Consider the unit 2-sphere and a smooth simple closed curve c embedded in it. I would guess that under the well studied parabolic equation which evolves the curve according to its curvature vector, that c shrinks through embedded sccs to a "round" point iff it bounds disks in S^2 of unequal area, and the "side" it shrinks to (which given the shrink can be unambiguously be identified as the "inside" or the "outside") is the one of smaller area. If the two sides have the same area, I would similarly guess that the curve shrinks to an equator. Does anyone know if these guesses are correct?
2 Answers
Actually, this result seems to be due to Mike Gage, and is so attributed by Grayson: @article {MR1046497, AUTHOR = {Gage, Michael E.}, TITLE = {Curve shortening on surfaces}, JOURNAL = {Ann. Sci. \'Ecole Norm. Sup. (4)}, FJOURNAL = {Annales Scientifiques de l'\'Ecole Normale Sup\'erieure. Quatri`eme S\'erie}, VOLUME = {23}, YEAR = {1990}, NUMBER = {2}, PAGES = {229--256}, ISSN = {0012-9593}, CODEN = {ASENAH}, MRCLASS = {53C22 (35K55 58G11)}, MRNUMBER = {1046497 (91a:53072)}, MRREVIEWER = {Dennis M. DeTurck}, URL = {http://www.numdam.org/item?id=ASENS_1990_4_23_2_229_0}, }
I believe that is correct and due to Matt Grayson. Under the curve shortening flow on the unit $S^2$, the evolution of the area $A(t)$ enclosed by the curve $c(t)$ is given by $dA/dt = - \int_c \kappa ds = A -2\pi$, where the last equality is by the Gauss--Bonnet formula. Solving this ODE, we get $A(t) =2\pi + e^t(A(0) -2\pi )$. If $A(0)=2\pi$, then $A(t)=2\pi$ for all $t$ and the curve limits to a great circle as $t\rightarrow \infty$. On the other hand, if $A(0) \neq 2\pi$, then the smaller of the two enclosed areas goes to zero at finite time $t_\max=\ln (2\pi /|A(0)-2\pi |)$ and the curve limits to a round point as $t\rightarrow t_\max$.
This existence and convergence to a round point or equator is due to Grayson's extension of the Gage--Hamilton theorem on convex plane curves to both the embedded case and the case of an ambient curved surface.