The Kaluza-Klein program and Supergravity and Sugra Talk van...The Kaluza-Klein program and Supergravity Lectures at Erice, June 2016, at the occasion of the 40th Anniversary of Supergravity

The Kaluza-Klein program

and Supergravity

Lectures at Erice, June 2016, at the occasion of the

40th Anniversary of Supergravity

Also: 50th Anniversary of my rst visit to Erice*

30th Anniversary of the paper I will discuss.

*1966 School on Strong and Weak Interactions, with lectures by Coleman onSU(3), Radicati, Cabibbo, Gell-Mann, Glashow, Phillips, Zumino(!), Gatto andothers. The stamp of Coleman.

The KK reduction of d = 10 IIB sugra on S5 (KRN)(Kim, Romans and vN)

S5AdS5

⊗

Contents:

1. History (fun)

2. The theoretical minimum of Sugra

3. General set-up of KK reductions

4. The nitty-gritty road: Explicit construction of spherical harmonics

5. The Royal Road: Coset theory and Young tableaux

6. New developments

References:

1. Du, Nilsson, Pope, Kaluza-Klein Supergravity, Phys. Rept. 130

(1986) 1. Focusses on S7 reduction of d = 11 sugra to d = 4.

2. Castellani, D'Auria, Fré, Supergravity and Superstrings:Textbook on geometric perspectives of sugra. Full of information(3 volumes).

3. Lectures by P. van Nieuwenhuizen at the Les Houches 1983 andTrieste 1984 (Coset Methods).

4. H.J. Kim, L.J. Romans and P. van Nieuwenhuizen, Mass spectrumof chiral ten-dimensional N = 2 supergravity on S5, Phys. Rev. D32 (1985) 389. This is the subject of these lectures. Has becomecentral for the AdS/CFT program.

History (fun)

History of dimensional reduction.

1912: Gunnar Nordström proposes 2φ = 4πGTµµ.Natural extension of ∇2φ = 4πGρ after SR in 1905.

1913: He proposes unication of EM and gravity

Aµ(x, z) = Aµ(x), φ(x) . Drops z

(d = 5 Maxwell) = (d = 4 Maxwell) + (d = 4 gravity)

Ruled out in 1919: no light bending. (And −16 precession

of Mercury.)1921: Theodor Kaluza interchanges Maxwell and gravity:

gµν(x, z) =

(gµν(x) Aµ(x)Aν(x) φ(x)

)(d = 5 Einstein) = (d = 4 Einstein) + (d = 4 Maxwell) + ?

Also drops z. Puts det gµν = −1.

Note: In both cases one decomposes µ = (µ, 5).

PvN (CNYITP) The KK program and Sugra Jun, 2016 1 / 29

History (fun)

History of dimensional reduction (continued).

1926: O. Klein and V. Fock: Keep z in

φ(x, z) =∑∞

k=−∞ φk(x)eιkzR .

Dφ(x, z) =

[2x +

(∂∂z

)2]φ =

∑k

[(2x − k2

R2

)φk(x)

]eιkzR

1938: Unication of gravity + strong + EM (+ weak)Klein proposes

(gµν)

5×5=

gµ5 =

(Aµ(x) Bµ(x)eιez

B∗µ(x)e−ιez Aµ(x)

)a matrix!

g55 = 1

Bµ is Yukawa eld (strong ints) and Aµ is photon.Note: e = e(EM) = e(strong). From Einstein (d = 5) onegets (part of) nonabelian theory for SU(2)!

1 2k

1

4

MR

1 2

2

−−→

−−→


History (fun)

History of dimensional reduction (continued).

1938: C. Möller: Charge independence of nuclear forces?Klein: gµ5 = Ba

µσa +Aµ : SU(2)× U(1) (weak ints)!

≥1940: P. Jordan, Y. Thiry, C. Brans, R. Dicke, · · · , strings, · · · :Keep the scalar(s) (dilaton, moduli).

1953: A. Pais and W. Pauli.

1954: Yang-Mills and Pauli.


History (fun)

(Non)Abelian symmetries from Einstein transformations.

Set

gµα(x, z) =∑I

BIµ(x)KIα(z)

α onMinternal

µ in Minkowski

The Killing vectors satisfy

DαKβ +DβKα = 0 ; [KαI ∂α,K

βJ ∂β] = fIJ

LKγL∂γ

Now equate

δgµα = δBIµKIα = (∂µξ

β)gβα + (∂αξβ)gµβ + ξβ∂βgµα .

