Two balls maximize the third Neumann eigenvalue in hyperbolic space
TTWO BALLS MAXIMIZE THE THIRD NEUMANN EIGENVALUEIN HYPERBOLIC SPACE
P. FREITAS AND R. S. LAUGESEN
Abstract.
We show that the third eigenvalue of the Neumann Laplacian in hyper-bolic space is maximal for the disjoint union of two geodesic balls, among domainsof given volume. This extends a recent result by Bucur and Henrot in Euclideanspace, while providing a new proof of a key step in their argument Introduction and results
Which shape maximizes the second eigenvalue of the Neumann Laplacian amongEuclidean regions of given volume? This question, which arises naturally in thewake of the Rayleigh–Faber–Krahn result for the Dirichlet Laplacian, was answeredin the 1950s by Szeg˝o [28] for two-dimensional simply–connected domains and byWeinberger [30] for the general case in all dimensions, the answer being the ball.Which shape maximizes the third eigenvalue of the Neumann Laplacian? Thistime the answer is a disjoint union of two equal balls. The proof of this result hadto wait for more than 50 years, and it was again proved first for simply–connecteddomains in the plane, by Girouard, Nadirashvili and Polterovich in 2009 [16], and infull generality by Bucur and Henrot in 2019 [10].1.1.
Hyperbolic results.
In this paper we extend the latter result to hyperbolicspace, showing that two disjoint geodesic balls of the same volume maximize thethird Neumann eigenvalue of the Laplace–Beltrami operator among regions of givenvolume. More precisely, our main result is the following, in hyperbolic spaces ofdimension n greater than or equal to 2. Theorem A (Third hyperbolic Neumann eigenvalue is maximal for two balls) . Let Ω be a bounded open set with Lipschitz boundary in the hyperbolic space H n of constantnegative curvature κ , and denote the eigenvalues of the Neumann Laplacian in Ω by η ≤ η ≤ η ≤ · · · . Then η (Ω) ≤ η ( B (cid:116) B ) , where B is a ball having half the hyperbolic volume of Ω . Equality holds if and onlyif Ω is a disjoint union of two balls in H n , of equal hyperbolic volume. As in Bucur and Henrot’s proof in the Euclidean setting, our approach builds onWeinberger’s method for the second eigenvalue. In his proof, the required orthog-onality is to the first (constant) eigenfunction, and thus amounts to obtaining trial
Mathematics Subject Classification.
Primary 35P15. Secondary 55M25.
Key words and phrases. degree theory, vibrating membrane, spectral theory, shape optimization. a r X i v : . [ m a t h . SP ] S e p THIRD NEUMANN EIGENVALUE functions with zero average. For the third eigenvalue, the trial functions must havezero average and also be orthogonal to the first excited state. This construction ismade possible, at its heart, by the fact that the eigenvalue η ( B ) of the ball hasmultiplicity n . More precisely, η ( B (cid:116) B ) has multiplicity 2 n because the spectrum ofa disjoint union is simply the union of the spectra and so the first three eigenvaluesof the disjoint union of two balls satisfy0 = η ( B (cid:116) B ) = η ( B (cid:116) B ) < η ( B (cid:116) B ) = η ( B ) . This multiplicity will be instrumental in constructing trial functions that satisfy therequired orthogonality conditions. Further, we provide a new way of identifying suchtrial functions. This method also works in Euclidean space, yielding a new proof ofthe central step in Bucur and Henrot’s result.Another key ingredient in our proof is a degree theory computation of Petrides [26]for maps between spheres, and here too we present what we believe to be a simpler,global proof. The proof of Theorem A is then covered in the second part of the paper,beginning in Section 7.We recall that the hyperbolic analogue of the Szeg˝o–Weinberger result for the sec-ond eigenvalue, asserting maximality of the ball under a hyperbolic volume constraint,was proved by Bandle [7, 8] for 2-dimensional simply connected surfaces, with the gen-eral higher dimensional case later mentioned by Chavel [11, p. 80], [12, pp. 43–44].Detailed expositions were provided by Ashbaugh and Benguria [5, §
6] and Xu [31,Theorem 1], and the requisite center of mass lemmas can be found in Benguria andLinde [9, Theorem 6.1] and Laugesen [23, Corollary 2].1.2.
Euclidean results.
Since our methods apply both to Euclidean and hyperbolicspaces, we first carry out the proof in the former case, as this corresponds to a simplerset-up and, we believe, makes the general approach clearer.Write B = B n for the unit ball in R n , where n ≥
2. Let Ω ⊂ R n be a bounded openset with Euclidean volume | Ω | and Lipschitz boundary. The Lipschitz requirementrestricts Ω to having finitely many components. The Neumann eigenvalue problemfor the Laplacian is − ∆ u = µu in Ω ∂u∂ν = 0 on ∂ Ωand, under the above conditions, its spectrum is discrete with eigenvalues satisfying0 = µ ≤ µ ≤ µ ≤ · · · → ∞ . The maximization result for µ , obtained by Bucur and Henrot in [10], is the following. Theorem B (Bucur and Henrot: third Neumann eigenvalue is maximal for twoballs) . If Ω ⊂ R n is a bounded open set with Lipschitz boundary then µ (Ω) | Ω | /n ≤ µ ( B (cid:116) B )(2 | B | ) /n . Equality holds if and only if Ω is a disjoint union of two balls of equal volume. HIRD NEUMANN EIGENVALUE 3
One of the goals of this paper is to present a proof of Theorem B that is differentat its homotopic core from Bucur and Henrot’s proof. That argument begins inSection 5, and a discussion of the similarities and differences between the two proofsis provided in Section 5.1.
Remarks on the literature.
Girouard, Nadirashvili and Polterovich [16, Theorem 1.1.3]first proved the result on the third eigenvalue for simply connected domains in theplane, and Girouard and Polterovich [17, Theorem 1.7] extended the argument tosurfaces with variable nonpositive curvature. Their Neumann trial functions in theunit disk were modified to Euclidean space by Bucur and Henrot [10], and adaptedto Robin eigenvalues in the planar case by Girouard and Laugesen [15].Girouard, Nadirashvili and Polterovich worked in terms of measures folded acrosshyperplanes (which in their context meant hyperbolic geodesics in the disk), andthey maximized over 2-dimensional spaces of trial functions. Girouard and Laugesenclarified the construction by composing trial functions with a fold map in order toobtain trial functions that are even across the hyperplane, and they used uniquenessof the center of mass point to reduce from a 2-dimensional trial space to a single trialfunction. This function depends on a parameter lying on the circle, and so the degreetheory required to finish the proof is a straightforward winding number argument.Bucur and Henrot worked in the Euclidean context in all dimensions, and employeda different parameterization for what is essentially the same family of trial functions.The degree theory required to finish their proof in higher dimensions is more difficult:they finished with an ingenious two-step homotopy argument for their vector field in R n .One contribution of the current paper is what we believe to be a conceptually sim-pler proof of Bucur and Henrot’s “two ball” Theorem B. We adapt the parameteriza-tion of Girouard, Nadirashvili and Polterovich to Euclidean space in all dimensions,and rely on uniqueness of the center of mass point together with a beautiful degreetheory result of Petrides [26] for maps between spheres (Theorem 2.1 below).A “relaxed” version of the eigenvalue maximization problem, in which the indicatorfunction of the region Ω is replaced by a weight function, was treated by Bucur andHenrot [10, Theorem 3]. Their weight could have unbounded support. We shall notpursue such generalizations here.1.3. Future directions.
