# Can spinors be explained or understood without group or representation theory?

Vectors, either as abstract mathematical elements of a vector space (in this case the definition of the vector is divorced from any notion of transformation), or as elements of tangent spaces on manifolds (in this case the definition of the vector implies a relationship between transformations on the manifold and transformations of the vector). Both cases can be understood without any references to group or representation theory. Is it possible to do the same for spinors? Or does every explanation of spinors require contact with group and representation theory? If so, then, (1) why? and (2) what does this mean for physical interpretation or intuition about spinors?

A bit more explication: Spinors are often described as: "The things that transform under the $$\text{Spin}(p, q)$$ (double cover $$SO(p, q)$$) group". Or, equivalently, Spinors are members of irreducible representations of the $$\text{Spin}(p, q)$$ group. Ok.. but we don't (or at least don't need to) define vectors in this way. That is, we don't say define vectors as "the things that transform under the $$SO(p, q)$$ group". In a manifold context we can define them as elements of the tangent space and then derive how the vectors transform. Mathematically, it seems like we should be able to do this for spinors (that is define them in some way and then observe based on that definition that they transform like representations of $$\text{Spin}(p, q)$$). If not, then I'm curious to know why not.

# IMPORTANT NOTE

This is a bit late in the game (I'm posting after the bounty has expired), but I would like to welcome new answers or edits to existing answers that address the question: "Can spinors be defined without group or representation theory". I think it's clear that a full understanding of spinors requires a discussion of group and representation theory. I would like to know if we can at least mathematically define them without reference to group/representation theory. If you think this warrants a new question then let me know in the comments.

• It is possible to conceptualise spinors as geometric objects, although some aspects of their behaviour may be less intuitive from this point of view. See math.stackexchange.com/q/2269714 Dec 8, 2023 at 10:35
• I'm not sure if you can even understand vectors without representation theory, so I'd be even more pessimistic about spinors. There are other ways of getting intuition about spinors, but they don't explain why we don't have additional kinds of objects, e.g., that only return to their original configuration after a rotation of $8\pi$ compared with $4\pi$ for spinors and $2\pi$ for vectors. Dec 8, 2023 at 22:47
• While it is possible to discuss spinors without explicitly mentioning representation theory, the fact is that all mathematics is interconnected; representation theory can be seen as a branch of group theory. So, would you like a description of spinors that makes no mention of group theory at all? Or a description of spinors that makes no mention of Lie groups and Lie algebras? What about such notions as homotopy and the fundamental group of a manifold? Dec 10, 2023 at 15:38
• @BrianBi A very valid point! Ordinary classical objects, of course, return to themselves when rotated by $2\pi$. (Imagine that you, the observer, rotate by $360^\circ$: you expect everything to be the same. It would be strange if you rotated through $360^\circ$ and everything changed sign! But that is exactly what happens with spinors, since they are quantum objects, and a spinor and its negative belong to the same ray.) Spinors return to themselves after a rotation by $4\pi$, but why? Why not $8\pi$, or something else? Group theory, and homotopy theory, must be invoked to explain that. Dec 10, 2023 at 15:52
• @printf yeah sure, an explanation without group theory would be great. Again, vectors can be understood without group theory (elements of the tangent space of a differentiable manifold). Dec 10, 2023 at 16:51

Yes. There is a subject called 'Geometric Algebra' which is an intuitive, geometric interpretation of Clifford algebras. Here, we extend vectors representing lines/lengths to higher dimensional entities: 'bivectors' representing planes/areas, trivectors representing volumes, and so on, along with scalars; we allow them to be added together, much like we add the real and imaginary parts of a complex number; and we introduce a 'geometric multiplication' which incorporates and unifies the scalar products, dot products, and cross products of vector algebra. The multiplication means that not only are these geometric shapes, they are also geometric transformations, like reflections and rotations.

It turns out that the spinors are equivalent to the even subalgebra of a geometric algebra - the elements corresponding to scalars, bivectors, quadvectors, etc. spanning an even number of dimensions, and leaving out the odd elements like vectors and trivectors. There are a couple of technical caveats about the relationship, but to begin with it is easiest to think of them that way.

Starting with 2D vectors, we get a geometric algebra with a basis consisting of one scalar, two vector, and one bivector element. The even subalgebra is the scalar and bivector components, and is identical to the complex numbers. These are 2D spinors (i.e. one complex dimension).

Starting with 3D vectors, we get a geometric algebra with a basis consisting of one scalar, three vector, three bivector, and one trivector element. (Note the 'Pascal's triangle' pattern in the numbers.) The even subalgebra is again the scalar and bivector components, and is identical to the quaternions, and also the Pauli algebra. These are 4D spinors (i.e. two complex dimensions) over a 3D vector space.

Starting with 4D vectors (with a mixed signature if we want to apply it to spacetime), we get a geometric algebra with a basis consisting of one scalar, four vector, six bivector, four trivector, and one quadvector element. The even subalgebra is the scalar, bivector, and quadvector components, and is identical to the biquaternions, and the Dirac algebra. These are 8D spinors (with four complex dimensions) over a 4D vector space.

Geometric algebra gives a much more geometrically intuitive picture, but isn't as popular as it probably deserves to be because it requires learning a new notation and formalism, and doesn't actually do anything the standard approaches can't do - they are mathematically equivalent. But it does let you handle spinors algebraically without requiring matrix representations, or coordinate systems. And if you are one of those people who are not content with abstract algebraic manipulation and want to know what it all means, it's an excellent approach.

For a fairly gentle introduction to geometric algebra, you could try Hestenes's lecture here. I'd advise skipping the first 7 pages of evangelism, but it gets better after that. Have a look at Francis and Kosowsky's The Construction of Spinors in Geometric Algebra for a discussion on spinors, or the copious literature on geometric algebra and spacetime algebra.

