这是一份imperial帝国理工大学 GG14作业代写的成功案例
In the NS-R sector the wavefunction can be written as
$$
|\phi\rangle_{e}=\left(b_{-1 / 2}^{\mu} \psi_{\mu e}+\cdots\right)|0\rangle
$$
and the physical state conditions imply that
$$
\partial \psi_{\mu}=\partial^{\mu} \psi_{\mu}=0, \ldots
$$
In the R-NS sector the wavefunction can be written as
$$
|\phi\rangle_{e}=\left(\bar{b}{-1 / 2}^{\mu} \psi{\mu \epsilon}^{\prime}+\cdots\right)|0\rangle
$$
$$
\partial \psi_{\mu}^{\prime}=\partial^{\mu} \psi_{\mu}^{\prime}=0, \ldots
$$
In the R-R sector the wavefunction can be written asand the resulting on-shell conditions are
$$
(\not{\partial}){e}^{\beta} S{\beta \delta}=0=(\not{\not}){\delta}^{\gamma} S{\epsilon \gamma}, \ldots .
$$
G1G3 COURSE NOTES ：
Raising the first index with the inverse metric and taking the trace we find that
$$
\Omega=\frac{1}{D} \partial_{\rho} f^{\mu} \partial_{\kappa} f^{v} \eta_{\mu \nu} \eta^{\rho \kappa} .
$$
Let us consider an infinitesimal transformation
$$
f^{\mu}=x^{\mu}+\varepsilon^{\mu}(x),
$$
then at lowest order, equation (8.1.4) becomes
$$
\partial_{\mu} \varepsilon_{v}+\partial_{v} \varepsilon_{\mu}=\frac{2}{D} \eta_{\mu \nu} \partial^{\rho} \varepsilon_{\rho},
$$
where
$$
\varepsilon_{v}=\varepsilon^{\mu} \eta_{\mu \nu} .
$$