text:
\begin{document}

\title
{Integrality Properties of Open-Closed Mirror Maps}
\author{Jian Zhou}
\address{Department of Mathematical Sciences\\Tsinghua University\\Beijing, 100084, China}
\email{jzhou@math.tsinghua.edu.cn}

\begin{abstract}
We propose a conjecture on integrality property of the open-closed mirror maps
of compact Calabi-Yau manifolds.
Some examples are presented.
\end{abstract}

\maketitle

In this paper we will propose a conjecture on integrality properties
of open-closed mirror maps for compact Calabi-Yau manifolds,
inspired by the Lian-Yau integrality for closed mirror map for compact Calabi-Yau manifold
\cite{Lia-Yau, Lia-Yau2, Zud, Kra-Riv, Kra-Riv2, Kra-Riv3, Del},
and its recent extension to the local Calabi-Yau case \cite{Zho1}.
The integrality of open-closed mirror map in the noncompact Calabi-Yau $3$-fold case
was observed in \cite{Zho1},
in this work we propose to extend it to the compact Calabi-Yau case and more noncompact examples, not necessarily three dimensional,
based on a related work \cite{Zho2}.

\section{Integrality Conjecture of Open-Closed Mirror Maps}

\subsection{The conjecture}

Suppose that we have a system of Picard-Fuchs equations in variables $x_0, x_1, \dots, x_N$
that comes from some charge vectors
describing some brane geometry in some Calabi-Yau geometry,
in the large volume phase.
Then it is expected that the system has a holomorphic solution
$g_0(x_0, x_1, \dots, x_N)$ such that $g_0(0, \dots, 0) = 1$
and $N$ logarithmic solutions of the form
$$g_1^{(i)}(x_0, \dots, x_N) = \log x_i \cdot g_0(x_0, \dots, x_N)
+ h_1^{(i)}(x_0, \dots, x_N), \qquad i =0, 1, \dots, N,$$
where $h_1^{(i)}$ are holomorphic functions such that
$h^{(i)}_1(0, \dots, 0) = 0$.
The open-closed mirror map is defined by:
\be
q_i = \exp(g_1^{(i)}/g_0) = x_i \exp (h_1^{(i)}/g_0), \qquad i=0, 1, \dots, N.
\ee
(For some examples,
see e.g. \cite{Ler-May}.)
On can find the inverse map by Lagrange-Good inversion formula \cite{Good} as in \cite{Zho1}.

Our conjecture is

\begin{Conjecture}
The Taylor series of $q_0, q_1, \dots, q_N$ in $x_0, x_1, \dots, x_N$ are integers,
and vice versa.
\end{Conjecture}

We will present some examples for which we verify this conjecture.
We will focus on some cases with one closed string moduli parameter.
Extensions to the case of multiple closed string moduli parameters
will be left to subsequent work.

\subsection{Some Picard-Fuchs systems related to open-closed mirror map}

Let $k_1,  \dots, k_n$ be positive integers such that
\be \label{eqn:Sum=1}
\frac{1}{k_1} + \cdots + \frac{1}{k_n} = 1.
\ee
See \cite{Zho2} for a list of solutions for $n=2,3,4,5$
under the assumption that $k_1 \leq  \dots \leq k_n$.
Here we do not require this assumption.
Let $k=l.c.m.(k_1, \dots, k_n)$ and $w_i = \frac{k}{k_i}$.
Then we have
\be
k_1 w_1 = \cdots = k_n w_n = w_1 + \cdots + w_n = k.
\ee
Consider charge vectors formed from such a solution:
\be \label{eqn:Charge0}
\begin{pmatrix}
\tilde{l}^{(1)} \\ \tilde{l}^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|cccc|cc}
-k & w_1 & w_2 & \cdots & w_n & 0 & 0 \\
-1 &  1 & 0 & \cdots & 0 & -1 & 1
\end{array} \biggr).
\ee
We will transform them to
\be
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|cccc|cc}
-k+w_1 & 0 & w_2 & \cdots & w_n & w_1 & -w_1 \\
-1 &  1 & 0 & \cdots & 0 & -1 & 1
\end{array} \biggr),
\ee
where $l^{(1)} = \tilde{l}^{(1)} - w_1 \tilde{l}^{(0)}$ and $l^{(0)} = \tilde{l}^{(0)}$.
We will consider the extended Picard-Fuchs system \cite{Ler-May} associated with three charge vectors
$l^{(1)}$, $l^{(0)}$ and $\tilde{l}_1 = l^{(1)} + w_1 l^{(0)}$:
\be \label{eqn:PF}
\cL_1 S(x_0,x_1) =0, \qquad \cL_0 S(x_0,x_1) = 0, \qquad \cL_1' S(x_0,x_1) = 0,
\ee
where
\bea
&& \cL_0 =  \big[ \theta_0 - x_0 (1 + \theta_0 + (k-w_1) \theta_1) \big] (\theta_0 - w_1 \theta_1), \\
&& \cL_1 = \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i \theta_1-j)
\prod_{j=0}^{w_1-1} (-\theta_0+w_1\theta_1-j) \\
&& \qquad - x_1 \prod_{j=1}^{k-w_1} (\theta_0+(k-w_1)\theta_1+j)
\prod_{j=0}^{w_1-1} (-\theta_0+w_1\theta_1+j), \nonumber \\
&& \cL_1'
= \prod_{j=0}^{w_1-1} (\theta_0-j) \cdot \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i \theta_1 -j)
- x_0^{w_1}x_1 \prod_{j=1}^{k} (\theta_0+(k-w_1)\theta_1+j),
\eea
where $\theta_i = x_i \frac{\pd}{\pd x_i}$, $i=0,1$.
When $n=4$,
some examples of this system correspond to the extended Picard-Fuchs system associated with
B-branes in compact Calabi-Yau $3$-folds in the large volume phase (see e.g. \cite{AHMM}).
See  \cite{Li-Lia-Yau} and its references for some recent work on extension of mirror symmetry to
the open string case.

\subsection{Derivation of the Picard-Fuchs operators}
Let us recall the well-known procedure of derivation of Picard-Fuchs operators from the charge vectors
for the convenience of the reader.
Given a vector $l = (l_0, l_1, \dots, l_{n+2}) \in \bZ^{n+3}$,
consider the affine complex space $\bC^{p+1}$ with linear coordinates $(a_0, \dots, a_{n+2})$,
define an operator
\be
\cD_l = \prod_{l_j > 0} \big( \frac{\pd}{\pd a_i} \big)^{l_j}
- \prod_{l_j < 0} \big( \frac{\pd}{\pd a_i} \big)^{-l_j}.
\ee
Introduce variables
\be
x_0 = \frac{a_1a_{n+2}}{a_0a_{n+1}}, \qquad
x_1 = (-1)^k \frac{a_2^{w_2} \cdots a_n^{w_n}a_{n+1}^{w_1}}{a_0^{k-w_1} a_{n+2}^{w_1}}.
\ee
Suppose that we have a function $\tilde{\Pi}(x_0,x_1)$ such that
\be
\Pi(a_0, \dots, a_{n+2}) = \frac{1}{a_0} \tilde{\Pi}(x_0,x_1)
\ee
satisfies the equations
\be
\cD_{l^{(i)}} \Pi(a_0, \dots, a_{n+2}) =0, \quad i =0, 1, \qquad
\cD_{\tilde{l}^{(1)}} \Pi(a_0, \dots, a_{n+2}) = 0.
\ee
For $i=0$,
\ben
&& \prod_{l_j^{(0)}>0} \big( \frac{\pd}{\pd a_i} \big)^{l^{(0)}_j} \Pi(a_0, \dots, a_{n+2})
= \frac{\pd}{\pd a_1} \frac{\pd}{\pd a_{n+2}} (\frac{1}{a_0} \tilde{\Pi}(x_0, x_1)) \\
& = & \frac{1}{a_0} \frac{\pd}{\pd a_1} (\frac{1}{x_0} \frac{\pd x_0}{\pd a_{n+2}} \theta_0 \tilde{\Pi}(x_0,x_1)
+ \frac{1}{x_1} \frac{\pd x_1}{\pd a_{n+2}} \theta_1 \tilde{\Pi}(x_0,x_1)) \\
& = & \frac{1}{a_0a_{n+2}} \frac{\pd}{\pd a_1} ( \theta_0 \tilde{\Pi}(x_0,x_1)
- w_1 \theta_1 \tilde{\Pi}(x_0,x_1)) \\
& = &  \frac{1}{a_0a_{n+2}}
(\frac{1}{x_0} \frac{\pd x_0}{\pd a_1} \theta_0
+ \frac{1}{x_1} \frac{\pd x_1}{\pd a_1} \theta_1) ( \theta_0 - w_1 \theta_1) \tilde{\Pi}(x_0,x_1)) \\
& = & \frac{1}{a_0a_1a_{n+2}} \theta_0 (\theta_0 - w_1 \theta_1) \tilde{\Pi}(x_0, x_1),
\een
and
\ben
&& \prod_{l_j < 0} \big( \frac{\pd}{\pd a_i} \big)^{-l^{(0)}_j}
\Pi(a_0, \dots, a_{n+2})
= \frac{\pd}{\pd a_0} \frac{\pd}{\pd a_{n+1}} (\frac{1}{a_0} \tilde{\Pi}(x_0,x_1)) \\
& = & \frac{\pd}{\pd a_0} (\frac{1}{a_0} \frac{1}{x_0}\frac{\pd x_0}{\pd a_{n+1}} \theta_0 \tilde{\Pi}(x_0,x_1)
+ \frac{1}{a_0} \frac{1}{x_1}\frac{\pd x_1}{\pd a_{n+1}} \theta_1 \tilde{\Pi}(x_0,x_1)) \\
& = & \frac{1}{a_{n+1}}\frac{\pd}{\pd a_0} (\frac{1}{a_0} \cdot
(-\theta_0 + w_1 \theta_1) \tilde{\Pi}(x_0,x_1)) \\
& = & \frac{1}{a_{n+1}} (-\frac{1}{a_0^2} + \frac{1}{a_0}  \frac{1}{x_0}\frac{\pd x_0}{\pd a_0} \theta_0
+ \frac{1}{a_0}  \frac{1}{x_1}\frac{\pd x_1}{\pd a_0} \theta_1)
(-\theta_0 + w_1 \theta_1) \tilde{\Pi}(x_0,x_1) \\
& = &  \frac{1}{a_0^2a_{n+1}} (-1 - \theta_0 - (k-w_1) \theta_1)
(-\theta_0 + w_1 \theta_1) \tilde{\Pi}(x_0,x_1).
\een
Therefore,
\ben
\theta_0 (\theta_0 - w_1 \theta_1) \tilde{\Pi}(x_0, x_1)
+ \frac{a_1a_{n+2}}{a_0a_{n+1}} (1 + \theta_0 + (k-w_1) \theta_1)
(-\theta_0 + w_1 \theta_1) \tilde{\Pi}(x_0,x_1) = 0.
\een
This is the equation $\cL_0 \tilde{\Pi}(x_0,x_1) = 0$.
The other two equations are derived in the same fashion.

\subsection{Holomorphic solutions}

In this subsection we will show that the extended Picard-Fuch system \eqref{eqn:PF}
has a holomorphic solution of the form
$$g_0 = 1 + \sum_{m_0+ m_1 > 0} a_{m_0,m_1} x_0^{m_0}x_1^{m_1}.$$
We will derive its explicit expression as follows.

The Picard-Fuchs equations for $g_0$ can be rewritten as follows:
\ben
&& \sum_{m_0+ m_1 > 0} m_0(m_0-w_1m_1) a_{m_0,m_1} x_0^{m_0}x_1^{m_1} \\
& = & x_0 \sum_{m_0+ m_1 > 0} (m_0-w_1m_1) (1+m_0+(k-w_1)m_1) a_{m_0,m_1} x_0^{m_0}x_1^{m_1}, \\
&& \sum_{m_0+ m_1 > 0} \prod_{j=0}^{w_1-1} (-m_0+w_1m_1+j) \prod_{i=2}^n \prod_{j=0}^{w_i-1}
(w_i m_1-j) a_{m_0,m_1} x_0^{m_0}x_1^{m_1} \\
& = &  x_1 \prod_{j=1}^{k-w_1} (m_0+(k-w_1)m_1+j)
\prod_{j=0}^{w_1-1} (-m_0+w_1m_1+j) a_{m_0,m_1} x_0^{m_0}x_1^{m_1}, \\
&& \sum_{m_0+ m_1 > 0} \prod_{j=0}^{w_1-1} (m_0-j) \cdot \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i m_1 -j) a_{m_0,m_1} x_0^{m_0}x_1^{m_1} \\
& = & x_0^{w_1}x_1(k!+ \sum_{m_0+ m_1 > 0} \prod_{j=1}^{k} (m_0+(k-w_1)m_1+j) a_{m_0,m_1} x_0^{m_0}x_1^{m_1}).
\een
These equations give rise to some recursions which can be solved directly to give us
the following solution:
\be
g_0(x_0,x_1) = \sum_{m \geq 0} \frac{(km)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m.
\ee
Clearly $g_0(x_0,x_1) \in \bZ[[x_0^{w_1}x_1]]$.

