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It seems to be well-known that in dimension 3, we can simultaneously diagonalize the metric and the Ricci tensor at any given point (see https://mathoverflow.net/questions/80452/diagonalizability-of-the-curvature-operator).

I am unable to prove this myself, and so I am hoping for a proof of this fact or a source which provides the proof. I think the link above has a proof by counting dimensions, but I am looking for a more straightforward proof if there is one.

Didier
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61plus
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1 Answers1

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This is not a particular statement about the dimension 3, but a general statement. As a symmetric bilinear form, the Ricci curvature admits an orthonormal basis of eigenvectors, which trivially diagonalizes the metric as well. See, e,g, this MSE post.

Didier
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  • I see, I was confused and the pointwise statement indeed follows from standard linear algebra. If we want this to be true locally, are you aware if that is still true in dimension 3? – 61plus Jan 06 '22 at 16:23
  • @61plus The MO posted you link consider the whole curvature operator, not only the Ricci tensor. – Arctic Char Jan 06 '22 at 16:38
  • With coordinates, this is certainly not true. Concerning just moving frames, I have to admit I can't convince myself right now – Didier Jan 06 '22 at 17:00
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    Actually I mentioned the curvature tensor in the title because the Ricci tensor and Riemannian curvature tensor determine each other in dimension 3. – 61plus Jan 06 '22 at 18:03
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    Note that the eigenvectors might not be smooth functions of the manifold. See https://mathoverflow.net/questions/60533/can-always-a-family-of-symmetric-real-matrices-depending-smoothly-on-a-real-para – Deane Jan 06 '22 at 18:06