这是一份imperial帝国理工大学 GG14作业代写的成功案例
$$
C=-B^{T} A
$$
then
$$
\gamma_{m}^{T}=-\rho(-1)^{t+1} C \gamma_{m} C^{-1} .
$$
Evaluating $\left(\gamma_{m}^{}\right)^{\dagger}=\left(\gamma^{\dagger}\right)^{}$ using equations we find that
$$
\gamma_{m} A^{-1} B^{\dagger} A^{} B=A^{-1} B^{\dagger} A^{} B \gamma_{m} .
$$
$$
A^{-1} B^{\dagger} A^{} B=\mu I . $$ However, using $$ A^{}=\gamma_{1}^{} \ldots \gamma_{t}^{}=\rho^{t} B A B^{-1}
$$
and equation then becomes
$$
B^{\dagger} B=\mu \rho^{t} I .
$$
G1F3 COURSE NOTES ：
Weyl symmetry:
$$
\begin{aligned}
\delta e_{\alpha}{ }^{a} &=\Lambda e_{\alpha}{ }^{a}, \delta \psi_{\alpha}=\frac{1}{2} \Lambda \psi_{\alpha}, \
\delta x^{\mu} &=0, \delta \chi^{\mu}=-\frac{1}{2} \Lambda \chi^{\mu}, \delta N^{\mu}=-\Lambda N^{\mu} .
\end{aligned}
$$
Local ‘S-supersymmetry’:
$$
\delta \psi_{\alpha}=\gamma_{\alpha} \zeta,
$$
where $\zeta_{A}$ is a Majorana spinor which depends on $\xi^{\alpha}$ in an arbitrary way. All the other fields are inert under $S$-supersymmetry.
The Poincaré group:
$$
\delta x^{\mu}=-\Lambda^{\mu}{ }{v} x^{v}+a^{\mu}, \delta \chi^{\mu}=-\Lambda^{\mu}{ }{v} \chi^{v} \text { etc. }
$$