Can we do better than "a spinor is something that transforms like a spinor"?

Question

It's common for students to be introduced to tensors as "things that transform like tensors" - that is, their components must transform in a certain way when we change coordinates. However, we can do better by defining a tensor as a multilinear map from $ V\times...\times V\times V^\ast\times ...\times V^\ast\to \mathbb{F} $, where $V$ is a vector space over $\mathbb{F}$ (often taken to be a tangent space). The transformation law then follows.

My current understanding of spinors feels like the first, dissatisfying definition: they're just "things that transform like spinors" - that is, they're elements of a vector space which transform according to a projective representation of $SO(n)$ which is genuinely multi-valued (i.e. it's not just a true representation of $SO(n)$). We could call this the "spinor transformation law". Note that this is something we've put in "by hand": the way a spinor transforms is not a property of some underlying object, but is built into our definition.

My question is: can we define spinors without reference to the way they transform, just as we did for tensors? Is there some object "underlying" the definition of spinors in terms of transformations, just as tensors are "really" multilinear maps?

\begin{align} \text{Tensor Transformation Law}&\to \text{Tensors as multilinear maps}\\ \text{Spinor Transformation Law}&\to \text{??? } \end{align}

Answer 1

The proper analogous formalization of spinors is not to view them as some sort of different functions from tensors on the same underlying vector space $V$, but instead to expand our idea of the underlying geometry: Where tensors are multilinear functions on vector spaces, tensors with "spinor" and "vector" parts are multilinear functions on super vector spaces $V = V_0\oplus V_1$ where the odd part $V_1$ is a spinorial representation of $\mathrm{Spin}(V_0)$. (nlab calls these spaces super-Minkowski spacetimes).

Via the dual representation, the linear functions on $V_1$ inherit a representation of the spin group. The (multi)linear functions also inherit the super-grading (a linear function that is zero on the odd part is even, and a linear function that is zero on the even part is odd), and purely even such functions are just ordinary tensors, and purely odd functiona are pure spinors.

Note that we still put in the spin representation $V_1$ by hand - the choice is not determined by the base space $V_0$. This is, in some way, not surprising - a notion of "spin" and spinor is genuinely more than just having a vector space: All (pseudo-Riemannian) manifolds (modeled on the vector spaces $\mathbb{R}^n$) have a notion of tensors built on tensor products of the (co)tangent spaces, but not all manifolds have spinors, i.e. the possibility to associate consistently a spinorial representation to every point of the manifold. For simple vector spaces the choice of a notion of spin is not obstructed, but it is still a choice.

That the supergeometric approach is nevertheless the "correct" (or at least a useful) one is seen when we turn to field theory, where one must represent fermionic/spinorial degrees of freedom by anti-commuting variables, and the $\mathbb{Z}/2$-grading of the underlying vector space then allows us to do this simply by declaring that the odd components anti-commute.

Answer 2

I think you're asking for intuition in the wrong direction here.

Suppose that somebody is already familiar with vectors, and wants to understand tensors. That's possible, because tensor representations are built out of vectors, i.e. the rank $2$ tensor representation is just the product of two vector representations, or equivalently a rank $2$ tensor is a bilinear map on two vectors.

But it's precisely the opposite with spinors. Spinors aren't built out of vectors, instead vectors are built out of spinors! Spinors are the simplest possible representations of the Lorentz group, and the vector representation is the product of a left-handed spinor and a right-handed spinor (or equivalently, a vector is a bilinear map on two spinors).

In other words, asking what underlies spinors is the wrong question. Spinors are the structure underlying everything you already knew. You need to rebuild your understanding with the spinors at the bottom, not at the top.

This happens a lot in physics: you can't ask for an intuitive derivation of a fundamental thing from a composite thing. What you're asking for is analogous to asking which atoms a proton is made of, or how many protons are inside a quark, or how to build a vector out of tensors. (Incidentally, learning fancier mathematics, as suggested in the other answers, never answers such questions, because these questions inherently don't have answers. What really happens is that in the process of learning the math, you get familiar with these new elementary objects. Once you can work with them fluently, you stop worrying about explaining them in terms of things you knew before, because you understand them on their own terms.)

Answer 3

Yes. Spinors are elements of representation spaces of objects known as a Clifford algebras.

A Clifford algebra is basically a vector space turned into an algebra via the product rule

$$ v\cdot w=2g(v,w)\Bbb{1} $$

where $g$ is some metric on the vector space itself. The most famous Clifford algebra is the Dirac algebra, i.e. the algebra of the Dirac matrices (for which the vector space is $\Bbb{R}^{4}$ and the metric is the Minkowski metric). If instead you use $\Bbb{R}^{3}$ as the base vector space, with the Euclidean metric, you obtain the Pauli algebra.

Once you have a Clifford algebra you can look for its representations (or "modules", as they are known in the literature). The elements of these representations are the spinors. The spinors corresponding to $\Bbb{R}^{4}$ with the Minkowski metric are the Dirac spinors, whereas those corresponding to $\Bbb{R}^{3}$ with the Euclidean metrics are the spinors of $SO(3)$/$SU(2)$.

