Exact solutions in supergravity theory
James T. Liu 26 July 2005
Lecture 1: Introduction and overview of supergravity Lecture 2: Conditions for unbroken supersymmetry Lecture 3: BPS black holes and branes
Lecture 4: The LLM bubbling AdS construction
Lecture 4 Outline
• Brief overview of the G-structure (Killing tensor) analysis
• Worked example: The LLM bubbling AdS construction
Reduction of IIB on S3 × S3 Obtaining the solution
The vacuum AdS5 × S5 background
G-structure (Killing tensor) analysis
• A powerful means of classifying and constructing new supersymmetric backgrounds was pioneered by Gauntlett, Gutowski, Martelli, Pakis, Sparks, Tod, Waldram. . .
Given a Killing spinor , construct all possible tensors T(n) = Γ(n)
Killing spinor → background isometries → specialized coordinates
• Outline of the procedure
0) Choose initial isometries of the background geometry 1) Construct all possible spinor bilinears
2) Derive the algebraic identities (Fierz relations) between bilinears 3) Obtain the differential identities and identify additional symmetries
4) Specialize the choice of coordinates and solve the appropriate equations
G-structures
• The resulting supersymmetric backgrounds may be classified according to G-structures
a principle sub-bundle of the frame bundle with fiber in GL(n, R) GL(6, R) → O(6) → SO(6) → SU (3)
metric orientable Calabi-Yau (J(2), Ω(3)) The failure of ∇ˆ to be compatible with the G-structure is measured by the intrinsic torsion (in this case dJ(2) and dΩ(3))
See e.g. Gauntlett et al., hep-th/0411194
• We will not focus on the classification, but will instead illustrate the construction with an example
Bubbling AdS5×S5 geometry [Lin, Lunin and Maldacena, hep-th/0409174] (LLM)
The LLM construction
• By focusing on the near-horizon geometry of a stack of N D3-branes
Strings on AdS5 × S5 ←→ N = 4 super-Yang Mills
• Both sides have the identical isometry group
SU (2, 2|4) ⊃ SO(2, 4) × SO(6)
• The AdS/CFT conjecture then relates
− States in N = 4 super-Yang Mills
− String configurations/giant gravitons in AdS5 × S5
− Exact supergravity backgrounds
• Investigate the 1/2 BPS sector of the theory
. . . given by operators, states or configurations with ∆ = J1
1/2 BPS states in N = 4 SYM
• Consider the super-Yang-Mills fields Aµ, 4χr, 6φi
1/2 BPS chiral primaries are built from operators with conformal dimensional equal to R-charge, ∆ = J1
X = φ1 + iφ2, Y = φ3 + iφ4, Z = φ5 + iφ6
Tr(XJ), Tr(Xn)Tr(XJ −n), etc.
• Reduce the system to matrix quantum mechanics X(xµ) → X(t)
Free fermion phase space
pλ
λ
1/2 BPS states in supergravity
• A ∆ = J1 state still preserves SO(4) × SO(4) symmetry
SO(2, 4) × SO(6) ⊃ SO(2)∆ × SO(4) × SO(2)J1 × SO(4)
Consider writing AdS5 × S5 as
ds210 = [− cosh2ρ dt2 + dρ2 + sinh2ρ dΩ23] + [cos2θ dφ2 + dθ2 + sin2θ deΩ23]
= [− cosh2ρ dt2 + cos2θ dφ2 + dρ2 + dθ2] + [sinh2ρ dΩ23 + sin2θ deΩ23] Giant gravitons rotate on S5 (t,φ) but may expand either in AdS5 (Ω3) or S5 (Ωe3)
• We thus seek a supergravity solution preserving SO(4) × SO(4) isometry
IIB sugra in D = 10 −→ Effective D = 4 model (breathing mode reduction on S3 × S3)
IIB supergravity on S
3× S
3• Start with IIB theory in ten dimensions
(gµν, B(2), φ, C(0), C(2), C(4))
• Focus on D3-branes dissolving into fluxes
ds210 = gµνdxµdxν + eH(eGdΩ23 + e−GdeΩ23) F(5) = F(2) ∧ dΩ3 − ∗4e−3GF(2) ∧ deΩ3
• The reduction on S3 × S3 yields an effective D = 4 Lagrangian
e−1L4 = e3H[R + 152 ∂H2 − 32∂G2 − 14e−3(H+G)Fµν2 + 12e−H cosh G]
• Use of symmetry reduces the system to a simpler one
gµν, H, G, F(2) in D = 4
Effective D = 4 supersymmetry
• We reduce the IIB gravitino variation (only non-trivial variation)
δψM = [∇M + 16·5!1 iFN P QRSΓN P QRSΓM]ε
With the reduction ansatz, this turns into
δψµ = [∇µ − 161 e− 32(H+G)FνλΓνλΓ(3)Γµ]ε
δψa = [ ˆ∇a + 14ΓaΓµ∂µ(H + G) − 161 e− 32(H+G)FµνΓµνΓ(3)Γa]ε δψ˜a = [ ˆ∇a˜ + 14Γ˜aΓµ∂µ(H − G) − 161 e− 32(H+G)FµνΓµνΓ(3)Γ˜a]ε
