Is $L^1$ strong convergence of Jacobians valid for maps between manifolds?

Question

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

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 $(2)$.

Answer 1

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 $\Vert Ju_k-Ju\Vert_1\geq\epsilon$.