• Truly, very nice! Two questions that I'm sure are addressed in the references (I've started poking at them). (1) you say that the existence of geometric multiplication means elements of Clifford algebras can be interpreted as geometric transformations. Can you elaborate on this? (2) Can you elaborate on why we might expect even subalgebras of a geometric algebra to be relevant in physics? Dec 8, 2023 at 16:14
• Oh... I'm reading the Hestene's lecture. I think the answer to my second question (why we might expect even subalgebras to be relevant) is that the full clifford algebra has a lot of redundancy. Each k-vector has a dual n-k vector. It seems like moving to the even subalgebra does work to eliminates this redundancy in the description. Dec 9, 2023 at 5:23
• In what sense is this notion of spinor equivalent to the usual notion of "spinor"? Your even subalgebras in dimension $d$ have complex dimension $2^{d - 2}$, while the usual notion of Dirac spinor has complex dimension $2^{\lfloor d/2 \rfloor}$, with the Weyl spinors in even dimension having complex dimension $2^{d/2 - 1}$. You give examples exactly for $d$ small enough that we can match it with the usual Dirac or Weyl spinor, but how is this supposed to work in arbitrary dimension? Dec 9, 2023 at 13:54
• Why is this so upvoted? As far as I can see this entire 'subject' is just a verbatim rewriting of Clifford algebras but using geometric terminology to make it appear different... Or am I mistaken? Dec 17, 2023 at 22:58
• As you say, but in normal language, the answer says "spinors form a subalgebra of Clifford algebra". But this is well known and adds nothing - the only thing that is "new" is the bizarre and frankly unhelpful terminology. I assume this is the definition of spinors you learnt in the first place. Moreover, the answer does not even say how the spin group acts on this subspace, thus begs the question ("why are these called spinors?"). Of course, you can describe this action (representation), but you are merely repeating the introduction of any book or chapter on spinors. Dec 20, 2023 at 8:56

Yes, you can define spinors as a minimal left ideal in a Clifford algebra, however this is really just representation theory in disguise.

As discussed in the history section of the wiki, the idea for doing this comes from noticing that a wave function solution of the Dirac equation say in 2D $$\psi(x) = \begin{bmatrix} \psi_1 \\ \psi_2 \end{bmatrix}$$ can be rewritten as the first column in a matrix (of the same size as the 2D Clifford algebra matrices) $$\begin{bmatrix} \psi_1 & 0 \\ \psi_2 & 0 \end{bmatrix}$$ and becomes a minimal left ideal in the two (and three) dimensional Clifford algebra (given by the Pauli matrices) $$\{ I ; \gamma^i ; \gamma^i \gamma^j \ \ (i < j) \ \} = \{I,\sigma_x,\sigma_y , \sigma_x \sigma_y \}.$$ This is obviously just another way to talk about applying gamma matrices, or an operator like the Dirac equation $$\gamma^i \partial_i = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \partial_x + \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \partial_y,$$ to a wave function $$\psi(x) = \begin{bmatrix} \psi_1 \\ \psi_2 \end{bmatrix}$$ that lives in a representation of the Clifford algebra and is acted on by the $$\gamma^{\mu}$$ matrices. Thus, even if you try to ignore representation theory, in reality it is staring you in the face.

The explicit construction of spinors as minimal left ideals can be motivated as follows. In 4D for example, our Clifford algebra $$\gamma_{n}$$, $$n=1,...,4$$, can be written as $$\gamma_i = a_i^{\dagger} + a_i$$ and $$\gamma_{2+i} = i (a_i^{\dagger} - a_i)$$, $$i=1,2$$. A wave function is then $$\psi(x) = \begin{bmatrix} \psi_1(x) \\ \psi_2(x) \\ \psi_3(x) \\ \psi_4(x) \end{bmatrix} = \psi_1(x) |0 \rangle + \psi_2(x) a_1^{\dagger}|0 \rangle + \psi_3(x) a_2^{\dagger}|0 \rangle + \psi_4(x) a_1^{\dagger} a_2^{\dagger}|0 \rangle$$ Instead of interpreting the vacuum $$|0 \rangle$$ as living in some abstract representation, we can set $$|0 \rangle = a_1 a_2$$ since this satisfies $$a_i|0 \rangle = 0$$ just like a vacuum does, so that $$\psi(x) = \psi_1(x) a_1 a_2 + \psi_2(x) a_1^{\dagger}a_1 a_2 + \psi_3(x) a_2^{\dagger}a_1 a_2 + \psi_4(x) a_1^{\dagger} a_2^{\dagger}a_1 a_2$$ The subspace of $$a_i$$ satisfy $$a_i^2 = 0$$, thus they form an 'isotropic subspace', as do the $$a_i^{\dagger}$$. Thus we can build everything up from the existence of an isotropic subspace in a Clifford algebra. It's clear this is just representation theory in disguise.

The question is why such weird objects arise in physics, e.g. in 3 and 4 dimensions. In other words, this begs the question of why a Clifford algebra even arises in physics in the first place.

The real reason is we want to preserve say rotational or Lorentz symmetry, and in quantum mechanics we have to work with wave functions living in a(n irreducible) representation of these orthogonal groups, so we are seeking irreducible representations of the orthogonal groups, which these provide.

For example, compare this to what Dirac does when he takes the Klein Gordon equation $$(\partial^2 + m^2)\psi(x) = 0$$, and assumes some $$(\gamma^{\mu} \partial_{\mu} + m) \Psi(x) = 0$$ exists which 'squares' to Klein-Gordon, forcing him to assume $$\{\gamma^{\mu},\gamma^{\nu}\} = 2 \eta^{\mu \nu} I.$$

Here he is implicitly assuming the existence of a linear representation of the Clifford algebra, where the spinor $$\Psi(x)$$ is thus acted on by the $$\gamma^{\mu}$$ matrices.

