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To be precise: given $\phi \in C^{\infty}(M)$, does there exist a (semi-)Riemannian metric $g$ such that $\Delta_{g}\phi = 0$?

I avoided fixing a coordinate system on purpose here; I'm not sure how that would come into play, but I expect it might.

Stańczyk
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    https://math.stackexchange.com/questions/3638437/why-is-there-no-harmonic-function-on-compact-riemannian-manifold – mfl Dec 14 '20 at 20:46
  • @mfl I am not sure how this answers the question – Stańczyk Dec 14 '20 at 23:39
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    @Stańczyk Choose a nonconstant function on a compact manifold. It will not be harmonic w.r.t. any metric. – Kajelad Dec 14 '20 at 23:50
  • @Kajelad Of course this true of a Riemannian metric. More generally? – Ted Shifrin Dec 15 '20 at 02:45
  • Even if one allows for a metric of any signature, there will still be counterexamples (e.g. $S^2$, which admits no indefinite nondegenerate metrics due to the hairy ball theorem). Determining if such a metric exists when $M$ does admit an indefinite metrics might be difficult, though. – Kajelad Dec 15 '20 at 03:39

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