1 Introduction

UB-ECM-PF/07/30 SISSA-76-2007-EP

New non-abelian effects on D branes

Roberto Iengo and Jorge G. Russo

International School for Advanced Studies (SISSA)

Via Beirut 2-4, I-34013 Trieste, Italy

INFN, Sezione di Trieste

Institució Catalana de Recerca i Estudis Avançats (ICREA)

Departament ECM, Facultat de Física, Universitat de Barcelona,

Diagonal 647, 08028 Barcelona, Spain

Abstract

We extend the Myers dielectric effect to configurations with angular momentum. The resulting time-dependent D0 brane bound states can be interpreted as describing rotating fuzzy ellipsoids. A similar solution exists also in the presence of a RR magnetic field, that we study in detail. We show that, for any finite , above a certain critical angular momentum is energetically more favorable for the bound state system to dissociate into an abelian configuration of D0 branes moving independently. We have investigated this problem in the low-energy expansion of the non-abelian D brane action for generic . In the case we find explicit solutions of the full non-abelian Born-Infeld D brane dynamics, which remarkably have the same structure and confirm the features of the low-energy approximation. We further study D string configurations representing fuzzy funnels deformed by the magnetic field and by the rotational motion.

## 1 Introduction

The study of the D brane dynamics led to many important results in the last years. When D branes are put together, a nonabelian gauge symmetry emerges [1]. The scalar fields describing the transverse displacements of the branes become matrix-valued in the adjoint representation of the gauge group. This leads to solutions describing non-commutative configurations where branes of lower dimensions can generate fuzzy higher dimensional branes. The basic example is the Myers “dielectric effect” [2]. In this case D0 branes moving in an external four-form field reduce their energies by forming a bound state, whose geometry can be recognized as the geometry of a fuzzy sphere. The system can be interpreted as a fuzzy version of the bound state between a spherical D2 brane and D0 branes, and a precise correspondence exists between the corresponding energies and radii in the large limit.

Another interesting example of non-abelian effects in D brane systems are the “funnel” solutions found in [3, 4], representing bound states of D1 and D3 branes. The non-abelian solutions appear in the D1 brane system even in the absence of external fields. Remarkably, the dynamics produces a fuzzy version of the BIon system, spike solutions where D strings extend out of the D3 brane [5, 6, 7]. A review of different non-abelian phenomena can be found in [8] and various interesting applications can be found e.g. in [9][19].

Finding more general non-abelian solutions is in general complicated due to the non-linear nature of the equations. In this paper we will find a class of time-dependent exact solutions, which is remarkably simple despite the non-linear effects. We will also turn on an external magnetic RR field, generalizing a study made in [20] for a single Dp brane moving in an external RR magnetic field to the case of branes. We will first consider the case of the D0 brane system, and then show that similar time-dependent solutions exist for the funnel solutions of [3, 4].

When the magnetic field is absent, our D0 brane system is equivalent to the one studied by Myers [2], where D0 branes move under the influence of an external RR four-form field strength . In our case it will be convenient to take and , in order to satisfy the Gauss constraint in the presence of angular momentum. The static system of Myers describes D0 branes forming a bound state representing a fuzzy 2-sphere geometry (for further discussions on these geometries see e.g. [21, 22]). Here we will have the same system to start with, in terms of , (in the static limit our system is not the fuzzy version of studied in [9]). We will first study what happens when angular momentum is given to the system and then add the magnetic field. The structure of the time-dependent solution is similar with or without the magnetic field, and the main features are as follows. At sufficiently small angular momentum, the energy of the non-abelian system is lower than the energy of D0 branes moving independently under the action of the magnetic field. The non-abelian system represents a fuzzy ellipsoid. As the angular momentum is increased, there is a critical point where the energy of the non-abelian bound state overcomes the energy of the abelian configuration of independent D0 branes. We interpret this as a signal that the D0-D2 brane bound state must break for angular momenta above a certain critical value, that we calculate.

Here we will use the non-abelian D brane action derived in [2]. We will consider the flat space background , and the world-volume gauge field strength . We will turn on RR gauge fields and neglect the back reaction on the metric. This is justified for weak fields. More precisely, one assumes that the fields are sufficiently weak so that the corresponding energy density multiplied by the Newton constant is much smaller than one.

