时空线元$\dd s^2=g_{\mu\nu}\dd x^\mu\dd x^\nu$，进行对偶基变换$\omega^a=e^a_{\; \mu}\dd x^\mu$，使得新基下度规为洛伦兹度规$\dd s^2=\eta_{ab}\omega^a\omega^b$，
度规变换$g_{\mu\nu}=e^a_{\; \mu} e^b_{\; \mu}\eta_{ab}$，对偶变换$\lbrace e_{a}^{\;\mu}\rbrace$满足$e_{a}^{\;\mu}e_{\;\nu}^a=\delta_\nu^\mu$，$e_{a}^{\;\mu}e_{\;\mu}^b=\delta_a^b$，基$v_a=e_a^{\;\mu}\partial_\mu$满足$v_a\omega^b=\delta_a^b$
现在得到了分别由希腊字母$\mu,\nu,\lambda,\cdots$ 和拉丁字母$a,b,c,\cdots$ 表示的两组基，将前者称为时空基，后者称为自旋基，时空基到自旋基的变换$e^a_{\; \mu}$的选取并不唯一，可以差一个自旋基中的变换，即洛伦兹矩阵$e’^{a}_{\; \mu}=L^a_{\; b}e^b_{\; \mu}$，变换后仍可得到洛伦兹度规，$L^a_{\; c}L^b_{\; d}\eta_{ab}=\eta_{cd}$，
在坐标$x$变化时，除了时空基$\partial_\mu$会产生改变外，也会使得自旋基$v_a$产生变化，对于基空间为时空基与自旋基直积空间的旋量而言，时空的切丛联络不能描述自旋基的变化，因此需引入能够描述混合基 $e^a_{\; \mu}$ 变化的联络
考虑自旋矢量场$A^a=e^a_{\; \mu} A^\mu$，引入自旋联络，其描述当时空坐标变化时自旋基的变化情况
$$\nabla_\mu A^a=\partial_\mu A^a+\Gamma_\mu A^a=\partial_\mu A^a+\omega_{\mu\; b}^{\;a} A^b \tag{1} \label{eq1}$$
注意现在将变换$e^a_{\; \mu}$视为基，
$$\nabla_\mu A^a=\nabla_\mu(e^a_{\; \nu} A^\nu)=(\nabla_\mu e^a_{\; \nu})A^\nu+e^a_{\; \nu}(\nabla_\mu A^\nu) \tag{2} \label{eq2}$$
$A^\nu$为时空矢量，因此其协变导数由奇维塔联络$\nabla^L_\mu$决定，
$$e^a_{\; \nu}(\nabla_\mu A^\nu)=e^a_{\; \nu}(\partial_\mu A^\nu+\Gamma_{\mu\lambda}^\nu A^\lambda)$$
代入$\eqref{eq2}$与$\eqref{eq1}$比较即可得到协变导数对基的作用，其应为零
$$\nabla_\mu e^a_{\; \nu}=\partial_\mu e^a_{\; \nu}-\Gamma_{\mu\nu}^\lambda e^a_{\; \lambda}+\omega_{\mu \; b}^{\; a} e^b_{\; \nu} \tag{3} \label{eq3}$$
对于混合基矩阵 $M^{\;\mu}_{b} =e_a^{\; \mu}A^aB_b $，
$$\nabla_\mu M=\nabla^L_\mu M+[\Gamma_\mu,M] \tag{4} \label{eq4}$$
要求联络与度规适配 $\nabla_\mu \eta_{ab}=0$
$$\nabla_\mu \eta_{ab}=\partial_\mu \eta_{ab}-\omega_{\mu \; a}^{\;c}\eta_{cb}-\omega_{\mu \; b}^{\;c} \eta_{ca}$$
可知$\omega$关于后两个指标 反对称，$\omega_{\mu ab}=-\omega_{\mu ba}$，
联络形式 $\omega^a_{\;b}=\omega_{\mu \; b}^{\; a}\dd x^\mu$，
对易系数 $[v_a,v_b]=f_{\;ab}^c v_c$，反对易系数 $\lbrace v_a,v_b\rbrace=h_{\;ab}^c v_c$
对$\omega^a$外微分，$\dd \omega^a=\nabla_\mu^\text{L} e^a_{\; \nu} \dd x^\mu\wedge\dd x^\nu=e^a_{\; [\nu,\mu]}\dd x^\mu\wedge\dd x^\nu$，
由于用$\dd x^\mu$表示，故$\nabla_\mu^\text{L}$为奇维塔联络，$\nabla_\mu^\text{L} e^a_{\; \nu}=\partial_\mu e^a_{\; \nu}-\Gamma_{\mu\nu}^\lambda e^a_{\; \lambda}$，