\subsection{Logarithmic solutions}

In this subsection we will show that the extended Picard-Fuch system \eqref{eqn:PF}
has two logarithmic solutions of the form
$$g_1^{(i)} = g_0 \log x_i + \sum_{m_0+ m_1 > 0} a^{(i)}_{m_0,m_1} x_0^{m_0}x_1^{m_1}$$ ($i=0,1$)
to the Picard-Fuchs system \eqref{eqn:PF}.
We will derive their explicit expressions as follows.
From $\cL_1'g_1^{(1)} =0$ we get
\ben
&& \sum_{m_0+ m_1 > 0}  \prod_{j=0}^{w_1-1} (m_0-j) \cdot \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i m_1 -j) a^{(1)}_{m_0,m_1} x_0^{m_0}x_1^{m_1} \\
& + & \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_i(m-1))!}
\sum_{i=2}^n \sum_{j=0}^{w_i-1} \frac{1}{w_im-j} (x_0^{w_1} x_1)^m\\
& = & x_0^{w_1}x_1 \sum_{m_0+ m_1 > 0} \prod_{j=1}^{k} (m_0+(k-w_1)m_1+j) a^{(1)}_{m_0,m_1} x_0^{m_0}x_1^{m_1} \\
& + & x_0^{w_1}x_1\sum_{m=0}^\infty \frac{(k(m+1))!}{\prod_{i=1}^n (w_im)!}
\sum_{j=1}^k \frac{k-w_1}{km+j} (x_0^{w_1}x_1)^m.
\een
Compare the coefficients of $x_0^{m_0}x_1^{m_1}$ on both sides.
When $w_0 = w_1 m$ and $w_2 = m$,
\ben
&& a^{(1)}_{w_1m, m}
+ \frac{(km)!}{\prod_{i=1}^n (w_im)!} \sum_{i=2}^n \sum_{j=0}^{w_i-1} \frac{1}{w_im-j}  \\
& = &  \frac{\prod_{j=0}^{k-1} (km-j)}{\prod_{i=1}^n \prod_{j=0}^{w_i-1} (w_i m -j)} a^{(1)}_{w_1(m-1),m-1}
+  \frac{(km)!}{\prod_{i=1}^n (w_im))!}\sum_{j=0}^{k-1} \frac{k-w_1}{km-j}.
\een
Hence we have
\be
a^{(1)}_{w_1m,m} =  \frac{(km)!}{\prod_{i=1}^n (w_im)!}
(\sum_{j=1}^{km} \frac{k-w_1}{j} - \sum_{i=2}^n \sum_{j=1}^{w_im} \frac{1}{j}).
\ee
For $m_1 \neq w_1m_0$,
we have
\ben
&& \prod_{j=0}^{w_1-1} (m_0-j) \cdot \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i m_1 -j) a^{(1)}_{m_0,m_1}  \\
& = & \prod_{j=0}^{k-1} (m_0+(k-w_1)m_1-j) a^{(1)}_{m_0-w_1,m_1-1},
\een
hence when $m_0 > w_1m_1$ we have
\be \label{eqn:>}
a^{(1)}_{m_0,m_1} = \frac{(m_0+(k-w_1)m_1)!}{m_0! \prod_{i=2}^n (w_im_1)!}
a^{(1)}_{m_0-w_1m_1, 0}.
\ee
We will determine $a^{(1)}_{m,0}$ from another equation.

From $\cL_0 g_1^{(1)} = 0$ we get
\ben
&& \sum_{m_0+ m_1 > 0} m_0(m_0-w_1m_1) a^{(1)}_{m_0,m_1} x_0^{m_0}x_1^{m_1}
- w_1  \sum_{m \geq 1} \frac{(km)!w_1m }{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m \\
& = & x_0 \sum_{m_0+ m_1 > 0} (m_0-w_1m_1) (1+m_0+(k-w_1)m_1) a^{(1)}_{m_0,m_1} x_0^{m_0}x_1^{m_1} \\
& - & w_1 x_0 \sum_{m \geq 0} \frac{(km+1)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m.
\een
Compare the coefficients of $x_0^{m_0}x_1^{m_1}$ on both sides.
For $m_0=m>0$ and $m_1 = 0$,
\ben
&& a_{1,0}^{(1)} = - w_1, \qquad
 m^2 a^{(1)}_{m,0} = m (m-1) a^{(1)}_{m-1,0},
\een
and so
\ben
a^{(1)}_{m,0} = - \frac{1}{m}w_1.
\een
Hence by combining with \eqref{eqn:>} for $m_0 > w_1m_1$,
\be
a^{(1)}_{m_0,m_1} = - w_1 \frac{(m_0+(k-w_1)m_1)!}{m_0! \prod_{i=2}^n (w_im_1)!} \frac{1}{m_0 - w_1m_1}.
\ee
When $m_0 \neq 0, w_1m_1$ or $w_1m_1+1$,
\be
a^{(1)}_{m_0,m_1}
=  \frac{(m_0-1-w_1m_1) (m_0+(k-w_1)m_1)}{m_0 (m_0-w_1m_1)} a^{(1)}_{m_0-1,m_1},
\ee
therefore,
for $m_0 < w_1m_1$,
we have
\be \label{eqn:<}
a^{(1)}_{m_0,m_1}
= \frac{(m_0+(k-w_1)m_1)!}{m_0!((k-w_1)m_1)!} \frac{w_1m_1}{w_1m_1-m_0} a^{(1)}_{0,m_1}.
\ee
We will determine $a^{(1)}_{0,m_1}$ from another equation.

From $\cL_1 g_1^{(1)} = 0$ we get:
\ben
&& \sum_{m_0+ m_1 > 0} \prod_{j=0}^{w_1-1} (-m_0+w_1m_1-j) \prod_{i=2}^n \prod_{j=0}^{w_i-1}
(w_i m_1-j) a^{(1)}_{m_0,m_1} x_0^{m_0}x_1^{m_1} \\
& + &  \sum_{m=1}^\infty
\frac{(-1)^{w_1-1} w_1! (km)!}{(w_1m)! \prod_{i=2}^n (w_i(m-1))!} (x_0^{w_1}x_1)^m \\
& = &   x_1 \sum_{m_0+ m_1 > 0} \prod_{j=1}^{k-w_1} (m_0+(k-w_1)m_1+j)
\prod_{j=0}^{w_1-1} (-m_0+w_1m_1+j) a^{(1)}_{m_0,m_1} x_0^{m_0}x_1^{m_1}\\
& + & x_1 \sum_{m=1}^\infty
\frac{w_1! (km+k-w_1)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m.
\een
By comparing the coefficients of $x_1^{m_1}$ we get:
\ben
&& \prod_{i=1}^n w_i! \cdot a^{(1)}_{0,1} =  w_1! (k-w_1)!, \\
&& \prod_{i=1}^n \prod_{j=0}^{w_i-1}
(w_i m_1-j) a^{(1)}_{0,m_1}  =  \prod_{j=0}^{k-w_1-1} ((k-w_1)m_1-j)
\prod_{j=1}^{w_1} (w_1m_1-j) a^{(1)}_{0,m_1-1},
\een
hence
\be
a^{(1)}_{0,m_1} =  \frac{((k-w_1)m_1)!}{\prod_{i=2}^n (w_im_1)!} \frac{1}{m_1}.
\ee
Combining with \eqref{eqn:<} we get:
\be
a^{(1)}_{m_0,m_1} =- w_1 \frac{(m_0+(k-w_1)m_1)!}{m_0!\prod_{i=2}^n (w_im)!} \frac{1}{m_0-w_1m_1}.
\ee

Therefore,
we get the following solution:
\bea
g_1^{(1)}(x_0,x_1) & = & \log x_1 \cdot \sum_{m \geq 0} \frac{(km)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m \\
& + & \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!}
(\sum_{j=1}^{km} \frac{k-w_1}{j} - \sum_{i=2}^n \sum_{j=1}^{w_im} \frac{1}{j})(x_0^{w_1}x_1)^m  \nonumber \\
& - & \sum_{m_0+m_1>0, m_0 \neq w_1m_1}
w_1 \frac{(m_0+(k-w_1)m_1)!}{m_0!\prod_{i=2}^n (w_im)!} \frac{x_0^{m_0} x_1^{m_1}}{m_0-w_1m_1}. \nonumber
\eea
In the same fashion one can get the following solution:
\bea
g_1^{(0)}(x_0,x_1) & = & \log x_0 \cdot \sum_{m \geq 0} \frac{(km)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m \\
& + & \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!}
(\sum_{j=1}^{km} \frac{1}{j} - \sum_{j=1}^{m} \frac{1}{w_1j}) (x_0^{w_1}x_1)^m \nonumber \\
& + & \sum_{m_0+m_1>0, m_0 \neq w_1m_1}
\frac{(m_0+(k-w_1)m_1)!}{m_0!\prod_{i=2}^n (w_im)!} \frac{x_0^{m_0} x_1^{m_1}}{m_0-w_1m_1}. \nonumber
\eea

\subsection{Open-closed mirror map and its integrality}

The open-closed mirror map is given by
\be
q_0 = \exp (g_1^{(0)}/g_0), \qquad
q_1 = \exp (g_1^{(1)} /g_0).
\ee
As a special case of Conjecture 1,
one should have
\be
q_0, q_1 \in \bZ[[x_0,x_1]].
\ee
Notice that in both cases we have some extra terms,
so the integrality property of open-closed mirror map is different from
the Lian-Yau integrality of closed mirror map,
because it involves the integrality of the following series:
\be
\exp (\sum_{m_0 \neq w_1m_1}
\frac{(m_0+(k-w_1)m_1)!}{m_0!\prod_{i=2}^n (w_im)!} \frac{x_0^{m_0} x_1^{m_1}}{m_0-w_1m_1}/
\sum_{m \geq 0} \frac{(km)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m).
\ee

Applying the Lagrange-Good inversion formula as in \cite{Zho1},
one can get the explicit expressions for the inverse open-closed mirror map.
Write
\be
q_i = \sum_{m_0+m_1>0} A_{m_0,m_1}^{(i)} x_0^{m_0}x_1^{m_1},
\ee
then $A_{m_0,m_1}^{(0)}$ and $A_{m_0,m_1}^{(1)}$ are the coefficients of $x_0^{m_0-1}x_1^{m_1}$
and $x_0^{m_0}x_1^{m_1-1}$ of
\be
(1+\theta_0 h_0 + \theta_1 h_1) \cdot \exp (-m_0h_0-m_1h_1),
\ee
where $h_i = g_1^{(i)}/g_0 - \log x_i$.

\subsection{The Picard-Fuchs system in a different phase}
One can also consider the Picard-Fuchs system associated with the charge vectors \eqref{eqn:Charge0}.
The associated variables are:
\be
\tilde{x}_0 = \frac{a_1a_{n+2}}{a_0a_{n+1}}, \qquad \tilde{x}_1 = (-1)^k \frac{a_1^{w_1} \cdots a_n^{w_n}}{a_0^k}.
\ee
I.e.,
\be \label{eqn:Change}
\tilde{x}_0 = x_0, \qquad \tilde{x}_1 = x_0^{w_1} x_1.
\ee
Let $\tilde{\theta}_i = \tilde{x}_i \frac{\pd}{\pd \tilde{x}_i}$.
Then we have:
\begin{align}
\theta_0 & = \tilde{\theta}_0 + w_1 \tilde{\theta}_1, &
 \theta_1 & = \tilde{\theta}_1, \\
\tilde{\theta}_0 & = \theta_0 - w_1 \theta_1, & \tilde{\theta}_1 & = \theta_1.
\end{align}
Under such change of variables,
the Picard-Fuchs operators become
\bea
&& \cL_0 =  \big[ \tilde{\theta}_0  + w_1 \tilde{\theta}_1- x_0 (1 + \tilde{\theta}_0 + k \tilde{\theta}_1) \big] \tilde{\theta}_0, \\
&& \cL_1 = \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i \tilde{\theta}_1-j)
\prod_{j=0}^{w_1-1} (-\tilde{\theta}_0-j) \\
&& \qquad - \tilde{x}_0^{-w_1} \tilde{x}_1 \prod_{j=1}^{k-w_1} (\tilde{\theta}_0+k\tilde{\theta}_1+j)
\prod_{j=0}^{w_1-1} (-\tilde{\theta}_0+j), \nonumber \\
&& \cL_1'
= \prod_{j=0}^{w_1-1} (\tilde{\theta}_0+w_1\tilde{\theta}_1-j) \cdot
\prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i \tilde{\theta}_1 -j)
- \tilde{x}_1 \prod_{j=1}^{k} (\tilde{\theta}_0+k\tilde{\theta}_1+j).
\eea
Consider the system given by
\be
\cL_0 S(\tilde{x}_0, \tilde{x}_1) = 0, \qquad \cL_1' S(\tilde{x}_0, \tilde{x}_1) = 0.
\ee
It has a solution holomorphic at $\tilde{x}_0 = \tilde{x}_1 = 0$:
\be
\tilde{g}_0(\tilde{x}_0, \tilde{x}_1)
= 1 + \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!} \tilde{x}_1^m,
\ee
and a logarithmic solution:
\be
\tilde{g}_1^{(1)}(\tilde{x}_0, \tilde{x}_1)
= \log \tilde{x}_1 \cdot \tilde{g}_0(\tilde{x}_0, \tilde{x}_1)
+ \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!}
(\sum_{j=1}^{km} \frac{k}{j} - \sum_{i=1}^n \sum_{j=1}^{w_im} \frac{1}{j}) \tilde{x}_1^m.
\ee
They are not obtained from $g_0(x_0,x_1)$ and $g_1^{(1)}(x_0,x_1)$ by change of variables \eqref{eqn:Change},
and they do not satisfy the equation
$\cL_1 S(\tilde{x}_0, \tilde{x}_1) = 0$.
Furthermore,
one can easily see that there is no solution of the form
$\tilde{g}_1^{(0)}
= \log \tilde{x}_0 \cdot \tilde{g}_0(\tilde{x}_0, \tilde{x}_1)
+ \tilde{h}_1^{(1)}(\tilde{x}_0, \tilde{x}_1)$,
with $\tilde{h}_1^{(1)}$ holomorphic in $\tilde{x}_0$ and $\tilde{x}_1$.
Indeed,
if we write $\tilde{h}_1^{(0)}(\tilde{x}_0, \tilde{x}_1)
= \sum_{m_0+m_1> 0} \tilde{a}_{m_0,m_1}^{(0)} \tilde{x}_0^{m_0} \tilde{x}_1^{m_1}$,
then from $\cL_1' \tilde{h}_1^{(0)} = 0$ we get:
\ben
&& \sum_{m_0+m_1> 0} \prod_{j=0}^{w_1-1} (m_0+w_1m_1-j) \cdot
\prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i m_1 -j) \tilde{a}_{m_0,m_1}^{(0)} \tilde{x}_0^{m_0} \tilde{x}_1^{m_1} \\
& + & \sum_{m=1}^\infty \sum_{j=0}^{w_1-1} \frac{1}{w_1m-j} \cdot
\frac{(km)!}{\prod_{i=1}^n (w_i(m-1))!} \tilde{x}_1^m \\
& = & \tilde{x}_1 \sum_{m_0+m_1 > 0} \prod_{j=1}^{k} (m_0+km_1+j) \tilde{a}_{m_0,m_1}^{(0)} \tilde{x}_0^{m_0} \tilde{x}_1^{m_1} \\
& + & \tilde{x}_1 \sum_{m=0}^\infty \sum_{j=1}^{k}\frac{1}{km+j}
\frac{(k(m+1))!}{\prod_{i=1}^n (w_im)!} \tilde{x}_1^m.
\een
Compare the coefficients of $\tilde{x}_1$ on both sides:
$$
0 = k! \sum_{j=1}^k \frac{1}{j},
$$
a contradiction.

