A condition for a manifold with boundary to be a hemisphere

In this short note I will give a characterization when a Riemannian manifold ${(M,g)}$ with boundary is isometric to a hemisphere. This result is due to Huang-Wu.

 Theorem 1 Let ${B}$ be the unit ball in ${\mathbb{R}^n}$. Suppose ${u}$ is function on ${\overline B}$ such that ${u\in C^2(B)\cap C(\overline B)}$, and let ${M}$ be it graph. Suppose ${M}$ has scalar curvature ${R\geq n(n-1)}$, then ${M}$ is isometric to the standard hemisphere ${\mathbb{S}^n_+}$.

 Remark 1 This is a special case of the Min-Oo conjecture, which states that for a compact Riemannian manifold ${M^n}$ with boundary, if ${R\geq n(n-1)}$, ${\partial M}$ is isometric to ${\mathbb{S}^{n-1}}$ and ${\partial M}$ is totally geodesic in ${M}$, then ${M}$ is isometric to the hemisphere. This conjecture is true when ${n=2}$ (by Eichmair), and is true under different variation of the assumptions, e.g. ${R\geq n(n-1)}$ strengthened to ${Ric\geq (n-1)g}$, the case which was proved by Hang-Wang. The conjecture turns out to be false for $n\geq 3$, as counterexamples were constructed by Brendle-Marques-Neves very recently, using deformation techniques.

We begin with two lemmas.

 Lemma 2 If $M^n$ is the graph of a function $u\in C(\overline B)\cap C^2(B)$, let ${H}$ be the mean curvature of ${M}$, then $\displaystyle \left|\int_B H\right|\leq n Vol(B).$

Proof: As the unit normal is ${\frac{(Du,-1)}{\sqrt{1+|Du|^2}}=\frac{(Du,-1)}{w}}$, it is easy to see that

$\displaystyle H= \sum_{i=1}^n \partial_i(\frac{\partial_i u}{w})=div (\frac{Du}{w}).$

Thus by divergence theorem,

$\displaystyle \int_{B_a} H= \int_{\partial {B_a}} \frac{Du}{w}\cdot \nu\leq \int_{\partial {B_a}}1= Vol(\partial {B_a})= \frac{n}{a}Vol(B_a)$

where ${B_a}$ is the ball of radius ${0 and ${\nu}$ is the normal of its boundary, taking ${a\rightarrow 1}$, we are done. $\Box$

 Lemma 3 Let ${A}$ be the shape operator of any hypersurface ${M^n\subset \mathbb{R}^{n+1}}$, and let ${\overset{\circ}A}$ be its trace-free part (i.e. ${tr \overset \circ A=0}$ and ${A-\overset \circ A\perp \overset \circ A}$), then $\displaystyle \left (\frac{H}{n}\right)^2 = \frac{ R}{ n(n-1)}+ \frac{|\overset\circ A|^2}{n(n-1)}.$

Proof: By Gauss equation,

$\displaystyle R= H^2 -|A|^2.$

Note that

$\displaystyle {\overset \circ A}_i ^j = A_i ^j- \frac{H}{n}\delta _i ^j,$

as ${H= tr(A)}$. So by Pythagoras theorem,

$\displaystyle |A|^2= |\overset \circ A|^2 +\left|\frac{H}{n}Id\right|^2= |\overset \circ A|^2 + \frac{H^2}{n}.$

Substitute it into the Gauss equation, we are done. $\Box$

Note that ${\stackrel \circ A=0}$ if and only if the shape operator is a multiple of the identity, in other words, the hypersurface is umbilical at that point (meaning all directions are principal directions. )

We now prove the main theorem.
Proof: As ${R\geq n(n-1)}$, Lemma 3 implies ${H\geq n}$. Combining this with Lemma 2, we conclude that ${H=n}$. But then this implies ${\stackrel \circ A=0}$, i.e. ${M}$ is umbilical of constant curvature ${1}$, so it is contained in a unit sphere. Together with the boundary condition ${\partial M=\mathbb{S}^{n-1}}$, we can get the result. $\Box$

This entry was posted in Geometry. Bookmark the permalink.