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10.3 Cosmological solutions with electric flux

Let us now make use of these considerations to construct some explicit supergravity solutions. We begin by analyzing the simplest configuration (31,13), of three points and one line. It is displayed in Figure 50View Image. This case is the only possible configuration for n = 3.
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Figure 50: (31,13): The only allowed configuration for n = 3.

This example also exhibits some subtleties associated with the Hamiltonian constraint and the ensuing need to extend ¯π”€ when the algebra dual to the geometric configuration is finite-dimensional. We will come back to this issue below.

10.3.1 General discussion

In light of our discussion, considering the geometric configuration (31,13) is equivalent to turning on only the component π’œ123 (t) of the 3-form that parametrizes the generator E123 in the coset representative 𝒱 (t) ∈ β„°10βˆ•π’¦ (β„°10). Moreover, in order to have the full coset description, we must also turn on the diagonal metric components corresponding to the Cartan generator 123 h = [E ,F123]. The algebra has thus basis {e,f,h} with

123 1 ∑ a 2 1 2 3 e ≡ E , f ≡ F123, h = [e,f] = −3- K a + 3(K 1 + K 2 + K 3), (10.14 ) a⁄=1,2,3
where the form of h followed directly from the general commutator between Eabc and F def in Section 8. The Cartan matrix is just (2) and is nondegenerate. It defines an A1 = 𝔰𝔩(2, ℝ) regular subalgebra. The Chevalley–Serre relations, which are guaranteed to hold according to the general argument, are easily verified. The configuration (31,13) is thus dual to A1,
𝔀 = A . (10.15 ) (31,31) 1
This A1 algebra is simply the 𝔰𝔩(2, ℝ)-algebra associated with the simple root α10. Because the Killing form of A1 restricted to the Cartan subalgebra π”₯A = ℝh 1 is positive definite, one cannot find a solution of the Hamiltonian constraint if one turns on only the fields corresponding to A1. One needs to enlarge A1 (at least) by a one-dimensional subalgebra ℝl of π”₯E10 that is timelike. As will be discussed further below, we take for l the Cartan element K44 + K55 + K66 + K77 + K88 + K99 + K1010, which ensures isotropy in the directions not supporting the electric field. Thus, the appropriate regular subalgebra of E10 in this case is A1 ⊕ ℝl.

The need to enlarge the algebra A1 was discussed in the paper [127Jump To The Next Citation Point] where a group theoretical interpretation of some cosmological solutions of eleven-dimensional supergravity was given. In that paper, it was also observed that ℝl can be viewed as the Cartan subalgebra of the (non-regularly embedded) subalgebra A1 associated with an imaginary root at level 21, but since the corresponding field is not excited, the relevant subalgebra is really ℝl.

10.3.2 The solution

In order to make the above discussion a little less abstract we now show how to obtain the relevant supergravity solution by solving the β„°10-sigma model equations of motion and then translating these, using the dictionary from Section 9, to supergravity solutions. For this particular example the analysis was done in [127Jump To The Next Citation Point].

In order to better understand the role of the timelike generator l ∈ π”₯E 10 we begin the analysis by omitting it. The truncation then amounts to considering the coset representative

123 𝒱(t) = eφ(t)heπ’œ123(t)E ∈ β„°10βˆ•π’¦ (β„°10). (10.16 )
The projection 𝒫 (t) onto the coset becomes
1 [ ( )𝒯] 𝒫(t) = -- ∂𝒱 (t) 𝒱(t)−1 + ∂𝒱 (t)𝒱 (t)−1 2 = ∂φ (t)h + 1-e2φ(t)∂ π’œ123(t)(E123 + F123), (10.17 ) 2
where the exponent is the linear form α(φ) = 2φ representing the exceptional simple root α123 of E10. More precisely, it is the linear form α acting on the Cartan generator φ(t)h, as follows:
α(φh ) = φ⟨α, h⟩ = φ ⟨α, α∨ ⟩ = α2 φ = 2φ. (10.18 )
The Lagrangian becomes
β„’ = 1(𝒫 (t)|𝒫(t)) 2 1-4φ(t) = ∂ φ(t)∂φ(t) + 4e ∂ π’œ123(t)∂π’œ123(t). (10.19 )
For convenience we have chosen the gauge n = 1 of the free parameter in the β„°10βˆ•π’¦ (β„°10)-Lagrangian (see Section 9). Recall that for the level one fields we have π’Ÿ π’œabc(t) = ∂π’œabc(t), which is why only the partial derivative of π’œ (t) 123 appears in the Lagrangian.

