# Geometrical point of view of the harmonic constraints ($\Delta g_{ij}=0$) in General Relativity

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What does it mean, from the geometrical point of view, use (in General Relativity) of the constraints on the metric tensor's coefficients such that $\Delta g_{ij}=0$? (where $\Delta$ is the Beltrami-Laplace Operator, $g_{ij}$ the metric tensor).
With $\Delta g_{ij}=0$, I mean the Laplace-Beltrami operator, applied componentwise to the components of the metric tensor.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini
retagged Jul 18, 2016
You say "in General Relativity," but there is no such constraint that is applied generically in general relativity. In the very early days of GR, it was thought that the metric's determinant needed to be constrained to be -1 everywhere. Today, it is often convenient to make a particular choice of gauge, such as harmonic coordinates en.wikipedia.org/wiki/Harmonic_coordinate_condition . But the only constraint on the metric that is absolutely mandatory in the standard modern formulation of GR is that it not be degenerate.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Ben Crowell
Another little question...in this particular kind of metric (for example in dimension 2) where $\Delta g=0$ the curvature isn't necessary zero, is correct?

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini
@AlexanderPigazzini: The condition $\Delta g_{ij} = 0$ does not constrain the curvature in any simple way. When $n=2$, there will be some very high order polynomial relation among $K$ and its first $m$ covariant derivatives that characterizes the existence of such a coordinate system, but I don't know what that is explicitly.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Robert Bryant
Thank you very much!

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini

Slightly off-topic, but the harmonic conditions  with $\square$ instead of $\Delta$ can be interpreted as the gravitation field equations (amongst others) in a flat Minkowsky space-time; see more details in https://arxiv.org/abs/0810.4393

thanks prof. Bryant, then from what I understand, there is no interest in a metric of this type, am I right? ...I mean that there isn't interest in $g=h(x,y)(dx^2+dy^2)$ where $h>0$ and satisfies $h_{xx}+h_{yy}=0$, or wrong?
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