Construction of four dimensional spaces with positive...

Construction of four dimensional spaces with positivecosmological constants from M theory

MIR MEHEDI FARUK

Department of Physics,McGill University.

[email protected]

March 25, 2019

MIR MEHEDI FARUK (McGill University) March 25, 2019 1 / 55

Overview

Maldacena Nunez no-go theorem

Type IIB string theory

status of de-sitter type solution from type IIB string theory(direct product space)

uplifting to M theory.

bound on quantum corrections from M theory

Hierarchy problem and breakdown of four dimensional EFT.

attempted solution

breaking explicit de sitter isometry in spacetime sector- Kasner-DeSitter type solutiontime dependent internal manifold

Some new results


Maldacena Nunez No-Go theorem

Considering direct product manifold, M = M4 ×MD−4

ds2D = ds2

4 + ds2D−4 ≡ gµνdx

µdxν + gmndxmdxn. (1)

M4, spanned by coordinates xµ, µ = 0, .., 3 and a transverse space MD−4,spanned by coordinates xm,m = 4, ..,D − 1.For a positively curved spacetime, i.e. R4 ≡ gµνRµν . > 0, we must satisfythe condition:

(D − 6)Tµµ > 4Tm

m . (2)


Einstein gravity in 10D

Einstein-Hilbert action coupled to matter in D spacetime dimensions:

Stotal =1

KD

∫dDx

√−GDRD +

∫dDxLint, (3)

where KD is the D-dimensional Newton constant, RD is the Ricci scalar inD dimensions, GD is the determinant of the D-dimensional metric. Finally,Lint is the Lagrangian for the local or global fields that couple to gravity.It can contain for example, global fluxes, scalar fields, local sources andterms that describe graviton self coupling.


Fluxes coupled to gravity

The flux Lagrangian

LFint = −√−GDFa1...aqF

a1...aq , (4)

where F is a q-form. The above Lagrangian leads to the followingstress-energy tensor:

T FMN = − 2√

−GD

δLintδgMN

(5)

= −gMNF2 + 2qFMa2..aqF

a2...aqN . (6)

The condition (2) now stands us,

4(1− q)F 2 > −Fµa2..aqFµa2..aqq (D − 2) . (7)


Fluxes coupled to gravity (Continued)

We can consider two types of fluxes:

1. the first type with legs only along the internal directions.2. the second type with legs in M4.For the first type of flux ai = m, n for all i and Fµa2..aqF

µa2..aq = 0. Nowas we know F 2 > 0, the only way (7) can be maintained is q < 1.

In case of second scenario-

For q ≥ 4 we will consider 4 out of q legs along M4 and the rest of its legsalong the internal directions. When q < 4, then all the legs will be alongMD−4, to protect De Sitter isometry. With this condition on the fluxes,one obtains the following identities:


Fluxes coupled to gravity (Continued)

F 2 = Fa1a2..aqFa1a2..aq = C (q, 4)Fµ1..µ4a5..aqF

µ1..µ4a5..aq

Fµa2..aqFµa2..aq = C (q − 1, 3)Fµ1..µ4a5..aqF

µ1..µ4a5..aq , (8)

where the coefficient C (q, k) is defined by

C (q, k) ≡ q!

(q − k)!k!. (9)

This in turn gives us

Fµa2..aqFµa2..aq =

4

qF 2. (10)

Using the above relation and the fact that F 2 < 0 for fluxes along leg intime direction, condition 2 will be satisfied if and only if

D < q + 1. (11)


Scalar field coupled to gravity

Lagrangian for a scalar field interacting with gravity is given by

Lφint = −√−GD

(∂Mφ∂

Mφ+ V (φ)). (12)

The stress-energy tensor is given by

TφMN = −gMN

(∂Kφ∂

Kφ+ V (φ))

+ 2∂Mφ∂Nφ. (13)

Then with the stress-energy tensor given above, the only way (2) issatisfied is if and only if

∂µφ∂µφ+ V (φ) > 0. (14)

Now demanding that the M4 is isotropic in space but dependent on time,we readily find ∂µφ∂

µφ = g tt∂tφ∂tφ < 0 since g tt < 0. Thus if V (φ) < 0,M4 will not have positive curvature. In type IIB string theory, the scalaraxio-dilaton field τ has no potential and thus will not aid in constructingpositive curvature.


