Scaler Curvature Equation

on Sn

Min Ji

Academy of Mathematics and System Sciences

Chinese Academy of Sciences

December 2008

1. Introduction

Let Sn = x ∈ Rn+1 : |x| = 1, n ≥ 3,

g0 = dx21 + dx2

2 + · · · + dx2n+1.

• Scalar Curvature Equation:

−cn u + R0u = R un+2n−2

u > 0,on Sn (1)

where

cn =4(n − 1)

(n − 2), R0 = n(n − 1),

R : Sn → R1 is a given positive, smooth function.

• Kazdan-Warner Problem: Given R : Sn → R1, find metric g

s.t.

– R = scalar curvature of g;

– g is pointwise conformal to g0.

• Equivalent Problem: Write

g = u4

n−2 g0, u > 0.

Then geometric problem ⇐⇒ solvability of (1).

• Remarks:

a) When n = 2, it is called the Nirenberg Problem which involves a

different equation.

b) Difficulties in two aspects

1) analysis: not compact, no minimun if R 6=constant;

2) topology: Kazdan-Warner necessary condition

c) Many works: Morser, Hong, Chen-Ding, Chang-Yang, Han,

Chen-Li, Chang-Liu, Ma, Xu-Yang, Escobar-Schoen, Hebey, Chen,

Zhang, Chang-Gursky-Yang, Y.Y. Li, Ambrosetti-Malchiodi,

Chen-Lin, Lin, Cao ...

• Minimizers in symmetric cases

Let H be a subgroup of the orthogonal transformations on Rn+1.

Let R be H-symmetric.

Consider

XH = u ∈ H1 : u(x) = u(hx), ∀h ∈ H.

Minimize the functional in XH

n = 2: Morser, Hong, Chen-Ding, · · ·

n ≥ 3: Escobar-Schoen, Hebey, Chen, · · ·

under some additional conditions

• Bahri-Coron Theorem (1991): Let n = 3.

Assumptions:

a) R is a Morse function;

b) |∇R| + |R| 6= 0 on Sn;

c) Index condition:

∑

∇R(x)=0,R(x)<0

(−1)ind(x) 6= (−1)n,

where ind(x)= Morse index.

Conclusion: Solution of (1) exists.

• Remark:

When n = 2, same result was given by Chang-Yang (1987).

Methods: Variational method + Minimax procedure.

• Generalized Bahri-Coron Theorem (Li, 1995):

Assumptions:

a) At every critical point p, under a geodesic normal coordinate

system centered at p,

R(y) = R(0) +n

∑

j=1

bj |yj |α + R(y), |y| ≪ 1,

where α ∈ (n − 2, n), bj 6= 0(∀j),∑n

j=1 bj 6= 0, and, R(y) = o(|y|α).

b) Index condition:∑

∇R(p)=0,Pn

j=1 bj<0

(−1)i(p) 6= (−1)n,

where i(p) = ♯bj : bj < 0.

Conclusion: Solutions of (1) exist.

• Remark: Allow α ∈ (1, n) if R ≈ R0.

Topological degree approach:

— Two steps:

1) Perturbation problem: R ≈ R0.

2) Deformation problem:

Rµ = µR + (1 − µ)R0, ∀ ǫ ≤ µ ≤ 1, (ǫ << 1).

Assume the a prior estimate

0 < C(R, n, ǫ)−1 ≤ uµ ≤ C(R, n, ǫ)

=⇒The Leray-Schauder degree associated with R1 = R and

Rǫ ≈ R0 are the same.

Perturbation Result:

— Chang-Yang (1991, 1993), Chang-Gursky-Yang(1993), Li (1995)

Assumptions:

1) ‖R − R0‖ ≤ ǫn;

2) Li’s uniform condition on R.

Conclusion: If deg(Gt, B, 0) 6= 0 as t ≫ 1, then solution of (1)

exists. Here

B = x ∈ Rn+1 : |x|2 = 1

the family of maps

Gt : Sn → Rn+1 (t ≥ 1) −−− Chang-Yang Map

Leray-Schauder degree = (−1)n deg(Gt, B, 0)

• Chang-Yang Map:

— the family of maps Gt : Sn → Rn+1, t ≥ 1

Gt(P ) =

∫

Sn

(

R φP,t

)

(x)x dx, ∀P ∈ Sn, t ≥ 1,

where φP,t : Sn → Sn is the conformal transformation, generated

by y 7→ ty (y : Sn \ P → Rn, the stereographic projection

when P is viewed as the north pole of Sn)

• Li’s uniform condition:

∃t0 > 0 and ω(t) with limt→∞ ω(t) = 0, s.t.