Choose ξβ(x, z) =∑

ΛI(x)KβI (z). Result (just substitute):

δBIµ(x) = ∂µΛI(x) + f IJLB

Jµ (x)ΛL(x) = (DµΛ)I .


History (fun)

General expansions into (spherical) harmonics.

1. gµα(x, z) = hµα(x, z) = BI5µ (x)Y I5

α (z) +((((((((

(BI1µ (x)DαY

I1(z).

For I5 = k = 1: Killing vectors. Degeneracy on Sn is 12n(n+ 1).

Gauge choice Dα(0)hµα = 0, eliminates longitudinal harmonics.

2. gαβ(x, z) = g(0)αβ (z) + hαβ(x, z)

hαβ(x, z) = h(αβ)(x, z) + 1ng

(0)αβ (z)hγγ

h(αβ)(x, z) = ΦI14(x)Y I14αβ (z) +

ΦI5(x)D(αY

I5β) (z)

+((((

(((((((

ΦI1(x)D(αDβ)YI1(z)

Gauge choice: Dα(0)h(αβ) = 0. Y I14

αβ is transversal and traceless.

3. For gravitinos: ψµ(x, z) =(ψµ(x, z), ψα(x, z)

).

ψµ(x, z) = ψI±Lµ (x)ΞI

±L (z) ; ψα(x, z) = ψ(α)(x, z) + χI(x)ταη

I+(z).The gauge Γαψα = 0 leaves τα-term.

ψ(α)(x, z) = ψI±L (x)Ξ

I±Lα (z) + ψI

±L (x)D(α)Ξ

I±L (z). The lowest of

ΞI±L (z) are Killing spinors η±(z)

(a√

of a Killing vector).


The theoretical minimum of Sugra

The theoretical minimum of Supergravity.

1. Rigid Susy:

δboson = fermion× εδfermion = ∂boson× ε

The derivative lls the gap in dimensions([fermion]− [boson] = 1

2 , [ε] = −12).

[δ(ε1), δ(ε2)]boson = ε1ε2∂︸︷︷︸translation

boson

[ε1Q, ε2Q] = (ε1γµε2)Pµ : “Q,Q = P” superalgebras

Local ε(x)→ local translation → g.c.t. (GR)

Local susy = Supergravity (Sugra).

2. Fields: gµν (or its square root eµm, eµ

meνnηmn = gµν : Cartan).

ψµa gravitino(s): spin 1 ⊗ spin 1

2=spin32 ⊕ · · ·

Other elds.

↑ Dirac's√

leads to Susy!




3.

δsusyψµa = 1

κ∂µεa + more (gauge eld of susy)

δsusyeµm = κεγmψµ (δboson = fermion× ε)

Newton constant: κ2 = 16πGc4

.

spin 3

2 + spin 2not

spin 32 + spin 1but · · ·

4. Vielbeins (Cartan: repères mobiles) from Dirac equation (Wigner).

Lat spaceDirac = −λγmδµm∂µλ with γm, γn = 2ηmn .

In curved space: γµ(x), γν(x) = 2gµν(x).Ansatz: γµ(x) = γmem

µ(x). Then emµen

νηmn = gµν (square root).If eµ

m xed by gµν : teleparallelism (Einstein).Arbitrary em

µ: local Lorentz rotation of frames (Cartan, Weyl).

δlLemµ = lm

n(x)enµ .




Dirac action is Einstein invariant if

LD,E = −(det eµm)λγµ(x)∂µλ

but also local Lorentz invariant if

LD,E,lL = −(det eµm)λγµ(x)Dµ(ω)λ

Dµ(ω)λ = ∂µλ+ 14ωµ

mnγmγnλ

δlLλ = 14λ

mnγmnλ

δlLωµmn = −Dµ(ω)λmn

Dµλmn = ∂µλ

mn + ωµmm′λ

m′n + ωµnn′λ

mn′ .

5. For free gravitinosOne derivative (Dirac)Real (Majorana)

LRS = −12 ψµγ

µνρ∂νψρ

(ψµ)D = (ψµ)M where (ψµ)D = ψ†µιγ0 and (ψµ)M = ψTµC.

