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}

  • 4
    $\begingroup$ Just to let you know, there's a really cool, and small, book by Cartan called The Theory of Spinors, maybe you'll find what you need there. $\endgroup$
    – Quiver
    May 23, 2020 at 17:47
  • $\begingroup$ What does bother you in the spinor transformation laws? I find much more diffcult for imagining the fact that the projections of the spin are disctere whatever spin vaue is - integer or half-integer. $\endgroup$ May 23, 2020 at 18:00
  • 4
    $\begingroup$ Saying a tensor is a multilinear map doesn't actually answer anything. That just means a tensor transforms like a multilinear map on vectors, which depends on how vectors transform. You can just as well say that a vector is a bilinear map on spinors, but I bet that wouldn't satisfy you. $\endgroup$
    – knzhou
    May 23, 2020 at 19:28
  • 2
    $\begingroup$ Seeking a spinor analog of "tensors as multilinear maps" might not be the path that leads most physicists to become comfortable with spinors. The path might be more like this: Quantum physics is expressed in terms of observables. If we only require the pattern of observables to be Poincaré-symmetric, without requiring that the scaffolding we use to construct them be Poincaré-symmetric, then we are led to consider scaffolding that uses the covering group of the Poincaré group -- spinor fields. (This perspective generalizes to curved spacetime, even though I expressed it in flat spacetime.) $\endgroup$ May 24, 2020 at 1:41
  • 3
    $\begingroup$ @knzhou I don't think I agree with that. Vectors are just maps $C^\infty (\mathcal{M}) \to \mathbb{F}$ satisfying linearity and Leibniz. No reference needed to transformations. This gives us a "foundation" on which to build tensors, again without reference to transformation properties. Indeed, you could do most differential geometry without ever even knowing how the components of these tensors transform. Spinors do not enjoy this property. $\endgroup$ May 24, 2020 at 9:59

5 Answers 5


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.

  • $\begingroup$ Any hints on understanding how a spinor can be (ab)used as a coordinate? I know this is pretty popular but I just don't get it. $\endgroup$
    – user257090
    May 23, 2020 at 18:36
  • 3
    $\begingroup$ This answer seems like exactly what I was looking for. Is it standard for physicists to treat spacetime as this suitably enlarged manifold ("super-Minkowski spacetime"), or is this only used in certain theories (e.g. supersymmetric theories)? $\endgroup$ May 23, 2020 at 18:47
  • $\begingroup$ @JacobDrori It is very common in supersymmetric theories, but you will see this more or less frequently also in all contexts that worry about quantization - you need a way to have "classical" fermionic variables in order to define a quantization process on them, and so they are usually added to the spacetime, phase space or whatever else the relevant configurational space in the respective context is. $\endgroup$
    – ACuriousMind
    May 23, 2020 at 18:51
  • $\begingroup$ @JacobDrori Sorry to interrupt, but he explicitly states that $V_1$ is a spinorial representation How is it now suddenly OK to use representations? $\endgroup$
    – user257090
    May 23, 2020 at 18:51
  • $\begingroup$ @DoctorNuu I have just now added a paragraph discussing that. $\endgroup$
    – ACuriousMind
    May 23, 2020 at 18:52

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.)

  • 1
    $\begingroup$ That's basically what I was trying to say (but apparently failed). Taken as building blocks, spinors are most fundamental. A point to note, however, is that when starting from a manifold, directions seem to come first and naturally, while for spinors one needs to consider the Geometric/Clifford algebra generated by those directions and find spinors in there. $\endgroup$
    – user257090
    May 23, 2020 at 20:00
  • 5
    $\begingroup$ I think I get the gist of what you're saying. The thing that is making it hard to accept is the fact that vectors really do seem to be the more "basic" objects on a manifold, as DoctorNuu said above. They're just maps on the tangent space, and we don't really care about how they transform unless we want to choose some coordinates to do calculations. But with spinors, it seems there is no definition which doesn't involve their transformation properties. Hence vectors seem genuinely more "natural". $\endgroup$ May 23, 2020 at 20:16
  • $\begingroup$ Yes, but we can talk about vector spaces axiomatically, with no mention of transformations, and then later derive the transformation properties from the axioms. Can we do the same thing with "spinor spaces"? $\endgroup$ May 26, 2020 at 2:53
  • $\begingroup$ @The_Sympathizer But that's not what we do for vectors. At the level of all vector spaces (whether they describe a spinor, vector, tensor, heatmap, inventory of a warehouse, stock price timeseries, or array in a C program), we can always derive the transformation rule for the components under an arbitrary change of basis, completely axiomatically. $\endgroup$
    – knzhou
    May 26, 2020 at 3:02
  • $\begingroup$ @The_Sympathizer In order to derive the transformation rule for the components under a physical rotation, you need to define the words "physical rotation". A priori the formalism of vector spaces knows nothing about physical space, which is why they can describe both warehouse inventories (where physical rotations are meaningless) and tensors. You put in what a rotation means by hand to get how a vector rotates, you can do the same for a spinor. $\endgroup$
    – knzhou
    May 26, 2020 at 3:03

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 $SO(3)$/$SU(2)$.

  • 1
    $\begingroup$ There we have it again. The old race between physics and maths. Physics was there first, but maths gets all the credit ;-) $\endgroup$
    – user257090
    May 23, 2020 at 18:23
  • 1
    $\begingroup$ All this is correct but I'm not sure it answers OP's question - they seem to be asking for a notion of spinors that is not just "spinors are elements of the following vector space". $\endgroup$
    – ACuriousMind
    May 23, 2020 at 18:27
  • 1
    $\begingroup$ Thanks for the explanation. I'm new to stackexchange and am pretty amazed at how quickly people respond. ACuriousMind is right, though: I was hoping for a definition which doesn't involve any "new" vector spaces other than what the base space already gives us. Defining spinors as elements of a representation space seems to be essentially defining them by their transformation properties, which is what I was trying to avoid. Apologies for my rather ambiguous question, and thanks for the answer regardless. $\endgroup$ May 23, 2020 at 18:35
  • $\begingroup$ Good answer +1, this is what I would have said. But how does this relate to tensors (as multilinear objects...)? $\endgroup$
    – lcv
    May 23, 2020 at 18:55
  • 1
    $\begingroup$ @DoctorNuu Actually, as far as I know, Clifford algebras were invented by Clifford himself. Who died in 1879. So let's give the mathematicians some credit! $\endgroup$ May 23, 2020 at 19:14

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.


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.

  • $\begingroup$ I would have written something along these lines if I had more time/patience. The $I_3$ makes it a little confusing, however. There are several other ways to write the Dirac equation in this framework, which I prefer. The nice part (in all of them) is that spinors have natural one-sided transformations, while vectors (like the derivative or A-fields after 'gauging') have two-sided transformation with the same $R$, no need for the usual awkward translations between representations. $\endgroup$
    – user257090
    May 25, 2020 at 13:35
  • $\begingroup$ @non-user38741 what's your preferred way of defining the Dirac eq in this framework? $\endgroup$
    – Timo
    May 25, 2020 at 14:02
  • $\begingroup$ My own, similar to older papers from Doran, Lasenby. The curious thing is that it never quite matches in D=1+3 because Cl(1,3) or Cl(3,1) has representations on $\mathbb{R}(4)$ or $\mathbb{H}(2)$. $\endgroup$
    – user257090
    May 26, 2020 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.