\section{Integrality of Local Open-closed Mirror Maps}

\subsection{Integrality of local open-closed mirror map: The outer brane case}
Consider the charge vectors
\be \label{eqn:Charge}
\begin{pmatrix}
\hat{l}^{(1)} \\ \hat{l}^{(0)}
\end{pmatrix}
= \begin{pmatrix}
\tilde{l}^{(1)} \\ -\tilde{l}^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|cccc|cc}
-k & w_1 & w_2 & \cdots & w_n & 0 & 0 \\
1 &  -1 & 0 & \cdots & 0 & 1 & -1
\end{array} \biggr).
\ee
The associated variables are
\bea
\hat{x}_0 = \frac{a_0a_{n+1}}{a_1a_{n+2}} = \frac{1}{\tilde{x}_0}, \qquad
\hat{x}_1 = (-1)^k \frac{a_1^{w_1} \cdots a_n^{w_n}}{a_0^k} = \tilde{x}_1.
\eea
Suppose that we have a function $\Pi(\hat{x}_0, \hat{x}_1)$ such that
the following equations are satisfied:
\be
\cD_{\hat{l}^{(0)}} \Pi(\hat{x}_0, \hat{x}_1) =0,  \qquad
\cD_{\hat{l}^{(1)}} \Pi(\hat{x}_0, \hat{x}_1) = 0.
\ee
They give rise to the following equations:
\bea
&& \hat{\theta}_0 (\hat{\theta}_0 - k \hat{\theta}_1) \Pi
= \hat{x}_0 \hat{\theta}_0 (\hat{\theta}_0 - w_1 \hat{\theta}_1) \Pi, \\
&& \prod_{j=0}^{w_1-1} (-\hat{\theta}_0 + w_1\hat{\theta}_1 - j) \cdot
\prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i\hat{\theta}_1 - j) \Pi
= (-1)^k \hat{x}_1 \prod_{j=0}^{k-1} (\hat{\theta}_0 - k \hat{\theta}_1 - j) \Pi.
\eea
We will refer to this system as the local extended Picard-Fuchs system associated with the charge
vectors $\hat{l}^{(0)}$ and $\hat{l}^{(1)}$.

Clearly $\Phi_0 = 1$ is a solution,
and we have the following logarithmic solutions:
\be
\Phi_1^{(0)} = \log \hat{x}_0 + \sum_{m\geq 1} \frac{(km)!}{\prod_{i=1}^n (w_im)!} \frac{\hat{x}_1^m}{m},
\quad \Phi_1^{(1)} = \log \hat{x}_1 - \sum_{m\geq 1} \frac{(km)!}{\prod_{i=1}^n (w_im)!} \frac{\hat{x}_1^m}{km}.
\ee
The local open-closed mirror map is given by
\be
\hat{X}_0 = \hat{x}_0 \exp \sum_{m\geq 1} \frac{(km)!}{\prod_{i=1}^n (w_im)!} \frac{\hat{x}_1^m}{m},
\qquad \hat{X}_1 = \hat{x}_1 \exp \sum_{m\geq 1} \frac{-(km)!}{\prod_{i=1}^n (w_im)!} \frac{\hat{x}_1^m}{km}.
\ee
By a result proved in \cite{Zho1},
we have
\ben
&& \hat{x}_0^{-1}\hat{X}_0, \hat{x}_1^{-1}\hat{X}_1 \in \bZ[[\hat{x}_1]].
\een

\subsection{Integrality of local open-closed mirror map: An inner brane case}
One can also consider the following charge vectors:
\bea
&& \begin{pmatrix} \tilde{\hat{l}}^{(1)} \\ \tilde{\hat{l}}^{(0)} \end{pmatrix}
= \begin{pmatrix} \tilde{l}^{(1)} + \tilde{l}^{(0)} \\ \tilde{l}^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|cccc|cc}
-k+1&  w_1 -1 & w_2 & \cdots & w_n & -1 & 1 \\
-1 & 1 & 0 & \cdots & 0 & 1 & -1
\end{array} \biggr).
\eea
The associated moduli parameters are
\be
\tilde{\hat{x}}_0 = \frac{a_1a_{n+1}}{a_0a_{n+2}}, \qquad
\tilde{\hat{x}}_1 = (-1)^{k-1} \frac{a_1^{w_1-1}a_2^{w_2} \cdots a_n^{w_n}a_{n+2}}{a_0^{k-1}a_{n+1}}.
\ee
The local extended Picard-Fuchs system associated with these charge vectors is given by:
\bea
&& (\tilde{\hat{\theta}}_0-\tilde{\hat{\theta}}_1)(\tilde{\hat{\theta}}_0 + (w_1-1) \tilde{\hat{\theta}}_1) \Pi
= \tilde{\hat{y}}_0 (-\tilde{\hat{\theta}}_0 + \tilde{\hat{\theta}}_1)
(-\tilde{\hat{\theta}}_0 - (k-1) \tilde{\hat{\theta}}_1) \Pi, \\
&& (\tilde{\hat{\theta}}_0 - \tilde{\hat{\theta}}_1) \cdot
\prod_{j=0}^{w_1-2} (\tilde{\hat{\theta}}_0 + (w_1-1) \tilde{\hat{\theta}}_1-j) \cdot
\prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i \tilde{\hat{\theta}}_1 - j) \Pi  \\
&& = \hat{y}_1 \cdot  (-\tilde{\hat{\theta}}_0 + \tilde{\hat{\theta}}_1)
\prod_{j=0}^{k-2} (\tilde{\hat{\theta}}_0 + (k-1)\tilde{\hat{\theta}}_1+j) \Pi. \nonumber
\eea
Clearly $\hat{\Phi}_0 = 1$ is a solution and we have two logarithmic solutions:
\bea
&& \hat{\Phi}_1^{(1)}
= \log \tilde{\hat{x}}_1 + \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!}
\frac{(k-1) (\tilde{\hat{x}}_0 \tilde{\hat{x}}_1)^m}{km}, \\
&&  \hat{\Phi}_1^{(0)}
= \log \tilde{\hat{x}}_0 - \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!}
\frac{(\tilde{\hat{x}}_0 \tilde{\hat{x}}_1)^m}{km}.
\eea
The local open-closed mirror map is given by
\bea
&& \tilde{\hat{X}}_1 = e^{\hat{\Phi}_1^{(1)}}
= \tilde{\hat{x}}_1 \exp \sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!}
\frac{(k-1) (\tilde{\hat{x}}_0\tilde{\hat{x}}_1)^m}{km}, \\
&& \tilde{\hat{X}}_0 = e^{ \hat{\Phi}_1^{(0) }}
= \tilde{\hat{x}}_0 \exp \sum_{m=1}^\infty \frac{-(km)!}{\prod_{i=1}^n (w_im)!}
\frac{(\tilde{\hat{x}}_0 \tilde{\hat{x}}_1)^m}{km}.
\eea
By a result in \cite{Zho1},
we have
$\tilde{\hat{x}}_0^{-1} \tilde{\hat{X}}_0, \tilde{\hat{x}}_1^{-1} \tilde{\hat{X}}_1 \in \bZ[[\tilde{\hat{x}}_0\tilde{\hat{x}}_1]]$.

\subsection{Integrality of local open-closed mirror map: Another inner brane case}
One can also consider the the charge vectors
\be
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|cccc|cc}
-k+w_1 & 0 & w_2 & \cdots & w_n & w_1 & -w_1 \\
-1 &  1 & 0 & \cdots & 0 & -1 & 1
\end{array} \biggr),
\ee
where $l^{(1)} = \tilde{l}^{(1)} - w_1 \tilde{l}^{(0)}$ and $l^{(0)} = \tilde{l}^{(0)}$.
The associated moduli parameters are
\be
x_0 = \frac{a_1a_{n+2}}{a_0a_{n+1}}, \qquad
x_1 = (-1)^k \frac{a_2^{w_2} \cdots a_n^{w_n}a_{n+1}^{w_1}}{a_0^{k-w_1} a_{n+2}^{w_1}}.
\ee
The associated local extended Picard-Fuchs system is given by:
\be \label{eqn:PF5}
\check{\cL}_1 S(x_0,x_1) =0, \qquad \check{\cL}_0 S(x_0,x_1) = 0, \qquad \check{\cL}_1' S(x_0,x_1) = 0,
\ee
where
\bea
&& \check{\cL}_0 =  \big[ \theta_0 - x_0 (\theta_0 + (k-w_1) \theta_1) \big] (\theta_0 - w_1 \theta_1), \\
&& \check{\cL}_1 = \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i \theta_1-j)
\prod_{j=0}^{w_1-1} (-\theta_0+w_1\theta_1-j) \\
&& \qquad - x_1 \prod_{j=0}^{k-w_1-1} (\theta_0+(k-w_1)\theta_1+j)
\prod_{j=0}^{w_1-1} (-\theta_0+w_1\theta_1+j), \nonumber \\
&& \check{\cL}_1'
= \prod_{j=0}^{w_1-1} (\theta_0-j) \cdot \prod_{i=2}^n \prod_{j=0}^{w_i-1} (w_i \theta_1 -j)
- x_0^{w_1}x_1 \prod_{j=0}^{k-1} (\theta_0+(k-w_1)\theta_1+j).
\eea
We have a holomorphic solution $\check{g}_0 = 1$
and two logarithmic solutions:
\bea
&& \check{g}_1^{(0)} = \log x_0 + \sum_{m=1}^\infty \frac{(km-1)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m, \\
&& \check{g}_1^{(1)} = \log x_1 + (k-w_1) \sum_{m=1}^\infty  \frac{(km-1)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m.
\eea
The local open-closed mirror map is given by
\bea
&& Q_0 = \exp( \check{g}_1^{(0)}/\check{g}_0) = x_0 \exp \sum_{m=1}^\infty \frac{(km-1)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m, \\
&& Q_1 = \exp (\check{g}_1^{(1)}/\check{g}_0) = x_1
\exp \sum_{m=1}^\infty  \frac{(k-w_1)\cdot (km-1)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m.
\eea
By a result in \cite{Zho1},
we have
\be
x_0^{-1}Q_0, x_1^{-1}Q_1 \in \bZ[[x_0^{w_1}x_1]].
\ee
Using Lagrange-Good inversion formula as in \cite{Zho1},
one can find explicit expressions for $x_0(Q_0,Q_1), x_1(Q_0,Q_1) \in \bZ[[Q_0,Q_1]]$.
Let $x_i = \sum_{m_0+m_1>0}  A_{m_0,m_1}^{(i)}Q_0^{m_0}Q_1^{m_1}$.
Then $A_{m_0,m_1}^{(0)}$ and $A_{m_0,m_1}^{(1)}$ are the coefficients of
$x_0^{m_0-1}x_1^{m_1}$ and $x_0^{m_0}x_1^{m_1-1}$ respectively of
\ben
\sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m
\cdot  \exp (-m_0-(k-w_1)m_1) \sum_{m=1}^\infty \frac{(km-1)!}{\prod_{i=1}^n (w_im)!} (x_0^{w_1}x_1)^m.
\een
It follows that $x_i = Q_i \sum_{a=0}^\infty C_a^{(i)} (Q_0^{w_1}Q_1)^a$,
where $C_a^{(0)}$ is the coefficient of $y^a$ in
\ben
\sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!} y^m
\cdot  \exp (-ka-1) \sum_{m=1}^\infty \frac{(km-1)!}{\prod_{i=1}^n (w_im)!} y^m
\een
and $C_a^{(1)}$ is the coefficient of $y^a$ in
\ben
\sum_{m=1}^\infty \frac{(km)!}{\prod_{i=1}^n (w_im)!} y^m
\cdot  \exp (-ka-(k-w_1)) \sum_{m=1}^\infty \frac{(km-1)!}{\prod_{i=1}^n (w_im)!} y^m.
\een

Now we generalize the discussions in \cite{Zho2} to the open-closed case.
Write
\be
g_0(z_0,z_1) = 1+ \sum_{m_0+m_1>0} c_{m_0,m_1} q_0^{m_0}q_1^{m_1}
=  1+ \sum_{m_0+m_1>0} C_{m_0,m_1} Q_0^{m_0}Q_1^{m_1}.
\ee
If Conjecture 1 holds,
then the coefficients $\{c_{m_0,m_1}\}_{m_0 + m_1\geq 1}$ and $\{C_{m_0,m_1}\}_{m_0+m_1 \geq 1}$  are integers.
One can express $q_0$ and $q_1$ as integral power series
in $Q_0$ and $Q_1$, and vice versa.
If Conjecture 1 holds,
then the following conjecture holds.

\begin{Conjecture}
There are integers $\alpha^{(i)}_{m_0,m_1}$ and $\beta_{m_0,m_1}^{(i)}$ such that
\be
Q_i=q_i\prod_{m_0+m_1\geq 1} (1-q_0^{m_0}q_1^{m_1})^{\alpha^{(i)}_{m_0,m_1}},
\ee
\be
q_i=Q_i\prod_{m_0+m_1\geq 1} (1-Q_0^{m_0}Q_1^{m_1})^{\beta^{(i)}_{m_0,m_1}}.
\end{equation}
for $i=0,1$.
\end{Conjecture}

\section{Examples}

We use Maple to compute some open-closed mirror maps and local open-closed mirror maps and
check their integrality properties for the examples presented below.

