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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?

CBBAM
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    https://math.stackexchange.com/questions/3295410/is-a-spinor-an-element-of-the-spin-group-or-an-object-that-transforms-under-t/3295484#3295484 – Moishe Kohan Nov 09 '23 at 00:05
  • @MoisheKohan Your post is actually one of the ones I saw and I found it very helpful! I did not comment on that post as it was old so I am happy to see your reply here. If I have understood correctly, an ordinary "vector" and a spinor are both vectors in that they belong to a vector space, but the difference is how they are acted upon by $SO(3)$? That is, a "vector" is a vector whose action can be exponentiated to give back $SO(3)$, whereas a spinor is a vector whose action does reduce to $SO(3)$ when exponentiated? – CBBAM Nov 09 '23 at 00:53
  • @MoisheKohan Also this would imply that just specifying the vector $v \in V$ is not enough to distinguish whether it is a "vector" or a spinor. We would also need extra information about the vector space such as the representation of $SO(3)$ it carries (or, alternatively, its dimension). Have I gotten this all right? – CBBAM Nov 09 '23 at 01:02
  • When you write SO(3), do you actually mean SO(4) or SO(3,1)? While mathematically speaking SO(3) will work just fine, it is an unusual choice from the physics standpoint. In any case, both spinors and vectors are elements of a vector space, just the group action is different. I am not sure what it has to do with exponentiation. In physics, one usually starts with SO(3,1) and its "standard" representation on $R^4$; then a physicist would call an element of $R^4$ a vector. The corresponding spinor group is $Spin(3,1)\cong SL(2,C)$ has a faithful action on $R^4\cong C^2$. – Moishe Kohan Nov 09 '23 at 01:40
  • Elements of $R^4$ (or $C^2$), which, of course, are (real or complex) vectors, will be also called spinors since they are acted upon via a spinor representation of $Spin(3,1)$. Just knowing the dimension is not enough, you need to know the group action. Yes, it is confusing and no wonder that in physics literature these things are not well-explained. I can add a proper answer once the issue of SO(3) is resolved... – Moishe Kohan Nov 09 '23 at 01:41
  • @MoisheKohan I used SO(3) as I thought this is the symmetry assumed in non-relativistic quantum mechanics. But the relativistic case of SO(3,1) is ok with me too, I'm more focused on the math. After reading Hall's book on quantum mechanics I thought the dimension determines whether it's a vector or a spinor, as Hall says "Each type of particle has spin $l$, which is a non-negative integer or half-integer. The Hilbert space for such a particle is $L^2 \hat{\otimes} V_l$, where $V_l$ is an irreducible projective representation of SO(3) of dimension $2l + 1". – CBBAM Nov 09 '23 at 03:35

2 Answers2

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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.

Gibbs
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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.

Holmes
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