Maximization of the second conformal eigenvalue of spheres

We establish in this paper an upper bound on the second eigenvalue of n-dimensional spheres in the conformal class of the round sphere. This upper bound holds in all dimensions and is asymptotically sharp as the dimension increases.

0 = λ 0 (M, g) < λ 1 (M, g) ≤ λ 2 (M, g) ≤ · · · ≤ λ k (M, g) ≤ . . . which goes to +∞ as k → +∞. The two first eigenvalues are simple, the eigenfunctions associated to λ 0 = 0 being the constant functions. A natural, and often adressed, question is to get estimates on the eigenvalues thanks to some geometric assumptions. In this paper, we discuss maximisation of eigenvalues for metrics in a given conformal class with fixed volume. We focus on the case of the standard sphere.
We let S n be the unit sphere of R n+1 for n ≥ 2. If g is a metric on S n , we are interested in the scale invariant quantity Λ n,k (g) = λ k (S n , g)V ol g (S n ) 2 n In dimension 2, we can maximize Λ 2,k on regular metrics. An inequality has been proved for k = 1 by Hersch [6] : with equality iff g is the round metric. He followed the proof of the maximization by Szegö [9] of the first non zero Neumann eigenvalue for planar domains, attained by discs. Nadirashvili found an optimal maximization for k = 2. He proved in [8] that where the supremum is attained in the degenerate case of the union of two identical spheres. His idea was used later in [5] to show that among simply connected planar domains, the second non zero Neumann eigenvalue is maximal in the degenerate case of two discs of the same area.
If we look for an analogous inequality in dimension n ≥ 3, we have to restrict our attention to some classes of metrics since Λ n,k is not bounded on the set of regular metrics (see [2]). It is natural, as suggested in [4] and [3], to consider the set of metrics in some conformal class. Indeed, in any given conformal class, Λ n,k (g) admits some upper bound (see [7]). Thus we define the conformal spectrum of (S n , [g 0 ]), where [g 0 ] is the class of metrics conformal to the round metric g 0 , by The theorem of Hersch was generalized in this framework in [4]. We have that where σ n is the volume of the unit n-dimensional sphere. We know almost nothing about λ c k (S n , [g 0 ]) for k ≥ 2. A lower bound was obtained by a method of conformal surgery in [3]. For all k, we have that Nadirashvili, Girouard and Polterovich conjectured in [5] that this inequality is an equality in all dimensions for k = 2, where the supremum is attained for the union of two identical spheres : In the way to this conjecture, the following theorem gives an "asymptotically sharp" upper bound : Theorem : Let n ≥ 2 and g ∈ [g 0 ] a metric on S n conformal to the round metric. Then where K n is a constant independant of g ∈ [g 0 ] given by Note that K 2 = 1, that 1 < K n ≤ 1.04 for all n ≥ 3 and that lim n→∞ K n = 1. The theorem is sharp in dimension 2 and was in fact already proved by Nadirashvili in [8]. In [5], Girouard, Nadirashvili and Polterovich established this inequality in odd dimensions.
We prove in this paper this theorem in all dimensions, unifying the previous proofs in dimension n = 2 and in odd dimensions and by the way extending it. The starting point of the proof is a construction, described in section 1 below, initiated by Nadirashvili [8] and used by Girouard, Nadirashvili and Polterovich [5] in odd dimension. However, our use of this construction differs from that of these two papers : we use the min-max characterisation of the second eigenvalue up to the end of the proof (see section 3), capitalizing on a new topological fact proved in section 2.

