 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.

Thank you in advance!

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini
asked Jul 4, 2016
retagged Jul 18, 2016
Since you are applying $\Delta$ to individual coordinate components, this seems like a strongly coordinate dependent condition, which might be hard to interpret geometrically. If you take $\Delta_g$ to be defined by $g$ itself, via its Levi-Civita connection, $\Delta_g g = 0$ is an identity, since $g$ itself is covariantly constant.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Igor Khavkine
Moreover, the equation $\Delta g_{ij}=0$ is an overdetermined system of equations for the coordinate system. For most metrics $g$, such coordinates don't exist, even locally.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Robert Bryant
Thanks for your answers! and if we had harmonic coordinates where $g_{ij}=\lambda * \delta_{ij}$?

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini
@AlexanderPigazzini: Well, there are almost none of those (i.e., harmonic, conformal coordinates). For example, in dimension $2$, this is equivalent to saying that the metric is what is called a Liouville metric, i.e., it admits a nontrivial quadratic first integral of its geodesic flow. Such metrics can be put in the local form $$g = h(x,y)(dx^2+dy^2)$$ where $h>0$ satisfies $h_{xx}+h_{yy}=0$ and, conversely, any such metric is a Liouville metric. (A nontrivial example is the metric on the general ellipsoid in $3$-space.)

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

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
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