In order for this to work he has to define the action of an orthogonal transformation $$S(\Lambda)$$ on $$\Psi(x)$$ which acts via direct multiplication of $$S(\Lambda)$$ on $$\Psi(x)$$, while it acts on the $$\gamma^{\mu}$$'s via conjugation $$\gamma'^{\mu} = S \gamma^{\mu} S^{-1}$$ (which reduces to a commutator in the Lie algebra), i.e. the $$\gamma^{\mu}$$ transform as a vector representation. Everything works very nicely and fits into place with little effort, but representation theory underlies it all.

Thus, spinors are inherently tied to representation theory, and on a fundamental level they arise from wanting to construct 'fundamental representations' of the orthogonal groups, and this is how Cartan discovered them in 1913, the idea of which is given in:

• 1: Ramond, 'Group Theory, A Physicists Survey'
• Your construction of $\gamma_i = a_i^{\dagger} + a_i$ and $\gamma_{2+i} = i (a_i^{\dagger} - a_i)$ in 4D seems to work only with Euclidian metric. How do you define $a_i^{\dagger}$ and $a_i$ in Minkowski space-time? Dec 20, 2023 at 18:28
• @MadMax is your concern addressed in this part of the Eigenchris video on how ideals can be constructed from idempotents that are constructed from nilpotents: youtu.be/bfuxhLnDzlQ?t=1264? Basically the idea is you may need $a_i = \gamma_i \pm \gamma_{i+1}$ or $a_i = \gamma_i \pm i \gamma_{i+1}$ where the relative factor of $i$ is needed to account for $\gamma_i = \pm 1$ depending on the metric signature? Dec 20, 2023 at 21:46
• @MadMax As an example, starting from ${\rm U}(1)$, with $a^{\dagger}$ and $a$, where $\{a,a^{\dagger}\} = 1$, if we set $\gamma^1 = a^{\dagger} + a$ and $\gamma^{1+1} = \gamma^2 = i (a^{\dagger} - a)$ we get ${\rm SO}(2)$, if we set $\gamma^1 = i (a^{\dagger} + a)$ and $\gamma^2 = i(a^{\dagger} - a)$ we get ${\rm SO}(1,1)$. Jagerber My answer addresses your edit, the minimal left ideal definition is technically independent of rep theory, I'm not sure what the problem with that is. I will remove this comment in a while to keep things clear/focused I ask you both to do the same, thank you. Dec 20, 2023 at 22:05

OP is asking why we need representation theory to understand spinor bundles but not for vector/tensor bundles.

This is similar to asking why we need the full machinery of Riemannian integration to understand the area below the function $$f(x)=x^2$$, but not for the function $$f(x)=x$$. The short answer is: linear functions are so simple that you can calculate their area without reference to Riemann partitions etc. But generically, areas do require these notions. For simple functions you can "get away" with this, only because they are so simple you can phrase everything using very primitive notions. But if you want to generalize and unify areas to arbitrary functions, you need to use the correct language.

The formal definition of a spinor is an element of a special type of principal bundle. And general associated vector bundles are defined using the language of representation theory. A tensor bundle is just another special type of vector bundle, just one so simple that you can "get away" with it and define it without (explicit) use of representation-theoretic language. But spinor, and more general, bundles, do require the full machinery of representation theory. Which brings me to the actual answer to the question in the OP:

## It is not possible to understand spinors without representation theory.

People will mumble things about a Dirac belt and the like, and you will be no closer to getting any useful information out of it.

You cannot understand spinors without representation theory. And, for that matter, you cannot understand vectors/tensors/etc without it either!

The problem is that, in physics, we use the word vector (and, more generally, tensor) with multiple different meanings. The most basic definition is that a vector is an element of a vector space, but this is rarely what we mean when we say vector in a physics context. For example, I could formally take any two variables, say, pressure and temperature, and define a formal vector space with coordinates $$(P,T)$$. This is, according to the basic definition, a vector. But a rather useless one.

(As a side note: One could take the rigorous approach to thermodynamics, in which case entropy is for example a one-form, an element of the (co)tangent bundle, as in your definition. In this formalism, entropy and other state variables are all (co)vectors. This is more useful than the silly definition above, but it is still not what we usually mean by vector in physics.)

In physics, when we say vector, we most likely mean "an object that transforms under the vector (also known as fundamental) representation of some group". For example, when we say that the electromagnetic potential $$A^\mu$$ is a vector, we don't just mean "an element of an abstract vector space" (which, any of its components would also be a vector in that sense), but instead, what we mean is that this object transforms as $$A\mapsto \Lambda A$$ under Lorentz transformations. In other words, when we say $$A^\mu$$ is a vector, we really mean that it is a Lorentz vector. Similarly, when we say that the electromagnetic field $$F^{\mu\nu}$$ is a tensor, what we really mean is that it is a Lorentz tensor, not just any tensor (e.g., a tensor is also an element of a formal vector space, so $$F^{\mu\nu}$$ is also a vector, just not a Lorentz vector).

In this sense, the abstract notion of vector does not require representation theory, but the way it is used in physics very much does. The reason is that, by "vector", we almost never mean "an element of an abstract vector space", instead we mean a specific representation of a certain group (usually Lorentz, but not always; for example you will also sometimes hear that quarks are vectors of color $$SU(3)$$).

Spinors are no different. There is an abstract definition (an element of a spinor bundle, some words about not factoring through a double cover, etc). But in physics, we specifically mean a Lorentz spinor, meaning an object that transforms as the spinor representation of the Lorentz group. Being a representation, spinors are also vectors. But they are a very specific type of vectors, they are elements of the spinor representation space of Lorentz.

