The Bethe/Gauge Correspondence

62 Chapter 4. Bethe/gaugeThe main result on the gauge side is of course the effective twisted superpotential (3.73) interms of the ground states σ n ,˜W eff (σ) = i τN∑n=1σ n + 12π+ 12π+ 12πN∑ ∑L f(σ n − ˜m l f )n=1 l=1N∑ ∑L¯f(−σ n − ˜m k¯f )n=1 k=1N∑ ∑L a(σ m − σ n − ˜m j a)m,n j=1m≠n(log σn − ˜m l fµ)− 1(log −σn − ˜m k¯fµ)− 1(log σm − σ n − ˜m j )a− 1 ,µ(4.3)and the resulting set of N coupled vacuum equations (3.74):∏ Lfl=1 (σn − ˜m l f∏ )L¯fk=1 (σn + ˜m k¯f ) = ∏NµL f−L¯f (−1) L¯f−L ae 2πiτ∏L am≠n j=1σ n − σ m + ˜m j aσ n − σ m − ˜m j a. (4.4)These equations describe the vacuum structure on the Coulomb branch of a two-dimensionalN = (2, 2) sym with massive chiral matter multiplets. The matter fields are in the fundamental,anti-fundamental or adjoint representation of the gauge group; we will continue to refer to theserepresentations with R ∈ {f,¯f, a}. These theories have the following parameters to play aroundwith:• the numbers L R of flavours of fields in the representation R (see Section 3.5);• the number N of colours;• the twisted masses ˜m l R ∈ C, for 1 ≤ l ≤ L R (Sections 3.2.3 and 3.3.3);• the complex gauge coupling τ = ir + ϑ/2π consisting of the FI-parameter r ∈ R and thevacuum angle ϑ ∈ S 1 (Sections 3.3.2 and 3.5).It’s time to match the two.4.1.1 Main exampleLooking at the above results, the leading observation for the Bethe/gauge correspondence fromSection 1.4 is clear:If we start with the appropriate field content, for suitable values of the parametersof the theory, the vacuum equation of two-dimensional gauge theory with N = (2, 2)supersymmetry coincides with the Bethe Ansatz equations of a quantum integrablemodel.The Bethe roots λ n are related to the supersymmetric vacua σ n . There is a small subtletyregarding the mass dimensions of λ n and σ n , but we will ignore this for now, and write λ n ≈ σ n .We will get back to this shortly and replace the ‘≈’ by equalities. (The difference in the placementof the index ‘n’ is immaterial.)To see what values we should take for the parameters to fit the vacuum equation (4.4) withthe bae (4.1) we start at the gauge side. The first observation is that we need L f = L¯f =: L andL a = 1, so that the vacuum equations (4.4) becomeL∏l=1σ n − ˜m l fσ n + ˜m l¯f= (−1) L−1 e iϑ−2πr N ∏m≠nσ n − σ m + ˜m aσ n − σ m − ˜m a. (4.5)Secondly, we can absorb the sign in front of the right-hand side by a shift in τ, passing on toτ ′ := τ + (L − 1)/2; we drop the prime. This parameter is related to the twist parameter ϑ of