\subsection{The $n=2$ case}
There is only one solution $k_1= k_2= 2$ to \eqref{eqn:Sum=1} and so $k=2$ and $w_1=w_2 = 1$.
To the charge vectors
\be \label{eqn:n=2}
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|cc|cc}
-1 & 0 & 1 & 1 & -1 \\
-1 &  1 & 0 & -1 & 1
\end{array} \biggr)
\ee
the extended Picard-Fuchs operators are:
\ben
&& \cL_0 =  \big[ \theta_0 - x_0 (1 + \theta_0 +  \theta_1) \big] (\theta_0 -  \theta_1), \\
&& \cL_1 =  \theta_1(-\theta_0+w_1\theta_1) - x_1 (\theta_0+\theta_1+1)(-\theta_0+w_1\theta_1),  \\
&& \cL_1' = \theta_0 \theta_1
- x_0 x_1 \prod_{j=1}^{2} (\theta_0+\theta_1+j).
\een
We have the following solutions:
\ben
g_0(x_0,x_1) & = & \sum_{m \geq 0} \frac{(2m)!}{(m!)^2} (x_0x_1)^2, \\
g_1^{(0)}(x_0,x_1) & = & \log x_0 \cdot \sum_{m \geq 0} \frac{(2m)!}{(m!)^2} (x_0x_1)^m
+ \sum_{m=1}^\infty \frac{(2m)!}{(m!)^2} \sum_{j=m+1}^{2m} \frac{1}{j} (x_0x_1)^m \\
& + & \sum_{m_0 \neq m_1}
\frac{(m_0+m_1)!}{m_0! m_1!} \frac{x_0^{m_0} x_1^{m_1}}{m_0-m_1}, \\
g_1^{(1)}(x_0,x_1) & = & \log x_1 \cdot \sum_{m \geq 0} \frac{(2m)!}{(m!)^2} (x_0x_1)^m
+ \sum_{m=1}^\infty \frac{(2m)!}{(m!)^2}
\sum_{j=m+1}^{2m} \frac{1}{j} (x_0x_1)^m  \\
& - & \sum_{m_0 \neq m_1}
\frac{(m_0+m_1)!}{m_0!m_1!} \frac{x_0^{m_0} x_1^{m_1}}{m_0-m_1}.
\een
The first few terms of the Taylor series of the mirror map is given by:
\ben
q_0 & = & x_0 + (x_0^2-x_0x_1) + x_0^3 + (x_0^4 + x_0^3x_1-2x_0^2x_1^2) \\
& + & (x_0^5+2x_0^4x_1-x_0^3x_1^2) +(x_0^6+ 3x_0^5x_1+x_0^4x_1^2-5x_0^3x_1^3) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1+x_1^2)
+x_1^3 + (-2x_0^2x_1^2 + x_0x_1^3+x_1^4) \\
& + & (2x_0x_1^4-x_0^2x_1^3+x_1^5) + (3x_0x_1^5+x_0^2x_1^4-5x_0^3x_1^3+x_1^6) + \cdots.
\een
The inverse open-closed mirror map is given by:
\ben
x_0 & = & q_0 +(- q_0^2 + q_0q_1) + (q_0^3-2q_0^2q_1) + (-q_0^4-3q_0^2q_1^2)  \\
& + & (q_0^5+2q_0^3q_1^2+2q_0^4q_1) + (4q_0^4q_1^2-q_0^6+5q_0^3q_1^3-4q_0^5q_1) + \cdots, \\
x_1 & = & q_1 + (-q_1^2+q_0q_1) + (-2q_0q_1^2+q_1^3) + (-q_1^4-3q_0^2q_1^2) \\
& + & (q_1^5+2q_0^2q_1^3+2q_0q_1^4) + (4q_0^2q_1^4+5q_0^3q_1^3-q_1^6-4q_0q_1^5) + \cdots.
\een

The local extended Picard-Fuchs operators associated with \eqref{eqn:n=2} are:
\ben
&& \check{\cL}_0 =  \big[ \theta_0 - x_0 (\theta_0 + \theta_1) \big] (\theta_0 - \theta_1), \\
&& \check{\cL}_1 = \theta_1 (-\theta_0+\theta_1) - x_1 (\theta_0+\theta_1) (-\theta_0+w_1\theta_1), \\
&& \check{\cL}_1' = \theta_0 \theta_1 - x_0x_1 \prod_{j=0}^{1} (\theta_0+\theta_1+j).
\een
We have a holomorphic solution $\hat{g}_0 = 1$
and two logarithmic solutions:
\ben
&& \check{g}_1^{(0)} = \log x_0 + \sum_{m=1}^\infty \frac{(2m-1)!}{(m!)^2} (x_0x_1)^m
= \log \frac{2x_0}{1+\sqrt{1-4x_0x_1}}, \\
&& \check{g}_1^{(1)} = \log x_1 + \sum_{m=1}^\infty  \frac{(2m-1)!}{(m!)^2} (x_0x_1)^m
= \log \frac{2x_1}{1+\sqrt{1-4x_0x_1}}.
\een
The local open-closed mirror map is given by
\ben
&& Q_0 = x_0 \exp \sum_{m=1}^\infty \frac{(2m-1)!}{(m!)^2} (x_0x_1)^m
= \frac{2x_0}{1 + \sqrt{1-4x_0x_1}}, \\
&& Q_1 = x_1 \exp \sum_{m=1}^\infty  \frac{(2m-1)!}{(m!)^2} (x_0x_1)^m
= \frac{2x_1}{1 + \sqrt{1-4x_0x_1}}.
\een
One can solve these equations to get:
\ben
x_0 = \frac{Q_0}{1+Q_0Q_1}, \qquad
x_1 = \frac{Q_1}{1-Q_0Q_1}.
\een
We have checked
\ben
q_0 = Q_0 \frac{1-Q_1}{1-Q_0}, \qquad q_1 = Q_1 \frac{1-Q_0}{1-Q_1},
\een
and
\ben
Q_0 = q_0 \frac{1+q_1}{1+q_0}, \qquad Q_1 = q_1 \frac{1+q_0}{1+q_1}.
\een

\subsection{The $n=3$ case}

There are three cases to consider:
$(k|w_1, w_2,w_3) = (3|1,1,1)$, $(4|2,1,1)$, $(6|3,2,1)$.
For the first case,
consider the charge vectors:
\ben
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|ccc|cc}
-2 & 0 & 1 & 1 & 1 & -1 \\
-1 & 1 & 0 & 0 & -1 & 1
\end{array} \biggr),
\een
The extended Picard-Fuchs operators are given by:
\ben
&& \cL_0 =  \big[ \theta_0 - x_0 (1 + \theta_0 + 2 \theta_1) \big] (\theta_0 - \theta_1), \\
&& \cL_1 = \theta_1^2 (-\theta_0+\theta_1)
- x_1 \prod_{j=1}^{2} (\theta_0+2\theta_1+j) (-\theta_0+\theta_1),  \\
&& \cL_1'
=  \theta_0 \theta_1^2
- x_0 x_1 \prod_{j=1}^{3} (\theta_0+2\theta_1+j),
\een
with solutions
\ben
g_0(x_0,x_1) & = & \sum_{m \geq 0} \frac{(3m)!}{(m!)^3} (x_0x_1)^m, \\
g_1^{(0)}(x_0,x_1) & = & \log x_0 \cdot \sum_{m \geq 0} \frac{(3m)!}{(m!)^3} (x_0x_1)^m
+ \sum_{m=1}^\infty \frac{(3m)!}{(m!)^3} \sum_{j=m+1}^{3m} \frac{1}{j}  (x_0x_1)^m \\
& + & \sum_{m_0 \neq m_1}
\frac{(m_0+2m)!}{m_0!(m_1!)^2} \frac{x_0^{m_0} x_1^{m_1}}{m_0-m_1}, \\
g_1^{(1)}(x_0,x_1) & = & \log x_1 \cdot \sum_{m \geq 0} \frac{(3m)!}{(m!)^3} (x_0x_1)^m
+ 2 \sum_{m=1}^\infty \frac{(3m)!}{(m!)^3} \sum_{j=m+1}^{3m} \frac{1}{j}(x_0x_1)^m  \\
& - & \sum_{m_0 \neq m_1}
w_1 \frac{(m_0+2m_1)!}{m_0!(m!)^2} \frac{x_0^{m_0} x_1^{m_1}}{m_0-m_1}.
\een
The open-closed mirror map is given by:
\ben
q_0 & = & x_0 +(x_0^2-2x_0x_1) + (x_0^3+3x_0^2x_1-x_0x_1^2) +
(x_0^4+9x_0^3x_1-29x_0^2x_1^2-2x_0x_1^3) \\
& + & (x_0^5+16x_0^4x_1+27x_0^3x_1^2-23x_0^2x_1^3-5x_0x_1^4) \\
& + & (x_0^6+24x_0^5x_1+127x_0^4x_1^2-527x_0^3x_1^3-63x_0^2x_1^4-14x_0x_1^5) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1 + 2x_1^2)
+ (8x_0x_1^2+5x_1^3)
+ (-16x_0^2x_1^2+33x_0x_1^3 + 14x_1^4) \\
& + & (-x_0^3x_1^2+111x_0^2x_1^3+124x_0x_1^4+42x_1^5) \\
& + & ( -x_0^4x_1^2-307x_0^3x_1^3 +606x_0^2x_1^4 +462x_0x_1^5 + 132 x_1^6) + \cdots.
\een
The local open-closed mirror map is given by:
\ben
&& Q_0 = x_0 \exp \sum_{m=1}^\infty \frac{(3m-1)!}{(m!)^3} (x_0x_1)^m, \\
&& Q_1 = x_1 \exp \sum_{m=1}^\infty  \frac{2 \cdot (3m-1)!}{(m!)^3} (x_0x_1)^m.
\een

For $(k|w_1, w_2,w_3) = (4|2,1,1)$,
one can consider the charge vectors:
\ben
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|ccc|cc}
-2 & 0 & 1 & 1 & 2 & -2 \\
-1 & 1 & 0 & 0 & -1 & 1
\end{array} \biggr),
\een
The open-closed mirror map is given by:
\ben q_0 & = & x_0 +(x_0^2-x_0x_1) + (x_0^3-7x_0^2x_1-x_0x_1^2) +
(x_0^4+12x_0^3x_1-5x_0^2x_1^2-2x_0x_1^3) \\
& + & (x_0^5+20x_0^4x_1-39x_0^3x_1^2-14x_0^2x_1^3-5x_0x_1^4) \\
& + & (x_0^6+29x_0^5x_1-299x_0^4x_1^2-80x_0^3x_1^3-45x_0^2x_1^4-14x_0x_1^5) + \cdots, \\
q_1 & = & x_1 + (-2x_0x_1 + 2x_1^2) + (x_0^2x_1+8x_0x_1^2+5x_1^3)
+ (4 x_0^2x_1^2+34x_0x_1^3 + 14x_1^4) \\
& + & (- 56x_0^3x_1^2+107x_0^2x_1^3+128x_0x_1^4+42x_1^5) \\
& + & (40 x_0^4x_1^2 +220x_0^3x_1^3 + 592 x_0^2x_1^4 + 476x_0x_1^5 + 132 x_1^6) + \cdots. \een
The inverse open-closed mirror map is given by:
\ben
x_0 & = & q_0 + (-q_0^2+q_0q_1) + (q_0^3+6q_0^2q_1) + (-q_0^2q_1^2-q_0^4-25q_0^3q_1) + \cdots, \\
x_1 & = & q_1 + (2q_0q_1-2q_1^2) + (q_0^2q_1-18q_0q_1^2+3q_1^3) + (50q_0q_1^3-56q_0^2q_1^2-4q_1^4) + \cdots.
\een
The local open-closed mirror map is given by:
\ben
&& Q_0 = x_0 \exp \sum_{m=1}^\infty \frac{(4m-1)!}{(2m)!(m!)^2} (x_0^2x_1)^m, \\
&& Q_1 = x_1 \exp \sum_{m=1}^\infty  \frac{2\cdot (4m-1)!}{(2m)!(m!)^2} (x_0^2x_1)^m.
\een
The inverse local open-closed mirror map is given by:
\ben
&& x_0 = Q_0(1-3Q_0^2Q_1 - 12 Q_0^4Q_1^2 - 253 Q_0^6Q_1^3 - 6033 Q_0^8Q_1^4+ \cdots), \\
&& x_1 = Q_1(1-6Q_0^2Q_1 -15 Q_0^4Q_1^2 -434 Q_0^6Q_1^3 -10404 Q_0^8Q_1^4+ \cdots).
\een
One has
\ben
q_0 & = & Q_0-Q_0Q_1+Q_0^2-Q_0Q_1^2-7Q_0^2Q_1+Q_0^3-2Q_0Q_1^3-5Q_0^2Q_1^2+9Q_0^3Q_1+Q_0^4\\
& & -5Q_0Q_1^4 -14 Q_0^2Q_1^3 -30 Q_0^3Q_1^2 + 14 Q_0^4Q_1 + Q_0^5 \\
& & - 14Q_0Q_1^5-45Q_0^2Q_1^4
-65Q_0^3Q_1^3-215Q_0^4Q_1^2+20Q_0^5Q_1+Q_0^6 + \cdots \\

q_1 & = & Q_1+2Q_1^2-2Q_0Q_1+5Q_1^3+8Q_0Q_1^2+Q_0^2Q_1+14Q_1^4+34Q_0Q_1^3-2Q_0^2Q_1^2\\
&& +42Q_1^5+128Q_0Q_1^4+83Q_0^2Q_1^3-38Q_0^3Q_1^2\\
&& +132Q_1^6+476Q_0Q_1^5+502Q_0^2Q_1^4+100Q_0^3Q_1^3
+28Q_0^4Q_1^2+1776Q_0Q_1^6 + \cdots.
\een
One can also consider the charge vectors corresponding to another phase of the $B$-brane:
\ben
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|ccc|cc}
-3 & 0 & 2 & 1 & 1 & -1 \\
-1 & 1 & 0 & 0 & -1 & 1
\end{array} \biggr),
\een
Then the open-closed mirror map is given by:
\ben
q_0 & = & x_0 + (x^2-3x_0x_1) + (x_0^3+10x_0^2x_1-3x_0x_1^2) \\
& + & (x^4+28x_0^3x_1-111x_0^2x_1^2-10x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1+3x_1^2) + (42x_0x_1^2+12x_1^3) \\
& + & (-63x_0^2x_1^2+192x_0x_1^3+55x_1^4) + \cdots.
\een

