Commutator of covariant derivatives acting on a vector density

Question

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.

Answer 1

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.