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5.1 topology of the conformal boundary Σ

5.1.2. Null Cone for Points in Σ d

0+√1

2 i+j

,0+√1

2 i+j

=gp(0,0) + 1

2 gp(i,i) +gp j,j

+gp i,j

=gp i,j ,

where in each step we used the results of the first calculation. This proves the remaining

equa-tions.

5.1.2. Null Cone for Points in Σ

d

The aim of this section is to show local equality of Σ and the geodesic null cones in a neigh-bourhood of pointsp∈Σd. In this section, it will be useful to interpret the gradient vector field gradσas a field onΣ. To shorten the notation, gradσ|Σ:Σ→TMwill just be denoted gradσif there is no risk of confusion.

We will first specify the structure ofΣthat is induced by the Morse lemma in a neighbourhood of points inΣd.

Lemma5.1.4. Let be p ∈ Σd, then there is a neighbourhood U of p such that Σ∩ U is a cone quadric over p.

Proof: Following the proof of Lemma 1.2.5, there is a neighbourhood U of p and a diffeomor-phism ϕ : U → Rn such that Σ∩ U is the preimage of 0 with respect to − ϕ02

+ ϕ12

+

· · ·+ ϕn−12

. The sign does not matter here. Finally Σ∩ U is homeomorphic to the (n−1)

dimensional double cone inRn.

Lemma5.1.5. The restricted map gradσ|Σcc→TM has the following properties.

(i) gradσp is an element of TpΣc ⊂ TpM for each p ∈ Σc and hence a tangent vector field of Σc. Consequently one has gradσ|Σ

c∈X(Σc). (ii) The(1, 1)tensor∇gradσfulfils onΣ

Xgradσ=|ΣρX. (5.1) (iii) Let X∈X(M)be a vector field that is tangent toΣc, i.e. g(X, gradσ) =0onΣc. Then∇gradσX

also is tangent toΣc. In other words

∀p∈Σc:gp Xp, gradσp

=0 ⇒ ∀p∈Σc:gp

gradσXp, gradσp

=0.

88 Chapter 5: almost einstein structures with vanishing almost scalar curvature

Proof: Considerp∈Σc. For (i) it suffices to show gradσp∈TpΣc. Letγ:I→Σcbe an arbitrary smooth curve withγ(0) =0 and ˙γ(0) =X∈TpΣc. Thenσγ≡0 alongγand therefore

0= d dt

t=0σγ=gp gradσp,Xp

=dσp Xp

. (5.2)

Since in addition we haveg(gradσp, gradσp) =0, gradσpmust be an element ofTpΣc. Assuming gradσp would be in the algebraic complement then any basis of TpΣ could be completed by gradσp to a basis of TpM, which on the other hand would be annihilated by dσp. This would contradictpto be an element ofΣc.

For the second claim we may use that by (1.5) and (1.6) on an almost Einstein structure (M,g,σ) we find

∇gradσ+σP]+ρid=0.

Consequently onΣthis gives∇gradσ = ρid.

By (i) gradσ is tangent toΣ. Hence the last claim is consequence of the following calculation g

gradσX, gradσ

=|Σ gradσ(g(X, gradσ) )−g

X,∇gradσgradσ

(i i)

= Σ

gradσ(0) +ρg(X, gradσ)

=|Σ0.

As gradσ identifies the isotropic direction on Σc in Lorentzian signature this implies that every tangent null vector in TΣc will be a multiple of gradσ. This has the following conse-quences.

Lemma5.1.6. Let p ∈ Σc be a point, X ∈ TpΣ a null vector andγ : I → Σc a smooth curve with non-vanishing null tangent vectorγ. Then˙

(i) there is a constantκRsuch that X = κ·gradσp. (ii) γis a null pregeodesic.

(iii) any integral curve ofgradσ starting onΣcstays withinΣc for all values of its parameter.

(iv) any integral curve of gradσonΣcis a null pregeodesic.

(v) γcan be reparametrised to an integral curve ofgradσ.

Proof: The statements will be proven separately.

