I'm reading "Mean curvature flow with surgeries of two–convex hypersurfaces" by Gerhard Huisken and Carlo Sinestrari and I didn't understand how to prove the proposition $3.2$ on page $148$. The authors indicate the paper "Four-Manifolds with Positive Isotropic Curvature" by Hamilton for the definition of intrinsic spherical and cylindrical curvature (see page $35$).

My doubts:

  1. What means that "$\overline{R}m$ is the identity curvature operator on the standard round sphere"?

  2. What is the standard frame of $S^{n-1} \times \mathbb{R}$? Is the product metric? If this is the case, then is the metric of $S^{n-1}$ induced by the immersion in $\mathbb{R}^n$?

  3. The authors state "We will add the word intrinsic to mean that we are referring to Hamilton’s definitions concerning the Riemann curvature tensor, applied to our induced metric." in the paragraph previous to the proposition $3.2$, but Hamilton considered a spherical parametrization $P$ (see page $27$ of Hamilton's work for the definition). Is the induced metric of Huisken and Sinestrari given by the immersion of $\mathcal{M}^n$ into $\mathbb{R}^{n+1}$?

  4. How to relate $\varepsilon$ and $\varepsilon'$? I can't see how "This is an immediate consequence of the Gauss equations".

Thanks in advance!

  1. You can think of the curvature tensor as a linear operator on 2-forms ($\mathfrak R:\Omega^2\to\Omega^2$), called the curvature operator. The round sphere has a curvature operator which is the identity map (up to a factor, depending on your conventions).

  2. In MCF, the sphere is always the round unit sphere (unless they are explicitly scaling it). They mean the product metric.

  3. Of course, see Lemma 2.1.

  4. The Gauss equations read $R_{ijkl}=h_{ij}h_{jl}-h_{il}h_{jk}$. The extrinsic spherical curvature condition means that the Weingarten map is close to that of the unit sphere. This means that, by the Gauss equations, the Riemann tensor will be close to that of the unit sphere. The exact way that $\varepsilon'$ depends on $\varepsilon$ and $n$ is a bit annoying and depends on exactly how you define these norms. At the end of the day, it doesn't matter what the exact constants are, just that $\varepsilon'=O(\varepsilon)$ as $\varepsilon\to 0$. If you're unsure about this, you need to sit down and estimate $|\mathfrak R-\mathrm{id}|$ given that $|W-\mathrm{id}|<\varepsilon$ (note that these are two different norms and identity maps).

As friendly advice, you're probably not read to read Huisken-Sinestrari if you're asking these questions. Unfortunately there isn't a better resource to learn this material but you should first understand Huisken's '84 paper on convex MCF. This would answer a couple of your questions already.

  • $\begingroup$ Firstly, I would like to thanks for the advice! As you said, I don't have the prerequisites to read Huisken-Sinestrari's paper, although I'm studying this paper by myself and I'm very motivated to understand the process of surgery in the MCF. I will certainly work in the Huisken's '84 paper on convex MCF. $\endgroup$ – George Jul 20 at 0:04
  • $\begingroup$ About your answer for my question $4$, I can see intuitively that the Riemann tensor will be close to that of the unit sphere as the Weingarten map is close to that of the unit sphere, but I will try do the computation, I just need more one thing to start the computation: you said the norms are differents. What norms are being used? I was thinking that it was used the norm of tensors or the operator norm, but I'm confuse about what norms are used. $\endgroup$ – George Jul 20 at 0:05
  • $\begingroup$ @George Well the point is that since the norms are on finite dimensional spaces (the norms are defined invariantly on the fibers of certain vectors bundles over $\mathcal M$), the actual choice of norm is irrelevant for this lemma. The usual norm for the second fundamental form/Weingarten map/shape operator is the Hilbert-Schmidt norm, so $|h|^2=h_{ij}h^{ij}$. But the usual norm on the curvature operator is actually the operator norm, because in 3 dimensions this corresponds to the largest (in absolute value) sectional curvature. This makes some 3-dimensional Ricci flow things mildly easier. $\endgroup$ – Ryan Unger Jul 20 at 2:37
  • $\begingroup$ The way I interpret this proposition is that $h_{ij}=\delta_{ij}+O(\varepsilon)$ so $$R_{ijkl}=\delta_{ij}\delta_{jl}-\delta_{il}\delta_{jk}+O(\varepsilon)$$ when $\varepsilon$ is small. That this is equivalent to the norm statement is just a general principle. $\endgroup$ – Ryan Unger Jul 20 at 2:39

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