I am reading the paper "Minimal Immersions of 2-Manifolds into Spheres". In (1.1), page 379, the author wrote that $\langle x_i,x_{jk}\rangle=0$. I don't know why this equality holds true but I think this is not a difficult one. Did I miss something? I would be grateful if someone could help me!
The link is behind a paywall, so I explain more about the notations here:
$(M,g)$ is a closed connected 2-dimensional Riemannian manifold.
Consider the isometric minimal immersions $\tilde{x}:M\rightarrow\mathbb{S}^N(1)$.
$j:\mathbb{S}^N(1)\rightarrow\mathbb{R}^{N+1}$ is the canonical embedding.
$x:=j\circ\tilde{x}$.
$x_i,x_{jk}$ is the covariant derivatives.
$\langle,\rangle$ is the inner product in $\mathbb{R}^{N+1}$.
So nothing specific was assumed about the local coordinates. This is just the usual cyclic permutation trick for solving for Christoffel symbols. Before we start, note that the covariant derivatives $x_{ij}$ and $x_{ji}$ are equal: \begin{align*} x_{ij} &= \frac{\partial^2 x}{\partial u^j\partial u^i} + \sum\Gamma^k_{ij}\frac{\partial x}{\partial u^k} \\ &= \frac{\partial^2 x}{\partial u^i\partial u^j} + \sum\Gamma^k_{ji}\frac{\partial x}{\partial u^k} \\ &= x_{ji}, \end{align*} since the connection is torsion-free.
Since the connection is compatible with the Riemannian metric, we have $$0 = \nabla_{\partial/\partial u^k} g_{ij} = \langle x_i,x_{jk}\rangle + \langle x_{ik},x_j\rangle.$$ We cyclically permute, and do the usual $+$ $-$ $+$ trick: \begin{alignat*}2 0 &= \langle x_i,x_{jk}\rangle + \langle x_{ik},x_j\rangle & \quad (+) \\ 0 &= \langle x_j,x_{ki}\rangle + \langle x_{ji},x_k\rangle & (-) \\ 0 &= \langle x_k,x_{ki}\rangle + \langle x_{kj},x_i\rangle & (+) \end{alignat*} Using the symmetry of the covariant derivatives we asserted at the beginning, we end up with $2\langle x_i,x_{jk}\rangle = 0$, as desired.