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I don't know much about this subject at all; I'm only just getting into it. As it turns out, a physicist friend of mine asked me a formulation of the following:

Suppose $M$ is a surface in $\mathbb{R}^3$ and $M_t$ is a variation of $M$ with $M_0=M$. Under Mean Curvature Flow, is it possible for each (or any) of the $M_t$ to be minimal, including $M$?

I looked at some articles and in one by Colding et al, it says that $\nabla Vol(M_t) = H \mathbf{n}$ where $H=k_1+k_2$ (principle curvatures) and $\mathbf{n}$ is a unit normal. Then we have

$$\frac{d}{dt}Vol(M_t) = - \langle\nabla Vol,\nabla Vol \rangle = -\int_{M_t} H^2$$

For minimal surfaces, $H=0$ so it seems that the change in the volume of $M_t$ has to be constant. What if the surface undergoes a different sort of flow? Can all the $M_t$ be minimal surfaces? Does this extend to higher dimensions?

Thanks in advance.

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    $\begingroup$ The mean curvature of a minimal surface is zero everywhere, so the surface does not move at all. Trivially, all $M_t$ are minimal surfaces because they are all identical to $M$. $\endgroup$ – Rahul Jun 2 '16 at 18:30
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    $\begingroup$ By definition, a closed surface is a compact surface without boundary. Such a surface must have at least one point with positive Gauss curvature, where both principal curvatures are non-negative, and for this reason cannot be a minimal surface. $\endgroup$ – Thomas Jun 2 '16 at 18:50

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