The reason why this simple looking model contains information about eleven-dimensional supergravity is that the A1 subalgebra represented by (e, f,h) is embedded in E10 through the level 1-generator E123, and hence this Lagrangian corresponds to a consistent subgroup truncation of the β„°10- sigma model.

Let us now study the dynamics of the Lagrangian in Equation (10.19View Equation). The equations of motion for π’œ123 (t) are

( ) 1 4φ(t) 1 4φ(t) ∂ 2-e ∂ π’œ123(t) = 0 = ⇒ 2e ∂π’œ123 (t) = a, (10.20 )
where a is a constant. The equations for the β„“ = 0 field φ may then be written as
2 2 −4φ(t) ∂ φ(t) = 2a e . (10.21 )
Integrating once yields
∂ φ(t)∂φ (t) + a2e−4φ(t) = E, (10.22 )
where E plays the role of the energy for the dynamics of φ (t). This equation can be solved exactly with the result [127Jump To The Next Citation Point]
1 [ 2a √ --] 1 φ (t) = --ln √---cosh Et ≡ -ln H (t). (10.23 ) 2 E 2
We must also take into account the Hamiltonian constraint
β„‹ = (𝒫 |𝒫 ) = 0, (10.24 )
arising from the variation of n(t) in the β„° 10-sigma model. The Hamiltonian becomes
1- 4φ(t) β„‹ = 2∂ φ(t)∂φ (t) + 2 e ∂ π’œ123(t)∂π’œ123(t) ( 2 −4φ(t)) = 2 ∂φ (t)∂φ (t) + a e = 2E. (10.25 )
It is therefore impossible to satisfy the Hamiltonian constraint unless E = 0. This is the problem which was discussed above, and the reason why we need to enlarge the choice of coset representative to include the timelike generator l ∈ π”₯E10. We choose l such that it commutes with h and E123,
123 [l,h] = [l,E ] = 0, (10.26 )
and such that isotropy in the directions not supported by the electric field is ensured. Most importantly, in order to solve the problem of the Hamiltonian constraint, l must be timelike,
2 l = (l|l) < 0, (10.27 )
where (⋅|⋅) is the scalar product in the Cartan subalgebra of E10. The subalgebra to which we truncate the sigma model is thus given by
¯π”€ = A1 ⊕ ℝl ⊂ E10, (10.28 )
and the corresponding coset representative is
𝒱&tidle;(t) = eφ(t)h+ &tidle;φ(t)leπ’œ123(t)E123. (10.29 )
The Lagrangian now splits into two disconnected parts, corresponding to the direct product SL (2,ℝ )βˆ•SO (2) × β„,
( ) 2 &tidle;β„’ = ∂φ (t) ∂φ(t) + 1e4φ(t)∂π’œ123(t)∂ π’œ123(t) + l-∂φ&tidle;(t) ∂&tidle;φ(t). (10.30 ) 4 2
The solution for &tidle; φ is therefore simply linear in time,
2 ∘ -- φ&tidle;= |l | &tidle;E t. (10.31 )
The new Hamiltonian now gets a contribution also from the Cartan generator l,
2 &tidle;β„‹ = 2E − |l | &tidle;E. (10.32 )
This contribution depends on the norm of l and since l2 < 0, it is possible to satisfy the Hamiltonian constraint, provided that we set
&tidle;E = -2-E. (10.33 ) |l2|