Now that the third Neumann eigenvalue is known to bemaximal for two disjoint balls in spaces of non-positive constant curvature (Euclideanand hyperbolic), it is natural to ask the same question in the positive curvature case,that is, for domains in the standard round sphere. The analogue of Weinberger’s resultfor the second eigenvalue is known to hold for domains contained in a hemisphere andcertain other domains [5], but is not known to hold in general. Thus some restrictionon the domain might be needed to get a result on the third eigenvalue.Another extension would be to higher eigenvalues: one might ask whether thefourth eigenvalue is maximal for three disjoint balls. This turns out not to be thecase. Numerical work by Antunes and Freitas [2, Section 4] (see also the chapter byAntunes and Oudet in [18]) suggests the maximizer for µ among Euclidean domains THIRD NEUMANN EIGENVALUE of given area is not a union of disks, but rather a domain that looks somewhat likethree touching disks with the joining regions smoothed out.Averaging the eigenvalues can improve their behavior and lead to a positive result.Notably, the harmonic mean of µ , . . . , µ n is maximal for the ball, among domainsin R n of given volume, by a recent result of Wang and Xia [29, Theorem 1.1] thatdirectly strengthens Weinberger’s theorem for µ . They also prove the analogousresult for domains in hyperbolic space [29, Theorem 1.2]. Thus, for example, Wangand Xia’s result implies the harmonic mean of µ and µ is maximal for the ball, amongdomains of given volume in n dimensions ( n ≥ µ by itself is maximal forthe disjoint union of two balls. Looking now to the future, it is an open problemwhether the spectral zeta function (cid:80) ∞ j =2 µ − sj is minimal for the ball, when s > n/ The Petrides theorem on the degree of a map with reflection symmetry
Our trial function construction for the third Neumann eigenvalue will depend on atopological theorem due to Petrides. His result, which is similar to the Borsuk–Ulamtheorem for odd mappings, says that if a map from the sphere to itself has a certainreflection symmetry property then its degree must be nonzero.This fact will play a role in the paper analogous to that of the Brouwer fixed pointtheorem (or no-retraction theorem) in Szeg˝o and Weinberger’s papers for the secondNeumann eigenvalue. Namely, the topological result will be used in Proposition 5.4to show existence of a trial function that is orthogonal to the first two Neumanneigenfunctions.We will give a new proof for Petrides’s theorem. His proof was local in nature,whereas the one below is global.Given a unit vector p in Euclidean space, write R p for the reflection with respectto the hyperplane through the origin that is perpendicular to p : R p ( y ) = y − y · p ) p. Theorem 2.1 (Reflection symmetry implies nonzero degree; Petrides [26, Claim 3]) . Assume φ : S m → S m is continuous, for some m ≥ . If φ satisfies the reflectionsymmetry property φ ( − p ) = R p (cid:0) φ ( p ) (cid:1) , p ∈ S m , (1) HIRD NEUMANN EIGENVALUE 5 then φ has nonzero degree, meaning φ is not homotopic to a constant map. Moreprecisely, if m is odd then deg( φ ) = 1 , and if m is even then deg( φ ) is odd. First we need an elementary result about maps that are homotopic to the identity.
Lemma 2.2.
Suppose ψ : S m → S m is continuous, where m ≥ . If ψ ( p ) · p ≥ forall p ∈ S m then deg( ψ ) = 1 . If ψ ( p ) · p ≤ for all p ∈ S m then deg( ψ ) = ( − m +1 . The intersection of the two cases, where ψ ( p ) · p = 0 for all p (“a hairy ball”), canobviously occur only for odd m , in which case the degree is 1. Proof.
Suppose ψ ( p ) · p ≥ p . Then ψ is homotopic to the identity viaΨ( p, t ) = (1 − t ) ψ ( p ) + tp | (1 − t ) ψ ( p ) + tp | , p ∈ S m , t ∈ [0 , , where Ψ( p,
0) = ψ ( p ) , Ψ( p,
1) = p , and the numerator is nonzero because its dotproduct with p is (1 − t ) ψ ( p ) · p + t , which is positive when t ∈ (0 , ψ .If ψ ( p ) · p ≤ p , then deg( − ψ ) = 1 by the case just proved, and so deg( ψ ) =( − m +1 . (cid:3) Proof of Theorem 2.1.
The case of the circle ( m = 1) admits a simple winding numberproof, and so we give that argument first. Regarding p ∈ S as a complex number, thereflection formula becomes R p ( z ) = − p z for complex numbers z . Thus the reflectionsymmetry hypothesis (1) implies that φ ( − p ) φ ( p ) = − p φ ( p ) φ ( p ) = − p . The argument of the right side increases by 4 π as p goes once around the circle. Onthe left side, the arguments of φ ( p ) and φ ( − p ) increase by the same amount as eachother, and hence must increase by 2 π , implying that φ has degree 1.Now we prove the theorem for all m ≥ Step 1 — Reducing to a smooth φ whose normal component changes sign. By anargument of Petrides [26, p. 2391] it is possible to reduce the theorem to the caseof φ smooth. This involves using successive smoothing on m + 1 almost-hemispherescentered on the coordinate axes, with the approximations extended to complementaryhemispheres via the reflection symmetry formula. So, from now on, φ is assumed tobe smooth.For s ∈ [0 ,
1) define the level sets A ( s ) = { p ∈ S m : φ ( p ) · p > s } , B ( s ) = { p ∈ S m : φ ( p ) · p < s } and Z ( s ) = { p ∈ S m : φ ( p ) · p = s } . These sets are invariant under the antipodal map, meaning A ( s ) = − A ( s ), B ( s ) = − B ( s ) and Z ( s ) = − Z ( s ), because the normal component is even: φ ( p ) · p = R p ( φ ( p )) · R p ( p ) = φ ( − p ) · ( − p ) THIRD NEUMANN EIGENVALUE by the reflection symmetry hypothesis. We may further suppose φ ( p ) · p is nonconstant,for otherwise Theorem 2.1 follows immediately from Lemma 2.2. Step 2 — Reducing Z to a smooth submanifold. Choose a regular value s ∈ [0 , φ ( p ) · p , so that the level set Z ( s ) is an imbedded submanifold in thesphere that forms the boundary of both the superlevel set A ( s ) and the sublevel set B ( s ). Define a homotopyΦ( p, t ) = φ ( p ) − tp | φ ( p ) − tp | , p ∈ S m , t ∈ [0 , s ] , and note that Φ( p,
0) = φ ( p ) and the numerator is nonzero for each t because φ ( p )is a unit vector while | tp | = t ≤ s <
1. Denote the right endpoint map of thehomotopy by ψ ( p ) = Φ( p, s ), so that ψ is a smooth map of the sphere to itself withdeg( ψ ) = deg( φ ). Observe also that ψ satisfies the reflection symmetry condition (1),because both φ ( p ) and p satisfy it. Thus it suffices to prove the theorem for ψ .The zero superlevel set for ψ is A ψ (0) = { p ∈ S m : ψ ( p ) · p > } = { p ∈ S m : φ ( p ) · p > s } = A ( s ) , and similarly for its sublevel and zero sets B ψ (0) = B ( s ) and Z ψ (0) = Z ( s ), respec-tively. Thus, by changing the name of ψ to φ and dropping the ψ index, we maysuppose from now on that φ is smooth and Z = Z (0) is an imbedded submanifold inthe sphere that forms the boundary of both A = A (0) and B = B (0). That is, Z islocally the graph of a smooth function with A on one side of the graph and B on theother side. Step 3 — Comparison map.
The degree of φ will be compared with the degree ofthe map (cid:101) φ : S m → S m defined by (cid:101) φ ( p ) = (cid:40) φ ( p ) φ ( − p ) = (cid:40) φ ( p ) when p ∈ A ∪ Z , R p φ ( p ) when p ∈ B ∪ Z ,where the reflection symmetry assumption (1) has been used on B ∪ Z ; notice thedefinition is consistent on Z since φ ( p ) · p = 0 there and so φ ( p ) = R p φ ( p ) on Z .This map (cid:101) φ is piecewise smooth, and deg( (cid:101) φ ) = 1 by Lemma 2.2, because (cid:101) φ ( p ) · p ≥ p ∈ S m . Step 4 — Odd dimensions.