Answer 4

Well, you should look at (irreducible) representations of the Lorentz group. Basically you want all of your ingredients to have correct and consistent transformations under the Lorentz group.

The Weyl and Dirac spinors are the most basic objects satisfying that requirement.

Starting from those you can build vectors as a (multiplicative) combos of two spinors. That's why in old texts you sometimes see spinors referred to as 'half-vectors'. Also, in this context, they use only 'half' of the transformation of a vector, i.e. one-sided vs two-sided.

In that sense its Spinors->Vectors->Tensors.

If you feel fancy, you can also look at things in the context of Geometric Algebra or Spacetime Algebra going back to David Hestenes. Here you can have spinors free of any matrix representation.

Two other references with different perspectives also come to mind: Spinors and space-time (Penrose) and GRAVITATION (Misner Thorne Wheeler)

The common theme of all approaches is however the special, fundamental transformation properties they have. You can't get around that.

Answer 5

I'm taking the Clifford algebra route as pointed out by non-user38741 and Giorgio Comitini, but I'll try to justify intuitively how to end up there and how the spinor transformation law appears inevitable. So I start with geometric algebra, which is simply another name for Clifford algebra when used in a physics context, and the vectors are taken to be elements of the algebra themselves (i.e. we're not imposing a separate matrix algebra). So take $\mathbb{R}^{n, m}$ with inner product $<\cdot,\cdot>$, and define the geometric algebra $\mathcal{G}(\mathbb{R}^{n, m})$ as the freest associative algebra over $\mathbb{R}^{n, m}$ which satisfies

\begin{equation} v^2 = <v, v>, \end{equation} where the square is, of course, the algebra multiplication. We will call the multiplication in this algebra the geometric product.

Admittedly, this does introduce another space, but that is an extremely natural one: the elements of the geometric algebra can be interpreted to consist of the scalars, the vectors of $\mathbb{R}^{n, m}$, the bivectors $u\wedge v$ where $u$ and $v$ are vectors and $u\wedge v := \frac{1}{2}(uv - vu)$, the 3-vectors $u\wedge v\wedge w$ and so on, up to (n + m) -vectors. The $n$ -vectors can be interpreted as directed area/volume/n-volume elements. For a whimsical introduction, see "Imaginary numbers are not real", or as a thorough introduction either Hestenes' "Clifford algebra to Geometric Calculus" or Doran and Lasenby's Geometric Algebra for Physicists.

Now, it turns out that a rotation of the vector $v$ in the plane defined by a simple bivector $\omega$ by $|\omega|$ radians (where the absolute value is $\sqrt{-\omega^2}$, since the square of $\omega$ is negative) can be expressed in geometric algebra (GA) as

\begin{equation} v \mapsto \exp(\omega) v \exp(-\omega), \end{equation} where the exponential is defined by the usual power series, with the multiplication being the geometric product, and a simple bivector is a bivector that can be written as the wedge product $a \wedge b$ for some vectors $a, b$. A general rotation is then given by the same formula, but with the $\omega$ not being necessarily simple (i.e. it may need to be a sum of several simple bivectors). The result of the exponential is then in the even subalgebra, i.e. built out of objects that can be expressed as a sum of products of an even number of vector factors. We call the result of the exponentiation a rotor, and often denote $R = \exp(\omega)$. Then the object on the right hand side of the transformation can also be written as $\tilde{R}$, where the tilde denotes reversion, which simply means taking each factor in a geometric product and reversing their order. Further, $R \tilde{R} = 1$ when $R$ is a rotor.

The first glimpse of a spinor-like transformation law appears: in general, we can rotate all elements of the space by the two-sided rotation law given above, and nothing changes. However, if we represent rotations by the rotor $\exp(\omega)$, then the composition of rotations is given by $\exp(\omega_1) \exp(\omega_2)$, which is also a rotor.

Now, let us stick specifically to $\mathbb{R}^{1, 3}$. Then we can write the free Dirac equation as \begin{equation} \nabla \psi I_3 + m \psi = 0, \end{equation} where $\nabla$ is the vector derivative $\nabla = e^\mu \partial_\mu$, and the $e^\mu$ are basis vectors acting via the geometric product (so that $\nabla$ itself is algebraically a vector). The Dirac field $\psi$ takes values in the even subalgebra of the geometric algebra. $I_3$ is a three-vector which appears to pick a preferred slice of spacetime, and therefore break Lorenz invariance. However, consider another choice given by $I'_3 = R I_3 \tilde{R}$. Then the corresponding new Dirac equation is

\begin{equation} \nabla \psi' R I_3 \tilde{R}+ m \psi' = 0. \end{equation} Now if $\psi$ solves the original Dirac equation, then clearly $\psi' = \psi \tilde{R}$ solves this new equation with $I_3'$. In other words, when the object $I_3$ transforms like a (three)-vector under rotations, then $\psi$ transforms like a spinor, and the transformation law has appeared.

Then note that the physical predictions of the theory only depend on the Dirac bilinears, which in this language can be written analogously to \begin{equation} \psi I_3 \tilde{\psi}, \end{equation} and that when $I_3$ transforms as a three-vector and $\psi$ as a spinor, the physical predictions remain unchanged. In other words, the spinor transformation law is required here to keep the physical predictions of the theory independent of the choice of the directed volume element $I_3$.

