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In Sec. 12.1 of 'Superstring Theory', by Green, Schwarz, and Witten, the minimal spin connection is motivated as follows:

If we are to avoid modifying the standard content of general relativity, the two notions of the covariant derivative of a vector $V$ must be equivalent. This will be so, in the sense that $D_{\mu}V^{a} = e^{a\nu}D_{\mu}V_{\nu}$, if we define the spin connection so that the covariant derivative of the veilbein is zero, $D_{\mu}e^{a}_{\nu}=0$.

Until recently, I have always thought that this argument was quite sensible. But now I have doubts, due to the following considerations: The covariant derivative of a vector field $A$, say, in local Lorentz coordinates is given by

$$D_{\mu }A_{a} \equiv\partial _{\mu }A_{a}-\frac{1}{2}\omega _{\mu cd}\left( V^{cd}\right) ^{b}{}_{a}A_{b} =\partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b},$$

where $V^{cd}$ are the generators of the vector representation of the Lorentz group, given by $(V^{cd})^{a}{}_{b} = \eta ^{cb}\delta _{a}^{d}-\eta ^{db}\delta _{a}^{c}$. But then, using as well the standard expression for $D_{\mu }e^{a}{}_{\nu }$,

\begin{align} D_{\mu }A_{\nu } & = D_{\mu }\left( e^{a}{}_{\nu }A_{a}\right) \\ &=\left( D_{\mu }e^{a}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }D_{\mu }A_{a} \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }+\omega _{\mu }{}^{a}{}_{b}e^{b}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }\left( \partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b}\right) \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }\right) A_{a}-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }A_{a}+e^{a}{}_{\nu } \partial _{\mu }A_{a} \\ &=\partial _{\mu }A_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }A_{\rho } \\ &\equiv \nabla _{\mu }A_{\nu }, \end{align}

where $\Gamma ^{\rho }{}_{\mu \nu}$ are the Christoffel symbols. The derivation depends only on the antisymmetry of the spin connection, not on it being minimal. But if $D_{\mu }A_{\nu }=\nabla _{\mu }A_{\nu }$, as thus seems to be the case, then the covariant derivative of any tensor field (as a tensor product of vector fields) is the standard GR one, and thus the standard content of GR would seem to be unaltered. Am I doing something fundamentally wrong?

In Sec. 12.1 of 'Superstring Theory', by Green, Schwarz, and Witten, the minimal spin connection is motivated as follows:

If we are to avoid modifying the standard content of general relativity, the two notions of the covariant derivative of a vector $V$ must be equivalent. This will be so, in the sense that $D_{\mu}V^{a} = e^{a\nu}D_{\mu}V_{\nu}$, if we define the spin connection so that the covariant derivative of the veilbein is zero, $D_{\mu}e^{a}_{\nu}=0$.

Until recently, I have always thought that this argument was quite sensible. But now I have doubts, due to the following considerations: The covariant derivative of a vector field $A$, say, in local Lorentz coordinates is given by

$$D_{\mu }A_{a} \equiv\partial _{\mu }A_{a}-\frac{1}{2}\omega _{\mu cd}\left( V^{cd}\right) ^{b}{}_{a}A_{b} =\partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b},$$

where $V^{cd}$ are the generators of the vector representation of the Lorentz group, given by $(V^{cd})^{a}{}_{b} = \eta ^{ca}\delta _{b}^{d}-\eta ^{da}\delta _{b}^{c}$. But then, using as well the standard expression for $D_{\mu }e^{a}{}_{\nu }$,

\begin{align} D_{\mu }A_{\nu } & = D_{\mu }\left( e^{a}{}_{\nu }A_{a}\right) \\ &=\left( D_{\mu }e^{a}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }D_{\mu }A_{a} \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }+\omega _{\mu }{}^{a}{}_{b}e^{b}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }\left( \partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b}\right) \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }\right) A_{a}-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }A_{a}+e^{a}{}_{\nu } \partial _{\mu }A_{a} \\ &=\partial _{\mu }A_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }A_{\rho } \\ &\equiv \nabla _{\mu }A_{\nu }, \end{align}

where $\Gamma ^{\rho }{}_{\mu \nu}$ are the Christoffel symbols. The derivation depends only on the antisymmetry of the spin connection, not on it being minimal. But if $D_{\mu }A_{\nu }=\nabla _{\mu }A_{\nu }$, as thus seems to be the case, then the covariant derivative of any tensor field (as a tensor product of vector fields) is the standard GR one, and thus the standard content of GR would seem to be unaltered. Am I doing something fundamentally wrong?