Suppose m is odd. Write J[ φ ] for the determinant ofthe derivative map of φ , so thatdeg( φ ) = S m J[ φ ]( p ) dS ( p )= 1 | S m | ˆ A J[ φ ]( p ) dS ( p ) + 1 | S m | ˆ B J[ φ ]( p ) dS ( p ) , (2)where we need not integrate over Z because that set has measure zero. In the integralover A , we may replace φ with (cid:101) φ . In the integral over B , write N ( p ) = − p for the HIRD NEUMANN EIGENVALUE 7 antipodal map, and change variable to obtain that ˆ B J[ φ ]( p ) dS ( p ) = ˆ N − ( B ) (J[ φ ] ◦ N )( p ) dS ( p )= ˆ N − ( B ) J[ φ ◦ N ]( p ) dS ( p )by the chain rule, because J[ N ] = ( − m +1 = 1 when m is odd. Next, since N − ( B ) = − B = B , and φ ◦ N = (cid:101) φ on B , we conclude ˆ B J[ φ ]( p ) dS ( p ) = ˆ B J[ (cid:101) φ ]( p ) dS ( p ) . Thus we may replace φ with (cid:101) φ in the second integral of (2) also, and so deg( φ ) =deg( (cid:101) φ ) = 1. Step 5 — Even dimensions.
Suppose m is even. By arguing as in the previousstep, except this time using that J[ N ] = ( − m +1 = − m is even, we finddeg( φ ) = 1 | S m | ˆ A J[ φ ]( p ) dS ( p ) − | S m | ˆ B J[ (cid:101) φ ]( p ) dS ( p ) . (3)Obviously alsodeg( (cid:101) φ ) = 1 | S m | ˆ A J[ φ ]( p ) dS ( p ) + 1 | S m | ˆ B J[ (cid:101) φ ]( p ) dS ( p ) . Since deg( (cid:101) φ ) = 1, adding the last two equations shows thatdeg( φ ) + 1 = 2 1 | S m | ˆ A J[ φ ]( p ) dS ( p ) . Thus to show φ has odd degree, we want to show the expression1 | S m | ˆ A J[ φ ]( p ) dS ( p ) (4)is an integer. We accomplish this by showing it equals the degree of a mapping.We regard the closure A = A ∪ Z as a compact manifold without boundary, byidentifying antipodal points in Z . In more detail, given a point p ∈ Z and any small ε >
0, one can form a neighborhood of p in A by gluing together the sets { q ∈ A : | q − p | < ε } and { q ∈ A : | q − ( − p ) | < ε } along their common interface, which is the smooth submanifold { q ∈ Z : | q − p | < ε } N ∼ { q ∈ Z : | q − ( − p ) | < ε } , where N ∼ denotes the antipodal equivalence relation on Z . A little thought shows thatthe resulting manifold A is orientable, because m is even. This manifold might notbe connected, but it can have only finitely many components, since Z is a smoothsubmanifold of the sphere. THIRD NEUMANN EIGENVALUE
Define a piecewise smooth map α : A → S m by restriction: let α = φ | A , notingthe restriction is consistent under the antipodal identification on Z because p ∈ Z = ⇒ φ ( p ) · p = 0 = ⇒ φ ( p ) = R p φ ( p ) = ⇒ φ ( p ) = φ ( − p ) , by the reflection symmetry hypothesis. The degree of the map α is exactly theexpression in (4) (by the de Rham approach to degree theory [25, Corollary III.2.4]),which means expression (4) must be an integer, completing the proof. Remark.
A more “symmetrical” formulation of Step 5, when m is even, arises fromregarding both A and B as compact orientable manifolds without boundary, via theantipodal identification on Z . Define β : B → S m by restriction as β = φ | B , so thatdeg( φ ) = deg( α ) − deg( β ◦ N ) by formula (3), where α : A → S m and β ◦ N : B → S m .Since also 1 = deg( (cid:101) φ ) = deg( α ) + deg( β ◦ N ) by definition of (cid:101) φ , adding the twoequations shows that deg( φ ) + 1 = 2 deg( α ), and hence deg( φ ) is odd. Remark.
The nonvanishing of deg( φ ) can be proved in a different way when m isodd. For suppose to the contrary that φ is homotopic to a constant map, say to p (cid:55)→ e where e is the unit vector in the first coordinate direction. Then φ ( − p )is also homotopic to the constant map, while R p (cid:0) φ ( p ) (cid:1) is homotopic to the map p (cid:55)→ R p ( e ). After joining these homotopies by the reflection symmetry hypothesis(1), we conclude that p (cid:55)→ R p ( e ) is homotopic to a constant. But R p ( e ) has nonzerodegree when m is odd, by an observation of Girouard, Nadirashvili and Polterovich[16, p. 656]. This contradiction shows that φ has nonzero degree. (cid:3) An elementary lemma related to trial functions
The following lemma is needed in both the Euclidean and hyperbolic parts of thepaper, when trial functions of the form g ( r ) x j /r are employed. The idea is due toWeinberger [30]. Write | x | = r . Lemma 3.1.
Suppose w ∗ and w ∗ are Borel measures on R n , n ≥ , and g ( r ) is C -smooth for r ≥ , with g (0) = 0 . If µ ∈ R satisfies µ ≤ ´ R n |∇ (cid:0) g ( r ) x j /r (cid:1) | dw ∗ ( x ) ´ R n (cid:0) g ( r ) x j /r (cid:1) dw ∗ ( x ) for j = 1 , . . . , n , and each integral is positive and finite, then µ ≤ ´ R n (cid:0) g (cid:48) ( r ) + n − r g ( r ) (cid:1) dw ∗ ( x ) ´ R n g ( r ) dw ∗ ( x ) . Proof.
Clearing the denominator and summing over j gives that µ ˆ R n g ( r ) n (cid:88) j =1 (cid:0) x j /r (cid:1) dw ∗ ( x ) ≤ ˆ R n n (cid:88) j =1 |∇ (cid:0) g ( r ) x j /r (cid:1) | dw ∗ ( x ) . HIRD NEUMANN EIGENVALUE 9
The sum on the left simplifies to 1, giving the desired denominator. The gradient onthe right can be computed straightforwardly, obtaining that n (cid:88) j =1 |∇ (cid:0) g ( r ) x j /r (cid:1) | = g (cid:48) ( r ) + n − r g ( r ) , (5)which is what we want for the numerator. (cid:3) Weighted mass transplantation
A minor adaptation of Weinberger’s mass transplantation method will be neededlater. Write V ω (Ω) = ˆ Ω ω ( r ) dx for the volume of Ω with respect to a radial weight ω ( r ). Lemma 4.1 (Mass transplantation) . Suppose ω ( r ) is a positive, Lebesgue measurable,radial weight function on R n , and Ω L and Ω U are measurable sets in R n with finiteweighted volume adding up to twice the weighted volume of a ball B centered at theorigin: V ω (Ω L ) + V ω (Ω U ) = 2 V ω ( B ) < ∞ . (6) If h : [0 , ∞ ) → R is decreasing then (cid:0) ˆ Ω L + ˆ Ω U (cid:1) h ( r ) ω ( r ) dx ≤ ˆ B h ( r ) ω ( r ) dx. (7) If h ( r ) is strictly decreasing then equality holds if and only if Ω L = Ω U = B up tosets of measure zero.Proof. The volume hypothesis (6) implies that V ω (Ω L \ B ) + V ω (Ω U \ B ) = V ω ( B \ Ω L ) + V ω ( B \ Ω U ) . (8)Since h is radially decreasing, we find (cid:0) ˆ Ω L + ˆ Ω U (cid:1) h ( r ) ω ( r ) dx = (cid:0) ˆ Ω L ∩ B + ˆ Ω U ∩ B (cid:1) h ( r ) ω ( r ) dx + (cid:0) ˆ Ω L \ B + ˆ Ω U \ B (cid:1) h ( r ) ω ( r ) dx ≤ (cid:0) ˆ Ω L ∩ B + ˆ Ω U ∩ B (cid:1) h ( r ) ω ( r ) dx + (cid:0) V ω (Ω L \ B ) + V ω (Ω U \ B ) (cid:1) h (1) (9)= (cid:0) ˆ B ∩ Ω L + ˆ B ∩ Ω U (cid:1) h ( r ) ω ( r ) dx + (cid:0) V ω ( B \ Ω L ) + V ω ( B \ Ω U ) (cid:1) h (1) by (8) ≤ (cid:0) ˆ B ∩ Ω L + ˆ B ∩ Ω U (cid:1) h ( r ) ω ( r ) dx + (cid:0) ˆ B \ Ω L + ˆ B \ Ω U (cid:1) h ( r ) ω ( r ) dx (10)= 2 ˆ B h ( r ) ω ( r ) dx < ∞ , which proves the inequality (7) in the proposition.Suppose h is strictly decreasing and equality holds in (7). Then equality musthold in (9), which forces the sets Ω L \ B and Ω U \ B to have weighted volume zero(using here that h is strictly decreasing), and hence to have measure zero, since ω > B \ Ω L and B \ Ω U have measure zero. (cid:3) Proof of Theorem B (Euclidean) — constructing the trial functions
The idea of constructing trial functions from the eigenfunctions of the ball goes backto Szeg˝o [28] and Weinberger [30], in their work on the second eigenvalue. Girouard,Nadirashvili and Polterovich [16] folded these trial functions across a hyperplane inorder to obtain trial functions for the third eigenvalue. (Technically, they folded thedomain rather than the trial functions, following an idea of Nadirashvili [24] on thesphere.) They worked in the disk in 2 dimensions with hyperbolic reflections, but theconstruction adapts to Euclidean space too, as developed below. Bucur and Henrot[10] constructed the same family of trial functions by gluing rather than folding,and they parameterized the trial functions somewhat differently. These matters arediscussed in more detail at the end of this section.
