Moduli fixing, flux vacua and landscape analysis

Lecture 1: Overview of challenges and approaches

Frederik Denef

KU Leuven

Tehran, April 13, 2007

Outline

The Landscape Problem

Challenges for string phenomenology

L

LHC

S

?

1. Fundamental mass scale Ms � Mew

2. Controlled vacua ↔ realistic vacua

3. Multitude of vacua

Controlled vacua ↔ realistic vacua: susy moduli

R1,3

X6

I String vacua ≡ solutions ∼ R1,3 × X6 to (corrected) 10dsupergravity.

I Susy most efficient to control corrections.

I But: susy solutions (e.g. type II with X6 CY) come with

geometric moduli massless scalars. ×I Quantum corrections can lift those, but...

Controlled vacua ↔ realistic vacua: weak coupling runaways

I Modulus ρ = Vol(X6) or 1/g2s or eVol(C)/gs or ...

I Example potential:

V (ρ) =1

ρ2− 1

ρ3+

1

4ρ4+O(

1

ρ5).

1 2 3 4 5

0.1

0.2

0.3

0.4

V

ρweakstrong

?

I Weak coupling ρ→∞ = flat 10d space ⇒ V (ρ) → 0.×I Minima “naturally” at strong coupling, beyond control...

Fluxes

= p-form field strengths F = Fµ1···µpdxµ1 ∧ · · · ∧ dxµp .

Nontrivial p-cycles Ci in X6 can carry quantized magnetic F -flux:∫Ci

F = Ni ∈ Z

Energy density in effective 4d theory:

V (φ,N) = V0(φ) +

∫X6(φ)

|F |2 = V0(φ) + gij(φ)N iN j

For suitable N, stabilized, reasonably controlled vacua! [KKLT]

But...

Multitude of vacua

Large degeneracy from fluxes

Large degeneracyfrom choice of topology

Large degeneracyfrom moduli potential

Low energy effective field theory parameters

How big is the set of string vacua?

I infinite: [AdS5 × S5]N ' N = 4 SU(N) SYM, N ∈ Z.I but, surprisingly, if rough observational constraints + control

are imposed, all evidence points (nontrivially) to finitenumber. E.g.:

I A priori infinitely many compactification topologies, butCheeger’s finiteness theorem “⇒” only finite set hasKaluza-Klein radii bounded above while keeping corrections togeometry under control. [Acharya-Douglas]

I # susy flux vacua for given X6 easily ∼ 10500, butI None of these 10100n-type models fully established as

metastable susy-breaking vacua (in KKLT scenario, largenumber of cycles typically leads to sub-string-size cycles henceloss of control).

I Not even roughly known how many vacua there are compatiblewith present experimental data (might even be zero!)

I So, perhaps we can just figure out our vacuum from exp data?

Low energy predictions from string theory in four easy steps

1. Construct/enumerate all vacua meeting rough observationalconstraints (4 huge dim, no massless scalars, tdecay > 10 Gyr, ...).(labeled by discrete microscopic data ~m: topology, flux, criticalpoints, ...)

Table 6 – continued from previous page

nr Total occ. MIPFs Chan-Paton Group Spectrum x Solved

411 31000 17 U(3)× U(2)× U(1)× U(1) AAVA 0 Y

417 30396 26 U(3)× U(2)× U(1)× U(1) AAVS 0 Y

495 23544 14 U(3)× U(2)× U(1)× U(1) AAVS 0

509 22156 17 U(3)× U(2)× U(1)× U(1) AAVS 0 Y

519 21468 13 U(3)× U(2)× U(1)× U(1) AAVA 0 Y

543 20176(*) 38 U(3)× U(2)× U(1)× U(1) VVVV 1/2 Y

617 16845 296 U(5)× O(1) AV 0 Y

671 14744(*) 29 U(3)× U(2)× U(1)× U(1) VVVV 1/2

761 12067 26 U(3)× U(2)× U(1) AAS 1/2 Y!

762 12067 26 U(3)× U(2)× U(1) AAS 0 Y!

