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Let $\{e_1,...,e_n\}$ be standard basis of $\mathbb R^n$, consider the cylinder $$ L=\{x\in \mathbb R^n : |x-\langle x, e_1\rangle e_1|=1 \} $$ Obviously, mean curvature of $L$ is $H=n-1$. Assume $\{x_1,..., x_{n-1}\}$ is a local coordinate of $ L$, then , the gradient of $H$ is $$ \nabla H = g^{ij} \frac{\partial H}{\partial x_i}\frac{\partial}{\partial x_j}=0 $$ On the other hand, the normal vector of $L$ is $$ \nu=x-\langle x, e_1\rangle e_1 $$ And $$ \langle x, \nu\rangle =1 $$ So, we have $$ H=(n-1)\langle x, \nu \rangle $$ Then, $$ \nabla \langle x, \nu \rangle =0 $$ Now I want to verify $\nabla \langle x, \nu \rangle =0$.
$$ \nabla \langle x, \nu \rangle =g^{ij}\frac{\partial \langle x, \nu \rangle }{\partial x_i}\frac{\partial}{\partial x_j} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g^{ij} (\langle \nabla_i x, \nu\rangle + \langle x, \nabla _i \nu \rangle) \frac{\partial}{\partial x_j} $$ It is equal to show $$ \langle \nabla_i x, \nu\rangle + \langle x, \nabla _i \nu \rangle =0 $$ Because $\nabla_i x$ is tangent vector, $\langle \nabla_i x, \nu\rangle=0$. Then $$ \langle x, \nabla _i \nu \rangle = \langle x, -h_{ij}g^{jk}\frac{\partial x}{\partial x_k} \rangle $$ If the local coordinate is $$ x_1=e_1, ~~~~\{x_2,...,x_n\} \text{ is the spherical coordinates } $$ Then, we have $$ g_{ij}=\delta_{ij} ~~~\text{ and }~~~ h_{ij}=\begin{equation} \left\{ \begin{aligned} &1 ~~~~~i=j\ne 1 \\ &0 ~~~~~\text{others}\\ \end{aligned} \right. \end{equation} $$ Then $$ \langle x, -h_{ij}g^{jk}\frac{\partial x}{\partial x_k} \rangle=0 $$ Since $$ \langle x, \frac{\partial x}{\partial x_i} \rangle =0 ~~~\text{when $i\ne 1$,} ~~~~~\text{and}~ h_{11}=0 $$

PS: At beginning , I don't know how to calculate $\nabla_i x$ and $\nabla _i \nu$. Then, I know , and add my calculate here.

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