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This is probably a very naive question, but so far I could not find an answer:

Let $(M,g)$ be a Riemannian manifold. Can we always find "orthogonal coordinates" locally?

More precisely, I am asking if for every $p \in M$ there exists a neighbourhood $U$ and a diffeomorphism $\phi:\mathbb{R}^n \to U$, such that $g_{ij}=g(d\phi(e_i),d\phi(e_j))=0$ for $i \neq j$.

Clarification: Note that I want $g_{ij}=0$ on all $U$, not just at $p$. Also, I allow $g_{ii} \neq g_{jj}$ for $i \neq j$ (the special case where $g_{ii}$ is independent of $i$ is called isothermal coordinates-and corresponds to conformal flatness of $U$).

Of course, this is weaker than requiring $M$ to be conformally flat, since a (linear) map which maps an orthogonal basis to an orthogonal basis does not need to be conformal.

Asaf Shachar
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  • This only makes $g_{ij}(p)=0$ but I want $g_{ij}=0$ on all $U$. – Asaf Shachar Nov 30 '16 at 11:19
  • No, because I only demand $g_{ij}=0$ for $i \neq j$. I do not assume $g_{ij}=\delta_{ij}$, so $\phi$ is not necessarily an isometry. – Asaf Shachar Nov 30 '16 at 12:08
  • I missed that, you want $g_{ij}(s) = 0$ for $i \ne j, s \in U$ and you allow $g_{ii}(s)$ to vary... Ok then say it clearly. And with a triangulation you should be able to see when the constraint is satisfiable. – reuns Nov 30 '16 at 12:12
  • I have edited the question, I hope it is clearer now. Can you please elaborate on the triangulation's idea? What is the obstruction you are thinking of? – Asaf Shachar Nov 30 '16 at 16:07
  • Say clearly that $g_{ii}(s)$ is allowed to vary ! And no sorry I can't help more. Maybe you can find a criterion making $g_{ii}(s_0)$ to diverge as the triangulation is refined ? – reuns Nov 30 '16 at 16:24

1 Answers1

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A Riemannian metric $g$ on an $n$-dimensional manifold is called locally diagonalizable if it is locally isometric to a Riemannian metric on a domain in $R^n$ with diagonal metric tensor. In dimension $n=2$ every Riemannian metric is locally diagonalizable due to existence of isothermal coordinates.

For $n\ge 3$ the problem of local diagonalizability was solved in

D. DeTurck, D. Yang, Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243–260.

They proved that for $n=3$ every Riemannian metric is indeed locally diagonalizable. This result fails for $n\ge 4$, for instance, for the standard Fubini-Study metric on the complex-projective plane $\mathbb C P^2$:

Gauduchon, Paul; Moroianu, Andrei, Non-existence of orthogonal coordinates on the complex and quaternionic projective spaces, J. Geom. Phys. 155, Article ID 103770, 6 p. (2020). ZBL1466.53019.

The following is an interesting open problem (already raised by Gauduchon and Moroianu):

Is there any topological obstruction for the existence of metrics with orthogonal coordinates, or does every smooth manifold carry such metrics?

Moishe Kohan
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  • I took another look at the paper you cite. I think the statements for $n > 3$ is something like this: If the metric is diagonal with respect to coordinates $(x^1, \dots, x^n)$, then for any distinct values of $1 \le i,j,k,l \le n$, $$W_{ijkl} = 0. $$ The condition that for any orthonormal frame $(e_1, \dots, e_n)$, $$W(e_i,e_j,e_k,e_l) = 0$$ is not necessary for the existence of diagonal coordinates. We in fact prove that this condition implies $W = 0$, i.e., the metric is conformally flat and therefore trivially diagonalizable. – Deane Oct 06 '24 at 15:35
  • @Deane: Sorry, this means that I misremembered your paper. I will correct this later during the week when I am over my flu and can think more clearly. Incidentally, do you know of topological obstructions to existence of Riemannian metrics with locally diagonalizable metric tensor? – Moishe Kohan Oct 06 '24 at 16:52
  • thanks. I know nothing about global questions. Since it's a system and, worse, nonlinear and nonelliptic, we simply don't have the tools to attack global questions such as topological obstructions. – Deane Oct 06 '24 at 18:47
  • For 4-dim manifold $M$ and $p\in M$, consider a 3-dim submanifold contains $p$. Then, near the $p$, give a 3-dim local coordinate such that its Riemannian metric is diagonalizable. Then, choose a geodesic through $p$ and perpendicular to the submanifold at $p$. Then, the local coordinate and the geodesic form a new geodesic. The new coordinate seemly be local diagonalizable near $p$. Just feel. – Enhao Lan Oct 07 '24 at 02:22
  • @EnhaoLan, I suggest you try to write a rigorous proof of your claim. – Deane Nov 10 '24 at 23:07
  • @Deane My ability is too far away from yours. I'm afraid to touch your questions. I am trying this (https://math.stackexchange.com/questions/4740602/the-length-of-shortest-curve-dividing-topological-sphere-into-two-parts-of-same). Do you have idea or recommended references ? Thanks. – Enhao Lan Nov 11 '24 at 01:14
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    It’s not ability. You have to learn how to write rigorous proofs. This is a good one to try. Sometimes you succeed. Sometimes you don’t. And it’s always a struggle. – Deane Nov 11 '24 at 01:52