CIMAT Guanajuato, México - IHP · 2012. 12. 6. · CIMAT Guanajuato, México Institut Henri...

Yamabe equationBifurcation

Yamabe equation on Sn × Sm

Jimmy Petean

CIMATGuanajuato, México

Institut Henri PoincaréNovember, 2012



IntroductionQuestionsExample

Yamabe equation: Introduction

M closed C∞ manifold [compact connected, ∂M 6= 0]

g = 〈 , 〉x Riemannian metric on M.

Look for metrics of constant scalar curvature in the conformal class

[g] = {f .g : f : M → R>0}.It amounts to solving the Yamabe equation:

−an∆gu + sgu = λ up−1

an = 4(n−1)n−2 , sg the scalar curvature, p = pn = 2n

n−2 ,

λ is the scalar curvature of up−2g.




Sign is determined by [g].Yamabe equation is the Euler-Lagrange equation forthe Hilbert-Einstein functional restricted to [g]:

S(h) =

∫M sh dvolh

Vol(M,h)n−2

n

Yamabe constant of (M, [g])

Y (M, [g]) = infh∈[g]

S(h)




1 Infimum in the definition of Y (M, [g]) is always achivedH. Yamabe-N. Trudinger-T. Aubin-R. Schoen. There isalways at least one (volume 1) solution of the Yamabeequation.

2 Solution is unique if Y (M, [g]) ≤ 0.

3 Solution is unique if g is Einstein (M. Obata).

4 Noncompact family of solutions on [gn0 ]

the conformal class of the round metric on Sn, conformaldiffeomorphisms.




Questions

1 Compute Y (M) = sup{[g]} Y (M, [g]) ≤ Y (Sn, [gn0 ])

(T. Aubin).

2 Find all solutions of the Yamabe equation on [g].

Can we solved the Yamabe equation on (Sn × Sm,gn0 + Tgm

0 )?(T > 0, n,m ≥ 2 )

Solution is not unique for T big (or small).




It is important to understand Y (Sm × TSn)

limT→∞

Y (Sn × Sm, [gn0 + Tgm

0 ])) = Y (Sn × Rm,gn0 + gE )

(K. Akutagawa-L. Florit-J. Petean)

It is fundamental for understanding the behavior of Y (M) undercodim k ≥ 3-surgery

Y (M) ≥ min {Y (M), c(n, k)}

(B. Ammann-M. Dahl-E. Humbert).




Example

Sn × TS1, n ≥ 2 (R. Schoen, O. Kobayashi)

Solutions give periodic solutions on Sn × R,then solutions on Rn+1 − {0},which must be radial (L. Caffarelli-B. Gidas-J. Spruck).Then solutions of the Yamabe equation only depend on theS1-variable: solutions of

u′′ − (n − 1)2

4(u − upn+1−1) = 0

(T-periodic for some T>0)




Or S1 × TSn, consider 2-periodic solutions of

u′′ − λ(u − up−1) = 0

u′(0) = 0, u′(1) = 0.

Linearize at u = 1: v ′′ − λ(2− p)v = 0

v ′′ + αv = 0

(for some α > 0), v ′(0) = 0, v ′(1) = 0.

→ v(t) = cos(√αt),√α = kπ.




Then we have a sequence

λ1 < λ2 < ...→∞

which corresponds to

T1 > T2 > ...→ 0

so that new solutions of the Yamabe equation appear.

Solutions close to (1,Ti) will have (i − 1) local extrema in (0,1).



Setting, radial solutionsIsoparametric functionsIsoparametric eigenfunctionsCounting

Bifurcation: construction of solutions

GT = gm0 + Tgn

0

It is Einstein if T = n−1m−1

We consider T < n−1m−1 .

sGT = m(m − 1) + (1/T )n(n − 1)Normalize the Yamabe equation

−∆GT u +sGT

an+mu −

sGT

an+mupm+n−1 = 0.

Linearization of the Yamabe equation at GT :

−∆GT v + (2− pm+n)sGT

an+mv = −∆GT v +

sGT

n + m − 1v = 0.




Let µ(T ) = 1n+m−1sGT

Compare with eigenvalues of −∆GT :

0, m, 2(m+1),...

n/T, 2(n+1)/T,...

m+n/T,2(m+1) + n/T,...

Computation : µ(T ) < n/T , for any T < nm−1 .

For T = n−1m we have µ(T ) = m.

