Some additions on cosmology

Signed-off-by: Riccardo Finotello <riccardo.finotello@gmail.com>
This commit is contained in:
2020-10-04 12:09:49 +02:00
parent 5123a92e78
commit 5a7583760e
6 changed files with 1574 additions and 636 deletions

View File

@@ -4,6 +4,7 @@
\RequirePackage{amsmath} %--------------------- math mode
\RequirePackage{amssymb} %--------------------- math symbols
\RequirePackage{amsfonts} %-------------------- math fonts
\RequirePackage{bm} %-------------------------- boldsymbols
\RequirePackage{mathtools} %------------------- mathematical tools
\RequirePackage{mathrsfs} %-------------------- better cal
\RequirePackage{slashed} %--------------------- slashed characters

View File

@@ -3,53 +3,53 @@ In this appendix we explain the conventions used for \SU{2} and show the details
\subsection{Conventions}
We parameterise \SU{2} matrices $U$ with a vector $\vb{n} \in \R^3$ such that:
We parameterise \SU{2} matrices $U$ with a vector $\vec{n} \in \R^3$ such that:
\begin{equation}
U(\vb{n})
U(\vec{n})
=
\cos(2 \pi n)\, \1_2
+
i\, \frac{\vb{n} \cdot \vb{\sigma}}{n}\, \sin(2 \pi n),
i\, \frac{\vec{n} \cdot \vec{\sigma}}{n}\, \sin(2 \pi n),
\label{eq:su2parametrisation}
\end{equation}
where $n = \norm{\vb{n}}$ and $0 \le n \le \frac{1}{2}$.
We also identify all $\vb{n}$ when $n=\frac{1}{2}$ since in this case $U(\vb{n})= -\1_2$.
where $n = \norm{\vec{n}}$ and $0 \le n \le \frac{1}{2}$.
We also identify all $\vec{n}$ when $n=\frac{1}{2}$ since in this case $U(\vec{n})= -\1_2$.
The parametrisation is such that:
\begin{eqnarray}
U^*(\vb{n})
U^*(\vec{n})
& = &
\sigma^2\, U(\vb{n})\, \sigma^2
\sigma^2\, U(\vec{n})\, \sigma^2
=
U(\widetilde{\vb{n}}),
U(\widetilde{\vec{n}}),
\\
U^{\dagger}(\vb{n})
U^{\dagger}(\vec{n})
& = &
U^T(\widetilde{\vb{n}})
U^T(\widetilde{\vec{n}})
=
U(-\vb{n}),
U(-\vec{n}),
\\
-U(\vb{n})
-U(\vec{n})
& = &
U(\widehat{\vb{n}})
U(\widehat{\vec{n}})
\label{eq:U_props}
\end{eqnarray}
where $\sigma^2$ is the second Pauli matrix, $\widetilde{\vb{n}} = \qty( -n^1, n^2, -n^3 )$ and $\widehat{\vb{n}} = - \qty(\frac{1}{2} -n )\, \frac{\vb{n}}{n}$.
where $\sigma^2$ is the second Pauli matrix, $\widetilde{\vec{n}} = \qty( -n^1, n^2, -n^3 )$ and $\widehat{\vec{n}} = - \qty(\frac{1}{2} -n )\, \frac{\vec{n}}{n}$.
