A. Particle-like solutions and their dual operators We can start by considering rotating particle solutions. When we endow only τ - dependence to the coordinates, the action and the constraint equations become equivalent to that of geodesic motion. More concretely, Eq. (20) becomes G$_{00}$ = 0, which is in fact the energy integral of the equations derived from the action Eq. (19). Eq. (21) becomes trivial. One can easily conclude that with uniform rotations as in Eq.(23), the equations for µ, ˜ θ, θ demand that µ = 0 or π $_{2}$. When µ = 0, the value of ˜ θ is irrelevant, since the S 2 embedded in CP 2 is collapsed. With µ = π $_{2}$, we can choose ˜ θ = 0 or π . With θ , we always have a choice between 0 and π . For instance, we can match the different particle-like solutions with vectors in some totally symmetrized tensor representations of SU (3) × SU (2) as follows. Let us start with SU (3), and assume the first two entries of the fundamental representations describe the position in S 2 parameterized by ˜ θ, ˜ φ . We can then assign where e$_{1}$ = (1 , 0 , 0) $^{T}$, e$_{2}$ = (0 , 1 , 0) $^{T}$, e$_{3}$ = (0 , 0 , 1) $^{T}$. The SU (2) part is easy. We can interpret orbits at θ = π (0) as direct products of spin-up (spin-down) states. We are now ready to give the dual operators for various particle-like solutions. For the lowest nontrivial chiral operators, Eqs. (6) and (7) explain how they constitute ( 10 , 3 ) representation. It is straightforward to generalize to long operators and match them against classical point-like solutions. First, This assignment (or the definitions given in Eqs. (26) to (29)) is justified when we calculate the conserved charges. As it is usual with supersymmetric solutions, one can verify that the conserved charges are given by the energy: we find R = Δ , Q$_{3}$ = 3Δ 4 , Q$_{8}$ = √ 3Δ 4 , J$_{3}$ = Δ 2 from the membrane calculation. And this is precisely what we can just read off from Eq. (33), when we for instance also make use of Eq. (24) and Eq. (30)! This duality mapping works also with other solutions: we have 9 µ = π $_{2}$, ˜ θ = π, θ = 0 ⇒ Tr $^{[}$( e − $^{ia}$X 2 $_{12}$X 2 $_{23}$X 2 $_{23}$) n $^{]}$, (34) µ = π $_{2}$, ˜ θ = 0 , θ = 0 ⇒ Tr $^{[}$( e − $^{ia}$X 1 $_{12}$X 1 $_{23}$X 1 $_{23}$) n $^{]}$. (33) µ = π $_{2}$, ˜ θ = 0 ⇒ e$_{1}$ ⊗ e$_{1}$ ⊗ · · · ⊗ e$_{1}$, (30) µ = π $_{2}$, ˜ θ = π ⇒ e$_{2}$ ⊗ e$_{2}$ ⊗ · · · ⊗ e$_{2}$, (31) µ = 0 ⇒ e$_{3}$ ⊗ e$_{3}$ ⊗ · · · ⊗ e$_{3}$, (32)