Friday, April 8, 2011 K¨ahlerModuliInﬂationandWMAP7

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Introduction Review of the type IIB flux compactification The K¨ahler moduli inflation Examples of the Swiss-cheese model Inflations with generic potentials Conclusions

K¨ahler Moduli Inflation and WMAP7

Soonkeon Nam

with Sunggeun Lee

IJMP A26 (2011) 1073 [arXiv:1006.2876]

Department of Physics, Kyung Hee University

April. 8, 2010 Chuo/Ochanomizu Univ.

Soonkeon Nam K¨ahler moduli inflation and WMAP7

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宇宙初期に

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태초에 .. 현대적 벽화

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Introduction

Inflation : very successful, but the potential is ad-hoc → desperately seeking a theory

String Theory : fundamental theory but → desperately seeking experimental test

It would be most satisfactory if we could derive inflation from string theory

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!  isotropic component

! dipole (motion of solar system) (!T T/ _CMB )_! ₌₁ "10^-3 # v = 371 km/s

-5

20 10 Large Scale Structure

(!T T/ _CMB )_{! "} " #

!  multipole components

=2.73 K TCMB

CMB Full Sky Map

-5

2 700 10

(!T T/ _CMB ) " "_! #

COBE-DMR (1990) WMAP (2003~)

!  WMAP 7th year data

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Angular resolution

WMAP 2years WMAP 8years Planck 1year

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Planck 1yr Map

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Table of contents

1 Introduction

2 Review of the type IIB flux compactification

3 The K¨ahler moduli inflation

4 Examples of the Swiss-cheese model Two K¨ahler moduli model: P_1,1,1,6,9⁴

Three K¨ahler moduli models: F₁₁ and P_1,3,3,3,5⁴

5 Inflations with generic potentials

A toy model :V = V₀(1 − αφe⁻^φ)

The Potential : V = V₀(1 − αφ^4/3e⁻^φ⁻^4/3)

6 Conclusions

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1. Why Inflation from string theory?

Slow-roll inflation:

V(!)

!

� ≡ ¹₂ �

M_PV^� V

�2

� 1

|η| ≡ �

�� ^M²^P_V^V^��

��

� � 1

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What we need to do

Find a scalar potential V from string theory which satisfies the slow roll conditions

� = M_Planck² 2

�V ^� V

�

� 1, η = M_P² V ^��

V � 1 Number of e-folding N > 60

N(t) =

� _t_e_nd

t_init

H(t^�)dt^� = 1

M_Planck²

� _φ_init

φ_end

V

V ^� dφ Density perturbations

η is sensitive to Planck suppressed operators

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Density Perturbation

δ_H = 2

5 P_R^1/2 = 1 5π√

3

V ^3/2

M_Planck³ V ^� = 1.91 × 10⁻⁵

n_s − 1 = ∂ log P_R

∂ log k � 2η − 6�

dn

d log k � 24�² − 16�η + 2ξ²

n_grav = d log P^grav (k)

d log k = −2�

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What we need to do

Find a scalar potential V from string theory which satisfies the slow roll conditions

� = M_Planck² 2

�V ^� V

�

� 1, η = M_P² V ^��

V � 1 Number of e-folding N > 60

Density perturbations

η is sensitive to Planck suppressed operators

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Eta-Problem

K = ! ! +!

( ) ( )

2 2

/ 1

2

p 3

K M F

p

V e W K W K W K W W

! ! !! ! ! M

"

# $

= % + + " &

% &

' (

( )⁽ ⁾

( )

2/ 2 2

2

1 3

Mp

p

e W W W W W

M

!

! ! ! !

+ " #

= % + + + $ &

% &

' (

! !

!2 2

2

0 0 2

3 _p p

M H

V V

M

! !

!

= =

= + +"

2

2 (1)

m O H !

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String Moduli

•

Some moduli are good inflaton candidates

•

Some moduli tend to interfere with inflation

•

Volume modules / Dilaton

•

Runaway directions

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Moduli Stabilization

•

Two Important Mechanisms

•

Fluxes of p-form field strength

•

Quantum Corrections

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History of String Cosmology

Before 1986 : Calabi-Yau Compactifications : many free Moduli (size and shape of CY3)

g_mn, B_mn, ϕ, A^I_m

Dilaton S, K¨ahler T , Complex Structure U, Wilson Lines W , Between 1986 and 1990 : Use geometric moduli as inflatons : Flat potential (V = 0)

Or too steep

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Open string moduli:

Brane separation Also Wilson Lines

See See AvgoustidisAvgoustidis talk talk

(t>1995) More moduli!

