What's the equivalent of the polarization vector of photons for gravitons? Spin-0, spin-1/2, and spin-1 particles have an associated quality expressing the spin structure of the field.
Spin-0 fields obviously have no extra quality. Spin-1/2 fields are expressed by spinors and gamma matrix algebra (the gamma matrices can be expressed in spin matrices). Spin-1 fields have a polarization vector.
If we extend this to massless spin-2 fields, in particular, the graviton, then what extra quality the field carries? I will be a tensor, but what will be the components of this tensor? Is it a metric tensor? Or, like the polarization vector for, say, photons, does it give the direction of momentum (=energy, in the case of photons) transfer?
 A: 
Spin-0 fields obviously have no extra quality.

When $\ell = 0$ we have $2\ell + 1 = 1$. This is why we say that spin-0 fields transform in the (trivial) 1-dimensional representation of the rotation group.

Spin-1/2 fields are expressed by spinors and gamma matrix algebra (the gamma matrices can be expressed in spin matrices).

When $\ell = 1/2$ we have $2\ell + 1 = 2$. This is why we say that spin-1/2 fields transform according to the two-dimensional irreducible representation of the rotation group.
Often we show the explicit representation using Pauli matrices. For electrons/positrons we actually package up two separate two-dimensional representations together into the four-dimensional gamma matrices.

Spin-1 fields have a polarization vector.

When $\ell = 1$ we have $2\ell + 1 = 3$. This is why we say that spin-1 fields transform according to the three-dimensional irreducible representation of the rotation group. Photons also have a bit of funny business due to the fact that they are massless, so we only end up with transverse polarization (only two polarizations to worry about).

If we extend this to massless spin-2 fields,

You're jumping over spin 3/2, but ok. Just want to point out real quick that we would expect there to be some spin 3/2 particles that would transform according to the irreducible 4-dimensional representation.

in particular, the graviton, then what extra quality the field carries?

It should transform according to a $2\ell + 1 = 5$ dimensional irreducible representation of the rotation group. So, yes, it will be more tensor-like than vector-like. The reason the 3-dimensional irreducible representation is vector-like is because we can repackage up the three $Y_{1,m}$ into something that looks like a vector. For the 5-dimensional representation, we will have components transforming like the five $Y_{2m}$.