由$\Gamma_{\mu\nu}^\lambda$对称性可进行反称化将其消去，$e^a_{\; [\nu,\mu]}=\displaystyle\frac{1}{2}(\partial_\mu e^a_{\; \nu}-\partial_\nu e^a_{\; \mu})=-\displaystyle\frac{1}{2}f_{\;\mu\nu}^a$
曲率形式 $R^a_{\; b}=\dd \omega^a_{\;b}+\omega^a_{\;c} \wedge \omega^c_{\;b}=(\partial\sideset{_{[\mu}}{_{\nu]\; b} ^{\;\;a}}{\omega}+\omega_{[\mu \; c}^{\;\;a}\omega_{\nu] \; b}^{\;\;c})\dd x^\mu\wedge\dd x^\nu$
挠率形式 $T^a=\dd \omega^a+\omega^a_{\;b}\wedge \omega^b=(e^a_{\; [\nu,\mu]}+\omega_{[\mu \;\nu]}^{\;\;a})\dd x^\mu\wedge\dd x^\nu$
施加无挠条件 $T^a=0$，可解出自旋联络系数
$$\omega_{\mu \;b}^{\;a}=-e_b^{\; \nu}(\partial_\mu e^a_{\; \nu}-\Gamma_{\mu\nu}^\lambda e^a_{\; \lambda}) \tag{5} \label{eq5}$$
代入$\eqref{eq3}$可发现无挠条件实际上等价于
$$\nabla_\mu e^a_{\; \nu}=0$$
联络系数可用对易系数表示为
$$\omega_{a\mu\nu}=\frac{1}{2}(f_{a\mu\nu}+f_{\nu\mu a}-f_{\mu\nu a})$$
$$\Gamma_{\mu\nu}^a=-\displaystyle\frac{1}{2}h_{\;\mu\nu}^a$$
旋量$\psi$在自旋基$v_a,\tilde{v}_a$下的转移函数为洛伦兹矩阵，$\tilde{\psi}=L\psi$，其应满足
$$\tilde{\Gamma}_\mu=L \Gamma_\mu L^{-1}- (\partial_\mu L) L^{-1} \tag{6}\label{eq6}$$
考虑无穷小洛伦兹变换$L^a_{\;b}=\delta^a_{\;b}+\delta\epsilon^a_{\;b}$，由$L^a_{\;b}L^c_{\;d}\eta_{ac}=\eta_{bd}$，可得$\delta\epsilon_{ab}$ 反称，$\delta\epsilon_{ab}=\eta_{ac}\delta\epsilon^c_{\; b}=-\delta\epsilon_{ba}$，
而洛伦兹群李代数由反称阵张成，因此$L^a_{\;b}$的群表示也可写为 $S(L)=I+i \delta\epsilon^{ab} S_{ab}$，$S_{ab}=-S_{ba} \in \mathfrak{so}(1,n-1)$ ，注意$a,b$固定时$\delta\epsilon^{ab}$为数，而$S_{ab}$为矩阵
两种表示之间关系 $\delta\epsilon^a_{\;b}=i \delta\epsilon^{cd} (S_{cd})^a_{\; b}$，可知$(S_{cd})^a_{\; b}=\displaystyle\frac{1}{2}(\delta_c^a\eta_{bd}-\delta_d^a\eta_{bc})$，将$S(L)=I+i \delta\epsilon^{ab} S_{ab}$代入$\eqref{eq6}$中
$$\tilde{\Gamma}_\mu=\Gamma_\mu+i\delta\epsilon^{ab}[S_{ab},\Gamma_\mu]-i\partial_\mu(\delta\epsilon^{ab})S_{ab}\tag{7}\label{eq7}$$
而联络$\Gamma_\mu$也应在李代数中取值，令其为$\Gamma_\mu=\Gamma_\mu^{\; ab} S_{ab} $，由$S_{ab}$反称性，$\Gamma_\mu^{\; ab}$关于$a,b$反称，现在考虑$S_{ab}$的对易关系
对于有限洛伦兹变换$S(L)=\exp(i \epsilon^{ab} S_{ab})$，$S(L)S(L’)S(L^{-1})=S(LL’L^{-1})$
先将$L’$取为无穷小变换，对右边进行反称化，
$$S(L)S_{ab}S(L^{-1})=\displaystyle\frac{1}{2}[L^c_{\; a}(L^{-1})^d_{\; b}-L^c_{\; b}(L^{-1})^d_{\; a}]S_{cd}$$
再将$L$取为无穷小变换可得对易关系
$$[S_{ab},S_{cd}]=\displaystyle\frac{i}{2}(\eta_{ac}S_{bd}+\eta_{bd}S_{ac}-\eta_{ad}S_{bc}-\eta_{bc}S_{ad}) \tag{8}\label{eq8}$$
代入$\eqref{eq7}$中得