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Since 1996 : Open String moduli : Brane separation, Wilson lines

1999 : Brane Inflation: V = 0 (Dvali-Tye)

2001 : Brane-Antibrane Inflation (Burgess et al, Dvali et al)

V (Y ) = A − B Y ^d^⊥⁻² A = 2T_pV_� = 2e^ϕ

(M_sr_⊥)^d^⊥ M_s²M_Planck² B = βe^2ϕ

M_s⁸ T_p²V_� = βe^ϕM_Planck² M^2(d^⊥⁻²⁾r_⊥^d^⊥

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An open string state becomes tachyonic at a critical inter brane separation : End of inflation

Intersecting Brane Inflation : Intersecting D4, D5 and D6 branes when two extra dimensions are products of three two tori

Orientifold Model

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History of Moduli Stabilization

Before 1986 : Closed string moduli

Between 1986 and 1990 : Gaugino Condensation

Between 1991 and 2002 : Emergence or more moduli (D-brane positions)

after 2002 : Fixing moduli using Fluxes (GKP and KKLT)

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The Problem

•

String theory/M-theory has many solutions or vacua

•

Degenerary : Discrete + Continuous

•

How to break SUSY and lift vacuum degeneracy

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KKLT

•

Fluxes fix moduli

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ALL MODULI STABILISED !

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Inflation in string theory

KKLMMT brane-anti-brane inflation!

Racetrack modular inflation!

D3/D7 brane inflation!

DBI inflation (non-minimal kinetic terms)!

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Universe Universe

D3 D3 BraneBrane or

or

D7 D7 BraneBrane

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KKLT Scenario

Type IIB String on Calabi-Yau Orientifold Turn on Fluxes

�

a

F₃ = n_a,

�

b

H₃ = m_b Superpotential

W =

�

G₃ ∧ Ω, G₃ = F₃ − iSH₃ Scalar Potential

V = e^K |D_aW |²

Minimum of Potential : D_aW = 0 fixes U_a and S.

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Why KKLT?

•

In general fluxes backreact on geometry

•

So Calabi Yau is not maintained

•

However here we have only mild backreaction

•

So the extra dimension is conformal to CY

•

4D effective action is well understood

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Fixing K¨ahler Moduli

Nonperturbative D7 Eﬀects

W = W₀ + �

i

A_ie⁻^aⁱ^Tⁱ To lift to de Sitter ; Add anti D3 branes

Kahler Potential

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Why KKLT? (2)

•

Separation of moduli

•

Complex Structure, Axion-Dilaton

•

Fixed at tree level

•

Kahler moduli

•

Potential from subleading effects

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Potential Problems

Cosmological Moduli Problem (Overclosing of the Universe or ruining of nuleosynthesis)

Coughlan et al 82, Banks et al 94, De Carlos et al 93 V=0 (Brane-Brane Inflation)

Fine tuning (Brane-anti-Brane case, η problem)

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Brane-Antibrane Inflation

So called η problem : Slow roll Inflation possible.

Need 10⁻³ fine tuning of parameters needed for 60 e-folding n_s � 1.05

Burgess, Stoica, Cline, Quevedo

D3-D7 on K 3 × T ² : Kallosh et al.

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Tachyonic Inflation

Sen et al

S =

�

d⁴x√

−g

�M_Planck²

2 R − AV (T )�

1 + B∂_µT ∂^µT

�

V (T ) = V₀sech(T /√

2α^�) Need large A, B for slow roll

No fine tuning needed but we need large fluxes.

Non-canonical Kinetic Term

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Review of the type IIB flux compactification

Compactify 10D string theory on CY3 orientifold.

For G₃ = F₃ − iτH₃ where we have RR 3-form flux F₃ and NS-NS 3 form flux H₃ turned on. (The back-reaction of the flux is mild.)

V = e^K (|D_iW |² − 3|W |²), K = −2 log[V^s] − log

�

−i

�

M

Ω ∧ Ω¯

�

− log[S + ¯S], W =

�

X

G₃ ∧ Ω + �

i

A_i(S, U)e⁻^aⁱ^Tⁱ .

Taking the minimum of the potential D_iW = 0 fixes U and S at tree level. Nonperturbative D7 brane eﬀects fixes K¨ahler moduli.

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Fixing the K¨ahler Moduli

Potential in the large volume limit V ∼

� 1

V^s a²_s |A_s|²(−k_sskt^k)e⁻^2a^s^τ^s

− 1 Vs²

a_sτ_se⁻^a^s^τ^s |A_sW | + ξ

Vs³ |W |²

� .

