Cosmological Constant, Near Brane Behavior and Singularities · Introduction Outline Introduction...

Cosmological Constant,Near Brane Behavior and Singularities

Daniel Junghans

Department of Physics & Institute for Advanced StudyHong Kong University of Science and Technology

arXiv: 1301.5647 (with F. F. Gautason and M. Zagermann)

Outline

Introduction

The cosmological constant as a sum of source terms

Singular anti-D3-branes in KKLT

Conclusion

Daniel Junghans Cosmological Constant, Near Brane Behavior and Singularities 2 / 34

Introduction

Outline

Introduction

Conclusion

Introduction

Localised sourcesI Localised sources (D-branes, O-planes) are important ingredients in

string theory/supergravity compactifications:SUSY breaking, tadpole cancelation, dS uplifts, ...

e.g. Kachru, Kallosh, Line, Trivedi 03

transverse space

profile

I Equations of motion (Einstein, dilaton, RR fields) include deltafunctions:

Sloc = µpep−3

4 φ∫

d10x√g δ(9−p)(x)− µp

∫Cp+1 ∧ δ(9−p)

Complicated dynamics in the compact dimensions...

Usually too hard to solve!

Introduction

transverse space

profile

Sloc = µpep−3

4 φ∫

∫Cp+1 ∧ δ(9−p)

Usually too hard to solve!

Introduction

transverse space

profile

Sloc = µpep−3

4 φ∫

∫Cp+1 ∧ δ(9−p)

Usually too hard to solve!Daniel Junghans Cosmological Constant, Near Brane Behavior and Singularities 4 / 34

Introduction

Smearing

I Common trick: only solve integrated equations of motion but neglectbackreaction in compact space

I equivalent to assuming δ(9−p) → const.

transverse space

profile

transverse space

profile

“smeared limit”, “probe approximation”, “satisfy tadpole condition”

Easier! But also justified?

Introduction

Smearing

transverse space

profile

transverse space

profile

Easier! But also justified?

Introduction

Smearing

transverse space

profile

transverse space

profile

Easier! But also justified?Daniel Junghans Cosmological Constant, Near Brane Behavior and Singularities 5 / 34

Introduction

Does this make any sense?I In general: no guarantee that smeared solutions approximate fully

backreacted ones! Observables could change, or solution could evencease to exist...

Douglas, Kallosh 09Blaback, Danielsson, DJ, Van Riet, Wrase, Zagermann 10, 11

I Intuitive argument: smearing conceals the non-trivial charge/tensionprofile in compact space. Could hide forces and lead to fake solutions.Balance of forces between sources and flux could be due to smearing!

I Exception: when ingredients (sources, fluxes) saturate BPS bound,smearing is guaranteed to be ok by no-force condition

Introduction

Does this make any sense?

I Taking into account backreaction is important but complicated...I Most constructions relevant for phenomenology/cosmology only

obtained in the smeared limit, effects of backreaction poorlyunderstood!

I Would be good to be able to compute observables (e.g. cosmologicalconstant) without resorting to questionable approximations likesmearing!

Often possible (for the cosmological constant)!

Introduction

Does this make any sense?

I Taking into account backreaction is important but complicated...I Most constructions relevant for phenomenology/cosmology only

obtained in the smeared limit, effects of backreaction poorlyunderstood!

I Would be good to be able to compute observables (e.g. cosmologicalconstant) without resorting to questionable approximations likesmearing!

Often possible (for the cosmological constant)!

Outline

Introduction

Conclusion

Scaling symmetries

Type II supergravity exhibits 2 global scaling symmetries, due to dilatonand mass scaling of the classical action

Witten 85; Burgess, Font, Quevedo 86

I Dilaton scaling:

e−φ → se−φ, gMN →√

sgMN , Cn−1 → sCn−1

S = Sbulk + Sloc → s2Sbulk + sSloc

I Mass scaling:

gMN → t−2gMN , Cn−1 → t−(n−1)Cn−1, B → t−2B

S = Sbulk + Sloc → t−8Sbulk +∑

pt−p−1S(p)

2 scaling symmetries, broken by localised sources!