$$\tilde{\Gamma}_\mu^{\; ab}=\Gamma_\mu^{\; ab}+\Gamma_{\mu\; c}^{\; b}\delta\epsilon^{ca}-\Gamma_{\mu\; c}^{\; a}\delta\epsilon^{cb}-i\partial_\mu(\delta\epsilon^{ab})$$
对$\omega_{\mu\; b}^{\;a}$作相似操作，$\tilde{\nabla}_\mu \tilde{\psi}=L \nabla_\mu \psi$，可得
$$\tilde{\omega}_\mu^{\; ab}=\omega_\mu^{\; ab}+\omega_{\mu\; c}^{\; b}\delta\epsilon^{ca}-\omega_{\mu\; c}^{\; a}\delta\epsilon^{cb}-\partial_\mu(\delta\epsilon^{ab})$$
比较即得$\Gamma_{\mu}^{\; ab}=i\omega_{\mu}^{\; ab}$，旋量的协变导数和其对易现在可写为
$$\nabla_\mu\psi=\partial_\mu\psi+i\omega_{\mu}^{\; ab}S_{ab}\psi$$
$$[\nabla_\mu,\nabla_\nu]\psi=-iR_{\mu\nu}^{\;\;\;ab}S_{ab}\psi$$
协变导数写成分量形式
$$\nabla_\mu\psi^a=\partial_\mu\psi^a+i\omega_{\mu}^{\; cd}(S_{cd})^a_{\; b}\psi^b$$
现在考虑平直时空狄拉克方程 $(i\gamma^a \partial_a-m)\psi=0$，$\lbrace \gamma^a,\gamma^b\rbrace=2\eta^{ab}I$ 在弯曲时空的推广，将普通导数换为协变导数 $\partial_a \to e_a^{\;\mu} \nabla_\mu$
$$(i\gamma^a e_a^{\;\mu} \nabla_\mu-m)\psi=(i\underline{\gamma}^\mu \nabla_\mu-m)\psi=0 $$
$\underline{\gamma}^\mu=\gamma^a e_a^{\;\mu}$ 为与时空有关的狄拉克矩阵，满足$\lbrace{\underline{\gamma}^\mu,\underline{\gamma}^\mu\rbrace}=2g^{\mu\nu}I$
现在考虑混合基变化$\psi’=S(L)\psi$，$e^{\prime \;\nu}_b=\displaystyle\frac{\partial x’^\nu}{\partial x^\mu}(L^{-1})^a_{\; b}e^{\;\mu}_{a}$，代入狄拉克方程
$$(iS(L)\gamma^a S^{-1}(L)L^b_{\;a} e^{\prime \;\nu}_b \nabla’_\nu -m)\psi’=0$$
可知 $S(L)\gamma^a S^{-1}(L)L^b_{\;a}=\gamma^b$，考虑无穷小变换$L^b_{\;a}=\delta^b_{\;a}+\delta\epsilon^b_{\;a}$
$$i\delta\epsilon_{cd}[S^{cd},\gamma^a]+\delta\epsilon^a_{\;b}\gamma^b=0$$
取$S^{cd}=\displaystyle-\frac{i}{8}[\gamma^c,\gamma^d]$ 可同时满足该式和对易关系$\eqref{eq8}$，协变导数和其对易成为
$$\nabla_\mu\psi=\partial_\mu\psi +\frac{1}{8}\omega_{\mu ab}[\gamma^a,\gamma^b]\psi \tag{9}\label{eq9}$$
$$[\nabla_\mu,\nabla_\nu]\psi=-\frac{1}{8}R_{\mu\nu ab}[\gamma^a,\gamma^b]\psi \tag{10}\label{eq10}$$
还原单位，狄拉克方程现在可写为
$$ i\hbar\gamma^a(v_a+\frac{1}{8}\omega_{abc}[\gamma^b,\gamma^c])\psi=mc \psi \tag{11}\label{eq11}$$
利用$\eqref{eq4}$和$\eqref{eq5}$ 可得
$$\nabla_\mu\underline{\gamma}^\nu=0$$
结合$\eqref{eq10}$和比安基恒等式 $R_{\mu\nu\lambda\rho}+R_{\mu\lambda\rho\nu}+R_{\mu\rho\nu\lambda}=0$ 可得
$$(\underline{\gamma}^\mu\nabla_\mu)^2=\square +\displaystyle\frac{R}{4} \tag{12}\label{eq12}$$
From Quantum Field Theory In Curved Spacetime by Leonard Parker and David Toms