Whether or notωµmn is anindependent eld.←↓




Gauge invariance: δψµ = ∂µε. Also tree unitarity xes this.Die Eleganz, die überlasz ich den Schneidern.Action for d = 4 N = 1 Sugra:

L =

LHE︷︸︸︷− 1

2κ2 eR(e, ω)

LRS︷︸︸︷− e

2 ψµγµνρDν(ω)ψρ

Minimal Einstein and local Lorentz covariantization.

e =√−g = det eµ

m ; γµνρ(x) = emµ(x)en

ν(x)erρ(x)γ[mγnγr]

R(ω, e) = Rµνmn(ω)em

νenµ ; Dν(ω)ψρ = ∂νψρ + 1

4ωνmnγmγnψρ

Rµνmn = ∂µων

mn − ∂νωµmn ; δsusyψµ = 1κDµ(ω)ε

+ ωµmkων

kn − ωνmkωµkn ; Dµ(ω)ε =(∂µ + 1

4ωµmnγmγn

)ε.

= gauge curvature for SO(3, 1).




6. Second- versus rst-order formalism. Requiring thatδψµ = 1

κDµ(ω)ε in LRS cancels δeµm in LHE xes

δeµm = κεγmψµ! The remaining variations factorize:

δL(remaining) = − 12κ2 ε

µνρσεmnrs(δωµ

mneνr + κ

6 εγmnrDµψν

)︸︷︷︸Deser & Zumino:

δωµmn=··· (rst order)

(Dρ(ω)eσ

s − κ2

4 ψργsψσ

)︸︷︷︸

FFN: ωµmn(e,ψ)(second order)

Superspace uses second order.

7. Superspace: xµ(µ = 0, 1, 2, 3); θα(α = 1, 2); θα(α = 1, 2)Superelds: V∗(x, θ) (* Lorentz indices)Pµ: translations x

µ → xµ + aµ. Local aµ(x) become g.c.t. (GR)Qα: translations θ

α → θα + εα. Local εα(x) become local susy!Mmn: local Lorentz rotations (frames, falling lifts)Superspace Supergravity is Supersymmetric General Relativity.


General set-up of KK reductions

General set-up of KK.

1. Find a solution of classical eld eqs. Our case: AdS5 ⊗ S5 (c = 1R)

AdS5 S5

g(0)µν (x) g

(0)αβ (z)

R(0)µνρσ = c2

(g

(0)µρ g

(0)νσ − g(0)

µσ g(0)νρ

)R

(0)αβγδ = −c2

(g

(0)αγ g

(0)βδ − g

(0)αδ g

(0)βγ

)F

(0)µνρστ = c

√−g(0)(x)εµνρστ F

(0)αβγδε = c

√g(0)(z)εαβγδε

2. Decompose all elds into background + quantum uctuations

Φ∗? = Φ(0)∗? (x or z) + ϕ∗?(x, z)

3. Expand ϕ∗?(x, z) =∑ϕI∗(x)Y I

? (z). The Y I? (z) are spherical

harmonics (Legendre; Green in EM; for scalars on S2 in QM; forvectors on S2 in Jackson).

4. Substitute into linearized eld equations for ϕ. Collect all termswith the same Y?. Mixing of ϕ∗(x) occurs: NOT discussed below.

10=5+5:

xµ zα⊗


The nitty-gritty road: Explicit construction of spherical harmonics

Details of KK: Scalar harmonics on Sn (n = 5).

One complex scalar eld B(x, z) in d = 9 + 1 IIB sugra, only quantum

B(x, z) =∑

Bk(x)Y k(z) , k = 0, 1, 2, · · ·

Spectrum: −2zYk(z) = λs(n, k)Y k(z) and degeneracy? (s for scalar)

Recall QM: Y 0 = 1 , Y 1 = x+ ιy, x− ιy, z , Y 2 = x2 + y2 − 2z2, xy, · · ·.Choose monomial P k(x) in xµ (coords of Rn+1). Impose 2P k(x) = 0⇒traceless. Go from xµ to yµ = (r, θα). (Polar coords, θα = zα.)

ds2 = dr2 + r2dΩ2

gµν(y) =

(1 0

0 r2gαβ(θ)

)2 = 1√

g∂∂yµ√ggµν ∂

∂yν

P k(x) = P k(y)

0 = 2P k(x) = 2P k(y). Use P k(y) = rkY k(θ).

0 = 2P k(y) =(

1rn∂rr

n∂r + 1r2 2S(θ)

)rkY k(θ)

⇒ −2S(θ)Y k(θ) = k(n+ k − 1)Y k(θ) .

chain rule



Young tableaux for scalars.

: symmetric and traceless irrep of SO(n+ 1).

ds(n, k) = #P k −#P k−2 (for the trace)

=

(n+ k

k

)−(n+ k − 2

k − 2

).

k = 0 : a constant

k = 1 : xµ

k = 2 : xµxν − 1n+1δ

µν x2

...

