On the symmetric scalar curvature problem on S^n

On the Symmetric Scalar Curvature Problem on S n Antonio AMBROSETTI - Andrea MALCHIODI

∗

Abstract We discuss some existence results dealing with the Scalar Curvature problem on S n in the presence of various symmetries.

1

Introduction

The purpose of this paper is to discuss some new existence results on the Scalar Curvature problem dealing with symmetric curvatures close to a positive constant. The paper can be seen as a natural completion of the recent work [2], where the non-symmetric case is studied. The problem of prescribing the Scalar Curvature on the standard sphere (S n , g0 ), n ≥ 3, consists in finding a metric g conformal to g0 such that its scalar curvature is a prescribed function R. Letting g = u4/(n−2) g0 (here and always int his paper we take n ≥ 3) and taking R = 1 + εK0 , ε small, the Scalar Curvature problem amounts to finding positive solutions of

(1)

−4

n−1 ∆g0 u + u = (1 + εK0 ) u(n+2)/(n−2) , n−2

u ∈ H 1 (S n ).

The Scalar Curvature problem has been broadly investigated: see e.g. [3, 5, 11, 12, 2]. In particular, the two papers [11, 12] contain a wide and updated bibliography. The case in which the curvature R is invariant under the action of a group Σ has also been studied. When Σ acts on S n without fixed points, Hebey has shown in [10] that there is no topological obstruction for the existence. When n = 2, the case of an even R has been handled by Moser [13]. Escobar/Schoen [9] have extended [13] to any n ≥ 3 under a suitable flatness condition. The case that Σ has fixed points is also handled in [8, 10] under some quantitative conditions which relate the value of R at the fixed points q such that R0 (q) = 0. For other recent results on the symmetric curvature problem see also [6, 7]. Our main results are given in sections 5 and 6. In the former we discuss the case in which the fixed point set F = Fix(Σ) = {±p∗ }. In Theorem 5.1 we show that a symmetric solution of (1) exists provided that K0 (p∗ ) = K0 (−p∗ ); we can also handle some cases in which K0 (p∗ ) 6= K0 (−p∗ ). Another type of result is Theorem 5.3 where we require a quantitative condition involving K0 (±p∗ ) and the mean of K0 on S n . Theorems 5.1 and 5.3 extend the result of section 4 of [2], where the specific case of a rotational symmetry is discussed. Theorem 5.3 is also related to some results of [10], see Remark 6.3. As a byproduct of our arguments we also recover (without any flatness condition) the case of a free action discussed in [9, 13], see Theorem 5.4. In section 6 we handle the remaining case that Σ has an S n−k of fixed points. First, we make a kind of assumptions like the preceding ones and prove a result, Theorem 6.1, which can be seen as an extension of Theorem 5.3. ∗ SISSA, via Beirut 2-4, Trieste 34014 Italy. Supported by M.U.R.S.T. under the national project Variational methods and nonlinear differential equations.

1

A different set of hypotheses are made in Theorem 6.4. Roughly, we show that in the presence of a symmetry it suffices to consider K0 restricted to the fixed points of the action and to make on such a restriction the same type of hypotheses as Bahri-Coron [5] and Yanyan Li [11, 12], see also [2]. This is a new kind of result that improves those of [5, 11, 12, 2]. An interesting question to be pursued is to extend Theorem 6.4 to curvatures not necessarly close to a constant. The first ingredient of the proofs is a perturbation result in critical point theory [1]. Roughly, through a finite dimensional reduction, one seeks solutions of (1) by studying a suitable functional Γ on R+ × Rn . This approach has been used in [2] to face equation (1) and has the remarkable advantage that it gives rise to results in a greater generality and that the arguments of the proofs are rather simple. The abstract approach is recalled in Section 2. Actually, the problems we handle need some improvement of the abstract setting and a new result in this direction is given Section 3. The second step relies on the remark that the group action on S n induces a suitable symmetry on Γ. More precisely, we show in Section 4 that there exists a group τ , related to Σ, under which Γ is invariant. As a consequence, we are led to study a symmetric finite dimensional functional and this allows us to prove our existence results in a quite simple way. require some situation in Notation R+ stands for [0, +∞). If h : V → R is a smooth functional, Crit(h) denotes the set of its critical points: Crit(h) = {x ∈ V : ∇h(x) = 0}. If p = (p1 , . . . , pn , pn+1 ) ∈ S n , we set p = (p0 , pn+1 ) with p0 = (p1 , . . . , pn ), and p = (p1 , p00 ) with p00 = (p2 , . . . , pn ). PN = (0, . . . , 0, 1) and PS = (0, . . . , 0, −1) denote the North and South poles on S n respectively. E = D1,2 (Rn ). πn : S n 7→ Rn denotes the stereographic projection. R dV is the volume element of g0 and ωn = S n dV .

