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\urldef\footurlb\url{cocalc.com/dfriedan/DM/SM}
\urldef\footurla\url{physics.rutgers.edu/~friedan}
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%
\setcounter{tocdepth}{2}
%
\begin{document}
%
\def\title{Calculations for\\ {``Thermodynamic stability of a 
cosmological\\[0.5ex]
SU(2)-weak gauge field''}}
%
\begin{center}
{\LARGE \title}
\vskip4ex
{\large Daniel Friedan}
\vskip2ex
{\it
New High Energy Theory Center
and Department of Physics and Astronomy,\\
Rutgers, The State University of New Jersey,\\
Piscataway, New Jersey 08854-8019 U.S.A. and
\vskip1ex
Science Institute, The University of Iceland,
Reykjavik, Iceland
\vskip1ex
\href{mailto:dfriedan@gmail.com}{dfriedan@gmail.com}
%\href{mailto:dfriedan@gmail.com}{\tt dfriedan@gmail.com}
\qquad
\href{https://physics.rutgers.edu/\textasciitilde friedan/}
%{physics.rutgers.edu/\texttt{\char`\~}friedan}
{physics.rutgers.edu/\texttilde friedan}
%{\tt \path{physics.rutgers.edu/~friedan/}}
}
\vskip2ex
\today
\end{center}
%
%
%	ABSTRACT
%
\begin{center}
\vskip3ex
{\sc Abstract}
\vskip2.5ex
%\parbox{0.9\linewidth}{
\parbox{0.96\linewidth}{
\hspace*{1.5em}
% arXiv.org limit is 1920 characters
This note shows calculations
for the paper {\it Thermodynamic stability of a cosmological SU(2)-weak gauge field}.
The section numbers and headings match those of the paper.
The numerical calculations and some algebra calculations 
are performed in two
SageMath notebooks.
The Supplemental Materials --- this note and the SageMath notebooks 
along with printouts as html and pdf
--- are available at \href{https://arXiv.org}{\tt arXiv.org}
as ancillary files associated with the paper, at
\href{https://cocalc.com/dfriedan/DM/SM}{\tt 
cocalc.com/dfriedan/DM/SM},
and at \href{https://physics.rutgers.edu/~friedan}{\tt physics.rutgers.edu/\texttilde friedan}.
}
\end{center}

\vspace*{-2ex}


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%	TABLE OF CONTENTS
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%


\setcounter{section}{1}

\section{Spin(4)-symmetric CGF}

$S^{3}$ is the unit 3-sphere in $\Reals^{4}$ with metric
$\hat g_{ij}(\hat x)$.
\eq
\hat x \in  \Reals^{4}
\qquad
\delta_{\mu\nu} \hat x^{\mu} \hat x^{\nu} = 1
\qquad
(\hat x^{\mu}) = (\hat x^{i},\hat x^{4}) 
\qquad
\hat g_{ij}(\hat x)d\hat x^{i}d\hat x^{j} = \delta_{\mu\nu} d\hat x^{\mu} d\hat x^{\nu}
\en
$S^{3}$ is identified  with $\SU(2)$ by
\eq
\hat x \in S^{3}
\:\:\:\longleftrightarrow\:\:\:
g_{\hat x} = \hat x^{4} \mathbf1 + \hat x^{k}  i^{-1} \sigma_{k} \;\in\SU(2)
\en
$\sigma_{k}$ are the Pauli matrices,
\eq
\sigma_{i}\sigma_{j} = \delta_{ij} + i \epsilon_{ij}{}^{k}\sigma_{k}
\en
$\Spin(4)$ is $\SU(2)_{L}{\ntimes} \SU(2)_{R}$.
It acts on $S^{3}$ as $\SO(4)$.
\eq
U=(g_{L},g_{R})
\qquad
\hat x \mapsto U \hat x
\qquad
g_{U\hat x} = g_{L} \,g_{\hat x}\, g_{R}^{-1}
\en
The metric is 
\eq
\hat g_{ij}(\hat x) = \frac12 \tr \,\left[\hat \partial_{i} g_{\hat x} \hat \partial_{j}  
(g_{\hat x}^{-1}) \right]
\qquad
\hat \partial_{i} = \frac{\partial}{\partial \hat x^{i}}
\en
The Dirac matrices at $\hat x$ are the components of an $\su(2)$-valued 1-form
\eq
\hat \gamma_{i}(\hat x)
=  \frac12 g_{\hat x} \hat \partial_{i} (g_{\hat x}^{-1})
\en
At the north pole,
\eq
\hat N = (0,0,0,1)
\qquad
\hat g_{ij}(\hat N) = \delta_{ij}
\qquad
\hat \gamma_{i}(\hat N) = \frac{i}{2}\sigma_{i}
\en
so everywhere on $S^{3}$
\eq
\hat \gamma_{i}(\hat x)\hat \gamma_{j}(\hat x)
=  -\frac14 \hat g_{ij}(\hat x) -\frac12 \hat\epsilon_{ij}{}^{k}(\hat x) 
\hat \gamma_{k}(\hat x)
\en
where $\hat\epsilon_{ijk}(\hat x)$ is the volume 3-form.
Also at $\hat N$,
\eqg
\hat \partial_{[i} \hat \gamma_{j]} =  \frac12 \hat \partial_{[i}g_{\hat x} \hat 
\partial_{j]} (g_{\hat x}^{-1})
 =  \frac12 i^{-1} \sigma_{[i} i \sigma_{j]}
 = i  \epsilon_{ij}{}^{k}\sigma_{k}
 = 2 \epsilon_{ij}{}^{k} \hat \gamma_{k}
 = -2 [\hat \gamma_{i},\, \hat \gamma_{j}]
 \\[1ex]
\hat \partial_{i} \hat \gamma_{j} + [\hat \gamma_{i},\, \hat \gamma_{j}]
- i\leftrightarrow j = 0
\eng
so the $\Spin(4)$-symmetric covariant derivative acting on spinors is
\eqg
\hat \nabla_{i} = \hat \partial_{i} + \hat \gamma_{i}
\qquad
\hat \nabla_{[i} \hat \gamma_{j]} = 0
\eng
with curvature
\eq
[\hat \nabla_{i},\,\hat \nabla_{j}] = \hat \partial_{[i} \hat 
\gamma_{j]} + [\hat \gamma_{i},\,\hat \gamma_{j}]
=\hat \epsilon_{ij}{}^{k} \hat \gamma_{k}
\en
Extending $\hat \nabla_{i}$ to be the metric covariant derivative on 
tensors as well,
\eq
\hat \nabla_{i} \hat \gamma_{j} = 0
\en
because $ \hat \nabla_{[i} \hat \gamma_{j]} = 0$ and $\hat \nabla_{i} \hat \gamma_{j} + \hat \nabla_{j} \hat 
\gamma_{i}$ would have to equal $A_{ij}^{k}\hat \gamma_{k}$ for 
$A_{ij}^{k}$ a 3-tensor symmetric in $i,j$.  There is no such 
$\SO(4)$-symmetric 3-tensor.