ω(t)|Gt(P )| ≥ ‖R φP,t − R(P )‖L2 ‖R φP,t − R(P )‖L2n/(n+2)

∀ t ≥ t0 and P ∈ Sn.

• Remarks:

a) Li’s uniform condition is verified for those special forms of R.

Especially, all critical points need to be non-degenerate;

b) Index condition ⇐⇒ deg(Gt, B, 0) 6= 0;

c) flatness condition: α ∈ (n − 2, n) =⇒ a priori estimate;

d) R is not allowed to be Cα+1 when α is odd.

• Questions:

Q1) How to verify Li’s uniform condition when R admits some

degenerate, or infinitely many, critical points?

Q2) How to verify the degree condition deg(Gt, B, 0) 6= 0 when R is

neither a Morse function nor of a special form?

Ji, Scaler Curvature Equation on Sn, Part I, · · ·

• Main Idea:

a) Establish asymptotic formulas for both Gt as t ≫ 1, and

‖R φP,t − R(P )‖L2 , ∀P ∈ Sn, as t ≫ 1.

(use to control both sides of Li’s condition)

b) Construct simple maps, G(β)t , having the same degree as Gt.

• Consequences:

1) Using these two asymptotic formulas, we can easily verify Li’s

uniform condition, by posing explicit conditions on R, without

assuming isolation and non-degeneracy of its critical points.

2) Leading to existence results using degree condition

deg(G(β)t , B, 0) 6= 0.

(the degree of G(β)t is easier to compute due to its simplicity)

The maps G(β)t :

• Definition: Let 2 ≤ β ≤ 2[n2 ] , an integer. Define, ∀x ∈ Sn,

G(β)t (x) : =

∇Re · ∇x − (Re)x when β is even,

∇Ro · ∇x − (Re)x when β is odd,

Re : =

R when β ≤ 3,

R + 1t2R + · · · + 1

t2(k−1)k−1R when β ≥ 4,

Ro : = Re +1

t2kkR, k = [

β

2].

• Special Cases:

— β = 2: G(2)t = ∇R · ∇x − (R)x := G independent of t;

— β = 3: G(3)t = G + 1

t2∇R · ∇x;

— β = 4: G(4)t = G + 1

t2(∇R · ∇x − (2R) x).

2. Analytical Characterizations

Ji, Scaler Curvature Equation on Sn, Part II, · · ·

Discuss

deg(G(β)t , B, 0) 6= 0 for an integer β

Let R be H-symmetric.

1) Existence of H-symmetric solutions:

deg(G(β)t , BH , 0) 6= 0

where

BH = x ∈ Rn+1 : |x|2 = 1, hx = x, ∀h ∈ H.

2) Existence of sub-H-symmetric solutions:

• It is possible that (1) do admit solutions, but does not admit

H-symmetric solutions(Chen-Li’s example). Necessarily

deg(G(β)t , BH , 0)= 0.

• If deg(G(β)t , BH , 0)= 0, find a subgroup Q ⊂ H, s.t.

deg(G(β)t , BQ, 0) 6= 0 (as t ≫ 1) ⇒ ∃Q-symmetric solution .

• General maps

F : Sm → Rm+1, defined by

F (x) = ∇f1(x) · ∇x + f2(x)x, x ∈ Sm,

where 0 ≤ m ≤ n, f1 ∈ C1(Sm), f2 ∈ C0(Sm), satisfying

|∇f1| + |f2| 6= 0, on Sm.

Taking m = dim(SH), SH = ∂BH ,

f1 =k

∑

i=0

1

t2iiR

∣

∣

∣

SH

, f2 = −l

∑

i=1

1

t2i−2iR

∣

∣

∣

SH

, t ≥ t0,

where 0 = 1, k = [β−12 ], l = [β

2 ].

3) Extension to cases where R is H-symmetric-like provided that

deg(G(β)t , B, 0) 6= 0 in H-symmetric cases.

• H-symmetric-like property:

∇f1(x) · h−1∇f1(hx) + f2(x)f2(hx) ≥ 0 x ∈ Sn, h ∈ H.

Remarks:

When R is H-symmetric, then

• f1, f2 are H-symmetric

left terms = |∇f1(x)|2 + |f2(x)|2 > 0.

