This is a follow-up of this question.

Let $M$ be a compact oriented Riemannian manifold, of dimension $d\ge2$. Let $f_n \in \text{Diff}^+(M)$ be a sequence of orientation-preserving conformal 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?).

As an element in a Sobolev space, $f$ is only defined up to a measure zero set. So, the question is whether or not there always exists a representative which is smooth and is either constant or a diffeomorphism.

In this answer, a sequence of diffeomorphisms $f_n:\mathbb{S}^1 \to \mathbb{S}^1$ is constructed; the limit of $f_n$ maps half of the circle onto $\mathbb{S}^1$, and the other half to a point. I tried to build from this sequence a higher-dimensional example, but couldn't do so in a conformal way. (e.g. $f_n \oplus f_n$ or $f_n \oplus \text{Id}$ are not conformal).

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    $\begingroup$ Do you have an example where the limit is a constant map? Even with $p>\dim M$? I haven't checked carefully but I guess dilations $x\mapsto nx$ in $\mathbb{CP}^1$ converge to the constant $\infty$ map for $p<2.$ $\endgroup$
    – Dap
    Feb 24, 2018 at 9:55
  • $\begingroup$ @Dap: The linked question contains a conformal example converging to a constant. It involves dilations in stereographic coordinates, so I think that it is essentially the same as yours. $\endgroup$ Feb 24, 2018 at 11:22
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    $\begingroup$ I suspect that the answer is affirmative if $M=\mathbb S^2$. This suspicion comes from complex analysis. If I am not mistaken, the only conformal diffeomorphisms of 𝕊2 onto 𝕊2 are Möbius transformations. These are essentially matrices with nonzero determinant. Weak convergence must translate into convergence of the matrix elements, perhaps to a singular matrix, corresponding to a constant map. $\endgroup$ Feb 24, 2018 at 11:48
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    $\begingroup$ The thing to have in mind is that conformal transformations satisfy the convergence property, i.e. every sequence of conformal transformations of a compact Riemannian manifold contains a subsequence which either converges to a conformal transformation or to a constant map uniformly on compacts away from one point. This should be discussed in Ferrand's papers (solution of the Lichnerovich conjecture). From this, you can probably get Sobolev convergence. $\endgroup$ Feb 24, 2018 at 17:16
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    $\begingroup$ Actually, in view of the fact that the Lichnerovich conjecture is true (proven by Ferrand), the group of conformal transformations of a compact Riemannian manifold is compact unless the manifold is $S^n$ with a conformally flat metric. In the former case, a sequence of conformal transformations will limit to a diffeomorphism. In the latter case, one would have to comb the literature on quasiconformal mappings. Check Iwanec and Martin's book (it is the most recent). Likely it will contain the result that you want. $\endgroup$ Feb 26, 2018 at 3:57

1 Answer 1


Let me convert my comment into an answer.

Theorem. Let $C(M)$ be the conformal group of a Riemannian manifold $M$ with $dim(M)=n\ge 2$. If $M$ is not conformally equivalent to $S^n$ or $E^n$, then $C(M)$ is inessential, i.e. can be reduced to a group of isometries by a conformal change of metric.

This theorem has a long and complicated history (in particular, a long history of incorrect proofs), you can find its proof and historic discussion in

J. Ferrand, The action of conformal transformations on a Riemannian manifold. Math. Ann. 304 (1996), no. 2, 277–291.

Now, given this theorem, if $M$ is compact, unless $M$ is conformal to $S^n$, any sequence of elements $f_i\in C(M)$ subconverges to a conformal diffeomorphism. This leaves out the case of a conformal metric on $S^n$. The elements of $C(M)$, in this case, are (smooth) quasiconformal transformations (in the sense of the standard round metric on $S^n$). As such, they satisfy the "convergence property", i.e. every sequence subconverges either to a constant (uniformly on compacts away from one point) or to a quasiconformal transformation. In the latter case, it is not hard to see that the limit is a diffeomorphism. In the former case, in order to get convergence in $W^{1,p}$ (most likely, $p=n$), one needs to work more and I do not have time for this. Check out books on quasiconformal maps, for instance

Iwaniec and Martin, "Geometric Function Theory and Non-linear Analysis", Oxford University Press, 2002.

Also, note that every sequence of qausiconformal mappings of $S^n$ converging to a constant can be written as a composition $f_n\circ h_n$ where the sequence $h_n$ converges to a quasiconformal mapping and $f_n$ is a sequence of Moebius transformations which converges to a constant map (away from one point). This effectively reduces the problem to the case of mappings of the form $x\mapsto \lambda_i x$, $x\in R^n$, $\lambda_i>0$ is a sequence diverging to infinity. (One also has to check how $W^{1,p}$-convergence behaves under the composition.)

  • $\begingroup$ Thanks for your efforts, really; this is a great answer. Two questions: (1) Do you know if the statement holds in the case of $\mathbb{E}^n$? (is limit of conformal diffeomorphisms a diffeomorphism or a point)? $\endgroup$ Feb 28, 2018 at 8:00
  • $\begingroup$ (2) How do you conclude after reducing the problem to the isometric case that the limit is again an isometry? (when $p>\dim M$ this implies uniform convergence, so it's clear. When $p < \dim M$ the only argument I know is this: Any sequence of isometries has a convergent subsequence in $W^{1,p}$ whose limit is an isometry. Thus, if such a sequence is convergent in $W^{1,p}$, then its limit must be again an isometry. This is theorem 1.4 here. $\endgroup$ Feb 28, 2018 at 8:00
  • $\begingroup$ @AsafShachar: Yes, in the case of $E^n$ also the limit of a sequence of conformal transformations (if it exists) is either a Moebius transformation or a constant map (but convergence is understood not in $W^{1,p}$-sense, only in the sense of uniform convergence on compacts (away from the exceptional point). As for isometries: Any sequence of isometries subconverges to an isometry in $C^\infty$-topology (uniformly on compacts). This is a corollary of the Myers-Steenrod theorem plus Arzela-Ascoli. $\endgroup$ Feb 28, 2018 at 12:58

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