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This statement shows in do carmo's differential geometry of curves and surface as

"Local conformality is easily seen to be an equivalence relation, that is , if S1 is locally conformal to S2 and S2 is locally conformal to S3, then S1 is locally conformal to S3. "

Well, the transitivity and reflectivity are obvious, but is symmetry satisfied here? In my intuition, local diffeomorphism is not a equivalent relation (does not satisfy symmetry, is it true?), and local isometric/conformal need local diffeomorphism. So I think local conformality is only transitive, not a equivalent relation. I guess I have something wrong here. Can someone help?

Thanks in advance .

J.Han
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  • If $g_1 = \lambda g_2$ and $g_2 = \mu g_3$ with $g_i$ metrics and $\lambda, \mu$ positive functions, then $g_1 = (\lambda \mu) g_3$. – Didier Mar 01 '21 at 17:43
  • I think I misread your question. A more relevant comment: if $g_1 = \lambda g_2$ with $\lambda$ positive, then $g_2 = (1/\lambda)g_1$ with $1/\lambda$ positive. – Didier Mar 01 '21 at 18:47
  • How does Do Carmo define "locally conformal"? – Kajelad Mar 01 '21 at 19:00
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    "In my intuition" could be the problem. I suggest you try to write rigorously what it means for a local diffeormorphism to be an equivalence relation and see if you can prove the statement to be true. If you don't succeed, I suggest you post your incomplete proof here and say where you got stuck. – Deane Mar 01 '21 at 21:50
  • No, it is not an equivalence relation, just like "locally isometric" is not. See here for some suggestions how to modify the definition to get an equivalence relation. – Moishe Kohan Mar 02 '21 at 08:33

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