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Action of Clifford-elements on vectors & spinors

  1. Sep 18, 2008 #1
    Hi all!

    I am currently preparing a talk about Clifford algebras and pin/spin-groups. Since half the audience will consist of physicians (as I am myself) I also want to get more into the connection of the mathematical definitions and derivations (as one may find in Baker, "Matrix groups" or, more for the physical liking, "Analysis, Manifolds and Physics (vol. 2)" of Choquet-Bruhat Y. & Dewitt-Morette) to the physical tools of everydays use, like Weyl-spinors, Majorana representation, behaviour of spinors under rotations.

    Especially the last point is unclear to me. The only somewhat good explanation I could find was in Wikipedia, article "Spinor" (http://en.wikipedia.org/wiki/Spinor" [Broken]). Under >> "Examples" >> "Two dimensions" it is written that the action of elements on vectors is

    [tex]\gamma\left(u\right) = \gamma u\gamma^*[/tex]

    whereas on spinors it is

    [tex]\gamma\left(\phi\right) = \gamma \phi[/tex].

    So the spinor shell be just a complex number. But where do these actions come from? What distinguishes, in this special case, vector and spinor? I am somewhat confused.

    Thanks everybody helping me out!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 18, 2008 #2
    I'm not sure what that article what that article means by action on a spinor, but as used, "the action on a vector" part is just a rotation.

    If u lies in the plane of the bivector [itex]i = \sigma_1\sigma_2[/itex], then yes, this is just a complex product, but this action also works as a rotation in higher dimensions. This is because both [itex]i[/itex] and the scalar component of the spinor will both commute with any component perpendicular to the plane.

    For example, if you consider the split of a vector into parts parallel to the plane and perpendicular to the plane

    u = u_\parallel + u_\perp = (u \cdot i) \frac{1}{i} + (u \wedge i) \frac{1}{i}

    and a spinor

    \gamma = \exp(i\theta/2) = \cos\theta/2 + i\sin\theta/2

    the action is linear, so both components can be considered separately. For the parallel to the plane part one has (and you can verify this by multiplying the bits out)

    \exp(i\theta/2) u_\parallel \exp(-i\theta/2) = \exp(i\theta) u_\parallel

    ... this takes the form of a normal complex rotation.

    for the component out of the plane one has:

    \exp(i\theta/2) u_\perp \exp(-i\theta/2) = \exp(i\theta/2) \exp(-i\theta/2) u_\perp = u_\perp

    Thus the action produces a rotation around a vector normal to the plane, but formulates this in a way that works in any dimension (like 4D where one can't express this normal in an unambigous fashion.)

    Geometrically, a spinor action of this sort can rotate higher grade elements such as planes in an intuitive fashion. I can't imagine geometrically what it would mean to apply such an action to a mixed grade object such as this 2,0 spinor, nor how that produces the "complex product" mentioned in the wiki article.

    fwiw. A reference that I'd recommend for this material is 'Geometric Algebra for Computer Science'. It doesn't have the applied to physics focus but is much easier to understand then the Doran/Lasenby text for example.
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