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\begin{abstract}
We show that for all $n \geq 2$, there exists a doubling linearly locally contractible metric space $X$ that is topologically a $n$-sphere such that every weak tangent is isometric to $\R^n$ but $X$ is not quasisymmetrically equivalent to the standard $n$-sphere. The same example shows that $2$-Ahlfors regularity in Theorem 1.1 of \cite{BK02} on quasisymmetric uniformization of metric $2$-spheres is optimal.
\end{abstract}

\maketitle

\section{Introduction}

Let $(X, d_X)$ and $(Y, d_Y)$ be two metric spaces. A homeomorphism $f: X \to Y$ is a quasisymmetry if there exists a homeomorphism $\eta:[0,\infty) \to [0,\infty)$ such that for all $x,y,z \in X$, with $x \neq z$, we have
\begin{align*}
\frac{d_Y(f(x), f(y))}{d_Y(f(x), f(z))} \leq \eta\br{ \frac{d_X(x,y)}{d_X(x,z)}}.
\end{align*}
In this case we say that $X$ and $Y$ are quasisymmetrically equivalent.

In \cite{Kinneberg17} Kinneberg characterized metric circles that are quasisymmetric to the standard circle in terms of weak tangents (defined in Section 4 below):
\begin{thm}[Kinneberg, \cite{Kinneberg17}]
A doubling metric circle $\mathcal{C}$ is quasisymmetrically equivalent to the standard circle $\mathbb{S}^1$ if and only if every weak tangent of $\mathcal{C}$ quasisymmetrically equivalent to the real line $\R$ based at $0$.
\end{thm}
\noindent Here a metric space is said to be doubling if for every $R > 0$, every ball $B$ of radius $2R$ can be covered by $N$ balls of radius $R$, where $N$ is a positive integer that does not depend on the choice of the ball $B$. In this paper we prove that Kinneberg's result cannot be extended to higher dimensions:
\begin{thm}\label{Theorem}
For every $n \geq 2$, there exists a doubling, linearly locally contractible metric space $X$ that is topologically an $n$-sphere such that every weak tangent is isometric to $\R^n$ but $X$ is not quasisymmetrically equivalent to the standard $n$-sphere.
\end{thm}

When $n = 2$, one can compare our result with the following Theorem:
\begin{thm}[Bonk and Kleiner, \cite{BK02}]\label{Bonk_Kleiner}
Let $Z$ be an 2-Ahlfors regular metric space homeomorphic to $\mathbb{S}^2$. Then $Z$ is quasisymmetric to $\mathbb{S}^2$ if and only if $Z$ is linearly locally contractible.
\end{thm}
\noindent Our Theorem \ref{Theorem} shows that the conclusion of Theorem \ref{Bonk_Kleiner} is false if we replace $2$-Ahlfors regularity with $Q$-Ahlfors regularity for $Q >2$.

Our study is also related to the following theorem, proven in \cite{Lindquist17}:
\begin{thm}\label{Intro_Thm1}
Let $(Z,d)$ be a doubling metric space homeomorphic to $\mathbb{S}^2$. The following are equivalent:
\begin{enumerate}[(i)]
\item $(Z, d)$ is quasisymmetrically equivalent to the standard $2$-sphere.
\item For every $x \in Z$, there exists an open neighborhood $U$ of $x$ in $Z$ such that $U$ is quasisymmetrically equivalent to $\D$.
\end{enumerate}
\end{thm}
\noindent An alternative proof of Theorem \ref{Intro_Thm1} can be given using ideas in \cite{GW18}. Roughly speaking, Theorem \ref{Intro_Thm1} says local geometry properties promote to global property. Since weak tangents are local, one could ask the following question:

\begin{question}
Suppose $(Z,d)$ is doubling and linearly locally connected. Are the following two statements equivalent?
\begin{enumerate}[(i)]
\item $Z$ is quasisymmetrically equivalent to the standard $d$-sphere $\mathbb{S}^d$
\item Every weak tangent of $Z$ is quasisymmetrically equivalent to $\R^d$.
\end{enumerate}
\end{question}
\noindent When $Z$ is a doubling and linearly locally connected metric sphere, statement (i) implies statement (ii). However, our construction shows that statement (ii) does not imply statement (i).

\begin{ack} The author would like to thank Mario Bonk and John Garnett for their support and many conversations. The author thanks Jeff Lindquist for helpful discussions and Wenbo Li for his comments and corrections.
\end{ack}

\section{The Modification of the Metric on $\R$}

We first construct a metric $\delta$ on $\R$ so that every weak tangent of $(\R, \delta)$ is isometric to $(\R, 0, d)$, where $d$ is the Euclidean metric on $\R$, and so that there are segments $[a_j, b_j]$ of $(\R, \delta)$ so that $\ell([a_j, b_j]) / \delta(a_j, b_j) \to +\infty$. Here, and in the remaining of the paper, $\ell(\gamma)$ denotes the length of the curve $\gamma$ with respect to the metric $\delta$. Once the metric $\delta$ is constructed, we show that for all $n \geq 2$, the weak tangents of the product space $(\R, \delta) \times \R^{n-1}$ are isometric to the Euclidean space $\R^n$, but because $\ell([a_j, b_j]) / \delta(a_j, b_j) \to +\infty$, the product space $(\R, \delta) \times \R^{n-1}$ cannot be quasisymmetric to the Euclidean space $\R^n$.

To construct $(\R, \delta)$, we glue together segments $[a_j, b_j]$ that look like snowflake curves of Hausdorff dimension $(\alpha_j)^{-1}$ i.e. intervals $I$ equipped with the metric $d^{\alpha_j}$. Such snowflake curves will have infinite length, so we modify the metric on the segment $[a_j, b_j]$ to get a metric $d_n$ so that $d_n$ looks like the Euclidean metric when two points are close together, but it looks like $d^{\alpha_j}$ when the two points are far apart.

Note that our construction of $(\R, \delta)$ only modifies the geometry of a bounded subset of $\R$, and therefore we can embed our $n$-dimensional construction into a topological $n$-sphere.

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To begin the construction let $\alpha \in (0,1)$ and $c \in (0,1)$ and define
\[ \varphi_{\alpha, c}(x) = \begin{cases}
L(\alpha, c) x,& x \in [0,c] \\
\br{ \frac{x - c(1-\alpha)}{1 - c(1-\alpha)}}^\alpha,& x \in [c,1],
\end{cases} \]
where
\[ L(\alpha, c) = \frac{1}{c} \br{\frac{c\alpha}{1 - c(1-\alpha)}}^\alpha. \]
The function $\varphi_{\alpha, c}$ is the only function that has the following properties:
\begin{enumerate}
\item $\varphi_{\alpha, c}(0) = 0$,
\item $\varphi_{\alpha, c}(1) = 1$,
\item $\varphi_{\alpha, c}$ is linear on $[0,c]$,
\item There exists $a,b \in \R$, with $a \neq 0$, such that $\varphi_{\alpha ,c}(x) = (ax - b)^\alpha$ when $x \in [c,1]$,
\item $\varphi_{\alpha, c}$ is continous and differentiable at $c$.
\end{enumerate}
In addition to the above properties that uniquely define $\varphi_{\alpha, c}$, we observe that $\varphi_{\alpha, c}$ has the following extra properties:
\begin{enumerate}
\item $\varphi_{\alpha, c} :[0,1] \to [0,1]$ is a homeomorphism,
\item $\varphi_{\alpha, c}$ is concave on $[0,1]$.
\end{enumerate}

\begin{lemma}\label{lemma_sup_multiplicative}
For any $\alpha \in (0,1)$, $c \in (0,1]$, and $a, b \in [0,1]$, we have
\[ \varphi_{\alpha, c}(ab) \varphi_{\alpha, c}(a) \varphi_{\alpha, c}(b). \]
\end{lemma}
\begin{proof}
The function $L(\alpha, c)$ is decreasing in $c$ to $1$, therefore if $0 < c_1 \leq c_2 \leq 1$, then
\[ \varphi_{\alpha, c_1} \geq \varphi_{\alpha, c_2}. \]
Let
\[ \varphi_\ast(x) = \frac{\varphi_{\alpha, c}(ax)}{\varphi_{\alpha, c}(a)}.  \]
The function $\varphi_\ast$ meets condition (1)-(5) listed above for some $c_\ast \in [c, 1]$, therefore $\varphi_{\ast} = \varphi_{\alpha, c_\ast}$. This implies that $\varphi_\ast \leq \varphi_{\alpha, c}(x)$ for all $x \in [0,1]$. Taking $x = b$, we have
\[ \varphi_{\alpha, c}(ab) \leq \varphi_{\alpha, c}(a)\varphi_{\alpha, c}(b). \]
\end{proof}

\begin{lemma}\label{lemma_rough_isometries_1}
For any fixed $\alpha \in (0,1)$, for any $c \in (0,1)$, and whenever $0 \leq t \leq x \leq 1$, we have
\[ 0 \leq \varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) -\varphi_{\alpha, c}(x) \leq \br{ 2 - 2^{\alpha}}\varphi_{\alpha, c}(x). \]
\end{lemma}
\begin{proof}
When $x\leq c$, for all $t \in [0,x]$, we have $\varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) -\varphi_{\alpha, c}(x) = 0$. When $x \geq 2c$ is fixed, $\varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) -\varphi_{\alpha, c}(x)$ is maximized when
\[ \varphi_{\alpha, c}'(t) - \varphi_{\alpha, c}'(x-t) = 0, \]
which is possible only if $x- t = t$ i.e. $2t = x$. Then we have
\begin{align*}
\varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) -\varphi_{\alpha, c}(x)
&\leq 2\varphi_{\alpha, c}(x/2) - \varphi_{\alpha, c}(x)  \\
&\leq 2\varphi_{\alpha, c}(x/2) - 2^{\alpha}\varphi_{\alpha, c}(x/2) \\
&= (2 - 2^{\alpha})\varphi_{\alpha, c}(x/2).
\end{align*}
When $c < x < 2c$, we have
\[ \varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) \leq L(\alpha, c)x = \frac{x}{c} \varphi_{\alpha, c}(c). \]
By the concavity of $\varphi_{\alpha, c}$, we have
\[ \frac{\varphi_{\alpha, c}(x)}{x} \geq \frac{\varphi_{\alpha, c}(2c)}{2c}. \]
Therefore
\begin{align*}
\varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) -  \varphi_{\alpha, c}(x)
&\leq \frac{x}{c} \varphi_{\alpha, c}(c)-  \frac{x \varphi_{\alpha, c}(2c)}{2c}
= \frac{x}{2c} \br{ 2 \varphi_{\alpha, c}(c) -\varphi_{\alpha, c}(2c)}.
\end{align*}
But
\[ 2 \varphi_{\alpha, c}(c) -\varphi_{\alpha, c}(2c) \leq  (2 - 2^{\alpha})\varphi_{\alpha, c}(c) = (2 - 2^{\alpha})cL(\alpha, c). \]
Therefore
\begin{align*}
\varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) -  \varphi_{\alpha, c}(x)
\leq \frac{x}{2c} (2 - 2^{\alpha})cL(\alpha, c)

= (2 - 2^{\alpha}) \varphi_{\alpha, c}(x/2).
\end{align*}
In any case, we have
\[ 0 \leq \varphi_{\alpha, c}(t) + \varphi_{\alpha, c}(x-t) -  \varphi_{\alpha, c}(x) \leq (2 - 2^{\alpha}) \varphi_{\alpha, c}(x/2) \leq (2 - 2^{\alpha}) \varphi_{\alpha, c}(x). \]
\end{proof}

\section{The One Dimensional Construction}

For any $\alpha \in (0,1)$, we have
\begin{equation}\label{equation_1}
\lim_{c \to 0^+} L(\alpha, c) = +\infty.
\end{equation}
Let $\alpha_n$ be an increasing sequence in $(0,1)$ such that $\lim_{n \to \infty} \alpha_n = 1$. By (\ref{equation_1}), we can choose $c_n \in (0,1)$ such that $c_n \to 0$, and
\[ \lim_{n \to \infty} L(\alpha_n, c_n) = +\infty. \]
Let $\varphi_n = \varphi_{\alpha_n, c_n}$. Choose a sequence $s_n$ so that for all $n \in \N$, $s_n < 2\br{\frac{1}{n} - \frac{1}{n+1}}$, such that $s_n L(\alpha_n, c_n)$ is decreasing and $\sum_{n \in\N} s_n L(\alpha_n, c_n) < \infty$.

For all $n \in \N$, let $a_n = \frac{1}{n} -s_n, b_n = \frac{1}{n}$.
Let $I_n = [a_n, b_n]$, and equip $I_n$ with the metric $\delta_n = s_n\varphi_n \circ (s_n^{-1} d)$, where $d$ is the usual Euclidean metric on $I_n$. Note that
\begin{enumerate}
\item The distance between the two endpoints of $I_n$ is $\delta_n(a_n, b_n) = s_n$.
\item $I_n$ is rectifiable and the length of $I_n$ is $\ell(I_n) = s_nL(\alpha_n, c_n)$.
\end{enumerate}
We construct a metric $\delta$ on $\R$ so that if $x \leq y$,
\begin{enumerate}
\item $\delta = \delta_n$ when restricted to $I_n \times I_n$,
\item $\delta(x,y) = d(x,y)$ when $x,y \in \R \bs \bigcup_{i \geq } I_i$,
\item $\delta(x,y) = d(x, a_n) + \delta_n(a_n, y)$ when $x \in \R \bs \bigcup_{i \in \N} I_i$ and $y \in I_n$,
\item $\delta(x,y) = \delta_n(x, b_n) + d(b_n, y)$ when $x \in I_n$ and $y \in  \R \bs \bigcup_{i \in \N} I_i$, and
\item $\delta(x,y) = \delta_n(x, b_n) + d(b_n, a_m) + d_m(a_m, y)$ when $x \in I_n$, $y \in I_m$.
\end{enumerate}

\section{The Weak Tangents of $(\R, \delta)$}

A \textit{pointed metric space} is a a triplet $(X, x_0, d_X)$, where $(X, d_X)$ is a metric space and $x_0$ is a point in $X$. Let $\vep > 0$. A map $\psi: (X, x_0, d_X) \to (Y, y_0, d_Y)$ between two pointed metric spaces $(X, x_0, d_X)$ and $(Y, y_0, d_Y)$ is a \textit{$\vep$-rough isometry} if
\begin{enumerate}
\item $\psi(x_0) = y_0$,
\item $d_Y(\psi(X), Y) \leq \vep$, and
\item for all $x_1, x_2 \in X$,
\[\abs{ d_Y(\psi(x_1), \psi(x_2)) - d_X(x_1, x_2)} \leq \vep. \]
\end{enumerate}
Note that a $\vep$-rough isometry may not be continuous. The \textit{pointed Gromov-Hausdorff distance} between 2 pointed metric spaces, denoted $d_{GH}\br{(X, x_0, d_X), (Y, y_0, d_Y)}$, is defined as the infimum of $\vep$ for which we can find a $\vep$-rough isometry $\psi_\delta: (X, x_0, d_X) \to (Y, y_0, d_Y)$.

A pointed metric space $(T, p, d_T)$ is called a \textit{weak tangent} of another metric space $(X, d_X)$ if there exist points $x_n \in X$ and positive integers $\lambda_n \to +\infty$ such that $(X, x_n, \lambda_n d_X)$ converges to $(T, p, d_T)$ in pointed Gromov-Hausdorff sense, i.e. for all $R > 0$, and for all $\vep > 0$ there exists $N > 0$ such that for all $n \geq N$,
\begin{align*}
d_{GH}\br{\wbar{B}_{\lambda_n d_X}(x_n, R + \vep), \wbar{B}_{d_T}(p, R)} \leq \vep.
\end{align*}
In particular, we get $(X, x_n, \lambda_n d_X) \to (T, p, d_T)$ if for all $R > 0$, and for all $\vep > 0$ there exists $N > 0$ such that for all $n \geq N$,
\begin{align*}
d_{GH}\br{\wbar{B}_{\lambda_n d_X}(x_n, R), \wbar{B}_{d_T}(p, R)} \leq \vep.
\end{align*}
Our notion of pointed Gromov-Hausdorff convergence is adopted from \cite[Definition 11.3.1]{The_Purple_Book}. Also see \cite{Burago} for detailed discussion on Gromov-Hausdorff distance and Gromov-Hausdorff convergence.