\setcounter{section}{5}

\section{Quadratic term in the action}

\subsection{Expand the action}

The $\Spin(4)$-symmetric classical gauge field has covariant derivative:
\eqg
D^{\cl}_{0}=\partial_{\that}\quad
D^{\cl}_{k} = \hat\nabla_{k} +\hat b(\that) \hat\gamma_{k}
\eng
The curvature is
\eqa
F^{\cl}_{0j}&amp; = [D^{\cl}_{0},\,D^{\cl}_{j}] 
= \partial_{\that}\hat b \,\hat \gamma_{j}
\\[2ex]
F^{\cl}_{ij} &amp;= [D^{\cl}_{i},\,D^{\cl}_{j}] 
=  [\hat\nabla_{i},\, \hat\nabla_{j} ] + \hat b^{2} 
[\hat\gamma_{i},\, \hat\gamma_{j} ] 
= (1-\hat b^{2})\hat\epsilon_{ij}{}^{k} \hat\gamma_{k}
\ena
The perturbed gauge field has covariant derivative
\eq
D_{0} = D^{\cl}_{0} = \partial_{\that}
\qquad
D_{i} = D^{\cl}_{i}+B_{i}
\en
and curvature
\eqa
F_{0j}&amp; = [D_{0},\,D_{j}] 
= F^{\cl}_{0j} + \partial_{\that}B_{j}
\\[1ex]
F_{ij} &amp;= [D_{i},\,D_{j}] 
=   F^{\cl}_{ij} + D^{\cl}_{[i}B_{j]} +  [B_{i},\,B_{j}]
\\[1ex]
D^{\cl}_{i}B_{j} &amp;= \hat\nabla_{i}B_{j} + \hat b  
[\hat\gamma_{i},\,B_{j}]
\ena
The Yang-Mills action is
\eq
\frac1\hbar S_{\gauge}
=\frac1{2 g ^{2}}\int  
\tr(- F_{\mu\nu} F^{\mu\nu})  \, \sqrt{-\hat g} \,d^{3}\hat x\, d\that
\en
\eqa
\tr( - F_{\mu\nu} F^{\mu\nu})
&amp;= \tr\left( 2F_{0j}F_{0}{}^{j} - F_{ij}F^{ij}\right)
\ena
The zeroth order term in $B$ is
\eqa
\tr(- F_{\mu\nu} F^{\mu\nu})_{0} 
&amp;= 
\tr\left( 2 (\partial_{\that}\hat b)^{2} \hat \gamma_{j}\hat 
\gamma^{j}
- (1-\hat b^{2})^{2}\hat\epsilon_{ij}{}^{k}\hat\epsilon^{ijk'} 
\gamma_{k}\gamma_{k'}\right)
\\[1ex]
&amp;= 3 \left[-(\partial_{\that}\hat b)^{2} +(1-\hat b^{2})^{2}\right]
\ena
The first order term is
\eqa
\tr( - F_{\mu\nu} F^{\mu\nu})_{1}
&amp;= \tr\left( 4F^{\cl}_{0j}\partial_{\that}B^{j} - 4F^{\cl}_{ij}
D^{\cl i}B^{j}\right)
\\[1ex]
&amp;= \tr\left( 4\partial_{\that}\hat b \,\hat \gamma_{j}
\partial_{\that}B^{j} - 4(1-\hat b^{2})\hat\epsilon^{ijk} \hat\gamma_{k}
( \hat\nabla_{i}B_{j} + \hat b [\hat\gamma_{i},\,B_{j}] )\right)
\\
&amp;= 
\partial_{\that}\left( 4\partial_{\that}\hat b 
\,\tr( \hat  \gamma_{j}B^{j}) \right)
-4 \left( \partial^{2}_{\that}\hat b 
+2 (\hat b^{2}-1) \hat b \right)\tr( \hat  \gamma_{j}B^{j}) 
\ena
The second order term is
\eqa
\tr( - F_{\mu\nu} F^{\mu\nu})_{2}
&amp;= \tr( - 2 \partial_{\that}B_{j} \partial_{\that}B^{j}
-D^{\cl}_{[i}B_{j]}D^{\cl\,[i}B^{j]}
-2 F^{\cl}_{ij} [B^{i},\,B^{j}])
\ena
The zeroth order term in the action is
\eq
\frac1\hbar S_{0}
=
\Vol(S^{3})\frac{3  }{g^{2}}
\int
\bigg[-\frac12 \bigg(\frac{d\hat b}{d\that}\bigg)^{2}+ \frac12 
(\hat b^{2}-1)^{2}\bigg]
d\that
\qquad
\Vol(S^{3}) = 2\pi^{2}
\en
The first order term is
\eqg
\frac1\hbar S_{1}
=
\frac{2  }{g^{2}}
\int
\bigg[
\partial_{\that}\left( \partial_{\that}\hat b 
\,B_{0}\right)
- \left( \partial^{2}_{\that}\hat b 
+2 (\hat b^{2}-1) \hat b \right)B_{0}
\bigg]
d\that
\\[1ex]
B_{0}(\that) =
\int
\tr( \hat  \gamma_{j}B^{j})
\, \sqrt{-\hat g} \,d^{3}\hat x
\eng
The second order term in the action is
\eqa
\frac1\hbar S_{2}
&amp;=
\frac1{2 g ^{2}}\int  
 \tr(  2 \partial_{\that}B_{j} \partial_{\that}B^{j}
-D^{\cl}_{[i}B_{j]}D^{\cl\,[i}B^{j]}
-2 F^{\cl}_{ij} [B^{i},\,B^{j}])
\, \sqrt{-\hat g} \,d^{3}\hat x\, d\that
\\[1ex]
&amp;= \frac1{g ^{2}}\int  
 \tr(  \partial_{\that}B_{j} \partial_{\that}B^{j}
- B_{i} \Box B^{i})
\, \sqrt{-\hat g} \,d^{3}\hat x\, d\that
\ena
\eq
\int   \tr(  B_{i} \Box B^{i}) \, \sqrt{-\hat g} \,d^{3}\hat x
= \int   \tr( D^{\cl}_{[i}B_{j]}D^{\cl\,i}B^{j}
+ F^{\cl}_{ij} [B^{i},\,B^{j}])
 ) \, \sqrt{-\hat g} \,d^{3}\hat x
\en
Changing variables
\eq
\that = \epsilon z
\qquad
\hat b(\that) = \frac1\epsilon b(z)
\en
the second order term becomes
\eq
\label{eq:quadratic1}
\frac1\hbar S_{2}
= 
\frac1{\epsilon g ^{2}}\int  
 \tr(  \partial_{z}B_{i} \partial_{z}B^{i}
- B_{i} \bK  B^{i})
\, \sqrt{-\hat g} \,d^{3}\hat x\, dz
\en
\eq
\bK = \epsilon^{2} \Box
\en
\eq
\int   \tr(  B_{i} \bK B^{i}) \, \sqrt{-\hat g} \,d^{3}\hat x
= \int   \tr( \epsilon^{2} D^{\cl}_{[i}B_{j]}D^{\cl\,i}B^{j}
+\epsilon^{2} F^{\cl}_{ij} [B^{i},\,B^{j}])
 ) \, \sqrt{-\hat g} \,d^{3}\hat x
\en

\subsection{Operators $\Gamma$ and $\starnabla $}
Define linear operators on the perturbations $B_{i}$.
\eqg
\label{eq:Gammastarnabladefs}
\Gamma B_{i} = \epsilon_{i}{}^{jk}[\gamma_{j},\, B_{k}]
\qquad
\starnabla  B_{i} = \epsilon_{i}{}^{jk}\hat\nabla_{j} B_{k}
\eng
so
\eqg
\starDcl  B_{i} =  \epsilon_{i}{}^{jk}D^{\cl}_{j} B_{k}
= \starnabla B_{i} + \hat b \,\Gamma B_{i}
\\[1ex]
D^{\cl}_{[i}B_{j]}D^{\cl\,i}B^{j}
=( \starDcl  B_{k} )( \starDcl  B^{k})
\\[1ex]
\tr(F^{\cl}_{ij} [B^{i},\,B^{j}])
= (1-\hat b^{2} )\tr(2\epsilon_{ij}{}^{k}\hat \gamma_{k} B^{i}\,B^{j})
= (1-\hat b^{2} )\tr (B_{k} \Gamma B^{k})
\eng
$\Gamma$ and $\starnabla $ are symmetric operators.
\eqa
\int \tr \left(B'{}^{i}\starnabla B_{i}\right) 
&amp;= \int \tr\left( B'{}^{i} \epsilon_{i}{}^{jk}\hat\nabla_{j} B_{k}\right)
= \int \tr \left(\epsilon_{k} {}^{ji}{} \hat\nabla_{j} B'{}_{i}  B^{k}\right)
\\[1ex]
&amp;=\int \tr \left(\starnabla  B'_{k} B^{k}\right)
\ena
\eqa
\int \tr \left(B'{}^{i}\Gamma B_{i}\right) &amp;= \int \tr\left( B'{}^{i} 
\epsilon_{i}{}^{jk}(\gamma_{j} B_{k}- B_{k}\gamma_{j})\right)
=  \int \tr \left(\epsilon_{i}{}^{jk} (B'{}^{i}\gamma_{j} B_{k} - 
\gamma_{j} B'{}^{i}B_{k})\right)
\\[1ex]
&amp;=  \int \tr \left(\epsilon_{k}{}^{ji} [\gamma_{j},\, B'{}_{i}] B^{k}\right)
= \int \tr \left( \Gamma B'_{k}B^{k}\right)
\ena
So
\eqa
\label{eq:bK}
\bK &amp;= \epsilon^{2}(\starnabla + \hat b \Gamma)^{2} 
+  \epsilon^{2}(1-\hat b^{2}) \Gamma
\\[1ex]
&amp;= (\Gamma^{2}-\Gamma) b^{2}
+(\starnabla \Gamma + \Gamma \starnabla)\epsilon b 
+ \left((\starnabla)^{2}+\Gamma\right)\epsilon^{2}
\ena




