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An introduction to the Penrose inequality III

Levi Lopes de Lima

Department of MathematicsFederal University of Ceará

Gelosp2013 - July, 2013

Levi Lopes de Lima (DM–UFC) A Penrose inequality São Paulo, July 2013 1 / 23

Joint work with Fred Girão (UFC/Fortaleza).

We use the inverse mean curvature flow to prove a sharp Alexandrov-Fenchel-type inequalityfor strictly mean convex hypersurfaces in a certain class of locally hyperbolic manifolds ofdimension n ≥ 3.

This provides natural generalizations of the classical Minkowski inequality for convexhypersufaces in Rn.

As an application we establish an optimal Penrose inequality for asymptotically locallyhyperbolic (ALH) graphs carrying a minimal horizon, with a precise description of whathappens in the equality case.

This provides a large class of examples of initial data sets (corresponding to time-symmetricsolutions of Einstein equations in General Relativity with a negative cosmological constant)for which an optimal Penrose inequality holds true.

In particular, in the physical dimension n = 3 we obtain, for this class of IDS, a proof of aPenrose-type inequality for exotic black holes solutions first conjectured by Gibbons andChrusciel-Simon.

The reference metrics (Chrusciel-Herzlich-Nagy)

Fix n ≥ 3, ε = 0,±1 and let (Nn−1, h) be a closed space form with curvature ε.

In the product manifold Pε = Iε × N, consider the metric

gε =dr2

ρε(r)2+ r2h, r ∈ Iε,

whereρε(r) =

√r2 + ε.

Here, I−1 = (1,+∞) and I0 = I1 = (0,+∞).

The metric gε is locally hyperbolic (Kgε ≡ −1).

For instance, if ε = 1 and (N, h) is a round sphere then (P1, g1) is hyperbolic space Hn.

Also, if ε = 0 and (N2, h) is a torus then (P0, g0) is a cusp manifold.

gε =dr2

whereρε(r) =

√r2 + ε.

Here, I−1 = (1,+∞) and I0 = I1 = (0,+∞).

gε =dr2

whereρε(r) =

√r2 + ε.

Here, I−1 = (1,+∞) and I0 = I1 = (0,+∞).

gε =dr2

whereρε(r) =

√r2 + ε.

Here, I−1 = (1,+∞) and I0 = I1 = (0,+∞).

gε =dr2

whereρε(r) =

√r2 + ε.

Here, I−1 = (1,+∞) and I0 = I1 = (0,+∞).

gε =dr2

whereρε(r) =

√r2 + ε.

Here, I−1 = (1,+∞) and I0 = I1 = (0,+∞).

gε =dr2

whereρε(r) =

√r2 + ε.

Here, I−1 = (1,+∞) and I0 = I1 = (0,+∞).

Asymptotically locally hyperbolic manifolds (Chrusciel-Herzlich-Nagy)

Definition

Fix ε and (N, h) as above. A complete n-dimensional manifold (M, g), possibly carrying a compactinner boundary Σ, is said to be asymptotically locally hyperbolic (ALH) if there exist subsetsK ⊂ M and K0 ⊂ Pε, with K compact, and a diffeomorphism Ψ : M − K → Pε − K0 such that

‖Ψ∗g − gε‖gε = O(r−τ ), ‖DΨ∗g‖gε = O(r−τ ), r → +∞,

for some τ > n/2. We also assume that Rg + n(n − 1) ∈ L1.

For this class of manifolds, a mass-like invariant m(M,g) ∈ R can be defined as

m(M,g) = limr→+∞

(ρε(divgεe − d trgεe)− i∇gερεe + (trgεdρε)

)(νr )dNr ,

where e = Ψ∗g − gε, Nr = {r} × N, νr is the outward unit vector to Nr and

2(n − 1)τn−1, τn−1 = arean−1(N, h).

This invariant measures the rate of the convergence g → g0,ε as r → +∞.

Here, we are leaving aside the cases where N = Sn/Γ, Γ 6= {Id}.

Definition

)(νr )dNr ,

2(n − 1)τn−1, τn−1 = arean−1(N, h).

Definition

)(νr )dNr ,

2(n − 1)τn−1, τn−1 = arean−1(N, h).

Definition

)(νr )dNr ,

2(n − 1)τn−1, τn−1 = arean−1(N, h).

Definition

)(νr )dNr ,

2(n − 1)τn−1, τn−1 = arean−1(N, h).

Definition

)(νr )dNr ,

2(n − 1)τn−1, τn−1 = arean−1(N, h).