The group product of two elements $U(\vb{n} \circ \vb{m} ) = U(\vb{n})\, U(\vb{m})$ has an explicit realisation as:
The group product of two elements $U(\vec{n} \circ \vec{m} ) = U(\vec{n})\, U(\vec{m})$ has an explicit realisation as:
\begin{equation}
\begin{split}
\cos(2 \pi \norm{\vb{n} \circ \vb{m}})
\cos(2 \pi \norm{\vec{n} \circ \vec{m}})
& =
\cos(2 \pi n)\, \cos(2 \pi m)
-
\sin(2 \pi n)\, \sin(2\pi m)\, \frac{\vb{n} \cdot \vb{m}}{n\, m},
\sin(2 \pi n)\, \sin(2\pi m)\, \frac{\vec{n} \cdot \vec{m}}{n\, m},
\\
\sin(2 \pi \norm{\vb{n} \circ \vb{m}})\,
\frac{\vb{n} \circ \vb{m}}{\norm{\vb{n} \circ \vb{m}}}
\sin(2 \pi \norm{\vec{n} \circ \vec{m}})\,
\frac{\vec{n} \circ \vec{m}}{\norm{\vec{n} \circ \vec{m}}}
& =
\cos(2 \pi n)\, \sin(2\pi m)\, \frac{\vb{m}}{m}
\cos(2 \pi n)\, \sin(2\pi m)\, \frac{\vec{m}}{m}
+
\sin(2 \pi n)\, \cos(2\pi m)\, \frac{\vb{n}}{n}.
\sin(2 \pi n)\, \cos(2\pi m)\, \frac{\vec{n}}{n}.
\end{split}
\label{eq:product_in_SU2}
\end{equation}
@@ -58,9 +58,9 @@ The group product of two elements $U(\vb{n} \circ \vb{m} ) = U(\vb{n})\, U(\vb{
Let $I = 1,\, 2,\, 3,\, 4$ and define:
\begin{equation}
\tau_I = \qty( i\, \1_2,\, \vb{\sigma} ),
\tau_I = \qty( i\, \1_2,\, \vec{\sigma} ),
\end{equation}
where $\vb{\sigma} = \qty( \sigma^1,\, \sigma^2,\, \sigma^3 )$ are the Pauli matrices.
where $\vec{\sigma} = \qty( \sigma^1,\, \sigma^2,\, \sigma^3 )$ are the Pauli matrices.
It is possible to show that:
\begin{equation}
\begin{split}
@@ -122,7 +122,7 @@ If the vector $X^I$ is real, using~\eqref{eq:tau_props} we have:
A rotation in spinor representation is defined as:
\begin{equation}
X'_{(s)} = U_{L}(\vb{n})\, X_{(s)}\, U_{R}^{\dagger}(\vb{m})
X'_{(s)} = U_{L}(\vec{n})\, X_{(s)}\, U_{R}^{\dagger}(\vec{m})
\end{equation}
and it is equivalent to:
\begin{equation}
@@ -138,9 +138,9 @@ through
\frac{1}{2}
\tr(
\qty( \tau_I )^{\dagger}\,
U_{L}(\vb{n})\,
U_{L}(\vec{n})\,
\tau_J\,
U_{R}^{\dagger}(\vb{m})
U_{R}^{\dagger}(\vec{m})
).
\end{equation}
The matrix $R$ is the $4$-dimensional rotation matrix we are looking for since:

View File

@@ -6,28 +6,28 @@ In this appendix we show the computation of the parameters of the hypergeometric
In the main text we set
\begin{equation}
D~
\rM_{\vb{\infty}}~
\rM_{\infty}~
D^{-1}
=
e^{-2\pi i \delta_{\vb{\infty}}}\,
\cL(\vb{n}_{\vb{\infty}}),
e^{-2\pi i \delta_{\infty}}\,
\cL(\vec{n}_{\infty}),
\end{equation}
where $\cL(\vb{n}_{\vb{\infty}}) \in \SU{2}$.
where $\cL(\vec{n}_{\infty}) \in \SU{2}$.
The previous equation implies
\begin{equation}
\qty( D\, \rM_{\vb{\infty}}\, D^{-1} )^\dagger
\qty( D\, \rM_{\infty}\, D^{-1} )^\dagger
=
\qty( D\, \rM_{\vb{\infty}}\, D^{-1} )^{-1},
\qty( D\, \rM_{\infty}\, D^{-1} )^{-1},
\end{equation}
which can be rewritten as
\begin{equation}
\widetilde{\rM}_{\vb{\infty}}^{-1}~
\widetilde{\rM}_{\infty}^{-1}~
\cC^{\dagger}\, D^{\dagger}\, D\, \cC
=
\cC^{\dagger}\, D^{\dagger}\, D\, \cC~
\widetilde{\rM}_{\vb{\infty}}^{-1}.