For $(k|w_1, w_2,w_3) = (6|3,2,1)$,
we check the following three cases corresponding to three different phases of the $B$-brane:
For the charge vectors
\ben
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|ccc|cc}
-3 & 0 & 2 & 1 & 3 & -3 \\
-1 & 1 & 0 & 0 & -1 & 1
\end{array} \biggr),
\een
we have
\ben
q_0 & = & x_0 + (x_0^2-x_0x_1) + (x_0^3-7x_0^2x_1-2x_0x_1) \\
& + & (x_0^4-37x_0^3x_1-17x_0^2x_1^2-7x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-3x_0x_1+3x_1^2) + (3x_0^2x_1+9x_0x_1^2+12x_1^3) \\
& + & (-x_0^3x_1+45x_0^2x_1^2+81x_0x_1^3+55x_1^4) + \cdots;
\een
for the charge vectors
\ben
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|ccc|cc}
-4 & 0 & 3 & 1 & 2 & -2 \\
-1 & 1 & 0 & 0 & -1 & 1
\end{array} \biggr),
\een
we have
\ben
q_0 & = & x_0 + (x_0^2-2x_0x_1) + (x_0^3-22x_0^2x_1-5x_0x_1^2) \\
& + & (x_0^4+212x_0^3x_1-49x_0^2x_1^2-24x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-2x_0x_1+4x_1^2) + (x_0^2x_1+32x_0x_1^2+22x_1^2) \\
& + & (342x_0^2x_1^2+284x_0x_1^3+140x_1^4) + \cdots;

\een
for the charge vectors
\ben
\begin{pmatrix}
l^{(1)} \\ l^{(0)}
\end{pmatrix}
= \biggl( \begin{array}{c|ccc|cc}
-5 & 0 & 2 & 3 & 1 & -1 \\
-1 & 1 & 0 & 0 & -1 & 1
\end{array} \biggr),
\een
we have
\ben
q_0 & = & x_0 + (x_0^2-10x_0x_1) + (x_0^3+77x_0^2x_1-55x_0x_1^2) \\
& + & (x_0^4+227x_0^3x_1-2635x_0^2x_1^2-785x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1 + 10x_1^2)
+(525x_0x_1^2+155x_1^3) \\
& + & (-685x_0^2x_1^2+6905x_0x_1^3+2885x_1^4) + \cdots.
\een

\subsection{Some $n>3$ cases}

For $n=4$,
there are thirteen cases to consider when $w_1 \geq \cdots \geq w_4$ listed in \cite{Zho2};
when $n=5$ there are 147 cases.
We have checked some of them using our Maple algorithm.
The following cases are particularly interesting because they correspond to the large volume phase
of $B$-brane geometries in one-moduli Calabi-Yau hypersurfaces in weighted
projective spaces:
$$(k|w_1, \dots, w_5) = (5|1,1,1,1,1), (6|2,1,1,1,1),
(8|4,1,1,1,1),(10|5,2,1,1,1).
$$
When $(k|w_1, \dots, w_5) = (5|1,1,1,1,1)$, for the charge vectors
\be
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-4 &  0 & 1 & 1 & 1 & 1&  -1 & 1 \\
-1 & 1 & 0 & 0 & 0 & 0 &1 & -1
\end{array} \biggr)
\ee
we have
\ben
q_0 & = & x_0 + (x_0^2-24x_0x_1) +(x_0^3-24x_0^2x_1-972x_0x_1^2) \\
& + & (x_0^4+216x_0^3x_1-20772x_0^2x_1^2-95264x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1+24x_1^2) + (746x_0x_1^2+1548x_1^3) \\
& + & (-1010 x_0^2x_1^2 + 36732 x_0x_1^3 + 155744 x_1^4) + \cdots.
\een

When $(k|w_1, \dots, w_5) = (6|2,1,1,1,1)$, we consider two phases.
For the charge vectors
\ben
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-4 & 0 & 1 & 1 & 1 & 1 & -2 & 2 \\
-1 & 1 & 0 & 0 & 0 & 0 & 1 & -1
\end{array} \biggr)
\een
we have
\ben
q_0 & = & x_0 + (x_0^2-12x_0x_1) + (x_0^3-132x_0^2x_1-558x_0x_1^2) \\
& + & (x_0^4+570x_0^3x_1-6678x_0^2x_1^2-54328x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-2x_0x_1+24x_1^2) + (x_0^2x_1+192x_0x_1^2+1548x_1^3) \\
& + & (1632x_0^2x_1^2+17784x_0x_1^3+155744x_1^4) + \cdots;
\een
for the charge vectors
\ben
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-5 &  0 & 2 & 1 & 1 & 1&  -1 & 1 \\
-1 & 1 & 0 & 0 & 0 & 0 &1 & -1
\end{array} \biggr)
\een
we have
\ben
q_0 & = & x_0 + (x_0^2-60x_0x_1) + (x_0^3+462x_0^2x_1-7650x_0x_1^2) \\
& + & (x_0^4+1362x_0^3x_1-225270x_0^2x_1^2-2271800x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1+60x_1^2)+(-60x_0x_1^2+11250x_1^3) \\
& + & (1890x_0^2x_1^2+175050x_0x_1^3+3405800x_1^4) + \cdots.
\een

When $(k|w_1, \dots, w_5) = (8|4,1,1,1,1)$,
for the charge vectors
\ben
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-4 &  0 & 1 & 1 & 1 & 1&  -4 & 4 \\
-1 & 1 & 0 & 0 & 0 & 0 &1 & -1
\end{array} \biggr)
\een
we have
\ben
q_0 & = & x_0 + (x_0^2-6x_0x_1) + (x_0^3-46x_0^2x_1-297x_0x_1^2) \\
& + & (x_0^4-226x_0^3x_1-3297x_0^2x_1^2-28946x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-4x_0x_1+24x_1^2) + (6x_0^2x_1+64x_0x_1^2+1548x_1^3) \\
& + & (-4x_0^3x_1+224 x_0^2x_1^2 + 10608 x_0x_1^3 + 155744 x_1^4) + \cdots;
\een
for the charge vectors
\be
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-7 &  0 & 4 & 1 & 1 & 1 &  -1 & 1 \\
-1 & 1 & 0 & 0 & 0 & 0 & 1 & -1
\end{array} \biggr)
\ee
we have
\ben
q_0 & = & x_0 + (x_0^2-210x_0x_1) + (x_0^3 + 2676x_0^2x_1 -113085x_0x_1^2) \\
& + & (x_0^4+8556x_0^3x_1-4420395x_0^2x_1^2-137765950x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1+210x_1^2) + (23212x_0x_1^2+157185x_1^3) \\
& + & (-29302x_0^2x_1^2+8462685x_0x_1^3+194522650x_1^4) + \cdots.
\een

When $(k|w_1, \dots, w_5) = (10|5,2,1,1,1)$,
for the charge vectors
\ben
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-5 &  0 & 2 & 1 & 1 & 1&  -5 & 5 \\
-1 & 1 & 0 & 0 & 0 & 0 &1 & -1
\end{array} \biggr)
\een
we have
\ben
q_0 & = & x_0 + (x_0^2-12x_0x_1) + (x_0^3-102x_0^2x_1-1818x_0x_1^2) \\
& + & (x_0^4-522x_0^3x_1-23838x_0^2x_1^2-538168x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-5x_0x_1+60x_1^2) + (10x_0^2x_1 + 150 x_0x_1^2 + 11250 x_1^3) \\
& + & (-10x_0^3x_1 + 450 x_0^2x_1^2 + 86250 x_0x_1^3 + 3405800 x_1^4) + \cdots;
\een
for the charge vectors
\be
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-8 &  0 & 5 & 1 & 1 & 1&  -2 & 2 \\
-1 & 1 & 0 & 0 & 0 & 0 &1 & -1
\end{array} \biggr)
\ee
we have
\ben
q_0 & = & x_0 + (x_0^2-168x_0x_1) + (x_0^3-3192x_0^2x_1-166068x_0x_1^2) \\
& + & (x_0^4+33534x_0^3x_1-3742116x_0^2x_1^2-336621152x_0x_1^3) + \cdots, \\
q_1 & = & x_1 + (-2x_0x_1 + 336x_1^2) + (x_0^2x_1 + 5376 x_0x_1^2 + 416808 x_1^3) \\
& + & (-11760 x_0^2x_1^2 + 9366672 x_0x_1^3 + 859605376 x_1^4) + \cdots;
\een
for the charge vectors
\be
\begin{pmatrix} l^{(1)} \\ l^{(0)} \end{pmatrix}
= \biggl( \begin{array}{c|ccccc|cc}
-9 & 0&  5 & 2  & 1 & 1&  -1 & 1 \\
-1 & 1 & 0 & 0 & 0 & 0 &1 & -1
\end{array} \biggr)
\ee
we have
\ben
q_0 & = & x_0 + (x_0^2-1512x_0x_1) + (x_0^3+ 27654 x_0^2x_1 - 8046108 x_0x_1^2) \\
& + & (x_0^4 + 95694 x_0^3 x_1 - 378472500 x_0^2x-1^2 - 93766645728 x_0 x_1^3) + \cdots, \\
q_1 & = & x_1 + (-x_0x_1 + 1512 x_1^2) + (265332x_0x_1^2 + 10332252 x_1^3) \\
& + & (121554726048x_1^4+719463276x_0x_1^3-334884x_0^2x_1^2) + \cdots.
\een
Some $n>4$ cases have also been checked.

\vspace{.1in}
{\em Acknowledgements}.
This research is partly supported by NSFC grants (10425101 and 10631050)
and a 973 project grant NKBRPC (2006cB805905).


TITLE: Polynomial factorization to irreducible factors with respect to field
QUESTION [3 upvotes]: I have a question, I think I don't understand this material very well and could use an explanation / some help.
Basically we are asked to decompose $x^5-x$ to irreducible factors over $R,F2,F5,C$
But I don't understand how the field has anything to do with it?
I would just say: $x^5-x = x(x^4-1) = x(x^2-1)(x^2+1) = x(x-1)(x+1)(x^2+1)$ and that's it, I can't factor it anymore. I don't see how fields are related.

REPLY [4 votes]: Over $\mathbb{C}$, $x^2+1$ factors as $(x+i)(x-i)$. Over $F_2$, it factors as $(x+1)(x+1)$. Over $F_5$, it factors as $(x+2)(x+3)$.
Remark: Polynomials of degree $1$ are irreducible, so it's over. Note that any polynomial with complex coefficients of degree $\ge 1$ can be expressed as a product of polynomials of degree $1$ and coefficients in $\mathbb{C}$. This is a version of the Fundamental Theorem of Algebra.
For $F_2$, note that $1^2+1=0$. So by a generalization of a familiar result, $x-1$ divides $x^2+1$ over $F_2$. We gave the polynomial $x-1$ the equivalent name $x+1$.
For $F_5$, note that $(2)^2+1=0$. So over $F_5$, the polynomial $x-2$, that is, $x+3$, divides $x^2+1$.


TITLE: A conceptual understanding of kg*m/s/s
QUESTION [0 upvotes]: Force is measured in kgm/s/s, or rather the rate in change of the rate in change of mass times length.
Conceptually, what might the product of mass and length be? If force is the second derivative of mass times length, that means force is the rate of change of the rate of change of mass times length; what, conceptually, really is mass times length? How would you represent it? Or is this a stupid question that arises out of a misunderstanding of our usage of units?

REPLY [1 votes]: Force is the mass times the rate of change of the rate of change of position (not length):
$$
F=m\frac{d}{dt}\left(\frac{dx}{dt}\right)=m\frac{d^2x}{dt^2},\tag{1}
$$
which shows also that the force is the mass times the second derivative of position, rather than,
$$
F=x\frac{d^2m}{dt^2}\tag{2}
$$
so you've got your relationship backwards. You could also remember this by knowing that (a) $a=dv/dt$ and (b) $F=ma$ (which is how Newton wrote it) to get
$$
F=m\frac{dv}{dt}=m\frac{d^2x}{dt^2}\,.\tag{3}
$$
So clearly (1) and (3) match while (2) doesn't seem to have any meaning, even though the units "work out" (NB: I would not put much stock in analyzing things in terms if their units, otherwise you might think torque and energy are the same thing because their units are the same.).
The National Institute of Standards and Technology (NIST) has a neat list of derived SI units. Perusing it, $\rm m\cdot kg$ does not appear together alone (there are terms that have m kg and factors of time in it).


TITLE: What is the difference between generalized momentum and ordinary momentum?
QUESTION [0 upvotes]: I'm studying about motion equation of charge in electromagnetic field.
Lagrangian of charge in E.M field is
$L=-mc^2\sqrt{1-v^2/c^2}+\frac{e}{c}\mathbf{A}\cdot \boldsymbol{v}-e\phi$ .
Thus generalized momentum is
$\boldsymbol{P}=\frac{\partial L}{\partial \boldsymbol{v}}=\frac{m \boldsymbol{v}}{\sqrt{1-v^2/c^2}}+\frac{e}{c}\boldsymbol{A}=\boldsymbol{p}+\frac{e}{c}\boldsymbol{A}$ ,
and it's obvious that generalized momentum has different value with ordinary momentum.
So here's my question. What is the difference between generalized momentum and ordinary momentum?

REPLY [2 votes]: They're (in principle) different quantities, that coincide for velocity-independent potentials in a Cartesian coordinate system. The Lagrangian and the generalized momenta in such a system are
$$
\begin{align}
L(x, y, z, v_x, v_y, v_z; t) &= -mc^2\sqrt{1-\frac{v^2}{c^2}} - V(x, y, z) \\
\frac{\partial L}{\partial v_i} &= \frac{mv_i}{\sqrt{1 - \frac{v^2}{c^2}}}
\end{align}
$$
The generalized momentum corresponds to the ordinary 3-momentum in this case.
The invariance of the Lagrangian under translation of a coordinate leads to conservation of the generalized momentum corresponding to that coordinate, through Lagrange's equation of motion. There is no conservation law for the ordinary momentum of a system.
The momentum conservation of the Newtonian formulation is a special case of this general conservation law, as is Newtonian angular momentum conservation.


