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 ?

Related Questions

Different equations for mean curvature flow

Updated January 02, 2018 23:20 PM

Small eigenvalue of $\Delta$ under mean curvature flow

Updated November 05, 2017 03:20 AM

Huisken's Distance Comparison principle for Ricci flow

Updated November 06, 2017 07:20 AM

Mean Curvature Flow equation, where does it come from?

Updated November 06, 2017 07:20 AM