# Are conformal maps between Riemannian manifolds real-analytic?

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 that the metrics are $$C^{\infty}$$. Note that every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with the smooth structure.

A reasonable starting point would be to know whether every $$C^{\infty}$$ conformal map between real-analytic manifolds with real-analytic Riemannian metrics is real-analytic. (but what I am asking seems much harder).

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

Motivation:

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

For the Euclidean case, I 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, the only non-obvious to handle is $$f(x)=\frac{A x}{|x|^2}, 0 \notin \Omega.$$

which is real-analytic as a multiplication of two analytic maps. ($$1/x^2$$ is analytic on $$\mathbb{R} \setminus \{0\}$$).

• Actually, the question doesn't make sense the way you phrased it. It's true that every smooth manifold has a real-analytic structure, but it's not unique in the sense that you need it to be. Given two real-analytic atlases on $M$, say $\mathscr A_1$ and $\mathscr A_2$, there's a real-analytic diffeomorphism from $(M,\mathscr A_1)$ to $(M,\mathscr A_2)$. But the identity map might not be analytic between these two structures. So a specific conformal diffeomorphism might be analytic with respect to one pair of analytic atlases, but not with respect to another. Commented Sep 14, 2018 at 17:36
• I'm pretty sure that what is true is that if $M$ and $N$ are real-analytic manifolds (i.e., with specific real-analytic atlases) endowed with real-analytic Riemannian metrics, then any conformal map from $M$ to $N$ is real-analytic. But I haven't found a reference. Commented Sep 14, 2018 at 17:37
• Thank you, I agree that the question does not make sense as I phrased it. Just to be sure: Given a specific smooth atlas $A$ on a manifold $M$, let $A_1$ be a compatible real-analytic atlas (this just means a real-analytic "sub-atlas" , right?). If we compose $A_1$ with a smooth diffeomorphism of $(M,A)$ which is not real-analytic w.r.t $A_1$, do we get another real-analytic atlas which is compatible with $(M,A)$ and different from $A_1$? Commented Sep 14, 2018 at 20:16
• (Also, I am actually interested in the case where the metrics are only $C^{\infty}$, but maybe this is too hopeful. I have edited the question according to your comments) Commented Sep 14, 2018 at 20:16
• You're right -- any real-analytic atlas compatible with $A$ is a "sub-atlas" of $A$. And your idea about how to get a different analytic atlas also compatible with $(M,A)$ is also right. Commented Sep 14, 2018 at 20:19