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Let $\Omega\subset \mathbb{R}^n$ be open and let $f:\Omega\times [0,T)\to \mathbb{R}$ be a smooth function. Consider the graph function $\phi:\Omega\times [0,T)\to \mathbb{R}^{n+1}$ given by $\phi(p,t)=(p,f(p,t))$. In general $\phi$ is said to flow in mean curvature if $$ \partial_t \phi=H\nu $$ Where $H$ is the mean curvature function for the pullback metric $g_{ij}=\delta_{ij}+f_if_j$ at each time, and $$\nu=\frac{(-\nabla f,1)}{1+|\nabla f|^2}$$ is the unit vector perpendicular to the immersed manifold $\phi|_t$.

This is a very common example and yet I get confused. In fact, $$\partial_t\phi=(0,...,0,\partial_t f)=\partial_t f\textbf{e}_{n+1}$$ and therefore $\nu||\textbf{e}_{n+1}$ i.e. $\nu=\textbf{e}_{n+1}$, and $\nabla f=0$.

This is clearly ridiculous and yet I cannot find my mistake.

Or Kedar
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1 Answers1

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It is because there is a tangential diffeomorphism that is involved.

In general for any family of immersions $\phi : M\times [0,T)\to \mathbb R^{n+1}$, the MCF equation is $\partial_t \phi=H\nu$. But one can also consider

$$\tag{1} (\partial_t \tilde \phi )^\perp = H\nu, \text{ or } \langle \partial_t \tilde \phi, \nu \rangle = H. $$

The difference is that $\partial _ t\tilde \phi$ might contain non-zero tangential part $X = (\partial_t \tilde \phi)^\top$. However, one can consider the following: let $Y$ be the time-dependent vector fields on $M$ such that $\tilde \phi_* Y = X$. Consider the one-parameter groups of diffeomorphism on $M$ generated by $-Y$:

$$\tag{2} y: M \times [0,T) \to M, \ \ \frac{\partial y}{\partial t} (y(x, t), t) = -Y(y(x, t)).$$

Then $\tilde \phi$ satisfies (1) if and only if the composition

$$ \phi (x, t) = \tilde \phi((y(x, t), t)$$

satisfies the usual MCF.

Going back to the graphical cases. We are really solving (1), which becomes

$$ \frac{\partial _t f }{ \sqrt{1+ |\nabla f|^2}} = H\Rightarrow \partial _t f = \sqrt{1+|\nabla f|^2} \operatorname{div} \left(\frac{\nabla f}{\sqrt{1+|\nabla f|^2}}\right) $$

Remark The trick to use (1) instead of the usual MCF is called the DeTurck trick and is orginated from Ricci flow. You can see that explained in here or p.548 here. (a related question)

Remark It seeems that whether or not (1) is really equivalent to the MCF depends on whether (2) as a ODE is solvable. When $M$ is compact this can always be done. In the non-compact cases, there might be some reason why (2) can always be solving, but I cannot find it in literature (As you can see in Ecker-Huisken's paper, they did not even bother to explain). In fact a doubt is raised in this MSE post

Arctic Char
  • 16,972
  • If $\partial_t\tilde{\phi}(x,t) = \tilde{H}(x,t)\tilde{\nu}(x,t) + X(x,t)$ then I believe $\partial_t y(y(x,t),t) = X(y(x,t))$ should be changed to $$D_y\tilde{\phi}(y(x,t),t) \cdot \partial_t y(x,t) = -X(y(x,t),t),$$ since $$\partial_t \phi(x,t) = \frac{d \tilde{\phi}(y(x,t),t)}{d t} = \partial_t \tilde{\phi}(y(x,t),t) + D_y \tilde{\phi}(y(x,t),t)\cdot \partial_t y(x,t) = \tilde{H}(y(x,t),t)\tilde{\nu}(y(x,t),t) + X(y(x,t),t) - X(y(x,t),t) = \tilde{H}(y(x,t),t)\tilde{\nu}(y(x,t),t)= H(x,t)\nu(x,t).$$ Am I misunderstanding something? – lightxbulb Jun 22 '25 at 12:28
  • Regarding the remark - if a graph function can "fold over" under mean curvature flow then I suppose that the solution of the ODE will fail there, i.e., you would have only local existence of a solution. I do not know whether there are theorems that state that if you start with a heightmap/graph it always remains a heightmap/graph under mean curvature flow. Even if this is true for the MCF, it could happen that this is not true for the wave equation using Laplace-Beltrami (e.g. resulting in waves that overhang). It's still probably the most reasonable thing to do if you are limited to graphs. – lightxbulb Jun 22 '25 at 16:20
  • @lightxbulb Thanks, I have fixed the error. – Arctic Char Jun 30 '25 at 12:21