Towards a fully stringy computation of Yukawa couplings on non factorized tori and non abelian twist correlators (I): the classical solution and action

We consider the simplest possible setting of non abelian twist fields which corresponds to $SU(2)$ monodromies. We first review the theory of hypergeometric function and of the solutions of the most general Fuchsian second order equation with three singularities. Then we solve the problem of writing the general solution with prescribed $U(2)$ monodromies. We use this result to compute the classical string solution corresponding to three $D2$ branes in $R^4$. Despite the fact the configuration is supersymmetric the classical string solution is not holomorphic. Using the equation of motion and not the KLT approach we give a very simple expression for the classical action of the string. We find that the classical action is not proportional to the area of the triangle determined by the branes intersection points since the solution is not holomorphic. Phenomenologically this means that the Yukawa couplings for these supersymmetric configurations on non factorized tori are suppressed with respect to the factorized case.


Introduction and conclusions
Since the beginning, D-branes have been very important in the formal development of string theory as well as in attempts to apply string theory to particle phenomenology and cosmology. However, the requirement of chirality in any physically realistic model leads to a somewhat restricted number of possible D-brane set-ups. An important class of models are intersecting brane models where chiral fermions can arise at the intersection of two branes at angles. Most of these computable models are based on D6 branes at angles in T 6 or its orbifolds.
To ascertain the phenomenological viability of a model the computation of Yukawa couplings and flavor changing neutral currents plays an important role. This kind of computations involves the computations of (excited) twist fields correlators. Besides the previous computations many other computations often involve correlators of twist fields and excited twist fields. It is therefore important and interesting in its own to be able to compute these correlators. The literature concerning orbifolds (see for example [1], [2], [3]) intersecting D-branes on factorized tori (see for example [4]), magnetic branes with commuting magnetic fluxes (see for example [5]) or involving "abelian" twist fields in various applications (see for example [6]) is very vast. These results are mainly based on the so called stress-tensor method [1] and concerns mainly non excited twists even if results for excited twists [7] were obtained. Some of the previous results were also obtained in the infinite charge formalism and boundary state formalism [8]. Within the Reggeon framework (see for example [9], [10]) the generating functions for the three points correlators were also obtained in a somewhat complex way. Finally in [11] and [12] based on previous results [13] and a mixture of the path integral approach with the reggeon approach the generating function of all the correlators with an arbitrary number of (excited) twist fields and usual vertices was given in the case of abelian twist fields. These computations boil down to the knowledge of the Green function in presence of twist fields and of the correlators of the plain twist fields. In this way the computations were made systematic differently from many previous papers where correlators with excited twisted fields have been computed on a case by case basis without a clear global picture. The same results were then recovered using the canonical quantization approach in [14].
Until now only the case of factorized tori has been considered at the stringy level. It is clear that the non factorized case is more generic and technically by far more complex. It concerns the so-called non abelian twists for which only a handful papers can be found in the literature of the last 30 years [15]. It is therefore interesting to try to understand how special the results from the factorized case are and to try to clarify the technical issues involved.
In this very technical paper we start the investigation of these configurations. We start considering the case of three D6 branes embedded in R 10 . The relevant configuration can be effectively described by three euclidean E2 branes in C 2 = R 4 . We can think of embedding the first E2 brane as Z 1 = Z 2 = 0. Then the second and third E2 branes are generically characterized by a SO(4) matrix (or more precisely by an equivalence class, i.e a point in the Grassmannian SO(4)/SO(2) × SO (2)) which describes how they are embedded with respect to the first one. However we limit our analysis to the simplest case where these matrices are characterized by an equivalence class of SU (2) . If these two matrices commute then we are in the abelian case if not we deal with the by far more difficult non abelian case. Even if we do not consider the most general case it is however interesting enough to start grasping the issues involved. Moreover this configuration is supersymmetric since there are spinors invariant under the other SU (2) of the "internal" rotation SO(4) ≡ SU (2) × SU (2).
Due to the technicality of the computations involved we have preferred to write down the details therefore the paper has grown in dimension making necessary to split it into different parts. In this part we recapitulate the mathematical tools necessary and we find the classical solution of the bosonic string. In a companion paper we deal with the Green function which is necessary to compute the correlators involving excited twist fields. We are nevertheless still not very close to determine from first principles the normalization of the three twist field correlator as it happens also for all the other papers on the subject [15]. The reason being the impossibility of writing explicitly the classical solution of the string with four branes with a non abelian configuration.
We can nevertheless draw some interesting conclusions. In particular using the path integral approach theN B point twist field correlator can be written roughly as where M t with 1 ≤ t ≤N B are the monodromies. Therefore the knowledge of the classical solution gives the main contribution e −S E,cl ({xt, Mt} 1≤t≤N B ) even if the quantum contribution N ({x t , M t } 1≤t≤N B ) is necessary for the complete result. It then follows that given three D6 t (1 ≤ t ≤ 3) branes the leading order of the Yukawa coupling in a truely stringy computation is given by Naively one could think that S E,cl is simply the area of the triangle determined by the three interaction points but it is not so. These interaction points always define a 2 dimensional real plane in R 4 but differently from the cases discussed before in the literature the embedding of the string worldsheet which follows from the equation of motion is not a flat triangle, i.e. a triangle which lies in the plane determined by the three interactions points. In fact figure 10 shows the actual line traced by the endpoint of the classical string while the naive path should be a segment. This implies that Yukawa couplings in non factorized models are supressed with respect to the factorized ones. The reason is that the classical string solution is not holomorphic (this must not be confused with the fact that the branes emdeddings are holomorphic in the proper set of coordinates). The paper is organized as follows. In section 2 we recapitulate the classical mathematics needed for the computation. In particular we consider the monodromies associated with the general solution of the second order Fuchsian equation with three singular points located at 0, 1 and ∞. Given the relation between the parameters of the equation and the monodromies we solve the inverse problem, i.e. given the monodromies in U (2) find the properly normalized combination of solutions which has the desired monodromy set. This solution is obviously expressed using the hypergeometric function as in eq.s (25,39,41). Another not so commonly appreciated result of the discussion is that monodromies depend on whether the base point is the upper or lower half plane.
In section 3 we consider the string action and the boundary conditions we have to impose. The boundary conditions are better expressed as a local problem for the monodromies on the double string coordinates and a global problem.
In section 4 we then proceed to find the actual classical solution. The upshot of this is more general than the three D-branes case. It turns out that for the U (2) case the bulk of most of the information is contained in the local behavior of the solution, the indeces of the Fuchsian equation, which are determined by the modulus of the vector n which parametrizes the SU (2) monodromy as M = exp (i2π n · σ). The normalization of the solution depends also on the other parameters. Nevertheless the indeces are not sufficient to completely fix the solution but in the simplest case we consider since there are the accessory parameters.
Finally, in section 5 we compute the classical action corresponding to the the solution found. We do this in a more general way which allows us express the action as a linear function of some coefficients opposed to the usual way of getting an expression quadratic. Moreover we clearly show that in the holomorphic case the action has a geometrical meaning.