Scaling symmetries

Type II supergravity exhibits 2 global scaling symmetries, due to dilatonand mass scaling of the classical action

Witten 85; Burgess, Font, Quevedo 86

I Dilaton scaling:

e−φ → se−φ, gMN →√

sgMN , Cn−1 → sCn−1

S = Sbulk + Sloc → s2Sbulk + sSloc

I Mass scaling:

gMN → t−2gMN , Cn−1 → t−(n−1)Cn−1, B → t−2B

S = Sbulk + Sloc → t−8Sbulk +∑

pt−p−1S(p)

2 scaling symmetries, broken by localised sources!

The MethodI Scaling symmetries useful for computing Λ! Shown before for models

in 6D SUGRA and for fluxless type II compactifications withcodimension 2 sources

Aghababaie, Burgess, Cline, Firouzjahi, Parameswaran, Quevedo, Tasinato, Zavala 03Burgess, Maharana, van Nierop, Nizami, Quevedo 11

I Basic idea: consider an action satisfying a scaling symmetry:

S[τkiψi ] = τkS[ψi ]

Take τ derivative:∫ ∑i

kiτki−1ψi

δS[τkiψi ]

δ(τkiψi )= kτk−1S[ψi ]

Evaluating at τ = 1, using δS[ψi ]/δψi = 0:

S[ψi ] = 0

The MethodI Scaling symmetries useful for computing Λ! Shown before for models

in 6D SUGRA and for fluxless type II compactifications withcodimension 2 sources

Aghababaie, Burgess, Cline, Firouzjahi, Parameswaran, Quevedo, Tasinato, Zavala 03Burgess, Maharana, van Nierop, Nizami, Quevedo 11

I Basic idea: consider an action satisfying a scaling symmetry:

S[τkiψi ] = τkS[ψi ]

Take τ derivative:∫ ∑i

kiτki−1ψi

δS[τkiψi ]

δ(τkiψi )= kτk−1S[ψi ]

Evaluating at τ = 1, using δS[ψi ]/δψi = 0:

S[ψi ] = 0

The Method

I Consider 4D Einstein equation:

Rd =d2L

I Substitute Ricci scalar for warped spacetime (ds210 = e2Ads2

d + ds210−d )

Rd =2d

d − 2e−2AΛ− e−dA∇2edA

and integrate over 10D:8vV

d − 2Λ = 2S

(v =∫?d 1, V =

∫?10−d e(d−2)A are volume factors)

I Thus, the scaling symmetries implyΛ = 0

The Method

Rd =d2L

d + ds210−d )

Rd =2d

d − 2Λ = 2S

(v =∫?d 1, V =

The Method

Rd =d2L

d + ds210−d )

Rd =2d

d − 2Λ = 2S

(v =∫?d 1, V =

The MethodIn string theory, this is usually more complicated...

I Localised sources explicitly break scaling symmetries:

S → s2Sbulk + sSloc, S → t−8Sbulk +∑

pt−p−1S(p)

I Flux compactifications: topologically non-trivial background fluxes!Field strengths not globally given by a gauge potential: F 6= dC(non-trivial gauge patches!)

ddτ S[τkiψi ] 6=

∫ ∑i

kiτki−1ψi

δS[τkiψi ]

δ(τkiψi )

Total derivatives need not integrate to zero...I Expect 2 contributions to Λ: localised source terms + flux

contribution

pt−p−1S(p)

ddτ S[τkiψi ] 6=

∫ ∑i

kiτki−1ψi

δS[τkiψi ]

δ(τkiψi )

Total derivatives need not integrate to zero...

I Expect 2 contributions to Λ: localised source terms + fluxcontribution

pt−p−1S(p)

ddτ S[τkiψi ] 6=

∫ ∑i

kiτki−1ψi

δS[τkiψi ]

δ(τkiψi )

Total derivatives need not integrate to zero...I Expect 2 contributions to Λ: localised source terms + flux

contributionDaniel Junghans Cosmological Constant, Near Brane Behavior and Singularities 12 / 34

How to take into account fluxI Convenient possibility: explicitly separate non-exact parts of NSNS

and RR field strengths (instead of dealing with globally not definedgauge potentials)

I For H, writeH = dB + Hb

Consistent with Bianchi identity dH = 0!I However, subtlety for RR field strengths: Bianchi identities are usually

more complicated (coupling to source terms!)

dFn − H ∧ Fn−2 = δn+1

AnsatzFn = dCn − H ∧ Cn−2 + 〈eB ∧ F b〉n

would therefore only solve Bianchi away from the sources... wrong!

