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

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.

• 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. – Igor Khavkine Jul 4 '16 at 12:13
• 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. – Robert Bryant Jul 4 '16 at 12:43
• @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.) – Robert Bryant Jul 4 '16 at 13:19
• @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. – Robert Bryant Jul 4 '16 at 17:29