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).

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).

*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).

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    $\begingroup$ If the map $\psi_\lambda$ you gave is well-defined, then the following should be well-defined also: $$\psi^{(1)}_\lambda(x_1, x_2\ldots x_{n+1})=(\lambda x_1, \lambda x_2, \ldots , \lambda x_n, x_{n+1}), $$ (modulo stereographic projection) and I would expect it to weakly converge to an immersion of the sphere onto a great circle. This should prove that the limit can be nonconstant. $\endgroup$ Feb 21, 2018 at 18:33
  • $\begingroup$ Thanks, this looks nice. Since the image is of positive codimension, this shows the limit is not an immersion, right? $\endgroup$ Feb 21, 2018 at 20:35
  • $\begingroup$ Oh sorry, yeah, I meant "submersion". Of course it is not an immersion. $\endgroup$ Feb 21, 2018 at 20:36
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    $\begingroup$ @GiuseppeNegro: It's not a submersion either - there is no submersion from $S^n$ to $S^1$ for $n>1$. Proof: Such a map lifts to a submersion $S^n\rightarrow \mathbb{R}$ because $S^n$ is simply connected. Submersions are open maps and this map is closed because $S^n$ is compact. It follows that the map $S^n\rightarrow \mathbb{R}$ is surjective, which is absurd since $\mathbb{R}$ is non-compact. $\endgroup$ Feb 24, 2018 at 1:05
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    $\begingroup$ @GiuseppeNegro By the way, your nice example, when looked from a "Sobolev point of view", is exactly like my original example: since only a set of measure zero is mapped onto a great circle (and the rest is mapped to a point), we can think of this map as the constant map. $\endgroup$ Feb 24, 2018 at 8:55

2 Answers 2


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.


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.


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