We have now found a consistent truncation of the 𝒦 (β„°10) × β„°10-invariant sigma model which exhibits SL (2,ℝ ) × SO (2) × β„-invariance. We want to translate the solution to this model, Equation (10.23View Equation), to a solution of eleven-dimensional supergravity. The embedding of 𝔰𝔩(2,ℝ ) ⊂ E10 in Equation (10.14View Equation) induces a natural “Freund–Rubin” type (1 + 3 + 7) split of the coordinates in the physical metric, where the 3-form is supported in the three spatial directions 1 2 3 x ,x ,x. We must also choose an embedding of the timelike generator l. In order to ensure isotropy in the directions x4, ⋅⋅⋅ ,x10, where the electric field has no support, it is natural to let l be extended only in the “transverse” directions and we take [127Jump To The Next Citation Point]

4 10 l = K 4 + ⋅⋅⋅ + K 10, (10.34 )
which has norm
(l|l) = (K4 + ⋅⋅⋅ + K10 |K4 + ⋅⋅⋅ + K10 ) = − 42. (10.35 ) 4 10 4 10
To find the metric solution corresponding to our sigma model, we first analyze the coset representative at β„“ = 0,
[ ] &tidle;𝒱(t)|| = Exp φ(t)h + &tidle;φ(t)l . (10.36 ) β„“=0
In order to make use of the dictionary from Section 9.3.6 it is necessary to rewrite this in a way more suitable for comparison, i.e., to express the Cartan generators h and l in terms of the 𝔀𝔩(10,ℝ )-generators Kab. We thus introduce parameters ξa (t) b and &tidle;ξab(t) representing, respectively, φ and &tidle;φ in the 𝔀𝔩(10, ℝ)-basis. The level zero coset representative may then be written as
[ 10 ( ) ] &tidle; || ∑ a &tidle;a a 𝒱(t)β„“=0 = Exp ξ a(t) + ξ a(t) K a [a=1 ] ∑10 ( a a ) a ( 1 1 2 2 3 3 ) = Exp ξ a(t) + ξ&tidle;a(t) K a + ξ 1(t)K 1 + ξ 2(t)K 2 + ξ 3(t)K 3 , (10.37 ) a=4
where in the second line we have split the sum in order to highlight the underlying spacetime structure, i.e., to emphasize that &tidle;a ξ b has no non-vanishing components in the directions 1 2 3 x ,x ,x. Comparing this to Equation (10.14View Equation) and Equation (10.34View Equation) gives the diagonal components of a ξ b and a &tidle;ξ b,
ξ11 = ξ22 = ξ33 = 2φ βˆ•3, ξ44 = ⋅⋅⋅ = ξ1010 = − φ βˆ•3, ξ&tidle;44 = ⋅⋅⋅ = &tidle;ξ1010 = &tidle;φ. (10.38 )
Now, the dictionary from Section 9 identifies the physical spatial metric as follows:
¯a ¯b ξ(t)+ &tidle;ξ(t) ¯a ξ(t)+&tidle;ξ(t) ¯b gab(t) = ea (t)eb (t)δ¯a¯b = (e )a (e )bδ¯a¯b (10.39 )
By observation of Equation (10.38View Equation) we find the components of the metric to be
g11 = g22 = g33 = e4φβˆ•3, −2φβˆ•3+2&tidle;φ (10.40 ) g44 = ⋅⋅⋅ = g(10)(10) = e .
This result shows clearly how the embedding of h and l into E10 is reflected in the coordinate split of the metric. The gauge fixing √ -- N = g (or n = 1) gives the gtt-component of the metric,
2 14&tidle;φ−2φβˆ•3 gtt = N = e . (10.41 )
Next we consider the generator E123. The dictionary tells us that the field strength of the 3-form in eleven-dimensional supergravity at some fixed spatial point x0 should be identified as
F (t,x ) = π’Ÿ π’œ (t) = ∂ π’œ (t). (10.42 ) t123 0 123 123
It is possible to eliminate the π’œ123(t) in favor of the Cartan field φ (t) using the first integral of its equations of motion, Equation (10.20View Equation),
1 -e− 4φ(t)∂ π’œ123(t) = a. (10.43 ) 2
In this way we may write the field strength in terms of a and the solution for φ,
4φ(t) −2 Ft123(t,x0) = 2ae = 2aH (t). (10.44 )
Finally, we write down the solution for the spacetime metric explicitly:
[ ] ∑10 ds2 = − e14&tidle;φ+2φβˆ•3dt2 + e4φβˆ•3 (dx1 )2 + (dx2 )2 + (dx3)2 + e2&tidle;φ−2φβˆ•3 (dx¯a)2 ¯a=4 √ -- √-- 10 1βˆ•3 13 &tidle;Et 2 −2βˆ•3 [ 12 2 2 3 2] 1βˆ•3 2&tidle;E1 t∑ ¯a 2 = − H (t)e dt + H (t) (dx ) + (dx ) + (dx ) + H (t)e (dx ) , ¯a=4 (10.45 )
where
2a √ -- H (t) = √---cosh Et. (10.46 ) E
This solution coincides with the cosmological solution first found in [61Jump To The Next Citation Point] for the geometric configuration (3 ,3 ) 1 1, and it is intriguing that it can be exactly reproduced from a manifestly β„° × π’¦ (β„° ) 10 10-invariant action, a priori unrelated to any physical model.

