Spinors Demystified

Rotations represented as vectors.

eben.cowley42@gmail.com


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Summary

Spinors very roughly correspond to rotations of a vector space with a quadratic form (a function for determining square magnitude of a vector). A certain type of spinor is used to represent matter in particle physics. That may sound strange, but it turns out that two instances of spinors are the familiar complex numbers and quaternions. The framework that they arise in, geometric algebra, allows for the concise representation of rotations and reflections in indefinitely many dimensions. The exact way that spinors relate to rotations allows us to deduce interesting topological facts about the rotation groups of 3D and 4D space.

Lie Groups

A set $S$ with an operation $*$ is a group if and only if for any $a,\, b,\, c \in S$ we have that $$a*b\in S$$ $$(a*b)*c=a*(b*c)$$ $$\exists e\in S \, : \, e*a=a*e=a$$ $$\exists a^{-1}\in S \, : \, a*a^{-1}=a^{-1}*a=e$$

These conditions are referred to as closure, associativity, identity, and inverse, respectively.

For a Lie group, the set $S$ is equipped with a geometric structure such that multiplication and inversion are smooth functions. Informally, if $a\in S$ then the functions $f,\, g,\, h:S\rightarrow S$ where $f(x)=a*x$, $\,g(x)=x*a$, and $\, h(x)=x^{-1}$ then $f$, $g$, and $h$ have derivatives of all orders.

If you're new to abstract algebra, you're probably thinking "What? The derivative of $ax$ is $a$ and the derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$." This is true, the non-zero reals with multiplication is our first example of a Lie group. The non-zero part is important because the inverse has to be defined for all elements of the group, $0^{-1}$ is undefined on the real numbers.

Another straightforward example of a Lie group is the real numbers with addition. That is, $S$ is $\mathbb{R}$ and $*$ is $+$. The identity in this case is $0$, so we have $x^{-1}=-x$, which has derivative $-1$, and for some real $a$ we have $f(x)=x+a$ which has derivative $1$.

Just as a smooth curve in the plane has a tangent line and a smooth surface in space has a tangent plane, a Lie group can be thought of as a tangent space. The tangent lines of a curve are all fundamentally the same vector space, as is with the tangent planes of a surface. The specific algebraic structure we are going to talk about can be thought of as the tangent space at the identity element of the group. This tangent space is referred to as the Lie algebra of the given Lie group. The essential connection between the Lie algebra and the Lie group is the existence of an exponential map, $\exp$, going from the Lie algebra to the component of the manifold containg the identity element and satisfying $\exp(x+y)=\exp(x)\exp(y)$. This map could be thought of as a function which takes a vector in the tangent space of the identity and bends it onto our manifold - this visualization makes it clear why the codomain of the exponential map must be the identity component.

For example, the non-zero reals under multiplication are disconnected. The group has two components: the negatives and positives. The positives are the identity component because they contain the identity, $1$. Thus, the Lie algebra is $\mathbb{R}$ and the exponential map is simply exponentiation because $e^x$ takes a real number, outputs a positive real, and satisfies the exponentiation property. We will also find that the natural extension of exponentiation, base $e$, to the complex numbers and qauternions functions as an exponential map in the Lie theoretic sense. In all the examples of matrix Lie groups that I'm aware of, it is true that the exponential map is the matrix exponential.

Geometric Algebra

The Lie groups that are of interest in this article can be represented with an under-rated concept in math called geometric algebra. Geometric algebra refers to the Clifford algebra of a Euclidean space. I lean towards the use of the term "geometric algebra" because most of my examples are Euclidean, but I talk about the Clifford algebra of spacetime at the end of the article.

Recall that for $u,v\in \mathbb{R}^3$ $$u\times v = |u||v|\sin(\theta)\hat{n}$$ where $|v|$ is the magnitude of $v$, $\theta$ is the angle between the vectors, and $\hat{n}$ is a unit vector determined by the right-hand rule. From this it can be derived that the magnitude of the cross product is the area of the parallelogram formed by the two vectors.