\section{Spin(4) decomposition}

The perturbations of the gauge field form the $\Spin(4)$ representation
\eqa
\left\{ B_{i}^{j}(\hat x)\right\} 
&amp;=  \mathop\oplus\limits_{j_{3}} \;(j_{1}\otimes j_{2}\otimes j_{3}, j_{3})
\qquad
j_{1}=j_{2}=1
\\[1ex]
&amp;= \mathop\oplus\limits_{j_{3},J}  \Complexes^{N(J,j_{3})} \otimes (J,j_{3})
\qquad
(J,j_{3}) = (j_{L},j_{R})
\ena
The infinitesimal gauge transformations form the representation
\eqg
\left\{ v^{i}(\hat x)\right\} 
=  \mathop\oplus\limits_{j_{3}} \;(1\otimes j_{3}, j_{3})
= \mathop\oplus\limits_{j_{3},J}  \Complexes^{N_{\gauge}(J,j_{3})} \otimes (J,j_{3})
\\[1ex]
\Complexes^{N_{\gauge}(J,j_{3})} \subseteq  \Complexes^{N(J,j_{3})}
\eng
The triple tensor product ${j_{1}} \otimes {j_{2}} \otimes {j_{3}} $
can be decomposed in two ways,
\eqa
\label{eq:decompositions}
({j_{1}} \otimes {j_{2}}) \otimes {j_{3}} 
&amp;= \left(\mathop\oplus\limits_{J_{12}} J_{12}\right)\otimes {j_{3}} 
=  \mathop\oplus\limits_{J}\left( \mathop\oplus\limits_{J_{12}} 
J\right)
= \mathop\oplus\limits_{J}\left( \Complexes^{N(J,j_{3})}\otimes J\right)
\\[1ex]
{j_{1}} \otimes (j_{2} \otimes {j_{3}} )
&amp;= j_{1}\otimes\left( \mathop\oplus\limits_{J_{23}} J_{23}\right)
=\mathop\oplus\limits_{J}\left(\mathop\oplus\limits_{J_{23}}  J\right)
= \mathop\oplus\limits_{J} \left(\Complexes^{N(J,j_{3})}\otimes 
J\right)
\ena
with
\eq
J_{12} \in \{0,1,2\}
\qquad
J_{23}-j_{3} \in \{-1,0,1\}
\qquad
J-j_{3} \in \{0,\pm 1,\pm 2\}
\en
The tensor product for the gauge symmetries decomposes
\eq
1 \otimes {j_{3}} = \mathop\oplus_{J} 
\left(\Complexes^{N_{\gauge}(J,j_{3})}\otimes \mathbf{J}\right)
\qquad
J -j_{3} \in \{-1,0,1\}
\en
The representations that occur in the decompositions
are listed in Table \ref{table:decomp}.
\begin{table}[ht]
$$
\begin{array}{c@{\quad}c@{\qquad}c@{\quad}c@{\qquad}ccc}
J-j_{3} &amp; j_{3} &amp; J_{12} &amp; J_{23}-j_{3} &amp; N &amp; N_{\mathrm{gauge}} &amp; 
N_{\phys}\\
\hline \\[-2ex]
0        &amp;  0   &amp; 0 &amp; 1 &amp; 1 &amp; 0&amp; 1\\[0.5ex]
0        &amp;  \frac12    &amp; 0,1 &amp; 0,1 &amp; 2 &amp; 1 &amp; 1\\[0.5ex]
0   &amp;  \ge 1   &amp; 0,1,2 &amp; -1,0,1 &amp; 3 &amp; 1 &amp; 2 \\[1ex]
1         &amp; 0     &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 0\\[0.5ex]
1 &amp; \ge\frac12  &amp; 1,2 &amp; 0,1 &amp; 2 &amp; 1&amp; 1 \\[1ex]
-1       &amp; 1  &amp; 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0\\[0.5ex]
-1  &amp; \ge\frac32  &amp; 1,2 &amp; -1,0 &amp; 2 &amp; 1 &amp; 1\\[1ex]
2 &amp; \ge 0  &amp; 2 &amp; 1 &amp; 1 &amp; 0 &amp; 1\\[0.5ex]
-2 &amp;\ge 2  &amp; 2 &amp; -1 &amp; 1 &amp; 0 &amp; 1\\[1ex]
\end{array}
$$
\caption{Decomposition of $\mathbf{1}\otimes\mathbf{1}\otimes\mathbf{j_{R}}$.}
\label{table:decomp}
\end{table}
The physical degrees of freedom are the reprentations with 
$N_{\phys}=N-N_{\gauge}&gt;0$.  The physical representations are listed in 
Table~\ref{table:decompphys}.
The infinite series are parametrized by $j$ rather than $j_{R}$ 
so that the symmetry $j_{L} \leftrightarrow j_{R}$ becomes 
$j\leftrightarrow -j$.


\begin{table}[ht]
$$
\begin{array}{c@{\quad}cc@{\quad}c@{\qquad}cc@{\qquad}ccc}
&amp;j_{L}&amp; j_{R} &amp;  &amp; J_{12} &amp; J_{23}&amp;   N &amp;  N_{\mathrm{gauge}} &amp;  N_{\mathrm{phys}} \\
\hline \\[-2ex]
\mathbf{1}&amp; 0&amp;0&amp;  &amp;0&amp;1&amp; 1 &amp; 0 &amp; 1\\[0.5ex]
\mathbf{2}&amp; \frac12&amp;\frac12&amp; &amp; 0,1 &amp; \frac12,\frac32 &amp; 2 &amp; 1 &amp; 1\\[1.5ex]
\mathbf{3}_{j} &amp; j-\frac12&amp;j-\frac12&amp;  \frac32\le j  &amp;0,1,2&amp; 
j-\frac32,j-\frac12,j+\frac12  &amp; 3 &amp;  1 &amp; 2\\[1.5ex]
\mathbf{2}_{j} &amp;j&amp; j-1&amp;   \frac32 \le j &amp;1,2 &amp;j-1,j &amp; 2 &amp;1 &amp; 1 \\[0.5ex]
&amp;-j-1&amp; -j&amp;   j\le -\frac32&amp; 1,2&amp;-j-1,-j &amp; 2 &amp;1 &amp; 1 \\[1.5ex]
\mathbf{1}_{j} &amp;j+\frac12&amp;j-\frac32 &amp; \frac32 \le j&amp;2 &amp;j-\frac12 &amp; 1 &amp;0 &amp; 1\\[0.5ex]
 &amp; -j-\frac32&amp;-j+\frac12 &amp; j\le -\frac32 &amp;2 &amp; -j-\frac12 &amp; 1 &amp;0 &amp; 1
\end{array}
$$
\caption{The physical representations $N_{\mathrm{phys}}=N-N_{\gauge}&gt;0$ .}
\label{table:decompphys}
\end{table}

\section{Five ODEs}

\subsection{$\su(2)$ notation}

The $\su(2)$ generators in the $j=1/2$ representation are
\eq
L^{(1/2)}_{a} =\mu_{a} = \frac{i\sigma_{a}}2 
\qquad
[\mu_{a},\, \mu_{b}] = - \epsilon_{ab}{}^{c}  \mu_{c}
\en
The generators $L_{a}$ of a general representation satisfy the same 
commutation relations,
\eq
 \qquad
[L_{a},\, L_{b}] = - \epsilon_{ab}{}^{c}  L_{c}
\en
The Casimir operator is
\eq
C = - \frac12 L^{a}L_{a}
\en
using $\delta_{ab}$ to raise/lower indices.  The Casimir for the 
irreducible representation $j$ is
\eq
C_{j} = \frac 12 j(j+1)
\en
The adjoint representation is $j=1$,
\eq
L^{(1)}_{a} v_{b} = \epsilon_{ab}{}^{c}v_{c}
\qquad
C_{1} = 1
\en


\subsection{The representation $\{B_{i}\}$}

The Dirac matrices on $S^{3}$ form an $\su(2)$-valued 1-form
which can be written
\eq
\hat \gamma_{i}(\hat x) = \hat \gamma_{i}^{a}(\hat x) \mu_{a}
\qquad
\hat \gamma_{i}^{a} \hat \gamma_{a}^{j} = \delta_{i}^{j}
\qquad
\hat \gamma_{i}^{a} \hat \gamma_{b}^{i} = \delta^{a}_{b}
\en
where indices are raised/lowered with $\hat g_{ij}$ and $\delta_{ab}$.
The gauge field perturbation $B_{i}$ can be written
\eq
B_{i}(\hat x) = B_{i}^{a}(\hat x) \mu_{a} = B_{ab}(\hat x)  \hat 
\gamma^{b}_{i} \mu^{a}
\en
making explicit the representation
\eq
\left\{ B_{ab}(\hat x)\right\} 
= \mathop\oplus\limits_{j_{3}} (j_{1}\otimes j_{2} \otimes j_{3},j_{3})
\qquad
j_{1}=j_{2}=1
\en
Let $L^{1}_{a}$, $L^{2}_{a}$, $L^{3}_{a}$ be the generators of 
the representations $j_{1}$, $j_{2}$, $j_{3}$.
\eq
L^{1}_{a} B_{bc} = \epsilon_{ab}{}^{d} B_{dc}
\qquad
L^{2}_{a} B_{bc} = \epsilon_{ac}{}^{d} B_{bd}
\qquad
L^{3}_{a} B_{bc} = \frac12 \hat \gamma_{a}^{i}(\hat x) \hat \partial_{i} B_{bc}
\en
The last formula is verified by noting that $L^{3}_{a}$ is the 
generator of left multiplication acting on functions so should act as 
a right-invariant vector field and by checking at the north pole.
\eq
\hat \gamma_{a}^{i}(\hat N) = \delta_{a}^{i}
\qquad
g_{\hat N + \epsilon^{a} \hat \gamma_{a}^{i}/2} =  1 - \epsilon^{a}\mu_{a}
\en
Tthe Casimirs of the individual factors are
\eq
C_{j_{1}}=C_{j_{2}} =1
\qquad
C_{j_{3}} = \frac12 j_{3}(j_{3}+1)
\en
Write
\eqg
L_{a}^{12} = L_{a}^{1} + L_{a}^{2}
\qquad
C_{12} = -\frac12 \delta^{ab}L_{a}^{12}L_{b}^{12} = -\frac12 \delta^{ab} ( L_{a}^{1} + L_{a}^{2})
( L_{b}^{1} + L_{b}^{2})
\\[1ex]
L_{a}^{23} = L_{a}^{2} + L_{a}^{3}
\qquad
C_{23} = -\frac12 \delta^{ab}L_{a}^{23}L_{b}^{23} = -\frac12 \delta^{ab} ( L_{a}^{2} + L_{a}^{3})
( L_{b}^{2} + L_{b}^{3})
\eng
In the decompositions (\ref{eq:decompositions})
\eq
C_{12} = \frac12 J_{12}(J_{12}+1)
\qquad
C_{23} = \frac12 J_{23}(J_{23}+1)
\en

