The weak limit of a sequence of diffeomorphisms is either a diffeomorphism or a constant?

Question

Let $M$ be a compact oriented manifold, and let $f_n \in \text{Diff}^+(M)$ be a sequence of orientation-preserving diffeomorphisms which weakly converges in the Sobolev space* $W^{1,p}(M;M)$ to a limit $f$. (for some $p>1$. I am ready to assume $p > \dim M$).

Is it true that $f$ must be either a constant map or a diffeomorphism?

(Or perhaps $f$ must be either constant or an immersion?).

More formally, we need some care: As an element in a Sobolev space, $f$ is only defined up to a measure zero set. So, the question is wether or not there is always a representative which is smooth and is either constant or a diffeomorphism).

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Note that indeed convergence to a constant can happen- here is the example I have in mind:

Take $M=\mathbb{S}^n$. Consider the following one-parameter family of diffeomorphisms $\psi_{\lambda}:\mathbb{S}^n \to \mathbb{S}^n$, $\lambda >0$: $\psi_{\lambda}$ is obtained by using the stereographic projection, then dilating by $\lambda$ (i.e. $x \to \lambda x$) and finally projecting back. $(\psi_{\lambda})_{\lambda >0}$ is a family of diffeomorphisms that weakly converge to the pole when $\lambda \to \infty$.

(Strictly speaking, when looking at the pointwise convergence of this sequence, the south pole is fixed, but everything else converges to the north pole. Since a point has a measure zero, we can ignore this when thinking on the limit as a function in a Sobolev space).

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*Strictly speaking, convergence in a Sobolev space requires a metric on the manifold. One embeds $M$ isometrically in $\mathbb{R}^D$, and then views maps $M \to M$ as maps $M \to \mathbb{R}^D$. Now, under this identification, the weak convergence in $W^{1,p}(M;M)$ is exactly the one taken from $W^{1,p}(M;\mathbb{R}^D)$. However, I think that for a compact manifold, the specific choice of a metric does not affect the convergence notion. (If it does, then just assume $M$ is equipped with a metric).

Answer 1

Just an attempt without thorough check: consider a sequence of maps $f_n$ which map $S^1 $ diffeomorphically to itself, such that $f_n$ maps the left half circle $\{3/2 \pi > \phi > \pi/2\} $ to $$\{\phi :(3/2 + (1/2-1/(2n))\,\pi >\phi > 1/(2n)\,\pi \}$$ and the rest of the circle to the small region $$ \{\phi :1/(2n)\pi > \phi \text{ or } \phi > (3/2 +(1/2-1/(2n))\,\pi \}$$

In other words: you stretch the left half of the circle so that, in the end, it covers the whole circle, while the right half shrinks to a point.

Assuming such a sequence, if constructed properly, converges weakly in $W^{1,p}$ (which I did not check, but I currently cannot see why it should not) the limit will map the left part of the circle to the whole circle minus one point and the right part of the circle to a point.

A similar construction should be possible in higher dimensions.

Answer 2

Partial result: The weak limit $f$ cannot be a point if $p>\dim M$.

If $f_n$ converges weakly to $f$ in $W^{1,p}$ and $p > \dim M$, then the $W^{1,p}$ norm of $f_n$ are uniformly bounded and thus $f_n\to f$ in $C^{0,\alpha}$ for some $\alpha>0$ (We used $p>\dim M$ here). Then $f$ has to be surjective: if not, there is an nonempty open set $U\subset M$ so that $f(x)\notin U$ for all $x\in M$. Since $f_n\to f$ uniformly, there is a nonempty open set $V\subset U$ so that $f_n(x) \notin V$ for all $x\in M$ and $n$ large. This contradicts that $f_n$ is surjective.