Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am unclear how the inverse mean curvature flow starts with a minimal surface. If there is some point $p$ with mean curvature $H(P)=0$, how should we treat this problem and let the flow start?

share|improve this question
    
Why do you believe that the flow starts at such a point or for a minimal surface? –  Ben McKay Mar 23 at 17:24
    
If the initial hypersurface for inverse mean curvature flow is the minimal surface, the penrose inequality could be proved with the monotonicity of the Hawking mass. I am a little confused about the beginning of the flow and want to understand that more clearly. –  Jer Mar 24 at 8:33
add comment

1 Answer 1

up vote 4 down vote accepted

It isn't clear exactly what sort of initial conditions you're requiring.

The difficulty with minimal initial conditions is part of the reason why it was an amazing result when Huisken--Ilmanen constructed a "weak inverse mean curvature flow" which

(1) Can start at a minimal surface (technically, for certain things to work nicely, it should be outer-minimizing)

(2) Exists for all time in an asymptotically flat manifold.

AND

(3) Still satisfies Geroch monotonicity, i.e. the Hawking mass is monotone along the flow.

That one could find a "flow" which satisfies (1) and (2) while still keeping (3) is incredible.

Their paper is very readable, although quite long, so I'd recommend that you take a look at it, rather than I try to explain the ideas here.


EDIT: I've added some more information below. Its not totally clear what your motivation for the question is; if you add more information perhaps I can answer your question better.


It is sometimes possible to define a classical flow which starts at a minimal surface. For example, in the Riemannian Schwarzschild metric $$ g = \left(1+\frac{2m}{r}\right)^4 \delta, $$ on $\mathbb{R}^3\setminus \{0\}$, there is a inverse mean curvauture flow $\Sigma_t = \{r(t)\}\times \mathbb{S}^2$ defined for $t>0$, where $\lim_{t\searrow 0} r(t) = m/2$. I'll leave it to you to compute the associated ODE (HINT: the easiest way is to use the fact that $|\Sigma_t|=e^t|\Sigma_0|$.

Be very careful with what I mean here for $t=0$. In particular, the PDE is not satisfied for $t=0$, just $t>0$.


Here are some results about the classical flow:

http://www.ams.org/mathscinet-getitem?mr=1064876

http://www.ams.org/mathscinet-getitem?mr=1082861

http://www.ams.org/mathscinet-getitem?mr=1753358

share|improve this answer
    
Does that mean that in the classic condition we have admitted the existence of the inverse mean curvature flow starting from a minimal surface? –  Jer Mar 24 at 8:40
    
Sorry, I don't understand your question, can you rephrase it? –  Otis Chodosh Mar 24 at 14:05
    
I am wondering that in classic case, we admitted this flow could be start from a minimal surface and using this initial condition to make other conclusions. But till Huisken and Ilmanen, people then actually make it clear with weak inverse mean curvature flow. is it right? –  Jer Mar 25 at 2:09
    
@Jer, I'm sorry but I am having difficulty understanding your question. Could you please clarify? In particular, by "we admit" do you mean "we assumed"? –  Otis Chodosh Mar 25 at 3:43
1  
Ok, I will compute the inverse mean curvature flow relevant to the Riemannian Schwarzschild metric and read the papers you recommended to me. Thank you very much! –  Jer Mar 28 at 11:56
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.