1 2k

5

12

MR

1

6

20

Scalars Y k(z) on S5



Details: Vector harmonics on Sn. (Jackson for S2)

Choose P kµ (x) =(

0, 0, P k(x), 0, · · · , 0). Impose again

0 = 2P kµ =∂yν

∂xµ2P kν (y)⇒

2P kr (y) = 0

2P kα(y) = 0

P kr (y) = ∂xµ

∂r Pkµ = xµ

r Pkµ = rkρk(θ)

P kα(y) = ∂xµ

∂θα Pkµ = rk+1V k

α (θ) .

0 = 2P kα =(∂r − 1

r

)(∂r − 1

r

)rk+1V k

α (θ) + rk−12S(θ)V kα (θ)

+ rk−1∂αρ(θ) + nlrk−1Vα(θ) + rk−1(∂αρ(θ)− Vα(θ)

).

BUT: Vα(θ) = ∂ασ(θ) + Y kα (θ) with DαYα = 0.

Then −2SYkα = λv(n, k)Y k

α with λv(n, k) = k(n+ k− 1)− 1, k = 1, 2, 3.

just substitute



Vector harmonics continued.

Young tableau:

⊗ = ⊕↓ k=1,2,3,···

⊕∂xµ

∂yν Pµ(x) ρ′ Yα σ′

dv(n, k) = (n+ 1)ds(n, k)− ds(n, k + 1)− ds(n, k − 1)

k = 1: Killing vectors:νµ

= Y k=1α = xµ∂αx

ν − xν∂αxµ

Degeneracy: 12(n+ 1)n.

k = 2:ν ρµ

= Y k=2α = 1

3

[∂αx

µ((xν xρ))− ∂αxν((xµxρ))]

+13

[∂αx

µ((xρxν))− ∂αxρ((xµxν))],

where ((xµxν)) = xµxν − 1nδµν x2.

1 2k

4

11

MR

15

64

Vectors Yα(z) on S5

Note the Young symmetryand the tracelessness



Details: Spinor harmonics.

First spin 12 : Ξ±,k(z) (k = 0 are Killing spinors η±). Later spin 3

2 .

Spin 12: Begin again in Rn+1 with massless (nonchiral) Dirac

equation in Cartesian coord: γmδmµ ∂∂xµ ψ

(k)(x) = 0. Theψ(x) have in one entry xµ1 · · · xµk and further traces.Examples:

k = 0 : χ

k = 1 :(xµ − 1

n+1/xγµ

)χ

k = 2 :[xµxν − 1

n+3

(/xγµxν + /xγν xµ + x2δµν

)]χ ≡ xµxνcα;µν

where χ is a constant spinor and c is gamma-traceless:

(γµ)αβcβ;µν = 0.

Then go to polar (or other) coords yν = (r, θα);ψ(k)(x) = ψ(k)(y)

0 = γm(δm

µ ∂yν

∂xµ

)∂∂yν ψ

(k)(y).



Spinor harmonics continued.

We need more! In Rn+1 with polar coordinates:(ds)2 = dr2 + r2dθ2 + r2 sin2 θdφ2 + · · ·. Then a diagonal vielbein:E1r = 1, E2

θ = r, E3φ = r sin θ + · · ·. We need in general a suitable local

Lorentz rotation Λ(θ): Λ−1ψk(y) ≡ ψk(y). Insert unity ΛΛ−1 = I:

Λ(

Λ−1γmΛ)(

δmµ ∂yν

∂xµ

)(∂∂yν +

(Λ−1 ∂

∂yν Λ))

ψk(y) = 0.

If Lmn = (eλ)mn, then Λ = e14λmnγmγn , and

Λ−1γmΛ = Lmnγn ; Λ−1 ∂

∂yν Λ = 14 ωµ

mn︸︷︷︸pure gauge

γmγn.

Finally,

γn(∂yν

∂xµ δmµLmn

)︸︷︷︸

Enν

Dνψk = 0.

Given Enν nd Lmn (or vice-versa for stereographic coords).




Use ψk(y) = Λ−1(θ)ψk(y) = Λ−1ψk(x) = rkψk(θ) (because Λ isindependent of r). Then[

γ1(∂r + n

21r︸︷︷︸

from ωa1a

)+ 1

r/DS(θ)

]rkψk(θ) = 0

For even n+ 1: Use γ1 =(

0 II 0

), γa+1 =

(0 −ισaισa 0

)and with

ψk =(

Ξ+,k

Ξ−,k

)one gets

∓ι /DSΞ∓,k = −(k + n

2

)Ξ∓,k ⇒ λspinor(n, k) = ±

(k + n

2

).