\widetilde{\rM}_{\infty}^{-1}.
\end{equation}
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.
As $\widetilde{\rM}_{\infty}$ is a generic diagonal matrix, the previous equation implies that the off-diagonal elements of $\cC^{\dagger}\, D^{\dagger}\, D\, \cC$ must vanish.
We therefore have
\begin{equation}
\begin{split}
@@ -76,71 +76,71 @@ 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 $\omega_{\bart-1} = 0$ is simpler to compute given that we choose $\cL(\vb{n}_{\vb{0}})$ and $\cR(\widetilde{\vb{m}}_{\vb{0}})$ to be diagonal.
The monodromy in $\omega_{\bart-1} = 0$ is simpler to compute given that we choose $\cL(\vec{n}_{0})$ and $\cR(\widetilde{\vec{m}}_{0})$ to be diagonal.
We impose:
\begin{eqnarray}
\mqty( \dmat{1, e^{-2\pi i c^{(L)}}} )
& = &
e^{-2\pi i \delta_{\vb{0}}^{(L)}}\,
\mqty( \dmat{e^{2\pi i n_{\vb{0}}}, e^{-2\pi i n_{\vb{0}}}} ),
e^{-2\pi i \delta_{0}^{(L)}}\,
\mqty( \dmat{e^{2\pi i n_{0}}, e^{-2\pi i n_{0}}} ),
\\
\mqty( \dmat{1, e^{-2\pi i c^{(R)}}} )
& = &
e^{-2\pi i \delta_{\vb{0}}^{(R)}}\,
\mqty( \dmat{e^{-2\pi i m_{\vb{0}}}, e^{2\pi i m_{\vb{0}}}} ),
e^{-2\pi i \delta_{0}^{(R)}}\,
\mqty( \dmat{e^{-2\pi i m_{0}}, e^{2\pi i m_{0}}} ),
\end{eqnarray}
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}.
where $n^3_{0} = \norm{\vec{n}_{0}} = n_{0}$ and $m^3_{0} = \norm{\vec{m}_{0}} = m_{0}$ with $0 \le n_{0},\, m_{0} < 1$ due to the conventions \eqref{eq:maximal_torus_left} and \eqref{eq:maximal_torus_right}.
We thus have:
\begin{equation}
\begin{split}
\delta_{\vb{0}}^{(L)}
\delta_{0}^{(L)}
& =
n_{\vb{0}} + k_{\delta^{(L)}_{\vb{0}}},
n_{0} + k_{\delta^{(L)}_{0}},
\qquad
k_{\delta^{(L)}_{\vb{0}}} \in \Z,
k_{\delta^{(L)}_{0}} \in \Z,
\\
c^{(L)}
& =
2 n_{\vb{0}} + k_c,
2 n_{0} + k_c,
\qquad
k_c \in \Z.
\end{split}
\label{eq:cL}
\end{equation}
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 $\alpha \le \delta_{\vb{0}}^{(L)} \le \alpha + \frac{1}{2}$.
Given that $0 \le n_{\vb{0}} < \frac{1}{2}$ we simply take $\alpha = 0$ and set $\delta_{\vb{0}}^{(L)} = n_{\vb{0}}$.
Since the determinant of the right hand side is $e^{-4 \pi i \delta_{0}^{(L)}}$, the range of definition of $\delta_{0}^{(L)}$ is $\alpha \le \delta_{0}^{(L)} \le \alpha + \frac{1}{2}$.
Given that $0 \le n_{0} < \frac{1}{2}$ we simply take $\alpha = 0$ and set $\delta_{0}^{(L)} = n_{0}$.
