TITLE: Finding the Closest Point to the Centre of a Rotating Vector Field
QUESTION [0 upvotes]: I am working on the analysis of a data set that resulted from a simulation of a Tornado. I need to be able to transform the vectors for the velocities at each measured point in the space from Cartesian to Polar coordinates. That requires me to be able to located the centre of the tornado at each time step to be able to locate my origin in the Polar coordinates.
I have ran into issues where the centre is not clearly defined by obvious things such as the highest velocities (being the tornado wall) having close proximity to the lowest velocity (the eye of the tornado) as this simulation is not a simple one, it attempts to be very true to real life including other funnel formations off to the side. This meant that simply searching for a point where the vectors have no velocity or even point in a circular pattern resulted in logical errors where my centre would be way off from where I can see it should be.
The vector field, the red dot representing where my current code located the eye to be
The gradient of the same vector field shown as vector arrows
I have looked around on here but haven't been able to find much, the closest thing I could find was a topic that did get a reply but I don't really know how to apply it to my data set as I have velocity in the x and velocity in the y directions for each x and y position on my plane (only looking at one plane of height at a time) for 110 x 190 points. So to find the dot product of this to find a direction I think wouldn't work on it's own. An idea I had was to use the gradients to find a "well" in the velocities, but actually implementing a search for this has proven quite difficult. Any suggestions on how to do this no matter the size of the eye and other smaller formations that mimic but are not a centre would be greatly appreciated.

REPLY [0 votes]: Using what Stefan said above, I was inspired by the idea of using the overlap of the transformed circles to make a "probability" of my circle centre. This method would have been quite the overkill though, so I looked deeper and (with help of a friend) found an AMAZING paper on the topic! I didn't quite use the method mentioned in this paper either though.
Instead I used the curl to make my "well" that should outline the vortex wall instead of the gradients. Looking at a surface plot of the curl it seemed that I should be able to use the maximum values (the yellow volcano-like peak) for the curl to locate my centre. To do this, I decided to make a value that would capture the x highest curl values (for either direction). The value was based on how large my dataset is so that should I have many many points and a large vortex I can choose more max values for better interpolation here is what the vortex looks like with the velocity vectors. From there, I simply found the mean of the points and eliminated any outliers that fall too far from the mean and indicate an error (resulting from an outside swirling).
Finally then, I could apply what the paper linked above describes where they take the normal to the tangential of the vectors (in my case only the ones that matched the max curl criterion) and then find the intersection of those lines. From this point, since there is not an exact intersection between all points, I simply picked a point right by most of the intersections and did a double linear interpolation between the nearest 4 points to get my centre!
I hope this is useful to someone out there with this specific problem!


TITLE: Combinatorial proof for $\sum_{k=0}^p {p+q\choose k} {p+q-k\choose p-k}=2^p {p+q \choose p}$
QUESTION [1 upvotes]: I'm looking for a combinatorial proof of the identity:
$$\sum_{k=0}^p {p+q\choose k} {p+q-k\choose p-k}=2^p {p+q \choose p} \text{   (1)}$$
I'm especially curious about its relationship with this other identity:
$$\sum_{k=0}^n {p\choose k} {q\choose n-k}= {p+q \choose n} \text{   (2)}$$
Formula (2) is obvious in the sense that choosing $n$ elements from two sets of $p$ and $q$ elements respectively could be done in $n$ distinct ways, but there surely is a nuance I'm missing because I don't see how this is any different from the LHS of (1).
Thanks in advance

REPLY [1 votes]: Consider the number of strings of length $p+q$ such that

at $p$ positions there is either $0$ or $1$ and
at the remaining $q$ positions there is $2$

Now,
RHS:

Choose $p$ positions from $p+q$ positions and fill these $p$ positions with $0$'s and $1$'s resp. (The remaining positions are filled with $2$'s): $\color{blue}{\binom{p+q}{p}\cdot 2^p}$

LHS: (split the counting according to the number of occurring $0$'s)

Choose $k$ from $p+q$ positions to place there $0$'s and choose $p+q-k$ positions to place there $1$'s. (The remaining positions are filled with $2$'s): $\color{blue}{\sum_{k=0}^p\binom{p+q}{k}\binom{p+q-k}{p-k}}$


TITLE: Write the $p$-sylow subgroup as a quotient of the group.
QUESTION [0 upvotes]: Let $G$ be an abelian finite group and let $G^{(p)}$ be its unique $p$-Sylow subgroup. Is there a way to write $G^{(p)}$ as a quotient of $G$, i.e. is there a subgroup $H\subseteq G$ with $G/H\cong G^{(p)}$?

