# The space of Sobolev maps between Riemannian manifolds

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 $$$$W^{1,2}(\mathcal{M}, \mathcal{N})=\{ u \in W^{1,2}(\mathcal{M}, \mathbb{R}^n)\mid u(x) \in \mathcal{N}\, a.e.\}.$$$$

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?

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

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.