Dynamic instability of solitons in dimensional gravity with negative cosmological constant

Transcription

1 Dynamic instaility of solitons in 4 + -dimensional gravity with negative cosmological constant Feruary 23, 2006 Astract We present heuristic arguments suggesting that perturations of Eguchi- Hanson solitons in 4 + -dimensional gravity with negative cosmological constant evolve into naked singularities, while perturations of anti-de Sitter space evolve into lack holes. In support of the first, we rigorously show that perturations of Eguchi-Hanson cannot evolve into spacetimes with horizons. Finally, we contrast these conjectures and results with an orital staility theorem for AdS-Schwarzschild. Introduction In this note, we shall show the following Theorem.. Consider asymptotically flat smooth initial data S, ḡ, K for the vacuum Einstein equations with cosmological constant Λ < 0, possessing iaxial Bianchi IX symmetry, sufficiently close to Eguchi-Hanson initial data with topology S 3 /Z p at infinity. Let M, g denote the maximal development of the initialoundary value prolem with standard oundary conditions at infinity preserving the mass, and let π : M Q denote the projection map to the 2-dimensional Lorentzian quotient Q. Then J I = Q. Theorem. can e paraphrased y the statement: Perturations of Eguchi-Hanson cannot form horizons. In addition to the aove Theorem, we have the following Theorem.2. For small perturations of Eguchi-Hanson, the conserved mass M at I + satisfies M p,λ < M < 0, where M p,λ denotes the mass of the Eguchi-Hanson soliton. On the other hand, from previous work? we have Theorem.3. There are no static regular solutions of the Einstein equations with cosmological constant Λ < 0 and with topology S 3 /Z p at infinity, with mass satisfying. One thus may conjecture:

2 Conjecture.. The Q of Theorem. has Penrose diagram: I + S Moreover, timelike infinity I + is incomplete. Conjecture. can e paraphrased y the statement: Eguchi-Hanson is dynamically unstale and perturations of it lead to naked singularities. We compare the situation with that of perturations of AdS. Here, we have again y a classical theorem of Boucher, Gions and Horowitz: Theorem.4. There are no static regular asympototically AdS vacuum spacetimes with mass M > 0. On the other hand, horizons may now form in this case. Thus we may conjecture: Conjecture.2. Consider now asymptotically AdS smooth initial data S, ḡ, K possessing iaxial Bianchi IX symmetry, sufficiently close to trivial initial data, and let Q e as efore. Then Q has a suset with Penrose diagram: H + I + S Moreover, timelike infinity I + is complete. Conjecture.2 can e paraphrased y the statement: Anti-de Sitter space is dynamically unstale and perturations of it lead to lack holes. Both instailities can e thought of as purely non-linear memory effects, necessitated y the conservation of mass at I +. In fact, somewhat ironically perhaps, their plausaility is related to the validity of linear staility. In contrast, in the case where trapped surfaces are present we have the following Theorem.5. Consider now asymptotically AdS data with Bianchi IX symmetry containing a trapped surface. Then Q contains a suset with Penrose diagram: H + I + In particular, this is true for data sufficiently close to Schwarzschild-AdS data. S 2

