$\newcommand{\M}{\mathcal{M}}$
$\newcommand{\N}{\mathcal{N}}$
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\Vol}{\operatorname{Vol}}$
$\newcommand{\Det}{\operatorname{Det}}$
$\newcommand{\Volm}{\operatorname{Vol}\_{\M}}$
$\newcommand{\Voln}{\operatorname{Vol}\_{\N}}$


Let $\M,\N$ be smooth, connected, oriented, compact $n$-dimensional Riemannian manifolds. Let $u\_k,u \in W^{1,n}(\M,\N)$ be **Lipschitz** and satisfy $u\_k \to u$ in $W^{1,n}(\M,\N)$. (strong convergence).



> 
> Is it true that $Ju\_k \to Ju$ strongly in $L^1(\M)$?
> 
> 
> 


I can prove that $|Ju\_k| \to |Ju|$ strongly in $L^1(\M)$ (see below), so if we can prove that $Ju\_k \to Ju$ a.e. we are done.


I tried to prove that $Ju\_n \to Ju$ a.e. by using local coordinates, but this doesn't seem trivial; $u\_k$ does not necessarily converge uniformly to $u$, so it is not clear how to do that. (Note that the values of $Ju\_k,Ju$ at a point $p$ depend upon the images $u\_k(p),u(p)$, unlike in the Euclidean case).




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I use the definition $W^{1,n}(\M,\N)=\{ u \in W^{1,n}(\M,\R^D) , u(x) \in \N a.e.\}$, where $\N$ is implicitly assumed to be isometrically embedded in $\R^D$ via some embedding $i$. $W^{1,n}(\M,\N)$ inherits the notion of strong convergence from the ambient space $W^{1,n}(\M,\R^D)$.


The Jacobians are defined via the Riemannian and orientation structures, i.e. by requiring $u\_k^\*\Voln=(Ju\_k) \Volm$ where $\Volm,\Voln$ are the Riemannian volume forms of $\M$ and $\N$ respectively.




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*Proof that $|Ju\_k| \to |Ju|$ strongly in $L^1$:*


$u\_k \to u$ in $W^{1,n}(\M,\N)$ means $i \circ u\_k \to i \circ u$ in $W^{1,n}(\M,\R^D)$, so in particular $d(i \circ u\_k) \to d(i \circ u)$ in $L^{n}$. (we regard $d(i \circ u\_k)$ as maps $T\M \to T\R^D$.)


A vector bundle map $L:T\M \to T\R^D$ have an associated notion of "absolute value Jacobian" defined by $\Det L=\sqrt{\det(L^TL)}=\det(\sqrt{L^TL})$. (we do not have a signed Jacobian since the dimension of the target fiber space is greater than that of the source.)


Specifying this to the maps $d(i \circ u\_k):T\M \to T\R^D$, we easily obtain $\Det d(i \circ u\_k) \to \Det d(i \circ u)$. Finally we note that $\Det d(i \circ u\_k)=|Ju\_k|$.




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**Edit:**


Let me explain why I don't think that $Ju\_n \to Ju$ a.e. is obvious: By definition, we have
$$
(\Voln)\_{u\_k(p)}\big( (du\_k)\_{p}(v\_1),\dots,(du\_k)\_{p}(v\_1) \big)=(u\_k^\*\Voln)\_p(v\_1,\dots,v\_n)=(Ju\_k)\_p (\Volm)\_p(v\_1,\dots,v\_i), \tag{1}
$$
where $v\_i \in T\_p\M$.


So, we need to show that
$$(\Voln)\_{u\_k(p)}\big( (du\_k)\_{p}(v\_1),\dots,(du\_k)\_{p}(v\_1) \big) \to (\Voln)\_{u(p)}\big( (du)\_{p}(v\_1),\dots,(du)\_{p}(v\_1) \big) \, \, \, \text{a.e,} \tag{2}$$  

and we may assume that $u\_k \to u$ and $d(i \circ u\_k) \to d(i \circ u)$ a.e. on $\M$. Thus $d(i \circ u\_k)\_p(v\_i) \to d(i \circ u)\_p(v\_i)$. The question is why does that imply the convergence $(du\_k)\_{p}(v\_i)\to du\_{p}(v\_i)$ in $T\N$, which is what I think that we need in order establish the limit $(2)$.



Is $L^1$ strong convergence of Jacobians valid for maps between manifolds?
You actually do not need to assume that the mappings are Lipschitz as it is true for general $W^{1,n}$ mappings



> 
> **Theorem.** If $\mathcal{M}$ and $\mathcal{N}$ are smooth compact and oriented manifolds, $\mathcal{N}\subset\mathbb{R}^D$, and $u,u\_k\in W^{1,n}(\mathcal{M},\mathcal{N})$, $u\_k\to u$ in $W^{1,n}$, then the Jacobians converge in $L^1$, $Ju\_k\to Ju$.
> 
> 
> 


*Proof*.
Suppose that $Ju\_k$ does not converge to $Ju$ in $L^1$. Then for a subsequence (still denoted by $u\_k$) we will have $\Vert Ju\_k-Ju\Vert\_1\geq\epsilon$. If we take a further subequence, we can also assume that $u\_k\to u$ and $Du\_k\to Du$ a.e.


Since $\mathcal{M}$ is compact, we can use a finite atlas which allows us to assume that $\mathcal{M}=B^n(0,1)$. Since the mappings need not be continuous, localization of the mappings in an atlas on $\mathcal{N}$ is not possible.


It is assumed that $\mathcal{N}$ is a submanifold of $\mathbb{R}^D$. Let $\omega$ be the volume form on $\mathcal{N}$. By extension, we can always assume that $\omega$ is a compactly supported smooth form on $\mathbb{R}^D$ so
$$
\omega=\sum\_{|I|=n}\omega\_I dy^I,
\quad
dy^I=dy^{i\_1}\wedge\ldots\wedge dy^{i\_n},
\quad
1\leq i\_1<\ldots<i\_n\leq D. 
$$


If $u\in W^{1,n}(B^n(0,1),\mathcal{N})$, then we can interpret the Jacobian as the $n$-form:
$$
u^\*\omega(x)=\sum\_{|I|}(\omega\_I\circ u)du^{i\_1}\wedge\ldots\wedge du^{i\_n}=Ju(x)dx^1\wedge\ldots\wedge dx^n
$$
Note that $\omega\_I\circ u\_k\to \omega\_I\circ u$ a.e. and the functions are uniformly bounded because $\omega$ is bounded as a function on $\mathbb{R}^D$.
Also it easily follows from the triangle inequality and Holder's inequality that
$$
du\_k^{i\_1}\wedge\ldots\wedge du\_k^{i\_n}\to
du^{i\_1}\wedge\ldots\wedge du^{i\_n}
\quad
\text{in $L^1$.}
$$
Therefore, it easily follows (triangle inequality plus dominated convergence theorem) that $u\_k^\*\omega\to u^\*\omega$ in $L^1$, but that contradicts
$\Vert Ju\_k-Ju\Vert\_1\geq\epsilon$.


