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Consider the highest weight vector of the Spinor rep of $SO(1,d-1)$ where $d=2m+1$. It can be shown that: $$\gamma_i \gamma_{m+i}v=v\tag{*}$$ I cannot see why this relation does not imply that $v=0$? My reasoning why it should be zero is as follows:

Multiply both sides of * by $\gamma_{m+i} \gamma_i$ and use that (for $i\ne $1): $$\gamma_i \gamma_i=I$$ then we get: $$v=\gamma_{m+i} \gamma_{i}v$$ using the fact that $\gamma_{m+i}$ and $\gamma_i$ anti-commute gives us: $$v=-\gamma_i \gamma_{m+i}v$$ now sub in *: $$v=-v$$ hence $$v=0$$

Can someone explain why this is wrong?

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  • $\begingroup$ Please prove (*), in particular, check all sign conventions. I suspect the correct relation in your case should be $\mathrm{i}\gamma_i\gamma_{m+i} v = v$, but this question is unanswerable without knowing how you got to (*). $\endgroup$ – ACuriousMind Dec 11 '17 at 13:50
  • $\begingroup$ Frankly, I'm not gonna read through a bunch of handwritten notes to find some sign/factor of $\mathrm{i}$ that either you or the author of these notes dropped, and I don't think it is reasonable to expect any other user of this site to do so. $\endgroup$ – ACuriousMind Dec 11 '17 at 14:11
  • $\begingroup$ @ACuriousMind In all fairness this is never what I asked anyone to do, and didn't know it was down to a factor of $i$ error until your comment. Any answer explaining why $\mathrm{i}\gamma_i\gamma_{m+i} v = v$ is correct would do. (p.s. they are not my notes) $\endgroup$ – Quantum spaghettification Dec 11 '17 at 14:18
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The operators $H_i = \gamma_i \gamma_{m+i}$ fulfill $$ H_i^2 = - \eta_{ii}\eta_{(m+i)(m+i)},$$ and therefore for $i$ not time-like we have $H_i^2 = -1$. This means that $H_i$ has eigenvalues of $\pm\mathrm{i}$ (just observe that $H_i$ is diagonalizable and the eigenvalues must square to $-1$), so your equation (*) cannot be correct when $i$ is not time-like, which I presume is what you mean by $i\neq 1$, and should instead be $$ \gamma_i \gamma_{m+i} v = \mathrm{i}v,$$ which leads to no contradiction.

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