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Note: This post is intended to be about Ricci-calculus. My definition of tensor densities in this post makes a scalar density of weight 1 essentially equivalent to a maximal-degree differential form, so answers along the line of modern differential geometry are almost useless to me here.

Given an $n$ dimensional real $C^\infty$ manifold $M$, I hereby define a scalar density of weight 1, $\rho$, at $p\in M$ as a rule that assigns to any local chart containing $p$ a real number with the understanding that during coordinate change, this number transforms as $$ \rho'=\det\frac{\partial x}{\partial x'}\ \rho. $$

Let $\nabla_\mu$ be a linear connection that acts on vector fields in a local trivialization as $$ \nabla_\mu X^\nu=\partial_\mu X^\nu+C^\nu_{\mu\sigma}X^\sigma. $$

Linear connections are induced by $\nabla$ on the cotangent bundle and the tensor bundles by the natural conditions that $\nabla$ should commute with contractions and obey the Leibniz rule with respect to tensor products.

Since scalar densities of weight 1 at different points, we also need to introduce a connection on the density bundle, given in a local trivialization as $$ \nabla_\mu\rho=\partial_\mu\rho+C_\mu\rho, $$ this $C_\mu$ is a priori independent of $C^\nu_{\mu\rho}$.

Now, we can check that during coordinate change, $C_\mu$ transforms as $$ C_\mu'=\frac{\partial x^\mu}{\partial x^{\mu'}}C_\mu-\frac{\partial^2x^\nu}{\partial x^{\mu'}\partial x^{\nu'}}\frac{\partial x^{\nu'}}{\partial x^\nu}, $$ and this is the same as the transformation rule for $-C^\nu_{\mu\nu}$, so one natural way to extend the connection in the tangent bundle to a connection in the density bundle is to define $C_\mu=-C_{\mu\nu}^{\nu}$, however this approach relies a lot on transformation rules, and I don't like it.

One other natural condition would be to abuse the relationship between scalar densities and differential forms, as the density $\rho$ corresponds to the differential form $\rho_{\mu_1,...,\mu_n}=\rho\pi_{\mu_1,...,\mu_n}$, where $\pi$ is Levi-Civita's symbol. Then the natural condition is to demand that $\nabla_\nu\rho_{\mu_1...\mu_n}=(\nabla_\nu\rho)\pi_{\mu_1,...,\mu_n}$.

I tried to check this, but this quickly became untractable as $$ \nabla_\nu\rho_{\mu_1...\mu_n}=(\partial_\nu\rho)\pi_{\mu_1...\mu_n}-\rho\left\{C^\sigma_{\nu\mu_1}\pi_{\sigma\mu_2...\mu_n}+...+C^\sigma_{\nu\mu_n}\pi_{\mu_1...\mu_{n-1}\sigma}\right\}, $$ and I have no clue how to evaluate the terms in the curly brackets.

Question: Would this latter definition of the covariant derivative of a scalar density of weight 1 reduce to the $C_\mu=-C_{\mu\nu}^\nu$ relation implied by the transformation properties?

If so, I would be terribly appreciative of any pointers as to how to perform the calculation that shows this. I tried using several identities and meanings related to the Levi-Civita symbol but nothing simplified the curly bracket.

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Since the space of scalar densities/top forms is one-dimensional, we can make things easier by just computing the component $\nabla_\nu \rho_{1\ldots n}$ instead of having all those extra indices flying around everywhere. Using your formula, this becomes

$$ \nabla_\nu \rho_{1 \ldots n} = \partial_\nu \rho - \rho \left( C^\sigma_{\nu 1} \pi_{\sigma 2 \ldots n} + \cdots + C^\sigma_{\nu n} \pi_{1\ldots (n-1)\sigma}\right).$$

Note that $\pi_{\sigma 2 \ldots n} = \delta_{\sigma}^{1}$, so the first term in the parentheses is $C^\sigma_{\nu 1} \delta_{\sigma}^{1} = C^1_{\nu 1}$. Similarly, if we replace some index $k \in \{ 1, \ldots, n\}$ in $\pi_{1\ldots n}$ with $\sigma$ and contract it with $C^\sigma_{\nu k}$, we get $C^k_{\nu k}$ (no sum); and thus adding these all up we get $$\nabla_\nu \rho_{1 \ldots n} = \partial_\nu\rho - C^\mu_{\nu \mu} \rho$$ as desired.

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  • $\begingroup$ Cool! I feel stupid for not thinking about this :/ . $\endgroup$ – Bence Racskó May 6 '17 at 10:56

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