\subsection{Motivation}

As seen in the previous sections, the study of viable phenomenological models in string theory involves the analysis of the properties of systems of D-branes.
The inclusion of the physical requirements deeply constrains the possible scenarios.
In particular the chiral spectrum of the \sm acts as a strong restriction on the possible setup.
In this section we study models based on \emph{intersecting branes}, which represent a relevant class of such models with interacting chiral matter.
We focus on the development of technical tools for the computation of Yukawa interactions for D-branes at angles~\cite{Chamoun:2004:FermionMassesMixing,Cremades:2003:YukawaCouplingsIntersecting,Cvetic:2010:BranesInstantonsIntersecting,Abel:2007:RealisticYukawaCouplings,Chen:2008:RealisticWorldIntersecting,Chen:2008:RealisticYukawaTextures,Abel:2005:OneloopYukawasIntersecting}.
The fermion--boson couplings and the study of flavour changing neutral currents~\cite{Abel:2003:FlavourChangingNeutral} are keys to the validity of the models.
These and many other computations require the ability to calculate correlation functions of (excited) twist and (excited) spin fields.
The goal of the section is therefore to address such challenges in specific scenarios.

The computation of the correlation functions of Abelian twist fields can be found in literature and plays a role in many scenarios, such as magnetic branes with commuting magnetic fluxes~\cite{Angelantonj:2000:TypeIStringsMagnetised,Bertolini:2006:BraneWorldEffective,Bianchi:2005:OpenStoryMagnetic,Pesando:2010:OpenClosedString,Forste:2018:YukawaCouplingsMagnetized}, strings in gravitational wave background~\cite{Kiritsis:1994:StringPropagationGravitational,DAppollonio:2003:StringInteractionsGravitational,Berkooz:2004:ClosedStringsMisner,DAppollonio:2005:DbranesBCFTHppwave}, bound states of D-branes~\cite{Gava:1997:BoundStatesBranes,Duo:2007:NewTwistField,David:2000:TachyonCondensationD0} and tachyon condensation in Superstring Field Theory~\cite{David:2000:TachyonCondensationD0,David:2001:TachyonCondensationUsing,David:2002:ClosedStringTachyon,Hashimoto:2003:RecombinationIntersectingDbranes}.
A similar analysis can be extended to excited twist fields even though they are more subtle to treat and hide more delicate aspects~\cite{Burwick:1991:GeneralYukawaCouplings,Stieberger:1992:YukawaCouplingsBosonic,Erler:1993:HigherTwistedSector,Anastasopoulos:2011:ClosedstringTwistfieldCorrelators,Anastasopoulos:2012:LightStringyStates,Anastasopoulos:2013:ThreeFourpointCorrelators}.
Results were however found starting from dual models up to modern interpretations of string theory~\cite{Sciuto:1969:GeneralVertexFunction,DellaSelva:1970:SimpleExpressionSciuto}.
Correlation functions involving arbitrary numbers of plain and excited twist fields were more recently studied~\cite{Pesando:2014:CorrelatorsArbitraryUntwisted,Pesando:2012:GreenFunctionsTwist,Pesando:2011:GeneratingFunctionAmplitudes} blending the CFT techniques with the path integral approach and the canonical quantization~\cite{Pesando:2008:MultibranesBoundaryStates,DiVecchia:2007:WrappedMagnetizedBranes,Pesando:2011:StringsArbitraryConstant,DiVecchia:2011:OpenStringsSystem,Pesando:2013:LightConeQuantization}.

We consider D6-branes intersecting at angles in the case of non Abelian relative rotations presenting non Abelian twist fields at the intersections.
We try to understand subtleties and technical issues arising from a scenario which has been studied only in the formulation of non Abelian orbifolds \cite{Inoue:1987:NonAbelianOrbifolds,Inoue:1990:StringInteractionsNonAbelian,Gato:1990:VertexOperatorsNonabelian,Frampton:2001:ClassificationConformalityModels} and for D-branes relative \SU{2} rotations~\cite{Pesando:2016:FullyStringyComputation}.
In this configuration we study three D6-branes in $10$-dimensional Minkowski space $\ccM^{1,9}$ with an internal space of the form $\R^4 \times \R^2$ before the compactification.
The D-branes are embedded as lines in $\R^2$ and as two-dimensional surfaces inside $\R^4$.
We focus on the relative rotations which characterise each D-brane in $\R^4$ with respect to the others.
In total generality, they are non commuting \SO{4} matrices.
We study the classical solution of the bosonic string which dominates the behaviour of the correlator of twist fields.
Using the path integral approach we can in fact separate the classical contribution from the quantum fluctuations and write the correlators of $N_B$ twist fields $\sigma_{\rM_{(t)}}( x_{(t)} )$ as:\footnotemark{}
\footnotetext{%
  Ultimately $N_B = 3$ in our case.
}
\begin{equation}
  \left\langle
    \finiteprod{t}{1}{N_B}
    \sigma_{\rM_{(t)}}(x_{(t)})
  \right\rangle
  =
  \cN
  \qty(
    \qty{ x_{(t)},\, \rM_{(t)} }_{1 \le t \le N_B}
  )\,
  e^{-S_E\qty( \qty{ x_{(t)}, \rM_{(t)} }_{1 \le t \le N_B} )},
\end{equation}
where $\rM_{(t)}$ (for $1 \le t \le N_B$) are the monodromies induced by the twist fields, $N_B$ is the number of D-branes and $x_{(t)}$ are the intersection points on the worldsheet.
Even though quantum corrections are crucial to the complete determination of the normalisation of the correlator, the classical contribution of the Euclidean action represents the leading term of the Yukawa couplings.
We focus on its contribution to better address the differences from the usual factorised case and generalise the results to non Abelian rotations of the D-branes.
We do not consider the quantum corrections as they cannot be computed with the actual techniques.
Their calculations requires the correlator of four twist fields which in turn requires knowledge of the connection formula for Heun functions which is not known.

We therefore study the boundary conditions for the open string describing the D-branes embedded in $\R^4$.
In particular we first address the issue connected to the global description of the embedding of the D-branes.
In conformal coordinates we rephrase such problem into the study of the monodromies acquired by the string coordinates.
These additional phase factors can then be specialised to \SO{4}, which can be studied in spinor representation as a double copy of \SU{2}.
We thus recast the issue of finding the solution as $4$-dimensional real vector to a tensor product of two solutions in the fundamental representation of \SU{2}.
We then see that these solutions are well represented by hypergeometric functions, up to integer factors.
Physical requirements finally restrict the number of possible solutions.


\subsection{D-brane Configuration and Boundary Conditions}

We focus on the bosonic string embedded in $\ccM^{1, d + 4}$.
The relevant configuration of the D-branes is seen as two-dimensional Euclidean planes in $\R^4$.
We specifically concentrate on the Euclidean solution for the classical bosonic string in this scenario.
The mathematical analysis is however more general and can be applied to any Dp-brane embedded in a generic Euclidean space $\R^q$.
The classical solution can in principle be defined in this case provided the ability to write the explicit form of the basis of functions with the proper boundary and monodromy conditions.
This is possible in the case of three intersecting D-branes but in general it is an open mathematical issue.
In the case of three D-branes with generic embedding we can in fact connect a local basis around one intersection point to another, the third depending on the first two intersections, by means of Mellin-Barnes integrals.
This way the solution can be explicitly and globally constructed.
With more than three D-branes the situation is by far more difficult since the explicit form of the connection formulas is not known.


\subsubsection{Intersecting D-branes at Angles}

Let $N_B$ be the total number of D-branes and $t = 1,\, 2,\, \dots,\, N_B$ be
an index defined modulo $N_B$ to label them.
We describe one of the D-branes in a well adapted system of coordinates
$X_{(t)}^I$, where $I = 1,\, 2,\, 3,\, 4$, as:
\begin{equation}
  X_{(t)}^3 = X_{(t)}^4 = 0.
  \label{eq:well-adapt-embed}
\end{equation}
We thus choose $X_{(t)}^1$ and $X_{(t)}^2$ to be the coordinates parallel to the D-brane $D_{(t)}$ while $X_{(t)}^3$ and $X_{(t)}^4$ are the coordinates orthogonal to it.

\begin{figure}[tbp]
  \centering
  \begin{subfigure}[b]{0.45\linewidth}
    \centering
    \def\svgwidth{\linewidth}
    \import{img}{branesangles.pdf_tex}
    \caption{D-branes as lines on $\R^2$.}
  \end{subfigure}
  \hfill
  \begin{subfigure}[b]{0.45\linewidth}
    \centering
    \def\svgwidth{\linewidth}
    \import{img}{welladapted.pdf_tex}
    \caption{Well adapted system of coordinates.}
  \end{subfigure}
  \caption{%
    Geometry of D-branes at angles.
  }
  \label{fig:branes_at_angles}
\end{figure}

The well adapted reference coordinates system is connected to the global $\R^{4}$ coordinates $X^I$ by a transformation:
\begin{equation}
  \tensor{(X_{(t)})}{^I}
  =
  \tensor{(R_{(t)})}{^I_J}\, \tensor{X}{^J}
  -
  \tensor{(g_{(t)})}{^I},
  \qquad
  I,\, J = 1,\, 2,\, 3,\, 4,
  \label{eq:brane_rotation}
\end{equation}
where $R_{(t)}$ represents the rotation of the D-brane $D_{(t)}$ and $g_{(t)} \in \R^4$ its translation with respect to the origin of the global set of coordinates (see \Cref{fig:branes_at_angles} for a two-dimensional example).

While we could naively consider $R_{(t)} \in \SO{4}$, rotating separately the subset of coordinates parallel and orthogonal to the D-brane does not affect the embedding.
In fact it just amounts to a trivial redefinition of the initial well adapted coordinates.
The rotation $R_{(t)}$ is actually defined in the Grassmannian:
\begin{equation}
  R_{(t)}
  \in
  \mathrm{Gr}(2, 4)
  =
  \frac{\SO{4}}{\rS\qty( \OO{2} \times \OO{2} )},
\end{equation}
that is we just need to consider the left coset where $R_{(t)}$ is a representative of an equivalence class
\begin{equation}
  \qty[ R_{(t)} ]
  =
  \qty{ R_{(t)} \sim \cO_{(t)} R_{(t)} },
\end{equation}
where $\cO_{(t)} = \rS\qty( \OO{2} \times \OO{2} )$ is defined as
\begin{equation}
  \cO_{(t)}
  =
  \mqty( \dmat{\cO^{\parallel}_{(t)}, \cO^{\perp}_{(t)}} )
\end{equation}
with $\cO^{\parallel}_{t} \in \OO{2}$, $\cO^{\perp}_{t} \in \OO{2}$ and $\det \cO_{(t)} = 1$.
The superscript $\parallel$ represents any of the coordinates parallel to the D-brane, while $\perp$ any of the orthogonal.
Notice that the additional $\Z_2$ factor in $\rS\qty( \OO{2} \times \OO{2} )$ with respect to $\SO{2} \times \SO{2}$ can be used to set $g_{(t)}^{\perp} \ge 0$.


\subsubsection{Boundary Conditions for Branes at Angles}

The peculiar embedding of the D-branes has natural consequences on the boundary conditions of the open strings.
Let $\tau_E = i \tau$ be the Wick rotated time direction.
We define the usual upper plane coordinates:
\begin{eqnarray}
  u
  =
  x + i y
  =
  e^{\tau_E + i \sigma}
  & \in &
  \ccH \cup \qty{ z \in \C \mid \Im z = 0 },
  \\
  \baru
  =
  x - i y
  =
  e^{\tau_E - i \sigma}
  & \in &
  \bccH \cup \qty{ z \in \C \mid \Im z = 0 },
\end{eqnarray}
where $\ccH = \qty{ z \in \C \mid \Im z > 0 }$ is the upper complex plane and $\bccH = \qty{ z \in \C \mid \Im z < 0 }$ is the lower complex plane.
In conformal coordinates $u$ and $\baru$, D-branes at $\sigma = 0$ and $\sigma = \pi$ are mapped onto the real axis $\Im z = 0$.
We use the symbol $D_{(t)}$ to label both the brane and the interval representing it on the real axis of the upper half plane:
\begin{equation}
  D_{(t)} = \qty[ x_{(t)}, x_{(t-1)} ],
  \qquad
  t = 2,\, 3,\, \dots,\, N_B,
  \qquad
  x_{(t)} < x_{(t-1)}.
\end{equation}
The points $x_{(t)}$ and $x_{(t-1)}$ represent the worldsheet intersection points of the brane $D_{(t)}$ with the branes $D_{(t+1)}$ and $D_{(t-1)}$ respectively.
The choice of the intervals must be carefully considered: since the D-branes are defined modulo $N_B$, the shorthand for the interval $D_{(1)} = \qty[ x_{(1)}, x_{(N_B)} ]$ should actually be:
\begin{equation}
  D_{(1)}
  =
  \left[ x_{(1)}, +\infty \right) \cup \left( -\infty, x_{(N_B)} \right].
\end{equation}

In the global coordinates system associated to the subspace $\R^4 \subset \ccM^{1, d+4}$ where D-branes intersect, the relevant part of the action in conformal gauge is:
\begin{equation}
  \begin{split}
    S_{\R^4}
    & =
    \frac{1}{2 \pi \ap}
    \iint\limits_{\ccH}
    \dd{u} \dd{\baru}\,
    \ipd{u} X^I\, \ipd{\baru} X^J\,
    \eta_{IJ}
    \\
    & =
    \frac{1}{4 \pi \ap}
    \iint\limits_{\R \times \R^+}
    \dd{x}\dd{y}\,
    \qty(
      \ipd{x} X^I\, \ipd{x} X^J
      +
      \ipd{y} X^I\, \ipd{y} X^J
    )\,
    \eta_{IJ},
  \end{split}
  \label{eq:string_action}
\end{equation}
where $2\, \ipd{u} = \ipd{x} - i\, \ipd{y}$ and $2\, \ipd{\baru} = \ipd{x} + i\, \ipd{y}$.
The \eom in these coordinates are:
\begin{equation}
  \ipd{u} \ipd{\baru} X^I( u, \baru )
  =
  \frac{1}{4}
  \qty( \ipd{x}^2 + \ipd{y}^2 ) X^I( x+iy, x-iy )
  =
  0.
  \label{eq:string_equation_of_motion}
\end{equation}
Their solution factorises as usual in holomorphic and anti-holomorphic components $X^I( u, \baru ) = X^I( u ) + \barX^I( \baru )$.

