Lecture 9: RR-sector and D-branes

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1 Lecture 9: RR-sector and D-branes José D. Edelstein University of Santiago de Compostela STRING THEORY Santiago de Compostela, March 6, 2013 José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

2 RR fields couple to extended objects Recall that the antisymmetric tensor B µν in the NS-NS sector couples directly to the string world-sheet: the string carries (electric) charge w.r.t. B µν, S = 1 4πα d 2 ξ ɛ αβ B µν (X) α X µ β X ν. The Lagrangian changes by a total derivative under the gauge symmetry, B µν B µν + µ Λ ν ν Λ µ In electromagnetism, the gauge invariant degrees of freedom are contained in the field strength, F = d A. Similarly, H = d B, H µνρ = µ B νρ + ν B ρµ + ρ B µν However, the situation for the R-R potentials C (n) is very different, because the vertex operators for the R-R states involve only the F (n+1). Thus, only the field strengths, not the potentials, would couple to the string. Thus elementary, perturbative string states cannot carry any charge with respect to the R-R gauge fields C (n). José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

3 RR fields couple to extended objects We are thereby forced to search for non-perturbative degrees of freedom which couple to these potentials. Clearly, they must be extended objects that sweep out a p + 1-dimensional world-volume as they propagate in time, generalizing the notion of a string, q d n ξ ɛ a µ1 0 a n 1 x x µn W n ξ a 0 ξ a C(n) µ n 1 1 µ n q C (n). W n in complete analogy with electromagnetic and B-field minimal couplings. The massless states of the R-R sectors are given by the CG decompositions: Type IIA: 16 s 16 c = [0] [2] [4]. Type IIB: 16 s 16 s = [1] [3] [5] +. They corresponds to completely antisymmetric tensors of rank n, or n-forms. The + subscript indicates a self-duality condition. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

4 Electric and magnetic coupling of extended objects Consider F (p+2), a p + 2-form representing an antisymmetric tensor field with p + 2 indices in D dimensions. It is the field strength of a potential, F (p+2) = dc (p+1), that electrically couples to a p + 1-dimensional object, W p+1, µ p W p+1 C (p+1), W p+1 being the world-volume of an extended object called p-brane. The Hodge dual of F (p+2) is F (D p 2) = F (p+2) = d C (D p 3). Its potential couples magnetically to the extended object W D p 3, µ D p 4 C (D p 3). W D p 3 José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

5 Extended objects in type II superstring theory In superstring theory, D = 10 thus we have two possible couplings F (n+2) couples to { electric n branes, magnetic 6 n branes. Recall that the F (n) forms resulted from the tensor product of two MW spinor representations in ten dimensions, F (n) µ 1 µ n = R ψ (0) +(±) Γ [µ 1 µ n] ψ (0) (±) R. Because of GSO projection, the states ψ (0) (±) R have definite Γ11 eigenvalue ±1. Thus, given that Γ 11 Γ [µ1 Γ µn] = ( 1)[ n 2] n! (10 n)! 2 ɛµ 1 µ n ν 1 ν 10 n Γ [ν1 Γ ν 10 n], there is an isomorphism (electric-magnetic duality ) F (n) µ 1 µ n ɛ µ 1 µ n ν 1 ν 10 n F (10 n) ν 1 ν 10 n. This identifies the representations [n] [10 n]; in particular, [5] is self-dual. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

6 Branes in type IIA theory Recall that, in type IIA: 16 s 16 c = [0] [2] [4]. Thus, there are even branes. In the NS-NS sector: H 3 couples to { electric 1 branes F1-string magnetic 5 branes NS5-brane In the R-R sector: F [2] couples to { electric 0 branes D0-brane magnetic 6 branes D6-brane F [4] couples to { electric 2 branes D2-brane magnetic 4 branes D4-brane José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

7 Branes in type IIB theory In type IIB: 16 s 16 s = [1] [3] [5] +. Thus, there are odd branes. In the NS-NS sector: H 3 couples to { electric 1 branes F1-string magnetic 5 branes NS5-brane In the R-R sector: F [1] F [3] couples to couples to { electric 1 branes D(-1)-brane magnetic 7 branes D7-brane { electric 1 branes D1-brane magnetic 5 branes D5-brane F [5] couples to { electric 3 branes D3-brane magnetic 3 branes D3-brane Since the F [5] is self-dual, full electromagnetic duality is in place. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