Construction of test functions
In this section, we describe the construction of Nadirashvili [8] (see also [5]) which is at the basis of our theorem as well as of the previous results. Let g be a metric on S n conformal to g 0 of volume 1. We denote by dv g the measure associated to g. We shall use in this paper the min-max characterization of the second eigenvalue of the Laplacian which tells us in particular that for all 2-dimensional subspaces E of functions in H 1 (S n ) with mean value 0. The aim is to find a suitable space E of test-functions such that (1) gives the estimate of the theorem.
On (S n , g 0 ), the eigenspace associated to λ 1 (S n , g 0 ) has dimension n + 1 : it is the set of linear forms of R n+1 written X s = (s, .) for s ∈ R n+1 . We will build E with these functions, and as Hersch did for λ 1 (S n , g), we proceed to a renormalisation of measures in order to keep the orthogonality to constants. For ξ ∈ B n+1 , we let d ξ : B n+1 → B n+1 be defined by which is a conformal transformation when restricted to the unit sphere.
We say that d ξ renormalizes a finite measure dν on the n-sphere if The Hersch lemma says that for all finite measures dν, such a ξ exists. Moreover it is unique and depends continuously on dν (the set of finite measures is considered as the topological dual of the continuous bounded functions) as proved in [5], Proposition 4.1.5. We call ξ the renormalization point of dν.
We also define families of measures parametrized by the set of caps of S n , denoted by C : We denote by dµ a the "lift" of the measure dv g by the cap a ∈ C : where a * = S n \ā and τ a is the conformal reflection with respect to the boundary circle of a, that is is the reflection of R n+1 with respect to the hyperplane orthogonal to p. Let ξ(a) be the renormalization point of dµ a . We set dν a = (d ξ(a) ) * dµ a . Thanks to this family of measures, we can define a new family of test functions orthogonal to the constants : By a Hölder inequality, the numerator of the Rayleigh quotient is less than a conformal invariant.
Let us define the multiplicity of a finite measure : Definition : The multiplicity of a finite measure dν on S n is the dimension of the eigenspace W associated to the maximal eigenvalue of the quadratic form : We say that dν is multiple if its multiplicity is greater than or equal to 2. Otherwise, we say that dν is simple.
As was noticed in [5], we know that if dv g is multiple, then we can choose (1). In this case, the theorem would be proved. In [5], it was proved that there necessarily exists such a multiple measure in odd dimensions (see below).
Let us now assume that all measures dv g and dν a , for a ∈ C, are simple. Up to a renormalisation and a rotation, we may assume that We denote by [s(a)] the unique direction of maximization of the quadratic form associated to dν a . With the parametrization (r, p) ∈ (−1, 1) × S n of C, the maps ξ : C → B n+1 and [s] : C → RP n are continuous. Moreover, one may prove that if r → −1, that is a → S n , we have :

Properties of the lift of the maximal direction
Let us study the maps ξ and [s] at the light of the links between a cap a ∈ C and its symmetrical cap a * = S n \ā. With the parameter (r, p) ∈ (−1, 1) × S n , notice that a * r,p = a −r,−p . and is an orthogonal map.
Proof : We set η = −τ a (−ξ). Let t ∈ S n , then . This is true for all t ∈ S n , and uniqueness of the renormalization point ensures that ξ * = η.
The same argument with the function Remark : Thanks to this claim 1, we can prove the theorem in odd dimensions. Indeed, when r → 1 that is a → {p}, we use (3) in order to obtain : (3). Therefore, following [5] in odd dimensions, the map [s] : [−1, 1] × S n → RP n defines a homotopy between the constant map [e 1 ] of degree 0 and φ(p) = R p [e 1 ] of degree 4. Thus, there is a contradiction and there exists a multiple measure among dv g and dν a for a ∈ C.
We do not prove that the assumption that all measures are simple lead to a contradiction. Indeed, it is not clear that in even dimensions, such a configuration can not happen. Instead, we look for suitable test functions like in Nadirashvili's proof in dimension 2 [8]. However, inspired by the method of [5], we use a topological argument to get symmetric properties of the lifts of the maximal directions.  Proof : We assume by contradiction that ǫ = 1. We set f (p) = s(0, p) for p ∈ S n . This function f is continuous on the sphere and satisfies ∀p ∈ S n , f (−p) = R p f (p) (4) Indeed, R a0,p = R p because τ a0,p = R p . Using claim 3 below, we know that such a map f can not have degree 0. However, the map s : [−1, 0] × S n → S n defines a homotopy between s 0 = f and s −1 = −e 1 of degree zero. Thus, there is a contradiction.
We have used the following topology result : Claim 3 : Let f : S n → S n a continuous map which satisfies (4). Then, if n is odd, deg(f ) = 1 and if n is even, deg(f ) ∈ 2Z + 1.
Proof : We first prove the claim for smooth functions which have a property of transversality (step 1) and we show that this case is generic (step 2).
Step 1 -Let f : S n → S n be a smooth function which satisfies (4). Let us assume that for all fix point x ∈ S n of f , T x f − I : T x S n → T x S n is an isomorphism. Then, if n is odd, deg(f ) = 1 and if n is even, deg(f ) ∈ 2Z + 1.