3 Theorem.5 can e paraphrased y the statement: Schwarzschild-AdS is oritally stale. One can conjecture that Conjecture.3. The solution in J I + suitaly approaches a Schwarzschild-AdS solution as advanced and retarded time go to infinity. The aove conjecture can e paraphrased y the statement: The Schwarzschild-AdS family is asymptotically stale. Indeed, the fact that Schwarzschild-AdS is stale while Eguchi-Hanson and AdS may not e is related to the fact that the former is emedded in a one-parameter family 2 : For now, the different conserved mass of the perturations does not deny them a static endstate. 2 Biaxial Bianchi IX AdS We will say that a spacetime M, g admits iaxial Bianchi IX symmetry if topologically, M = Q SU2, for Q a 2-dimensional manifold possily with oundary, and where gloal coordinates u and v can e chosen on Q such that g = u, vdu dv + 4 r2 u, v e 2Bu,v σ 2 + σ2 2 + e 4Bu,v σ3 2 2 where B, Ω, and r are functions Q R, and the σ i are a standard asis of left invariant one-forms on SU2, i.e. such that coordinates θ, φ, ψ can e chosen on SU2 with σ = sin θ sin ψdφ + cos ψdθ, σ 2 = sin θ cos ψdφ sin ψdθ, σ 3 = cos θdφ + dψ. 3 Geometrically, the function B measures the squashing of the three sphere. From the five-dimensional Einstein equations with a negative cosmological constant Λ = 6 we can derive the two contraint equations R µν = 2 3 Λg µν = 4 g µν 4 u Ω 2 u r = B,u 2, 5 v Ω 2 v r = B,v 2, 6 and a system of non-linear wave equations for the quantities r, Ω and B: r,uv = R,ur,v Ω2 r 3 r r, 7 u v log Ω = Ω2 R r 2 r,ur,v 3B,u B,v + Ω2 2, 8 2 Of course, one can think also of AdS as a special memer of the Schwarzschild-AdS family. And Conjecture.3 may still e true for perturations of AdS. 3

4 B,uv = 3 r,u 2 r B,v 3 r,v 2 r B,u Ω2 e 2B 3r 2 e 8B 9 Here R is the scalar curvature of the group orit, given y R = 2e 2B 2 e 8B 0 We can define a renormalized Hawking mass m = r2 + 4r,ur,v 2 + r4 2 = m are + r4 2 Note that m > m are. Using the equations of motion one estalishes the following monotonicity properties of the renormalized Hawking mass. u m 0 2 v m 0 3 Note that in the asymptotically flat case, l, the renormalized Hawking mass reduces to the standard Hawking mass. 3 Eguchi-Hanson AdS The Eguchi-Hanson metric is a memer of the iaxial class discussed in the last section. If we define r = ρ a4 6 4 ρ 4 = + ρ2 5 B = 6 ln a4 ρ 4 6 du = dt 2 dρ 7 + ρ2 l a4 2 ρ 4 dv = dt + 2 dρ 8 + ρ2 l a4 2 ρ 4 we recover the Eguchi-Hanson AdS metric in the perhaps more familiar t, ρ coordinates: + ρ2 dt 2 + dρ ρ2 l a4 2 ρ 4 ρ 2 a4 ρ 4 4 dψ + cos θdφ 2 + ρ2 4 Regularity at ρ = a, the center, enforces the coordinate ψ to have period dθ 2 + sin 2 θdφ 2 9 q 2π. + a2 On the other hand, ψ has to have period 4π p for an integer p to remove the string singularities at the poles. The two conditions lead to the relation p a 2 = for p 3. Hence the topology of the orital space is S 3 /Z p. Note that there are no regular AdS Eguchi-Hanson solitons with p = 2. 4

5 4 Proof of Theorem. 4. Non-Existence of marginally trapped surfaces It is straightforward to compute the Hawking mass of Eguchi-Hanson at infinity m = lim m = a 4 ρ 6 2 We also note that the renormalized mass diverges to for ρ a. These little facts immedeatly lead to the following Lemma : Eguchi-Hanson AdS will not form marginally trapped surfaces under perturations. Proof : Under sufficiently small perturations the renormalized mass will still e negative at infinity. By monotonicity of m, the renormalized mass is negative everywhere in the spacetime. But m = r r,ur,v + r can t hold together with r,v 0, r,u < 0, the conditions for marginally trapped surfaces. Hence such surfaces must e asent. To prove Theorem. we need to estalish the stronger statement that no horizons will form. For this we will need a rigorous setup of the initial value prolem in asymptotically locally AdS spaces within the symmetry class considered to prove an extension principle analogous to the one used in the old paper. 4.2 Initial Value Formulation for asymptotically locally AdS In this section we construct the initial value for asymptotically AdS spaces within the iaxial Bianchi-IX class of metrics Local existence in the interior Can e done in the standard way Mihalis notes. Moreover, we have an extension principle for these diamonds from our old paper Local existence near timelike null-infinity On the AdS-oundary, where r := r = 0, we choose coordinates such that u = v. The situation is indicated in the figure. We will now specify initial data on the u u = v u, v 0 û, û v û, v o ɛ u 0, v 0 v = v 0 ray satisfying the u-constraint and enforce the function Ω or equivalently, 5