Massless spectrum of Type II superstring theory

Type IIB-

NS-NS Sector- gMN , BMN , φ

NS-R Sector- ψµ, χ

R-R sector- C (0), C(2)MN , C

(4)MNPQ

Dp branes, p = odd

For the type IIA theory the left- and right-moving R-sector ground statesare chosen to have the opposite chirality. And as a result now the NS-Rfermionic states come with two different chiralities.

NS-NS Sector- gMN , BMN , φ

NS-R Sector- ψµ, χ

R-R sector- C(1)M , C

(3)MNP

Dp branes, p = even


Type IIB superstring theory

The low-energy effective action for type IIB superstring theory in Eisnteinframe,

Stotal = SSUGRA + Sloc, (15)

where,

SSUGRA =1

2κ210

∫d10x

√−G10

(R − ∂Mτ∂

M τ

2|Imτ |2− |F5|2

4 · 5!− G3 · G3

12Imτ

)

+1

8iκ210

∫C4 ∧ G3 ∧ G3

Imτ. (16)

Here τ = C0 + ie−φ; G10 = det gMN ,M,N = 0, .., 9; gMN is the metric inEinstein frame and G3 = F3 − τH3. F5 is defined by

F5 = F5 −1

2C2 ∧ H3 +

1

2B2 ∧ F3. (17)

as usual.MIR MEHEDI FARUK (McGill University) March 25, 2019 10 / 55

Direct product space with Branes

The action for a Dp-brane is given by

SDp = −∫

dp+1σ Tp eφ(p+1)

4

√−f + µp

∫ (C ∧ e F

)p+1

. (18)

where f is the determinant of the metric fab, defined in the following way:

fab = fab + Fab, fab = gMN∂XM

∂σa∂XN

∂σband Fab = Fab + Bab.(19)

and Cp+1 is the RR flux. As above, Fab is raised or lowered with thepullback metric fab. Note that the sign of µp determines whether we havea brane or an anti-brane. However both branes and anti-branes havepositive tension Tp > 0. The topological term cannot enter the

stress-energy tensor since δSCS

δgMN = 0 where SCS = µp∫ (

C ∧ e F)p+1

is the

Chern-Simons action.


Further conditions on forms

To evaluate the trace of the stress-energy tensor, we will restrict the formof the fields to ensure Poincare invariance in the non-compact spacetime .These conditions are the following:

•The fluxes H3 and F3 only have legs along M6, and τ depends only onxm, the coordinates of M6.

• F5 will have legs in the xµ directions. Then by imposing self duality andPoincare invariance, one obtains the general form

F5 = (1 + ∗10) dα ∧ dt ∧ dx ∧ dy ∧ dz , (20)

where α(xM) is a scalar field which is a function of all coordinatesxM ,M = 0, .., 9.

Having laid down the required conditions, we will now analyze theindividual cases with branes, anti-branes and orientifold planes.


Direct product space with D3 Branes

Re-writing the Lagrangian,

LDp = −Tpeφ(p+1)

4

√−f√gD−p−1δ

10−p−1(x − x) (21)

For Dp brane with p = 3, we have D3 or anti-D3 branes which fill up M4.Thus the induced metric is

fab = gab, for a, b = µ, ν

fab = 0 for a, b 6= µ, ν. (22)

Therefore we find,

Tµ

µ (D3/D3)= −T3e

φ

√−f√g6√−G10

(4 + Fµµ

)δ6(x − x)

Tmm (D3/D3)

= 0. (23)

However, since the flux F is anti-symmetric while the metric is symmetric,Fµµ = 0. Thus neither the D3 nor the anti-D3 brane tensor satisfies theconstraint (2).


Direct product space with D5 and D7 Branes

For Poincare invariance in the noncompact dimensions, we will fill up M4

with the Dp or anti-Dp branes and the remaining worldvolume will fill upsome Sp−3 cycle inside the transverse space MD−p−1. If xm, xn denotecoordinates of the cycle Sp−3, then we have

fab = gab, for a, b = µ, ν,m, n

fab = 0 for a, b 6= µ, ν,m, n. (24)

And we obtain

Tµ

µ (Dp/Dp)= −Tpe

φ(p+1)4

√−f√gD−p−1√−G10

(4 + Fµµ

)δD−p−1(x − x)

Tmm (Dp/Dp)

= −Tpeφ(p+1)

4

√−f√gD−p−1√−G10

(p − 3 + F u

u

)δD−p−1(x − x)

(25)

A Using the form above, we can readily see that neither the Dp nor antiDp-brane stress-energy tensor satisfies the constraint (2) for p = 5, 7.


