5.3 Intersecting M2 and M5-branes

As a second example of BPS excitation, consider the 1/4 BPS configuration M5 ⊥ M2 (1 ) corresponding to the brane array
M5: 1 2 3 4 5 x x x x x (266 ) M2: x x x x 5 6 x x x x .
The idea is to describe an infinite M5-brane by the static gauge and to turn on a transverse scalar field 6 X to account for the M2-brane excitation. However, 6 X is not enough to support an M2-brane interpretation, since the latter is electrically charged under the 11-dimensional supergravity three form A3. Thus, the sought M5-brane soliton must source the A056 components. From the Wess–Zumino coupling
∫ dV2 ∧ π’œ3 , (267 )
one learns that the magnetic (dV )ˆaˆbˆc components, where hatted indices stand for world space directions different from σ5, i.e., ˆa ⁄= 5, must also be excited.

The full ansatz will assume delocalisation along the σ5 direction, so that the string-like excitation in the 6 X direction can be viewed as a membrane:

μ μ i i X = σ , X = c , X6 (σˆa) = y(σˆa), β„‹5 ˆaˆb = 0 . (268 )

Supersymmetry analysis:
The M5-brane kappa symmetry matrix (157View Equation) in the temporal gauge a = τ reduces to

[ -------- ] Γ = ∘------1-------- 1-πœ–a1...a5Γ γ − 1√ − det𝒒 Γ γ H&tidle;ab − Γ γ ta . (269 ) κ − det(𝒒 + H&tidle;) 5! 0 a1...a5 2 0 ab 0 a
For the subset of configurations described by the ansatz (268View Equation), it follows
ta = 0 , H&tidle;ˆaˆb = 0 &tidle;5ˆa √-Πˆa--- ˆa 1-ˆaˆa1ˆa2ˆa3 H = − det𝒒 , Π = 3!πœ– β„‹ ˆa1ˆa2ˆa3 . (270 )
This reduces Eq. (269View Equation) to
1 [ ] Γ κ = ∘---------------Γ 012345 + ∂ ˆayΓ 05yΓ 05Γ ˆaΓ 012345 − Γ 05Γ ˆaΠaˆ− ∂ˆay ΠˆaΓ 05y . (271 ) − det(𝒒 + H&tidle;)
To solve the kappa symmetry preserving condition (214View Equation), I impose two projection conditions
Γ 012345πœ– = πœ–, Γ 05yπœ– = πœ– (272 )
on the constant Killing spinors πœ–. The eight supercharges satisfying them match the ones preserved by M5 ⊥ M2 (1). Using Eq. (272View Equation) in Eq. (271View Equation), Γ κ keeps a non-trivial dependence on Γ Γ 05 ˆa. Requiring its coefficient to vanish gives rise to the BPS condition
Π = − ∂ y. (273 ) ˆa ˆa
Overall, the kappa symmetry preserving condition (214View Equation) reduces to the purely algebraic condition
∘ -------------- &tidle; ( ˆaˆb ) − det(𝒒 + H )πœ– = 1 + δ ∂ˆay∂ˆby πœ–. (274 )
To check this holds, notice the only non-vanishing components of &tidle;H μν are H&tidle; 5ˆa
ˆb H&tidle; = 𝒒 𝒒 ˆ√---Π-----. (275 ) 5ˆa 55 aˆb − det 𝒒
This allows us to compute the determinant
− det(𝒒 + H&tidle;) = det(𝒒 + &tidle;H H&tidle; ) = det(𝒒 )(1 + 𝒒ˆaˆbH&tidle; &tidle;H ) , (276 ) ˆaˆb 5ˆa 5ˆb ˆaˆb 5ˆa 5ˆb
which becomes a perfect square once the BPS equation (273View Equation) is used
− det(𝒒 + H&tidle;) = (1 + δˆaˆb∂ y∂ y )2 . (277 ) ˆa ˆb
This shows that Eq. (274View Equation) holds automatically. Thus, the solution to the kappa symmetry preserving condition (214View Equation) for the ansatz (268View Equation) on an M5-brane action is solved by the supersymmetry projection conditions (272View Equation) and the BPS equation (273View Equation). Since the soliton involves a non-trivial world volume gauge field, the Bianchi identity dβ„‹3 = 0 must still be imposed. This determines the harmonic character for the excited transverse scalar in the four dimensional world space ω4
∂ˆa∂ y = 0. (278 ) ˆa

Hamiltonian analysis:
The Hamiltonian analysis for this system was studied in [225Jump To The Next Citation Point] following the M5-brane phase space formulation given in Eq. (229View Equation). For static configurations, the Hamiltonian constraint can be solved by the energy density β„° as

β„°2 1 --2- = 1 + (∂y)2 + -|β„‹&tidle;|2 + |β„‹&tidle;⋅ ∂y |2 + |V |2 (279 ) TM5 2
1 |β„‹ &tidle;|2 = &tidle;β„‹abβ„‹&tidle;cd δacδbd, β„‹&tidle;ab = --πœ€abcdeβ„‹cde, 6 |β„‹&tidle;⋅ ∂y|2 = &tidle;β„‹abβ„‹&tidle;cd ∂by∂dyδac, |V |2 = V V δab, (280 ) a b
and world space indices were denoted by latin indices σa a = 1,...,5. It was noted in [225Jump To The Next Citation Point] that by introducing a unit length world space 5-vector ζ, i.e., ζaζbδab = 1, the energy density could be written in the suggestive form
β„° 2 || 1 ||2 -2--= |ζa ± β„‹&tidle;ab ∂by|2 + 2||∂[ayζb] ±-δacδbd &tidle;β„‹cd|| TM5 2 + (ζa∂ay)2 + |V |2. (281 )
The unit vector provides a covariant way of introducing a preferred direction in the 5-dimensional world space. Choosing ζ5 = 1 and ζˆa = 0, to match the delocalisation direction in our bosonic ansatz, one derives the inequality
-β„°--≥ 1 ± 1Π ˆa∂ˆay. (282 ) TM5 6
The latter is saturated if and only if
∂5y = 0 β„‹5ˆaˆb = 0 (283 )
β„‹3 = ± ⋆ dy (284 )
where β„‹3 is only defined on the 4-dimensional subspace ω4, orthogonal to ζ, and ⋆ is its Hodge dual. This confirms the BPS nature of Eq. (273View Equation). Since β„‹3 is closed, y is harmonic in ω4. To regulate the divergent energy, one imposes periodic boundary conditions in the 5-direction making the orbits of the vector field ζ have finite length L. Then, the total energy satisfies
E ≥ E + L ⋅ |Z |, (285 ) 0
where Z is the topological charge
∫ Z = β„‹ ∧ dy . (286 ) w4 3
The tension of the soliton, i.e., energy per unit of length, equals T = E − E βˆ•L 0. It is bounded by Z. It only equals the latter for configurations satisfying Eq. (284View Equation). Singularities in the harmonic function match the strings found in [301Jump To The Next Citation Point]. To check this interpretation, consider a solution with a single isolated point singularity at the origin. Its energy can be rewritten as the small radius limit of a surface integral over a 3-sphere surrounding the origin. Since y is constant on this integration surface, one derives the string tension [225Jump To The Next Citation Point]
∫ T = μ lim y (δ) where μ = β„‹ (287 ) δ→0 S3 3
is the string charge. Even though this tension diverges, it does so consistently, being the boundary of a semi-infinite membrane.
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