\subsection{$\Gamma $ and $\starnabla$ in terms of Casimirs}

The operator $\Gamma$ defined in (\ref{eq:Gammastarnabladefs})
is
\eq
\Gamma B_{i} = \epsilon_{i}{}^{jk}[\hat\gamma_{j},\, B_{k}]
= \epsilon_{i}{}^{jk}[\hat\gamma_{j}^{c}\mu_{c},\, 
B_{a'b'}\hat\gamma^{b'}_{k}\mu^{a'}]
= - \epsilon_{i}{}^{jk}\hat\gamma_{j}^{c}\hat\gamma^{b'}_{k}
B_{a'b'}\epsilon_{c}{}^{a'}{}_{a} \mu^{a}
\en
Now use
\eq
\epsilon_{abc} = 
\epsilon_{ijk}\hat\gamma^{i}_{a}\hat\gamma^{j}_{b}\hat\gamma^{k}_{c}
\en
to get
\eqg
\Gamma(B_{ab} \mu^{a}\hat \gamma ^{b}_{i})
=- \epsilon_{i'}{}^{jk}
\hat\gamma_{b}^{i'}
\gamma_{j}^{c}\hat\gamma^{b'}_{k}
\epsilon_{c}{}^{a'}{}_{a}
B_{a'b'} \mu^{a}\hat\gamma_{i}^{b}
=- \epsilonb{}^{cb'}
\epsilon_{c}{}^{a'}{}_{a}
B_{a'b'} \mu^{a}\hat\gamma_{i}^{b}
\\[1ex]
\Gamma B_{ab} = - \delta^{cd}\epsilon_{ca}{}^{a'}\epsilon_{db}{}^{b'}
B_{a'b'}
\eng

\eq
\Gamma = - \delta^{cd} L^{1}_{c} L^{2}_{d} 
= C_{12} -C_{j_{1}}-C_{j_{2}}
= C_{12}-2
\en
Next, define
\eq
\stard = \starnabla - \Gamma
\en
\eqa
\stard(B_{ab}\hat\gamma_{i}^{b}\mu^{a}) 
&amp;= \epsilon_{i}{}^{jk}\hat\partial_{j}(B_{ab}\hat\gamma_{k}^{b}\mu^{a})
= \epsilon_{i}{}^{jk}(\hat\partial_{j}B_{ab})\hat\gamma_{k}^{b}\mu^{a}
+ \epsilon_{i}{}^{jk}B_{ab}\hat\partial_{j}\hat\gamma_{k}^{b}\mu^{a}
\\[1ex]
&amp;= 
-\epsilon^{c}{}_{b}{}^{b'}(\hat\gamma_{c}^{j}\hat\partial_{j}B_{ab'})\hat\gamma_{i}^{b}\mu^{a}
+ \epsilon_{i'}{}^{jk}\hat \gamma^{i'}_{b} \hat\partial_{j}\hat\gamma_{k}^{b'}
B_{ab'}\hat\gamma^{b}_{i}\mu^{a}
\\[1ex]
\stard B_{ab} &amp;= -2 \delta^{cd} L^{2}_{c}L^{3}_{d}
+ A^{b'}_{b} B_{ab'}
\qquad
A^{b'}_{b} = \epsilon_{i'}{}^{jk}\hat \gamma^{i'}_{b} \hat\partial_{j}\hat\gamma_{k}^{b'}
\\[2ex]
A^{b'}_{b}\mu_{b'} &amp;= \epsilon_{i'}{}^{jk}\hat \gamma^{i'}_{b} \hat\partial_{j}\hat\gamma_{k}
= \epsilon_{i'}{}^{jk}\hat \gamma^{i'}_{b}
\frac12\hat\partial_{[j}\hat\gamma_{k]}
= - \epsilon_{i'}{}^{jk}\hat \gamma^{i'}_{b}
[\hat\gamma_{j},\,\hat\gamma_{k}]
\\[1ex]
&amp;
=  \epsilon_{i'}{}^{jk}\hat \gamma^{i'}_{b}
\epsilon_{jk}{}^{i}\hat\gamma_{i}
= 2\gamma^{i}_{b}\hat\gamma_{i} = 
2\gamma^{i}_{b}\hat\gamma_{i}^{c}\mu_{c} = 2\mu_{b}
\\[1ex]
A^{b'}_{b} &amp;=2 \delta^{b'}_{b}
\\[2ex]
\stard &amp;= 2(C_{23}-C_{j_{2}}-C_{j_{3}}) +2
= 2(C_{23}-C_{j_{3}})
\ena

\subsection{Wigner 6-j symbol and Racah W-coefficient}

The two decompositions (\ref{eq:decompositions}) give two canonical 
bases for $\Complexes^{N(J,j_{3})}$,
\eq
\Complexes^{N(J,j_{3})} = \mathop\oplus\limits_{J_{12}} 
\hat e_{12}(J_{12})\otimes\Complexes 
=  \mathop\oplus\limits_{J_{23}} 
\hat e_{23}(J_{23})\otimes\Complexes 
\en
$C_{12}$ is diagonal in the first basis, $C_{23}$ in the second.
\eq
C_{12} \,\hat e_{12}(J_{12}) = \frac12 J_{12} (J_{12}+1) \,\hat e_{12}(J_{12})
\qquad
C_{23} \,\hat e_{23}(J_{23}) = \frac12 J_{23} (J_{23}+1) \,\hat 
e_{23}(J_{23})
\en
But they cannot be diagonalized simultaneously unless $N=1$.
The two bases are related by a matrix,
\eq
\hat e_{12}(J_{12}) = \sum_{J_{23}} U(J_{12},J_{23}) \,\hat e_{23}(J_{23})
\en
The matrix elements $U(J_{12},J_{23})$ are called the {\it recoupling 
coefficients}.  Traditionally the recoupling coefficients are written
\eq
U(J_{12},J_{23}) = 
\inprod{\,(j_{1},(j_{2}j_{3})J_{23})J\,}{\,((j_{1}j_{2})J_{12}j_{3})J\,}
\en
They are related to the Wigner 6-j symbol
and the Racah W-coefficient,
\eq
\frac{U(J_{12},J_{23})) }
{\sqrt{(2J_{12}+1)(2J_{23}+1)}}
=
(-1)^{j_{1}+j_{2}+j_{3}+J}
\left\{
\begin{array}{ccc}
j_{1} &amp; j_{2}&amp; J_{12}\\
j_{3}&amp; J &amp;J_{23}
\end{array}
\right\}
=W(j_{1}j_{2}Jj_{3};J_{12}J_{23})
\en
Given $U$,
the two Casimirs
can be written as matrices in the same basis,
\eq
C_{12} 
= \frac12 J_{12}(J_{12}+1)
\qquad
C_{23} = U \,\frac12 
J_{23}(J_{23}+1) \,U^{t}
\en
Racah's algorithm for the W-coefficient \cite{PhysRev.62.438} is:

\eqg
 W(j_{1}j_{2}Jj_{3};J_{12}J_{23}) =W(abcd;ef) 
\\[1ex]
a = j_{1}\quad
b = j_{2}\quad
c = J \quad
d = j_{3}\quad
e = J_{12}\quad
f = J_{23}\quad
\eng
\vskip1ex

\eqg
W(abcd;ef) = \Delta(a,b,e) \,\Delta(c,d,e) \,\Delta(a,c,f) 
\,\Delta(b,d,f) \; w(abcd;ef)
\\[1ex]
\Delta(a,b,c) = \left[
\frac{(a+b-c)! (a-b+c)! (-a+b+c)! }
{(a+b+c+1)!}
\right]^{1/2}
\eng
\vskip1ex

\eq
w(abcd;ef) = \sum_{z}\;
\frac{(-1)^{z+\beta_{1}}(z+1)!}
{\prod\limits_{i=1}^{4}(z-\alpha_{i})!\; \prod\limits_{i=1}^{3}  (\beta_{i}-z)!}
\en

The sum is over integers $z$ in the range
\eqg
\max(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})
\le z \le
\min(\beta_{1},\beta_{2},\beta_{3})
\\[1ex]
\alpha_{1}=a+b+e
\qquad 
\alpha_{2}=c+d+e
\qquad 
\alpha_{3}=a+c+f
\qquad 
\alpha_{4}=b+d+f
\\[1ex]
\beta_{1}=a+b+c+d
\qquad 
\beta_{2}=a+d+e+f
\qquad 
\beta_{3}=b+c+e+f
\eng

