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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.

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  • $\begingroup$ 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. $\endgroup$
    – K.defaoite
    Feb 15, 2022 at 11:25
  • $\begingroup$ 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. $\endgroup$
    – K.defaoite
    Feb 15, 2022 at 11:25
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    $\begingroup$ @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 $\endgroup$
    – Didier
    Feb 15, 2022 at 11:30
  • $\begingroup$ @OP What is your definition of divergence? What do you mean by $\nabla X(e_i,e_i)$? $\endgroup$
    – Didier
    Feb 15, 2022 at 11:36
  • $\begingroup$ @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. $\endgroup$ Feb 15, 2022 at 11:46

1 Answer 1

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$\DeclareMathOperator{\div}{div} \DeclareMathOperator{\tr}{tr}$

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}

There are plenty of other proofs (on this website also) relying on the coordinate expression of the Riemannian volume form. This method is known as the method of moving frames (referring to the local orthonormal (co-)frame). It surprisingly turns out that it is less known whilst it is (in my opinion) simpler. Some people (including me) prefer to work in a coordinate-free manner, and to that end, moving frames are good friends.

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  • $\begingroup$ An elegant proof!Thanks a lot. $\endgroup$ Feb 15, 2022 at 14:11

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