Show that $\hbox{div}(X)=\sum_{j=1}^{n}\langle\nabla_{E_j}X,E_j\rangle$

Question

Let $M$ a riemannian manifold. Let $X\in\chi(M)$ and $f$ a function $C^{\infty}$ in $M$. Define the divergence of $X$ as a function $div X:M\to\mathbb{R}$ given by $\operatorname{div}X(p)=\{\mbox{trace of the linear application }\} Y(p)\to\triangledown_{Y}X(p)$, $p\in M$, show that $$\hbox{div}(X)=\sum_{j=1}^{n}\langle\nabla_{E_j}X,E_j\rangle$$

Where $X=\sum_{i}{f_{i}E_{i}}$ and $E_{i}$ is geodesic frame ($\nabla_{E_{i}}{E_{j}}=0$).

Well, if $X=\sum_{i}{E_{i}}$ and $Y=\sum_{j}{f_{j}E_{j}}$, reemplace in the equation $\nabla_{X}{Y}=\sum_{ij}{x_{i}y_{j}\nabla_{X_{i}}X_{j}}+\sum_{ij}{x_{i}X_{i}(y_{j})X_{j}}$, but I don't see the relation, (maybe the relation is more simple) any hint is appreciated. Thanks!

Answer 1

$\{e_i\}$ is basis of $T_pM$ ,and $\{\omega^i\}$ is dual basis. Under some suitable frame $(\nabla _{e_i}e_j=0,\langle e_i,e_j\rangle=0(i\ne j))$, $$ div(X)=tr(Y\rightarrow \nabla _Y X)=\omega^i(\nabla_i X)=\omega^i(e_i(X^j)e_j)=e_i(X^i) $$ Besides $$ \langle \nabla_j(X^ie_i) ,e_j \rangle =\langle e_j(X^i)e_i ,e_j\rangle=e_i(X^i) $$

Answer 2

The easiest way for me to think of this problem is to write $Y(p) \to \nabla_Y X(p)$ as a matrix. Let $X = f^iE_i$ and $Y = y^jE_j$. Then

\begin{eqnarray*} \nabla_YX & = & y^i \nabla_{E_i} (f^jE_j) \\ & = & y^jE_i(f^j) E_j + y^i f^j \nabla_{E_i}E_j \\ & = & y^iE_i(f^j)E_j \\ & = & \left [ \begin{array}{cccc} E_1(f^1) & E_2(f^1) & \ldots & E_n(f^1) \\ E_1(f^2) & E_2(f^2) & \ldots & E_n(f^2) \\ \vdots & \vdots & \ddots & \vdots \\ E_1(f^n) & E_2(f^n) & \ldots & E_n(f^n) \\ \end{array} \right ] \left [ \begin{array}{c} y^1 \\ y^2 \\ \vdots \\ y^n \\ \end{array} \right ] \end{eqnarray*}

where we used the fact that $\nabla_{E_i}E_j$ can be chosen to be zero at the point $p$. Does this seem a little clearer now?