\subsection{The five $\mathbf{K}(z)$ matrices}

The quadratic action is (\ref{eq:quadratic1})
\eqg
\frac1\hbar S_{2}
= 
\frac1{\epsilon g ^{2}}\int  
 \tr(  \partial_{z}B_{i} \partial_{z}B^{i}
- B_{i} \bK  B^{i})
\, \sqrt{-\hat g} \,d^{3}\hat x\, dz
\eng
with equation of motion
\eq
\left(\frac{d^{2}}{dz^{2}} + \bK(z) \right ) B_{i} = 0
\en
Equation (\ref{eq:bK}) gives $\bK(z)$ as
\eq
\bK=   \left(\Gamma^{2}-\Gamma\right)b^{2}
+\left(\starnabla \Gamma + \Gamma \starnabla\right)\epsilon b 
+ \big[(\starnabla)^{2}+\Gamma\big]\epsilon^{2}
\en
now expressed in terms of Casimirs,
\eq
\Gamma = C_{12}-2
\qquad
\starnabla-\Gamma =2C_{23}-j_{3}(j_{3}+1)
\en
For each irreducible $(j_{L},j_{R})$ occuring in the space 
$\{B_{i}\}$ of gauge field perturbations,
$\bK(z)$ is an $N{\times}N$ matrix acting on
$\Complexes^{N}$, $N = N(j_{L},j_{R})$.

The $\bK(z)$ matrix for {\bf Case $\mathbf{1}$}
$(j_{L},j_{R}) = (0,0)$ is calculated by hand
below.
The $\bK(z)$ matrices 
for the other four  sets of representations
listed in Table~\ref{table:decompphys}
are calculated in Sagemath in the notebook
{\it $K(z)$ matrices}.
For the series of representations indexed by $j$,
the parameter $p$ is defined as
\eq
p^{2} = 4 \epsilonb^{2} C_{(j_{L},j_{R})} 
= 2\epsilonb^{2}  \left[j_{L}(j_{L}+1)+ j_{R}(j_{R}+1)
\right]
\en
The representations are indexed by
$
 (j_{L},j_{R}) = (J,j_{3})
$.

\vskip1ex
\paragraph{Case $\mathbf{1}$}
$$
\begin{array}{c@{\quad}cc@{\quad}c@{\qquad}cc@{\qquad}ccc}
&amp;j_{L}&amp; j_{R} &amp;  &amp; J_{12} &amp; J_{23}-j_{3}&amp;   N &amp;  N_{\mathrm{gauge}} &amp;  N_{\mathrm{phys}} \\
\hline \\[-2ex]
\mathbf{1}&amp; 0&amp;0&amp;  &amp;0&amp;1&amp; 1 &amp; 0 &amp; 1
\end{array}
$$

\vskip1ex
\noindent
\eqg
C_{12}=0\quad C_{23}=1
\qquad
\Gamma = -2\quad \starnabla=0
\\[1ex]
\bK = 6 b^{2} -2\epsilonb^{2}
\eng

\vskip1ex
\paragraph{Case $\mathbf{2}$}

$$
\begin{array}{c@{\quad}cc@{\quad}c@{\qquad}cc@{\qquad}ccc}
&amp;j_{L}&amp; j_{R} &amp;  &amp; J_{12} &amp; J_{23}-j_{3}&amp;   N &amp;  N_{\mathrm{gauge}} &amp;  N_{\mathrm{phys}} \\
\hline \\[-2ex]
\mathbf{2}&amp; \frac12&amp;\frac12&amp; &amp; 0,1 &amp; 0,1 &amp; 2 &amp; 1 &amp; 1
\end{array}
$$

\vskip2ex

\noindent
\eqg
\mathbf{K} =
\begin{pmatrix}
6 &amp; 0 \\
0 &amp; 2
\end{pmatrix}
b^{2}
+
\begin{pmatrix}
0 &amp; -3\sqrt{2}\\
-3\sqrt{2} &amp; 0
\end{pmatrix}
\epsilonb b
+
\begin{pmatrix}
0 &amp; 0 \\
0 &amp; 1
\end{pmatrix}
\epsilonb^{2}
\eng


\vskip1ex
\paragraph{Case $\mathbf{3_{j}}$}

$$
\begin{array}{c@{\quad}cc@{\quad}c@{\qquad}cc@{\qquad}ccc}
&amp;j_{L}&amp; j_{R} &amp;  &amp; J_{12} &amp; J_{23}-j_{3}&amp;   N &amp;  N_{\mathrm{gauge}} &amp;  N_{\mathrm{phys}} \\
\hline \\[-2ex]
\mathbf{3}_{j} &amp; j-\frac12&amp;j-\frac12&amp;  \frac32\le j  &amp;0,1,2&amp; 
-1,0,1  &amp; 3 &amp;  1 &amp; 2
\end{array}
$$

\vskip2ex

\noindent
\eqg
\mathbf{K}_{2} =
\begin{pmatrix}
6 &amp; 0  &amp; 0\\
0 &amp; 2 &amp; 0 \\
0 &amp; 0 &amp; 0
\end{pmatrix}
b^{2}
+
\begin{pmatrix}
0 &amp; -6 &amp; 0 \\
-6 &amp; 0 &amp; 0 \\
0  &amp; 0 &amp; 0
\end{pmatrix}
\sigma  b
+
\begin{pmatrix}
\alpha^{2}  &amp; 0 &amp;  \alpha  \\
0 &amp; \alpha^{2}+  1 &amp; 0 \\
\alpha &amp; 0 &amp;  1
\end{pmatrix}
2 \sigma^{2} 
\\[2ex]
p^{2} 
= 4  \epsilonb^{2} \left(j^{2}-\frac14\right)
\qquad
\sigma =   \sqrt{\frac{2}{3}} \sqrt{j^{2}-\frac1{4}} 
\,\epsilonb 
=\frac{p}{\sqrt{6}}
\qquad
\alpha =  \sqrt2 \sqrt{\frac{j^{2}-1}{j^{2}-\frac1{4}}}
\eng



\paragraph{Case $\mathbf{2_{j}}$}
$$
\begin{array}{c@{\quad}cc@{\quad}c@{\qquad}cc@{\qquad}ccc}
&amp;j_{L}&amp; j_{R} &amp;  &amp; J_{12} &amp; J_{23}-j_{3}&amp;   N &amp;  N_{\mathrm{gauge}} &amp;  N_{\mathrm{phys}} \\
\hline \\[-2ex]
\mathbf{2}_{j} &amp;j&amp; j-1&amp;   \frac32 \le j &amp;1,2 &amp;0,1 &amp; 2 &amp;1 &amp; 1 \\[0.5ex]
&amp;-j-1&amp; -j&amp;   j\le -\frac32&amp; 1,2&amp;-1,0 &amp; 2 &amp;1 &amp; 1
\end{array}
$$
\vskip2ex

\noindent
\eqg
\mathbf{K} =
\begin{pmatrix}
2 &amp; 0 \\
0 &amp; 0
\end{pmatrix}
b^{2}
+
\begin{pmatrix}
-1 &amp; 0 \\
0 &amp; 1
\end{pmatrix}
2 \sigma b
+
\begin{pmatrix}
\alpha^{2} &amp; \alpha \\
\alpha &amp; 1
\end{pmatrix}
2 \sigma^{2}
\\[2ex]
p^{2} = 4  \epsilonb^{2} j^{2}
\qquad
\sigma= \epsilonb j = \frac12 p
\qquad
\alpha = \sqrt{1-\frac1{j^{2}}}
\eng




\paragraph{Case $\mathbf{1_{j}}$}
$$
\begin{array}{c@{\quad}cc@{\quad}c@{\qquad}cc@{\qquad}ccc}
&amp;j_{L}&amp; j_{R} &amp;  &amp; J_{12} &amp; J_{23}-j_{3}&amp;   N &amp;  N_{\mathrm{gauge}} &amp;  N_{\mathrm{phys}} \\
\hline \\[-2ex]
\mathbf{1}_{j} &amp;j+\frac12&amp;j-\frac32 &amp; \frac32 \le j&amp;2 &amp;1 &amp; 1 &amp;0 &amp; 1\\[0.5ex]
 &amp; -j-\frac32&amp;-j+\frac12 &amp; j\le -\frac32 &amp;2 &amp; -1 &amp; 1 &amp;0 &amp; 1
\end{array}
$$
\vskip2ex

\noindent
\eqg
\mathbf{K} =
4\sigma b
+4\alpha^{2} \sigma^{2}
\\[1ex]
p^{2} = 4 \epsilonb^{2} \left(j^{2}+\frac34\right)
\qquad
\sigma = \epsilonb j = \frac12 \sqrt{p^{2} - 3 \epsilonb^{2}}
\qquad
\alpha = \sqrt{1+\frac1{4j^{2}}}
\eng