In the well adapted frame~\eqref{eq:well-adapt-embed} we describe an open string with one of the endpoints on $D_{(t)}$ through the relations:
\begin{eqnarray}
  \eval{\ipd{\sigma} X^i_{(t)}( \tau, \sigma )}_{\sigma = 0}
  =
  \eval{\ipd{y} X^i_{(t)}( u, \baru )}_{y = 0}
  & = &
  0,
  \qquad
  i = 1,\, 2,
  \label{eq:neumann_bc}
  \\
  X^m_{(t)}( \tau, 0 )
  =
  X^m_{(t)}( x, x )
  & = &
  0,
  \qquad
  m = 3,\, 4,
  \label{eq:dirichlet_bc}
\end{eqnarray}
where $x \in D_{(t)} = \qty[ x_t, x_{t-1} ]$ and the index $i$ labels the Neumann boundary conditions while $m$ labels the Dirichlet coordinates associated to the direction orthogonal to the D-branes.

As the presence of $g_{(t)}^m$ in \eqref{eq:brane_rotation} and \eqref{eq:dirichlet_bc} may complicate the analysis, we consider the derivative along the boundary direction of \eqref{eq:dirichlet_bc} to remove the dependence on the translation vector.
This procedure produces simpler boundary conditions which are nevertheless not equivalent to the original~\eqref{eq:neumann_bc} and \eqref{eq:dirichlet_bc}: they will be recovered later by adding further constraints.
The simpler boundary conditions we consider in the global coordinates are:
\begin{eqnarray}
  \tensor{\qty( R_{(t)} )}{^i_J}
  \eval{\ipd{\sigma} X^J( \tau, \sigma )}_{\sigma = 0}
  & = &
  i\, \tensor{\qty( R_{(t)} )}{^i_J}
  \qty(
    \ipd{u} X^J( x + i\, 0^+ ) - \ipd{\baru} \barX^J( x - i\, 0^+ )
  )
  =
  0,
  \\
  \tensor{\qty( R_{(t)} )}{^m_J}
  \eval{\ipd{\tau} X^J( \tau, \sigma )}_{\sigma = 0}
  & = &
  i\, \tensor{\qty( R_{(t)} )}{^m_J}
  \qty(
    \ipd{u} X^J( x + i\, 0^+ ) + \ipd{\baru} \barX^J( x - i\, 0^+ )
  )
  =
  0,
\end{eqnarray}
where $i = 1,\, 2$, $m = 3,\, 4$ and $x \in D_{(t)}$.

With the introduction of the target space embedding of the worldsheet interaction point between D-branes $D_{(t)}$ and $D_{(t+1)}$, $f_{(t)}$, we recover the full boundary conditions in terms of discontinuities on the real axis:
\begin{equation}
  \begin{cases}
    \ipd{u} X^I( x + i\, 0^+ )
    & =
    \tensor{\qty( U_{(t)} )}{^I_J}
    \ipd{\baru} \barX^J( x - i\, 0^+ ),
    \qquad
    x \in D_{(t)}
    \\
    X^I( x_{(t)}, x_{(t)} )
    & =
    f_{(t)}
  \end{cases}.
  \label{eq:discontinuity_bc}
\end{equation}
In the last expression we introduced the matrix
\begin{equation}
  U_{(t)}
  =
  \qty( R_{(t)} )^{-1}\,
  \cS\,
  R_{(t)}
  \in
  \frac{\SO{4}}{\rS\qty( \OO{2} \times \OO{2} )},
  \label{eq:Umatrices}
\end{equation}
where
\begin{equation}
  \cS
  =
  \mqty( \dmat{ 1, 1, -1, -1 } )
  \label{eq:reflection_S}
\end{equation}
embeds the difference between Neumann and Dirichlet conditions.
Given its definition $U_{(t)}$ is such that $U_{(t)} = \qty( U_{(t)} )^{-1} = \qty( U_{(t)} )^T$.

The target space vector $f_{(t)}$ recovers the apparent loss of information suffered when losing $g_{(t)}$.
Consider for instance the embedding equations~\eqref{eq:dirichlet_bc} for any two intersecting D-branes $D_{(t)}$ and $D_{(t+1)}$.
Introducing the auxiliary quantities
\begin{eqnarray}
  \cR_{(t,\, t+1)}
  =
  \mqty( R_{(t)}^m \\ R_{(t+1)}^n )
  & \in &
  \GL{4}{\R},
  \qquad
  m, n = 3, 4,
  \\
  \cG_{(t,\, t+1)}
  =
  \mqty( g_{(t)}^m \\ g_{(t+1)}^n )
  & \in &
  \R^4,
  \qquad
  m, n = 3, 4,
\end{eqnarray}
we can compute the intersection point as:
\begin{equation}
  f_{(t)}
  =
  \qty( \cR_{(t,\, t+1)} )^{-1}\,
  \cG_{(t,\, t+1)}.
\end{equation}
Information on $g_{(t)}$ is thus recovered through the global boundary conditions in the second equation in \eqref{eq:discontinuity_bc}.


\subsubsection{Doubling Trick and Branch Cut Structure}

In conformal coordinates we thus introduced the discontinuities~\eqref{eq:discontinuity_bc} across each D-brane which define a non trivial cut structure on the plane.
One way to deal with them is to introduce the \emph{doubling trick} by gluing the relations along an arbitrary but fixed D-brane $D_{(\bart)}$:
\begin{equation}
  \ipd{z} \cX(z) =
  \begin{cases}
    \ipd{u} X(u)
    &
    \qif
    z = u \qand \Im z > 0 \qor z \in D_{(\bart)}
    \\
    U_{(\bart)}\,
    \ipd{\baru} \barX(\baru)
    & \qif z = \baru \qand \Im z < 0 \qor z \in D_{(\bart)}
  \end{cases}.
  \label{eq:real_doubling_trick}
\end{equation}
Let then $\cU_{(t,\, t+1)} = U_{(t+1)}\, U_{(t)}$ and $\tcU_{(t,\, t+1)} = U_{(\bart)}\, U_{(t)}\, U_{(t+1)}\, U_{(\bart)}$.
The boundary conditions in terms of the doubling field are:
\begin{eqnarray}
  \ipd{z} \cX( x_t + e^{2 \pi i}( \eta + i\, 0^+ ) )
  & = &
  \cU_{(t,\, t+1)}
  \ipd{z} \cX( x_t + \eta + i\, 0^+ ),
  \label{eq:top_monodromy}
  \\
  \partial \cX( x_t + e^{2 \pi i}( \eta - i\, 0^+ ) )
  & = &
  \tcU_{(t,\, t+1)}
  \ipd{z} \cX( x_t + \eta - i\, 0^+ ),
  \label{eq:bottom_monodromy}
\end{eqnarray}
for $0 < \eta < \min\qty( \abs{x_{(t-1)} - x_{(t)}}, \abs{x_{(t)} - x_{(t+1)}} )$ in order to consider only the two adjacent D-branes $D_{(t)}$ and $D_{(t+1)}$.
Matrices $\cU_{(t,\, t+1)}$ and $\tcU_{(t,\, t+1)}$ are the non trivial monodromies arising from the rotation of the D-branes.

Since the relative rotations between consecutive D-branes are non Abelian, for each interaction point there are two monodromies \cU and $\tcU$ depending on the location of the base point of the closed loop: one for paths starting in the upper plane \ccH and one for paths starting in $\bccH$.
As a consequence of the geometry of the rotations of the D-branes, a path on the complex plane enclosing all of them does not present a monodromy:
\begin{equation}
  \finiteprod{t}{1}{N_B}\,
  \cU_{(\bart - t, \bart + 1 - t)}
  =
  \finiteprod{t}{1}{N_B}\,
  \tcU_{(\bart + t, \bart + 1 + t)}
  =
  \1_4.
  \label{eq:homotopy_rep}
\end{equation}
The complex plane has therefore branch cuts running between the D-branes at finite as shown in \Cref{fig:finite_cuts}.
We thus translated the rotations of the D-branes encoded in the matrices $R_{(t)}$ in terms of $\cU_{(t,\, t+1)}$ and $\tcU_{(t,\, t+1)}$ which are matrix representations of the homotopy group of the complex plane with the described branch cut structure.

\begin{figure}[tbp]
  \centering
  \def\svgwidth{0.5\textwidth}
  \import{img}{branchcuts.pdf_tex}
  \caption{%
    Branch cut structure of the complex plane with $N_B = 4$.
    Cuts are pictured as solid coloured blocks running from one intersection point to another at finite.
  }
  \label{fig:finite_cuts}
\end{figure}

As a consistency check, the action~\eqref{eq:string_action} can be computed in terms of the doubling field \cX.
The map
\begin{equation}
  x_{(t)} + \eta \pm i\, 0^+
  \quad
  \mapsto
  \quad
  x_{(t)} + e^{2 \pi i}( \eta \pm i\, 0^+)
\end{equation}
must leave the action untouched since it does not depend on the branch cut structure.
In fact we can show that
\begin{equation}
  S_{\R^4}
  =
  \frac{1}{4 \pi \ap}
  \iint\limits_{\C}
  \dd{z} \dd{\barz}\,
  \ipd{z} \cX^T(z)\,
  U_{(\bart)}\,
  \ipd{\barz} \cX(\barz).
\end{equation}
As a matter of fact the action does not depend on the branch structure of the complex plane.


\subsection{D-branes at Angles in Spinor Representation}

In the previous section we showed that it is possible to map the information on the rotations of the D-branes to non trivial monodromies of the doubling field.
We thus recast the issue of solving the \eom of the string in the presence of rotated boundary conditions to the search for an explicit solution $\ipd{z} \cX( z )$ reproducing the non trivial monodromies in~\eqref{eq:top_monodromy} and \eqref{eq:bottom_monodromy}.

The field $\ipd{z} \cX( z )$ is technically a $4$-dimensional \emph{real} vector which has $N_B$ non trivial monodromy factors represented by $4
\times 4$ real matrices, one for each interaction point $x_{(t)}$.
A solution of the \eom is encoded in four linearly independent functions with $N_B$ branch points.
In principle we could try to write it as a solution to fourth order differential equations with $N_B$ finite Fuchsian points.
This is however an open mathematical debate.
In fact the basis of such functions around each branch point are usually complicated and defined up to several free parameters.
Moreover the explicit connection formulae between any two of them is an unsolved mathematical problem.
Using contour integrals and writing the functions as Mellin--Barnes integrals it might be possible to solve the issue in the case $N_B = 3$ but it is certainly not the best course of action.

On the other hand $N_B = 3$ is exactly the case we are investigating.
In what follows we use the isomorphism
\begin{equation}
  \SO{4}
  \cong
  \frac{\SU{2} \times \SU{2}}{\Z_2}
  \label{eq:su2isomorphism}
\end{equation}
to map the problem of finding a $4$-dimensional real solution to the \eom to a quest for a $2 \times 2$ complex matrix.
Such matrix is a linear superposition of tensor products of vectors in the fundamental representation of two different \SU{2} groups.
These vectors are solutions to second order differential equations with three Fuchsian points, possibly the hypergeometric equation.
The task is then to find the parameters of the hypergeometric functions producing the spinor representation of the monodromies in~\eqref{eq:top_monodromy} and \eqref{eq:bottom_monodromy}.


\subsubsection{Doubling Trick and Rotations in Spinor Representation}

We recall some of the properties of the isomorphism~\eqref{eq:su2isomorphism} in~\Cref{sec:isomorphism}.
We define the spinor representation of $X$ as:
\begin{equation}
  X_{(s)}( u, \baru ) = X^I( u, \baru )\, \tau_I,
\end{equation}
where $\tau = \qty( i\, \1_2,\, \vec{\sigma} )$ and $\vec{\sigma}$ is the vector of the Pauli matrices.
Consider then:
\begin{equation}
  \ipd{z} \cX_{(s)}( z )
  =
  \begin{cases}
    \ipd{u} X_{(s)}(u)
    & \qif
    z \in \ccH \qor z \in D_{(\bart)}
    \\
    U_{L}(\vec{n}_{(\bart)})\,
    \ipd{\baru} X_{(s)}(\baru)\,
    U_{R}^{\dagger}(\vec{m}_{(\bart)})
    & \qif z \in \bccH \qor z \in D_{(\bart)}
  \end{cases}.
  \label{eq:spinor_doubling_trick}
\end{equation}

As in the real representation the discontinuities on the D-branes can be cast into monodromy factors with respect to the D-brane $D_{(\bart)}$. Branch cut structure and considerations on the homotopy group are left unchanged as long as we consider both left and right sectors of $\SU{2}_L \times \SU{2}_R$ at the same time.
Let $0 < \eta < \min\qty( \abs{x_{(t)} - x_{(t-1)}}, \abs{x_{(t+1)} - x_{(t)}} )$.
We find:
\begin{eqnarray}
  \ipd{z} \cX_{(s)}( x_t + e^{2 \pi i}( \eta + i\, 0^+) )
  & = &
  \cL_{(t,\, t+1)} \ipd{z}\,
  \cX_{(s)}( x_t + \eta + i\, 0^+ )\,
  \cR_{(t,\, t+1)}^{\dagger},
  \label{eq:top_spinor_monodromy}
  \\
  \ipd{z} \cX_{(s)}( x_t + e^{2 \pi i}( \eta - i\, 0^+) )
  & = &
  \tcL_{(t,\, t+1)}\,
  \ipd{z} \cX_{(s)}( x_t + \eta - i\, 0^+ )\,
  \tcR_{(t,\, t+1)}^{\dagger},
  \label{eq:bottom_spinor_monodromy}
\end{eqnarray}
  where:
\begin{eqnarray}
  \cL_{(t,\, t+1)}
  & = &
  U_{L}(\vec{n}_{(t+1)})\,
  U_{L}^{\dagger}(\vec{n}_{(t)}),
  \\
  \tcL_{(t,\, t+1)}
  & = &
  U_{L}(\vec{n}_{(\bart)})\,
  U_{L}^{\dagger}(\vec{n}_{(t)})\,
  U_{L}(\vec{n}_{(t+1)})\,
  U_{L}^{\dagger}(\vec{n}_{(\bart)}),
  \\
  \cR_{(t,\, t+1)}
  & = &
  U_{R}(\vec{m}_{(t+1)})\,
  U_{R}^{\dagger}(\vec{m}_{(t)}),
  \\
  \tcR_{(t,\, t+1)}
  & = &
  U_{R}(\vec{m}_{(\bart)})\,
  U_{R}^{\dagger}(\vec{m}_{(t)})\,
  U_{R}(\vec{m}_{(t+1)})\,
  U_{R}^{\dagger}(\vec{m}_{(\bart)}).
\end{eqnarray}

In spinor representation the action~\eqref{eq:string_action} becomes
\begin{equation}
  \begin{split}
    S_{\R^4}
    & =
    \frac{1}{4 \pi \ap}
    \iint\limits_{\ccH}
    \dd{u} \dd{\baru}\,
    \tr(\ipd{u} X_{(s)}(u, \baru) \cdot \ipd{\baru} X^{\dagger}_{(s)}(u, \baru))
    \\
    & =
    \frac{1}{8 \pi \ap}
    \iint\limits_{\C}
    \dd{z} \dd{\barz}\,
    \tr(
      U_{L}(\vec{n}_{(\bart)})\,
      \ipd{z} \cX_{(s)}(z, \barz)\,
      U_{R}^{\dagger}(\vec{m}_{(\bart)})\,
      \ipd{\barz} \cX_{(s)}^{\dagger}(z, \barz)
    ).
  \end{split}
  \label{eq:action_doubling_fields_spinor_representation}
\end{equation}
It is possible to show that the closed loop $x_t + \eta \pm i\, 0^+ \mapsto x_t + e^{2 \pi i}( \eta \pm i\, 0^+ )$ does not generate additional contributions in the action.