The black hole solutions I

Fix ε = 0,±1, m > 0 and consider the interval

Im,ε = {r > rm,ε},

where rm,ε is the positive root of

r2 + ε−2m

rn−2= 0.

If (Nn−1, h) is a compact space form with curvature ε, in the product manifoldPm,ε = Im,ε × N define the metric

gm,ε =dr2

ρm,ε(r)2+ r2h,

ρm,ε(r) =

√r2 + ε−

2mrn−2

We note that gm,ε extends smoothly to Pm,ε = [rm,ε,+∞)× N and the slice defined byr = rm,ε is called the horizon.

Im,ε = {r > rm,ε},

r2 + ε−2m

rn−2= 0.

gm,ε =dr2

ρm,ε(r)2+ r2h,

ρm,ε(r) =

√r2 + ε−

2mrn−2

Im,ε = {r > rm,ε},

r2 + ε−2m

rn−2= 0.

gm,ε =dr2

ρm,ε(r)2+ r2h,

ρm,ε(r) =

√r2 + ε−

2mrn−2

Im,ε = {r > rm,ε},

r2 + ε−2m

rn−2= 0.

gm,ε =dr2

ρm,ε(r)2+ r2h,

ρm,ε(r) =

√r2 + ε−

2mrn−2

Im,ε = {r > rm,ε},

r2 + ε−2m

rn−2= 0.

gm,ε =dr2

ρm,ε(r)2+ r2h,

ρm,ε(r) =

√r2 + ε−

2mrn−2

The black hole solutions II

If (θ1, · · · , θn−1) are orthonormal coordinates in N then the sectional curvatures of gm,ε are

Kgm,ε (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,ε (∂θi , ∂θj ) = −1 +

so that the scalar curvature of gm,ε is Rgm,ε = −n(n − 1).

Moreover, each gm,ε is a static metric in the sense that ρm,ε satisfies

(∆ρm,ε)gm,ε − Hessgm,ερm,ε + ρm,εRicgm,ε = 0,

which means that the Lorentzian metric

gm,ε = −ρ2m,εdt2 + gm,ε,

defined on Qm,ε = R× Pm,ε, is a solution to the vacuum Einstein field equations withnegative cosmological constant:

Ricgm,ε= −ngm,ε.

Thus, gm,ε defines an initial data set for a time-symmetric (actually, static) vacuum solution ofEinstein equations carrying a black hole.

Kgm,ε (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,ε (∂θi , ∂θj ) = −1 +

Kgm,ε (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,ε (∂θi , ∂θj ) = −1 +

Kgm,ε (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,ε (∂θi , ∂θj ) = −1 +

The black hole solutions III

One easily verifies that, as r → +∞,

‖gm,ε − gε‖gε = O(mr−n) ,

where gε is the corresponding reference metric.

Thus, each gm,ε, m > 0, is asymptotically locally hyperbolic (ALH).

Physical reasoning allows us to interpret m as the total mass of the black hole solution gm,ε.

Indeed, a computation shows that m(Pm,ε,gm,ε) = m.

The black hole solutions IV

It turns out that each gm,ε can be isometrically embedded as a graph in (Qε, gε), whereQε = R× Pε and

gε = ρε(r)2dt2 +dr2

ρε(r)2+ r2dθ2.

Notice that (Qε, gε) is locally hyperbolic!

The radial function defining the graph, u = um,ε(r), satisfies u(rm,ε) = 0 and

ρε(r)2(

1ρm,ε(r)2

ρε(r)2, r ≥ rm,ε.

It follows that the graph realization of the black hole solution meets the slice t = 0orthogonally along the minimal ‘horizon’ H defined by r = rm,ε.

Notice also that the mass m relates to the area |H| of the black hole horizon by

( |H|τn−1

) nn−1

(|H|τn−1

) n−2n−1

, τn−1 = arean−1(N).

ρε(r)2+ r2dθ2.

ρε(r)2(

1ρm,ε(r)2

( |H|τn−1

) nn−1

(|H|τn−1

) n−2n−1

ρε(r)2+ r2dθ2.

ρε(r)2(

1ρm,ε(r)2

( |H|τn−1

) nn−1

(|H|τn−1

) n−2n−1

ρε(r)2+ r2dθ2.

ρε(r)2(

1ρm,ε(r)2

( |H|τn−1

) nn−1

(|H|τn−1

) n−2n−1

ρε(r)2+ r2dθ2.

