Why aren't gravitons spin 1? Expressing the metric as $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, assuming $h_{\mu \nu} \ll 1$ we can write the Einstein Hilbert action to leading order in $h_{\mu \nu}$ and quantize the linearized Einstein Hilbert action to construct the graviton field. Gravitons are spin 2 particles, which is easiest to see by noting that $h_{\mu \nu}$ has two indices. These enjoy a "gauge symmetry" corresponding to diffeomorphisms.
However, classically, gravity can be understood to be largely analogous to a gauge theory. The Christoffel symbol $\Gamma^{\alpha}_{\beta \mu}$ takes the place of $A^a_\mu T^a$ as the gauge field.
Note that $\Gamma^{\alpha}_{\beta \mu}$ has three indices, however the $\alpha,\beta$ indices can be understood as a matrix, much like the Lie algebra elements $T^a$ in Yang Mills theory.
If we quantize this field instead shouldn't we not be able to realize gravity as a theory mediated by spin 1 gauge particles?
(This should be especially true if we take the action to be the Kretschmann scalar, which seems to be equivalent to the Yang Mills Lagrangian.
$$
\mathrm{Tr}(F_{\mu \nu} F^{\mu \nu}) \leftrightarrow R^a_{b \mu \nu} {R^b_{a}}^{\mu \nu}
$$
However, this would obviously give a different theory than Einstein gravity.)
 A: Imagine a gravitational wave traveling along the $z$ axis. We can choose coordinates (aka fix a gauge if you prefer that language) where the components of the metric perturbation are
\begin{equation}
h_{\mu\nu} = \left(
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & h_+ & h_\times & 0 \\
0 & - h_\times & h_+ & 0 \\
0 & 0 & 0 & 1
\end{matrix}
\right)
\end{equation}
where $h_+$ and $h_\times$ are the amplitudes of the $+$ and $\times$ polarizations.
Under a rotation about the $z$ axis by an angle $\psi$, the time parts of the above metric perturbation $h_{00}$ and $h_{0i}$ are invariant, while the spatial part $h_{ij}$ transforms as
\begin{equation}
h_{ij} \rightarrow R(\psi)_{ia} h_{ab} R_{jb}(\psi)
\end{equation}
where $R(\psi)_{ij}$ is a rotation matrix
\begin{equation}
R(\psi)_{ij} = \left(
\begin{matrix}
\cos\psi & \sin \psi & 0 \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\end{equation}
Working through the algebra and using various trig identities, you can show this implies that $h_{ij}$ is invariant under a rotation by $\pi$, which is characteristic of a spin-2 particle. In general after a rotation you will find that the components of $h_{ij}$ are multiplied by functions of $2\psi$, another characteristic sign of a spin-2 particle.
The precise definition is in terms of helicity. An eigenstate of the helicity operator corresponds to a circularly polarized gravitational wave. A circularly polarized wave with $h = h_+=e^{\pm i\pi/2}h_\times$ will transform as $h \rightarrow e^{\pm i 2\psi} h$. The $\pm 2$ here are the eigenvalues of the helicity operator; a massless particle with a $\pm 2$ helicity eigenstates is a massless spin-2 particle.