6 m to e constant on the oundary. We also impose B = 0 on the oundary to make the spacetime asymptotically locally AdS. We shall e aler to infer local existence in at least a small region as indicated in the figure.all assumptions and the local existence theorem is made precise in the following Proposition: Let Ω init, r init, B init e C 2 -functions defined on [u 0, u ] v 0 satisfying the constraint r,u init 2 u r init 2 Ω init 2 = Ω init 2 rinit B,u init 2 23 Assume B init, v 0 C C 24 r init, v 0 C 25 r init,u r init, v 0 C 26 2 log Ω init, v 0 C C 27 Then there exists an ɛ > 0, dependent on the choice of C only, such that, defining û = minu 0 + ɛ, u there exist unique C 2 -functions Ω, r, B on the region R = {u, v R u v} where R = [u 0, û] [v 0, û] 28 coinciding with Ω init, r init, B init on [u 0, û] v 0, satisfying B = const, Ω = const and hence m = const on the oundary and solving the evolution equations r,uv = 3 Ω2 r 3 R + 3 r r r,u r,v r 2 29 log Ω,uv = 2 Ω2 r 2 R + 3 r,u r,v r 2 3B,u B,v + Ω B,uv = 3 2 r,u rb,v r,v rb,u 3 Ω2 r 2 e 2B e 8B 3 Note that we have reformulated the r equation in terms of r = r. as well as the constraints r,u u r 2 = 2 r B,u 2 32 r,u v r 2 = 2 r B,v Proof: The proof proceeds y reformulating the evolution in terms of a constraction mapping etween Banach spaces. We define the Banach space X := {Ω, r, B functions on R with Ω C 0 R and r, B C R } 34 with distance function d [Ω, r, B,, r 2, B 2 ] = max log Ω log, r r 2, u log r u log r 2, v log r v log r 2, B B 2 C 35 6

7 Comment: Triangle inequality needs to e checked. Motivation for this comes from the finiteness of the Hawking mass m = 2 r r,u r,v r r 4 36 We see that for m to e finite at r 0 where Ω is constant and r,u = r,v we need that r,u r is finite for r 0 to cancel the cosmological term. Let us define the suspace and consider the map given y log Ω u, v = X E = {Ω, r, B X d [Ω, r, B, 0, 0]}. 37 v u v 0 v 0 v v Φ : X E X E 38 Ω, r, B Ω, r, B 39 [ 2 Ω2 r 2 R + 3 r ],u r,v r 2 3B,u B,v + Ω2 2 ū, v dūd v + log Ω u, v 0 log Ω v, v 0 + log Ω u 0, v 0 40 v u [ r u, v = 3 Ω2 r 3 R + 3 r r r ],u r,v r 2 ū, v dūd v + r u, v 0 r v, v 0 + r u 0, v 0 4 v u [ 3 B r,u u, v = 2 r B,v + 3 r,v 2 r B,u 3 Ω2 r 2 e 2B e 8B ] ū, v dūd v One checks that v 0 v + B u, v 0 B v, v 0 + B u 0, v 0 42 u Bu, v u = v Bu, v 0 Bv, v v Bv, v o Bv, v o Bu 0, v 0 Ω u, v 0 = Ωu, v 0, 43 r u, v 0 = ru, v 0, 44 B u, v 0 = Bu, v 0, 45 Ω v, v = Ωu 0, v 0, 46 r v, v = ru 0, v 0, 47 B v, v = Bu 0, v