Note that in modern terminology, this solution is an SM 2-brane solution (see, e.g., [143] for a review) since it can be interpreted as a spacelike (i.e., time-dependent) version of the M 2-brane solution. From this point of view the world volume of the SM 2-brane is extended in the directions x1,x2 and x3, and so is Euclidean.

In the BKL-limit this solution describes two asymptotic Kasner regimes, at t → ∞ and at t → − ∞. These are separated by a collision against an electric wall, corresponding to the blow-up of the electric field Ft123(t) ∼ H − 2(t) at t = 0. In the billiard picture the dynamics in the BKL-limit is thus given by free-flight motion interrupted by one geometric reflection against the electric wall,

e (β) = β1 + β2 + β3, (10.47 ) 123
which is the exceptional simple root of E10. This indicates that in the strict BKL-limit, electric walls and SM 2-branes are actually equivalent.

10.3.3 Intersecting spacelike branes from geometric configurations

Let us now examine a slightly more complicated example. We consider the configuration (62,43), shown in Figure 51View Image. This configuration has four lines and six points. As such the associated supergravity model describes a cosmological solution with four components of the electric field turned on, or, equivalently, it describes a set of four intersecting SM 2-branes [96Jump To The Next Citation Point].

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Figure 51: The configuration (62,43), dual to the Lie algebra A1 ⊕ A1 ⊕ A1 ⊕ A1.

From the configuration we read off the Chevalley–Serre generators associated to the simple roots of the dual algebra:

e1 = E123, e2 = E145, e3 = E246, e4 = E356. (10.48 )
The first thing to note is that all generators have one index in common since in the graph any two lines share one node. This implies that the four lines in (62,43) define four commuting A1 subalgebras,
(62,43) ⇐ ⇒ 𝔀(62,43) = A1 ⊕ A1 ⊕ A1 ⊕ A1. (10.49 )
One can make sure that the Chevalley–Serre relations are indeed fulfilled for this embedding. For instance, the Cartan element b1b2b3 h = [E ,Fb1b2b3] (no summation on the fixed, distinct indices b1,b2,b3) reads
1- ∑ a 2- b1 b2 b3 h = − 3 K a + 3 (K b1 + K b2 + K b3). (10.50 ) a⁄=b1,b2,b3
Hence, the commutator [h,Ebicd] vanishes whenever Ebicd has only one b-index,
[h, Ebicd] = − 1[(Kcc + Kdd ),Ebicd] + 2-[(Kb1b + Kb2b + Kb3b ),Ebicd] ( 3 ) 3 1 2 3 1- 1- 2- bicd = − 3 − 3 + 3 E = 0 (i = 1,2,3). (10.51 )
Furthermore, multiple commutators of the step operators are immediately killed at level 2 whenever they have one index or more in common, e.g.,
123 145 123145 [E ,E ] = E = 0. (10.52 )
To fulfill the Hamiltonian constraint, one must extend the algebra by taking a direct sum with ℝl, 7 8 9 10 l = K 7 + K 8 + K 9 + K 10. Accordingly, the final algebra is A1 ⊕ A1 ⊕ A1 ⊕ A1 ⊕ ℝl. Because there is no magnetic field, the momentum constraint and Gauss’ law are identically satisfied.