\subsubsection{Special Form of Matrices for D-Branes at Angles} 
\label{sec:special_SO4}

The \SU{2} matrices involved in this scenario with D-branes intersecting at angles have a particular form.
In the left sector (i.e.\ $\SU{2}_L$ matrices) we have:
\begin{equation}
  \cL_{(t,\, t+1)}
  =
  U_{L}(\vec{n}_{(t+1)})\,
  U_{L}^{\dagger}(\vec{n}_{(t)})\,
  =
  -\vec{v}_{(t+1)} \cdot \vec{v}_{(t)}
  +
  i\, (\vec{v}_{(t+1)} \times \vec{v}_{(t)}) \cdot \vec{\sigma} ,
\end{equation}
with $\norm{\vec{v}_{(t)}}^2 = 1$.
This is a consequence of the peculiar properties of the \SO{4} matrices $U_{(t)}$ defined in \eqref{eq:Umatrices}.
Hence the corresponding $\SU{2}_L \times \SU{2}_R$ element $(U_{L}(\vec{n}_{(t)}),\, U_{R}(\vec{m}_{(t)}))$ reflects such characteristics.
In particular for the left part we have
\begin{equation}
  U_{L}(\vec{n}_{(t)})
  =
  i\, \vec{v}_{(t)} \cdot \vec{\sigma},
  \qquad
  \norm{\vec{v}_{(t)}}^2 = 1,
  \label{eq:special_UL_brane_t}
\end{equation}
since $U_{(t)}^2 = \1_4$ implies that $U_{L}^2 = \pm \1_2$.
The right sector clearly follows the same discussion.

In fact \cS in~\eqref{eq:reflection_S} can be represented as $U_{L} = U_{R} = i\, \sigma_1$.
Then any matrix $U_{L}(\vec{n}_{(t)})$ is of the form $U_{L}(\vec{n}_{(t)}) = i\, U(\vec{r}_{(t)}) \cdot \sigma_1 \cdot U^\dagger(\vec{r}_{(t)})$, for some $\vec{r}_{(t)}$ as follows from~\eqref{eq:Umatrices}.
Such matrix has vanishing trace and squares to $-\1_2$ hence the term  proportional to two-dimensional unit matrix in the expression of the generic \SU{2} element given in \Cref{sec:isomorphism} vanishes.
As a consequence $n_{(t)} = \frac{1}{4}$ such that~\eqref{eq:special_UL_brane_t} follows.


\subsection{The Classical Solution}

In the previous sections we defined the principal tools to study the non Abelian embedding of the D-branes.
In what follows we start the investigation of the relation between the hypergeometric solutions and the monodromies due to the geometry of the D-branes.


\subsubsection{The Choice of Hypergeometric Functions}

We build the spinorial representation with \SU{2} matrices and solutions of Fuchsian equations with $N_B$ regular singular points.
We are specifically interested in a solution with $N_B = 3$.
We fix the usual \SL{2}{\R} invariance by mapping the three intersection points $x_{(\bart-1)}$, $x_{(\bart+1)}$ and $x_{(\bart)}$ to $\omega_{\bart-1} = \omega_{x_{(\bart-1)}} = 0$, $\omega_{\bart+1} = \omega_{x_{(\bart+1)}} = 1$ and $\omega_{\bart} = \omega_{x_{(\bart)}} = \infty$ respectively through:
\begin{equation}
  \omega_{u}
  =
  \frac{u - x_{(\bart-1)}}{u - x_{(\bart)}}
  \cdot
  \frac{x_{(\bart+1)} - x_{(\bart-1)}}{x_{(\bart+1)} - x_{(\bart)}}
  \label{eq:def_omega}
\end{equation}
The cut structure for this choice is presented in~\Cref{fig:hypergeometric_cuts}.
The map also defines $\arg(\omega_t - \omega_z) \in \left[ 0,\, 2\pi \right)$ for $t = \bart-1,\, \bart+1$.
We choose $\bart = 1$ in what follows.

\begin{figure}[tbp]
  \centering
  \def\svgwidth{0.35\linewidth}
  \import{img}{threebranes_plane.pdf_tex}
  \caption{%
    Fixing the \SL{2}{\R} invariance for $N_B = 3$ and $\bart = 1$ leads to a cut structure with all the cuts defined on the real axis towards $\omega_{\bart} = \infty$.}
  \label{fig:hypergeometric_cuts}
\end{figure}

The map~\eqref{eq:def_omega} moves the generic Fuchsian singularities to known points on the complex plane.
The functions reproducing the necessary monodromies are basis of hypergeometric functions.
We define:
\begin{equation}
  \hyp{a}{b}{c}{z}
  =
  \zeroinfsum{k}\,
  \frac{\poch{a}{k}\poch{b}{k}}{\gfun{c+k}}~
  \frac{z^k}{k!}
  =
  \frac{1}{\gfun{c}}~
  \tensor[_2]{F}{_1}(a,\, b;\, c;\, z),
\end{equation}
where $\tensor[_2]{F}{_1}(a,\, b;\, c;\, z)$ is the Gauss hypergeometric function and $\gfun{s}$ is the Euler Gamma function.
The function \hyp{a}{b}{c}{z} is well defined for any value of its parameters.\footnotemark{}
\footnotetext{%
  It is not necessary to require $c \in \Z_+$ as in the definition of the Gauss hypergeometric function.
}
We define a vector of independent hypergeometric functions:
\begin{equation}
  B_{0}(z)
  =
  \mqty(
    \hyp{a}{b}{c}{z}
    \\
    (-z)^{1-c}~\hyp{a+1-c}{b+1-c}{2-c}{z}
  )
  \label{eq:basis_0}
\end{equation}
as basis of functions around $z = 0$ with a branch cut on the interval $\left[ 0, +\infty \right)$.
The choice of the branch cuts follows from the cut on $\left[ 1, +\infty \right)$ coming from $\hyp{a}{b}{c}{z}$ which has a singularity at $z = 1$ and the cut on $\left[ 0, +\infty \right)$ from $(-z)^{1-c}$.

As argued in~\eqref{eq:homotopy_rep}, the homotopy group of the complex plane with the branch cut structure of~\Cref{fig:hypergeometric_cuts} is such that a closed loop around all the singularities is homotopically trivial.
The corresponding product of the monodromy matrices~\eqref{eq:homotopy_rep} is the unit matrix.
Let for instance $\cM_{\omega_z}^{\pm}$ be the monodromy matrix which represents a closed loop around $\omega_z$ (the $+$ sign represents a path starting in $\ccH$, while $-$ is a path with base point in $\bccH$).
The triviality property is realised through:
\begin{equation}
  \cM_{0}^+\,
  \cM_{1}^+\,
  \cM_{\infty}^+
  =
  \cM_{\infty}^-\,
  \cM_{1}^-\,
  \cM_{0}^-
  =
  \1_2
  \label{eq:monodromy_relations}
\end{equation}
The monodromy matrix $\omega_{\bart+1} = 1$ can thus be recovered as a product of monodromies around $0$ and $\infty$ given the properties
\begin{equation}
  \begin{split}
    \cM_{0}^+
    & =
    \cM_{0}^-
    =
    \cM_{0},
    \\
    \cM_{\infty}^+
    & =
    \cM_{\infty}^-
    =
    \cM_{\infty},
  \end{split}
\end{equation}
which encode the peculiar branch cut structure due to the doubling trick gluing the intervals on one arbitrary D-brane.
These matrices are an abstract representation of the monodromy group since they are in an arbitrary basis.

Using the basis in $z = 0$~\eqref{eq:basis_0} it is straightforward to find the explicit representation $\rM_{0}$ of the abstract monodromy $\cM_{0}$:
\begin{equation}
  \rM_{0}( c ) = \mqty( \dmat{1, e^{-2\pi i c}} ).
  \label{eq:monodromy_zero}
\end{equation}
The computation of the monodromy matrix $\rM_{\infty}$ representing the monodromy in $\omega_z  =\infty$ in the basis \eqref{eq:basis_0} requires to first compute the monodromy representation $\trM_{\infty}$ of the abstract monodromy $\cM_{\infty}$ in the basis of hypergeometric functions around $z = \infty$:
\begin{equation}
  B_{\infty}(z)
  =
  \mqty(
    (-z)^{-a}~\hyp{a}{a+1-c}{a+1-b}{z^{-1}}
    \\
    (-z)^{-b}~\hyp{b}{b+1-c}{b+1-a}{z^{-1}}
  ).
\end{equation}
This basis is connected to~\eqref{eq:basis_0} through the transition matrix
\begin{equation}
  \cC( a,\, b,\, c )
  =
  \frac{\pi}{\sin(\pi(a-b))}
  \mqty(
    \frac{1}{\gfun{b}\gfun{c-a}}
    &
    -\frac{1}{\gfun{a} \gfun{c-b}}
    \\
    \frac{1}{\gfun{1-a}\gfun{b+1-c}}
    &
    -\frac{1}{\gfun{1-b}\gfun{a+1-c}}
  ),
  \label{eq:transition_matrix}
\end{equation}
as $B_{0}(z) = \cC(a,\, b,\, c)~B_{\infty}(z)$.
Through the loop $z \mapsto z e^{-2\pi i}$ we find:
\begin{equation}
  \trM_{\infty}( a,\, b )
  =
  \mqty( \dmat{e^{2\pi i a}, e^{2\pi i b}} ).
\end{equation}
Finally we can build the desired monodromy:
\begin{equation}
  \rM_{\infty}
  =
  \cC(a,\, b,\, c)\,
  \trM_{\infty}(a,\, b)\,
  \cC^{-1}(a,\, b,\, c).
  \label{eq:monodromy_infty}
\end{equation}


\subsubsection{The Monodromy Factors}

With the previous definitions we reproduce the monodromies of the doubling field in its spinor representation~\eqref{eq:top_spinor_monodromy}.\footnotemark{}
\footnotetext{%
  In general we do not need to consider~\eqref{eq:bottom_spinor_monodromy} since they are the same monodromies.
}
These monodromies are tensor products of two basis of hypergeometric functions: the first basis reproduces the monodromies defined as $\cL$ and the second one those defined as $\cR$ in~\eqref{eq:top_spinor_monodromy}.
Since in principle there can be several combinations of parameters of the hypergeometric function yielding the same monodromy, we consider the full solution to be a linear superposition of all possible contributions:
\begin{equation}
  \ipd{z} \cX(z)
  =
  \pdv{\omega_z}{z}\,
  \sum\limits_{l,\, r}
  c_{lr}\,
  \ipd{z} \cX_{l,r}(\omega_z),
  \label{eq:formal_solution}
\end{equation}
where we drop the index representing the spinorial representation to lighten the notation.
We write any possible solution in a factorised form as
\begin{equation}
  \ipd{z} \cX_{l,\,r}(\omega_z)
  =
  (-\omega_z)^{A_{lr}}\,
  (1-\omega_z)^{B_{lr}}\,
  \cB_{0,\, l}^{(L)}(\omega_z)
  \qty( \cB_{0,\, r}^{(R)}(\omega_z) )^T,
  \label{eq:formal_solution_lr}
\end{equation}
where  $l$ and $r$ label the parameters associates with the left and right sectors of the hypergeometric function.
We introduce the left basis element
\begin{equation}
  \begin{split}
    \cB_{0,\, l}^{(L)}(\omega_z)
    & =
    D^{(L)}_l~
    B_{0,\,l}^{(L)}(\omega_z)
    \\
    & =
    \mqty( 1 & 0 \\ 0 & K_l^{(L)} )\,
    \mqty(
      \hyp{a_l}{b_l}{c_l}{\omega_z}
      \\
      (-z)^{(1-c_l)}\,
      \hyp{a_l+1-c_l}{b_l+1-c_l}{2-c_l}{\omega_z}
    )
  \end{split}
\end{equation}
where $D_l^{(L)} \in \GL{2}{\C}$ is a relative normalisation matrix weighting differently the components of the basis.\footnotemark{}
\footnotetext{%
  In general they can be different for each solution.
}
The right sector follows in a similar way.
Notice that the matrices $D^{(L)}_l$ do not fix the absolute normalisation contained in $c_{lr}$.


\subsubsection{Parameters of the Trivial Monodromy}

Using the previous relations we can determine the possible $\ipd{z} \cX_{l,r}(\omega_z)$ with the desired monodromies.
In this section we study the case of the most general \SU{2} matrices despite the fact that in~\Cref{sec:special_SO4} we argued that they have a specific form.