8 Hence the quantities Ω, r, B satisfy the correct oundary conditions. In particular, since Ω and r are constant on timelike null-infinity, the Hawking mass is also constant. We will now show that Φ maps indeed to X E. We estimate log Ω 3C + ɛ 2 2 e2e E E2 + 3E 2 + e2e 2l 2 r 2C + ɛ 2 3 e2e E e2e E + 4E E2 r 0 ɛ 2 3 e2e E e2e E + 4E E B 3C + ɛ E E E E + 3 e2e E 2 e 2E + e 8E 52 Comment: Do we need a lower positive ound on r? We can similarly estimate the u- and v-derivatives of the quantities Ω, r, B, e.g. B,u 3 C + ɛ 2 E E E E + 3 e2e E 2 e 2E + e 8E 53 B 3,v C + 2ɛ 2 E E E E + 3 e2e E 2 e 2E + e 8E 54 Thus if ɛ is small enough depending on C and E only Φ indeed maps to X E. To show that Φ is a contraction, we have to similarly ound differences, which can e done in a completely analogous fashion. Result: For sufficiently small ɛ the map Φ is a contraction. Banach s fixed point theorem then assures a fixed point Ω, r, B X E. Clearly the Ω, r, B satisfy the evolution equations and the right oundary conditions. The only thing we finally need to ensure is that the two constraint equations are also satisfied. Using the evolution equations we compute r,u 2Ω v u r 2 = v 2 r B,u 2 55 from which we conclude that the u-constraint equation will e propagated: If it is satisfied on the initial v = v 0 ray, it will e satisfied everywhere in R. In particular, it will hold on the oundary r = 0. However, from the identity u + v m = 0 56 holding on the oundary, we easily see that the v-constraint equation is identical to the u-constraint equation on the oundary. Since r,v 2Ω u v r 2 = u 2 r B,v 2, 57 the v-constraint is propagated in the u-direction from the r = 0 oundary and will e satisfied in all of R. 4.3 Extension Principle near timelike null-infinity 5 Proof of Theorem.2 To estalish Theorem.4 we will prove that the Eguchi-Hanson-AdS metric minimizes the mass in the class of iaxial Bianchi IX metrics which have topology S 3 /Z p 8

9 at infinity. This can e done in the following way. Consider the initial value pro- I + I+ EH initial data EH initial data lem for asymptotically locally AdS-spaces illustrated aove. We assume that the initial data are prescried oth on a spacelike hypersurface extending to and on a null-hypersurface extending from to the timelike null-infinity of Eguchi-Hanson. Let us prescrie exactly Eguchi-Hanson initial data on the spacelike part. On the dotted null-line we can prescrie B freely. Since the mass is assumed to e constant on null-infinity for asymptotically locally AdS spaces minimizing the mass of the space time corresponds to finding a function B on the initial data slice that minimizes the mass at infinity. 5. Gauge conditions We fix coordinates on the dotted null-initial data surface in the diagram y choosing ρ = v 58 to hold on the line. Here = 2 + ρ p a4 2 ρ 4 p 59 is a constant. At the Eguchi-Hanson functions take the following values = + ρ2 p 60 r,u = ρ,u = 5 2 a4 6 ρ 4 a4 p 3ρ 4 6 p r,v = r,u 62 Using these initial conditions at and the equations of motion 6, 7 we can derive the expressions for the quantities Ω, r,u, r,v on the whole line: ρ = r,u = r ρ 2 ρ + ρ2 p [ r ρ 2 r = ρ a4 ρ exp ρ 4 ρ 6 3 RB ρ r ρ2 63 ρ ρ B ρ 2 d ρ > 0 64 ] ρ d ρ + r ρ C r ρ 2 < r,v = a4 6 ρ 4 a4 3ρ 4 >