\subsection{Gauge symmetries}

A gauge variation of the classical solution is
a solution of the equation of motion 
that is first order in $b(z)$.
\eq
w_{\gauge}(z) =  w_{1}b(z) + w_{0}
\en
Use
\eq
b'' = 2\epsilonb^{2} b -2 b^{3}
\en
and write
\eq
\bK = \bK_{2}b^{2} +\bK_{1}b +\bK_{0}
\en
to calculate
\eqa
0 &amp;= \frac{d^{2} w}{dz^{2}} +\mathbf{K}(z) w
\\[1ex]
&amp;= w_{1} (2\epsilonb^{2} b -2 b^{3})
+ 
\left(\mathbf{K}_{2} b^{2} + \mathbf{K}_{1}  b
+ \mathbf{K}_{0}  \right)
\left(w_{1}b+  w_{0}\right)
\\[1ex]
&amp;= (\bK_{2}-2)w_{1} b^{3}
+(\bK_{2}w_{0}+\bK_{1}w_{1}) b^{2}
\\[1ex]
&amp;\qquad\qquad
{}+(2\epsilonb^{2}w_{1}+\bK_{0}w_{1}+\bK_{1}w_{0})b
+ \bK_{0}w_{0}
\ena
So the coefficients $w_{0}$, $w_{1}$ are the solutions of
\eqa
0&amp;  =\bK_{0}w_{0}
\\[1ex]
0&amp;  =(\bK_{2}-2)w_{1}
\\[1ex]
0&amp;  =\bK_{2}w_{0}+\bK_{1}w_{1}
\\[1ex]
0&amp;  = \bK_{1}w_{0}+(\bK_{0}+2\epsilonb^{2})w_{1}
\ena

\paragraph{Case $\mathbf{2}$}

\eqg
\mathbf{K}_{2} =
\begin{pmatrix}
6 &amp; 0 \\
0 &amp; 2
\end{pmatrix}
\qquad
\mathbf{K}_{1} =
\begin{pmatrix}
0 &amp; -3\sqrt{2}\\
-3\sqrt{2} &amp; 0
\end{pmatrix}
\epsilonb
\qquad
\mathbf{K}_{0} =
\begin{pmatrix}
0 &amp; 0 \\
0 &amp; 1
\end{pmatrix}
\epsilonb^{2}
\eng

\eqg
w_{1} = 
B
\begin{pmatrix}
0
\\
1
\end{pmatrix}
\qquad
w_{0} = A
\begin{pmatrix}
1
\\
0
\end{pmatrix}
\qquad
0 = A
\begin{pmatrix}
6
\\
0
\end{pmatrix}
+
B
\begin{pmatrix}
-3\sqrt2
\\
0
\end{pmatrix}
\epsilonb
\\[1ex]
A= \frac{\epsilonb}{\sqrt2} B
\qquad
0
=
A
\begin{pmatrix}
0
\\
-3\sqrt2
\end{pmatrix}
\epsilonb
+
B
\begin{pmatrix}
0
\\
3
\end{pmatrix}
\epsilonb^{2}
=0
\eng
\eqg
w_{\gauge}
= 
\begin{pmatrix}
0
\\
1
\end{pmatrix}
b
+
\begin{pmatrix}
1
\\
0
\end{pmatrix}
\frac{\epsilonb }{\sqrt2}
\eng

\paragraph{Case $\mathbf{3_{j}}$}

\eqg
\mathbf{K}_{2} =
\begin{pmatrix}
6 &amp; 0  &amp; 0\\
0 &amp; 2 &amp; 0 \\
0 &amp; 0 &amp; 0
\end{pmatrix}
\qquad
\mathbf{K}_{1} =
\begin{pmatrix}
0 &amp; -6  &amp; 0 \\
-6  &amp; 0 &amp; 0 \\
0  &amp; 0 &amp; 0
\end{pmatrix}
\sigma_{3}
\qquad
\mathbf{K}_{0} =
\begin{pmatrix}
\alpha_{3}^{2}  &amp; 0 &amp;  \alpha_{3}  \\
0 &amp; \alpha_{3}^{2}+  1 &amp; 0 \\
\alpha_{3} &amp; 0 &amp;  1
\end{pmatrix}
2 \sigma_{3}^{2}
\\[2ex]
\sigma_{3} =   \sqrt{\frac{2}{3}} \sqrt{j^{2}-\frac1{4}} \,\epsilonb
\qquad
\alpha_{3} =  \sqrt2 \sqrt{\frac{j^{2}-1}{j^{2}-\frac1{4}}}
\eng

\eqg
w_{1} = B
\begin{pmatrix}
0
\\
1
\\
0
\end{pmatrix}
\qquad
w_{0} = A
\begin{pmatrix}
1
\\
0
\\
-\alpha_{3}
\end{pmatrix}
\\[1ex]
0=
 A
\begin{pmatrix}
6
\\
0
\\
0
\end{pmatrix}
+
B
\begin{pmatrix}
-6 \sigma_{3}
\\
0
\\
0
\end{pmatrix}
\qquad
A =  \sigma_{3} B
\eng

\eqg
0 =A
\begin{pmatrix}
0
\\
-6 \sigma_{3}
\\
0
\end{pmatrix}
+B
\begin{pmatrix}
0
\\
2\epsilonb^{2}+ (\alpha_{3}^{2}+1) 2\sigma_{3}^{2} 
\\
0
\end{pmatrix}
\eng

\eqg
\begin{aligned}
0 &amp;= B
\left[
-6\sigma_{3}^{2}
+2\epsilonb^{2}+  2\sigma_{3}^{2} (\alpha_{3}^{2}+1)
\right]
=
2 B
\left[
\epsilonb^{2}+   \sigma_{3} (\alpha_{3}^{2}-2)
\right]
\\[1ex]
&amp;=
2 B\epsilonb^{2}
\left[
1+  \frac23 \left(j^{2}-\frac14\right) 
2 \left(
\frac{-\frac34}{j^{2}-\frac14}
\right)
\right]
\\[1ex]
&amp;=0
\end{aligned}
\eng

\eq
w_{\gauge}
=
\begin{pmatrix}
0
\\
1
\\
0
\end{pmatrix}
b
+\begin{pmatrix}
1
\\
0
\\
-\alpha_{3}
\end{pmatrix}
\sigma_{3}
\en




\paragraph{Case $\mathbf{2_{j}}$}

\eqg
\mathbf{K}_{2} =
\begin{pmatrix}
2 &amp; 0 \\
0 &amp; 0
\end{pmatrix}
\qquad
\mathbf{K}_{1} =
\begin{pmatrix}
-1 &amp; 0 \\
0 &amp; 1
\end{pmatrix}
2 \sigma_{2}
\qquad
\mathbf{K}_{0} =
\begin{pmatrix}
\alpha_{2}^{2} &amp; \alpha_{2} \\
\alpha_{2} &amp; 1
\end{pmatrix}
2 \sigma_{2}^{2}
\\[2ex]
\sigma_{2} = \epsilonb j
\qquad
\alpha_{2} = \sqrt{1-\frac1{j^{2}}}
\eng
\eqg
w_{1} = B
\begin{pmatrix}
1
\\
0
\end{pmatrix}
\qquad
w_{0} = A
\begin{pmatrix}
1
\\
-\alpha_{2} 
\end{pmatrix}
\\[1ex]
0 = 
 A
\begin{pmatrix}
2
\\
0
\end{pmatrix}
+
2\sigma_{2}
\begin{pmatrix}
-B
\\
0
\end{pmatrix}
\qquad
A= \sigma_{2}B
\\[1ex]
0 = B\begin{pmatrix}
2 \epsilonb^{2}+2\alpha_{2}^{2}\sigma_{2}^{2}
\\
2\alpha_{2}\sigma_{2}^{2}
\end{pmatrix}
+\sigma_{2}B
\begin{pmatrix}
-2\sigma_{2}
\\
-2\alpha_{2} \sigma_{2}
\end{pmatrix}
=
2 B
\begin{pmatrix}
\epsilonb^{2}+\alpha_{2}^{2}\sigma_{2}^{2}
-\sigma_{2}^{2}
\\
0
\end{pmatrix}
=0
\eng
\eq
w_{\gauge} =   
\begin{pmatrix}
1
\\
0
\end{pmatrix}
b
+
\begin{pmatrix}
1
\\
-\alpha_{2}
\end{pmatrix}
\sigma_{2}
\en



\section{Time-translation zero-mode}

The zero-mode is written as a free particle in the reparametrized time 
$\tilde z$ given by
\eq
\frac{d\tilde z}{dz} = \frac{1}{\cn'(z) ^{2}}
\en
The following calculates the periodicities of $\tilde z(z)$.  The 
results are not used in the paper,
but will be relevant for further investigation of the zero-mode 
integral.