Monodromies of the hypergeometric function
To solve the problem of finding the string classical solution, we are interested in finding complex functions with a given set of singular points and monodromies. A good starting point to construct these functions is to consider the Fuchsian linear differential equations (which are reviewed in appendix A) since the solutions come naturally in vectors with given monodromies.
Since we are interested in the case with three singular points and U (2) monodromies we would now like to summarize some basics facts on the hypergeometric function which we need in the following.
Our main interest is the derivation of monodromies. In particular we discuss one point which seems to be overlooked or implicit in the literature, i.e. the monodromies do depend on the point we start the loop 1 . It is in fact well known that the homotopy group is defined starting from a base point and that all of these groups are isomorphic. This does not however mean that their representations in the vector space of the solutions of the hypergeometric equation are equal. And actually they are not. In appendix B we show this point in a local setup and at the end of section 3.2 we explicitly show how this fact is needed to demonstrate that the string action is well defined when it is written using the string coordinates obtained by the doubling trick.

Paths
We consider the loops 2 γ (+) [0] , γ (+) [1] , γ [∞] having a base point in the upper half plane H + ≡ H 3 and looping in counterclockwise direction around the marked points z 0 = 0, 1, ∞ respectively as shown in figure 1. We consider also γ (+) [0] γ (+) [1] γ (+) [∞] z 0 More explicitly we define We denote γ [a] * γ [b] the loop formed by first going around γ [a] and then γ [b] , i.e e have then that 1 Historically we notice that even the first paper on the monodromies for the hypergeometric function by Riemann in 1857 [17] seems not to consider the two cases. 2 Here and in the following we denote the singular point by a subscript in square parenthesis, e.g. γ [1] this is to avoid confusion with the index associated with the brane. Indices associated with the branes are put in round parenthesis, e.g. f (1) . Moreover we use {1} as subscript to denote well adapted objects for the singular point z = 1, see for example eq. (14). 3 We define H = {z ∈ C; z ≥ 0} and z 0 Figure 2: The three different paths around the marked points 0, 1, ∞ starting in the lower half plane.
These equations can be generalized to N points with coordinates z i such that The first equation in eq. (3) is shown in figure 3 and and equivalent version of the second one, i.e. γ (−) [0] * γ (+) [1] The reason why we find two different products in the upper and lower half plane is simple. Not all the paths γ (+) do transform into the γ (−) ones when we move the base point from the upper half plane to the lower one. Moreover there are three different ways of moving a point from the upper half plane to the lower one. These ways are characterized by the two marked points between which we move the base point. There are therefore three different possibilities.
In figure 5 we show what happens when we move the base point from the upper half plane to the lower half plane between 0 and 1. Both γ (+) [0] and γ (+) [1] are transformed into the corresponding paths γ (−) [0] and γ (−) [1] [1] as it is shown in figure 6 or equivalently to γ when we consider the sphere and we move the path on the sphere. Both these expressions are compatible with eq.s (3).  [1] .
If we move the base point between 0 and ∞ then γ [0] when we move counterclockwise around the cuts as shown in figure 7 and γ (−) when we move clockwise. The two expressions are nevertheless equal because of the second equation in (3).
Finally in figure 8 we show what happens when we move the base point between 1 and ∞. z 0 z 0 . z 0 z 0 .

Hypergeometric equation and its solutions
As discussed before in order to find a basis of solutions with U (2) monodromies we start considering the most general Fuchsian differential equation of order n = 2 with N = 3 singularities. Specializing what described in app.
A we can write it as where ρ i a (i = 1, 2, N = 3, a = 1, n = 2) are called the indices and give the possible behaviors of the solutions around the singular points, i.e generically we have a mixture as Its general solution can be formally written as by using the Papperitz-Riemann P -symbol .
Using the last property we can limit ourselves to consider the P -symbol which is associated with the hypergeometric equation which has singular points z = 0, 1 and z = ∞ where the respective indices are 0, 1 − c, 0, c − a − b and a, b. This equation has the obvious perturbative solution around the z = 0 singular point given by 4 4 While we use the same notation used by NIST the normalization differs. The reason of this choice is to have simpler monodromy matrices as eq. (22) shows.
Hence F a b c ; z is among the solutions represented by the P -symbol (10). Notice also that F a b c ; z is a function defined on the whole complex plane minus the cut and not only for |z| < 1 where the series converges. The other independent solution around z = 0 can be found using the P -symbol properties which yield Now in order to derive the monodromies matrices we need to understand how the natural basis of the solutions at z = 0 to the hypergeometric equation behaves away from the singular point z = 0 and around the other singular points. In the fundamental sheet this basis (in which it has a diagonal monodromy matrix and has a simple power expansion around {0}) is given by Again these are the functions defined over all the complex plane minus cuts in the fundamental sheet. The cuts arise from both from the F a b c ; z and from the (−z) 1−c . While the cut from F a b c ; z is naturally (but non compulsory) between 1 and +∞ the cut from (−z) 1−c can be set either from 0 to −∞ or from 0 to +∞. The former situation is the one depicted in figures 5 and 6 while the latter is the one in figure 7. Our choice is to set both cuts along the real positive axis, i.e. what it is depicted in figure 7.
Using again the properties of the P-symbol we realize that the basis of solutions in 1/z (which has a diagonal monodromy matrix and have a simple power expansion around {∞}) is given by The starting point to connect the two basis is the Barnes integral repre-sentation of a solution of the hypergeometric equation given by where a, b, c ∈ C and Γ is the path from −i∞ to +i∞ which has all the s = −a − 1 − n, s = −b − 1 − n with n ∈ N poles to the left and s = n ones to the right. Notice that the integrand up to the factor (−z) s is a function with only isolated singularities at finite while the F a b c ; z has both isolated singularities and one cut which originates from the logarithm in (−z) s = exp(log(−z)s). The cut starts apparently at z = 0 but the expansion (12) shows at it actually starts at z = 1.
Notice also that F a b c ; z is a function defined on the whole complex plane minus the cut and not only for |z| < 1.
The presence of the logarithmic cut means that we must specify the value of the logarithm in order to compute the integral and establish its existence. Moreover for seeing in a clear way the connection of the previous function with the usual hypergeometric 2 F 1 and therefore the absence of the cut between z = 0 and z = 1 we can expand it for |z| < 1. To do so we close the path Γ on the left but this can be done only if −π < arg(−z) < π because the modulus of the integrand behaves as |z| Re(s) e Im(s) [−π sign(arg(s)−arg(−z)] for large |s|. When we close on the left we find the original perturbative solution (12) In a similar way for |z| > 1 and still with −π < arg(−z) < π we can close the Γ path on the right and get for where the second equality is obtained using the perturbative definition of Once we have decided where the cuts are we can compute the series expansion of the basis of solutions (14) around z = 0 as in eq. (12) or around z = ∞ as in eq. (18) by expanding the Barnes integral for |z| > 1 and closing the path to the left.
We can then relate the basis of solutions (14) in z with the basis of solutions in 1/z as (20) It is then immediate to compute the monodromies for the basis of solutions in z (14) as where the second expression comes from the obvious monodromy at z = ∞ for the basis of solutions in 1/z (15) along with eq. (20). These monodromy matrices are the same whether we start in the upper of lower half-plane because of our choice of cuts. This is not however true for the monodromy matrices M (±) {0} [1] around z = 1 which do generically 5 depend on the base point. From the fact that monodromy matrices are a representation of the homotopy group (3) they can be derived as A naive way of understanding why this happens is to look at figure 7 and realize that the neighborhood of z = 1 is cut into two disconnected pieces by the cuts.
Obviously the same relations hold also for the matrices M {∞} which are obtained starting from the good basis at z = ∞ (20) since the two sets of matrices are connected by a conjugation by the C matrix.