Analogous results hold in the right sector.
Furthermore from the third equation in \eqref{eq:parameters_equality_zero} and from the first equation in \eqref{eq:cL} we can restrict:
\begin{equation}
n_{\vb{0}} + m_{\vb{0}} - A \in \Z.
n_{0} + m_{0} - A \in \Z.
\end{equation}
We then need to find $3$ equations to determine $a^{(L)}$, $b^{(L)}$ and $\delta^{(L)}_{\vb{\infty}}$.
We then need to find $3$ equations to determine $a^{(L)}$, $b^{(L)}$ and $\delta^{(L)}_{\infty}$.
After that we then fix the remaining factors in $B$ and $\abs{K^{(L)}}$.
The equations follow from~\eqref{eq:parameters_equality_infty}.
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}:
The first two equations for $a^{(L)}$, $b^{(L)}$ and $\delta^{(L)}_{\infty}$ follow by considering the trace of~\eqref{eq:parameters_equality_infty}:
\begin{equation}
e^{\pi i ( a^{(L)} + b^{(L)} )} \cos(\pi( a^{(L)} - b^{(L)} ) )
=
e^{-2\pi i \delta^{(L)}_{\infty}} \cos(2\pi n_{\vb{\infty}}),
e^{-2\pi i \delta^{(L)}_{\infty}} \cos(2\pi n_{\infty}),
\end{equation}
which is satisfied by:
\begin{equation}
\begin{split}
\delta^{(L)}_{\vb{\infty}}
\delta^{(L)}_{\infty}
& =
-
\frac{1}{2}(a^{(L)} + b^{(L)})
+
\frac{1}{2} k_{\delta^{(L)}_{\vb{\infty}}},
\frac{1}{2} k_{\delta^{(L)}_{\infty}},
\qquad
k_{\delta_{\vb{\infty}}} \in \Z,
k_{\delta_{\infty}} \in \Z,
\\
a^{(L)} - b^{(L)}
& =
2\, (-1)^{p^{(L)}}\, n_{\vb{\infty}}
2\, (-1)^{p^{(L)}}\, n_{\infty}
+
(-1)^{q^{(L)}}\, k_{\delta^{(L)}_{\vb{\infty}}}
(-1)^{q^{(L)}}\, k_{\delta^{(L)}_{\infty}}
+
2\, k'_{a b},
\qquad
@@ -154,31 +154,31 @@ We therefore have:
\begin{equation}
a^{(L)} - b^{(L)}
=
2\, n_{\vb{\infty}}
2\, n_{\infty}
+
k_{\delta^{(L)}_{\vb{\infty}}}
k_{\delta^{(L)}_{\infty}}
+
2 k_{ab},
\qquad
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 $\omega_{\bart-1} = 0$.
The allowed values for $k_{\delta^{(L)}_{\infty}}$ 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 $k_{\delta^{(L)}_{\infty}} = 0,\, 1$.
As in the previous case we have $\alpha \le \delta_{\infty}^{(L)} \le \alpha + \frac{1}{2}$ with $\alpha$ technically arbitrary.
We cannot thus choose a vanishing $k_{\delta^{(L)}_{\infty}}$ but we have to consider $k_{\delta^{(L)}_{\infty}} = 0,\, 1$.
We find a third relation by considering the entry
\begin{equation}
\Im\qty(
e^{+2\pi i \delta_{\vb{\infty}}^{(L)}}\,
e^{+2\pi i \delta_{\infty}^{(L)}}\,
D^{(L)}\,
\rM_{\vb{\infty}}^{(L)}\,
\rM_{\infty}^{(L)}\,
\qty( D^{(L)} )^{-1}
)_{11}
=
\Im\qty(
\cL(n_{\vb{\infty}})
\cL(n_{\infty})
)_{11}.