REPLY [2 votes]: Yes. A finite abelian group is always the direct product of its Sylow subgroups. I.e.
$\begin{align}G\cong G^{(p_1)}\times G^{(p_2)}\times...\times G^{(p_n)}\end{align}$
where $p_i$ are the distinct primes in the factorization of $|G|$. This follows from the structure theorem for finite abelian groups. So you can simply choose $H$ as the product of all other Sylow subgroups.


TITLE: How to show that gcd(|a|,|b|) = gcd(a,b)?
QUESTION [0 upvotes]: How to show that gcd(|a|,|b|) = gcd(a,b) ?
I tried many ways but not convinced that im doing it right. i tried to look at different scenarios where :
a,b>0
a,b<0
a>0,b<0
but I know I am doing it totally wrong.

REPLY [1 votes]: Show that every common divisor of $a$ and $b$ is also a common divisor of $|a|$ and $|b|$.  Then vice verse.  Since the set of common divisors of $a$ and $b$ is the same as the set of common divisors of $|a|$ and $|b|$, the largest element in each set has to be the same.
Another way is to invoke Bezout.  If $d=\gcd(a,b)$ then there exist integers $x$ and $y$ so that $d=ax+by$.  By adjusting the signs on $x$ and $y$ you can get a similar expression for $\gcd(|a|,|b|).$


TITLE: Local form of Ampere's law
QUESTION [1 upvotes]: How can I find the magnetic field created by an infinite straight wire of radius $R$ and current density $\vec{j}$ using only the local form of Ampere's law : $\vec{\nabla}\times \vec{B} = \mu_0 \vec{j}$ ?

REPLY [1 votes]: Outside of the wire where $r>R$, the curl of the B-field is zero (no current density) and inside, it is non-zero. Using symmetry, we know that if $\vec{j}$ only depends on r, which is often the case, then the B-field must also only depend on r. $\vec{B}=\vec{B}(r)$. If we look at the curl in cylindrical coordinates, we find that $\nabla \times \vec{B}=-\frac{dB_z(r)}{dr}\hat{\phi}+\frac{1}{r}\frac{d(rB_{\phi}(r))}{dr}\hat{z}$. Inside, we have $\vec{j}=j(r)\hat{z}$ for a current running along the length of the wire. So that means $\frac{1}{r}\frac{d(rB_{\phi}(r))}{dr}=\mu_0j(r)$. Integrate this to get your answer. Outside of the wire, we do the same thing but with $\vec{j}=0$, and impose the boundary condition at $r=R$ to fix the integration constant. It's definitely easier to use the integral form for this type of problem though.


TITLE: Changing the scalar curvature (k = 0,+1,-1) with coordinate transformations?
QUESTION [0 upvotes]: I would like to prove that I can (or can't) change curvature of space, k = 0,+1,-1, via general coordinate transformations, which in principle can mix space and time coordinates together.

REPLY [0 votes]: By the term "space" I meant this: If 4D metric is given by
G_μν = diag[1,-a(t)^2/1-kr^2,-a^2(t)r^2,-a(t)^2r^2sin(Θ)^2],
in some coordinates (t,r,Θ,ϕ), then a 3D "subspace" is simply
G_ij = diag[-a(t)^2/1-kr^2,-a^2(t)r^2,-a(t)^2r^2sin(Θ)^2],
which seems like your other case, surface of t=const.


text:
\section{Three-strand pretzel knots}\label{sec:pretzels}
\begin{proof}[Proof of Theorem~\ref{thm:not left}]
Let $K = P (2k+1,2\ell+1,2m+1)$ where $k, \ell, m \geq 1$.  That $\Sigma_3(K)$ is an L-space is proved in \cite{Ter}.  We will construct an explicit presentation of $\pi_1(\Sigma_3(K))$.  Let $M$ be the exterior of $K$.  It is shown in \cite{Tr} that $\pi_1 (M)$ has a presentation with generators $x,y,z$ and relators
\begin{equation}\label{eq:not left1}
(xy^{-1})^m x(xy^{-1})^{-m} (yz^{-1})^{k+1} z^{-1} (yz^{-1})^{-(k+1)} \end{equation}
\begin{equation}\label{eq:not left2}
(yz^{-1})^k y(yz^{-1})^{-k} (zx^{-1})^{\ell+1} x^{-1} (zx^{-1})^{-(\ell+1)}
\end{equation}
\begin{equation}\label{eq:not left3}
(zx^{-1})^\ell z(zx^{-1})^{-\ell} (xy^{-1})^{m+1} y^{-1} (xy^{-1})^{-(m+1)}
\end{equation}
where $x,y,z$ are meridians of $K$.