First of all consider the matrices in \eqref{eq:monodromy_zero} and \eqref{eq:monodromy_infty}.
We impose:
\begin{eqnarray}
  &&\begin{cases}
    D^{(L)}\,
    \rM_{0}^{(L)}\,
    \qty( D^{(L)} )^{-1}
    =
    e^{-2\pi i \delta_{0}^{(L)}}\,
    \cL(\vec{n}_{0})
    \\
    D^{(R)}\,
    \rM_{0}^{(R)}\,
    \qty( D^{(R)} )^{-1}
    =
    e^{-2\pi i \delta_{0}^{(R)}}\,
    \cR^*(\vec{m}_{0})
    =
    e^{-2\pi i \delta_{0}^{(R)}}\,
    \cR(\widetilde{\vec{m}}_{0})
    \\
    e^{2\pi i ( A_{lr} - \delta_{0}^{(L)} -
    \delta_{0}^{(R)} )}
    =
    1
    \end{cases},
  \label{eq:parameters_equality_zero}
  \\
  &&\begin{cases}
    D^{(L)},
    \rM_{\infty}^{(L)}\,
    \qty( D^{(L)} )^{-1}
    =
    e^{-2\pi i \delta_{\infty}^{(L)}}\,
    \cL(\vec{n}_{\infty})
    \\
    D^{(R)}\,
    \rM_{\infty}^{(R)}\,
    \qty( D^{(R)} )^{-1}
    =
    e^{-2\pi i \delta_{\infty}^{(R)}}\,
    \cR^*(\vec{m}_{\infty})
    =
    e^{-2\pi i \delta_{\infty}^{(R)}}\,
    \cR(\widetilde{\vec{m}}_{\infty})
    \\
    e^{2\pi i ( A_{lr} + B_{lr} - \delta_{\infty}^{(L)} -
    \delta_{\infty}^{(R)} )}
    =
    1
  \end{cases},
  \label{eq:parameters_equality_infty}
\end{eqnarray}
where we defined
\begin{eqnarray}
  \cL(\vec{n}_{0})
  & = &
  \cL_{(\bart-1,\,\bart)}
  =
  U_L(\vec{n}_{(\bart)})\,
  U_L^{\dagger}(\vec{n}_{(\bart-1)}),
  \\
  \cL(\vec{n}_{\infty})
  & = &
  \cL_{(\bart,\, \bart+1)}
  =
  U_L(\vec{n}_{(\bart+1)})
  U_L^{\dagger}(\vec{n}_{(\bart)}),
  \\
  \cR(\vec{m}_{0})
  & = &
  \cR_{(\bart-1,\, \bart)}
  =
  U_R(\vec{n}_{(\bart)})
  U_R^{\dagger}(\vec{n}_{(\bart-1)}),
  \\
  \cR(\vec{m}_{\infty})
  & = &
  \cR_{(\bart,\, \bart+1)}
  =
  U_R(\vec{n}_{(\bart+1)})
  U_R^{\dagger}(\vec{n}_{(\bart)}).
\end{eqnarray}
The range of $\delta_{0}^{(L)}$ is
\begin{equation}
  \alpha \le \delta_{0}^{(L)} \le  \alpha + \frac{1}{2},
\end{equation}
that is the width of the range is only $\frac{1}{2}$ and not $1$ as one would naively expect.
This is a consequence of the fact that $e^{- 4 \pi i \delta_{0}^{(L)}}$ is the determinant of the right hand side of the first equation in \eqref{eq:parameters_equality_zero}.
We then choose $\alpha = 0$ for simplicity.
The same considerations hold true for all the other additional parameters $\delta_{0}^{(R)}$ and $\delta_{\infty}^{(L,\,R)}$.

Since we are interested in relative rotations of the D-branes, we choose the
rotation in $\omega_{\bart-1} = 0$ in the maximal torus of $\SU{2}_L \times \SU{2}_R$ without loss of generality: as we have two independent groups, we can in fact fix the orientation of both vectors $\vec{n}_{0}$ and $\vec{m}_{0}$.
In particular we set:
\begin{eqnarray}
  \vec{n}_{0}
  =
  ( 0,\, 0,\, n_{0}^3 ) \in \R^3,
  & \qquad &
  0 < n_{0}^3 < \frac{1}{2},
  \label{eq:maximal_torus_left}
  \\
  \widetilde{\vec{m}}_{0}
  =
  ( 0,\, 0,\, -m_{0}^3 ) \in \R^3,
  & \qquad &
  0 < m_{0}^3 < \frac{1}{2},
  \label{eq:maximal_torus_right}
\end{eqnarray}
where $n_{0}^3 = 0$ is excluded to avoid considering a trivial rotation.
We then define the parameters of the rotation in $\omega_{\bart} = \infty$ to be the most general
\begin{equation}
  \begin{split}
    \vec{n}_{\infty}
    & =
    ( n_{\infty}^1,\, n_{\infty}^2,\, n_{\infty}^3 ),
    \\
    \widetilde{\vec{m}}_{\infty}
    & =
    ( -m_{\infty}^1,\, m_{\infty}^2,\, -m_{\infty}^3 ),
\end{split}
\end{equation}
We could actually set $n_{\infty}^2 = m_{\infty}^2 = 0$ since the choice of the ``gauge''~\eqref{eq:maximal_torus_left} and~\eqref{eq:maximal_torus_right} is preserved by \U{1} rotations mixing $n_{\infty}^1$ and $n_{\infty}^2$.
We nevertheless keep the general expression in order to check the computations.

Solving~\eqref{eq:parameters_equality_zero} and~\eqref{eq:parameters_equality_infty} connects the parameters of the hypergeometric function to the parameter of the rotations (see \Cref{sec:parameters}) thus reproducing the boundary conditions of the intersecting D-branes through the non trivial monodromies of the basis of hypergeometric functions.
We find:
\begin{eqnarray}
  a_l^{(L)}
  =
  n_{0}
  +
  (-1)^{f^{(L)}}\, n_{1}
  +
  n_{\infty}
  +
  \ffa^{(L)}_l,
  & \qquad &
  \ffa^{(L)}_l \in \Z,
  \\
  b_l^{(L)}
  =
  n_{0}
  +
  (-1)^{f^{(L)}}\, n_{1}
  -
  n_{\infty}
  +
  \ffb^{(L)}_l,
  & \qquad &
  \ffb^{(L)}_l \in \Z,
  \\
  c_l^{(L)}
  =
  2\, n_{0}
  +
  \ffc^{(L)}_l,
  & \qquad &
  \ffc^{(L)}_l \in \Z,
  \\
  \delta_{0}^{(L)}
  =
  n_{0},
  \\
  \delta_{\infty}^{(L)}
  =
  -
  n_{0}
  -
  (-1)^{f^{(L)}}\, n_{1},
  \\
  K^{(L)}_l
  =
  -\frac{1}{2 \pi^2}\,
  \cG(a_l^{(L)},\, b_l^{(L)},\, c_l^{(L)})\,
  \cF(a_l^{(L)},\, b_l^{(L)},\, c_l^{(L)})\,
  \frac{n^1_{\infty}+ i\, n^2_{\infty}}{n_{\infty}},
\label{eq:K_factor_value}
\end{eqnarray}
where $f^{(L)} \in \qty{ 0,\, 1 }$.
For the sake of brevity we defined two auxiliary functions, namely $\cG(a,\, b,\, c) = \gfun{1-a}\, \gfun{1-b}\, \gfun{a+1-c}\, \gfun{b+1-c}$ and $\cF(a,\, b,\, c) = \sin(\pi c)\, \sin(\pi(a-b))$.
We also introduced the norm $n_{1} = \norm{\vec{n}_{1}}$ of the rotation vector around $\omega_{\bart+1} = 1$.
Its dependence on the other parameters is encoded in~\eqref{eq:monodromy_relations}, where $\rM^+_{1} = \rM^{-1}_{0}\, \rM^{-1}_{\infty}$, and the composition rule~\eqref{eq:product_in_SU2}:
\begin{equation}
  \cos(2\pi n_{1})
  =
  \cos(2\pi n_{0})\,
  \cos(2\pi n_{\infty})
  -
  \sin(2\pi n_{0})\,
  \sin(2\pi n_{\infty})\,
  \frac{n_{\infty}^3}{n_{\infty}}.
  \label{eq:dependent_monodromy_main_text}
\end{equation}
Relations for the right sector follow under the interchange of $(L)$ with $(R)$ and $\vec{n} \leftrightarrow \vec{m}$.

Parameters $A_{lr}$ and $B_{lr}$ follow the previous results and equations~\eqref{eq:parameters_equality_zero} and \eqref{eq:parameters_equality_infty}:
\begin{eqnarray}
  A_{lr}
  =
  n_{0} + m_{0} + \ffA_{lr},
  & \qquad &
  \ffA_{lr} \in \Z,
  \\
  B_{lr}
  (-1)^{f^{(L)}}\, n_{1} + (-1)^{f^{(R)}}\, m_{1} + \ffB_{lr},
  & \qquad &
  \ffB_{lr} \in \Z.
\end{eqnarray}


\subsubsection{Equivalent Solutions and Necessary Parameters}

There are ambiguities in the equations presented in the previous section.
In fact the choice of $f^{(L)}$ and $f^{(R)}$ looks arbitrary and leading to an undefined solution.
We can use properties of the hypergeometric functions to show that any choice does not affect the final result.
Specifically we can start with certain values but we can recover the others through:
\begin{equation}
  \rP
  \qty{
    \mqty{
      0   & 1     & \infty &   \\
      0   & 0     & a      & z \\
      1-c & c-a-b & b      &
    }
  }
  =
  (1-z)^{c-a-b}\,
  \rP
  \qty{
    \mqty{
      0   & 1     & \infty &   \\
      0   & 0     & c-b    & z \\
      1-c & a+b-c & c-a    &
    }
  },
\end{equation}
where \rP is the Papperitz-Riemann symbol for the hypergeometric functions.
We can then assign any admissible value to $f^{(L)}$ and $f^{(R)}$ and then recover the other through the identification:
\begin{eqnarray}
  f^{(L)}{}' & = & \qty( 1 + f^{(L)} )~\text{mod}~2,
  \\
  \ffa_l' & = & \ffc_l - \ffb_l,
  \\
  \ffb_l' & = & \ffc_l - \ffa_l,
  \\
  \ffc_l' & = &\ffc_l.
\end{eqnarray}
A similar procedure applies also for the right sector.
Finally we also identified the ``free'' parameters:
\begin{eqnarray}
  \ffA_{lr}'
  & = &
  \ffA_{lr},
  \\
  \ffB_{lr}'
  & = &
  \ffB_{lr} - \ffa^{(L)}_l - \ffa^{(R)}_r - \ffb^{(L)}_l - \ffb^{(R)}_r + \ffc^{(L)}_l + \ffc^{(R)}_r.
\end{eqnarray}
The choice of $f^{(L,\,R)}$ is thus simply a convenient relabeling of parameters.
We choose $f^{(L)} = f^{(R)} = 0$ for simplicity.
Moreover in order to get a well defined solution we must impose constraints on the hypergeometric parameters.
We require:
\begin{eqnarray}
  c_l^{(L)} & \not\in & \Z,
  \\
  a_l^{(L)} + b_l^{(L)} & \not\in & \Z + \frac{1}{2}.
\end{eqnarray}

The relations between the parameters of the hypergeometric functions and the monodromies associated to the rotation of the intersecting D-brane are more general than needed.
The number of parameters necessary to fix the configuration is $6$, that is the amount of parameters to uniquely determine $n_{0}^3$, $n_{\infty}^1$, $n_{\infty}^3$ and $m_{0}^3$, $m_{\infty}^1$, $m_{\infty}^3$.
As noticed before we can in fact fix $n_{\infty}^2 = m_{\infty}^2 = 0$.
This is a consequence of the fact that all parameters depend on the norm of the rotation vectors exception made for $K^{(L)}$ and $K^{(R)}$.
They depend on $n_{\infty}^1 + i n_{\infty}^2$ and $m_{\infty}^1 + i m_{\infty}^2$.
Performing a $\SU{2}_L$ and $\SU{2}_R$ rotation around the third axis and a shift of the parameters $\delta_{\infty}$, the phases of the normalisation factors $K$ can vanish.


\subsubsection{The Importance of the Normalization Factors}

Using the \rP symbol the solutions can be symbolically written as
\begin{equation}
  \begin{split}
    &
    (-\omega)^{\ffA}\, (1-\omega)^{\ffB}
    \times
    \\
    &
    \times
    \rP \qty{
      \mqty{
          0
          &
          1
          &
          \infty
          &
          \\
          n_{0}
          &
          n_{1}
          &
          n_{\infty} + \ffa^{(L)}
          &
          \omega
          \\
          -n_{0} + 1 - \ffc^{(L)}
          &
          -n_{1} - \ffa^{(L)} - \ffb^{(L)} + \ffc^{(L)}
          &
          -n_{\infty} + \ffb^{(L)}
          &
      }
    }
    \\
    &
    \times
    \rP \qty{
      \mqty{
          0
          &
          1
          &
          \infty
          &
          \\
          m_{0}
          &
          m_{1}
          &
          m_{\infty} + \ffa^{(R)}
          &
          \omega
          \\
          -m_{0} + 1 - \ffc^{(R)}
          &
          -m_{1} - \ffa^{(R)} - \ffb^{(R)} + \ffc^{(R)}
          &
          -m_{\infty} + \ffb^{(R)}
          &
      }
      }.
  \end{split}
  \label{eq:symbolic_solutions_using_P}
\end{equation}
The normalisation parameters $K$ cannot however be guessed from the \rP symbol.

As we are interested in finding the truly independent solutions to the original problem, we can use properties of the hypergeometric functions to reduce the number of possible choices of the integer factors in the definition of the parameters.
It is possible to show that any hypergeometric function $\hyp{a + \ffa}{b + \ffb}{c + \ffc}{z}$ can be written as a combination of \hyp{a}{b}{c}{z} and any of its contiguous functions~\cite{::NISTDigitalLibrary}.
For instance we can choose:
\begin{equation}
  \hyp{a + \ffa}{b + \ffb}{c + \ffc}{z}
  =
  h_1(a,\, b,\, c;\, z)\,
  \hyp{a+1}{b}{c}{z}
  +
  h_2(a,\, b,\, c;\, z)\,
  \hyp{a}{b}{c}{z},
  \label{eq:reduction_F_F+}
\end{equation}
where $h_1$ and $h_2$ are finite sums of integer (both positive and negative) powers of $z$ and negative powers of $1-z$.
For simplicity let:
\begin{equation}
  \begin{split}
    F & = \hyp{a}{b}{c}{z},
    \\
    F(a+k) & = \hyp{a+k}{b}{c}{z},
    \\
    F(b+k) & = \hyp{a}{b+k}{c}{z},
    \\
    & \dots
  \end{split}
\end{equation}
Similarly we use a shorthand notation for the basis of the hypergeometric functions:\footnotemark{}
\footnotetext{%
  In this expression we introduce a slight abuse of notation since $K_{a,b,c}$ depends on a phase which is not a function of $a$, $b$ or $c$.
  See for instance~\eqref{eq:K_factor_value} and~\eqref{eq:n12+n22}.
}
\begin{equation}
  \cB_{0}(a,\, b,\, c;\, z)
  =
  \mqty(
    \hyp{a}{b}{c}{z}
    \\
    K_{a,b,c}~
    (-z)^{(1-c)}~
    \hyp{a+1-c}{b+1-c}{2-c}{z}
  ).
\end{equation}
We can then algorithmically apply the following relations
\begin{equation}
  \begin{split}
    (c-a)\, F(a-1) + (2a-c+(b-a)z)\, F - a(1-z)\, F(a+1) & = 0.
    \\
    (b-a)\, F + a\, F(a+1) - b\, F(b+1) & = 0,
    \\
    (c-a-b)\, F + a (1-z)\, F(a+1) - (c-b)\, F(b-1) & = 0,
    \\
    (a+(b-c)z)\, F - a (1-z)\, F(a+1) + (c-a)(c-b)z\, F(c+1) & = 0,
    \\
    (c-a-1)\, F + a\, F(a+1) - F(c-1) & = 0,
  \end{split}
  \label{eq:contiguous_functions}
\end{equation}
to eliminate unwanted integer factors and keep only \hyp{a}{b}{c}{z} and any of its contiguous functions.

