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I'm not very familiar with the concepts in Riemannian geometry, so I'll be very glad if you could point out if my question is naive or meaningless!

For a Riemannian manifold, after diagonalizing the metric, what kind of manifold has $g_{11}$ being a constant? I know that for sphere and hypersphere we have $g_{11}=r^2$ where $r$ is the radius of the sphere, see reference here, so I wonder if there are any other manifolds equipped with metric whose $g_{11}$ is constant.

Elio Li
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    What do you mean by "after diagonalising the metric"? In general, there is no coordinate frame in which the metric is diagonal. – Didier Oct 05 '24 at 09:53
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    In general, I wouldn't expect much. If I recall correctly, all $3$-manifolds (and thus $2$ and $1$ manifolds as special cases) admit local coordinates making the metric tensor diagonal locally (rather than pointwise; we can then cover the manifold with finitelt many of these local charts. This is exactly what happens in the sphere case you mentioned). Thus, the only meaningful thing here in low dimensions will be a single component of your metric tensor being constant, which I suppose says curvature only occurs along restricted directions (in reference to the given local coordinates) – Brevan Ellefsen Oct 05 '24 at 10:14
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    See the reference in my answer here for local diagonalizability of Riemannian metrics. – Moishe Kohan Oct 05 '24 at 14:39
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    If the metric tensor is of the form $g_{11} = c$ and $g_{1j} = 0$ for $j \ne 1$, then the coordinate curves in the direction $\partial_1$ are constant speed geodesics. – Deane Oct 06 '24 at 15:47

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