For flat backgrounds, the action of [2] reduces to , where

 SBI=−Tp∫dp+1σSTr(√det(Qij)√−det(P[Gαβ+Gαi(Q−1−δ)ijGjβ])), (1.1)
 Qij≡δij+iλ[Φi,Φk]Gkj

and the Wess-Zumino term is

 SWZ=Tp∫STr(P[eiλiΦiΦ∑C(n)]) . (1.2)

As usual, represents the transverse displacements, . For further details we refer to [2, 8].

## 2 D0 brane bound states

### 2.1 Time-dependent non-abelian solutions

The low-energy Lagrangian for in the presence of an external RR field strength is given by

 L=−NT0+T0λ2Tr(12˙Φ2+14[Φi,Φj][Φi,Φj]+i3ΦiΦjΦkF(4)tijk(t))+O(λ4) , (2.1)

where the the non-zero components of the field strength are , i.e.

 F(4)tijk=−2fεijkfor {i,j,k}=1,2,3  or  {i,j,k}=4,5,6 . (2.2)

The equations of motion are

 −¨Φi+[Φj,[Φi,Φj]]−ifεijk[Φj,Φk]=0 ,i,j,k=1,2,3 , (2.3)

and similarly for . They have to be supplemented by the Gauss constraint

 9∑i=1[˙Φj,Φj]=0 , (2.4)

coming from the equation of motion of the gauge field , which was set to zero. Having a field strength with non-vanishing components in both directions 123 and 456 will permit to solve the constraint due to a balance between the plane 12 and the plane 45. With this choice of field strength components, the Myers solution [2] is given by

 Φk=Φk+3=f2 αk , k=1,2,3 , (2.5)

where are two matrix representations of the SU(2) algebra

 [αi,αj]=2iεijkαk , (2.6)

Introducing new variables , the satisfy the commutation relations defining a fuzzy two-sphere:

 [Φi+,Φj+]=2ir0εijkΦk+ ,r0=f . (2.7)

while other commutators vanish (he gauge field configuration also admits a solution representing a fuzzy configuration, where the are given in terms , commuting with the [9]).

We now look for time-dependent solutions to this system. We consider the following ansatz:

 Φ1 = k1(t)α1+k2(t)α2 ,Φ4=k1(t)α1−k2(t)α2 , Φ2 = q1(t)α1+q2(t)α2 ,Φ5=−q1(t)α1+q2(t)α2 , Φ3 = Φ6=m0α3 . (2.8)

The equations of motion reduce to

 −¨k1−8(k22+q22+m20)k1+4fm0q2=0 , −¨k2−8(k21+q21+m20)k2−4fm0q1=0 , −¨q1−8(k22+q22+m20)q1−4fm0k2=0 , −¨q2−8(k21+q21+m20)q2+4fm0k1=0 , −2(k21+k22+q21+q22)m0+f(k1q2−k2q1)=0 . (2.9)

Although the system is non-linear, it admits a very simple solution:

 k1 = r0 cos(ωt) ,k2=r0 sin(ωt) , q1 = −r0 sin(ωt) ,q2=r0 cos(ωt) , m0 = f4 , (2.10)

with defined by the equation

 ω2=8r20−12f2 . (2.11)

The Gauss constraint (2.4) is identically satisfied.

Introducing , we find

 Φ1+ = 2r0cos(tω)α1 ,Φ2−=−2r0sin(tω)α1 , Φ2+ = 2r0cos(tω)α2 ,Φ1−=2r0sin(tω)α2 , Φ3+ = f2 α3 ,Φ3−=0 , (2.12)

They describe the algebra of a rotating fuzzy ellipsoid. To visualize this, at any given time one can make a rotation of coordinates so that the algebra is the same as the one at , where vanish and satisfy the following commutation relations:

 t=0:   [Φ1+,Φ2+]=16ir20fΦ3+ ,[Φ3+,Φ1+]=ifΦ2+ ,[Φ2+,Φ3+]=ifΦ1+ . (2.13)

At generic , the ellipsoid rotates penetrating into the part of the space. The eccentricity of the ellipsoid is always the same and determined by the ratio . When this is equal to one, one gets from (2.11) and we recover the Myers static solution. From (2.11) we see that must be greater than one in order to have oscillatory solutions.

In the large limit it should describe a bound state of D0 branes and a D2 brane. The classical configuration can be visualized by replacing the matrices with spherical harmonics , i.e.