It is straightforward to verify that the product of the relators \eqref{eq:not left1}, \eqref{eq:not left2} and \eqref{eq:not left3} is the identity in the free group on $x,y$ and $z$,  and so relator \eqref{eq:not left3} may be eliminated.  Let $X$ be the 2-complex corresponding to the resulting presentation; thus $X$ has a single 0-cell $c$, three 1-cells $e_x, e_y, e_z$ corresponding to $x,y,z$, respectively, and two 2-cells $D_1,D_2$ corresponding to the relators \eqref{eq:not left1} and \eqref{eq:not left2}.  Let $p : X_3 \to X$ be the 3-fold cyclic cover.  Then $p^{-1}(c) = \{c_0, c_1,c_2\}$, say, and $p^{-1} (e_x) = e_x^{(0)}\cup e_x^{(1)} \cup e_x^{(2)}$, where $e_x^{(i)}$ is a path in $X_3$ from $c_i$ to $c_{i+1}$, $i\in \zed/3$; similarly for $p^{-1} (e_y)$ and $p^{-1} (e_z)$.  Each 2-cell $D_j$, $j\in \{1,2\}$, lifts to three 2-cells $D_j^{(i)}$, $i\in \zed/3$.

To get a presentation for $\pi_1(X_3)$ we choose a maximal tree in the 1-skeleton $X_3^{(1)}$; we take this to be $e_z^{(0)}\cup e_z^{(1)}$.
The path $e_x^{(i)}$ now represents an element $x_i \in \pi_1 (X_3)$, and similarly for $e_y^{(i)}, e_z^{(i)}$.
Note that $z_0 = z_1 =1 \in \pi_1 (X_3)$.
The 2-cells $D_j^{(i)}$ give the following relations, where $i\in \zed/3$:
\begin{equation}\label{eqn:pretz1}
(x_i y_i^{-1})^m x_i (x_{i+1} y_{i+1}^{-1} )^{-m} (y_{i+1} z_{i+1}^{-1})^{k+1}
z_i^{-1}(y_iz_i^{-1})^{-(k+1)} =1
\end{equation}
\begin{equation}\label{eqn:pretz2}
(y_i z_i^{-1})^k y_i (y_{i+1} z_{i+1}^{-1})^{-k} (z_{i+1} x_{i+1}^{-1})^{\ell+1}
x_i ^{-1} (z_i x_i^{-1})^{-(\ell+1)} =1.
\end{equation}
Note that $\pi_1(X_3) \cong \pi_1 (M_3)$, where $M_3$ is the three-fold cyclic cover of $M$.  Let $\mu$ be a meridian of $K$ and $\pi:M_3 \to M$ the covering map.  Then, recall that $\Sigma_3(K)$ is given by $M_3(\pi^{-1}(\mu^3))$.   Thus, to get a presentation of $\pi_1 (\Sigma_3 (K))$ we must adjoin the branching relations $x_0 x_1 x_2 = y_0 y_1 y_2 = z_0 z_1 z_2 =1$.  Since we have that $z_0 = z_1 = 1$, we must have that $z_2 = 1$ as well.  Eliminating $z_0, z_1,$ and $z_2$ from \eqref{eqn:pretz1} and \eqref{eqn:pretz2}, we obtain
\begin{equation}\label{eqn:pretz1new}
(x_i y_i^{-1})^m x_i (x_{i+1} y_{i+1}^{-1})^{-m} y_{i+1}^{k+1} y_i ^{- (k+1)} =1
\end{equation}
\begin{equation}\label{eqn:pretz2new}
y_i^{k+1} y_{i+1}^{-k} x_{i+1}^{-(\ell+1)} x_i^\ell =1.
\end{equation}

As in the proof of Theorem~\ref{thm:twobridge}, the relations \eqref{eqn:pretz1new} and \eqref{eqn:pretz2new} have the property that for each generator, the exponents of all the occurrences of that generator in the relator have the same sign.  These signs are given in Tables~\ref{table:pretzrel1} and \ref{table:pretzrel2}, for relations \eqref{eqn:pretz1new} and \eqref{eqn:pretz2new} respectively.