For odd n+ 1: Multiply by (1± ιγ1). Then (1∓ ιγ1)ψk = Ξ±,k and/DSΞ±,k = −ι

(k + n

2

)Ξ±,k. Same result.




1 2k

- 92

- 72

- 52

52

72

92

MR

4

20

60

4*

20*

60*

Spinors Ξ±(z) on S5

Young Tableau • • • • • :

dspinor(n, k) =[(n+ k

k

)︸︷︷︸

#P (k)

−(n+ k − 1

k − 1

)︸︷︷︸Dirac equation

]2[n

2]︸︷︷︸

factor 1/2for ψ+,ψ−




For polar coordinates: Λ−1 = e12θγ1γ2

e12φγ2γ3

e12χγ3γ4

e12ζγ4γ5

. (Eulerangles: L = · · · e−φL23e−θL12)

k = 0: Killing spinors: Dαη± = ∓ ι

2γαη±. On S2:

Ξ±,k=0 = Λ−1(1∓ιγ1)χ =

η±I = eι

φ/2(

cos θ/2(1∓ι)− sin θ/2(1±ι)

)η±II = e−ι

φ/2(

sin θ/2(1∓ι)cos θ/2(1±ι)

)

2 η+

and2 η−

k = 1:

Ξ±,k=1 = xµ︸︷︷︸Y k=1

(Λ−1χ

)︸︷︷︸

η±

− 1n+1 x

ν(

Λ−1γνΛ)

︸︷︷︸rγ1

(Λ−1γµΛ

)︸︷︷︸

usedΛ−1γµΛ= ∂xµ

∂yνEνnγn

= xµ

rγ1+ 1

r( /DS x

µ)

Λ−1χ

= 1n+1

[nY k=1 ∓

(/DSY

1)]η±. (KRN!)




Spin 32(gravitino on S5): Combine vectors and spinors.

ΞL,±,ka,α = DaΞ±,kα ; ΞT,±,ka,α = (Dax

µ)xρ1 · · · xρk (Σ±)αβ︸︷︷︸

Λ−1(θ)

ψ±

β;ρ1 ρkµ

·

The constant spinors ψ are(anti) chiral if n+ 1 = even(anti) symmetric in µ, ρ1, · · · , ρkgamma-traceless γµ∓ψ

±µρ1···ρk = γρ1

∓ ψ±µρ1···ρk = 0.

Γa =(

0 γaγa 0

); Γn+1 =

(0 ιI−ιI 0

); γ±A = (γa,±ιI) .

Identity (needed for spectrum):

Σ±γ∓AΣ± = YABγ∓B , A,B = 1, · · · , n+ 1.

In stereographic coordinates easy to check:

YAB =

(δab − 2zazb

4R2+z2−2za

4R2+z2

2zb

4R2+z24R2−z2

4R2+z2

); Σ± =

2R± ι/z√4R2 + z2

·

Pauli matrices



Mass Spectrum on S5

(e = 1

R

).


The Royal Road: Coset theory and Young tableaux

The Royal Road: Cosets and Young tableaux.

S5 : G/H = SO(6)/SO(5). Coset representatives: L(z) ∈ SO(6) in anyirrep. L−1dL = eaKa + 1

2ωabHab (Cartan-Maurer). Dene:

Y = L−1 ⇒ dY = −(L−1dL

)L−1 =

(−eaKa − 1

2ωabHab

)Y .

DαY = −KαY or Da︸︷︷︸di. op.

Y = − Ka︸︷︷︸constant matrix

Y←sph. harm.

DbDaY = −[Db,Ka]Y −KaDbY

⇓−ωbca KcY = −ωbacDcY

DbDaY = KaKbY

2SY =(δabKaKb

)Y = δABTATB − 1

2HabHab

−2SY =[C2(SO(6))− C2(SO(5))

]Y

Degeneracy of Y : dimension of Young tableau.

∣∣∣∣∣ In any irrep R of G.Decomposes into irrepsof H, on which 2S acts.



Examples of Young tableau.

TensorsR :

g1g2g3f1f2f3

# boxes = r

C2(R of SO(n)) = −r(n− 1)−∑f2i +

∑g2i .

Scalars:

k︷︸︸︷of SO(6); • of SO(5); k = 0, 1, 2, · · ·.