U (2) monodromies: constraints on the Papperitz equation
As sketched in the introduction and better explained in the following sections we want to use the doublet of solutions as the key element to build solutions to the string e.o.m with monodromies in U (2). Therefore we are interested in finding a doublet of functions whose monodromies belong to U (2). As it is clear from eq.s (21) does not generically belong to U (2). Therefore the doublet of solutions of the hypergeometric equation (11) given by the basis (14) is not what we are looking for but it is a close relative.
To fix the problem around z = 0 and z = ∞ we can consider which is a solution of a Fuchsian equation with three singular points at z = 0, 1 and ∞ and where we have allowed for arbitrary complex rescaling Comparing the indices we see therefore that our solution is within the general solution represented by the Papperitz-Riemann symbol as Notice that given the previous symbol we can easily find two independent solutions but generically these solutions will not generate the desired monodromies. In order to get the desired we need to normalize and recombine the solutions to get the solution in eq. (25) back. We parametrize a U (2) matrix as where n = (n 1 , n 2 , n 3 ) and n = | n|. In appendix C we report some useful formula such as the effect on the parameters N, n given by the product of two elements or the opposite of an element. We are then interested in the relation among the parameters a, b, c, d, f , d {0}1 and d {0}2 and the parameters N [0] , n [0] and N [∞] , n [∞] which parametrize the U (2) monodromy matrices by a rescaling) around z = ∞ for the basis around z = 0 given by E [0] (z) in eq. (25).
As discussed in the appendix D the monodromy around z = 0 can only be in the maximal torus of U (2) and we get U (2) monodromies when the parameters are real a, b, c, d, f ∈ R. (30) Moreover they must satisfy the constraint which is necessary in order to be able to find a value for . In fact the ratio of the moduli of parameters d {0}1 and d {0}2 is fixed in order to have U (2) monodromies as Their relative phase e i2πδ [0] defined by is arbitrary. Nevertheless it fixes part of the information on the versor associated to n [∞] as it enters the last of eq.s (38). It is then immediate to see that the monodromy around z = 0 has U (2) parameters where the integers k N [0] ∈ Z and k n [0] ∈ Z are uniquely fixed by the range in which N [0] , n [0]3 can vary and by Similarly from the trace of the monodromy around z = ∞ we find that U (2) has parameters where the integers k N [∞] ∈ Z and k n [∞] ∈ Z and the "sign" s n[∞] ∈ {0, 1} are again uniquely fixed by the range in which N [∞] , n [∞] can vary and by Moreover we have also

From U (2) monodromies to parameters of Papperitz equation
The previous equations can be also inverted. In this way we can find the parameters a, b, c, d, f and d {0}2 /d {0}1 given the U (2) monodromies at z = 0 and z = ∞. The former must be in the maximal torus of U (2) generated by the Cartan subalgebra and is characterized by N [0] and n 3 [0] . The latter is a generic U (2) matrix and is fixed by giving N [∞] and n [∞] .
With a simple algebra we find for ( where all ks are arbitrary integers, (−) s A ∈ {±1} and the quantity A with 0 ≤ A < 1/2 is defined as 7 Because of the relation eq. (24) among the monodromies we can expect that A is connected with n [1] . Using eq. (150) we can establish in a more precise way this relation.
is in the proper definition range the quantity A is actually n [1] the modulus of the vector n [1] which parametrizes the SU (2) part of the upper half plane monodromy matrix M (+) [0] [1] in z = 1 as defined in eq. (24). Moreover N [1] , the quantity A is actually 1 2 −n [1] and N [1] [1] and N [1] In order to fix almost completely the solution we need also We then fix completely our definition of the solution by choosing Using the explicit values of the parameters it is possible to verify that and therefore that the previous expression for is always meaningful. At first sight the presence of the sign ambiguity (−) s A ∈ {±1} would hint to the existence of two different families of solutions where each member of any class is labeled by the integers ks. It is not so. This can be seen using Euler relation Denoting byã,b, . . . the quantities for (−) s A = −1 it is easy to verify that The analogous relation for the second component of E is then also satisfied. This equivalence can be seen in a less precise way by comparing the Papperitz-Riemann symbols associated to the two solutions. Using eq. (27) we write for the (−) The two symbols coincide again when we use the previous identifications therefore the two families are actually the same.
As we have discussed above there are actually two possible monodromies at z = 1 (24) depending on whether our base point is in the upper or lower half plane, nevertheless since the SU (2) in z = 0 is in the maximal torus, i.e. n [0] = n [0]3 k there is no difference in the modulus of n [1] . Actually also n [1]3 is invariant and only n [1]1 and n [1]2 are different.