Let M be a set of separable, uniformly doubling, and uniformly linearly locally contractible pointed metric spaces. The pointed Gromov Hausdorff convergence on the set $M$ induces a topology on $M$ that is metrizable\cite{LeDonne}. In particular, if a sequence of pointed metric space converges to a Gromov-Hausdorff limit, then the limit is unique.

The goal of this section is to prove the following proposition that describe all the weak tangents of $(\R, \delta)$.
\begin{prop}\label{Prop_1}
For all $a_n \in \R$, and for all positive integers $\lambda_n \to +\infty$, $(\R, a_n, \lambda_n \delta)$ converges in pointed Gromov-Hausdorff sense to $(\R, 0, d)$.
\end{prop}
Note that Proposition \ref{Prop_1} guarentees the existence of weak tangents of $(\R, \delta)$. To prove the above proposition we will use the following three lemmas:

\begin{lemma}\label{lemma_rough_isometries_2}
Suppose $x, y, z \in \R$ are three points so that $x \leq y \leq z$. Suppose $N = \inf\{n \in \N: \{x,y,z\} \cap I_N \neq \emptyset\} < \infty$. Then
\[ 0 \leq \delta(x,y) + \delta(y,z) - \delta(z,x) \leq \br{2 - 2^{\alpha_N}} \min\{ s_N, \delta(x,y) \} \]
\end{lemma}
\begin{proof}
If $N = +\infty$, then $\delta(x,y) + \delta(y,z) - \delta(z,x) = 0$. Otherwise, let
\begin{align*}
a &= \sup\{ a_n \leq y: n \in \N\} \vee \sup\{b_n \leq y: n \in \N\} \vee x \\
b &= \inf\{ a_n \geq y: n \in \N\} \wedge \inf\{b_n \geq y: n \in \N\} \wedge z
\end{align*}
We have $x \leq a \leq y \leq b \leq z$. By definition of $\delta$, we have
\begin{align*}
\delta(x,y) + \delta(y,z) - \delta(x,z)
&= \br{ \delta(x,a) + \delta(a,y)} + \br{ \delta(y,b) + \delta(b,z)} \\
&\hspace{30pt} - \br{ \delta(x,a) + \delta(a,b) + \delta(b,z)} \\
&= \delta(a,y) + \delta(y,b) -\delta(a,b).
\end{align*}
Bu our choice of $a$ and $b$, either $a,y,b \in I_n$ for some $n \geq N$, or $(a, b) \cap \bigcup_{n \in \N} I_n = \emptyset$. In the latter case, we have
\[ \delta(x,y) + \delta(y,z) - \delta(z,x) = \delta(a,y) + \delta(y,b) -\delta(a,b) = 0. \]
In the former case, Lemma \ref{lemma_rough_isometries_1} gives
\begin{align*}
0 &\leq \delta(x,y) + \delta(y,z) - \delta(z,x)
= \delta(a,y) + \delta(y,b) -\delta(a,b)  \\
&\leq \br{2 - 2^{\alpha_n}} \delta(a,b)
= \br{2 - 2^{\alpha_N}} \min\{ s_N, \delta(x,y) \}.
\end{align*}
Since $\alpha_N \leq \alpha_n \leq 1$, we have our desired conclusion.
\end{proof}

\begin{lemma}\label{lemma_rough_isometries_3'}
Let $p \in \R$ and $r > 0$ be arbitrary. Suppose $N = \inf\{n \in \N: \wbar{B}_{\delta}(p,r)\cap I_n \neq \emptyset\} < \infty$. Let $a = \inf\{x \in \R: \delta(x,p) \leq r\}$ and $b = \sup\{x \in \R: \delta(x,p) \leq r\}$. The map
\begin{align*}
\psi: ([a,b], p, \delta) \to ([-r, r],0,d)\\
\psi(x) = \begin{cases}
-\delta(p,x) ,& x \leq p \\
\delta(p,x),& x > p.
\end{cases}
\end{align*}
is a $\vep$-rough isometry, where $\vep = \br{2 - 2^{\alpha_N}} \min\{ s_N, 2r\}$.
\end{lemma}
\begin{proof}
Since $\psi$ is surjective and fixes $p$, it remains to check that for all $x,y \in [a,b]$, we have
\[ \br{d(\psi(x), \psi(y)) - \delta(x,y)} \leq \vep. \]
Suppose $x \leq y$. If $x \leq y \leq p$, then
\begin{align*}
\abs{d(\psi(x), \psi(y)) - \delta(x,y)}
&= \abs{ \abs{ \delta(p,x) - \delta(p,y)} - \delta(x,y)} \\
&= \abs{\delta(p,y) - \delta(p, x) - \delta(x,y)} \\
&\leq \br{2 - 2^{\alpha_N}} \min\{ s_N, \delta(p,x)\}  \\
&= \br{2 - 2^{\alpha_N}} \min\{ s_N, 2r\} .
\end{align*}
The second to last inequality is a consequence of Lemma \ref{lemma_rough_isometries_2}. If $x \leq p \leq y$, then
\begin{align*}
\abs{d(\psi(x), \psi(y)) - \delta(x,y)}
&= \abs{ \abs{ \delta(p,x) - \delta(p,y)} - \delta(x,y)} \\
&= \abs{\delta(p,y) + \delta(p, x) - \delta(x,y)} \\
&\leq \br{2 - 2^{\alpha_N}} \min\{ s_N, \delta(x,y)\}  \\
&= \br{2 - 2^{\alpha_N}} \min\{ s_N, 2r\}.
\end{align*}
If $p \leq x \leq y$, then following a similar arguement as when $x \leq y \leq p$, we get
\begin{align*}
\abs{d(\psi(x), \psi(y)) - \delta(x,y)}
\leq \br{2 - 2^{\alpha_N}} \min\{ s_N, 2r\}
\end{align*}
This verifies that $\psi$ is a $\vep$-rough isometry.
\end{proof}

\begin{lemma}\label{lemma_rough_isometries_3}
For $r \in (0,1)$, we have
\[ \sup_{p \in \R} d_{GH}((\wbar{B}_{\delta}(p,r), p, \delta),(\wbar{B}_{d}(0,r), 0, d)) = o(r) \]
as $r \to 0$.
\end{lemma}
\begin{proof}
Let $p \in \R$ and $r > 0$ be arbitrary. Let $N = \inf\{n \in \N: \wbar{B}_{\delta}(p,r)\cap I_n \neq \emptyset\}$. If $N = +\infty$, then $\delta = d$ on $\wbar{B}_{\delta}(p,r)$. We have
\[ d_{GH}((\wbar{B}_{\delta}(p,r), p, \delta),(\wbar{B}_{d}(0,r), 0, d)) = 0. \]
If $N < \infty$, we consider 2 cases:

Case 1: $r \geq \frac{s_N}{2}\varphi_N(c_N)$. As $r \to 0$, $N \to +\infty$, therefore $\br{2 - 2^{\alpha_N}} \to 0$. In this case Lemma \ref{lemma_rough_isometries_3'} implies
\begin{align*}
d_{GH}((\wbar{B}_{\delta}(p,r), p, \delta),(\wbar{B}_{d}(0,r), 0, d)) \leq 2\br{2 - 2^{\alpha_N}} r = o(r).
\end{align*}

Case 2: $r < \frac{s_N}{2}\varphi_N(c_N)$. In this case, $\delta$ is a length metric on $[a,b]$, and $\psi$ is an isometry between two length spaces. We have
\[ d_{GH}((\wbar{B}_{\delta}(p,r), p, \delta),(\wbar{B}_{d}(0,r), 0, d)) = 0. \]
\end{proof}

\begin{proof}[Proof of Proposition \ref{Prop_1}]
Let $\{a_n\}_{n \in \N}$ be a sequence in $\R$ and $\{\lambda_n\}_{n \in \N}$ be a sequence of positive numbers that diverges to $+\infty$. By Lemma \ref{lemma_rough_isometries_3},
\begin{align*}
&d_{GH}\br{(\wbar{B}_{\lambda_n \delta}(a_n, R ), a_n, \lambda_n \delta),(\wbar{B}_{\lambda_n d}(0, R), 0, \lambda_n d)} \\
&\hspace{30pt}= \lambda_n d_{GH}\br{(\wbar{B}_{\delta}(a_n, \lambda_n^{-1}R ), a_n, \delta),( \wbar{B}_{d}(0, \lambda_n^{-1}R), 0, d)} = \lambda_n o(\lambda_n^{-1}R).
\end{align*}
But $(\wbar{B}_{\lambda_n d}(0, R), 0, \lambda_n d) = \wbar{B}_{d}(0, R), 0, d))$ by symmetry of $\R$. As $n \to \infty$, $\lambda_n^{-1}R \to 0$. We have
\[ (\wbar{B}_{\lambda_n \delta}(a_n, R), a_n, \delta ) \to \wbar{B}_{d}(0, R), 0, d). \]
This is true for all $R > 0$. We conclude that $(\R, a_n, \lambda_n \delta) \to (\R, 0, d)$.
\end{proof}

\section{Linear Local Contractibility and Assouad Dimension of $(\R, \delta)$}

In this section we establish two properties of the space $(\R, \delta)$. These properties often appear in the study of quasisymmetry classes of metric spheres. Both properties are discussed in detail in \cite{Heinonen}.

\begin{defn}
Let $C > 1$ be a constant. A metric space is \textit{$C$-linearly locally contractible} if every small ball is contractible inside a ball whose radius is $C$ times larger.  A metric space is \textit{linearly locally contractible} if it is $C$-linearly locally contractible for some $C > 0$.
\end{defn}

\begin{defn}
Let $N > 0$. A metric space is \textit{$N$-doubling} if for all $R > 0$, every open ball of $2R$ can be covered by $N$ balls of radius $R$. A metric space is \textit{doubling} if it is $N$-doubling for some $N > 0$.
\end{defn}

Both doubling and linear local contractibility are preserved under quasisymmetry. The Euclidean spaces like $\R^n$ or $\mathbb{S}^n$ are doubling and linearly locally contractible. The doubling property also ensures the existence of weak tangents.

\begin{prop}\label{Prop_3}
The space $(\R, \delta)$ is $1$-linearly locally contractible.
\end{prop}
\begin{proof}
Any open ball $B$ in $(\R, \delta)$ is an open interval $(a,b)$. Denote $p$ the center of $B$ (in $(\R, \delta)$.) Note that the map $x \mapsto \delta(p, x)$ is increasing on $\{x \in \R: x \geq p$, and decreasing on $\{x \in \R: x \leq p\}$. Therefore the map $H(x,t) = tx + (1-t)p$ is a homotopy of $(a,b)$ to $\{p\}$ in $B$. This proves that $(\R, \delta)$ is $1$-linearly locally contractible.
\end{proof}

\begin{prop}\label{Prop_2}
The space $(\R, \delta)$ is doubling.
\end{prop}

If a metric space $X$ is doubling, then there exists $\beta > 0$ and $C > 0$ such that for all $\vep \in (0, 1/2)$ and $r > 0$, any set of diameter $r$ in $X$ can be covered by at most $C \vep^{-\beta}$ subsets of diameter at most $\vep r$. The function $\vep \mapsto C \vep^{-\beta}$ is called the \textit{covering function} of $X$. The Assouad dimension of $X$ is defined to be the infimum of all $\beta$ so that a covering function of the form $\vep \mapsto C \vep^{-\beta}$ of $X$ exists. Conversely, any metric space of finite Assouad dimension is doubling. Proposition \ref{Prop_2} will follow from the stronger proposition below.

\begin{lemma}
For each $n \in \N$, the function $f_n(\vep) = 2\vep^{-\alpha_n^{-1}}$ is a covering function of $(I_n, \delta)$.
\end{lemma}
\begin{proof}
Every subinterval of $I_n$ of $\delta$-diameter $r \in [0, s_n]$ has $d$-diameter $s_n\varphi_n^{-1}(s_n^{-1}r)$. Thus our goal is to show that for every $r \in [0, s_n]$, and every $\vep \in (0, 1/2)$, every subset of $I_n$ of $d$-diameter $s_n\varphi_n^{-1}(s_n^{-1}r)$ can be covered by no more than $(\vep^{-\alpha_n}+1)$-many subintervals of $I_n$ of $d$-diameter at most $s_n\varphi_n^{-1}(s_n^{-1}\vep r)$. The number of subintervals we need can be bounded from above by
\begin{align*}
\frac{s_n\varphi_n^{-1}(s_n^{-1}r)}{s_n\varphi_n^{-1}(s_n^{-1}\vep r)} + 1
\leq \sup_{y \in (0,1]} \frac{\varphi_n^{-1}(y)}{\varphi_n^{-1}(\vep y)} + 1
= \sup_{y \in [\varphi_n(c_n),1]} \frac{\varphi_n^{-1}(y)}{\varphi_n^{-1}(\vep y)} + 1.
\end{align*}
We claim that the last supremum is attained when $y = 1$. This is equivalent to
\begin{equation}\label{equation_2}
\frac{\varphi_n(\varphi_n^{-1}(\vep y))}{\varphi_n(\varphi_n^{-1}(y))} \leq \varphi_n\br{ \frac{\varphi_n^{-1}(\vep y)}{\varphi_n^{-1}(y)}}
\end{equation}
Inequality (\ref{equation_2}) is true by Lemma \ref{lemma_sup_multiplicative}.

Suppose $\varphi_n(x_0) = \vep$. When $\vep > \varphi_n(c_n)$, we have $x_0 > c_n$, and
\begin{align*}
\vep^{\alpha_n^{-1}} = \frac{x_0 - c(1-\alpha_n)}{1-c(1-\alpha_n)} \leq x_0.
\end{align*}
When $0 < \vep < \varphi_n(c_n)$, we have $x_0 < c_n$ and $\vep = \varphi_n(x_0) = \frac{x_0}{c} \varphi_n(c_n)$. Since $ \varphi_n(c_n)^{\alpha_n^{-1}} \leq c$, we have
\begin{align*}
\frac{1}{x_0}
= \frac{\varphi(c)}{c\vep}
= \vep^{-\alpha_n^{-1}} \br{ \frac{\vep}{\varphi(c)}}^{\alpha_n^{-1}-1} \frac{\varphi(c)^{\alpha_n^{-1}}}{c} \leq \vep^{-\alpha_n^{-1}}.
\end{align*}
In any case, we can take the covering function of $I_n$ to be
\begin{align*}
\vep^{-\alpha_n^{-1}}+1 \leq 2\vep^{-\alpha_n^{-1}}.
\end{align*}
\end{proof}

\begin{prop}\label{Prop_4}
The Assouad dimension of $(\R, \delta)$ is $1$.
\end{prop}
\begin{proof}
Let $\beta > 1$ be arbitrary. There exists $N \in \N$ such that when $n \geq N$, $\alpha^{-1} <\beta$. Let $C = \max_{n < N}\{2^{\alpha_n^{-1} - \beta}\} \geq 1$. Then the function $\vep \mapsto 2C \vep^{-\beta}$, where $\vep \in (0, 1/2]$, is a covering function of $(I_n, \delta)$ for all $n \in \N$. Thus $\vep \mapsto 4C \vep^{-\beta}$ is a covering function of $(\R, \delta)$.