O Planes

On the other hand, for an orientifold, we have the action

SOp = −∫

dp+1σ TOpeφ(p+1)

4

√−f + µOp

∫Cp+1, (26)

where the orientifold has negative tension, i.e. TOp < 0. Here µp is thecharge of the Op-plane and we have the relation |TOp| = e−φ|µOp|. Alsonote that since the Op plane has negative charge, we haveµp = eφTOp = −eφ|TOp|.


Direct product space with O planes

The tension of O3-planes taken to lie in M4 is given by

Tµµ (O3) = 4|TO3|eφ

√−f√g6√−G10

δ6(x − x)

Tmm (O3) = 0, (27)

In case of Op-planes with p = 5, 7, assuming as above that the spacetimedirections M4 are filled, we find

Tµµ (Op) = 4|TOp|e

φ(p+1)4

√−f√gD−p−1√−G10

δD−p−1(x − x)

Tmm (Op) = |TOp|e

φ(p+1)4

√−f√gD−p−1√−G10

(p − 3) δD−p−1(x − x). (28)

Orientifolds have negative tension, so there is a possibility that theconstraint (2) might be satisfied when O-planes are included. However wewill see that this does not lead to positive curvature in four dimensions.



The Einstein equations arising from variation of the action (??) withrespect to the metric:

Rµν = −gµν

[G3 · G3

48 Imτ+

F 25

8 · 5!

]+

Fµabcd Fabcdν

4 · 4!+ κ2

10Nf

(T locµν −

1

8gµνT

loc

),

Rmn = −gmn

[G3 · G3

48 Imτ+

F 25

8 · 5!

]+

Fmabcd Fabcdn

4 · 4!+

G bcm Gnbc

4 Imτ+∂mτ∂nτ

2 |Imτ |2

+ κ210Nf

(T locmn −

1

8gmnT

loc

), (29)

where Nf is the number of localized objects contributing to Sloc.



Considering product manifold, taking the trace of the first equation in (29)gives

R4(xµ) = −G3 · G3

12 Imτ+

Fµabcd Fµabcd

4 · 4!+κ2

10Nf

2

(Tµ locµ − Tm loc

m

). (30)

The left-hand side is independent of xm, and hence the right-hand sideshould be as well. It follows that we can evaluate the right-hand side atany value of xm, and so we are free to consider xm away from the localizedOp-planes, where the local O-plane stress-energy tensor gives zero. As wehave already studied, the flux and local or smeared Dp or anti-Dp branecontributions to R4 are negative definite. Thus we obtain

R4 ≤ 0. (31)


Warped product manifolds with O planes


M theory uplift from type IIB

The type IIB metric we want to study,

ds2 =1

Λ(t)√h

(−dt2 + ηijdzidzj + dx23 ) +

√hgmndy

mdyn, (32)

where, Λ(t) = Λ|t|2. For this to be a dS solution, we demand that Λ bestrictly positive.The corresponding metric in the M theory is

ds2 = e2A(y ,t)(−dt2 + dx21 + dx2

2 )

+ e2B(y ,t)gmndymdyn + e2C(y ,t)(dx2

3 + dx211) (33)

where they are defined as,

e2A = Λ(t)−43 h(y)−

23 (34)

e2B = Λ(t)−13 h(y)

13 (35)

e2C = Λ(t)23 h(y)

13 (36)


M theory action

We begin by setting up the M-theory uplift of the IIB system we areinterested in. The action for M-theory is given by

S = Sbulk + Sbrane + Scorr , (37)

where Sbulk is the standard supergravity action for M-theory with a 3-formflux C and corresponding field strength G4, Sbrane is the contribution fromM2-branes, and Scorr is a curvature correction to the action. Scorr is acurvature correction to the action. The supergravity and brane actions aregiven by

Sbulk =1

2κ2

∫d11x

√−g[R − 1

48G 2

]− 1

12κ2

∫C ∧ G ∧ G , (38)

Sbrane = −T2

2

∫d3σ√−γ[γµν∂µX

M∂νXNgMN − 1 (39)

+1

3!εµνρ∂µX

M∂νXN∂ρX

PCMNP ], (40)


M theory action

In the action we have assumed a minimal coupling of the brane to thefluxes and,

T2 is the tension of the M2-brane.