Notice that $\cB_{0}$ is a basis element of the possible solutions of the classical and quantum string \eom
Using any relation in~\eqref{eq:contiguous_functions} we can change $a$, $b$ or $c$ by one unit coherently in both hypergeometric functions contained in $\cB_{0}$.
For example from the first equation in~\eqref{eq:contiguous_functions} we expect:
\begin{equation}
  (c-a)\, \cB_{0}(a-1)
  +
  (2a-c+(b-a)z)\, \cB_{0}
  -
  a(1-z)\, \cB_{0}(a+1)
  =
  0,
\end{equation}
which can be used to lower and rise the integer factors in $a$.
The relation holds only because the normalisation factor $K$ is present.
In fact coefficients in this equation equal those in the relation for the first component of $\cB_{0}$.
It is not trivial for the second component where the factor $K$ is key to the consistency.
Similarly the relation needed to lower $c$ reads:
\begin{equation}
  (a-c)(b-c)\, \cB_{0}(c+1)
  +
  (a+(b-c)z)\, \cB_{0}
  -
  a(1-z) \cB_{0}(a+1)
  =
  0.
\end{equation}


\subsubsection{Constraints from the Finite Euclidean Action}

In previous sections we present a general procedure to write all possible independent solutions to the classical string \eom
However not all of them are physically acceptable.
In fact we require the finiteness of the Euclidean action~\eqref{eq:action_doubling_fields_spinor_representation}.

In principle it could appear obvious to use~\eqref{eq:contiguous_functions} to restrict the possible arbitrary integers to:
\begin{eqnarray}
  \ffa^{(L)} \in \qty{ -1,\, 0 },
  & \qquad &
  \ffa^{(R)} \in \qty{ -1,\, 0 },
  \\
  \ffb^{(L)} = 0,
  & \qquad &
  \ffb^{(R)} = 0,
  \\
  \ffc^{(L)} = 0,
  & \qquad &
  \ffc^{(R)} = 0.
\end{eqnarray}
We could then use~\eqref{eq:reduction_F_F+} to write the possible solution as
\begin{equation}
  \begin{split}
    \ipd{z} \cX(z)
    & =
    \pdv{\omega_z}{z}\,
    (-\omega_z)^{n_{0} + m_{0}}\,
    (1-\omega_z)^{n_{1} + m_{1}}
    \\
    & \times
    \sum\limits_{\ffa^{(L,\,R)} \in \qty{ -1, 0 }} 
    h(\omega_z,\, \ffa^{(L,R)})
    \times
    \\
    & \times
    \cB_{0}^{(L)}(a^{(L)} + \ffa^{(L)},\, b,\, c;\, \omega_z)
    \qty(
      \cB_{0}^{(R)}(a^{(R)} + \ffa^{(R)},\, b,\, c;\, \omega_z)
    )^T.
  \end{split}
  \label{eq:doubling_field_expansion}
\end{equation}
The issue is therefore to find an explicit form for $h(\omega_z,\, \ffa^{(L,R)})$ yielding a finite action.

We could however use the symbolic solution~\eqref{eq:symbolic_solutions_using_P} to find basis of solutions with finite action.
As a matter of fact, finding the possible solutions with finite action can be recast to finding conditions such that the field $\ipd{z} \cX(z)$ is finite by itself.
Linearity of this condition ensures a simpler approach with respect to the quadratic action of the string.
From~\eqref{eq:action_doubling_fields_spinor_representation} it is clear that the action can be expressed as the sum of the product of any possible couple of elements of the expansion~\eqref{eq:formal_solution}.
We thus need to take into examination all possible pairs of contributions $\ipd{z} \cX_{l_1 r_1}(z)~ \ipd{\barz} \cX_{l_2 r_2}(\barz)$.
Near its singular points, the behavior of any element of solution~\eqref{eq:formal_solution} can be easily read from its symbolic representation~\eqref{eq:symbolic_solutions_using_P}:
\begin{equation}
  \ipd{z} \cX(z)
  \stackrel{\omega_z \to \omega_t}{\sim}
  \omega_t^{C_t}
  \mqty( \omega_t^{k_{t_1}} \\ \omega_t^{k_{t_2}} )
  \mqty( \omega_t^{h_{t_1}} & \omega_t^{h_{t_2}} ).
\end{equation}
It can be verified that the convergence of the action both at finite and infinite intersection points is ensured by the same constraints found when imposing the convergence at any point of the classical solution
\begin{equation}
  X_{(s)}(u,\, \baru)
  =
  f_{(s)\, (\bart-1)}
  +
  \finiteint{u'}{x_{(\bart-1)}}{u}
  \ipd{u'} \cX_{(s)}(u')
  +
  U_L^{\dagger}(\vec{n}_{{\bart}})
  \qty[
    \finiteint{\baru'}{x_{(\bart-1)}}{\baru}
    \ipd{\baru'} \cX_{(s)}(\baru')
  ]
  U_R(\vec{m}_{{\bart}}),
  \label{eq:classical_solution}
\end{equation}
which follows in spinor representation from~\eqref{eq:spinor_doubling_trick} and where $f_{(s),\, (\bart-1)} = f^I_{(\bart-1)}\, \tau_I$.
We specifically find:
\begin{equation}
  \begin{split}
    C_t + k_{t_i} + h_{t_j} > -1,
    \qquad
    i,\, j \in \{1,2\},
    & \qquad
    \omega_t \in \{0, 1\},
    \\
    C_t + k_{t_i} + h_{t_j} < -1,
    \qquad
    i,\, j \in \{1,2\},
    & \qquad
    \omega_t = \infty.
  \end{split}
  \label{eq:constraints_finite_X}
\end{equation}

For simplicity first consider the case of a trivial right rotation.\footnotemark{}
\footnotetext{%
  That is require that $U_R$ is proportional to the identity.
}
In this case~\eqref{eq:symbolic_solutions_using_P} becomes
\begin{equation}
  (-\omega)^\ffA\,
  (1-\omega)^\ffB\,
  \rP\qty{
  \mqty{
    0
    &
    1
    &
    \infty
    &
    \\
    n_{0}
    &
    n_{1}
    &
    n_{\infty} + \ffa^{(L)}
    &
    \omega
    \\
    -n_{0} + 1 -\ffc^{(L)}
    &
    -n_{1} - \ffa^{(L)} - \ffb^{(L)} + \ffc^{(L)}
    &
    -n_{\infty} + \ffb^{(L)}
    &
  }
  }.
\end{equation}
The only possible solution compatible with~\eqref{eq:constraints_finite_X} is
\begin{equation}
  \rP\qty{
  \mqty{
    0
    &
    1
    &
    \infty
    &
    \\
    n_{0} - 1
    &
    n_{1} - 1
    &
    n_{\infty} + 1
    &
    \omega
    \\
    -n_{0}
    &
    -n_{1}
    &
    -n_{\infty} + 2
    &
  }
  },
  \label{eq:X_solution_pure_L}
\end{equation}
that is $\ffa^{(L)} = -1$, $\ffb^{(L)} = 0$, $\ffc^{(L)} = 0$, $\ffA = -1$ and  $\ffB = -1$.

In the general the solution is more complicated and it depends on the relation between the rotation vectors $\vec{n}_{0,\, 1,\, \infty}$, $\vec{m}_{0,\, 1,\, \infty}$.
For each possible case the solution is however unique and it is given by
\begin{enumerate}
  \item $n_{0} > m_{0}$ and $n_{1} > m_{1}$:
    \begin{equation}
      \rP\qty{
        \mqty{
          0
          &
          1
          &
          \infty
          &
          \\
          n_{0} - 1
          &
          n_{1} - 1
          &
          n_{\infty} + 1
          &
          \omega
          \\
          -n_{0}
          &
          -n_{1}
          &
          -n_{\infty} + 2
          &
        }
      }
      \rP\qty{
        \mqty{
          0
          &
          1
          &
          \infty
          &
          \\
          m_{0}
          &
          m_{1}
          &
          m_{\infty}
          &
          \omega
          \\
          -m_{0} + 1
          &
          -m_{1}
          &
          -m_{\infty} + 1
          &
        }
      },
      \label{eq:X_solution>>}
   \end{equation}

   \item $n_{0} > m_{0}$, $n_{1} < m_{1}$ and $n_{\infty} > m_{\infty}$:
     \begin{equation}
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           n_{0} - 1
           &
           n_{1}
           &
           n_{\infty}
           &
           \omega
           \\
           -n_{0}
           &
           -n_{1}
           &
           -n_{\infty} + 2
           &
         }
       }
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           m_{0}
           &
           m_{1} - 1
           &
           m_{\infty} + 1
           &
           \omega
           \\
           -m_{0}
           &
           -m_{1}
           &
           -m_{\infty} + 1
           &
         }
       },
       \label{eq:X_solution><>}
     \end{equation}

   \item $n_{0} > m_{0}$, $n_{1} < m_{1}$ and $n_{\infty} < m_{\infty}$:
     \begin{equation}
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           n_{0} - 1
           &
           n_{1}
           &
           n_{\infty} + 1
           &
           \omega
           \\
           -n_{0}
           &
           -n_{1}
           &
           -n_{\infty} + 1
           &
         }
       }
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           m_{0}
           &
           m_{1} - 1
           &
           m_{\infty}
           &
           \omega
           \\
           -m_{0}
           &
           -m_{1}
           &
           -m_{\infty} + 2
           &
         }
       },
       \label{eq:X_solution><<}
     \end{equation}

   \item $n_{0} < m_{0}$, $n_{1} > m_{1}$ and $n_{\infty} > m_{\infty}$:
     \begin{equation}
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           n_{0}
           &
           n_{1} - 1
           &
           n_{\infty}
           &
           \omega
           \\
           -n_{0}
           &
           -n_{1}
           &
           -n_{\infty} + 2
           &
         }
       }
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           m_{0} - 1
           &
           m_{1}
           &
           m_{\infty} + 1
           &
           \omega
           \\
           -m_{0}
           &
           -m_{1}
           &
           -m_{\infty} + 1
           &
         }
       },
       \label{eq:X_solution<>>}
     \end{equation}

   \item $n_{0} < m_{0}$, $n_{1} > m_{1}$ and $n_{\infty} < m_{\infty}$:
     \begin{equation}
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           n_{0}
           &
           n_{1} - 1
           &
           n_{\infty} + 1
           &
           \omega
           \\
           -n_{0}
           &
           -n_{1}
           &
           -n_{\infty} + 1
           &
         }
       }
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           m_{0} - 1
           &
           m_{1}
           &
           m_{\infty}
           &
           \omega
           \\
           -m_{0}
           &
           -m_{1}
           &
           -m_{\infty} + 2
           &
         }
       },
       \label{eq:X_solution<><}
     \end{equation}

   \item $n_{0} < m_{0}$, $n_{1} < m_{1}$:
     \begin{equation}
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           n_{0}
           &
           n_{1}
           &
           n_{\infty}
           &
           \omega
           \\
           -n_{0}
           &
           -n_{1}
           &
           -n_{\infty} + 1
           &
         }
       }
       \rP\qty{
         \mqty{
           0
           &
           1
           &
           \infty
           &
           \\
           m_{0} - 1
           &
           m_{1} - 1
           &
           m_{\infty} + 1
           &
           \omega
           \\
           -m_{0}
           &
           -m_{1}
           &
           -m_{\infty} + 2
           &
         }
       }.
       \label{eq:X_solution<<}
     \end{equation}
\end{enumerate}
The parameters associated to this list of solutions are summarised in~\Cref{tab:coeffs_k}, where the symmetry under the exchange of $n$ and $m$ becomes evident.

\begin{table}[tbp]
  \centering
  \begin{tabular}{@{}ccc|rr|rrr|rrr@{}}
    \toprule
    & & & \ffA & \ffB & $\ffa^{(L)}$ & $\ffb^{(L)}$ & $\ffc^{(L)}$ & $\ffa^{(R)}$ & $\ffb^{(R)}$ & $\ffc^{(R)}$ \\
    \midrule
    $n_{0} > m_{0}$ & $n_{1} > m_{1}$ & $n_{\infty} \lessgtr m_{\infty}$ & -1 & -1 & -1 & 0 & 0 & 0 & +1 & +1 \\
    $n_{0} > m_{0}$ & $n_{1} < m_{1}$ & $n_{\infty} > m_{\infty}$ & -1 & -1 & -1 & +1 & 0 & 0 & 0 & +1 \\
    $n_{0} > m_{0}$ & $n_{1} < m_{1}$ & $n_{\infty} < m_{\infty}$ & -1 & -1 & 0 & 0 & 0 & -1 & +1 & +1 \\
    $n_{0} < m_{0}$ & $n_{1} > m_{1}$ & $n_{\infty} > m_{\infty}$ & -1 & -1 & -1 & +1 & +1 & 0 & 0 & 0 \\
    $n_{0} < m_{0}$ & $n_{1} > m_{1}$ & $n_{\infty} < m_{\infty}$ & -1 & -1 & 0 & 0 & +1 & -1 & +1 & 0 \\
    $n_{0} < m_{0}$ & $n_{1} < m_{1}$ & $n_{\infty} \lessgtr m_{\infty}$ & -1 & -1 & 0 & +1 & +1 & -1 & 0 & 0 \\
    \bottomrule
  \end{tabular}
  \caption{Integer shifts in the parameters of the hypergeometric function.}
  \label{tab:coeffs_k}
\end{table}