Note that, unlike vectors, spinors are an object that only makes sense for the orthogonal group. The unitary group $$SU(N)$$, for example, does not have spinor representations (modulo, perhaps, low-rank isomorphisms if you want to get pedantic). This is why spinors require a Riemannian metric, otherwise the structure group of the manifold would be $$GL(n)$$, which doesn't have spinor representations.

The very definition of (Lorentz) spinor requires representation theory, any claim to the contrary is useless wordplay.

## But you can rephrase everything in a more geometric language, if you will.

If you take a generic manifold, you can define the tangent bundle and various Cartesian powers thereof. You can even restrict to smaller bundles in a consistent way by imposing symmetry/anti-symmetry constraints. For example, a rank-2 tensor can be decomposed into its symmetric and anti-symmetric parts in a way consistent with changes of coordinates.

In rep theory language, a manifold has a natural $$GL(n)$$ structure and therefore there are natural bundles which transform as representations of $$GL(n)$$. The symmetry/anti-symmetry of tensors is just the statement that the irreducible representations of $$GL(n)$$ can be obtained by applying suitable symmetrizers to the fundamental representation, this is the old Schur correspondence between irreps of $$S_n$$ and $$GL_n$$. You don't need to phrase this in rep theoretic language, but why would you not? This language makes it clear that there are no further constraints you could impose in a way consistent with generic changes of coordinates, since these representations are all the irreps of $$GL(n)$$.

If you have a non-degenerate metric, the $$GL(n)$$ structure group can be restricted to an $$O(n)$$ group. You can now further decompose your tensors into its trace, symmetric-traceless, and anti-symmetric parts, and this is still compatible with metric-preserving changes of coordinates.

You can again rephrase this in rep theoretic language: all irreps of $$O(n)$$ can be recovered from irreps of $$GL(n)$$ by imposing suitable tracelessness constraints.

A key point of $$O(n)$$ is that, unlike $$GL(n)$$, this group is not simply-connected, so there are more options for the various bundles you can consider. In particular, $$O(n)$$ has a (pair of, actually) canonical universal-cover, $$Pin(n)$$. A spinor field is an element of the corresponding $$Pin(n)$$ bundle, i.e., it is a section of a double cover of your structure group.

This is the definition of spinor in geometric language. You can phrase this in rep theoretic language by saying that $$Pin(n)$$ has representations that do not give representations of $$O(n)$$, they are projective representations instead. This is the exact same claim as above, just with different words. You can define spinors without explicitly using representation theory language, but the theory is 100% there, you are just using different words.

## Just don't reinvent the wheel.

At the end of the day, the common property of any definition of spinor is that it picks up a minus sign under a $$2\pi$$ rotation. This is a statement about how spinors transform. Representation theory is precisely the language of "how things transform". You can try to use a different set of words in your definition of spinor, of course. But we already have a theory that describes how things transform, representation theory, so every definition of spinor will, at least implicitly, use rep-theoretic notions, even if you decide to use a different terminology.

• I don't buy that we need rep theory to understand physics vectors. Take a spacetime manifold with a Reimannian Minkowski metric. If you perform any change of coordinates on this metric the tangent vectors (physics vectors) on the manifold will transform as contravariant vectors. The Lorentz coordinate changes are distinguished because they are metric preserving in this manifold, so the vectors on this manifold are Lorentz vectors. Dec 13, 2023 at 14:42
• That is, we don't define Lorentz vectors as objects transform $\boldsymbol{A} \to \boldsymbol{\Lambda} \boldsymbol{A}$ under Lorentz transformations, we define Lorentz vectors as elements of tangent spaces to Reimannian manifolds with Minkowski metrics and then this transformation rule is derived. Maybe I should say Pseudo-Reimannian manifold if we're talking about Minkowski metrics. Dec 13, 2023 at 14:43
• @Jagerber48 There is representation theory all over your two comments. You may choose not to use the words representation theory, but it is 100% there. If you want to use geometric language, you could say: a metric gives you a canonical restriction of the structure group $GL(n)\to O(n)$. And $O(n)$ is not simply connected so there is a canonical cover $Pin(n)$ which is simply connected. A spinor is an element of the $Pin(n)$ bundle, i.e., there is a 2-to-1 canonical map to $TM$. This is of course just a rephrasing of the rep theory statement about reps of O(n) vs Pin(n). It is equivalent. Dec 13, 2023 at 14:50
• Sorry, I don't see how my definition of a physics vector is requiring representation theory. I just define a physics vector as a tangent vector on a differentiable manifold. No representation theory there right? I now make the observation about my physics vectors which is that if the manifold has a Minkowski metric then Lorentz transformations are metric preserving and transform vectors contravariantly. The fact that Lorentz transformations are metric preserving is maybe considered a group theoretic statement, but the contravariant transformation is again just differential geometry. Dec 13, 2023 at 15:00
• But again, the Minkowski and Lorentz stuff (1) arguably doesn't involve group/rep theory and (2) was not required for the definition of a vector. Now, of course, I only give a definition for a vector here. I don't give a definition for a spinor yet, hence my question. Dec 13, 2023 at 15:01

There is this way of introducing spinors geometrically and another way (mine) :

On the sphere of radius $$R$$ , we take a large circle (geodesic) and two others which are orthogonal ($$S_{1}\lor S_{2}$$) see diagram Each point on the great circle corresponds to two points on the blue and red circles by 'spherical' projection', the rays of the red and blue circles are $$\rho_{1}=R \cos(\theta),\rho_{2}=R \sin(\theta)$$ which can be represented mathematically by: $$\;\psi_{1}=\rho_{1}e^{i\alpha} ,\psi_{2}=\rho_{2}e^{i\beta}$$,therefore each point on the sphere (great circle) can be represented by two complex numbers which are the components of a spinor $$\phi=(\psi_{1},\psi_{2})$$.