2

Some results from [1] and [2]

To keep our notation as close as possible to the one used in [2], we prefer to deal with the equation obtained from (1) by applying the stereographic projection π = πn . Setting K(x) = K0 (π −1 x), one is led to the problem (apart from an unimportant constant)

(2)

(

−∆u = (1 + εK(x)) u(n+2)/(n−2) u > 0, u ∈ E.

in Rn

The Functional Setting. Let us recall some results of [1], Section 2 and [2], Sections 2 and 3, to which we refer for more details. Let us consider the functional Z Z Z ∗ 1 1 ε 2 2∗ (3) fε (u) = |∇u| − ∗ |u| − ∗ K(x)|u|2 , 2∗ = 2n/(n − 2). 2 Rn 2 Rn 2 Rn The non trivial critical points of fε give rise to a solution of (2). Let
(2−n)/2 (n−2)/4 z0 (x) = Cn · 1 + |x|2 , Cn = [n(n − 2)] 2

denote the positive radially symmetric solution of the unperturbed problem −∆u = u(n+2)/(n−2)

(4)

u ∈ E.

and set Z = {zµ,ξ (x) = µ

2−n 2

µ > 0, ξ ∈ Rn }.

z0 ((x − ξ)/µ) ,

Every z ∈ Z is a critical point of the unperturbed functional f0 . Moreover D2 f0 is Fredholm of index zero and Ker D2 f0 = Tz Z, the tangent space to Z at z. According to Proposition 3.1, this allows us to define, for ε small, a smooth function wε (z) : Z 7→ (Tz Z)⊥ such that the critical points z ∗ ∈ Z of the functional ϕε : Z 7→ R,

ϕε (z) := fε (z + wε (z))

give rise to critical points uε = z ∗ + wε (z ∗ ) of fε . The study of ϕε is accomplished using the fact that

(5)

   (i) ϕε (z) = b −  

(ii) ϕ0ε (z)

ε 2∗ Γ(z)

+ o(ε),

= − 2ε∗ Γ0 (z) + o(ε),

where b = f0 (z) and Γ, in the present case, has the expression 1 : Z 2n/(n−2) Γ(µ, ξ) = K(µy + ξ)z0 (y)dy. Rn

Hereafter we will use the co-ordinate (µ, ξ) for points of Z. To be short, we will also write wε (µ, ξ) instead of wε (zµ,ξ ). It is convenient to recall a result which will be often used in the sequel. (AB) Let X ∗ = (µ∗ , ξ ∗ ) ∈ Crit (Γ) (with µ∗ > 0) be such that either (i) X ∗ is a proper local minimum or maximum of Γ, or (ii) the local degree of ∇Γ at X ∗ is different from 0. Then near uε = z(µ∗ , ξ ∗ ) + wε (µ∗ , ξ ∗ ) there is a critical point of fε which gives rise to a solution of (2).

Behaviour of Γ. Concerning the behaviour of Γ, let us recall the following facts, see [2]. First of all, Γ can be smoothly extended to µ = 0 by setting (6)

Γ(0, ξ) = lim Γ(µ, ξ) = c0 K(ξ), µ→0

c0 > 0.

Furthermore there results ∂µ Γ(0, ξ) = 0 and hence ξ is a critical point of K iff (0, ξ) is a critical point of Γ. Finally one has (7)

2 ∂µµ Γ(0, ξ) = c1 ∆K(ξ),

c1 > 0.

In the sequel we will use the same symbol for Γ and the above extension defined on all of R+ ×Rn . The context will make it clear which one we are dealing with. However, in order to apply (AB) to a critical point (µ, ξ) of this extended Γ, one must obviously have µ > 0. 1 The reader should remark that we are using a different notation from [2]: the present Γ equals 2∗ times the Γ in [2].

3

A result of [2]. In Section 5 we will improve the results of [2] in the case of symmetric curvatures. For brevity, we will only indicate how to handle the case covered in the first part of Corollary 3.9 of [2]. For the reader’s convenience, we report below the statement we are interested in, rephrased in the present setting. Theorem 2.1 Suppose that (K0 )

PN is a non-degenerate minimum for K0 on S n ;

(K1 )

Crit(K) is finite and all ξ ∈ Crit(K) are non degenerate;

(K2 )

∆K(ξ) 6= 0 for all ξ ∈ Crit(K) and X (−1)m 6= (−1)n .

