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<a<1} 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

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