XM denotes the worldsheet coordinates of the brane.

γµν is the induced metric on the brane,

The corrections to the action are of the form Rn or Gn (or a combinationthereof) and can come from several sources, such as- instantoncorrections, tree level α′ corrections, and loop corrections.


Transformation parameters under T duality

Transformation of gs and ls under T duality

l IIBs −→ l IIAs = l IIBs (41)

g IIBs −→ g IIA

s =l IIBs

R3g IIBs (42)

the type IIA coupling gs is proportional to:

gs ∝(Λ|t|2

)1/2h1/4, (43)

Dimensional reduction on a circle of radius R11 gives a relation betweenNewton’s constant in ten and 11 dimensions

G11 = 2R11G10 (44)


General Form of The Stress-Energy Tensor

The stress-energy tensor has 3 contributions:

TMN = TMNG + TMN

corr + TMNB , (45)

The stress-energy contributions are then given by

TMNG =

1

12

[GMPQRGN

PQR −1

8gMNGPQRSGPQRS

](46)

TMNB (x) = −κ

2T2n3√−g

∫d3σ√−γγµν∂µXM∂νX

Nδ11(x − xb) (47)

TMNcorr ≡

∑i

(gs)2αi CMN, i , (48)

where again xb is the spacetime position of the brane. CMN, i istime-neutral functions of the curvature tensors R, RMN and RMNPQ andthe GMNPQ components.


Einstein Tensors

Now we are in a position to attempt to search for solutions, by separatelyexamining the mn, ab, and µν components of the Einstein equation. Wecan now calculate the Einstein tensors from our M theory metric (eq.(33)). In different components they are -

Gmn = Gmn −∂mh∂nh

2h2+ gmn

[∂kh∂

kh

4h2− 6Λh

](49)

Gab = δabΛ(t)

[− R

2− 9hΛ +

gpk∂ph∂kh

4h2

], (50)

Gµν = − ηµνΛ(t)

[R

2h+

gmk∂kh∂mh

4h3− �h

2h2+ 3Λ

], (51)


The fluxes we switch on are-Gmnpq ≡ 4∂[mCnpq], Gmnpa ≡ 4∂[mCnpa]. Gmnab ≡ 4∂[mCnab] and

Gmµνρ = ∂m

(εµνρhΛ(t)2

).

To study the stress-energy tensor from the G -fluxes we have to firstexpress the various components of the G -fluxes GMNPQ in terms of their

unwarped components GMNPQ as:

G 012m = G 012m[Λ(t)]13/3h5/3,

Gmnpa = Gmnpa[Λ(t)]1/3h−4/3,

Gmnab = Gmnab[Λ(t)]−2/3h−4/3

Gmnpq = Gmnpq[Λ(t)]4/3h−4/3

where what we have done here is to simply isolate the warp factordependences of GMNPQ and express its components in terms of GMNPQ .This also means that GMNPQ ≡ GMNPQ by definition.


Classical contribution to energy momentum tensors for G4fluxes

T (G)mn = gmn

∂kh∂kh

4h2− ∂mh∂nh

2h2+

1

4h

[GmlkaG

lkan − 1

6gmnGpklaG

pkla

]+

Λ(t)

12h

[Gmlkr G

lkrn − 1

8gmnGpklr G

pklr

]+

1

4hΛ(t)

[GmlabG

labn − 1

4gmnGpkabG

pkab

]. (52)

T (G)ab =

Λ(t)

12h

[GamnpG

mnpb − δab

Gmnpc Gmnpc

2+ δab

3gmp∂mh∂ph

h

]

+1

4h

[GacmnG

cmnb − 1

4δabGmncd G

mncd

]− δab

[Λ(t)]2

4 · 4!hGmnpqG

mnpq. (53)


Classical contribution to energy momentum tensors for G4fluxes Cont.

