Volume preserving mean curvature flow preserving uniformly convex

Question

Let $(M_t,g_t)$ be a Riemannian manifold evolve by volume preserving mean curvature flow. So , for second fundamental form, we have $$ \partial_t h_{ij}=\Delta h_{ij}-2H h_{im}h^m_j+hh_{im}h^m_j + |A|^2 h_{ij} $$ $H=g^{ij}h_{ij}$ is mean curvature, $|A|^2=g^{ij}g^{kl}h_{ik}h_{jl}$ is inner product of second fundamental form. If the initial manifold is uniformly convex : the eigenvalues of its second fundamental form are strictly positive everywhere. Then, how to show $M_t$ still be uniformly convex for all $t\ge 0$ where the solution exists ?

Answer 1

Use Theorem 9.1 of Hamilton's THREE-MANIFOLDS WITH POSITIVE RICCI CURVATURE, which tells you that if a time-dependent symmetric tensor field $h$ satisfies $$\partial_t h_{ij} = \Delta h_{ij} + N_{ij}$$

with the reaction term $N$ satisfying the null-eigenvector condition $$h_{ij} v^i = 0 \implies N_{ij}v^iv^j \ge 0,$$

then positive-definiteness $h \ge 0$ is preserved in time. You can derive this from the scalar maximum principle by studying the scalar function $v \mapsto h(v,v)$ on the unit tangent bundle.

In this case we have $N_{ij} = -2H h_{im}h^m_j+hh_{im}h^m_j + |A|^2 h_{ij}$, so assuming $h_{ij}v^i = 0$ we see $N_{ij} v^i = 0$. Thus the inequality is preserved.