ξ = −^ζ^(3)χ(M_2(2π)³ ⁾. χ(M) = is Euler number of M. We require ξ > 0 so h^2,1 > h^1,1.

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4. Type IIB: Closed string models

Inflaton = a Kähler modulus (or its axionic partner) (i) Racetrack inflation BBCEGKLQ (2004)

Lalak, Ross, Sarkar (2005) Greene, Weltman (2005) BBCEGKLQ (2006)

K = tree level

W = W_flux + A₁e⁻^a¹^T¹ + A₂e⁻^a²^T²

Careful fine-tuning (e.g. very large gauge group ranks) Inflation near a saddle point

n_s ≈ 0.95, r unobservably small

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(ii) Blow-up modulus inflation Conlon, Quevedo (2005) Cicoli’s talk

Setup: Large volume scenario

Balasubramanian, Berglund (2004)

Balasubramanian, Berglund, Conlon, Quevedo (2005)

K = tree level + α^� W = W_flux + �

i A_ie⁻^aⁱ^Tⁱ

Becker, Becker, Haack, Louis (2002)

Inflaton = a blow-up modulus

Potential problem: loop corrections to K

(iii) Fiber inflation Cicoli, Burgess, Quevedo (2008) Cicoli’s talk

Inflaton = Kähler mod. affected only by loop corrections Example in K3-fibered CY’s Tensor modes?

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Racetrack Potentials

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The K¨ahler moduli inflation

Calabi Yau volume V^s = α

� τ

3

12 −

�n

i=2

λ_iτ

3

i2

�

= α

2√ 2

�

(T₁ + ¯T₁)³² −

�n

i=1

λ_i(T_i + ¯T_i)³²

�

The condition for other moduli to be stable ρ =

λ_n a^3/2_n

�_n

i=2 λ_i a^3/2_i

< 1.

Therefore, since λ_i > 0 and a_i > 0, the stability condition is satisfied for h^1,1 > 2.

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� τ

3

12 −

�n

i=2

λ_iτ

3

i2

�

= α

2√ 2

�

(T₁ + ¯T₁)³² −

�n

i=1

λ_i(T_i + ¯T_i)³²

�

λ_n a^3/2_n

�_n

i=2 λ_i a^3/2_i

< 1.

Large Volume

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� τ

3

12 −

�n

i=2

λ_iτ

3

i2

�

= α

2√ 2

�

(T₁ + ¯T₁)³² −

�n

i=1

λ_i(T_i + ¯T_i)³²

�

λ_n a^3/2_n

�_n

i=2 λ_i a^3/2_i

< 1.

Large Volume Small holes

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Two K¨ahler moduli model: P_1,1,1,6,9⁴

Examples of the Swiss-cheese model

Use Calabi-Yau 3 : Weighted projective space with 4-cycles τ_a One of 4-cycles is large and controls the size of the ‘cheese’

and others are small and denote the holes

Figure: Swiss Cheese

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h^1,1 = 2

The overall volume in terms of 2-cycle volumes t’s is given by V^s = 1

6(3t₁²t₅ + 18t₁t₅² + 36t₅³). (1) With the four-cycle volumes such as τ₄ = ^t₂¹² , τ₅ = ^(t¹^+6t₂ ⁵⁾² , the of volume Calabi-Yau manifold takes the form

V^s = 1 9√

2

� τ

3

52 − τ

3

42

�

. (2)

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The large volume claims that τ₅ → ∞ and τ₄ remains small. The superpotential is then given by

W = W₀ + A₄e⁻^a⁴^τ⁴. (3) We take the limit V^s → ∞(τ₅ → ∞) with a₄τ₄ = log V^s. Then,

the potential becomes V ∼ 1

V^s a²₄|Aˆ₄|²√

τ₄e⁻^2a⁴^τ⁴ − 1 Vs²

a₄τ₄e⁻^a⁴^τ⁴|Aˆ₄Wˆ ₀|+ ξ

Vs³ |Wˆ ₀|². (4)

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By taking the limit e^aⁿ^τⁿ � Vs², the above potential is simplified to V_inf = V₀ − 4τ_nWˆ ₀a_nAˆ_ne⁻^aⁿ^τⁿ

V_s² , (5)

The canonically normalized field is obtained through straightforward calculation:

τ_n^c =

� 4λ 3V^s τ

3

n4 . (6)

In terms of τ_n^c, the inflationary potential becomes V_infl = V₀ − 4 ˆW₀a_nAˆ_n

Vs²

� 3V^s 4λ

�²₃

(τ_n^c)⁴³ e⁻^aⁿ(³_4λ^Vs )²³ ^(τn^c)⁴³ . (7)