Since $(-\infty, 0)$ has Assouad dimension $1$, the proposition follows.
\end{proof}

\section{Higher Dimension Construction}

Let $d \geq 2$. We will denote by $d_{Euclid}$ the Euclidean metric on $\R^d$. Let
\[ X_d = \br{\R \times \R^{d-1}, \sqrt{\delta^2 + d_{Euclid}^2}}\]
be the product of $(\R, \delta)$ and $(\R^{d-1}, d_{Euclid})$. Write $\rho_n = \sqrt{\delta^2 + d_{Euclid}^2}$. Here are some facts about $X_d$.

\begin{prop}\label{HD_Prop_1}
\begin{enumerate}[(a)]

\item Every weak tangent of $X_d$ is isometric to $(\R^d, 0, d_{Euclid})$.
\item $X_d$ is doubling and linearly locally contractible.
\end{enumerate}
\end{prop}
\begin{proof}
\begin{enumerate}[(a)]
\item Every weak tangent of $X_d$ is of the form $(T \times \R^{n-1}, (x, 0), d_T \times d_{Euclid})$, where $(T, x, d_T)$ is a weak tangent of $(\R, \delta)$.
\item Recall that $(\R, \delta)$ is doubling (Proposition \ref{Prop_2}). $X_d$ is product of doubling metric space, hence doubling. By Proposition \ref{Prop_3}, $(\R, \delta)$ is $C$-linearly localy contractible for some $C > 1$. Let $x = (x_1, x_2)$ be any point in $\R \times \R^{d-1}$, and $r >0$ be arbitrary. The ball $B(x,r)$ in $X_d$ can first be contracted to $\{x_1\} \times B(x_2, r)$ within a $B(x, Cr)$, which can then be contracted to the point $\{x_1, x_2\}$.
\end{enumerate}
\end{proof}

Every finite segment in $(\R, \delta)$ is rectifiable. Let $\mu_1$ be a the measure on $(\R, \delta)$ given by length. For $d \geq 2$, let $\mu_d$ be the product measure $\mu_1 \times \lambda_{d-1}$ on $X_d$, where $\lambda_{d-1}$ is the $(d-1)$-dimensional Lebesgue measure on $\R^{d-1}$.

In the remaining of this section we show that $X_d$ is not quasisymmetrically equivalent to $\R^d$. To do that we consider a geometric quantity that is roughly preserved under quasisymmetry called modulus. Given a family $\Gamma$ of curves in a measured metric space $(X, d_X, \mu)$, we say that a function $\rho:X \to [0,\infty)$ is admissible if for all $\gamma \in \Gamma$,
\begin{align*}
\int_\gamma \rho(x) \,ds \geq 1.
\end{align*}
Let $Q > 0$. We define the \textit{$Q$-modulus} of $\Gamma$ as
\begin{align*}
\mod_Q(\Gamma) = \inf\{ \int_X \rho^Q \,d\mu : \rho \text{ admissible }\}.
\end{align*}
Let $E, F \subset X$ be two disjoint nondegerate continua in $X$. Let $\Gamma_{E,F}$ to be the collection of all rectifiable curves joining $E$ and $F$. We write $\mod_Q(E,F) = \mod_Q(\Gamma_{E,F})$.

Moduli behave nicely under quasisymmetry, as illustrated by the following theorem.
\begin{thm}[Tyson, \cite{Tyson98}]\label{Tyson_theorem}
Let $X, Y$ be locally compact, connected, $Q$-Ahlfors regular metric spaces, where $Q > 1$, and $f:X \to Y$ be a quasisymmetric homeomorphism. Then there exists $C > 1$ such that for all curve family $\Gamma \subset X$, we have
\[ \frac{1}{C} \mod_Q(\Gamma) \leq \mod_Q(f(\Gamma)) \leq C\mod_Q(\Gamma). \]
\end{thm}
See the next section for the definition of $Q$-Ahlfors regularity.

We can now prove that $X_d$ is not quasisymmetrically equivalent to $\R^d$. The idea is that if the two spaces are quasisymmetrically equivalent, then for any curve family $\Gamma$, $\Gamma$ and $f(\Gamma)$ should have comparable moduli. We know that $\R^d$ has the property that any disjoint nondegerate continua $E, F$ in $\R^d$ satisfy
\begin{equation}\label{modulus}
\mod_d(E,F) \leq \phi\br{ \frac{d(E,F)}{\diam(E) \wedge \diam(F)}}.
\end{equation}
for some non-increasing function $\phi:[0,\infty) \to (0,\infty)$. However, Proposition \ref{HD_Prop_3} shows that some sequence disjoint nondegerate continua $E_n, F_n$ in $X_d$ do not behaviour as in (\ref{modulus}). The only problem is that $X_d$ is not $d$-Ahlfors regular, so we cannot apply Theorem \ref{Tyson_theorem} directly. In Proposition \ref{HD_Prop_2}, however, we will show that the inequality
\[  \mod_d(E_n,F_n) \leq C \mod_d(f(E_n), f(F_n)) \]
holds for some $C$ independent of $C$.

From now on, we will denote $\Delta(E,F) = \frac{d(E,F)}{\diam(E) \wedge \diam(F)}$.

\begin{prop}\label{HD_Prop_3}
There exists $E_n, F_n \subset X_d$ such that $\Delta(E_n, F_n) = \frac{d(E_n, F_n)}{\diam E_n \wedge \diam F_n} = 1$ and $\mod_d(E_n, F_n) \to +\infty$.
\end{prop}
\begin{proof}
Take
$E_n = I_n \times \{0\} \times [0,s_n]^{d-2},
 F_n = I_n \times \{s_n\} \times [0,s_n]^{d-2}$.
Then $d(E_n, F_n) = s_n$, and $\diam E_n = \diam F_n = s_n$, therefore
\[ \Delta(E_n, F_n) = \frac{d(E_n, F_n)}{\diam E_n \wedge \diam F_n} = 1. \]
For each $n \in \N$, and for $x \in I_n$ and $(v_2, \ldots, v_{d-1}) \in [0,s_n]^{d-2}$, let $\gamma_{x, v_2, \ldots, v_{d-1}}$ be the path
\[ t \mapsto \br{x, t, v_2, \ldots, v_{d-1}}, t \in [0,s_n]. \]
Let
\[ \Gamma_n = \{ \gamma_{x, v_2, \ldots, v_{d-1}} : x \in I_n, (v_2, \ldots, v_{d-1}) \in [0,s_n]^{d-2}\}\]
be the family of straight lines joining $E_n$ and $F_n$ that meet $E_n$ orthogonally. Then
\[ \mod_d(E_n, F_n) \geq \mod_d(\Gamma_n). \]

Let $\rho: X_d \to \R_{\geq 0}$ be an admissible function for $\Gamma_n$. This means for all $\gamma_{x, v_2, v_3, \ldots, v_{d-1}} \in \Gamma_n$, we have
\[ \int_{\gamma_{x, v_2, v_3, \ldots, v_{d-1}}} \rho(t) \,dt
= \int_0^{s_n} \rho(x,t,v_2, \ldots, v_{d-1}) \,dt \geq 1. \]
Integrating over $I_n \times [0,s_n]^{d-2}$ and applying Fubini's theorem, we get
\begin{align*}
\int_{I_n \times [0,s_n]^{d-2}} \,d\mu_1 \times \lambda_{d-2}
&\leq \int_{I_n \times [0,s_n]^{d-2}} \int_{\gamma_{x, v_2, v_3, \ldots, v_{d-1}}} \rho(t) \,dt \,d(\mu_1 \times \lambda_{d-2}) \\
&= \int_{I_n \times [0,s_n]^{d-1}} \rho \,d\mu_d.
\end{align*}
Applying H\"older's inequality, we get
\begin{align*}
\int_{I_n \times [0,s_n]^{d-1}} \rho \,d\mu_d
&\leq \br{ \int_{I_n \times [0,s_n]^{d-1}} \,d\mu_d}^{\frac{1}{\delta}} \br{ \int_{I_n \times [0,s_n]^{d-1}} \rho^d \,d\mu_d}^{\frac{1}{d}}
\end{align*}
where $\delta$ is the conjugate exponent of $d$ (so $\frac{1}{\delta} + \frac{1}{d} = 1$). We have
\begin{align*}
\int_{I_n \times [0,s_n]^{d-1}} \rho^d \,d\mu_d
&\geq \br{ \int_{I_n \times [0,s_n]^{d-1}} \,d\mu_d}^{-\frac{d}{\delta}}\br{\int_{I_n \times [0,s_n]^{d-2}} \,d\mu_1 \times \lambda_{d-2}}^d  \\
&=\br{ (s_n)^{(d-1)} \ell(I_n)}^{-\frac{d}{\delta}} \br{(s_n)^{d-2} \ell(I_n)}^d \\
&= 2^{n} \ell(I_n).
\end{align*}
This is true for all admissible function $\rho$, therefore
\[ \mod_d(\Gamma_n) \geq  s_n^{-1} \ell(I_n). \]
As $n \to \infty$, $s_n^{-1} \ell(I_n) = L(\alpha_n, c_n) \to +\infty$. We have
\[ \mod_d(E_n, F_n) \geq \mod_d(\Gamma_n) \to +\infty. \]
\end{proof}

\begin{prop}\label{HD_Prop_2}
$X_d$ is not quasisymmetrically equivalent to any subset of $\R^d$.
\end{prop}
\begin{proof}
For each $n \in \N$, and for $x \in I_n, (v_2, \ldots, v_{d-1}) \in [0,s_n]^{d-2}\}$, let $\gamma_{x, v_2, \ldots, v_{d-1}}$ be the path
\[ t \mapsto \br{x, t, v_2, \ldots, v_{d-1}}, t \in [0,s_n]. \]
Let
\[ \Gamma_n = \{ \gamma_{x, v_2, \ldots, v_{d-1}} : x \in I_n, (v_2, \ldots, v_{d-1}) \in [0,s_n]^{d-2}\}\]
be the family of straight lines joining $E_n$ and $F_n$ that meet $E_n$ orthogonally.
Suppose $f: X_d \to \R^d$ is a $\eta$-quasisymmetric embedding. We denote by $\gamma^*$ the image of the curve $\gamma$. For a path family $\Gamma$, define $\Gamma^* = \bigcup_{\gamma \in \Gamma} \gamma*$, and define $\diam(\Gamma) = \inf\{\diam \gamma: \gamma \in \Gamma\}$.

Choose $E_n$ and $F_n$ as in Proposition \ref{HD_Prop_3}. We know from $\diam(E_n) \wedge \diam(F_n) = d(E_n, F_n)$ that
\[ \frac{d(f(E_n), f(F_n))}{\diam(f(E_n)) \wedge \diam(f(F_n))} \sim 1, \]
where the implicit constant for $\sim$ depends only on $\eta$. For the same reason, there exists $\alpha > 1$, depending only on $\alpha,\beta > 1$, depending only on $\eta$, so that $f(\Gamma_n)^* \subset B_n$ for a ball $B_n$ with diameter $r_n \leq \alpha \,d(E_n, F_n) \leq \beta \diam(f(\Gamma))$.

For each $n \in \N$, we can cover $\Gamma^*$ by squares $\{R_i\}_{i \in I_n}$ of diameter $s_n\varphi_n(c_n)$ so that their sides are either parallel to the paths in $\Gamma$ or orthogonal to the paths in $\Gamma$. We can choose $\{R_i\}$ so that these $R_i$'s don't overlap and their union is precisely $\Gamma^*$. For each rectifiable path $\gamma$, denote by $\ell(\gamma)$ its length.
Let
\[ \rho_n = (\diam f(\Gamma_n))^{-1} \sum_{i \in I_n} \frac{\diam(f(R_i))}{\diam(R_i)} \mathbbm{1}_{f^{-1}(B_n) \cap R_i}. \]
be a function on $X_d$.
For all $\gamma \in \Gamma_n$,

 We have
\[ \int_{\gamma} \rho_n(s) \,ds
= (\diam f(\Gamma_n))^{-1} \sum_{i \in I_n} \frac{\diam(f(R_i))}{\diam(R_i)} \ell( f^{-1}(B_n) \cap R_i \cap \gamma). \]
For each $i$, $f^{-1}(B_n) \cap R_i \cap \gamma = R_i \cap \gamma$. When $R_i \cap \gamma \neq \emptyset$, $\ell(R_i \cap \gamma) = \diam(R_i)$. As $\{R_i\}_{i \in I_n}$ covers $\Gamma^*$, we have
\begin{align*}
 \int_{\gamma} \rho_n(s) \,ds
\geq (\diam f(\Gamma_n))^{-1}  \sum_{i \in I_n, R_i \cap \gamma \neq \emptyset } \diam(f(R_i))
\geq  (\diam f(\Gamma_n))^{-1}  \diam (f(\gamma))
\geq 1.
\end{align*}
This means $\rho_n$ is admissible for $\Gamma_n$. We have
\begin{align*}
\mod_d \Gamma_n
&\leq \int \rho_n^d \,d\mu_d
= (\diam f(\Gamma_n))^{-d} \sum_{i \in I_n} \br{ \frac{\diam(f(R_i))}{\diam(R_i)}}^d \mu_d(f^{-1}(B_n) \cap R_i) \\
&= (\diam f(\Gamma_n))^{-d} \sum_{i \in I_n} \br{ \frac{\diam(f(R_i))}{\diam(R_i)}}^d \mu_d(R_i) \\
&= (\diam f(\Gamma_n))^{-d} \sum_{i \in I_n} \br{ \frac{\diam(f(R_i))}{\diam(R_i)}}^d \diam(R_i)^d.
\end{align*}
For the last equality, we make use of the fact that $R_i$ are chosen so small that $\mu_d(R_i) = \diam(R_i)^d$. We get
\begin{align*}
\mod_d \Gamma_n
\lesssim (\diam f(\Gamma_n))^{-d} \sum_{i \in I_n} \br{ \diam(f(R_i)) }^d
\lesssim (\diam f(\Gamma_n))^{-d} \sum_{i \in I_n} \lambda_d(f(R_i)).
\end{align*}
Here we use the fact that the Lebesgue $\lambda_d$ on $\R^d$ is $d$-Ahlfors regular and that $f(R_i)$ are uniform quaisdisks. Since $\{f(R_i)\}_{n\in\N}$ are disjoint subsets of $B_n$, we have
\begin{align*}
\mod_d \Gamma_n \lesssim (\diam f(\Gamma_n))^{-d} \lambda_d(B_n) \lesssim 1.
\end{align*}
where all the implicit constants for $\lesssim$ depends only on $\eta$. But this is a contradiction to Proposition \ref{HD_Prop_3}.
\end{proof}

From Prop \ref{HD_Prop_1} and Prop \ref{HD_Prop_2}, $X_d$ is homeomorphic to $\R^d$, doubling and linearly localy connected, it is not quasisymmetric to $\R^d$. Our proof shows that the any ball in $X_d$ centered at $0$ cannot be quasisymmetrically embedded into $\R^d$.