10 where we defined the constant C to e C = 6 a 4 3ρ 4 p ρ 2 p + ρ2 p Note that at all functions take their Eguchi-Hanson-AdS values. < The first variation We now want to extremize the functional [ I [B + tξ] = 2 4r3 B + tξ 2 Ω [B + tξ] 2 r,u [B + tξ] + rr 2 ] 3 R [B + tξ] dρ which is the mass difference etween infinity and, y varying B. Here Ω [B + tξ] = + ρ2 ρ p r exp r B + tξ 2 d ρ r,u [B + tξ] = r 2 ρ We want r r2 Ω [B + tξ] R [B + tξ] ρ d ρ + C r r 2 70 R [B + tξ] = 2e 2B+tξ 2 e 8B+tξ 7 0 = 2 d I [B + tξ] = dt t=0 A B C D + dρ 8r3 Ω 3 dρ 4r3 dρ 8r3 r,ub ρ ξ dρ 8 3 d Ω r,u B 2 dt t=0 d r,u B 2 dt t=0 r e 2B e 8B ξ We will treat the four terms seperately and refer to them y their corresponding letters given on the left. From 69 we compute d ρ Ω [B + tξ] = Ω B ξ ρ B ξ ρ d ρ 73 dt ρ t=0 p and using 73 and 70 we otain d ρ r,u [B + tξ] =,u B dt ρ t=0 p + [ ρ 4 e 2B r 2 e 8B 4Ω2 B ρ 3 r r ρ r 2 ξ ρ d ρ 3 R + r2 72 ] + 2r,u B ξ ρ d ρ 74 0

11 We will refer to the expression in square rackets as Q [B ρ] in the following. It will turn out to e a quantity that vanishes y the equation of motion. To otain the equation of motion we now have to isolate the ξ-terms in 72 using integration y parts. The A -term will leave us with 8r 3 Ω 3 A A2 d Ω r,u B 2 = dt t=0 [ 6r 4 r,u B 3 ] ξ ρ dρ 8r 3 r,u B 2 ρ Let us consider the B term. 4r 3 d r,u B 2 dt dρ = t=0 B B2 8r 3 2 r,u B 4r B 2 B ξ ρd ρ dρ ρ B ξ ρ d ρ dρ ρ r 2 Q [B ρ] ξ ρ d ρ dρ The term B cancels A2. The term B2 can e simplified y an equation y parts, using the constraint 6: 2 B2 = Q [B ρ] ξ ρ d ρ 76 The terms C and D of 72 simply read [ 8r 3 r,u B ρ C + D = r 75 e 2B e 8B ] ξ ρ dρ 77 The term C can e comined with the term A again using the constraint 6 to yield 8r 3 r,u B A + C = ξ ρ dρ 78 Ω2 We now add up the terms A, B2, C and D to arrive at the equations of motion. 8r 3 r,u B 0 = + 8 r e 2B e 8B + 3 [ r 2 Ω2 + Ω2 8 e 2B e 8B ] 8Ω2 B 3 r r ρ 3 R + r2 + 4r,u B 79 This expression can e simplified consideraly, if we use equation 7, which in our gauge can e written as ρ r 2 r 3 r2 r,u = R 80

12 The result is [ 4 3 r Since e 2B e 8B 4Ω2 B r ρ ] 3 R + r2 + 2r,u B = 0 8 0, 82 which is a consequence of equation 64, the equation of motion is equal to the term in square rackets and can e written as B ρ = B fρ Ω2 R rr,u 3 + r2 e 2B 3r 2 e 8B 83 r,u where fρ = ρ4 + a 4 3ρ 4 + a 4 ρρ 4 a 4 3ρ 4 a 4 There is also a neat way of writing this equation: ρ r 3 r,u B = Ω2 r 3 which upon integration leads to the result r 3 e 2B e 8B dρ = a4 3ρ 2 p 84 e 2B e 8B 85 The mass is extremized if B satisfies 83. Of course the equation is supplemented y oundary conditions. At the function B has to assume its Eguchi-Hanson value whereas at it has to vanish to satisfy the criterion of asymptotically locally AdS. It is straightforward to check that B EH = 6 log a4 ρ 4 87 indeed satisfies 83 and the oundary condtions. To prove Theorem.4 it suffices to show that we are in a local minimum. This can e done y calculating the second variation, which we are going to do in the next susection. 86 Comment: However, one could e more amitious and prove that Eguchi-Hanson is the gloal minimizer. One way to achieve this would e to show uniquess of the solution to 83 via the maximum principle. This work is currently in progress. An important property of extremizers follows directly from the maximum principle applied to the differential equation 83: Lemma : everywhere in [,. Any extremizer satisfies B 0 B 0 B ρ 0 88 Proof: At an extremum of a solution B the differential equation 83 reads B ρ = 3r 2 r,u e 2B e 8B. 89 2

