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Let $\Sigma, M$ be compact Riemannian manifolds. By embedding $M$ isometrically into $\mathbb{R}^N$, one can define the Sobolev spaces $W^{k,p}(\Sigma, M)$ by $$W^{k,p}(\Sigma,M) = \{ u \in W^{k,p}(\Sigma,\mathbb{R}^N) \,\, | \,\, u(z) \in M \, \mathrm{a.e} \}.$$ 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 plane gets mapped a.e to the tangent plane of the image of $M$ in $\mathbb{R}^N$?

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I've also posted this question on mathoverflow, with some additional details.… – levap Oct 4 '12 at 11:51
I expect the answer is yes and you ought to consider the approximative derivative of your map, which equals the weak derivatives almost everywhere and has much of the properties of the total derivative. – user141904 Apr 9 '14 at 10:24

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