The identities in section~\ref{sect:Jacobi} below
are used to integrate.
\eqa
\tilde z &amp;= \int \frac{dz}{\sn(z)^{2}\dn(z)^{2}} 
= \int 
\frac{k^{2}\sn(z)^{2}+\dn(z)^{2}}{\sn(z)^{2}\dn(z)^{2}}\,dz
\\[1ex]
 &amp;= k^{2}\int \nd^{2}\,dz +\int\ns^{2}\,dz
\\[1ex]
 &amp;=\frac{ k^{2}}{k'{}^{2}}(\cE-k^{2}\sn\cd)
+ z-\dn\cs-\cE
\\[1ex]
 &amp;=
 \left(\frac{ k^{2}}{k'{}^{2}}- 1\right)\cE
 + z
- \frac{ k^{4}}{k'{}^{2}} \frac{\sn\cn}{\dn}- \frac{\dn\cn}{\sn}
\\[1ex]
 &amp;=
\left(\frac{ k^{2}}{k'{}^{2}}- 1\right)\cE
 + z
- \frac{\cn}{ k'{}^{2}\sn\dn}
\left( { k^{4}} \sn^{2}+k'{}^{2} \dn^{2}\right)
\ena

\eqa
k^{4} \sn^{2}+k'{}^{2} \dn^{2}
&amp;= k^{4} \sn^{2}+k'{}^{2} (1-k^{2} \sn^{2})
= k'{}^{2} + k^{2} (k^{2}-k'{}^{2})\sn^{2}
\ena

\eqa
\tilde z &amp;=
\left(\frac{ k^{2}}{k'{}^{2}}- 1\right)\cE
 + z
- \frac{\cn}{ \sn\dn}
- \frac{\cn}{ k'{}^{2}\sn\dn}k^{2}(k^{2}-k'{}^{2}) \sn^{2}
\\[1ex]
&amp;=
- \frac{\cn}{ \sn\dn}
+ z
+\left(\frac{ k^{2}}{k'{}^{2}}- 1\right)
\left(\cE
- k^{2}\frac{\cn\sn}{\dn}
\right)
\ena
So $\tilde z$ has periodicities
\eqa
\tilde z(z+2K) - \tilde z(z) 
&amp;= 2K
+ \left(\frac{ k^{2}}{k'{}^{2}}- 1\right)(\cE(z+2K)-\cE(z))
\\[1ex]
&amp;= 2K
+ \left(\frac{ k^{2}}{k'{}^{2}}- 1\right)2E
\ena
\eqa
\tilde z(z+2K'i) - \tilde z(z) 
&amp;= 2K'i
+ \left(\frac{ k^{2}}{k'{}^{2}}- 1\right)(\cE(z+2K'i)-\cE(z))
\\[1ex]
&amp;= 2K'i
+ \left(\frac{ k^{2}}{k'{}^{2}}- 1\right)(2K'-2E')i
\ena


\subsection{More on Jacobi elliptic functions}
\label{sect:Jacobi}

These identities are from the DLMF \cite[Chapters 22 and 
19]{DLMF2022}.


\paragraph{$\sn(z,k)$ and $\dn(z,k)$}

\eqg
\sn^{2}+\cn^{2} =1
\qquad
k^{2}\sn^{2}+\dn^{2}=1
\\[1ex]
\sn(z) \sim  z + O(z^{3})
\qquad
\dn(z) \sim 1+ O(z^{2})
\eng
\qquad
\eqg
\cn' = -\sn\dn
\qquad
\dn' = -k^{2}\sn\cn
\\[1ex]
\dn'' = (1+k'{}^{2}) \dn-2 \dn^{3}
\qquad
(\dn')^{2} = (1-\dn^{2})(\dn^{2}-k'{}^{2})
\\[1ex]
\eng

\paragraph{half-periods}

\eqg
\cn(z+2K) = -\cn(z)
\qquad
\sn(z+2K) = -\sn(z)
\qquad
\dn(z+2K) = \dn(z)
\\[1ex]
\cn(z+2K'i) = -\cn(z)
\qquad
\sn(z+2K'i) = \sn(z)
\qquad
\dn(z+2K'i) = -\dn(z)
\eng

\paragraph{the other Jacobi elliptic functions}

\eqg
\nc = \frac1{\cn}
\qquad
\ns = \frac1{\sn}
\qquad
\nd = \frac1{\dn}
\qquad
\cd = \frac{\cn}{\dn}
\qquad
\cs = \frac{\cn}{\sn}
\eng

\paragraph{Jacobi's amplitude function $\phi=\am(z,k)$}

\eqg
\phi= \am(z) = \int_{0}^{z} \dn
\qquad
\sn(z) = \sin(\phi)
\qquad
\cn(z) = \cos(\phi)
\\[1ex]
\am(z+2K) = \am(z) +\pi
\qquad
\am(2K) = \pi
\eng

\paragraph{Jacobi epsilon function $\cE(z,k)$ and zeta function 
$\cZ(z,k)$}

\eqg
\cE(z) = \int_{0}^{z} \dn^{2}
= k^{2}\sn\cd + k'{}^{2} \int \nd^{2}
= z - \dn\cs -\int \ns^{2}
\eng
\eq
\cZ(z) = \cE(z) - \frac{E(k)}{K(k)} z
= \frac{\theta_{0}'}{\theta_{0}}
\en
$E(k)$ is the complete elliptic integral of the second kind,
\eq
E(k) = \cE(K) = E(\pi/2,k)
\qquad
E(\phi,k) = \cE(z,k)
\en
satisfying
\eq
EK'+E'K-K K' =\frac{\pi}{2}
\en


\eq
\cE(-z)=-\cE(z)
\qquad
\cE(z) = z + O(z^{3})
\en

\eqa
\cE(u+v) &amp;= \cE(u) +\cE(v)-k^{2} \sn(u)\sn(v) \sn(u+v)
\\[1ex]
\cZ(u+v) &amp;= \cZ(u) +\cZ(v)-k^{2} \sn(u)\sn(v) \sn(u+v)
\ena

\eq
\cZ(z+2K) = \cZ(z)
\qquad
\cZ(z+2K'i) = \cZ(z) - \frac{\pi}{K} i
\en

\eq
\cE(z+2K) =\cE(z) + 2E
\qquad
\cE(z+2K'i) = \cE(z) + 2\left( K'-E'\right)i
\en

\eqa
\cE(z+2K'i)  &amp;= \cE(z) + \cZ(z+2K'i) - \cZ(z) 
+ \frac{E}{K} 2K'i
\\[1ex]
&amp;= \cE(z) - \frac{\pi}{K} i+ \frac{E}{K} 2K'i
\\[1ex]
&amp;= \cE(z)  + \frac{2i}{K} \left( - \frac{\pi}{2} + E K'\right)
\\[1ex]
&amp; = \cE(z) +2\left( K'-E'\right)i
\ena


\setcounter{section}{9}

\section{Classical mechanics analytic in complex time $z$}


\eqg
\cA(z) =  
\begin{pmatrix}
0 &amp; - 1 \\[1ex]
\bK(z) &amp; 0
\end{pmatrix}
\qquad
\Omega =
i \begin{pmatrix}
0 &amp; 1 \\[1ex]
-1 &amp; 0
\end{pmatrix}
\eng

\eqa
\cA(z)^{t} \Omega +\Omega \cA(z) &amp;= 
\begin{pmatrix}
0 &amp;  \bK(z)^{t}\\[1ex]
- 1 &amp; 0
\end{pmatrix}
i \begin{pmatrix}
0 &amp; 1 \\[1ex]
-1 &amp; 0
\end{pmatrix}
+
i \begin{pmatrix}
0 &amp; 1 \\[1ex]
-1 &amp; 0
\end{pmatrix}
\begin{pmatrix}
0 &amp; - 1 \\[1ex]
\bK(z) &amp; 0
\end{pmatrix}
\\[1ex]
&amp;=
i
\begin{pmatrix}
\bK(z)-\bK(z)^{t} &amp; 0 \\[1ex]
0  &amp; 0
\end{pmatrix}
\ena

\eq
\bK(z) = \bK(z)^{t}
\quad
\Longleftrightarrow
\quad
\cA(z)^{t} \Omega +\Omega \cA(z) = 0
\en

\section{Quantum mechanics analytic in complex time $z$}

\paragraph{Canonical commutation relations}
\eqg
\cQ = \begin{pmatrix}
q\\
p
\end{pmatrix}
\qquad
\cQ^{t} = \begin{pmatrix}
q^{t} &amp;
p^{t}
\end{pmatrix}
\qquad
\cQ \cQ^{t} 
=\begin{pmatrix}
q q^{t} &amp; qp^{t}
\\
pq^{t} &amp; p p^{t}
\end{pmatrix}
\\[1ex]
\left( \cQ \cQ^{t}\right)^{t}- \cQ \cQ^{t}  =
\begin{pmatrix}
(q q^{t})^{t} - q q^{t} &amp; (pq^{t})^{t}- qp^{t}
\\
(qp^{t})^{t}-pq^{t}&amp; (p p^{t})^{t}-p p^{t}
\end{pmatrix}
\eng
\vskip2ex

\eq
\left( \cQ \cQ^{t}\right)^{t}- \cQ \cQ^{t}  = \Omega
\quad
\Longleftrightarrow
\quad
\begin{aligned}
(q q^{t})^{t} - q q^{t} &amp;= 0\\
(p p^{t})^{t}-p p^{t} &amp;= 0\\
(pq^{t})^{t}- qp^{t} &amp;= i
\end{aligned}
\en