We conclud this section by a proof of Theorem \ref{Theorem}.
\begin{proof}[Proof of Thoerem \ref{Theorem}]
The $d$-dimensional unit ball $B(0,1)$ in $\R^d$, equipped with the metric
\[ \wtil{\rho}(x, y) = \frac{\rho\br{ \frac{x}{1-\abs{x}}, \frac{y}{1-\abs{y}}}}{1+ \rho\br{ \frac{x}{1-\abs{x}}, \frac{y}{1-\abs{y}}}}. \]
The completion of the space $(B(0,1), \wtil{\rho})$ is $\wbar{B}(0,1)$, and the metric on the boundary is same as the Euclidean metric. Glue the space $\wbar{B}(0,1)$ with another hemisphere to form a topological $d$-sphere. This $d$-sphere is doubling, locally linearly contractible, and every weak tangent is isometric to $\R^d$, but it cannot be a quasisphere.
\end{proof}

\section{Ahlfors Regularity}
Let $Q > 0$. A measured metric space $(X, d, \mu)$ is said to be $Q$-Ahlfors regular if for all $x \in X$ and $r \leq \diam X$, we have
\[ \mu(B(x,r)) \sim r^Q. \]
Ahlfors regularity and doubling property of metric spaces are related notions. We record a result from \cite{Heinonen}:

\begin{thm} \label{Heinonen_AR_2}\cite[Theorem 14.6]{Heinonen}
Let $X$ be a complete, connected metric space of finite Assouad dimension $\beta$. Then for each $Q > \beta$, there exists a quasisymmetric homeomorphism of $X$ onto a closed $Q$-Ahlfors regular subset of some $\R^N$.
\end{thm}
With these facts our example gives:
\begin{thm}\label{Q-AR-counter-example}
For every $Q > 2$, there exists a $Q$-Ahlfors regular and linearly locally contractible metric space $X$ that is topologically an $2$-sphere such that every weak tangent is uniformly quasisymmetric to $\R^2$ but $X$ is not quasisymmetric to the standard $2$-sphere.
\end{thm}
\begin{proof}
Let $Q > 2$. By Proposition \ref{Prop_4}, the Assouad dimension of $X_2$ is $2$. Proposition \ref{Heinonen_AR_2} says that there exist a distortion function $\eta:[0,\infty) \to [0,\infty)$ and a $\eta$-quasisymmetry $\varphi:X_2 \to X'$, where $X'$ is a closed $Q$-Ahlfors regular subset of $\R^N$. By Proposition \ref{HD_Prop_2} and Proposition \ref{HD_Prop_1}(b), $X'$ is not quasisymmetric to the standard $2$-sphere, but every weak tangent of $X'$ is $\eta$-quasisymmetric to $(\R^2, 0)$.
\end{proof}

\begin{comment}
What happens in higher dimensions? In
\begin{thm}
For each $n \geq 3$ there exists an almost smooth $n$-Ahlfors regular and linearly locally contractible length metric $d$ in $\mathbb{S}^n$ with a singular set consisting of a single point $x_0 \in \mathbb{S}^n$ for which there is no quasiconformal homeomorphism $(\mathbb{S}^n, d) \to \mathbb{S}^n$.
\end{thm}
\end{comment}

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TITLE: Find all rings with 2 elements
QUESTION [2 upvotes]: How to find all rings (up to isomorphism) with two elements?
Is it only $(\mathbb{Z}_2, +, *)$?

REPLY [2 votes]: Hint: draw the addition and multiplication tables.
There is only one group of 2 elements. That fixes the addition table. Try to prove that the neutral element with respect to addition must be an absorbing element with respect to multiplication: that fixes three quarters of the multiplication table.


TITLE: How can I find the P matrix in $P^{-1}AP $= D?
QUESTION [1 upvotes]: I have the following exercise:
Let a matrix A = $\begin{pmatrix}
4 & 0  & 1\\
 2 & 2  & 1\\
 0 & 4   & 1
\end{pmatrix}$
1) Determine its eigenvalues and their multiplicity.
2) Give a basis of the eigen spaces associated with each of the
distinct eigenvalues of A.
3) find P and D such that $P^{-1}AP$ = D is diagonal.
I've done 1) and 2) and found the basis
B ={ $\begin{pmatrix}
-1 \\
-1 \\
4
\end{pmatrix}$ , $\begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}$ , $\begin{pmatrix}
-1 \\
1 \\
1
\end{pmatrix}$ }
Now, in my notes they say that the matrix P is constituted of the vectors of the basis found.
So, P = $\begin{pmatrix}
-1 & -2  & -1\\
 -1 & 1  & 1\\
 4 & 4   & 1
\end{pmatrix}$
Also, because we have a 3x3 matrix and we've found three distinct eigen values, the matrix D exists and as such we put the eigen values on the diagonal line of the matrix. ($E_(\lambda_1) = 0$, $E_(\lambda_2) = 2$, $E_(\lambda_3) = 5$ )
D = $\begin{pmatrix}
0 & 0  & 0\\
 0 & 2  & 0\\
 0 & 0   & 5
\end{pmatrix}$
Finally to verify that this is all true I compute $AP = PD$. However I didn't find the same result on the left side and the right side of the equation. Where did I go wrong?
$AP \ne PD$
$\begin{pmatrix}
    0 & -4 &-3\\
 0 & 2  & 1 \\
 0 & 8  & 5 \\
\end{pmatrix} \ne \begin{pmatrix}
    0 & -4 &-5 \\
 0 & 2 & 5 \\
 0 & 8 & 5 \\
\end{pmatrix} $

REPLY [2 votes]: A quick look at your calculations, I think your last eigen vector is wrong. The last eigen vector is (1, 1, 1) and not (-1, 1, 1). I think that would solve your problem.


TITLE: Question regarding the formulation of a proof using the mean value theorem for integrals
QUESTION [0 upvotes]: Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous, and suppose that $\int_c^df=0$ for every interval $[c,d]\subset[a,b]$. Show that $f(x)=0$ for every $x\in[a,b]$.

I just need assistance writing a formal proof of this. I believe I have the general idea:
By the mean value theorem, there exists an $z\in[c,d]$ such that $f(z)(d-c)=0$. Because $\int_c^df=0$ for $\textit{every}$ subinterval $[c,d]\subset[a,b]$, it follows that $f(x)=0$ for every $x\in[a,b]$.
How do I make this rigorous?

REPLY [0 votes]: You proved correctly that every interval $[c,d]\subset[a,b]$ (with more than one point) has a zero of $f$. Therefore, $\{x\in[a,b]\,|\,f(x)=0\}$ is dense in $[a,b]$. Since $f$ is countinuous, it follows from this that $f$ is the null function.


TITLE: How to not feel bad for doing math?
QUESTION [4 upvotes]: I have a MsC and want to take a PhD in algebraic topology. Probably very few people in the world will have any interest of my thesis.  They will pay me for doing my hobby. Its the only job I can think of which doesnt contribute to bettering the world somehow. I feel like I should get a real job like doctor or garbageman. I  get bad conscience. How can I feel better for pursuing my dream?

REPLY [4 votes]: Perhaps you need to read  G. H. Hardy's A Mathematician's Apology.

One of the main themes of the book is the beauty that mathematics
  possesses, which Hardy compares to painting and poetry. For Hardy, the
  most beautiful mathematics was that which had no practical
  applications in the outside world (pure mathematics) and, in
  particular, his own special field of number theory. Hardy contends
  that if useful knowledge is defined as knowledge which is likely to
  contribute to the material comfort of mankind in the near future (if
  not right now), so that mere intellectual satisfaction is irrelevant,
  then the great bulk of higher mathematics is useless.
  He justifies the pursuit of pure mathematics with the argument that
  its very "uselessness" on the whole meant that it could not be misused
  to cause harm.
  On the other hand, Hardy denigrates much of the applied mathematics as
  either being "trivial", "ugly", or "dull", and contrasts it with "real
  mathematics", which is how he ranks the higher, pure mathematics.

[bolding mine]
Somewhat ironically, some topics that were in Hardy's day the domain of pure mathematics now have very practical importance, eg, algorithms used in public key cryptography rely on things like the difficulty of factorization of large numbers, properties of discrete logarithms, elliptic curve theory, etc.


TITLE: Finding a homeomorphism between a part of ball and part of euclidean plane
QUESTION [1 upvotes]: So i have two topological spaces on euclidean plane that are defined as:
$X=\{(x,y)\in\mathbb{R^2}|x<0,y<0, x^2+y^2<4\}$
$Y=\{(x,y)\in\mathbb{R^2}|x>1,y<-1\}$
So i have one quarter of ball in the third quadrant and one is just an infinite rectangle if I may call it like that.
So i wonder how can i find a homeomorphism between these two spaces.
Thank you in advance.

REPLY [0 votes]: Hint: Start with an easier setting: $X'= \{(x,y):x> 0,y>0, x^2+y^2 < 1 \},$ $Y' = \{(x,y): x> 0, y>0\}.$ The map
$$(x,y) \to \frac{(x,y)}{1-(x^2+y^2)}$$
is then a homeomorphism of $X'$ onto $Y'.$ You're just a few simple maps away from this.


text:
\begin{document}

\setcounter{page}{1}

\title[$\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$]{$\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$}

\author[S. KABBAJ, M. ROSSAFI]{S. KABBAJ$^1$ \MakeLowercase{and} M. ROSSAFI$^2$$^{*}$}

\address{$^{1}$Department of Mathematics, University of Ibn Tofail, B.P. 133, Kenitra, Morocco}
\email{\textcolor[rgb]{0.00,0.00,0.84}{samkabbaj@yahoo.fr}}

\address{$^{2}$Department of Mathematics, University of Ibn Tofail, B.P. 133, Kenitra, Morocco}
\email{\textcolor[rgb]{0.00,0.00,0.84}{rossafimohamed@gmail.com}}

\subjclass[2010]{42C15, 46L06}

\keywords{$\ast$-frame, operator frame, $\ast$-operator frame, $C^{\ast}$-algebra, Hilbert $\mathcal{A}$-modules.}

\date{12/12/2017;
\newline \indent $^{*}$Corresponding author}

\begin{abstract}
In this paper, a new notion of frames is introduced: $\ast$-operator frame as generalization of $\ast$-frames in Hilbert $C^{\ast}$-modules introduced by A. Alijani and M. A. Dehghan \cite{Ali} and we establish some results.
\end{abstract} \maketitle

\section{Introduction}
Frames were first introduced in 1952 by Duffin and Schaeffer \cite{Duf} in the study of nonharmonic fourier series. Frames possess many nice properties which make them very useful in wavelet analysis, irregular sampling theory, signal processing and many other fields. The frame theory was quickly generalized.
There were several kinds of generalization. For example: g-frames, $\ast$-frames, $\ast$-g-frames and $\ast$-$K$-g-frames in Hilbert $A$-modules \cite{Ros2}.

In this article, a new notion of frames is introduced: $\ast$-operator frame as generalization of $\ast$-frames in Hilbert $C^{\ast}$-modules introduced by A. Alijani and M. A. Dehghan \cite{Ali} and we establish some results.

The paper is organized as follows:

 in section 2, we briefly recall the definitions and basic properties of $C^{\ast}$-algebra, Hilbert $C^{\ast}$-modules, frames and $\ast$-frames in Hilbert $C^{\ast}$-modules;

 in section 3, we introduce the $\ast$-operator frame, the $\ast$-operator frame transform and the $\ast$-frame operator;

 in section 4, we investigate tensor product of Hilbert $C^{\ast}$-modules, we show that tensor product of $\ast$-operator frames for Hilbert $C^{\ast}$-modules $\mathcal{H}$ and $\mathcal{K}$, present an $\ast$-operator frames for $\mathcal{H}\otimes\mathcal{K}$, and tensor product of their $\ast$-frame operators is the $\ast$-frame operator of their tensor product of $\ast$-operator frames.
\section{Preliminaries}
Let $I$ be a countable index set. In this section we briefly recall the definitions and basic properties of $C^{\ast}$-algebra, Hilbert $C^{\ast}$-modules, frame, $\ast$-frame in Hilbert $C^{\ast}$-modules. For information about frames in Hilbert spaces we refer to \cite{Ch}. Our reference for $C^{\ast}$-algebras is \cite{Dav,Con}. For a $C^{\ast}$-algebra $\mathcal{A}$, an element $a\in\mathcal{A}$ is positive ($a\geq 0$) if $a=a^{\ast}$ and $sp(a)\subset\mathbf{R^{+}}$. $\mathcal{A}^{+}$ denotes the set of positive elements of $\mathcal{A}$.
\begin{definition}
	\cite{BA}. Let $ \mathcal{A} $ be a unital $C^{\ast}$-algebra and $\mathcal{H}$ be a left $ \mathcal{A} $-module, such that the linear structures of $\mathcal{A}$ and $ \mathcal{H} $ are compatible. $\mathcal{H}$ is a pre-Hilbert $\mathcal{A}$-module if $\mathcal{H}$ is equipped with an $\mathcal{A}$-valued inner product $\langle.,.\rangle_{\mathcal{A}} :\mathcal{H}\times\mathcal{H}\rightarrow\mathcal{A}$, such that is sesquilinear, positive definite and respects the module action. In the other words,
	\begin{itemize}
		\item [(i)] $ \langle x,x\rangle_{\mathcal{A}}\geq0 $ for all $ x\in\mathcal{H} $ and $ \langle x,x\rangle_{\mathcal{A}}=0$ if and only if $x=0$.
		\item [(ii)] $\langle ax+y,z\rangle_{\mathcal{A}}=a\langle x,y\rangle_{\mathcal{A}}+\langle y,z\rangle_{\mathcal{A}}$ for all $a\in\mathcal{A}$ and $x,y,z\in\mathcal{H}$.
		\item[(iii)] $ \langle x,y\rangle_{\mathcal{A}}=\langle y,x\rangle_{\mathcal{A}}^{\ast} $ for all $x,y\in\mathcal{H}$.
	\end{itemize}
	For $x\in\mathcal{H}, $ we define $||x||=||\langle x,x\rangle_{\mathcal{A}}||^{\frac{1}{2}}$. If $\mathcal{H}$ is complete with $||.||$, it is called a Hilbert $\mathcal{A}$-module or a Hilbert $C^{\ast}$-module over $\mathcal{A}$. For every $a$ in $C^{\ast}$-algebra $\mathcal{A}$, we have $|a|=(a^{\ast}a)^{\frac{1}{2}}$ and the $\mathcal{A}$-valued norm on $\mathcal{H}$ is defined by $|x|=\langle x, x\rangle_{\mathcal{A}}^{\frac{1}{2}}$ for $x\in\mathcal{H}$.
	\begin{example} \cite{Tro}
		If $ \{\mathcal{H}_{k}\}_{k\in\mathbf{N}} $ is a countable set of Hilbert $\mathcal{A}$-modules, then one can define their direct sum $ \oplus_{k\in\mathbb{N}}\mathcal{H}_{k} $. On the $\mathcal{A}$-module $ \oplus_{k\in\mathbb{N}}\mathcal{H}_{k} $ of all sequences $x=(x_{k})_{k\in\mathbb{N}}: x_{k}\in\mathcal{H}_{k}$, such that the series $ \sum_{k\in\mathbb{N}}\langle x_{k}, x_{k}\rangle_{\mathcal{A}} $ is norm-convergent in the $\mathcal{C}^{\ast}$-algebra $\mathcal{A}$, we define the inner product by
		\begin{equation*}
		\langle x, y\rangle:=\sum_{k\in\mathbb{N}}\langle x_{k}, y_{k}\rangle_{\mathcal{A}}
		\end{equation*}
		for $x, y\in\oplus_{k\in\mathbb{N}}\mathcal{H}_{k} $.

		Then $\oplus_{k\in\mathbb{N}}\mathcal{H}_{k}$ is a Hilbert $\mathcal{A}$-module.

		The direct sum of a countable number of copies of a Hilbert $\mathcal{C}^{\ast}$-module $\mathcal{H}$ is denoted by $l^{2}(\mathcal{H})$.
	\end{example}
\end{definition}
	Let $\mathcal{H}$ and $\mathcal{K}$ be two Hilbert $\mathcal{A}$-modules. A map $T:\mathcal{H}\rightarrow\mathcal{K}$ is said to be adjointable if there exists a map $T^{\ast}:\mathcal{K}\rightarrow\mathcal{H}$ such that $\langle Tx,y\rangle_{\mathcal{A}}=\langle x,T^{\ast}y\rangle_{\mathcal{A}}$ for all $x\in\mathcal{H}$ and $y\in\mathcal{K}$.

