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

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.

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.