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*.