(8)

∆K(ξ)<0

Then (2) has a solution. Here m = m(K, ξ) denotes the Morse index of ξ. Let us recall that a condition like (8) was first introduced in [5] . Remark 2.2 The condition (K0 ) corresponds to assumption (K1.a) of [2]. As we will show below, see (14), it implies that ∇Γ(X) · X < 0 for all |X| >> 1, which is nothing but Lemma 3.3 of [2]. Dependence of Γ on K0 . For future reference let us show how Γ depends upon K0 . By the change of variables x = π(p) one finds (up to a positive constant which is not important) ∗ dV = z02 dx and hence Z (9) Γ(µ, ξ) = K(µπ(p) + ξ)dV. Sn

In particular, we get: (10)

Z

Γ(1, 0) =

K0 dV

Sn

(11) (12)

Γ(0, 0) = ωn · K(0) = ωn · K0 (PS ) Γ(0, ξ) = ωn · K0 (π −1 (ξ)).

Moreover, letting pµ,ξ ∈ S n be such that π(pµ,ξ ) = µπ(p) + ξ, (9) becomes Z Γ(µ, ξ) = K0 (pµ,ξ )dV. Sn

Since pµ,ξ → PN as µ + |ξ| → ∞ we infer (13)

lim µ+|ξ|→∞

Γ(µ, ξ) = ωn · K0 (PN ).

Finally, we set X = (µ, ξ) and show, as anticipated in Remark 2.2, that there exists R > 0 such that (14)

∇Γ(X) · X < 0 ∀ |X| ≥ R, 4

provided (K0 ) holds. A straight calculation yields  Z  ∂K0 (p) 0 2 0 ∇Γ(X) · X = −pn+1 ∇p0 K0 (p) · p + |p | dV. ∂pn+1 Sn Above we have understood the dependence of p, p0 , pn+1 on (µ, ξ). Condition (K0 ) implies that the preceding integrand is negative provided pµ,ξ is close enough to PN . Since pµ,ξ → PN as µ2 + |ξ|2 → ∞, (14) follows. Further remarks. There is a natural isomorphism ι between E and H 1 (S n ) given by ι

u 7−→ u b,

u b(p) = u(π(p))/z0 (π(p)).

This map ι induces a correspondence between the fundamental solutions zµ,ξ of (4) and the fundamental solutions Uα of −4

(15)

n−1 ∆g0 v + v = v (n+2)/(n−2) , n−2

v ∈ H 1 (S n ).

n+1 Precisely, we choose to parametrize Uα with α = (α1 , α00 ) ∈ S+ := {α ∈ S n+1 : α1 > 0} in such a way that

(16)

n+1 α = αµ,ξ = π −1 (µ, ξ) ∈ S+ .

Uα = zbµ,ξ ,

If (α1 , α00 ) = (1, 0) then U1,0 (q) ≡ 1; otherwise Uα is characterized by its maximum Mα and by the point Qα = (0, (1 − α12 )−1/2 α00 ) where Mα is achieved. Moreover, setting  Z  ∗ 1 1 1 fbε (v) = |∇g0 v|2 + |v|2 − ∗ (1 + εK0 )v 2 dV, cn 2 2 Sn there results (17)

fε (u) = fbε (b u).

b Finally, as for the auxiliary functionals Γ, b Γ(α) =

Z

∗

Sn

K0 Uα2 dV,

one finds (18)

3

b µ,ξ ). Γ(µ, ξ) = Γ(α

A preliminary result

As we will see in Theorem 5.1-(a) below, there are cases in which (AB) cannot be used, and one needs to study ϕε . The main goal in this section is to discuss a result in this direction, proving Proposition 3.1 ϕε ∈ C 2 (Z, R) and there results  − n−2  = b · (1 + εK0 (PS )) 2 ,  limµ→0 ϕε (µ, 0) (19)   − n−2 limµ→+∞ ϕε (µ, 0) = b · (1 + εK0 (PN )) 2 . 5

Proof. First, a preliminary discussion on the definition of w is in order. As in [1], letting  0  fε (z + w) − αz˙ (20) H(α, w, z, ε) = . (w, z) ˙ one has H = 0 if and only if w ∈ (Tz )⊥ and ∇fε (z + w) ∈ Tz Z. The unknown function (α, w) is implicitly defined as the solution of the equation H = 0 and is found via the Implicit Function Theorem. Unlike [1], here it is convenient to use the contraction theorem as in [4]. There holds H(α, w, z, ε) = 0

⇔

H(0, 0, z, 0) +

where R(α, w, z, ε) = H(α, w, z, ε) − Rz,ε (α, w) = −

∂H (0, 0, z, 0)[α, w] + R(α, w, z, ε) = 0, ∂(α, w)

∂H ∂(α,w) (0, 0, z, 0)[α, w].