• G(β)t is equivariant w.r.t. H

⇒ the degree has ”dimension reduction” property:

deg(G(β)t , B, 0) ↔ deg(G

(β)t , BH , 0)

DEGREE FORMULAS for

F (x) = ∇f1(x) · ∇x + f2(x)x, x ∈ Sm

Let N = (0, · · · , 0, 1) be the north pole of Sm.

Identify Sm\N with Rm via the stereographic projection:

yi(x) =xi

1 − xm+1, 1 ≤ i ≤ m, x ∈ Sm.

Without loss of generality,

∇f1(N) = 0.

Let f∗i : Rm → R1 (i = 1, 2), induced by fi, i.e.,

f∗i (y) := fi(x(y)), y ∈ Rm,

Ω− = y : f∗2 (y) < 0, Ω+ = y : f∗

2 (y) > 0.

•Theorem 1.

a) If m = 0, then

deg(F, B1, 0) =

±1, as f2(N)f2(−N) > 0;

0, as f2(N)f2(−N) < 0.

b) If m ≥ 1, then

deg(F, Bm+1, 0)

=

1 − deg(∇f∗1 , Ω−, 0), as f2(N) > 0;

(−1)m+1(

1 − (−1)m deg(∇f∗1 , Ω+, 0)

)

, as f2(N) < 0.

where ∇ is the gradient on the Euclidean spaces.

3. Existence results:

Case 1 dim(SH) = 0;

Case 2 dim(SH) = m ≥ 1: Denote

Σ := x ∈ Sn : ∇R(x) = R(x) = 0, ΣH := Σ⋂

SH ,

K = x ∈ Sn : ∇R(x) = 0, KH := K⋂

SH .

• Corollary (β = 2). Let R ∈ C2,γ .

Assumptions:

ΣH = ∅, and

deg(∇R, R < 0 ∩ Rm, 0) 6= (−1)m as R(N) > 0

deg(∇R, R > 0 ∩ Rm, 0) 6= 1 as R(N) < 0,

Conclusion: H-symmetric solution of (1) exists.

• Remark: R(N) 6= 0 (N ∈ KH).

• Cases where ΣH 6= ∅ are allowed

Di =

∇m, if i = 2m + 1 (m ≥ 0),

m, if i = 2m (m ≥ 1).

L(p) = mini : DiR(p) 6= 0, ∀p ∈ KH

Condition O(L)

—involves mainly conditions on DiR near p for odd i ≤ L.

—Example:

The condition O(3) at p : for some ǫ > 0

∇R · ∇R ≥ −(1 − ǫ)|∇R||∇R|, near p.

Theorem 2. Let R ∈ Cβ,γ .

Assumptions:

a) the condition O(L) at any p ∈ ΣH , w.r.t. H, with L = L(p) ≤ β;

b) E+ (resp. E−) is empty, or is a finite set consisting of isolated

critical points of R|SH and

∑

p∈E+

Indp ∇(R|SH ) 6= 1 ( resp.∑

p∈E−

Indp ∇(R|SH ) 6= (−1)m).

where

E− = p ∈ KH : L = L(p) is even, DLR(p) < 0,

E+ = p ∈ KH : L = L(p) is even, DLR(p) > 0.

Conclusion: H-symmetric solution of (1) exists.

• Remarks:

a) Critical points p of odd L(p) make no contribution to the index

counting formula.

b) Isolation of critical points are only assumed either in E+ or E−.

c) E+ and E− are open sets. So isolation of critical points does not

imply the finiteness of these sets in general.

(in fact this is true if Indp ∇R 6= 0 is further assumed ∀p ∈ E+

(resp. ∀p ∈ E−)).

d) If p is a non-degenerate critical point of R|SH in the sense of

Morse, then Indp ∇R|SH = (−1)ind(p), ind(p) =Morse index of

R|SH at p .

Corollary(β = 2, 3). Let n ≥ 4.

Assumptions:

a) When ΣH 6= ∅, assume ∇R 6= 0 on ΣH , and for pj ⊂ SH ,

pj → p ∈ ΣH , ∃ǫ > 0 s.t.

∇R · ∇R ≥ −(1 − ǫ)|∇R||∇R| at pj .

b) One of the following holds:

(1) R ≥ 0 on SH ;

(2) R ≤ 0 on SH ;

(3) Critical points of R|SH in R < 0 ∩ SH are non-degenerate

∑

x∈KH ,R(x)<0

(−1)Ind(x) 6= (−1)m;

(4) Critical points of R|SH in R > 0 ∩ SH are non-degenerate

and∑

x∈KH ,R(x)>0

(−1)Ind(x) 6= 1.