TITLE: Determine all subspaces $U$ of $\mathbb{R}^3$ such that $\phi (U)=U$.
QUESTION [1 upvotes]: We have the matrix $A=\begin{pmatrix}1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & -1\end{pmatrix}$ and let $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ x\mapsto Ax$ be the endomorphism to $A$.
I have found the eigenvalues which are $1,2,-1$. Since we have three dinstinct eigenvalues it follows that $A$ and so also $\phi$ is diagonalizable.
Is that correct?
Then for each eigenvalue I calculated the eigenspaces. The set of the three eigenvectors forms a basis of $\mathbb{R}^3$, right?
I want to determine all subspaces $U$ of $\mathbb{R}^3$ such that $\phi (U)=U$.
How can we do that? Is this maybe related to the eigenvectors?

REPLY [5 votes]: What you did is fine. And now, if $v_1$, $v_2$ and $v_{-1}$ are eigenvalues corresponding to the eigenvalues $1$, $2$, and $-1$ respectively, then the subspaces $U$ of $\Bbb R^3$ such that $\phi(U)=U$ are:

$\{0\}$;
$\Bbb Rv_1$;
$\Bbb Rv_2$;
$\Bbb Rv_{-1}$;
$\Bbb Rv_1\bigoplus\Bbb Rv_{-1}$;
$\Bbb Rv_1\bigoplus\Bbb Rv_2$;
$\Bbb Rv_{-1}\bigoplus\Bbb Rv_2$;
$\Bbb R^3$.

It is clear that each of these spaces is invariant. On the other hand, let $U\subset\Bbb R^3$ be such that $\phi(U)\subset U$. If $\dim U$ is $0$, $1$ or $3$, it's clear that $U$ belongs to the list from above. If $\dim U=2$, let $\psi\colon U\longrightarrow U$ be the restriction of $\phi$ to $U$. So, $\psi$ has $2$ complex eigenvalues, which must be a subset of $\{1,2,-1\}$. But then $U$ has to be one of the $3$ $2$-dimensional spaces from above.


TITLE: Ice cube in a thermally isolated system ... Will any of it melt?
QUESTION [12 upvotes]: If an ice cube at $0$ °C is in a thermally isolated system on its own, will any of it melt?
The chemistry teacher says it will reach a state of equilibrium of half ice and half water due to variation in kinetic energy of the particles but I don't understand where the latent heat energy would come from, unless half of the ice cube ends up as several degrees below zero to supply the energy to melt the other half, which goes against thermal equilibrium principles.

REPLY [4 votes]: Your teacher is wrong.
The ice will stay at the same ice-water fraction it was when you sealed the container.
If the system wasn't in thermal equilibrium when you closed it, then you could have a phase change since the energy would already be there. But in that case you're not talking about thermodynamics since thermo only deals with systems in equilibrium.


TITLE: Finding probability of earning points
QUESTION [2 upvotes]: A professor asks a true/false question with ten individual questions. Suppose the professor assigns grades of: $10$ points for each correct response, $0$ points for each absent response, and $-10$ for each incorrect response, but where negative totals are always replaced by zero. Les Guess, a student who understands nothing of the professor's class, flips a fair coin to decide if he will answer each question true or false.
a) In how many different ways might Les answer the question?
b) What is the probability that Les Guess will earn $n$ points? (This is just the number of ways to answer the question that will result in n points divided by the total number of ways of answering the question.)

REPLY [2 votes]: I will (for the moment) assume that he picks true, false, or abstain equally likely. (otherwise, why mention the abstain possibility?)
(a) There are $3$ possibilities for each question, so there are $3^{10}$ ways to take the test.
(b) To score $10n$, he needs to answer $k$ questions right and $k-n$ answers wrong. So the number of ways is
$${10 \choose k} {{10-k} \choose {k-n}}$$
for $0 \le k \le 10$, where ${ x \choose 0}=1$, ${x \choose {j<0}} = 0$, and ${0 \choose 10\ge j\ge0} = 1$.
Then the total probability is
$$P(10n) = \frac{1}{3^{10}}\sum_{k=0}^{10} {10 \choose k} {{10-k} \choose {k-n}}$$
for $n > 0$. This is symmetric with $n < 0$ and the probability of having equal numbers of rights and wrongs is given by the above definition of $P(10n)$ by setting $n=0$.
Other interpretation
Now I will assume that he chooses true or false equally likely and never abstains.
(a) There are $2^{10}$ ways to take the test.
(b) Let $r$ denote the number of correct answers and $w$ the number of wrong answers. Say he scores $10n$. We have that
$$r+w = 10 \\ r - w = n$$
So if $n$ is odd, there are no ways. If $n$ is even, then $r=5+n/2$ and $w = 5-n/2$. We now choose $r$ of the $10$ questions to be correct, so there are ${10 \choose {5+n/2}}$ ways to take the test while scoring $n$. So the probability is:
$$
   P(10n) = \left\{
     \begin{array}{lr}
       \frac{1}{2^{10}}{10 \choose {5+n/2}} & : \text{n even}\\
       0 & : \text{n odd}
     \end{array}
   \right.
$$
for $1 \le n \le 10$. For $n=0$, we can score negative or $0$. Negative n is symmetric with the positive $n$, and ${10 \choose 5}$ is the number of ways to get exactly $0$, so
$$P(0)=\frac{1}{2^{10}}{10 \choose 5} + \sum_{n=1}^{10} P(10n)$$


TITLE: Random variables definition
QUESTION [1 upvotes]: I am looking at my lecture notes and would like to ask for a clarification of the following property. So I have a definition of a random variable:
" $X(w) \in \mathbb{R}$, i.e. $X:\Omega \rightarrow \mathbb{R}$ and $\omega \in \Omega$ with the property that $\forall x \in \mathbb{R}, \left\{\omega \mid X(\omega)\leq x  \right\} \in \Sigma$"
where $\Sigma$ is the sigma-field.
I don't quite follow that last part. What does it mean?

REPLY [2 votes]: The last part implies that $X$ is a measurable function from $(Ω,Σ)$ to $(\mathbb R,\mathcal B(\mathbb R))$

$$ X^{-1}((-\infty,x)):=\{ω\in Ω\mid X(ω)\le x\}\in \Sigma ,\;\;\forall x\in \mathbb R $$

Now,

Why only sets of the form $(-\infty, x)$? Because they generate the Borel σ-algebra on $\mathbb R$ (so is suffices to look at them, i.e. to check measurability on the generating set.). And
Why has $X$ to be measurable? Because we want to calculate (or measure) probabilities of events (which are subsets of $Σ$.)


TITLE: Shifting Parabolas Question about equation
QUESTION [0 upvotes]: I have a question about shifting parabolas. The equation is to shift  $$f(x) = x^2$$
$$to$$ $$g(x)=(x-6)^2$$
Now I know that the formula is $(x-h)+k$ so that the $6$ moves to the right as in $-(-6)$.
My question is, on the first equation $x$ is raised to the power of $2$, while on the second equation not only $x$, but also $-6$ is raised to the power of $2$. Why is the correct answer "to get the function $g$, shift $f$ right by $6$ units" instead of $36$ as in raised to the power of $2$.

REPLY [1 votes]: Fix some $z$. When we send $z$ through $f$, we get $f(z)=z^2$. To get the same "output" from g, we can input $z+6$, since $g(z+6)=((z+6)-6)^2=(z+6-6)^2=z^2=f(z)$.
Since $z$ was just some arbitrary number, what we did above should work for other inputs.  So to get the same result from $g$ as we do from $f$, we input a value to $g$ that is 6 greater than that which we put through $f$.
If $g$ were $f$ shifted 36 to the right (which it is not), we would have
$f(z)=z^2=g(z+36)=((z+36)-6)^2=(z+30)^2$ for all $z$.


TITLE: Normally distributed from 0 to 99th percentile
QUESTION [1 upvotes]: Assume the length of waiting at supermarket is approximately normally distributed with mean 6 minutes and standard deviation 1.5 minutes.
(1) What length of the waiting time constitutes the 99th percentile ($Φ_.99$)?
My way is
For 99th percentile, find p(z > 2.327) = .01
find (y-6)/1.5 = 2.237 <=> y = 9.4905. For 0th percentile, y should be negative infinity. However, it does not make sense since we talk about time here. We should take 0 instead.
The waiting time length is between 0 and 9.4905
Could anyone please check (1) for me?

REPLY [2 votes]: I assume (1) and (2) mean the same thing here.
What you have done is reasonable, except you have rounded at the wrong time: if the normal approximation was exact, the value would be closer to $9.4895$ then to $9.4905$, but there is in any case no good reason for being any more precise than $9.49$ minutes or perhaps $9$ minutes and $29$ seconds if you prefer sexagesimal.
As you say, a normal approximation cannot be exact if negative values are impossible.  Your approach of putting this very small amount (about $0.00003$) at $0$ rather than negative is probably the most sensible thing to do.
In real life, queuing times are often zero and can have a right skewed distribution and so a normal approximation is probably inappropriate.


text:
\section{Introduction}
Optimal transport (OT) theory~\cite{Villani09,figalli2017monge} has recently inspired several works in data science, where dealing with and comparing probability distributions, and more generally positive measures, is an important staple (see~\citep{Peyre2019computational} and references therein). For these applications of OT to be successful, a belief now widely shared in the community is that some form of regularization is needed for OT to be both scalable and avoid the curse of dimensionality~\cite{dereich2013constructive,fournier2015rate}. Two approaches have emerged in recent years to achieve these goals: either regularize directly the measures themselves, by looking at them through a simplified lens; or regularize the original OT problem using various modifications. The first approach exploits well-known closed-form identities for OT when comparing two univariate measures or two multivariate Gaussian measures. In this approach, one exploits those formulas and operates by summarizing complex measures as one or possibly many univariate or multivariate Gaussian measures.
The second approach builds on the fact that for arbitrary measures, regularizing the OT problem, either in its primal or dual form, can result in simpler computations and possibly improved sample complexity. The latter approach can offer additional benefits for data science: because the original marginal constraints of the OT problem can also be relaxed, regularized OT can also yield useful tools to compare measures with different total mass --- the so-called ``unbalanced'' case~\citep{benamou03}--- which provides a useful additional degree of freedom. Our work in this paper stands at the intersection of these two approaches. To our knowledge, that intersection was so far empty: no meaningful closed-form formulation was known for regularized optimal transport. We provide closed-form formulas of entropic (OT) of two Gaussian measures for balanced and unbalanced cases.

\paragraph{Summarizing measures \emph{vs.} regularizing OT.}
Closed-form identities to compute OT distances (or more generally recover Monge maps) are known when either (1) both measures are univariate and the ground cost is submodular~\cite[\S2]{SantambrogioBook}: in that case evaluating OT only requires integrating that submodular cost w.r.t. the quantile distributions of both measures; or (2) both measures are Gaussian, in a Hilbert space, and the ground cost is the squared Euclidean metric~\cite{dowson82frechet,gelbrich1990}, in which case the OT cost is given by the Wasserstein-\citeauthor{bures1969extension} metric~\citep{bhatia2018bures,malago2018wasserstein}. These two formulas have inspired several works in which data measures are either projected onto 1D lines~\citep{rabin-ssvm-11,2013-Bonneel-barycenter}, with further developments in~\cite{paty2019subspace,kolouri2019generalized,titouan2019sliced}; or represented by Gaussians, to take advantage of the simpler computational possibilities offered by the Wasserstein-Bures metric~\cite{heusel2017gans,muzellec2018elliptical,chen2018optimal}.

Various schemes have been proposed to regularize the OT problem in the primal~\cite{cuturi2013sinkhorn,FrognerNIPS} or the dual~\citep{shirdhonkar2008approximate,arjovsky2017wasserstein,2016-Cuturi-siims}. We focus in this work on the formulation obtained by~\cite{chizat17}, which combines entropic regularization~\cite{cuturi2013sinkhorn} with a more general formulation for unbalanced transport~\cite{2017-chizat-focm,LieroMielkeSavareShort,LieroMielkeSavareLong}. The advantages of unbalanced entropic transport are numerous: it comes with favorable sample complexity regimes compared to unregularized OT~\citep{genevay19sample}, can be cast as a loss with favorable properties~\citep{genevay18a,feydy2019}, and can be evaluated using variations of the Sinkhorn algorithm~\cite{genevay2016stochastic}.

\paragraph{On the absence of closed-form formulas for regularized OT.}
Despite its appeal, one of the shortcomings of entropic regularized OT lies in the absence of simple test-cases that admit closed-form formulas. While it is known that regularized OT can be related, in the limit of infinite regularization, to the energy distance~\cite{ramdas2017wasserstein}, the absence of closed-form formulas for a fixed regularization strength poses an important practical problem to evaluate the performance of stochastic algorithms that try to approximate regularized OT: we do not know of any setup for which the ground truth value of entropic OT between continuous densities is known. The purpose of this paper is to fill this gap, and provide closed form expressions for balanced and unbalanced OT for Gaussian measures. We hope these formulas will prove useful in two different ways: as a solution to the problem outlined above, to facilitate the evaluation of new methodologies building on entropic OT, and more generally to propose a more robust yet well-grounded replacement to the Bures-Wasserstein metric.

\paragraph{Related work.} From an economics theory perspective, \citet{galichon} provided a closed form for an ``equilibrium 2-sided matching problem'' which is equivalent to entropy-regularized optimal transport. Second, a sequence of works in optimal control theory \cite{chen16, chen18, chen20} studied stochastic systems, of which entropy regularized optimal transport between Gaussians can be seen as a special case, and found a closed form of the optimal dual potentials. Finally, a few recent concurrent works provided a closed form of entropy regularized OT between Gaussians: first \citet{gerolin20} found a closed form in the univariate case, then \citet{mallasto20} and \citet{delbarrio20} generalized the formula for multivariate Gaussians. The closest works to this paper are certainly those of \citet{mallasto20} and \citet{delbarrio20} where the authors solved the balanced entropy regularized OT and studied the Gaussian barycenters problem. To the best of our knowledge, the closed form formula we provide for unbalanced OT is novel. Other differences between this paper and the aforementioned papers are highlighted below.