By investigating the sigma model solution corresponding to the algebra 𝔀(6,4 ) 2 3, augmented with the timelike generator l,

¯π”€ = A1 ⊕ A1 ⊕ A1 ⊕ A1 ⊕ ℝl, (10.53 )
we find a supergravity solution which generalizes the one found in [61Jump To The Next Citation Point]. The solution describes a set of four intersecting SM 2-branes, with a five-dimensional transverse spacetime in the directions t,x7,x8,x9, x10.

Let us write down also this solution explicitly. The full set of generators for 𝔀 (62,43) is

123 145 246 356 e1 = E , e2 = E , e3 = E , e4 = E f1 = F123, f2 = F145, f3 = F246, f4 = F356 ∑ h1 = − 1- Kaa + 2(K11 + K22 + K33 ), 3 3 1a⁄=∑1,2,3 2 h2 = − -- Kaa + -(K11 + K44 + K55 ), (10.54 ) 3a⁄=1,4,5 3 1 ∑ a 2 2 4 6 h3 = − 3- K a + 3(K 2 + K 4 + K 6), a⁄=2,4,6 1-∑ a 2- 3 5 6 h4 = − 3 K a + 3(K 3 + K 5 + K 6). a⁄=3,5,6
The coset element for this configuration then becomes
φ(t)h +φ (t)h +φ (t)h +φ (t)h +&tidle;φ(t)l π’œ (t)E123+π’œ (t)E145+π’œ (t)E246+π’œ (t)E356 𝒱(t) = e 1 1 2 2 3 3 4 4 e 123 145 246 356 . (10.55 )
We must further choose the timelike Cartan generator, l ∈ π”₯ E10, appropriately. Examination of Equation (10.54View Equation) reveals that the four electric fields are supported only in the spatial directions 1 6 x ,⋅⋅⋅ ,x so, again, in order to ensure isotropy in the directions transverse to the S-branes, we choose the timelike Cartan generator as follows:
l = K77 + K88 + K99 + K1010, (10.56 )
which implies
( ) l2 = (l|l) = K77 + K88 + K99 + K1010 |K77 + K88 + K99 + K1010 = − 12. (10.57 )
The Lagrangian for this system becomes
l2 β„’(62,43) = β„’1 + β„’2 + β„’3 + β„’4 + --∂&tidle;φ(t)∂φ&tidle;(t), (10.58 ) 2
where β„’1,β„’2, β„’3 and β„’4 represent the SL (2, ℝ) × SO (2)-invariant Lagrangians corresponding to the four A1-algebras. The solutions for φ1 (t),⋅⋅⋅ ,φ4(t) and &tidle;φ(t) are separately identical to the ones for φ (t) and &tidle;φ(t), respectively, in Section 10.3.2. From the embedding into E 10, provided in Equation (10.54View Equation), we may read off the solution for the spacetime metric,
√ --- ds2 = − (H H H H )1βˆ•3e23 E−tdt2 + (H H )−2βˆ•3(H H )1βˆ•3(dx1)2 (62,43) 1 2 3 4 1 4 2 3 +(H1H3 )−2βˆ•3(H2H4 )1βˆ•3(dx2)2 + (H1H2 )− 2βˆ•3(H3H4 )1βˆ•3(dx3)2 −2βˆ•3 1βˆ•3 42 − 2βˆ•3 1βˆ•3 5 2 +(H3H4 ) (H1H2 ) (dx ) + (H2H4 ) (H1H3 ) (dx ) 1√--- ∑10 +(H2H3 )−2βˆ•3(H1H4 )1βˆ•3(dx6)2 + (H1H2H3H4 )1βˆ•3e 6 E −t (dx¯a)2. (10.59 ) ¯a=7
As announced, this describes four intersecting SM 2-branes with a 1 + 4-dimensional transverse spacetime. For example the brane that couples to the field associated with the first Cartan generator is extended in the directions x1, x2,x3. By restricting to the case φ1 = φ2 = φ3 = φ4 ≡ φ the metric simplifies to
( 2a )4 βˆ•3 √ -- 2√ -- ( 2a ) −2βˆ•3 √-- ∑ 6 ′ ds2(62,43) = − √--- cosh4βˆ•3 Et e3 &tidle;Etdt2 + √--- cosh− 2βˆ•3 Et (dxa )2 E E a′=1 ( )4 βˆ•3 √ -- 10 -2a- 4βˆ•3√ -- 16 &tidle;Et∑ ¯a 2 + √ E cosh Et e (dx ), (10.60 ) ¯a=7
which coincides with the cosmological solution found in [61Jump To The Next Citation Point] for the configuration (62,43). We can therefore conclude that the algebraic interpretation of the geometric configurations found in this paper generalizes the solutions given in the aforementioned reference.