Eigenfunctions depend only on Sm (the ‘big’ variable)




Bifurcation: Pick Ti such that µ(Ti) is an eigenvalue of −∆gm0

,µ(Ti) = i(m + i − 1) =λi,m .Consider the Yamabe equation as an operator

C2,α(Sm × Sn)× R→ C0,α(Sm × Sn)

The linearization is Fredholm, index 0, the kernel Vi hasdimension

D(m,i) =(

m + i − 1m − 1

)-(

m + i − 3m − 1

)On a neighborhood U of (0,Ti) ∈ Vi ⊕ R we have defined amap F : U → V⊥i such that if u = 1 + v + F (v ,T )

(Id − Π)

(−∆GT (u) +

sGT

an+m(u)−

sGT

an+m(u)pm+n−1

)= 0.

where Π is the L2-orthogonal projection on Vi .Yamabe equation on Sn × Sm



Then around (1,Ti) the space of solutions of the GT -Yamabeequation is the set of functions u = 1 + v + F (v ,T ) such that

Π

(−∆GT (u) +

sGT

an+m(u)−

sGT

an+m(u)pm+n−1

)= 0.

Uniqueness implies that for each v ∈ Vi , F (v ,T ) ∈ C2,α(Sm).Therefore always u ∈ C2,α(Sm).

Reduce to solutions u : Sm → R>0 which solve

−∆gm0

u +sGT

an+m(u − up−1) = 0

p = pn+m < pm




Results, radial solutions

Let λ =sGTan+m

, we are interested in

λ >(m + n)(m − 1)(m + n − 2)

4(m + n − 1)

(corresponds to T = (n − 1)/(m − 1)).There are no (non-constant) solutions ifλ ≤ λ1 = m/(p − 2) = m(m + n − 2)/4, which corresponds toT ≥ T1 (M-F. Bidaut-Veron-L. Veron)

But there are non-constant solutions for T close to T1, T > T1.

Look for radial solutions with respect to some axis, functionsinvariant under the action of SO(n − 1).

With this restriction the space of solutions of the linearizedequation has dimension one, use local bifurcation.




How does the space of solutions of the bifurcation equationlook like?

Moreover (Q. Jin, YanYan Li, H. Hu), the solution appearingafter λi which corresponds to a solution of the ODE

u′′ + (m − 1)cos(t)sin(t)

u′ + λ(up−1 − u) = 0

has (i − 1) local extrema (besides the poles) andfor each λ > λi there is at least one such solution.

Moving the axis one is obtaining m-dimensional families ofsolutions.




Argument

The linearized equation (restricted to radial functions) is

ϕ′′ + (m − 1)cos(t)sin(t)

ϕ′ + λ(p − 2)ϕ = 0

ϕ′(0) = ϕ′(π) = 0.It has dimension 0 or one. The solutions at the bifurcationpoints can be computed explicitly, solution at the λi will have izeros ((i − 1) local extrema in (0, π) ).

Therefore radial solutions near λi will have (i − 1) extrema.

Look for the connected component Ci of the solutionsappearing at λi , in the space of non-trivial solutions (6= (1, λ)).

All elements in Ci must have exactly (i − 1) local extrema (sincethey solve a second order ODE).




Then since Ci only contains one trivial solution, which is abifurcation point of odd multiplicity, Ci is not compact (by P.Rabinowitz Theorem).

But since p < pm the space of solutions (u,T ) with T moving ina compact interval is compact.

Therefore Ci must pass over every λ > λi .

Theorem(G. Henry-Petean) Let S be an isoparametric hypersurface ofSm of degree k. Then for each positive integer i there exist atleast i solutions of the equation on Sm forλ ∈ (λik ,m/(p − 2) , λ(i+1)k ,m/(p − 2)] which are constant alongS.




The simplest generalization of the above is to consider anothercodimension 1 action on Sm.

But also...a function f : (M,g)→ R is called isoparametric if

∆f = a(f ), ‖∇f‖2 = b(f )

where a is continuous, b is smooth.

Then for an isoparametric function f let Sf be {ϕ ◦ f}.

Example: x1 on Sm is an isoparametric function, a(t) = −mt ,b(t) = 1− t2. In this case Sf are the radial functions.x2

1 is also an isoparametric function: a(t) = 2(1−m)t − 2t2,b(t) = 4(t − t2).

If f : (M,g)→ [0,1] is an isoparametric function, then 0 and 1are the only critical values (Q-M. Wang). Regular level sets arecalled isoparametric hypersurfaces.




Let (M,g) be a closed Riemannian manifold and f : M → R anisoparametric function. Then there exist eigenfunctions of theLaplace operator of (M,g) which belong to Sf . Let us firstobserve the following:

LemmaIf u ∈ Sf is smooth then ∆gu ∈ Sf .

Proof.Let u = ϕ ◦ f , for a smooth function ϕ. Then by a directcomputation ∆gu = ϕ′′b(f ) + ϕ′a(f ) ∈ Sf .