 X1 = k1(t) cos(φ)sin(θ)+k2(t) sin(φ)sin(θ) , X2 = q1(t) cos(φ)sin(θ)+q2(t) sin(φ)sin(θ) , X3 = f4 cos(θ) , (2.14)

and a similar configuration in . This surface should represent the dual D2 brane, though we were unable to prove it from the dual D2 brane system, which for time-dependent configurations becomes complicated.

The energy of the solution is given by

 E=NT0+132λ2T0cN(−f4+4f2ω2+12ω4) , (2.15)

where we used

 Tr[αiαj]=cNδij , cN=N3(N2−1) ,  i,j=1,2,3 , (2.16)

which holds for irreducible representations (for reducible representations one gets a coefficient less than ). The formula (2.15) shows that the rotational motion increases the energy of the system, as expected.

### 2.2 Adding a RR magnetic field

Consider an external RR field , with non-vanishing components and similar ones in the plane 4,5, i.e. . This can be produced by D6 brane and anti D6 brane configurations. From eq. (1.2), one can read the additional term in the lagrangian

 ΔL=T0λTr[C(1)i˙Φi]=T0λ212bTr[Φ2˙Φ1−Φ1˙Φ2−Φ5˙Φ4+Φ4˙Φ5] . (2.17)

The equations of motion now become

 −¨Φi+[Φj,[Φi,Φj]]−ifεijk[Φj,Φk]−εij3b˙Φj=0 ,   i=1,2 , −¨Φ3+[Φj,[Φ3,Φj]]−ifε3jk[Φj,Φk]=0 . (2.18)

The equations for are similar with . The constraint is

 9∑i=1[Pi,Φi]=0 , (2.19)

where for and , and otherwise . The simplest solution is obtained by considering all to be diagonal, with entries , representing the positions of D0 branes. In this case they move independently in the magnetic field, giving rise to the usual Landau motion of the form , , .

Here we will be interested in non-abelian solutions. Once we turn on non-diagonal components of the , then only the center-of-mass coordinate –proportional to the identity– follows the Landau motion, where the D0 branes move collectively under the magnetic force. This motion decouples, since the identity commutes with all other matrices of , so in what follows we consider the part only, i.e. traceless .

We consider the same ansatz (2.8), which now leads to the equations

 −¨k1−8(k22+q22+m20)k1+4fm0q2−b˙q1=0 , −¨k2−8(k21+q21+m20)k2−4fm0q1−b˙q2=0 , −¨q1−8(k22+q22+m20)q1−4fm0k2+b˙k1=0 , −¨q2−8(k21+q21+m20)q2+4fm0k1+b˙k2=0 , −2(k21+k22+q21+q22)m0+f(k1q2−k2q1)=0 . (2.20)

A solution is given again by (2.10), where now

 ω2+bω−8r20+12f2=0 , (2.21)

i.e.

 ω=−12(b±√b2+32r20−2f2 ) . (2.22)

The constraint (2.19) is identically satisfied.

The commutation relations of the are the same as in (2.13). We see that a sufficiently large magnetic field permits to have oscillatory solutions for arbitrarily small values of the eccentricity .

Another interesting feature is that there are two solutions that represent a spherical configuration . One is the obvious solution . This is the Myers solution, which survives even in the presence of the magnetic field, since a static configuration does not feel the magnetic force. A second solution is . This is an oscillatory solution which remarkably remains spherical.

The energy of the general solution (2.10), (2.22) is given by

 E=NT0+λ2T0cN(2r20ω2+8r40−r20f2) , (2.23)

where we used (2.16).

Let us compare the energy of the spherical configuration for the two cases: i) and ii) . In these cases the energy formula (2.23) gives

 E(i)=NT0−132λ2T0cNf4 ,E(ii)=E(i)+18λ2T0cNb2f2 . (2.24)

The first case with energy is the same as the result of [2]. We see that the second case has more energy. More generally, using that and one can see that the solution with less energy is the Myers static solution and energy . To see this, one can rewrite the energy (2.23) in the form

 E=NT0+18λ2T0cN(−14f4+ω2(b+ω)2+16r20ω2) , (2.25)

where we used the relation (2.21).

The energy can also be expressed in terms of the conserved angular momenta defined by

 J12=T0λ2Tr[˙Φ1Φ2−˙Φ2Φ1+12b(Φ21+Φ22)] , (2.26)
 J45=T0λ2Tr[˙Φ4Φ5−˙Φ5Φ4−12b(Φ24+Φ25)] . (2.27)

The non-abelian properties of the dynamics provides a contribution to an “intrinsic spin” of the system, to be distinguished from the contribution of the center-of-mass, which represents the orbital part.