Setting

−1 ∂H (0, 0, z, 0) R(α, w, z, ε). ∂(α, w)

one finds that H(α, w, z, ε) = 0

⇔

(α, w) = Rz,ε (α, w).

The function Rz,ε turns out to be a contraction mapping from some ball Bρ(ε) into itself. In fact, −1  ∂H (0, 0, z, 0) has a norm smaller than some constant C0 , and if (α, w) ∈ the operator ∂(α,w) n+2

Bρ (0), then kRk ≤ C1 · (ε + ρ n−2 ), for some fixed constant C1 > 0. Hence if ρ(ε) = 2C0 C1 ε, for |ε| sufficiently small, Rz,ε maps Bρ(ε) into Bρ(ε) . Moreover, there holds   4 4 (21) kRz,ε (α0 , w0 ) − Rz,ε (α, w)k ≤ C2 ε + kwk n−2 + kw0 k n−2 · kw − w0 k, where also C2 is a fixed positive constant. Therefore, if |ε| is sufficiently small, Rz,ε is a contraction in Bρ(ε) . This proves the existence of (α, w) for |ε| sufficiently small, uniformly for z ∈ Z. Hence ϕε is well defined on Z and is clearly of class C 2 there. Next, let us consider the functional fεµ : E → R defined by setting Z Z ∗ 1 1 |∇u|2 − ∗ (1 + εK(µx))|u|2 (µ > 0). fεµ (u) = 2 Rn 2 Rn There results fεµ (u) = fε (µ−(n−2)/2 u e),

(µ > 0)

fεµ ,

where u e(x) = u(x/µ). Corresponding to the preceding arguments give rise to the existence of wεµ ∈ (Tz0 Z)⊥ such that ∇fεµ (z0 + wεµ ) ∈ Tz0 Z. Of course there results (22)

wεµ (x) = µ

n−2 2

wε (zµ,0 )(µx).

The functional fεµ can be trivially extended to µ = 0 by setting Z Z ∗ 1 1 fε0 = |∇u|2 − ∗ (1 + εK(0)) |u|2 . 2 Rn 2 Rn Next we show: 6

Lemma 3.2 Set tε = (1 + εK(0))−

n−2 4

, and wε0 = (tε − 1)z0 . Then there results

lim wεµ = wε0 = (tε − 1)z0 .

(23)

µ→0

Proof. ¿From elementary computations it follows that ∇fε0 (z0 + wε0 ) = 0 and that fε0 (z0 + n−2 wε0 ) = (1 + εK(0))− 2 . We denote by Rεµ the functions Rz,ε (see above) corresponding to fεµ and for z = z0 . The H¨ older inequality and the dominated convergence theorem yield Z 1/q ∗ (24) kRεµ (αε0 , wε0 ) − Rε0 (αε0 , wε0 )k ≤ εC3 |K(0) − K(µx)|q |wε0 |2 → 0 (µ → 0), Rn

where C3 > 0 and q = 2n/(n + 2) is the conjugate exponent of 2∗ . Furthermore, from (21), it follows that Rµ is a contraction in Bρ(ε) for ε sufficiently small. So we have k(αεµ , wεµ ) − (αε0 , wε0 )k = kRεµ (αεµ , wεµ ) − Rε0 (αε0 , wε0 )k
≤ C4 kRεµ (αεµ , wεµ ) − Rεµ (αε0 , wε0 )k + kRεµ (αε0 , wε0 ) − Rε0 (αε0 , wε0 )k . Using (24) and the fact that Rµ is a contraction (recall that both (αεµ , wεµ ) and (αε0 , wε0 ) belong to Bρ(ε) ), we deduce that wεµ → wε0 in E. Proof of Proposition 3.1 Completed. Let us prove the first of (19). Using (22) a straight calculation yields ϕε (µ, 0) = fε (zµ,0 + wε (µ, 0)) = fεµ (z0 + wεµ ).

(25) Furthermore

fεµ (z0 + wεµ )

= fεµ (z0 + wεµ ) − fε0 (z0 + wεµ ) + fε0 (z0 + wεµ ) Z ∗ 1 = ε (K(0) − K(µx))|z0 + wεµ |2 + fε0 (z0 + wεµ ). ∗ 2 Rn

Hence Lemma 3.2 implies (26)

fεµ (z0 + wεµ ) → fε0 (z0 + wε0 ) = f0 (tε z0 ) as µ → 0.