where Γ(3) = −iΓ456 lives on the first S3 and we have used the fact that IIB spinors have definite Γ11 chirality
Effective D = 4 supersymmetry
• There is one final step in obtaining effective four-dimensional Killing spinor equations
Choose a Dirac matrix decomposition
Γµ = γµ × 1 × 1 × σ1 Γa = 1 × σa × 1 × σ2 Γ˜a = γ5 × 1 × σa˜ × σ1
along with ε = × χ × χ ×e » 1
0 –
• We also demand that η and ηe are Killing spinors on S3 × S3
[ ˆ∇a + 12iη ˆσa]χ = 0 [ ˆ∇a˜ + 12iη ˆeσa˜]η = 0e (η,η = ±1)e
The Killing spinor equations
• The result may be written as
δψµ = [∇µ + 161 ie− 32(H+G)Fνλγνλγµ]
δχH = [γµ∂µH + e− 12H(ηe− 12G − iηγe 5e12G)]
δχG = [γµ∂µG − 14ie− 32(H+G)Fµνγµν + e−
1 2H
(ηe− 12G + iηγe 5e12G)]
This looks vaguely like D = 4 gauged supergravity coupled to matter (but this is not a consistent truncation)
• Our goal is to solve these Killing spinor equations in order to obtain all 1/2 BPS solutions
Supersymmetry analysis
• We now apply the G-structure (Killing tensor) analysis
Gauntlett et al. CMP 247, 421 (2004); CQG 20, 4587 (2003);
CQG 20, 5049 (2003); CQG 21, 4335 (2004)
• Start with the spinor bilinears
For a four-dimensional Dirac spinor, we define the real quantities
f1 = γ5 f2 = i Kµ = γµ Lµ = γµγ5 Y µν = iγµνγ5 In addition, we may also consider expressions involving (c · · · ) where c = TC
• The algebraic identities give relations between these tensors
L2 = −K2 = f12 + f22 etc.
⇒ Kµ is timelike and Lµ is spacelike
The differential identities
• Start with the gravitino variation
∇µ = −161 ie− 32(H+G)Fνλγνλγµ
∇µ = 161 ie− 32(H+G)Fνλγµγνλ
• For Kν = γν we find
∇µKν = 14e− 32(H+G)(f2Fµν − f1 ∗ Fµν) → ∇(µKν) = 0 so Kµ is a (timelike) Killing vector (take K = ∂/∂t)
• For Lν = γνγ5 we find
∇µLν = 14e− 32(H+G)(12gµνFλρY λρ − 2F(µλYν)λ) → dL(1) = 0 so L(1) is a (spacelike) closed 1-form (take L(1) = dy)
Pinning down f
1and f
2• The gravitino variation also yields
∂µf1 = 14e− 32(H+G) ∗ FµνKν ∂µf2 = −14e− 32(H+G)FµνKν
magnetic electric
• This may be combined with additional ‘differential’ identities obtained from δχH and δχG
f1∂µ(H − G) = 12e− 32(H+G) ∗ FµνKν f2∂µ(H + G) = −12e− 32(H+G)FµνKν
→ d[e− 12(H−G)f1] = 0 d[e− 12(H+G)f2] = 0
→ f1 = be12(H−G) f2 = ae12(H+G) where a and b are constants
Specializing the metric
• We consider one more constraint from the δχH equation
: γµ∂µH = −e− 12H(ηe− 12G − iηγe 5e
1 2G
)
→ Kµ∂µH = i “
ηe− 12(H+G)f2 + ηee − 12(H−G)f1”
→ Kµ∂µH = 0 and aη + bη = 0e
Choose, e.g., a = b = η = −η = 1e or a = b = η = −η = −1e This gives a 1/4 + 1/4 = 1/2 BPS configuration
• This normalization yields
L2 = −K2 = f12 + f22 = eH(eG + e−G) = 2eH cosh G
Specializing the metric
• We now have enough information to write down the metric in a convenient coordinate system (where K = ∂/∂t and L(1) = dy)
ds24 = −h−2(dt + Vidxi)2 + h2(e2γ(dxi)2 + dy2)
Combining L(1) = dy with L(1) = deH obtained from δχH, we obtain eH = y h−2 = 2y cosh G
• We may also perform a similar analysis on the 1-form ωµ = cγµ to find
dω = 0 and that it has normalized components along x1 and x2
→ we are allowed to choose coordinates such that γ = 0
The form field F
(2)• What remains is a determination of F(2) and dV Recall the differential identities
∂µf1 = 14e− 32(H+G) ∗ FµνKν ∂µf2 = −14e− 32(H+G)FµνKν
magnetic electric
• With f1 = e12(H−G) and f2 = e12(H+G), we obtain
F(2) = −de2(H+G) ∧ (dt + V ) − h2e3G ∗3 de2(H−G)
where we may substitute in eH = y and h−2 = 2y cosh G
The metric vector V
(1)• Finally, note that the antisymmetric part of the differential identity ∇µKν
yields
dK = 12e− 32(H+G)(f2F(2) − f1 ∗ F(2))
This may be combined with K = −h−2(dt + V ) to give
dV = −2h4eH ∗3 dG or dV = −12y−1 ∗3 d tanh G
• Define z = 12 tanh G so that dV = −y−1 ∗3 dz
• We now obtain the consistency condition d2V = 0 or
d „ 1
y ∗3 dz
«
= 0
Interpretation of the solution
• What have we learned?