LemmaThere exist an infinite sequence 0 < λ1 < λ2 < λ3 < · · · <∞such that there exists an eigenfunction fi ∈ Sf of ∆g witheigenvalue −λi .

The sequence 0 > −λ1 > −λ2 > . . . is of course asubsequence of the spectrum of ∆g .

LemmaFor each eigenvalue λi the space of Sf -eigenfunctions hasdimension one.




If f is an isoparametric function and u : R→ R is smooth,u′ ◦ f 6= 0, then u ◦ f is also an isoparametric function.In the case of Sm one has a certain normalization:if M ⊂ Sn is an isoparametric hypersurface then H. Münznerproved that there exists a homogeneous polynomialF : Rn+1 → R of degree k which satisfies theCartan-Münzner equations:

< ∇F ,∇F >= k2‖x‖2k−2 (1)

∆F =12

ck2‖x‖k−2, (2)

(c is an integer) such that M is a regular level set of f = F|Sn ,which is an isoparametric function.In this situation we will say that f is an isoparametric function ofdegree k and similarly that M is an isoparametric hypersurfaceof degree k .




Moreover, k can only take the values 1, 2, 3, 4 or 6 andcoincides with the number of distinct principal curvatures of M.In case k = 3 (or 1) the principal curvatures have the samemultiplicities and in case k = 2,4 or 6 there are integers m1 andm2 (which might be equal) such that half of the principalcurvatures have multiplicity m1 and the other half m2; inparticular (k/2)(m1 + m2) = m − 1. Then the constant c ism2 −m1. The polynomial F is called a Cartan-Münznerpolynomial. Note that interchanging m1 and m2 corresponds toreplacing F with −F .




Consider the obvious O(i)×O(j)-action on Si+j−1 ⊂ Ri+j. For(x , y) ∈ Ri+j write x2 = x2

1 + ...+ x2i and y2 = y2

1 + ...+ y2j .

Then x2 and y2 are homogeneous polynomials of degree 2invariant under the action. Let F = x2 − y2. Then F is aCartan-Münzner polynomial with k = 2, m1 = i − 1, m2 = j − 1.The corresponding isoparametric hypersurfaces have twoprincipal curvatures and are diffeomorphic to Si−1 × Sj−1. Asbefore we denote by f = F|Si+j−1

; then Sf is the family ofO(i)×O(j)-invariant functions.A classification of homogeneous isoparametric hypersurfacesin the sphere was given by W. Y. Hsiang and H. B. Lawson. Butthere are families of examples of non-homogeneousisoparametric hypersurfaces. The first examples were found byH. Ozeki and M. Takeuchi.




Let H be the real quaternion algebra and let u −→ u be thenatural involution. We can identify R16 with H2 × H2, we notex ∈ R16 as x = (u0,u1, v0, v1) where ui , vi ∈ H for i = {0,1}. LetF : R16 −→ R defined by

F = ‖x‖4 − 2F0(x) (3)

where

F0(x) = 4(

u0v0 + u1v1

)(u0v0 + u1v1

)−

(u0v0 + u1v1 + v0u0 + v1u1

)2(4)

+[u1u1 − v1v1 + u0v0 + v0u0

]2

Ozeki and Takeuchi showed that the function F (that satisfiedthe equations with k = 4 and c = 1) gives non homogeneousisoparametric hypersurfaces in S15.




LemmaLet f : Sm → [−1,1] be an isoparametric function obtained asthe restriction to Sm of a solution F of de Cartan-Munznerequations. Let k be the degree of F . Then for each i = 0,1, ...there is an eigenfunction fi ∈ Sf of the Laplacian ∆Sm witheigenvalue λik ,m. The space of such eigenfunctions hasdimension 1 and is generated by pi ◦ f where pi is a monicpolynomial of degree i which has i distinct simple roots in theinterval (−1,1). Moreover, if λj,m is an eigenvalue of ∆Sm |Sf

then j = ik for some i.




F is a homogeneous polynomial of degree k . LetU = F − (c/(m + 1))(x2

1 + ...+ x2m+1)k/2. Then U is a harmonic

homogeneous polynomial of degree k .(c = 0 if k is odd)u = U|Sm ∈ Sf is an eigenfunction of −∆Sm with eigenvalueλk ,m.u = p1(f ) with p1(t) = t − (c/(m + 1)).u verifies that Su = Sf ,

∆u = −λk ,m u

‖∇u‖2 = ‖∇f‖2 = k2(1− f 2) = k2(1− (u + c/(m + 1))2)




Now α ◦ f is an eingenfunction of −∆Sm with eigenvalue λi,m ifand only if α solves the second order ODE, Oi(α) =

k2α′′(t)(1− t2) + α′(t)ck2 − k(m + k − 1)t

2+ λi,mα(t) = 0.

