User:Jheald/sandbox/GA/Spinors -- sui generis, or more general?

Trying to build on earlier thread Talk:Spinor#Spinors_in_the_Geometric_Algebra_literature, and references therein.

• Possible new article target: Spinors in Geometrical Algebra.

• Classically, define a spinor as something corresponding to a (usually complex) column vector, that a matrix from the matrix representation of the Clifford algebra could operate on:

${\begin{pmatrix}.\\M\\.\end{pmatrix}}{\begin{pmatrix}.\\.\\.\end{pmatrix}}={\begin{pmatrix}.\\.\\.\end{pmatrix}}$

• Then, recognise that such matrix elements are isomorphic to elements of the algebra right-multiplied by a projector element

${\begin{pmatrix}.\\.\\.\end{pmatrix}}\simeq {\begin{pmatrix}.&0&0&\cdots \\.&0&0&\cdots \\.&0&0&\dots \end{pmatrix}}={\begin{pmatrix}.\\M'\\.\end{pmatrix}}{\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}}$

In 2D this corresponds to post-multiplying by ½(1 + e₁), in 3D to ½(1 + e₃) etc.

• Transformations in the smaller "spinor space" can be spanned by a sub-algebra of the original Clifford algebra (since the projector is an idempotent, so absorbs anything in the direction of itself):

For example, Cl_1,0(R) in 2D; Cl_2,0(R) in 3D

• So compared to a general multivector equation, a spinor equation is an equation in a projection of the original algebra

For example, for an equation on a sphere (spherical harmonics), one might want to project out the radial dependence

• But

Does the spinor equation really depend on the choice of the original projection vector?

Projecting with a different vector, eg ½(1+e₂) would be equivalent to projecting a different column. But does that give the same equation?

• Guess: it depends, on whether the Clifford operator that is being applied to the spinor is invariant under the exchange e₃ ↔ e₂

More generally, need to look at how things transform under rotations and other transformations of the reference axes.

But. That can't be a pre-condition, because if we're operating on a general multivector, doesn't choosing any column to project give the same equation?

Indeed. Therefore the projected equation can't just apply in one projection, it has to apply in all projections (or, at least, all similar projections).

This must be what it means that the projected space Δ is called isotropic.

• So the projection is the way to get a vector-like thing out of the matrix.

If we were dealing with matrices, we could then look at each basis element of the vector-like thing, and look at how that transforms; then putting those together would give a complete description of the transformation -- would it make sense to try to do something similar?

So what you do is to see how the multivector operates on a general spinor; then to operate on a general operand multivector, you would project that multivector out orthogonally into different spinor components, apply the operator, then put them all back together again. That might be quite neat.

Doran et al hit the Pauli equation with something like that, and get out helicity; and then say it's always like bivectors...

Idempotent projectors

So the key thing appears to be idempotent projectors like

$(1+e_{3})\;$

(at least in a space where e₃² = 1; if e₃² = -1, then may need the pseudoscalar in there to do something similar - cf Bott periodicity & Lounesto: Tilt to the opposite metric).

We can adjust our matrix representation so that (1+e₃) is represented by the block matrix $\left({\begin{smallmatrix}1&0\\0&0\end{smallmatrix}}\right)$

The opposite block matrix $\left({\begin{smallmatrix}0&0\\0&1\end{smallmatrix}}\right)$ then represents the element (1-e₃).

These two block matrices are orthogonal, just as

$(1+e_{3})(1-e_{3})=0\;$

Right-multiplying a general matrix M by the matrix representing (1+e₃) zeroes the right-hand half of the columns:

$\left({\begin{smallmatrix}*&*\\{*}&*\end{smallmatrix}}\right)\left({\begin{smallmatrix}1&0\\0&0\end{smallmatrix}}\right)=\left({\begin{smallmatrix}*&0\\{*}&0\end{smallmatrix}}\right)$.

Correspondingly, we can break a general multivector A into two parts,

$A=A(1+e_{3})+A(1-e_{3})\;$

If the original A had 2ⁿ degrees of freedom, each of the smaller ones has 2^n-1 degrees of freedom, and is spanned by Cl_n-1 (1+e₃).