\subsubsection{The Basis of Solutions}
\label{sec:true_basis}

In the previous section we produced one solution for each ordering of the $n_{\omega_z}$ with respect to $m_{\omega_z}$.
There are however other solutions connected to the $\Z_2$ equivalence class in the isomorphism between \SO{4} its double cover.
Given a solution $(\vec{n}_{0},\, \vec{n}_{1},\, \vec{n}_{\infty}) \oplus (\vec{m}_{0},\, \vec{m}_{1},\, \vec{m}_{\infty})$, we can in fact replace any couple of $\vec{n}$ and $\vec{m}$ by $\widehat{\vec{n}}$ and $\widehat{\vec{m}}$ and get an apparently new solution.\footnotemark{}
\footnotetext{%
  We need to change two rotation vectors because the monodromies are constrained by~\eqref{eq:monodromy_relations}.
}
For instance we could consider $(\widehat{\vec{n}}_{0},\, \widehat{\vec{n}}_{1},\, \vec{n}_{\infty}) \oplus (\vec{m}_{0},\, \widehat{\vec{m}}_{1},\, \widehat{\vec{m}}_{\infty})$.
On the other hand the previous substitution would change the \SO{4} in both $\omega = 0$ and $\omega = \infty$: it does not represent a new solution.
We are left therefore with three possibilities besides the original one:
\begin{equation}
  \begin{split}
    (\widehat{\vec{n}}_{0},\, \widehat{\vec{n}}_{1},\, \vec{n}_{\infty})
    & \oplus
    (\widehat{\vec{m}}_{0},\, \widehat{\vec{m}}_{1},\, \vec{m}_{\infty}),
    \\
    (\widehat{\vec{n}}_{0},\, \vec{n}_{1},\, \widehat{\vec{n}}_{\infty})
    & \oplus
    (\vec{m}_{0},\, \widehat{\vec{m}}_{1},\, \widehat{\vec{m}}_{\infty}),
    \\
    (\widehat{\vec{n}}_{0},\, \vec{n}_{1},\, \widehat{\vec{n}}_{\infty})
    & \oplus
    (\vec{m}_{0},\, \widehat{\vec{m}}_{1},\, \widehat{\vec{m}}_{\infty}).
  \end{split}
\end{equation}
We can gauge fix the $\Z_2$ choice by letting $\vec{n}_{0}^3,\, \vec{m}_{0}^3 > 0$ as required by~\eqref{eq:maximal_torus_left} and~\eqref{eq:maximal_torus_right}.
We are thus left with two possible solutions
\begin{align}
  (\vec{n}_{0},\, \vec{n}_{1},\, \vec{n}_{\infty})
  & \oplus
  (\vec{m}_{0},\, \vec{m}_{1},\, \vec{m}_{\infty}),
  \\
  (\vec{n}_{0},\, \widehat{\vec{n}}_{1},\, \widehat{\vec{n}}_{\infty})
  & \oplus
  (\vec{m}_{0},\, \widehat{\vec{m}}_{1},\, \widehat{\vec{m}}_{\infty}).
\end{align}
These are the original and a modified solution obtained by acting with a parity operator $P_2$ on the rotation parameters at $\omega = 1,\, \infty$ on both left and right sector at the same time.
We then need to ensure its independence in order to accept it as a possible solution.

As shown in~\Cref{tab:coeffs_k}, there are only two different cases up to left-right symmetry.
The first is
\begin{equation}
  \Big\lbrace
    (n_{0} > m_{0},\, n_{1} > m_{1},\, n_{\infty} >  m_{\infty} ),~
    (n_{0} > m_{0},\, \hat{n}_{1} < \hat{m}_{1},\, \hat{n}_{\infty} < \hat{m}_{\infty})
  \Big\rbrace,
\end{equation}
which is mapped to
\begin{equation}
  \Big\lbrace
    (n_{0} < m_{0},\, n_{1} < m_{1},\, n_{\infty} <  m_{\infty} ),~
    (n_{0} < m_{0},\, \hat{n}_{1} > \hat{m}_{1},\, \hat{n}_{\infty} > \hat{m}_{\infty})
  \Big\rbrace
\end{equation}
by the left-right symmetry.
The second is
\begin{equation}
  \Big\lbrace
    (n_{0} > m_{0},\, n_{1} > m_{1},\, n_{\infty} <  m_{\infty} ),~
    (n_{0} > m_{0},\, \hat{n}_{1} < \hat{m}_{1},\, \hat{n}_{\infty} > \hat{m}_{\infty})
  \Big\rbrace,
\end{equation}
which is mapped to
\begin{align}
  \Big\lbrace
    (n_{0} < m_{0},\, n_{1} < m_{1},\, n_{\infty} >  m_{\infty} ),~
    (n_{0} < m_{0},\, \hat{n}_{1} > \hat{m}_{1},\, \hat{n}_{\infty} < \hat{m}_{\infty})
  \Big\rbrace
\end{align}
by the same symmetry.

We can then study the two solutions in the two cases.
We first perform the computations common to both cases and then we explicitly specialise the calculations.
Computing the parameters of the hypergeometric functions of the first solution leads to:
\begin{equation}
  \begin{cases}
    a^{(L)} & = n_{0} + n_{1} + n_{\infty} + \ffa^{(L)}
    \\
    b^{(L)} & = n_{0} + n_{1} - n_{\infty} + \ffb^{(L)}
    \\
    c^{(L)} & = 2\, n_{0} + \ffc^{(L)}
  \end{cases},
  \qquad
  \begin{cases}
    a^{(R)} & = m_{0} + m_{1} + m_{\infty} + \ffa^{(R)}
    \\
    b^{(R)} & = m_{0} + m_{1} - m_{\infty} + \ffb^{(R)}
    \\
    c^{(R)} & = 2\, m_{0} + 1 + \ffc^{(R)}
  \end{cases}.
\end{equation}
The values of the constants are in \Cref{tab:coeffs_k}.
We then derive the factors $K^{(L)}$ and $K^{(R)}$ using~\eqref{eq:K_factor_value}.
The first solution reads:
\begin{equation}
  \begin{split}
    \ipd{\omega} \cX_1
    & =
    (-\omega)^{n_{0} + m_{0} - 1}\,
    (1-\omega)^{n_{1} + m_{1} - 1}\,
    \\
    & \times
    \mqty(
      \hyp{a^{(L)}}{b^{(L)}}{c^{(L)}}{\omega}
      \\
      K^{(L)}\, (-\omega)^{1 - c^{(L)}}\,
      \hyp{a^{(L)} + 1 - c^{(L)}}{b^{(L)} + 1 - c^{(L)}}{2 - c^{(L)}}{\omega}
    )
    \\
    & \times
    \mqty(
      \hyp{a^{(R)}}{b^{(R)}}{c^{(R)}}{\omega}
      \\
      K^{(R)}\, (-\omega)^{1 - c^{(R)}}\,
      \hyp{a^{(R)} + 1 - c^{(R)}}{b^{(R)} + 1 - c^{(R)}}{2 - c^{(R)}}{\omega}
    )
  \end{split}
  \label{eq:first_solution}
\end{equation}

The parameters of the second solution read
\begin{equation}
  \begin{split}
    &
    \begin{cases}
      \hat{a}^{(L)}
      & =
      n_{0} + \hat{n}_{1} + \hat{n}_{\infty} + \hat{\ffa}^{(L)}
      =
      c^{(L)} - a^{(L)} + \ffa^{(L)} - \ffc^{(L)} + \hat{\ffa}^{(L)} + 1
      \\
      \hat{b}^{(L)}
      & =
      n_{0} + \hat{n}_{1} - \hat{n}_{\infty} + \hat{\ffb}^{(L)}
      =
      c^{(L)} - b^{(L)} + \ffb^{(L)} - \ffc^{(L)} + \hat{\ffb}^{(L)}
      \\
      \hat{c}^{(L)}
      & =
      2\, n_{0} + \hat{\ffc}^{(L)}
      =
      c^{(L)} - \ffc^{(L)} + \hat{\ffc}^{(L)}
    \end{cases}
    \\
    &
    \begin{cases}
      \hat{a}^{(R)}
      & =
      m_{0} + \hat{n}_{1} + \hat{n}_{\infty} + \hat{\ffa}^{(R)}
      =
      c^{(R)} - a^{(R)} + \ffa^{(R)} - \ffc^{(R)} + \hat{\ffa}^{(R)} + 1
      \\
      \hat{b}^{(R)}
      & =
      m_{0} + \hat{n}_{1} - \hat{n}_{\infty} + \hat{\ffb}^{(R)}
      =
      c^{(R)} - b^{(R)} + \ffb^{(R)} - \ffc^{(R)} + \hat{\ffb}^{(R)}
      \\
      \hat{c}^{(R)}
      & =
      2\, m_{0} + \hat{\ffc}^{(R)}
      =
      c^{(R)} - \ffc^{(R)} + \hat{\ffc}^{(R)}
    \end{cases}
  \end{split}
\end{equation}
The two cases differ only for constant factors and not in structure.


\paragraph{Case 1}

Consider $n_{0} > m_{0}$, $n_{1} > m_{1}$ and $n_{\infty} > m_{\infty}$.
The associated second solution is $n_{0} > m_{0}$, $\hat{n}_{1} < \hat{m}_{1}$ and $\hat{n}_{\infty} < \hat{m}_{\infty}$.
Its parameters are:
\begin{equation}
  \begin{cases}
    \hat{a}^{(L)} & = c^{(L)} - a^{(L)}
    \\
    \hat{b}^{(L)} & = c^{(L)} - b^{(L)}
    \\
    \hat{c}^{(L)} & = c^{(L)}
  \end{cases},
  \qquad
  \begin{cases}
    \hat{a}^{(R)} & = c^{(R)} - a^{(R)}
    \\
    \hat{b}^{(R)} & = c^{(R)} - b^{(R)} + 1
    \\
    \hat{c}^{(R)} & = c^{(R)} + 1
  \end{cases},
\end{equation}
The normalisation factors are
\begin{equation}
  \hat{K}^{(L)} = K^{(L)},
  \qquad
  \hat{K}^{(R)} = \frac{K^{(R)}}{a^{(R)} (c^{(R)} - b^{(R)})}.
\end{equation}
Using Euler relation
\begin{equation}
  \hyp{a}{b}{c}{\omega} = (1-\omega)^{c-a-b}\, \hyp{c-a}{c-b}{c}{\omega},
\end{equation}
we can write the second solution as
\begin{equation}
  \begin{split}
    \ipd{\omega} \cX_2
    & =
    (-\omega)^{n_{0} + m_{0} - 1}\,
    (1-\omega)^{n_{1} + m_{1}}\,
    \\
    & \times
    \mqty(
      \hyp{a^{(L)}}{b^{(L)}}{c^{(L)}}{\omega}
      \\
      K^{(L)}\, (-\omega)^{1 - c^{(L)}}\,
      \hyp{a^{(L)} + 1 - c^{(L)}}{b^{(L)} + 1 - c^{(L)}}{2 - c^{(L)}}{\omega}
    )
    \\
    & \times
    \mqty(
      \hyp{a^{(R)} + 1}{b^{(R)}}{c^{(R)} + 1}{\omega}
      \\
      \hat{K}^{(R)}\, (-\omega)^{- c^{(R)}}\,
      \hyp{a^{(R)} + 1 - c^{(R)}}{b^{(R)} - c^{(R)}}{1 - c^{(R)}}{\omega}
    )
  \end{split}.
\end{equation}
In this solution the left basis is exactly the same as in the first solution~\eqref{eq:first_solution} while the right basis differs for $a^{(R)} \mapsto a^{(R)} + 1$ and $c^{(R)} \mapsto c^{(R)} + 1$.


\paragraph{Case 2}

Consider now the second option $n_{0} > m_{0}$, $n_{1} > m_{1}$ and $n_{\infty} < m_{\infty}$.
For the second solution we have $n_{0} > m_{0}$, $\hat{n}_{1} < \hat{m}_{1}$ and $\hat{n}_{\infty} > \hat{m}_{\infty}$ and the parameters are explicitly:
\begin{equation}
  \begin{cases}
    \hat{a}^{(L)} & = c^{(L)} - a^{(L)} - 1
    \\
    \hat{b}^{(L)} & = c^{(L)} - b^{(L)} - 1
    \\
    \hat{c}^{(L)} & = c^{(L)}
  \end{cases},
  \qquad
  \begin{cases}
    \hat{a}^{(R)} & = c^{(R)} - a^{(R)}
    \\
    \hat{b}^{(R)} & = c^{(R)} - b^{(R)}
    \\
    \hat{c}^{(R)} & = c^{(R)}
  \end{cases},
\end{equation}
The normalisation factors $K$ are:
\begin{equation}
  \hat{K}^{(L)}
  =
  K^{(L)}\,
  \frac{(b^{(L)} - 1)(c^{(L)} - a^{(L)} - 1)}{a^{(L)} (c^{(L)} - b^{(L)})},
  \qquad
  \hat{K}^{(R)}
  =
  K^{(R)}.
\end{equation}
Using Euler relation we write the second solution for the second case as
\begin{equation}
  \begin{split}
    \ipd{\omega} \cX_2
    & =
    (-\omega)^{n_{0} + m_{0} - 1}\,
    (1-\omega)^{n_{1} + m_{1}}\,
    \\
    & \times
    \mqty(
      \hyp{a^{(L)} + 1}{b^{(L)} - 1}{c^{(L)}}{\omega}
      \\
      \hat{K}^{(L)}\, (-\omega)^{1 - c^{(L)}}\,
      \hyp{a^{(L)} + 2 - c^{(L)}}{b^{(L)} - c^{(L)}}{2 - c^{(L)}}{\omega}
    )
    \\
    & \times
    \mqty(
      \hyp{a^{(R)}}{b^{(R)}}{c^{(R)}}{\omega}
      \\
      \hat{K}^{(R)}\, (-\omega)^{- c^{(R)}}\,
      \hyp{a^{(R)} + 1 - c^{(R)}}{b^{(R)} + 1 - c^{(R)}}{2 - c^{(R)}}{\omega}
    )
  \end{split}.
\end{equation}
The right basis is the same as in the first solution while the left basis differs for $a^{(L)} \mapsto a^{(L)} + 1$ and $b^{(L)} \mapsto b^{(L)} - 1$.


\subsubsection{The Solution}
  
We showed that there are two independent solutions.
The general solution for $\ipd{\omega} \cX$ is therefore:
\begin{equation}
  \ipd{\omega} \cX
  =
  C_1\, \ipd{\omega} \cX_1 + C_2\, \ipd{\omega} \cX_2.
  \label{eq:general_solution}
\end{equation}
The final solution depends only on two complex constants, $C_1$ and $C_2$, which we can fix imposing the global conditions in \eqref{eq:discontinuity_bc}, that is the second equation in the solution \eqref{eq:classical_solution}.
As the three intersection points in target space always define a triangle on a 2-dimensional plane, we impose the boundary conditions knowing two angles formed by the sides of the triangle (i.e.\ the branes between two intersections) and the length of one of them.
Since we already fixed the parameters associated to the rotations, we need to compute the length of one of the sides.
Consider for instance the length of $X(x_{\bart+1},\, x_{\bart+1}) - X(x_{\bart-1},\, x_{\bart-1})$.
Explicitly we impose the four real equations in spinorial formalism
\begin{equation}
  \finiteint{\omega}{0}{1}
  \ipd{\omega} \cX(\omega)
  +
  U_L^{\dagger}(\vec{n}_{{\bart}})
  \qty[
    \finiteint{\bomega}{0}{1} \ipd{\bomega} \cX(\bomega)
  ]
  U_R(\vec{m}_{{\bart}})
  =
  f_{{\bart+1}\, (s)} - f_{{\bart-1}\, (s)},
\end{equation}
where we used the mapping~\eqref{eq:def_omega} to write the integrals in the $\omega$ variables.
This equation has enough \dof to fix completely the two complex parameters $C_1$ and $C_2$.
The final generic solution is thus uniquely determined.


