Add appendix for branes at angles
Signed-off-by: Riccardo Finotello <riccardo.finotello@gmail.com>
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351
sec/app/parameters.tex
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351
sec/app/parameters.tex
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In this appendix we show the computation of the parameters of the hypergeometric functions and their relation with the rotation parameters.
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\subsection{Consistency Conditions of the Monodromy Matrices}
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In the main text we set
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\begin{equation}
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D~
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\rM_{\vb{\infty}}~
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D^{-1}
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=
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e^{-2\pi i \delta_{\vb{\infty}}}\,
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\cL(\vb{n}_{\vb{\infty}}),
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\end{equation}
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where $\cL(\vb{n}_{\vb{\infty}}) \in \SU{2}$.
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The previous equation implies
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\begin{equation}
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\left( D\, \rM_{\vb{\infty}}\, D^{-1} \right)^\dagger
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=
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\left( D\, \rM_{\vb{\infty}}\, D^{-1} \right)^{-1},
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\end{equation}
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which can be rewritten as
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\begin{equation}
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\widetilde{\rM}_{\vb{\infty}}^{-1}~
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\cC^{\dagger}\, D^{\dagger}\, D\, \cC
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=
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\cC^{\dagger}\, D^{\dagger}\, D\, \cC~
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\widetilde{\rM}_{\vb{\infty}}^{-1}.
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\end{equation}
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As $\widetilde{\rM}_{\vb{\infty}}$ is a generic diagonal matrix, the previous equation implies that the off-diagonal elements of $\cC^{\dagger}\, D^{\dagger}\, D\, \cC$ must vanish.
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We therefore have
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\begin{equation}
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\begin{split}
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\abs{K}^{-2}
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& =
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-\frac{\cC_{21}\, \cC^*_{22}}{\cC_{11}\, \cC^*_{12}}
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\\
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& =
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-\frac{1}{\pi^4}\,
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\abs{\gfun{a} \gfun{b} \gfun{c-a} \gfun{c-b}}^2 \times
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\\
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& \times
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\sin(\pi a)\, \sin^*(\pi (c-a))\, (\sin(\pi b)\, \sin^*(\pi (c-b)))^*.
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\end{split}
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\end{equation}
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When $a,\, b,\, c \in \R$ this ultimately means that
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\begin{equation}
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\sin(\pi a)\, \sin(\pi (c-a))\, \sin(\pi b)\, \sin(\pi (c-b)) < 0.
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\label{eq:constraint_from_K^2}
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\end{equation}
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Since the previous equation is invariant under integer shift of any of its parameters, we can consider just the fractional parts $0 \le \{a\},\, \{b\},\, \{c\} < 1$.
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In order to have \U{2} monodromies finally requires
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\begin{equation}
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0 \le \{b\} < \{c\} < \{a\} < 1
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\qq{or}
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0 \le \{a\} < \{c\} < \{b\} <1.
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\label{eq:K_consistency_condition}
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\end{equation}
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Should we request \U{1,1} monodromies as in moving rotated branes then we get:
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\begin{equation}
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\abs{K}^{-2}
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=
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\frac{\cC_{21}\, \cC^*_{22}}{\cC_{11}\, \cC^*_{12}}.
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\end{equation}
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This would then imply
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\begin{equation}
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0 \le \{c\} < \{a\},\, \{b\} < 1
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\qq{or}
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0 \le \{a\},\, \{b\} < \{c\} < 1.
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\end{equation}
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\subsection{Fixing the Parameters}
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We can finally show in details the computation of the parameters of the basis of hypergeometric functions used in the main text.
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The relation between these and the \SU{2} matrices can be computed requiring that the monodromies induced by the choice of the parameters equal the monodromies produced by the rotations of the D-branes.
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The monodromy in $\omega_{\bt-1} = 0$ is simpler to compute given that we choose $\cL(\vb{n}_{\vb{0}})$ and $\cR(\widetilde{\vb{m}}_{\vb{0}})$ to be diagonal.