ρε(r)2(

1ρm,ε(r)2

( |H|τn−1

) nn−1

(|H|τn−1

) n−2n−1

ρε(r)2+ r2dθ2.

ρε(r)2(

1ρm,ε(r)2

( |H|τn−1

) nn−1

(|H|τn−1

) n−2n−1

The Penrose conjecture for ALH manifolds

Let (M, g) be an ALH manifold (relative to the reference metric gε). Assume thatRg ≥ −n(n − 1) and that M carries an outermost minimal horizon Σ. Then,

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

with the equality occurring if and only if (M, g) is (isometric to) the corresponding black holesolution.

In the physical dimension n = 3, this appears as a conjectured Penrose-type inequality inpapers by Gibbons and Chrusciel-Simon.

In the following we establish this inequality for ALH graphs in any dimension n ≥ 3, includingthe corresponding rigidity statement.

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

ALH hypersurfaces in Qε

DefinitionA complete, isometrically immersed hypersurface (M, g)# (Qε, gε), possibly with an innerboundary Σ, is asymptotically locally hyperbolic (ALH) if there exist subsets K ⊂ M, K0 ⊂ Pε suchthat M − K , the end of M, can be written as a vertical graph over Pε − K0, with the graph beingassociated to a smooth function u : Pε − K0 → R such the previous asymptotic conditions holdsfor the nonparametric chart Ψu(x , u(x)) = x , x ∈ K0. Moreover, we assume thatRΨu∗g + n(n − 1) is integrable.

Under these conditions, the mass of (M, g) is well defined and can be computed by using Ψu .

The integral formula for the mass

For any hypersurface M ⊂ Qε = R× Pε endowed with a unit normal N, an old formula byReilly says that

divM (G(A)X) = 2σ2(A)Θ,

where G(A) = σ1(A)I − A is the Newton tensor of the shape operator A, X is the tangentialcomponent of ∂/∂t and Θ = 〈N, ∂/∂t〉. This uses that ∂/∂t is Killing and that Kgε

≡ −1.

Assume from now on that M ⊂ Qε is ALH and its inner boundary Σ lies on a horizontal totallygeodesic hypersurface, say P ' Pε. Assume further that M meets P orthogonally along Σ(which implies that Σ ⊂ M is minimal and hence a horizon).

TheoremUnder the above conditions,

m(M,g) = cn

Θ (Rg + n(n − 1)) dM + cn

ˆΣρεHdΣ,

where H is the mean curvature of Σ ⊂ P and ρε(r) =√

r2 + ε. In particular, if Rg ≥ −n(n − 1)and M is a graph (Θ > 0) then

m(M,g) ≥ cn

ˆΣρεHdΣ.

≡ −1.

m(M,g) = cn

Θ (Rg + n(n − 1)) dM + cn

ˆΣρεHdΣ,

m(M,g) ≥ cn

ˆΣρεHdΣ.

≡ −1.

m(M,g) = cn

Θ (Rg + n(n − 1)) dM + cn

ˆΣρεHdΣ,

m(M,g) ≥ cn

ˆΣρεHdΣ.

≡ −1.

m(M,g) = cn

Θ (Rg + n(n − 1)) dM + cn

ˆΣρεHdΣ,

m(M,g) ≥ cn

ˆΣρεHdΣ.

≡ −1.

m(M,g) = cn

Θ (Rg + n(n − 1)) dM + cn

ˆΣρεHdΣ,

m(M,g) ≥ cn

ˆΣρεHdΣ.

The Alexandrov-Fenchel inequality

We have seen thatm(M,g) ≥ cn

ˆΣρ0,εHdΣ.

In order to proceed, we need a new Alexandrov-Fenchel inequality for a class ofhypersurfaces in (Pε, gε)!

TheoremIf Σ ⊂ Pε is star-shaped and strictly mean convex (H > 0) then

ˆΣρεHdΣ ≥

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

with the equality holding if and only if Σ is a slice.

ˆΣρ0,εHdΣ.

ˆΣρεHdΣ ≥

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

ˆΣρ0,εHdΣ.

ˆΣρεHdΣ ≥

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

ˆΣρ0,εHdΣ.

ˆΣρεHdΣ ≥

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

ˆΣρ0,εHdΣ.

ˆΣρεHdΣ ≥

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

The optimal Penrose inequality

This proves the first part of our main result.