Conclusion: H-symmetric solution of (1) exists.

• Remarks.

a) This extends Bahri-Coron Theorem (which assumes n = 3,

H = I, Σ = ∅; and R is a Morse function).

b) This extends result of Ambrosetti-Malchiodi (2001), which

assumes ΣH = ∅; and R|SH is a Morse function

• Special form:

(H = I)At each critical point p, under the stereographic

coordinates centered at p,

R(y) = R(0) +n

∑

i=1

biyαii + o(

∑

i

|biyi|αi), |y| ≪ 1.

• Remark:

Comparing with that considered by Li, this is more general by

— allowing some of the bi’s to vanish;

— allowing the power αi’s to be different;

— allowing R to be smooth (it involves powers of yi rather than

those of |yi| )

• Corollary. Let R ∈ C2[ n2 ],γ and denote

α = minαi : bi 6= 0,

± = p ∈ K : α is even, ±∑

αi=α bi > 0,

E = p ∈ K : all αi are even.

Assumptions:

a) For each p ∈ K, α ≤ 2[n2 ] is either odd, or even with

∑

αi=α bi 6= 0.

b) + (resp. − ) is finite, and∑

p∈E∩+

(−1)i(p) 6= 1 (resp.∑

p∈E∩−

(−1)i(p) 6= (−1)n).

Conclusion: Solution of (1) exists.

• Remark: Critical points p with some odd αi(p) make no

contribution to the index counting formula.

Existence of sub-H-symmetric solutions

Typical case: axisymmetric case

R(x) = ζ(xn+1) for some ζ : [−1, 1] → R1

— symmetric w.r.t. H = O(n)

— Xu-Yang (n=2, 1993), Chen-Li(n ≥ 3, 2001):

Under a non-degenerate condition,

not monotone ⇔ existence (no any symmetry)

— Observation

deg(G, B, 0) 6= 0 ⇔

R(N)R(−N) > 0, if n is even.

R is not monotone, if n is odd.

Our Conclusion:

not monotone ⇐⇒ deg(G(β)t , BQ, 0) 6= 0 with Q = O(n − 1).

• Remarks:

a) The non-monotonicity is necessary for solvability.

b) Possibly deg(G(β)t , BH , 0) = 0 for H = O(n).

—This shows that the most suitable symmetry group to use is

O(n − 1) rather than O(n).

In other words, an axisymmetric R should be regarded as a

functions of 2-variables rather than a 1-variable function.

Use the latitude as the variable:

R = R(θ) = ζ(sin θ) −π

2≤ θ ≤

π

2.

• Theorem. Let R ∈ C2[ n2 ],γ be axisymmetric.

Assumption:

|R′(θ)| + |R′′(θ)| + · · · + |R(2[ n2 ])(θ)| 6= 0, ∀θ ∈ [−π/2, π/2].

Conclusion:

R is not monotone iff (1) has O(n − 1)-symmetric solutions.

Remarks:

• The Theorem improves the result of Xu-Yang (n=2), Chen-Li

(n ≥ 3, flatness conditions, and R only admits local extrema type

of critical points, but allows R < 0 somewhere)

Near every positive critical point r0:

R = R(r) = R(r0) + a|r − r0|α + h(|r − r0|),

with a 6= 0, n − 2 < α < n, h′ = o(sα−1).

• Our result for the sub-symmetric solution is given for the general

functions of (m + 1) variables (m < n):

R(x) = ζ(u), ∀x = (x1, · · · , xn−m, u), |x|2 = 1

where ζ is a smooth function on the closed ball Bm+1 ⊂ Rm+1.

— R is Hl-symmetric ∀l ∈ [m, n],

Hl := O(n − l), with the action: ∀h ∈ Hl,

hx = (hx′, xn−l+1, · · · ), ∀x = (x′, xn−l+1, · · · ) ∈ Rn+1.

— m = 0 corresponds to the axisymmetric R.

• Our general assumption:

either |R − R0| ≤ ǫn,

or satisfies ”flatness conditions (if n ≥ 4):

(∗)α with α > n − 2 near c. pts. of R

e.g.

assume α > n − 2 in last Corollary;

assume |R′(θ0)| + · · · + |R(n−2)(θ0)| = 0 in last Theorem.