Levi-Civita connection acting on tensor density

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 $$\mathrm{vol}_{g}$$...

• I don't know who is Mr or Mrs Leipniz, but the differentiation rule is named after Leibniz! Commented Nov 17, 2022 at 10:34

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 $$\nabla_X\mathrm{vol}_g = 0$$.
Let $$vol_g=\sqrt{\det g}\,dx^1\wedge\cdots\wedge dx^n$$, where $$g$$ denotes the Gram matrix $$\langle\tfrac\partial{\partial x_i},\frac\partial{\partial x_j}\rangle$$, and $$n=$$dimension of the Riemannian manifold.
Then, for any vector field $$X$$, \begin{align*} \nabla_X vol_g={}&X(\sqrt{\det g})\, dx^1\wedge\cdots\wedge dx^n \\ &+\sqrt{\det g}\,(\nabla_X dx^1)\wedge\cdots\wedge dx^n \\ &\cdots \\ &+\sqrt{\det g}\,dx^1\wedge\cdots\wedge(\nabla_X dx^n). \end{align*} We have $$\nabla_X dx^i=\Gamma^i_{jk}d x^jX^k$$ hence $$(\nabla_X dx^1)\wedge\cdots\wedge dx^n=\Gamma^1_{1k}X^kdx^1\wedge\cdots\wedge dx^n \\ \cdots \\ dx^1\wedge\cdots\wedge (\nabla_X dx^n)=\Gamma^n_{nk}X^kdx^1\wedge\cdots\wedge dx^n$$ so it remains to check that $$\frac{X(\sqrt{\det g})}{\sqrt{\det g}}=\sum_{i=1}^n\Gamma^{i}_{ik}X^k.$$ This follows from the following identity for Christoffel symbols of the Levi-Civita connection: $$\sum_{i=1}^n\Gamma^i_{ik}=\frac{\partial}{\partial x^k}\log\sqrt{\det g}.$$ The latter can be proved as follows: $$\sum_{i=1}^n\Gamma^i_{ik}=\frac 12g^{il}\left(\frac{\partial g_{lk}}{\partial x^i}+\frac{\partial g_{li}}{\partial x^k}-\frac{\partial g_{ki}}{\partial x^l}\right)=\frac 12g^{il}\frac{\partial g_{li}}{\partial x^k}=\frac{\partial}{\partial x^k}\log\sqrt{\det g}.$$