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What is the transformation law for spinors (pinors) under parity in an odd number of spacetime dimensions?

I know how to derive the transformation properties of spinors (pinors) under parity in an even number of spacetime dimensions. Let $$\eta^{ab} = \mathrm{diag} (1, 1\ldots 1, -1, -1 \ldots -1) $$ where there are $p$ entries of $1$ and $q$ entries of $-1$, and $p+q=n$. Let the gamma matrices $\gamma^a$ generate the real Clifford algebra $\mathrm{Cl}(p,q)$, $$ \{ \gamma^a, \gamma^b \} = 2 \eta^{ab} $$
The Pin group $\mathrm{Pin}(p,q)$ is defined as the set of invertible elements $S_{\Lambda}$ of $\mathrm{Cl}(p,q)$ that satisfy $$ S_{\Lambda} \gamma^a S_{\Lambda}^{-1} = {\Lambda^a}_b \gamma^b $$ for some element ${\Lambda^a}_b$ of the orthogonal group $\mathrm{O}(p,q)$, and also $S_{\Lambda}S_{\Lambda}^{\tau} = \pm 1$, where the superscript $\tau$ denotes a linear operator that reverses the order of products, e.g. $(\gamma_0 \gamma_1 \gamma_2)^{\tau} = \gamma_2 \gamma_1 \gamma_0$. For each $\Lambda$, there are two solutions for $S_{\Lambda}$ that differ by a minus sign, and the the map that sends these two solutions to $\Lambda$ is a $2-1$, homomorphism from $\mathrm{Pin}(p,q)$ to $\mathrm{O}(p,q)$.

A parity transformation in the orthogonal group $\mathrm{O}(p,q)$ that inverts the $i$-th spatial axis is given by $P_i = \mathrm{diag}(1, 1 \ldots 1, -1, 1, 1, \ldots 1)$, with the entry of $-1$ acting on the $i$-th spatial axis. To find a parity transformation on a spinor (pinor), one solves the above equations for $\Lambda = P_i$. In an even number of spacetime dimensions, the solution is $$ S_{P_i} = \pm \gamma_i \gamma_n $$ where $\gamma_n = \prod_{a=0}^{n-1} \gamma_a$.

Crucial to make this work is the fact that $\gamma_n$ anti-commutes with all of the gamma matrices $\gamma_a$ in an even number $n$ of spacetime dimensions. However, in an odd number of spacetime dimensions, the operator $\gamma_n$ is proportional to the identity matrix, and thus commutes with everything. And indeed I believe there is no operator in an odd number of spacetime dimensions that anticommutes with all the gamma matrices $\gamma_a$. As a result, I cannot see that there is a solution to the above equations for a parity transformation on spinors (pinors) in odd dimensions.

The main reference I have been using is In section 5, page 65, a similar conclusion is reached. Then it is said that the 2-1 homomorphism/covering map from $\mathrm{Pin}(p,q)$ to $\mathrm{O}(p,q)$ given above is not surjective in an odd number of spacetime dimensions, and in particular it does not 'hit' axis reflections in $\mathrm{O}(p,q)$.

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Note that "spinors" correspond to the even part of the Clifford algebra, or, equivalently, to the algebra generated by the commutators $[\gamma_\mu, \gamma_\nu]$ (and the identity). So we work with $Spin (p,q)$ and $SO(p,q)$ rather than $Pin (p,q)$ and $O(p,q)$. For instance, $Spin(p,q) = Spin(q,p)$, but this isomorphism is not necessary true for $Pin$ – Trimok Feb 6 '14 at 14:24
Yes, I should have used the word "pinor", question edited. – Steven Feb 6 '14 at 16:03
up vote 0 down vote accepted

I believe I'm ready to answer my own question.

The pin group can alternately be defined as the set of all invertible elements $S_{\Lambda} \in \mathrm{Cl}(p,q)$ satisfying $S_{\Lambda} S_{\Lambda}^{\tau} = \pm 1$ and $$ \alpha(S_{\Lambda}) \gamma^a S_{\Lambda}^{-1} = {\Lambda^a}_b \gamma^b $$ for some element ${\Lambda^a}_b \in \mathrm{O}(p,q)$. The map $\alpha: \mathrm{Cl}(p,q) \rightarrow \mathrm{Cl}(p,q)$ sends odd elements of $\mathrm{Cl}(p,q)$ to minus themselves, and even elements to themselves. It is an algebra automorphism.

This defines a second $2-1$ homomorphism called the twisted map from $\mathrm{Pin}(p,q)$ to $\mathrm{O}(p,q)$ that is surjective in any number of spacetime dimensions. In particular, the elements that get mapped to a reflection of the $i$-th spatial axis are $\pm \gamma_i$.

The major difference when using the twisted map to define parity transformations is that $\gamma^a$ is now a pseudovector since it transforms with a minus sign under reflections.

The twisted map is the only surjective homomophrism from $\mathrm{Pin}(p,q)$ to $\mathrm{O}(p,q)$ in odd dimensions and therefore it must be used to define the parity transform for spinors. In an even number of spacetime dimensions, there is a choice to be made. There is further ambiguity in the sign of the parity operator, and the spacetime metric signature, which can lead to different parity operators. Which parity operator is 'correct' is a matter that is determined by experiment.

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protected by Qmechanic Apr 30 '14 at 14:02

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