Equivalence of the Dirac representation of the Lorentz algebra and its conjugate in even dimensions - Polchinski

Question

In Polchinski's String Theory, Appendix B.1 we look at the smallest irreps of the Clifford algebra in even-dimensional spacetimes $d=2k+2$. In (B.1.16) he defines two operators from the gamma matrices $\Gamma^{\mu}$ $$ B_1 = \Gamma^3 \Gamma^5\ldots \Gamma^{d-1} \quad , \quad B_2 = \Gamma B_1 $$ where $ \Gamma = i^{-k} \Gamma^0 \Gamma^1 \ldots \Gamma^{d-1}$ is the chirality operator.

In (B.1.18) Polchinski then writes that for either $B_1$ or $ B_2$ and only for these two matrices the following holds: $$ B \Sigma^{\mu \nu} B^{-1} = - \Sigma^{\mu \nu *} \tag{B.1.18} \label{1}$$ where $\Sigma^{\mu \nu} = - \tfrac{i}{4} [\Gamma^{\mu},\Gamma^{\nu}] $ are the Lorentz generators and $*$ denotes complex conjugation. In other words this signifies the equivalence with the conjugate of the Dirac representation of the Lorentz algebra.

Question: How do you show that \eqref{1} indeed holds only for $B_1$ or $B_2$?

Any help much appreciated.

Answer 1

I have found my answer. It requires a little bit of knowledge of gamma matrices. Of course, as the question is stated, we can find more than $2$ matrices which satisfy $~(\rm B.1.18)~$, as for example $~\lambda B~$ for any $~0\neq\lambda~$ also works and most linear combinations of these matrices work as well. What is meant here is that $~B_{1,2}~$ are the independent generators. We will show this now.

Assume we have some $ B_1 $ and $B_2$ such that $~(\rm B.1.18)~$ holds. Defining $~X= B_2 B_1^{-1} ~$, this implies that $$ [X,\Gamma^{\mu \nu}]=0 \tag{1}$$ where $~\Gamma^{\mu \nu}=\Gamma^{[\mu} \Gamma^{\nu]} \equiv \frac{1}{2}[\Gamma^{\mu},\Gamma^{\nu}] ~$. Below we show that this fact implies that $$ X=a\mathbf{1}+b\Gamma ~. \tag{2} $$ This then answers our question, as $~B_2=aB_1+b\Gamma B_1~$ so the generators indeed are $~B_1~$ and $~\Gamma B_1~$.

Proof of $(2)$ from $(1)$

This assumes basic knowledge of gamma matrices which can be found e.g. in Chapter 3 of the book Supergravity by Freedman and van Proeyen. First define $~\Gamma^{\mu_1\ldots\mu_r}=\Gamma^{[\mu_1}\ldots\Gamma^{\mu_r]}~$ to be the completely antisymmetrized products of gamma matrices. It can be shown that the set $$ \{\mathbf{1},\Gamma^{\mu},\Gamma^{\mu_1\mu_2},\ldots,\Gamma^{\mu_1\ldots\mu_d} \} $$ where the index values satisfy $~\mu_1<\ldots<\mu_r~$ forms a basis of the space of (complex) matrices of dimension $~2^{k+1} \times 2^{k+1}~$ where $~d=2k+2~$. Any such matrix $~M~$ can thus be expanded as $$ M=\sum_{k=0}^{d} \frac{1}{k!} m_{\mu_1\ldots\mu_k} \Gamma^{\mu_1\ldots\mu_k} ~.$$ Expanding in this way our matrix $~X=B_2 B_1^{-1}~$ the condition $(1)$ becomes $$0= \sum_{k=0}^{d} \frac{1}{k!} x_{\mu_1\ldots\mu_k} [\Gamma^{\mu_1\ldots\mu_k}, \Gamma^{\nu \rho}] \tag{3} $$ Using the algorithm explained in Chapter 3 of Supergravity we can show that $$ [\Gamma^{\mu_1\ldots\mu_k}, \Gamma_{\nu \rho}] = 4k \Gamma^{[\mu_1\ldots\mu_{k-1}}_{\phantom{[\mu_1\ldots\mu_{k-1}}[\rho} \delta^{\mu_k]}_{\nu]}~. \tag{4}$$ Note that $~[\mathbf{1},\Gamma_{\nu \rho}]=0~$ as well as $~[\Gamma^{\mu_1 \ldots \mu_d} ,\Gamma_{\nu \rho}]=0~$, which also follows from $(4)$ because we have only $d$ distinct indices. Note that $~\Gamma^{\mu_1 \ldots \mu_d} \propto \varepsilon^{\mu_1\ldots\mu_d}\Gamma ~$ where $~\varepsilon^{\mu_1\ldots\mu_d}~$ is the Levi-Civita tensor. Altogether $(3)$ and $(4)$ imply that $$ x_{\mu_1\ldots\mu_k}=0 \qquad \text{for} \quad 1\leq k < d ~.$$ This gives $(2)$, quod erat demonstrandum.