text:
\begin{document}
\title{Doubling measures on uniform Cantor sets\thanks{Supported
by the NSFC (Nos. 11271114 and 10971056).}}

\author{Chun WEI,
\quad  Shengyou WEN,\thanks{Corresponding author. E-mail:
sywen\_65@163.com}\quad and Zhixiong WEN}

\maketitle

\begin{abstract}
We obtain a complete description for a probability measure to be doubling on an arbitrarily
given uniform Cantor set. The question of which doubling measures on such a Cantor set can be
extended to a doubling measure on $[0,1]$ is also considered.

\medskip

\noindent{\bf Keywords}\, Uniform Cantor Set, Doubling Measure, Extension

\noindent{\bf 2010 MSC:}   28A12 (Primary), 30L10 (Secondary)
\end{abstract}

\section{Introduction}

It is well known that every complete doubling metric space carries a doubling measure
\cite{LS,KRS,VK} and that doubling measures are rich
when the underlying doubling metric space has no dense isolated points \cite{KW}.
In the Euclidean $n$-space, a measure, defined by integrating an $A_\infty$
weight, is doubling \cite{G}. These measures form a subclass of absolutely
continuous doubling measures on $\mathbb{R}^n$.
Examples of singular doubling measures can be obtained from self-similar measures
on the unit square \cite{Fa,Y}.
The existence of singular doubling measures have also been studied extensively
in general metric spaces \cite{KW,Tu,Wu}. Recall that a Borel regular measure $\mu$ on a
metric space $X$ is doubling, if there is a constant $C\geq 1$ such that
\begin{equation}\label{eq1}
0<\mu(B(x,2r))\leq C\mu(B(x,r))<\infty\,\,\mbox{for all $x\in X$ and $r>0$.}
\end{equation}

Doubling measures naturally arise in the study of the gradients of
convex functions on the Euclidean $n$-space. See \cite{KM,KMW}, where it is showed that,
in many cases, the gradient of
a convex function is a quasisymmetric self-homeomorphism of $\mathbb{R}^n$, by which
the pullback measure of the Lebesgue measure is doubling. Doubling measures can be
applied to classify
subsets of a metric space, in which, for example,
a subset is called very thin, if it is null for all doubling measures on the metric space.
This leads to the study on fat and thin sets \cite{B,hww,ORS}.
Doubling measures can also be used to construct quasisymmetric deformations of the
underlying metric space, so that the objective space is in some sense better than the original space,
for example, it can be proved that, if a metric space is uniformly perfect and carries a doubling
measure, then it is quasisymmetrically
equivalent to an Ahlfors regular space \cite{H1}.

Though examples of doubling measures are rich, a complete description for doubling measures
on a given metric space is usually difficult. It is clear that a measure on the real line
$\mathbb{R}$ is doubling if and only if it is a pullback measure of a quasisymmetric self-homeomorphism
of $\mathbb{R}$. However, this is not true for higher dimensional Euclidean
space. In the present paper, we study doubling measures on uniform Cantor sets.
We shall give a sufficient and necessary condition
for the doubling property of a measure on an arbitrarily given uniform Cantor set.
The question of which doubling measure on such a Cantor set can be extended to a doubling
measure on $[0,1]$ is also discussed. For the related papers on Cantor sets and doubling measures
on $[0,1]$, we refer to \cite{CS,PW,SW,WW}.

We begin with uniform Cantor sets. Let
$\mathbf{n}=\{n_k\}_{k=1}^\infty$ be a sequence of
integers, where $n_k\geq2$. Let $\mathbf{c}=\{c_k\}_{k=1}^\infty$ be a sequence of real
numbers in $(0,1)$ such that $(n_k-1)c_k<1$ for all $k\geq1$. The
uniform Cantor set $E(\mathbf{n},\mathbf{c})$ is defined by
\begin{equation}\label{eq2}
E(\mathbf{n},\mathbf{c})=\bigcap_{k=0}^\infty E_k,
\end{equation}
where $\{E_k\}$ is a sequence of nested compact sets in $[0,1]$,
$E_0=[0,1]$, and $E_k$ is obtained by deleting from every
component $I$ of $E_{k-1}$ $(n_k-1)$ open intervals of equal length
$c_k|I|$, such that the remaining $n_k$ closed intervals are of
equal length. A component of $E_k$ will be called a component of
level $k$ and a component of $E_{k-1}\setminus E_k$ will be called
a gap of level $k$. Denote by $N_k$, $\delta_k$, and
$\varepsilon_k$, respectively, the number of components of level
$k$, the length of a component of level $k$, and the length of a
gap of level $k$. Then, from the definition
\begin{equation}\label{eq3}
N_k=\prod_{i=1}^kn_i,\ \
\delta_k=\prod_{i=1}^k\frac{1-(n_i-1)c_i}{n_i},\ \
\varepsilon_k=c_k\delta_{k-1}.
\end{equation}
Note that the uniform Cantor set $E(\mathbf{n},\mathbf{c})$
is of Lebesgue measure
$$\mathcal{L}(E(\mathbf{n},\mathbf{c}))=\prod_{i=1}^\infty(1-(n_i-1)c_i).$$
Therefore
\begin{equation}\label{eq4}
\mathcal{L}(E(\mathbf{n},\mathbf{c}))>0\Leftrightarrow\{n_kc_k\}\in\ell^1,
\end{equation}
where $\ell^p$ denotes the set of real sequences $\{a_k\}_{k=1}^\infty$ with $\sum_{k=1}^\infty
|a_k|^p<\infty$.

For a uniform Cantor set
$E(\mathbf{n},\mathbf{c})$ we denote by $\mathcal{I}_k$ the family of its components of level $k$
and by $\mathcal{G}_k$ the family of its gaps of level $k$. Then $\sharp\mathcal{I}_k=N_k$
and $\sharp\mathcal{G}_k=N_{k-1}(n_k-1)$, where $\sharp$ denotes the cardinality.
Let $\mathcal{I}=\cup_{k=1}^\infty\mathcal{I}_k$ and $\mathcal{G}=\cup_{k=1}^\infty\mathcal{G}_k$.
To label these components and gaps, we relate a word to each of them as
follows: Let
\begin{equation}\label{eq5}
W_k=\{i_1i_2\cdots i_k : 1\leq i_j\leq n_j, 1\leq j\leq
k\}\,\,\mbox{and}\,\, W=\cup_{k=1}^\infty W_k.
\end{equation}
A member in $W_k$ is called
a word of length $k$ and a member in $W$ is called a word of finite
length. Let the words in $W$ be ordered in the lexicographic order.
Then the first $n_1$ words are $1,2,\cdots,n_1$. We denote, from left to right, by $I_1,I_2,\cdots,I_{n_1}$
the $n_1$ components in $\mathcal{I}_1$ and by
$G_1,G_2,\cdots,G_{n_1-1}$ the $n_1-1$ gaps in $\mathcal{G}_1$. Inductively,
supposing that $I_w$ has been defined for a word $w\in W_{k-1}$, we denote, from left to right, by
$I_{w1},I_{w2},\cdots,I_{wn_k}$ the $n_k$ members of
$\mathcal{I}_k$ in $I_w$ and by $G_{w1},G_{w2},\cdots,G_{w(n_k-1)}$ the
$n_k-1$ members of $\mathcal{G}_k$ in $I_w$. With the above notation, once a word $w\in W_k$ is
specified, the relative position of the component $I_w$ of level $k$ is
determined. For example, the minimal word $11\cdots 1$ of length $k$ corresponds to the
leftmost component in $\mathcal{I}_k$ and the maximal word $n_1n_2\cdots n_k$ of length $k$
corresponds to the rightmost component in $\mathcal{I}_k$.
We say that two components $I_{w}$ and $I_{u}$ are adjacent, if $w,u\in W$ are adjacent in the sense
of the lexicographic order. For example, given $w\in W_{k-1}$ and $1\leq i<n_k$,
the maximal word $win_{k+1}\cdots n_{k+t}$ of length $k+t$ with prefix $wi$ and the minimal word $w(i+1)1\cdots 1$ of
length $k+t$ with prefix $w(i+1)$ are adjacent. They correspond to two adjacent components in $\mathcal{I}_{k+t}$. Note that
$$E(\mathbf{n},\mathbf{c})=\bigcap_{k=1}^\infty
\bigcup_{\omega\in W_k}I_\omega$$ and that
$$I_\omega =I_{w1}\cup G_{w1}\cup I_{w2}\cup G_{w2}\cup\cdots\cup I_{w(n_k-1)}\cup G_{w(n_k-1)}\cup I_{wn_k}$$
for all $w\in W_{k-1}$, where, if $k=1$, then $I_w=[0,1]$ and $wi=i$.

We also need some terminology on vectors. A vector $P=(p_1,p_2,\cdots,p_k)$ is called a positive probability,
if its components are all positive and $\sum_{i=1}^kp_i=1$.
A positive probability vector $P=(p_1,p_2,\cdots,p_k)$ is called $C$-uniform,
where $C\geq 1$ is a constant, if
\begin{equation}\label{jia}
C^{-1}\leq \frac{
p_{i+1}+\cdots+p_{i+l}}{p_{j+1}+\cdots+p_{j+l}}
\leq C\,\,\mbox{whenever $0\leq i\leq j\leq i+l\leq j+l\leq k$.}
\end{equation}
The condition means that two of
sums of consecutive $l$ components of $P$ are comparable, if they are adjacent.
Clearly, the vector $P$ is $1$-uniform if and only if $p_1=p_2=\cdots=p_k$.
We say that $P$ is $(C,s)$-uniform, where $1\leq s\leq k$, if
the condition (\ref{jia}) holds only for $s\leq l\leq k$. Thus, the
$(C,1)$-uniformness and the $C$-uniformness are the same. We say that a
sequence $\mathbb{P}=\{P_k\}$ of positive probability vectors
is $C$-uniform, if each $P_k$ is $C$-uniform. We say that $\mathbb{P}$ is ultimately $C$-uniform,
if there is an integer $k_0\geq1$ such that $P_k$ is $C$-uniform for each $k\geq k_0$.

Let $E(\mathbf{n},\mathbf{c})$ be a uniform Cantor set.
Let $\mathbb{P}=\{P_k\}_{k=1}^\infty$ be a sequence of positive probability vectors, where
$$P_k=(p_{k,1},p_{k,2},\cdots,p_{k, n_k})$$
has $n_k$ components.
Such a sequence of vectors will be called $\mathbf{n}$-matching. Given an $\mathbf{n}$-matching
sequence $\mathbb{P}$ of vectors, there is a unique probability measure on
$E(\mathbf{n},\mathbf{c})$, denoted by $\mu_\mathbb{P}$, satisfying
\begin{equation}\label{eq6}
\mu_\mathbb{P}(I_{wi})=p_{k,i}\mu_\mathbb{P}(I_w)\mbox{ for all $w\in W_{k-1}$ and $1\leq i\leq n_k$.}
\end{equation}
The positivity assumption of probability vectors is reasonable. Otherwise, the corresponding measure would support on
a proper compact subset of $E(\mathbf{n},\mathbf{c})$.

The following data of the uniform Cantor set $E(\mathbf{n},\mathbf{c})$ are crucial in stating our results.
Denote
\begin{equation}\label{eq7}
\Lambda=\{k\geq1: \varepsilon_k<\delta_k\}.
\end{equation}
Thus, a gap in $\mathcal{G}_k$ is shorter than a component in $\mathcal{I}_k$ at every level $k\in\Lambda$.
For each $k\in\Lambda$ let $m_k$ be the unique positive integer such
that
\begin{equation}\label{eq8}
\delta_{k+m_k}\leq\varepsilon_k<\delta_{k+m_k-1}.
\end{equation}
and let $s_k$ be the unique positive
integer such that
\begin{equation}\label{eq9}
s_k\delta_{k+m_k}+(s_k-1)\varepsilon_{k+m_k}\leq\varepsilon_k<(s_k+1)\delta_{k+m_k}+s_k\varepsilon_{k+m_k}.
\end{equation}
Write $E(\mathbf{n},\mathbf{c},\Lambda, \{m_k\}_{k\in\Lambda}, \{s_k\}_{k\in\Lambda})$
for $E(\mathbf{n},\mathbf{c})$, when these related data are emphasized.

\begin{thm}\label{1}
Let $E=E(\mathbf{n},\mathbf{c},\Lambda, \{m_k\}_{k\in\Lambda}, \{s_k\}_{k\in\Lambda})$ be a uniform
Cantor set. Let $\mu_\mathbb{P}$ be a measure determined by an $\mathbf{n}$-matching sequence
$\mathbb{P}$ of positive probability vectors. Then $\mu_\mathbb{P}$ is doubling on $E$ if and only
if there is a constant $C>1$ such that $\mathbb{P}$ is $C$-uniform and satisfies, for each $k\in\Lambda$,
the following conditions:

$(a)$ $(P_{k+t},P_{k+t})$ is $C$-uniform and
\begin{equation}\label{eq11}
C^{-1}\prod_{j=1}^tp_{k+j,1}\leq\prod_{j=1}^t{p_{k+j, n_{k+j}}}\leq C\prod_{j=1}^tp_{k+j,1}\,\,\mbox{for all\,\, $1\leq t<m_k$.}
\end{equation}

$(b)$ $(P_{k+m_k},P_{k+m_k})$ is $(C,s_k)$-uniform.

