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.


1 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.

  • $\begingroup$ 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
    Commented May 31, 2020 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
    Commented May 31, 2020 at 1:48

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