On the proof of Cheng-Yau gradient estimate

Question

I am reading Schoen-Yau's Differential Geometry book and having a question in the proof of Cheng-Yau gradient estimate. We consider a complete $n$-dimensional Riemannain manifold $M$ and let $u$ be a positive harmonic function on $M$. Now fix a point $p$ in $M$ and we consider $|\nabla (|\nabla u|)|^2$ at the point $p$. The book said "since the computation is local, we can choose a normal coordinate in a neighborhood of $p$ such that $u_i(p)=0$ for $i\geq 2$ and $u_1(p)=|\nabla u|_{(p)}$." Here $u_i$ denotes the covariant derivative of $u$ w.r.t. $\frac{\partial}{\partial x_i}$. I am confused about this assumption. I know the definition of normal coordinates but I don't see how to get this assumption by choosing a normal coordinate. Any comments are welcome!

Answer 1

Given any basis $v_1,\ldots,v_n$ of $T_pM$, one can define normal coordinates $F:B_{\varepsilon}^{\mathbb{R}^n}(0)\to M$ by $F(t_1,\ldots,t_n)=\exp_p(t_1v_1+\cdots+t_nv_n).$ Let $x$ denote $F^{-1}$; then the coordinate vector fields at the point $p$ are $\frac{\partial}{\partial x^1}|_p=v_1$ and $\frac{\partial}{\partial x^2}|_p=v_2$ and so on.

For this problem, take any $g_p$-orthonormal basis with $v_1=\frac{\nabla u(p)}{|\nabla u(p)|}$. Then for any $i=1,\ldots,n$ you have $$\frac{\partial u}{\partial x^i}(p)=\Big\langle v_i,\nabla u(p)\Big\rangle=\Big\langle v_i,v_1\Big\rangle|\nabla u(p)|$$ which by the $g_p$-orthonormality of $v_1,\ldots,v_n$ is $0$ if $i\geq 2$ and is $|\nabla u(p)|$ if $i=1.$