There results f0 (tε z0 )

Z Z ∗ 1 2 1 ∗ tε |∇z0 |2 − ∗ t2ε (1 + εK(0)) |z0 |2 2 2 n Rn  RZ  Z ∗ 1 1 2 2 2 −2 2∗ = tε |∇z0 | − ∗ tε (1 + εK(0)) |z0 | . 2 Rn 2 Rn =

Then the definition of tε yields (27)

f0 (tε z0 ) = t2ε

 Z  Z ∗ 1 1 |∇z0 |2 − ∗ |z0 |2 = b · t2ε . 2 Rn 2 Rn

Putting together (25), (26) and (27) we finally find lim ϕε (µ, 0) = fε0 (z0 + wε0 ) = b · t2ε .

µ→0

This proves the first of (19). The second of (19) can be proved repeating the preceding arguments, with obvious changes. Remark 3.3 More generally, one can show that limµ→0 ϕε (µ, ξ) = b(1 + εK(ξ))−(n−2)/2 . 7

4

About Symmetry.

We will suppose there is a group of isometries Σ ⊂ O(n + 1) which acts on S n and that K0 is Σ-invariant: K0 (σp) = K0 (p) for all σ ∈ Σ. These conditions will always be assumed in the sequel and understood. The action of Σ induces in a natural way an action σ 0 on Rn as well as an action τ on R+ ×Rn , under which K, resp. ϕε and Γ, are invariant. The former is defined by σ 0 = πn ◦ σ ◦ πn−1 ; let Σ0 denote the corresponding group. Plainly, K(σ 0 (x)) = K(x) for all x ∈ Rn . The latter map is obtained as follows: τ : R+ × Rn (µ, ξ)

π −1

−→ S n+1 −→ P

σ e

−→ −→

S n+1 σ e(P )

π

−→ R+ × Rn −→ τ (µ, ξ),

where the map σ e : S n+1 → S n+1 is given by Id × σ, namely is defined on P ∈ S n+1 by setting P = (P1 , P 00 ) 7−→ (P1 , σ(P 00 )).

Above we understand that the map σ has been extended to Rn+1 by homogeneity. With an abuse of notation we will also denote the corresponding group by τ . Examples. (σ1 ) If σ : (p0 , pn+1 ) 7→ (−p0 , pn+1 ) then σ 0 : x 7→ −x and τ : (µ, ξ) 7→ (µ, −ξ); (σ2 ) If σ : (p0 , pn+1 ) 7→ (p0 , −pn+1 ) then σ 0 : x 7→ x/|x|2 and   µ ξ , ; τ : (µ, ξ) 7→ µ2 + |ξ|2 µ2 + |ξ|2 (σ3 ) If σ : (p0 , pn+1 ) 7→ (−p0 , −pn+1 ) then σ 0 : x 7→ −x/|x|2 and   µ −ξ τ : (µ, ξ) 7→ , . µ2 + |ξ|2 µ2 + |ξ|2 We will show, see Proposition 4.2, that Γ is τ -invariant. It is also convenient to indicate the relationships between the fixed point sets Fix(Σ), Fix(Σ0 ) and Fix(τ ). First of all one has (28)

ξ ∈ Fix(Σ0 ) ⇐⇒ π −1 ξ ∈ Fix(Σ) \ {PN }.

Next, let Cξ denote the half circle in R+ × Rn through (1, 0), (0, ξ) and (0, −ξ/|ξ|2 ) if ξ 6= 0; otherwise we set C0 = {ξ = 0}. Then one finds (29)

[

Fix(τ ) =

Cξ .

ξ∈Fix

(Σ0 )

Remark 4.1 It is worth pointing out that, in any case, the set Fix(τ ) contains the point (1, 0). We end this section with a Proposition that will be constantly used in the sequel. Proposition 4.2 If K0 is Σ-invariant, then Γ and ϕε are invariant under the action τ . Therefore, at any X ∈ Fix(τ ), the vectors ∇Γ(X), ∇ϕε are tangent to Fix(τ ) (if X is an isolated point of Fix(τ ) we mean that ∇Γ(X) = ∇ϕε (X) = 0). 8

For the proof, some preliminary lemmas are in order. b b σ (α)). Lemma 4.3 There results Γ(α) = Γ(e b and since K0 is σ-invariant we infer Proof. ¿From the definition of Γ Z Z ∗ ∗ b Γ(α) = K0 (p)Uα2 (p)dV = K0 (σ(p))Uα2 (p)dV Sn

Sn

Changing the variable q = σ(p) we find Z b Γ(α) =

Sn

∗

K0 (q)Uα2 (σ −1 (q))dV.

