# Question about Neumann eigenvalues on manifolds

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 $$\Omega.$$
• Here's a heuristic. Call the smallest Neumann eigenvalue the "zeroth" so we can just say "first." The first eignvalue for the hemisphere is no smaller than $2$ (as that's the first nontrivial Neumann eigenvalue of the sphere). A long narrow convex subset of diameter $L$ will look like a geodesic of length $L$, so its first eigenvalue is close to the first eigenvalue of the interval of length $L$, namely $\frac{\pi^2}{L^2}$. Since the longest possible diameter is $L=\pi$, the smallest first eigenvalue of a long narrow interval should be $1$ and $1 < 2$
– Neal
Commented Apr 5 at 17:54
• @Neal that is what I had in mind, but I am having trouble showing that the eigenvalues of a thin tube around the half-geodesic converge to the eigenvalues on the half-geodesic with the flat metric as the thickness of the tube tends to zero. Commented Apr 5 at 20:42
• Do you need to show convergence? Just estimate the first eigenvalue of the thin tube: Take tubular coordinates $(s, r, \theta)$ where $s$ is the long dimension and $r,\theta$ vary in the thin cross-section disk. Let $u(s, r, \theta) = \cos(\pi s / L)$. Maybe I'm wrong about the complexity of the computation, but I think one should be able to show $\int u = 0$ and get a good estimate its Rayleigh quotient from above.
– Neal
Commented Apr 5 at 21:37
• (Also, I think convergence is delicate: the eigenfunctions that have radial variation across the tube will have their eigenvalues blow up, only the eigenfunctions that have minimal/no radial variation will have their eigenvalues converge.)
– Neal
Commented Apr 5 at 21:40

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.

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 $$\mu_1(\Omega) < \mu_1(\mathbb{S}_+^2)$$.

• Thank you for your answer, but I am not sure if the thin strip is a geodesically convex subset of the hemisphere, what do you think? Commented Apr 9 at 21:30
• Good catch. I think this is fixed by considering a "banana" between two geodesics that intersect at the equator on the boundary of $\mathbb{S}^+$, andhave an angle of $2\epsilon$ between them. This domain's convex (a geodesic connecting two points must be the "long way round" since a shorter path along the bdry can be found), and the estimate is modified with inner integral limits dependent on $s$. I think since integrands are all positive, the upper bound will carry through. If I have time later this week I'll check this and update my answer more carefully. But lmk if you see a flaw here.
– Neal
Commented Apr 11 at 3:46
• The first eigenvalue of this lune is $\frac{\pi}{2\epsilon}(\frac{\pi}{2\epsilon}+1)$ which blows up as $\epsilon\to 0,$ so I am not sure if this argument would work. Commented Apr 12 at 3:30
• @Student you sure that's not the first Dirichlet eigenvalue?
– Neal
Commented Apr 12 at 17:26
• that is also the first Dirichlet eigenvalue, see Section 2 here link.springer.com/article/10.1007/s10455-021-09797-y Commented Apr 12 at 18:09