$M$ is a $n$-dimensional smooth manifold without boundary . $F: M \rightarrow \mathbb R^{n+1}$ is a smooth embedding. $A$ is the second foundamental form , and $H$ is mean curvature. $\nu$ is the normal vector. $x$ is position vector. If $$ H=\langle x, \nu \rangle \tag{1} $$ it is easy to see hyperplane, sphere, cylinder satisfy the equation. But I don't know whether there are other manifold satisfy this equation, especially , manifold with negative mean curvature?

In the Huisken's Asymptotic behavior for singularities, he prove that if $M$, $n\ge 2$, is compact with nonnegative mean curvature $H$ and satisfy (1), then $M$ is sphere of radius $\sqrt n$.

Parts of this proof is to prove $\frac{|A|^2}{H^2}$ is constant. I don't know why he know to calculate this quantity. Exactly , this quantity has maximum principle. Whether there are any geometric view ?

Last, I guess the geometric essence of (1) is the principle curvature of must be constant or zero, right ?

  • $\begingroup$ Just a TeX note: \tag{1} sets an equation number without the hard-coded spaces. :) $\endgroup$ – Andrew D. Hwang Jun 18 '17 at 14:32
  • $\begingroup$ Instead of $(1)$ you meant this formula ? $\endgroup$ – reuns Jun 19 '17 at 0:47
  • $\begingroup$ @user1952009 This formula is local present. But not all manifold can be presented as entire graph. About the entire graph self shrinker in $\mathbb R^{n+1}$, Lu Wang's A Bernstein type theorem for self-similar shinkers asserts they are hyperplanes. But I don't find paper about the cylindic graph self-shrinker. $\endgroup$ – lanse7pty Jun 20 '17 at 2:59
  • $\begingroup$ I'm not sure if this is what you are looking for, but if $F:M\to\mathbf{R}^{m+1}$ is an isometric immersion, where $\mathrm{dim}M=m$, then you can show $\Delta F = mH_{\nu}$, where the Laplace-Beltrami operator acts on $F$ coordinatewise, and $H_{\nu}$ is the mean curvature vector of $F(M)$. Note that in the case of a sphere, $F=\mathrm{id}$ is an isometric immersion and a direct computation shows $\Delta F = 2 x$, and so by taking the dot product with respect to the unit normal you find that $H=\langle{H_\nu,\nu}\rangle=\langle x,\nu\rangle$, which is exactly the above equality. $\endgroup$ – yousuf soliman Jun 20 '17 at 17:48
  • $\begingroup$ @YousufSoliman Could you detail talk about how to show $\Delta F=mH_\nu$ ? Thanks very much. $\endgroup$ – lanse7pty Jun 23 '17 at 13:47

A submanifold $M$ which satisfies the equation (1) is called a self-shrinker. The name comes from the fact that if $M$ is a submanifold which satisfies one, when the MCF starting at $M$ is given by

$$M_t = \sqrt{1-2t} M, t\in [0, 1/2) $$

up to diffeomorphism.

Self-shrinkers arise as a singularity model of mean curvature flow. In the type I singularity case, Huisken shows in his paper that under the type I rescaling, the MCF converges subsequentially, locally smoothly to a submanifold satisfying the self-shrinking equation (1), see related question here.

Using again Huisken's monotonicity formula and use instead the parabolic rescaling, Ilmanen (Singularity of mean curvature flow of surfaces) and White show independently that the limit of the parabolic rescaling converges weakly to a self-shrinking flow, which is modelled by a (weak) self-shrinker.

The fact that a compact self-shrinker with $H\ge 0$ must be a sphere is generalized to the non-compact case here: if $M$ is an properly embedded self-shrinkers with $H\ge 0$, then it is either the sphere or the generalized cylinders $\mathbb S^k \times \mathbb R^{n-k}$. Actually they defined the entropy stability condition and show that all entropy stable self-shrinkers must have $H\ge 0$. See this question for more details.