8 Electric-magnetic duality Let us recall how the story goes in Maxwell theory. In the absence of charges and currents, d F = 0 and d F = 0, where F is the 2-form field strength describing electric and magnetic fields. The equations are symmetric under the interchange of F and F. Assuming that sources can be added in a symmetric fashion, d F = J m and d F = J e, we face Dirac s quantization condition: the wave function of an electrically charged particle moving in the field of a monopole is uniquely defined if e g 2π Z. We have seen that our branes, D-branes, can both couple electrically or magnetically. Their charges are measured using Gauss law. The Dirac quantization condition has a straightforward generalization, µ p µ 6 p 2π Z. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

9 Type II low energy effective actions Similarly to the bosonic case, the vanishing of the Weyl anomaly demands, β µν (g) = β µν (B) = β µν (Φ) = 0, where these equations are covariant complicated expressions of the massless fields. In type II superstrings we have, in addition, we should include the RR-forms in a way compatible with supersymmetry. Thus, these equations coincide with those arising in ten dimensional theories of supergravity. The number of supersymmetries is 32. Whereas type IIB theory is chiral, type IIA is not. We will present their Lagrangians next. Higher order α corrections would lead to higher powers of the curvature, as in the bosonic string. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

10 Type IIA effective action The Lagrangian of type IIA supergravity in the string frame reads ( 1 S IIA = 2κ 2 e 2Φ d 10 x g R + 4 dφ dφ ! H [3] H [3] 1 [ 1 2 2! F [2] F [2] + 1 ]) 4! F [4] F [4] + B [2] F [4] F [4], where F [2] = dc [1] F [4] = dc [3] C [1] H [3] H [3] = db [2] We can go to the Einstein frame by (g µν ) string = g 1/2 s e Φ/2 (g µν ) Einstein where g s = e Φ(r ) is the string coupling constant. Then, g string = (g 1/2 s e Φ/2 ) 5 g Einstein = g 5/2 s e 5Φ/2 g Einstein. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

11 Type IIA effective action In general, for any p form (F [p] F [p] ) string = g p/2 s e p Φ/2 (F [p] F [p] ) Einstein. The resulting action in the Einstein frame reads: ( SIIA E 1 = 2κ 2 d 10 x g E R E 1 2 dφ dφ ! e Φ H [3] H [3] [ 1 2! e 3 2 Φ F [2] F [2] + 1 4! e 1 2 Φ F [4] F [4] + B [2] F [4] F [4] ]). Gravity is now canonically normalized, as well as the dilaton kinetic term, but the coupling with the R-R forms is more involved. Notice that to compute the solution corresponding to a specific D-brane, S = 1 2κ 2 d 10 x { g R 1 2 µφ µ φ 1 } 2 e an φ 2 F n. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

12 Type IIB effective action The Lagrangian of type IIB supergravity in the string frame reads S IIB = ( 1 2κ 2 e 2Φ d 10 x g R + 4 dφ dφ 1 ) 2 3! H [3] H [3] 1 ( 4κ 2 F [1] F [1] + 1 3! F [3] F [3] ! F [5] F [5] ) C [4] F [3] H [3] where F [1] = dc [0] F [3] = dc [2] C [0] H [3] H [3] = db [2] F [5] = dc [4] 1 2 C [2] H [3] B [2] F [3] supplemented by the additional on-shell constraint F [5] = F [5]. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

13 Type IIB effective action It can also be driven to its Einstein frame form, S E IIB = 1 2κ ( d 10 x g E R E 1 1 dφ dφ 2 [ e 2Φ F [1] F [1] + 1 3! eφ F [3] F [3] ) C [4] F [3] H [3]. 2 3! e Φ H [3] H [3] 1 5! F [5] F [5] Again, to compute the solution corresponding to a specific D-brane, S = 1 2κ 2 d 10 x { g R 1 2 µφ µ φ 1 } 2 e an φ 2 F n. We have a 3 = 1 for the NS-NS 3-form, H [3], and a n = 5 n 2 for any RR n-form, F [n]. Notice that for n = 5, i.e., the self-dual D3-brane, the dilaton decouples. Even the remaining string theories fit into this quite simple action. ] José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