	We also reserve the notation $End_{\mathcal{A}}^{\ast}(\mathcal{H},\mathcal{K})$ for the set of all adjointable operators from $\mathcal{H}$ to $\mathcal{K}$ and $End_{\mathcal{A}}^{\ast}(\mathcal{H},\mathcal{H})$ is abbreviated to $End_{\mathcal{A}}^{\ast}(\mathcal{H})$.

\begin{definition}
	\cite{F4}. Let $ \mathcal{H} $ be a Hilbert $\mathcal{A}$-module. A family $\{x_{i}\}_{i\in I}$ of elements of $\mathcal{H}$ is a frame for $ \mathcal{H} $, if there exist two positive constants $A$ and $B$ such that for all $x\in\mathcal{H}$,
	\begin{equation}\label{1}
		A\langle x,x\rangle_{\mathcal{A}}\leq\sum_{i\in I}\langle x,x_{i}\rangle_{\mathcal{A}}\langle x_{i},x\rangle_{\mathcal{A}}\leq B\langle x,x\rangle_{\mathcal{A}}.
	\end{equation}
	The numbers $A$ and $B$ are called lower and upper bound of the frame, respectively. If $A=B=\lambda$, the frame is $\lambda$-tight. If $A = B = 1$, it is called a normalized tight frame or a Parseval frame. If the sum in the middle of \eqref{1} is convergent in norm, the frame is called standard.
\end{definition}
\begin{definition}
	\cite{Ali}. Let $ \mathcal{H} $ be a Hilbert $\mathcal{A}$-module. A family $\{x_{i}\}_{i\in I}$ of elements of $\mathcal{H}$ is an $\ast$-frame for $ \mathcal{H} $, if there	exist strictly nonzero elements $A$ , $B$ in $\mathcal{A}$, such that for all $x\in\mathcal{H}$,
	\begin{equation}\label{3}
		A\langle x,x\rangle_{\mathcal{A}} A^{\ast}\leq\sum_{i\in I}\langle x,x_{i}\rangle_{\mathcal{A}}\langle x_{i},x\rangle_{\mathcal{A}}\leq B\langle x,x\rangle_{\mathcal{A}} B^{\ast}.
	\end{equation}
	The elements $A$ and $B$ are called lower and upper bound of the $\ast$-frame, respectively. If $A=B=\lambda$, the $\ast$-frame is $\lambda$-tight. If $A = B = 1$, it is called a normalized tight $\ast$-frame or a Parseval $\ast$-frame. If the sum in the middle of \eqref{3} is convergent in norm, the $\ast$-frame is called standard.
\end{definition}
\begin{lemma}\label{2.7}
	\cite{Ali}. If $\varphi:\mathcal{A}\longrightarrow\mathcal{B}$ is an $\ast$-homomorphism between $\mathcal{C}^{\ast}$-algebras, then $\varphi$ is increasing, that is, if $a\leq b$, then $\varphi(a)\leq\varphi(b)$.
\end{lemma}
\begin{lemma}\label{2.8}
	\cite{Ali}. Let $\mathcal{H}$ and $\mathcal{K}$ be two Hilbert $\mathcal{A}$-modules and $T\in End_{\mathcal{A}}^{\ast}(\mathcal{H},\mathcal{K})$.
	\begin{itemize}
		\item [(i)] If $T$ is injective and $T$ has closed range, then the adjointable map $T^{\ast}T$ is invertible and $$\|(T^{\ast}T)^{-1}\|^{-1}I_{\mathcal{H}}\leq T^{\ast}T\leq\|T\|^{2}I_{\mathcal{H}}.$$
		\item  [(ii)]	If $T$ is surjective, then the adjointable map $TT^{\ast}$ is invertible and $$\|(TT^{\ast})^{-1}\|^{-1}I_{\mathcal{K}}\leq TT^{\ast}\leq\|T\|^{2}I_{\mathcal{K}}.$$
\end{itemize}
\end{lemma}
\section{$\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$}
We begin this section with the following definition.
\begin{definition}
	A family of adjointable operators $\{T_{i}\}_{i\in I}$ on a Hilbert $\mathcal{A}$-module $\mathcal{H}$ over a unital $C^{\ast}$-algebra is said to be an operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$, if there exist positive constants $A, B > 0$ such that
	\begin{equation}\label{eq3}
		A\langle x,x\rangle\leq\sum_{i\in I}\langle T_{i}x,T_{i}x\rangle\leq B\langle x,x\rangle, \forall x\in\mathcal{H}.
	\end{equation}
	The numbers $A$ and $B$ are called lower and upper bound of the operator frame, respectively. If $A=B=\lambda$, the operator frame is $\lambda$-tight. If $A = B = 1$, it is called a normalized tight operator frame or a Parseval operator frame. If only upper inequality of \eqref{eq3} hold, then $\{T_{i}\}_{i\in I}$ is called an operator Bessel sequence for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$.

	If the sum in the middle of \eqref{eq3} is convergent in norm, the operator frame is called standard.
\end{definition}
Throughout the paper, series like \eqref{eq3} are assumed to be convergent in the norm sense.
\begin{example}
	Let $\mathcal{A}$ be a Hilbert $C^{\ast}$-module over itself with the inner product $\langle a,b\rangle=ab^{\ast}$.
		Let $\{x_{i}\}_{i\in I}$ be a frame for $\mathcal{A}$ with bounds $A$ and $B$. For each $i\in I$, we define $T_{i}:\mathcal{A}\to\mathcal{A}$ by $T_{i}x=\langle x,x_{i}\rangle,\;\; \forall x\in\mathcal{A}$. $T_{i}$ is adjointable and $T_{i}^{\ast}a=ax_{i}$ for each $a\in\mathcal{A}$. And we have
	\begin{equation*}
	A\langle x,x\rangle\leq\sum_{i\in I}\langle x,x_{i}\rangle\langle x_{i},x\rangle\leq B\langle x,x\rangle, \forall x\in\mathcal{A}.
	\end{equation*}
	Then
	\begin{equation*}
	A\langle x,x\rangle\leq\sum_{i\in I}\langle T_{i}x,T_{i}x\rangle\leq B\langle x,x\rangle, \forall x\in\mathcal{A}.
	\end{equation*}
	So $\{T_{i}\}_{i\in I}$ is an operator frame in $\mathcal{A}$ with bounds $A$ and $B$.
\end{example}
\begin{example}
	Let $\mathcal{H}$ and $\mathcal{K}$ be separable Hilbert spaces and let $B(\mathcal{H,K)}$ be the set of all bounded linear operators from $\mathcal{H}$ into $\mathcal{K}$. $B(\mathcal{H,K)}$ is a Hilbert $B(\mathcal{K})$-module with a $B(\mathcal{K})$-valued inner product $\langle S, T\rangle= ST^{\ast}$ for all $S, T\in B(\mathcal{H,K)}$, and with a linear operation of $B(\mathcal{K})$ on $B(\mathcal{H,K)}$ by composition of operators.

	Let $I=\mathbf{N}$ and fix $(a_{i})_{i\in\mathbf{N}}\in l^{2}(\mathbf{C})$.

	Define: $T_{i}(X)=a_{i}X, \forall X\in B(\mathcal{H,K)}, \forall i\in\mathbb{N}$. We have
	\begin{equation*}
	\sum_{i\in\mathbb{N}}\langle T_{i}x,T_{i}x\rangle=\sum_{i\in\mathbb{N}}|a_{i}|^{2}\langle X, X\rangle, \forall X\in B(\mathcal{H,K)}.
	\end{equation*}
	Then $\{T_{i}\}_{i\in\mathbb{N}}$ is $\sum_{i\in\mathbb{N}}|a_{i}|^{2}$-tight operator frame.
\end{example}
Now we define the $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$.
\begin{definition}
	A family of adjointable operators $\{T_{i}\}_{i\in I}$ on a Hilbert $\mathcal{A}$-module $\mathcal{H}$ over a unital $C^{\ast}$-algebra is said to be an $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$, if there exists two strictly nonzero elements $A$ and $B$ in $\mathcal{A}$ such that
	\begin{equation}\label{eqq33}
		A\langle x,x\rangle A^{\ast}\leq\sum_{i\in I}\langle T_{i}x,T_{i}x\rangle\leq B\langle x,x\rangle B^{\ast}, \forall x\in\mathcal{H}.
	\end{equation}
	The elements $A$ and $B$ are called lower and upper bounds of the $\ast$-operator frame, respectively. If $A=B=\lambda$, the $\ast$-operator frame is $\lambda$-tight. If $A = B = 1_{\mathcal{A}}$, it is called a normalized tight $\ast$-operator frame or a Parseval $\ast$-operator frame. If only upper inequality of \eqref{eqq33} hold, then $\{T_{i}\}_{i\in i}$ is called an $\ast$-operator Bessel sequence for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$.
\end{definition}
We mentioned that the set of all of operator frames for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ can be considered
as a subset of $\ast$-operator frame. To illustrate this, let $\{T_{j}\}_{i\in I}$ be an operator frame for Hilbert $\mathcal{A}$-module $\mathcal{H}$
with operator frame real bounds $A$ and $B$. Note that for $x\in\mathcal{H}$,
\begin{equation*}
(\sqrt{A})1_{\mathcal{A}}\langle x,x\rangle_{\mathcal{A}}(\sqrt{A})1_{\mathcal{A}}\leq\sum_{i\in I}\langle T_{i}x,T_{i}x\rangle\leq(\sqrt{B})1_{\mathcal{A}}\langle x,x\rangle_{\mathcal{A}}(\sqrt{B})1_{\mathcal{A}}.
\end{equation*}
Therefore, every operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ with real bounds $A$ and $B$ is an $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ with $\mathcal{A}$-valued $\ast$-operator frame bounds $(\sqrt{A})1_{\mathcal{A}}$ and $(\sqrt{B})1_{\mathcal{B}}$.
\begin{example}
	Let $\mathcal{A}$ be a Hilbert $C^{\ast}$-module over itself with the inner product $\langle a,b\rangle=ab^{\ast}$.
	Let $\{x_{i}\}_{i\in I}$ be an $\ast$-frame for $\mathcal{A}$ with bounds $A$ and $B$, respectively. For each $i\in I$, we define $T_{i}:\mathcal{A}\to\mathcal{A}$ by $T_{i}x=\langle x,x_{i}\rangle,\;\; \forall x\in\mathcal{A}$. $T_{i}$ is adjointable and $T_{i}^{\ast}a=ax_{i}$ for each $a\in\mathcal{A}$. And we have
	\begin{equation*}
	A\langle x,x\rangle A^{\ast}\leq\sum_{i\in I}\langle x,x_{i}\rangle\langle x_{i},x\rangle\leq B\langle x,x\rangle B^{\ast}, \forall x\in\mathcal{A}.
	\end{equation*}
	Then
	\begin{equation*}
	A\langle x,x\rangle A^{\ast}\leq\sum_{i\in I}\langle T_{i}x,T_{i}x\rangle\leq B\langle x,x\rangle B^{\ast}, \forall x\in\mathcal{A}.
	\end{equation*}
	So $\{T_{i}\}_{i\in I}$ is an $\ast$-operator frame in $\mathcal{A}$ with bounds $A$ and $B$, respectively.
\end{example}
\begin{remark}
The examples $3.3$ and $3.4$ in \cite{A} are examples of $\ast$-operator frame.
\end{remark}
Similar to $\ast$-frames, we introduce the $\ast$-operator frame transform and $\ast$-frame operator and establish some properties.
\begin{theorem} \label{2.3}
	Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame with lower and upper bounds $A$ and $B$, respectively. The $\ast$-operator frame transform $R:\mathcal{H}\rightarrow l^{2}(\mathcal{H})$ defined by $Rx=\{T_{i}x\}_{i\in I}$ is injective and closed range adjointable $\mathcal{A}$-module map and $\|R\|\leq\|B\|$. The adjoint operator $R^{\ast}$ is surjective and it is given by $R^{\ast}(\{x_{i}\}_{i\in I})=\sum_{i\in I}T_{i}^{\ast}x_{i}$ for all $\{x_{i}\}_{i\in I}$ in $l^{2}(\mathcal{H})$.
\end{theorem}
\begin{proof}
	By the definition of norm in $l^{2}(\mathcal{H})$
	\begin{equation}\label{3.3}
	\|Rx\|^{2}=\|\sum_{i\in I}\langle T_{i}x,T_{i}x\rangle\|\leq\|B\|^{2}\|\langle x,x\rangle\|, \forall x\in\mathcal{H}.
	\end{equation}
	This inequality implies that $R$ is well defined and $\|R\|\leq\|B\|$. Clearly, $R$ is a linear $\mathcal{A}$-module map. We now show that the range of $R$ is closed. Let $\{Rx_{n}\}_{n\in\mathbb{N}}$ be a sequence in the range of $R$ such that $\lim_{n\to\infty}Rx_{n}=y$. For $n, m\in\mathbb{N}$, we have
	\begin{equation*}
	\|A\langle x_{n}-x_{m},x_{n}-x_{m}\rangle A^{\ast}\|\leq\|\langle R(x_{n}-x_{m}),R(x_{n}-x_{m})\rangle\|=\|R(x_{n}-x_{m})\|^{2}.
	\end{equation*}
		Since $\{Rx_{n}\}_{n\in\mathbb{N}}$ is Cauchy sequence in $\mathcal{H}$, then

	$\|A\langle x_{n}-x_{m},x_{n}-x_{m}\rangle A^{\ast}\|\rightarrow0$, as $n,m\rightarrow\infty.$

	Note that for $n, m\in\mathbb{N}$,
	\begin{align*}
\|\langle x_{n}-x_{m},x_{n}-x_{m}\rangle\|&=\|A^{-1}A\langle x_{n}-x_{m},x_{n}-x_{m}\rangle A^{\ast}(A^{\ast})^{-1}\|\\
&\leq\|A^{-1}\|^{2}\|A\langle x_{n}-x_{m},x_{n}-x_{m}\rangle A^{\ast}\|.
	\end{align*}
	Therefore the sequence $\{x_{n}\}_{n\in\mathbb{N}}$ is Cauchy and hence there exists $x\in U$ such that $x_{n}\rightarrow x$ as $n\rightarrow\infty$. Again by \eqref{3.3} we have $\|R(x_{n}-x)\|^{2}\leq\|B\|^{2}\|\langle x_{n}-x,x_{n}-x\rangle\|$.

	Thus $\|Rx_{n}-Rx\|\rightarrow0$ as $n\rightarrow\infty$ implies that $Rx=y$. It concludes that the range of $R$ is closed. Next we show that $R$ is injective. Suppose that $x\in\mathcal{H}$ and $Rx=0$. Note that $A\langle x,x\rangle A^{\ast}\leq\langle Rx,Rx\rangle$ then $\langle x,x\rangle=0$ so $x=0$ i.e. $R$ is injective.

	For $x\in\mathcal{H}$ and $\{x_{i}\}_{i\in I}\in l^{2}(\mathcal{H})$ we have $$\langle Rx, \{x_{i}\}_{i\in I}\rangle=\langle \{T_{i}x\}_{i\in I}, \{x_{i}\}_{i\in I}\rangle=\sum_{i\in I}\langle T_{i}x, x_{i}\rangle=\sum_{i\in I}\langle x, T_{i}^{\ast}x_{i}\rangle=\langle x, \sum_{i\in I}T_{i}^{\ast}x_{i}\rangle.$$
	Then $R^{\ast}(\{x_{i}\}_{i\in I})=\sum_{i\in I}T_{i}^{\ast}x_{i}$. By injectivity of $R$, the operator $R^{\ast}$ has closed range and $\mathcal{H}=range(R^{\ast})$, which completes the proof.
\end{proof}
Now we define $\ast$-frame operator and studies some of its properties.
\begin{definition}
Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame with $\ast$-operator frame transform $R$ and lower and upper bounds $A$ and $B$, respectively. The $\ast$-frame operator $S:\mathcal{H}\to\mathcal{H}$ is defined by $Sx=R^{\ast}Rx=\sum_{i\in I}T_{i}^{\ast}T_{i}x,\;\;\forall x\in\mathcal{H}$.
\end{definition}
\begin{theorem}
	The $\ast$-operator frame $S$ is bounded, positive, self-adjoint, invertible and $\|A^{-1}\|^{-2}\leq\|S\|\leq\|B\|^{2}$.
\end{theorem}
\begin{proof}
	By definition we have, $\forall x, y\in\mathcal{H}$:
	\begin{align*}
	\langle Sx,y\rangle&=\left\langle\sum_{i\in I}T_{i}^{\ast}T_{i}x,y\right\rangle\\
	&=\sum_{i\in I}\langle T_{i}^{\ast}T_{i}x,y\rangle\\
	&=\sum_{i\in I}\langle x,T_{i}^{\ast}T_{i}y\rangle \\
	&=\left\langle x,\sum_{i\in I}T_{i}^{\ast}T_{i}y\right\rangle\\
	&=\langle x,Sy\rangle.
	\end{align*}
	Then $S$ is a selfadjoint.

