Consider an arbitrary $d$-dimensional (pseudo)-Riemannian manifold $(\mathcal{M},g)$ with Levi-Civita connection and some covariant $2$-tensor density $T$. More precisely, we consider $T$ to be given by
$$T=T^{\prime}\otimes\mathrm{vol}_{g}\in\Gamma^{\infty}(T^{\ast}\mathcal{M}^{\otimes 2}\otimes{\bigwedge}^{d}T^{\ast}\mathcal{M})\cong\Gamma^{\infty}(T^{\ast}\mathcal{M}^{\otimes 2})\otimes_{C^{\infty}(\mathcal{M})}\Omega^{d}(\mathcal{M}),$$
where $\mathrm{vol}_{g}$ denotes the volume $d$-form of $(\mathcal{M},g)$ and $\Omega^{d}(\mathcal{M})$ denotes the space of top-degree differential forms on $\mathcal{M}$. In other words, we consider a density $T$ with tensor part $T^{\prime}$ and density part $\mathrm{vol}_{g}$.
Now, I try to figure out what $\nabla_{X}T$ is for some vector field $X$. Using the Leibniz rule, this can be written as
$$\nabla_{X}T=(\nabla_{X}T^{\prime})\otimes\mathrm{vol}_{g}+T^{\prime}\otimes\nabla_{X}\mathrm{vol}_{g}.$$
Maybe its trivial, but is there are general way to compute $\nabla_{X}\mathrm{vol}_{g}$? By definition of the Levi-Civita connection, $\nabla g=0$. Does this imply $\nabla_{X}\mathrm{vol}_{g}=0$? I am not sure, since there is the square root of $g$ appearing ing $\mathrm{vol}_{g}$…
Best Answer
The Riemannian volume form of an oriented Riemann manifold is parallel: $\nabla \mathrm{vol}_g=0$.
Indeed, let $\{E_1,\ldots,E_n\}$ be a local positively oriented orthonormal frame. For $j\in \{1,\ldots,n\}$, let $\theta^j = g(\cdot,E_j)$. Then $\theta^1\wedge\cdots\wedge \theta^n$ is a $n$-form, and agrees with $\mathrm{vol}_g$ on $\{E_1,\ldots,E_n\}$: it follows that locally, $$ \mathrm{vol}_g = \theta^1\wedge\cdots\wedge \theta^n. $$ Therefore, for any $X$, locally, $$ \label{star} \nabla_X\mathrm{vol}_g = \sum_{j=1}^n \theta^1\wedge\cdots\wedge \nabla_X\theta^j \wedge \cdots \wedge \theta^n. \tag{$\star$} $$ Fix $j\in \{1,\ldots,n\}$. From $\|E_j\|^2 = 1$, one has $g(\nabla_XE_j,E_j)=0$, and then, $\nabla_XE_j \in \mathrm{span}\left\{ E_k \mid k\neq j\right\}$. It follows that $$ \nabla_X\theta^j = g(\cdot,\nabla_XE_j) \in \mathrm{span}\left\{\theta^k \mid k\neq j\right\}. $$ Hence, any term in the sum of equation \eqref{star} is then zero, and $\nabla_X\mathrm{vol}_g = 0$.