Consistent with Bianchi identity dH = 0!

I However, subtlety for RR field strengths: Bianchi identities are usuallymore complicated (coupling to source terms!)

dFn − H ∧ Fn−2 = δn+1

Consistent with Bianchi identity dH = 0!I However, subtlety for RR field strengths: Bianchi identities are usually

more complicated (coupling to source terms!)

dFn − H ∧ Fn−2 = δn+1

How to take into account fluxI Way out: express all internal (components of) RR field strengths by

dual field strengths using the duality relation

e5−n

2 φF intn = (−1)

(n−1)(n−2)2 ?10 F ext

10−n

(“ext”: dual RR field strength, which is spacetime-filling)

Dual field strengths do not have source terms in Bianchi (would bepoint-like in spacetime)

I Then we can write

F ext10−n = dC ext

9−n − H ∧ C ext7−n + 〈eB ∧ F b〉10−n

Can check that Bianchis are solved for this ansatz...

Example: F3 can have D5-brane term in Bianchi: dF3 − H ∧ F1 = δ4,but we can write F3 = −e−φ ?10 F7 and then defineF7 = dC6 −H ∧ C4 + 〈eB ∧ F b〉7 such that Bianchi dF7 −H ∧ F5 = 0is solved.

How to take into account fluxI Way out: express all internal (components of) RR field strengths by

dual field strengths using the duality relation

e5−n

2 φF intn = (−1)

(n−1)(n−2)2 ?10 F ext

10−n

(“ext”: dual RR field strength, which is spacetime-filling)

Dual field strengths do not have source terms in Bianchi (would bepoint-like in spacetime)

I Then we can write

F ext10−n = dC ext

9−n − H ∧ C ext7−n + 〈eB ∧ F b〉10−n

Can check that Bianchis are solved for this ansatz...

Example: F3 can have D5-brane term in Bianchi: dF3 − H ∧ F1 = δ4,but we can write F3 = −e−φ ?10 F7 and then defineF7 = dC6 −H ∧ C4 + 〈eB ∧ F b〉7 such that Bianchi dF7 −H ∧ F5 = 0is solved.

General resultI Lengthy calculation yields

8vVd − 2Λ =

p − 32 c

) [S(p)

DBI + S(p)CS

∫F(c)

v ,V: volume factors, c: free parameter (determines which linearcombination of 2 symmetries is exploited)

I Flux term

F(c) = −∑n≥d

n − 52 c

n ∧⟨eB ∧ σ(F int)

⟩10−n

− cHb ∧(e−φ ?10 H −

⟨σ(F int) ∧ C ext

)σ: ±1 (depends on form degree), 〈. . .〉n: projects to n-form part

I Choose c to set to zero first line, H eom often allows to gauge to zerosecond line

8vVd − 2Λ =

p − 32 c

) [S(p)

DBI + S(p)CS

∫F(c)

I Flux term

F(c) = −∑n≥d

n − 52 c

⟩10−n

− cHb ∧(e−φ ?10 H −

8vVd − 2Λ =

p − 32 c

) [S(p)

DBI + S(p)CS

∫F(c)

I Flux term

F(c) = −∑n≥d

n − 52 c

⟩10−n

− cHb ∧(e−φ ?10 H −

Example: the GKP solution

I 4D Minkowski vacua from type IIB compactifications on warped CYGiddings, Kachru, Polchinski 01

I Simple example: setup with H and F3 fluxes and O3-planes

ds2 = e2Ads24 + e−2Ads2

6 , F3 = −e−φ ?6 H,

F5 = −(1 + ?10)e−4A ?6 de4A, C ext4 = ?4(e4A + a)

I The flux term F(c) in our expression for Λ is zero for a = 0, c = −1!

F(c) = −c Hb ∧[e−φ ?10 H + F3 ∧ C ext

]+ (1 + c) F b

7 ∧ F3 = 0.