TheoremIf M ⊂ Q0,ε is an ALH graph as above, with Σ ⊂ P = P0,ε being mean convex (H ≥ 0) andstar-shaped, then

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

with the equality holding if and only if (M, g) is (congruent to) the graph realization of thecorresponding black hole solution.

For ε = 1, this sharpens previous results by Dahl-Gicquaud-Sakovich.

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

m(M,g) ≥12

( |Σ|τn−1

) nn−1

(|Σ|τn−1

) n−2n−1

The proof of AF I

The proof uses the IMCF:∂X∂t

= −ξ

where ξ is the inward unit normal to Σ.

It is convenient to use the parameter s satisfying ds = dr/ρε(r), which gives

arcsinh r ε = 1

log r , ε = 0log(2

√r2 − 1 + 2r), ε = −1

In terms of this parameter,gε = ds2 + λε(s)2h,

λε(s) =

sinh s ε = 1

es, ε = 0es

4 + e−s, ε = −1

Notice that λ2ε = λ2

ε + ε.

The proof of AF I

= −ξ

arcsinh r ε = 1

log r , ε = 0log(2

√r2 − 1 + 2r), ε = −1

λε(s) =

sinh s ε = 1

es, ε = 0es

4 + e−s, ε = −1

ε + ε.

The proof of AF I

= −ξ

arcsinh r ε = 1

log r , ε = 0log(2

√r2 − 1 + 2r), ε = −1

λε(s) =

sinh s ε = 1

es, ε = 0es

4 + e−s, ε = −1

ε + ε.

The proof of AF I

= −ξ

arcsinh r ε = 1

log r , ε = 0log(2

√r2 − 1 + 2r), ε = −1

λε(s) =

sinh s ε = 1

es, ε = 0es

4 + e−s, ε = −1

ε + ε.

The proof of AF I

= −ξ

arcsinh r ε = 1

log r , ε = 0log(2

√r2 − 1 + 2r), ε = −1

λε(s) =

sinh s ε = 1

es, ε = 0es

4 + e−s, ε = −1

ε + ε.

The proof of AF I

= −ξ

arcsinh r ε = 1

log r , ε = 0log(2

√r2 − 1 + 2r), ε = −1

λε(s) =

sinh s ε = 1

es, ε = 0es

4 + e−s, ε = −1

ε + ε.

The proof of AF II

It is shown that if the initial hypersurface Σ0 ⊂ P0,ε is star-shaped and strictly mean convex(H > 0) then the evolving hypersurface Σt is defined for all t > 0, remains star-shaped andstrictly mean convex and expands to infinity in the sense that the principal curvaturesconverge exponentially to 1 as t → +∞.

Moreover, there exists α ∈ R so that if u = u(t , θ) is the graphing function then the rescaling

u(t , θ) = u(t , θ)−t

n − 1

converges to α in the sense that

|∇u|+ |∇2u| = o(1).

In particular,

λε(u) ∼ λε(u) ∼ et

n−1 .

The proof of AF II

u(t , θ) = u(t , θ)−t

n − 1

|∇u|+ |∇2u| = o(1).

In particular,

n−1 .

The proof of AF II

u(t , θ) = u(t , θ)−t

n − 1

|∇u|+ |∇2u| = o(1).

In particular,

n−1 .

The proof of AF II

u(t , θ) = u(t , θ)−t

n − 1

|∇u|+ |∇2u| = o(1).

In particular,

n−1 .

The proof of hyperbolic AF III

DefineJ (Σ) = −

pdΣ, p = 〈Dρε, ξ〉,

andK(Σ) = τn−1A(Σ)

nn−1 , A(Σ) = A/τn−1.

These quantities appear in the following preliminary result.

TheoremIf Σ ⊂ Pε is star-shaped and strictly mean convex then

J (Σ) ≤ K(Σ),

with the equality holding if and only if Σ is totally umbilical.

DefineJ (Σ) = −

nn−1 , A(Σ) = A/τn−1.

J (Σ) ≤ K(Σ),

DefineJ (Σ) = −

nn−1 , A(Σ) = A/τn−1.

J (Σ) ≤ K(Σ),

DefineJ (Σ) = −

nn−1 , A(Σ) = A/τn−1.

J (Σ) ≤ K(Σ),

DefineJ (Σ) = −

nn−1 , A(Σ) = A/τn−1.

J (Σ) ≤ K(Σ),

The proof of hyperbolic AF IV

Letting Σ flow under the IMCF, we have

n − 1J ,

where (∗) is a recent inequality by Brendle.

On the other hand,dAdt

= A ⇒dKdt

n − 1K,

and this immediately yieldsddtJ −K

n−1≥ 0.

But the asymptotics gives

limt→+∞

J −K

n−1= 0,

and the theorem follows.

n − 1J ,

= A ⇒dKdt

n − 1K,

n−1≥ 0.

limt→+∞

J −K

n−1= 0,

n − 1J ,

= A ⇒dKdt

n − 1K,

n−1≥ 0.

limt→+∞

J −K

n−1= 0,

n − 1J ,

= A ⇒dKdt

n − 1K,

n−1≥ 0.

limt→+∞

J −K

n−1= 0,

n − 1J ,

= A ⇒dKdt

n − 1K,

n−1≥ 0.

limt→+∞

J −K

n−1= 0,

The proof of hyperbolic AF V

We now considerI(Σ) =

ˆΣρεHdΣ.

If K is the extrinsic scalar curvalure of Σ, then

ρεKH

dΣ + 2J

≤n − 2n − 1

I + 2J ,

so that the previous theorem gives

(I − (n − 1)K) ≤n − 2n − 1

(I − (n − 1)K) + 2 (J −K)

≤n − 2n − 1

(I − (n − 1)K) .

ˆΣρεHdΣ.

ρεKH

dΣ + 2J

≤n − 2n − 1

I + 2J ,

(I − (n − 1)K) ≤n − 2n − 1

(I − (n − 1)K) + 2 (J −K)

≤n − 2n − 1

(I − (n − 1)K) .

ˆΣρεHdΣ.

ρεKH

dΣ + 2J

≤n − 2n − 1

I + 2J ,

(I − (n − 1)K) ≤n − 2n − 1

(I − (n − 1)K) + 2 (J −K)

≤n − 2n − 1

(I − (n − 1)K) .

ˆΣρεHdΣ.

ρεKH

dΣ + 2J

≤n − 2n − 1

I + 2J ,

(I − (n − 1)K) ≤n − 2n − 1

(I − (n − 1)K) + 2 (J −K)

≤n − 2n − 1

(I − (n − 1)K) .

The proof of hyperbolic AF VI

The previous inequality can be rewritten as

dLdt≤ 0,

whereL(Σ) = A(Σ)

− n−2n−1 (I(Σ)− (n − 1)K(Σ)) .

But, as we shall see below, the asymptotics also gives

lim inft→+∞

L(t) ≥ (n − 1)τn−1ε,

so thatL(0) ≥ (n − 1)τn−1ε,

as desired.

dLdt≤ 0,

whereL(Σ) = A(Σ)

− n−2n−1 (I(Σ)− (n − 1)K(Σ)) .

lim inft→+∞

L(t) ≥ (n − 1)τn−1ε,

as desired.

dLdt≤ 0,

whereL(Σ) = A(Σ)

− n−2n−1 (I(Σ)− (n − 1)K(Σ)) .

lim inft→+∞

L(t) ≥ (n − 1)τn−1ε,

as desired.

dLdt≤ 0,

whereL(Σ) = A(Σ)

− n−2n−1 (I(Σ)− (n − 1)K(Σ)) .

lim inft→+∞

L(t) ≥ (n − 1)τn−1ε,

as desired.

The proof of hyperbolic AF VII (the lower bound for L)

A computation using the asymptotics of the flow gives

A(Σt ) =

λn−1ε + o(e

(n−3)tn−1 ),

and ˆΣt

ρεHdΣt = (n − 1)

ˆλ2ελ

n−2ε + o(e

(n−2)t(n−1) ).

Hence, if we use the characteristic equation λ2ε = λ2

ε + ε,

lim inft→+∞

L(Σt ) = (n − 1)τn−1 lim inft→+∞

fflλ2ελ

n−2ε −

(fflλn−1ε

) nn−1

+ o(e(n−2)t

n−1 )(fflλn−1ε )

) n−2n−1

+ o(e(n−4)t

n−1 )

≥ (n − 1)τn−1ε lim inft→+∞

fflλn−2ε(ffl

λn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

+(n − 1)τn−1 lim inft→+∞

fflλnε −

(fflλn−1ε

) nn−1

(fflλn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

o(e(n−2)t

n−1 )(fflλn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

The proof of hyperbolic AF VII (the lower bound for L)A computation using the asymptotics of the flow gives

A(Σt ) =

λn−1ε + o(e

(n−3)tn−1 ),

and ˆΣt

ˆλ2ελ

n−2ε + o(e

(n−2)t(n−1) ).

