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Let $(M^3,g)$ be a 3-dim. Riemannian manifold. Suppose there is an atlas of coordinate charts for $M$ such that, in each chart, the metric components can be written as

$$ g_{ij}(x)=R(\theta(x),\phi(x))\,\text{diag}\left(\lambda_1(x),\lambda_2(x),\lambda_3(x)\right)R^\mathsf{T}(\theta(x),\phi(x)), $$

with $R(\theta,\phi)$ the rotation matrix that (in Eucl. space) takes the vector $\hat{\mathbf{k}}=(0,0,1)$ to $(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$, and $\theta(x),\;\phi(x)$ general (smooth) functions. ($R(\theta,\phi)$ may be written as a $\theta$ rotation around the $y$ the axis followed by a $\phi$ rotation around $z$).

Consider now the metric $\tilde{g}_{ij}(x)=\text{diag}\left(\lambda_1(x),\lambda_2(x),\lambda_3(x)\right)$.

Question Is there a relation between the curvature tensors $\mathsf{Ric}(g_{ij})$ and $\mathsf{Ric}(\tilde{g}_{ij})$ ? In particular, for the case $\lambda_1(x)=\lambda_2(x)=$const. and $\lambda_3(x)$=const., can we factor out the $\lambda_1,\;\lambda_3$ dependence of $\mathsf{Ric}(g_{ij})$ ? In two-dimensions the curvature scalar of $g_{ij}(x)=R(\theta(x))\,\text{diag}\left(\lambda_1,\lambda_2\right)R^\mathsf{T}(\theta(x))$ doesn't depend on the constants, up to a multiplicative factor, so I wonder if something similar happens with one dimension more. A naïve computation of the components of the Ricci tensor seems to indicate that the answer is no, but maybe one can conclude something from more general arguments.

DanielKatzner
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  • (1) $R$ isn't specified uniquely by just sending $e_z$ to some vector. It is, however, uniquely specified if you take it to be rotation by $\theta$ about $y$-axis followed by rotation by $\varphi$ about $z$-axis. (2) in the simplest case you have a warped product – user10354138 Mar 03 '21 at 18:46
  • @user10354138 Yes, thank you. I could rephrase the questions saying that $g=\text{SO}(3)\cdot\text{diag}(\lambda_1,\lambda_2,\lambda_3)\cdot\text{SO}(3)$, but I don't think the analysis would be that different if we change the representation of $R$. – DanielKatzner Mar 03 '21 at 19:35
  • @user10354138 As for the warped product, I don't see how this metric (or which particular case) fits in that definition. $\theta$ and $\phi$ are functions of the coordinates $(x_1,x_2,x_3)$ and there's no separation of variables in none of the elements of $g_{ij}$. – DanielKatzner Mar 03 '21 at 19:41
  • Every Riemannian metric on a 3-dimensional manifold is locally isometric to a "diagonal" Riemannian metric. See my answer here. – Moishe Kohan Mar 04 '21 at 14:51
  • @MoisheKohan That's true, one knows that already from simple counting of degrees of freedom. But it happens the same as in two dimensions with isothermal coordinates: the explicit change of coordinates is very hard or impossible to solve (in 2d one has to solve the Beltrami equation, in 3d I guess some similar system of PDEs). But maybe I'm wrong, do you know of a recipe to explicitly construct the diagonal coordinates ? Thank you. – DanielKatzner Mar 04 '21 at 15:04
  • No, one cannot prove such results by counting dimensions. And computations of such isometries can only be done numerically. – Moishe Kohan Mar 04 '21 at 15:05

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