We claim: Uα (σ −1 (q)) = Uσe(α) (q).

(30)

If (30) holds the Lemma follows immediately, because Z Z ∗ 2∗ −1 b σ (α)). b Γ(α) = K0 (q)Uα (σ (q))dV = K0 (q)Uσe2(α) (q)dV = Γ(e Sn

Sn

It remains to prove the claim. As anticipated in Section 2, when α 6= (1, 0) the functions Uα are characterized by their maxima Mα = max Uα and by the points where Mα is achieved, namely Qα = (0, sα00 ) where α = (α1 , α00 ) and s = (1 − α12 )−1/2 . It is easy to see that (31)

Mα = Mβ

⇐⇒

α1 = β1 .

Observe that the maximum Mα0 of Uα (σ −1 (q)) is equal to Mα and is achieved at the point Q0α = (0, σ(sα00 )) = (0, sσ(α00 )). On the other hand, Uα (σ −1 (q)) is a solution of (15) and thus there exists β ∈ S n+1 such that Uα (σ −1 (q)) = Uβ (q). ¿From Mβ = Mα0 = Mα and (31) it follows that α1 = β1 . Moreover, Q0α = Qβ implies (0, sσ(α00 )) = (0, s0 β 00 ), with s0 = (1−β12 )−1/2 , namely sσ(α00 ) = s0 β 00 . Since α1 = β1 then s = s0 and thus β 00 = σ(α00 ). In conclusion, taking into account the definition of σ e, there results β = (α1 , σ(α00 )) = σ e(α). This shows the claim (30) is true and completes the proof of the lemma. Let σ ∗ (v) be the pull-back of v ∈ H 1 (S n ) through σ, σ ∗ (v)(p) = v(σ(p)),

p ∈ S n , σ ∈ Σ.

It is easy to see that fbε (v) = fε (ι−1 v), see (17), is invariant through σ ∗ . Define Aσ : E → E by setting Aσ = ι−1 ◦ σ ∗ ◦ ι. Since ι and σ ∗ are linear isometries, also Aσ does. Moreover, the functional fε is invariant through the group AΣ = ∪σ∈Σ Aσ , because there results fε (u) ≡ fbε (ιu) = fbε (σ ∗ ιu) ≡ fε (ι−1 σ ∗ ιu) ≡ fε (Aσ u).

We also remark that from the definition of τ (µ, ξ), there results Aσ zµ,ξ = zτ (µ,ξ) . Lemma 4.4 There results Aσ wε (µ, ξ) = wε (τ (µ, ξ)). 9

Proof. According to the definition of w, it suffices to prove:
⊥ (i) Aσ wε (µ, ξ) ∈ Tzτ (µ,ξ) Z , (ii) fε0 (zτ (µ,ξ) + Aσ wε (µ, ξ)) ∈ Tzτ (µ,ξ) Z. As for (i), it is an easy consequence of the fact that Aσ is an isometry. Actually there results Tzτ (µ,ξ) Z = Aσ (Tzµ,ξ Z). To show that (ii) holds we argue as follows. Since Aσ is linear and fε is Aσ -invariant, one has 1 1 [fε (Aσ u + tAσ v) − fε (Aσ u)] = [fε (u + tv) − fε (u)] , u, v ∈ E, t ∈ R. t t This implies fε0 (Aσ u)[Aσ v] = fε0 (u)[v], u, v ∈ E. Hence fε0 (zτ (µ,ξ) +Aσ wε (µ, ξ)), which equals fε0 (Aσ (zµ,ξ +wε (µ, ξ))), belongs to Tzτ (µ,ξ) Z because, once again, Aσ is an isometry. This proves (ii) and the Lemma follows. Proof of Proposition 4.2. 1) Let us first show that Γ is τ -invariant. From (18) and Lemma 4.3, there results (32) Using (16) it follows that

b µ,ξ ) = Γ(e b σ (α Γ(µ, ξ) = Γ(α

mu,ξ )).

b σ (αµ,ξ )) = Γ(e b σ π −1 (µ, ξ)). Γ(e

Using again (18) and recalling the definition of τ , we infer

b σ π −1 (µ, ξ)) = Γ(πe Γ(e σ π −1 (µ, ξ)) = Γ(τ (µ, ξ)).

