There was a question about it here http://math.stackexchange.com/questions/12894/distinction-between-adjoint-and-formal-adjoint)
12. Jack - February 26, 2011: By the way, it’s ‘Hermann’, not ‘Herman’.
13. Emmanuel Kowalski - February 26, 2011: The Laplace operator mentioned in Comment #8 is already an unbounded essentially self-adjoint operator on the standard L^2 space (if the manifold is compact without boundary), and one can do a fair amount of work without needing to characterize the domain of the self-adjoint extension too precisely. Indeed, to think of a Sobolev space as the domain of the Laplace operator, the exponent s has to be just right [s=2] and the norm has to be defined very precisely. (Not with arbitrary positive constants for the contribution of derivatives of such and such order, as in the definition of N_s(f) in the post).

Indeed, one can understand that a very precise definition is needed because one must retain the property that eigenfunctions with distinct eigenvalues are orthogonal — thus for the domain of the Laplace operator, the orthogonality is very important.
14. Todd Trimble - March 5, 2011: The horse may be dead, but with regard to “smaller subspaces, larger duals”, it might help to observe that the subspace inclusions are typically dense, and therefore epimorphisms. Dualizing gives you monomorphisms, hence bigger duals.

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