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We impose:
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\begin{eqnarray}
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\mqty( \dmat{1, e^{-2\pi i c^{(L)}}} )
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& = &
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e^{-2\pi i \delta_{\vb{0}}^{(L)}}\,
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\mqty( \dmat{e^{2\pi i n_{\vb{0}}}, e^{-2\pi i n_{\vb{0}}}} ),
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\\
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\mqty( \dmat{1, e^{-2\pi i c^{(R)}}} )
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& = &
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e^{-2\pi i \delta_{\vb{0}}^{(R)}}\,
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\mqty( \dmat{e^{-2\pi i m_{\vb{0}}}, e^{2\pi i m_{\vb{0}}}} ),
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\end{eqnarray}
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where $n^3_{\vb{0}} = \norm{\vb{n}_{\vb{0}}} = n_{\vb{0}}$ and $m^3_{\vb{0}} = \norm{\vb{m}_{\vb{0}}} = m_{\vb{0}}$ with $0 \le n_{\vb{0}},\, m_{\vb{0}} < 1$ due to the conventions \eqref{eq:maximal_torus_left} and \eqref{eq:maximal_torus_right}.
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We thus have:
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\begin{equation}
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\begin{split}
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\delta_{\vb{0}}^{(L)}
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& =
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n_{\vb{0}} + k_{\delta^{(L)}_{\vb{0}}},
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\qquad
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k_{\delta^{(L)}_{\vb{0}}} \in \Z,
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\\
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c^{(L)}
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& =
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2 n_{\vb{0}} + k_c,
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\qquad
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k_c \in \Z.
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\end{split}
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\label{eq:cL}
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\end{equation}
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Since the determinant of the right hand side is $e^{-4 \pi i \delta_{\vb{0}}^{(L)}}$, the range of definition of $\delta_{\vb{0}}^{(L)}$ is $\alpha \le \delta_{\vb{0}}^{(L)} \le \alpha + \frac{1}{2}$.
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Given that $0 \le n_{\vb{0}} < \frac{1}{2}$ we simply take $\alpha = 0$ and set $\delta_{\vb{0}}^{(L)} = n_{\vb{0}}$.
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Analogous results hold in the right sector.
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Furthermore from the third equation in \eqref{eq:parameters_equality_zero} and from the first equation in \eqref{eq:cL} we can restrict:
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\begin{equation}
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n_{\vb{0}} + m_{\vb{0}} - A \in \Z.
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\end{equation}
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We then need to find $3$ equations to determine $a^{(L)}$, $b^{(L)}$ and $\delta^{(L)}_{\vb{\infty}}$.
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After that we then fix the remaining factors in $B$ and $\abs{K^{(L)}}$.
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The equations follow from~\eqref{eq:parameters_equality_infty}.
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The first two equations for $a^{(L)}$, $b^{(L)}$ and $\delta^{(L)}_{\vb{\infty}}$ follow by considering the trace of~\eqref{eq:parameters_equality_infty}:
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\begin{equation}
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e^{\pi i ( a^{(L)} + b^{(L)} )} \cos(\pi( a^{(L)} - b^{(L)} ) )
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=
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e^{-2\pi i \delta^{(L)}_{\infty}} \cos(2\pi n_{\vb{\infty}}),
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\end{equation}
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which is satisfied by:
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\begin{equation}
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\begin{split}
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\delta^{(L)}_{\vb{\infty}}
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& =
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-
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\frac{1}{2}(a^{(L)} + b^{(L)})
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+
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\frac{1}{2} k_{\delta^{(L)}_{\vb{\infty}}},
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\qquad
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k_{\delta_{\vb{\infty}}} \in \Z,
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\\
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a^{(L)} - b^{(L)}
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& =
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2\, (-1)^{p^{(L)}}\, n_{\vb{\infty}}
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+
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(-1)^{q^{(L)}}\, k_{\delta^{(L)}_{\vb{\infty}}}
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+
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2\, k'_{a b},
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\qquad
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k'_{ab} \in \Z,
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\end{split}
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\end{equation}
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where $p^{(L)},\, q^{(L)} \in \left\lbrace 0, 1 \right\rbrace$.
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Notice that changing the value of $p^{(L)}$ corresponds to swapping $a$ and $b$: since the hypergeometric function is symmetric in those parameters we can fix $p^{(L)}=0$.
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Redefining $k'$ we can always set $q^{(L)}=0$.
