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:
This can be achieved with the assignments (Lounesto, p. 14):
Reversion is achieved by transposing.
- 2. Now identify Δ, the space of spinors, as R2, the space of column vectors on which the matrices act
- 3. These spinors can be related to elements of the algebra if we make the assignments
These elements are a sub-algebra of the original Clifford algebra, spanned by
- A Clifford algebra on this subspace would be Cl1,0(R)
- Comments
The significance of the element ½ (1 + e1) is clear if we consider its corresponding matrix element,
This makes clear that ½ (1 + e1) is an
and why it annuls elements of the Clifford algebra that correspond to the projection out of other columns
- Nilpotent route
Alternatively, one starts by finding a nilpotent element (which will be represented by a nilpotent matrix)...
"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 ??
- Isotropic subspace
So we get an
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:
Left handed Weyl spinors:
Can both be drived from the real spinors
More explicitly
We start with the representation of the algebra using the
The spinors are then the space of column vectors on which these matrices act:
We now look for eigenvectors of the block-matrix .
This gives one eigenspace spanned by , which we shall call the right-handed Weyl spinor,
and one eigenspace spanned by , 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
a right-handed Weyl spinor corresponds to a space spanned by the elements
and a left-handed Weyl spinor corresponds to a member of the space spanned by the elements
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...)