Conflicting definitions of a spinor

Question

I'm moving my question from math.stackexchange over here because it got no attention over there, even after 3 weeks and a 50-point bounty, and also because this is a very physics-oriented math question.

I've come across two definitions of "spinors" that I'm having a hard time reconciling:

1. Spinors are the "square root" of a null vector (see here, and also Cartan's book "The Theory of Spinors")
2. Spinors are minimal ideals in a Clifford algebra (see here, and several other texts like "Clifford Algebras and Spinors" by Lounesto)

I'll give a run-down of these definitions, and then I have some questions at the end.

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For definition #1, we take a vector $\vec{v} = x \vec{e_x} + y \vec{e_y} + z \vec{e_z}$ and write it as a linear combination of the sigma matrices to get a $2 \times 2$ hermitian matrix. This gives us:

$$\vec{v} \rightarrow V = x\sigma_x + y\sigma_y + z\sigma_z$$ $$ = x \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix} + y \begin{bmatrix} 0 & -i \\ i & 0\end{bmatrix} + z \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix} $$ $$ = \begin{bmatrix} z & x-iy \\ x+iy & -z\end{bmatrix} $$

We can check that $det(V) = - (length(\vec{v}))^2 = - (x^2 + y^2 + z^2) $. If the vector is "null", this quantity is zero. If the determinant of the matrix is zero, we can "factor" it as the multiplication of a column and row of complex numbers, which are the left and right spinors:

$$ V = \begin{bmatrix} z & x-iy \\ x+iy & -z\end{bmatrix} = 2 \begin{bmatrix} \xi_1 \\ \xi_2 \end{bmatrix} \begin{bmatrix} -\xi_2 & \xi_1 \end{bmatrix} $$

In this representation, when we want to rotate the components of a vector, we use a double-sided transformation $V \rightarrow UVU^\dagger$ with $U \in SU(2)$ being a $2 \times 2$ unitary matrix. So if a null vector is "split" into a (left) column spinor and a (right) row spinor, each spinor gets acted on by a single-sided transformation involving $U$ (for left) or $U^\dagger$ (for right).

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Now for Definition #2, we look for minimal ideals of a Clifford Algebra. For those unfamiliar, a "left ideal" is a term from algebra for a set where any left-multiplication brings you back into the set. "Right ideal" means the same thing, but for right-multiplication. A "minimal ideal" means an ideal with no sub-ideals inside it.

We can start with the Clifford Algebra $Cl(3,0)$ with the 3 basis vectors $\{ \sigma_x, \sigma_y, \sigma_z \}$ with $(\sigma_i)^2=+1$, and all anti-commuting with each other, $\{\sigma_i, \sigma_j\} = \sigma_i \sigma_j + \sigma_j \sigma_i = 0$ for $i\neq j$.

My understanding is that every (minimal) ideal has a (minimal) projector element $(p^2 = p)$ associated with it. So we can use the projector $p_+ = \frac{1}{2}(1 + \sigma_z) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $ to get a "left ideal", by left-multiplying the algebra on it, $Cl(3,0)p_+$. The basically gives us the set of $2 \times 2$ matrices with only the left column being non-zero. There is also the projector $p_- = \frac{1}{2}(1 - \sigma_z) = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $, which gives us another left ideal: the set of $2 \times 2$ matrices with only the right column being non-zero.

So in this interpretation, some examples of spinors as members of a minimal ideal would be:

$$\sigma_x p_+ =\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$

$$\sigma_x p_- =\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$

In this case, spinors actually add together to give us a vector... for example $(\sigma_x p_+) + (\sigma_x p_-) = \sigma_x$.

We could also act on projectors from the right to get $p_\pm Cl(3,0)$, which would give right ideals (the sets of $2 \times 2$ matrices with either the top row non-zero or bottom row non-zero).

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I don't understand the connection between these two definitions. I realize we can write

$$\begin{bmatrix} \xi_1 \\ \xi_2 \end{bmatrix} \begin{bmatrix} -\xi_2 & \xi_1 \end{bmatrix} = \begin{bmatrix} \xi_1 & 0 \\ \xi_2 & 0 \end{bmatrix} \begin{bmatrix} -\xi_2 & \xi_1 \\ 0 & 0 \end{bmatrix} $$

which makes column/row spinors look a bit more like members of left and right ideals, using the projectors I used in definition #2.