There is an operator, $\wedge$, called the wedge product, that generalizes the cross product to any dimension. In $\mathbb{R}^2$ we can think of $u\wedge v$ as an oriented parallelogram - one whose sides are given by each of the vectors and whose orientation is clockwise if $v$ is clockwise from $u$ and anticlockwise otherwise. This kind of object is called a bivector. If we pick an orthonormal basis for the plane, $e_1$ and $e_2$, then all bivectors are spanned simply by $e_1 \wedge e_2$. To see this, all we need is the following simple set of rules:

$$e_i \wedge e_i = 0$$ $$e_i \wedge e_j = -e_j \wedge e_i$$ $$e_i \wedge (e_j \wedge e_k) = e_i \wedge e_j + e_i \wedge e_k$$ $$e_i\wedge (e_j\wedge e_k)=(e_i\wedge e_j)\wedge e_k$$ And multiplication of anything with any scalar behaves nicely, to be concise. Note that these rules can be applied to an orthonormal basis for any Euclidean space. Now let's continue with the 2-dimensional example and wedge $ae_1 + be_2$ with $ce_1+de_2$: $$(ae_1 + be_2) \wedge (ce_1+de_2)$$ $$=ae_1\wedge (ce_1 + de_2) + be_2\wedge(ce_1+de_2)$$ $$=ae_1\wedge ce_1 + ae_1\wedge de_2 + be_2\wedge ce_1 + be_2\wedge de_2$$ $$=0 + ade_1 \wedge e_2 + bce_2\wedge e_1 + 0$$ $$=ade_1\wedge e_2 - bce_1\wedge e_2$$ $$=(ad-bc)e_1\wedge e_2$$ Thus, we see that when we wedge 2 vectors in $\mathbb{R}^2$ we get the signed area of the parallelogram that they form times $e_1\wedge e_2$ - the orientation that I mentioned earlier is encapsulated by the sign.

For $v_1,v_2,...,v_n\in\mathbb{R}^n$, $v_1\wedge v_2,\wedge . . . \wedge v_n$ will be $Ve_1\wedge e_2\wedge . . .\wedge e_n$ where $V$ is the hypervolume of the n-parallelotope formed by the vectors and the $e_i$ form an orthonormal basis for $\mathbb{R}^n$.

The geometric product was invented by William Kingdon Clifford and is denoted $uv$ for $u,v\in\mathbb{R}^n$. It is defined by $$uv=u \wedge v + u \cdot v$$ You're probably thinking "but $\wedge$ yields a bivector and $\cdot$ yields a scalar," which is true. In a geometric algebra, you can add objects together even if they don't have the same grade. For example, $3+e_2+2e_1e_3-5e_4e_5e_6$ is a valid element of the geometric algebra of $\mathbb{R}^6$. $3$ is of grade 0, a scalar, $e_2$ is of grade 1, a vector, $2e_1e_3$ is of grade 2, a bivector, etc.

This is all probably starting to seem very contrived, so let's see an example of the geometric product in action: $$(e_1e_2)^2=e_1e_2e_1e_2=-e_1e_2e_2e_1=-e_2e_2=-1$$ Notice that I've used the fact that the geometric product of a unit vector with itself is $1$ (because the wedge product is zero and the dot product is 1). So why did I choose this example? Since $e_1e_2$ squares to $-1$, it can be identified as the imaginary unit! Indeed, it can be verified by following the rules for geometric multiplication that $$(a+be_1e_2)(c+de_1e_2)=ad - bc + (ac + bd)e_1e_2$$ Which is complex multiplication. Notice that this expression uses only objects of even grade. The complex numbers are therefore what's referred to as the even sub-algebra of the geometric algebra of $\mathbb{R}^2$. We are allowed to say sub-algebra instead of just sub-space because the sub-set isn't just closed under addition, but also geometric multiplication. That is, the product of two elements of an even grade will always have an even grade (this is easily verified).

Elements of even sub-algebras of geometric algebras (or more generally, Clifford algebras) are referred to as spinors.