In Sec. 12.1 of 'Superstring Theory', by Green, Schwarz, and Witten, the minimal spin connection is motivated as follows:

If we are to avoid modifying the standard content of general relativity, the two notions of the covariant derivative of a vector $V$ must be equivalent. This will be so, in the sense that $D_{\mu}V^{a} = e^{a\nu}D_{\mu}V_{\nu}$, if we define the spin connection so that the covariant derivative of the veilbein is zero, $D_{\mu}e^{a}_{\nu}=0$.

Until recently, I have always thought that this argument was quite sensible. But now I have doubts, due to the following considerations: The covariant derivative of a vector field $A$, say, in local Lorentz coordinates is given by

$$D_{\mu }A_{a} \equiv\partial _{\mu }A_{a}-\frac{1}{2}\omega _{\mu cd}\left( V^{cd}\right) ^{b}{}_{a}A_{b} =\partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b},$$

where $V^{cd}$ are the generators of the vector representation of the Lorentz group, given by $(V^{cd})^{a}{}_{b} = \eta ^{cb}\delta _{a}^{d}-\eta ^{db}\delta _{a}^{c}$. But then, using as well the standard expression for $D_{\mu }e^{a}{}_{\nu }$,

\begin{align} D_{\mu }A_{\nu } & = D_{\mu }\left( e^{a}{}_{\nu }A_{a}\right) \\ &=\left( D_{\mu }e^{a}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }D_{\mu }A_{a} \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }+\omega _{\mu }{}^{a}{}_{b}e^{b}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }\left( \partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b}\right) \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }\right) A_{a}-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }A_{a}+e^{a}{}_{\nu } \partial _{\mu }A_{a} \\ &=\partial _{\mu }A_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }A_{\rho } \\ &\equiv \nabla _{\mu }A_{\nu }, \end{align}

where $\Gamma ^{\rho }{}_{\mu \nu}$ are the Christoffel symbols. The derivation depends only on the antisymmetry of the spin connection, not on it being minimal. But if $D_{\mu }A_{\nu }=\nabla _{\mu }A_{\nu }$, as seems to be suggested, then the covariant derivative of any tensor field (as a tensor product of vector fields) is the standard GR one, and thus the standard content of GR would seem to be unaltered. Am I doing something fundamentally wrong?

In Sec. 12.1 of 'Superstring Theory', by Green, Schwarz, and Witten, the minimal spin connection is motivated as follows:

If we are to avoid modifying the standard content of general relativity, the two notions of the covariant derivative of a vector $V$ must be equivalent. This will be so, in the sense that $D_{\mu}V^{a} = e^{a\nu}D_{\mu}V_{\nu}$, if we define the spin connection so that the covariant derivative of the veilbein is zero, $D_{\mu}e^{a}_{\nu}=0$.

Until recently, I have always thought that this argument was quite sensible. But now I have doubts, due to the following considerations: The covariant derivative of a vector field $A$, say, in local Lorentz coordinates is given by

$$D_{\mu }A_{a} \equiv\partial _{\mu }A_{a}-\frac{1}{2}\omega _{\mu cd}\left( V^{cd}\right) ^{b}{}_{a}A_{b} =\partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b},$$

where $V^{cd}$ are the generators of the vector representation of the Lorentz group, given by $(V^{cd})^{a}{}_{b} = \eta ^{cb}\delta _{a}^{d}-\eta ^{db}\delta _{a}^{c}$. But then, using as well the standard expression for $D_{\mu }e^{a}{}_{\nu }$,

\begin{align} D_{\mu }A_{\nu } & = D_{\mu }\left( e^{a}{}_{\nu }A_{a}\right) \\ &=\left( D_{\mu }e^{a}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }D_{\mu }A_{a} \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }+\omega _{\mu }{}^{a}{}_{b}e^{b}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }\left( \partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b}\right) \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }\right) A_{a}-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }A_{a}+e^{a}{}_{\nu } \partial _{\mu }A_{a} \\ &=\partial _{\mu }A_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }A_{\rho } \\ &\equiv \nabla _{\mu }A_{\nu }, \end{align}

where $\Gamma ^{\rho }{}_{\mu \nu}$ are the Christoffel symbols. The derivation depends only on the antisymmetry of the spin connection, not on it being minimal. But if $D_{\mu }A_{\nu }=\nabla _{\mu }A_{\nu }$, as thus seems to be the case, then the covariant derivative of any tensor field (as a tensor product of vector fields) is the standard GR one, and thus the standard content of GR would seem to be unaltered. Am I doing something fundamentally wrong?