### Spinors

In §5, we argued that spacetime vectors are better represented as matrices than as column vectors. What, then, do column vectors represent?

In §6, we viewed 2-component column vectors as vectors in $\RR^n$ for appropriate $n$; imposing a normalization condition led us to regard these as points in $\SS^{n-1}$. But this is really nothing more than counting the (real) degrees of freedom.

A better indication of what these column vectors represent is given by the fact that they “square” to null vectors, in the sense of (15) of §6. These are *spinors*, more precisely *Weyl* or *Penrose spinors* [ 13 ].

How do we “rotate” a spinor? How do we rotate the square root of a vector? Comparing the Lorentz transformation (7) of §5 with the relationship (17) of §6 between spinors and (null) vectors yields the answer: A Lorentz transformation $\MM$ acts on spinors via \begin{equation} v \longmapsto \MM v \end{equation} How does the corresponding null vector transform? Equation (7) of §5 tells us that \begin{equation} vv^\dagger \longmapsto \MM (vv^\dagger) \MM^\dagger \end{equation} Consistency would require that this vector be the square of $\MM v$, so that \begin{equation} \MM (vv^\dagger) \MM^\dagger = (\MM v) (v^\dagger \MM^\dagger) \equiv (\MM v) (\MM v)^\dagger \label{compat} \end{equation} We refer to this condition as *compatibility*. Over $\HH$, it is automatically satisfied due to associativity, but over $\OO$ it is a nontrivial condition. However, all of the generators given in §5, do satisfy this condition. *We will always
assume that Lorentz transformations are described in terms of a sequence of
compatible generators.*

Recall that a generator such as $\RRR_z$, given in (12) of §5. in terms of a parameter $\theta$, describes a rotation through an angle $2\theta$. If $\theta=\pi$, corresponding to a rotation through a full circle, then a vector $\XX$ is unchanged. However, $\RRR_z$ reduces to $-\II$, rather than $\II$. In particular, a spinor is *not* left unchanged by this transformation; it requires a rotation by $4\pi$ (corresponding to $\theta=2\pi$) to do that! This makes sense; since the spinor $v$ is the square root of the vector $vv^\dagger$, it “rotates” half as fast.

But what good are spinors? It turns out that spinors represent particles with half-integer spin. But what does this mean?