The complete abelian solution cannot be recovered
It is then interesting to take the abelian limit of the U (2) monodromies, i.e. consider the case where all monodromies are in the maximal torus U (1) 2 . This in order to make contact with the previous papers dealing with the factorized cases. Naively this seems to be possible since we have monodromies in U (2) but this is not actually the case. One intuitive reason is that we are not considering the general monodromies in R 4 . We can nevertheless obviously obtain a U (1) monodromy at the price of setting to zero one of the two solutions of the U (2) case.
More technically the reason why we cannot obtain solutions with U (1) 2 ⊂ U (2) monodromy is the following. Suppose we want to find a second order equation which has as solutions the following two abelian solutions then the indices would be It follows immediately that the sum of the indices is zero and therefore there cannot exist a Fuchsian second order equation with these solutions since for any Fuchsian second order equation with three singularities the sum of indices is one. This can also be confirmed directly computing the associated second order differential equation for w as and checking that the leading behavior for z → ∞ for the coefficient of w is O 1 z 2 while it should be O 1 z 4 in order to be Fuchsian. In any case it is interesting to consider how far we can go trying to recover the abelian solutions. We can actually recover one of the two abelian solutions while the other is set to zero because of the scaling coefficients d {0}1 and d {0}2 .
Explicitly the abelian case corresponds to n [0] = n [0]3 k and n [∞] = n [∞]3 k. Then from eq. (40) we read immediately that In any of these cases one of the quantities a, b, a − c and b − c is an integer and therefore the ratio is either zero or infinity. This means that either d {0}1 or d {0}2 is zero and hence that one of the components of the vector of the basis solutions E [0] (z) is zero. Another way of explaining this it is to notice that the monodromy matrix at z = ∞ cannot ever become diagonal but at most triangular.

String action and branes configuration
Our aim is to describe the configuration of three D2 branes in R 4 with global monodromies in U (2) ⊂ SO(4). Here we discuss the simplest possible setting.
The part of our interest of the Euclidean action for the string in conformal gauge is given by where u, v, · · · ∈ H belong to the upper half plane H ( u ≥ 0), I = 1, . . . 2N (for the case of our interest N = 2) are the labels of flat coordinates and The complex string coordinate is a map from the upper half plane to a real surface with boundaries in R 4 .
In the cases considered previously the map was from the upper half plane to a polygon Σ in C ≡ R 2 , i.e. X : H → Σ ⊂ C. For example in fig. 9 we have pictured the interaction ofN B = 4 branes at angles D (t) with t = 1, . . .N B . The interaction between brane D (t) and D (t+1) is in f (t) ∈ C. We use the conventions that index t is defined moduloN B and that In the case we consider we have only three branes and therefore only three interaction points. These interaction points always define a 2 dimensional real plane in R 4 but differently from the cases discussed before in the literature the embedding of the string worldsheet which follows from the equation of motion is not a flat triangle, i.e. a triangle which lies in the plane determined by the three interactions points. In fact figure 10 shows the actual line traced by the endpoint of the classical string. This line should be compared with the naive path which is a segment.
Using the notation described in the next subsection we can describe more precisely the setup. The endpoint is shown in good local coordinates for Figure 9: Map from the upper half plane to the target polygon Σ with untwisted in and out strings. The map X(u,ū) folds the boundary of the upper half plane starting from x = −∞ in a counterclockwise direction and preserves the orientation.

Local and global branes configuration
Any brane D (t) t = 1, . . .N B can be described in locally adapted real coordinates X I (Dt) as X where N (t) runs over the normal directions. For a D2 in C 2 ≡ R 4 we have I = 1, 2, 3, 4 and the normal directions can be taken to be in two classes either N (t) = 3, 4 (more generally The former leads to an embedding in locally adapted complex coordinates as Z 1 (Dt) = 0 while the former leads to We want to make contact with what usually done for D1 embedded into C where the embedding is described Z 1 (Dt) = 0 therefore we use this former embedding given in eq.s (53) even if it is apparently less elegant. Because of this choice the tangent directions index T runs over the complementary coordinates which in the D2 in R 4 case are T = 3, 4.
The locally adapted complex coordinates are connected to the global complex coordinates used in defining the string action by a roto-translation as where the rotation is restricted to U (N ) ⊂ O(N ). This is shown in figure  11 in the case of R 2 where we have set U (t) = e −iπαt (0 ≤ α t < 1) to make contact with the notation used in previous papers [?]. Figure 11: The relation between local and global coordinates in the simplest case of R 2 . The bold line is the brane D t with local coordinates Z (Dt) = (X (Dt) +iY (Dt) )/ √ 2. It has a distance | √ 2 g z (t) | from the origin.
Our choice is dictated by the fact that we want to find the simplest configurations which lead to U (2) monodromies of our interest .
This means that the local embedding conditions for D (t) can be written in global coordinates as  We conclude therefore that our configuration withN B branes requirê N B N (N + 3)/2 real dof.s to be specified. This number will be the same when we count it in a different way the next section.

String boundary conditions
It is immediate to write down the eom associated with the action (50) as along with their general solution as with u ∈ H. On this solution we must impose the boundary conditions. In local real adapted coordinates they read where N (t) runs over the normal directions and P (t) runs over the parallel directions to the brane D t . The same conditions can be expressed using the well adapted complex coordinates as where we have supposed that the string boundary lies on D (t) when x t < x < x t−1 and that the normal and tangent directions in local coordinates are always labeled by the same indexes. These boundary conditions imply (but are not equivalent because of the necessity of taking a derivative) where the reflection matrix R (Dt)(Z)(t) in local adapted complex coordinates Z (Dt) is given by and is idempotent R 2 (Dt)(Z)(t) = I 2N . In global coordinates the previous equations become Then the global boundary conditions are equivalent to the following conditions where R (Z)(t) is an idempotent matrix, i.e. R 2 (Z)(t) = I 2N and f (t) is the intersection point between D (t) and D (t+1) . The matrix R (Z)(t) is idempotent since it is conjugated to R (Dt)(Z)(t) . Explicitly R (Z)(t) is given by The intersection point 8 between D (t) and D (t+1) f (t) is given by The matrices R (Z)(t) are somewhat trivial since they are conjugate to a reflection matrix and therefore we have Nevertheless we need to specifyN B N (N + 3)/2 real dof.s in order to fix the boundary conditions completely. N (N + 1)/2 of these dof.s come from the parameters of the unitary symmetric matrices U (t) (t = 1 . . .N B ) as it can be easily seen by writing U = exp(iH). The otherN B N real dof.s come from the complex vectors f (t) (t = 1 . . .N B ). These are subject to the following constraintsN where the first one is actually a consequence of the second set of constraints.
Actually the second set of constraints is simply stating the following geometrical fact: . Therefore for our computation of the dof.s we need to consider the second set of constraints only which halves the real dof.s in the set {f (t) } from 2N B N real dof.s toN B N . The meaning of these constraints is roughly to say that we need the "rotation" matrices U (t) (t = 1 . . .N B ), one corner f (t) for a fixed t and the "lengths" ofN B − 2 sides to describe the string configuration.

