How does the spin connection transform under a linear perturbation to the metric tensor?

Question

I want to know how the spin connection

\begin{equation} \begin{aligned} \omega_{\mu ab}&=(e_a)_\rho\nabla_\mu (e_b)^\rho\\ &=(e_a)_\rho\partial_\mu (e_b)^\rho+(e_a)_\rho\Gamma^\rho_{\mu\sigma} (e_b)^\sigma \end{aligned} \end{equation}

transform under a linear transformation of the metric tensor of the form

\begin{equation} g_{\mu\nu}\rightarrow g_{\mu\nu}+\epsilon h_{\mu\nu} \end{equation}

where $h_{\mu\nu}$ is an arbitrary perturbation and $\epsilon$ is a small parameter. I know that the Levi-Civita connection is

\begin{equation} \Gamma^\rho_{\mu\sigma}=\frac{1}{2}g^{\rho\alpha}\big(g_{\alpha\sigma,\mu}+g_{\mu\alpha,\sigma}-g_{\mu\sigma,\alpha}\big) \end{equation}

However, I'm not sure how I would write the tetrad basis $(e^\rho)_a$ in terms of the metric, so I'm not sure how to expand the spin connection up to linear order in $h_{\mu\nu}$.

Attempted answer:

One of the comments gave a link to a thesis where I found an expression for the tetrad basis to linear order in $h_{\mu\nu}$

\begin{equation} e_{A\rho}\rightarrow \tilde{e}_{A\rho}=e_{A\rho}+\frac{1}{2}h_{\rho\sigma}e_A^\sigma \end{equation}

and

\begin{equation} e_{A}^\rho\rightarrow \tilde{e}_{A}^\rho=e_{A}^\rho-\frac{1}{2}h^{\rho\sigma}e_{A\sigma} \end{equation}

which checks out when we calculate the full metric as $\tilde{g}_{\mu\nu}=e_{A\mu}e^A_\nu$.

With this expression we can calculate the linear perturbation to the spin connection as

\begin{equation} \begin{aligned} \tilde{\omega}_{\mu ab}&=\tilde{e}_{a\rho}\tilde{\nabla}_\mu \big(\tilde{e}_b)^\rho \\ &=\big(e_{a\rho}+\frac{1}{2}h_{\rho\sigma}e_a^\sigma\big)\tilde{\nabla}_\mu\big(e_b^\rho -\frac{1}{2}h^{\rho\sigma} e_{b\sigma}\big) \\ &=e_{a\rho}\tilde{\nabla}_\mu e_b^\rho+\frac{1}{2}h_{\rho\sigma}e_a^\sigma\tilde{\nabla}_\mu e_b^\rho -e_{a\rho}\frac{1}{2}\tilde{\nabla}_{\mu}\big(h^{\rho\sigma} e_{b\sigma}\big)\\ &=e_{a\rho}\nabla_\mu e_b^\rho+e_{a\rho}\delta \Gamma^\rho_{\mu\alpha}e_b^\alpha+\frac{1}{2}h_{\rho\sigma}e_a^\sigma\nabla_\mu e_b^\rho -e_{a\rho}\frac{1}{2}\nabla_{\mu}\big(h^{\rho\sigma} e_{b\sigma}\big)\\ &=\omega_{\mu ab}+e_{a\rho}\delta \Gamma^\rho_{\mu\alpha} e_b^\alpha+\frac{1}{2}h_{\rho\sigma}e_a^\sigma\nabla_\mu e_b^\rho -\frac{1}{2}e_{a\rho}h^{\rho\sigma}\nabla_{\mu}e_{b\sigma}-\frac{1}{2}e_{a\rho}e_{b\sigma}\nabla_{\mu}h^{\rho\sigma} \\ &=\omega_{\mu ab}+e_{a\rho}\delta \Gamma^\rho_{\mu\alpha} e_b^\alpha-\frac{1}{2}e_{a\rho}e_{b\sigma}\nabla_{\mu}h^{\rho\sigma} \\ &=\omega_{\mu ab}+\frac{1}{2}e_{a\rho}e_b^\alpha g^{\rho\sigma}\big(h_{\sigma\alpha;\mu}+h_{\mu\sigma;\alpha}-h_{\mu\alpha;\sigma}\big)-\frac{1}{2}e_{a\rho}e_{b\sigma}\nabla_{\mu}h^{\rho\sigma} \end{aligned} \end{equation}

so we get that

\begin{equation} \tilde{\omega}_{\mu ab}=\omega_{\mu a b}+ e_a^\alpha e_b^\beta h_{\mu[\alpha;\beta]} \end{equation}

Is this correct?

Answer 1

In my notes I have the variation of the torsion-free spin connection due to the variation of the vierbein as being $$ \delta \omega_{ijk} =-\frac 12\left\{( \eta_{ib}( \nabla_j [e^{*b}_\alpha \delta e^\alpha_k]- \nabla_k [e^{*b}_\alpha \delta e^\alpha_j])\right.\nonumber\\ \quad+\eta_{jb}( \nabla_k [e^{*b}_\alpha \delta e^\alpha_i]- \nabla_i [e^{*b}_\alpha \delta e^\alpha_k])\nonumber\\ \quad \left.-\eta_{kb}( \nabla_i [e^{*b}_\alpha \delta e^\alpha_j]- \nabla_j [e^{*b}_\alpha \delta e^\alpha_i])\right\}+ \omega_{ij\mu} \delta e^\mu_k.\nonumber $$ The very last term spoils the frame-bundle tensor character of $ \delta \omega_{ijk}$. Now $$ \omega_{ijk}= \omega_{ij\mu}e^\mu_k $$ and so $$ \delta \omega_{ijk}= (\delta \omega_{ij\mu}) e^\mu_k+ \omega_{ij\mu}\delta e^\mu_k. $$ Thus $\delta\omega_{ij\mu}$ is a frame-bundle tensor, and $$ (\delta \omega_{ij\mu}) e^\mu_k =-\frac 12\left\{( \eta_{ib}( \nabla_j [e^{*b}_\alpha \delta e^\alpha_k]- \nabla_k [e^{*b}_\alpha \delta e^\alpha_j])\right.\nonumber\\ \quad+\eta_{jb}( \nabla_k [e^{*b}_\alpha \delta e^\alpha_i]- \nabla_i [e^{*b}_\alpha \delta e^\alpha_k])\nonumber\\ \quad \left.-\eta_{kb}( \nabla_i [e^{*b}_\alpha \delta e^\alpha_j]- \nabla_j [e^{*b}_\alpha \delta e^\alpha_i])\right\}. \nonumber $$

I'm not sure whether this helps your particlular question, but it is useful for showing that the Belinfante tensor coincides with the Hilbert E&M tensor.