Let $\mathfrak n^\alpha$ be a vector density of weight 1. Define the covariant derivative $\nabla$ such that under a coordinate transformation $x^\mu \to \bar x^\mu$ $$ \nabla_\rho \mathfrak n^\alpha \to \left\lvert \frac{\mathrm d \bar x^\mu}{\mathrm d x^\nu} \right\rvert \frac{\partial x^\sigma}{\partial \bar x^\rho} \frac{\partial \bar x^\alpha}{\partial x^\beta} \nabla_\sigma \mathfrak n^\beta $$ Is this the correct form of the covariant derivative?: $$ \mathfrak q_\nu^\alpha \equiv \nabla_\nu \mathfrak n^\alpha = \partial_\nu \mathfrak n^\alpha + \Gamma^\alpha_{\nu\beta} \mathfrak n^\beta - \Gamma^\rho_{\nu\rho} \mathfrak n^\alpha $$ I'm trying to calculate the action of a commutator of covariant derivatives on $\mathfrak n^\alpha$, and ultimately the analogue to what the Ricci tensor means for vectors. Here's what I have so far: $$ \nabla_\mu \mathfrak q_\nu^\alpha - \nabla_\nu \mathfrak q_\mu^\alpha = (\partial_\mu \mathfrak q_\nu^\alpha + \Gamma^\alpha_{\mu\beta} \mathfrak q_\nu^\beta - \Gamma^\sigma_{\mu\nu} \mathfrak q_\sigma^\alpha - \Gamma^\rho_{\mu\rho} \mathfrak q_\nu^\alpha) - (\partial_\nu \mathfrak q_\mu^\alpha + \Gamma^\alpha_{\nu\beta} \mathfrak q_\mu^\beta - \Gamma^\sigma_{\nu\mu} \mathfrak q_\sigma^\alpha - \Gamma^\rho_{\nu\rho} \mathfrak q_\mu^\alpha) $$

\begin{multline} {}=(\partial_\mu (\Gamma^\alpha_{\nu\beta} \mathfrak n^\beta - \Gamma^\rho_{\nu\rho} \mathfrak n^\alpha ) + \Gamma^\alpha_{\mu\beta} \mathfrak q_\nu^\beta - \Gamma^\rho_{\mu\rho} (\partial_\nu \mathfrak n^\alpha + \Gamma^\alpha_{\nu\beta} \mathfrak n^\beta)) - {} \\ (\partial_\nu (\Gamma^\alpha_{\mu\beta} \mathfrak n^\beta - \Gamma^\rho_{\mu\rho} \mathfrak n^\alpha ) + \Gamma^\alpha_{\nu\beta} \mathfrak q_\mu^\beta - \Gamma^\rho_{\nu\rho} (\partial_\mu \mathfrak n^\alpha + \Gamma^\alpha_{\mu\beta} \mathfrak n^\beta)) \end{multline}

\begin{multline} {}=(\partial_\mu \Gamma^\alpha_{\nu\beta} \mathfrak n^\beta - \partial_\mu \Gamma^\rho_{\nu\rho} \mathfrak n^\alpha + \Gamma^\alpha_{\mu\beta} (\Gamma^\beta_{\nu\gamma} \mathfrak n^\gamma - \Gamma^\rho_{\nu\rho} \mathfrak n^\beta ) - \Gamma^\rho_{\mu\rho} \Gamma^\alpha_{\nu\beta} \mathfrak n^\beta) - {} \\ (\partial_\nu \Gamma^\alpha_{\mu\beta} \mathfrak n^\beta - \partial_\nu \Gamma^\rho_{\mu\rho} \mathfrak n^\alpha + \Gamma^\alpha_{\nu\beta} (\Gamma^\beta_{\mu\gamma} \mathfrak n^\gamma - \Gamma^\rho_{\mu\rho} \mathfrak n^\beta ) - \Gamma^\rho_{\nu\rho} \Gamma^\alpha_{\mu\beta} \mathfrak n^\beta) \end{multline}