\noindent Hereafter, if $P=(p_1,p_2,\cdots,p_j)$, then $(P,P)=(p_{1},p_2,\cdots,p_{j},p_{1},p_2,\cdots,p_{j})$.
\end{thm}

From theorem \ref{1} some easier conditions for
the doubling property of the measure $\mu_{\mathbb{P}}$ on $E$ can be formulated.

\begin{cor}\label{c1}
If $\sup _{k\geq1}n_k<\infty$, then the measure $\mu_\mathbb{P}$ is
doubling on $E$ if and only
if there is a constant $C>1$ such that

(a)\, $C^{-1}p_{k,i}\leq p_{k,i+1}\leq Cp_{k,i}$ for all $1\leq i<n_k$ and for all $k$, and

(b)\, the condition (\ref{eq11}) is satisfied for all $1\leq t<m_k$ and for all $k\in\Lambda$.
\end{cor}

\begin{cor}\label{c2}
If $\sup_{k\in\Lambda}m_k<\infty$, then the measure $\mu_\mathbb{P}$
is doubling on $E$ if and only
if there is a constant $C>1$ such that

(a)\, $\mathbb{P}$ is $C$-uniform, and

(b)\, for every integer $k\in\Lambda$, $(P_{k+t},P_{k+t})$ is $C$-uniform for all $1\leq t<m_k$
and $(P_{k+m_k},P_{k+m_k})$ is $(C,s_k)$-uniform.
\end{cor}

\begin{cor}\label{c0}
If $\mathbb{P}$ is ultimately $1$-uniform,
then $\mu_{\mathbb{P}}$ is doubling on $E$.
\end{cor}

Theorem \ref{1} can be generalized. Let $\mu$ be a Borel probability measure on
$E$ with $\mu(I_w)>0$ for all $w\in
W$. For each $k\geq 1$ and for each $w\in W_{k-1}$ let
\begin{equation}\label{eq66}
p_{wi}=\frac{\mu(I_{wi})}{\mu(I_w)},\,\,
1\leq i\leq n_k.
\end{equation}
Then there is a one-to-one
correspondence between the set of Borel probability measures $\mu$
on $E$, with $\mu(I_w)>0$ for all $w\in W$,
and the set of sequences $\{p_w\}_{w\in W}$, with
$\sum_{i=1}^{n_k}p_{wi}=1$ and $0<p_{wi}<1$ for all $w\in W_{k-1}$ and $k\geq 1$. Write
\begin{equation}\label{cr}
P_w=(p_{w1},p_{w2},\cdots,p_{wn_k})
\end{equation}
for each $w\in W_{k-1}$. Then $\mu$ is a measure considered in theorem \ref{1}, provided that
for every fixed integer $k\geq 1$, the vectors $P_w$ are the same for all $w\in W_{k-1}$.
Write $wiu_t$ for the maximal word in $W_{k+t}$ with prefix $wi$
and $wi1^t$ the minimal word in $W_{k+t}$ with prefix $wi$. Our next theorem gives a complete
description of doubling measures on a uniform Cantor set.

\begin{thm}\label{t11}
Let $E=E(\mathbf{n},\mathbf{c},\Lambda, \{m_k\}_{k\in\Lambda}, \{s_k\}_{k\in\Lambda})$ be a
uniform Cantor set. Let $\mu$ be a Borel probability measure on $E$ with $\mu(I_w)>0$ for all
$w\in W$. Let $\{p_w\}_{w\in W}$ be the corresponding sequence.
Then $\mu$ is doubling on $E$ if and only
if there is a constant $C>1$ such that $P_w$ is $C$-uniform for all ${w\in W}$ and satisfies,
for each $k\in\Lambda$, the following conditions:

$(a)$ $(P_{wiu_{t-1}},P_{w(i+1)1^{t-1}})$ is $C$-uniform and
\begin{equation}\label{eq13}
C^{-1}\prod_{j=1}^{t}p_{wiu_j}\leq\prod_{j=1}^{t}{p_{w(i+1)1^j}}\leq C\prod_{j=1}^{t}p_{wiu_j}
\end{equation}
for all $w\in W_{k-1}$, $1\leq i<n_k$, and $1\leq t<m_k$.

$(b)$ $(P_{wiu_{m_k-1}},P_{w(i+1)1^{m_k-1}})$ is $(C,s_k)$-uniform.
\end{thm}

For a uniform Cantor set $E(\mathbf{n},\mathbf{c})$ it is known that
$\nu(E(\mathbf{n},\mathbf{c}))=0$ for all doubling measures $\nu$ on $[0,1]$
if and only if
$\{n_kc_k\}_{k=1}^\infty\not\in\cup_{q\geq1}\ell^q$ \cite{hww}. As a consequence,
if $E(\mathbf{n},\mathbf{c})$ carries a doubling measure which can be extended to a
doubling measure on $[0,1]$, then $\{n_kc_k\}_{k=1}^\infty\in\cup_{q\geq1}\ell^q$.
Our main result on the extension question of doubling measures on a uniform Cantor
set is the following theorem.

\begin{thm}\label{t2}
Let $E=E(\mathbf{n},\mathbf{c})$ be a uniform Cantor set.
Let $\mu_{\mathbb{P}}$ be a measure determined by an ultimately $1$-uniform $\mathbf{n}$-matching
sequence $\mathbb{P}$ of positive probability vectors. Then $\mu_{\mathbb{P}}$ can be
extended to a doubling measure on $[0,1]$ if and only if $\{n_kc_k\}_{k=1}^\infty\in\ell^1$.
\end{thm}

\section{Proof of Theorem \ref{1}}

In what follows
$C$ will denote a constant depending only on the constants in question and it
may be different in every appearance. Write $A\sim B$
for $C^{-1}B\leq A\leq CB$, and $A\preceq B$
for $A\leq CB$, when $A, B$ are quantities. For an interval $I$, denote by
$x^-(I)$ and $x^+(I)$,
respectively, the left and the right endpoints of $I$.

\medskip

\noindent{\bf Proof of the `only if' part.\,}
Let $E=E(\mathbf{n},\mathbf{c},\Lambda, \{m_k\}_{k\in\Lambda}, \{s_k\}_{k\in\Lambda})$ be a uniform
Cantor set. Let $\mu_\mathbb{P}$ be a measure determined by an $\mathbf{n}$-matching sequence
$\mathbb{P}$ of positive probability vectors. Suppose that $\mu_\mathbb{P}$
is doubling on $E$. We are going to show that the sequence $\mathbb{P}$ satisfies
the conditions of theorem \ref{1}. The doubling property of the measure $\mu_\mathbb{P}$ will be used frequently in the following equivalent form:
$$\mu(B(x, r))\sim\mu(B(y,t))\mbox{ for all $x,y\in E$ and $r, s>0$ with $|x-y|\preceq r\sim s$}.$$
Note that, for all $k\geq1$, the endpoints of components of level $k$ of $E$ belong to $E$. The proof consists of the following claims.

\medskip

{\bf{Claim 1.}} $p_{k,2}\leq Cp_{k,1}$ and $p_{k,n_k-1}\leq Cp_{k,n_k}$ for all $k\geq 1$.

\medskip

Proof of Claim 1. Let $w\in W_{k-1}$ be the minimal word of length $k-1$.
Then the corresponding component $I_w$
is the leftmost component of level $k-1$. We easily see that
$$I_{w1}=E_k\cap B(x^-(I_{w1}),\delta_k+\varepsilon_k)\,\,\mbox{and}
\,\, I_{w2}\subseteq B(x^-(I_{w2}),\delta_k+\varepsilon_k).$$
Observing that $x^-(I_{w1}),x^-(I_{w2})\in E$
and $|x^-(I_{w1})-x^-(I_{w2})|=\delta_k+\varepsilon_k$, it follows from the
doubling property of $\mu_\mathbb{P}$ that
$$\mu_\mathbb{P}(I_{w2})\leq\mu_\mathbb{P}(B(x^-(I_{w2}),\delta_k+\varepsilon_k))\leq C\mu_\mathbb{P}(B(x^-(I_{w1}),\delta_k+\varepsilon_k))=C\mu_\mathbb{P}(I_{w1}),$$
giving $p_{k,2}\leq Cp_{k,1}$ by (\ref{eq6}). Similarly, we have $p_{k,n_k-1}\leq Cp_{k,n_k}$
by considering the maximal word $w\in W_{k-1}$ and the rightmost two components of level $k$.
This proves claim 1.

\medskip

{\bf{Claim 2.}} $p_{k,i}\sim p_{k,i+1}$ for all $1\leq i<n_k$ and $k\geq 1$.

\medskip

Proof of Claim 2. Let $k$ be given. When $n_k=2$, claim 2 follows from claim 1 directly.
Next we assume that $n_k>2$. Consider the components $I_{wi}$ and $I_{w(i+1)}$ of level $k$,
where $w\in W_{k-1}$ and $1\leq i< n_k$. Then $G_{wi}$ is the gap of level $k$ between them.
Three possible cases may happen.

\medskip

Case 1. $\varepsilon_k\geq\delta_k$.

\medskip

Since $\varepsilon_k\geq\delta_k$, we easily see that
\begin{equation}\label{peq2}
I_{wi}=E_k\cap B(x^+(I_{wi}),\varepsilon_k)\,\,\mbox{and}\,\, I_{w(i+1)}=E_k\cap B(x^-(I_{w(i+1)}),\varepsilon_k)
\end{equation}
for $1<i<n_k-1$. Observing that $x^+(I_{wi}),x^-(I_{w(i+1)})\in E$ and $|x^+(I_{wi})-x^-(I_{w(i+1)})|=\varepsilon_k$,
we get from doubling property of $\mu_\mathbb{P}$ that
$$\mu_\mathbb{P}(I_{wi})\sim\mu_\mathbb{P}(I_{w(i+1)}),$$
which yields $p_{k,i}\sim p_{k,i+1}$ by (\ref{eq6}). For $i=1$, letting $w$ be the minimal word in $W_{k-1}$,
the relationship (\ref{peq2}) remains to be true, and so
$p_{k,1}\sim p_{k,2}$. For $i=n_k-1$, considering the maximal word in $W_{k-1}$,
we get $p_{k,n_k-1}\sim p_{k,n_k}$ in the same way.

\medskip

Case 2. $\varepsilon_k<\delta_k$ and $n_{k+1}>2$.

\medskip

Consider balls $B(x,r)$ and $B(y,r)$, where
$$x=x^-(G_{wi})-r,\,\, y=x^+(G_{wi})+r,\,\, r=[\frac{n_{k+1}-1}{2}](\delta_{k+1}+\varepsilon_{k+1}).$$
We see that $x$ is an endpoint of a component of level $k+1$ in $I_{wi}$ and that $y$ is
an endpoint of a component of level $k+1$ in $I_{w(i+1)}$, so $x,y\in E$.
Note also that
$$B(x,r)\subseteq I_{wi}\,\,\mbox{and}\,\, B(y,r)\subseteq I_{w(i+1)}.$$
Since $n_{k+1}\geq 3$, we have
$4[\frac{n_{k+1}-1}{2}]\geq n_{k+1}$, so $4r>\delta_k$, giving
$$I_{wi}\subseteq B(x,4r)\,\,\mbox{and}\,\, I_{w(i+1)}\subseteq B(y,4r).$$ Therefore
$$\mu_{\mathbb{P}}(I_{wi})\sim\mu_{\mathbb{P}}(B(x,r))\,\,\mbox{and}\,\,\mu_{\mathbb{P}}(I_{w(i+1)})\sim\mu_{\mathbb{P}}(B(y,r)).$$
Additionally, since $\varepsilon_k<\delta_k$ has been assumed, we have $|x-y|=2r+\varepsilon_k\leq6r$, so
$\mu_{\mathbb{P}}(B(x,r))\sim\mu_{\mathbb{P}}(B(y,r))$ by the doubling property of $\mu_{\mathbb{P}}$.
It then follows that $\mu_{\mathbb{P}}(I_{wi})\sim\mu_{\mathbb{P}}(I_{w(i+1)})$, giving
$p_{k,i}\sim p_{k,i+1}$ by (\ref{eq6}).

\medskip
Case 3. $\varepsilon_k<\delta_k$ and $n_{k+1}=2$.
\medskip

In the case of $\varepsilon_{k+1}\geq\delta_{k+1}$, consider $B(x^-(I_{wi2}),r)$ and $B(x^+(I_{w(i+1)1}),r)$,
where $r=\delta_{k+1}+\frac{\varepsilon_k}{3}$. Since $r<\delta_{k+1}+\frac{\delta_k}{3}\leq\delta_{k+1}+\varepsilon_{k+1}$,
we see that
$$I_{wi2}\subseteq E_{k}\cap B(x^-(I_{wi2}),r)\subseteq I_{wi}$$ and
$$I_{w(i+1)1}\subseteq
E_{k}\cap B(x^+(I_{w(i+1)1}),r)\subseteq I_{w(i+1)}.$$
Observing that
$|x^+(I_{w(i+1)1}-x^-(I_{wi2})|=2\delta_{k+1}+\varepsilon_k\leq3r$,
we get from the doubling property of $\mu_{\mathbb{P}}$ that
$$\mu_{\mathbb{P}}(B(x^-(I_{wi2}),r))\sim\mu_{\mathbb{P}}(B(x^+(I_{w(i+1)1}),r)).$$ Since $n_{k+1}=2$, we have $\mu_{\mathbb{P}}(I_{wi2})\sim\mu_{\mathbb{P}}(I_{wi})$
and $\mu_{\mathbb{P}}(I_{w(i+1)1})\sim\mu_{\mathbb{P}}(I_{w(i+1)})$ by claim 1.
Therefore $\mu_{\mathbb{P}}(I_{wi})\sim\mu_{\mathbb{P}}(I_{w(i+1)})$, and so $p_{k,i}\sim p_{k,i+1}$ by (\ref{eq6}).

\medskip

In the case of $\varepsilon_{k+1}<\delta_{k+1}$, since $n_{k+1}=2$, we have
$$B(x^-(I_{wi2}),\delta_{k+1})\subseteq I_{wi}\subseteq B(x^-(I_{wi2}),3\delta_{k+1})$$
and
$$B(x^+(I_{w(i+1)1}),\delta_{k+1})\subseteq I_{w(i+1)}\subseteq B(x^+(I_{w(i+1)1}),3\delta_{k+1}).$$
Since $\varepsilon_k<\delta_k$, we get
$|x^+(I_{w(i+1)1}-x^-(I_{wi2})|=2\delta_{k+1}+\varepsilon_k\leq5\delta_{k+1}$.
Then $\mu_{\mathbb{P}}(I_{wi})\sim\mu_{\mathbb{P}}(I_{w(i+1)})$ by the doubling property of $\mu_{\mathbb{P}}$, giving
$p_{k,i}\sim p_{k,i+1}$.
This completes the proof of claim 2.