By a straightforward computation Oik (t i) =k2i(i−1)t i−2(1−t2)+it i−1(ck2/2−k(m+k−1)t)+ik(m+ik−1)t i

= Ct i−1 + Dt i−2

for some C,D ∈ R. Moreover for j < iOik (t j) = Et j + Ft j−1 + Gt j−2 for some E ,F ,G ∈ R and E 6= 0.We set p0 = 1, p1(t) = t − (c/(m + 1)) (as before) and it followsthat there is exactly one monic polynomial of degree i , pi whichsolves Oik (pi) = 0.




Assume pi has i roots −1 < t1 < ... < ti < 1. If we call t0 = −1and ti+1 = 1 then it is enough to prove that pi+1 has at leastone root in each interval (tj , tj+1). Sturm comparison theorem:

p′′i +((1/2)ck2 − k(n + k − 1)t)

k2(1− t2)p′i +

λik ,n

k2(1− t2)pi = 0,

p′′i+1 +((1/2)ck2 − k(n + k − 1)t)

k2(1− t2)p′i+1 +

λ(i+1)k ,n

k2(1− t2)pi+1 = 0,

with λ(i+1)k ,n > λik ,n.To prove the last statement in the lemma, pick j ,ik < j < (i + 1)k . Let ϕ be a non-trivial solution of Oj(ϕ) = 0.Then applying the same Sturm comparison to pi and ϕ wouldprove that ϕ has at least i + 1 zeros. And then applying thesame Sturm comparison to ϕ and pi+1 would prove that pi+1has at least i + 2 zeros, which is false.




Then restrict the equation to SfThe linearized equation has a 1-dimesional kernel on the λik ,i = 1,2, ....Then these are bifurcation points. Solutions appearing aroundλik are given by solutions of a second order ODE with(i − 1)-local extrema.

The argument for radial solutions (Jin-Li-Xu) works to prove thatthere are solutions in Sf with (i − 1)-local extrema for anyλ > λik .




Isoparametric hypersurfaces of degree 1 in Sm arehomogeneous, the corresponding solutions to the Yamabeequation are the radial solutions.

Similarly isoparametric hypersurfaces of degree 2 areSi × Sm−i−1, the orbits of the isometric actions ofO(i + 1)×O(m − i), with 1 ≤ i ≤ [m − 1/2].

Isoparametric hypersurfaces of degree 3 were classified by E.Cartan: the three principal curvatures have the samemultiplicity which can be 1, 2, 4 or 8.The corresponding isoparametric hypersurfaces in S4, S7, S13

and S25 are called Cartan hypersurfaces and they are tubesover the canonical embeddings of the projective planes FP2

(where F are the real, complex, quaternionic or Cayleynumbers).




There are essentially two isoparametric hypersurfaces ofdegree 6 in the sphere, one in S7 and the other in S13. In bothcases all the principal curvature have the same multiplicities, 1and 2 respectively.

Isoparametric hypersurfaces of degree 4 provide the richestexamples. They are not completely classified.It is known that new examples of isoparametric hypersurfacesof degree 4 can only appear in S31.

Let

T m,ni =

n(n − 1)

(m + n − 1)λi,m −m(m − 1)




ThenLet Nm,n

Y (T ) be the number of isometrically distinct unit volumemetrics of constant scalar curvature in the conformal class of(Sm × Sn,gm

0 + Tgn0 ) and suppose that T ∈ [T m,n

i+1 ,Tm,ni ). Then:

a) If m = 2j with j 6= 2 then Nm,nY (T ) ≥ i + [(2j − 1)/2][i/2].

b) If m = 2j + 1 with j 6= 1,3,4,6,7,9,12 thenNm,n

Y (T ) ≥ i + j[i/2] + (γ(j) + β(j))[i/4].c) 1 If m = 3 then Nm,n

Y (T ) ≥ i + [i/2].2 If m = 4 then Nn,k

Y (T ) ≥ i + [i/2] + [i/3].3 If m = 7 then Nn,k

Y (T ) ≥ i + 3[i/2] + [i/3] + [i/4] + [i/6].4 If m = 9 then Nn,k

Y (T ) ≥ i + 4[i/2] + 2[i/4].5 If m = 13 then Nn,k

Y (T ) ≥ i + 6[i/2] + [i/3] + [i/4] + [i/6].6 If m = 15 then Nn,k



Y (T ) ≥ i + 12[i/2] + [i/3] + [i/4].




Thank you !!


CIMAT Guanajuato, México - IHP · 2012. 12. 6. · CIMAT Guanajuato, México Institut Henri...

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