# Smoothness of conformal transformations

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?

• If $M=\mathbb{R}$ with the standard metric, any $C^1$-function $f$ with $f'(t)>0$ for all $t$ is conformal in your sense. – abx Jan 28 at 6:04
• I believe one dimension is usually excluded in conformal geometry, since any two metrics are in the same conformal class – Quarto Bendir Jan 28 at 6:08

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.