(i) The first claim is a consequence of Σcbeing a null submanifold. Assume X ∈ TpΣ to be a null vector in TΣc. From (5.2) we get g(gradσ,X) = 0 such that span{gradσ,X} is a totally isotropic vector subspace of TpM. As the signature of M is Lorentzian, the subspace has dimension one and the claim follows.

(ii) From part(i)we conclude that there is a smooth map c : I →R\ {0}such that ˙γ(t) = c(t)gradσγ(t). We now showγ to fulfil Equation (1.78)

(∇γ˙γ˙) (t) = (∇γ˙c·gradσ) (t)

= c˙(t)gradσγ(t) +c(t) (∇γ˙ gradσ) (t)

(5.1)

=

c˙(t)

c(t) −c(t)ργ(t)

˙ γ(t). Consequently it is a pregeodesic.

5.1 topology of the conformal boundary Σ 89

(iii) As stated before the restriction of gradσ to Σc is a vector field in X(Σc). Therefore the differential equation ˙η(t) = gradση(t) on Σc with initial data η(0) = p ∈ Σc and

˙

η(0) = gradσp ∈ TpΣc has a unique solution η : (−e,e) → Σc. The claim follows directly.

(iv) Letη : I →Σcbe an integral curve of gradσ. Then by definition ˙η(t) = gradση(t) is a non-vanishing null vector for all t ∈ I. Hence η satisfies the requirements of (i i) and so is a pregeodesic.

(v) Letη : J →Σcbe the smooth null curve with non-vanishing tangent vector, parametrised by the interval J = (t0,t1). By (i) there is a smooth non-vanishing map f : J → R such that ˙η(t) = f(t)gradση(t). Reparametrisation with a function h : ˜J := (s0,s1) → J leads to the required differential equation:

(η◦h)0 =h0 · f ◦h·gradσγ◦h

=! gradσγ◦h

where we denote the derivative with respect to the parameter on ˜J with a prime. The resulting ordinary differential equationh0· f ◦h = 1 is implicitly solved by separation of variables to

Z h(s) h(s0)

d x

f(x) = s−s0.

The integral on the left-hand side is strictly monotonic as a function of h(s) due to non-vanishing of f. Hence a solution h(s) to this equation exists. It is strictly monotonic too and has an inverse. As a resultηcan be reparametrised to an integral curve of gradσon Σc.

Lemma5.1.7. Letγ :(α,β) → Σc be a maximal integral curve ofgradσ. If the limitlimt→αγ(t) orlimt→βγ(t)exists, then it is an element ofΣd

Proof: The gradient vector field is non-vanishing onΣc. Hence for each point in Σc we can find a chart neighbourhood such that at least one component of gradσ is nowhere vanishing on this neighbourhood. By Lemma1.1.16every maximal integral curve entering that neighbourhood will leave it within finite values of the parameter. Therefore no point in Σc can be the limit of a maximal integral curve. SinceΣ is a closed subset of M and γ(t) ∈ Σfor all t ∈ (α,β), the limit must be in Σand by the previous considerations in Σ\Σc = Σd if existent. This proves

the claim.

Lemma5.1.8. Every point p∈ Σd is an attractor or repeller ofgradσin M.

Proof: Let be(Up,x)a normal coordinate chart atp, withx = (x0, . . . ,xn−1) :UpRn. The intention in this proof is to make use of the theorem of Poincaré and Lyapunov1.1.17.

In coordinates we have Hessgσ =

n−1 i,

j=0

ijσ

d xi⊗d xj

n−1 i,j,k

=0

Γki jkσ

d xi⊗d xj

for the Hessian ofσ and

gradσ =

n−1 i,j

=0

gi j(jσ)i

=

n−1 i

=0

gradσii

90 Chapter 5: almost einstein structures with vanishing almost scalar curvature

for the gradient. Here gi j are the components of the inverse matrix of g at y ∈ Up. The coordinate representationgrad^σ := (gradσ0◦x−1, . . . , gradσn−1◦x−1) is a smooth map fromx(Up) ⊂RntoRn. We will drop the tilde in the following considerations and identify the components of gradσ with this representation. In normal coordinates the Christoffel symbols Γki j vanish at p and hence

Hessσp =

n−1 i,j

=0

ijσ

(p)d xi⊗d xj

p. Corollary1.4.12(i) then gives

ijσ

(p) =−ρ(p)gi j(p). (5.3) We now calculate the coordinate Jacobian gradσ0. Using Einstein notation its components are

(gradσ0)ik = kgi jjσ

= (kgi j)jσ+g(i j∂kjσ.