\medskip

{\bf{Claim 3.}} $\mathbb{P}$ is $C$-uniform.

\medskip

Proof of Claim 3.  Since
the $C$-uniformness of $\mathbb{P}$ is equivalent to that every $P_k$ satisfies
\begin{equation}\label{peq1}
p_{k,i+1}+\cdots+p_{k,i+l}\sim p_{k,i+l+1}+\cdots+p_{k,i+2l}
\end{equation}
for all $i\geq0$ and $l\geq1$ with $i+2l\leq n_k$, it suffices to show (\ref{peq1}).
Fix $k$. When $l=1$ or $n_k\leq 3$, the relationship (\ref{peq1}) follows from claim 2 directly.
Next we assume that $l\geq 2$ and $n_k>3$. Let $w\in W_{k-1}$ and $i\geq 0$ be given such that
$i+2l\leq n_k$. Observing that there are balls $B(x,r)$ and $B(y,r)$
with $x,y\in E$ and $|x-y|\leq 3r$ such that
$$E_k\cap B(x,r)\subseteq I_{w(i+1)}\cup\cdots\cup I_{w(i+j)}\subseteq B(x,3r)$$
and
$$E_k\cap B(y,r)\subseteq I_{w(i+j+1)}\cup\cdots\cup I_{w(i+2j)}\subseteq B(y,3r),$$
we immediately get
$$\mu_{\mathbb{P}}(I_{w(i+1)}\cup\cdots\cup I_{w(i+j)})\sim\mu_{\mathbb{P}}(I_{w(i+j+1)}\cup\cdots\cup I_{w(i+2j)}),$$
giving the relationship (\ref{peq1}) by (\ref{eq6}). This completes the proof of claim 3.

\medskip

{\bf{Claim 4.}} $\prod_{j=1}^t{p_{k+j,1}}\sim\prod_{j=1}^t{p_{k+j, n_{k+j}}}$ for all $k\in\Lambda$
and $1\leq t<m_k$.

\medskip

Proof of Claim 4. Let $k\in\Lambda$ be fixed. Without loss of generality, assume that $m_k>1$.
Let $1\leq t<m_k$ be given.
Then $\varepsilon_k<\delta_{k+t}$ by the definition of $m_k$.
Let $w\in W_{k-1}$ and $1\leq i<n_k$ be given. Let $wiu\in W_{k+t}$ be the maximal word
with prefix $wi$ and $w(i+1)v\in W_{k+t}$ be the minimal word with prefix $w(i+1)$.
Then, by (\ref{eq6}) $$\mu_{\mathbb{P}}(I_{wiu})=\mu_{\mathbb{P}}(I_{w})p_{k,i}\prod_{j=1}^t{p_{k+j, n_{k+j}}}$$ and $$\mu_{\mathbb{P}}(I_{w(i+1)v})=\mu_{\mathbb{P}}(I_{w})p_{k,i+1}\prod_{j=1}^t{p_{k+j,1}}.$$
Note that the gap between $I_{wiu}$ and $I_{w(i+1)v}$ is $G_{wi}$. Since $\varepsilon_k<\delta_{k+t}$,
one has $|G_{wi}|<|I_{wiu}|=|I_{w(i+1)v}|$. To complete this proof, we consider two cases:

\medskip

Case 1. $n_{k+t+1}>2$.

\medskip

Arguing as case 2 of claim 2, we have two balls, $B(x,r)$ and $B(y,r)$, with $x,y\in E$ and $|x-y|\leq 6r$, such that
$$B(x,r)\subseteq I_{wiu}\subseteq B(x,4r)\,\,\mbox{and}\,\, B(y,r)\subseteq I_{w(i+1)v}\subseteq B(y,4r),$$
which, combined with the doubling property of $\mu_{\mathbb{P}}$, yields $\mu_{\mathbb{P}}(I_{wiu})\sim\mu_{\mathbb{P}}(I_{w(i+1)v})$.

\medskip

Case 2. $n_{k+t+1}=2$.

\medskip

Consider two subcases: $\varepsilon_{k+t+1}\geq\delta_{k+t+1}$ and $\varepsilon_{k+t+1}<\delta_{k+t+1}$.
In the first subcase, arguing as case 3 of claim 2,
we have two balls, $B(x,r)$ and $B(y,r)$, with $x,y\in E$ and $|x-y|\leq 3r$, such that
$$I_{wiu2}\subseteq E_{k+t}\cap B(x,r)\subseteq I_{wiu}\mbox{ and }I_{w(i+1)v1}\subseteq E_{k+t}\cap B(y,r)\subseteq I_{w(i+1)v}.$$
In the second subcase, arguing as case 3 of claim 2, we have two balls, $B(x,r)$ and $B(y,r)$,
with $x,y\in E$ and $|x-y|\leq 5r$, such that
$$B(x,r)\subseteq I_{wiu}\subseteq B(x,3r)\mbox{ and }B(y,r)\subseteq I_{w(i+1)v}\subseteq B(y,3r).$$
Therefore, in both subcases, we have
$\mu_{\mathbb{P}}(I_{wiu})\sim\mu_{\mathbb{P}}(I_{w(i+1)v})$.

Now $\mu_{\mathbb{P}}(I_{wiu})\sim\mu_{\mathbb{P}}(I_{w(i+1)v})$ is proved.
Since $p_{k,i}\sim p_{k,i+1}$ has been proved in claim 2, we get
$\prod_{j=1}^t{p_{k+j,1}}\sim\prod_{j=1}^t{p_{k+j, n_{k+j}}}$. This completes the proof of claim 4.

\medskip

{\bf{Claim 5.}} $(P_{k+t},P_{k+t})$ is $C$-uniform for all $k\in\Lambda$
and $1\leq t<m_k$.

\medskip

Proof of Claim 5. Since $\mathbb{P}$ has been proved to be $C$-uniform, it suffices to show that
the sum of the first $j$ terms of $P_{k+t}$ is comparable to the sum of the last $j$ terms of $P_{k+t}$
for all $k\in\Lambda$, $1\leq t<m_k$, and $1\leq j\leq n_{k+t}$. Let these integers be given.
Let $w\in W_{k-1}$ and $1\leq i<n_k$ be given. Let $u\in W_{k+t-1}$ be the maximal word
with prefix $wi$ and $v\in W_{k+t-1}$ be the minimal word with prefix $w(i+1)$. Then $G_{wi}$ is
the gap between $I_u$ and $I_v$.
Consider components $I_{un_{k+t}}$, $I_{u(n_{k+t}-1)}$,$\cdots$, $I_{u(n_{k+t}+1-j)}$ and
$I_{v1}$, $I_{v2}$, $\cdots$, $I_{vj}$ of level $k+t$. Since $\varepsilon_k<\delta_{k+t}$,
by the same argument as that of claim 3 we get
$$\mu_{\mathbb{P}}(I_{un_{k+t}}\cup\cdots\cup I_{u(n_{k+t}+1-j)})
\sim\mu_{\mathbb{P}}(I_{v1}\cup\cdots\cup I_{vj}),$$
which, together with (\ref{eq6}), yields
$$p_{k+t,n_{k+t}}+\cdots+p_{k+t,n_{k+t}+1-j}
\sim p_{k+t,1}+\cdots+p_{k+t,j}.$$
This completes the proof of claim 5.

\medskip

{\bf{Claim 6.}} $(P_{k+m_k},P_{k+m_k})$ is $(C,s_k)$-uniform for all $k\in\Lambda$.

\medskip

Proof of Claim 6. Since $\mathbb{P}$ has been proved to be $C$-uniform, it suffices to show that
the sum of the first $j$ terms of $P_{k+m_k}$ is comparable to the sum of the last $j$ terms of $P_{k+m_k}$
for all $k\in\Lambda$ and $s_k\leq j\leq n_{k+m_k}$, where, by the definition, $s_k$ satisfies
$$s_k\delta_{k+m_k}+(s_k-1)\varepsilon_{k+m_k}\leq\varepsilon_k<(s_k+1)\delta_{k+m_k}+s_k\varepsilon_{k+m_k}.$$
Let $k$ and $j$ be given. Let $w\in W_{k-1}$ and $1\leq i<n_k$.
Let $u\in W_{k+m_k-1}$ be the maximal word
with prefix $wi$ and let $v\in W_{k+m_k-1}$ be the minimal word with prefix $w(i+1)$.
Then $G_{wi}$ is the gap between $I_{un_{k+m_k}}$ and $I_{v1}$.

\medskip
Case 1. $s_k=j=1$. In this case, $\delta_{k+m_k}\leq\varepsilon_k\leq2\delta_{k+m_k}+\varepsilon_{k+m_k}$, so we have
$$I_{un_{k+m_k}}\subseteq E\cap B(x^-(G_{wi}),\varepsilon_k)\subseteq I_{un_{k+m_k}}\cup I_{u(n_{k+m_k}-1)}$$
and $$I_{v1}\subseteq E\cap B(x^+(G_{wi}),\varepsilon_k)\subseteq I_{v1}\cup I_{v2}.$$
From claim 2 we get
$$\mu_{\mathbb{P}}(I_{un_{k+m_k}})\sim\mu_{\mathbb{P}}(I_{un_{k+m_k}}\cup I_{u(n_{k+m_k}-1)})\,\,\mbox{and}\,\,\mu_{\mathbb{P}}(I_{v1})\sim\mu_{\mathbb{P}}(I_{v1}\cup I_{v2)}).$$
Since $x^-(G_{wi}), x^+(G_{wi})\in E$ and $|x^-(G_{wi})-x^+(G_{wi})|=\varepsilon_k$, it follows from the doubling
property of $\mu_{\mathbb{P}}$ that
$\mu_{\mathbb{P}}(B(x^-(G_{wi}),\varepsilon_k))\sim\mu_{\mathbb{P}}(B(x^+(G_{wi}),\varepsilon_k))$, so
$$\mu_{\mathbb{P}}(I_{un_{k+m_k}})\sim\mu_{\mathbb{P}}(I_{v1}),$$ which, together with claim 4, yields
$p_{k+m_k, n_{k+m_k}}\sim p_{k+m_k, 1}$.

\medskip
Case 2. $s_k>1$ or $j>1$. In this case, arguing as we did in claim 3, we get
$$\mu_{\mathbb{P}}(I_{un_{k+m_k}}\cup\cdots\cup I_{u(n_{k+m_k}+1-j)})\sim\mu_{\mathbb{P}}(I_{v1}\cup\cdots\cup I_{vj}),$$
which, combined with claim 4, yields $$p_{k+m_k, n_{k+m_k}}+\cdots+p_{k+m_k, n_{k+m_k}+1-j}\sim p_{k+m_k, 1}+\cdots+p_{k+m_k, j}.$$
This completes the proof of claim 6 and the `only if' part is thus proved.

\medskip

\noindent{\bf Proof of the `if' part.} Suppose that $\mathbb{P}$ meets the conditions of Theorem \ref{1}.
We are going to show that $\mu_\mathbb{P}$ is doubling on $E$.

Let $B(x,r)$ be a ball of the real line with $x\in E$ and $r\in(0,1)$.
Let $I$ be the smallest component such that $I\supseteq E\cap B(x,2r)$.
Suppose that $I$ is at the level $k-1$. Then $B(x,2r)$ intersects at least two components of level $k$, so
$2r\geq \varepsilon_k.$
Denote
$$\mathcal{A}=\{L:L\in\mathcal{I}_k,\,\, L\cap B(x,2r)\neq\emptyset\}.$$
Then $\cup_{L\in\mathcal{A}}L\subseteq I$ and $4r\geq(\sharp\mathcal{A}-2)\delta_k+(\sharp\mathcal{A}-1)\varepsilon_k$,
where $\sharp$ denotes the cardinality. We consider four cases as follows.

\medskip

Case 1. $\sharp\mathcal{A}\geq 4$.
\medskip

In this case, we see that $\sharp\mathcal{A}\sim\sharp\{L\in\mathcal{A}: L\subseteq B(x,r)\}$.
Since $P_k$ is $C$-uniform, we get $\mu_{\mathbb{P}}(B(x,r))\sim\mu_{\mathbb{P}}(B(x,2r))$.

\medskip

Case 2. $\sharp \mathcal{A}=3$.

\medskip

In this case, $4r>{\delta_k}$. When $n_{k+1}\geq 4$,
we see that $B(x,r)$ contains at least $[\frac{n_{k+1}}{4}]$
components of level $k+1$ in $L(x)$, where $L(x)\in\mathcal{A}$ is the component
of level $k$ containing $x$. Since $P_{k+1}$ and $P_k$ are $C$-uniform,
we have $$\mu_{\mathbb{P}}(B(x,r))\sim\mu_{\mathbb{P}}(L(x))\sim\mu_{\mathbb{P}}(B(x,2r)).$$
When $n_{k+1}=2$ or $3$, we see from $\sharp \mathcal{A}=3$ that $B(x,r)$ contains at least a
component of level $k+1$. Therefore the last relationship remains true.

\medskip

Case 3. $\sharp\mathcal{A}=2$ and $\varepsilon_{k}\geq\delta_{k}$.

\medskip

In this case, since $2r\geq{\varepsilon_k}$, we have $2r\geq{\delta_k}$. By the same argument
as that of Case 2 we get $\mu_{\mathbb{P}}(B(x,r))\sim\mu_{\mathbb{P}}(B(x,2r))$.