String boundary conditions for double fields
Non trivial rotation matrices arise when we use the doubling trick. In particular we can glue the upper and lower half planes along the segment (xt, xt −1 ) which corresponds to thet brane as , withŨ (t) = U * (t) and whereH is the interior of the upper half plane. Then the boundary conditions become the discontinuities and the boundary values Notice that the two fields are independent and not connected by a complex conjugation even if their monodromies are complex conjugate. In fact we find withŨ (t) = U * (t) . As discussed in appendix B where we pay attention to the ordering of the generically non commuting matrices U −1 (u) U (t) the previous discontinuities can be rewritten as monodromies as ( ∈ R, 0 < , min( if we start in the upper half plane and if we start in the lower half plane. The previous monodromy matrices are not completely arbitrary U (N ) matrices since for exampleM T (t,t) = U (t+1)M T (t,t) U * (t+1) . Moreover they satisfy a constraint which follows from the fact they are a representation of the homotopy group given in eq.s. (4), for example in the case While the existence of two sets of monodromies matrices may seem weird it is necessary to verify that the Euclidean action written using the double fields has not any cut. In the previous expression Z j (t) (w)| w=z means that Z j (t) (w) is evaluated for w →z. Because of this when we go around x t counterclockwise with z ∈ H we havez ∈ H − and we go round clockwise inz hence the first factor in the first addend contributes [U −1 (t+1) U t) ] T and the second factor in the first addend contributes Similarly for the tilded fields.

The classical solution
We are now ready to explicitly compute the classical solution. We try to be as general as possible as far as possible but then we apply the general procedure to the case of interest, i.e. the SU (2) monodromies.

Summary of the previous steps
Let us summarize the what done up to now and see how we proceed further on.
• We start with the embedding data given by U (t) and g (t) as follows from the embedding of the D2 (t) is given in eq. (55).
• Using these data we can compute the intersection points f (t) between D (t) and D (t+1) as in eq. (67) and the symmetric matrices U (t) = U T (t) U (t) . • We choose a brane D (t) which we use for the doubling trick as in eq.s (69).
• We can compute the monodromies matrices M (t) = U −1 (t+1) U (t) as in eq.s (73). These data are nevertheless independent of the way we perform the doubling trick and are all what it is needed to compute the derivatives ∂Z and ∂Z. However these matrices do depend on the order of the interaction points x t and these can be changed by changing g (t) as it is easy to see in the simplest abelian case withN B = 3.
• The world sheet interaction points are x t (t = 1, . . .N B ) where the twists are inserted. They are singular points which need to be remapped to the singular points of any possible solution 9 . Moreover the brane D (t) for which xt < x < xt −1 where the doubled solution (69) has no cut must be remapped to the interval where solutions have no cut. In analogy with what happens for the hypergeometric function (and for the fixed singularities of Heun function) we can assume that the three canonical singular are 0, 1 and ∞ and that the interval without cut is the real axes interval (−∞, 0) as discussed after eq. (14). This remapping can then be done with the SL(2, R) transformation In particular we have called ω (t) the complex variable on which any solution depends in order not to confuse it with the original doubled complex variable z. In our caseN B = 3 and the possible solutions are given by any E {0} compatible with all constraints given in eq. (39). If we chooset = 1 then the SL(2, R) transformation maps x 3 , x 2 and x ± 1 to 0, 1 and ∓∞ respectively. The general mapping is