\subsection{Recovering the \texorpdfstring{\SU{2}}{SU(2)} and the Abelian Solution}

In this section we show how this general procedure includes both the solution with pure \SU{2} rotation matrices and the solution with Abelian rotations of the D-branes.
The Abelian solution emerges from this construction as a limit and produces the known result for Abelian $\SO{2} \times \SO{2} \subset \SO{4}$ rotations in the case of a factorised space $\R^4 = \R^2 \times \R^2$.

\subsubsection{Abelian Limit of the \texorpdfstring{\SU{2}}{SU(2)} Monodromies}

\begin{table}[tbp]
  \centering
  \begin{tabular}{@{}rr|cc|cr|c@{}}
    \toprule
    $\vec{n}_{0}$                 &
    $\vec{n}_{\infty}$            &
    \multicolumn{2}{c|}{relations}    &
    $n_{1}$                      &
    $\vec{n}_{1}$                 &
    $\sum\limits_{t} \vec{n}_{\vec{t}}$
    \\
    \midrule
    $n_{0}\, \vec{k}$                        &
    $n_{\infty}\, \vec{k}$                   &
    $n_{0} + n_{\infty} < \frac{1}{2}$ &
    $n_{0} \lessgtr n_{\infty}$        &
    $n_{0} + n_{\infty}$               &
    $-n_{1}\, \vec{k}$                       &
    $0$
    \\
    $n_{0}\, \vec{k}$                        &
    $n_{\infty}\, \vec{k}$                   &
    $n_{0} + n_{\infty} > \frac{1}{2}$ &
    $n_{0} \lessgtr n_{\infty}$        &
    $1 - (n_{0} + n_{\infty})$         &
    $n_{1}\, \vec{k}$                        &
    $\vec{k}$
    \\
    $n_{0}\, \vec{k}$                               &
    $-n_{\infty}\, \vec{k}$                         &
    $n_{0} + n_{\infty} \lessgtr \frac{1}{2}$ &
    $n_{0} > n_{\infty}$                      &
    $n_{0} - n_{\infty}$                      &
    $-n_{1}\, \vec{k}$                              &
    $0$
    \\
    $n_{0}\, \vec{k}$                               &
    $-n_{\infty}\, \vec{k}$                         &
    $n_{0} + n_{\infty} \lessgtr \frac{1}{2}$ &
    $n_{0} < n_{\infty}$                      &
    $-n_{0} + n_{\infty}$                     &
    $n_{1}\, \vec{k}$                               &
    $0$
    \\
    \bottomrule
  \end{tabular}
  \caption{Abelian limit of \SU{2} monodromies}
  \label{tab:Abelian_composition}
\end{table}
Here we compute the parameter $\vec{n}_{1}$ given two Abelian rotation in $\omega = 0$ and $\omega = \infty$ using the standard expression for two \SU{2} element multiplication given in~\eqref{eq:product_in_SU2} in~\Cref{sec:isomorphism}.
Results are shown in~\Cref{tab:Abelian_composition}.

\begin{figure}[tbp]
    \centering
    \def\svgwidth{0.8\textwidth}
    \import{img}{abelian_angles_case1.pdf_tex}
    \caption{%
      The Abelian limit when the triangle has all acute angles.
      This corresponds to the cases  $n_{0} + n_{\infty}< \frac{1}{2}$ and $n_{0}< n_{\infty}$ which are exchanged under the parity $P_2$.}
    \label{fig:Abelian_angles_1}
\end{figure}

\begin{figure}[tbp]
    \centering
    \def\svgwidth{0.8\textwidth}
    \import{img}{abelian_angles_case2.pdf_tex}
    \caption{%
      The Abelian limit when the triangle has one obtuse angle.
      This corresponds to the cases $n_{0} + n_{\infty}> \frac{1}{2}$ and $n_{0}> n_{\infty}$ which are exchanged under the parity $P_2$.}
    \label{fig:Abelian_angles_2}
\end{figure}

Under the parity transformation $P_2$ the previous four cases are grouped
into two sets $\{ n_{1} = n_{0} + n_{\infty},\, \hat{n}_{1} = -n_{0} + \hat{n}_{\infty} \}$ and $\{ n_{1} = 1 - (n_{0} + n_{\infty}),\, \hat{n}_{1} = n_{0} - \hat{n}_{\infty} \}$.
Geometrically the first group corresponds to the same geometry which is depicted in~\Cref{fig:Abelian_angles_1} while the second in~\Cref{fig:Abelian_angles_2}.
We can in fact arbitrarily fix the orientation of $D_{(3)}$ to obtain these geometrical interpretations.
Since $n^3_{0} > 0$ we can then fix the orientation of $D_{{1}}$.
$D_{{2}}$ is then fixed relatively to $D_{{1}}$ by the sign of $n^3_{\infty}$.
The sign of $n^3_{1}$ then follows.

Differently from the usual geometric Abelian case, this group analytical approach distinguishes between the possible orientations of the D-branes.
In fact we can compare all possible D-brane orientation and the group parameter $n^3$ with the angles in the Abelian configuration in~\Cref{fig:usual_Abelian_angles}.
The relation between the usual Abelian paramter $\epsilon_{\vec{t}}$ and $n_{\vec{t}}^3$ is
\begin{equation}
  \varepsilon_{\vec{t}}
  =
  n_{\vec{t}}^3 + \theta(-n^3_{\vec{t}})
  \label{eq:Abelian_vs_n_simple_case},
\end{equation}
when all $m = 0$.

\begin{figure}[tbp]
  \centering
  \def\svgwidth{0.8\textwidth}
  \import{img}{usual_abelian_angles.pdf_tex}
  \caption{%
    The geometrical angles used in the usual geometrical approach to the Abelian configuration do not distinguish among the possible branes orientations.
    In fact we have $0 \le \alpha < 1$ and $0 < \varepsilon < 1$.
  }
  \label{fig:usual_Abelian_angles}
\end{figure}


\subsubsection{The Abelian Limit of the Left Solutions}

We can then compute the basis element for the entries of~\Cref{tab:coeffs_k} for any value of $n_{1}$ given in~\Cref{tab:Abelian_composition}.
For simplicity we consider the left sector of the solution and drop the notation identifying it to avoid cluttering the equations.
The right sector follows in a similar way.

In the Abelian limit either $K = 0$ or $K = \infty$.
We can absorb the infinite divergence in a constant term globally multiplying the solution and use:
\begin{eqnarray}
  \eval{D}_{K = 0} & = & \mqty( 1 & \\ & 0 ),
  \\
  \eval{D}_{K = \infty} & = & \mqty( 0 & \\ & 1 ).
\end{eqnarray}
Results are summarised in~\Cref{tab:Left_Abelian_solutions} where we left some hypergeometric functions in their symbolic form for compactness even though they are in fact elementary functions since either $a$ or $c - b$ equal $-1$.
\begin{table}[tbp]
  \centering
  \begin{tabular}{@{}ccc@{}}
    \toprule
    $\qty( \ffa^{(L)},\, \ffb^{(L)},\, \ffc^{(L)} )$ &
    $n_{1}$                                            &
    $\qty( \cB^{(L)}( z ) )^T$
    \\
    \midrule
    \multirow{4}{*}{$(-1,\, 0,\, 0)$} &
    $n_{0} + n_{\infty}$    &
    $\mqty( (1 - z)^{-2\, n_{\infty} - 2\, n_{0} + 1} & 0 )$
    \\
                                                      &
    $1 - \qty( n_{0} + n_{\infty} )$ &
    $\mqty( 1 & 0 )$
    \\
                                   &
    $n_{0} - n_{\infty}$ &
    $\mqty( 1 & (-z)^{1 - 2\, n_{0}} )$
    \\
                                     &
    $- n_{0} + n_{\infty}$ &
    $\mqty( 1 & 0 )$
    \\
    \midrule
    \multirow{4}{*}{$(-1,\, 1,\, 0)$} &
    $n_{0} + n_{\infty}$    &
    $\mqty( \hyp{2\, n_{\infty} + 2\, n_{0} - 1}
                {2\, n_{0} + 1}
                {2\, n_{0}}
                {z}
            & 0 )$
    \\
                                                      &
    $1 - \qty( n_{0} + n_{\infty} )$ &
    $\mqty( 1 & 0 )$
    \\
                                   &
    $n_{0} - n_{\infty}$ &
    $\mqty( 0 & (-z)^{1 - 2\, n_{0}} )$
    \\
                                     &
    $- n_{0} + n_{\infty}$ &
    $\mqty( 0 & (1-z)^{2\, n_{0} - 2\, n_{\infty}}\, (-z)^{1 - 2\, n_{0}} )$
    \\
    \midrule
    \multirow{4}{*}{$(0,\, 0,\, 0)$} &
    $n_{0} + n_{\infty}$    &
    $\mqty( (1-z)^{-2\, n_{\infty} - 2\, n_{0}} & 0 )$
    \\
                                                      &
    $1 - \qty( n_{0} + n_{\infty} )$ &
    $\mqty( 0 & (1-z)^{2\, n_{\infty} + 2\, n_{0} - 2}\, (-z)^{1 - 2\, n_{0}} )$
    \\
                                   &
    $n_{0} - n_{\infty}$ &
    $\mqty( (1-z)^{2\, n_{\infty} - 2\, n_{0}} & 0 )$
    \\
                                     &
    $- n_{0} + n_{\infty}$ &
    $\mqty( 1 & 0 )$
    \\
    \midrule
    \multirow{4}{*}{$(-1,\, 1,\, 1)$} &
    $n_{0} + n_{\infty}$    &
    $\mqty( (1-z)^{-2\, n_{\infty} - 2\, n_{0}} + 1 & 0 )$
    \\
                                                      &
    $1 - \qty( n_{0} + n_{\infty} )$ &
    $\mqty( 1 & 0 )$
    \\
                                   &
    $n_{0} - n_{\infty}$ &
    $\mqty( 0 & (-z)^{-2\, n_{0}})\, \hyp{-1}{1 - 2\, n_{\infty}}{1 - 2\, n_{0}}{z} )$
    \\
                                     &
    $- n_{0} + n_{\infty}$ &
    $\mqty( 0 & (1-z)^{-2\, n_{\infty} + 2\, n_{0} + 1}\, (-z)^{-2\, n_{0}} )$
    \\
    \midrule
    \multirow{4}{*}{$(0,\, 0,\, 1)$} &
    $n_{0} + n_{\infty}$    &
    $\mqty( 0 & (-z)^{-2\, n_{0}} )$
    \\
                                                      &
    $1 - \qty( n_{0} + n_{\infty} )$ &
    $\mqty( 0 & (1-z)^{2\, n_{\infty} + 2\, n_{0} - 1}\, (-z)^{-2\, n_{0}} )$
    \\
                                   &
    $n_{0} - n_{\infty}$ &
    $\mqty( 0 & (-z)^{-2\, n_{0}}) )$
    \\
                                     &
    $- n_{0} + n_{\infty}$ &
    $\mqty( 1 & 0 )$
    \\
    \midrule
    \multirow{4}{*}{$(0,\, 1,\, 1)$} &
    $n_{0} + n_{\infty}$    &
    $\mqty( (1-z)^{-2\, n_{\infty} -2\, n_{0}} & 0 )$
    \\
                                                      &
    $1 - \qty( n_{0} + n_{\infty} )$ &
    $\mqty( 0 & (1-z)^{2\, n_{\infty} + 2\, n_{0} - 2}\, (-z)^{-2\, n_{0}} )$
    \\
                                   &
    $n_{0} - n_{\infty}$ &
    $\mqty( 0 & (-z)^{-2\, n_{0}}) )$
    \\
                                     &
    $- n_{0} + n_{\infty}$ &
    $\mqty( 0 & (1-z)^{2\, n_{0} - 2\, n_{\infty}}\, (-z)^{-2\, n_{0}} )$
    \\
    \bottomrule
  \end{tabular}
  \caption{Abelian limit of the solutions}
  \label{tab:Left_Abelian_solutions}
\end{table}


\subsubsection{The \texorpdfstring{$\SU{2}_L$}{Left SU(2)} Limit}

We recover the non Abelian \SU{2} solution by considering $m_{\vec{t}}\sim 0$.
This is the first specific case shown in~\Cref{sec:true_basis}.
In this scenario the left solution $\cB^{(L)}$ is always the same and matches the previous computation, however the right sector seems to give different solutions when different Abelian limits are taken.
Studying all possible solutions we find that all of them give the same answer in the limit $m_{\vec{t}} \to 0$, i.e.\ both $\cB^{(R)} = \mqty(1 & 0)^T$ and $\cB^{(R)} = \mqty(0 & 1)^T$.\footnotemark{}
\footnotetext{%
  We write ``possible solutions'' because $m_{1} = 1 - \qty( m_{0} + m_{\infty} )$ is not.
}
The difference is the solution obtained from $n_{0} > m_{0}$, $n_{1} > m_{1}$ and $n_{\infty} > m_{\infty}$ or $n_{0} > m_{0}$, $\hat{n}_{1} < \hat{m}_{1}$ and $\hat{n}_{\infty} < \hat{m}_{\infty}$.
In any case the solution is factorised in the form $\cB^{(L)}(z) \otimes \mqty(C & C')^T$ which is expected since the right sector plays no role.