\eqg
\left[(q q^{t})^{t} - q q^{t} \right]^{ab} = q^{b}q^{a} -q^{a}q^{b}
\qquad
\left[(p p^{t})^{t}-p p^{t}\right]_{ab} = p_{b}p_{a} -p_{a}p_{b}
\\[1ex]
[(pq^{t})^{t}- qp^{t}]^{a}_{b} = p_{b}q^{a} - q^{a}p_{b}
\eng
\vskip2ex

\eq
\left( \cQ \cQ^{t}\right)^{t}- \cQ \cQ^{t}  = \Omega
\quad
\Longleftrightarrow
\quad
\begin{aligned}
[ q^{a},\,q^{b}] &amp;= 0\\
[p_{a},\,p_{b}]  &amp;= 0\\
[p_{b},\,q^{a}] &amp;= i\delta_{b}^{a}
\end{aligned}
\en

\paragraph{Phase-space action}
\eqa
\frac1{2i}\cQ^{t} \Omega \left( \frac{d}{dz} + \cA \right)\cQ
&amp;= \frac1{2}
\begin{pmatrix}
q^{t} &amp;
p^{t}
\end{pmatrix}
\begin{pmatrix}
0 &amp; 1 \\[1ex]
-1 &amp; 0
\end{pmatrix}
\left( \frac{d}{dz} + \begin{pmatrix}
0 &amp; - 1 \\[1ex]
\bK(z) &amp; 0
\end{pmatrix}
 \right)
\begin{pmatrix}
q\\
p
\end{pmatrix}
\\[1ex]
&amp;= \frac1{2}
\begin{pmatrix}
q^{t} &amp;
p^{t}
\end{pmatrix}
\begin{pmatrix}
0 &amp; 1 \\[1ex]
-1 &amp; 0
\end{pmatrix}
\begin{pmatrix}
\frac{dq}{dz} -p\\
\frac{dp}{dz} +Kq
\end{pmatrix}
\\[1ex]
&amp;= -\frac12 p^{t}\frac{dq}{dz}+ \frac12 q^{t}\frac{dp}{dz} 
 +\frac12 p^{t} p+\frac12 q^{t}Kq
\ena

\setcounter{section}{17}


\section{Numerical evidence for Property {\bf P}}

\subsection{Imaginary period monodromy $\Moni$ at $t=K$}

Identities for Jacobi elliptic functions.  For $\tau\in \Reals$,
\eqg
\cn(K+\tau i,k) = -k' \sd(\tau i,k) = -k' \sd(\tau,k') i
\\[2ex]
\sd(\tau,k') = \overline{\sd(\tau,k')}
\\[1ex]
\sd(-\tau,k') = - \sd(\tau,k')
\qquad
\sd(\tau+2K,k') = - \sd(\tau,k')
\eng

\noindent
Let $t= K$ and let $C_{K}$ be the vertical path passing through $K$.
\eqg
z = K+\tau i
\qquad
b(z) = k \cn(z,k) = - k k' \sd(\tau,k') i =  F(\tau)i
\eng

\eqg
F(\tau) = - k k' \sd(\tau,k')
\qquad
\bK(z) = - \bK_{2} F(\tau)^{2}+ \bK_{0}
+ \bK_{1} F(\tau) i
\eng

\eqg
F(\tau) = \overline{F(\tau)}
\qquad
F(-\tau) = - F(\tau)
\qquad
F(\tau+2K') = - F(\tau)
\eng


\noindent
Write the propagator along $C_{K}$
\eq
\cP(\tau_{2},\tau_{1}) = \cP_{C_{K}}(K+\tau_{2}i,K+\tau_{2}i)
\en
The symmetries of $F(\tau)$ give
\eq
\cP(4K'-\tau_{2},4K'-\tau_{1} )= \overline{\cP(\tau_{2} ,\tau_{1})}
\qquad
\cP(2K'-\tau_{2} ,2K'-\tau_{1})  = 
\cR\cP(\tau_{2} ,\tau_{1})\cR 
\en


\noindent
The imaginary period monodromy matrix is
\eq
\Moni = \cP(4K',0)
\en
Define
\eq
\Monifour = \cP(K',0) \qquad
\Monitwo = \cP(2K',0) 
\en
Then
\eqa
\Monitwo &amp;= \cP(2K',K') \cP(K',0) = \cR \cP(0,K') \cR \cP(K',0)
\\[1ex]
&amp;= \cR \Monifour^{-1} \cR\Monifour = \cR \Omega \Monifour^{t} \Omega \cR \Monifour
\\[2ex]
\Moni &amp;= \cP(4K',2K') \cP(2K',0) = \overline{\cP(0,2K')} \cP(2K',0) 
\\[1ex]
&amp;= \overline{\Monitwo}^{-1} \Monitwo= \Omega \Monitwo^{\dagger}\Omega \Monitwo
\ena
so to calculate $\Moni$ it is enough to calculate $\Monifour$.
It is enough to integrate the ode from $K$ to $K+K'i$.


\subsection{$\cV_{\phys}(t)$}

For an ode with $N_{\gauge}&gt;0$, the gauge solution is
$\cW_{\gauge}(z)$.
The physical phase-space $\cV_{\phys}(t)$ is 
the quotient $\cW_{\gauge}^{\perp} /\Complexes \cW_{\gauge}$
where  $\cW_{\gauge}^{\perp}\Complexes^{N}$ is the $\Omega$-complement
of $\cW_{\gauge}(z)$.
For the numerical computations it is useful to represent $\cV_{\phys}(t)$
as a subspace of $\cW_{\gauge}^{\perp}$.
Write
\eq
\cV= \cV
\qquad
\cV_{\phys}= \cV_{\phys}(t)
\qquad
\cW_{\gauge} = \cW_{\gauge}(t)
\en
Define the vector
\eqg
\tilde\cW_{\gauge} = i \Omega \cW_{\gauge}
\eng
satisfying
\eqg
\cW_{\gauge}^{t} \tilde \cW_{\gauge} = 0
\qquad
\cW_{\gauge}^{t}\cW_{\gauge} = \tilde \cW_{\gauge}^{t}\tilde \cW_{\gauge}
\\[1ex]
\cW_{\gauge}^{t} \Omega \tilde \cW_{\gauge} =
i \cW_{\gauge}^{t} \cW_{\gauge}
\eng
$\cW_{\gauge}$ is real so $\tilde\cW_{\gauge}$ is also real.
The orthogonal complement $\tilde\cW_{\gauge}^{\perp}$ is the 
$\Omega$-orthogonal complement of $\cW_{\gauge}$.
\eqg
\tilde\cW_{\gauge}^{t} \cW = 0
\quad
\Leftrightarrow
\quad
\cW^{t}_{\gauge} \Omega \cW = 0
\eng
$\tilde \cW$ is not a natural vector since $\Omega$ is a bilinear form
on $\cV$, not a linear operator.  $\tilde \cW$ depends on a choice of 
bilinear form on $\cV$.
Choosing $\tilde \cW_{\gauge}$ give the  decomposition
\eq
\cV = \Complexes \tilde \cW_{\gauge} \oplus \Complexes  \cW_{\gauge} 
\oplus \cV_{\phys}
\en
representing $\cV_{\phys}$ as a codimension two subspace of $\cV$.
Define 
\eq
P = \cW_{\gauge} (\cW_{\gauge}^{t}\cW_{\gauge})^{-1}
\cW_{\gauge}^{t}
\qquad
\tilde P = \tilde\cW_{\gauge} (\tilde\cW_{\gauge}^{t}\tilde\cW_{\gauge})^{-1}
\tilde\cW_{\gauge}^{t}
\en
which are commuting projections
\eqg
P^{2} = P
\qquad
\tilde P^{2} = \tilde P
\qquad
P\tilde P = \tilde P P = 0
\qquad
P^{t} = P
\qquad
\tilde P^{t} =\tilde  P
\eng
Then define
\eq
P_{\gauge} = 1-\tilde P
\qquad
P_{\phys} = P_{\gauge} - P = 1-\tilde P -P
\en
so
\eqg
P_{\gauge}\cV  = (W_{\gauge})^{\perp_{\Omega}}
\qquad
\cV = \Complexes \tilde \cW_{\gauge} \oplus  (W_{\gauge})^{\perp_{\Omega}}
\\[1ex]
P_{\phys}\cV =  \cV_{\phys}
\qquad
(W_{\gauge})^{\perp_{\Omega}}  =  \Complexes  \cW_{\gauge} 
\oplus \cV_{\phys}
\\[1ex]
\cV =\Complexes \tilde \cW_{\gauge} \oplus \Complexes  \cW_{\gauge} 
\oplus \cV_{\phys}
\eng

\eqg
i\Omega \cP(i\Omega)^{t}  = 
\tilde\cW_{\gauge} (\cW_{\gauge}^{t}\cW_{\gauge})^{-1} \tilde\cW_{\gauge}^{t}
= \tilde \cP
\\[1ex]
\Omega \cP \Omega = \tilde \cP
\qquad
\Omega \cP = \tilde \cP \Omega
\qquad
\Omega = \tilde \cP \Omega \cP +  \cP \Omega \tilde \cP
\eng

%\newpage

\bibliographystyle{ytphys}

\raggedright
\phantomsection
\bibliography{Literature}
\addcontentsline{toc}{section}{\numberline{}References}

\end{document}

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