We can note that

$AB=A(1+e_{3})B+A(1-e_{3})B\;$

$\qquad =A(1+e_{3})[(1+e_{3})B]+A(1-e_{3})[(1-e_{3})B]$

so each half of A pulls out a similar half of multivectors it acts on.

Repeating this process, with each idempotent projector we apply we can halve the width of the active part of the representation matrix, until eventually we reach something one (or perhaps two; or just possibly four) column(s) wide. This corresponds to what has classically been called a spinor.

• One could then look at how particular parts of the algebra (eg rotations) act on such a column, with a view to identifing transformations to reduce the algebra into non-mixing part.

The effect of the projector is to flatten the e₃ direction out of the Clifford algebra; this is because

$e_{3}(1+e_{3})=e_{3}+1=1(1+e_{3})\;$

-- we can imagine breaking the multivector into blades, permuting any e₃ dependence to the right, then replacing that factor of e₃ with a factor of (+1).

Similarly, for the other projector

$e_{3}(1-e_{3})=e_{3}-1=-1(1-e_{3})\;$

so this operator is equivalent to flattening the multivector by replacing right-factors of e₃ with a factor of (-1).

(so (1 ± e₃) has eigen-elements 1 (trivial) and e₃ (eigenvalue ±1)... relevant?)

The symmetry operation

$e_{3}\mapsto -e_{3}$

would switch the two projectors over. Cf the Charge Conjugation operator, as discussed in Lounesto. Consequences of the physical world being invariant under such re-labelling ?

But. We can also left multiply; doing both gives a projection

$(1+e_{3})A(1+e_{3})\;$

corresponding to the submatrix $\left({\begin{smallmatrix}*&0\\0&0\end{smallmatrix}}\right)$

... this is going to keep only the even-graded part of A (excluding e₃) -- exactly what we need for analysing rotations ...except that rotations in the e_ie₃ plane are going to now show up in the "odd" sub-half.

which is different from

$(1-e_{3})A(1+e_{3})\;$

corresponding to the submatrix $\left({\begin{smallmatrix}0&*\\0&0\end{smallmatrix}}\right)$

... this is going to keep only the odd-graded part of A -- which will be zero for rotations

If the interpretation about zapping all of the e₃ factors out of the multivector were correct, how would this be possible? Needs to be looked at.

Aside: what happens to a rotation like $R=\cos(\theta /2)+e_{31}\sin(\theta /2)=\exp(e_{31}\;\theta /2)$ ?

The rotation sends

$e_{1}\mapsto Re_{1}{\tilde {R}}=e_{1}\cos(\theta )+e_{3}\sin(\theta )$

This on the face of it is going to be flattened to

$sin(\theta )+e_{1}\cos(\theta )$

... which is a scalar plus a vector, i.e. weird.

... compare this to the usual projector operator, which would give $e_{1}\cos(\theta )$ <-- check this; & examine the difference further

A thought: e₃ represents a reflection. The projection (1 + e₃) therefore corresponds to symmetrising with respect to this reflection (cf a symmetric function); (1 - e₃) corresponds to anti-symmetrising with respect to this reflection.

Projective geometry

Talking about the effect of rotations, Weyl–Brauer matrices has: "the action of the orthogonal group on spinors is not single-valued, but instead descends to an action on the projective space associated to the space of spinors."

Dorst et al, p. 308: "Projective transformations are the general linear transformations in the space ℝⁿ⁺¹, interpreted in the space ℝⁿ. Such general transformations can transform finite points to infinite points (e.g., the rotation in ℝⁿ⁺¹ that turns e₀ into e₁ is among them)."

Here we have exactly a rotation in ℝⁿ⁺¹ (our original space), that is turning e₀ into e₁ in our flattened space --> so ℝⁿ⁺¹ would be the projected space over our flattened space. More to think about on this.

Resources

• Lounesto: ch.10: Dirac eqn, esp pp.146-149; & ch. 12; & especially ch.17 (pp.225--229)

• Brauer & Weyl, "Spinors in n dimensions". JSTOR, only p.1 available