13 The term e 2B e 8B has the sign of B, the term rω2 3r 2 r,u is always positive ecause r,u is always negative. By the maximum principle, there can only e maxima B ρ < 0 if B itself is negative and minima if B is positive. Since B starts at with a positive value namely its Eguchi-Hanson value and is zero at infinity, the function B must go to zero monotonically. Thus we conclude B < 0 and B > 0. The statement aout the second derivative then follows from applying these facts to 83. Corollary: Different extremizers can only intersect at their oundary points. Furthermore, using an asymptotic expansion, one can show Lemma 2: Nontrivial solutions near infinity ehave like B ρ 4 90 Proof: From equation 63 we can derive that r = ρ + O ρ 3 In our gauge, the Hawking mass can e written as m = r2 r + r4 2 κ 2 92 with κ = Ω2 4r,u. Consider first the flat case, l. To ensure finite mass at infinity, asymptotically ehave like Hence κ = + O ρ 2 κ = + O ρ 2 κ 9 should Using that R 3 2 near infinity, the differential equation 83 written out to lowest order in ρ reads Bρ = 3 ρ B + 8 ρ 2 B 95 This Euler-type equation has a unique solution that decays at infinity, B = ρ. 4 For the AdS-case we need a slightly different argument. In this case the ehaviour of κ to ensure finite Hawking mass at infinity is changed to κ = ρ O ρ 2 96 κ = l2 ρ 2 + O ρ 4 97 Hence, to lowest order, the diffential equation 83 in the AdS-case reads B ρ = 5 ρ B 98 Again this equation has a unique solution that decays, namely B = ρ 4. 3

14 5.3 The second variation The second variation is d 2 dt 2 I [B + tξ] = t=0 24r 3 Ω 4 r,u B 2 d 2 Ω [B + tξ] dρ dt t=0 8r Ω 3 r,ub 2 d 2 dt 2 Ω [B + tξ] dρ t=0 6r Ω 3 B 2 d d Ω [B + tξ] r,u [B + tξ] dρ dt dt t=0 t= Ω 3 r d,u Ω [B + tξ] B ξ dρ dt t=0 4r 3 d 2 5 dt 2 r,u [B + tξ] B 2 dρ t=0 6r 3 d 6 r,u [B + tξ] B ξ dρ dt t= r 3 r,u ξ 2 dρ 6 rr 3 ξ2 e 2B 4e 8B dρ We will refer to these terms on the right y their corresponding oxed numers given on the left. To evaluate this expression we recall the formulae 73 and 74, where we can simplify the latter using the vanishing of the square racket y the equation of motion. Note that we then have the relation d r,u [B + tξ] = 2 r,u dt Ω t=0 d Ω [B + tξ] dt t=0 Furthermore, we will need the expressions d 2 dt 2 Ω [B + tξ] = d Ω [B + tξ] Ω dt t=0 t= rr,u B ξρ ρ + Ω ξ 2 ρ d ρ 0 and d 2 dt 2 r,u [B + tξ] = t=0 [ ρ r 2 r ξ2 e 2B 4e 8B r ξ e 2B e 8B Ω d Ω [B + tξ] r dt t= r2 d R d 2 ] r dt 2 Ω [B + tξ] + Ω [B + tξ] d ρ dt t=0 t=0 02 4