By requirements of the lemma,pis a critical point with gradσ(p) = 0 and hence the first term vanishes at p. Combining the result with Equation (5.3) gives

(gradσ0)ik(p) = −ρ(p)

j

gi j(p)gk j(p)

= −ρ(p)δik, (5.4)

where δik are the components of the identity matrix. By Corollary 1.4.12(ii) this implies that gradσ0 does not vanish at p and moreover is positive or negative definite depending on the sign ofρ. Now the theorem of Poincaré and Lyapunov can be applied, which gives the claim. In particular p is an attractor forρ(p) < 0 and an repeller forρ(p) > 0.

Corollary 5.1.9. Let be p ∈ Σd. Then with respect to gradσ, there is an attracting or repelling neighbourhood Up of p that also is a normal neighbourhood of p. In particularU := expp1(Up)is a convex neighbourhood of0in TpM.

Proof: Due to Lemma5.1.8, pis an attractor or repeller. Following the calculations therein, Equa-tion (5.4) gives(gradσ0)ik(p) = −ρ(p)δikin normal coordinates(U,ϕ). Therefore gradσ0(0) clearly is positive or negative definite and hence using Proposition 1.1.18 there is an open ball Br(0) ⊂ ϕ(U), which is an attracting or repelling neighbourhood of 0 with respect to ϕgradσ. ConsequentlyUp := ϕ−1(Br(0) )is an attracting or repelling neighbourhood. The ball is taken with respect to the canonical vector space metric inRn.

Moreover, Br(0) corresponds to an open ball in TpM in a suitable basis as we started with normal coordinatesϕ. Specifically there is a basis{ei}ofTpMsuch thatY ∈ Br(0)is mapped

to exp−1ϕ−1(Y) =Y0e0+· · ·+Yn−1en−1 inU.

This corollary gives rise to the following definition.

Definition 5.1.10. Let be p ∈ M an attractor or repeller of X ∈ X(M). A neighbourhood U ⊂ TpM of 0 is calledstable neighbourhoodwith respect to X at p, if (i) expp(U)is a normal neighbourhood of p and (i i) expp(U) is an attracting or repelling neighbourhood of p with respect toX.

Lemma5.1.11. For p ∈ Σd there are convex sets U ⊂ K ⊂ U˜ ⊂ TpM containing the origin 0 such thatKis compact andU,U˜ are stable neighbourhoods with respect togradσ at p.

A proof can be found in the appendix.

5.1 topology of the conformal boundary Σ 91

Remark. As a matter of fact ˜U can be chosen such that its image under expp is a subset of some convex neighbourhood of p. First choose some convex neighbourhood V of p. Then ϕ

exp−1p (V)is well defined and open inRn. Hence using the construction of the proof above r can be scaled down such that the ball Br(0) is a subset of expp1(V) and still is attracting or repelling. The last claim holds, since every ball inside an attracting ball sharing its centre is attracting or repelling itself.

Proposition5.1.12. For p ∈ Σdthe singularity setΣlocally coincides with the geodesic null cone in p.

In particular there is a neighbourhood Upof p such that UpΣ = Cp(Up).

Proof: First using Lemma 5.1.11we choose convex subsets U ⊂ K ⊂ U˜ ⊂ TpM such that K is compact andU, ˜Uare stable neighbourhoods of the origin. We will in advance require ˜Uto be small enough such that expp(U˜)is subset of some convex neighbourhood of p. We then define the following sets

K := expp(K) U˜ := expp(U˜) C:= CpM∪ {0}∩K C˜ :=CpM∩U˜

S := expp1(Σ∩K) ⊂ K S˜ := exp−1p Σc∩U˜⊂ U˜

It suffices to show that S is an open and closed subset of C. We remark that p ∈ C∩S by definition such that it suffices to show thatS\ {0}is open and closed inC\ {0}.