\paragraph{Contributions.} Our contributions can be summarized as follows:

\begin{itemize}
\item \Cref{thm:otclosedform} provides a closed form expression of the entropic (OT) plan $\pi$, which is shown to be a Gaussian measure itself (also shown in \citep{galichon, chen20, mallasto20, delbarrio20}). Here, we furthermore study the properties of the OT loss function: it remains well defined, convex and differentiable even for singular covariance matrices unlike the Bures metric.
\item Using the definition of debiased Sinkhorn barycenters~\citep{luise19,janati20}, \Cref{thm:barycenters} shows that the entropic barycenter of Gaussians is Gaussian and its covariance verifies a fixed point equation similar to that of \citet{agueh11}. \citet{mallasto20} and \citet{delbarrio20} provided similar fix point equations however by restricting the barycenter problem to the set of Gaussian measures whereas we consider the larger set of sub-Gaussian measures.

\item As in the balanced case,  \Cref{thm:unbalanced} provides a closed form expression of the unbalanced Gaussian transport plan. The obtained formula sheds some light on the link between mass destruction and the distance between the means of $\alpha, \beta$ in Unbalanced OT.
\end{itemize}

\paragraph{Notations.}

$\Scal^d$ denotes the set of square symmetric matrices in $\RR^{d\times d}$. $\Scal^d_{++}$ and $\Scal^d_+$ denote the cones of positive definite and positive semi-definite matrices in $\Scal^d$ respectively.
Let $\Ncal(\ba, \bA)$ denote the multivariate Gaussian distribution with mean $\ba \in \RR^d$ and variance $\bA \in \Scal^d_{++}$.
$f = \Qcal(\ba, \bA)$ denotes the quadratic form $f: x\mapsto -\frac{1}{2}(x^\top \bA x - 2\ba^\top x)$ with $\bA \in \Scal^d$. For short, we denote $\Qcal(\bA) = \Qcal(0, \bA)$. Whenever relevant, we follow the convention $0\log 0 = 0$. $\Mcal^+_p$ denotes the set of non-negative measures in $\RR^d$ with a finite p-th order moment and its subset of probablity measures $\Pcal_p$. For a non-negative measure $\alpha \in \Mcal^+_p(\RR^d)$, $\Lcal_2(\alpha)$ denotes the set of functions  $f: \RR^d\to\RR$ such that $\Esp_\alpha(|f|^2)=\int_{\RR^d} |f|^2 \dd \alpha < +\infty$. With $\bC\in\Scal^d_{++}$ and $\ba,\bb\in\RR^d$, we denote the squared Mahalanobis distance: $\|\ba-\bb\|_{\bC}^2 = (\ba-\bb)^\top\bC(\ba-\bb)$.

config: arxiv


TITLE: How to show that the limit of a function does not exist, while $\limsup$ and $\liminf$ do?
QUESTION [0 upvotes]: It does seem like a very simple question, but I don't see how to do it properly.
How can we show that the limit of
$$3^{\lfloor\frac{-\log(j)}{\log(2)}\rfloor}j^{\frac{\log(3)}{\log(2)}}$$
as $j\rightarrow\infty$does not exist, while $\limsup$ and $\liminf$ exist and are equal to $1$ and $\frac{1}{3}$?
I can see it intuitively but don't see how to prove it. Thank you very much for any help.

REPLY [1 votes]: Every term of the sequence is in $(\frac13,1]$ since $x-1\lt\lfloor x\rfloor\leqslant x$ for every $x$. If $j=2^n$, the result is $1$. If $j=2^n+1$, the result is $\frac13\,\left(1+\frac1{2^n}\right)$. Hence the limsup is $1$ and the liminf is $\frac13$.


text:
\section{Contextual stochastic block model}\label{KPCA-sec-CSBM}

\subsection{Problem setup}

Contextual network analysis concerns discovering interesting structures such as communities in a network with the help of node attributes. Large-scale applications call for computationally efficient procedures incorporating the information from both sources. For community detection in the contextual setting, various models and algorithms have been proposed and analyzed \citep{ZLZ16, WFe16, BVR17, MMa17, DSM18, MHC19, YSa20}. How to quantify the benefits of aggregation is a fundamental and challenging question.
We study community detection under a canonical model for contextual network data and prove the optimality of a simple spectral method.

To begin with, we present a binary version of the stochastic block model \citep{HLL83} that plays a central role in statistical network analysis \citep{Abb17}.
We use a label vector $\by = (y_1,\cdots,y_n)^{\top} \in \{\pm 1\}^n$ to encode the block (community) memberships of nodes. For any pair of nodes $i$ and $j$, we connect them with probability $\alpha$ if they are from the same block. Otherwise, the connection probability is $\beta$.

\begin{definition}[Stochastic Block Model]\label{KPCA-defn-SBM}
	For $n \in \ZZ_+$, $\by \in \{ \pm 1 \}^n$ and $0 < \alpha,  \beta < 1$, we write $\bA \sim \mathrm{SBM} ( \by, \alpha, \beta)$ if $\bA \in \{ 0, 1 \}^{n\times n}$ is symmetric, $A_{ii} = 0$ for all $i\in[n]$, $\{ A_{ij} \}_{1 \leq i < j \leq n}$ are independent, and
	\begin{align*}
	\PP ( A_{ij} = 1 ) = \begin{cases}
	\alpha & \mbox{ if } y_i  = y_j \\
	\beta & \mbox{ if } y_i \neq  y_j
	\end{cases},\qquad
	\forall i \neq j.
	\end{align*}
\end{definition}

In addition to the network, we also observe an attribute vector $\bx_i \in \RR^d$ of each node $i$ and postulate the Gaussian mixture model in Definition \ref{KPCA-defn-GMM}. Given the labels and other parameters, the network $\bA$ and node attributes $\{ \bx_i \}_{i=1}^n$ are assumed to be independent.
We borrow the name ``contextual stochastic block model'' from \cite{DSM18}.
More general versions can be found in \cite{BVR17}, \cite{DSM18} and \cite{YSa20}. In another line of research, the network $\bA$ is generated based on the covariates $\{ \bx_i \}_{i=1}^n$ \citep{WFe16, MMa17, MHC19}.

For simplicity, we impose uniform priors on the label vector $\by$ and the direction of separation vector $\bmu$. The two blocks are then approximately balanced.

\begin{definition}[Contextual Stochastic Block Model]\label{KPCA-defn-CSBM}
For $n \in \ZZ_+$, $0 <  \alpha , \beta < 1$, $d \geq 2$ and $R>0$, we write $( \by, \bmu, \bA, \{ \bx_i \}_{i=1}^n ) \sim \mathrm{CSBM} ( n, d, \alpha, \beta, R )$ if
\begin{enumerate}
\item the label vector $\by$ and separation vector $\bmu$ are independently generated from the uniform distributions over $\{ \pm 1 \}^n$ and $\{ \bu \in \RR^d :~ \| \bu \|_2 = R \}$, respectively;
\item given $\by$ and $\bmu$, the network $\bA$ and attributes $\{ \bx_i \}_{i=1}^n $ are independently generated from $\mathrm{SBM} ( \by, \alpha, \beta)$ and $\mathrm{GMM} ( \bmu, \by)$, respectively.
\end{enumerate}
\end{definition}

The goal of contextual community detection is to reconstruct $\by$ based on $\bA$ and $\{ \bx_i \}_{i=1}^n$.
We consider a commonly-used regime of the network where the connection probabilities $\alpha$, $\beta$ scale like $q_n/n$ for some $1 \ll q_n \leq \log n$ and differ by a constant factor. When $q_n \gg \log n$, one can easily recover the communities perfectly from $\bA$ \citep{Abb17}. When $q_n = O(1)$, it is not possible to achieve vanishingly small misclassification error \citep{ZZh16}. We are interested in the intermediate regime $1 \ll q_n \leq \log n$.
Meanwhile, recall that $\mathrm{SNR} = R^4 / (R^2 + d / n)$ in (\ref{KPCA-snr-gmm}) is the signal-to-noise ratio of the Gaussian mixture model. We take $\mathrm{SNR} \asymp q_n$ to ensure that the signal strengths of $\bA$ and $\{ \bx_i \}_{i=1}^n$ are comparable. There is no specific assumption on the dimension $d=d_n$. It may be bounded or diverge at any rate.

\begin{assumption}[Asymptotics]\label{KPCA-assumption-CSBM-q}
	Let $a$, $b$ and $c$ be positive constants. $( \by, \bmu, \bA, \{ \bx_i \}_{i=1}^n ) \sim \mathrm{CSBM} ( n, d, \alpha, \beta, R )$ with $1 \ll q_n \leq \log n$, $\alpha = \frac{a q_n}{n}$, $\beta = \frac{b q_n}{n}$ and $R^4 / (R^2 + d / n) = c q_n $.
\end{assumption}

\subsection{An aggregated spectral estimator}

On the one hand, Section \ref{KPCA-sec-GMM-GMM} shows that the leading eigenvector $\bu_1$ of the hollowed Gram matrix $\bG = \cH( \bX \bX^{\top} )$ is optimal for the Gaussian mixture model. From now on we rename it as $\bu_1(\bG)$ to avoid ambiguity. On the other hand, the second eigenvector $\bu_2(\bA)$ of $\bA$ estimates the labels under the stochastic block model \citep{AFW17}. To get some intuition, suppose that half of the entries in $\{ y_i \}_{i=1}^n$ are $+1$'s and the others are $-1$'s so that $ \mathbf{1}_n^{\top} \by = 0$. For such $\by$, it is easy to see from
\begin{align} \label{fan1}
\EE ( \bA | \by ) = \frac{\alpha+ \beta}{2} \mathbf{1}_n \mathbf{1}_n^{\top} + \frac{\alpha- \beta}{2} \by \by^{\top}
\end{align}
that its second eigenvector $\by/\sqrt{n}$ reveals the community structure. We propose an estimator for the integrated problem by aggregating the two individual spectral estimators $\bu_2(\bA)$ and $\bu_1(\bG)$. Without loss of generality, we assume $\langle \bu_2(\bA), \bu_1(\bG) \rangle \geq 0$ to avoid cancellation.

Let us begin the construction. The ideal `estimator'
\[
\hat y_i^{\mathrm{genie}} = \argmax_{ y = \pm 1 }  \PP ( y_i =  y | \bA , \bX, \by_{-i} ).
\]
is the best guess of $y_i$ given the network, attributes, and labels of all nodes (assisted by  Genie) except the $i$-th one. It is referred to as a genie-aided estimator or oracle estimator in the literature and is closely related to fundamental limits in clustering \citep{ABH16, ZZh16}, see Theorem \ref{KPCA-lemma-fundamental}. To mimic $\hat y_i^{\mathrm{genie}}$, we first approximate its associated odds ratio.

\begin{lemma}\label{KPCA-lemma-CSBM-linearization}
Under Assumption \ref{KPCA-assumption-CSBM-q}, for each given $i$, we have
\begin{align*}
\bigg| \log \bigg( \frac{ \PP ( y_i =  1 | \bA , \bX, \by_{-i} ) }{ \PP ( y_i = -1 | \bA , \bX, \by_{-i} ) } \bigg) -
\bigg[ \bigg(
\log (a/b) \bA + \frac{2}{n + d / R^2 } \bG
\bigg) \by \bigg]_i
\bigg| = o_{\PP} (q_n;~q_n).
\end{align*}
\end{lemma}

The $i$-th coordinate of $\bA \by$ corresponds to the log odds ratio $ \log [ \PP ( y_i =  1 | \bA , \by_{-i} ) / \PP ( y_i = -1 | \bA , \by_{-i} ) ]$ for the stochastic block model \citep{ABH16}. From $A_{ii} = 0$ we see that $(\bA \by)_i = \sum_{j \neq i} A_{ij} y_j$ tries to predict the label $y_i$ via majority voting among the neighbors of node $i$. Similarly, $(\bG \by)_i$ relates to the log odds ratio $ \log [ \PP ( y_i =  1 | \bX , \by_{-i} ) / \PP ( y_i = -1 | \bX , \by_{-i} ) ]$ for the Gaussian mixture model. The overall log odds ratio is linked to a linear combination of $\bA \by$ and $\bG \by$ thanks to the conditional independence between $\bA$ and $\bX$ in Definition \ref{KPCA-defn-CSBM}. The proof of Lemma \ref{KPCA-lemma-CSBM-linearization} can be found in Appendix \ref{KPCA-lemma-CSBM-linearization-proof}.

Intuitively, Lemma \ref{KPCA-lemma-CSBM-linearization} reveals that
\begin{align*}
\sgn \bigg( \log (a/b) \bA \by + \frac{2}{n + d / R^2 } \bG \by \bigg) \approx (\hat y_1^{\mathrm{genie}}, \cdots, \hat y_n^{\mathrm{genie}})^{\top}
\end{align*}
The left-hand side still involves unknown parameters $a/b$, $R$ and $\by$.  Once these unknowns are consistently estimated, the substitution version of the left-hand side provides a valid estimator that mimics well the genie-aided estimator and hence is optimal.
Heuristics of linear approximation in Theorem \ref{KPCA-thm-GMM-Lp} above and \cite{AFW17} suggest
\begin{align*}
\bu_2(\bA) \approx \bA \bar\bu / \bar\lambda_A \qquad \text{and} \qquad \bu_1(\bG) \approx \bG \bar\bu / \bar\lambda_G.
\end{align*}
Here $\bar\bu = \by / \sqrt{n}$, $\bar\lambda_A = n(\alpha-\beta)/2$ is the second largest (in absolute value) eigenvalue of $\EE(\bA | \by)$ when $\alpha \neq \beta$ and the two blocks are equally-sized, and $\bar\lambda_G = n R^2$ is the leading eigenvalue of $\bar\bG = \bar\bX \bar\bX^{\top}$. Hence
\begin{align}
~~& \log (a/b) \bA \by + \frac{2}{n + d / R^2 } \bG \by \notag\\
& \approx \log (a/b) \sqrt{n} \bar\lambda_A \bu_2(\bA) + \frac{2}{n + d / R^2 } \sqrt{n} \bar\lambda_G \bu_1(\bG)  \notag \\
& \propto  \frac{n(\alpha-\beta)}{2} \log \bigg( \frac{\alpha}{\beta} \bigg) \bu_2(\bA) + \frac{2 R^4}{R^2 + d / n} \bu_1(\bG),
\label{eqn-CSBM-aggregation}
\end{align}
which yields a linear combination of $\bu_2(\bA)$ and $\bu_1(\bG)$. The coefficient in front of $\bu_1(\bG)$ is twice the $\mathrm{SNR}$ in (\ref{KPCA-snr-gmm}) for Gaussian mixture model.  Analogously, we may regard $\frac{n(\alpha-\beta)}{4} \log ( \alpha / \beta ) $ as a signal-to-noise ratio for the stochastic block model.