(solution implies: e−φ ?10 H = −e4A?4 ∧ F3 = −F3 ∧ C ext4 )

Example: the GKP solution

I 4D Minkowski vacua from type IIB compactifications on warped CYGiddings, Kachru, Polchinski 01

I Simple example: setup with H and F3 fluxes and O3-planes

ds2 = e2Ads24 + e−2Ads2

6 , F3 = −e−φ ?6 H,

F5 = −(1 + ?10)e−4A ?6 de4A, C ext4 = ?4(e4A + a)

I The flux term F(c) in our expression for Λ is zero for a = 0, c = −1!

F(c) = −c Hb ∧[e−φ ?10 H + F3 ∧ C ext

]+ (1 + c) F b

7 ∧ F3 = 0.

(solution implies: e−φ ?10 H = −e4A?4 ∧ F3 = −F3 ∧ C ext4 )

Example: the GKP solutionI Reminder:

8vVd − 2Λ =

p − 32 c

) [S(p)

DBI + S(p)CS

∫F(c)

This yields (p = 3, d = 4, F = 0):

4vV[S(3)

DBI + S(3)CS

I Evaluate this:

4vV µ3

∫ (?4e4A − C ext

)∧ δ6 =

14V NO3 µ3

(e4A0 − e4A0

)We thus recover the result

Λ = 0of the GKP solution!

I Works for several other examples of type IIA/IIB SUGRAcompactifications as well (see paper)...

8vVd − 2Λ =

p − 32 c

) [S(p)

DBI + S(p)CS

∫F(c)

This yields (p = 3, d = 4, F = 0):

4vV[S(3)

DBI + S(3)CS

]I Evaluate this:

4vV µ3

∫ (?4e4A − C ext

)∧ δ6 =

14V NO3 µ3

(e4A0 − e4A0

8vVd − 2Λ =

p − 32 c

) [S(p)

DBI + S(p)CS

∫F(c)

This yields (p = 3, d = 4, F = 0):

4vV[S(3)

DBI + S(3)CS

]I Evaluate this:

4vV µ3

∫ (?4e4A − C ext

)∧ δ6 =

14V NO3 µ3

(e4A0 − e4A0

Outline

Introduction

Conclusion

KKLT Review

I Construction of meta-stable dS vacua in string theoryKachru, Kallosh, Line, Trivedi 03

I 3 ingredients:1. no-scale Minkowski solution of the GKP type with warped throatregion2. non-perturbative corrections3. uplift by anti-D3 branes

= meta-stable dS (?)

Klebanov-Strassler solutionI Non-compact solution of type IIB supergravity with H and F3 flux

Klebanov, Strassler 00

Deformed conifold: topologically a cone over S2 × S3, with singularapex replaced by finite S3 (smooth)

Candelas, de la Ossa 90

I Put F3 flux through A-cycle and H flux through B-cycle

∫A F3 = M,

∫B H = K

Fluxes can be shown to generate large (but finite) warping in the tipregion eA ∼ e−K/M

I can be embedded as a local region into a compact manifoldGiddings, Kachru, Polchinski 01

∫A F3 = M,

∫B H = K

∫A F3 = M,

∫B H = K

Non-perturbative correctionsI Only complex structure moduli stabilised by fluxes at tree-level,

Kahler moduli unstabilised (simplest case: only volume modulus ρ)

I Consider non-perturbative corrections to the effective potential tobreak no-scale structure, e.g. from gaugino condensation onD7-branes or Euclidean D3-brane instantons

Kachru, Kallosh, Line, Trivedi 03; Witten 96

I stabilises all moduli in supersymmetric AdS vacuum

Kahler moduli unstabilised (simplest case: only volume modulus ρ)I Consider non-perturbative corrections to the effective potential to

break no-scale structure, e.g. from gaugino condensation onD7-branes or Euclidean D3-brane instantons

Kahler moduli unstabilised (simplest case: only volume modulus ρ)I Consider non-perturbative corrections to the effective potential to

break no-scale structure, e.g. from gaugino condensation onD7-branes or Euclidean D3-brane instantons

anti-D3 brane upliftI add a small number of anti-D3 branes to the setup and adjust flux

numbers accordingly to solve the tadpole condition

I adds a positive term to the effective potential and uplifts the CC to apositive value, still (meta-)stable in suitable parameter range

