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Let $\mathcal{M}, \mathcal{N}$ be two Riemannian manifods. Suppose that $\mathcal{N}$ is properly and isometrically embedded in $\mathbb{R}^n$. The space of Sobolev maps between $\mathcal{M}$ and $\mathcal{N}$ is defined as follows \begin{equation} W^{1,2}(\mathcal{M}, \mathcal{N})=\{ u \in W^{1,2}(\mathcal{M}, \mathbb{R}^n)\mid u(x) \in \mathcal{N}\, a.e.\}. \end{equation}

Many papers say that $W^{1,2}(\mathcal{M}, \mathcal{N})$ is a Banach manifold, but no one gives a reference about how to prove this result. Can anybody give me some references?

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  • $\begingroup$ You might find useful references in the review cited in this answer. $\endgroup$ Dec 5, 2022 at 10:34
  • $\begingroup$ @IgorKhavkine Thanks. However, it doesn't contain much information about the manifold structure of the space of sobolev maps. $\endgroup$
    – gaoqiang
    Dec 14, 2022 at 8:02
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    $\begingroup$ "Many papers say..." Can you give references to these papers? I haven't seen such a statement. $\endgroup$ Dec 17, 2022 at 15:43

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This is too long for a comment so I am posting it as an answer. I believe, I have seen a statement in the literature that $W^{1,p}(\mathcal{M},\mathcal{N})$ is a Banach manifold if and if $p> \dim\mathcal{M}$ (although I am not sure about the case $p=\dim\mathcal{M}$). However, there was no proof of this fact neither a reference. Unforlunately, I do not remember where I saw this statement. It was probably in one of the papers of Karen Uhlenbeck, but I am not entirely sure. In any case, there is no natural manifold structure in $W^{1,p}(\mathcal{M},\mathcal{N})$ if $p\leq\dim\mathcal{M}$, because the Sobolev maps cannot be localized in the coordinate charts and this is why the space of Sobolev mappings between manifolds is defined through the embedding into $\mathbb{R}^n$. In fact the topological structure of the space $W^{1,p}(\mathcal{M},\mathcal{N})$ is very complicated.

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