T (G)µν = −ηµν

[(∂h)2

4Λ(t)h3+

GmnpaGmnpa

4!Λ(t)h2+

GmnpqGmnpq

4 · 4!h2+

GmnabGmnab

16h2[Λ(t)]2

]. (54)


Classical contribution to energy momentum tensors for M2brane

T (B)mn = 0 (55)

T (B)ab = 0 (56)

T (B)µν = − κ2T2n3

h2Λ(t)√gδ8(x − X )ηµν , (57)


Quantum correction towards energy momentum tensor

T corrmn = h1/3

∑i

[Λ(t)]αi C imn. (58)

T corrab = h1/3

∑i

[Λ(t)]αi+1C iab. (59)

T corrµν = h−2/3

∑i

[Λ(t)]αi−1C iµν . (60)


Analysis of the equation of motion

Remembering the stress-energy tensor has 3 contributions:

TMN = TMNG + TMN

corr + TMNB , (61)

The Einstein equation,

GMN = TMN (62)

Time-independent pieces,

Gmn − gmn6Λh =1

4h

[GmlkaG

lkan − 1

6gmnGpklaG

pkla

]+ h1/3

∑αi=0

C imn(63)

(R

2+ 9hΛ

)δab +

1

12h

[GamnpG

mnpb − δab

Gmnpc Gmnpc

2

]+h1/3

∑αi=0

C iab = 0 (64)


Analysis of the equation of motion

and contribution from (µ, ν) componenets,(R

2h− �h

2h2+ 3Λ

)=

GmnpaGmnpa

4!h2+κ2n3T2δ

8(x − X )

h2√g

− 1

3h2/3

∑{αi}=0

Cµ, iµ ,(65)

Taking trace for the internal components, it is,∑{αi}=0

Cm, im = − 2

h1/3(R + 18hΛ), (66)

∑{αi}=0

Ca, ia = − 1

h1/3(R + 18hΛ), (67)

which, in turns tells us,

R = −18hΛ− h1/3

1

2

∑{αi}=0

Ca, ia +1

4

∑{αi}=0

Cm, im

, (68)


Analysis of equation of motion

rewrite the constraint (65) as

−�h =GmnpaG

mnpa

12+ 12h2Λ +

2κ2n3T2δ8(x − X )√g

+h4/3

1

2

∑{αi}=0

C a, ia +

1

4

∑{αi}=0

C m, im − 2

3

∑{αi}=0

C µ, iµ

. (69)

We now integrate out equation (69) over the compact eight-dimensionalmanifold, we see that the LHS integrates to zero as the warp factor h is aglobally defined quantity, and we get

1

12

∫d8x

√g GmnpaG

mnpa + 12Λ

∫d8x

√g h2 + 2κ2T2n3

+

∫d8x

√gh4/3

1

2

∑{αi}=0

Ca, ia +1

4

∑{αi}=0

Cm, im − 2

3

∑{αi}=0

Cµ, iµ

= 0.

(70)


Final remark

In the absence of fluxes and higher-curvature corrections the aboveequation implies that the simplest solution will be Λ = 0, a fourdimensional Minkowski space.

In the presence of fluxes, and in the presence or absence of thehigher-curvature corrections, it is not difficult to see that the Λ < 0solution is favored.

for a solution to exist we must have the following condition

1

2

∑{αi}=0

〈Ca, ia 〉+1

4

∑{αi}=0

〈Cm, im 〉 − 2

3

∑{αi}=0

〈Cµ, iµ 〉 < 0. (71)


No Hierarchy in quantum corrections and loss of EFTdescription

Lets take a look at the correction contribution to energy momentumtensors,

TMN ≡∑i

h1/3(Λ|t|2

)αi C(i)MN , (72)

Clearly if we arrange the series (72) with decreasing αi , i.e if we make thefollowing arrangements:

αi > αi+1, (73)

then there is some hierarchy between the various quantum terms, at leastperturbatively. This hierarchy will be lost if αi = 0. The logic behind suchan arrangement is to note that the type IIA coupling gs is proportional to:

gs ∝(Λ|t|2

)1/2h1/4, (74)

which decreases slowly with time towards weak coupling, but is stronglycoupled at early times and therefore our choice of gs does indeed provide


No Hierarchy in quantum corrections and loss of EFTdescription -II

some hierarchy between the various quantum pieces with g < 1. Lets takea look at some time-neutral terms in the quantum sum:

Λ(1) ≡GmnpqG ab

mn Gabpq

M3p

, Λ(2) ≡R2RabRabG

mnabGpqabGpqcdGcdmn

M12p

Λ(3) ≡GrsabG

rsabR[mn][pq]R[mn]GpqcdR[cd ]

M9p

, Λ(4) ≡RRmnpqGmn

abGpqab

M6p

Λ(5) ≡R2GmnabG

mnab

M6p

, Λ(6) ≡�2GmnabG

mnab

M6p

,

Λ(7) ≡(�R)GmnabG

mnab

M6p

, (75)

and � here is defined with respect to the six-dimensional base only. Notethat Λ(i)’s do not scale with respect to time but appear with differentorders in Mp.