C2(SO(6)) = −k(5)− k2 + k

C2(SO(5)) = 0

λs(n, k) = k(k − 4).

Vectors:

k︷︸︸︷of SO(6); of SO(5); k = 1, 2, 3, · · ·.

YAB =

↑Yn+1

B

+ YaB

↓



Examples continued.

Antisym. tensors:

k+1︷︸︸︷of SO(6); of SO(5); k = 1, 2, 3.

Symmetric tensors:

k︷︸︸︷of SO(6); of SO(5);

k = 2, 3, · · ·.



Examples continued.

SpinorsR :

• • • •• •• •

C2(R of SO(n)) = −rn− 18n(n+ 1)−

∑f2i +

∑g2i .

Spin 12: • • • • of SO(6); • of SO(5).

λspinor(n, k) =[k × 6 + 1

8 × 6× 5 + k2 − k]−[

18 × 5× 4

]= k2 + 5k + 10

8 k = 0, 1, 2.

λ = −2S = − /D /D + 14R = − /D /D − 1

420r2

ι /DΞk,± = ±(k + 12)Ξk,± .


New developments

Exceptional Field Theory. (O. Hohm, H. Samtleben, arXiv:1312.0614)

Idea: Introduce new coordinates YM ⇒ extended sugra which containsboth N = 1 d = 11 and IIB.

The YM are in 27 of E6 (recall: scalars from torus compactication formcoset G/H = E6/USp(8)).

Action: S =∫d5xd27Y e

[R+ gµν(DµMMN )

(DνMMN

)− 1

4MMNFµν,MFµνN + more].

Symmetries: External gen. di. (EGD) and Internal gen. di. (IGD).

δEGD(ξ)eµm(x, Y ) = ξν(x, Y )Dνeµ

m + (Dµξν)eν

m

Dµeνm = ∂µeν

m −ANµ ∂Neνm − 13∂MA

Mµ eν

m

δIGD(Λ)VM (x, Y ) = “LΛ”VM = ΛK(x, Y )∂KVM

+ 6(PNM )KL(∂KΛL)VN + λ(∂LΛL)VM

(if VA

M∈E6 then

also δ(Λ)VAM∈E6

)(PNM )KL = (tα)NM (tα)KL (α = 1, · · · , 78)

= 118δM

NδLK︸︷︷︸

needed for P2=P

+ 16δM

KδLN︸︷︷︸

usual term

− 53dNKP dMLP︸︷︷︸

inv. tensors of E6


http://arxiv.org/abs/1312.0614

New developments

Exceptional Field Theory continued.

Closure: [LΛ1 ,LΛ2 ] = L[Λ1,Λ2]

Denes E bracket.

Requires section constraint:

dMNK∂N∂KA = 0;

dMNK∂NA∂KB = 0.

KK reduction: Make ansatz (sph. harm.)

gµν(x, Y ) = ρ−2(Y )gµν(x)

AµM (x, Y ) = ρ−1(Y )U−1N

M (Y )AµN (x)

MMN (x, Y ) = UMK(Y )UNL(Y )MKL(x)

Consistency: EMN (x, Y ) ≡ ρ−1(Y )U−1(Y )MN must satisfy

LEMEN = −XMNKEK

(EM = EMN ∂

∂Y N

).


New developments

Exceptional Field Theory continued.

Consistency: for BM (x, Y ) = ENM (Y )BN (x);ΛM (x, Y ) = ENM (Y )ΛN (x).

One gets δBM = ENM (Y )δBN (x) (denition)

= LΛBM = LEK(Y )ΛK(x)

(ENM (Y )BN (x)

)= ΛK(x)

(LEKEN

M)︸︷︷︸

require: −XKNLELM then

δBM (x)=−ΛK(x)XKNMBN (x).

BN (x)

Two solutions of the section constraint yield

N = 1 d = 11 if E6(6) → Sl(6)×Gl(1); (Note: E6(6) is real.)27→ (ym,

Ymn,ym6 + 15 + 6

)

IIB if E6(6) → Sl(5)× Sl(2).

27→ (ym,ymα,ymα,y

α

5 + 10 + 10 + 2)

Big Question: Is this Bookkeeping or Deep Physics?PvN (CNYITP) The KK program and Sugra Jun, 2016 29 / 29

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The Kaluza-Klein program and Supergravity and Sugra Talk van...The Kaluza-Klein program and Supergravity Lectures at Erice, June 2016, at the occasion of the 40th Anniversary of Supergravity