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  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 in Theoretical Physics by Alexander Pigazzini (30 points) [ no revision ]
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
@AlexanderPigazzini: I wouldn't say that there is no interest, it's just of specialized interest. There was a time when Liouville metrics were of great interest indeed, but now they are mainly of interest to the integrable systems folks, not so much in general relativity.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Robert Bryant
Thank prof.! my question arises because I was looking for a possible interest in the study of totally umbilical immersions with this particular type of metric and thought to find was something in GR.

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

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