Riemannian Curvature Functionals: Lecture II

Jeff Viaclovsky

Park City Mathematics Institute

July 17, 2013

Lecture Outline

Today we will discuss the following:

• To understand the local behavior of the functional near anEinstein metric, we only need to look at conformal andtransverse-traceless directions, the other directions do notmatter.

• Ebin-Palais slice theorem; consider the functional as amapping from M/D, the space of Riemannian metricsmodulo diffeomorphisms.

• Saddle point structure and the Yamabe invariant.

• Local description of the moduli space of Einstein metrics.

Splitting

We let K : T ∗M → S20(T ∗M) be the conformal Killing operator

(Kα)ij = ∇iαj +∇jαi −1

2(δα)gij .

Lemma

The space of symmetric 2-tensors admits the following orthogonaldecomposition:

S2(T ∗M) = {f · g} ⊕ {K(α)} ⊕ {δh = 0, trg(h) = 0}.

To prove this, consider the operator � = δK.

Exercise

(i) Show that this operator is elliptic and self-adjoint.(ii) Prove that the kernel of � is exactly the space of conformalKilling forms.

Proof of Lemma

Given h ∈ S20(T ∗M), consider the 1-form δh. By Fredholm theory,

the equation �α = δh has a solution if and only if δh is orthogonalto the kernel of the adjoint operator, which is exactly the space ofconformal Killing 1-forms (by the exercise). If κ is any conformalKilling 1-form, then∫

M〈δh, κ〉 =

1

2

∫M〈h,Kκ〉 = 0.

So the equation �α = δh has a solution, which proves thath−Kα is divergence-free.

Another decomposition

We note that the orthogonal decomposition

S2(T ∗M) = {f · g} ⊕ {K(α)} ⊕ {δh = 0, trg(h) = 0}

implies the decomposition

S2(T ∗M) = {f · g} ⊕ {L(α)} ⊕ {δh = 0, trg(h) = 0}.

Proposition

If (M, g) is Einstein, with Ric = λ · g, then this latterdecomposition is a direct sum, unless (M, g) is isometric to(S4, gS).

Proof of proposition

We need to show that the spaces {f · g} and {L(α)} haveintersection {0}. So if L(α) = f · g, then taking a trace, we have

2δα = 4f,

which implies that K(α) = 0. Taking a divergence, we have

∇i(∇iαj +∇jαi − (1/2)(δα)gij) = ∆αj +∇i∇jαi − (1/2)∇j(δα)

= ∆αj + (1/2)∇j(δα) + λαj .

However, we have that

∆α = −∆Hα+Ric ∗ α = (dδ + δd)α+ λα.

Putting these together, we obtain

�α = (3/2)dδα+ δdα+ 2λα.

Proof of proposition continued

Pairing the equation

(3/2)dδα+ δdα+ 2λα = 0,

with α and integrating,

−3

2

∫M|δα|2 −

∫M|dα|2 + 2λ

∫M|α|2 = 0.

This implies that α = 0 if λ < 0 (so any conformal Killing fieldvanishes on a negative Einstein metric).If λ = 0, we see that δα = 0 and dα = 0. In particular, α is aKilling 1-form, and we are done.In the case λ > 0, applying a divergence to the top equation yields

(3/2)∆(δα) + 2λ(δα) = 0.

By Lichnerowicz’ Theorem, this implies that δα = 0 unless (M, g)is isometric to (S4, gS), so α is Killing.

Second variation as a bilinear form

Let us recall the second variation is

E ′′(h, h) = V ol(g)−1/2∫M〈h, Jh〉dV,

where J is the operator

J = λh+G′h.

Using polarization, the Hessian of E is the bilinear form given by

E ′′(h1, h2) = V ol(g)−1/2∫M〈h1, Jh2〉dV.

Proposition

The decomposition

S2(T ∗M) = {f · g} ⊕ {L(α)} ⊕ {δh = 0, trg(h) = 0}

is orthogonal with respect to E ′′(·, ·)

First, E ′′(L(α), ·) = 0 from diffeomorphism invariance. So we justneed to check that

E ′′(f · g, z) = 0

if z is TT. To see this,

E ′′(f · g, z) = V ol(g)−1/2∫M〈f · g, Jz〉dV

= V ol(g)−1/2∫M〈f · g, 1

2∆z +Rm ∗ z〉dV

= V ol(g)−1/2∫MfRijipzjpdV = 0.

Summary

Summarizing the above discussion:

If h is any symmetric 2-tensor, then decompose h as

h = f · g + Lα+ z,

where z is TT. Then

E ′′(h, h) = E ′′(f · g, f · g) + E ′′(z, z).