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We therefore have:
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\begin{equation}
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a^{(L)} - b^{(L)}
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=
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2\, n_{\vb{\infty}}
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+
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k_{\delta^{(L)}_{\vb{\infty}}}
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+
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2 k_{ab},
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\qquad
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k_{a b}\in \Z.
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\label{eq:aL-bL}
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\end{equation}
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The allowed values for $k_{\delta^{(L)}_{\vb{\infty}}}$ follow a construction similar to the monodromy around $\omega_{\bt-1} = 0$.
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The main difference is given by the fact that $\frac{1}{2}(a^{(L)} + b^{(L)})$ may a priori take values in an interval of width $1$.
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As in the previous case we have $\alpha \le \delta_{\vb{\infty}}^{(L)} \le \alpha + \frac{1}{2}$ with $\alpha$ technically arbitrary.
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We cannot thus choose a vanishing $k_{\delta^{(L)}_{\vb{\infty}}}$ but we have to consider $k_{\delta^{(L)}_{\infty}} = 0,\, 1$.
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We find a third relation by considering the entry
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\begin{equation}
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\Im\left(
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e^{+2\pi i \delta_{\vb{\infty}}^{(L)}}\,
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D^{(L)}\,
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\rM_{\vb{\infty}}^{(L)}\,
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\left( D^{(L)} \right)^{-1}
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\right)_{11}
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=
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\Im\left(
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\cL(n_{\vb{\infty}})
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\right)_{11}.
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\end{equation}
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Using
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\begin{equation}
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\det \cC
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=
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\frac{\sin(\pi c^{(L)})}{\sin(\pi(a^{(L)}-b^{(L)}))},
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\end{equation}
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and the second equation in~\eqref{eq:cL} and~\eqref{eq:aL-bL} leads to:
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\begin{equation}
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\cos(\pi( a^{(L)} + b^{(L)} - c^{(L)} ))
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=
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(-1)^{k_c+k_{\delta^{(L)}_{\vb{\infty}}} }\, \cos(2\pi \cA^{(L)}),
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\end{equation}
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where
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\begin{equation}
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\cos(2\pi \cA^{(L)})
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=
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\cos(2\pi n_{\vb{0}})\,
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\cos(2\pi n_{\vb{\infty}})
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-
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\sin(2\pi n_{\vb{0}})\,
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\sin(2\pi n_{\vb{\infty}})\,
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\frac{n_{\vb{\infty}}^3}{n_{\vb{\infty}}}.
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\label{eq:cos_n1}
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\end{equation}
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This expression is connected with rotation parameter in the third interaction point $\omega_{\bt+1} = 1$.
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In fact $\cos(2\pi \cA^{(L)}) = \cos(2\pi {n}_{\vb{1}})$.
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We then write
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\begin{equation}
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a^{(L)} + b^{(L)} - c^{(L)}
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=
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2\, (-1)^{f^{(L)}}\, n_{\vb{1}}
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+
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k_c
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+
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k_{\delta^{(L)}_{\vb{\infty}}}
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+
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2\, k_{abc},
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\qquad
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k_{abc}\in \Z,
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\end{equation}
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with $f^{(L)} \in \left\lbrace 0, 1 \right\rbrace$.
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The request
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\begin{equation}
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A
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+
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B
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-
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n_{\vb{0}}
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-
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m_{\vb{0}}
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-
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(-1)^{f^{(L)}}\, n_{\vb{1}}
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-
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(-1)^{f^{(R)}}\, m_{\vb{1}}
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\in \Z
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\end{equation}
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finally fixes the $B$ parameter in the third equation of~\eqref{eq:parameters_equality_infty}.
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So far we can summarise the results in
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\begin{eqnarray}
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a
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=
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n_{\vb{0}} + (-1)^{f^{(L)}} n_{\vb{1}} + n_{\vb{\infty}} + m_a,
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& \qquad &
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m_a \in \Z,
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\\
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b
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=
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n_{\vb{0}} + (-1)^{f^{(L)}} n_{\vb{1}} - n_{\vb{\infty}} + m_b,
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& \qquad &
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m_b \in \Z,
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\\
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c
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=
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2\, n_{\vb{0}} + m_c,
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& \qquad &
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m_c \in \Z,
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\\
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\delta_{\vb{0}}^{(L)}
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=
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n_{\vb{0}},
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\\
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\delta_{\vb{\infty}}^{(L)}
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=
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- n_{\vb{0}} - (-1)^{f^{(L)}} n_{\vb{1}} + m_c + 2\, m_\delta,
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& \qquad &
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m_{\delta} \in \Z,
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\\
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A
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=
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n_{\vb{0}} + m_{\vb{0}} + m_A,
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& \qquad &
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m_A \in \Z,
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\\
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B
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=
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(-1)^{f^{(L)}}\, n_{\vb{1}} + (-1)^{f^{(R)}}\, m_{\vb{1}} + m_B,
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& \qquad &
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m_B \in \Z.
