A field of metric tensors fully characterises the curvature of a vacuum space-time. (For example, the spacetime between some single point masses which are themself not part of the manifold) The metric tensor $g_{\mu\nu}$ is a $n$ x $n$ tensor in $n$ dimensions.
With $g_{\mu\nu} = g_{\nu\mu}$, the number of obviously independent functions reduces to:
4D - 10 (spacetime)
3D - 6
2D - 3
However, the curvature of a 2D plane, which is described in this list by a field of metric tensors with three independent functions, can be fully characterised by using only one single parameter: the altitude above ground, as it is done in colormaps of geographical/topological maps.
How comes that two degrees of freedom vanish by changing the form of the plot?
Is this somehow generalizable, can the number of independent functions be reduced also for the metric tensor of spacetime? How and to which number?
Does it help, therein, that the speed of light in vacuum is constant regardless of the curvature of spacetime?
EDIT: This question can be read as "How many scalar fields are necessary to fully characterise any possible tensorfield of metric tensors in 2, 3, 4... dimensions?"