Before talking about why geometric algebra is so useful I also need to introduce the geometric inverse. If $A$ is a multivector, a geometric product of vectors, $a_1a_2...a_i$ then its reverse, $A^\dagger$, is $a_ia_{i-1}...a_1$. For example, if $$A=e_1(e_2+e_3)(e_4-e_5)$$ then, $$A^\dagger = (e_4-e_5)(e_2+e_3)e_1=-e_1(e_2+e_3)(e_4-e_5)$$ The reverse can be extended to any element of the algebra by declaring linearity: $(a+b)^\dagger=a^\dagger + b^\dagger$. The point of the reverse is that $AA^\dagger$ is a scalar. This means that the quantity $$A^{-1}=\frac{A^\dagger}{AA^\dagger}$$ is well-defined, and that it is the right multiplicative inverse of $A$. This means $AA^{-1}=1$. The expression for the inverse may seem familiar because it is the same as the inverse of a complex number with the exception that the conjugate, $A^*$, is used instead of the reverse, $A^\dagger$ (The definitions are the same if the complex numbers are considered as the even sub-algebra of the geometric algebra of $\mathbb{R}^2$).

The geometric inverse of a vector is useful because it allows us to easily write down the parallel and perpendicular components of a vectors with respect to each other. If we have a vector $v$ and a non-zero vector $m$ then $v$ can be written in the form $v_{\perp m}+v_{||m}$. That is, the sum of the perpendicular and parallel components. The equations for these components are as follows: $$v_{\perp m}=(v \wedge m)m^{-1}$$ $$v_{||m}=(v\cdot m)m^{-1}$$ I'll leave the full verification to the reader but it should be intuitive because the bivector components of the wedge product correspond to the vector components of the cross product, whose magnitude is related to the sine of the angle between the vectors, and $v\cdot m= |v||m|\cos\theta$, as usual.

If $v'$ is the reflection of $v$ across the hyperplane orthogonal to $m$ then $$v'=v_{\perp m} - v_{|| m}$$ $$= (v \wedge m)m^{-1} - (v\cdot m)m^{-1}$$ $$=-(m \wedge v + m \cdot v)m^{-1}$$ $$=-mvm^{-1}$$ Which is a very nice, concise way of representing a reflection.

Rotations are where things start to get really interesting. If you have two vectors, reflecting across both of the hyperplanes orthogonal to each of the vectors is equivalent to a rotation through the plane spanned by the two vectors by twice the angle between the vectors. Say the vectors are unit and named $r$ and $s$. Then reflecting $v$ across $r$'s hyperplane followed by $s$'s is the following: $$v'=-s(-rvr^{-1})s^{-1}$$ $$=s(rvr^{-1})s^{-1}$$ $$=(sr)v(sr)^{-1}$$ $$=(sr)v(rs)$$ Note that the last step here is justified by the fact that the vectors are unit, their inverse is equal to their reverse. Continuing with the derivation: $$=(s\cdot r + s\wedge r)v(r\cdot s + r\wedge s)$$ $$=(s\cdot r + s\wedge r)v(s\cdot r - s\wedge r)$$ Now, let's let $e_1$ and $e_2$ be orthonormal vectors that span the same plane as $r$ and $s$ and are oriented the same way. Continuing: $$=(\cos\theta + \sin\theta e_1e_2)v(\cos\theta - \sin\theta e_1e_2)$$ Where $\theta$ is the angle between $r$ and $s$. And now the good part: $$v'=e^{\theta e_1e_2}ve^{-\theta e_1e_2}$$ This is simply Euler's formula, and it works because $(e_1e_2)^2=-1$! Note that this is a rotation of $v$ by $2\theta$, because it is a composition of two reflections. If we want to rotate $v$ by $\theta$ through a plane given by a unit bivector $I$, the formula is: $$v'=e^{I\frac{\theta}{2}}ve^{-I\frac{\theta}{2}}$$ And that's a rotation!

Some important terms that I want to introduce here: For any invertible element of a geometric algebra $r$ and arbitrary element $v$, $rvr^{-1}$ is referred to as conjugation of $v$ by $r$. The expression $e^{I\frac{\theta}{2}}$ is refferred to as a rotor, and this term can be used to mean any multivector conjugating a vector.

If you're familiar with the quaternions, you probably noticed that the rotation formula here is exactly the same for one with quaternions, except that $I$ should represent a unit vector in the direction of the axis of rotation. Remember that in 3D space a unit vector corresponds to a unit bivector (because every line through the origin corresponds to a plane through the origin). This means the quaternions are the even sub-algebra of the geometric algebra of $\mathbb{R}^3$. The real part is the scalar part and the imaginary part is the bivector part. This means that quaternions could be referred to as the spinors of three dimensional space.