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\end{eqnarray}
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$K^{(L)}$ is finally determined from
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\begin{equation}
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\left( D^{(L)}\, \rM_{\vb{\infty}}\, \left( D^{(L)} \right)^{-1} \right)_{21}
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=
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e^{-2\pi i \delta_{\vb{\infty}}^{(L)}}\,
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\left( \cL(n_{\vb{\infty}}) \right)_{21},
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\label{eq:fixing_K_21}
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\end{equation}
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and get:
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\begin{equation}
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K^{(L)}
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=
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-\frac{(-1)^{m_a + m_b + m_c}}{2 \pi^2}\,
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\cG( a^{(L)},\, b^{(L)},\, c^{(L)} )\,
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\sin(2 \pi n_{\vb{0}})
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\sin(2 \pi n_{\vb{\infty}})
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\frac{n^1_{\vb{\infty}} + i\, n^2_{\vb{\infty}}}{n_{\vb{\infty}}},
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\label{eq:app_B_K21}
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\end{equation}
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where $\cG( a,\, b,\, c ) = \gfun{1-a}\, \gfun{1-b}\, \gfun{a+1-c}\, \gfun{b+1-c}$.
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\subsection{Checking the Consistency of the Solution}
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We check the consistency condition \eqref{eq:K_consistency_condition} using~\eqref{eq:product_in_SU2}.
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The result is
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\begin{equation}
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\begin{split}
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\left( K^{(L)} \right)^{-1}
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& =
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\frac{(-1)^{m_a + m_b + m_c}}{2 \pi^2}\,
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\cG(1 - a^{(L)},\, 1 - b^{(L)},\, 2 - c^{(L)})\,
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\\
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& \times
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\sin(2 \pi n_{\vb{0}})\,
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\sin(2 \pi n_{\vb{\infty}})\,
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\frac{n^1_{\vb{\infty}} -i n^2_{\vb{\infty}}}{n_{\vb{\infty}}},
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\end{split}
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\label{eq:app_B_K12}
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\end{equation}
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where the function $\cG( a,\, b,\, c )$ was defined at the end of the previous section.
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Compatibility with~\eqref{eq:app_B_K21} requires
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\begin{equation}
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\frac{(n^1_{\vb{\infty}})^2 + (n^2_{\vb{\infty}})^2}{n^2_{\vb{\infty}}}
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=
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-4 \frac{\sin(\pi a) \sin(\pi(c-a))\sin(\pi b) \sin(\pi(c-b))}
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{\sin^2(\pi c) \sin^2(\pi(a-b))}.
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\label{eq:n12+n22}
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\end{equation}
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We can then rewrite~\eqref{eq:cos_n1} as
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\begin{equation}
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\frac{(n^3_{\vb{\infty}})^2}{n^2_{\vb{\infty}}}
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=
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\frac{(\cos(\pi (a-b)) \cos(\pi c)- \cos(\pi(a+b-c)))^2}
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{\sin^2(\pi c) \sin^2(\pi(a-b))}.
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\end{equation}
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It is then possible to verify that the sum of the left and right hand sides of~\eqref{eq:n12+n22} and the last equation are equal to $1$.
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The same consistency check can also be performed by computing $K^{(L)}$ from
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\begin{equation}
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\left( D^{(L)}\, \rM_{\vb{\infty}}\, \left( D^{(L)} \right)^{-1} \right)_{12}
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=
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e^{-2\pi i \delta_{\vb{\infty}}^{(L)}}\,
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\left( \cL(n_{\vb{\infty}}) \right)_{12},
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\end{equation}
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instead of \eqref{eq:fixing_K_21}.
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% vim: ft=tex
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