Up to NS fermions
Signed-off-by: Riccardo Finotello <riccardo.finotello@gmail.com>
This commit is contained in:
@@ -33,7 +33,7 @@ The parametrisation is such that:
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U(\widehat{\vb{n}})
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\label{eq:U_props}
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\end{eqnarray}
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where $\sigma^2$ is the second Pauli matrix, $\widetilde{\vb{n}} = \left( -n^1, n^2, -n^3 \right)$ and $\widehat{\vb{n}} = - \left(\frac{1}{2} -n \right)\, \frac{\vb{n}}{n}$.
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where $\sigma^2$ is the second Pauli matrix, $\widetilde{\vb{n}} = \qty( -n^1, n^2, -n^3 )$ and $\widehat{\vb{n}} = - \qty(\frac{1}{2} -n )\, \frac{\vb{n}}{n}$.
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The group product of two elements $U(\vb{n} \circ \vb{m} ) = U(\vb{n})\, U(\vb{m})$ has an explicit realisation as:
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\begin{equation}
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@@ -58,17 +58,17 @@ The group product of two elements $U(\vb{n} \circ \vb{m} ) = U(\vb{n})\, U(\vb{
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Let $I = 1,\, 2,\, 3,\, 4$ and define:
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\begin{equation}
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\tau_I = \left( i\, \1_2,\, \vb{\sigma} \right),
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\tau_I = \qty( i\, \1_2,\, \vb{\sigma} ),
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\end{equation}
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where $\vb{\sigma} = \left( \sigma^1,\, \sigma^2,\, \sigma^3 \right)$ are the Pauli matrices.
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where $\vb{\sigma} = \qty( \sigma^1,\, \sigma^2,\, \sigma^3 )$ are the Pauli matrices.
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It is possible to show that:
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\begin{equation}
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\begin{split}
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\left( \tau_I \right)^{\dagger}
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\qty( \tau_I )^{\dagger}
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& =
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\eta_{IJ}\, {\tau}^I,
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\\
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\left( \tau^I \right)^*
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\qty( \tau^I )^*
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& =
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-\sigma_2\, \tau_I\, \sigma_2,
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\end{split}
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@@ -85,7 +85,7 @@ The following relations are then a natural consequence:
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& = &
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2\, \eta_{IJ},
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\\
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\tr(\tau_I \left( \tau_J \right)^{\dagger})
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\tr(\tau_I \qty( \tau_J )^{\dagger})
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& = &
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2\, \delta_{IJ}.
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\end{eqnarray}
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@@ -99,7 +99,7 @@ We can recover the components using the previous properties:
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X^I
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=
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\frac{1}{2}\, \delta^{IJ}\,
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\tr(X_{(s)} \left( \tau_J \right)^{\dagger})
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\tr(X_{(s)} \qty( \tau_J )^{\dagger})
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=
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\frac{1}{2}\, \eta^{IJ}\, \tr(X_{(s)} \tau_J),
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\end{equation}
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@@ -126,7 +126,7 @@ A rotation in spinor representation is defined as:
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\end{equation}
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and it is equivalent to:
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\begin{equation}
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\left( X' \right)^I
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\qty( X' )^I
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=
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\tensor{R}{^I_J}\,
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X^J
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@@ -137,7 +137,7 @@ through
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=
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\frac{1}{2}
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\tr(
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\left( \tau_I \right)^{\dagger}\,
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\qty( \tau_I )^{\dagger}\,
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U_{L}(\vb{n})\,
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\tau_J\,
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U_{R}^{\dagger}(\vb{m})
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@@ -167,7 +167,7 @@ From the second equation in \eqref{eq:tau_props} and the first equation in \eqre
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\end{equation}
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Furthermore the direct computation of the determinant of $R$ using the parametrisation~\eqref{eq:su2parametrisation} shows that $\det R = 1$.
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Finally the explicit choice of the basis $\tau$ ensures $R$ to be a real matrix which ensures $R \in \SO{4}$.
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Since $\left\lbrace U_{L},\, U_{R} \right\rbrace$ and $\left\lbrace -U_{L},\, -U_{R} \right\rbrace$ generate the same \SO{4} matrix then the correct isomorphism takes the form:
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Since $\qty{ U_{L},\, U_{R} }$ and $\qty{ -U_{L},\, -U_{R} }$ generate the same \SO{4} matrix then the correct isomorphism takes the form:
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\begin{equation}
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\SO{4}
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\cong
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@@ -15,9 +15,9 @@ In the main text we set
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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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\qty( D\, \rM_{\vb{\infty}}\, D^{-1} )^\dagger
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=
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\left( D\, \rM_{\vb{\infty}}\, D^{-1} \right)^{-1},
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\qty( D\, \rM_{\vb{\infty}}\, D^{-1} )^{-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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@@ -147,7 +147,7 @@ which is satisfied by:
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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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where $p^{(L)},\, q^{(L)} \in \qty{ 0, 1 }$.
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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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@@ -170,16 +170,16 @@ We cannot thus choose a vanishing $k_{\delta^{(L)}_{\vb{\infty}}}$ but we have t
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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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\Im\qty(
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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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\qty( D^{(L)} )^{-1}
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)_{11}
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=
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\Im\left(
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\Im\qty(
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\cL(n_{\vb{\infty}})
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\right)_{11}.
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)_{11}.
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\end{equation}
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Using
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\begin{equation}
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@@ -221,7 +221,7 @@ We then write
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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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with $f^{(L)} \in \qty{ 0, 1 }$.
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The request
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\begin{equation}
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A
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@@ -284,10 +284,10 @@ So far we can summarise the results in
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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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\qty( D^{(L)}\, \rM_{\vb{\infty}}\, \qty( D^{(L)} )^{-1} )_{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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\qty( \cL(n_{\vb{\infty}}) )_{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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@@ -310,7 +310,7 @@ We check the consistency condition \eqref{eq:K_consistency_condition} using~\eqr
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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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\qty( K^{(L)} )^{-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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@@ -341,10 +341,10 @@ We can then rewrite~\eqref{eq:cos_n1} as
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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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\qty( D^{(L)}\, \rM_{\vb{\infty}}\, \qty( D^{(L)} )^{-1} )_{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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\qty( \cL(n_{\vb{\infty}}) )_{12},
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\end{equation}
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instead of \eqref{eq:fixing_K_21}.
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