A first naive look
After having summarized what done until now we can proceed further without paying attention to some subtleties which will be addressed in the next subsection.
• In general the monodromy matrix M (t−1) = U −1 (t) U (t−1) at ω (t)t−1 = ω (t) (xt −1 ) = 0 is not diagonal therefore we need a U (N ) (N = 2 in the case at hand) transformation V (t) to diagonalize it since we want to use the previous results where the monodromy matrix U [0] at ω (t) = 0 is diagonal. In particular we want Naively we could think that given whichever solution of the previous equation would work. It is not so as we discuss in the next subsection.
• Given the matrix V (t) we can map the monodromies matrices M (t) = U (t+1,t) into the monodromies matrices of any solution E {0} given in eq.
(25) as Correspondingly the monodromies matricesM (t,t) = U * (t,t+1) for ∂Z (t) are mapped intoŨ We can takeṼ () = V * () so thatŨ [] = U * [] . When we specialize the previous generic discussion to our case with N = 2 we can parametrize the monodromies as Notice that the existence of solutions depends on the possibility of fixing these integers so that we get a finite Euclidean action.
• We can now look for the general solution for the derivative of the classical string e.o.m (which depend only on the monodromy matrices) among the linear combinations , where r labels the possible independent solutions E {0}r which have the required monodromies and finite action as necessary for a classical solution. We have also allowed for a different V (t) r for any possible solution.
Because of this we could in principle believe that also the corresponding monodromies matrices U [] depend on r. As we show in the next subsection for the case at hand it is not the case and that the only dependence of V (t) r on r is through a phase. Similarly for s with the tilded quantities.
• Finally we can determine the classical solution by fixing the constants as and bs from the global conditions. In fact we can write the classical solution as and then impose the further global conditions (71) in order to fix the constants.
We can summarize what we have got until now by saying that twist conditions are local and therefore knowing the modulus of the twists is enough to determine up to integers the indices. The sentence "up to integer" is important since the sum of all indices for the Papperitz-Riemann equation, i.e. the general Fuchsian second order equation with three singularities must add to 1 forN B = 3 twists. In fact we can write the previous statement for the case of monodromies in SU (2),N B = 3 twist fields in a sketchy way as where all indices sum to 0 while they should sum to 1. In a more precise way we can write 10 s∈{±} k (t)s = 1. Now requiring that the action be finite implies that we must set where we have introduced for notation simplicity which has however a different range w.r.t. n (0 ≤ n < 1 2 ). Similarly we havẽ Notice that generically the behavior of the solution around any singular point is given by the sum of the two possible behaviors as for example in eq. (19) which shows that even if we start from a function having a unique index in a singularity we end up with a mixture of indices in other singularities. This means that we must require the all combinations of the two indices at any singular point must give a finite action. For example at ω (t) = ∞ we require 2(n (t) +k (t)+ ) > 2, 2(−n (t) +k (t)− ) > 2 and (n (t) +k (t)+ )+(−n (t) +k (t)− ) > 2.
In particular the last equation means that k (t)+ + k (t)− ≥ 4. As it is obvious form the explicit expressions there is an asymmetry among points and this is disturbing since the P symbol is invariant under SL (2, C) transformations. This asymmetry is however apparent since the P symbol is directly connected to ∂ ω (t) Z and not to ∂ z Z. To see how this solve the asymmetry issue consider one of the simplest cases where we useω (t) = 1/ω (t) as new variable then t) . This restores completely the symmetry among the points. Another point on which is worth noticing and commenting is the appearance of an antiholomorphic part in Z while in the corresponding case with abelian monodromies this does not happen. In fact the previous discussion shows that there is one solution for ∂Z and one for ∂Z. The reason why we get two solutions can be easily understood by noticing that we have more equations to fix the coefficients a r and b s than in the abelian case. In fact according to the discussion after eq.s (68) the configuration is determined bŷ N B N (N + 3)/2 = 15 real parameters and the proposed solution (84) depends on f (t−1) , U (t) and M (t) (t = 1, . . .N B ) for a total of 2N +N B N (N + 1)/2 real parameters. This happens since given U (t) and M (t) we can compute all U (t) and any symmetric unitary U (t) is specified by N (N + 1)/2 real dof.s. Hence we still need (N B − 2)N = 2 real equations to fix the vector the remaining quantities f (t) (t =t − 1), in particular we can simply determine f (t) − f (t−1) . On the other side the reason why we get both a holomorphic and antiholomorphic contribution to Z is less obvious and may be traced back to the fact that minimal area surface is not anymore drawable on a plane despite we have only three interaction points which uniquely fix a plane in R 4 . This happens because the rotations of the D2 branes are not abelian.
This can give the impression that anything can be done easily also for more complex cases. Unfortunately, it is not so. Let us see why.
The same approach can be generalized to theN B = 4 case with SU (2) global symmetry. So we can write using a generalized P-symbol in the casē t = 4 in a sketchy way where a is the location of the fourth singularity whence we have fixed the other three and with The question could be solved by a direct computation if it were not because of two issues. The first one is that the general solution of the general Fuchsian second order equation with four singularities is not uniquely determined by the indices as it happens for the hypergeometric function but it has an accessory parameter q. In fact, as reviewed in appendix A, a Fuchsian equation of order N withN B singular points hasN B N (N −1)/2−N 2 +1 free accessory parameters. The fixing of the accessory parameters is therefore the first issue it is necessary to solve if we want to consider more complex cases than the actual one.
Another issue and more fundamental is that in order to write the actual solution we need to normalize the solutions in order to get the desired monodromies. This is what done in eq. (25) for the caseN B = 3 and N = 2 which discuss in this paper. However as it is clear from the discussion in the present case this normalization does depend on the continuation formulas for the different basis of solutions around the different singular points. Unfortunately this problem is not solved in the general case even in the simplest case of Heun function whose P symbol is given in eq. (92) and which corresponds to the second order Fuchsian equation with four singularities.
At least a couple of possible ways forward can be imagined to try to solve these issues in this case: • consider special M (t) values which correspond to algebraic solutions to the differential equation. This amounts to say that we are actually working a higher genus Riemann surface as done in the paper by Inoue [15]; • try to use CFT factorization.
Another point is worth discussing. In the usual factorized case where we consider only R 2 there are two possible cases forN B = 4 as discussed in [?] and they are labeled by an integer M which in this case can be either 1 or 2. The situation is pictured in figure (12). As long as we limit ourselves to the R 2 case we cannot move from one case to the other by moving the point B without going through the straight line, i.e. the case of no twist. This explains why the two cases are different in R 2 . At first sight this should be not true in R 3 since we can rotate the curve ABC around the AP C axis in a third dimension without going through the straight line. Hence we would expect to have only one case. Actually it is no. The reason is that while rotating the curve ABC in the third dimension in order to deform the left configuration to the right one the minimal area bounded by ABCD starts increasing and a certain point a second configuration bounded by AP CD and ABCP of equal area appears 11 .
The same issues are present also for theN B = 3 case with SU (3) global monodromies where we can expect to write in a sketchy way 11 An easy model to see what is going on is to approximate the minimal area configurations by a sum of triangles. We consider the simplest case with A ≡ (0, 0, 0), P ≡ (B, 0, 0), C ≡ (2B, 0, 0), D ≡ (B, H, 0) and B ≡ (B, h cos θ, h sin θ). This case corresponds to the case where both the triangles ACD and ACB are isosceles. Then the approximated area for the M = 1 case is twice the area of the triangle ABD and is A M =1 = H 2 h 2 + B 2 H 2 + B 2 h 2 − (H 2 h 2 cos 2 θ + 2B 2 cos θ) and the approximated area for the M = 2 case is the sum of the two triangles ACD and ABC and is A M =2 = B(H + h). It is the immediate to see that for the case B 2 < Hh the maximum of where n (i) and m (i) are the independent local SU (3) rotation parameters and again we have one accessory parameter q. Nevertheless we can consider special cases where the monodromies are known ( [18]).

