User:Jheald/sandbox/GA/Spinors in two dimensions

Per the recipes in article spinor, let us construct respresentations of spinors in two dimensions:

Component spinors: real form

 * 1. Create an explicit representation of the Clifford algebra. (ie, a Clifford module).

Basis:
 * $$e_1^2 = e_2^2 = +1; \qquad (e_1 e_2)^2 = -1$$

This can be achieved with the assignments (Lounesto, p. 14):



1 \simeq \begin{pmatrix} 1&0\\0&1 \end{pmatrix};\; e_1 \simeq \begin{pmatrix} 1&0\\0&-1 \end{pmatrix};\; e_2 \simeq \begin{pmatrix} 0&1\\1&0 \end{pmatrix};\; e_1 e_2 \simeq \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} $$

Reversion is achieved by transposing.


 * 2. Now identify &Delta;, the space of spinors, as R2, the space of column vectors on which the matrices act


 * $$M \begin{pmatrix}u\\v\end{pmatrix} = \begin{pmatrix}u'\\v'\end{pmatrix}$$


 * 3. These spinors can be related to elements of the algebra if we make the assignments


 * $$\begin{pmatrix}u\\v\end{pmatrix} \simeq \begin{pmatrix}u&0\\v&0\end{pmatrix}$$

These elements are a sub-algebra of the original Clifford algebra, spanned by


 * $$f_0 = \tfrac{1}{2} (1 + e_1)$$
 * $$f_1 = \tfrac{1}{2} {e_1 - e_1 e_2} = \tfrac{1}{2} e_2 (1 + e_1)$$


 * A Clifford algebra on this subspace would be Cl1,0(R)

The significance of the element ½ (1 + e1) is clear if we consider its corresponding matrix element,
 * Comments
 * $$\tfrac{1}{2} (1 + e_1) \simeq \begin{pmatrix}

1&0\\ 0&0 \end{pmatrix};$$ This makes clear that ½ (1 + e1) is an idempotent
 * $$\tfrac{1}{2} (1 + e_1) \tfrac{1}{2} (1 + e_1) = \tfrac{1}{2} (1 + e_1)$$

and why it annuls elements of the Clifford algebra that correspond to the projection out of other columns
 * $$\tfrac{1}{2} (1 - e_1) \tfrac{1}{2} (1 + e_1) = 0$$

Alternatively, one starts by finding a nilpotent element (which will be represented by a nilpotent matrix)...
 * Nilpotent route
 * $$\tfrac{1}{2} e_1 (1 - e_2) \simeq \begin{pmatrix}

0&0\\ 1&0 \end{pmatrix};$$ "The construction via nilpotent elements is more fundamental in the sense that an idempotent may then be produced from it" -- don't understand this. ... ?? maybe to do with iterating a nilpotent element to build up a flag ??

So we get an isotropic subspace
 * Isotropic subspace
 * $$W = \mathrm{span}(f_0,f_1)\;$$

is one because it contains the nilpotent above.

Why "isotropic" ? -- no clear derivation I can see yet from the more ordinary sense of "equal in all directions"

"maximal isotropic subspace" --> pull out the whole off-diagonal part of the column? No, not quite right.

Weyl spinors
The action of γ ∈ Cℓ02,0 on a spinor φ ∈ C is given by ordinary complex multiplication:

Right handed Weyl spinors: $$\gamma(\phi) = \gamma\phi$$

Left handed Weyl spinors: $$\gamma(\phi)=\bar{\gamma}\phi$$

Can both be drived from the real spinors

More explicitly
We start with the representation of the algebra using the Weyl-Brauer matrices:



1 \simeq \begin{pmatrix} 1&0\\0&1 \end{pmatrix};\; e_1 \simeq \begin{pmatrix} 0&1\\1&0 \end{pmatrix};\; e_2 \simeq \begin{pmatrix} 0&i\\-i&0 \end{pmatrix};\; e_1 e_2 \simeq \begin{pmatrix} -i&0\\0&i \end{pmatrix} $$

The spinors are then the space of column vectors on which these matrices act:


 * $$M \begin{pmatrix}u\\v\end{pmatrix} = \begin{pmatrix}u'\\v'\end{pmatrix}$$

We now look for eigenvectors of the block-matrix $$U = \left(\begin{smallmatrix}1 & 0 \\ 0 & -1\end{smallmatrix}\right)$$.

This gives one eigenspace spanned by $$\left(\begin{smallmatrix}1 \\ 0\end{smallmatrix}\right)$$, which we shall call the right-handed Weyl spinor,

and one eigenspace spanned by $$\left(\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right)$$, which we shall call the left-handed Weyl spinor.

In terms of elements of the algebra

 * There is no element of Cl2(R) that we can identify with M.
 * Nor can we construct either Weyl spinor from elements of Cl2(R)
 * However, we can recognise M as the what could correspond to e3 in a representation of Cl3(R) that contained this representation of Cl2(R)

Taking this route, the spinor space would be a space spanned by the elements
 * $$x \; \tfrac{1}{2}(1 + e_3)$$

a right-handed Weyl spinor corresponds to a space spanned by the elements
 * $$\tfrac{1}{2}(1 + e_3) \; x \; \tfrac{1}{2}(1 + e_3)$$

and a left-handed Weyl spinor corresponds to a member of the space spanned by the elements
 * $$\tfrac{1}{2}(1 - e_3) \; x \; \tfrac{1}{2}(1 + e_3)$$

where x is a general element of Cl3(R).


 * if x is a general element of Cl2(R) -- i.e. no e3 factors -- then the right spinor will represent the even part of x, and the left spinor the odd part.

Interpretation


"How the hell do I add a scalar to a vector ?"

(I know how to add a scalar to a bivector, and what it means...)