Question about Neumann eigenvalues on manifolds

Question

Question: Let $\Omega\subset \mathbb{S}^2_+$ be any geodesically convex subset of the hemisphere $\mathbb{S}^2_+$. Then is it true that $\mu_1(\Omega)\geq \mu_1(\mathbb{S}^2_+)$ where $\mu_1$ is the first (non-trivial) Neumann eigenvalue of the Laplacian?

The above statement clearly fails when $\mathbb{S}^2_{+}$ is replaced by a rectangle in the plane and $\Omega$ is a thin strip around the diagonal of the rectangle. Still, I am wondering if such a construction generalizes to the hemisphere as well.

Update: Surprisingly this problem seems harder than it looks.

• So far the best estimates on $\mu_1(\Omega)$ that I found here: $$\mu_1(\Omega)\geq \frac{\pi^2}{D^2} + \frac{1}{2}$$ where $D=\operatorname{diameter}(\Omega).$ Thus if $\mu_1(\Omega)<2$ then $D> \frac{\pi}{\sqrt{3/2}}.$ This also gives a lower bound on the perimeter of $\Omega$ since $P(\Omega)\geq 2 D \geq \frac{2\sqrt{2}}{\sqrt{3}} \pi.$
• On the other hand, it is also known that $\mu_1(\Omega)\leq \mu_1(B)$ where $B$ is a geodesic ball of the same volume as $\Omega.$

Answer 1

The original inequality indicated in the question appears to be true -- for various instances of such a result see:

• Theorem 4.3 in J.F. Escobar, Uniqueness theorems on conformal deformation of metrics, Sobolev inequalities, and an eigenvalue estimate. Comm. Pure Appl. Math.43(1990), no.7, 857–883, and
• Theorem 4.15 in V. Bayle and C. Rosales, Some isoperimetric comparison theorems for convex bodies in Riemannian manifolds, Indiana Univ. Math. J. 54 No. 5 (2005), 1371–1394.

Answer 2

Let $\Omega$ be a connected domain and denote its the Neumann eigenvalues by $ 0 = \mu_0(\Omega) < \mu_1(\Omega) \leq \cdots $

Let $\mathbb{S}_+^2 = \{(x,y,z)\in\mathbb{S}^2\ |\ y \geq 0\}$ be the hemisphere facing the positive $y$ direction. Because every Neumann eigenfunction on $H$ extends to an eigenfunction on $\mathbb{S}^2$ by reflection, $\mu_1(H) \geq 2 = \mu_1(\mathbb{S}^2)$.

We will formalize the intuition that a long, thin convex subset should have a low-frequency spectrum very similar to that of a line segment. We do this by defining a convex subset $\Omega\subset \mathbb{S}_+^2$ and a test function $u$ on $\Omega$, and estimate its Rayleigh quotient $R(u)$ to show $$ \mu_1(\mathbb{S}_+^2) \geq 2 \geq R(u) \geq \mu_1(\Omega).$$

Define $\Omega$ as the image of $$ [0,\pi]\times[-\epsilon,\epsilon] \ni (s,r)\to \left( \cos(s)\cos(r), \sin(s)\cos(r), \sin(r)\right) \in \mathbb{S}_+^2$$ This defines geodesic normal coordinates for a very thin strip around the equator $H\cap\mbox{$xy$-plane}$. The $s$ coordinate measures distance around the equator, and the $r$ coordinate measures distance away from the equator.

Figure that the first Neumann eigenfunction on the strip should be basically the same as on the line, so pick a test function $u(s,r) = \cos(s)$. Calculating in the $(s,r)$ coordinates, the $L^2$ norm of $u$ is $\pi\sin(\epsilon) $ and the $L^2$ norm of its gradient is $$ \frac{\pi}{2}\int_{-\epsilon}^\epsilon \frac{dr}{\cos(r)}. $$ The integral factor is bounded by $2\epsilon \leq \int_{\epsilon}^\epsilon dr/\cos(r) \leq 2\epsilon/\cos(\epsilon)$

So the Rayleigh quotient of $u$ is estimated above and below by $$ \frac{\epsilon}{\sin(\epsilon)} \leq R(u) \leq \frac{\epsilon}{\cos(\epsilon)\sin(\epsilon)} $$

By choosing $\epsilon$ small enough, the upper bound is less than $2$. The function $u$ is mean-free, and $\mu_1(\Omega)$ minimizes the Rayleigh quotient over all mean-free functions. So the estimate on $R(u)$ shows that $\mu_1(\Omega) < \mu_1(\mathbb{S}_+^2)$.