15 We insert the relation 00 into term 3 and 6 of 99 and equation 0 into term 2 leading to some cancellations. Finally, we insert 73 and use partial integration to arrive at the following simplified expression for the first three terms of r 5 r,u +2+3 = 2 B 4 ξ 2 dρ 6r 3 + r,u B 2 8r 3 + r,u B 2 ρ ρ 2 B ξ ρ d ρ dρ 2 ξ 2 d ρ dρ 03 The terms 4 and 6 together simplify to 4+6 = 64r 4 r,uξ B 2 ξ dρ 04 Term 5 is the most painful. We have to insert 02 and use several partial integrations, thery applying 6 continuously. Let us do this separately for each term in the square racket of 02 and refer to the results as 5a, 5, 5c. 5a = 6 rξ 2 3 e 2B 4e 8B dρ + 6rξ 2 3 e 2B 4e 8B dρ 05 The second term oviously cancels the term 8 in 99. 5a + 8 = r 6 rξ 2 e 2B 4e 8B dρ 06 3 The 5 -term will consist of two terms ecause we have to insert 73. We refer to these two terms as 5I and 5II. 5I = 64r 2 B 3 e 2B e 8B ξ 2 dρ 07 5II = r + 3 e 2B e 8B ρ ξ B ξ ρ d ρ dρ 3 ρ p 08 The 5c term will me made of four terms ecause we have to insert the relation 0 and then the square of 73. We will refer to these four terms as 5cI... 5cIV. 5cII = 5cI = 32 r 3 B r 2 B ξ 3 ρ r2 R 3 r2 R ξ 2 dρ 09 B ξ ρ d ρ dρ 0 5

16 5cIII = 5cIV = 8 r 4 r 3 R r2 ρ 2 B ξ ρ d ρ dρ 3 ρ r2 R ξ 2 ρ d ρ dρ 2 We now find that the second term of 03 cancels 5II + 5cII + 5cIII using the equation of motion 83: +2+3 second + 5II + 5cII + 5cIII = 0 3 Using the equation 80 we can susume the term 5cIV together with the last term of 03 and 7 to give +2+3 last + 5cIV + 7 = r 8r 3 r,u ξ 2 dρ 4 Finally, the sum of 5I, 5cI, the first term of 03 and the term 04 simplifies to give 5I + 5cI first = r 64r 4 2 r,u B 2 ξξ dρ 5 We are ready to state the final result: d 2 dt 2 I [B + tξ] = t=0 8r [ ] r 3 r,u dρ ξ 2 + 8r4 r,u B 2 ρ p 2 ξξ 2Ω2 r e 2B 4e 8B ξ 2 3 = 8r [ r 3 2 r,u dρ ξ ρ r 2 B r r,u 2Ω2 r ] e 2B 4e 8B ξ Positivity of the second variation 6 We want to show that the integral on the right hand side is positive. Unfortunately, the integrand is not manifestly positive. However, for large enough we can show the following Proposition : For any extremizer, we can choose large enough such that the integrand of 6 is manifestly positive. Proof sketched: The ξ 2 in 6 is already manifestly positive. Using the asymptotic ehaviour of an extremizer Lemma 2 aove and of r,u, Ω, it is straightforward to show that for large enough ρ the ξ 2 term in 6 will also ecome positive. More precisely, let us compute the weight w of the ξ 2 term in 6. Using the equation of motion 83 we otain w = 8 r 2 B e 2B e 8B 4 r3 B 2 R r2 2Ω2 r e 2B 4e 8B 3 7 6

17 For the general case, we will have to work harder. Let us plug in the Eguchi- Hanson values into 6 and use Hardy inequalities to prove positivity of 6 at least for the Eguchi-Hanson case. The Eguchi-Hanson values are r = ρ a4 ρ 4 6 Ω ρ = ρ 2 ρ 4 a 4 4 r,u = + ρ2 a 4 3ρ 4 6 ρ 4 a4 ρ and the positivity of 6 at Eguchi-Hanson is equivalent to the positivity of [ dρ + ρ2 ρ 4 a 4 ] 2 ξ 2 ρ [ 2ρ 3ρ 8 6a 4 ρ 4 a 8 ] 2 32a 8 ρ 3 + a 4 3ρ 4 2 a 4 3ρ 4 2 ξ 2 > 0 This is an inequality of Hardy-type. We are going to prove it seperately for the flat case and the additional l -terms in case of a cosmological constant. Since ρ 2 p > a we will set the left oundary equal to a and prove positivity for all functions ξ vanishing at x = a and infinity. The flat case: For the flat terms, we need to prove that [ 2ρ 3ρ 8 + 6a 4 ρ 4 + a 8 ] [ dρ a 4 3ρ 4 2 ξ 2 ρ 4 a 4 ] < dρ ξ ρ a holds for any ξ vanishing at a and at infinity. We can scale out the parameter a y setting ρ = x a: [ 2x 3x 8 + 6x 4 + ] [ dx 3x 4 2 ξ 2 x 4 ] < dx ξ,x x Note that for x c = a the inequality is trivially satisfied. Step : We show that the inequality 23 holds for any ξ, which vanishes at x = and takes aritrary values at the point x =.04. [.04 2x 3x 8 + 6x 4 + ].04 2 x dx 3x 4 2 ξ 2 = 5ξ x 6.04 dx 2 6x ξξ,x.04 2x 3x 8 + 6x x 4 2 ξ x dx 2 3x 4 + x 6 2 x + 6x 4 3x 8 ξ,x 2 dx 24 using Hoelder s inequality and therefore [ 2x 3x 8 + 6x 4 + ].04 dx 3x 4 2 ξ x 2 3x 4 + x 6 2 x + 6x 4 3x 8 ξ,x 2 dx 25 7

