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 $divX(p)=\mbox{trace of the linear application } Y(p)\to\triangledown_{Y}X(p) $, $p\in M$ and gradient of $f$ as the vector field $grad f$ in $M$ define by $\langle grad f(p),v\rangle=df_{p}(v)$, $p\in M$, $v\in T_{p}M$. Let $E_{i}$ a geodesic frame in $p\in M$. Show that $$\triangledown(f)=\sum_{i=1}^{n}{(E_{i}(f))E_{i}(p)}\quad \mbox{ div}X(p)=\sum_{i=1}^{n}{E_{i}(f_{i})(p)}$$
Where $X=\sum_{i}{f_{i}E_{i}}$.
My approach: Note that, $\langle\mbox{grad} (f)(p),V\rangle=df_{p}(v)$, $p\in M$, $v\in T_{p}M$. So, $\langle\mbox{grad} (f)(p),E_{i}(p)\rangle=df_{p}(E_{i})=E_{i}(f)(p)$, then $$\langle\sum_{i=1}^{n}{a_{i}E_{i}(p)},E_{j}(p)\rangle=\sum_{i=1}^{n}{a_{i}}\langle E_{i}(p),E_{j}(p)\rangle=\sum_{i=1}^{n}{a_{i}\delta_{i,j}}=a_{j}$$
I don't know this is right or not. For the second equation, suppose that $Y(p)=\sum_{i=1}^{n}{a_{i}E_{i}(p)}$ and $X(p)=\sum_{j=1}^{n}{b_{j}E_{j}(p)}$, then
$$\triangledown_{Y}X\vert_p=\sum_{i=1}^{n}{a_{i}\triangledown_{E_{i}}(X)}\vert_{p}=\sum_{i=1}^{n} a_i \left({\sum_{j=1}^{n}{b_{j}\triangledown_{E_{i}}E_{j}}}+E_{i}(b_{j})E_{j}\right)\vert_{p}=\sum_{j=1}^{n}\left(\sum_{i=1}^{n}{a_{i}E_{i}(b_{j})}\right)E_{j}\vert_{p}$$
Now, how I see that $\mbox{div }X=\sum_{i=1}^{n}{E_{i}(b_{i})(p)}$. Thanks!
Best Answer
First of all you have to note that $$\hbox{div}(X)=\sum_{j=1}^{n}\langle\nabla_{E_j}X,E_j\rangle$$ taking in account that $E_i$ are orthonormal. There fore by putting $X=\sum_{i=1}^{n}f_iE_i$, we can easily get $$\hbox{div}(X)=\sum_{i,j=1}^{n}\langle\nabla_{E_j}(f_iE_i),E_j\rangle =\sum_{i,j=1}^{n}\bigl(E_j(f_i)\langle E_i,E_j\rangle-f_i\langle\nabla_{E_j}E_i,E_j\rangle\bigr)=\sum_{i=1}^{n}E_i(f_i)$$