The sphere as a surface is 2d, the large circles as straight lines (1d), we want to provide the sphere with a specific reference and we know that the complex numbers are 1d, each straight line of the 2d sphere can be represented by two 1d coordinate vectors, i.e. $$(\vec{i} \equiv e^{i\alpha}, \vec{j} \equiv e^{i\beta})$$.

Each point on the 'line' of a sphere of radius 1 is written in this reference frame $$(cos\theta e^{i\alpha}, sin\theta e^{i\beta} )=\phi$$,and we can define spherical vectors.

To establish their transformations A course of higher mathematics III part 1, linear algebra,V.I.Smirnov, from an algebraic point see: Geometry lessons, Vth semester, Group and Lie Algebras, M.Postnikov,....

• For more details see: The Theory of Spinors - Cartan, Elie Dec 10, 2023 at 9:12

I am a math student just learning QFT, so I will try to explain from my mathematical point of view as to why spinors should exist. First I would like to say that a vector space and what physicists call vectors are two different things: a vector space is something like $$\mathbb{R}^n$$, while what physicists call vectors is the tangent bundle of a manifold. To clarify this, let's begin by reviewing what (and why) vector bundles are. In fact spinors are also just vector bundles, and thus mathematically aren't more special than any other vector bundle.

A rank $$n$$ vector bundle $$E$$ on a manifold $$M$$ is a vector space of dimension $$n$$ attached to every point on a manifold. A good way to think of this is a parametrized family of vector spaces. For later use, a section of a vector bundle is an assignment of a vector at the vector space of every point that "varies continuously". This was a confusing thing for me to get used to because not all vector bundles are trivial (i.e. $$E$$ is a product $$M\times \mathbb{R}^n$$). Some trivial examples are a(n infinite) cylinder (rank 1 vector bundle over a circle) and $$S^2\times\mathbb{R}^2$$ (a trivial rank 2 bundle over a sphere; harder to visualize). Nontrivial examples are the open mobius strip (a "twisted" family of rank 1 vector spaces over a circle; each fiber should technically be extended infinitely) and the tangent bundle to the sphere (a "twisted" family of rank 2 vector spaces over a sphere). The latter example can be visualized as the tangent plane attached to each point of $$S^2$$ embedded in $$\mathbb{R}^3$$. This twisting is harder to imagine, but is guaranteed by the hairy ball theorem, which states that a sphere has no nonvanishing global sections (vector fields) while (you should check) every trivial rank n vector bundle admits n everywhere linearly independent sections. In fact, it can be easily shown (try it) that there is no nonvanishing global section of the mobius bundle as well. In minkowski space (the context of basic QFT), only trivial vector bundles are needed (any vector bundle over a contractible manifold can be trivialized), but it is good to have nontrivial examples of vector bundles to understand that they are richer than vector spaces.

First question: why should we care at all about vector bundles? We learn from quantum mechanics that particle states are described by vectors in some vector space which describes the space of states. In QFT, we have quantum fields which associate a particle state to every point in space, so the space of all states is described by a vector space at every point i.e. a vector bundle! A particular state is described by a section of said vector bundle (a state at every point). The way to think of it is that sections are essentially $$\mathbb{R}^n$$ (or $$\mathbb{C}^n$$) valued functions, but more precisely, this is only true locally. If it was true globally, the tangent bundle of the sphere would admit a nontrivial tangent vector field: e.g. the field "(1,0)". It is true globally iff the vector bundle is trivial, so a vector bundle is to be thought of as a parametrized function where the codomain may change point to point, but every codomain is isomorphic.

Now, we have not actually stated what vector bundle we're working with. One example of a vector bundle that is canonically associated to any manifold is the tangent bundle. This is the bundle that "locally looks like the manifold"; I won't go into details of defining this because people prefer their own of a multitude of definitions which are at the end of the day equivalent. I have no personal qualms to picturing a literal tangent plane of the manifold embedded in some high dimensional $$\mathbb{R}^N$$, but I've found the best way to think of the tangent vectors at a point is the possible "infinitesimal deformations" of the point. You can skip this next part of the paragraph, but I'll explain vector bundles some more. There are ways to get new bundles out of old through operations which give a new vector space out of old ones such as the dual of a vector space or the direct sum of vector spaces or the tensor product of vector spaces. Performing these operations "fiberwise" on vector bundles gives new vector bundles. In particular taking tensor powers of the tangent bundle give what physicists call "tensors" and taking duals give "covectors", and combining give arbitrary $$(k,l)$$ tensors. Tensor products correspond to keeping track of the states of multiple particles. As an example, on the sphere, a generic section of the tangent bundle vanishes at 2 points, a generic section of the tensor square of the tangent bundle vanishes at 4 points, while the dual of the tangent bundle has no nonvanishing global sections (follows from the Chern class)!

Now, on $$\mathbb{R}^4$$, I mentioned that any vector bundle can be trivialized, so really they are classified by their dimension, and you don't have this weird "twisting" phenomenon. However, we have more structure at hand, namely, symmetries! We postulate that whatever physics we have (whatever vector bundle we choose) should be invariant under Lorentz symmetry i.e. sections which are solutions to EOM should be sent to other solutions. To give a very simple example of why this is useful, take the differential equation for a hydrogen atom. This equation is rotationally symmetric, and thus if $$\psi(x)$$ is a solution and $$R$$ is a rotation, $$\psi(Rx)$$ is also a solution. This means that $$SO(3)$$ acts of the space of solutions i.e. the solutions form a representation. Then, the irreducible representations ("states") can be understood by studying the abstract representation theory of $$SO(3)$$ or its Lie algebra $$so(3)$$ - these complex representations are classified and allow you to quantize the states (or at least understand the possibilities) without ever having to solve the differential equation!