\medskip

Case 4. $\sharp\mathcal{A}=2$ and $\varepsilon_{k}<\delta_{k}$.

\medskip

Let $I_{wi}$ and $I_{w(i+1)}$ be members of $\mathcal{A}$, where $w\in W_{k-1}$ and $1\leq i<n_k$.
Without loss of generality, assume that $x\in I_{wi}$. Let $J(x)$ be the biggest component such that
$x\in J(x)\subseteq B(x,r)$. Then $J(x)=I$ or $J(x)\subseteq I_{wi}$. It is obvious that
$\mu_{\mathbb{P}}(B(x,r))=\mu_{\mathbb{P}}(B(x,2r))$ for $J(x)=I$. Also, if $J(x)=I_{wi}$,
we easily see from the $C$-uniformness of $P_k$ that
$$\mu_{\mathbb{P}}(B(x,r))\geq\mu_{\mathbb{P}}(I_{wi})\geq C\mu_{\mathbb{P}}(I_{wi}\cup I_{w(i+1)})
\geq C\mu_{\mathbb{P}}(B(x,2r)).$$
Next assume $J(x)\neq I_{wi}$. Thus $J(x)$ is at the level $k+t$ for some $t\geq1$.
Let $I_u$ be the component of level $k+t-1$ containing $J(x)$, where $u\in W_{k+t-1}$.
Then we have $B(x,r)\subset I_u\cup G\cup I_v$ and $r<\delta_{k+t-1}$ by the maximality of $J(x)$,
where $I_v$ is the nearest component of level $k+t-1$ on the right of $I_u$
and $G$ is the gap between $I_u$ and $I_v$. Therefore, $B(x,2r)$ meets at most four components of level $k+t-1$,
that is, $\sharp\mathcal{B}\leq 4$, where
$$\mathcal{B}=\{L:L\mbox{ is a component of $E_{k+t-1}$ and}\,\, L\cap B(x,2r)\neq\emptyset\}.$$
Clearly, $I_u\in\mathcal{B}$. We claim that $\mu_{\mathbb{P}}(L)\sim\mu_{\mathbb{P}}(I_u)$ for all $L\in\mathcal{B}$.
In fact, let $L$ and $K$ be two adjacent components in $\mathcal{B}$,
then the gap between $L$ and $K$ is at the level $k+j$ for some $0\leq j\leq t-1$.
When the gap is at the level $k+t-1$, we immediately get
$\mu_{\mathbb{P}}(L)\sim\mu_{\mathbb{P}}(K)$ from the $C$-uniformness of $P_{k+t-1}$.
The gap being at the level $k+j$ for some $1\leq j\leq t-2$ is possible only when
$$\varepsilon_{k+j}\leq2r<2\delta_{k+t-1}<\delta_{k+j},$$  and if it happened, we would have $m_{k+j}\geq t-1$.
If $m_{k+j}= t-1$, then $s_{k+j}=1$, so the $(C, s_{k+j})$-uniformness
of $(P_{k+j+m_{k+j}},P_{k+j+m_{k+j}})$ implies $\mu_{\mathbb{P}}(L)\sim\mu_{\mathbb{P}}(K)$.
If $m_{k+j}> t-1$, we may use the condition (\ref{eq11}) to
get $\mu_{\mathbb{P}}(L)\sim\mu_{\mathbb{P}}(K)$. Finally, we consider the case where the gap is at the level $k$.
Note that $t\leq m_k+1$, in fact, if not, we would get from the definition of $m_k$
that $2r\leq2\delta_{k+m_k+1}<\delta_{k+m_k}\leq\varepsilon_k$,
contradicting $2r\geq\varepsilon_k$. Also, if $t=m_k+1$ happened, $(P_{k+m_k},P_{k+m_k})$ would have been $C$-uniform. In fact,
if $t=m_k+1$ then $\delta_{k+m_k}\leq{\varepsilon_k}\leq 2r\leq2\delta_{k+m_k}$
by the definition of $m_k$, so $s_k=1$, and hence $(P_{k+m_k},P_{k+m_k})$ is $C$-uniform,
because it has been assumed to be $(C, s_k)$-uniform. Then $\mu_{\mathbb{P}}(L)\sim\mu_{\mathbb{P}}(K)$
follows from the above discussion and the assumptions of theorem \ref{1}. This proves the claim.

Now let
$$\mathcal{C}=\{L:L\mbox{ is a component of $E_{k+t}$ and}\,\, L\subseteq I_u\cap B(x,r)\}.$$
Two possible subcases may happen.

\medskip
Subcase 1. $B(x,2r)\supseteq I_u$.
\medskip

In this subcase, we have
$\sharp\mathcal{C}\geq[\frac{n_{k+t}}{2}]$. It follows from the $C$-uniformness of $P_{k+t}$ that
$$\mu_{\mathbb{P}}(B(x,r))\geq C\mu_{\mathbb{P}}(I_u).$$
Since $\sharp\mathcal{B}\leq 4$, we get from the above claim that
$\mu_{\mathbb{P}}(B(x,r))\sim\mu_{\mathbb{P}}(B(x,2r))$.

\medskip
Subcase 2. $B(x,2r)\subset I_u\cup G\cup I_v$.
\medskip

In this subcase, it is clear that $2r<|I_u|=\delta_{k+t-1}$,
so we have $t\leq m_k$ by the definition of $m_k$. Observing that $I_u\subseteq I_{wi}$
and $I_v\subseteq I_{w(i+1)}$, with $u\in W_{k+t-1}$ being the maximal word with
prefix $wi$ and $v\in W_{k+t-1}$ being the minimal word with
prefix $w(i+1)$, we see that $G$ is a gap of level $k$.
When $1\leq t<m_k$, we use the assumption (a) to get $\mu_{\mathbb{P}}(B(x,r))\sim\mu_{\mathbb{P}}(B(x,2r))$.
When $t=m_k$, remembering that $s_k$ has been defined to satisfy
$$s_k\delta_{k+m_k}+(s_k-1)\varepsilon_{k+m_k}\leq\varepsilon_k<(s_k+1)\delta_{k+m_k}+s_k\varepsilon_{k+m_k},$$
which, together with $2r\geq\varepsilon_{k}$, yields
$2r\geq s_k\delta_{k+m_k}+(s_k-1)\varepsilon_{k+m_k}$, we may use the $C$-uniformness of $P_{k+m_k}$,
the $(C, s_k)$-uniformness of $(P_{k+m_k},P_{k+m_k})$, and the assumption (\ref{eq11}) to get
$\mu_{\mathbb{P}}(B(x,r))\sim\mu_{\mathbb{P}}(B(x,2r))$.
This completes the proof of the `if' part.

\section{Proof of Theorem \ref{t11}}

The proof of theorem 2 may go in step as that of theorem 1. For the proof of the `only if' part, suppose
that $\mu$ is doubling on the uniform Cantor set $E$.
We shall prove that the sequence $\{P_w\}_{w\in W}$ satisfies the conditions of theorem 2.
As we did in theorem 1, the proof consists of the following claims.

\medskip

{\bf{Claim A.}} $p_{w2}\leq Cp_{w1}$ and $p_{w(n_k-1)}\leq Cp_{wn_k}$ for all $w\in W_{k-1}$ and $k\geq 1$.

\medskip

{\bf{Claim B.}} $p_{wi}\sim p_{w(i+1)}$
for all $w\in W_{k-1}$, $1\leq i<n_k$, and $k\geq 1$.

\medskip

{\bf{Claim C.}} $P_w$ is $C$-uniform for all ${w\in W}$.

\medskip

{\bf{Claim D.}} $\prod_{j=1}^{t}{p_{w(i+1)1^j}}\sim\prod_{j=1}^{t}p_{wiu_j}$ for all $w\in W_{k-1}$,
$1\leq i<n_k$, $1\leq t<m_k$, and $k\in \Lambda$.

\medskip

{\bf{Claim E.}} $(P_{wiu_{t-1}},P_{w(i+1)1^{t-1}})$ is $C$-uniform for all $w\in W_{k-1}$, $1\leq i<n_k$,
$1\leq t<m_k$, and $k\in \Lambda$.

\medskip

{\bf{Claim F.}} $(P_{wiu_{m_k-1}},P_{w(i+1)1^{m_k-1}})$ is $(C,s_k)$-uniform for all $k\in\Lambda$.

\medskip

If, for every fixed integer $k\geq 1$, the vectors $P_w$ are the same for all $w\in W_{k-1}$,
then the above claims are exactly those in theorem 1. Under this condition, {\it claim A} has a
simpler proof, because the word $w\in W_{k-1}$ in question may be assumed to be the minimal or
maximal words. Now, without assuming this condition, we are going to prove {\it claim A}. Instead,
we will show the following

\begin{lemma}\label{m}
$p_{w2}\sim p_{w1}$ and $p_{w(n_k-1)}\sim p_{wn_k}$ for all $w\in W_{k-1}$ and $k\geq 1$.
\end{lemma}

Proof of Lemma \ref{m}. We only prove $p_{w2}\sim p_{w1}$. The proof of the other relationship
is obviously similar.
Let $k\geq 1$ and $w\in W_{k-1}$ be given. To prove $p_{w2}\sim p_{w1}$, it suffices to show
$\mu(I_{w2})\sim\mu(I_{w1})$. Without loss of generality, assume that
$w\in W_{k-1}$ is neither minimal nor maximal.

\medskip

Case 1. $\varepsilon_k\geq\delta_k$.

\medskip

Let $I_u$ be the component of level $k$ next to the left of $I_{w1}$. Let $I_v$ be the component of
level $k$ next to the right of $I_{w2}$.
Let $G$ be the gap
between $I_u$ and $I_{w1}$. Let $G'$ be the gap between $I_{w2}$ and $I_v$. Then $G'$ is a gap
of level $k$ if and only if $n_k>2$. Remember
that $\mu$ is doubling on $E$ and that $\varepsilon_k\geq\delta_k$ has been assumed.

Noting that, if $\delta_k\leq|G|<\varepsilon_k$ then
$$I_{w1}=E_k\cap B(x^+(I_{w1}),|G|)\mbox{ and } I_{u}=E_k\cap B(x^-(I_{u}),|G|),$$
and that, if $|G|<\delta_k$ then
$$I_{w1}=E_k\cap B(x^+(I_{w1}),\delta_k)\mbox{ and }I_{u}=E_k\cap B(x^-(I_{u}),\delta_k),$$
we find that, if $|G|<\varepsilon_k$, then
\begin{equation}\label{rr}
\mu(I_{w1})\sim \mu(I_u).
\end{equation}
Arguing as we just did, if $|G'|<\varepsilon_k$, then
\begin{equation}\label{rrr}
\mu(I_{w2})\sim \mu(I_v).
\end{equation}

Now we apply (\ref{rr}) and (\ref{rrr}) to prove $\mu(I_{w1})\sim\mu(I_{w2})$. We consider nine subcases.

\medskip

Subcase 1. $|G|\geq\varepsilon_k$ and $|G'|\geq\varepsilon_k$.

Since $\varepsilon_k\geq\delta_k$ has been assumed,
we have $$I_{w1}=E_k\cap B(x^+(I_{w1}),\varepsilon_k)\,\,\mbox{and}\,\, I_{w2}=E_k\cap B(x^-(I_{w2}),\varepsilon_k),$$ and so
$\mu(I_{w1})\sim\mu(I_{w2}).$

\medskip

Subcase 2. $|G|<\delta_k$ and $|G'|<\delta_k$.

In view of (\ref{rr}) and (\ref{rrr}),
$\mu(I_{u})\sim\mu(I_{w1})$
and $\mu(I_{v})\sim\mu(I_{w2})$.
Since $|G'|<\delta_k\leq\varepsilon_k$, we have $n_k=2$, as we have said.
We see that the gap next to the right of $I_v$ belongs to $\mathcal{G}_k$. Then we have
$$
\mu(I_{w1})+\mu(I_{u})=\mu(B(x^+(I_{u}),\varepsilon_k+\delta_k))
$$
and
$$
\mu(I_{w2})+\mu(I_{v})=\mu(B(x^-(I_{v}),\varepsilon_k+\delta_k)),
$$
and so $\mu(I_{w1})\sim\mu(I_{w2})$.

\medskip

Subcase 3. $\delta_k\leq|G|<\varepsilon_k$ and $|G'|\geq\varepsilon_k$.

In view of (\ref{rr}), $\mu(I_{w1})\sim\mu(I_{u})$. Observing that
$$I_{w1}\subseteq E_k\cap B(x^+(I_{w1}),\varepsilon_k)\subseteq I_u\cup I_{w1}\,\,\mbox{and}\,
\,I_{w2}=E_k\cap B(x^-(I_{w2}),\varepsilon_k),$$
we get $\mu(I_{w1})\sim\mu(I_{w2})$.

\medskip

Subcase 4. $\delta_k\leq|G'|<\varepsilon_k$ and $|G|\geq\varepsilon_k$.

Arguing as we just did in subcase 3, we have $\mu(I_{w1})\sim\mu(I_{w2})$.

\medskip

Subcase 5. $\delta_k\leq|G|<\varepsilon_k$ and $|G'|<\delta_k$.

We have
$\mu(I_{u})\sim\mu(I_{w1})$ and $\mu(I_{v})\sim\mu(I_{w2})$. We also have
$$I_{w1}\subseteq E_k\cap B(x^-(I_{w1}),\varepsilon_k+\delta_k)\subseteq I_u\cup I_{w1}$$
and
$$\mu(I_{w2})+\mu(I_{v})=\mu(B(x^-(I_{v}),\varepsilon_k+\delta_k)),$$
and so
$\mu(I_{w1})\sim\mu(I_{w2}).$

\medskip

Subcase 6. $\delta_k\leq|G'|<\varepsilon_k$ and $|G|<\delta_k$.

We get $\mu(I_{w1})\sim\mu(I_{w2})$ by an argument analogous to case 5.

