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?

• 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

• The $4$ Bianchi identities reduce the number of independent Einstein equations to $6$ from $10$ and this is in correspondence with the reduction of the number of independent degrees of freedom of the metric from $10$ to $6$ due to the diffeomorphism invariance of Einstein equations. Bianchi identities and diffeomorphism invariance represent the same gauge redundancy---so they should simply be verified to be taking down the same number of degrees of freedom, should not be added. Correct me if I am misunderstanding. I am basing my understanding on S. Weinberg, Chapter 7, Section 4, pp. 161. – Feynmans Out for Grumpy Cat Dec 13 '18 at 4:24
As i can remember, for the case of classical electrodynamic the d.o.f counting start after the assumption of the Bianchi Identity, and at the end the desired result came all from the gauge freedom. In fact $\partial_{[\mu}F_{\nu\alpha]}=0$ only choose a suitable form for $F_{\mu\nu}$, for example $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$ After this choise from the four components of $A_\mu$ (our starting point for d.o.f counting) we must take into account the gauge freedom in choosing the four-potential, i.e. $$A_\mu\sim A_\mu+\partial_\mu\Lambda.$$ For example if we want $\partial_\mu A^\mu=0$, we must perform a gauge transformation with $\Lambda$ such that $$\color{red}{\Box\Lambda=-\partial_\mu A^\mu}.$$ At this point we are left with another possible gauge transformation (residual gauge freedom) such that $\partial_\mu A^\mu=0$ still holds, i.e. we must choose another $\Lambda$ with $$\color{blue}{\Box\Lambda=0}.$$ It is the gauge freedom that fix the correct numbers of degree of freedom for the photon $$4-\color{red}{1}-\color{blue}{1}=2.$$ In the case of GR the path is the same: Einstein equation, differential Bianchi equation for $R_{\mu\nu\alpha\beta}$ and the gauge freedom (space-time diffeomorphisms). Again you can perform two of these transformation and you have as you say $10-4-4=2$ d.o.f.