Do spinors form a vector space?

His notes sound a bit confused. He says correctly that the group SU(2) is not a vector space, but seems to think that that this means that the column vectors that the SU(2) matrices acts on do not form a vector space. I think that he might be thinking of the orbit that you get when you choose a fixed starting spinor ${bf v}_0$ and consider the set you get by acting on it by all possible SU(2) matrices. This subset of spinors (the orbit of ${bf v}_0$) is not a vector space, but that does not mean that the set of all spinors spinors is not a vector space.

The question is tricky, because we must answer the following question: "can we make superpositions of different spinors of the same rank?". Mathematically - yes, we can, why not?, it is just a matter of definition of a spinor space, but physically we may not because the energy of a system may depend on the particular spin configuration, so different spinors belong to different Hamilton eigenvectors, and we must make superpositions of the total time-dependent wave functions rather than just their spinor parts, I guess.

H. Ohanian (Am. J. Phys. 54 (6) 1986) tried to give a "classical" meaning to spin as an angular momentum of a spinor wave packet. The most essential part of his calculations is fading the wave packet in the transverse directions. But this fading is just a simplified (mathematical) description of influence of the particle source and the surrounding matter. I can safely say that spin is a property of an elementary excitation of a complex system (world).

I've only had a quick look at the Coddens paper, but if I understand it correctly what he is doing is the equivalent of saying that the set of unit vectors are not vectors, because they don't form a vector space. You can't 'add' two unit vectors together and always get another unit vector.

He mentions Hestenes and geometric algebra in his introduction, which in my view gives a much clearer intuitive geometric picture of spinors. (I'm not clear on why he rejects this.) In geometric algebra, the spinors are just the even sub-algebra - the linear combinations of all products of an even number of basis vectors. Since each basis vector represents an (oriented) reflection, pairs of basis vectors represent rotations (i.e. pairs of reflections).

In 2D, the geometric algebra consists of linear combinations of a scalar (1), two basis vectors (x and y) and a bivector representing the plane (xy). The even sub-algebra is the linear combinations of $1, xy$ where $(xy)^2=-1$. The 2D spinors are thus just the complex numbers.

In 3D, the geometric algebra consists of a scalar (1), three basis vectors (x, y, z), three bivectors representing coordinate planes (yz, xz, xy), and a trivector (xyz) representing the oriented volume element. The even sub-algebra is the linear combinations of $1, yz, xz, xy$ where $(yz)^2=(xz)^2=(xy)^2=-1$. The 3D spinors are thus just the quaternions.

If we have two generic vectors $a$ and $b$, then $ab=acdot b+awedge b=|a||b|(textrm{cos }theta+B textrm{sin }theta)$ where $B$ is the unit bivector in the plane of $a$ and $b$ and $theta$ is the angle between them. If the vectors being multiplied are parallel, the result is a pure scalar, if perpendicular then the result is a pure bivector. The dot product is familiar from vector algebra, and is just the scalar part of the geometric product. The wedge product is the bivector part, and is dual to vector algebra's cross product. (We have to use the dual to turn it into another vector because vector algebra can't cope with bivectors. However, this doesn't work perfectly because it transforms wrongly under reflections - i.e. the cross product gives a 'pseudovector', not a vector. Pseudovectors are actually bivectors.) Thus both products from vector algebra are unified in a single product in geometric algebra, and correspond to finding the 'real' and 'imaginary' components of a spinor.

The complex column vector representation of spinors arises from noting that $xy$ acts like the imaginary unit $i$, so $alpha(1)+beta(xy)=alpha+beta i$ and $gamma(yz)+delta(xz)=(gamma+delta(xy))(yz)=(gamma+delta i)(yz)$ so the quaternion can be encoded as two complex coefficients of a vector space over the basis $1, yz$.

In geometric algebra, spinors implement rotation by conjugation. To rotate a vector $v$ by the rotation represented by the spinor $S$, we find $SvS^{-1}$. If $S$ has a norm different from 1, then $S^{-1}$ has the reciprocal norm, and they cancel. This is why a (non-zero) scalar multiple of a spinor corresponds to the same rotation, and why we can add spinors together without giving rise to any problems. We commonly use unit spinors (called rotors) to represent rotations so that we have an unambiguous representation, like we commonly use unit vectors to represent things like surface normals, where the length doesn't matter. (That is to say, as a complex column vector, these are 'unitary' and transform as SU(2).) But using unit spinors to represent rotations doesn't mean they're not elements of a vector space any more than using unit vectors does. It's also the case that the sign cancels too, so $-S$ gives the same rotation as $S$, but is not the same spinor.

I find that geometrically, the best way of visualising a 3D spinor is as the angle between a pair of oriented reflection planes. ('Oriented' means the plane has a well-defined front and back.) If the planes are identical, the reflections cancel, and the identity results. (i.e. with parallel vectors we get a pure scalar 1.) As you rotate one plane with respect to the other, the results is a rotation through twice the angle between the planes, about the axis along which the planes intersect. When the angle between the planes reaches 180 degrees, the planes are parallel, but with normals in opposing directions. Again, the reflections cancel, so this represents a 360 degree rotation, but this is not the same spinor! The 180 degree angle means one plane is the negative of the other. If you keep on rotating the reflection planes, eventually a 360 degree angle between reflections corresponds to a 720 degree rotation, and we have identity again. It is not true that rotating a spinor 360 degrees gives the opposite sign, and you have to rotate 720 degrees to return to where you started. What happens is that rotating the spinor through an angle increases the corresponding rotation by twice the angle, so rotating a spinor 180 degrees negates the spinor but turns the rotation right around, and rotating the spinor 360 degrees corresponds to a rotation of 720 degrees. The spinor is not the rotation - it is the angle between the pair of (oriented) reflection planes that make it up.