So we have shown that to check the second variation, we reallyonly need to consider conformal variations and TT variationsseparately.

Ebin-Palais slice theorem

The above discussion was at the level of the “tangent space to thespace of Riemannian metrics at g”. We will next transfer thisstatement directly to the space of Riemannian metrics near gmodulo diffeomorphism.

Theorem

For each metric g1 in a sufficiently small C`+1,α-neighborhood of g(` ≥ 1), there is a C`+2,α-diffeomorphism ϕ : M →M such that

θ ≡ ϕ∗g1 − g

satisfies

δg

(θ − 1

ntrg(θ)g

)= 0.

Proof of Ebin-Palais

Let {ω1, . . . , ωκ} denote a basis of the space of conformal Killingforms with respect to g. Consider the map

N : C`+2,α(TM)× Rκ × C`+1,α(S2(T ∗M))→ C`,α(T ∗M)

given by

N (X, v, θ) = Nθ(X, v)

=(δg[ ◦︷ ︸︸ ︷ϕ∗X,1(g + θ)

]+∑i

viωi),

where ϕX,1 denotes the diffeomorphism obtained by following theflow generated by the vector field X for unit time, and ◦ denotesthe traceless part with respect to g.

Proof of Ebin-Palais continued

linearizing in (X, v) at (X, v, θ) = (0, 0, 0), we find

N ′0(Y, a) =d

dε

(δg[ ◦︷ ︸︸ ︷ϕ∗εY,1(g)

]+∑i

(εai)ωi)∣∣∣ε=0

=(δg[

◦︷ ︸︸ ︷LgY [] +

∑i

aiωi)

=(�Y [ +

∑i

aiωi),

where Y [ is the dual one-form to Y .

Proof of Ebin-Palais continued

The adjoint map (N ′0)∗ : Cm+2,α(T ∗M)→ Cm,α(TM)× Rκ isgiven by

(N ′0)∗(η) =((�η)],

∫〈η, ωi〉 dV

),

where (�η)] is the vector field dual to �η.

If η is in the kernel of the adjoint, the first equation implies that ηis a conformal Killing form, while the second implies that η isorthogonal (in L2) to the space of conformal Killing forms. Itfollows that η = 0, so the map N ′0 is surjective.

Proof of Ebin-Palais continued

Omitting a few technical details for simplicity, applying aninfinite-dimensional version of the implicit function theorem, givenθ1 ∈ C`+1,α(S2(T ∗M)) small enough we can solve the equationNθ1 = 0; i.e., there is a vector field X ∈ C`+2,α(TM), a v ∈ Rκ,such that

δg[

◦︷︸︸︷ϕ∗g1] +

∑i

viωi = 0,

where ϕ = ϕX,1. Letting θ = ϕ∗g1 − g, then θ satisfies

δg[◦θ] +

∑i

viωi = 0,

Pairing with ωj , for j = 1 . . . κ, and integrating by parts, we seethat vj = 0, and we are done.

Some exercises

• Verify the above formula for (N ′0)∗

• By adding a scaling factor to the map N , modify the aboveargument to show that we can find a constant c (dependingupon g1), and find

θ ≡ ecϕ∗g1 − g,

so that in addition to the traceless part of θ being TT, θ alsosatisfies ∫

trg θ dVg = 0.

That is, we can also “gauge away” the scale-invariance of thefunctional. Equivalently, we can look at a slice of unit-volumemetrics modulo diffeomorphism.

Summary

Combining the above discussions, given any g1 sufficiently near g,we can write

ϕ∗g1 = g + θ,

with θ = f · g + z, and z is TT. Then

E(g1) = E(ϕ∗g1) (from diffeomorphism invariance)

= E(g + θ)

= E(g) + E ′g(θ) + E ′′g (f · g + z, f · g + z) + remainder

= E(g) + E ′′g (f · g, f · g) + E ′′g (z, z) + remainder.

Theorem

The local behavior of E , when considered as a map on M/D (thespace of Riemannian metrics modulo diffeomorphism), isdetermined by the conformal and TT directions (to second order).

Saddle point structure and the Yamabe invariant.

We have seen that the functional E is minimizing in the conformaldirections, but maximizing (modulo a finite-dimensional subspace)in the TT directions. So an Einstein metric is always a saddle pointfor E . This suggests defining the following min-max type invariant.

First, we minimize in the conformal direction:

Y (M, [g]) = infg∈[g]E(g).

This is called the conformal Yamabe invariant.

From Aubin’s estimate,

Y (M, [g]) ≤ E(Sn, gS).

It is known that this inequality is strict, unless (M, g) is conformalto (Sn, gS). Thus:

The smooth Yamabe invariant

Theorem

(The Yamabe problem) If (Mn, g) is compact, then there exists aconformal metric g ∈ [g] which has constant scalar curvature, andwhich minimizes E in its conformal class.