1. Do the two definitions describe the exact set same of objects? Or is one more general?
2. Is there a procedure for factoring null elements of a clifford algebra, like $\sigma_x + i \sigma_y$ into a product of spinors?
3. What motivates the notion of a "single-sided multiplication" on spinors in definition #2? There's no mention of "factoring a vector into spinors" anywhere, so I don't understand why we'd be motivated to do single-sided transformations on spinors. Is it just the fact that the spinors form an ideal, and therefore multiplication by any matrix gives another spinor?

Answer 1

I'll address the first question:

Do the two definitions describe the exact set same of objects? Or is one more general?

The definition in terms of a minimal left ideal is standard, for any dimension and signature. Given any minimal left ideal, the action of the Clifford algebra on a basis for that ideal defines an irreducible matrix representation of the Clifford algebra, which is how we usually define spinors in physics. See section 2.5 in ref 1.

The definition in terms of null vectors is different: it's the definition of a so-called pure spinor. That's what the rest of this answer is about.

Pure spinors: definition

Let $g_{jk}$ be a flat metric with signature $(p,q)$. The definition of pure spinor that I've found in the mathematical literature (refs 2 and 3) either allows vectors to have complex components, which includes the case illustrated in the question, or focuses on signatures for which $|p-q|\leq 1$. But in the physics literature, I've seen the same idea applied to Clifford algebras over the real numbers with Lorentzian signature instead. Ref 4 is an example. The following definition covers both cases. This definition seems to be consistent with the definition in chapter 3 of ref 1, but I'm not a mathematician, so please consult the references to double-check what I've written here.

Let $\gamma_1,\gamma_2,...,\gamma_N$ be a set of generators for the Clifford algebra Cliff($p,q$), satisfying $$ \newcommand{\bfx}{\mathbf{x}} \gamma_j \gamma_k + \gamma_k \gamma_j=2g_{jk} $$ as usual. A linear combination $\bfx = \sum_k x_k\gamma_k$ with real coefficients $x_k$ will be called a vector. Or, we can allow the coefficients to be complex, in which case the signature $(p,q)$ doesn't matter except for the total number of dimensions $N\equiv p+q$. In either case, a subspace $S$ such that $\bfx^2=0$ for all $\bfx\in S$ will be called a totally isotropic subspace, and it will be called a maximal totally isotropic subspace if it has the largest possible number of dimensions among all such subspaces. A spinor $\psi$ will be called pure if $\bfx \psi=0$ for all $\bfx\in S$ for some maximal totally isotropic subspace $S$. This assumes that spinors — not necessarily pure — are already defined in the standard way, as in the first paragraph above.

When are all spinors pure?

Now the question can be reframed like this: For what dimensions (in the complex case) and what signatures (in the real case) are all spinors pure? I don't know the complete answer, but here are a couple of excerpts regarding special cases:

• According to ref 4, regarding the special case of Lorentzian signature: "In 3,4,6 and 10 dimensions, a general spinor in the smallest irreducible spinor representation of the Lorentz group is automatically pure." The fact that normed division algebras exist only in dimensions 1, 2, 4, and 8 seems to be related to this. The relationship is reviewed in ref 5 and the appendix of ref 6.

Regarding the cases typically considered in the mathematical literature (complex coefficients, or real coefficients with $|p-q|\leq 1$):

• According to section 109 in ref 2: "It is obvious that any spinor can be taken, in an infinity of ways, as a sum of pure spinors..."

• From page 109 in ref 1: "In general not all spinors will be pure; whereas we can always choose a basis of pure spinors, linear combinations of pure spinors will not in general be pure."

• From page 113 in ref 1: "Eight dimensions are interesting as the lowest number of dimensions in which not all semi-spinors are pure." (What they call semi-spinors is what physicists would call chiral or Weyl spinors in the even-dimensional case.)

Altogether, there are dimensions/signatures where all spinors are pure (like the case shown in the question), and dimensions/signatures where not all spinors are pure.

Why are pure spinors interesting?

I haven't studied this subject enough to know all of the reasons, but one reason is that their association with maximal totally isotropic subspaces provides pure spinors an appealing geometric flavor. This seems to be useful for proving some properties of general spinors, at least in the cases of complex coefficients or $|p-q|\leq 1$ that are usually studied in the mathematical literature.