So spinors do have a perfectly understandable geometric interpretation.

Let me pick up from DanielC's post. The idea that a column matrix could represent a group element can also be found in any regular representation of a finite group. Let us consider e.g. the permutation group $S_{n}$. We can label arbitrarily each of the $n!$ permutations $p_{j} in S_{n}$ with a number $j in [1,n!]cap{mathbb{N}}$. The order the group elements acquire this way has no importance. Any order will do. The regular representation is then given by $n! times n!$ representation matrices and each group element $p_{j}$ is also represented by a $n! times 1$ matrix ${mathbf{p}}^{(j)}$, whose entries are $[,{mathbf{p}}^{(j)},]_{k} = delta_{kj}$. That is, all entries take the value $0$ except the one on line $j$, which takes the value $1$. The square matrix representing a group element $p_{j}$ would just represent $p_{j}$ by the group automorphism: $T_{p_{j}}: qin S_{n} rightarrow T_{p_{j}}(q) = p_{j} circ q in S_{n}$. We could call the $n! times 1$ column matrices ${mathbf{p}}^{(j)}$ "column vectors" and they would span a "vector space" $(V,{mathbb{K}})$ over some number field ${mathbb{K}}$ that could be ${mathbb{R}}$ or ${mathbb{C}}$. The group $S_{n}$ would then be a finite discrete subset (of $n!$ points) of the vector space $(V,{mathbb{K}}) = ({mathbb{K}}^{n!},{mathbb{K}})$. It would constitute an orthonormal basis for $(V,{mathbb{K}})$. But this is just shallow nonsense, because in the representation a sum ${mathbf{p}} + {mathbf{q}}$ of two such $n! times 1$ column matrices ${mathbf{p}}$ and ${mathbf{q}}$ would by isomorphism correspond to the sum of two permutations as in Eq.1 of Symmetry, 13, 659 (2021).

It is very obvious that this operation is just not defined, and the same applies for any other linear combination $sum_{j=1}^{n!} c_{j} {mathbf{p}}^{(j)}$, with $c_{j} in {mathbb{K}}, forall j in [1,n!] cap{mathbb{N}}$, that does not belong to $S_{n}$. All points of ${mathbb{K}}^{n!}backslash S_{n}$ are a priori meaningless. Only one operation is defined by the axioms of a group.

It is for the same reason that in general relativity the curved space-time manifold should not be considered as embedded in a vector space, but described intrinsically. The points you would have to add to obtain an extension in the form of a vector space wherein the curved manifold could be embedded do not exist physically. That vector space would have to be ${mathbb{R}}^{5}$, just like the two-dimensional surface of a sphere is embedded in ${mathbb{R}}^{3}$. The points of the extension to ${mathbb{R}}^{5}$ that do not belong to space-time would just be physically meaningless.

Consider now the complex vector space ${mathbb{C}}^{2}$ and a general point $(zeta_{1},zeta_{2}) in {mathbb{C}}^{2}$ of it. Until something is done about it, such a point has no obvious physical meaning. A first step consists in pointing out that the spinors $psi = [,zeta_{1},zeta_{2},]^{top}$ of SU(2), which can be given the meaning of rotations around the origin in ${mathbb{R}}^{3}$, belong to the set ${mathscr{C}} = { (zeta_{1},zeta_{2})parallel zeta^{*}_{1}zeta_{1} + zeta^{*}_{2}zeta_{2} = 1} subset {mathbb{C}}^{2}$. These spinors are isometries, i.e. special elements of the vector space $L({mathbb{R}}^{3},{mathbb{R}}^{3})$. They are geometrical/physical operators. Each rotation corresponds to two spinors which are identical up to a factor $pm 1$. Each rotation defines therefore two points of the curved manifold ${mathscr{C}} $, but the converse is also true: each such pair of points of ${mathscr{C}} $ corresponds to a rotation, i.e. a pair of spinors of SU(2), which is a double covering of SO(3) due to the very existence of these two possible factors $pm 1$. The simplest way to prove this is to identify $(zeta_{1},zeta_{2})$ with the expression for a spinor that corresponds to the rotation $R(alpha,beta,gamma)$ where $(alpha,beta,gamma)$ are its Euler angles, as e.g. given by the first column of the matrix in eq. 1.2.29 of the book of J. Hladik. Therefore ${mathscr{C}}equiv$ SU(2). We can consider the manifold ${mathscr{C}}$ as embedded in ${mathbb{C}}^{2}$. In terms of real numbers, the rotation group, represented by ${mathscr{C}}$, is then a three-dimensional manifold embedded in the four-dimensional vector space ${mathbb{C}}^{2} equiv {mathbb{R}}^{4}$. The situation is completely analogous to the one of four-dimensional space-time embedded in ${mathbb{R}}^{5}$ as described above.

It is therefore algebraically feasible to calculate linear combinations $c_{1}psi_{1} + c_{2}psi_{2}$, where $(c_{1},c_{2}) in {mathbb{C}}^2$, or to consider elements of ${mathbb{C}}^{2} backslash {mathscr{C}}$ but this is purely formal and a priori devoid of any geometrical meaning in terms of some element of $L({mathbb{R}}^{3},{mathbb{R}}^{3})$, which is the natural embedding for the rotation group SO(3) $subset L({mathbb{R}}^{3},{mathbb{R}}^{3})$. What other kind of embedding of SO(3) could we else imagine to give $(zeta_{1},zeta_{2}) in {mathbb{C}}^{2} backslash {mathscr{C}}$ meaning?

This argument is similar to that made for $S_{n}$, but this time the group is a Lie group and therefore no longer a discrete finite set but a differentiable manifold.