Proof of step 1 -Let F be defined by
We notice that if F never vanishes, F |F | defines a homotopy between f and σ, the antipodal map and deg(f ) = deg(σ) = (−1) n+1 . Now, F (x, t) = 0 if and only if t = 0 and x is a fix point of f and then, Thus, DF (x, 0) is an isomorphism, and 0 is a regular value. We write (x 1 , 0), · · · , (x r , 0) the regular points of F −1 (0). Let's approximate F by its differential in the neighborhood of its zeros. Let α > 0 and, set for 1 ≤ i ≤ r, φ i : B xi (α) → B 0 (α) ⊂ T xi S n the exponential chart at We define a cut-off function 0 ≤ ψ ≤ 1 such that ψ = 1 on One may choose α > 0 small enough so that G t s is well defined for all t ∈ (−α, α) \ {0}. Then, for 0 < t < α, G t 1 is homotopic to G t 0 = Ft |Ft| , so to f , and G −t 1 is homotopic to σ. We now write, for t ∈ (−α, α), g t = G t 1 .
Let us look at the behaviour of g t = Lt |Lt| in the balls B xi ( α 2 ) when t → 0. We recall that Therefore, the image I t xi = g t (B xi ( α 2 )) blows up to the half-sphere D xi = {x ∈ S n ; (x, x i ) > 0} when t → 0.
Thanks to (4), x is a fix point of f if and only if −x is a fix point too. Moreover, by differentiating (4) at a fix point x, we obtain Let's renumber the fix points x 1 , · · · , x k , −x 1 , · · · , −x k (with r = 2k), so that x 1 , · · · , x k are in a same half sphere D p = {(x, p) > 0}. We choose ǫ < α small enough so that k i=1 I ǫ xi has a non-empty interior I. Then, for z ∈ I, there is a unique point in . Then by definition of degree and homotopy, In odd dimensions, deg(f ) = deg(σ) = 1 and in even dimensions, deg(f ) ∈ 2Z + 1. This ends the proof of step 1.
Step 2 -Let f : S n → S n be a continuous map which satisfies (4). Then there exists a map, homotopic to f , which satisfies the assumptions of step 1.
Proof of step 2 -Denote by (e 0 , e 1 , · · · , e n ) the canonical basis of R n+1 and B α k ⊂ D e k = {(x, e k ) > 0} the ball centered at e k such that d(B α k , D −e k ) = α > 0. Choose α small enough so that Let ǫ > 0. We build by induction maps g k : S n → S n such that g 0 = f and, for 0 ≤ k ≤ n, By density of smooth maps S n → R n+1 , choose h k such that h k − g k C 0 < ǫ. Let 0 ≤ φ ≤ 1 be a smooth cut-off function such that φ = 1 on B 2α i and φ = 0 on S n \ B α i . We let g k+1 be defined, provided ǫ is small enough, by for x ∈ D e k . Therefore g = g n+1 is smooth, satisfies (4) and g − f C 0 < Cǫ. If ǫ is small enough, g is homotopic to f . Let's now tackle the transversality condition. We write g in the following way where X is a tangent vector field of the sphere and |X| 2 + λ 2 = 1. Then, g satisfies (4) if and only if X and λ are even maps. By differentiating these equalities at a fix point x (with λ(x) = 1 and X(x) = 0), one may find T x g − I = T x X. Then, T x g − I is an isomorphism for all fix points x if and only if X is transverse to the zero vector field. Then, one may build by induction, with Sard's theorem in n-dimensional charts on D e k , smooth tangent vector fields X k such that X 0 = X and for 0 ≤ k ≤ n : If ǫ is small enough, thenf is well defined, satisfies the assumptions of step 1 and is homotopic to f . This ends the proof of step 2.
These two steps clearly end the proof of the claim.

Choice of test functions
Thanks to claim 2, one may easily deduce that where we have set, for this section u a = u s(a) a . Let r ∈ (−1, 1). We look at the space E generated by φ = X e1 and ψ r = u ar,e 1 .
By the claim 4, we know that α r = − S n ψ r (∆ g φ) dv g S n |∇ g φ| n g dv g By continuity, (claim 4), there exists r ∈ (−1, 1) such that α r − 2 2 n β r = 0. As already said, this completes the proof of the theorem.