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.