factor of $1/n!$. So, for instance, $\Gamma_{(0)}$ is the unit matrix, $\Gamma_{(0)}$ $=$ $I$, and $\sigma^{\mu\nu}$ $=$ $\Gamma_{(2)}^{\mu\nu}$. Though we will invariably retain $\gamma^\mu$ for $n$ $=$ $1$. One advantage of this basis is that $$\mbox{tr} \left( \Gamma_{(m)}^{\mu_1 \ldots \mu_m} 
\Gamma_{(n)}^{\nu_1 \ldots \nu_n} \right) ~ \propto ~ \delta_{mn}
I^{\mu_1 \ldots \mu_m \nu_1 \ldots \nu_n} 
\label{gamtr}$$ where $I^{\mu_1 \ldots \mu_m \nu_1 \ldots \nu_n}$ is the unit matrix in this $\Gamma$-space, [@37; @38; @39; @40; @41]. As a result of this there is a partitioning of the projection matrix into sectors with different $\Gamma_{(n)}$ as is evident in the examples in Appendix A. Two final points concerning the projectors are worth emphasising. First, our choice of basis tensors, ${\cal P}^i_{(k) \, \mu_1 \ldots \mu_{n_i} }(p,q)$, only has meaning strictly at the symmetric subtraction point. Away from this point there will be a bigger basis set of tensors since then we would have $p^2$ $\neq$ $q^2$, $p^2$ $\neq$ $(p+q)^2$ and $q^2$ $\neq$ $(p+q)^2$ as is evident from the explicit forms given in Appendix A. Second, at the symmetric point the number of independent tensors in each case is clearly larger than that of the asymmetric forward subtraction point considered in the RI${}^\prime$ scheme. 