# Is there a generalization of ${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$ for arbitrary connection?

I am studying general relativity and I am trying to understand how to perform variation of the Einstein–Hilbert action with respect to the metric $${{g}_{\mu \nu }}$$ and an arbitrary connection $${{\omega }^{\kappa }}_{\lambda \mu }$$ treating them as independent variables.
At some point I do have to perform integration by parts, where the relation $${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$$ could be useful. The problem is that I am convinced that this relation is not valid for an arbitrary connection. For the case of an arbitrary symmetric connection I have managed to prove the relation $${{\left( \sqrt{-g}{{V}^{a}} \right)}_{,\alpha }}=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}+\sqrt{-g}\frac{1}{2}{{g}^{\sigma \tau }}{{g}_{\sigma \tau }}_{;\alpha }{{V}^{\alpha }}$$ which gives $${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$$ if the connection is metric compatible.

Therefore I would like to ask if there is a way to obtain a generalization of $${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$$ that is valid for an arbitrary connection (non-symmetric and/or not metric-compatible)?

• Any two connections $\nabla_a$ and $\nabla'_a$ different from each other by $\nabla'_a\omega_b=\nabla_a\omega_b-C^c{}_{ab}\omega_c$ for an arbitrary $\omega_a$. Here, $C^c{}_{ab}$ may not possess any symmetry, depending on the properties of $\nabla_a$ and $\nabla'_a$. If you let $\nabla_a$ be the Levi-Civita connection, and further use $\nabla_a\omega_b=\partial_a\omega_b-\Gamma^c{}_{ab}\omega_c$ with $\Gamma^c{}_{ab}$ the Christoffel symbol, you can find a general result, which might not be beautiful. – Drake Marquis Aug 2 at 0:25

For the purposes of this answer, to adapt to the level of mathematics presented in OP, I will use (signed) densities, rather than differential forms.

Let $$\rho$$ be a nowhere vanishing scalar density of weight 1 (eg. of the same type as $$\sqrt{-g}$$, some people consider this to be weight -1 I guess). Such an object shall be called a volume density.

The covariant derivative of scalar densities of weight 1 (with respect to any connection) is defined as $$\nabla_\mu\rho=\partial_\mu\rho -\Gamma_{\mu\nu}^\nu\rho.$$

Here $$\Gamma$$ doesn't denote the Christoffel symbols, but the coefficients of an arbitrary linear connection.

The connection $$\nabla$$ is said to be volume-presering if there is such a $$\rho$$ such that $$\nabla_\mu\rho=0$$. We make some statements regarding volume-preserving connections.

• If $$\nabla$$ is volume-preserving then $$\Gamma_\mu\equiv\Gamma_{\mu\nu}^\nu$$ is a gradient: $$0=\nabla_\mu\rho=\partial_\mu\rho-\Gamma_\mu\rho \\ \partial_\mu\rho=\Gamma_\mu\rho \\ \partial_\mu\ln\rho=\Gamma_\mu.$$ Dividing by $$\rho$$ was possible because we assumed it is nowhere vanishing.
• If $$\nabla$$ is volume-preserving and torsionless, then for any vector field $$X^\mu$$ we have $$\nabla_\mu X^\mu\rho=\partial_\mu(X^\mu\rho):$$ $$\nabla_\mu X^\mu=\partial_\mu X^\mu+\Gamma_{\mu\nu}^\mu X^\nu=\partial_\mu X^\mu+\Gamma_{\nu\mu}^\mu X^\nu+T^\mu_{\ \mu\nu}X^\nu=\partial_\mu X^\mu+\partial_\mu\ln\rho X^\mu+T^\mu_{\ \mu\nu}X^\nu \\ =\frac{1}{\rho}\partial_\mu(\rho X^\mu)+T^\mu_{\ \mu\nu}X^\nu.$$ Hence if the connection is torsionless, the identity is proven.

• The connection $$\nabla$$ is locally volume-preserving if and only if the curvature tensor is $$\mathfrak{sl}(n,\mathbb R)$$-valued, which in this context means that $$R^\kappa_{\ \kappa\mu\nu}=0$$: $$\$$ First we prove that if the connection is volume preserving, then the curvature is traceless: $$R^\kappa_{\ \kappa\mu\nu}=\partial_\mu\Gamma^\kappa_{\nu\kappa}-\partial_\nu\Gamma^\kappa_{\mu\kappa}+\Gamma_{\mu\sigma}^\kappa\Gamma_{\nu\kappa}^\sigma-\Gamma_{\nu\sigma}^\kappa\Gamma_{\mu\kappa}^\sigma=\partial_\mu\partial_\nu\ln\rho-\partial_\nu\partial_\mu\ln\rho=0,$$ where the first two terms cancel because of the equality of mixed partials, and the cancellation in the second two terms is evident if the summation indices are renamed. $$\$$ We now illustrate that if the curvature tensor is traceless then there at least locally exists a volume density that is preserved by the connection. We have $$0=R^\kappa_{\ \kappa\mu\nu}=\partial_\mu\Gamma_\nu-\partial_\nu\Gamma_\mu.$$ Note that the cancellation of the last two terms doesn't depend on the specific form of the connection (it is essentially based on the identity $$\text{Tr}(AB)=\text{Tr}(BA)$$). By Poincaré's lemma we have then $$\Gamma_\mu=\partial_\mu\phi$$ for some function $$\phi$$, at least locally. But then one may check that $$\rho=e^\phi$$ transforms as a density, and is preserved by the connection, since $$\nabla_\mu\rho=\partial_\mu\rho-\Gamma_\mu\rho=\partial_\mu e^\phi-\partial_\mu\phi e^\phi=0.$$ An alternative line of thought is that on densities, the curvature is $$-R^\kappa_{\ \kappa\mu\nu}$$, so if this vanishes, then we can at least locally (on contractible domains) integrate the partial differential equation $$\nabla_\mu\rho=0$$.

TLDR:

The equation $$\nabla_\mu X^\mu\sqrt{-g}=\partial_\mu(X^\mu\sqrt{-g})$$ does not generalize to arbitrary connection. However if the connection is such that its curvature tensor is traceless, then one may locally find a volume density $$\rho$$ such that if $$\sqrt{-g}$$ is replaced with $$\rho$$, and $$\nabla$$ is torsionless, the equation will keep holding.

• Thank you very much for your reply. – Nikos Aug 5 at 7:58