I anti-branes feel force due to flux background and are driven towardsthe tip of the warped throat, where their tension is highly redshifted

Kachru, Pearson, Verlinde 02

Uplift to small, positive CC!

numbers accordingly to solve the tadpole conditionI adds a positive term to the effective potential and uplifts the CC to a

positive value, still (meta-)stable in suitable parameter range

Uplift to small, positive CC!Daniel Junghans Cosmological Constant, Near Brane Behavior and Singularities 22 / 34

Caveats

I Hard to estimate non-perturbative effects, exact form not known

I anti-D3 branes only considered in probe approximation, backreactioneffects not taken into account

I Does solution persist when backreaction is included? A lot of work inthe literature...

DeWolfe, Kachru, Mulligan 08; McGuirk, Shiu, Sumitomo 09Bena, Grana, Halmagyi 09; Bena, Giecold, Grana, Halmagyi, Massai 11

Blaback, Danielsson, DJ, Van Riet, Wrase, Zagermann 11Dymarsky 11; Massai 12; Bena, Grana, Kuperstein, Massai 12

Result: unexpected singularity in energy densities of H and F3 (do notcouple to branes)...

Caveats

I Hard to estimate non-perturbative effects, exact form not knownI anti-D3 branes only considered in probe approximation, backreaction

effects not taken into account

I Does solution persist when backreaction is included? A lot of work inthe literature...

DeWolfe, Kachru, Mulligan 08; McGuirk, Shiu, Sumitomo 09Bena, Grana, Halmagyi 09; Bena, Giecold, Grana, Halmagyi, Massai 11

Blaback, Danielsson, DJ, Van Riet, Wrase, Zagermann 11Dymarsky 11; Massai 12; Bena, Grana, Kuperstein, Massai 12

Caveats

I Hard to estimate non-perturbative effects, exact form not knownI anti-D3 branes only considered in probe approximation, backreaction

effects not taken into accountI Does solution persist when backreaction is included? A lot of work in

the literature...DeWolfe, Kachru, Mulligan 08; McGuirk, Shiu, Sumitomo 09

Bena, Grana, Halmagyi 09; Bena, Giecold, Grana, Halmagyi, Massai 11Blaback, Danielsson, DJ, Van Riet, Wrase, Zagermann 11

Dymarsky 11; Massai 12; Bena, Grana, Kuperstein, Massai 12

Caveats

I However, partial smearing of anti-D3 branes on S3 necessary!(in earlier works also linearisation around BPS background)

Does singularity remain in the full solution? Or is it an artifact ofpartial smearing? Is there a simple reason that forces the singularityto be there?

Our result for the CC might help...

Caveats

I However, partial smearing of anti-D3 branes on S3 necessary!(in earlier works also linearisation around BPS background)

Does singularity remain in the full solution? Or is it an artifact ofpartial smearing? Is there a simple reason that forces the singularityto be there?

Our result for the CC might help...

AnsatzI Split the CC into classical part and contribution due to

non-perturbative effects: Λ = Λclass + Λnp

Classical part given by

Λclass =1

4vV(S(3)

DBI + S(3)CS

∫F(−1)

I Most general ansatz for fields (use maximal symmetry and H eom):C ext

4 = ?4(α + a), H = eφ−4A ?6 (αF3 + X3) , F1 = 0(in simplest setup, where no p = 7 sources are present in the UV)X3: closed 3-form; α,A, φ: functions, a: gauge freedom

I Substituting into our expression yields

Λ = − 14V ND3 µ3

(e4A0 + α0

116V NO3 µ3

(e4A∗ − α∗

)− 1

Hb ∧ X3 + Λnp

Λclass =1

4vV(S(3)

DBI + S(3)CS

∫F(−1)

Λ = − 14V ND3 µ3

(e4A0 + α0

116V NO3 µ3

(e4A∗ − α∗

)− 1

Hb ∧ X3 + Λnp

Λclass =1

4vV(S(3)

DBI + S(3)CS

∫F(−1)