Spin and Pin

The expression above for the rotation of a vector descibes a simple rotation, that is, a rotation through a plane. In two and three dimensions, all rotations are simple rotations, this is not so in higher dimensions. If $r$ and $s$ are rotors through different planes, $$rsvs^{-1}r^{-1}=(rs)v(rs)^{-1}$$ is a compound rotation with rotor $rs$ that can not be described as a rotation through a single plane. Note that since a simple rotor is a scalar plus a bivector, an element of the even sub-algebra, every rotor, a product of simple rotors, is an element of the even sub-algebra.

A multivector $s$ in the geometric algebra of $\mathbb{R}^n$ such that $ss^\dagger=\pm 1$ is an element of the Pin(n) group. Conjugation by an element in the Pin group will be a number of reflections. If $s$ has an odd number of vector factors, conjugation is a reflection, and otherwise it is a rotation. We can also consider a subgroup of Pin, called Spin, in which all elements are even. Conjugation by elements of this group will always correspond to a rotation.

You're probably familiar with the groups O(n) and SO(n). If not, O(n) is the group of all matrices, $A$, such that $AA^T=I$. This is equivalent to saying that the columns of $A$ form an orthonormal basis. This tells us that the determinant of the matrix is always $\pm 1$, because the determinant is the wedge of all the columns. The determinant will be $1$ if the corresponding transformation preserves orientation (rotations) and will be $-1$ if the transformation flips orientation (reflections). SO(n) is the subgroup of O(n) such that all matrices have determinant $1$. All elements oh O(n) correspond to a unique rotation/reflection and all the elements of SO(n) correspond to a unique rotation. However, the rotor $r$, in Spin or Pin, corresponds to the same transformation as $-r$, it's simple to verify. This means that there are continuous maps $$Pin(n) \rightarrow O(n)$$ $$Spin(n) \rightarrow SO(n)$$ such that there are always two elements in the groups on the left that get mapped to the same element in the groups on the right. This kind of continuous map is called a double covering.

It's important to note that for n=2 there is not just a double covering but also an isomorphism. This can be verified by algebraically manipulating the rotation formula for two dimensions to get it to be the rotation of a complex number. This is important because the rotation group of complex numbers is the group of unit complex numbers which is, geometrically, a circle. No non-trivial loop on a circle can be continuously contracted to a point.

Due to some topological considerations somewhat beyond the scope of this article, every loop in Spin(n) can be continuously contracted to a point, this property is known as simple connectedness. Thus, Spin(n) for n>2 is simply connected. We know that it is also a covering space of SO(n), and the additional property of simple connectedness means that Spin(n) is referred to as a universal cover of SO(n).

We know that quaternions are the spinors of $\mathbb{R}^3$, so Spin(3) is the group of unit quaternions. This tells us something interesting about the topology of SO(3), the group of rotations in 3D space. The unit quaternions are geometrically a 3-sphere, and the double covering takes antipodes of a sphere and glues them together. All lines through the origin of four dimensional space intersect this 3-sphere at exactly one pair of antipodes. This means that the rotation group SO(3) is topologically equivalent to the space of lines through the origin of 4D space, which is known as real projective space.

Remember that when we composed two simple rotations in the expression $(rs)v(rs)^{-1}$, $r$ and $s$ were both of the form $e^{\frac{\theta}{2}I}$ where $I$ is the wedge of two orthonormal vectors spanning the plane of rotation. When the rotors are multiplied together, the bivector exponents will be added. This tells us is that the Lie algebra of the spin group is the space of all bivectors and that the exponential map is ordinary exponentiation. Equivalently, the space of all imaginary quaternions is the Lie algebra of the group of unit quaterions; just as the imaginary numbers are the Lie algebra of the group of unit complex numbers.

