I am reading some context from computer graphics about using eikonal equation to compute geodesics on triangular mesh, I find some reference e.g text
The authors state that if $\| \text{grad} \: \textit{f} \ \|_g=1$ then we get $$\nabla_{\text{grad} \: \textit{f}} \ \text{grad} \: \textit{f}=0$$
Could anyone provide a proof of this statement or point me to a reference where this is proven?
When $X$ is the gradient of a function $f$ then $X$ is rotation free: $$\tag{1} \nabla_\ell X_k=\nabla_k X_\ell. $$ The eikonal equation can be written as $$ (\nabla_\ell f)(\nabla^\ell f)=g^{\ell i}(\nabla_\ell f)(\nabla_if)=1\,. $$ Assuming that $\nabla$ is metric compatible we can differentiate and get \begin{align} 0&=g^{\ell i}(\nabla_k\nabla_\ell f)(\nabla_i f)+g^{\ell i}(\nabla_\ell f)(\nabla_k\nabla_if) \stackrel{(1)}=g^{\ell i}(\nabla_\ell\nabla_k f)(\nabla_i f)+g^{\ell i}(\nabla_\ell f)(\nabla_i\nabla_kf)\\[2mm] &=(\nabla^i\nabla_kf)(\nabla_if)+(\nabla_\ell f)(\nabla^\ell\nabla_kf)=2(\nabla_if)\nabla^i\nabla_kf\,. \end{align} which can be written as the directional derivative of $X=\nabla f$ with respect to itself: $$ 0=\nabla_XX\,. $$ This shows that $\nabla f$ is geodesic when it is eikonal.