\medskip

Subcase 7. $\delta_k\leq|G|<\varepsilon_k$ and $\delta_k\leq|G'|<\varepsilon_k$.

We have
$\mu(I_{u})\sim\mu(I_{w1})$ and $\mu(I_{v})\sim\mu(I_{w2})$. We also have
$$I_{w1}\subseteq E_k\cap B(x^-(I_{w1}),\varepsilon_k+\delta_k)\subseteq I_u\cup I_{w1}$$
and
$$I_{w2}\subseteq E_k\cap B(x^+(I_{w2}),\varepsilon_k+\delta_k)\subseteq I_v\cup I_{w2},$$
giving $\mu(I_{w1})\sim\mu(I_{w2}).$

\medskip

Subcase 8. $|G|<\delta_k$ and $|G'|\geq\varepsilon_k$.

We have
$\mu(I_{u})\sim\mu(I_{w1})$. Noting that
$$I_{w2}=E_k\cap B(x^-(I_{w2}),\varepsilon_k)\,\,\mbox{and}\,\,I_{w1}\subseteq E_k\cap B(x^+(I_{w1}),
\varepsilon_k)\subseteq I_u\cup I_{w1},$$
we get $\mu(I_{w1})\sim\mu(I_{w2})$.

\medskip

Subcase 9. $|G'|<\delta_k$ and $|G|\geq\varepsilon_k$.

Arguing as in case 8, we have $\mu(I_{w1})\sim\mu(I_{w2})$.

\medskip
Case 2. $\varepsilon_k<\delta_k$, $n_{k+1}=2$, and $\varepsilon_{k+1}\geq\delta_{k+1}$.
\medskip

We consider two subcases: $\varepsilon_k<\delta_{k+1}$ and $\varepsilon_k\geq\delta_{k+1}$. When $\varepsilon_k<\delta_{k+1}$,
one has $\mu(I_{w12})\sim\mu(I_{w21})$ by comparing the sizes of
$B(x^-(I_{w12}),\delta_{k+1})$ and $B(x^+(I_{w21}),\delta_{k+1})$ in measure $\mu$, which, combined with an observation that  $I_{w1}\subset B(x^-(I_{w12}),2\varepsilon_{k+1})$, yields $$\mu(I_{w1})\leq C\mu(B(x^-(I_{w12}),\varepsilon_{k+1}))\leq C(\mu(I_{w12})+\mu(I_{w21}))\leq C\mu(I_{w2}).$$
As $\mu(I_{w2})\leq C\mu(I_{w1})$ may be obtained similarly, we have $\mu(I_{w1})\sim\mu(I_{w2})$. When $\varepsilon_k\geq\delta_{k+1}$, since $3\varepsilon_{k+1}\geq\delta_k>\varepsilon_k$ by the assumptions of case 2,
one also has $\mu(I_{w12})\sim\mu(I_{w21})$ by comparing
$B(x^-(I_{w12}),r_{k})$ and $B(x^+(I_{w21}),r_{k})$, where $r_{k}=\min\{\varepsilon_k,\varepsilon_{k+1}\}$. Therefore, arguing as in the previous subcase gives
$\mu(I_{w1})\sim\mu(I_{w2})$.

\medskip
Case 3. $\varepsilon_k<\delta_k$, $n_{k+1}=2$, and $\varepsilon_{k+1}<\delta_{k+1}$.
\medskip

Case 4. $\varepsilon_k<\delta_k$ and $n_{k+1}>2$.

\medskip

For cases 3 and 4, the proof of $\mu(I_{w1})\sim\mu(I_{w2})$
is the same as that of claim 2 in section 2. We omit it. $\hfill\Box$

\medskip

Based on lemma \ref{m}, {\it claims B - F} can be proved in the same way as {\it claims 2 - 6}
of section 2. This completes the proof of the `only if' part.

\medskip

For the proof of the `if' part, suppose that the sequence $\{P_w\}_{w\in W}$ satisfies the conditions of theorem 2.
We have to prove that $\mu$ is doubling on the uniform Cantor set $E$. Arguing as we did in the proof of theorem 1,
the doubling property of $\mu$ can be proved similarly, and so it is omitted here.

\section{Proof of Theorem \ref{t2}}

Let $E=E(\mathbf{n},\mathbf{c},\Lambda, \{m_k\}_{k\in\Lambda}, \{s_k\}_{k\in\Lambda})$ be a uniform Cantor set.
The proof of Theorem \ref{t2} is based on the following lemmas.

\begin{lemma}\label{lem1}
If $E$ carries a doubling measure which can be extended to a doubling measure on $[0,1]$, then
$\{n_kc_k\}_{k=1}^\infty\in\cup_{q\geq1}\ell^q$.
\end{lemma}
\noindent{\bf Proof.\,} If $\{n_kc_k\}_{k=1}^\infty\not\in\cup_{q\geq1}\ell^q$, then we have $\nu(E)=0$
for all doubling measures $\nu$ on $[0,1]$ (see \cite{hww}), a contradiction.
$\hfill\Box$

\begin{lemma}\label{lem2}
Suppose that $E$ carries a doubling measure which can be extended to a doubling measure
on $[0,1]$ and that $\sup_{k\geq1}n_k<\infty$. Then $\sup_{k\in\Lambda}m_k=\infty$.
\end{lemma}
\noindent{\bf Proof.\,} If $\sup_{k\in\Lambda}m_k<\infty$, then there is an integer $N$ such
that $\varepsilon_k\geq\delta_{k+N}$ for all $k\geq 1$, i.e.
$$\prod_{i=k}^{k+N}\frac{1-(n_i-1)c_i}{n_i}\leq c_k.$$
It follows that
\begin{equation}\label{jj}
(\frac{1-\sup _{i\geq1}(n_i-1)c_i}{\sup _{i\geq1}n_i})^{N+1}\leq c_k
\end{equation}
for all $k\geq 1$. Since $E$ carries a doubling measure which can be extended
to a doubling measure
on $[0,1]$, we have from Lemma \ref{lem1} that $\{n_kc_k\}_{k=1}^\infty\in\cup_{q\geq1}\ell^q$.
This implies $\sup_{k\geq 1}(n_k-1)c_k<1$, which, together with (\ref{jj}) and the assumption
$\sup_{k\geq1}n_k<\infty$, gives $\inf_{k\geq 1}c_k>0$. This implies that $E$ is porous, so
we have $\nu(E)=0$ for all doubling measures $\nu$ on $[0,1]$ (see \cite{Wu1}), a contradiction. $\hfill\Box$

\medskip

\noindent{\bf Proof of Theorem \ref{t2}.} Suppose $\{n_kc_k\}_{k=1}^\infty\in\ell^1$.
Then $\mathcal{L}(E)>0$ by (\ref{eq4}). Since $\mathbb{P}$ is an ultimately $1$-uniform $\mathbf{n}$-matching
probability sequence, there is an integer $k_0\geq 1$ such that
$p_{k,i}=\frac{1}{n_k}$ for all $k\geq k_0$ and $1\leq i\leq n_k$. By this, we have
$$\mu_{\mathbb{P}}=\sum_{I\in\mathcal{I}_{k_0-1}}\frac{\mu_{\mathbb{P}}(I)}{\mathcal{L}(E\cap I)}\mathcal{L}|_{E\cap I}.$$
Let
$$\nu=\sum_{I\in\mathcal{I}_{k_0-1}}\frac{\mu_{\mathbb{P}}(I)}{\mathcal{L}(E\cap I)}\mathcal{L}|_{I}
+\sum_{J\in\mathcal{G}_{k_0-1}}\mathcal{L}|_J.$$
Since $\mathcal{I}_{k_0-1}\cup\mathcal{G}_{k_0-1}$ forms a partition of $[0,1]$ by finite intervals
and the restrictions of the measure $\nu$ to these intervals are comparable to Lebesgue measure, we easily see that $\nu$ is doubling on $[0,1]$.
Moreover, since $E\cap(\cup_{G\in\mathcal{G}_{k_0-1}}G)=\emptyset$, we have
$\mu_{\mathbb{P}}=\nu|_E.$  This proves that $\mu_{\mathbb{P}}$ can be extended to a doubling measure on $[0,1]$.

Conversely, suppose that $\mu_{\mathbb{P}}$ can be extended to a doubling measure $\nu$
on $[0,1]$.
Then we have from Lemma \ref{lem1} that
$\{n_kc_k\}_{k=1}^\infty\in\ell^q$ for some $q\geq1$,
which yields $\lim_{k\rightarrow\infty}n_kc_k=0$.
We are going to show $\{n_kc_k\}_{k=1}^\infty\in\ell^1$.

Since $\lim_{k\rightarrow\infty}n_kc_k=0$ and $\mathbb{P}$ is ultimately $1$-uniform,
we may choose an integer $k_0\geq 1$ such that
\begin{equation}\label{jia2}
\mbox{$n_kc_k<1/3$\, for all $k\geq k_0$}
\end{equation}
and that
\begin{equation}\label{jia3}
\mbox{$p_{k,i}=\frac{1}{n_k}$ for all $k\geq k_0$ and $1\leq i\leq n_k$.}
\end{equation}
Therefore $\{k:k\geq k_0\}\subseteq\Lambda$, where $\Lambda$ is as in (\ref{eq7}).
Let $k\geq k_0$ be fixed. By the definitions of integers $m_k$ and $s_k$, we have
\begin{equation}\label{jia1}
s_k\delta_{k+m_k}+(s_k-1)\varepsilon_{k+m_k}\leq\varepsilon_k<(s_k+1)\delta_{k+m_k}+s_k\varepsilon_{k+m_k}.
\end{equation}
Since $\varepsilon_k=c_k\delta_{k-1}$ by the construction of the uniform Cantor set $E$, we see from (\ref{eq3}) that
the right-hand inequality of (\ref{jia1}) can be rewritten as
$$c_k<\frac{(s_k+1)(1-(n_{k+m_k}-1)c_{k+m_k})+s_kn_{k+m_k}c_{k+m_k}}{n_{k+m_k}}\prod_{i=k}^{k+m_k-1}\frac{1-(n_i-1)c_i}{n_i}.$$
Therefore
\begin{equation}\label{p21}
n_kc_k\leq\frac{3s_k}{n_{k+1}n_{k+2}\cdots n_{k+m_k}}.
\end{equation}
On the other hand, let $w\in W_{k-1}$, $1\leq i<n_k$, and let $G_{wi}\in\mathcal{G}_k$ be
the corresponding gap of level $k$. Then $|G_{wi}|=\varepsilon_k$. Let $J=[x^-(G_{wi})-\varepsilon_k,x^-(G_{wi})]$ be an interval,
where $x^-(G_{wi})$ is the left endpoint. We see from the left-hand inequality of (\ref{jia1}) that
$J$ contains at least $s_k$ components of level $k+m_k$ in $I_{wi}$. Since $\nu$ is an extension
of $\mu_{\mathbb{P}}$ and $\nu$ is doubling on $[0,1]$, we get from (\ref{jia3}) that
$$\nu(G_{wi})\geq C\nu(J)\geq\frac{Cs_k\mu_{\mathbb{P}}(I_{wi})}{n_{k+1}\cdots n_{k+m_k}}.$$
Now, summing over all $1\leq i<n_k$ yields
\begin{equation}\label{gp}
\nu(\bigcup_{i=1}^{n_{k}-1}G_{wi})\geq\frac{Cs_k}{n_{k+1}\cdots n_{k+m_k}}\mu_{\mathbb{P}}(\bigcup_{i=1}^{n_{k}-1}I_{wi}).
\end{equation}
Since ${\mathbb{P}}$ is ultimately $1$-uniform, we have $\mu_{\mathbb{P}}(\cup_{i=1}^{n_{k}-1}I_{wi})\sim\mu_{\mathbb{P}}(I_w)$.
Summing over all $w\in W_{k-1}$, we get from (\ref{p21}) that
$$
\nu(\bigcup_{J\in\mathcal{G}_k}J)\geq\frac{Cs_k}{n_{k+1}\cdots n_{k+m_k}}
\geq Cn_kc_k
$$
Finally, summing over all $k\geq k_0$ yields
$$\infty>\nu([0,1])\geq \sum_{k=k_0}^\infty\nu(\bigcup_{J\in\mathcal{G}_k}J)
\geq C\sum_{k=k_0}^\infty n_kc_k$$
This proves $\{n_kc_k\}\in\ell^1$.

\medskip

\noindent{\bf Acknowledgement}\, The authors wish to thank the referee for
their valuable comments.


TITLE: correct typesetting for quantifiers
QUESTION [7 upvotes]: For years I have been typing and writing quantifiers in a certain way. Now that I am writing my thesis, my adviser is taking issue with some of these things. Since he is my adviser I'm going to do what he says, but I am curious about the general consensus on this.
As an example, let's say I wanted to write symbolically "There exists an element $a$ of $A$ such that $a$ is positive." My habitual way of doing this would be
$$\exists a\in A: a>0.$$
My adviser has 2 problems with this. Firstly he says there should be a space between the $\exists$ and the $a$. Secondly he says I cannot assume people will read the colon as "such that." So he would have me change this to:
$$\exists\ a\in A\text{ such that }a>0.$$
Which seems correct to you?
As for the space after the $\exists$,
it looks funny to me.
It also seems significant that $\LaTeX$
does not automatically put a space after the $\exists$
and I need to write \exists\ instead of \exists.
As for the colon, it's been a few years but I used to study logic, and I think in the conventions there my usage is fine. I'm not sure of the grammatical terminology but there is a sense in which the colon indicates that we are done quantifying things and are now going to indicate the property the quantified things have. Not only have I been writing this way for years, I have been teaching students to write this way.
I was never quite sure though about the colon with a universal quantifier? Like, is it conventional at least for some people, if I write "For all elements $a$ of $A$, $f(a)=c$" as follows?
$$\forall a\in A: f(a)=c.$$
A lot of people don't write the universal quantifier at the beginning either. I feel that my more hardcore logic professors would never do this, but a topologist would have no problem writing:
$$f(a)=c, \forall a\in A.$$
The predicate logic conventions, if I'm remembering them properly, seem just way more ... logical. But I want my writing to be familiar to my audience, which is something my adviser has the best feel for. I guess once I am more established I will have more freedom in how I write. In the meantime I'd like to hear which of these things look correct or incorrect, and please also mention what area of math you work in because that seems to matter.
UPDATE: I appreciate the many insightful comments this question has received, but would someone please post their answer as an answer?

