Understanding's Wikipedia's definition of a spinor

Question

I am trying to understand spinors from a mathematical view. I've seen similar questions on this website but I'm still unclear on what they are exactly. On Wikipedia they state:

Although spinors can be defined purely as elements of a representation space of the spin group (or its Lie algebra of infinitesimal rotations), they are typically defined as elements of a vector space that carries a linear representation of the Clifford algebra.

What confuses me is say we are working on space time so that $\mathbb{R}^4$ is our vector space. Going by the above passage, a spinor would be an element of $\mathbb{R}^4$ that carries a representation of the spin group, let us denote this pair as $(x, \rho)$ where $\rho$ is the representation. However $x$ is itself also a vector since $x \in \mathbb{R}^4$. So why are vectors and spinors referred to as two different objects? In other words, how does the representation play any role in describing the spinor itself (which as I understand is simply a vector in the underlying vector space)?

My current guess is that spinors are always to be taken as a pair consisting of a vector and a representation, for example $(x, \rho)$ above. Thus, spinors are actually also vectors, but the difference is that when you talk about rotations then spinors "rotate" differently than an ordinary vector in $\mathbb{R}^4$. However I am not sure if this is right since most textbooks do not describe anything along these lines. For example, no physics textbook describes a spinor as a pair $(x, \rho)$. Is this implicitly assumed?

Answer 1

Spinors are defined as vectors of certain representation spaces. Take $\mathbb R^n$ and its Clifford algebra $Cl(\mathbb R^n)$. Then its complexification $Cl(\mathbb R^n) \otimes \mathbb C$ is isomorphic (as an algebra) to either $\mathrm{End}(\mathbb C^{2^m})$, if $n=2m$, or to $\mathrm{End}(\mathbb C^{2^m})\oplus \mathrm{End}(\mathbb C^{2^m})$, if $n=2m+1$. Elements of any of the above $\mathbb C^{2^m}$ are called spinors when you restrict the action of $Cl(\mathbb R^n) \otimes \mathbb C$ to $\mathrm{Spin}(n)$ (which you can realize as a subgroup of $Cl(\mathbb R^n)$). I guess this is the definition Wikipedia is telling you about. All in all spinors are elements of very specific vector spaces with certain group (or algebra) actions.

Answer 2

If you are familiar with vector bundles then I think a way to answer this question is to consider vector bundles on a manifold that isn't $\mathbb R^n$.

Roughly speaking, a rank $k$ vector bundle over a manifold $X$ is a family of $k$ dimensional vector spaces $E_x$ for each $x \in X$, plus some extra conditions to get a nice topology on the disjoint union $E = \coprod E_x$. A section of a vector bundle is a map $s: X \to E$ where $s(x) \in E_x$.

An example is the trivial bundle where we just consider the cartesian product $X \times \mathbb R^k$. Sections of trivial bundles are equivalent to maps $s: X \to \mathbb R^k$. Another example is the tangent bundle. Sections of the tangent bundle are vector fields. In general, the tangent bundle of a manifold is not isomorphic to a trivial bundle.

A spinor on a manifold is a section of a certain vector bundle. One way of defining a spinor bundle is as a vector bundle with a map from the tangent bundle to its endomorphisms, which locally is Clifford multiplication.

It turns out that any vector bundle over $\mathbb R^n$ is trivial, and so the vector bundles over $\mathbb R^n$, up to isomorphism, are just classified by their dimension. What this means is that for any vector bundle $E \to \mathbb R^n$, we can find a bundle isomorphism

$$\phi: E \to \mathbb R^n \times \mathbb R^k.$$

Hence, if you have two rank $4$ vector bundles over $\mathbb R^4$ then they are isomorphic by the above, but this is not a particularly natural isomorphism.

So, over $\mathbb R^4$, vector fields and spinors are sections of isomorphic vector bundles and so in this sense you can say that they are the same. Indeed this is sometimes helpful.

For example, if we make the identification $\mathbb R^4 = \mathbb H$, then the positive spinor bundle $(S^+)$, the negative spinor bundle $(S^-)$ and the tangent bundle of $\mathbb H$ can each be identified with $\mathbb H \times \mathbb H$. Then the Clifford multiplication

$$ \gamma : T\mathbb R^4 = \mathbb R^4 \times \mathbb H \to \text{Hom}(S^+, S^-) $$

can be defined as

$\gamma(x, q)(x, p) = (x, qp)$. This map has the property that it is equivariant with respect to the appropriate $\text{Spin}(4) = Sp(1) \times Sp(1)$ representations. With this definition one can define things like the Dirac operator very explicitly.

For a general manifold, it won't be the case that all these bundles are isomorphic and so the spinors and vectors can't be identified, it is perhaps just a coincidence that on $\mathbb R^n$ it is possible to make such an identification.