# Equivalence class of the normal flow

by lanse7pty   Last Updated March 13, 2018 07:20 AM

Consider a compact manifold $M^n$ embedding in $\mathbb R^{n+1}$. We can deform it by some geometric flow. A kind of all, we only move it in the normal direct, for example, mean curvature $$\partial_tF=-H\nu$$ Another kind, we move it in the normal direct and position vector direct, for example, the area preserving mean curvature flow $$\partial_t F= -H\nu + \frac{h}{n}F$$ From this question, I know the moving in non-normal direct can be decomposed into normal moving and tangent moving. So I feel some flow is equal to normal flow. Assume I have a flow $$\partial_t F = \varphi F+ \psi \nu$$ and $F: M^n\times [0,T)\rightarrow \mathbb R^{n+1}$ is solution of it. Then $$\partial_t F = \varphi F^\top + (\varphi\langle F ,\nu\rangle +\psi)\nu$$ where $F^\top$ is tangential component of $F$. Then, finding a vector field $X_t\in\mathfrak{X}(M^n)$ such that $$\partial_t F - (\varphi\langle F ,\nu\rangle +\psi)\nu = d(F_t)(X_t)$$ Let $x_t:M^n\times [0,T)\rightarrow M^n$ be the flow of $X_t$, and $\widetilde F_t = F_t \circ x_t$, according to Amitai Yuval's answer, I guess I have $$\partial_t \widetilde F =(\varphi\langle \widetilde F ,\widetilde \nu\rangle +\psi)\widetilde \nu$$ In fact, I don't know how to calculate the above equality. Then, any flow liking $$\partial_t F = \varphi F+ \psi \nu$$ is equal to a normal flow which has form $$\partial_t \widetilde F =(\varphi\langle \widetilde F ,\widetilde \nu\rangle +\psi)\widetilde \nu$$ whether I am right ?

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