\end{equation}
Using
@@ -191,31 +191,31 @@ and the second equation in~\eqref{eq:cL} and~\eqref{eq:aL-bL} leads to:
\begin{equation}
\cos(\pi( a^{(L)} + b^{(L)} - c^{(L)} ))
=
(-1)^{k_c+k_{\delta^{(L)}_{\vb{\infty}}} }\, \cos(2\pi \cA^{(L)}),
(-1)^{k_c+k_{\delta^{(L)}_{\infty}} }\, \cos(2\pi \cA^{(L)}),
\end{equation}
where
\begin{equation}
\cos(2\pi \cA^{(L)})
=
\cos(2\pi n_{\vb{0}})\,
\cos(2\pi n_{\vb{\infty}})
\cos(2\pi n_{0})\,
\cos(2\pi n_{\infty})
-
\sin(2\pi n_{\vb{0}})\,
\sin(2\pi n_{\vb{\infty}})\,
\frac{n_{\vb{\infty}}^3}{n_{\vb{\infty}}}.
\sin(2\pi n_{0})\,
\sin(2\pi n_{\infty})\,
\frac{n_{\infty}^3}{n_{\infty}}.
\label{eq:cos_n1}
\end{equation}
This expression is connected with rotation parameter in the third interaction point $\omega_{\bart+1} = 1$.
In fact $\cos(2\pi \cA^{(L)}) = \cos(2\pi {n}_{\vb{1}})$.
In fact $\cos(2\pi \cA^{(L)}) = \cos(2\pi {n}_{1})$.
We then write
\begin{equation}
a^{(L)} + b^{(L)} - c^{(L)}
=
2\, (-1)^{f^{(L)}}\, n_{\vb{1}}
2\, (-1)^{f^{(L)}}\, n_{1}
+
k_c
+
k_{\delta^{(L)}_{\vb{\infty}}}
k_{\delta^{(L)}_{\infty}}
+
2\, k_{abc},
\qquad
@@ -228,13 +228,13 @@ The request
+
B
-
n_{\vb{0}}
n_{0}
-
m_{\vb{0}}
m_{0}
-
(-1)^{f^{(L)}}\, n_{\vb{1}}
(-1)^{f^{(L)}}\, n_{1}
-
(-1)^{f^{(R)}}\, m_{\vb{1}}
(-1)^{f^{(R)}}\, m_{1}
\in \Z
\end{equation}
finally fixes the $B$ parameter in the third equation of~\eqref{eq:parameters_equality_infty}.
@@ -243,51 +243,51 @@ So far we can summarise the results in
\begin{eqnarray}
a
=
n_{\vb{0}} + (-1)^{f^{(L)}} n_{\vb{1}} + n_{\vb{\infty}} + m_a,
n_{0} + (-1)^{f^{(L)}} n_{1} + n_{\infty} + m_a,
& \qquad &
m_a \in \Z,
\\
b
=
n_{\vb{0}} + (-1)^{f^{(L)}} n_{\vb{1}} - n_{\vb{\infty}} + m_b,
n_{0} + (-1)^{f^{(L)}} n_{1} - n_{\infty} + m_b,
& \qquad &
m_b \in \Z,
\\
c
=
2\, n_{\vb{0}} + m_c,
2\, n_{0} + m_c,
& \qquad &
m_c \in \Z,
\\
\delta_{\vb{0}}^{(L)}
\delta_{0}^{(L)}
=
n_{\vb{0}},
n_{0},
\\
\delta_{\vb{\infty}}^{(L)}
\delta_{\infty}^{(L)}
=
- n_{\vb{0}} - (-1)^{f^{(L)}} n_{\vb{1}} + m_c + 2\, m_\delta,
- n_{0} - (-1)^{f^{(L)}} n_{1} + m_c + 2\, m_\delta,
& \qquad &
m_{\delta} \in \Z,
\\
A
=
n_{\vb{0}} + m_{\vb{0}} + m_A,
n_{0} + m_{0} + m_A,
& \qquad &
m_A \in \Z,
\\
B
=
(-1)^{f^{(L)}}\, n_{\vb{1}} + (-1)^{f^{(R)}}\, m_{\vb{1}} + m_B,
(-1)^{f^{(L)}}\, n_{1} + (-1)^{f^{(R)}}\, m_{1} + m_B,
& \qquad &
m_B \in \Z.