With Lorentzian signature, pure spinors play an important role in the study of supersymmetry, as illustrated in refs 4,5,6.

By the way, ref 4 defines a pure spinor to be one that satisfies the Cartan-Penrose equation $$ \sum_{j,k}g^{jk}(\overline\psi\gamma_j\psi)\gamma_k\psi=0. $$ This looks different than the definition I showed above, but we can see the relationship between the definitions by writing $\bfx\equiv \sum_{j,k}g^{jk}(\overline\psi\gamma_j\psi)\gamma_k$, so that the preceding equation is $\bfx\psi=0$. Multiply this on the left by $\bfx$ so see that it is consistent with $\bfx^2=0$, and note that a maximal totally isotropic subspace is one-dimensional when the signature is Lorentzian (that is, when $p$ or $q$ equals $1$).

Other uses of the name "pure spinor"

Beware that other, inequivalent definitions of pure spinor may also occur in the literature. Ref 7 seems to be one example. Page 2 in that paper says, "Pure spinors (in the sense of Cartan) can be defined in any dimension, but that definition turns out to be not quite what we want." Page 4 says, "[This] is not the conventional pure-spinor condition... However, as confusion is unlikely to arise we refer to the conditions [imposed here] as pure-spinor conditions throughout this note."

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References:

1. Benn and Tucker (1987), An Introduction to Spinors and Geometry with Applications in Physics, Adam Hilger

2. Cartan, The Theory of Spinors, reprinted in english by Dover in 1981

3. Chevalley (1954), The Algebraic Theory of Spinors and Clifford Algebras: Part 2, reprinted by Springer in 1997

4. Banks, Fischler, and Mannelli (2004), Microscopic Quantum Mechanics of the $p=\rho$ Universe" (https://arxiv.org/abs/hep-th/0408076)

5. Baez and Huerta (2009), "Division Algebras and Supersymmetry I" (https://arxiv.org/abs/0909.0551)

6. Dray, Janesky, and Manogue (2000), "Octonionic hermitian matrices with non-real eigenvalues," (https://arxiv.org/abs/math/0006069)

7. Grassi and Wyllard (2005), "Lower-dimensional pure-spinor superstrings," (https://arxiv.org/abs/hep-th/0509140)

Answer 2

Only answering your 3rd question

What motivates the notion of a "single-sided multiplication" on spinors in definition #2? There's no mention of "factoring a vector into spinors" anywhere, so I don't understand why we'd be motivated to do single-sided transformations on spinors.

The second definition of Spinors permits transformations from both sides: $$ \Psi \rightarrow U_L\Psi U_R $$ where $U_L$ and $U_R$ are independent.

Let's say $\Psi$ is a minimal left ideal of a Clifford Algebra. Then $U_L$ transforms within the same ideal, meaning $U_L$ is Lorentz-type transformation (e.g. a right-handed electron after Lorentz transformation is still a right-handed electron). On the other hand, $U_R$ transforms between different ideals, meaning $U_R$ is internal-gauge-type transformation (e.g. a green quark can be transformed into a red guark under $SU(3)_C$).

Consequently, type-A vector (in external space) is product like $$ \Psi \Psi^{\dagger} $$ which transforms as $$ \Psi \Psi^{\dagger} \rightarrow U_L\Psi U_R U_R^{\dagger} \Psi^{\dagger} U_L^{\dagger} = U_L\Psi \Psi^{\dagger} U_L^{\dagger} $$ while type-B vector (in internal space) is product like $$ \Psi^{\dagger} \Psi $$ which transforms as $$ \Psi^{\dagger} \Psi \rightarrow U_R^{\dagger}\Psi^{\dagger} U_L^{\dagger} U_L \Psi U_R = U_R^{\dagger}\Psi^{\dagger} \Psi U_R $$ assuming that $U_L$ and $U_R$ are unitary (in reality, we deal with $\Psi \bar{\Psi}$ or $\bar{\Psi}\Psi $, therefore $U_L$ is not unitory).

As you can see, if your force: $$ U_R = U_L^{\dagger} $$ and then you get vector-like transformations for spinor $$ \Psi \rightarrow U_L\Psi U_L^{\dagger} $$ However, you forfeits the otherwise much richer symmetry structure baked into the spinor.