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


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

  • 2
    $\begingroup$ I don't know who is Mr or Mrs Leipniz, but the differentiation rule is named after Leibniz! $\endgroup$
    – Didier
    Commented Nov 17, 2022 at 10:34

2 Answers 2


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 $\nabla_X\mathrm{vol}_g = 0$.


Just for completeness, a proof in coordinates.

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: $$ \sum_{i=1}^n\Gamma^i_{ik}=\frac 12g^{il}\left(\frac{\partial g_{lk}}{\partial x^i}+\frac{\partial g_{li}}{\partial x^k}-\frac{\partial g_{ki}}{\partial x^l}\right)=\frac 12g^{il}\frac{\partial g_{li}}{\partial x^k}=\frac{\partial}{\partial x^k}\log\sqrt{\det g}. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .