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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 andChruściel-Simon.

The reference metrics (Chruściel-Herzlich-Nagy)

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

In the product manifold P� = I� × N, consider the metric

g� =dr2

ρ�(r)2+ r2h, r ∈ I�,

whereρ�(r) =

√r2 + �.

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

The metric g� is locally hyperbolic (Kg� ≡ −1).

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

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




g� =dr2

ρ�(r)2+ r2h, r ∈ I�,

whereρ�(r) =

√r2 + �.

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







g� =dr2

ρ�(r)2+ r2h, r ∈ I�,

whereρ�(r) =

√r2 + �.

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







g� =dr2

ρ�(r)2+ r2h, r ∈ I�,

whereρ�(r) =

√r2 + �.

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







g� =dr2

ρ�(r)2+ r2h, r ∈ I�,

whereρ�(r) =

√r2 + �.

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







g� =dr2

ρ�(r)2+ r2h, r ∈ I�,

whereρ�(r) =

√r2 + �.

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







g� =dr2

ρ�(r)2+ r2h, r ∈ I�,

whereρ�(r) =

√r2 + �.

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




Asymptotically locally hyperbolic manifolds (Chruściel-Herzlich-Nagy)

Definition

Fix � 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 ⊂ P�, with K compact, and a diffeomorphism Ψ : M − K → P� − K0 such that

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

ˆNr

(ρ�(divg�e − d trg�e)− i∇g�ρ�e + (trg�dρ�)

)(νr )dNr ,

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

cn =1

2(n − 1)τn−1, τn−1 = arean−1(N, h).

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

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


Definition






ˆNr


)(νr )dNr ,


cn =1

2(n − 1)τn−1, τn−1 = arean−1(N, h).




Definition






ˆNr


)(νr )dNr ,


cn =1

2(n − 1)τn−1, τn−1 = arean−1(N, h).




Definition






ˆNr


)(νr )dNr ,


cn =1

2(n − 1)τn−1, τn−1 = arean−1(N, h).




Definition






ˆNr


)(νr )dNr ,


cn =1

2(n − 1)τn−1, τn−1 = arean−1(N, h).




Definition






ˆNr


)(νr )dNr ,


cn =1

2(n − 1)τn−1, τn−1 = arean−1(N, h).



The black hole solutions I

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

Im,� = {r > rm,�},

where rm,� is the positive root of

r2 + �−2m

rn−2= 0.

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

gm,� =dr2

ρm,�(r)2+ r2h,

where

ρm,�(r) =

√r2 + �−

2mrn−2

.

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



Im,� = {r > rm,�},


r2 + �−2m

rn−2= 0.


gm,� =dr2

ρm,�(r)2+ r2h,

where

ρm,�(r) =

√r2 + �−

2mrn−2

.




Im,� = {r > rm,�},


r2 + �−2m

rn−2= 0.


gm,� =dr2

ρm,�(r)2+ r2h,

where

ρm,�(r) =

√r2 + �−

2mrn−2

.




Im,� = {r > rm,�},


r2 + �−2m

rn−2= 0.


gm,� =dr2

ρm,�(r)2+ r2h,

where

ρm,�(r) =

√r2 + �−

2mrn−2

.




Im,� = {r > rm,�},


r2 + �−2m

rn−2= 0.


gm,� =dr2

ρm,�(r)2+ r2h,

where

ρm,�(r) =

√r2 + �−

2mrn−2

.


The black hole solutions II

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

Kgm,� (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,� (∂θi , ∂θj ) = −1 +

2mrn,

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

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

(∆ρm,�)gm,� − Hessgm,�ρm,� + ρm,�Ricgm,� = 0,

which means that the Lorentzian metric

gm,� = −ρ2m,�dt

2 + gm,�,

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

Ricgm,� = −ngm,�.

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



Kgm,� (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,� (∂θi , ∂θj ) = −1 +

2mrn,






2 + gm,�,






Kgm,� (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,� (∂θi , ∂θj ) = −1 +

2mrn,






2 + gm,�,






Kgm,� (∂r , ∂θi ) = −1− (n − 2)mrn

andKgm,� (∂θi , ∂θj ) = −1 +

2mrn,






2 + gm,�,




The black hole solutions III

One easily verifies that, as r → +∞,

‖gm,� − g�‖g� = O(mr−n

),

where g� is the corresponding reference metric.

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

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

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



‖gm,� − g�‖g� = O(mr−n

),







‖gm,� − g�‖g� = O(mr−n

),







‖gm,� − g�‖g� = O(mr−n

),







‖gm,� − g�‖g� = O(mr−n

),







‖gm,� − g�‖g� = O(mr−n

),





The black hole solutions IV

It turns out that each gm,� can be isometrically embedded as a graph in (Q�, g�), whereQ� = R× P� and

g� = ρ�(r)2dt2 +

dr2

ρ�(r)2+ r2dθ2.

Notice that (Q�, g�) is locally hyperbolic!

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

ρ�(r)2(

dudr

)2=

1ρm,�(r)2

−1

ρ�(r)2, r ≥ rm,�.

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,�.

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

m =12

( |H|τn−1

) nn−1

+ �

(|H|τn−1

) n−2n−1

, τn−1 = arean−1(N).



g� = ρ�(r)2dt2 +

dr2

ρ�(r)2+ r2dθ2.



ρ�(r)2(

dudr

)2=

1ρm,�(r)2

−1

ρ�(r)2, r ≥ rm,�.



m =12

( |H|τn−1

) nn−1

+ �

(|H|τn−1

) n−2n−1




g� = ρ�(r)2dt2 +

dr2

ρ�(r)2+ r2dθ2.



ρ�(r)2(

dudr

)2=

1ρm,�(r)2

−1

ρ�(r)2, r ≥ rm,�.



m =12

( |H|τn−1

) nn−1

+ �

(|H|τn−1

) n−2n−1




g� = ρ�(r)2dt2 +

dr2

ρ�(r)2+ r2dθ2.



ρ�(r)2(

dudr

)2=

1ρm,�(r)2

−1

ρ�(r)2, r ≥ rm,�.



m =12

( |H|τn−1

) nn−1

+ �

(|H|τn−1

) n−2n−1




g� = ρ�(r)2dt2 +

dr2

ρ�(r)2+ r2dθ2.



ρ�(r)2(

dudr

)2=

1ρm,�(r)2

−1

ρ�(r)2, r ≥ rm,�.



m =12

( |H|τn−1

) nn−1

+ �

(|H|τn−1

) n−2n−1




g� = ρ�(r)2dt2 +

dr2

ρ�(r)2+ r2dθ2.



ρ�(r)2(

dudr

)2=

1ρm,�(r)2

−1

ρ�(r)2, r ≥ rm,�.



m =12

( |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 g�). 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 Chruściel-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 Q�

DefinitionA complete, isometrically immersed hypersurface (M, g)# (Q�, g�), possibly with an innerboundary Σ, is asymptotically locally hyperbolic (ALH) if there exist subsets K ⊂ M, K0 ⊂ P� suchthat M − K , the end of M, can be written as a vertical graph over P� − K0, with the graph beingassociated to a smooth function u : P� − 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 ⊂ Q� = R× P� 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 Kg� ≡ −1.

Assume from now on that M ⊂ Q� is ALH and its inner boundary Σ lies on a horizontal totallygeodesic hypersurface, say P ' P�. 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) = cnˆ

MΘ (Rg + n(n − 1)) dM + cn

ˆΣρ�HdΣ,

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

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

m(M,g) ≥ cnˆ

Σρ�HdΣ.







m(M,g) = cnˆ

MΘ (Rg + n(n − 1)) dM + cn

ˆΣρ�HdΣ,



m(M,g) ≥ cnˆ

Σρ�HdΣ.







m(M,g) = cnˆ

MΘ (Rg + n(n − 1)) dM + cn

ˆΣρ�HdΣ,



m(M,g) ≥ cnˆ

Σρ�HdΣ.







m(M,g) = cnˆ

MΘ (Rg + n(n − 1)) dM + cn

ˆΣρ�HdΣ,



m(M,g) ≥ cnˆ

Σρ�HdΣ.







m(M,g) = cnˆ

MΘ (Rg + n(n − 1)) dM + cn

ˆΣρ�HdΣ,



m(M,g) ≥ cnˆ

Σρ�HdΣ.

The Alexandrov-Fenchel inequality

We have seen thatm(M,g) ≥ cn

ˆΣρ0,�HdΣ.

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

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

cnˆ

Σρ�HdΣ ≥

12

( |Σ|τn−1

) nn−1

+ �

(|Σ|τn−1

) n−2n−1

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



ˆΣρ0,�HdΣ.



cnˆ

Σρ�HdΣ ≥

12

( |Σ|τn−1

) nn−1

+ �

(|Σ|τn−1

) n−2n−1




ˆΣρ0,�HdΣ.



cnˆ

Σρ�HdΣ ≥

12

( |Σ|τn−1

) nn−1

+ �

(|Σ|τn−1

) n−2n−1




ˆΣρ0,�HdΣ.



cnˆ

Σρ�HdΣ ≥

12

( |Σ|τn−1

) nn−1

+ �

(|Σ|τn−1

) n−2n−1




ˆΣρ0,�HdΣ.



cnˆ

Σρ�HdΣ ≥

12

( |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,� is an ALH graph as above, with Σ ⊂ P = P0,� 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 � = 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

H,

where ξ is the inward unit normal to Σ.

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

s =

arcsinh r � = 1

log r , � = 0log(2

√r2 − 1 + 2r), � = −1

In terms of this parameter,g� = ds2 + λ�(s)2h,

where

λ�(s) =

sinh s � = 1

es, � = 0es4 + e

−s, � = −1

Notice that λ̇2� = λ2� + �.

The proof of AF I


= −ξ

H,



s =

arcsinh r � = 1

log r , � = 0log(2

√r2 − 1 + 2r), � = −1


where

λ�(s) =

sinh s � = 1

es, � = 0es4 + e

−s, � = −1


The proof of AF I


= −ξ

H,



s =

arcsinh r � = 1

log r , � = 0log(2

√r2 − 1 + 2r), � = −1


where

λ�(s) =

sinh s � = 1

es, � = 0es4 + e

−s, � = −1


The proof of AF I


= −ξ

H,



s =

arcsinh r � = 1

log r , � = 0log(2

√r2 − 1 + 2r), � = −1


where

λ�(s) =

sinh s � = 1

es, � = 0es4 + e

−s, � = −1


The proof of AF I


= −ξ

H,



s =

arcsinh r � = 1

log r , � = 0log(2

√r2 − 1 + 2r), � = −1


where

λ�(s) =

sinh s � = 1

es, � = 0es4 + e

−s, � = −1


The proof of AF I


= −ξ

H,



s =

arcsinh r � = 1

log r , � = 0log(2

√r2 − 1 + 2r), � = −1


where

λ�(s) =

sinh s � = 1

es, � = 0es4 + e

−s, � = −1


The proof of AF II

It is shown that if the initial hypersurface Σ0 ⊂ P0,� 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

ũ(t , θ) = u(t , θ)−t

n − 1

converges to α in the sense that

|∇ũ|+ |∇2ũ| = o(1).

In particular,

λ�(u) ∼ λ̇�(u) ∼ et

n−1 .

The proof of AF II



ũ(t , θ) = u(t , θ)−t

n − 1


|∇ũ|+ |∇2ũ| = o(1).

In particular,

λ�(u) ∼ λ̇�(u) ∼ et

n−1 .

The proof of AF II



ũ(t , θ) = u(t , θ)−t

n − 1


|∇ũ|+ |∇2ũ| = o(1).

In particular,

λ�(u) ∼ λ̇�(u) ∼ et

n−1 .

The proof of AF II



ũ(t , θ) = u(t , θ)−t

n − 1


|∇ũ|+ |∇2ũ| = o(1).

In particular,

λ�(u) ∼ λ̇�(u) ∼ et

n−1 .

The proof of hyperbolic AF III

DefineJ (Σ) = −

ˆΣ

pdΣ, p = 〈Dρ�, ξ〉,

andK(Σ) = τn−1A(Σ)

nn−1 , A(Σ) = A/τn−1.

These quantities appear in the following preliminary result.

TheoremIf Σ ⊂ P� 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) = −

ˆΣ

pdΣ, p = 〈Dρ�, ξ〉,


nn−1 , A(Σ) = A/τn−1.



J (Σ) ≤ K(Σ),



DefineJ (Σ) = −

ˆΣ

pdΣ, p = 〈Dρ�, ξ〉,


nn−1 , A(Σ) = A/τn−1.



J (Σ) ≤ K(Σ),



DefineJ (Σ) = −

ˆΣ

pdΣ, p = 〈Dρ�, ξ〉,


nn−1 , A(Σ) = A/τn−1.



J (Σ) ≤ K(Σ),



DefineJ (Σ) = −

ˆΣ

pdΣ, p = 〈Dρ�, ξ〉,


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

dJdt

= nˆ

Σ

ρ�

HdΣ

(∗)≥

nn − 1

J ,

where (∗) is a recent inequality by Brendle.

On the other hand,dAdt

= A ⇒dKdt

=n

n − 1K,

and this immediately yieldsddtJ −K

An

n−1≥ 0.

But the asymptotics gives

limt→+∞

J −K

An

n−1= 0,

and the theorem follows.



dJdt

= nˆ

Σ

ρ�

HdΣ

(∗)≥

nn − 1

J ,



= A ⇒dKdt

=n

n − 1K,


An

n−1≥ 0.


limt→+∞

J −K

An

n−1= 0,




dJdt

= nˆ

Σ

ρ�

HdΣ

(∗)≥

nn − 1

J ,



= A ⇒dKdt

=n

n − 1K,


An

n−1≥ 0.


limt→+∞

J −K

An

n−1= 0,




dJdt

= nˆ

Σ

ρ�

HdΣ

(∗)≥

nn − 1

J ,



= A ⇒dKdt

=n

n − 1K,


An

n−1≥ 0.


limt→+∞

J −K

An

n−1= 0,




dJdt

= nˆ

Σ

ρ�

HdΣ

(∗)≥

nn − 1

J ,



= A ⇒dKdt

=n

n − 1K,


An

n−1≥ 0.


limt→+∞

J −K

An

n−1= 0,


The proof of hyperbolic AF V

We now considerI(Σ) =

ˆΣρ�HdΣ.

If K is the extrinsic scalar curvalure of Σ, then

dIdt

= 2ˆ

Σ

ρ�KH

dΣ + 2J

≤n − 2n − 1

I + 2J ,

so that the previous theorem gives

ddt

(I − (n − 1)K) ≤n − 2n − 1

(I − (n − 1)K) + 2 (J −K)

≤n − 2n − 1

(I − (n − 1)K) .



ˆΣρ�HdΣ.


dIdt

= 2ˆ

Σ

ρ�KH

dΣ + 2J

≤n − 2n − 1

I + 2J ,


ddt

(I − (n − 1)K) ≤n − 2n − 1

(I − (n − 1)K) + 2 (J −K)

≤n − 2n − 1

(I − (n − 1)K) .



ˆΣρ�HdΣ.


dIdt

= 2ˆ

Σ

ρ�KH

dΣ + 2J

≤n − 2n − 1

I + 2J ,


ddt

(I − (n − 1)K) ≤n − 2n − 1

(I − (n − 1)K) + 2 (J −K)

≤n − 2n − 1

(I − (n − 1)K) .



ˆΣρ�HdΣ.


dIdt

= 2ˆ

Σ

ρ�KH

dΣ + 2J

≤n − 2n − 1

I + 2J ,


ddt

(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−1�,

so thatL(0) ≥ (n − 1)τn−1�,

as desired.



dLdt≤ 0,


n−2n−1 (I(Σ)− (n − 1)K(Σ)) .


lim inft→+∞

L(t) ≥ (n − 1)τn−1�,

so thatL(0) ≥ (n − 1)τn−1�,

as desired.



dLdt≤ 0,


n−2n−1 (I(Σ)− (n − 1)K(Σ)) .


lim inft→+∞

L(t) ≥ (n − 1)τn−1�,

so thatL(0) ≥ (n − 1)τn−1�,

as desired.



dLdt≤ 0,


n−2n−1 (I(Σ)− (n − 1)K(Σ)) .


lim inft→+∞

L(t) ≥ (n − 1)τn−1�,

so thatL(0) ≥ (n − 1)τn−1�,

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−1� + o(e

(n−3)tn−1 ),

and ˆΣt

ρ�HdΣt = (n − 1)ˆλ̇2�λ

n−2� + o(e

(n−2)t(n−1) ).

Hence, if we use the characteristic equation λ̇2� = λ2� + �,

lim inft→+∞

L(Σt ) = (n − 1)τn−1 lim inft→+∞

fflλ̇2�λ

n−2� −

(fflλn−1�

) nn−1

+ o(e(n−2)t

n−1 )(fflλn−1� )

) n−2n−1

+ o(e(n−4)t

n−1 )

≥ (n − 1)τn−1� lim inft→+∞

fflλn−2�(ffl

λn−1�) n−2

n−1+ o(e

(n−4)tn−1 )

+

+(n − 1)τn−1 lim inft→+∞

fflλn� −

(fflλn−1�

) nn−1

(fflλn−1�

) n−2n−1

+ o(e(n−4)t

n−1 )

+


o(e(n−2)t

n−1 )(fflλn−1�

) 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−1� + o(e

(n−3)tn−1 ),

and ˆΣt


n−2� + o(e

(n−2)t(n−1) ).


lim inft→+∞


fflλ̇2�λ

n−2� −

(fflλn−1�

) nn−1

+ o(e(n−2)t

n−1 )(fflλn−1� )

) n−2n−1

+ o(e(n−4)t

n−1 )

≥ (n − 1)τn−1� lim inft→+∞

fflλn−2�(ffl

λn−1�) n−2

n−1+ o(e

(n−4)tn−1 )

+


fflλn� −

(fflλn−1�

) nn−1

(fflλn−1�

) 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−1� + o(e

(n−3)tn−1 ),

and ˆΣt


n−2� + o(e

(n−2)t(n−1) ).


lim inft→+∞


fflλ̇2�λ

n−2� −

(fflλn−1�

) nn−1

+ o(e(n−2)t

n−1 )(fflλn−1� )

) n−2n−1

+ o(e(n−4)t

n−1 )

≥ (n − 1)τn−1� lim inft→+∞

fflλn−2�(ffl

λn−1�) n−2

n−1+ o(e

(n−4)tn−1 )

+


fflλn� −

(fflλn−1�

) nn−1

(fflλn−1�

) 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−1� + o(e

(n−3)tn−1 ),

and ˆΣt


n−2� + o(e

(n−2)t(n−1) ).


lim inft→+∞


fflλ̇2�λ

n−2� −

(fflλn−1�

) nn−1

+ o(e(n−2)t

n−1 )(fflλn−1� )

) n−2n−1

+ o(e(n−4)t

n−1 )

≥ (n − 1)τn−1� lim inft→+∞

fflλn−2�(ffl

λn−1�) n−2

n−1+ o(e

(n−4)tn−1 )

+


fflλn� −

(fflλn−1�

) nn−1

(fflλn−1�

) 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,







G(A) := σ1(A)I − A ≥ 0,







G(A) := σ1(A)I − A ≥ 0,







G(A) := σ1(A)I − A ≥ 0,







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 Chruściel-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:

cnˆ

ΣHdΣ ≥

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

),




cnˆ

ΣHdΣ ≥

12

(|Σ|ωn−1

) n−2n−1

.


Further comments


m(M,g) ≥(

4πτ2

)3/2√ |Σ|16π

(1− γ +

|Σ|4π

),




cnˆ

ΣHdΣ ≥

12

(|Σ|ωn−1

) n−2n−1

.


Further comments


m(M,g) ≥(

4πτ2

)3/2√ |Σ|16π

(1− γ +

|Σ|4π

),




cnˆ

ΣHdΣ ≥

12

(|Σ|ωn−1

) n−2n−1

.


Further comments


m(M,g) ≥(

4πτ2

)3/2√ |Σ|16π

(1− γ +

|Σ|4π

),




cnˆ

ΣHdΣ ≥

12

(|Σ|ωn−1

) n−2n−1

.


Further comments


m(M,g) ≥(

4πτ2

)3/2√ |Σ|16π

(1− γ +

|Σ|4π

),




cnˆ

ΣHdΣ ≥

12

(|Σ|ωn−1

) n−2n−1

.


THANKS FOR YOUR ATTENTION!!!

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