An interesting note that doesn't really have to do with the quaternions as a spinor algebra is that rotations in 4D can always be represented using left and right quaternion multiplication. The first step is to define an isoclinic rotation as a rotation of $\mathbb{R}^4$ that can be decomposed into two rotations of equal angle across perpendicular 2-planes. If we assign an orientation to each of the hyperplanes then a rotation by $\theta$ through the first plane and by $-\theta$ through the second hyperplane is referred to as right-isoclinic; if both the angles have the same sign then the rotation is referred to as left-isoclinic. It turns out that all 4D rotations can be decomposed into a right and left isoclinic rotation. In fact, if we take a vector $v\in\mathbb{R}^4$ and consider it as a quaternion, and $p$ and $q$ are unit quaternions, $pv$ is a left-sioclinic rotation of $v$ and $vq$ is a right-isoclinic rotation of $v$. We can obtain the pair of orthogonal hyperplanes from the quaternion, $q$, by making one plane out of the real line and the line given by the imaginary part. We can also obtain the rotation angle by taking the inverse cosine of the real part. I encourage you to check out the wikipedia for more information on the algebra behind this decomposition.

Note that the rotation $(-p)v(-q)$ is the same as $pvq$. Since the group of unit quaternions is homeomorphic to $S^3$, the 3-sphere, this means that SO(4) is topologically the same as $S^3 \times S^3 /\mathbb{Z}_2$ where $\mathbb{Z}_2$ is $\{(1,1),(-1,-1)\}$ under component-wise multiplication. Since, as discussed above, $SO(3)$ is homeomorphic to $S^3/\mathbb{Z}_2'$ where $\mathbb{Z}_2'=\{1,-1\}$, we can perform a substitution: $$SO(4)=(SO(3)\times \mathbb{Z}_2')\times (SO(3)\times \mathbb{Z}_2')/\mathbb{Z}_2$$ Performing a somewhat informal cancellation with the isomorphic $\mathbb{Z}_2$ and $\mathbb{Z}_2'$: $$SO(4)=SO(3)\times SO(3)\times \mathbb{Z}_2$$ Revealing an unexpectedly simple connection between 3D and 4D rotations! Note that I went from talking about topological to algebraic isomorphisms without really saying much about it. I encourage you to verify that the above is indeed a group isomorphism.

Relation to Physics

In the summary I said that all of this can be done for any vector space with a quadratic form. Spacetime is $\mathbb{R}^4$ equipped with the following quadratic form: $$Q(x)=x_0^2-x_1^2-x_2^2-x_3^2$$ Where $x_0$ is time and $x_{i>0}$ is space. See my special relativity article for a justification of this. The procedure for setting up spacetime's Clifford algebra is as follows, choose an orthogonal basis of $\mathbb{R}^4$: $\gamma^0$, $\gamma^1$, $\gamma^2$, and $\gamma^3$. The multiplication works like so, $$(\gamma^i)^2=Q(\gamma^i)$$ $$\gamma^i\gamma^j=-\gamma^j\gamma^i$$ If we make the basis orthonormal and align it with the coordinates for the quadratic form then we have $\gamma_0^2=1$ and $\gamma_i^2=-1$. These are exactly the relations imposed on the Dirac $\gamma$ matrices! In fact, the Dirac equation: $$(i\partial_\mu\gamma^\mu-m)\psi=0$$ has the same meaning regardless of whether the $\gamma$'s are thought of as vectors in the Clifford algebra of spacetime or as matrices. The difference is that if we think of them as matrices, the solution will be in the form: $$\psi= \begin{pmatrix} w \\ x \\ y \\ z \end{pmatrix}$$ Where all the entries are complex (meaning there is a total of 8 real numbers encoded in the solution). On the other hand, if we think of the $\gamma$'s as vectors, the solution will be in the form: $$\psi=a+b\gamma^0\gamma^1+c\gamma^0\gamma^2+d\gamma^0\gamma^3+e\gamma^1\gamma^2+f\gamma^1\gamma^3+g\gamma^2\gamma^3+h\gamma_0\gamma_1\gamma_2\gamma_3$$ Notice that, either way, we get the same number of reals defining the solution. You may think that the first notation is better because its more concise, but keep in mind that, the same way conjugating by spinors of Euclidean space gives us rotations and reflections, conjugating by $\gamma$'s conveniently gives us Lorentz transformations. It also makes referring to fermion fields as "spinors" much more intuitive.

On the topic of physics, it's also worth noting that the symmetry groups of the electromagnetic and weak forces are U(1)=Spin(2) and SU(2)=Spin(3), respectively. One might be tempted to draw a link between these forces and 2D and 3D rotations; don't. Thanks to the Coleman-Mandula theorem we know that the connection is merely coincidental.