\begin{multline} {}=R^\alpha_{\beta\mu\nu} \mathfrak n^\beta + (- \partial_\mu \Gamma^\rho_{\nu\rho} \mathfrak n^\alpha - \Gamma^\alpha_{\mu\beta} \Gamma^\rho_{\nu\rho} \mathfrak n^\beta - \Gamma^\rho_{\mu\rho} \Gamma^\alpha_{\nu\beta} \mathfrak n^\beta) - ( - \partial_\nu \Gamma^\rho_{\mu\rho} \mathfrak n^\alpha - \Gamma^\alpha_{\nu\beta} \Gamma^\rho_{\mu\rho} \mathfrak n^\beta - \Gamma^\rho_{\nu\rho} \Gamma^\alpha_{\mu\beta} \mathfrak n^\beta) \end{multline}

$$ \nabla_\mu \nabla_\nu \mathfrak n^\alpha - \nabla_\nu \nabla_\mu \mathfrak n^\alpha = R^\alpha_{\beta\mu\nu} \mathfrak n^\beta - ( \partial_\mu \Gamma^\rho_{\nu\rho} - \partial_\nu \Gamma^\rho_{\mu\rho} ) \mathfrak n^\alpha $$ $$ \nabla_\mu \nabla_\nu \mathfrak n^\mu - \nabla_\nu \nabla_\mu \mathfrak n^\mu = \left[ R_{\beta\nu} - ( \partial_\beta \Gamma^\rho_{\nu\rho} - \partial_\nu \Gamma^\rho_{\beta\rho} ) \right] \mathfrak n^\beta $$ Could this be right? I'm suspicious that the tensor in brackets on the RHS has an antisymmetric part.


My calculation is that you are obtaining extra terms because you started with the wrong expression. If indeed $n^{\mu}$ is a vector density of weight 1, then with the Levi-Civita connection it could be written as $$n^{\mu} = \sqrt{-g}\,V^{\mu}$$ with $V$ an ordinary vector. Then, the covariant derivative of $n$ could be calculated to be $$\nabla_{\nu} n^{\mu} = \sqrt{-g} \; \nabla_{\nu} V^{\mu}\\ \quad \quad \quad \quad \quad \quad= \sqrt{-g}(\partial_{\nu} V^{\mu} + \Gamma^{\mu}_{\nu \rho} V^{\rho}) $$ then $$\nabla_{\zeta} \nabla_{\nu} n^{\mu} = \nabla_{\zeta} \, (\sqrt{-g} \; \nabla_{\nu} V^{\mu})\\ \quad \quad \quad= \sqrt{-g} \, \nabla_{\zeta}\nabla_{\nu} V^{\mu}.$$ Taking the commutator would lead to $$\nabla_{\zeta}\nabla_{\nu} n^{\mu} - \nabla_{\nu} \nabla_{\zeta} n^{\mu} = \sqrt{-g} \; ( \nabla_{\zeta}\nabla_{\nu} V^{\mu} - \nabla_{\nu} \nabla_{\zeta} V^{\mu}) \\ \quad \quad \quad = \sqrt{-g} \;(R^{\mu}_{\; \rho \zeta \nu} V^{\rho}) \\ \quad \quad \quad = R^{\mu}_{\; \rho \zeta \nu} \sqrt{-g} \; V^{\rho} = R^{\mu}_{\; \rho \zeta \nu} \, n^{\rho}.$$ You can check if this makes sense to you.

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  • $\begingroup$ If the the metric varies with $x^\mu$, then wouldn't the partial derivative of the vector density produce non-tensor-density terms beyond those in the partial derivative of an ordinary 4-vector? Thus the need from extra coupling terms to the Levi-Civita connection $\endgroup$ – rossng May 31 at 1:45
  • $\begingroup$ $ \partial_\rho \mathfrak n^\alpha \to \left\lvert \frac{\partial x^\mu}{\partial x^\nu} \right\rvert \frac{\partial x^\sigma}{\partial \bar x^\rho} \frac{\partial \bar x^\alpha}{\partial x^\beta} \left( \frac{\partial \mathfrak n^\beta}{\partial x^\sigma} + \frac{ \partial x^\beta}{\partial \bar x^\lambda} \frac{\partial^2\bar x^\lambda}{\partial x^\sigma \partial x^\gamma} \mathfrak n^\gamma + \frac{\partial x^\xi}{\partial\bar x^\lambda} \frac{\partial\bar x^\lambda}{\partial x^\sigma \partial x^\xi} \delta^\beta_\gamma \mathfrak n^\gamma \right) $ $\endgroup$ – rossng May 31 at 1:48

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