(33)

Putting together (32) and (33) we deduce that

Γ(µ, ξ) = Γ(τ (µ, ξ)), namely that Γ is τ -invariant. 2) Since fε is Aσ -invariant and Aσ is linear, we find fε (zµ,ξ + wε (µ, ξ)) = fε (Aσ (zµ,ξ + wε (µ, ξ))) = fε (Aσ zµ,ξ + Aσ wε (µ, ξ)) = fε (zτ (µ,ξ) + Aσ wε (µ, ξ)).
Using Lemma 4.4 we infer that fε (zµ,ξ + wε (µ, ξ)) = fε zτ (µ,ξ) + wε (τ (µ, ξ)) , namely that ϕε (µ, ξ) = ϕε (τ (µ, ξ)). This completes the proof.

5

Main existence results I

We will distinguish several cases depending on the nature of the set F := Fix (Σ). Let us recall that, since Σ ⊂ O(n), then either F = ∅, or F is apair of antipodal points, or else F = S n−k for some 0 < k < n. In this section we deal with the case that F is finite. The first two results deal with the case that F consists of two antipodal points. Theorem 5.1 Suppose there exists p∗ ∈ S n such that F = {p∗ , −p∗ }. Then (1) has a solution provided one of the following two conditions hold: (a)

K0 (p∗ ) = K0 (−p∗ );

(b)

K0 (p∗ ) < K0 (−p∗ ), resp. K0 (p∗ ) > K0 (−p∗ ), and ∆K0 (p∗ ) < 0, resp. ∆K0 (p∗ ) > 0. 10

Proof. We can take PN = p∗ and PS = −p∗ . Then (28)-(29) immediately yield that Fix(τ ) = {(µ, 0) : µ ∈ R+ }. From Proposition 4.2 it follows ∀ µ ∈ R+ .

∇ξ ϕε (µ, 0) = 0

As a consequence, it suffices to study µ 7→ ϕε (µ, 0). From K0 (PN ) = K0 (PS ), Proposition 3.1 implies that limµ→0 ϕε (µ, 0) = limµ→+∞ ϕε (µ, 0). Then, the real valued function µ 7→ ϕε (µ, 0) achieves the minimum or maximum at some µ0 > 0 (unless it is identically constant). In any case, we infer that ϕε has a critical point at some (µ, 0) with µ > 0 and (a) follows. (b) Let, for example, K0 (PS ) < K0 (PN ) and ∆K0 (PS ) < 0. In such a case (7) implies that 2 ∂µµ Γ(0, 0) < 0. On the other hand, as before, the inequality K0 (PS ) < K0 (PN ) jointly (11) and (13) implies that Γ(0, 0) < lim Γ(µ, 0). µ→+∞

Then Γ(µ, 0) achieves the minimum at some (µ∗ , 0) with µ∗ > 0, and the result follows from (AB). Remark 5.2 The solution we found is, by construction, a symmetric one. The same is true for the following Theorems. Instead of the condition K0 (p∗ ) = K0 (−p∗ ) we can use an assumption like that introduced in [10]. Theorem 5.3 Let p∗ ∈ S n be such that F = {p∗ , −p∗ } and suppose that either Z (H1 ) ωn K0 (±p∗ ) < K0 dV, Sn

or ∗

(H2 )

ωn K0 (±p ) >

Z

K0 dV.

Sn

Then (1) has a solution. Proof. As in the preceding Theorem, one is led to study Γ(µ, 0). Recall that (10) yields Z Γ(1, 0) = K0 dV, Sn

while (11)-(13) yield Γ(0, 0) = ωn · K0 (−p∗ ),

lim Γ(µ, 0) = ωn · K0 (p∗ ).

µ→∞

Then condition (H1 ) implies that Γ achieves the maximum in a point (µ∗ , 0) with µ∗ > 0. Applying (AB) the result follows. If (H2 ) holds, the preceding arguments can be repeated taking the minimum of Γ instead of the maximum. For completeness, let us consider the case in which the action of Σ is free and we recover, for curvatures close to a positive constant, the following known result. Theorem 5.4 Suppose that F = ∅. Then (1) has a solution. Proof. If F = ∅, then Fix(τ ) = {(1, 0)}, see Remark 4.1, and Proposition 4.2 implies that ∇ϕε (1, 0) = 0.

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6

Main existence results II

The results of this section deal with the case that F consists of an S n−k for some 0 < k < n. The first one is the counterpart of Theorem 5.3. Theorem 6.1 Let F = S n−k and let us suppose that either Z 0 (H1 ) ωn K0 (p) < K0 dV, ∀p ∈ S n−k , Sn

or (H20 )

ωn K0 (p) >

Z

K0 dV,

∀p ∈ S n−k .