\end{eqnarray}
$K^{(L)}$ is finally determined from
\begin{equation}
\qty( D^{(L)}\, \rM_{\vb{\infty}}\, \qty( D^{(L)} )^{-1} )_{21}
\qty( D^{(L)}\, \rM_{\infty}\, \qty( D^{(L)} )^{-1} )_{21}
=
e^{-2\pi i \delta_{\vb{\infty}}^{(L)}}\,
\qty( \cL(n_{\vb{\infty}}) )_{21},
e^{-2\pi i \delta_{\infty}^{(L)}}\,
\qty( \cL(n_{\infty}) )_{21},
\label{eq:fixing_K_21}
\end{equation}
and get:
@@ -296,9 +296,9 @@ and get:
=
-\frac{(-1)^{m_a + m_b + m_c}}{2 \pi^2}\,
\cG( a^{(L)},\, b^{(L)},\, c^{(L)} )\,
\sin(2 \pi n_{\vb{0}})
\sin(2 \pi n_{\vb{\infty}})
\frac{n^1_{\vb{\infty}} + i\, n^2_{\vb{\infty}}}{n_{\vb{\infty}}},
\sin(2 \pi n_{0})
\sin(2 \pi n_{\infty})
\frac{n^1_{\infty} + i\, n^2_{\infty}}{n_{\infty}},
\label{eq:app_B_K21}
\end{equation}
where $\cG( a,\, b,\, c ) = \gfun{1-a}\, \gfun{1-b}\, \gfun{a+1-c}\, \gfun{b+1-c}$.
@@ -316,16 +316,16 @@ The result is
\cG(1 - a^{(L)},\, 1 - b^{(L)},\, 2 - c^{(L)})\,
\\
& \times
\sin(2 \pi n_{\vb{0}})\,
\sin(2 \pi n_{\vb{\infty}})\,
\frac{n^1_{\vb{\infty}} -i n^2_{\vb{\infty}}}{n_{\vb{\infty}}},
\sin(2 \pi n_{0})\,
\sin(2 \pi n_{\infty})\,
\frac{n^1_{\infty} -i n^2_{\infty}}{n_{\infty}},
\end{split}
\label{eq:app_B_K12}
\end{equation}
where the function $\cG( a,\, b,\, c )$ was defined at the end of the previous section.
Compatibility with~\eqref{eq:app_B_K21} requires
\begin{equation}
\frac{(n^1_{\vb{\infty}})^2 + (n^2_{\vb{\infty}})^2}{n^2_{\vb{\infty}}}
\frac{(n^1_{\infty})^2 + (n^2_{\infty})^2}{n^2_{\infty}}
=
-4 \frac{\sin(\pi a) \sin(\pi(c-a))\sin(\pi b) \sin(\pi(c-b))}
{\sin^2(\pi c) \sin^2(\pi(a-b))}.
@@ -333,7 +333,7 @@ Compatibility with~\eqref{eq:app_B_K21} requires
\end{equation}
We can then rewrite~\eqref{eq:cos_n1} as
\begin{equation}
\frac{(n^3_{\vb{\infty}})^2}{n^2_{\vb{\infty}}}
\frac{(n^3_{\infty})^2}{n^2_{\infty}}
=
\frac{(\cos(\pi (a-b)) \cos(\pi c)- \cos(\pi(a+b-c)))^2}
{\sin^2(\pi c) \sin^2(\pi(a-b))}.
@@ -341,10 +341,10 @@ We can then rewrite~\eqref{eq:cos_n1} as
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 $1$.