No Hierarchy in quantum corrections and loss ofEFTdescription -III

let us consider any of the Λ(k) pieces in (75). They are all time-neutral,and for our purpose we can choose Λ(1) as a representative. Using this letus define the following function:

M6p

∫ y1

0

∫ y2

0....

∫ y8

0d8y ′√g8 �Λ(1)(y ′1, ..., y

′8) ≡ M3

pΓ(5)(y1, ..., y8)

which is by construction a time-neutral function also, but now appearswith a positive power of Mp. We can raise this to arbitrary powers togenerate positive powers of M3

p . By construction they are non-localfunctions and may therefore contribute to the non-local counter-terms.


No Hierarchy in quantum corrections and loss of EFTdescription -III

Looking at the series,

∑k

(Λ|t|2

)αk C(k)MN =

∑k

g2αks

(∑b

(ckb)MN

Mβkbp

)(76)

= g2α1s

[(c11)MN

Mβ11p

+(c12)MN

Mβ12p

+ ..

]+ g2α2

s

[(c21)MN

Mβ21p

+(c22)MN

Mβ22p

+ ..

]+ ......

where βkb ∈ Z; the gs provides a strong hierarchy because it can bepartially tuned by changing t that (ckb)MN are all time independentfunctions).


No Hierarchy in quantum corrections and loss of EFTdescription -IV

But the series when {αk} = 0, which are precisely the time-neutral seriesthat would contribute to the quantum corrections here.

∑{αk}=0

C(k)MN =

∑k

(∑b

(ckb)MN

Mβkbp

)(77)

=

[(c11)MN

Mβ11p

+(c12)MN

Mβ12p

+ ......

]+

[(c21)MN

Mβ21p

+(c22)MN

Mβ22p

+ ......

]+ .....,

as we mentioned earlier, there are also non-local corrections with positivepowers of Mp, so the series dont have a leading order term, potentiallydestroying this hierarchy. Preserving the Mp hierarchy in the presence ofthese corrections would require that the higher order non-local correctionsare sufficiently suppressed by the coefficients (ckb)MN , which cannot beguaranteed without a more detailed analysis of these terms.


EFT description

Each of the series in the brackets, which are basically C(k)MN , come

with all powers of Mp and now contribute equally to the sum!

The hierarchies provided by (ckb)MN , as mentioned above, are prettyweak to allow for any controlled approximation for the sum.

Thus our argument here implies that there is at least no simpleeffective field theory description for the background with full de Sitterisometries.

The QCD action is exact, analogous to the string world sheet action.The supergravity action we play with emerges only as a limit of stringtheory, somewhat analogous to chiral perturbation theory as an EFTof the pion

When quantum corrections to the pion EFT become large, it indicatesthat the pions split into quarks, in which case one must use the fullQCD action. We refer to this phenomena as the breakdown ofeffective field theory


EFT description: analogy with QCD

In more quantitative terms, for QCD the quantum corrections have theform:

Q1 ≡∑a

(g2YM)aCa, (78)

where Ca are computed from loops and for a < 0 we have thenon-perturbative (NP) effects. They eventually may be expressed aspowers of e−1/g2

YM . On the other hand, for the de Sitter case the series weexpect is:

Q2 ≡∑a

gas Ca, (79)

where Ca =∑

b DabMbp and b is summed for all positive and negative

integers in a way discussed in details earlier. If this is the case, EFT isdefined and would match with the EFT of QCD mentioned above.


EFT description: analogy with QCD - II

Unfortunately what we actually get is:

Q3 ≡∑a

Ca, (80)

with no gs dependences! Thus there is at least no simple EFT descriptionhere, evident from the loss of gs and Mp hierarchy, and therefore differsfrom QCD where we do expect an EFT description. All in all this showsthat to allow for a four-dimensional de Sitter in string theory we willrequire an infinite number of degrees of freedom at every scale resulting ina loss of EFT description. This is then the key point of difference betweenthe two theories


Attempted solution: breaking explicit de sitter isometry

Let us start with Kasner De Sitter metric,

ds2 =1

Λ(t)√h

(−dt2 + ef1(t)dx21 + ef2(t)dx2

2 + ef3(t)dx23 ) +

√hgmndy

mdyn(81)

