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I'm reading a lecture note on mean curvature flows (by C. Mantegazza) and studying about analysis of type-I singularities.

Let $M$ be a smooth closed $n$-dimensional manifold and $\varphi:M\times[a,\infty)\stackrel{C^\infty}\to\mathbb{R}^{n+1}$ be a mapping such that for each $s\in[a,\infty)$ the map $\varphi_s:=\varphi(\underline\ ,s):M\to\mathbb{R}^{n+1}$ is an immersion.

Then what does the statement "a sequence of hypersurfaces $\left(\varphi_{s_i}\right)_i$ locally smoothly converges (up to reparametrization) to some limit hypersurface $M_\infty$" mean?

Reading the following context in my textbook, the above statement seems to be strictly rougher than that "the sequence of maps $(\varphi_{s_i})$ converges to an immersion $\varphi_\infty$ in $C^\infty(M,\mathbb{R}^{n+1})$". It seems that $M_\infty$ can be unbounded subset of $\mathbb{R}^{n+1}$ while $M$ is assumed to be closed.

The figure below is what I am reading in the textbook. Here we set $\tilde\varphi$ to be the rescaling of a smooth solution of the mean curvature flow $\dfrac{\partial \varphi_t}{\partial t}=\Delta^t\varphi_t$ which develops a type I singularities at t=T.

a

I am glad if you show me an example of nontrivial convergence of hypersurfaces. Here the term "trivial" means the convergence of the immersions in $C^\infty(M,\mathbb{R}^{n+1})$.

Thank you.

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This is quiet unclear and (interestingly) has been used (without proof) as early as the work in Huisken' 1990 paper when he defined type I singularity. The point is that when you have a type I singularity, the second fundamental form satisfies a bound

$$\max _{\varphi_t(\cdot)} |A|^2 \le \frac{C_0}{T-t}$$

for some fixed constant $C_0$. Then one can define the rescaling (I am following the book by XP Zhu, so the normalizing constant might be different)

$$\tilde \varphi (x, \tau) = \frac{1}{\sqrt{2(T-t)}} (\varphi_t(x) -P), \ \ \tau = -\frac 12 \log\left( \frac{T-t}{t}\right)$$

This scaling factor $\frac{1}{\sqrt{2(T-t)}}$ has the advantage that $$\sup_\tau |\tilde A|^2 \le C$$

for all $\tau$, where $\tilde A$ is the second fundamental form of $\tilde \varphi$. Now using the (parabolic) equation for $|\tilde A|^2$, one can show that there are unform bound on all higher derivatives:

$$\sup_\tau |\tilde\nabla^m \tilde A|^2 \le C(m),\ \ \ \forall m.$$

And this condition (together with the fact that $\tilde\varphi_\tau$ do not excape to infinity) will forces that whenever you pick $\tau_i\to \infty$, there is a subsequence so that $\tilde \varphi_{\tau_i}$ converges locally smoothly to a hypersurfaces $\phi_\infty : M_\infty \to \mathbb R^{n+1}$.

You are right that $M_\infty$ will not be the topologically the same as $M$ in general. For example, you can start with the standard torus in $\mathbb R^3$ which is a surface of revolution. Then when you apply the MCF, when the torus will shrinks to a $\mathbb S^1$. Now if you choose any points $P$ in this $\mathbb S^1$ and apply the blow up at $P$, your get actually a cylinder $M_\infty$, which is not the torus.

Now go back to the claimed convergence. One can actually find a proof in the paper by Patrick Breuning (Theorem 1.3) in 2012. There are also some previous works on hypersurfaces, you may find that in the reference. So you are right that it is not a convergence as a mapping $\varphi_i \to \varphi_\infty$, but it is still quite strong as a convergence (locally smoothly)

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