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

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

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 $K \subset \subset \operatorname{Int}(\M)$.?

I am fine with assuming that $f_n$ are Lipschits and injective, and that $V(f_n(\M)) \to V(\N) $.

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 $K \subset \subset \M$.

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.

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

  • $\begingroup$ How about examples of bubbling where the body map $f$ is not constant? $\endgroup$ – Leo Moos Jan 14 at 15:08
  • $\begingroup$ Thanks, you might be right. I am not very familiar with bubbling phenomena. Do you have a specific example in mind? (can you please elaborate?) $\endgroup$ – Asaf Shachar Jan 14 at 16:51
  • 1
    $\begingroup$ Bubbling is kind of exactly what you do in your example, an image of the whole sphere concentrates in a single point, so that it cannot be seen in the usual "almost everywhere" sense, somewhat like a "bubbling off". In fact doesn't your example already contradict your assumption? You don't even have to calculate $Jf_n$ (if you do, I think it's easier if you scale the other way and have the Dirac at $0$), just by noting that the limit is a Dirac measure, you know that $Jf_n \to 0$ a.e. else and then using Egorov's theorem uniformly on large enough sets, thus $V(\{Jf_n \leq r\}) \to 0$. $\endgroup$ – mlk Jan 15 at 9:55
  • $\begingroup$ @mlk I'm a bit confused by the second part of your comment. Aren't you in fact showing that $V(\{ J f_n \geq r \}) \to 0$, with the inequality going in the other direction than what you claim? $\endgroup$ – Leo Moos Jan 15 at 17:08
  • $\begingroup$ @LeoMoos Sorry, you are right, I got confused by the inequality there. $\endgroup$ – mlk Jan 15 at 18:41

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.

  • $\begingroup$ Thank you very much! This is very interesting, and I will try to digest it . For now I have a question: Do you have any reference for these 'results in the theory of bubbling' you refer to? (I will point that if we require $f_n$ to be Lipschitz and bijective, then the area formula implies that $\int_{\mathcal{M}} Jf_n dx = V(f_n(\mathcal{M}))=V(\mathcal{N})$ so the condition you suggested holds). I would be very interested in any reference that states that under such suitable conditions you have $Jf_n dx \rightharpoonup Jf dx + V(\mathcal{N})\sum_{i\in I} a_i \delta_{x_i}$, as you mentioned. $\endgroup$ – Asaf Shachar Jan 16 at 19:09
  • $\begingroup$ @AsafShachar My personal go to reference for stuff like this would be Giaquita et al. "Cartesian currents in the calculus of variations I", but that might be a bit much and sometimes has a bit of a non-standard treatment of the topic. In general this is connected to the theory of harmonic maps and to degree theory and the topology of Sobolev mappings. There is a survey paper by Helein and Wood, "Harmonic maps", which has a lot of the ideas. $\endgroup$ – mlk Jan 16 at 21:32
  • $\begingroup$ Thank you! The more I study your answer the more impressed I become. And I think that your idea regarding the "bubbling theory" seems entirely right! I think that "Theorem 1 (Structure Theorem)" of "Cartesian currents in the calculus of variations" (Volume II, pg 363) should do the job. Unfortunately, I am not fluent in the language of currents. Here is a question you might be able to answer: If we assume $f_n$ are Lipschitz and injective, and $ \int_{M} Jf_n = V(f_n(M)) \to V(N)$. Does $Jf_n \ge 0$ converge in the sense of measures? (what is the relevant compactness theorem here?). $\endgroup$ – Asaf Shachar Jan 17 at 11:35
  • $\begingroup$ I guess that if it does converge, then it must converge to a current, which must be of the form you have described (by the theorem I mentioned). Thank you again for all your help! $\endgroup$ – Asaf Shachar Jan 17 at 11:35
  • $\begingroup$ @AsafShachar The basic compactness should be the usual one for measures, i.e. as $V(f_n(M))$ is bounded and $Jf_n$ has a sign, $Jf_ndx$ is a bounded sequence of non-negative measures (and since $M$ is compact, there are no problems with tightness). $\endgroup$ – mlk Jan 17 at 12:01

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