# Degrees of freedom of the graviton versus classical degrees of freedom

I have a puzzle I can not even understand. A graviton is generally understood in $D$ dimensions as a field with some independent components or degrees of freedom (DOF), from a traceless symmetric tensor minus constraints, we get:

1. A massless graviton has $D(D-3)/2$ d.o.f. in $D$-dimensional spacetime.

2. A massive graviton has $D(D-1)/2-1$ d.o.f. in $D$-dimensional spacetime.

Issue: In classical gravity, given by General Relativity, we have a metric (a symmetric tensor) and the Einstein Field Equations(EFE) provide its dynamics. The metric has 10 independent components, and EFE provide 10 equations. Bianchi identities reduce the number of independent components by 4. Hence, we have 6 independent components. However, for $D=4$, we get

1. 2 independent components.

2. 5 independent components.

Is the mismatch between "independent" components of gravitational degrees of freedom (graviton components) one of the reasons why General Relativity can not be understood as a quantum theory for the graviton?

Of course, a massive graviton is a different thing that GR but even a naive counting of graviton d.o.f. is not compatible with GR and it should, should't it? At least from the perturbative approach. Where did I make the mistake?

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Does this help? -- en.wikipedia.org/wiki/… . I don't see any quantum mechanics in the question at all. It seems purely classical. –  Ben Crowell Aug 16 '13 at 0:11
This question (v3) is also addressed in e.g this and this Phys.SE answers. –  Qmechanic Aug 16 '13 at 8:22
@BenCrowell Well, I have certainly some confussion, that is why I asked. GR is a classical field theory for the metric (without torsion). Gravitational field is provided with the aid of a metric. Therefore, I am interested in the number of independent components of the "graviton" due to the Weinberg's formulae I wrote above. However counting independent d.o.f. does not match what I believed to. –  riemannium Aug 16 '13 at 15:37