general relativity - What is the relation between the metric tensor and the graviton?

In Zee's quantum theory in a nutshell, at the end of chapter I.10, he states that

the graviton is of course the particle associated with the field $g_{\mu\nu}$.

My understanding of quantum field theory is fuzzy, but I think I understood that each field assumes its values in an irreducible respresentation of the Lorentz group, the internal state space (or a representation of its universal covering, which is equivalent to a projective representation). Such representations are classified by their restriction to the $\text{SO}(3)$ or $\text{Spin}(3) = \text{SU}(2)$ subgroup, which is a spin representation and is classified by a single number called the spin. The graviton field has spin 2.

The metric tensor $g_{\mu\nu}$ is a section of the symmetric square of the (real) tangent bundle, which can be seen as a representation of the Lorentz group by the action $g\mapsto \Lambda g\Lambda^{T}$.

1. My first question would be: is this essentially the graviton field (as Zee seems to say, just to double check)?

The metric tensor is 6-dimensional, whereas a spin 2 field is 5-dimensional. Since for the Minkowski metric $\eta$ we have $\Lambda\eta\Lambda^T = \eta$, multiples of $\eta$ form a 1-dimensional subrepresentation, whose complement/quotient will be a 5-dimensional representation.

My other questions would be:

2. 
   

   Is this complement indeed the actual graviton field?

   
   


3. 
   

   What is its meaning or interpretation? Is it something like the non-flat part of the metric?

   
   



4. 
   

   If the preceding two were correct, does the fact that it is factored out of the graviton field mean that the graviton doesn't see unform dilations of spacetime? If so (or if not), what does that mean?

   
   

If my question is misguided, please point out why!

Answer

A spin-2 field in 4D is not five-dimensional. The standard spin-2 object transforms in the $(1,1)$-representations of the Lorentz group where the $(m,n)$-labels are half-integers labelling an equivalent representation of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, see the Wikipedia article on representations of the Lorentz group for more information. The spin of the $(m,n)$-representation is $m+n$ because the actual rotation algebra embeds diagonally into the two copies of $\mathfrak{su}(2)$ we are representing. A pure spin two field $h_{\mu\nu}$ is given by traceless and symmetric 2-tensor, which has 9 d.o.f. in 4D.

If we want to conceive of GR as a QFT (on ordinary flat space) in which we can speak of gravitons (an approach usually called Pauli-Fierz theory, we are essentially quantizing linearized gravity. I will not discuss the various inconsistencies and their fixes this approach needs to actually yield a somewhat consistent QFT), then you just write $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ and treat $h$ as the dynamical field whose associated particles are gravitons. However, not all components of $h$ are physical - it enjoys a gauge symmetry $h_{\mu\nu}\mapsto h_{\mu\nu} + \partial_\mu X_\nu + \partial_\nu X_\mu$ for an arbitrary vector field $X$ - this is the infinitesimal version of the diffeomorphism invariance of GR.

At the end of the day, the gauge symmetry means that two d.o.f. of $h$ remain as physical - corresponding to the actual graviton and its possible polarizations.

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JamalS's answer to a similar question explicitly counts the d.o.f. of $h$ and describes an alternative way to deduce the spin of the particle associated to $h$ in the quantum theory by looking at the Noether currents for the Lorentz symmetry applied to the part of the Lagrangian containing $h$.