Construction of trial functions.
By scale invariance we may assume | Ω | = 2 | B | , so that Ω has twice the volume of the unit ball.Let g ( r ) be the radial part of the second Neumann eigenfunction for the unit ball,extended to be constant for r ≥
1. That is, g (0) = 0 and g ( r ) = (cid:40) r − n/ J n/ ( k (cid:48) n/ r ) , < r ≤ ,g (1) , r > , where k (cid:48) n/ is the first positive zero of ( r − n/ J n/ ( r )) (cid:48) and J n/ is the Bessel functionof order n/
2. The eigenvalue of the ball is µ ( B ) = ( k (cid:48) n/ ) . For example, in dimension n = 2 one gets k (cid:48) (cid:39) .
84. Notice g (cid:48) = 0 at r = 1, from both the left and right.What we need to know about the ball eigenfunction g ( r ) y j /r (where r = | y | and j = 1 , . . . , n ) is that g is continuous and increasing for r ≥
0, and positive for r > g (cid:48) is continuous with g (cid:48) (1) = 0. Proposition A.1 at the end of the paper providesa precise statement and references for these properties.Define v : R n → R n to be the vector field v ( y ) = g ( | y | ) y | y | , y ∈ R n \ { } , (11)with v (0) = 0. Note v is continuous everywhere, including at the origin since g (0) = 0.Each component v j ( y ) = g ( r ) y j /r of the vector field is a Neumann eigenfunctionon the unit ball. The eigenfunction equation − ∆ v j = µ ( B ) v j implies that g satisfiesthe Bessel-type equation − g (cid:48)(cid:48) ( r ) = n − r g (cid:48) ( r ) + (cid:18) µ ( B ) − n − r (cid:19) g ( r ) . (12) HIRD NEUMANN EIGENVALUE 11
A useful normalization is that by translating Ω we may suppose the “Weinbergerpoint” (or g -center of mass) of Ω lies at the origin: ˆ Ω v ( y ) dy = 0 . (13)The existence of such a translation was proved by Weinberger [30], using Brouwer’sfixed point theorem, although note the set-up is slightly different here because ourΩ has twice the volume of the ball to which g is adapted. Incidentally, an existenceproof by energy minimization was explored recently by Laugesen [22, Corollary 2](take f ≡ v across hyperplanes,thus constructing functions that are even with respect to the hyperplane. Let H = H p,t = { y ∈ R n : y · p < t } , p ∈ S n − , t ≥ , to be the open halfspace with normal vector p and “height” t , and let R H ( y ) ≡ R p,t ( y ) = y + 2( t − y · p ) p be reflection in the hyperplane ∂H . Define the “fold map” onto the closed halfspace H by F H ( y ) ≡ F p,t ( y ) = (cid:40) y if y ∈ H,R H ( y ) if y ∈ R n \ H. We will use the notation R H and F H whenever this is unambiguous, and R p,t and F p,t whenever it is necessary to refer to any of the arguments explicitly.Our trial functions on Ω will be the n components of the vector field y (cid:55)→ v ( F H ( y ) − c ) , where c ∈ R n . To visualize these trial functions, imagine centering the vector fieldat c and then replace its values in the complement of H with the even extension ofthe vector field across ∂H . The resulting vector field belongs to the Sobolev space H (Ω; R n ), since it is continuously differentiable on each side of the hyperplane ∂H and is continuous across the hyperplane. Remark.
The number of parameters matches the number of conditions to be satisfied,because the parameters ( p, t, c ) lie in the 2 n -dimensional space S n − × R × R n , andwe aim to satisfy 2 n orthogonality conditions, namely we want each of the n trialfunctions to be orthogonal to the first and second eigenfunctions on Ω. Thus theset-up considered is dimensionally compatible.The different possibilities for the trial functions are illustrated in Figure 1, for thespecial case in 2 dimensions where the hyperplane is the vertical axis and H is theleft halfspace. Orthogonality of trial functions to the constant.
We claim that there exists aunique point c H = c p,t ∈ R n such that each component of the vector field v H ( y ) = v ( F H ( y ) − c H ) - - - - - - - - - - - - - - - - - - Figure 1.
Trial function contour plots when H is the left halfplane(meaning p = (1 , , t = 0). Centering points where the trial functionsvanish: top c = ( − / , c = ( − , c = ( − , v ( F H ( y ) − c ). Right side: sine mode, from second component of thetrial vector field. The white circles sit at radius 1 from the point c andits reflected point. All trial functions are even with respect to H .is orthogonal to the constant function (the Neumann ground state) on Ω, meaning ˆ Ω v H ( y ) dy = ˆ Ω v (cid:0) F H ( y ) − c H (cid:1) dy = 0 , (14)and that this unique point c H depends continuously on the parameters ( p, t ) of thehalfspace H . These claims follow from [22, Corollary 3] (with f ≡ c H = − x H , and notice the hypotheses of the corollaryare satisfied because g is increasing with g ( r ) > r >
0. Incidentally, resultsin [22] can also handle unbounded sets of finite volume, if desired.We call c H the “center of mass point” corresponding to the halfspace H and, as inthe case of the fold map defined above, will use the notation c H or c p,t depending onthe context. HIRD NEUMANN EIGENVALUE 13
Lemma 5.1 (Location of the center of mass) . The point c H defined by condition (14) lies in the halfspace H .Proof. We prove the contrapositive. Suppose c H does not lie in H , so that c H · p ≥ t .Also note if y / ∈ ∂H then F H ( y ) ∈ H and so F H ( y ) · p < t , which implies F H ( y ) · p The hyperplane ∂H p, passes through the origin, and forms the common bound-ary of the complementary halfspaces H p, and H − p, . The reflection R p = R p, inter-changes the halfspaces, and so their fold maps are related by F − p, = R p ◦ F p, . (16)To prove c − p, = R p ( c p, ), we must show that R p ( c p, ) satisfies the condition de-termining the unique point c − p, , namely condition (14) with p and t replaced by − p and 0. That is, we must show ˆ Ω v (cid:0) F − p, ( y ) − R p ( c p, ) (cid:1) dy = 0 . By relations (15) and (16), the left-hand side is equal to ˆ Ω v (cid:0) R p ( F p, ( y ) − c p, ) (cid:1) dy = ˆ Ω R p v (cid:0) F p, ( y ) − c p, (cid:1) dy = R p ˆ Ω v (cid:0) F p, ( y ) − c p, (cid:1) dy. The last integral indeed equals 0, by condition (14) for the center of mass point c p, . (cid:3) Orthogonality of trial functions to the first excited state. Write f for a firstexcited state on Ω, that is, a Neumann eigenfunction of Ω corresponding to eigenvalue µ (Ω). We will prove that for some halfspace H , the components of the trial vector v H are orthogonal to f , meaning ˆ Ω v H ( y ) f ( y ) dy = ˆ Ω v (cid:0) F H ( y ) − c H (cid:1) f ( y ) dy = 0 . Define a vector field W ( p, t ) = ˆ Ω v (cid:0) F H ( y ) − c H (cid:1) f ( y ) dy, p ∈ S n − , t ≥ , where H = H p,t . It is easy to see W is continuous, since F H and c H depend contin-uously on the parameters p, t of the halfspace. The task is to show W vanishes forsome ( p, t ), which we do in Proposition 5.4 below.Let us investigate the vector field for large t . Choose τ > ⊂ H p,τ for all p ∈ S n − , which can be done since Ω is bounded. The next lemmashows that when t = τ , the vector field W can be evaluated explicitly, and it doesnot depend on p . Lemma 5.3 (Large positive t ) . For all p ∈ S n − , c p,τ = 0 and W ( p, τ ) = w, where w = ´ Ω vf dy is a constant vector (independent of p ).Proof. Since Ω ⊂ H p,τ , the fold map fixes Ω, so that F p,τ ( y ) = y for all y ∈ Ω. Hence ˆ Ω v (cid:0) F p,τ ( y ) − (cid:1) dy = ˆ Ω v ( y ) dy = 0by the normalization (13). Therefore the center of mass relation (14) holds with c p,τ = 0. Hence W ( p, τ ) = ´ Ω v ( y ) f ( y ) dy = w . (cid:3) Proposition 5.4 (Vanishing of the vector field) . W ( p, t ) = 0 for some p ∈ S n − and t ∈ [0 , τ ] .Proof. Suppose W ( p, t ) (cid:54) = 0 for all p, t , and define φ = W/ | W | , so that φ : S n − × [0 , τ ] → S n − . For t = τ the map is constant, with φ ( p, τ ) = w | w | by Lemma 5.3. Thus φ ( p, τ ) has degree 0 as a map from the sphere to itself.When t = 0, we find v (cid:0) F − p, ( y ) − c − p, (cid:1) = R p v (cid:0) F p, ( y ) − c p, (cid:1) by (15), (16) and Lemma 5.2. Multiplying the last equation by f ( y ) and integratingover Ω implies W ( − p, 0) = R p ( W ( p, φ ( − p, 0) = R p ( φ ( p, , p ∈ S n − . HIRD NEUMANN EIGENVALUE 15 That is, φ ( p, 0) satisfies the reflection symmetry hypothesis of Petrides’s Theorem 2.1.Hence φ ( p, 0) has nonzero degree, which is impossible because degree is a homotopyinvariant. Therefore W ( p, t ) = 0 for some p, t . (cid:3) The proof of Theorem B will be completed in the next section.5.1. Similarities and differences with the trial functions and homotopy ar-gument of Bucur and Henrot. As explained in the Introduction and at the begin-ning of this section, the trial function construction we are using developed in stagesthrough work of Szeg˝o [28], Weinberger [30], Nadirashvili [24], Girouard, Nadirashviliand Polterovich [16], Bucur and Henrot [10], Girouard and Laugesen [15], and nowthe current paper.The family of trial functions v ( F H ( y ) − c ) in this paper is the same as employedby Bucur and Henrot [10], although rather than parameterizing the trial functions asthey did in terms of points ( A, B ) ∈ R n × R n , we parameterize in terms of ( p, t, c ) ∈ S n − × R × R n , following the earlier work by Girouard, Nadirashvili and Polterovich inthe disk. The connection between the parameterizations is that A = c is the “center”for the trial function and B = R p,t ( c ) is the reflection of that point in the hyperplane ∂H p,t . Bucur and Henrot restrict A to lie in H , whereas we allow c to lie anywherein R n , but in practice this additional flexibility brings us no advantage because whenorthogonality to the constant function is imposed, one finds by Lemma 5.1 that theunique center of mass point c H must lie in H .A significant difference between their method and ours is that we reduce from a2 n -parameter family of trial functions to an n -parameter family. We do so by usinguniqueness of the center of mass point (that is, orthogonality to the constant) todetermine c in terms of p and t ; see condition (14). Another difference is that theymodify their 2 n -dimensional vector field by an explicit two-step homotopy in order toreduce to a vector field whose degree they can compute explicitly. We rely instead ona reflection symmetry observation (Lemma 5.2) that allows us to invoke the degreetheory result of Petrides (Theorem 2.1).6. Proof of Theorem B, continued — deploying the trial functions Fix the halfspace H = H p,t found in Proposition 5.4 for which the componentsof the vector field v H ( y ) = v ( F H ( y ) − c H ) are orthogonal to the first and secondeigenfunctions of the Neumann Laplacian on Ω, that is, to the constant functionand f . In this section, we estimate the third Neumann eigenvalue by inserting thecomponents of the vector field as trial functions into the Rayleigh quotient Q [ u ] = ´ Ω |∇ u | dy ´ Ω u dy for the Laplacian, and applying mass transplantation to arrive at a two-ball situation.These arguments go essentially as for Bucur and Henrot [10, pp. 344–345], who werein turn adapting the original techniques of Weinberger [30]. Rayleigh quotient estimate. Substituting each component v H, , . . . , v H,n of thevector field v H into the Rayleigh characterization µ (Ω) = min (cid:8) Q [ u ] : u ∈ H (Ω) , u ⊥ , u ⊥ f (cid:9) gives that µ (Ω) ≤ ´ Ω |∇ v H,j | dy ´ Ω ( v H,j ) dy , j = 1 , . . . , n. Write Ω L = H ∩ Ω − c H , Ω U = H ∩ R H (Ω) − c H , so that the region decomposes as Ω = (Ω L + c H ) ∪ R H (Ω U + c H ) ∪ (Ω ∩ ∂H ). Here “ L ”and “ U ” label the portions of the region corresponding to the “lower” and “upper”halfspaces, except the reflection in the definition of Ω U moves that piece to the lowerhalfspace. The sets Ω L and Ω U are not assumed to be disjoint, and indeed mightcoincide. The set Ω ∩ ∂H has measure zero, and so can be neglected in what follows.By evenness of v H,j with respect to reflection across ∂H , we may decompose theRayleigh quotient as µ (Ω) ≤ (cid:0) ´ H ∩ Ω + ´ H ∩ R H (Ω) (cid:1) |∇ v H,j | dy (cid:0) ´ H ∩ Ω + ´ H ∩ R H (Ω) (cid:1) ( v H,j ) dy . On H one has F H ( y ) = y and so v H ( y ) = v ( x ) = g ( r ) x/ | x | where x = y − c H and r = | x | . Making that change of variable gives µ (Ω) ≤ (cid:0) ´ Ω L + ´ Ω U (cid:1) |∇ v j | dx (cid:0) ´ Ω L + ´ Ω U (cid:1) v j dx , j = 1 , . . . , n. Applying Lemma 3.1 with the measures w ∗ and w ∗ being Lebesgue measure timesthe sum of indicators 1 Ω L + 1 Ω U now implies µ (Ω) (cid:0) ˆ Ω L + ˆ Ω U (cid:1) g ( r ) dx ≤ (cid:0) ˆ Ω L + ˆ Ω U (cid:1)(cid:0) g (cid:48) ( r ) + ( n − r − g ( r ) (cid:1) dx. At this stage we will introduce a minor simplification to the usual method, whichleads to needing monotonicity of only one function, rather than the usual two. (Thissimplification was exploited more seriously by Freitas and Laugesen [14, formula(14)].) The technique is to subtract µ ( B (cid:116) B ) ´ g dx from the left side of the inequalityand the equal value µ ( B ) ´ g dx from the right side, to get (cid:0) µ (Ω) − µ ( B (cid:116) B ) (cid:1)(cid:0) ˆ Ω L + ˆ Ω U (cid:1) g dx ≤ (cid:0) ˆ Ω L + ˆ Ω U (cid:1) h dx (17)where h ( r ) = g (cid:48) ( r ) + ( n − r − g ( r ) − µ ( B ) g ( r ) . Note h is continuous, since by construction g and g (cid:48) are continuous for all r ≥ 0, with g (0) = 0. Further, h has zero average over the unit ball, since by summing as in the HIRD NEUMANN EIGENVALUE 17 proof of Lemma 3.1 one finds ˆ B h ( r ) dx = n (cid:88) j =1 (cid:18) ˆ B |∇ v j ( x ) | dx − µ ( B ) ˆ B v j ( x ) dx (cid:19) = 0 , using that v j was constructed to be a second Neumann eigenfunction of the unit ball.This function h is strictly decreasing, by the following calculation due to Weinberger[30, formula (2.15)]. For 0 < r < h (cid:48) ( r ) = − n − r (cid:18) g (cid:48) ( r ) − g ( r ) r (cid:19) − µ ( B ) g ( r ) g (cid:48) ( r ) < , where g (cid:48)(cid:48) was eliminated from the formula with the help of the Bessel-type equation(12). For r > g is constant and g (cid:48) = 0 on thatrange, so that h (cid:48) ( r ) = − n − r g (1) < . Hence h is strictly decreasing.Since | Ω | = 2 | B | , mass transplantatiion as in Lemma 4.1 with ω ≡ ´ B h ( r ) dx = 0. From the left side of(17) we conclude µ (Ω) − µ ( B (cid:116) B ) ≤ 0. Equality obviously holds if Ω is a unionof two disjoint balls of equal volume. Finally, if equality holds then the equalitystatement of Lemma 4.1 implies that Ω L and Ω U equal B up to sets of measure zero,and so Ω ∩ H and Ω ∩ H c are each translates of B , up to sets of measure zero. TheLipschitz boundary assumption on the open set Ω then forces Ω ∩ H and Ω ∩ H c toactually equal those translates of B , and hence Ω is the union of two disjoint balls.This finishes the proof of Theorem B.7. Proof of Theorem A (hyperbolic) — constructing the trial functions The Euclidean proof in the preceding sections adapts robustly to the hyperbolicsetting. First we prepare the eigenvalue problem for the hyperbolic Laplacian, usingthe Poincar´e ball model. Construction of the hyperbolic Laplacian. The Laplacian or Laplace–Beltramioperator on a Riemannian manifold with metric g is1 √ det g n (cid:88) i,j =1 ∂ i ( (cid:112) det g g ij ∂ j u ) . In the Poincar´e ball model the metric is given by (1 − | x | ) − times the Euclideanmetric, that is, g ij ( x ) = (1 − | x | ) − δ ij , with inverse g ij ( x ) = (1 − | x | ) δ ij . Hence √ det g = (1 − | x | ) − n and the hyperbolic Laplacian becomes∆ hyp u = (1 − | x | ) n ∇ · (cid:0) (1 − | x | ) − n ∇ u (cid:1) . (18) Note. Our choice of metric gives a hyperbolic space with sectional curvature κ = − curvature one simply multiplies the metric by a constant factor, which results in theLaplacian and its eigenvalues also getting multiplied by a constant.Consider an open set Ω (cid:98) B that is compactly contained in the unit ball and hasLipschitz boundary. The Neumann eigenvalue problem for the hyperbolic Laplacianis − ∆ hyp u = ηu in Ω, ∂u∂ν = 0 on ∂ Ω. (19)The associated Rayleigh quotient is Q h [ u ] = ´ Ω |∇ u ( x ) | (1 − | x | ) dγ ( x ) ´ Ω u ( x ) dγ ( x )where the hyperbolic volume element is dγ ( x ) = 1(1 − | x | ) n dx. The weight function (1 − | x | ) − n is positive and bounded on Ω, because Ω is assumedto have compact closure in the unit ball, and so the Rayleigh quotient is well definedfor u in the unweighted Sobolev space H (Ω). Since ∂ Ω is assumed to be Lipschitz, H (Ω) imbeds compactly into L (Ω). Hence the Neumann spectrum of the hyperbolicLaplacian is well defined and discrete, with eigenvalues0 = η ≤ η ≤ η ≤ · · · → ∞ that are characterized by the usual minimax variational principle in terms of theRayleigh quotient. The first eigenvalue is η = 0, with constant eigenfunction, andthe eigenfunctions satisfy the natural boundary condition ∂u/∂ν = 0. Thus we havearrived at the eigenvalue problem (19) studied in Theorem A.The eigenvalues are invariant under hyperbolic isometries applied to Ω. Eigenfunctions of a ball. Consider the hyperbolic eigenvalue problem (19) on aball B = B ( a ) of radius a < r, θ ) ∈ [0 , × S n − we may separate variables in the form u ( r, θ ) = g ( r )Θ( θ )and substitute into the eigenfunction equation − ∆ hyp u = ηu. By expressing the hyperbolic Laplacian (18) in spherical coordinates, we obtain thatthe angular part Θ satisfies∆ θ Θ( θ ) + (cid:96) ( (cid:96) + n − θ ) = 0where (cid:96) ≥ θ denotes the spherical Laplacian. When (cid:96) = 0, givinga constant function Θ, the eigenfunctions on the ball are purely radial. For positiveintegers (cid:96) , the angular function Θ is a spherical harmonic and the eigenvalues havemultiplicity greater than 1. HIRD NEUMANN EIGENVALUE 19 The separation of variables shows that the radial part g satisfies − g (cid:48)(cid:48) ( r ) = (cid:18) n − r + 2( n − r − r (cid:19) g (cid:48) ( r ) + (cid:18) η (1 − r ) − (cid:96) ( (cid:96) + n − r (cid:19) g ( r ) (20)for 0 < r < a , with Neumann boundary condition g (cid:48) ( a ) = 0 . The key facts about the second eigenvalue and its eigenfunctions are given in Propo-sition A.2, which states that:the second eigenfunctions of the ball B ( a ) have angular dependence ofthe form g ( r ) x j /r for j = 1 , . . . , n , and the eigenvalue has multiplicity n . The radial part g has g (0) = 0 and g (cid:48) ( r ) > r ∈ (0 , a ), andsatisfies the boundary condition g (cid:48) ( a ) = 0.For example, in 2 dimensions, the eigenfunctions have the form g ( r ) cos θ and g ( r ) sin θ ,and, in general, the angular parts x /r, . . . , x n /r come from the spherical harmonicswith (cid:96) = 1. Construction of trial functions. Write B for a ball centered at the origin whosehyperbolic volume equals half the hyperbolic volume of Ω. Denote by a ∈ (0 , B . Let g ( r ) be the radial part of the second Neumanneigenfunction on B , as above, and extend it to the exterior of the ball by letting g ( r ) = g ( a ) , a < r < . Notice g is continuous and increasing for r ∈ [0 , r > 0, and g (cid:48) ( r )is continuous for r ∈ [0 , v : B → R n to be the vector field v ( y ) = g ( | y | ) y | y | , y ∈ B \ { } , with v (0) = 0. This v is continuous everywhere, including at the origin since g (0) =0. Each component v j ( y ) = g ( r ) y j /r of the vector field is a hyperbolic Neumanneigenfunction on the ball B with eigenvalue η ( B ). M¨obius isometries of the ball. The trial functions will depend on a family of M¨obiustransformations T x : B → B that are parameterized by x ∈ B and have the following properties: T ( y ) = y is theidentity, and when x (cid:54) = 0 the map T x ( · ) is a M¨obius self-map of the ball such that T x (0) = x and T x fixes the points ± x/ | x | on the unit sphere. In 2 dimensions themaps can be written in complex notation as T x ( y ) = x + y xy , x ∈ D , y ∈ D , where D (cid:39) B is the unit disk in the complex plane. In all dimensions [1, eq. (26)]: T x ( y ) = (1 + 2 x · y + | y | ) x + (1 − | x | ) y x · y + | x | | y | , x ∈ B , y ∈ B . Observe T x ( y ) is a continuous function mapping ( x, y ) ∈ B × B to T x ( y ) ∈ B , and T x ( · ) maps B to itself and ∂ B to itself, with T x (0) = x and inverse ( T x ) − = T − x .Each mapping T x is a hyperbolic isometry, and its derivative matrix at y is( DT x )( y ) = 1 − | T x ( y ) | − | y | × (orthogonal matrix) , by [1, Section 2.7]. Taking the determinant shows T x has Jacobian (1 −| T x ( y ) | ) n / (1 −| y | ) n .Hence the numerator of the hyperbolic Rayleigh quotient is invariant under M¨obiustransformations, because a straightforward change of variable reveals that ˆ E |∇ u ( y ) | (1 − | y | ) dγ ( y ) = ˆ T x ( E ) |∇ ( u ◦ T − x )( y ) | (1 − | y | ) dγ ( y )whenever E