\subsubsection{Relating the Abelian Angles with the Group Parameters}

Using the explicit form of the \SO{4} and $\SU{2} \times \SU{2}$ matrices, we can verify that when the left and right $\SU{2}$ parameters are $\vec{n} = n^3\, \vec{k}$ and $\vec{m} = m^3\, \vec{k}$ the rotation of the D-branes in the plane $14$ is a \SO{2} element
\begin{equation}
  \mqty( \cos(\theta) & \sin(\theta) \\ -\sin(\theta)&  \cos(\theta) ),
  \qquad
  \theta = n^3 - m^3,
\end{equation}
while in plane $23$ the angle is $\theta = n^3 + m^3$.
Comparing with the case $m = 0$ given in~\eqref{eq:Abelian_vs_n_simple_case}, we then guess that the general relation between the group parameters and the usual Abelian angles is:
\begin{equation}
  \begin{split}
    \varepsilon_{\vec{t}}
    & =
    n_{\vec{t}}^3 - m_{\vec{t}}^3 + \theta( -(n^3_{\vec{t}} - m_{\vec{t}}^3) ),
    \\
    \varphi_{\vec{t}}
    & =
    n_{\vec{t}}^3 + m_{\vec{t}}^3 + \theta( -(n^3_{\vec{t}} + m_{\vec{t}}^3) ).
  \end{split}
  \label{eq:Abelian_vs_n_general_case}
\end{equation}


\subsubsection{Recovering the Abelian Result: an Example}

To show that the construction of the Abelian limit is indeed working, we consider the first case in~\Cref{sec:true_basis} with $n_{1} = 1 - \qty( n_{0} + n_{\infty} )$ and $m_{1} = -m_{0} + m_{\infty}$.
This leads to two independent rational functions of $\omega_z$:
\begin{equation}
  \begin{split}
    \ipd{\omega_z} \cX(\omega_z)
    & =
    \mqty( i \ipd{\omega_z} \bar{\cZ}^1(\omega_z) &
           \ipd{\omega_z} \cZ^{2}(\omega_z)
           \\
           \ipd{\omega_z} \bar{\cZ}^{2}(\omega_z) &
           i \ipd{\omega_z} \cZ^{1}(\omega_z)
    )
    \\
    & =
    \mqty( i \ipd{\omega_z} ( \cX^1(\omega_z) - i \cX^4(\omega_z) ) &
           \ipd{\omega_z} ( \cX^2(\omega_z) + i \cX^3(\omega_z) )
           \\
           \ipd{\omega_z} ( \cX^2(\omega_z) - i \cX^3(\omega_z) )   &
           i \ipd{\omega_z} ( \cX^1(\omega_z) + i \cX^4(\omega_z) )
    )
    \\
    & =
    \mqty( 0 &
           C_1\,
           (1-\omega_z)^{\varepsilon_{1}-1}\,
           (-\omega_z)^{\varepsilon_{0}-1}
           \\
           0 &
           C_2\,
           (1-\omega_z)^{-\varphi_{1}}\,
           (-\omega_z)^{-\varphi_{0}}
    ),
  \end{split}
  \label{eq:Abelian_sol_example}
\end{equation}
where $C_1$ and $C_2$ are constants as in~\eqref{eq:general_solution}.
This is the known result in the presence of Abelian rotations of the D-branes: we have two different \U{1} sectors undergoing two different rotations
$\U{1}_1 \times \U{1}_2 \subset \SU{2}_L \times \SU{2}_R$.
In particular we used~\eqref{eq:Abelian_vs_n_general_case} to write the relation between the usual Abelian angles and the group parameters as
\begin{equation}
  \varepsilon_{0} = n_{0} - m_{0},
  \qquad
  \varepsilon_{1} = n_{1} - m_{1},
  \qquad
  \varepsilon_{\infty} = n_{\infty} + m_{\infty}
\end{equation}
such that $\sum\limits_{t} \varepsilon_{\vec{t}} = 1$, and
\begin{equation}
  \varphi_{0} = n_{0} + m_{0},
  \qquad
  \varphi_{1} = n_{1} + m_{1},
  \qquad
  \varphi_{\infty} = n_{\infty} - m_{\infty},
  \label{eq:Abelian_rotation_second}
\end{equation}
where $\sum\limits_{t} \varphi_{\vec{t}} = 2$, in order to approach the usual notation in the literature.
As usual we have $\ipd{\omega_z} \cZ^1( \omega_z ) \neq \qty[ \ipd{\omega_z} \bcZ^1( \omega_z ) ]^*$.

We can now build the Abelian solution to show the analytical structure of the limit.
We have
\begin{equation}
  \mqty( i \barZ^1( u,\, \baru ) & Z^2( u,\, \baru )
         \\
         \barZ^2( u,\, \baru ) & i Z^1( u,\, \baru )
  )
  =
  \mqty( i \barf^1_{(\bart - 1)} + i \finiteint{\omega}{0}{\bomega_{\baru}}\, \ipd{\omega} \cZ^1
         &
         f^2_{(\bart -1)} + \finiteint{\omega}{0}{\omega_u}\, \ipd{\omega} \cZ^2
         \\
         \barf^2_{(\bart - 1)} + \finiteint{\omega}{0}{\bomega_{\baru}}\, \ipd{\omega} \cZ^2
         &
         i f^1_{(\bart-1)} + i \finiteint{\omega}{0}{\omega_u}\, \ipd{\omega_z} \cZ^1
  )
\end{equation}
where we chose $R_{(\bart)} = \1_4$ so that $U_{(\bart)}$ in~\eqref{eq:Umatrices} is mapped to $(i \sigma_1, i \sigma_1) \in \SU{2} \times \SU{2}$.
Notice however that $\vec{n}_{\vec{t}} = n_{\vec{t}}^3\, \vec{k}$ implies that $v^3_{(t)} = 0$ in~\eqref{eq:special_UL_brane_t}.
Hence $U_L$ and $U_R$ are always off diagonal and their action on~\eqref{eq:Abelian_sol_example} is to fill the first column.
From the previous relations we can also recover the usual holomorphicity $\barZ^1(\baru) = \qty[ Z^1(u) ]^*$ of the sector with $\sum\limits_t \varepsilon_{\vec{t}} = 1$ and $\barZ^2(\baru) = \qty[ Z^2(u) ]^*$ of the sector with $\sum\limits_t \varphi_{\vec{t}} = 2$.


\subsubsection{Abelian Limits}

From the example in the previous section it is possible to consider both cases given in~\Cref{sec:true_basis} and all possible combinations of the expression of $n_{1}$ and $m_{1}$ for a total of $2 \times 4 \times 4 = 32$ possible combinations.
In almost all cases (in fact all but six) the solution in spinorial formalism is a $2 \times 2$ matrix which has two non vanishing entries, hence two independent Abelian solutions.
In the remaining cases the matrix has only one non vanishing entry but the constraints on $n$ and $m$ are not compatible, thus they should not be considered.
In the first case encountered in~\Cref{sec:true_basis} the inconsistent combinations are $\{ n_{1} = n_{0} + n_{\infty},~ m_{1} = 1 - (m_{0} + m_{\infty}) \}$ and $\{ n_{1} = 1 - (n_{0} + n_{\infty}),~ m_{1} = 1 - (m_{0} + m_{\infty}) \}$.
In the second case in~\Cref{sec:true_basis} the incompatible constraints appear when $n_{1} = -n_{0} + n_{\infty}$.


\subsection{The Physical Interpretation}

In this section we show some consequences of the explicit classical solution for the phenomenology of models involving D-branes intersecting at angles.
In particular we focus on the value of the action which plays a fundamental role in the hierarchy of the Yukawa couplings.


\subsubsection{Rewriting the Action}

Using the classical solution previously computed, it is possible to compute the classical action of the bosonic string and show its contribution to the correlation functions of twist fields and Yukawa couplings.
We use the equations of motion~\eqref{eq:string_equation_of_motion} to simplify the action~\eqref{eq:string_action}.
We get:
\begin{equation}
  4 \pi \ap \eval{S_{\R^4}}_{\text{on-shell}}
  =
  i \finitesum{t}{1}{3}\,
  \sum\limits_{m \in \{3, 4\}}
  g_{(t)\, m}\,
  \finiteint{x}{x_{(t)}}{x_{(t-1)}}
  \tensor{\qty( R_{(t)} )}{_{mI}}
  \eval{\qty( X_L'(x) - X_R'(x) )^I}_{y=0^+},
  \label{eq:area_tmp}
\end{equation}
where indices $I = 1,\, 2,\, 3,\, 4$ are summed over and $m = 3,\, 4$ are the transverse directions in the well adapted frame with respect to the D-brane.
As the number of D-branes is defined modulo $N_B = 3$, $D_{(1)}$ is split on two separate intervals:
\begin{equation}
  \qty[ x_{(1)},\, x_{(3)} ]
  =
  \left[ x_{(1)},\, +\infty \right)
  \cup
  \left( -\infty,\, x_{(3)} \right],
\end{equation}
as it is visually shown in~\Cref{fig:finite_cuts}.
For $x_{(t)} < x < x_{(t-1)}$ we have:
\begin{equation}
  X(x+iy,\, x-iy)
  =
  X^*(x+iy,\, x-iy)
  \quad
  \Rightarrow
  \quad
  X_L^*(x-iy)
  =
  X_R(x-iy) + Y,
\end{equation}
where $Y \in \R$ is a constant factor which cannot depend on the particular D-brane $D_{(t)}$.
In fact the continuity of $X_L(u)$ and $X_R(\baru)$ on the worldsheet intersection point ensures that
\begin{equation}
  \lim\limits_{x \to x_{(t)}^+} X(x, x)
  =
  \lim\limits_{x \to x_{(t)}^-} X(x, x),
\end{equation}
which does not allow $Y$ to depend on the specific D-brane while the reality of $X(u,\baru)$ implies that $\Im Y = 0$.
Now~\eqref{eq:area_tmp} becomes:
\begin{equation}
  \begin{split}
    4 \pi \ap \eval{S_{\R^4}}_{\text{on-shell}}
    & =
    -2 \finitesum{t}{1}{3}\,
    \sum\limits_{m \in \{3, 4\}}
    \eval{%
      g_{(t)\, m}\,
      \tensor{\Im \qty( R_{(t)} )}{_{mI}}
      X_L^I(x+i0^+)
    }^{x = x_{(t-1)}}_{x = x_{(t)}}
    \\
    & =
    -2 \finitesum{t}{1}{3}\,
    g^{(\perp)}_{(t)\, I}\,
    \eval{%
      \Im X_L^I(x+i0^+)
    }^{x = x_{(t-1)}}_{x = x_{(t)}} \in \R,
  \end{split}
  \label{eq:action_with_imaginary_part}
\end{equation}
where $g^{(\perp)}_{(t)\, I} = \sum_{m \in \{3, 4\}} \tensor{\qty( R_{(t)}^{-1} )}{_{mI}}\, g_{(t)\, m}$ is the transverse shift of $D_{(t)}$ in the global coordinates of the target space:
\begin{equation}
  g^{(\perp)}_{(t)\, I}\, \qty(f_{(t-1)} - f_{(t)})^I = 0.
  \label{eq:g_perp_Delta_f}
\end{equation}


\subsubsection{Holomorphic Case}

In this case there are global complex coordinates for which the string solution is holomorphic:
\begin{equation}
  Z^i(u, \baru) = Z^i_L(u),
  \qquad
  \barZ^i(u, \baru) = \bar{Z}^i(\baru) = \qty( Z^i_L(u) )^*,
\end{equation}
where $i = 1$ in the Abelian case and $i=1,\, 2$ in the \SU{2} case.
We also have $f^i_{(t)} = Z^i_L(x_{(t)} + i\, 0^+)$.
Equations~\eqref{eq:g_perp_Delta_f} and~\eqref{eq:action_with_imaginary_part} then become
\begin{eqnarray}
  \Re( g_{(t)\, i}^{(\perp)}\, \qty( f_{(t-1)} - f_{(t)} )^i )
  & = &
  0,
  \\
  4 \pi \ap \eval{S_{\R^4}}_{\text{on-shell}}
  & = &
  -2 \finitesum{t}{1}{3}\,
  \Im( g_{(t)\, i}^{(\perp)}\, \qty( f_{(t-1)} - f_{(t)} )^i ),
\end{eqnarray}
where the last equation shows that the action can be expressed using just the global data.

In the Abelian scenario we can further simplify the action and give a clear geometrical meaning.
Given to complex numbers $a, b \in \C$ such that $\Re(a^* b)=0$ then $\Im(a^* b) = \pm \abs{a} \abs{b}$.
This can be seen either by direct computation or by using a \U{1} rotation to set $b$ equal to $\abs{b}$.
Since the action is positive then we can write
\begin{equation}
  \eval{S_{\R^4}}_{\text{on-shell}}
  =
  \frac{1}{2 \pi \ap}
  \finitesum{t}{1}{3}\,
  \qty(
    \frac{1}{2} \abs{g^{(\perp)}_{(t)}} \,
    \abs{f_{(t-1)} - f_{(t)}}
  ),
\end{equation}
where a factor $\frac{1}{2}$ comes from raising the complex index in $g^{(\perp)}_{(t)\, 1}$.
The right hand side of the previous expression is the sum of the areas of the triangles having the interval between two intersection points on a given D-brane $D_{(t)}$ as base and the distance between the D-brane and the origin as height.
A visual reference can be found in~\Cref{fig:branes_at_angles}.

For the \SU{2} case we can use a rotation to map $(f_{(t-1)} - f_{(t)})^i$ to the form $\norm{f_{(t-1)} - f_{(t)}} \delta^i_1$.
Each term of the action can be interpreted again as an area of a triangle where the distance between the interaction points is the base.


\subsubsection{Generalisation and Summary}

\begin{figure}[tbp]
  \centering
  \def\svgwidth{0.35\textwidth}
  \import{img/}{brane3d.pdf_tex}
  \caption{%
    Pictorial $3$-dimensional representation of two D2-branes intersecting in the Euclidean space $\R^3$ along a line (in $\R^4$ the intersection is a point since the co-dimension of each D-brane is 2): since it is no longer constrained on a bi-dimensional plane, the string must be deformed in order to stretch between two consecutive D-branes.
    Its action can be larger than the planar area.
  }
  \label{fig:brane3d}
\end{figure}

In the general case there does not seem to be any possible way of computing the action~\eqref{eq:action_with_imaginary_part} in term of the global data.
Most probably the value of the action is larger than in the holomorphic case since the string is no longer confined to a plane.
Given the nature of the rotation its worldsheet has to bend in order to be attached to the D-brane as pictorially shown in~\Cref{fig:brane3d} in the case of a $3$-dimensional space.
The general case we considered then differs from the known factorized case by an additional contribution in the on-shell action which can be intuitively understood as a small ``bump'' of the string worldsheet in proximity of the boundary.

We thus showed that the specific geometry of the intersecting D-branes leads to different results when computing the value of the classical action, that is the leading contribution to the Yukawa couplings in string theory.
In particular in the Abelian case the value of the action is exactly the area formed by the intersecting D-branes in the $\R^2$ plane, i.e.\ the string worldsheet is completely contained in the polygon on the plane.
The difference between the \SO{4} case and \SU{2} is more subtle as in the latter there are complex coordinates in $\R^4$ for which the classical string solution is holomorphic in the upper half plane.
In the generic case presented so far this is in general no longer true.
The reason can probably be traced back to supersymmetry, even though we only dealt with the bosonic string.
In fact when considering \SU{2} rotated D-branes part of the spacetime supersymmetry is preserved, while this is not the case for \SO{4} rotations.

% vim: ft=tex
