Let $\psi$ denote a Dirac spinor then Weyl spinors are defined by: $$\psi_{L,R}=\frac{1}{2} (I\pm \gamma)\psi$$ on even dimensions $\gamma$ commutes with $\sigma_{\mu \nu}$ (generators used to define spinor rep) and as such this is clearly a subrep of the dirac spinors. But I don't understand why it is then claimed what we can't define Weyl spinors for odd dimensions. Might it also be true that $\psi_{L,R}$ form a subrep in this case - even if less explicitly?

On a related note Majorana-Weyl spinors are said to only exist in $2 \mod 8$ dimensions which is derived on the condition that: $$[\gamma, C]=0$$ but I cannot see why this needs to be true. Since surely we only need them to share a single eigenvector rather then all of them (c.f. this post).

Please can someone explain these two points?

Note on Notation: $\gamma=i^{-k}\gamma_0 \ldots \gamma_{d-1}$ and $C$ is the operator for which Majorana spinors are defined: $\psi =C\psi$


migrated from math.stackexchange.com Dec 26 '17 at 20:16

This question came from our site for people studying math at any level and professionals in related fields.

  • $\begingroup$ I think I have found the answer to the first of these, i.e. in the odd case $\gamma\propto 1 $ so the projector simply projects back onto the Dirac spinors or $0$ and in either case a trivial subrep. I am still confused about the second part. $\endgroup$ – Quantum spaghettification Dec 26 '17 at 13:18
  • 1
    $\begingroup$ I have found H. Murayama’s notes: hitoshi.berkeley.edu/230A/clifford.pdf to be a good and clear intro to the eightfold periodicty of Majorana and Weyl fermions. You want to look at, and understand the origin of, his table 4 for Minkowski signature. $\endgroup$ – mike stone Dec 26 '17 at 21:44
  • $\begingroup$ Appendix E of de Wit, B. and Smith, J. (1986), Field Theory in Particle Physics (North-Holland Personal Library), V1, ISBN 978-0444869999 should do the trick. $\endgroup$ – Cosmas Zachos Dec 27 '17 at 14:24
  • $\begingroup$ If mathematically inclined, peruse the Clifford algebra clock. $\endgroup$ – Cosmas Zachos Dec 27 '17 at 16:38
  • $\begingroup$ You seem to have taken an alarming and mysterious left turn with your oracular "surely we only need them to share a single eigenvector rather than all of them". I doubt your question is meaningful, unless you explain this preposterous misconception. $\endgroup$ – Cosmas Zachos Dec 27 '17 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.