	By Lemma \ref{2.8} and Theorem \ref{2.3}, $S$ is invertible. Clearly $S$ is positive.

	By definition of an $\ast$-operator frame we have
	\begin{equation*}
	A\langle x,x\rangle A^{\ast}\leq\sum_{i\in I}\langle T_{i}x,T_{i}x\rangle\leq B\langle x,x\rangle B^{\ast}.
	\end{equation*}
	So
	\begin{equation*}
	A\langle x,x\rangle A^{\ast}\leq\langle Sx,x\rangle\leq B\langle x,x\rangle B^{\ast}.
	\end{equation*}
	This give
	\begin{equation*}
	\|A^{-1}\|^{-2}\|x\|^{2}\leq\|\langle Sx,x\rangle\|\leq\|B\|^{2}\|x\|^{2}, \forall x\in\mathcal{H}.
	\end{equation*}
	If we take supremum on all $x\in\mathcal{H}$, where $\|x\|\leq1$, then $\|A^{-1}\|^{-2}\leq\|S\|\leq\|B\|^{2}$.
\end{proof}
\begin{corollary}
Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame with $\ast$-operator frame transform $R$ and lower and upper bounds $A$ and $B$, respectively. Then $\{T_{i}\}_{i\in I}$ is an operator frame for $\mathcal{H}$ with lower and upper bounds $\|(R^{\ast}R)^{-1}\|^{-1}$ and $\|R\|^{2}$, respectively.
\end{corollary}
\begin{proof}
	By Theorem \ref{2.3}, $R$ is injective and has closed range and by Lemma \ref{2.8}
	$$\|(R^{\ast}R)^{-1}\|^{-1}I_{\mathcal{H}}\leq R^{\ast}R\leq\|R\|^{2}I_{\mathcal{H}}.$$
	So$$\|(R^{\ast}R)^{-1}\|^{-1}\langle x, x\rangle\leq \sum_{i\in I}\langle T_{i}x, T_{i}x\rangle\leq\|R\|^{2}\langle x, x\rangle,\;\;\forall x\in\mathcal{H}.$$
	Then $\{T_{i}\}_{i\in I}$ is an operator frame for $\mathcal{H}$ with lower and upper bounds $\|(R^{\ast}R)^{-1}\|^{-1}$ and $\|R\|^{2}$, respectively.
\end{proof}
The following theorem is similar to theorem 2.3 in \cite{Ali}.
\begin{theorem} \label{Th3.9}
	Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame for $\mathcal{H}$, with lower and upper bounds $A$ and $B$, respectively and with $\ast$-frame operator $S$. Let $\theta\in End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be injective and has a closed range. Then $\{T_{i}\theta\}_{i\in I}$ is an $\ast$-operator frame for $\mathcal{H}$ with $\ast$-frame operator $\theta^{\ast}S\theta$ with bounds $\|(\theta^{\ast}\theta)^{-1}\|^{-\frac{1}{2}}A$, $\|\theta\|B$.
\end{theorem}
\begin{proof}
	We have
	\begin{equation}\label{eq11}
		A\langle\theta x, \theta x\rangle A^{\ast}\leq\sum_{i\in I}\langle T_{i}\theta x, T_{i}\theta x\rangle\leq B\langle\theta x, \theta x\rangle B^{\ast}, \forall x\in\mathcal{H}.
	\end{equation}
	Using Lemma \ref{2.8}, we have $\|(\theta^{\ast}\theta)^{-1}\|^{-1}\langle x,x\rangle\leq\langle \theta x,\theta x\rangle$, $\forall x\in\mathcal{H}$. This implies
	\begin{equation}\label{eq22}
		\|(\theta^{\ast}\theta)^{-1}\|^{-\frac{1}{2}}A\langle x,x\rangle(\|(\theta^{\ast}\theta)^{-1}\|^{-\frac{1}{2}}A)^{\ast}\leq A\langle \theta x,\theta x\rangle A^{\ast}, \forall x\in\mathcal{H}.
	\end{equation}
	And we know that $\langle \theta x,\theta x\rangle\leq\|\theta\|^{2}\langle x,x\rangle$, $\forall x\in\mathcal{H}$. This implies that
	\begin{equation}\label{eq33}
		B\langle \theta x,\theta x\rangle B^{\ast}\leq\|\theta\|B\langle x,x\rangle(\|\theta\|B)^{\ast}, \forall x\in\mathcal{H}.
	\end{equation}
	Using \eqref{eq11}, \eqref{eq22}, \eqref{eq33} we have
\begin{align*}
\|(\theta^{\ast}\theta)^{-1}\|^{-\frac{1}{2}}A\langle x,x\rangle(\|(\theta^{\ast}\theta)^{-1}\|^{-\frac{1}{2}}A)^{\ast}&\leq\sum_{i\in I}\langle T_{i}\theta x, T_{i}\theta x\rangle\\&\leq\|\theta\|B\langle x,x\rangle(\|\theta\|B)^{\ast}, \forall x\in\mathcal{H}.
\end{align*}
	So $\{T_{i}\theta\}_{i\in\mathbb{J}}$ is an $\ast$-operator frame for $\mathcal{H}$.

	Moreover for every $x\in\mathcal{H}$, we have
	$$\theta^{\ast}S\theta x=\theta^{\ast}\sum_{i\in I}T_{i}^{\ast}T_{i}\theta x=\sum_{i\in I}\theta^{\ast}T_{i}^{\ast}T_{i}\theta x=\sum_{i\in I}(T_{i}\theta)^{\ast}(T_{i}\theta)x.$$ This completes the proof.
\end{proof}
\begin{corollary}
Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame for $\mathcal{H}$, with $\ast$-frame operator $S$. Then $\{T_{i}S^{-1}\}_{i\in I}$ is an $\ast$-operator frame for $\mathcal{H}$.
\end{corollary}
\begin{proof}
	Result from theorem \ref{Th3.9} by taking $\theta=S^{-1}$.
\end{proof}
\begin{corollary}
	Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame for $\mathcal{H}$, with $\ast$-frame operator $S$. Then $\{T_{i}S^{-\frac{1}{2}}\}_{i\in I}$ is a Parseval $\ast$-operator frame for $\mathcal{H}$.
\end{corollary}
\begin{proof}
	Result from theorem \ref{Th3.9} by taking $\theta=S^{-\frac{1}{2}}$.
\end{proof}
The following theorem is a generalization of theorem 2.3 in \cite{Ali}.
\begin{theorem} \label{.}
	Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame for $\mathcal{H}$, with lower and upper bounds $A$ and $B$, respectively. Let $\theta\in End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be surjective. Then $\{\theta T_{i}\}_{i\in I}$ is an $\ast$-operator frame for $\mathcal{H}$ with bounds $A\|(\theta\theta^{\ast})^{-1}\|^{-\frac{1}{2}}$, $B\|\theta\|$.
\end{theorem}
\begin{proof}
	By the definition of $\ast$-operator frame, we have
	\begin{equation}\label{eqq11}
	A\langle x, x\rangle A^{\ast}\leq\sum_{i\in I}\langle T_{i}x, T_{i}x\rangle\leq B\langle x, x\rangle B^{\ast}, \forall x\in\mathcal{H}.
	\end{equation}
	Using Lemma \ref{2.8}, we have
	\begin{equation}\label{eqq111} \|(\theta\theta^{\ast})^{-1}\|^{-1}\langle T_{i}x, T_{i}x\rangle\leq\langle \theta T_{i}x,\theta T_{i}x\rangle\leq\|\theta\|^{2}\langle T_{i}x, T_{i}x\rangle, \forall x\in\mathcal{H}.
	\end{equation}
	 	Using \eqref{eqq11}, \eqref{eqq111}, we have
	 \begin{align*}
	 	\|(\theta\theta^{\ast})^{-1}\|^{-\frac{1}{2}}A\langle x,x\rangle(\|(\theta\theta^{\ast})^{-1}\|^{-\frac{1}{2}}A)^{\ast}&\leq\sum_{i\in I}\langle\theta T_{i}x, \theta T_{i}x\rangle\\&\leq B\|\theta\|\langle x,x\rangle(B\|\theta\|)^{\ast}, \forall x\in\mathcal{H}.
	 \end{align*}
	 So $\{\theta T_{i}\}_{i\in\mathbb{J}}$ is an $\ast$-operator frame for $\mathcal{H}$.
\end{proof}
The following theorem is similar of the theorem $3.5$ in \cite{Ros1}.
\begin{theorem}
	Let $(\mathcal{H},\mathcal{A},\langle.,.\rangle_{\mathcal{A}})$ and $(\mathcal{H},\mathcal{B},\langle.,.\rangle_{\mathcal{B}})$ be two Hilbert $\mathcal{C^{\ast}}$-modules and let $\varphi :\mathcal{A}\longrightarrow \mathcal{B}$ be a $\ast$-homomorphism and $\theta$ be a map on $\mathcal{H}$ such that $\langle \theta x,\theta y\rangle_{\mathcal{B}}=\varphi(\langle x, y\rangle_{\mathcal{A}})$ for all $x,y\in\mathcal{H}$. Also, suppose that $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ is an $\ast$-operator frame for $(\mathcal{H},\mathcal{A},\langle.,.\rangle_{\mathcal{A}})$ with $\ast$-frame operator $S_{\mathcal{A}} $ and lower and upper $\ast$-operator frame bounds $A$, $B$  respectively. If $\theta$ is surjective and $\theta T_{i}= T_{i}\theta$ for each $i$ in $I$, then $\{T_{i}\}_{i\in I}$ is an $\ast$-operator frame for $(\mathcal{H},\mathcal{B},\langle.,.\rangle_{\mathcal{B}})$ with $\ast$-frame operator $S_{\mathcal{B}} $ and lower and upper $\ast$-operator frame bounds $\varphi(A)$, $\varphi(B)$ respectively, and $\langle S_{\mathcal{B}}\theta x,\theta y\rangle_{\mathcal{B}}=\varphi(\langle S_{\mathcal{A}}x, y\rangle_{\mathcal{A}})$.
\end{theorem}
\begin{proof} Let $y\in\mathcal{H}$ then there exists $x\in\mathcal{H}$ such that $\theta x=y$ ($\theta$ is surjective). By the definition of $\ast$-operator frames we have
	$$A\langle x,x\rangle_{\mathcal{A}} A^{\ast}\leq\sum_{i\in I}\langle T_{i}x, T_{i}x\rangle_{\mathcal{A}}\leq B\langle x,x\rangle_{\mathcal{A}} B^{\ast}.$$
	By lemma \ref{2.7} we have
	$$\varphi(A\langle x,x\rangle_{\mathcal{A}} A^{\ast})\leq\varphi(\sum_{i\in I}\langle T_{i}x, T_{i}x\rangle_{\mathcal{A}})\leq\varphi( B\langle x,x\rangle_{\mathcal{A}} B^{\ast}).$$
	By the definition of $\ast$-homomorphism we have
	$$\varphi(A)\varphi(\langle x,x\rangle_{\mathcal{A}}) \varphi(A^{\ast})\leq\sum_{i\in I}\varphi(\langle T_{i}x, T_{i}x\rangle_{\mathcal{A}})\leq\varphi( B)\varphi(\langle x,x\rangle_{\mathcal{A}}) \varphi(B^{\ast}).$$
	By the relation betwen $\theta$ and $\varphi$ we get
	$$\varphi(A)\langle \theta x,\theta x\rangle_{\mathcal{B}} \varphi(A)^{\ast}\leq\sum_{i\in I}\langle \theta T_{i}x,\theta T_{i}x\rangle_{\mathcal{B}}\leq\varphi( B)\langle\theta x,\theta x\rangle_{\mathcal{B}} \varphi(B)^{\ast}.$$
	By the relation betwen $\theta$ and $T_{i}$ we have
	$$\varphi(A)\langle \theta x,\theta x\rangle_{\mathcal{B}} \varphi(A)^{\ast}\leq\sum_{i\in I}\langle T_{i}\theta x, T_{i}\theta x\rangle_{\mathcal{B}}\leq\varphi( B)\langle\theta x,\theta x\rangle_{\mathcal{B}} \varphi(B)^{\ast}.$$

	Then
	$$
	\varphi(A)\langle  y, y\rangle_{\mathcal{B}} (\varphi(A))^{\ast}\leq\sum_{i\in I}\langle T_{i}y, T_{i}y\rangle_{\mathcal{B}}
	\leq\varphi( B)\langle y,y\rangle_{\mathcal{B}} (\varphi(B))^{\ast} , \forall y\in\mathcal{H}.
	$$
	On the other hand we have
	$$
	\aligned
	\varphi(\langle S_{\mathcal{A}}x, y\rangle_{\mathcal{A}})&=\varphi(\langle\sum_{i\in I}T_{i}^{\ast}T_{i}x,y\rangle_{\mathcal{A}})\\
	&=\sum_{i\in I}\varphi(\langle T_{i}x, T_{i}y\rangle_{\mathcal{A}})\\
	&=\sum_{i\in I}\langle\theta T_{i}x,\theta T_{i}y\rangle_{\mathcal{B}}
	\\
	&=\sum_{i\in I}\langle T_{i}\theta x, T_{i}\theta y\rangle_{\mathcal{B}}\\
	&=\langle\sum_{i\in I}T_{i}^{\ast}T_{i}\theta x,\theta y\rangle_{\mathcal{B}}\\
	&=\langle S_{\mathcal{B}}\theta x,\theta y\rangle_{\mathcal{B}}.
	\endaligned
	$$
	Which completes the proof.
\end{proof}
\section{Tensor Product}
 Suppose that $\mathcal{A}, \mathcal{B}$ are unital $C^{\ast}$-algebras and  $\mathcal{A}\otimes\mathcal{B}$ is the completion of $\mathcal{A}\otimes_{alg}\mathcal{B}$ with the spatial norm. $\mathcal{A}\otimes\mathcal{B}$ is the spatial tensor product of $\mathcal{A}$ and $\mathcal{B}$, also suppose that $\mathcal{H}$ is a Hilbert $\mathcal{A}$-module and $\mathcal{K}$ is a Hilbert $\mathcal{B}$-module. We want to define $\mathcal{H}\otimes\mathcal{K}$ as a Hilbert $(\mathcal{A}\otimes\mathcal{B})$-module. Start by forming the algebraic tensor product $\mathcal{H}\otimes_{alg}\mathcal{K}$ of the vector spaces $\mathcal{H}$, $\mathcal{K}$ (over $\mathbb{C}$). This is a left module over $(\mathcal{A}\otimes_{alg}\mathcal{B})$ (the module action being given by $(a\otimes b)(x\otimes y)=ax\otimes by$ $(a\in\mathcal{A},b\in\mathcal{B},x\in\mathcal{H},y\in\mathcal{K})$). For $(x_{1},x_{2}\in\mathcal{H},y_{1},y_{2}\in\mathcal{K})$ we define $ \langle x_{1}\otimes y_{1},x_{2}\otimes y_{2}\rangle_{\mathcal{A}\otimes\mathcal{B}}=\langle x_{1},x_{2}\rangle_{\mathcal{A}}\otimes\langle y_{1},y_{2}\rangle_{\mathcal{B}} $. We also know that for $ z=\sum_{i=1}^{n}x_{i}\otimes y_{i} $ in $\mathcal{H}\otimes_{alg}\mathcal{K}$ we have $ \langle z,z\rangle_{\mathcal{A}\otimes\mathcal{B}}=\sum_{i,j}\langle x_{i},x_{j}\rangle_{\mathcal{A}}\otimes\langle y_{i},y_{j}\rangle_{\mathcal{B}}\geq0 $ and $ \langle z,z\rangle_{\mathcal{A}\otimes\mathcal{B}}=0 $ iff $z=0$.
This extends by linearity to an $(\mathcal{A}\otimes_{alg}\mathcal{B})$-valued sesquilinear form on $\mathcal{H}\otimes_{alg}\mathcal{K}$, which makes $\mathcal{H}\otimes_{alg}\mathcal{K}$ into a semi-inner-product module over the pre-$\mathcal{C}^{\ast}$-algebra $(\mathcal{A}\otimes_{alg}\mathcal{B})$.
The semi-inner-product on $\mathcal{H}\otimes_{alg}\mathcal{K}$ is actually an inner product, see \cite{Lan}. Then $\mathcal{H}\otimes_{alg}\mathcal{K}$ is an inner-product module over the pre-$\mathcal{C}^{\ast}$-algebra $(\mathcal{A}\otimes_{alg}\mathcal{B})$, and we can perform the double completion discussed in chapter 1 of \cite{Lan} to conclude that the completion $\mathcal{H}\otimes\mathcal{K}$ of $\mathcal{H}\otimes_{alg}\mathcal{K}$ is a Hilbert $(\mathcal{A}\otimes\mathcal{B})$-module. We call $ \mathcal{H}\otimes\mathcal{K} $ the exterior tensor product of $\mathcal{H}$ and $\mathcal{K}$. With $\mathcal{H}$ , $\mathcal{K}$ as above, we wish to investigate the adjointable operators on $ \mathcal{H}\otimes\mathcal{K} $. Suppose that $S\in End_{\mathcal{A}}^{\ast}(\mathcal{H})$ and $T\in End_{\mathcal{B}}^{\ast}(\mathcal{K})$. We define a linear operator $S\otimes T$ on $ \mathcal{H}\otimes\mathcal{K} $ by $S\otimes T(x\otimes y)=Sx\otimes Ty  (x\in\mathcal{H} ,y\in\mathcal{K})$. It is a routine verification that  is $S^{\ast}\otimes T^{\ast}$ is the adjoint of $S\otimes T$, so in fact $S\otimes T\in End_{\mathcal{A\otimes B}}^{\ast}(\mathcal{H}\otimes\mathcal{K})$. For more details see \cite{Dav,Lan}. We note that if $a\in\mathcal{A}^{+}$ and $b\in\mathcal{B}^{+}$, then $a\otimes b\in(\mathcal{A}\otimes\mathcal{B})^{+}$. Plainly if $a$, $b$ are Hermitian elements of $\mathcal{A}$ and $a\geq b$, then for every positive element $x$ of $\mathcal{B}$, we have $a\otimes x\geq b\otimes x$.