A more detailed look
From the discussion of the previous subsection it seems that any V (t) would do the job since all of them fix the same indices. However it is not so. The reason is that there are further constraints from the action of complex conjugation on ∂Z (t) (z) and ∂Z (t) (z). In fact eq.s (72) (or eq.s (84)) imply for any r and s since E ∼ D −1 B and B is a pair of hypergeometric functions with real parameters. These constraints are fundamental to get a solutions which satisfies the required boundary conditions, i.e. with the string boundaries on the branes. They are however not satisfied by a random solution V (t) r of eq. (79) or the equivalent forṼ (t) s . The problem in implementing them at this stage is that D {0} does depend on n [∞] as in eq.s (41). On the other side n [∞] in turn depends on V (t) because n [∞] is the parameter associated with the conjugation of U (t+1,t) by V (t) as in the last of eq.s (80). Therefore we must solve eq.s (94) and (80) together.
Let us now solve the previous constraints. This is done by comparing the previous equations (94) from the behavior of the doubled solution under complex complex conjugation with the definition U (t) = U T (t) U (t) as given in eq. (66). This comparison suggests to choose the "gauge" 12 so that we have then we can make the ansatz moreover R (t) ∈ U (2) since both U and V are in U (2). This implies R (t) = U (0, r (t) 2 j) 14 . Then we get for any unknown V (t) r We are now left with the problem of computing R (t) and U [∞] 15 . These are the only unknowns since, as noticed before, U [0] is completely determined and independent on r. This happens because it is in the Cartan, i.e. n [0] = n [0]3 k and M (t−1) and U [0] are conjugate, therefore if we write as before M (t−1) = U (0, n (t−1) ) we can always set n [0]3 = n (t−1) > 0.
Because of the same reason U [∞]r is partially determined and we must only fix the ratio n [∞]1 /n [∞]3 . We have therefore two unknowns r (t) 2 and n [∞]1 /n [∞]3 . They can be fixed using the first and second equation in the group of eq.s (80). The first one can be rewritten as Since 16 both the rhs and the lhs are symmetric matrices we can write U * (t) U (t−1) U † (t) = U (0,m (t−1) 1 i +m (t−1) 3 k) (withm (t−1) = n (t−1) ) and then get easily 12 Notice that we can always choose the gauge (95) by a "rotation" in the Cartan group of SU (2) since this does not change U [0] . 13 In principle R (t) should be written as R (t) r but as we now show it is actually independent on r.
14 The parametrization given in the main text is such that R (t) ∈ SU (2) but we could for example have R (t) = U (0, r (t) 2 j)σ 1 ∈ U (2). We can exploit this fact to choose turns out to be independent on r. 16 Since U [0] is diagonal we can easily compute U 1 2 [0] and then naively get R (t) = ±(U (t−1) U † (t) ) T U 1 2 [0] . This is however wrong since this would-be solution does not generically satisfy R T (t) R (t) = I.
Since − 1 2 ≤ r (t) 2 < 1 2 the matrix R (t) is completely determined up to a sign. This explicit solution shows that R (t) is independent on r.
The second equation of eq.s (80) can be used to fix completely U [∞] since it can be rewritten as All the previous eq.s can be solved since they are consistent with the various properties of the involved matrices, i.e.
. In particular this last property is valid only because of the choice of "gauge" (95).
Let us now consider what happens to the tilded quantities associated with ∂Z. All the previous equations remain unchanged with the substitution of the untilded quantities with the tilded ones. In particular the given quantities M (t) and U (t) are connected with the tilded ones by complex conjugation. This means thatṼ is the same of V * up to a phase. It also follows thatR = ±R since R is a real matrix. Finally,Ũ [] are the complex conjugate of the corresponding U [] . The important consequence is then that ñ () are the parameters associated with the complex conjugate of U [] , i.e ñ = (−n 1 , +n 2 , −n 3 ). However this does not mean that there exists oneẼ {0}s for each E {0}r since the determination of the possible solutions requires fixing all integers ks andks. In fact we can determine the parameters ks andks of the solutions E {0}r and E {0}s by requiring a finite classical action.
Finally, the fact that the solution for R is unique up to a sign has as a consequence for the way we write the conditions which can be used to fix the coefficients a and b. In fact we write the global boundary conditions (71) as Now where we have defined the coefficients In the previous expression (102) we can take a, b ∈ R and we are not obliged to consider the previous expression with a → |a| and b → |b| as it would follow from the direct application of eq. (98) where the phase would cancel the corresponding phase of a (and similarly for b).
Finally notice that the previous equation is simply asserting that there exists well adapted coordinates where the computations are more straightforward. In facts when we have decided which branes D (t) to use for the doubling trick and then we have mapped in the proper way the original worldsheet coordinate u into ω (t) as in eq. (77) then the space coordinates are good local coordinates at the interaction point xt −1 , i.e. coordinates for which the monodromy at ω (t)t−1 = 0 and are specially well suited to perform the explicit computations.

The classical solution with SU (2) monodromy
We are now ready to compute the classical solution in the simplest of all non abelian cases, i.e. when the global monodromy group is SU (2). We leave for future publications more complex cases. As we have discussed before in section 4.2 there is only one solution of the hypergeometric which is needed for and only one solution of the hypergeometric for ∂Z (t) (z) For determining the values of the constants a, b, c, d and f we have to make use of the discussion done in section 2.4. Would we not we could make many different associations between the previous P symbols and the P symbol in eq. (27) because of the permutation property of the P symbol (8). Moreover we would miss the proper normalizations. Now comparing the previous P symbols with the P symbol in eq. (27) and using eq.s (39) we get the following values since we can identify For the theẼ solution we have to remember thatñ (t−1)3 = −n (t−1)3 and n (t)1 = −n (t)1 which stems fromM = M * . Therefore we get We are now ready to write the E {0} andẼ {0} as with 17 N = −sign(n (t)1 ) sin[π(n (t−1) + n (t) − n (t+1) )] sin[π(n (t−1) + n (t) + n (t+1) )] sin[π(−(n (t−1) − n (t) ) + n (t+1) )] sin[π((n (t−1) − n (t) ) + n (t+1) )] .
Because ot this we reabsorb N in the definition of E {0}r=1 (ω) and from now on we useẼ Then we can write We are left with the task of fixing the two real coefficients a r=1 and b s=1 . This is done using any t in the equation (102). All of them are equivalent and must be consistent because the counting of the dof.s For example we can write One can in principle doubt that this system is solvable since we have two real unknowns a and b while we have to match four real numbers In order to see whether it is consistent and solvable we can however consider the following case where we do not need to distinguish whether we are integrating in the upper or lower half plane since we are working in the principal sheet in a region where the solution has not any cut which are those depicted in figure 7. In this case the rhs is real. Therefore we need to verify that also the lhs is real. This is however true because of eq. (68). The integrals in the previous equation can be expressed with the help of generalized hypergeometric functions as Explicitly we find where in accordance with eq. (103) we have defined

The classical action
The next task is to compute the classical action of the classical configuration we have found. Many of the recent papers use to express the classical action using the KLT formalism, i.e. they express the double integral on the complex plane as a sum of products of two line integrals. In the abelian case win N B = 3 the final answer is then simply the area of a triangle. In view of the fact that we expect the classical action must have something to do with an area we should suspect that an easier approach should be available. In facts it turns out that using the eom in the case of holomorphic solutions, which is unfortunately not our case, we show that the classical action can be easily expressed using the embedding data. We start from the classical action (50) and we use the eom along with the finitness of the action to immediately write 4πα S then we use the existence of different boundary conditions and the boundary conditions (59) along with (56) to write This expression can be written using complex coordinates as

The holomorphic case
It is now immediate to finish the computation when X i is holomorphic (or antiholomorphic) since in this case In the case of R 2 it is not difficult to see that the previous expression is actually the area of the polygon bounding the string. Few observations are needed to show this. The quantity √ 2 | g z (t) | is the distance of the side from the origin and | is the length of the side. Given the equation of the line through the side Y = aX +b the sign of g z (t) is the same of the sign of the product ab. The sign of on the Y axis. Then using the previous rules it is easy to see that each term of the sum computes the (signed) area of a subtriangle of the original polygon and that the sum is the area of the polygon In figure (13) we give a very simple example of how this works and in figure (14) a less trivial one.