An legitimate estimator for $\by$ is obtained by replacing the unknown parameters $\alpha$, $\beta$ and $R$ in (\ref{eqn-CSBM-aggregation}) by their estimates.  When the two classes are balanced, i.e. $\by^{\top} \mathbf{1}_n = 0$,  \eqref{fan1} yields $\lambda_{1}[ \EE(\bA | \by) ] = n(\alpha + \beta) / 2$ and $\lambda_{2}[ \EE(\bA | \by) ] = n(\alpha - \beta) / 2$. Here $\lambda_j(\cdot)$ denotes the $j$-th largest (in absolute value) eigenvalue of a real symmetric matrix. Hence,
\begin{align*}
\frac{n(\alpha-\beta)}{2} \log \bigg( \frac{\alpha}{\beta} \bigg)
& = \lambda_{2}[ \EE(\bA | \by) ] \log \bigg( \frac{ \lambda_{1}[ \EE(\bA | \by) ] + \lambda_{2}[ \EE(\bA | \by) ]}{ \lambda_{1}[ \EE(\bA | \by) ] - \lambda_{2}[ \EE(\bA | \by) ]} \bigg) \\
& \approx \lambda_2(\bA) \log \bigg( \frac{\lambda_1(\bA) + \lambda_2(\bA)}{\lambda_1(\bA) - \lambda_2(\bA)} \bigg) .
\end{align*}
It can be consistently estimated using the plug-in method.
Similarly, using $\lambda_1(\bar\bG) = n R^2$, we have
\begin{align*}
\frac{2 R^4}{R^2 + d / n} = \frac{2 [\lambda_1(\bar\bG)/n]^2 }{ \lambda_1(\bar\bG) / n + d / n } \approx \frac{2 \lambda_1^2(\bG) }{n \lambda_1(\bG) + n d }.
\end{align*}

Based on these, we get an aggregated spectral estimator $\sgn(\hat\bu)$ with
\begin{align}
\hat\bu = \log \bigg( \frac{\lambda_1(\bA) + \lambda_2(\bA)}{\lambda_1(\bA) - \lambda_2(\bA)} \bigg)  \lambda_2(\bA) \bu_2(\bA) +  \frac{2 \lambda_1^2(\bG) }{n \lambda_1(\bG) + n d } \bu_1(\bG).
\label{eqn-CSBM-aggregation-1}
\end{align}
Our estimator uses a weighted sum of two individual estimators without any tuning parameter.
When there are $K>2$ communities, it is natural to compute spectral embeddings using $(K-1)$ eigenvectors of $\bA$ and $\bG$, respectively. One may apply Procrustes analysis \citep{Wah65} to align the two embeddings and then use their linear combination for clustering. It would be nice to develop a tuning-free procedure similar to the above.

\cite{BVR17} propose a spectral method based on a weighted sum of the graph Laplacian matrix and $\bX \bX^{\top}$. \cite{YSa20} develop an SDP using a weighted sum of $\bA$ and a kernel matrix of $\{ \bx_i \}_{i=1}^n$. \cite{DSM18} study a belief propagation algorithm. Their settings are different from ours.

\subsection{Analysis of the estimator when $q_n = \log n$}

There are very few theoretical results on the information gain in combining the network and node attributes. \cite{BVR17} and \cite{YSa20} derive upper bounds for the misclassification error that depend on both sources of information. However, those bounds are not tight and cannot rigorously justify the benefits. \cite{DSM18} use techniques from statistical physics to derive an information threshold for weak recovery (i.e. better than random guessing) in some regimes. The threshold is smaller than those for the stochastic block model and the Gaussian mixture model. Their calculation is under the sparse regime where the maximum expected degree $n(\alpha + \beta) / 2$ of the network remains bounded as $n$ goes to infinity. They obtain a formal proof by taking certain large-degree limits.
To our best knowledge, the result below gives the first characterization of the information threshold for exact recovery and provides an efficient method achieving it by aggregating the two pieces of information.

We now investigate the aggregated spectral estimator (\ref{eqn-CSBM-aggregation-1}) under the Assumption \ref{KPCA-assumption-CSBM-q} with $q_n = \log n$. Our study shows that $\sgn(\hat\bu)$ achieves the information threshold for exact recovery as well as the optimal misclassification rate, both of which are better than
those based on a single form of data in terms of the mismatch $\cM$ in (\ref{eqn-gmm-mismatch}). To state the results, define
\begin{align}
I^*( a, b, c ) = \frac{(\sqrt{a} - \sqrt{b})^2 + c}{2}.
\label{KPCA-eqn-CSBM-h}
\end{align}

\begin{theorem}\label{KPCA-theorem-CSBM-eaxct}
	Let Assumption \ref{KPCA-assumption-CSBM-q} hold with $q_n = \log n$ and $a \neq b$.
\begin{enumerate}
\item When $I^*( a, b, c )  > 1$, we have $\lim_{n \to \infty} \PP [\cM ( \sgn( \hat\bu ) , \by ) = 0] = 1$.
\item When $I^*( a, b, c ) < 1$, we have $\liminf_{n \to \infty} \PP [\cM ( \hat\by , \by ) > 0] > 0$ for any sequence of estimators $\hat\by = \hat\by_n ( \bA, \{ \bx_i \}_{i=1}^n )$.
\end{enumerate}
\end{theorem}

Theorem \ref{KPCA-theorem-CSBM-eaxct} asserts that $I^*(a,b,c)$ quantifies the signal-to-noise ratio and the phase transition of exact recovery takes place at $I^*( a, b, c ) = 1$.  When $c = 0$ (node attributes are uninformative), we have
$I^*(a, b, 0) = ( \sqrt{a} - \sqrt{b} )^2 / 2 $; the threshold reduces to that for the stochastic block model [$|\sqrt{a} - \sqrt{b} | = \sqrt{2}$ by \cite{ABH16}].  Similarly, when $a = b$ (the network is uninformative), we have
$I^*(a, a, c) = c/2$; the threshold reduces to that for the Gaussian mixture model [$c = 2$ by \cite{Nda18}]. The relation (\ref{KPCA-eqn-CSBM-h}) indicates that combining two sources of information adds up the powers of each part. The proof of Theorem \ref{KPCA-theorem-CSBM-eaxct} is deferred to Appendix \ref{KPCA-theorem-CSBM-eaxct-proof}.

Figure \ref{fig-CSBM} demonstrates the efficacy of our aggregated estimator $\sgn( \hat\bu )$. The two experiments use $c = 0.5$ and $c = 1.5$ respectively. We fix $n = 500$, $d = 2000$ and vary $a$ ($y$-axis), $b$ ($x$-axis) from $0$ to $8$. For each parameter configuration $(a,b,c)$, we compute the frequency of exact recovery (i.e. $\sgn(\hat\bu) = \pm \by$) over 100 independent runs. Light color represents high chance of success. The red curves $(\sqrt{a} - \sqrt{b})^2 + c = 2$ correspond to theoretical boundaries for phase transitions, which match the empirical results pretty well. Also, larger $c$ implies stronger signal in node attributes and makes exact recovery easier.

\begin{figure}[ht]
	\begin{minipage}[b]{0.45\linewidth}
		\centering
		\includegraphics[width=0.7\textwidth]{1.png}
	\end{minipage}
	\hspace{0.5cm}
	\begin{minipage}[b]{0.45\linewidth}
		\centering
		\includegraphics[width=0.7\textwidth]{2.png}
	\end{minipage}
		\caption{Exact recovery for CSBM: $c = 0.5$ (left) and $c = 1.5$ (right).}
				\label{fig-CSBM}
\end{figure}

When $I^*(a,b,c) < 1$, exact recovery of $\by$ with high probability is no longer possible. In that case, we justify the benefits of aggregation using misclassification rates, by presenting an upper bound for $\sgn(\hat\bu)$ as well as a matching lower bound for all possible estimators. Their proofs can be found in Appendices \ref{KPCA-theorem-CSBM-error-rate-proof} and \ref{KPCA-thm-CSBM-lower-proof}.

\begin{theorem}\label{KPCA-theorem-CSBM-error-rate}
Let Assumption \ref{KPCA-assumption-CSBM-q} hold, $q_n = \log n$, $a \neq b$ and $I^*(a,b,c) \leq 1$. Then
	\begin{align*}
	\limsup_{n \to \infty} q_n^{-1} \log \EE \cM ( \sgn( \hat\bu ) , \by )  \leq - I^*( a, b, c ) .
	\end{align*}
\end{theorem}
\begin{theorem}\label{KPCA-thm-CSBM-lower}
Let Assumption \ref{KPCA-assumption-CSBM-q} hold. For any sequence of estimators $\hat\by = \hat\by_n ( \bA, \{ \bx_i \}_{i=1}^n )$,
	\begin{align*}
	\liminf_{n \to \infty} q_n^{-1} \log \EE \cM ( \hat{ \by } , \by ) \geq - I^*(a, b, c).
	\end{align*}
\end{theorem}

Theorems \ref{KPCA-theorem-CSBM-error-rate} and \ref{KPCA-thm-CSBM-lower} imply that in the $\log n$-regime, the aggregated spectral estimator $\sgn(\hat\bu)$ achieves the optimal misclassification rate:
\[
\EE \cM ( \sgn(\hat\bu) , \by ) = n^{-I^*(a,b,c)+o(1)}.
\]
When $c = 0$, it reduces to the optimal rate $n^{ - ( \sqrt{a} - \sqrt{b} )^2 / 2 + o(1) }$ for the stochastic block model (Definition \ref{KPCA-defn-SBM}) and when  $a=b$, the result reduces $n^{ - c / 2 + o(1) }$ for the Gaussian mixture model (Definition \ref{KPCA-defn-GMM}), respectively. It is easy to show that they are achieved by $\bu_2(\bA)$ \citep{AFW17} and $\bu_1(\bG)$ (Theorem \ref{KPCA-thm-GMM}), which are asymptotically equivalent to our aggregated estimator $\hat\bu$ in extreme cases $c \to 0$ and $a \to b$, respectively. In other words, our result and procedure encompass those for the stochastic block model and Gaussian mixture model as two specific examples.

\subsection{A modified estimator for the general case}

While \Cref{KPCA-thm-CSBM-lower} establishes a lower bound $e^{- q_n [ I^*(a,b,c) + o(1) ] }$ for misclassification under Assumption \ref{KPCA-assumption-CSBM-q} without restricting $q_n = \log n$, our aggregated spectral estimator $\sgn(\hat\bu)$ is only analyzed for the $\log n$-regime. If the network becomes sparser ($q_n \ll \log n$), the empirical eigenvalues $\lambda_1(\bA)$ and $\lambda_2(\bA)$ no longer concentrate around  $\frac{\alpha + \beta}{2}$ and $\frac{\alpha - \beta}{2}$ \citep{FOf05}. The eigenvector analysis of $\bu_2(\bA)$ in \cite{AFW17} breaks down. Consequently, the estimator (\ref{eqn-CSBM-aggregation-1}) fails to approximate the (scaled) vector of log odds $n^{-1/2} \log (a/b) \bA \by + n^{-1/2}  \frac{2}{n + d / R^2 } \bG \by  $.

Fortunately, the $\ell_p$ results for $\bu_1(\bG)$ in the current paper continue to hold, and $\hat{\by}_G = \sgn[\bu_1(\bG)]$ faithfully recovers $\by$. Hence, we need only to modify the first term in (\ref{eqn-CSBM-aggregation-1}) concerning the network $\bA$, which aims to approximate $n^{-1/2} \log (a/b) \bA \by $. To approximate $\log(a/b)$, we resort to $\bm{1}^{\top} \bA \bm{1} / n \approx \frac{\alpha + \beta}{2} = \frac{( a+b ) q_n}{n}$ and $\by \bA \by / n \approx \frac{\alpha - \beta}{2} = \frac{( a-b ) q_n}{n}$ so that $a/b \approx \frac{\bm{1}^{\top} \bA \bm{1} + \by \bA \by}{\bm{1}^{\top} \bA \bm{1} - \by \bA \by}$.  Thus, we propose a new estimator $\sgn(\tilde\bu)$ with
\begin{align}
\tilde\bu = \frac{1}{\sqrt{n}} \log \bigg( \frac{
\bm{1}^{\top} \bA \bm{1} + \hat{\by}_G^{\top} \bA \hat{\by}_G
}{
\bm{1}^{\top} \bA \bm{1} - \hat{\by}_G^{\top} \bA \hat{\by}_G
} \bigg) \bA  \hat{\by}_G
 \ +  \frac{2 \lambda_1^2(\bG) }{n \lambda_1(\bG) + n d } \bu_1(\bG),
\label{eqn-CSBM-aggregation-2}
\end{align}
where $\hat{\by}_G $ estimates $\by$.

The new estimator $\sgn(\tilde\bu)$ achieves the fundamental limit $e^{- q_n [ I^*(a,b,c) + o(1) ] }$ even if $q_n \ll \log n$. See \Cref{thm-CSBM-q} below and its proof in \Cref{sec-proof-thm-CSBM-q}.

\begin{theorem}\label{thm-CSBM-q}
Let Assumption \ref{KPCA-assumption-CSBM-q} hold and $a \neq b$. We have
	\begin{align*}
	\limsup_{n \to \infty} q_n^{-1} \log \EE \cM ( \sgn( \tilde\bu ) , \by )  \leq - I^*( a, b, c ) .
	\end{align*}
In addition, if $q_n = \log n$ and $ I^*( a, b, c ) > 1$ then $\lim_{n \to \infty} \PP [\cM ( \sgn( \tilde\bu ) , \by ) = 0] = 1$.
\end{theorem}