REPLY [5 votes]: First of all, I would avoid logical symbols as much as possible. For instance, you initial example could be phrased as "There exists a positive element $a$ in $A$". Now, if you really need to use quantifiers, my advice would be the following:

Avoid any unnecessary symbols like ":" or "."
Don't hesitate to add spaces, and possibly parentheses and brackets to improve readability. $\LaTeX$ offers a large panel of possibilities to do so. For instance, $\exists x\ \forall y\ \varphi(x,y)$ looks better than $\exists x \forall y \varphi(x,y)$.
Put quantifiers in the front, not at the end. Although it is acceptable to write "$f(n) > 0$ holds for every integer $n$", if you really need to use quantifiers, it is preferable to write
$$
\forall n \in \mathbb{Z}\quad f(n) > 0
$$
or, as suggested by Brian M. Scott,
$$
\forall n \in \mathbb{Z}\quad (f(n) > 0)
$$
Double check again. Do you really need quantifiers? You will not find a single quantifier in Rudin's Real and Complex Analysis. You will no find quantifiers in Bourbaki's chapters on topology either, although Bourbaki's style is usually very formal.


TITLE: Need help understanding the proof: if v is a left singular vector of A then v is a unit eigenvector of $AA^{T}$
QUESTION [1 upvotes]: This is the proof in my textbook:

What I don't understand it why " $AA^{T}u = 0u $ means that u is an eigenvector.
Is this a theorem that I don't know? That if you multiply a matrix by a vector and it's 0, then that vector is an eigenvector?
Further, we are asked to prove that u is a unit eigenvector. The answer only proves that it is an eigenvector. How do we show that it is a unit eigenvector?

REPLY [0 votes]: I'll write my SVD notation as $A=U \Sigma V^T$, assuming that $A \in \mathbb{R}^{m \times n}$, $U \in \mathbb{R}^{m \times m}$, $\Sigma \in \mathbb{R}^{m \times n}$, and $V \in \mathbb{R}^{n \times n}$. So we have a full SVD, $U^T U = I$, $V^T V = I$, and $\Sigma$ is diagonal with nonnegative entries. I mention this because everything is slightly different in the case of a reduced SVD.
Now if $u_i$ is the $i$th column of $U$, then $A^T u_i = V \Sigma^T U^T u_i$. $U^T u_i = e_i$ (check it). Then $\Sigma^T e_i$ is the $i$th column of $\Sigma^T$, which is $\sigma_i e_i$ (check it). Finally $V e_i$ is the $i$th column of $V$, so we have $A^T u_i = \sigma_i v_i$.
Doing the same thing for $A$ and $v_i$, we find that $A v_i = \sigma_i u_i$, so $A A^T u_i = A \left ( \sigma_i v_i \right ) = \sigma_i^2 u_i$.
Finally, $u_i$ are unit by the definition of the SVD, so there's no need to separately prove anything there.


TITLE: Is a solid sphere with a "bubble" in the middle topologically the same as a torus?
QUESTION [0 upvotes]: Suppose there was a solid sphere or ball with radius 2 such that a sphere with radius 1 was removed from the center making a hollow cavity. A sphere with a "bubble in it" if that's easier to visualize.
Is this topologically equivalent to a torus? to a sphere? to a whole other class of shapes I'm unaware of?

REPLY [2 votes]: First, Note that A sphere $\cong$ $S^n$ which is the set of points of distance equals to 1 from the origin. While a Ball $\cong$ $D^n$ which is the set of all points of distance $\le 1$ from the origin. They are different, so be careful while describing them.
I think the following argument answers your problem.
Let $X$ denotes the space you're talking about.

This might be a clearer argument to show that $X\not\cong T_1$, where $T_1$ denotes the torus, but will require some knowledge from algebraic topology.
Choose $x_0\in X$ and a continuous map $\alpha:I\to X$ s.t. $\alpha(0)=\alpha(1)=x_0$ (loop). You can always find a way to continuously deform it to a single point, so the fundamental group $\pi(X,x_0)$ is trivial, suggesting that $X$ is simply connected
However, a loop in $T_1$ that enclose the center hole cannot be deformed to a point, suggesting that $T_1$ is not simply connected. Therefore $X\not\cong T_1$.
More information, please see here and here.

But, $X$ can be deformation retract to a sphere $S^2$ through a continuous map $h_t:X\to A\cong S^2$ and the inverse is an inclusion map. So  $X$ is homotopy equivalent to $S^2$ i.e. $X\simeq S^2$. This concept is different from homeomorphism.
Definition of homotopy

to have a better understanding about those concept mentioned above, I recommend you to read Algebraic Topology by Allen Hatcher. You could found it on website http://gen.lib.rus.ec/


text:
\section{Kudla-Millson theory}
\label{sec:kudlamillson}

In the previous section, we studied certain cohomological representations
for $G=\O(p,q)$ with $(p,q)=(4,2)$ or $(0,6)$. In this section,
we recall the explicit construction of $(\fg,K)$-cohomology classes attached to these representations
using the Weil representation \`{a} la Kudla-Millson. While the original
papers of Kudla and Millson considered the case of the trivial local system,
 the case of more general local systems was discussed in Funke-Millson.
 We also study the restriction of these explicit $(\fg, K)$-cohomology classes
 to the subgroup $\O(2,2) \times \O(2,0)$ and $\O(0,4) \times \O(0,2)$ of
 $\O(4,2)$ and $\O(0,6)$ respectively.

\subsection{Groups and Lie algebras}
\label{ss:groups}

Let $V$ be an $m$-dimensional quadratic space over $\R$ of signature $(p,q)$, where $p$ and $q$ are non-negative integers such that $p+q=m$.
Namely, $V$ is equipped with a non-degenerate symmetric bilinear form
$\langle \cdot, \cdot \rangle : V \times V \rightarrow \R$ and an orthogonal basis $\{ e_i \, | \, 1 \le i \le m \}$ such that
\[
 \langle e_i, e_i \rangle =
 \begin{cases}
  +1 & \text{if $1 \le i \le p$,} \\
  -1 & \text{if $p+1 \le i \le m$.}
 \end{cases}
\]
We assume that $p$ and $q$ are even.
Let $G = \O(V) \simeq \O(p,q)$ be the orthogonal group of $V$.
Put
\[
 V_+ = \R e_1 + \dots + \R e_p, \qquad
 V_- = \R e_{p+1} + \dots + \R e_m,
\]
so that $V = V_+ \oplus V_-$.
We define a Cartan involution $\theta$ of $G$ by
\[
 \theta(g) = I_V \cdot g \cdot I_V,
\]
where $I_V = \id_{V_+} \oplus (-\id_{V_-})$.
Let $K$ be the maximal compact subgroup of $G$ with respect to $\theta$.
Then $K = \O(V_+) \times \O(V_-) \simeq \O(p) \times \O(q)$.
We define a maximal torus $T$ of $G$ by
\[
 T = \SO(V_1) \times \dots \times \SO(V_r) \simeq \SO(2)^r,
\]
where $V_i = \R e_{2i-1} + \R e_{2i}$ and $r = \frac{m}{2}$.

Let $\fg_0$ be the Lie algebra of $G$.
Then we have a $G$-equivariant isomorphism $\rho: \wedge^2 V \rightarrow \fg_0$ given by
\[
 \rho(u \wedge v)(w) = \langle u, w \rangle v - \langle v, w \rangle u.
\]
We take a basis $\{ X_{ij} \, | \, 1 \le i < j \le m \}$ of $\fg_0$ given by $X_{ij} = \rho(e_i \wedge e_j)$.
Let $\{ \omega_{ij} \, | \, 1 \le i < j \le m \}$ be its dual basis of $\fg_0^*$.
We have a Cartan decomposition
\[
 \fg_0 = \fk_0 \oplus \fp_0,
\]
where $\fk_0$ and $\fp_0$ are the $(+1)$-eigenspace and $(-1)$-eigenspace of $\theta$, respectively.
Note that $\fk_0$ is the Lie algebra of $K$.
Via the isomorphism $\rho$, we have
\[
 \fk_0 \simeq \wedge^2 V_+ \oplus \wedge^2 V_-, \qquad
 \fp_0 \simeq V_+ \otimes V_-.
\]
Let $\ft_0$ be the Lie algebra of $T$.
Let $\fg$, $\fk$, $\fp$, $\ft$ be the complexifications of $\fg_0$, $\fk_0$, $\fp_0$, $\ft_0$, respectively.
If $q=2$, then we have a complex structure $\id_{V_+} \otimes J_{V_-}$ on $\fp_0$, where $J_{V_-}$ is defined by
\[
 J_{V_-}(e_{m-1}) = -e_m, \qquad
 J_{V_-}(e_m) = e_{m-1}.
\]
Let $\fp^+$ and $\fp^-$ be the $(+i)$-eigenspace and $(-i)$-eigenspace of this complex structure in $\fp$, respectively.
Then
\[
 \fg = \fk \oplus \fp^+ \oplus \fp^-.
\]

Let $W$ be a $2$-dimensional symplectic space over $\R$.
Namely, $W$ is equipped with a non-degenerate skew-symmetric bilinear form $\langle \cdot, \cdot \rangle : W \times W \rightarrow \R$ and a basis $\{ e, f \}$ such that
\[
 \langle e, e \rangle = \langle f, f \rangle = 0, \qquad
 \langle e, f \rangle = 1.
\]
Let $G' = \Sp(W) \simeq \SL_2(\R)$ be the symplectic group of $W$.
Let $K' \simeq \U(1)$ be the standard maximal compact subgroup of $G'$, where $\U(1)$ is embedded into $\SL_2(\R)$ by $a + bi \mapsto \smat{a}{b}{-b}{a}$.
Let $\fg_0'$ be the Lie algebra of $G'$ and $\fg'$ its complexification.

\subsection{Finite-dimensional representations of $G$}
\label{ss:fin-dim-rep-SO}

Let $\{ \varepsilon_i \, | \, 1 \le i \le r \}$ be the basis of $\ft_0^*$ given by $\varepsilon_i(t) = t_i$ for
\[
 t = \left( \mat{}{t_1}{-t_1}{}, \dots, \mat{}{t_r}{-t_r}{} \right).
\]
We identify $\ft^*$ with $\C^r$ via this basis.
Under this identification, the weight lattice is given by $\Z^r$.
We take
\[
 \{ \varepsilon_i - \varepsilon_{i+1} \, | \, 1 \le i < r \} \cup \{ \varepsilon_{r-1} + \varepsilon_r \}
\]
as a set of simple roots.
Then $\lambda = (\lambda_1,\dots,\lambda_r) \in \Z^r$ is dominant if and only if
\[
 \lambda_1 \ge \dots \ge \lambda_{r-1} \ge |\lambda_r|.
\]

Now assume that $r \ge 2$.
We only consider dominant weights $\lambda$ of the form
\[
 \lambda = (\ell,0,\dots,0)
\]
for some non-negative integer $\ell$.
In particular, we have $\Ss_\lambda V = \Sym^\ell V$, where $\Ss_\lambda$ is the Schur functor associated to $\lambda$.
Put $\sss^\ell V = \Sym^\ell V \otimes \C$ and equip it with a non-degenerate $G$-invariant bilinear pairing $\langle \cdot, \cdot \rangle : \sss^\ell V \times \sss^\ell V \rightarrow \C$ given by
\[
 \langle v_1 \cdots v_\ell, w_1 \cdots w_\ell \rangle = \sum_{\sigma \in \mathfrak{S}_\ell} \langle v_1, w_{\sigma(1)} \rangle \cdots \langle v_\ell, w_{\sigma(\ell)} \rangle,
\]
where $\mathfrak{S}_\ell$ is the symmetric group of degree $\ell$.
We denote by $\Hs^\ell V$ the kernel of the contraction $\sss^{\ell} V \rightarrow \sss^{\ell-2} V$ given by
\[
 v_1 \cdots v_\ell \longmapsto
 \sum_{i<j} \langle v_i, v_j \rangle \cdot
 v_1 \cdots \hat{v}_i \cdots \hat{v}_j \cdots v_\ell.
\]
Then, by \cite[\S 19.5]{fulton-harris}, $\Hs^\ell V$ is the irreducible finite-dimensional representation of $G$ with highest weight $\lambda$.
Moreover, its highest weight vector is given by
\[
 (e_1+ie_2)^\ell.
\]

\subsection{Weil representations}

We recall the Schr\"odinger model $\cS(V)$ of the Weil representation $\omega$ of $G \times G'$ (relative to the character $x \mapsto e^{2 \pi i x}$ of $\R$), where $\cS(V)$ is the space of Schwartz functions on $V$.
By \cite[\S 5]{kudla-splitting}, the action of $G$ is given by
\[
 \omega(g)\varphi(x) = \varphi(g^{-1} x),
\]
and the action of $G'$ is given by
\begin{align*}
 \omega \mat{a}{}{}{a^{-1}} \varphi(x) & = a^{\frac{m}{2}} \varphi(ax), \\
 \omega \mat{1}{b}{}{1} \varphi(x) & = \varphi(v) e^{\pi i b \langle x, x \rangle}, \\
 \omega \mat{}{-1}{1}{} \varphi(x) & = i^{\frac{q-p}{2}} \int_V \varphi(y) e^{-2 \pi i \langle x, y \rangle} \, dy.
\end{align*}
Let $x_1, \dots, x_m$ be the coordinates on $V$ with respect to the basis $\{ e_1, \dots, e_m \}$.
We denote by $S(V)$ the subspace of $\cS(V)$ consisting of functions of the form $p \cdot \varphi_0$, where $p$ is a polynomial function and $\varphi_0$ is the Gaussian defined by
\[
 \varphi_0(x_1, \dots, x_m) = e^{- \pi (x_1^2 + \dots + x_m^2)}.
\]