Sn

Then (1) has a solution. Proof. We can take PN in such a way that F = {p ∈ S n : pn−k+2 = . . . = pn+1 = 0}. Now we find that Fix(Σ0 ) = Sn−k = {ξ ∈ Rn : |ξ| = 1, ξn−k+2 = . . . = ξn = 0} and Fix(τ ) =

[

Cξ .

ξ∈Sn−k

Repeating the arguments carried out in the proof of Theorem 5.3, one infers that the maximum, resp. minimum, of Γ on Fix(τ ) is achieved at some (µ∗ , ξ ∗ ) such that µ∗ > 0, and the result follows. Remark 6.2 (a) The arguments of Theorem 5.1-(a) jointly with Remark 3.3 can be used to handle the case that K0 = const. on F . (b) As in Theorem 5.1-(b) we can show that a solution exists provided there exists p¯ ∈ F such that K0 (¯ p) = maxF K0 , resp minF K0 , and ∆K0 (¯ p) > 0, resp. < 0. This improves, in the perturbative case, a result of [8]. Remark 6.3 Theorem 1 (and Corollary 1) of [10] deal with a general curvature function R, invariant for the action of a finite group G of isometries on S n . A solution of (1) is found under the assumption that Z 2/(n−2) (H3 ) ωn R(q) ≤ (card O(q)) · R dV, ∀ q ∈ Crit(R). Sn

Here O(q) is the orbit of q. When R = 1 + εK0 and q is not fixed (namely card O(q) ≥ 2) then (H3 ) is automatically verified. If q is a fixed point, as in our Theorems 5.3 and 6.1, our condition (H1 ), resp. (H10 ) is equivalent (apart from the fact that we require a strict inequality) to (H3 ). The new feature of our result is that we also allow the reversed inequality (H2 ) or (H20 ). This case cannot be handled by the approach of [10] because in that paper the solutions are sought as symmetric minima of the Sobolev quotient and have Morse index = 1. On the contrary, when (H2 ) or (H20 ) holds, the solution we find corresponds to a minimum of Γ. Since fε = f0 − εG, one can show that uε , as a critical point of fε , in general has Morse index greater than 1. Concerning the Morse index of the solutions of (2), see e.g. Remark 3.10-(iii) of [2].

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Our last result is related to Theorem 2.1 and hence we will make assumptions on K directly. Here we will take PN to belong to the fixed point set F . Precisely, we suppose there exists k, 1 ≤ k ≤ n, such that F = {p ∈ S n , p1 = . . . = pk = 0}.

(34)

An example of an action satisfying (34) is given by (p1 , . . . , pk , pk+1 , . . . , pn+1 ) 7→ (−p1 , . . . , −pk , pk+1 , . . . , pn+1 ). According to (29), if (34) holds then there results Fix (τ ) = {(µ, ξ) ∈ R+ × Rn : ξ1 = . . . = ξk = 0}. If we set Vn−k = {ξ ∈ Rn : ξ1 = . . . = ξk = 0} we can write Fix(τ ) = R+ × Vn−k . We will assume that K satisfies: (K3 )

Yk :=Crit(K) ∩ Vn−k is finite and all ξ ∈ Yk are non-degenerate for K|Vn−k ;

(K4 )

∆K(ξ) 6= 0 for all ξ ∈ Yk and X

˜ (−1)m 6= (−1)n−k .

ξ∈Yk , ∆K(ξ)<0

Here m ˜ denotes the Morse index of ξ as a critical point of K constrained on Vn−k . Let us point out that, since ∇K0 is tangent to S n−k and achieves the maximum and the minimum there, then K|Vn−k has its maximum or the minimum on Vn−k . As a consequence Yk 6= ∅. Theorem 6.4 Let Σ satisfies (34) and let (K0 ) hold. Moreover, suppose that K satisfies conditions (K3 − K4 ). Then (1) has a solution. Proof. Arguing as before, one finds that Fix(τ ) = R+ × Vn−k implies ∇ξ Γ(µ, ξ) = η ∈ Vn−k

∀ (µ, ξ) ∈ R+ × Vn−k .

Thus it suffices to seek the critical points of the functional Γ constrained on R × Vn−k . Then we can apply the results of [2] with n − k instead of n and the Theorem follows. Remark 6.5 (1) If k = n we recover Theorem 5.1. When k = 0 we would have the trivial group Σ = {e} and hence there is no symmetry. In that case Theorem 6.4 is nothing but Theorem 2.1 recalled in Section 2. When 1 ≤ k < n the new feature of our result is that we do not make any assumption on the critical points that K possesses off of Vn−k . (2) The preceding proof makes it clear that we can extend all the results of [2] to the case of curvatures which are invariant under the action of a group Σ satisfying (34).

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