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... = R(yi)L(yi)L −1(yf )R −1(yf ) . (3.46) Since we are only interested in traces of M , i.e. its eigenvalues, we will conjugate M by R(yi) to get M → M = L(yi)L−1(yf )R−1(yf )R(yi) . (3.47) We are interested in connections (3.22) of the form R(xµ) = exp (∫ x 0 a ) b−1(ρ) , L(xµ) = b−1(ρ) exp ( − ∫ x 0 a ) , (3.48) where b(ρ) ≡ exp(ρL0), a = atdt + axdx and a = atdt + axdx. The limit ρ → ∞ defines the boundary of the space. All the connections will as well satisfy Ax = e ρL1 +O(1) , Ax = e −ρL−1 +O(1) , (3.49) which means that the backgrounds are asymptotically AdS3 in accordance with e.g. [19, 20]. This guarantees that all backgrounds have a well defined SL(2,R) × SL(2,R) limit. Our curve satisfies boundary conditions ρ(yf ) = ρ(yi) ≡ ρ0 , x(yf )− x(yi) ≡ ∆x , t(y) = constant , (3.50) and we take ρ0 → ∞. From (3.46) and (3.48) we have M = e−2L0ρ0e∆axe2L0ρ0e−∆ax , (3.51) where ∆ax is the integral of ax with boundary conditions (3.50), and an analogous definition for ∆ax. It is rather clear here that divergent piece as ρ0 → ∞ is governed by L0. If we solve for (3.36) while enforcing the primary ordering of eigenvalues, we find that the leading order solution in eρ0 is e∆α2 ∼ e4ρ0 , an...