Λ = − 14V ND3 µ3

(e4A0 + α0

116V NO3 µ3

(e4A∗ − α∗

)− 1

Hb ∧ X3 + Λnp

Assumptions

I Topological flux: In conifold region, F3 carries flux on A cycle and Hcarries flux on B cycle (KS solution)

I IR boundary conditions: anti-D3 branes locally deform geometry likein flat space, i.e.

e2A → 0, gmn ≈ e−2Agmn

very close to the anti-branes. For unperturbed metric to benon-singular: assume that, at the tip,

eφ|F A3 |2 6= 0

F A3 : part of F3 along the A-cycle (finite S3)

Motivated by unperturbed case, where nonzero F A3 is necessary to

deform the conifold so that S3 remains finite at the tip

Assumptions

I Topological flux: In conifold region, F3 carries flux on A cycle and Hcarries flux on B cycle (KS solution)

I IR boundary conditions: anti-D3 branes locally deform geometry likein flat space, i.e.

e2A → 0, gmn ≈ e−2Agmn

very close to the anti-branes. For unperturbed metric to benon-singular: assume that, at the tip,

eφ|F A3 |2 6= 0

F A3 : part of F3 along the A-cycle (finite S3)

Motivated by unperturbed case, where nonzero F A3 is necessary to

deform the conifold so that S3 remains finite at the tip

Assumptions

I UV boundary conditions: solution approaches unperturbed one in theUV, i.e. fluxes and O3-plane boundary conditions are BPS:

α∗ ≈ e4A∗ , X UV3 ≈ 0 (1)

(reminder: in our ansatz H = eφ−4A ?6 (αF3 + X3), but in GKPα = e4A, H = eφ ?6 F3)

I Non-perturbative corrections: are captured by negative contribution,i.e.

Λ = Λclass − |Λnp|.

I Cosmological constant: uplifted to positive value by anti-D3 branes,as proposed in KKLT

Λ > 0

Assumptions

α∗ ≈ e4A∗ , X UV3 ≈ 0 (1)

Λ > 0

Assumptions

α∗ ≈ e4A∗ , X UV3 ≈ 0 (1)

Λ > 0

The flux termI What about the flux term in our expression for Λ?∫

M6Hb ∧ X3 =?

I By definition, X3 is closed. We can therefore make the ansatz:X3 = βωA

3 + dω2

(β: function, ωA3 : harmonic 3-form along A cycle, ω2: 2-form)

I Using dX3 = dωA3 = 0, we find

dβ ∧ ωA3 = 0,

which implies that β is constant over directions orthogonal to the S3.We can therefore set β = βUV = 0 without loss of generality.

I We can therefore write∫M6

Hb ∧ X3 =

Hb ∧ X UV3 = 0

Flux term vanishes due to UV boundary conditions!

M6Hb ∧ X3 =?

3 + dω2

dβ ∧ ωA3 = 0,

Hb ∧ X3 =

Hb ∧ X UV3 = 0

M6Hb ∧ X3 =?

3 + dω2

dβ ∧ ωA3 = 0,

Hb ∧ X3 =

Hb ∧ X UV3 = 0

M6Hb ∧ X3 =?

3 + dω2

dβ ∧ ωA3 = 0,

Hb ∧ X3 =

Hb ∧ X UV3 = 0

Flux term vanishes due to UV boundary conditions!Daniel Junghans Cosmological Constant, Near Brane Behavior and Singularities 28 / 34

The Singularity

I Since flux term is zero, Λ becomes:

Λ = − 14V ND3 µ3

(e4A0 + α0

116V NO3 µ3

(e4A∗ − α∗

)+ Λnp

I With assumptions e4A0 = 0, e4A∗ = α∗, Λnp < 0, this simplifies to

Λ = − 14V ND3 µ3α0 − |Λnp|

I With assumption Λ > 0, it then follows

|α0| > 0

α0 must be finite at the brane position for uplift to be possible!