We also recall the Fock model $\Ps(\C^m)$ of the Weil representation $\omega$ of $\fg \times \fg'$ (relative to $\lambda = 2 \pi i$), where $\Ps(\C^m)$ is the space of polynomial functions on $\C^m$.
We refer the reader to \cite[\S 7]{km-ihes}, \cite[Appendix A]{funke-millson} for details.
Let $z_1, \dots, z_m$ be the coordinates on $\C^m$.
Then, by \cite[Lemma A.3]{funke-millson}, we have a $\fg \times \fg'$-equivariant isomorphism $\iota : S(V) \simeq \Ps(\C^m)$ such that $\iota(\varphi_0) = 1$ and such that
\begin{equation}
\label{eq:weil-fock-isom}
\begin{aligned}
 \iota \left( x_\alpha - \frac{1}{2 \pi} \frac{\partial}{\partial x_\alpha} \right) \iota^{-1} & = \frac{1}{2 \pi i} z_\alpha, &
 \iota \left( x_\alpha + \frac{1}{2 \pi} \frac{\partial}{\partial x_\alpha} \right) \iota^{-1} & = 2i \frac{\partial}{\partial z_\alpha}, \\
 \iota \left( x_\mu - \frac{1}{2 \pi} \frac{\partial}{\partial x_\mu} \right) \iota^{-1} & = - \frac{1}{2 \pi i} z_\mu, &
 \iota \left( x_\mu + \frac{1}{2 \pi} \frac{\partial}{\partial x_\mu} \right) \iota^{-1} & = - 2i \frac{\partial}{\partial z_\mu}
\end{aligned}
\end{equation}
for $1 \le \alpha \le p$ and $p+1 \le \mu \le m$.

\subsection{Schwartz forms}
\label{ss:schwartzforms}

We now recall the Schwartz forms constructed by Kudla--Millson \cite{km1} in the case of trivial coefficients and Funke--Millson \cite{funke-millson} in general.
Let $\ell$ be a non-negative integer.
If $p \ge 1$, then as in \cite[\S 6.2]{funke-millson}, we define
\[
 \varphi_{q,\ell} \in \Ps(\C^m) \otimes \wedge^q \fp^* \otimes \sss^{\ell} V
\]
by
\[
 \varphi_{q,\ell} = \left( \frac{1}{4 \pi i} \right)^{\ell+q}
 \sum_{\alpha} \sum_{\beta}
 z_\alpha z_\beta \otimes \omega_\alpha \otimes e_\beta,
\]
where the sums run over $\alpha = (\alpha_1, \dots, \alpha_q) \in \{ 1, \dots, p \}^q$ and $\beta = (\beta_1, \dots, \beta_\ell) \in \{ 1, \dots, p \}^\ell$, and
\begin{align*}
 z_\alpha & = z_{\alpha_1} \cdots z_{\alpha_q}, &
 z_\beta & = z_{\beta_1} \cdots z_{\beta_\ell}, \\
 \omega_\alpha & = \omega_{\alpha_1 p+1} \wedge \cdots \wedge \omega_{\alpha_q p+q}, &
 e_\beta & = e_{\beta_1} \cdots e_{\beta_\ell}.
\end{align*}
(Note that we scale the Schwartz form given in \cite[\S 6.2]{funke-millson} by $2^{-\frac{q}{2}}$ and take its image under the projection $V^{\otimes \ell} \otimes \C \rightarrow \sss^\ell V$.)
Then we define
\[
 \varphi'_{q, \ell} \in \Ps(\C^m) \otimes \wedge^q \fp^* \otimes \Hs^\ell V
\]
as the image of $\varphi_{q,\ell}$ under the $G$-equivariant projection $\sss^\ell V \rightarrow \Hs^\ell V$.
Via the isomorphism $\iota$, we also regard $\varphi'_{q, \ell}$ as an element in $S(V) \otimes \wedge^q \fp^* \otimes \Hs^\ell V$.
By \cite[Theorem 5.6]{funke-millson}, $\varphi'_{q, \ell}$ is invariant under the diagonal action of $K$ and
\[
 (\omega(t) \otimes 1 \otimes 1) \varphi'_{q, \ell} = t^{\ell + \frac{m}{2}} \varphi'_{q, \ell}
\]
for $t \in K' \simeq \U(1)$.
By \cite[Theorem 5.7]{funke-millson}, $\varphi'_{q, \ell}$ defines a closed differential form on $G/K$.

Similarly, if $p=0$, then we define
\[
 \varphi_\ell \in \Ps(\C^m) \otimes \sss^{\ell} V
\]
by
\[
 \varphi_\ell = \left( \frac{1}{4 \pi i} \right)^{\ell}
 \sum_{\beta} z_\beta \otimes e_\beta,
\]
where the sum runs over $\beta = (\beta_1, \dots, \beta_\ell) \in \{ 1, \dots, m \}^\ell$.
Then we define
\[
 \varphi_\ell' \in \Ps(\C^m) \otimes \Hs^{\ell} V
\]
as the image of $\varphi_\ell$ under the $G$-equivariant projection $\sss^\ell V \rightarrow \Hs^\ell V$.
Via the isomorphism $\iota$, we also regard $\varphi'_\ell$ as an element in $S(V) \otimes \Hs^\ell V$.
Then $\varphi'_\ell$ is invariant under the diagonal action of $G$ and
\[
 (\omega(t) \otimes 1) \varphi'_\ell = t^{-\ell-\frac{m}{2}} \varphi'_\ell
\]
for $t \in K' \simeq \U(1)$.

\subsection{Restrictions and contractions}
\label{ss:Res-C}

For our applications, we consider a $6$-dimensional quadratic space $\tilde{V}$ over $\R$ of signature $(p,q)$, where
\[
 (p,q) = (4,2) \text{ or } (0,6).
\]
Let $\tilde{G} = \O(\tilde{V})$ be the orthogonal group of $\tilde{V}$.
As in \S \ref{ss:groups}, we take a basis $\{e_1, \dots, e_6\}$ of $\tilde{V}$ and define a Cartan involution $\theta$ of $\tilde{G}$.
Let $\tilde{K}$ be the maximal compact subgroup of $\tilde{G}$ with respect to $\theta$.
Let $\tilde\fg = \tilde\fk \oplus \tilde\fp$ be the complexified Lie algebra of $\tilde{G}$, where $\tilde\fk$ and $\tilde\fp$ are $(+1)$-eigenspace and $(-1)$-eigenspace of $\theta$, respectively.

Put
\[
 V = \R e_1 + \R e_2 + \R e_5 + \R e_6, \qquad
 V_0 = \R e_3 + \R e_4,
\]
so that $\tilde{V} = V \oplus V_0$.
Then we have a natural inclusion $G \hookrightarrow \tilde{G}$.
Also, the natural projection $\tilde{V} \twoheadrightarrow V$ induces a projection $\sss^\ell \tilde{V} \twoheadrightarrow \sss^\ell V$.
Composing this with the inclusion $\Hs^\ell \tilde{V} \hookrightarrow \sss^\ell \tilde{V}$ and the projection $\sss^\ell V \twoheadrightarrow \Hs^\ell V$, we obtain a $G$-equivariant projection
\begin{equation}
\label{eq:proj-Hs-ell}
 \Hs^\ell \tilde{V} \twoheadrightarrow \Hs^\ell V.
\end{equation}

\subsubsection{The case $(p,q)=(4,2)$}
\label{sss:Res-C-1}

In this case, $\tilde\fp^*$ is equipped with a basis $\{\omega_{ij} \, | \, 1 \le i \le 4, \, 5 \le j \le 6 \}$ as in \S \ref{ss:groups}.
Let $\tilde{\fp}^* \twoheadrightarrow \fp^*$ be the $K$-equivariant projection induced by the natural inclusion $\fp \hookrightarrow \tilde{\fp}$.
This together with \eqref{eq:proj-Hs-ell} gives rise to a $K \times G'$-equivariant map
\[
 \Res:
 S(\tilde{V}) \otimes \wedge^2 \tilde{\fp}^* \otimes \Hs^\ell \tilde{V}
 \longrightarrow S(\tilde{V}) \otimes \wedge^2 \fp^* \otimes \Hs^\ell V.
\]
Choose an isomorphism $\wedge^4 \fp^* \simeq \C$ so that $\omega_{15} \wedge \omega_{25} \wedge \omega_{16} \wedge \omega_{26} \mapsto 1$.
This induces a non-degenerate $K$-invariant bilinear pairing $\cdot \wedge \cdot : \wedge^2 \fp^* \times \wedge^2 \fp^* \rightarrow \wedge^4 \fp^* \simeq \C$.
For $\bo \in \wedge^2 \fp^*$ and $\vvv \in \sss^\ell V$, we define a contraction
\[
 \ccc_{\bo,\vvv} : S(\tilde{V}) \otimes \wedge^2 \fp^* \otimes \sss^\ell V \longrightarrow S(\tilde{V})
\]
by $\ccc_{\bo,\vvv} = 1 \otimes ( \cdot \wedge \bo) \otimes \langle \cdot, \vvv \rangle$.

Let $\varphi'_{2,\ell} \in S(\tilde{V}) \otimes \wedge^2 \tilde{\fp}^* \otimes \Hs^\ell \tilde{V}$ be the Schwartz form as in \S \ref{ss:schwartzforms}.
We shall compute $\ccc_{\bo,\vvv}(\Res(\varphi'_{2,\ell}))$ for $\bo$ and $\vvv$ given as follows.
Put
\begin{align*}
 \omega^{++} & = \omega_{15} + i \omega_{25} + i \omega_{16} - \omega_{26}, \\
 \omega^{+-} & = \omega_{15} + i \omega_{25} - i \omega_{16} + \omega_{26}, \\
 \omega^{-+} & = \omega_{15} - i \omega_{25} + i \omega_{16} + \omega_{26}, \\
 \omega^{--} & = \omega_{15} - i \omega_{25} - i \omega_{16} - \omega_{26}.
\end{align*}
Then for $t = (t_1, t_2) \in T \simeq \U(1)^2$ (where we identify $\U(1)$ with $\SO(2)$ by $a + b i \mapsto \smat{a}{b}{-b}{a}$), we have
\[
 t \cdot \omega^{\epsilon_1\epsilon_2} = t_1^{\epsilon_1} t_2^{\epsilon_2} \omega^{\epsilon_1\epsilon_2}.
\]
In particular,
\[
 (\fp^+)^* = \C \omega^{++} + \C \omega^{-+}, \qquad
 (\fp^-)^* = \C \omega^{+-} + \C \omega^{--}.
\]
Hence we obtain a basis of $\wedge^2 \fp^*$ given by
\begin{align*}
 \omega^{++} \wedge \omega^{-+} & \in \wedge^2 (\fp^+)^*, &
 \omega^{++} \wedge \omega^{--}, \, \omega^{-+} \wedge \omega^{+-}
 & \in (\fp^+)^* \wedge (\fp^-)^*, \\
 \omega^{+-} \wedge \omega^{--} & \in \wedge^2 (\fp^-)^*, &
 \omega^{++} \wedge \omega^{+-}, \, \omega^{-+} \wedge \omega^{--}
 & \in (\fp^+)^* \wedge (\fp^-)^*.
\end{align*}
Note that the above elements in $\wedge^2 (\fp^+)^*$, $(\fp^+)^* \wedge (\fp^-)^*$, $\wedge^2 (\fp^-)^*$ lie in
\[
 H^{2,0}(X_1 \times X_2), \qquad
 H^{1,1}(X_1 \times X_2), \qquad
 H^{0,2}(X_1 \times X_2),
\]
respectively.
Also, the elements in $(\fp^+)^* \wedge (\fp^-)^*$ in the first row lie in
\[
 H^{1,1}(X_1) \otimes H^{0,0}(X_2), \qquad
 H^{0,0}(X_1) \otimes H^{1,1}(X_2),
\]
respectively, whereas the elements in $(\fp^+)^* \wedge (\fp^-)^*$ in the second row (which are the most relevant for us) lie in
\[
 H^{1,0}(X_1) \otimes H^{0,1}(X_2), \qquad
 H^{0,1}(X_1) \otimes H^{1,0}(X_2),
\]
respectively.
From the representation-theoretic viewpoint, the former corresponds to the contribution of the trivial representation, whereas the latter corresponds to the contribution of holomorphic and anti-holomorphic vectors in the discrete series representation.
Put
\begin{align*}
 \bo^+ & = - \frac{1}{2i} \cdot \omega^{++} \wedge \omega^{+-} = (\omega_{15} + i \omega_{25}) \wedge (\omega_{16} + i \omega_{26}), \\
 \bo^- & = - \frac{1}{2i} \cdot \omega^{-+} \wedge \omega^{--} = (\omega_{15} - i \omega_{25}) \wedge (\omega_{16} - i \omega_{26}),
\end{align*}
and
\[
 \vvv^+ = \frac{1}{\ell!} \cdot (e_1 + i e_2)^\ell, \qquad
 \vvv^- = \frac{1}{\ell!} \cdot (e_1 - i e_2)^\ell.
\]

\begin{prop}
\label{p:schwartz-rc1}
For $\epsilon = \pm$, we have
\[
 \ccc_{\bo^\epsilon,\vvv^\epsilon}(\Res(\varphi'_{2,\ell}))(x) = (x_1 + \epsilon ix_2)^{\ell+2} \cdot \varphi_0(x).
\]
\end{prop}