It is interesting to compare the energy of the abelian solution where the D0 branes move independently under the action of the magnetic field with the present non-abelian solution, for the same value of the angular momentum. As mentioned above, the abelian solution is given by . The angular momentum of the abelian solution is given by

 Jabel12=−Jabel45=−12NT0λ2ρ20b ,ρ20≡1N(ρ21+...+ρ2N) , (2.28)

and the energy has the expected gyromagnetic contribution

 Eabel=NT0+NT0λ2ρ20b2=NT0+|bJ12|+|bJ45| . (2.29)

In the case of the non-abelian solution, the angular momentum is given by

 Jnonabel12=−Jnonabel45=T0λ2cN j ,j=r20(2ω+b) . (2.30)

To simplify the discussion, we assume that the center of mass does not move, i.e. that in the non-abelian case all the angular momentum comes from the intrinsic spin.

Using (2.22) and (2.30) we find the relation

 j=±r20√b2+32r20−2f2 . (2.31)

The energy can then be written in the form

 Enonabel=NT0+λ2T0cN(−8r40+j2r20−bj) . (2.32)

The first term represents as usual the contribution from the rest masses of the D0 branes. The second term proportional to is the binding energy. The next term is a kinetic contribution due to the rotation and finally the term corresponds, interestingly, to a gyromagnetic interaction with gyromagnetic factor equal to 1 (so that the magnetic moments of the non-abelian system are equal to ).

The energy (2.32) of the non-abelian solution has a complicated expression in terms of the angular momentum. The reason is that also depends on the angular momentum through (2.31) and it is determined by a cubic equation. Since increases monotonically with , there is a one-to-one correspondence between and and therefore the value of univocally determines the value of the energy.

Using (2.31) one finds that at large the energy has the expansion:

 Enonabel=NT0+λ2T0cN(321/3|j|4/3−125/3|j|2/3(f2−b22)−bj−196(f2−b22)2+...) (2.33)

In the special case we get and and one can write a simple expression for the complete energy

 Enonabel∣∣∣f2=b22=NT0+cNT0λ2(−bj+3⋅2−1/3|j|4/3) . (2.34)

Since the energy (2.33) for the non-abelian solution increases as for large (recall ), there must be a point where it overcomes the linearly-increasing energy (2.29) of the abelian configuration. This indicates the remarkable fact that there is a critical value of the angular momentum after which the Myers D2-D0 bound state breaks and the system prefers to behave as individual D0 branes.

The critical can be found as follows. Writing , the abelian energy can be written as

 Eabel=NT0+2T0λ2cN|bj| . (2.35)

Expressing in terms of as in (2.31), we see from (2.32) that for

 24r20−2f2+b2=b√32r20−2f2+b2 , j<0 24r20−2f2+b2=3b√32r20−2f2+b2 , j>0 (2.36)

This gives a unique solution for . Substituting into eq. (2.31) we find the critical angular momentum after which the bound state should become unstable:

 jcr = 1216(−b2+6f2+b√b2+6f2)√5b2+6f2+4b√b2+6f2 ,j<0 jcr = 124√3(5b2+2f2+b√33b2+6f2)√23b2+2f2+4b√33b2+6f2 , j>0. (2.37)

Note that the grows like for large . This suggests that in the dual D2 system this transition may not be seen, since the critical point moves to infinity as .

We stress that this transition exists also in the Myers system with , i.e. without the RR magnetic field. In this case and the energy at the critical point goes to zero. For , the abelian configuration describes D0 branes with uniform motion. After the transition, the D0 branes should carry the energy and angular momentum of the original system in terms of linear momenta , .

Let us now examine the behavior of the energy as . From eq.(2.31) we see that corresponds to if or if . Therefore, for the energy (2.32) takes the following form

 Enonabel → NT0−132cNT0λ2(f2−b22)2 ,    f2≥b22 , Enonabel → NT0 ,    f2

In Figures 1a, 1b, 1c we compare the energies for each solution as a function of the angular momentum, including the abelian case. The figures exhibit the point of the transition after which .

Finally, it is interesting to note that the spherical case , , with energy given in (2.24), can be attained for a particular .