The Singularity

Λ = − 14V ND3 µ3

(e4A0 + α0

116V NO3 µ3

(e4A∗ − α∗

)+ Λnp

Λ = − 14V ND3 µ3α0 − |Λnp|

|α0| > 0

The Singularity

Λ = − 14V ND3 µ3

(e4A0 + α0

116V NO3 µ3

(e4A∗ − α∗

)+ Λnp

Λ = − 14V ND3 µ3α0 − |Λnp|

|α0| > 0

The SingularityI Energy density of H = eφ−4A ?6 (αF3 + X3) in near-brane region:

e−φ|H|2 = eφ−8A|αF3 + X3|2 ≥ α2e−8Aeφ|F A3 |2

At leading order (gmn ≈ e−2Agmn):e−φ|H|2 ≥ α2e−2Aeφ|F A

I For finite eφ|F A3 |2 (assumption), this gives

e−φ|H|2 ∼ e−2A

Energy density is singular!I With dilaton equation ∇2φ = 1

2eφ|F3|2 − 12e−φ|H|2 and finite dilaton,

also F3 energy density must be singular!eφ|F3|2 ∼ e−2A

Anti-D3 uplift yields singular 3-form field strengths!

e−φ|H|2 ∼ e−2A

Energy density is singular!

I With dilaton equation ∇2φ = 12eφ|F3|2 − 1

2e−φ|H|2 and finite dilaton,also F3 energy density must be singular!

eφ|F3|2 ∼ e−2A

e−φ|H|2 ∼ e−2A

Anti-D3 uplift yields singular 3-form field strengths!Daniel Junghans Cosmological Constant, Near Brane Behavior and Singularities 30 / 34

Interpretation of singularity?

Several proposals...I Singularity might be due to linear perturbation around BPS

backgroundDymarsky 11

Now excluded!

I Myers effect: in flux background, anti-D3 branes can polarise into D5and/or NS5-branes

Myers 99; Polchinski, Strassler 00; Kachru, Pearson, Verlinde 02

Could resolve singularity... However, argued not to happen inBena, DJ, Kuperstein, Van Riet, Wrase, Zagermann 12

Bena, Grana, Kuperstein, Massai 12

Polarisation potential for worldvolume scalars does not admitpolarisation into D5-branes!

Several proposals...I Singularity might be due to linear perturbation around BPS

backgroundDymarsky 11

Now excluded!I Myers effect: in flux background, anti-D3 branes can polarise into D5

and/or NS5-branesMyers 99; Polchinski, Strassler 00; Kachru, Pearson, Verlinde 02

Could resolve singularity... However, argued not to happen inBena, DJ, Kuperstein, Van Riet, Wrase, Zagermann 12

Bena, Grana, Kuperstein, Massai 12

Polarisation potential for worldvolume scalars does not admitpolarisation into D5-branes!

I Is singularity resolved by stringy effects? Recent works suggest no...Bena, Buchel, Dias 12; Bena, Blaback, Danielsson, Van Riet 13

No BH solution cloaking the singularity! Suggests that singularity isnot ok in string theory

Gubser 00

I Singularity may indicate that solution has perturbative instability anddecays via brane-flux annihilation

Blaback, Danielsson, Van Riet 12Bena, Buchel, Dias 12; Bena, Blaback, Danielsson, Van Riet 13

No meta-stable de Sitter vacua from anti-D3 braneuplifts?

Gubser 00

Conclusion

Outline

Introduction

Conclusion

I Understanding backreaction effects is important for stringphenomenology/cosmology

I Classical cosmological constant in type II supergravity often fullydetermined by boundary behavior of the fields at the source positionsbut independent of bulk dynamics, for fully backreacted solutions

I Result also shows the existence of a singularity in the energy densitiesof H and F3 generated by anti-D3 branes in the KKLT scenario

I Future work: clarify meaning of singularity; better understand otheruplifting mechanisms such as Kahler uplifting, non-geometry, ...

Thank you!

Conclusion

I Understanding backreaction effects is important for stringphenomenology/cosmology

I Classical cosmological constant in type II supergravity often fullydetermined by boundary behavior of the fields at the source positionsbut independent of bulk dynamics, for fully backreacted solutions

I Result also shows the existence of a singularity in the energy densitiesof H and F3 generated by anti-D3 branes in the KKLT scenario

I Future work: clarify meaning of singularity; better understand otheruplifting mechanisms such as Kahler uplifting, non-geometry, ...

Thank you!

Cosmological Constant, Near Brane Behavior and Singularities · Introduction Outline Introduction...

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