Examples Indeed there are more examples than the standard one. Soon after Huisken's result, Angenent constructs using ODE the famous rotational symmetric self-shrinking doughnuts, which is an embedded self-shrinking surface of genus one. Chopp constructs numerically a lot more examples (compact or noncompact, higher genus). Some of the example there are recently constructed using desingularization techniques here, here. It is generally believed that the classification of self-shrinkers is impossible. Colding-Minicozzi have a compactness result though.

The study of self-shrinkers has been the central objects. I don't think I can give a fair review here.


To simplify some notation, I only consider the following case: Let $M$ be an $m$-dimensional Riemannian submanifold of $\mathbf{R}^{m+1}$. Fix any vector $\vec{v}\in\mathbf{R}^{m+1}$; we will consider the real valued function $f=\langle\mathrm{id},\vec{v}\rangle$. Now fix $p\in M$ and let $\{e_1,\dots,e_m\}$ be a geodesic frame in a neighborhood of $p$. Let $\vec{N}$ be the local normal frame in the same neighborhood of $p\in\mathbf{R}^{m+1}$. Now recall that the Laplace-Beltrami operator in geodesic coordinates is given by $$\Delta g = \sum_{i=1}^{m}e_i(e_i(g)),\qquad g\in\mathscr{C}^\infty(M).$$ Note that for some vector $x\in T_pM$, that $df_p(x)=\langle x,v\rangle$. We use this to compute $$\Delta f = \Delta\langle\mathrm{id},\vec{v}\rangle = \sum_{i=1}^{m}e_i(e_i\langle \mathrm{id},\vec{v}\rangle) = \sum_{i=1}^{m}e_i(df(e_i)) = \sum_{i=1}^{m}e_i\langle e_i,\vec{v}\rangle.$$ Now we use the compatibility of the Levi-Civita connection, $\overline\nabla$, of $\mathbf{R}^{m}$ with the Euclidean metric $\langle\cdot,\cdot\rangle$ to obtain $$\sum_{i=1}^{m}e_i\langle e_i,\vec{v}\rangle = \sum_{i=1}^{m}\langle\overline\nabla_{e_i}e_i,\vec{v}\rangle + \langle e_i,\overline\nabla_{e_i}\vec{v}\rangle = \sum_{i=1}^{m}\langle\overline\nabla_{e_i}e_i,\vec{v}\rangle=\sum_{i=1}^{m}\langle\nabla_{e_i}e_i,\vec{v}\rangle+\langle \mathsf{II}(e_i,e_i),\vec{v}\rangle,$$ where $\mathsf{II}$ is the vector valued second-fundamental form. Now since $e_i$ was chosen to be a geodesic frame at $p$ we have $\nabla_{e_i}e_i(p)=0$, and so $$\sum_{i=1}^{m}\langle\nabla_{e_i}e_i,\vec{v}\rangle+\langle \mathsf{II}(e_i,e_i),\vec{v}\rangle=\sum_{i=1}^{m}\langle \mathsf{II}(e_i,e_i),\vec{v}\rangle=mH\langle{N,\vec{v}}\rangle.$$ So we have shown that $$\Delta\langle{\mathrm{id},\vec{v}}\rangle=mH\langle{N,\vec{v}}\rangle.$$ Now by applying this result where $\vec{v}=x_i$ are the standard unit vectors of $\mathbf{R}^{m+1}$ we find that $$\Delta\mathrm{id}=mH\vec{N}.$$ To generalize this result to an isometrically immersed submanifold isn't much more work. This provides a very general relationship between the mean curvature vector and a submanifold of Euclidean space. In particular, we recover the relationship between the mean curvature and $\langle x,\nu\rangle$ of the sphere presented in your problem.

  • $\begingroup$ I like undergraduate student :) $\endgroup$ – ParaH2 Nov 5 '17 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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