# Some calculations about divergence of vector field on Riemannian manifold

Assume first that we have a Riemannian manifold $$(M,g)$$. Furthermore, $$X$$ is a vector field on M and $$\nabla$$ is the Levi-Civita connection as usual. Let $$\{e_i\}$$ be an orthonormal basis on M.

Then how can we get $$\operatorname{div} X =\sum_i\langle e_i,\nabla_{e_i}X\rangle$$? Where $$\operatorname{div}$$ represents the divergence of $$X$$. In other words, $$\operatorname{div} X=\operatorname{tr}(\nabla X)$$.

I try to use the definition of covariant differential and we have $$\operatorname{div} X = \sum_{i}\nabla X(e_i,e_i)=\sum_{i}\nabla_{e_i}X(e_i)$$ but what’s next? I have found many books and they just ignore the detail so can someone help me ? Many thanks to you.

• Firstly, requiring $e_i$ be an orthonormal basis is the wrong condition. You should instead require it to be a coordinate basis. If you don't know the difference, see here. Feb 15, 2022 at 11:25
• Second, I'm not exactly sure what you're trying to do here. Typically $\operatorname{div}=\operatorname{tr}\nabla$ is taken as a definition, not something that you are supposed to show. Feb 15, 2022 at 11:25
• @K.defaoite I usually define the divergence of a vector field $X$ to be the unique smooth function such that $L_X \mathrm{d}vol_g = (\mathrm{div} X) \mathrm{d}vol_g$ ($M$ should be orientable), and it makes sense trying to show the equality $\mathrm{div}X = \mathrm{tr} \nabla X$. Moreover, it totally makes sense to work with an orthonormal frame $\{e_1,\ldots,e_n\}$ since the definition is tensorial. No need to consider coordinate basis. In fact, it is the good condition since the trace of an endomorphism $A$ is given by $\sum_i \langle Ae_i,e_i\rangle$ only in an orthonormal basis Feb 15, 2022 at 11:30
• @OP What is your definition of divergence? What do you mean by $\nabla X(e_i,e_i)$? Feb 15, 2022 at 11:36
• @K.defaoite I want to prove how the divergence of a vector field X can be $\sum_i\langle e_i, \nabla_{e_i} X\rangle$. And I think i can just take an orthonormal frame $\{e_1,…, e_n\}$ as Didier said. Feb 15, 2022 at 11:46

$$\DeclareMathOperator{\div}{div} \DeclareMathOperator{\tr}{tr}$$
We consider the following definition of the divergence. Let $$(M,g)$$ be an orientable Riemannian manifold, with Riemannian volume form $$\mathrm{d}vol_g$$, and $$X$$ be a vector field. The divergence of $$X$$ is the unique smooth function $$\div X$$ such that $$L_X\mathrm{d}vol_g = (\div X) \mathrm{d}vol_g$$.
Let $$\{e_1,\ldots,e_n\}$$ be a local orthonormal frame and consider its dual frame $$\{\theta^1,\ldots,\theta^n\}$$, that is $$\theta^i = g(\cdot,e_i)$$. Then we have the equality $$\mathrm{d}vol_g = \theta^1\wedge\cdots\wedge \theta^n.$$ Indeed, these two volume forms are proportional and coincide on the orthonormal frame $$\{e_1,\ldots,e_n\}$$. Now, since $$L_X$$ is a derivation of the exterior algebra, it follows that $$L_X\mathrm{d}vol_g = \sum_{i=1}^n \theta^1\wedge\cdots\wedge L_X\theta^i \wedge \cdots \wedge \theta^n.$$ For any vector field $$Y$$, we have \begin{align} (L_X\theta^i)(Y) &= X\theta^i(Y) - \theta^i([X,Y]) & \text{by Leibniz rule}\\ &= Xg(Y,e_i) - g([X,Y],e_i)\\ &= g(\nabla_XY-[X,Y],e_i) + g(Y,\nabla_Xe_i) &\text{from the compatibility of g and \nabla}\\ &= g(\nabla_YX,e_i) + g(Y,\nabla_Xe_i) & \text{since \nabla is torsion-free}\\ &= (\theta^i\circ \nabla X)(Y) + g(Y,\nabla_Xe_i). \end{align} Applying $$\nabla_X$$ to the equality $$\|e_i\|^2 = 1$$ gives $$\nabla_Xe_i\perp e_i$$, so that the 1-form $$g(\cdot,\nabla_Xe_i)$$ vanishes on $$e_i$$, and is then a linear combination of $$\{\theta^j\}_{j\neq i}$$. It then disappears in the wedge product, and it follows that $$L_X\mathrm{d}vol_g = \sum_{i=1}^n \theta^1\wedge\cdots\wedge (\theta^i\circ \nabla X) \wedge \cdots \wedge \theta^n.$$ Finally, evaluating on the orthonormal frame $$\{e_1,\ldots,e_n\}$$ yields \begin{align} \div X &= \sum_{i=1}^n \theta^1(e_1)\times \cdots \times \theta^i(\nabla_{e_i}X) \times \cdots \theta^n(e_n)\\ &= \sum_{i=1}^n \theta^i(\nabla_{e_i}X)\\ &= \sum_{i=1}^n g(\nabla_{e_i}X,e_i)\\ &= \tr \nabla X. \end{align}