Now what do we mean by this invariance in general? One guess would be if a section $$\psi$$ (essentially a function) satisfies the equations of motion, then if $$x$$ gets sent to $$x'$$ under a Lorentz symmetry, we should have $$\psi(x)=\psi(x')$$. However, there is a big problem with this! $$\psi(x)$$ lives in the vector space over $$x$$ while $$\psi(x')$$ lives in the vector space over $$x'$$ and these cannot a priori be compared! Perhaps you think this is not a problem - as our vector bundle is trivial, each fiber is isomorphic and can be compared. Unfortunately, it is not so simple since the trivialization is not canonical. In other words, we can "twist around" the fibers while keeping the vector bundle trivial, but the comparisons of fibers are different. Imagine a trivial rank 2 vector bundle over a circle (an "infinite square torus") and "twisting it without tearing", this changes the trivialization. Also, if we could always compare fibers, we could find a nonvanishing tangent vector field on the sphere - take some vector at a single point and move it around by the transitive (takes any point to any other) action of rotations on the sphere - clearly this doesn't work!

So what is the solution? We need a way to compare vector spaces over different points which plays well with our group acting on our space in some way. Imagine a 1 parameter subgroup e.g. some rotation parametrized by angle or a boost parametrized by whatever you call the parameter. A point $$x_0$$ will trace out some path $$C(t)$$ where $$t$$ is the parameter and $$C(0)=x_0$$ (for a rotation, a part of a circle; for a boost, a part of a hyperbola). Now this part may get a bit confusing. Fix some trivialization of the vector bundle under consideration (as yet unspecified) i.e. a way to compare vector spaces at different points. Now, there is a "nice" comparison which plays well with our group action, but as explained, this may not be the trivialization that is chosen. So suppose that in our initial trivialization, we have a section (i.e. function to $$\mathbb{C}^n$$ for some $$n$$) $$\psi(x)$$. We said that perhaps $$\psi(C(t))\neq \psi(x_0)$$ (in this trivialization), but at the very least, $$\psi(C(t))=A(t)\psi(x_0)$$ where $$A(t)$$ is an $$n\times n$$ complex matrix depending on $$t$$ (as any two vectors are related as such). Interpretation: imagine the vector space over $$x_0$$ "twisting" along $$C(t)$$ as $$t$$ varies and our point traces the path; $$A(t)$$ measures this twist. Now technically $$A(t)$$ is not fully specified as we only know the output of $$\psi(x_0)$$, but by choosing different linearly independent sections, by imposing that the comparisons should be linear, $$A(t)$$ can be fully determined.

Here's the important point! $$A(s)A(t)=A(s+t)$$ as moving (and twisting) $$s$$ along and then $$t$$ along should be the same as moving (and twisting) $$s+t$$. Thus, $$A(t)$$ forms a representation of the one parameter subgroup on the fiber $$\mathbb{C}^n$$ of $$x_0$$ (which is identified with the fibers of $$C(t)$$ as it moves around)! This actually holds for all $$x\in \mathbb{R^4}$$ simultaneously, and the representation must be the same on each fiber by continuity. More generally, we are allowed to rotate and boost in some direction and then rotate and boost in another direction, and still maintain symmetry. Tracing through, the fiber $$\mathbb{C}^n$$ furnishes a representation of the Lorentz group!

As explained, the representation is essentially found by "dragging along the vector space" i.e. if we know how it works infitesimally, we can recover global behavior (the semigroup property above comes into play here). Thus, we can study a representation of a Lie group by the representations of its Lie algebra, which in this case turns out to be isomorphic to $$su(2)\bigoplus su(2)$$. This can be classified, and it gives you integral spin representations, and half spin representations - I consider this to be something you just "have to compute" though, I don't have much intuition as to why these are all the representations. Then any physics we want to do MUST arise as one of these. Hence, in this light, the spinor representation is, as a vector bundle (vector space), nothing more than $$\mathbb{C}^4$$, the same as the tangent bundle (tensored with $$\mathbb{C}$$)! What is different is the way that the Lorentz group acts on it.

Now, integral spin representations (I believe) correspond to the tangent bundle "vectors", its tensor powers "tensors", and duals (if $$G$$ has a representation on vector spaces, it induces representations on tensor products, duals, and direct sums). Note that these bundles actually naturally furnish a representation of the full diffeomorphism group of the manifold. The half integer representations correspond to spinors.

Now I finally get to the question that I think you're mainly wondering, why do these spinors not return to their initial position after rotating around $$2\pi$$? The answer (a bad one) is why should it! Explained better, while the vector bundle itself has no twisting (like the mobius band), the representation does. As we move along the circle traced out by $$x_0$$, the fiber vector space twists around, and when we get back, the vector space isn't the same as it started. This is a phenomenon known as holonomy - we can only control infinitesimal symmetry (by studying the Lie algebra), but global symmetry is not guaranteed. This is the same phenomenon as when you carry an orthonormal basis along a Mobius band, it reverses orientation when you get back, even though it always varied continuously. In fact, it's a miracle that it even does get back to where it started when we rotate $$4\pi$$. Mathematicians call "miracles" theorems.

It is a theorem (I believe) that representations of real Lie algebras are in 1-1 correspondence with representations of the corresponding simply connected Lie group. To explain, the Lie algebra structure on the tangent space to the identity of a Lie group determines the group structure locally, but global structure may differ via the fundamental group. A simply connected Lie group $$G$$ with the same Lie algebra as $$G'$$ is a covering space of $$G'$$, and a representation of $$G$$ is a representation of $$G'$$ iff it is trivial on "deck transformations". It is a theorem (again I believe) that every Lie algebra has a unique simply connected Lie group. Now, let's analyze $$su(2)=so(3)$$ (one of the factors of the decomposition of the Lorentz group comes from the subgroup of rotations). We know $$SO(3)$$ is a Lie group with Lie algebra $$so(3)$$, but is it simply connected? No! It turns out $$SU(2)$$, which is topologically isomorphic to a 3 sphere, and hence simply connected, forms a double cover ($$2:1$$ surjective map) of $$SO(3)$$. Thus, $$\pi_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$$. Thus, the representations computed from the Lie algebra correspond to $$SU(2)$$ (or $$Spin(3)$$ if you like), and this also explains why you get back to the identity after $$4\pi$$ instead of $$2\pi$$ - the half spin representations correspond to representations of $$SU(2)$$ which are not representations of $$SO(3)$$, it also explains why $$4\pi=2\times 2\pi$$ is the only phenomenon - $$|\pi_1(SO(3))|=2$$, and not, say, 6. So at most, any nontriviality (holonomy) must be resolved after 2 rotations.

Now why is physics only invariant under local symmetry and not global? I don't have a good answer other than observation I guess. You could postulate symmetry under $$SO(3)$$ rather than $$so(3)$$, but this would not yield accurate physics. Finally, there are many ways to get at the representations of $$su(2)$$. As explained, the Clifford algebra is one such way. Another way is a lovely argument with the eigenvectors of a set of generators of $$su(2)$$. I believe a third way is to have $$SU(2)$$ act on $$\mathbb{C}^2$$ with unitary coordinates $$(x,y)$$ and look at the induced homogenous polynomials of degree $$n$$ in $$x$$ and $$y$$ (which turn out to form irreps). The second way is the only way I can see that guarantees you have everyting, the first is neat, and the third has the advent of explicitly giving the group matrices instead of the Lie algebra matrices.

To summarize, why spinors exist is perhaps the wrong question; the answer at the end of the day is representation theory. The question that should be asked is why does representation theory appear in physics? This is because things are linear (QM), giving us vector spaces/bundles, and we have symmetries, thus giving us representations. As a good mathematician, let's change parameters: suppose we lived on a torus and had rotational symmetry along each axis - then we would instead care about representations of $$S^1\times S^1$$ or its Lie algebra, which is $$\mathbb{R}^2$$ (with trivial Lie bracket). If instead we lived on a sphere and demanded rotational symmetry, we would study representations of $$SO(3)$$ or $$so(3)$$ rather than the Lorentz group. Rather, if we consider the sphere as $$\mathbb{CP}^1$$ and we lived in a funky world where physics was left invariant by conformal transformations, we would study the representations of $$PGL_2(\mathbb{C})$$ and its associated Lie algebra. All of these topological spaces have nontrivial vector bundles as well which adds another level of complexity! Interestingly, we certainly don't know the topology of our universe, so these considerations may actually be fruitful one day.

• This answer is offering a complicated bundle example in order to find a reason for us to care about wanting reps of an orthogonal group (like Lorentz). Once we want reps of the ortho group, the answer resorts to the usual: 'look, for 4D Lorentz reps, we see $SU(2) \times SU(2)$, thus spinors!', and pointing out global aspects of this. This is not going to work so easy in 26D for example. Obviously this is just saying 'rep theory leads to spinors' and is begging the question as to why any orthogonal group admits these new crazy 'spinor' reps in any dimension, and my answer explains that. Dec 13, 2023 at 7:49
• Well yes, I am precisely trying to answer why we care about representation theory. After that, it's just a matter of actually doing the representation theory, which will involve some level of algebra at the end of the day, but is not the conceptual core of why it arises in physics imo. At the very least, you can always expect "spinor-like" reps without doing anything because $|\pi_1(SO(n))|=2$ for $n$ at least 3 (fiber bundle argument), so you can expect reps of the Lie algebra which require two rotations to trivialize without any work. The details of this rep will obviously involve more work. Dec 13, 2023 at 15:07

Not an expert on the topic, but here's an apparent difficulty with "construct and derive spinor transformations", maybe someone can resolve it or elaborate, interested to see.

First, some notes about the "local" picture might make the geometric algebra construction more concrete. Some of the answers imply the connection between rep theory and the Clifford/geometric bundle is a coincidence or trick. Actually it is natural and straightforward.

The reason for the connection is the bivectors: the set of bivectors is precisely the Lie algebra of rotations. Obviously the exponential of the Lie algebra is locally equivalent to the group, in this case locally Spin(p,q) ie locally SO(p,q). Spinors as even multivectors are the completion of the exponentials of bivectors. (This already shows why the multivector approach makes some sense and must have valid dimensionality.)

In particular suppose $$G$$ is the geometric (real Clifford) algebra built on vector space $$V=\mathbb{R}_{p,q}$$ with inner product of signature $$(p,q)$$. The space of bivectors is closed under the commutator of the geometric product, $$[B,B'] = B B' - B' B$$, and forms the Lie algebra $$so(p,q)$$.

It is with reps of this $$so(p,q)$$ that we're concerned. It has two natural ways to act on $$G$$: the "two-sided" action that rotates vectors (rotation in the plane of $$B$$ by angle $$|B|$$ for simple bivectors), $$v' = e^{-B/2} \; v \; e^{B/2},$$ and the "one-sided" one that corresponds to spinors, $$s' = e^{-B/2} s .$$ This second is natural when one identifies the vector rotation as $$v' = s v s^{-1}$$, the spinor rule is rotating a rotation.

Two-sided extends to a group action of $$SO(p,q)$$. One-sided doesn't, eg because $$e^{-2\pi B/2}=-1$$ (unit $$B$$). There's the $$2\pi$$ vs $$4\pi$$ vs $$8\pi$$... it's because the vector rotation has a half angle. One-sided extends to the Spin(p,q) rep.

