Recently, I received an innocent-looking question about spinors and fields. My correspondent noted that given a spin-2 particle represented by a spinor, it can be acted upon by the Pauli matrices; and a spin-3 particle represented by a vector can be acted upon by what he believed to be the Gell-Mann matrices. So what is being acted upon by the 5×5 matrices of a spin-2 theory, how do these matrices operate on tensors?

The following is a slightly edited version of what I wrote in reply.

The spin of a particle is an internal state. A spin-N particle has an internal state that can be represented by $2N+1$ base states: $N=1/2$ means 2 base states, $N=1$ means 3 base states, $N=3/2$ means four, $N=2$ means five, etc. So the internal state of a spin-$N$ particle is $(2N+1)$-dimensional (regardless of the number of dimensions in the geometric space of coordinates and perhaps time.)

The question is, what happens to a particle that has an internal spin state if I rotate that particle in geometric space. The group of rotations in $n$-dimensional geometric space is ${\rm O}(n)$, but this can be decomposed into more elementary operations described by "rotors". This is beautifully explained in Doran and Lasenby's Geometric Algebra for Physicists (Cambridge, 2003). The group of rotors in $n$ dimensions is called ${\rm Spin}(n)$, and this forms a double cover of ${\rm O}(n)$: for every rotation in ${\rm O}(n)$ there exist two different rotors in ${\rm Spin}(n)$.

The group of rotations in three dimensions is ${\rm O}(3)$. The corresponding group ${\rm Spin}(3)$ happens to be isomorphic to ${\rm SU}(2)$, but this is a coincidence; it is one of the so-called "accidental isomorphisms". This ONLY happens in three dimensions, where it is also true that ${\rm SU}(2)$ is isomorphic to the symplectic group ${\rm Sp}(1)$ (the group of unit quaternions).

(Also, in ONLY four dimensions, ${\rm Spin}(4)$ happens to be isomorphic to ${\rm Spin}(3)\times{\rm Spin}(3)$. Hence the fact that relativistic 4-component Dirac spinors in four dimensions can be decomposed into a pair of 2-component Weyl spinors, which is why the four-dimensional relativistic case can be constructed relatively easily once the three-dimensional non-relativistic case is dealt with. But I digress.)

EVERYTHING in my correspondent's e-mail is about doing things in three dimensions (i.e., nonrelativistic spin). We have particles with internal states that can be represented as two-component (for spin-1/2), three-component (for spin-1), etc. vectors. These are NOT spatial vectors but vectors in the abstract space of internal states (i.e., "quantum isospace"). Nonetheless, as we perform a rotation in actual, three-dimensional space, the internal state of the particle also changes.

Rotations in three dimensions can be written up using infinitesimal rotations around the $x$, $y$, $z$ axes. What happens is that when you rotate a spin-1/2 particle around the $x$, $y$, $z$ axes, its internal state (represented by two numbers) transforms under the corresponding Pauli matrix $\sigma_x$, $\sigma_y$, and $\sigma_z$, which also happen to be the infinitesimal generators of how rotors (i.e., spinors) themselves transform.

If you rotate a spin-1 particle, its internal state transforms under $J_x$, $J_y$, $J_z$, which also happen to be the infinitesimal angular momentum operators, which define how a spatial vector transforms under a rotation.

If the particle happens to be a spin-2 particle, its internal state is represented by a 5-component vector, and this vector transforms under a set of three 5×5 matrices $F_x$, $F_y$, $F_z$.

Note also that $J_x$, $J_y$, and $J_z$ are NOT the Gell-Mann matrices, nor does ${\rm SU}(3)$ have anything to do with the above. ${\rm SU}(3)$ does not correspond to any particular rotation; it has eight generators, acting on a three-component state vector. In quantum physics, ${\rm SU}(3)$ is used to represent quark "color", and the eight generators correspond to the eight gluons. This is the business of color charge, and it has nothing to do with spin. Also, ${\rm SU}(4)$ has nothing to do with spin-2. We remain firmly in three dimensions, with the rotation group ${\rm O}(3)$ and its double cover, ${\rm Spin}(3)={\rm Sp}(1)={\rm SU}(2)$.

So why do we call a spin-1/2 field a spinor field? Spin-1/2 means two base states; a spinor has two independent components. So a spin-1/2 field can be imagined as attaching a spinor to each point in spacetime. But this spinor "lives" in the "isospace" of internal states, nothing to do with spatial rotations. Same goes for spin-1: it means three base states, which happen to be the number of independent components of a 3-vector (or a unit 4-vector, or a 4-vector otherwise constrained, e.g., by a gauge condition). So we can imagine attaching a 3-vector or unit 4-vector to each point in spacetime, but once again, this vector "lives" in the isospace of internal states. For spin-2, the number of base states is 5. A symmetric rank-2 tensor in 3 dimensions has 6 independent components, one of which can be constrained by a gauge condition. A symmetric rank-2 tensor in 4 dimensions has 10 independent components, but this can be cut down to 5 by choosing a coordinate system and by enforcing a gauge condition. So we can imagine attaching a rank-2 tensor to each point in spacetime (again, the tensor "lives" in isospace).

I realize that this is rather difficult, and I'm not sure if my explanation clears things up or makes things even more confusing. I apologize if it's the latter; it only means that my own understanding of this topic is not as solid as I'd like to believe. In any case, I find the book I mentioned above, by Doran and Lasenby, a very enlightening read, along with Frankel's The Geometry of Physics.