is an open subset of the unit ball and u is C -smooth on E . (Thedenominator of the Rayleigh quotient is similarly invariant.) From the differentialgeometry perspective, the reason for the invariance is that the integrand on the leftcan be written intrinsically in terms of the metric as |∇ u ( y ) | (1 − | y | ) = |∇ g u ( y ) | g ,and the M¨obius transformation is an isometry with respect to the hyperbolic metric g . Translational centering. The M¨obius transformation enables us to impose a usefulnormalization. By replacing Ω with its hyperbolic translation T x (Ω), for some x , wemay require ˆ Ω v ( y ) dγ ( y ) = 0 , (21)as we now explain. The existence of such a “hyperbolic Weinberger center of mass”was known to Chavel [11, p. 80]. A detailed proof using Brouwer’s fixed point theo-rem appeared later in Benguria and Linde [9, Theorem 6.1]. For a proof by energyminimization, yielding also uniqueness and continuous dependence of the translationwith respect to Ω, see Laugesen [23, Corollary 2], noting that the hypotheses thereare satisfied with f ≡ 1, since ´ g ( r )(1 − r ) − dr = ∞ for our function g . Hyperbolic folding, and the trial functions. Next we develop the hyperbolic fold map.Let H p = { y ∈ B : y · p < } , p ∈ S n − , be the halfball with normal vector p . Its boundary relative to the ball is the set ∂H p = { y ∈ B : y · p = 0 } , which is the intersection of the ball with a hyperplanepassing through the origin. Define H ≡ H p,t = T pt ( H p ) , p ∈ S n − , t ∈ [0 , , which is the image of the halfball under the M¨obius translation T pt . (Taking t = 0gives H p, = H p .) The boundary relative to the ball is the hyperbolic hyperplane ∂H p,t = T pt ( ∂H p ). After writing R p ( y ) = y − y · p ) p HIRD NEUMANN EIGENVALUE 21 for the reflection map across ∂H p , we may define the hyperbolic reflection across ∂H = ∂H p,t by conjugation, as R H ≡ R p,t = T pt ◦ R p ◦ ( T pt ) − : B → B . This reflection is a hyperbolic isometry. Clearly R p, = R p . Define the hyperbolic“fold map” onto H byΦ H ( y ) ≡ Φ p,t ( y ) = (cid:40) y if y ∈ H,R H ( y ) if y ∈ B \ H, so that the fold map fixes each point in H and maps each point in B \ H to itshyperbolic reflection across ∂H .Our trial functions will be the components of the vector field v ◦ T − c ◦ Φ H , where c ∈ B is fixed. To visualize these trial functions, imagine “centering” thevector field at the point c with the help of the transformation T − c , and then replaceits values on B \ H by evenly extending the vector field from H via hyperbolic reflectionacross ∂H . The resulting vector field is continuously differentiable on each side of thehyperplane ∂H , and is continuous across the hyperplane.Contour plots for these trial functions can be visualized similar to the Euclideancase in Figure 1, except instead of spreading over the whole plane, the functions arecrammed into the unit disk. Orthogonality of trial functions to the constant. We claim that a unique “cen-ter of mass point” c H ∈ B exists such that each component of the vector field v H = v ◦ T − c H ◦ Φ H is orthogonal to the constant function (the Neumann ground state) on Ω, meaning ˆ Ω v H ( y ) dγ ( y ) = ˆ Ω v (cid:0) T − c H ◦ Φ H ( y ) (cid:1) dγ ( y ) = 0 . (22)(Recall dγ ( y ) is the volume element with respect to the hyperbolic metric.) Theexistence and uniqueness of c H follow directly from Laugesen [23, Corollary 3] with f ≡ 1, where the hypotheses of that result are satisfied because ´ g ( r )(1 − r ) − dr = ∞ . Furthermore, that work shows that c H depends continuously on the parameters( p, t ) of H p,t . Write c p,t for the center of mass point c H that corresponds to thehyperbolic halfspace H = H p,t .As in the Euclidean case, reflection commutes with v : v ◦ R p = R p ◦ v. (23)Reflection also conjugates with the M¨obius transformations, in the sense that( T R p x ◦ R p )( y ) = ( R p ◦ T x )( y ) , x ∈ B , y ∈ B , p ∈ S n − , (24)as can be verified using the definitions. To understand the last formula, rememberthat T x represent a hyperbolic translation by x , and so the analogous formula in Euclidean space simply says that R p x + R p y = R p ( x + y ), which is geometricallyobvious.Now we can show how the center of mass point behaves under reflection. Lemma 7.1 (Reflection invariance of the center of mass when t = 0) . For p ∈ S n − , c − p, = R p ( c p, ) . Proof. The hyperplane ∂H p, passes through the origin, and forms the common bound-ary of the complementary halfballs H p, and H − p, . The reflection R p interchangesthe halfballs, and so their fold maps are related byΦ − p, = R p ◦ Φ p, . (25)To prove c − p, = R p ( c p, ), we must show that R p ( c p, ) satisfies the condition de-termining the unique point c − p, , namely condition (22) with p and t replaced by − p and 0. That is, we must show ˆ Ω v (cid:0) T − R p ( c p, ) ◦ Φ − p, ( y ) (cid:1) dγ ( y ) = 0 . The integrand on the left is v ◦ T − R p ( c p, ) ◦ Φ − p, = R p ◦ v ◦ T − c p, ◦ Φ p, = R p ◦ v H p, by (23), (24) and (25) . Integrating over Ω gives zero (as desired) on the right side,thanks to condition (22) for the center of mass c p, . (cid:3) Orthogonality of trial functions to the first excited state. Write f for a firstexcited state on Ω, that is, a hyperbolic Neumann eigenfunction of Ω corresponding toeigenvalue η (Ω). We will prove that for some hyperbolic halfspace H , the componentsof the trial vector v H are orthogonal to f , meaning ˆ Ω v (cid:0) T − c H ◦ Φ H ( y ) (cid:1) f ( y ) dγ ( y ) = 0 . Define a vector field W ( p, t ) = ˆ Ω v (cid:0) T − c p,t ◦ Φ p,t ( y ) (cid:1) f ( y ) dγ ( y ) , p ∈ S n − , t ∈ [0 , . We want to show W vanishes for some ( p, t ). Note W is continuous, since the centerof mass point c p,t and the fold map Φ p,t depend continuously on p, t .First we investigate the vector field for t near 1. Choose τ > ⊂ H p,τ for all p ∈ S n − , which can be done since Ω is compactly containedin B . The next lemma shows that when t = τ , the vector field W can be evaluatedexplicitly, and it does not depend on p . Lemma 7.2 ( t near 1) . For all p ∈ S n − , c p,τ = 0 and W ( p, τ ) = w, where w = ´ Ω v ( y ) f ( y ) dγ ( y ) is a constant vector (independent of p ). HIRD NEUMANN EIGENVALUE 23 Proof. Since Ω ⊂ H p,τ , the fold map fixes Ω, so that Φ p,τ ( y ) = y for all y ∈ Ω. Hence ˆ Ω v (cid:0) τ ◦ Φ p,τ ( y ) (cid:1) dγ ( y ) = ˆ Ω v ( y ) dγ ( y ) = 0by the normalization (21). Therefore the center of mass relation (22) holds with c p,τ = 0. Hence W ( p, τ ) = ˆ Ω v ( y ) f ( y ) dγ ( y ) = w. (cid:3) It follows that W vanishes at some point: Proposition 7.3 (Vanishing of the vector field) . W ( p, t ) = 0 for some p ∈ S n − and t ∈ [0 , τ ] .Proof. When t = 0, we find v ◦ T − c − p, ◦ Φ − p, = R p ◦ v ◦ T − c p, ◦ Φ p, by (23), (24), (25) and Lemma 7.1. Multiplying the last equation by f ( y ) dγ ( y ) andintegrating over Ω implies W ( − p, 0) = R p ( W ( p, W satisfies the reflectionsymmetry condition. The rest of the proof goes exactly as in the Euclidean case inProposition 5.4, except using Lemma 7.2 instead of Lemma 5.3 (cid:3) Estimating the Rayleigh quotient. Fix the parameters ( p, t ) found in Proposi-tion 7.3, and write H = H p,t for the hyperbolic halfspace, so that by the propositionand the earlier formula (22), the vector field v H ( y ) = v (cid:0) T − c H ◦ Φ H ( y ) (cid:1) is orthogonalin L (Ω; dγ ) to both the constant function and the first excited state f .Substituting each component v H, , . . . , v H,n of the vector field as a trial functioninto the Rayleigh characterization of the third eigenvalue gives that η (Ω) ≤ Q h [ v H,j ] = ´ Ω |∇ v H,j | (1 − | y | ) dγ ( y ) ´ Ω ( v H,j ) dγ ( y ) , j = 1 , . . . , n. By evenness of v H,j with respect to hyperbolic reflection across ∂H , and using theinvariance of the numerator and denominator integrals under hyperbolic reflection,we find η (Ω) ≤ (cid:0) ´ H ∩ Ω + ´ H ∩ R H (Ω) (cid:1) |∇ v H,j | (1 − | y | ) dγ ( y ) (cid:0) ´ H ∩ Ω + ´ H ∩ R H (Ω) (cid:1) ( v H,j ) dγ ( y ) . Write Ω L = T − c H ( H ∩ Ω) , Ω U = T − c H ( H ∩ R H (Ω)) , so that the region decomposes as Ω = T c H (Ω L ) ∪ R H ( T c H (Ω U )) ∪ (Ω ∩ ∂H ). Theanalogy with the Euclidean case earlier in the paper should at this point be clear. On H one has Φ H ( y ) = y and so v H ( y ) = v ( x ) where x = T − c H ( y ). Changing variable from y to x with the help of M¨obius invariance of the numerator and denominatorintegrals, we obtain the inequality η (Ω) ≤ (cid:0) ´ Ω L + ´ Ω U (cid:1) |∇ v j | (1 − | x | ) dγ ( x ) (cid:0) ´ Ω L + ´ Ω U (cid:1) v j dγ ( x ) , j = 1 , . . . , n. Applying Lemma 3.1 with dw ∗ ( x ) = (1 − | x | ) (1 Ω L + 1 Ω U )( x ) dγ ( x ) , dw ∗ ( x ) = (1 Ω L + 1 Ω U )( x ) dγ ( x ) , we deduce that η (Ω) (cid:0) ˆ Ω L + ˆ Ω U (cid:1) g ( r ) dγ ( x ) ≤ (cid:0) ˆ Ω L + ˆ Ω U (cid:1)(cid:0) g (cid:48) ( r ) + ( n − r − g ( r ) (cid:1) (1 − r ) dγ ( x )where r = | x | .Subtracting η ( B (cid:116) B ) ´ g dγ ( x ) from the left side of the inequality and the equalvalue η ( B ) ´ g dγ ( x ) from the right side gives that (cid:0) η (Ω) − η ( B (cid:116) B ) (cid:1)(cid:0) ˆ Ω L + ˆ Ω U (cid:1) g ( r ) dγ ( x ) ≤ (cid:0) ˆ Ω L + ˆ Ω U (cid:1) h ( r ) dγ ( x ) (26)where h ( r ) = (cid:0) g (cid:48) ( r ) + ( n − r − g ( r ) (cid:1) (1 − r ) − η ( B ) g ( r ) . Note h is continuous, since by construction g and g (cid:48) are continuous for 0 ≤ r < g (0) = 0. Further, h has integral zero over the ball B , since by (5) one finds ˆ B h ( r ) dγ ( x ) = n (cid:88) j =1 (cid:18) ˆ B |∇ v j ( x ) | (1 − | x | ) dγ ( x ) − η ( B ) ˆ B v j ( x ) dγ ( x ) (cid:19) = 0 , using here that each v j is a hyperbolic eigenfunction on B with eigenvalue η ( B ).The function h is strictly decreasing for 0 < r < a , as one sees by differentiatingdirectly to find h (cid:48) ( r ) = − n − r (1 − r ) (cid:34)(cid:18) g (cid:48) ( r ) − − r r g ( r ) r (cid:19) + 4 r (1 + r ) g ( r ) r (cid:35) − η ( B ) g ( r ) g (cid:48) ( r ) < , where g (cid:48)(cid:48) was eliminated from the formula using the Bessel-type equation (20) with (cid:96) = 1. (Equivalent derivative calculations were given by Ashbaugh and Benguria [5,formula (6.1)] and Xu [31, p. 158].) For a < r < g is constant and g (cid:48) = 0 on that range, and so h (cid:48) ( r ) = ( n − g ( a ) ddr r − (1 − r ) < . Hence h is strictly decreasing for 0 < r < HIRD NEUMANN EIGENVALUE 25 Since Ω has twice the hyperbolic volume of B , the mass transplantation Lemma 4.1applied with the hyperbolic volume weight ω ( r ) = (1 − r ) − n implies that the rightside of (26) is less than or equal to 2 ´ B h ( r ) dγ ( x ) = 0. (This weight ω is definedonly for 0 ≤ r < 1, but that is enough for applying Lemma 4.1 since Ω L , Ω U and B all lie in the unit disk.) Equality certainly holds if Ω is a union of two disjoint ballsof equal hyperbolic volume. In the other direction, if the right side of (26) equals 0then the equality statement of Lemma 4.1 implies that Ω L and Ω U equal B up to setsof measure zero, and so Ω ∩ H and Ω ∩ H c are each hyperbolic translates of B , upto sets of measure zero. The Lipschitz boundary assumption on the open set Ω thenrequires Ω ∩ H and Ω ∩ H c to actually equal those translates of B , and hence Ω isthe union of two disjoint balls of equal hyperbolic volume. The proof of Theorem Ais complete. Acknowledgments This research was supported by the Funda¸c˜ao para a Ciˆencia e a Tecnologia (Portu-gal) through project UIDB/00208/2020 (Pedro Freitas), and a grant from the SimonsFoundation ( Appendix A. Second Neumann eigenfunction of ball is not radial Separation of variables reveals the form of the eigenfunctions for the ball, but doesnot say whether the second eigenfunction is purely radial or has angular dependence.The next result shows it is not radial. Proposition A.1 (Euclidean Laplacian) . The second eigenspace of the NeumannLaplacian on the ball B ( a ) ⊂ R n has a basis of the form { g ( r ) x j /r : j = 1 , . . . , n } ,where the radial part g has g (0) = 0 and g (cid:48) ( r ) > for r ∈ (0 , a ) , and satisfies theboundary condition g (cid:48) ( a ) = 0 . The result is well known, and was used by Weinberger [30]. It can be proved in avariety of ways that either emphasize or de-emphasize the role of special functions.For example, an approach using Bessel functions is given by Ashbaugh and Benguria[4, p. 562].The hyperbolic Laplacian for the Poincar´e ball model with curvature − hyp u = (1 − | x | ) n ∇ · (cid:0) (1 − | x | ) − n ∇ u (cid:1) . The second Neumann eigenfunction of this operator too has the form g ( r ) x j /r . Proposition A.2 (Hyperbolic Laplacian) . The second eigenspace of the hyperbolicNeumann Laplacian on the ball B ( a ) , < a < , has a basis of the form { g ( r ) x j /r : j = 1 , . . . , n } , where the radial part g has g (0) = 0 and g (cid:48) ( r ) > for r ∈ (0 , a ) , andsatisfies the boundary condition g (cid:48) ( a ) = 0 . For a proof using an ODE comparison method, see Ashbaugh and Benguria [5,Sections 3 and 6], which is based on arguments of Bandle [7], [8, pp. 122–128] forgeneral weighted eigenvalue problems. Bandle stated her result in 2 dimensions, andobserved that the method extends easily to higher dimensions [8, p. 153]. To connectAshbaugh and Benguria’s notation to ours, first note that the hyperbolic distance s from the origin to a point at Euclidean radius r satisfies ds/dr = (1 − r ) − , and so s = arctanh r . In their paper, the hyperbolic distance θ from the origin is twice asgreat, due to a different normalization, and so the relation between the variables is θ = 2 s = 2 arctanh r .A proof along similar lines is given by Xu [31, Lemma 2]. 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