This is a cross-post.

Let $M,N$ be oriented smooth ($C^{\infty}$) $n$-dimensional Riemannian manifolds, and let $f:M \to N$ be a smooth orientation-preserving weakly* conformal map.

Do there exist real-analytic structures on $M,N$ that make $f$ real-analytic?

I only assume the metrics are $C^{\infty}$. Every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with its smooth structure.

A reasonable starting point would be to know whether every $C^{\infty}$ conformal map between real-analytic manifolds with real-analytic metrics is real-analytic. (I don't know a reference for that; anyway, what I am asking seems harder).

*A weakly conformal map is a map whose differential at every point is either conformal or zero. (This is equivalent to $df^Tdf =(\det df)^{\frac{2}{n}} \, \text{Id}_{TM}$).


I am trying to understand if smooth weakly conformal maps whose differential vanishes at a point are constant (for dimensions $n \ge 3$). This seems to be the case for analytic maps, hence my interest in the possible analyticity of such maps.

For the Euclidean case, this follows directly by Liouville's theorem:

For $n=2$, every such map is complex-analytic. Let $\Omega \subseteq \mathbb{R}^n$ be an open subset, $n \ge 3$, and let $f:\Omega \to \mathbb{R}^n$ be a smooth conformal map. By Liouville's theorem, $f$ is of the form $$ f(x)=b+\alpha\frac{1}{|x-a|^\epsilon}A(x-a),$$

where $A$ is an orthogonal matrix, and $\epsilon \in \{0,2\}, b \in \mathbb{R}^n,\alpha \in \mathbb{R},a \in \mathbb{R}^n \setminus \Omega$.

So, up to translations and dilations, $ f(x)=\frac{A x}{|x|^2}$ (where $ 0 \notin \Omega$) which is real-analytic as a multiplication of two analytic maps. ($1/x^2$ is analytic on $\mathbb{R} \setminus \{0\}$).

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    $\begingroup$ While there is an analytic structure on the manifold, the Riemannian metric will not necessarily be real analytic. Does it bother you? $\endgroup$ Commented Oct 22, 2018 at 14:42
  • $\begingroup$ @PiotrHajlasz No, I am OK with the metrics to be only $C^{\infty}$ but not real analytic. (Although the case where the metrics are analytic is also interesting I guess. I assume that in that case, conformal maps-in high dimensions, do need to be analytic, but I don't know a reference for that). $\endgroup$ Commented Oct 23, 2018 at 9:11

2 Answers 2


As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $f\colon M\to N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could just pick any analytic structure on $N$, pull it back to $M$ by $f$ and make $f$ trivially analytic.

With that in mind, one can think of counterexamples already in 1-dimension. For instance let $(N,g_N) = (M,g_M) = (\mathbb{R},dx^2)$, where $dx^2$ is the metric that appears flat with respect to some global coordinate $x$. Let $$ f(x) = \begin{cases} (x+\epsilon) e^{-1/(x+\epsilon)^2} & x<-\epsilon \\ 0 & -\epsilon\le x \le \epsilon \\ (x-\epsilon) e^{-1/(x-\epsilon)^2} & \epsilon < x \end{cases} . $$ It is locally a diffeomorphism if you consider it as a map from $\mathbb{R}\setminus [-\epsilon,\epsilon]$ to $\mathbb{R}\setminus \{0\}$. But there is no pair of analytic structures on $\mathbb{R}$ that would make $f\colon \mathbb{R} \to \mathbb{R}$ analytic. If $f$ were analytic (with respect to any analytic atlas), then it could not vanish on any open set, while not being globally zero. But since we are in 1-dimension, any map is weakly conformal. So then $f\colon M \to N$ is weakly conformal, but can never be made analytic.

Higher dimensional examples could be generated from this one by using rotational symmetry and basically reducing to the 1-dimensional case.

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    $\begingroup$ Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question. $\endgroup$ Commented Oct 22, 2018 at 14:38
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    $\begingroup$ For example, I considered maps of the form $\vec{x} \mapsto \phi(|\vec{x}|) \vec{x}$ where $\phi : \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ is $0$ on $[0,1]$ and $>0$ thereafter. But I got that you need $\phi'(s)/\phi(s)$ to be bounded near $s=1$, so $\log \phi(s)$ is bounded, so $\phi$ can't go to $0$. $\endgroup$ Commented Oct 23, 2018 at 9:42
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    $\begingroup$ @DavidESpeyer, hmm, I think that I was overly optimistic when I wrote that. I basically had in mind the same idea as you, and now see that it doesn't easily work, just as you say. Sorry! $\endgroup$ Commented Oct 23, 2018 at 14:33

This is not an answer. Just a possible vector of attack for this problem.

According to the Theorem 11.4.6 of Harmonic Morphisms Between Riemannian Manifolds by Paul Baird and John C. Wood, every non-constant weakly conformal map between manifolds of the same dimension $n\geq 3$ is a conformal local diffeomorphism. The automorphism group of $(M, [g])$ is a Lie group whose Lie algebra is the algebra of conformal Killing vector fields.

According to the Theorem 3.8 of Normal BGG solutions and polynomials every conformal Killing vector field is given by a polynomial expression (of degree at most 2) in normal coordinates and normal frame. Given a Cartan geometry $(\pi: \mathcal{G} \to M, \omega \in \Omega^1(\mathcal{G}, \mathfrak{g})$, these normal coordinates and frames are given by flow flow of $\omega^{-1}(X)$ and by projection of that flow through $\pi.$

  • $\begingroup$ Thanks. Does your second comment (regarding the conformal vector fields) imply that such conformal maps are real-analytic? (It seems to me that the normal coordinates themselves could be only $C^{\infty}$ but not real-analytic, since the metrics are not assumed to be real-analytic, but maybe I am confused here). $\endgroup$ Commented Oct 23, 2018 at 13:21
  • $\begingroup$ @AsafShachar I don't know. These coordinates and frames are given by flows of certain vector fields comming from the Cartan connection. $\endgroup$ Commented Oct 23, 2018 at 13:54
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    $\begingroup$ Why doesn't your first comment answer the question in dimension $n > 3$? It seems you can pick a real analytic structure on the base and pull back sufficiently small charts via the local diffeomorphism to give a real analytic structure on the domain for which (in these charts) the map $f$ is the identity. I must be missing something. $\endgroup$
    – mme
    Commented Oct 23, 2018 at 14:21
  • $\begingroup$ @MikeMiller Because I didn't have enough coffee today. $\endgroup$ Commented Oct 23, 2018 at 15:05
  • $\begingroup$ You at least had enough to find these lovely references :) $\endgroup$
    – mme
    Commented Oct 23, 2018 at 16:28

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