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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 ?

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    $\begingroup$ Just apply Hamilton's tensor maximum principle to that evolution equation - at a null eigenvector of $A$ it reduces to $\partial_t A - \Delta A = |A|^2A\ge0.$ $\endgroup$ – Anthony Carapetis Mar 18 '17 at 2:46
  • $\begingroup$ @AnthonyCarapetis How to get the $|A|^2A$? Choice a normal coordinate, such that $g_{ij}=\delta_{ij}$ and $h_{ij}$ is diagonal. Then the above evolution equation is $\partial_t h_{ii}=\Delta h_{ii}-2Hh_{ii}^2+hh_{ii}^2+|A|^2h_{ii}$, then $\partial_t A =\Delta A +(h-H)|A|^2$. $\endgroup$ – lanse7pty Mar 18 '17 at 7:45
  • $\begingroup$ By $A$ I mean the full second fundamental form tensor. What I wrote is not quite true - I really mean that $(\partial_t h_{ij}-\Delta h_{ij}) v^i v^j \ge 0$ whenever $h_{ij} v^j = 0$, which is the "null eigenvector condition" required for the tensor maximum principle. $\endgroup$ – Anthony Carapetis Mar 18 '17 at 8:50
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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.

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  • $\begingroup$ Thanks, what is the relation of uniformly convex and second fundamental form ? I just Google a uniformly convex space. en.wikipedia.org/wiki/Uniformly_convex_space $\endgroup$ – lanse7pty Mar 18 '17 at 11:31
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    $\begingroup$ In this context I assume uniformly convex means $h \ge \epsilon g$ for some $\epsilon > 0$; i.e. the eigenvalues of the second fundamental form are bounded below by a uniform constant. Of course, since the manifold is compact, this uniformity doesn't really mean anything - it's equivalent to ordinary convexity $h \ge 0$. This maximum principle argument doesn't rule out the convexity constant $\epsilon$ decreasing in time - you need to study something like $h - \epsilon g$ (or I think Huisken uses $h - \epsilon H g$?) for that, which will be a little more subtle. $\endgroup$ – Anthony Carapetis Mar 18 '17 at 11:38
  • $\begingroup$ Thanks, seemly , Huisken is to prove $h\ge 0$ , then to prove $h \ge \epsilon Hg$, maybe , there is a typo on Huisken's paper. Besides, do you know any about this qustion math.stackexchange.com/questions/2184477/… . I fail to calculate it . $\endgroup$ – lanse7pty Mar 19 '17 at 6:46
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    $\begingroup$ @lanse7pty: that's a different notion for totally different spaces - we are talking about classical convexity of a domain in $\mathbb R^n$, not local convexity of a Banach space. Strictly speaking, uniform convexity is $h > 0$ and I've only proven convexity $h \ge 0$; but using a stronger maximum principle should preserve $h > 0$ using the same condition. $\endgroup$ – Anthony Carapetis Mar 19 '17 at 7:16
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    $\begingroup$ @lanse7pty: See e.g. Prop 12.47 of Chow & Knopf volume 2 (Ricci flow techniques and applications). $\endgroup$ – Anthony Carapetis Sep 27 '17 at 4:07

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