If one choose f1 = f2 = f3 = 0 we will get back the metric ofarXiv : 1402.5112. The M theory metric is

ds2 = e2A(y ,t)(−dt2 + ef1(t)dx21 + ef2(t)dx2

2 ) + e2B(y ,t)gmndymdyn +

e2C(y ,t)(dx23 + dx2

11)

where,

A(y , t) = A(y , t) +1

6f3(t) (82)

B(y , t) = B(y , t) +1

6f3(t) (83)

C (y , t) = C (y , t)− 1

3f3(t) (84)


Kasner De sitter type solution

where they are defined as,

e2A = Λ(t)−43 h(y)−

23 (85)

e2B = Λ(t)−13 h(y)

13 (86)

e2C = Λ(t)23 h(y)

13 (87)

withΛ(t) = Λt2 (88)


Einstein tensors


2h2+ gmn

[∂kh∂kh4h2

− 6hΛ

+ hΛ(t)(− 1

4f1f2 −

1

4f2f3 −

1

4f3f1 −

1

4f 21 −

1

4f 22

+ −1

4f 23 −

1

2

..f1 −

1

2

..f2 −

1

2

..f3

1

2t(3f1 + 3f2 + 3f3)

)](89)

Gab = e−f3δab

[Λ(t)(− R

2− 9hΛ +

gpk∂ph∂kh

4h2) +

Λ2(t) [ h(−1

4f1

2 − 1

4f2

2 − 1

2f3

2 − 1

4f1f2 −

1

2f1f3 −

1

2f2f3

−1

2

..f1 −

1

2

..f2 −

1

2

..f3 +

h

t(2f1 + 2f2 + 3f3)]

](90)


Einstein tensors

G00 = − η00

Λ(t)(R

2h+

gmk∂mh∂kh

4h3− �h

2h2+ 3Λ)

− f1t− f2

t− f3

t+

1

4f1f2 +

1

4f2f3 +

1

4f1f3

G11 = −ef1[ η11

Λ(t)(R

2h+

gmk∂mh∂kh

4h3− �h

2h2+ 3Λ)

− f2t− f3

t+

1

4f2f3 +

1

4f2

2 +1

4f3

2 +1

2

..f2 +

1

2

..f3]

G22 = −ef2[ η22

Λ(t)(R

2h+

gmk∂mh∂kh

4h3− �h

2h2+ 3Λ)

− f1t− f3

t+

1

4f1f3 +

1

4f1

2 +1

4f3

2 +1

2

..f1 +

1

2

..f3]


Fluxes and their scaling

Now,

G 012m = G 012m[λ(t)]13/3h5/3e−43f3e−(f1+f2) (91)

G 0mna = G 0mna[λ(t)]4/3h−1/3e−13f3 (92)

Gmnpa = Gmnpa[λ(t)]1/3h−4/3e−13f3 (93)

G 0mnp = G 0mnp[λ(t)]7/3h−1/3e−43f3 (94)

G 012a = G 012a[λ(t)]]10/3h5/3e−(f1+f2)e−13f3 (95)

G 0mab = G 0mab[λ(t)1]1/3h−1/3e23f3 (96)

Gmnab = Gmnab[λ(t)]−2/3h−4/3e23f3 (97)

Gmnpq = Gmnpq[λ(t)]4/3h−4/3e−43f3 (98)

(99)


Newly added fluxes and their scaling

The newly added fluxes and their scaling. we have defined fµ such as,

f0 = 0 (100)

Gµmnp = Gµmnp[λ(t)]7/3h−1/3e−43f3e−fµ (101)

Gµmna = Gµmna[λ(t)]4/3h−1/3e−13f3e−fµ (102)

Gµmab = Gµmab[λ(t)]1/3h−1/3e−fµe23f3 (103)

Gµνmn = Gµνmn[λ(t)]10/3h2/3e−fµ−fνe−43f3 (104)

Gµνma = Gµνmn[λ(t)]7/3h2/3e−fµ−fνe−13f3 (105)

Gµνab = Gµνab[λ(t)]4/3h2/3e−fµ−fνe23f3 (106)

(107)


Energy momentum tensors

TGmn = −(

∂mh∂nh

2h2− gmn

∂kh∂kh

4h2) +

1

4h(GmlkaG

lkan −

1

6gmnGpklaG

pkla)

+e−f3Λ(t)

12h(Gmlkr G

lkrn −

1

8gmnGpklr G

pklr ) +

ef3

4hΛ(t)(GmlabG

labn − 1

4gmnGpkabG

pkab)

