Metric tensor operator Please excuse me if the question doesn't make any sense. I'm only an interested layman.
One of the main principles of quantum physics is that a measurable is given by an operator $H$ acting over a vector space $V$. The measured quantity then would be an eigenvalue of $H$. If we consider the components $g_{ij}$ of the metric tensor $g$ as measurables, then it would be natural to assume the existence of some operators $H_{ij}$ such that the measurement gives
$$H_{ij}(x)v_x=g_{ij}(x)v_x$$
where $x$ runs over our universe manifold $M$, and $v_x$ belongs to a vector space $E_x$. Of course, $g_{ij}$ would be quantized and no longer interpreted as a metric tensor, but we can assume that the average value $\langle g_{ij}(x)\rangle$ behaves as a metric tensor (that is, $\langle g_{ij}(x)\rangle$ is governed by Einstein equations).
My questions are:

*

*Is there any research in quantum gravity following this direction or having an alike consequence?

*If there is, can we describe $H$ coordinate-freely? And what is $g_{ij}$'s interpretation?

 A: I believe what you are trying to do is not correct.
Thr values of $g_{\mu \nu}$ are not observables nor measurable.
Metric gives you Christoffel symbols and the Christoffel symbols give you the geodesic equations. Christoffel symbols are called connection.
This is not the only way to the to the geodesic equations. You can find the equation of motion with a totally different connect which is not related to any metric at all.
The connections which are related to the same geodesic corresponds to a class of connections which are related by projective transformation.
Metric is just a convenient tool to calculate the geodesic but not the only way. This means that it is not measurable or observable.
Quantization of gravity has already been done in string theory. The hard part about quantization of gravity is that the cross section of the interaction goes to infinity and it is not an easy fix. String theory provides one way to do it mathematically.
Finally, what you defined as H may not exist at all because GR equations are highly non linear while Schrodinger equation is linear. So you cannot use the same trick.
