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Progress on fermions with defects

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Signed-off-by: Riccardo Finotello <riccardo.finotello@gmail.com>
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Riccardo
2020-09-29 18:48:42 +02:00
parent 28c749e568
commit c81838154a
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6
sec/app/parameters.tex
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@@ -76,7 +76,7 @@ This would then imply
We can finally show in details the computation of the parameters of the basis of hypergeometric functions used in the main text.
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.
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.
The monodromy in $\omega_{\bart-1} = 0$ is simpler to compute given that we choose $\cL(\vb{n}_{\vb{0}})$ and $\cR(\widetilde{\vb{m}}_{\vb{0}})$ to be diagonal.
We impose:
\begin{eqnarray}
\mqty( \dmat{1, e^{-2\pi i c^{(L)}}} )
@@ -163,7 +163,7 @@ We therefore have:
k_{a b}\in \Z.
\label{eq:aL-bL}
\end{equation}
The allowed values for $k_{\delta^{(L)}_{\vb{\infty}}}$ follow a construction similar to the monodromy around $\omega_{\bt-1} = 0$.
The allowed values for $k_{\delta^{(L)}_{\vb{\infty}}}$ follow a construction similar to the monodromy around $\omega_{\bart-1} = 0$.
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 $1$.
As in the previous case we have $\alpha \le \delta_{\vb{\infty}}^{(L)} \le \alpha + \frac{1}{2}$ with $\alpha$ technically arbitrary.
We cannot thus choose a vanishing $k_{\delta^{(L)}_{\vb{\infty}}}$ but we have to consider $k_{\delta^{(L)}_{\infty}} = 0,\, 1$.
@@ -205,7 +205,7 @@ where
\frac{n_{\vb{\infty}}^3}{n_{\vb{\infty}}}.
\label{eq:cos_n1}
\end{equation}
This expression is connected with rotation parameter in the third interaction point $\omega_{\bt+1} = 1$.
This expression is connected with rotation parameter in the third interaction point $\omega_{\bart+1} = 1$.
In fact $\cos(2\pi \cA^{(L)}) = \cos(2\pi {n}_{\vb{1}})$.
We then write
\begin{equation}