# Can a Riemannian metric be analytic in non-analytically different coordinates?

Suppose I have two coordinates on the same (subset of a) Riemannian manifold. If the metric tensor is analytic in both coordinates, is the change of variables between them necessarily analytic?

In other (and perhaps clearer) words, is this conjecture true?

Let $U,V\subset\mathbb R^n$ be domains. Let $g$ be a real-analytic Riemannian metric on $U$ and let $\phi\colon V\to U$ be a smooth diffeomorphism. If $\phi^*g$ is a real-analytic metric on $V$, then $\phi$ is real analytic.

I have not been able to prove this or find a counterexample.

The difficulty in using the formula for the metric pulled back over $\phi$, $$\phi^*g_{ij}(x) = \partial_i\phi^k(x)g_{kl}(\phi(x))\partial_j\phi^l(x),$$ is that $\phi$ appears several times and in different ways in the right-hand side. Taking determinants yields $\det(\phi^*g(x))=\det(g(\phi(x)))\det(D\phi(x))^2$, which might help if one could change coordinates analytically on $U$ so that $\det(g(y))$ is constant and show that analyticity of $\det(D\phi)$ implies analyticity of $\phi$. (Compare this with the one-dimensional proof below.)

The claim is true if $n=1$, but I do not see how to generalize this argument: A Riemannian metric on $\mathbb R$ is just a positive scalar function. After a real-analytic change of coordinates on $U$ one can assume that $g\equiv1$ (Euclidean metric). Then $\phi^*g(x)=\phi'(x)^2$. Since this is analytic and $\phi'$ never vanishes, the function $\phi$ is analytic.

• Incidentally, the title seems strange, and not just because it asks the opposite question to the body of the text: since coordinates that can be related by an analytic change could be called "analytically identical" or similar, it might make more sense to say that those that cannot are "analytically different", rather than "non-analytically different". Commented Apr 6, 2017 at 22:47
• @LSpice I would parse "analytically different coordinates" as different coordinate systems related by an analytic diffeomorphism. I know the title asks the opposite question, but I could not come up with an equally concise and descriptive title otherwise. Asking "$P$?" and "$\neg P$?" is effectively equivalent. I'm open to suggestions for a new title if you have ideas, but I quite like the present one. Commented Apr 6, 2017 at 22:54
• It's probably not a big deal, since the question was answered decisively 6 minutes after being asked. I did make a suggestion, but obviously we parse it differently, so it seems reasonably to conclude that there is no universal opinion. (As to keeping a concise title, you could change the question in the body to match it—but then the Boolean in AntonPetrunin's nice answer would have the opposite of its intended sense, so that's probably not a good idea either.) Commented Apr 6, 2017 at 22:56