### 2.3 Corrections from Born-Infeld

We will now show that the same ansatz promotes to a solution of the full Born-Infeld Lagragian with only a correction in the equation defining the frequency and the constant value of , parametrized by . Here for simplicity we will consider the case of , so in what follows. The non-abelian Born-Infeld action is a non-linear functional of the non-abelian fields and a complete definition requires a prescription for the ordering of the fields under the trace. The action (1.1) was defined with the symmetric trace prescription [23], that we also adopt. For practical matters, this prescription amounts to treat and as commuting objects in the entries of the determinant. For the D0 brane system, the BI part of the Lagrangian can be written as

 (2.39)

We now consider the following ansatz:

 Φ1 = r0cos(ωt)α1+r0sin(ωt)α2 ,Φ4=¯r0cos(¯ωt)α1+¯r0sin(¯ωt) , Φ2 = −r0sin(ωt)α1+r0cos(ωt)α2 ,Φ5=−¯r0sin(¯ωt)α1+¯r0cos(¯ωt) , Φ3 = ¯m0α3 ,Φ6=¯m0α3 . (2.40)

We will compute the Lagrangian on this ansatz and find the values of and by variational principle with respect to and by imposing the Gauss constraint, which in these variables reads

 (∂∂ω+∂∂¯ω)Ltot=0 (2.41)

where includes the same couplings to and given in sects. 2.1 and 2.2. We have performed several checks that this procedure reproduces the correct equations.

Invariance under rotation of the Lagrangian independently in the planes 12 and 45 implies that we can compute at . For we find that the matrix is proportional to the identity, so taking the square root is straightforward (in the case of there would be terms proportional to inside the square root which are not necessarily proportional to the identity). The complete Lagrangian is given by

 Ltot=−2T0√−det~D+2λ2T0((4fm0+bω)r20+¯r20(4f¯m0−b¯ω)) , (2.42)
 −det~D = (2.43) − 8λ4r20¯r20(2m20+2¯m20+r20+¯r20)(ω−¯ω)2

The equations of motion and the constraint are solved by

 ¯r0=r0 ,¯m0=m0 ,¯ω=−ω . (2.44)

 0=f−4m0√1−4λ2r20ω2√1+16λ2r20(2m20+r20) , (2.45)
 0=4bm0ω(1−4λ2r20ω2)+f(8m20−8r20+ω2+48λ2r40ω2) . (2.46)

Solving (2.45) for we find

 m0=f4 √1+16λ2r40√1−2λ2f2r20−4λ2ω2r20 . (2.47)

Combining (2.45) and (2.46) one can get an algebraic equation determining the frequencies . One can then substitute in eq. (2.47) and find . Expanding in powers of we recover the same solution of section 2.1, as expected:

 m0=f4+O(λ2) ,ω2+bω−8r20+12f2=O(λ2) , (2.48)

where we used (2.47) and (2.46).

#### 2.3.1 Angular momentum and energy

The angular momenta are:

 J12 = ∂Ltot∂ω=2T0λ2r20(b+8ωm0f) . J45 = ∂Ltot∂¯ω=−2T0λ2r20(b+8ωm0f)=−J12 . (2.49)

The energy is:

 E = ω∂Ltot∂ω+¯ω∂Ltot∂¯ω−Ltot (2.50) = T0f(1+16λ2r40(1+8λ2m20ω2))2m0(1−4λ2r20ω2) .

where we used (2.45). Expanding in powers of , to order we reproduce the previous energy (2.23) with .

#### 2.3.2 System without magnetic field

In the particular case , the equations simplify. We get

 m0=f4 √1+48λ2r40 , ω2=8r20−12f2−24λ2f2r401+48λ2r40 . (2.51)

and the angular momentum and the energy are given by

 J ≡ J12=−J45=2T0√2λ2r20√16r20−f2(1+48λ2r40) , E = 2T0(1−2f2λ2r20)√1+48λ2r40 . (2.52)

The energy (2.52) reproduces, as expected, the formula (2.15), with given by (2.11), modulo terms . The possible values of and are constrained by the condition that and must be real. For the quantity under the square root, is negative and therefore becomes complex, so in what follows we assume . The minimum and the maximum values of are given by the value at which , i.e. defined by the condition . One can see that has a maximum value; for larger values of the angular momentum, the solution does not exist and the bound state must break.

For less than the maximum value,