\begin{proof}
Consider the diagram
\[
 \xymatrix{
 \sss^\ell \tilde{V} \ar[d]_-p \ar[r]^-r &
 \sss^\ell V \ar[d]^-q \\
 \Hs^\ell \tilde{V} \ar[r]_-s &
 \Hs^\ell V
 },
\]
where $p,q,r,s$ are the projections.
By \cite[\S 19.5]{fulton-harris}, we have
\begin{align*}
 \sss^\ell \tilde{V} & \simeq \tau_{(\ell,0,0)} \oplus \tau_{(\ell - 2,0,0)} \oplus \cdots \oplus \tau_{(\ell-2k,0,0)}, &
 \sss^\ell V & \simeq \tau_{(\ell,0)} \oplus \tau_{(\ell - 2,0)} \oplus \cdots \oplus \tau_{(\ell-2k,0)}, \\
 \Hs^\ell \tilde{V} & \simeq \tau_{(\ell,0,0)}, &
 \Hs^\ell V & \simeq \tau_{(\ell,0)},
\end{align*}
where $\tau_\lambda$ denotes the irreducible representation with highest weight $\lambda$ and $k = [\frac{\ell}{2}]$.
Also, by \cite[\S 25.3]{fulton-harris}, we have
\[
 \tau_{(j,0,0)}|_{G} \simeq \tau_{(j,0)} \oplus \tau^{\oplus 2}_{(j-1,0)} \oplus \cdots \oplus \tau_{(0,0)}^{\oplus j+1}.
\]
Hence if $p(x) = 0$ for $x \in \sss^\ell \tilde{V}$, then $(q \circ r)(x) = 0$.
This implies that the above diagram is commutative.
Since $\vvv^\epsilon \in \Hs^\ell V$, we have
\[
 \langle (s \circ p)(x), \vvv^\epsilon \rangle = \langle (q \circ r)(x), \vvv^\epsilon \rangle = \langle r(x), \vvv^\epsilon \rangle
\]
and hence
\[
 \ccc_{\bo^\epsilon,\vvv^\epsilon}(\Res(\varphi'_{2,\ell})) = \ccc_{\bo^\epsilon,\vvv^\epsilon}(\widetilde{\Res}(\varphi_{2,\ell})),
\]
where
\[
 \widetilde{\Res}:
 S(\tilde{V}) \otimes \wedge^2 \tilde{\fp}^* \otimes \sss^\ell \tilde{V}
 \longrightarrow
 S(\tilde{V}) \otimes \wedge^2 \fp^* \otimes \sss^\ell V
\]
is the natural projection.
By definition, we have
\[
 \ccc_{\bo^\epsilon,\vvv^\epsilon}(\widetilde{\Res}(\varphi_{2,\ell})) =
 \left( \frac{1}{4 \pi i} \right)^{\ell+2}
 \sum_{\alpha} \sum_{\beta} z_\alpha z_\beta (\omega_\alpha \wedge \bo^\epsilon)
 \langle e_\beta, \vvv^\epsilon \rangle
\]
in $\Ps(\C^6)$, where the sums run over $\alpha = (\alpha_1, \alpha_2) \in \{ 1, 2 \}^2$ and $\beta = (\beta_1, \dots, \beta_\ell) \in \{ 1, 2 \}^\ell$.
It is easy to see that
\[
 \sum_{\alpha} z_\alpha (\omega_\alpha \wedge \bo^\epsilon) = (z_1 + \epsilon i z_2)^2,
 \qquad
 \sum_{\beta} z_\beta \langle e_\beta, \vvv^\epsilon \rangle = (z_1 + \epsilon iz_2)^\ell,
\]
so that
\[
 \ccc_{\bo^\epsilon,\vvv^\epsilon}(\widetilde{\Res}(\varphi_{2,\ell})) =
 \left( \frac{1}{4 \pi i} \right)^{\ell+2}
 (z_1 + \epsilon iz_2)^{\ell+2}.
\]
This combined with \eqref{eq:weil-fock-isom} gives the desired formula.
\end{proof}

\subsubsection{The case $(p,q)=(0,6)$}
\label{sss:Res-C-2}

In this case, we have $\tilde{G} = \tilde{K}$ and $\tilde{\fp}^* = \{ 0 \}$.
As above, \eqref{eq:proj-Hs-ell} gives rise to a $G \times G'$-equivariant map
\[
 \Res: S(\tilde{V}) \otimes \Hs^\ell \tilde{V}
 \longrightarrow S(\tilde{V}) \otimes \Hs^\ell V.
\]
For $\vvv \in \sss^\ell V$, we define a contraction
\[
 \ccc_{\vvv} : S(\tilde{V}) \otimes \sss^\ell V \longrightarrow S(\tilde{V})
\]
by $\ccc_{\vvv} = 1 \otimes \langle \cdot, \vvv \rangle$.

Let $\varphi'_\ell \in S(\tilde{V}) \otimes \Hs^\ell \tilde{V}$ be the Schwartz form as in \S \ref{ss:schwartzforms}.
Then, as in Proposition \ref{p:schwartz-rc1}, we have:

\begin{prop}
\label{p:schwartz-rc2}
For $\epsilon = \pm$, we have
\[
 \ccc_{\vvv^\epsilon}(\Res(\varphi'_\ell))(x) = (-1)^\ell \cdot (x_1 + \epsilon ix_2)^\ell \cdot \varphi_0(x),
\]
where $\vvv^\epsilon = \frac{1}{\ell!} \cdot (e_1 + \epsilon i e_2)^\ell$.
\end{prop}

config: arxiv


TITLE: Finding the mode of a distribution
QUESTION [1 upvotes]: I've been trying to get a better understanding of distributions.  So far I understand how we get the formulas for mean and variance (by looking at the derivative of the moment generating function).  However, I'm confused as to the general process of finding the mode.
I understand the process laid out here, but I'm wondering why one would even look at the $k+1$ and $k$ terms in the first place.

REPLY [2 votes]: The question you refer to in the link seems to have been the subject of some bickering. Anyway, I gather that you're asking how to find the mode of a distribution in general, not just for the binomial distribution.
The mode is the outcome(s) which arises most frequently. This is easy to understand in a discrete random variable. If $X$ is a random variable which takes values in $\Omega$ then the $mode$ is the value $x \in \Omega$ for which $Pr[X= x]$ is maximised, in other words the point $x$ at which the p.m.f. $p(x)$ is a maximum. There may be many such points, in which case there is more than one $mode$. These points are not necessarily always next to each other.
With a continuous random variable, the $mode$ is the point(s) $x$ at which the density function $f(x)$ is a maximum. This can usually be found by differentiating the density function to find the points where the derivative is zero and then, importantly, also checking whether such points are actually maxima.


TITLE: Show that a graph cannot be a Cayley graph
QUESTION [1 upvotes]: How can I show that the following graph cannot be a Cayley graph:
I think it's not a cayley graph because it is not vertex transitive, I just don't know how to show it (or explain it more formally).

REPLY [1 votes]: *This has to come as a comment to the previous answer,I don't have enough reputation to make a comment to answers.
Please look at the definition of vertex-transitivity, it says for any two vertices say $v_{1},v_2\text{ }\exists \text{ } f\in Aut(X)$ such that $f(v_1)=v_2$.An automorphism is an isomorphism from a graph to itself, so it has to preserve subgraph structures.So suppose vertex $v_1$ is incident to three triangles and $v_{2}$ is incident to 2 triangles , we can't map $v_1$ to $v_{2}$ since it wont be preserving subgraph structures hence it's not vertex-transitive.
There is another way to prove this, since your graph has $|V(X)=7|$,so $V(X)=\mathbb{Z_{7}}$ and it's a degree 4 connected graph,you have to choose a $C,$with $|C|=4$ and $<C>=\mathbb{Z_7}$.For any $C$ which satisfies the above condition we will fail to find an isomorphism between $X(G,C)$ and this graph hence graph cannot be cayley.


TITLE: Proof that $\log_a b \cdot \log_b a = 1$
QUESTION [1 upvotes]: Prove that $\log_a b \cdot \log_b a = 1$

I could be totally off here but feel that I have at least a clue.
My proof is:
Suppose that $a = b$, then $a^{1} = b$ and $b^{1} = a$ and we are done.
Suppose now that $a \neq b$. We wish to show that: $\log_b a = \frac{1}{\log_a b}$.
\begin{align*}
1 &= \frac{1}{\log_a b} \cdot \frac{1}{\log_b a} \cdot \left(\log_a{b} \cdot \log_b a \right)    \\
   &=  \left(\log_a{b} \cdot \frac{1}{\log_a b} \right) \cdot  \left(\log_b a \cdot \frac{1}{\log_b a}\right)  \\
   &=  \left(\log_b a \cdot \frac{1}{\log_b a}\right) \cdot 1    \\
   &=1 \cdot 1    \\
  &= 1
\end{align*}

REPLY [2 votes]: Your argument does not prove the identity. I can substitute any two values for $\log_a b$ and $\log_b a$ and get the "proof"
\begin{align}
1 &= \frac1x \cdot \frac1y \cdot \left( x \cdot y \right) \\
&= \left( x \cdot \frac1x \right)\left( y \cdot \frac1y \right) \\
&= \left( x \cdot \frac1x \right) \cdot 1 \\
&= 1 \cdot 1 \\
&= 1
\end{align}
But we certainly don't have $xy = 1$ for all values of $x$ and $y$. Your manipulations aren't wrong, but you've only shown that $1 = 1$ not that $\log_a b \log_b a = 1$.


TITLE: Has this series for $\ln(2)$ been discovered yet?
QUESTION [4 upvotes]: $$\ln(2)=\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{9}+\frac{1}{10}-\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+...$$
This is something I came up with and was intrigued, and no this isn't random; there's a pattern to this.
Since you guys are asking for a relation, here it is;
$$\ln(2)=1+\sum_{n=1}^∞ \frac{(-1)^{S_n-1}}{S_n-1} $$
Where $S_n$ is the $n$-th number that is not a perfect power (A007916, and $S_1$ is $2$). Proof to this is also somewhat easy to derive.

REPLY [3 votes]: Equivalently, the problem is to compute $L=\sum_{s\in S}(-1)^s/(s-1)$ where $S$ is the set of all perfect powers. A similar problem of computing $\sum_{s\in S}1/(s-1)\color{gray}{[{}=1]}$ goes back to Goldbach as far as I know.
The idea is pretty much the same. Let $N$ be the set of all integers (strictly) greater than $1$. Then each $s\in S$ has a representation $s=a^n$ with unique $a\in N\setminus S$ and $n\in N$; the same is true for $s\in N$ if we also allow $n=1$. Hence (each of the series below is absolutely convergent)
\begin{align}
L&=\sum_{a\in N\setminus S}(-1)^a\sum_{n=2}^{\infty}\frac{1}{a^n-1}\\&=\sum_{a\in N\setminus S}(-1)^a\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}a^{-nk}\\&=\sum_{a\in N\setminus S}(-1)^a\sum_{k=1}^{\infty}\frac{1}{a^k(a^k-1)}\\&=\sum_{s\in N}\frac{(-1)^s}{s(s-1)}=2\ln 2-1.
\end{align}
[The sum in the question is $1+\sum_{s\in N\setminus S}(-1)^{s-1}/(s-1)=1-(\ln 2-L)$.]


TITLE: Should the identity labelled by red line be $\overline{f(Z)}=X$?
QUESTION [2 upvotes]: The above picture is from Milne's Etale Cohomology.
Suppose $A=\Bbb Z, \mathfrak q=(2T+3)$, consider $Z=\operatorname{Spec} \Bbb Z[T]/(2T+3)\to \operatorname{Spec} \Bbb Z[T]\to\operatorname{Spec} \Bbb Z$.  Then $f(Z)$ doesn't contain the prime ideal $(2)$.
Should the identity labelled by red line be $\overline{f(Z)}=X$?

REPLY [5 votes]: Yes, it should be $\overline{f(Z)} = X$. In fact, $\mathfrak{q} \cap A = 0$ literally means that the image in $X$ of the generic point of the irreducible subset $Z \subseteq \mathbb{A}^1_X$ --which corresponds to the prime ideal $\mathfrak{q} \subseteq A[T]$-- is equal to the generic point of $X = {\rm Spec}(A)$ (recall that $A$ is an integral domain). This implies that the composition $Z \hookrightarrow \mathbb{A}^1_X \rightarrow X$ is dominant (i.e., the image is dense), which literally means that $\overline{f(Z)} = X$.
And as your example correctly shows, one does not need to have $f(Z) = X$ in general.


TITLE: Proving an inequality using field order axioms only.
QUESTION [1 upvotes]: In a very recent post I was trying to get guidance on how to prove that $x^2+x>0$ whenever $x<0$.
I didn't consider this doesn't hold if $x \in (-1,0)$.
However, what I'm really trying to prove is that $x^2+x+1>0$ for $x\in R$
I have considered the following:
i) If $x\geq0$, then $x^2\geq0$. This means $x^2 +x \geq0$, from which we get $x^2+x+1\geq1$. Hence $x^2+x+1\geq0$.
ii) If $x<0$, then $-x>0$, so $(-x)(-x)>0$, which means $x^2>0$. As $-x>0>x$, we also have that $x^2-x+1>x^2+1>x^2+x+1$
In this ii) part is where I'm struggling. I know that $(x+\frac{1}{2})= x^2+x+\frac{1}{4}\geq0$, so I could pretty much conclude from here $x^2+x+1>x^2+x+\frac{1}{4}\geq0$. However, I'm trying to prove this only with algebraic manipulation for the case $x<0$. Can someone help?

REPLY [3 votes]: Why insist on treating $x<0$ separately? It isn’t natural for this problem.
Just note that no matter what $x$ is, you have
$$x^2+x+1=x^2+x+\tfrac14-\tfrac14+1$$
$$=(x+\tfrac12)^2 +\tfrac34\geq\tfrac34>0$$