ds2 = −h−2(dt + V )2 + h2(dx21 + dx22 + dy2) + y(eGdΩ23 + e−GdeΩ23) F(2) = −h
d(y2e2G) ∧ (dt + V ) + h2e3G ∗3 d(y2e−2G)i with
h−2 = 2y cosh G, z ≡ 12 tanh G, dV = −y−1 ∗3 dz where
h
∂12 + ∂22 + y ∂y1 y∂yi
z(x1, x2, y) = 0 harmonic in H3
or
h
∂12 + ∂22 + 3
y∂y + ∂y2i „ z y2
«
= 0 harmonic in R2 × R4
Linear superposition and bubbling
• Underlying the 1/2 BPS solution is a linear system
6 „ z(x1, x2, y) y2
«
= 0
• This admits a Green’s function solution
z(~x, y) = y2 π
Z 1
(|~x − ~x0|2 + y2)2z(~x0, 0)d2~x0 where boundary conditions are imposed at y = 0
• Because ds2 = · · · + y(eGdΩ23 + e−GdeΩ23) boundary conditions must be chosen to ensure a regular solution
Regularity of the geometry at y = 0
• When y → 0 the volume of ds26 = eGdΩ23 + e−GdeΩ23 goes to zero
• To be smooth, only a single S3 can collapse
Either eG → 0 or e−G → 0
⇒ G = ±∞ or z ≡ 12 tanh G = ±12 as y → 0
Suppose z → 12 as y → 0
Solving the harmonic equation gives an expansion eG ∼ y−1
so that h2dy2 + y(eGdΩ23 + e−GdeΩ23) ∼ dy2 + y2dΩ23 + deΩ23
• Boundary conditions: z(x1, x2, 0) = ±12 z= −1/2 x1
x z
2
= 1/2
Example: The AdS
5× S
5background
• We recover AdS5 × S5 by filling the ‘Fermi sea’
z ≡ 12 tanh G = r2 + y2 − `2
2p(r2 + y2 − `2)2 + 4y2`2 z = −1/2
1
x 2
x = r cosθ
= r sin z
θ
= 1/2
V = − r2 + y2 + `2
2p(r2 + y2 − `2)2 + 4y2`2dφ
• Make the change of coordinates y = ` sinh ρ sin θ r = ` cosh ρ cos θ
→ z = 1 2
sinh2ρ − sin2θ
sinh2ρ + sin2θ eG = sinh ρ sin θ and h−2 = `(cosh2ρ − cos2θ) V = −1
2
cosh2ρ + cos2θ cosh2ρ − cos2θ
Example: The AdS
5× S
5background
• Look at the metric
ds2 = −h−2(dt + V )2 + h2(dr2 + r2dφ2 + dy2) + y(eGdΩ23 + e−GdeΩ23)
= `
−(cosh2ρ − cos2θ)
„
dt − 1 2
cosh2ρ + cos2θ cosh2ρ − cos2θdφ
«2
+ dρ2 + dθ2
+ cosh2ρ cos2θ
cosh2ρ − cos2θdφ2 + sinh2ρ dΩ23 + sin2θ deΩ23 ff
= `
− cosh2ρ(dt − 12dφ)2 + dρ2 + sinh2ρ dΩ23 AdS5 + cos2θ(dt + 12dφ)2 + dθ2 + sin2θdeΩ23
ff
S5
• Note the mixing between t and φ
motion of the giant gravitons along the equator of S5
Some additional resources
• A rather incomplete list. . .
M. Duff, B. Nilsson and C. Pope, Kaluza-Klein supergravity, Phys. Rep. 130, 1 (1986).
P. Candelas and X. de la Ossa, Comments on Conifolds, Nucl. Phys. B342, 246 (1990).
M. Duff, R. Khuri and J. Lu, String solitons, Phys. Rep. 259, 213 (1995).
P. Aspinwall, K3 surfaces and string duality, hep-th/9611137.
K. Stelle, BPS branes in supergravity, hep-th/9803116.
J. Gauntlett, D. Martelli, J. Sparks and D. Waldram, Supersymmetric AdS Backgrounds in String and M-theory, hep-th/0411194.