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The three slow parameters �, η, and ξ goes to

� � 32V_s³

3β²W₀² a_n²A²_n√

τ_n(a_nτ_n)²e⁻^2aⁿ^τⁿ, η � 4a_nA_nV_s²

3λ_n√τ_nβW₀ × 4(a_nτ_n)²e⁻^aⁿ^τⁿ, ξ � 32(a_nA_n)²V_s⁴

9β²λ²W₀²τ_n × 8(a_nτ_n)⁴e⁻^2aⁿ^τⁿ. (8) The volume for 47 ≤ N_e ≤ 61 is roughly V^s ∼ 3.1 × 10⁶ (τ_n ∼ 7.0).

� < 10⁻¹², η � − 1

N_e , ξ � 1

N_e² . (9)

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the stabilization

∂V

∂V^s = ∂V

∂τ₄ = 0, (10)

and by taking the limit a₄τ₄ � 1 we get the following:

τ₄ ∼ (4ξ)²³ , and V^s ∼ ξ ¹³ |W₀|

a₄A₄ e^a⁴^τ⁴. (11)

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Three K¨ahler Moduli Cases

F₁₁

P_1,3,3,3,5⁴

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Fano threefold F₁₁ case

Z₂ quotient of real six dimensional CY with, h_1,1 = 3 and h_2,1 = 111

we can rewrite the volume as V^s = 1

3√ 2

� 2τ

3

a2 − τ

3

b2 − τ

3

c2

�

. (12)

The potentials were calculated by Denef, M. Douglas, and Floera (2004).

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P_1,3,3,3,5⁴ case

h_1,1 = 3 and h_1,2 = 75

Two stacks of D7-branes wrapping rigid 4-cycles D_A and D_B. Poetential calculated by Blumenhagen, et al (2008)

Phenomenology

The volume in terms of τ ’s V^s =

� 2 45

�

(5τ₁ + 3τ₂ + τ₃)³² − 1

3 (5τ₁ + 3τ₂)³² −

√5 3 τ

3

12

� .

(13) In terms of diagonal basis

τ_a = 5τ₁ + 3τ₂ + τ₃, τ_b = 5τ₁ + 3τ₂, τ_c = τ₁, (14) V^s =

� 2 45

� τ

3

a2 − 1 3τ

3

b2 −

√5 3 τ

3

c2

�

. (15)

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Changing of coordinates, using Euclidean D3-brane cycle τ_E₃ and standard model cycle τ_SM with the relations τ_b = 2τ_E₃ + τ_SM and τ_c = τ_E₃ − τ_SM, gives the potential:

V =

λ₁ ��

5(2τ_E₃ + τ_SM) + √

τ_E₃ − τ_SM�

e⁻^4πτ^E³

V^s −3λ₂τ_E₃e⁻^2πτ^E³

V_s² + λ₃ V_s³ . The potential has a critical point at τ_E₃ = 2τ_SM but this is not a

minimum but a saddle point along τ_SM at fixed τ_E₃ and V^s.

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The one loop corrected scalar potential has the form

V_F =

λ₁ ��

5(2τ_E₃ + τ_SM) + √

τ_E₃ − τ_SM�

e⁻^4πτ^E³

V^s − 3λ₂τ_E₃e⁻^2πτ^E³ Vs²

+ λ₃ Vs³

+

� 5

√τ_E₃ − τ_SM + 13√

√ 5

2τ_E₃ + τ_SM

� 1 15√

2V_s³ . (16)

The inflationary potential:

V = λ₃

Vs³ − 3λ₂τ_E₃e⁻^2πτ^E³ Vs²

, (17)

This has the same form as the previous section!

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Four Modulus Case

Collinucci, Kreuzer, Mayrhofer, Walliser (2009) P1,1,1,10,15⁴

R1 resolution of P_1,1,2,2,6⁴ /Z₂ R2 resolution of P1,1,1,10,15⁴

P_1,1,1,3,3⁴ /Z₃

We have studied the second case.

We have a similar potential.

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A toy model :V = V₀(1 − αφe⁻^φ)

The Potential : V = V₀(1 − αφ^4/3e⁻^φ⁻^4/3)

Inflations with generic potentials

For the cases with two, three, and four K¨ahler moduli fields, we have the following form of generic Inflaton Potential

V = V₀(1 − αφ^4/3e⁻^φ⁻^4/3)

Let us first consider the the case with a slightly simpler model with V = V₀(1 − αφe⁻^φ)

although this case is diﬀerent from the above case, since the field redefinition changes the kinetic part.