Indeed, there is a natural interpretation of the object $\psi$ as a product of a rotor, scaling and a transform between scalars and pseudoscalars in $\mathbb{R}^{1,3}$. In this way, the spinor transformation law appears naturally as the composition of rotors (or rotor-like objects). Of course, since there is no treatment of quantum field theory in the geometric algebra language, it's not clear how far or seriously this can be taken as an interpretation of the physical Dirac equation, but nonetheless it at least provides an example where spinors appear naturally, without manually imposing the transformation law. Rather, it comes from transformations of the solutions of the Dirac equation when the choice of the constant $I_3$ transforms by rotations.

I'm sure that this flash-introduction to the subject leaves many questions unanswered and it may be a bit confusing, but if I piqued your interest I suggest you follow some of the links here and continue further that way.

Answer 6

I was wrong here {

One can generalize spinors and create infinitely many kinds of objects by 'mimicking' a tensor. It is known that Metvv=v^2. So it would be logical to do va^2=a^2. If a spinor is the square root in some sense, one then can do va^2=v2;

So one must define the map vaa->v, which can be recovered through QM, or whatever the situation is. So spinors are likely not the most fundamental, because this can be applied as many times as one wants. It is seen that the square of a vector is a scalar due to the metric which is a vector squared. If the square of a spinor is a vector, then a vector times two spinors which is a spinor squared which is a vector. So this kind of context where baa=b2=a2^2 supports the existence of a spinor when going this route (and infinitely many other things). A spinor keeps the train of thought rolling past the assignment of real numbers in the metric, to create such objects. This is a bunch of careful choices to implement the square root of an object. But it is consistent in the result, and it works. Time evolution should also be determined. If Uvaa=v2, (where U defines the type of object such as a spinor) then d^n(Uvaa)=d^n(v2) determines the time evolution and an even larger whole new set of objects to explore.

In Short:So one does Uvss=v (s=spinor like object, v=vector).

For a general nth root one does Uva^n=v;

Another way to define a spinor is to 'slow down' the generators of the lie algebra by half.

Then one can restrict the values of U, or use that idea as another whole set of objects.

A spinor seems to act like both of these. The generators are slowed down and two of those objects can change a vector. } Doing more research and attempting to make it abelian to abelian, I found that {,}^(-1)(2*Met)=>Tensor_Product_Power(Met,1/2);

Simply put, it maps the undoing of the averaging (divide by 2) anticommutator binary operation (typically with the metric being the result of the anti-commutator of each object) to the half tensor product of that object (typically the metric (again)).

It goes like this: If {a,b}=2*Met, then A2XB2=Met;

A2 is the transformation of A, and a is in A, same deal with b,B,B2. The map is consistent (both ends are abelian so no need to worry yet).

The dimension for these items a and b is typically 2^k, while the dimensions for A2 and B2 is likely k.

I first thought today, that It maps the anti...-anti-commutator (look above to see what I mean, it's crazier than just a commutator. I didn't stutter nor am I using double negatives to make either positive or negative clauses. ), to the anti-product (gets a set).

That is obviously wrong, since one is abelian the other is not.

So I found this. I need more prefixes than the english language offers to make this sentence. Anti2-Anti-Average-Like-Commutator of the Metric goes to Anti2-Tensor-Product-Square of the Metric. That is the way to remember it (At least for me). The gamma matrices are the anti2-anti-commutator of the metric. I put a 2 there to warn you that this means undoing a binary operation to recover a set of items, rather than the anti as in anti-commutator kind of anti.

The best way to write it I found is,

Make_Spinor({,}^(-1)(Met))=(X)^(1/2)(Met).

The dimensions are likely either a matter of combinatorics, or quantum mechanics.

Another way to think about it quantum mechanically, is SMAT_{universal}(sqrt((...)a^2+(...)b^2+...),a,b,...) maps to MetX^(1/2).

What I mean by that, is solve a constant matrix for some square root of the sum of squares (or another combination via the metric) and map it to the vector that it's square tensor product is the metric. So that becomes even simpler. One does SMAT(sqrt(Met(s,s)),s)->sqrt_{Tensor_Product)(Met);

That becomes SMAT(sqrt O Met*s^2,...)->sqrt(SMAT(ds,v^2)).

That becomes SMATsqrt(...)->sqrtSMAT(...) (the simplest it can be).

SMAT(OUT,IN) is basically 'solve matrix' (like neural networking). I find that is the best way to notate it. Partial derivatives aren't always clear on what it doesn't include, while a matrix problem is 100% clear and one can execute it the same way every time.

My notation is different (I develop my own notation like newton,.... (long list)), so beware. But it is well suited to all of these concepts.

Hopefully I simplified it.

The main result (in very short form) is: sqrt*SMAT(...)<->SMAT * sqrt(..)...;

That is the main rule that recovers it all. To me it is currently the most important. I don't know about others (they might just call me crazy), but the work on spinors cares in itself.