Let I and J be countable index sets. Our next theorem is a generalization of theorem 2.2 in \cite{Ali}.
\begin{theorem}
	Let $\mathcal{H}$ and $\mathcal{K}$ be two Hilbert $C^{\ast}$-modules over unitary $C^{\ast}$-algebras $\mathcal{A}$ and $\mathcal{B}$, respectively. Let $\{\Lambda_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ and $\{\Gamma_{j}\}_{j\in J}\subset End_{\mathcal{B}}^{\ast}(\mathcal{K})$ be two $\ast$-operator frames for $\mathcal{H}$ and $\mathcal{K}$ with $\ast$-frame operators $S_{\Lambda}$ and $S_{\Gamma}$ and $\ast$-operator frame bounds $(A,B)$ and $(C,D)$ respectively. Then $\{\Lambda_{i}\otimes\Gamma_{j}\}_{i\in I,j\in J}  $ is an $\ast$-operator frame for Hibert $\mathcal{A}\otimes\mathcal{B}$-module $\mathcal{H}\otimes\mathcal{K}$ with $\ast$-frame operator $ S_{\Lambda}\otimes S_{\Gamma}$ and lower and upper $\ast$-operator frame bounds $A\otimes C$ and $ B\otimes D $, respectively.
\end{theorem}
\begin{proof}
By the definition of $\ast$-operator frames $\{\Lambda_{i}\}_{i\in I} $ and $\{\Gamma_{j}\}_{j\in J}$ we have
$$A\langle x,x\rangle_{\mathcal{A}} A^{\ast}\leq\sum_{i\in I}\langle \Lambda_{i}x,\Lambda_{i}x\rangle_{\mathcal{A}}\leq B\langle x,x\rangle_{\mathcal{A}} B^{\ast} , \forall x\in\mathcal{H},$$
and
$$C\langle y,y\rangle_{\mathcal{B}} C^{\ast}\leq\sum_{j\in J}\langle \Gamma_{j}y,\Gamma_{j}y\rangle_{\mathcal{B}}\leq D\langle y,y\rangle_{\mathcal{B}} D^{\ast} , \forall y\in\mathcal{K}.$$
Therefore
$$
\aligned
&(A\langle x,x\rangle_{\mathcal{A}} A^{\ast})\otimes (C\langle y,y\rangle_{\mathcal{B}} C^{\ast})\\&\leq\sum_{i\in I}\langle \Lambda_{i}x,\Lambda_{i}x\rangle_{\mathcal{A}}\otimes\sum_{j\in J}\langle \Gamma_{j}y,\Gamma_{j}y\rangle_{\mathcal{B}}\\
&\leq (B\langle x,x\rangle_{\mathcal{A}} B^{\ast})\otimes (D\langle y,y\rangle_{\mathcal{B}} D^{\ast}) , \forall x\in\mathcal{H} ,\forall y\in\mathcal{K}.
\endaligned
$$
Then
$$
\aligned
&(A\otimes C)(\langle x,x\rangle_{\mathcal{A}}\otimes\langle y,y\rangle_{\mathcal{B}}) (A^{\ast}\otimes C^{\ast})\\&\leq\sum_{i\in I,j\in J}\langle \Lambda_{i}x,\Lambda_{i}x\rangle_{\mathcal{A}}\otimes\langle \Gamma_{j}y,\Gamma_{j}y\rangle_{\mathcal{B}}\\
&\leq (B\otimes D)(\langle x,x\rangle_{\mathcal{A}}\otimes\langle y,y\rangle_{\mathcal{B}}) (B^{\ast}\otimes D^{\ast}) , \forall x\in\mathcal{H} ,\forall y\in\mathcal{K}.
\endaligned
$$
Consequently we have
$$
\aligned
&(A\otimes C)\langle x\otimes y,x\otimes y\rangle_{\mathcal{A\otimes B}} (A\otimes C)^{\ast}\\&\leq\sum_{i\in I,j\in J}\langle\Lambda_{i}x\otimes\Gamma_{j}y,\Lambda_{i}x\otimes\Gamma_{j}y\rangle_{\mathcal{A\otimes B}} \\
&\leq (B\otimes D)\langle x\otimes y,x\otimes y\rangle_{\mathcal{A\otimes B}} (B\otimes D)^{\ast}, \forall x\in\mathcal{H}, \forall y\in\mathcal{K}.
\endaligned
$$
Then for all $x\otimes y\in\mathcal{H\otimes K}$ we have
$$
\aligned
&(A\otimes C)\langle x\otimes y,x\otimes y\rangle_{\mathcal{A\otimes B}} (A\otimes C)^{\ast}\\&\leq\sum_{i\in I,j\in J}\langle(\Lambda_{i}\otimes\Gamma_{j})(x\otimes y),(\Lambda_{i}\otimes\Gamma_{j})(x\otimes y)\rangle_{\mathcal{A\otimes B}} \\
&\leq (B\otimes D)\langle x\otimes y,x\otimes y\rangle_{\mathcal{A\otimes B}} (B\otimes D)^{\ast}.
\endaligned
$$
The last inequality is satisfied for every finite sum of elements in $\mathcal{H}\otimes_{alg}\mathcal{K}$ and then it's satisfied for all $z\in\mathcal{H\otimes K}$. It shows that $\{\Lambda_{i}\otimes\Gamma_{j}\}_{i\in I,j\in J}  $ is $\ast$-operator frame for Hibert $\mathcal{A}\otimes\mathcal{B}$-module $\mathcal{H}\otimes\mathcal{K}$ with lower and upper $\ast$-operator frame bounds $A\otimes C$ and $ B\otimes D $, respectively.\\
By the definition of $\ast$-frame operator $S_{\Lambda}$ and $S_{\Gamma}$ we have:$$S_{\Lambda}x=\sum_{i\in I}\Lambda_{i}^{\ast}\Lambda_{i}x, \forall x\in\mathcal{H},$$
and
$$S_{\Gamma}y=\sum_{j\in J}\Gamma_{j}^{\ast}\Gamma_{j}y, \forall y\in\mathcal{K}.$$
Therefore
$$
\aligned
(S_{\Lambda}\otimes S_{\Gamma})(x\otimes y)&=S_{\Lambda}x\otimes S_{\Gamma}y\\
&=\sum_{i\in I}\Lambda_{i}^{\ast}\Lambda_{i}x\otimes\sum_{j\in J}\Gamma_{j}^{\ast}\Gamma_{j}y\\
&=\sum_{i\in I,j\in J}\Lambda_{i}^{\ast}\Lambda_{i}x\otimes\Gamma_{j}^{\ast}\Gamma_{j}y\\
&=\sum_{i\in I,j\in J}(\Lambda_{i}^{\ast}\otimes\Gamma_{j}^{\ast})(\Lambda_{i}x\otimes\Gamma_{j}y)\\
&=\sum_{i\in I,j\in J}(\Lambda_{i}^{\ast}\otimes\Gamma_{j}^{\ast})(\Lambda_{i}\otimes\Gamma_{j})(x\otimes y)\\
&=\sum_{i\in I,j\in J}(\Lambda_{i}\otimes\Gamma_{j})^{\ast})(\Lambda_{i}\otimes\Gamma_{j})(x\otimes y).
\endaligned
$$
Now by the uniqueness of $\ast$-frame operator, the last expression is equal to $S_{\Lambda\otimes\Gamma}(x\otimes y)$. Consequently we have $ (S_{\Lambda}\otimes S_{\Gamma})(x\otimes y)=S_{\Lambda\otimes\Gamma}(x\otimes y)$. The last equality is satisfied for every finite sum of elements in $\mathcal{H}\otimes_{alg}\mathcal{K}$ and then it's satisfied for all $z\in\mathcal{H\otimes K}$. It shows that $ (S_{\Lambda}\otimes S_{\Gamma})(z)=S_{\Lambda\otimes\Gamma}(z)$. So $S_{\Lambda\otimes\Gamma}=S_{\Lambda}\otimes S_{\Gamma}$.
\end{proof}
The two following theorems are a generalization of Theorem 3.5 in \cite{BA}.
\begin{theorem} \label{Th4.2}
	If $Q\in End_{\mathcal{A}}^{\ast}(\mathcal{H})$ is invertible and $\{\Lambda_{i}\}_{i\in I}\subset End_{\mathcal{A\otimes B}}^{\ast}(\mathcal{H\otimes K})$ is an $\ast$-operator frame for $\mathcal{H\otimes K}$ with lower and upper  $\ast$-operator frame bounds $A$ and $B$ respectively and $\ast$-frame operator $S$, then $\{\Lambda_{i}(Q^{\ast}\otimes I)\}_{i\in I} $ is an $\ast$-operator frame for $\mathcal{H\otimes K}$ with lower and upper $\ast$-operator frame bounds $\|Q^{\ast-1}\|^{-1}A$ and $\|Q\|B$ respectively and $\ast$-frame operator $(Q\otimes I)S(Q^{\ast}\otimes I)$.
\end{theorem}
\begin{proof}
	Since $Q\in End_{\mathcal{A}}^{\ast}(\mathcal{H})$, $Q\otimes I\in End_{\mathcal{A\otimes B}}^{\ast}(\mathcal{H\otimes K})$ with inverse $Q^{-1}\otimes I$. It is obvious that the adjoint of $Q\otimes I$ is $Q^{\ast}\otimes I$. An easy calculation shows that for every elementary tensor $x\otimes y$,
	$$
	\aligned
	\|(Q\otimes I)(x\otimes y)\|^{2}&=\|Q(x)\otimes y\|^{2}\\
	&=\|Q(x)\|^{2}\|y\|^{2}\\
	&\leq\|Q\|^{2}\|x\|^{2}\|y\|^{2}\\
	&=\|Q\|^{2}\|x\otimes y\|^{2}.
	\endaligned
	$$
	So $Q\otimes I$ is bounded, and therefore it can be extended to $\mathcal{H\otimes K}$. Similarly for $Q^{\ast}\otimes I$, hence $Q\otimes I$ is $\mathcal{A\otimes B}$-linear, adjointable with adjoint $Q^{\ast}\otimes I$. Hence for every $z\in\mathcal{H\otimes K}$ we have  $$\|Q^{\ast-1}\|^{-1}.|z|\leq|(Q^{\ast}\otimes I)z|\leq\|Q\|.|z|.$$
	By the definition of $\ast$-operator frames we have
	$$A\langle z,z\rangle_{\mathcal{A\otimes B}} A^{\ast}\leq\sum_{i\in I}\langle \Lambda_{i}z,\Lambda_{i}z\rangle_{\mathcal{A\otimes B}}\leq B\langle z,z\rangle_{\mathcal{A\otimes B}} B^{\ast}.$$
	Then
	$$
	\aligned
	&A\langle (Q^{\ast}\otimes I)z,(Q^{\ast}\otimes I)z\rangle_{\mathcal{A\otimes B}} A^{\ast}\\&\leq\sum_{i\in I}\langle \Lambda_{i}(Q^{\ast}\otimes I)z,\Lambda_{i}(Q^{\ast}\otimes I)z\rangle_{\mathcal{A\otimes B}}\\
	&\leq B\langle (Q^{\ast}\otimes I)z,(Q^{\ast}\otimes I)z\rangle_{\mathcal{A\otimes B}} B^{\ast}.
	\endaligned
	$$
	So$$\aligned&\|Q^{\ast-1}\|^{-1}A\langle z,z\rangle_{\mathcal{A\otimes B}}(\|Q^{\ast-1}\|^{-1}A)^{\ast}\\&\leq\sum_{i\in I}\langle \Lambda_{i}(Q^{\ast}\otimes I)z,\Lambda_{i}(Q^{\ast}\otimes I)z\rangle_{\mathcal{A\otimes B}}\\&\leq\|Q\|B\langle z,z\rangle_{\mathcal{A\otimes B}}(\|Q\|B)^{\ast}.\endaligned$$
	Now
	$$
	\aligned
	(Q\otimes I)S(Q^{\ast}\otimes I)&=(Q\otimes I)(\sum_{i\in I}\Lambda_{i}^{\ast}\Lambda_{i})(Q^{\ast}\otimes I)\\
	&=\sum_{i\in I}(Q\otimes I)\Lambda_{i}^{\ast}\Lambda_{i}(Q^{\ast}\otimes I)\\
	&=\sum_{i\in I}(\Lambda_{i}(Q^{\ast}\otimes I))^{\ast}\Lambda_{i}(Q^{\ast}\otimes I).
	\endaligned
	$$
	Which completes the proof.
\end{proof}
\begin{corollary}
	If $Q\in End_{\mathcal{B}}^{\ast}(\mathcal{K})$ is invertible and $\{\Lambda_{i}\}_{i\in I}\subset End_{\mathcal{A\otimes B}}^{\ast}(\mathcal{H\otimes K})$ is an $\ast$-operator frame for $\mathcal{H\otimes K}$ with lower and upper  $\ast$-operator frame bounds $A$ and $B$ respectively and $\ast$-frame operator $S$, then $\{\Lambda_{i}(I\otimes Q^{\ast})\}_{i\in I} $ is an $\ast$-operator frame for $\mathcal{H\otimes K}$ with lower and upper $\ast$-operator frame bounds $\|Q^{\ast-1}\|^{-1}A$ and $\|Q\|B$ respectively and $\ast$-frame operator $(I\otimes Q)S(I\otimes Q^{\ast})$.
\end{corollary}
\begin{proof}
Similar to the proof of the theorem \ref{Th4.2}.
\end{proof}
\bibliographystyle{amsplain}

config: arxiv


TITLE: Find the value of P[X>Y<Z>U] from a given joint distribution
QUESTION [2 upvotes]: Suppose the joint distribution is given by some function $f(x, y, z, u)$. Then find the value of $P[X>Y<Z>U]$ assuming that $f$ is non-zero only for $x > 0, y>0, z>0, u>0$
My approach: Simply integrate the joint distribution function, taking care of appropriate limits for all the possible cases.
However, I keep getting some form of "circular limits" on the integral.
For example, one arrangement of the values for $x, y, z, u$ can be $y < u < x < z$
and the integral looks like $\int_{0}^{z}\int_{u}^{\infty}\int_{0}^{x}\int_{y}^{\infty}f(x, y, z, u)dx dy dz du$
Here, the limits for $z$ and $u$ are dependent on each other's value. Similarly for $x$ and $y$.
What is the way to deal with such integral limits? Forgive me if this a silly doubt, but I've been out of practice for a while. Please note that the RV are not independent.
Thanks

