Smoothness of conformal transformations

Question

Given a smooth pseudo-Riemannian manifold $(M,g)$ one can define the conformal group as the set of smooth diffeomorphisms $\varphi:M\to M$ such that there is a positive smooth function $u$ with $\varphi^\ast g=ug$. One could also define it as the set of all $C^1$-diffeomorphisms $\varphi:M\to M$ such that there is a positive continuous function $u$ with $\varphi^\ast g=ug$. One could also define intermediate cases. I would expect these to be the same but it does not seem obvious, can someone clarify? Do any such answers rely on $M$ being Hausdorff or second-countable/paracompact?

Answer 1

Answers to your question are somewhat dimension and signature dependent.

For example, in dimension $2$, if $g$ is definite, then every $C^1$ conformal diffeomorphism is, in fact, real-analytic, because, locally, $g$ can be written in the form $g = F\,\mathrm{d}z{\circ}\mathrm{d}\bar z$ for some complex-valued coordinate $z$ and nonzero $F$. In such a coordinate, a $C^1$ conformal transformation is actually either $z$-holomorphic or conjugate $z$-holomorphic.

Also, in dimension $2$, if $g$ has split signature, then, locally $g$ is of the form $g = F(x,y)\,\mathrm{d}x{\circ}\mathrm{d}y$ for some coordinates $(x,y)$, and a $C^1$ $g$-conformal transformation is either of the form $\phi(x,y) = (p(x),q(y))$ or of the form $\phi(x,y) = (q(y),p(x))$, where $p$ and $q$ are $C^1$, and you don't get any better regularity than that since all such $C^1$-maps are $g$-conformal.

In dimensions higher than $2$, in the positive definite case, if $g$ is smooth, then any $C^1$ conformal diffeomorphism is smooth, though I think this is not an easy result, even when $g$ is flat. (See the proofs of Liouville's Theorem in the case of low regularity.) If you assume, say, $C^3$, though, then it's not too hard.

In dimensions higher than $2$, independent of signature, if you assume $g$ is smooth and $\phi$ is at least $C^4$ (though probably $C^3$ is OK), then $\phi$ must be smooth. About lower regularity, I don't know.