In a more general setting where we excite more roots of E10, the solutions of course become more complex. However, as long as we consider commuting subalgebras there will naturally be no coupling in the Lagrangian between fields parametrizing different subalgebras. This implies that if we excite a direct sum of m A 1-algebras the total Lagrangian will split according to

∑m β„’ = β„’k + &tidle;β„’, (10.61 ) k=1
where β„’k is of the same form as Equation (10.19View Equation), and &tidle; β„’ is the Lagrangian for the timelike Cartan element, needed in order to satisfy the Hamiltonian constraint. It follows that the associated solutions are
[ ] 1 ak ∘ --- φk(t) = 2-ln E-- cosh Ekt (k = 1,⋅⋅⋅ ,m ), ∘ --k (10.62 ) &tidle;φ(t) = |l2| &tidle;Et.
Furthermore, the resulting structure of the metric depends on the embedding of the A1-algebras into E10, i.e., which level 1-generators we choose to realize the step-operators and hence which Cartan elements that are associated to the φ k’s. Each excited A 1-subalgebra will turn on an electric 3-form that couples to an SM 2-brane and hence the solution for the metric will describe a set of m intersecting SM 2-branes.

As an additional nice example, we mention here the configuration (73,73), also known as the Fano plane, which consists of 7 lines and 7 points (see Figure 52View Image). This configuration is well known for its relation to the octonionic multiplication table [8]. For our purposes, it is interesting because none of the lines in the configuration are parallell. Thus, the algebra dual to the Fano plane is a direct sum of seven A1-algebras and the supergravity solution derived from the sigma model describes a set of seven intersecting SM 2-branes.

View Image

Figure 52: The Fano Plane, (73,73), dual to the Lie algebra A1 ⊕ A1 ⊕ A1 ⊕ A1 ⊕ A1 ⊕ A1 ⊕ A1.

10.3.4 Intersection rules for spacelike branes

For multiple brane solutions, there are rules for how these branes may intersect in order to describe allowed BPS-solutions [6]. These intersection rules also apply to spacelike branes [144] and hence they apply to the solutions considered here. In this section we will show that the intersection rules for multiple S-brane solutions are encoded in the associated geometric configurations [96Jump To The Next Citation Point].

For two spacelike q-branes, A and B, in M-theory the rules are

(qA + 1)(qB + 1) SM qA ∩ SM qB = ----------------− 1. (10.63 ) 9
So, for example, if we have two SM 2-branes the result is
SM 2 ∩ SM 2 = 0, (10.64 )
which means that they are allowed to intersect on a 0-brane. Note that since we are dealing with spacelike branes, a zero-brane is extended in one spatial direction, so the two SM 2-branes may therefore intersect in one spatial direction only. We see from Equation (10.59View Equation) that these rules are indeed fulfilled for the configuration (62,43).

In [72] it was found in the context of 𝔀+++-algebras that the intersection rules for extremal branes are encoded in orthogonality conditions between the various roots from which the branes arise. This is equivalent to saying that the subalgebras that we excite are commuting, and hence the same result applies to 𝔀++-algebras in the cosmological context39. From this point of view, the intersection rules can also be read off from the geometric configurations in the sense that the configurations encode information about whether or not the algebras commute.

The next case of interest is the Fano plane, (73,73). As mentioned above, this configuration corresponds to the direct sum of 7 commuting A1 algebras and so the gravitational solution describes a set of 7 intersecting SM 2-branes. The intersection rules are guaranteed to be satisfied for the same reason as before.


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