14 D-branes as classical solutions The equations of motion are: R µν 1 2 g µνr = 1 ( µ φ ν φ 1 ) 2 2 g µν λ φ λ 1 φ + φ 2(n 1)! ean τ µν, 1 ( µ g g µν ν φ ) 1 a n φ g 2 n! ean Fn 2 = 0, 1 1 ( µ g e a n φ F µν 2...ν n ) = 0, (n 1)! g where the electromagnetic stress-energy tensor reads τ µν = F µλ2...λ ni F ν λ 2...λ ni 1 2n g µν F 2 n. The most general metric that incorporates all the symmetries is: ds 2 = B 2 dt 2 + C 2 δ ij dy i dy j + F 2 dr 2 + G 2 r 2 dω 2 d 1. with all the functions depending only on r. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

15 D-branes as classical solutions There is an extremal solution ds 2 = H 2(7 p) with = (p + 1)(7 p) + 4 a 2 n and ( dt 2 + δ ij dy i dy j) + H 2(p+1) H = 1 + 4(7 p) Q r 7 p. ( dr 2 + r 2 dω 2 ) d 1. Now, electric solutions are those with p = n 2 whereas magnetic solutions have p = 8 n. The dilaton reads whereas the R-R form e φ = H 8 a p+2 or e φ = H 8 a 8 p, F ty 1 y p r = d dr H 1 or F θ1 θ 8 p = Q ω 8 p, the latter being proportional to the volume element of S 8 p. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

16 The D3-brane The solutions look simpler in the string frame, ds 2 = H ( 1 2 dt 2 + δ ij dy i dy j) + H 1 2 for any p. The mass of these solutions can be computed ( dr 2 + r 2 dω 2 ) 5. M = q they all saturate the BPS bound. q = Lp Ω 8 p 2κ 2 Q ; As mentioned earlier, the case n = 5 (that is, p = 3) is special. If we plug it into previous expressions, a 5 = 0, = 16, and ds 2 = ( ) 1 2 Q ( dt 2 r 4 + δ ij dy i dy j) ( ) 1 2 Q (dr 2 r 4 + r 2 dω 2 ) 5. If we focus in the region close to the throat, r 0, the metric behaves as ds 2 r 2 ( dt 2 + δ ij dy i dy j) + r 2 ( dr 2 + r 2 dω 2 5). This is AdS 5 S 5 with equal radii of curvature. José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

17 D-branes and boundary contributions Let us come back to the gauge symmetry of (we set 2πα = 1) S = 1 d 2 ξ ɛ αβ B µν (X) α X µ β X ν. 2 If we perform the gauge symmetry transformation, δb µν = µ Λ ν ν Λ µ then the action transforms as δs = d 2 ξ ɛ αβ µ Λ ν α X µ β X ν = d 2 ξ ɛ αβ α Λ ν β X ν. = = dτ dσ ( τ Λ ν σ X ν σ Λ ν τ X ν ) dτ dσ ( τ [Λ ν σ X ν ] σ [Λ ν τ X ν ]) the total derivative τ gives no boundary contribution but σ does! José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18

18 D-branes and boundary contributions From the point of view of the open strings, the D-branes are hypersurfaces where their end-points can lie, σ=π δs = dτ dσ ( σ [Λ ν τ X ν ]) = dτ [Λ ν τ X ν ] Now, we have to distinguish between X µ = (X m, X a ), where m = 0, 1,..., p, σ=π σ=π δs = dτ [Λ m τ X m + Λ a τ X a ] = dτ [Λ m τ X m ] since τ X a = 0 at both end-points. Gauge invariance fails at the end-points of the string! To restore it, we must add a couple of terms that give electric charge to the string end-points, S = 1 d 2 ξ ɛ αβ B µν (X) α X µ β X ν + dτ A m (X) dx m σ=π 2 dτ. The string end-points are oppositely charged and F mn F mn = F mn + B mn. σ=0 σ=0 σ=0 σ=0 José D. Edelstein (USC) Lecture 9: RR-sector and D-branes 6-mar / 18