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?

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