1024 7466 7 U(3)× U(2)× U(2)× U(1) VAAV 1

1125 6432 87 U(3)× U(3)× U(3) VVV * Y

1201 5764(*) 20 U(3)× U(2)× U(1)× U(1) VVVV 1/2

1356 5856(*) 10 U(3)× U(2)× U(1)× U(1) VVVV 1/2 Y

1725 2864 14 U(3)× U(2)× U(1)× U(1) VVVV 1/2 Y

1886 2381 115 U(6)× Sp(2) AV 1/2 Y!

1887 2381 115 U(6)× Sp(2) AV 0 Y!

1888 2381 115 U(6)× Sp(2) AV 1/2 Y!

2624 1248 3 U(3)× U(2)× U(2)× U(3) VAAV 1

2880 1049 34 U(5)× U(1) AS 1/2 Y!

2881 1049 34 U(5)× U(1) AS 0 Y!

2807 1096(*) 8 U(3)× U(2)× U(1)× U(1) VVVV 1/2

2919 1024 2 U(3)× U(2)× U(2)× O(3) VAAV 1

4485 400(*) 2 U(3)× U(2)× U(1)× U(1) VVVV 1/2

4727 352 3 U(3)× U(2)× U(1)× U(1) VVVV 1/2

4825 332 20 U(4)× U(2)× U(2) VAS 1/2 Y!

4902 320(*) 1 U(3)× U(2)× U(1)× U(1) VVVV 1/2 Y

4996 304 30 U(3)× Sp(2)× U(1)× U(1) VVVV 1/2 Y

6993 128(**) 1 U(3)× U(2)× U(2)× U(1) VVVV 1/2

7053 124 4 U(3)× U(2)× U(2)× U(1) VASV 1/2 Y!

7241 116(**) 4 U(3)× U(2)× U(2)× U(1) VVVV 1/2

7280 114 3 U(3)× Sp(2)× U(1) AVS 1/2

7464 108 1 U(3)× Sp(2)× U(1) VVT 1/2

7905 96(*) 1 U(3)× U(2)× U(1)× U(1) VVVV 1/2

8747 68(**) 3 U(3)× U(2)× U(1)× U(1) VVVV 1/2

8773 68 4 U(3)× U(2)× U(1)× U(1) VVVV 1/2

11347 32(**) 1 U(3)× U(2)× U(1)× U(1) VVVV 1/2

Continued on next page

44

Low energy predictions from string theory in four easy steps

2. Compute presently measurable low energy parameters Φ(continuous and discrete) of vacua with high accuracy.

# Generationsg

g

1

2

rk G

= computing map ~m 7→ Φ(~m).

Low energy predictions from string theory in four easy steps

3. Find unique vacuum compatible with experiment.

= find ~m such that Φ(~m) = Φexp.

Low energy predictions from string theory in four easy steps

4. Use this vacuum to predict everything we ever wanted to know.

m = ...

g = ...

G = ...*

*

*

= compute Φ̃everything we everwanted to know(~m).

Is this in principle a tractable problem?

?

Computational complexity theory

2027: string theory under full control!

Imagine we have systematic classification of all vacua, and that wecan compute for each vacuum every low energy quantity toarbitrarily high accuracy.

Now what?

Model for matching observed data with microscopic data

ε

Λ0

E.g. cosmological constant in Bousso-Polchinski model:

Λ(N) = −Λ0 +∑ij

gijNiN j

with flux N ∈ ZK . Example question: ∃N : 0 ≤ Λ(N) < ε ?

Can be extended to more complicated models, other parameters, ...

Computational complexity

I Bousso-Polchinski problem can be shown to be NP-complete.(Essentially due to high “nonlocality” of map N 7→ Λ(N).)

I ⇒ If you find polynomial time algorithm to solve this,everything we ever thought is difficult is actually easy, and youwin a $106 Clay prize.

I But more likely: no general polynomial time algorithm for BP(not even on quantum computer)

Caveats

many