File talk:Spinor on the circle.pdf


 * Misrepresentation:

The arrows should not change at the join, if the indicated direction maintains the original orientation (without the inversion), the nature of the mobius strip would automatically invert the opposing side. In other words. One "side"(sic) would be UP, and the other "side"(sic) would be DOWN, by nature. By flipping at the join, we do not see the process described. I don't know who the original author is; but, I would suggest modifying this image. I do not have the tools needed, or I would do it myself. --Cyberchip (talk) 17:19, 28 January 2015 (UTC)


 * I'm not exactly clear what your objection is, but the image does illustrate what it is supposed to (see the caption and technical footnote ate Spinor, where the image is used). The space of spinors associated to the Lorentz group SO(1,2) in three dimensions is a two-dimensional real vector space V, with an action of the group SL(2,R) (which is just the spin group in this signature).  This goes over to an action on the projective line PV over V, with kernel $$\pm 1$$, and so this action factors through the group SO(1,2) (and in particular the copy of SO(2) sitting inside there, which acts on the projective line by rotations).  The projective line PV has a tautological bundle (the Mobius bundle) and an non-zero element of V can, in a natural way, be identified with a point of that bundle.  The rotation group SO(2) (inside SO(1,2)) acts in the usual way on the circle.  But this action fails to globally lift to an action on the Mobius bundle.  In fact, when one integrates this action, we get a non-trivial holonomy of $$-1$$.   Sławomir Biały  (talk) 20:47, 28 January 2015 (UTC)


 * I admit, Lorentz is one of my weaker areas; I'm trying to see if I understand. You're saying in one way, this fits descriptively with the particular Lorentz group, but it's missing another point of inversion (π?), or doesn't need the extra inversion if we're talking projective lines. And, almost hits the mark, but with the mobius group, it misses because it has the one point, so the corollary by using the mobius strip doesn't conform.  So, it's wrong on for both frames of reference or mobius, or partly right on both frames but still fails as an analogy with mobius.  Is this close? I'm testing my concept (and I'm keenly interested in holography ; I don't often pick up on the terminology. Perhaps the other inversion is hidden. We're talking phase conjugation (matrix transformations), yes? maybe non-trivial zero @ 0,π, and 2π=0 crossovers, sign inversions due to harmonics?  Or am I all confused? lol sorry. I hope you don't mind if I pick your brain a bit; I"m self taught, and still working on Lie groups and such. It's a translation problem (a transformation problem :) I hope to see it with your words some day; I'm high on concept, weak on math symbolism. Looking forward to your reply, I have no peers to talk to.(I will look up every word you use, and eventually understand it.)  After studying the terminology you used, I think I'm beginning to see (not a Monodromy either?! (too soon?)) Cyberchip (talk) 09:09, 29 January 2015 (UTC)


 * Almost there. Let me try to explain more concretely.  A spinor for the three-dimensional Lorentz group is a real two-component vector $$\pi_A$$.  A null vector in three dimensions can be written as a product $$x_a=\pi_A\pi_{A'}$$ (Notation is roughly that of Penrose and Rindler.)  Null vectors form the usual cone in $$\mathbb R^3$$.  The set of null rays is the "circle" in the Mobius strip, and the points of the Mobius strip itself are the spinors $$\pi_A$$ lying over each null ray $$x_a$$ such that a factorization $$x_a=\pi_A\pi_{A'}$$ exists.  Clearly, for any fixed value of $$x_a$$, there will be two such factorizations though: one with $$\pi_A$$ and one with $$-\pi_A$$.  Having chosen one or the other factorization, we can subject the null vector $$x_a$$ to an infinitesimal rotation (or more general infinitesimal Mobius transformation), and we can still manage to choose the spinor $$\pi_A$$ in a consistent (continuous) way.  But, when we subject the vector $$x_a$$ to a full rotation, it is impossible to choose $$\pi_A$$ consistently, because it will have transformed to its negative.   Sławomir Biały  (talk) 12:21, 29 January 2015 (UTC)