18 holds for any ξ vanishing at x =. It is easily shown that holds in the interval [,.04]. 2 + x 2 3x 4 + x 6 2 x + 6x 4 3x 8 < x 4 2 x 26 Step 2: We show that 23 holds for any function ξ supported in [.04, c+0.] with ξc + 0. = 0 and ξ.04 aritrary. Since any ξ vanishing at x = and x = can e decomposed into a ξ vanishing at x = and x = c + 0. and a ξ 2 having support only in [c, and therefore trivially satisfying 23, this step will prove that 23 holds for all ξ vanishing at x = and x =. Define the weight functions and We have to show wx = { 2x 3x 8 +6x 4 + 3x 4 2 if x [.04, c] 0 if x [c, c + 0.] vx = x 4 2 x c wxξ 2 dx < K c vx ξ,x 2 dx 29 holds for functions ξ vanishing at the right oundary and a constant K. For such weighted Hardy-inequalities there is the following Theorem: cf Theorem 6.2 in [?] Let wx and vx e positive, measurale functions in the interval [a, ] and ξ e an asolutely continuous function also defined in the interval [a, ] and vanishing at the point. Then the inequality a wxξ 2 xdx C 2 holds for some constant C if and only if x B := sup a<x< F x := sup a<x< a a vx ξ,x 2 xdx 30 wxdx x Furthermore, the optimal constant C in 30 satisfies dx < 3 vx B C 2B 32 In our case, we can compute the function F x explictly from the given weights 28 and 27. The plot of F x in [.04, c + 0.] is given in Figure. Clearly, F x < 2 everywhere. By the theorem, the ideal constant C in 30 is smaller than one for the weights considered. Hence 29 holds even for some K smaller than. The AdS terms For the additional AdS terms in 2 we proceed along the same lines. We have to show 32a 8 ρ 3 dρ a 4 3ρ 4 2 ξ2 < 2 ρ ρ 4 a 4 ξ 2 dρ 33 a a 8

19 Figure : The function Fx or equivalently, using rescaled variales 32x 3 3x 4 2 ξ2 dx < 2 x x 4 ξ,x 2 dx 34 The latter inequality should hold for any function ξ vanishing at and infinity. Step : We show that 34 holds in [,.07] for functions ξ vanishing at x = and admitting aritrary values at x =.07:.07 32x x 4 2 ξ2 dx 0.9ξ x ξξ,x dx x x 3x 4 2 ξ2 dx 4 2 x 3 ξ,x 2 dx 36 and therefore Now since.07 32x x 4 2 3x 4 2 ξ2 dx x 3 ξ,x 2 dx 37 2 x 4 2 holds for x [,.07] the first step is complete. x 3 < 2 x x 4 38 Step 2: We show that 34 holds in [.07, M] for functions ξ, aritrary at x =.07 and vanishing at x = M, where M is an aritrary large numer. To do this we again construct the critical function in the theorem aove and show that it is everywhere smaller than 2. Figure 2 shows a plot of the critical function F for the case M = 0. 9

20 Figure 2: The function F x 20