The same consistency check can also be performed by computing $K^{(L)}$ from
\begin{equation}
\qty( D^{(L)}\, \rM_{\vb{\infty}}\, \qty( D^{(L)} )^{-1} )_{12}
\qty( D^{(L)}\, \rM_{\infty}\, \qty( D^{(L)} )^{-1} )_{12}
=
e^{-2\pi i \delta_{\vb{\infty}}^{(L)}}\,
\qty( \cL(n_{\vb{\infty}}) )_{12},
e^{-2\pi i \delta_{\infty}^{(L)}}\,
\qty( \cL(n_{\infty}) )_{12},
\end{equation}
instead of \eqref{eq:fixing_K_21}.

File diff suppressed because it is too large Load Diff

View File

@@ -1308,12 +1308,12 @@ These stacks would separately lead to a $\U{3} \times \U{2}$ gauge theory.
It would however be theory of pure force, without matter content.
Moreover we should also worry about the extra \U{1} groups appearing: these need careful consideration but go beyond the necessary analysis for what follows.
Matter fields are notoriously fermions transforming in the bi-fundamental representation $(\vb{N}, \vb{M})$ of the \sm gauge group~\eqref{eq:intro:smgroup}.
For example left handed quarks in the \sm transform under the $(\vb{3}, \vb{2})$ representation of the group $\SU{3}_C \otimes \SU{2}_L$.
Matter fields are notoriously fermions transforming in the bi-fundamental representation $(\vec{N}, \vec{M})$ of the \sm gauge group~\eqref{eq:intro:smgroup}.
For example left handed quarks in the \sm transform under the $(\vec{3}, \vec{2})$ representation of the group $\SU{3}_C \otimes \SU{2}_L$.
This is realised in string theory by a string stretched across two stacks of $3$ and $2$ D-branes as in the right of~\Cref{fig:dbranes:chanpaton}.
The fermion would then be characterised by the charge under the gauge bosons living on the D-branes.
The corresponding anti-particle would then simply be a string oriented in the opposite direction.
Things get complicated when introducing also left handed leptons transforming in the $(\vb{1}, \vb{2})$ representation: they cannot have endpoints on the same stack of D-branes as quarks since they do not have colour charge.
Things get complicated when introducing also left handed leptons transforming in the $(\vec{1}, \vec{2})$ representation: they cannot have endpoints on the same stack of D-branes as quarks since they do not have colour charge.
We therefore need to introduce more D-branes to account for all the possible combinations.
An additional issue comes from the requirement of chirality.
@@ -1338,7 +1338,7 @@ The light spectrum is thus composed of the desired matter content alongside with
\end{figure}
It is therefore possible to recover a \sm-like construction using multiple D-branes at angles as in~\Cref{fig:dbranes:smbranes}, where the angles have been drawn perpendicular but can in principle be arbitrary~\cite{Ibanez:2001:GettingJustStandard,Grimm:2005:EffectiveActionType,Sheikh-Jabbari:1998:ClassificationDifferentBranes,Berkooz:1996:BranesIntersectingAngles}.
For instance quarks are localised at the intersection of the \emph{baryonic} stack of D-branes, yielding the colour symmetry generators, with the \emph{left} and \emph{right} stacks, leading to the $( \vb{3}, \vb{2} )$ and $( \vb{3}, \vb{1})$ representations.
For instance quarks are localised at the intersection of the \emph{baryonic} stack of D-branes, yielding the colour symmetry generators, with the \emph{left} and \emph{right} stacks, leading to the $( \vec{3}, \vec{2} )$ and $( \vec{3}, \vec{1})$ representations.
The same applies to leptons created by strings attached to the \emph{leptonic} stack.
Combinations of the additional \U{1} factors in the resulting gauge group finally lead to the definition of the hypercharge $Y$.

File diff suppressed because it is too large Load Diff