Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\Sigma, M$ be smooth compact Riemannian manifolds. By embedding $M$ isometrically into $\mathbb{R}^N$, one can define the Sobolev spaces $W^{k,p}(\Sigma, M)$ by

\begin{equation} W^{k,p}(\Sigma,M) = \{ u \in W^{k,p}(\Sigma,\mathbb{R}^N) \,\, | \,\, u(z) \in M \,\, \mathrm{a.e} \}. \end{equation}

For a $u \in W^{k,p}(\Sigma, M)$, $k \geq 1$, is there some sense in which $u$ has a differential $du$ that maps $T\Sigma$ into $TM$?

By working in coordinates on $\Sigma$, one can think of $du$ locally as the matrix of the (weak) partial derivatives of the local representation of $u$. One can even show using the chain rule that the local definition extends to a global definition (a.e) of $du : T\Sigma \rightarrow \mathbb{R}^N \times \mathbb{R}^N$. But does each tangent space gets mapped a.e to the tangent space of the image of $M$ in $\mathbb{R}^N$?

(This question was asked on math.stackexchange, but didn't get a respond so I'm posting it here.)

I was thinking about this question while trying to interpret a weak formulation for a specific problem. Namely, if $u : (\Sigma, j) \rightarrow (M,J)$ is a smooth map between almost complex manifolds, we say that $u$ is $(j-J)$ holomorphic if it satisfies the generalized Cauchy-Riemann equation \begin{equation} du \circ j = J \circ du. \end{equation} If $u \in W^{1,p}(\Sigma, M)$ for $p > \dim \Sigma$, then $u$ is continuous and so necessarily localizable and we can define than $u$ is $(j-J)$ holomorphic if its local representations in coordinate systems in the domain and range $\tilde{u}$ satisfy \begin{equation} d\tilde{u} \circ \tilde{j} = \tilde{J} \circ d\tilde{u} \end{equation} where $d\tilde{u}$ is the matrix of (weak) partial derivatives. However, if $p \leq \dim \Sigma$, one usually uses an isometric embedding to define the Sobolev spaces. Now, I've seen definitions which say that in this case $u$ is $(j-J)$ holomorphic if it satisfies \begin{equation} du \circ j = J \circ du \,\,\mathrm{a.e} \end{equation} but I'm not sure how to interpret $J \circ du$ in the right side. The almost complex structure $J$ is only defined on $TM$. I can try to extend $J$ to be defined on $\left. T\mathbb{R}^N \right|_{M}$ (identifying $M$ with its image in $\mathbb{R}^N$) but I'm not sure if the resulting equation doesn't depend on the extension, hence the question.

share|improve this question
Why do you need to consider $p\leq\dim\Sigma$? For holomorphic curves, most treatments (by which I mean McDuff-Salamon) take $p>2$ (not least because $W^{1,2}\to L^2$ is not compact). –  Jonny Evans Oct 4 '12 at 12:08
If if the functions in $W^{k,p}$ are not continuous, i.e., k-\frac{\dim \Sigma}{p} <0$, then the condition $u(z)\in M$ a.e., is difficult to interpret. –  Liviu Nicolaescu Oct 4 '12 at 12:47
Jonny - I was just wondering if you can talk about $J$-holomorphic curves which has less regularity. I've found quite a few treatments in which W^{1,2} curves are discussed. For example, in ams.org/journals/tran/1994-342-02/S0002-9947-1994-1176088-1/… it is shown that W^{1,2} curves are smooth. In fim.math.ethz.ch/~triviere/papers/tiri.ps, there is a discussion of $J$-holomorphic "surfaces" (maps from $\Sigma$ of real dimension 4), for which (unsurprisingly) W^{1,2} is not enough to guarantee regularity. But I'm not sure how to interpret –  levap Oct 4 '12 at 14:12
"being a W^{1,2} J-holomorphic map". Liviu, what do you mean? –  levap Oct 4 '12 at 14:13
I think that the space of smooth holomorphic maps $\Sigma\longrightarrow M$ is closed in $W^{1,2}$. My (slightly biased) opinion is that if you use a Sobolov space setup to study holomorphic curves and you want to allow the domain of $\Sigma$ to vary, you should understand that you are doing something unnatural, and proceed with caution. There are several mistakes in the published J-holomorphic curves literature coming from people using Sobolev spaces inappropriately. –  Brett Parker Oct 4 '12 at 23:58

4 Answers 4

If we use the extrinsic definition, viewing $u$ as a vector valued Sobolev function, they are approximately differentiable a.e. (just replace the $L_\infty$ norm in the usual definition of differentiation by the $L_p$ norm, see Evans and Gariepy, Measure Theory and Fine Properties of Functions). Then at points where it's approximately differentiable, $du$ maps $TM$ to $TN$.

share|improve this answer

If you are interested, we have given an intrinsic definition of a weak derivative for maps between manifolds A. Convent et J. Van Schaftingen, emphasized Intrinsic colocal weak derivatives and Sobolev spaces between manifolds, [arXiv:1312.5858]. This weak derivative is map between the tangent bundles and allows to define consistently with previous definitions Sobolev spaces $W^{1, p} (\Sigma, M)$ between manifolds $\Sigma$ and $M$.

In particular, in you situation, $du$ is almost everywhere in $TM$.

share|improve this answer

You have to extend J to be defined on your embedding space. Thus, your equation a priori depends on your chosen embedding. Of course the point here is to prove certain properties of your solutions that make it a posteriori independent of the embedding. For instance, Riviere often works with such embeddings and then shows regularity properties (such as harmonic maps from surfaces which are a priori only $W^{1,2},$ but then are shown to be Hölder).

In my opinion the main point here is the bigger bag principle. Classically for instance to solve the Dirichlet problem you use minimization of the Dirichlet functional. To capture the limit however requires you to use a complete space. Then to recover a usable solution you have to use regularity results. Notice however that you cannot use these regularity results if you do not have your bigger bag already.

It is similar in the holomorphic curve setting, where sometimes it is useful to consider suitable embeddings of the target manifold and work with Sobolev spaces with low regularity (such spaces are not intrinsically defined if the Sobolev space in question does not embed into the space of continuous functions). Usually in holomorphic curve theory one considers moduli spaces involving a fixed target space. Then it is feasible to work with target-depending moduli spaces as long as you can prove your wanted statement. Sometimes this is not possible since at some point you need more regularity (for instance in gluing).

share|improve this answer

If $W^{k,p}$ is the space of all $f$ with $ D^\alpha f \in L^p$ for all $|\alpha| \le k$ on $\mathbb R^n$ (so that I know which index is what), then the Sobolev inequality says $W^{k,p} \subset C^m$ if $k> m+\frac{\dim \Sigma}{p}$. So if $m\ge 1$, the answer is yes.

share|improve this answer
Thanks for your answer, but I'm interested exactly in the cases for which the Sobolev embedding theorem does not guarantee that functions are $C^1$. –  levap Oct 4 '12 at 20:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.