+e−f3−fµΛ(t)2

4(GmpqµG

pqµn − 1

6gmnGpqrµG

pqrµ) +

e−fµΛ(t)1

4(GmpµaG

pµan − 1

2gmnGpqµaG

pqµa)

+ ef3e−fµ1

2(GmµabG

µabn − 1

4gmnGpµabG

pµab) + e−f3e−(fµ+fν)Λ(t)3h1

4(GmpµνG

pµνn − 1

4gmnGpqµνG

pqµν)

+1

2e−(fµ+fν)Λ(t)2h(GmaµνG

aµνn − 1

4gmnGpaµνG

paµν)

+ e−(fµ+fν)ef3Λ(t)h(− 1

16gmnGabµνG

abµν) (108)


FInal attempt: time dependent internal manifold

The type IIB metric is-

ds2 =1

Λ(t)√h

(−dt2 + dx21 + dx2

2 + dx23 ) + F1(t)

√hgαβ(y)dyαdyβ +

F2(t)√hgmn(y)dymdyn (109)

with α, β = 4, 5; m, n = 6, 7, 8, 9 and F1(t) = w(t)2 and F2(t) = 1w(t) . In

the M theory the metric takes the form,

ds2 = e2A(y ,t)(−dt2 + dx21 + dx2

2 ) + e2B1(y ,t)gαβdyαdyβ +

e2B2(y ,t)gmndymdyn + e2C(y ,t)(dx2

3 + dx211) (110)

where,

e2A = Λ(t)−43 h(y)−

23 (111)

e2B1 = F1(t)Λ(t)−13 h(y)

13 (112)

e2B2 = F2(t)Λ(t)−13 h(y)

13 (113)

e2C = Λ(t)23 h(y)

13 (114)


Fluxes and scaling

G 012α = G 012α[Λ(t)]13/3h5/3F−11 (115)

G 012m = G 012m[Λ(t)]13/3h5/3F−12 (116)

Gmnpa = Gmnpa[Λ(t)]1/3h−4/3F−32 (117)

Gmnαa = Gmnαa[Λ(t)]1/3h−4/3F−22 F−1

1 (118)

Gmαβa = Gmαβa[Λ(t)]1/3h−4/3F−21 F−1

2 (119)

Gαβγa = Gαβγa[Λ(t)]1/3h−4/3F−31 (120)

Gmαab = Gmnab[Λ(t)]−2/3h−4/3F−11 F−1

2 (121)

Gαβab = Gmnab[Λ(t)]−2/3h−4/3F−21 (122)

Gmnab = Gmnab[Λ(t)]−2/3h−4/3F−22 (123)


Fluxes and scaling

Gmnpq = Gmnpq[Λ(t)]4/3h−4/3F−42 (124)

Gmnpα = Gmnpα[Λ(t)]4/3h−4/3F−32 F−1

1 (125)

Gmnαβ = Gmnαβ[Λ(t)]4/3h−4/3F−22 F−2

1 (126)

Gmαβγ = Gmαβγ [Λ(t)]4/3h−4/3F−12 F−3

1 (127)

Gαβγδ = Gαβγδ[Λ(t)]4/3h−4/3F−41 (128)

(129)


Einstein tensor

The Einstein tensors are,

Gαβ = Gαβ −∂αh∂βh

2h2+ gαβ

[∂αh∂αh4h2

+F1

F2

∂mh∂mh

4h2− 6hΛF1 + htΛF1

−1

2ht2ΛF1 +

ht2ΛF 21

4F1− ht2ΛF 2

2 F1

2F 22

− ht2ΛF2F1

F2+

4htΛF2F1

F2− 2ht2ΛF2F1

F2

](130)

and,


2h2+ gmn

[F2

F1

∂αh∂αh

4h2+∂mh∂

mh

4h2− 6hΛF2 + 3htΛF2

−3

2ht2ΛF2 +

ht2ΛF 21 F2

4F 21

− 3ht2ΛF2F1

2F1+

2htΛF1F2

F1− ht2ΛF1F2

F1

](131)


The End


Things to do

try to find out carefully if the second attempt has an EFT description.

What is the form of four dimensional cosmological constant. is it timeindependent or dependent?

Constrain according to 2014. anything analogous to it? as there is notime independent piece in quantum corrections.

What would be the four dimensional lagrangian?

check out the anomaly condition.


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