The proof of this is involved, and we will not discuss here, but isdue to Aubin, Trudinger, Yamabe, with the most difficult casessolved by Schoen and Schoen-Yau (positive mass theorem).

The min-max invariant is then defined by

Y (M) = supg∈M

Y (M, [g]),

which we will call the smooth Yamabe invariant of M , also knownas the σ-invariant of M . Defined independently by O. Kobayashiand R. Schoen.

Some known cases

We will not focus on smooth Yamabe invariants in this lecture, butonly list a few known cases:

• Y (S4) = Y (S4, [gS ]) = 8π√

6.

• Y (S1 × S3) = Y (S4, [gS ]) = 8π√

6.

• Y (RP3) = Y (RP3, [gS ]) (proved by Bray-Neves).

• If (M3, gH) is compact hyperbolic, thenY (M3) = Y (M3, [gH ]) (proved by Perelman).

• Y (CP2) = Y (CP2, [gFS ]) = 12π√

2, where gFS is theFubini-Study metric (proved by LeBrun).

Some unkown cases

It is a very difficult problem to determine Yamabe invariants ingeneral. Here are a few prominent unknown cases:

• Y (S3/Γ), where S3/Γ is a spherical space form with |Γ| > 2.Achieved by the round metric?

• Y (CP2#CP2) = ? The only result known is due to O.Kobayashi: Y (CP2#CP2) ≥ Y (CP2).

• Y (CP2#CP2) = ? Again, the only result known is

Y (CP2#CP2) ≥ Y (CP2).

• Y (S2 × S2) = ? The only known result is thatY (S2 × S2) > Y (S2 × S2, gS2 + gS2) (strict inequality!). Thisfollows from the above exercise on S2×S2 and an observationof Wang-Ziller that CSC metrics sufficiently near an Einsteinmetric are also Yamabe minimizers in their conformal class.

Moduli space of Einstein metricsNext, given an Einstein metric g with Ric = λ · g, we would likeunderstand the space of solutions of the equation

Ric(g) = λ · g (1)

with g near g. This will be infinite-dimensional since ϕ∗g will alsobe a solution for any diffeomorphism ϕ : M →M .Therefore, we need to look at the space of Einstein metrics modulodiffeomorphism. Our goal is to prove:

Theorem

Assume g is Einstein with Ric(g) = λ · g and λ < 0. Then thespace of Einstein metrics near g modulo diffeomorphism is locallyisomorphic to the zero set of a map

Ψ : H1 → H1,

where

H1 = {h ∈ S2(T ∗M) : δgh = 0, trgh = 0,∆h+ 2Rm ∗ h = 0}.

Ellipticity

The diffeomorphism invariance also means that the above equationcannot be elliptic. Indeed, differentiating

Ric(ϕ∗t g) = ϕ∗t (Ric(g))

yields

Ric′(LXg) = LX(Ric(g)) = λ · LXg.

Exercise

Show that this implies that the symbol of Ric′ is not elliptic.

We will next describe a procedure called “gauging” which shows ineffect, that the diffeomorphism directions are the only obstructionto ellipticity. This is somewhat analogous to the “Coulomb gauge”in electrodynamics.

A gauge choiceRecall that, at an Einstein metric

(Ric′)ij =1

2

(−∆hij +∇i(δh)j +∇j(δh)i −∇i∇j(trh)

− 2Riljphlp + 2λhij

).

Define

βgh = δgh−1

2d(trgh)

Exercise

Show that

1

2Lβgh =

1

2

(∇i(δh)j +∇j(δh)i −∇i∇j(trh)

).

Therefore

(Ric′ − 1

2Lβg)h =

1

2

(−∆h− 2Rm ∗ h+ 2λh

).

The nonlinear map

Given θ ∈ C2,α(S2T ∗M), consider the map

P : C2,α(S2(T ∗M))→ C0,α(S2(T ∗M))

by

Pg(θ) = Ric(g + θ)− λ · (g + θ)− 1

2Lg+θβgθ.

Proposition

The operator P is elliptic.

This is immediate: from the previous slide, the linearized operatorat θ = 0 is

P ′h =1

2

(−∆h− 2Rm ∗ h

),

which is clearly elliptic.

Next lecture

Next time we will discuss the following:

• Complete the local description of the moduli space of Einsteinmetrics.

• Rigidity of Einstein metrics and discussion of some basicexamples.

• Quadratic functionals.

• Hitchin-Thorpe Inequality and examples of Einstein metrics indimension 4.

• Self-dual metrics and their deformation theory.