SinceΣ is a closed set, we see immediately thatSis closed. We will show it to be a subset of C. Considerx ∈ Σ∩K ⊂expp

. Since ˜Uis stable, there is a null integral curveγ : I →Σc

with γ(0) = x and either I = (−∞, 0] or I = [0,∞) such that γ(t) → p for t → ±∞.

Due to Lemma 5.1.6(ii) γ is a pregeodesic and can be reparametrised to a segment of a null geodesicη : [0,t0) →Σcwith η(0) = x and η → p fort →t0. Since expp

is subset of a convex neighbourhood of pand there is no geodesic ending inside M(see Lemma1.2.4),η can be extended to the value t0 by η(t0) = p. Reparametrising η to a radial geodesic in p gives x = expp(−t0η˙(t0) )for the null vector ˙η(t0). Therefore we have−t0η˙(t0) ∈ CpM∩K⊂ C and we conclude thatS is a closed subset ofCand consequentlyS\ {0} is a closed subset of C\ {0}.

Using the same arguments as above we get ˜Sto be a subset of ˜C. At this point we summarise the results gained so far as follows

S\ {0} ⊂closedC\ {0} S˜ ⊂ C˜.

Now we will showS\ {0}to be open inC\ {0}. First we assumex ∈ Σc∩U˜ and hence by the previous considerations exp−1p (x) ∈ S˜ ⊂ C˜. It suffices to show that there is a neighbour-hoodUofx such that

exp−1p (U∩(Σ∩K) ) = exp−1p (U)∩C,

since then exp−1p (U)∩exp−1p ( (Σ∩K) ) = expp1(U)∩C and therefore the set on the left-hand side contains expp1(x) and is open in C\ {0}. According to the first part of the proof we already have

exp−1p (U∩(Σ∩K) ) = exp−1p (U)∩S ⊂ exp−1p (U)∩C independent of the choice ofU.

For a concrete choice we assert Σc to be a submanifold of M. Therefore we can choose a chart(U,ϕ)of x such that ϕ(U∩Σc) = V× {0} ⊂ Rn and V ⊂ Rn−1 is an open set. We

92 Chapter 5: almost einstein structures with vanishing almost scalar curvature

assumeU to be small enough such that U ⊂ expp U˜ \ {0}. In particular (exp−1p (U), ˜ϕ) with ˜ϕ := ϕ◦exppis a chart for a neighbourhood of exp−1p (x)inTpM. Moreover we have

˜ ϕ

exp−1p (U)∩C˜

ϕ˜

expp1(U)∩S˜

= ϕ˜

expp1(U∩Σc)

= ϕ(U∩Σc)

= V× {0}.

C = CpM∩U˜ topologically is an open subset ofCpM and hence an (n−1)-dimensional sub-manifold. Consequently(exp−1p (U)∩S˜, ˜ϕ)must be a chart for a neighbourhood of expp1(x) in ˜C. In particular exp−1p (U)∩S˜ is a open neighbourhood of exp−1p (x)in ˜C. By arbitrariness of x ∈ Σc∩U˜ we get that ˜S is open in ˜C. Hence S\ {0} = S˜ ∩K is an open subset of C\ {0} =C˜ ∩K. Therefore we haveC\ {0} =S\ {0}and hence

C= S.

We now choose Up := expp(U). Then sinceU⊂ K, we get CpM∩U= exp−1p Σc∩Up

. (5.5)

Proposition1.2.8(Equation (1.85)) states that in a normal neighbourhood the geodesic null cone Cp(Up)in p is the image of the tangent null cone under the exponential map. Hence

Cp(Up) = expp(CpM∩U)

(5.5)

= Σc∩Up.

This proves the claim.