REPLY [1 votes]: Writing the integrals in alternate form to make the manipulations easier to see:
$$\begin{align}&P[X>Y<Z>U]\\[2ex]
&=P[Y<\min(X,Z),\ \ U<Z]\\[2ex]
&=\int_0^\infty dz \int_0^\infty dx \int_0^{\min(x,z)} dy \int_0^z du\, f(x,y,z,u)\\[2ex]
&=\int_0^\infty dz \left\{\int_0^z dx +\int_z^\infty dx\right\}\int_0^{\min(x,z)} dy \int_0^z du\, f(x,y,z,u)\\[2ex]
&=\left\{\int_0^\infty dz \int_0^z dx \int_0^x dy \int_0^z du\, + \int_0^\infty dz \int_z^\infty dx \int_0^z dy \int_0^z du\right\}\,f(x,y,z,u)\\[2ex]
\end{align}$$


TITLE: On integer solutions of $2\sqrt{\sqrt{2x}+\sqrt{y}}=\sqrt{x}+\sqrt{2y}$
QUESTION [4 upvotes]: $\textsf{Background}$
From the double-angle formula $\cos2\alpha=2\cos^2\alpha-1$, we can get $$\cos15^\circ=\sqrt{\frac{1+\cos30^\circ}2}=\frac{\sqrt{2+\sqrt3}}2$$ but we also know that it is equivalent to $\dfrac{\sqrt6+\sqrt2}4$.

This is an example of an equality such that $$2\sqrt{\sqrt{2x}+\sqrt{y}}=\sqrt{x}+\sqrt{2y}$$ after some rearranging. We can write $y$ in terms of $x$ without much bother: $$y=2+\frac x2+2\sqrt{1+\sqrt{2x}}-\sqrt{2x(1+\sqrt{2x})}.$$ But when are $x$ and $y$ integers? Here is a plot of the curve:
Some obvious solutions are $(0,0)$, $(0,4)$, $(2,3)$ and $(32,0)$. Of course, $x=2k$ for some integer $k$, leaving us with $$y=2+k+2\sqrt{1+2\sqrt k}-2\sqrt{k(1+2\sqrt k)}$$ Hence this boils down to finding $k$ such that $$(1-\sqrt k)\sqrt{1+2\sqrt k}$$ is an integer. Any advances on this?

REPLY [0 votes]: The stated condition leads easily to
$$(\sqrt x + \sqrt{2y} - 2\sqrt2)^2 = 8 - 4\sqrt y.$$
Hence $0 \le y \le 4.$ Trying each value of $y$ in turn, one finds that the only solutions with $x$ also an integer are the four already stated.


TITLE: Why does intergral of a double power only work when there is an another x in the intergral
QUESTION [1 upvotes]: why does this work and give a nice answer but this gives a weird answer.
For context, I was asked this question and didn't understand how the answer was explained because I don't understand why how the above pics work.

REPLY [2 votes]: Hint:
You can substitute $u=x^9$ then $du=9x^8dx$.
So that the integral becomes:
$$I=\int 15 x^{17}e^{x^9}dx=\dfrac {5}{3}\int ue^udu$$
Integrate by part.


TITLE: How prove this inequality $\sum\limits_{cyc}\frac{x^a\ln{x}}{(x^a+y+z)^2}\ge 0$
QUESTION [10 upvotes]: Question:

let $x,y,z$ be postive numbers,and such $xyz\ge 1$,and such $a$ is real numbers.show that
  $$\dfrac{x^a\ln{x}}{(x^a+y+z)^2}+\dfrac{y^a\ln{y}}{(y^a+x+z)^2}+\dfrac{z^a\ln{z}}{(
z^a+x+y)^2}\ge 0$$

I think this can use Integral to solve it.because
$$(x^a)'_{a}=x^a\ln{x}$$
so
$$\left(-\dfrac{1}{x^a+y+z}\right)'_{a}=\dfrac{x^a\ln{x}}{(x^a+y+z)^2}$$
so we  only prove this
$$-\left(\dfrac{1}{x^a+y+z}+\dfrac{1}{y^a+z+x}+\dfrac{1}{z^a+x+y}\right)'_{a}\ge 0,a\in R$$
then I can't use this condition $xyz\ge 1$,
Thank you

REPLY [2 votes]: I have a proof that applies for $a\geq -1$.
We start by making the change of variables $X = \exp(x)$ etc. The condition $xyz\geq 1$ becomes $X+Y+Z\geq 0$ and the ineqality we want to prove reads
$\sum_{\rm cyc}\frac{X\exp(aX)}{(\exp(aX) + \exp(Y) + \exp(Z))^2} \geq 0$
We combined the three terms to form a single fraction. The denominator is a product of squares and always positive. The nominator will consist of the three terms $Xf_1 + Yf_2 + Zf_3$
where
$f_1 = \exp(aX)(\exp(X)+\exp(aY)+\exp(Z))^2(\exp(X)+\exp(Y)+\exp(aZ))^2$
and $f_2,f_3$ are given by the same expression with a cyclic permutation of the variables.
To proceed we want to expand this product into single exponential terms and then use the approximation $\exp(x) > 1 + x$ on each of the terms. To avoid doing this manually we can use a trick: the approximation $\exp(x) > 1 + x$ implies $f_1 > T(f_1)$ where $T(f_1)$ is the first order Taylor series of $f_1$ around $(X,Y,Z)=(0,0,0)$
$T(f_1) = f_1(0,0,0) + \left.\frac{df_1}{dX}\right|_{(0,0,0)}X + \left.\frac{df_1}{dY}\right|_{(0,0,0)}Y + \left.\frac{df_1}{dZ}\right|_{(0,0,0)}Z$
which reads
$T(f_1) = 27(3 + (3a+4)X + 2(a+1)(Y+Z))$
and similar for $f_2,f_3$. Summing up for the three terms we find that the numerator satisfies
$Xf_1 + Yf_2 + Zf_3 \geq 27(2\alpha^2(a+1) + 3\alpha + (a+2)(X^2+Y^2+Z^2)) \geq 0$
at least for $a\geq -1/2$ since $\alpha \equiv X+Y+Z \geq 0$.
Equality is found only for $X=Y=Z=0$ or in terms of the original variables: $x=y=z=1$.


TITLE: Power series of $f(z)^8$
QUESTION [6 upvotes]: Goal: If $f(z)^8$ is analytic on some domain D and if $f(z)$ is continuous on domain $D$ with $f(0) = 0$, then the power series $f(z)^8 = \sum a_nz^n$ will begin with $n$ divisible by $8$
My attempt:
Allow $g(z) = f(z)^8$ which is analytic on some domain D. Then, we use Cauchy integral to calculate our $a_n$.
$$g^{(n)}(z) = a_n = \frac{n!}{2\pi i}\int_{D} \frac{g(w)}{(w-z)^{n+1}} dw$$
Now, I notice that we can split the integrand up, but I'm not quite sure if it's helpful
$$\frac{n!}{2\pi i} \int_D \frac{1}{(w-z)^{j}} \frac{f(w)^8}{(w-z)^{8k+1}} dw$$.
Where $j + 8k + 1 = n + 1$
Can anyone tell me if this is going in the right direction or any hints?

REPLY [2 votes]: Your assumption of continuity of $f$ along with the analyticity of $f^8$ is enough to conclude $f$ is analytic. For example, see the following:

If $z\mapsto f(z)^n$ is analytic then $f$ is analytic
If $f^2$ is an analytic function then so is $f$

The former gives the result immediately but is a tad tougher to prove, while the latter gives the result after noting $f$ being continuous implies $f^2$ and  $f^4$ are continuous, at which point we apply the latter result three times.
We thus know $f$ is analytic, so $$f(z) = \sum_{k=0}^\infty c_kz^k$$
Applying your edit that $f(0)=0,$ we must have $c_0 = 0.$ Let us first assume $c_1 \neq 0,$ so $$f(z) = \sum_{k=1}^\infty c_kz^k = c_1 z + \mathcal{O}(z^2)$$
Thus, by the Binomial Theorem, we have that
$$f(z)^8 = (c_1 z + \mathcal{O}(z^2))^8 = (c_1 z)^8 + \mathcal{O}(z^9)$$
Which proves the claim whenever $c_1 \neq 0.$
We now use this idea to prove the general case. Suppose $c_k = 0$ for $0 \le k \le n-1,$ but $c_n \neq 0.$ Then:
$$f(z)^8 = \left(\sum_{k=n}^\infty c_kz^k\right)^8 = (c_n z^n + \mathcal{O}(z^{n+1}))^8 = (c_n z^n)^8 + \mathcal{O}(z^{8(n+1)})$$
Which proves that the leading nonzero term will have a power divisible by $8.$ This of course extends readily to other powers of $f.$


TITLE: Is Bayes' theorem, is the denominator always assumed to be non-zero?
QUESTION [0 upvotes]: Is Bayes' theorem, is the denominator always assumed to be non-zero or, in certain circumstances, you can prove that it is actually never zero?

REPLY [0 votes]: Let $A$ and $B$ be two random variables taking values in $\Omega_X$ and $\Omega_Y$. Then:
\begin{equation}
P(A=a | B=b) = \frac{P(A=a, B=b)}{\sum\limits_{a' \in \Omega_X} P(A=a', B=b)}
\end{equation}
In this form I think it is easier to see why the denominator shouldn't be zero: it can be interpreted as a normalization factor. If it is zero it means that every term in the sum is zero since probabilities are non negative, that is in no circumstances the events $A=a$ and $B=b$ can occur together. If this is the case, the probability of $A=a$ given $B=b$ is ill defined


TITLE: Reference Request: Structure constants for G2
QUESTION [9 upvotes]: Let $G$ be a split semisimple real Lie group in characteristic zero, and let $B=TU$ be a Borel subgroup with unipotent radical $U$ and Levi $T$.  Fix an ordering on the roots $\Phi^+$ of $T$ in $U$, and for each root subgroup $U_{\alpha}$ of $U$, let $u_{\alpha}: \mathbb R \rightarrow U_{\alpha}$ be an isomorphism.
For all $\alpha, \beta \in \Phi^+$, there exist unique real numbers $C_{\alpha,\beta,i,j}$ (depending on the $u_{\alpha}$ and the ordering) such that for all $x, y \in \mathbb R$,
$$u_{\alpha}(x) u_{\beta}(y) u_{\alpha}(x)^{-1} = u_{\beta}(y) \prod\limits_{\substack{i,j>0\\ i\alpha + j \beta \in \Phi^+}} u_{i\alpha+j\beta}(C_{\alpha,\beta,i,j}x^iy^j)$$
I want to work out some examples with unipotent groups of exceptional semisimple groups, and am looking for table of structure constants for the root system G2.  Does anyone know a reference where an ordering on the roots is chosen and these constants are written down?

REPLY [5 votes]: Probably the earliest reference is the 1956-58 Chevalley seminar, available online in typed format, which has been republished in 2005 as a typeset book edited by P. Cartier: see Chapter 21. (No special assumption about the characteristic of the field is needed.)   My own later treatment of the classification of simple algebraic groups followed the same method in GTM 21 (Linear Algebraic Groups, Springer, 1975, 33.5).   A similar approach was taken in SGA3, as indicated by Peter McNamara.   (The later more elegant approach to the classification due to Takeuchi was worked out in Jantzen's book as well as Springer's textbook.)
Note too that the Lie algebra calculations were done in my earlier book GTM 9, first in the characteristic 0 setting: see 19.3.


TITLE: Generalizing Big O notation to arbitrary vector spaces
QUESTION [5 upvotes]: I'm constructing a Coq library for Big-O notation. Naturally, I'd like it to be as general as possible. The Wikipedia page on Big-O notation says

The generalization to functions taking values in any normed vector
  space is straightforward (replacing absolute values by norms)

Of course, there's no inline citation. My attempt at doing this yielded the following:
Definition big_O (f g : V -> V) : Prop :=
    ∃ k : K, 0 < k ∧ exists n0 : K, 0 < n0 ∧ ∀ n : V, n0 ≤ ∥n∥ -> ∥f n∥ ≤ k * ∥g n∥.

Which translates to the following in informal language:
Definition 1. Given a vector space $V$ with (semi)norm $\lVert-\rVert$ over a totally ordered field $(K,\leq)$ and functions $f,g:V\to V$, we say that  $f\in O(g)$ iff there exists some positive $k$ and $n_0$ in $K$ such that for all vectors $n\in V$ with $n_0 \leq \lVert n\rVert$, $\rVert f(n)\rVert\leq k\cdot\lVert g(n)\rVert$.
However, I feel like the following definition is somewhat more elegant and general:
Definition big_O (f g : V -> V) : Prop :=
    ∃ k : K, k ≠ 0 ∧ ∃ n0 : V, ∀ n : V, ∥n0∥ ≤ ∥n∥ -> ∥f n∥ ≤ ∥k · g n∥.

Which translates to the following:
Definition 2. Given a vector space $V$ with (semi)norm $\lVert-\rVert$ over a totally ordered field $(K,\leq)$ and functions $f,g:V\to V$, we say that  $f\in O(g)$ iff there exists some nonzero $k$ and some $n_0$ in $V$ such that for all vectors $n\in V$ with $\lVert n_0\rVert \leq \lVert n\rVert$, $\lVert f(n)\rVert\leq \lVert k\cdot g(n)\rVert$.
So I guess I have a few questions:

Does anyone have a good reference for the generalization to vector spaces?
Do you think these definitions are equivalent? Any proofs or counterexamples?
This is somewhat opinion-based, but any arguments for using one or the other?

REPLY [4 votes]: Here is a straightforward generalization of the asymptotic notation to  functions taking values in normed vector spaces:
Let $X$ be a topological space, $\overline X\supseteq X$ a superspace in which $X$ is dense, and $a\in\overline X$. Let $Y$ be a normed vector space, and $f,g\colon X\to Y$. Then we say

$f(x)=O(g(x))$ as $x\to a$ if there is a neighbourhood $U\ni a$ and $q>0$ such that $\|f(x)\|\le q\left\|g(x)\right\|$ for all $x\in U\cap X$;
$f(x)=o(g(x))$ as $x\to a$ if for every $q>0$, there is a neighbourhood $U\ni a$ such that $\|f(x)\|\le q\left\|g(x)\right\|$ for all $x\in U\cap X$.

This is for standard, real-valued norms. You seem to be interested also in norms valued in nonarchimedean ordered fields $K$. In this case, there are at least two useful possibilities how to cast the definition, namely (1) by taking $q\in\mathbb Q$, or (2) by taking $q\in K$. Which one is more suitable depends on the intended use case.
Further generalizations are possible, seeing as we didn’t actually use most of the structure of $Y$.
You are apparently trying to define only the special case with $a=\infty$. Thus, you need to decide what are appropriate neighbourhoods of infinity in your spaces (complements of bounded sets? complements of compact sets?).