In light of that it seems fair to say that embedding Spin(p,q) in Clifford(p,q) is not a trick but pretty fundamental. References eg Lie groups as spin groups or Doran+Lasenby. Especially D+L 3.3.2 for how these type of spinors appear in physics, 2.7.4 for rotations and $$2\pi$$ $$4\pi$$, 11.3 for spin groups. And 3.7 of Hall.

So can you construct spinor fields without rep theory and derive their transformations? I think the answer to the second part is basically no.

You can definitely construct them. (On a $$p,q$$ semi-Riemannian manifold.) Vector fields are sections of the tangent bundle. Tensor fields are linear functions of vectors (and dual vectors). In both cases the fact that these objects are geometrically well-defined (coordinate independent) determines the "transformation laws" of their components under change of basis, no rep theory needed.

Similarly, each tangent vector space can be extended to a geometric algebra. Multivectors of this "tangent Clifford bundle" have the same status as tangent vectors and tensors -- they are coordinate independent, which completely determines their (lack of) "transformations" under coordinate change. Spinors can be viewed as elements of the even subalgebra.

But there's a catch when it comes to the spinors. "Tangent spinors" in the sense above are coordinate independent, and therefore transform like the rest of the multivectors (ie like vectors) under coordinate change. So the one-sided spinor rep transformation law doesn't seem to naturally derive from coordinate independence. It's something postulated separately. You could see it as deriving from $$svs^{-1}$$ being coordinate independent.

But if its a separate postulate anyway, maybe you should just define spinors as sections of an arbitrary Clifford module bundle over the manifold --- going back to the pure rep theory and scrapping the tangent Clifford bundle.

Not sure the best way to look at it. Maybe you could outright define even-multivector-style spinors by a factorization with respect to a fixed reference vector, then they should inherit a transformation.

About the disparate dimensionality.. The even-multivector construction gives one particular representation, other reps of the same $$so(p,q)$$ are possible. Plus there's the question of choosing some particular matrix rep of the Clifford algebra. Isn't this just a case of comparing different dimensional reps of the same algebra? Eventually we bump against the question of what exactly "spinor" is supposed to mean... what representations and generalizations does physics actually care about?

• I'm not sure whether this answers goes with the geometric algebra definition of spinors as elements of the even subalgebra of the Clifford algebra or as minimal left ideals of the Clifford algebra. In either case, you discuss the transformation properties of spinors defined as elements of the "tangent Clifford bundle". I've assumed that these would actually transform as spinors under SO(3) rotations of the coordinates. I suspect the expected transformation properties would arise from the fact that spinors are superpositions of the right multivectors in the Clifford algebra... Dec 20, 2023 at 15:36
• despite each multivector on its own transforming more or less like a regular tensor. Dec 20, 2023 at 15:37
• Had in mind the even subalgebra, which more straightforwardly completes the bivector exponentials. I assume the left ideals definition is essentially reducing the even subalgebra into irreps? (But haven't scoured that connection too closely) Dec 20, 2023 at 17:03
• Any comment on your claim that the Clifford bundle transforms like vectors/tensors (because it's derived from a tensor algebra) compared to my hypothesis that it naturally transforms like spinors because it is a special superposition of multivectors? Dec 20, 2023 at 19:31
• I'd be much happier if it worked out as you say, but I don't think it can, it's too linear. If the basis goes $e'_i = R e_i R^{-1}$, then all strings of basis vectors, and hence also wedges, and then also sums of different ranks, all go like $A' = R A R^{-1}$ Dec 20, 2023 at 20:04

Let me offer a treatment of spinors in a specific case of Dirac spinors $$\psi$$, rather than in a general case. I mostly follow E. T. Whittaker, Proc. Roy. Soc. A 158, 38 (1937) (see also E. Cartan, The theory of spinors (Dover Publications, Inc., 1981), section 154).

The Dirac spinor can be represented as a sum of two chiral spinors: $$\psi=\psi_++\psi_-$$, where $$\psi_\pm=\frac{1}{2}(1\pm\gamma^5)\psi$$. Then the chiral spinors $$\psi_\pm$$ can be represented by antisymmetric second-rank tensors $$\left(\psi_\pm^{\mu\nu}\right) =\left(\overline{\psi_\pm^c} \sigma^{\mu\nu}\psi_\pm\right) =\left(\psi_\pm^T C \sigma^{\mu\nu}\psi_\pm\right)$$, where $$C$$ is the charge conjugation matrix, $$\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$$. The tensors define the chiral spinors up to a sign.

One can check that $$\psi_\pm^{\mu\nu}\psi_{\pm\mu\nu}=0$$ and $$\left(\psi^{\mu\nu}_\pm\right)=\left(\pm i\star\psi^{\mu\nu}_\pm\right)$$, where the Hodge dual of a second-rank antisymmetric tensor is defined as $$\label{eq:dual1b} \star F^{\alpha\beta}=\frac{1}{2}\epsilon^{\alpha\beta\gamma\delta} F_{\gamma\delta}$$. As a result, the tensors are fully defined by 3D vectors $$(\mathbf{\psi}^i_{\pm})=(\psi^{0 1}_\pm,\psi^{0 2}_\pm,\psi^{0 3}_\pm)$$.

One can check that the $$\mathbf{\psi}^i_{\pm}\mathbf{\psi}^i_{\pm}=0$$, so $$(\mathbf{\psi}^i_{\pm})$$ is a (complex) isotropic vector. Four-dimensional rotations in the 4D Minkowski space for the tensors correspond to rotations through complex angles in a 3D complex space for the 3D vectors (cf. the Weber vector (Riemann-Silberstein vector) $$\mathbf{E}+i \mathbf{H}$$, which fully defines the real antisymmetric second-rank electromagnetic field tensor $$F^{\mu\nu}$$).

Thus, the chiral spinors are pretty much equivalent to antisymmetric second-rank tensors with some special properties or to 3D isotropic vectors.