ε + ε,

lim inft→+∞

fflλ2ελ

n−2ε −

(fflλn−1ε

) nn−1

+ o(e(n−2)t

) n−2n−1

+ o(e(n−4)t

n−1 )

fflλn−2ε(ffl

λn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

fflλnε −

(fflλn−1ε

) nn−1

(fflλn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

o(e(n−2)t

) n−2n−1

+ o(e(n−4)t

n−1 )

A(Σt ) =

λn−1ε + o(e

(n−3)tn−1 ),

and ˆΣt

ˆλ2ελ

n−2ε + o(e

(n−2)t(n−1) ).

ε + ε,

lim inft→+∞

fflλ2ελ

n−2ε −

(fflλn−1ε

) nn−1

+ o(e(n−2)t

) n−2n−1

+ o(e(n−4)t

n−1 )

fflλn−2ε(ffl

λn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

fflλnε −

(fflλn−1ε

) nn−1

(fflλn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

o(e(n−2)t

) n−2n−1

+ o(e(n−4)t

n−1 )

A(Σt ) =

λn−1ε + o(e

(n−3)tn−1 ),

and ˆΣt

ˆλ2ελ

n−2ε + o(e

(n−2)t(n−1) ).

ε + ε,

lim inft→+∞

fflλ2ελ

n−2ε −

(fflλn−1ε

) nn−1

+ o(e(n−2)t

) n−2n−1

+ o(e(n−4)t

n−1 )

fflλn−2ε(ffl

λn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

fflλnε −

(fflλn−1ε

) nn−1

(fflλn−1ε

) n−2n−1

+ o(e(n−4)t

n−1 )

o(e(n−2)t

) n−2n−1

+ o(e(n−4)t

n−1 )

The proof of hyperbolic AF VI (rigidity)

The analysis here is based on recent work by Huang and Wu.

If the equality holds in Penrose then σ2(A) = 0, which implies that the graph M ⊂ Hn+1 ismean convex (σ1(A) ≥ 0).

An elementary algebraic inequality then implies that

G(A) := σ1(A)I − A ≥ 0,

which means by a classical computation that the graph M is an elliptic solution of σ2(A) = 0.

Since the black hole solution is elliptic as well, the rigidity follows by applying a suitableversion of the Maximum Principle.

G(A) := σ1(A)I − A ≥ 0,

Further comments

If N is a surface of genus γ ≥ 1, we obtain

m(M,g) ≥(

4πτ2

)3/2√|Σ|16π

(1− γ +

|Σ|4π

where we should take τ2 = 4π if γ = 1. This appears as a conjectured ineuqality in papers byGibbons and Chrusciel-Simon. Also, it is related to recent work by Lee-Neves.

Also, our AF inequality is related to recent work by Brendle-Hung-Wang, where a similarinequality is proved for hypersurfaces in adSS-space by essentially the same method.

If we take Λ→ 0, we recover the Minkowski inequality in Rn, first proved by Guan-Li:

HdΣ ≥12

(|Σ|ωn−1

) n−2n−1

There is by now a lot of activity on ‘higher order’ AF inequalities in space forms with potentialapplications to Penrose-type inequalities. See, for instance, papers by Ge-Wang-Wu, dealingwith the hyperbolic case and by Makowski-Scheuer, dealing with the spherical case andposted last tuesday in the arXiv!

Further comments

m(M,g) ≥(

4πτ2

)3/2√|Σ|16π

(1− γ +

|Σ|4π

HdΣ ≥12

(|Σ|ωn−1

) n−2n−1

Further comments

m(M,g) ≥(

4πτ2

)3/2√|Σ|16π

(1− γ +

|Σ|4π

HdΣ ≥12

(|Σ|ωn−1

) n−2n−1

Further comments

m(M,g) ≥(

4πτ2

)3/2√|Σ|16π

(1− γ +

|Σ|4π

HdΣ ≥12

(|Σ|ωn−1

) n−2n−1

Further comments

m(M,g) ≥(

4πτ2

)3/2√|Σ|16π

(1− γ +

|Σ|4π

HdΣ ≥12

(|Σ|ωn−1

) n−2n−1

Further comments

m(M,g) ≥(

4πτ2

)3/2√|Σ|16π

(1− γ +

|Σ|4π

HdΣ ≥12

(|Σ|ωn−1

) n−2n−1

THANKS FOR YOUR ATTENTION!!!

levi lopes de lima - institute of mathematics and ...gelosp2013/files/levi3.pdf · an introduction...

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