The general case
In the case where the solution is neither holomoprhic nor antiholomorphic the computation is more complex and requires the explicit knowledge of the D (1) (+, +, −) O Y X Figure 13: It is shown as the area of the triangle P 1 P 2 P 3 is decomposed into three subtriangles P 1 P 3 O, P 3 P 2 O and P 2 P 1 O. Along each side the three signs are the signs of is the equation of the line through the side.
solution. To simplify the expression (122) we start noticing that eq.s (63) imply that i.e. both U (t) Z L (x) and U (t) Z R (x) are step functions with discontinuities in the set of the interaction points {x t } t=1,...N B . In particular, from eq.s (124) it follows that U i (t) j ∂ x Z j L (x) and U i (t) j ∂ x Z j R (x) are real vectors hence eq. (122) can be written without the real projection as 4πα S The advantage of this expression is that we can directly compare with the global boundary conditions which can be written as to see that once we have solved for the global contition we need not to do more efforts to compute the classical action. The classical action can be written also as 4πα S D (1) (+, +, +) Figure 14: It is shown as the area of the polygon P 1 P 2 P 3 P 4 is decomposed into four subtriangles P 1 P 4 O, P 4 P 3 O and P 3 P 2 O, P 2 P 1 O. The contributions to S classical E of the first two triangles are positive while the ones from the last two is negative thus yielding minus the area of the interior of the polygon. Along each side the three signs are the signs of is the equation of the line through the side.
in such a way to show the deviation from the holomorphic case and that the real projection in eq. (123) is not really necessary. We can also make contact with the more explicit expression in eq. (102) by writing If we suppose that z = z i at finite is a singular point of the meromorphic coefficients and we require to have n solutions at this point with behavior like y ∼ (z − z i ) ρ i we deduce immediately that the previous equation can be written as where R i are regular functions at z = z i . The possible values of ρ i are the indeces at z = z i . If we consider N − 1 singular points at finite we get therefore where Q N −1 (z) = N −1 j=1 (z − z j ) andP i are polynomials. If we now want the infinity to be a regular singual point, i.e. we require to have n solutions with behavior like y ∼ 1 z ρ∞ we deduce that the most general Fuchsian equation with N − 1 regular singular points at finite and one at the infinite is given by (132) where Q N −1 (z) = N −1 j=1 (z − z j ) when and P k (z) are abitrary polynomials of order k. We consider the case where the infinity is a singular point since it is not immediate to write down the conditions for the regularity a infinity.
The previous equation for the special case n = 2 can be written in a more explicit form as with i γ i = 0. It is also easy to prove Fuchs' result according to which the sum of all indeces must be equal to (N − 2)n(n − 1)/2, i.e. n a=1 N i=1 ρ i a + ρ ∞ a = (N − 2)n(n − 1)/2.
(136) Using the residue theorem we get from which the theorem follows. Now a simple parameters counting gives n i=1 [i(N − 2) + 1] parameters in the polynomials P and nN indeces of which only nN − 1 are arbitrary by virtue of Fuchs' result. Therefore we have N n(n − 1)/2 − n 2 + 1 free accessory parameters. Only for n = 1 and n = 2, N = 3 there are no accesory parameters.
The general solution of the Fuchsian differential equation of order n with N singularities can be represented by a generalized P -symbol as where q ∈ C N n(n−1)/2−n 2 +1 . This symbol represents the space of all ∞ n solutions.
Since the equation is defined on P 1 the symbol is invariant under a SL(2, C transformationẑ = (az + b)/(cz + d), i.e.

B From discontinuities to monodromies
In this appendix we would like to show explicitely in a local setup that the monodromies depend on the base point. This discussion is useful for the derivation of the monodromies of the non abelian twists. Suppose we are given a vector of analytic functions f (u) with discontinuities U − U + Figure 15: The array of functions f (z) with their discontinuities.
U −1 + U − Figure 16: The array of functions F + (z) with their discontinuity.
with x ∈ R and x > 0 as shown in figure 15. We want to compute the monodromies associated with f (e i2π (x + i0 + )) and f (e i2π (x − i0 + )). From figure 15 it is clear that when we compute f (e i2π (x + i0 + )) we first cross the U − discontinuity and then the U + discontinuity in the opposite direction with respect to the definition and therefore we have a contribution U −1 + . What it is not obvious is whether the final contribution is U − U −1 + or U −1 + U − . To solve this issue we define a new function F + (z) by the doubling trick as This new function has only one discontinuity as shown in figure 16 It follows then that the monodromy is simply In particular using the definition it follows that This last results shows again that the monodromies depend on the base point and it is in accordance with the fact that to compute f (e i2π (x − i0 + )) we first cross the U −1 + discontinuity in the opposite direction and then the U − discontinuity.
Finally, notice that the same result can be obtained usong a diferent glueing, i.e.

D Details on U (2) monodromies
In this appendix we would like to give the details of the derivation of the relation between the indeces and and the parameters of the U (2) monodromies.
Since the monodromy at z = 1 is fixed once we have the monodromies at z = 0 and z = ∞ we will not deal with it.
Our strategy is first to find the constraints on the solution parameters so that the monodromies matrices are in U (2) and then find the relation between them and the U (2) parameters.

D.1 Constraints from the monodromy at z = 0
If we start from the basis of solutions E {0} (z) in eq. (25) the monodromy at is the monodromy of the Barnes basis at z = 0 (14) times (−z) d (1 − z) f −d around z = 0 and is the matrix used to rescale the two independent solutions of the Barnes basis at z = 0 (14) with P = C † (D {0} D † {0} ) −1 C. Since P must be invertible we have P 11 , P 22 > 0 therefore we get (a) = (b) = (f ) while the constraints involving P 12 , P 21 imply This constraint can be implemented as −4sin(πa * ) sin(πb)sin[π(a * − c)] sin[π(b − c)] ∈ R + .