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Picture below is from 242th page of

Huisken, Gerhard, Flow by mean curvature of convex surfaces into spheres, J. Differ. Geom. 20, 237-266 (1984). ZBL0556.53001.

enter image description here

$H$ is the mean curvature. I don't know how to get the red line under this frame. In my view, $$ \nabla H =(\frac{\partial H}{\partial x_1},...,\frac{\partial H}{\partial x_n}) $$ and $$ \frac{\partial }{\partial x_1}=e_1 =\frac{\nabla H}{|\nabla H|} $$

Enhao Lan
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1 Answers1

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"A frame" is not a global coordinate system, but rather a collection of bases for tangent spaces, say near a point as in the above text. If one introduces a frame where $e_1(p)=\frac{\nabla H(p)}{|\nabla H(p)|}$ then when using the basis that is given by the frame the vector $\nabla H(p)$ is written as a linear combination of basis vectors as $$\nabla H(p)= |\nabla H(p)| e_1(p) +0 e_2(p)+\ldots +0 e_n(p)$$ Thus, since the frame is orthonormal at each point, $\nabla_i H(p)= \nabla H(p) \cdot e_i(p)$ is as in the "red line".

Max
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  • Thanks, do you know how to understand $\nabla H=\frac{\partial H}{\partial x_i}\frac{\partial F}{\partial x_j}g^{ij}$ ? – Enhao Lan Apr 19 '20 at 12:57
  • What is the context for this second equation? I presume it's the same paper, but what section? – Max Apr 19 '20 at 17:24
  • @lanse7pty, $\nabla H$ is a tangent vector field, so the right-hand side is just decomposition to n-linear independent tangent vectors at a point. – STUDENT Oct 24 '20 at 07:24