$\newcommand{\M}{\mathcal{M}}$
$\newcommand{\N}{\mathcal{N}}$


Let $\M,\N$ be two-dimensional smooth, compact, connected, oriented Riemannian manifolds. (with or without boundaries).



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> Let $f\_n \rightharpoonup f$ in $W^{1,2}(\M,\N) $ with $Jf\_n > 0$ a.e., and suppose that the volume $V(\{x \in \M \, | \, Jf\_n \le r\}) \to 0$ when $n \to \infty$, for some $0<r<1$. Does $ Jf\_n \rightharpoonup Jf $ in $L^1(K)$ for every $K \subset \subset \operatorname{Int}(\M)$.?
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I am fine with assuming that $f\_n$ are Lipschits and injective, and that $V(f\_n(\M)) \to V(\N) $.




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The "higher integrability property of determinants" implies that if $\M,\N$ are open Euclidean domains, then $ Jf\_n \rightharpoonup Jf $ in $L^1(K)$ for any compact $K \subset \subset \M$.




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Without the assumption $V(Jf\_n \le r) \to 0$, the answer can be negative even when $f\_n$ are diffeomorphisms:


Take $\M=\N=\mathbb{S}^2$. Let $s: \mathbb{S}^2 \to \mathbb{R}^2 \cup \{\infty\}$ be the stereographic projection, and let $g\_k(x) = k x$ for $x \in R^2$ (and $g\_n(\infty) = \infty$.).


Set $ f\_n = s^{-1} \circ g\_n \circ s$. $f\_k$ are conformal, orientation preserving, smooth diffeomorphisms
and thus $ \int\_{\mathbb{S}^2 }Jf\_n=V(\mathbb{S}^2 )$. By conformality $\int\_{\mathbb{S}^2 } |Df\_n|^2 =2\int\_{\mathbb{S}^2 }Jf\_n$ is uniformly bounded, so $f\_n$ is bounded in $W^{1,2}$, and converges to a constant function. (asymptotically we squeeze bigger and bigger parts of the sphere to a small region around the pole).


So, we do not have weak convergence of $Jf\_n$ to $Jf=0$. (the
Jacobians converge as measures to a Dirac mass at the pole.) The question is if by adding the non-degeneracy constraint $V(Jf\_n \le r) \to 0$ we recover this 'Jacobian continuity' under weak convergence.




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\*(In my case of application $r=\frac{1}{4}$ but I don't think it matters).



Does weak continuity of Jacobians hold for non nondegenerate maps?
There is a counterexample, however there might be ways to avoid it.


Take $\mathcal{M} = \mathcal{N} =\mathbb{S}^2$, but now consider sequence of maps that cover the sphere twice, where you shrink the preimage of one of them to a point. Specifically consider using the stereographic projection as you did, consider $g\_n: \mathbb{C} \cup \{\infty\} \to\mathbb{C} \cup \{\infty\}$, that map $0$ to $0$, $\partial B\_{1/n}(0)$ to $\infty$ and $\infty$ to $0$ again, with positive Jacobian in between, e.g. something like
$$g\_n(x) := \begin{cases} \frac{x}{|x|}\tan(\pi nx/2)& \text{ for } |x| < \frac{1}{n} \\ \frac{\overline{x}}{|x|} \frac{1}{|x|-\frac{1}{n}} & \text{ for } |x| \geq \frac{1}{n} \end{cases}.$$


This fulfills all your conditions (however with $Jf\_n =0$ on a circle, which is a set of measure $0$), and after a slightly tedious calculation you should get $g\_n \rightharpoonup \frac{1}{x}$ in $W^{1,2}$, $Jf\_n \rightharpoonup 1 + 4\pi\delta\_0$ in the sense of distributions and $Jf\_n \geq \frac{1}{2}$ anywhere except on the small shrinking ball that corresponds to $B\_{1/n}(0)$. In fact $Jf\_n \to 1$ almost everywhere and uniformly on any set excluding a neighborhood of $0$.


For manifolds without boundaries, a good way to avoid this might be to require $\int\_{\mathcal{M}} Jf\_n dx = V(\mathcal{N})$. There are results in the theory of bubbling that roughly tell you that the above is the only thing that happens, i.e. if $f\_n \rightharpoonup f$ and $Jf\_n$ converges in the sense of measures, then $Jf\_n dx \rightharpoonup Jf dx + V(\mathcal{N})\sum\_{i\in I} a\_i \delta\_{x\_i}$, where $I$ is a finite set, $a\_i \in \mathbb{Z} \setminus \{0\}$ and $x\_i \in \mathcal{M}$, i.e. the only thing that can happen is whole copies of the target manifold "bubbling off" in single points. So if you only have one copy available, and your condition on $Jf\_n > r$ requires you to keep one, there are none to do that.


Specifically, since $Jf\_n$ is non-negative, the same is true for the limit as a measure, so $Jf \geq 0$ and $a\_i > 0$, but by testing with the constant function $1$ you get
$$V(\mathcal{N}) = \int\_{\mathcal{M}} Jf\_n dx \to \int\_{\mathcal{M}} Jf dx + V(\mathcal{N})\sum\_{i\in I} a\_i.$$
So either $I = 0$, which gives you the $L^1$-convergence, or $Jf=0$, which implies convergence a.e. and thus contradicts the assumption.