\begin{table}
\begin{center}
\begin{tabular}{| c | c | c | c | c | c | c |}
\hline
$i$  & $x_0$ & $x_1$ & $x_2$ & $y_0$ & $y_1$ & $y_2$ \\
\hline
$0$ & + & $\--$ & $\circ$ & $\--$ & + & $\circ$ \\
\hline
1 & $\circ$ & + & $\--$ & $\circ$ & $\--$ & + \\
\hline
2 & $\--$ & $\circ$ & + & + & $\circ$ & $\--$ \\
\hline
\end{tabular}
\vspace{.1in}
\caption{Signs of the exponents of the generators as they appear in the relation \eqref{eqn:pretz1new} for $i \in \mathbb{Z}/3$.  The notation $\circ$ indicates that the generator does not appear in the corresponding relation.}\label{table:pretzrel1}
\end{center}
\end{table}

\begin{table}
\begin{center}
\begin{tabular}{| c | c | c | c | c | c | c |}
\hline
$i$  & $x_0$ & $x_1$ & $x_2$ & $y_0$ & $y_1$ & $y_2$ \\
\hline
$0$ & + & $\--$ & $\circ$ & $+$ & $\--$ & $\circ$ \\
\hline
1 & $\circ$ & + & $\--$ & $\circ$ & $+$ & $\--$ \\
\hline
2 & $\--$ & $\circ$ & + & $\--$ & $\circ$ & $+$ \\
\hline
\end{tabular}
\vspace{.1in}
\caption{Signs of the exponents of the generators as they appear in the relation \eqref{eqn:pretz2new} for $i \in \mathbb{Z}/3$. The notation $\circ$ indicates that the generator does not appear in the corresponding relation.}\label{table:pretzrel2}
\end{center}
\end{table}
Suppose that there exists a left-invariant order, $<$, on $\pi_1(\Sigma_3 (K))$.  Since there is an automorphism of $\pi_1 (\Sigma_3 (K))$ sending $x_i$ to $x_{i+1}$, if some $x_i =1$ then $x_0 = x_1 =x_2 =1$.  Relation \eqref{eqn:pretz2new}
 then gives $y_i^{k+1} = y_{i+1}^k$, $i\in \zed/3$, which implies that each $y_i$ has finite order. Since we are assuming that $\pi_1(\Sigma_3 (K))$ is left-orderable, $y_0 = y_1 = y_2 =1$ and so $\pi_1 (\Sigma_3(K)) =1$, a contradiction.  Similarly, it follows that no $y_i$ is trivial in $\pi_1 (\Sigma_3 (K))$.

We say that an element $g \in \pi_1 (\Sigma_3 (K))$ is {\em positive} (respectively {\em negative}) if $g>1$ (respectively $g<1$).  Note that a product of elements with the same sign cannot be the identity.  Applying this to the relations \eqref{eqn:pretz1new} and \eqref{eqn:pretz2new}, Tables~\ref{table:pretzrel1} and \ref{table:pretzrel2} show that, for $i\in \zed/3$, if $x_i$ and $x_{i+1}$ have opposite sign then $y_i$ and $y_{i+1}$ have the same sign.  On the other hand, the relations $x_0 x_1 x_2 =1$, $y_0 y_1 y_2 =1$ show that the $x_i$ cannot all have the same sign, and the $y_i$ cannot all have the same sign.  This is clearly a contradiction.
\end{proof}

config: arxiv


TITLE: What is $\sqrt{(-1)^2}$
QUESTION [1 upvotes]: This question is primarily terminology based. In that $\sqrt{}$ denotes the principal square root.
Here are two reasoning

$\sqrt{(-1)^2}=1$  since $\sqrt{(-1)^2}=\sqrt{1}$ which we know has a principal square root of $1$.
Or  $\sqrt{(-1)^2}=-1$ since $\sqrt{(e^{\pi i})^2}=\sqrt{e^{2\pi i}}=e^{\pi i}=-1$

Which reasoning is correct and why? Also if possible can you leave a source.

REPLY [0 votes]: according to the absolute definition
$$\sqrt{x^2}=|x|=|-1|=1$$


TITLE: Related rates - Melting snowball
QUESTION [2 upvotes]: Assume that a snowball melts in such a way that its volume decreases at a rate proportional to its surface area. If half the original snowball has melted away after 2 hours, how much longer will it take for the snowball to disappear completely?
My try:
$$\frac{dV}{dt}=K4\pi r^2$$
and
$$V(0) = \frac{4}{3}\pi [r(0)]^3$$
$$V(2) = \frac{2}{3}\pi [r(0)]^3$$
but that's about it. I don't know how to find $t$ such that $V(t) = 0$ Also, it seems to me that this has some relationship to exponential decay, is that right or not?

REPLY [7 votes]: $V  = \frac {4}{3} \pi r^3\\
\frac {dV}{dr} = 4\pi r^2\\
\frac {dV}{dt} = \frac {dV}{dr}\frac {dr}{dt} = -4K\pi r^2\\
\frac {dr}{dt} = -K$
$r(t) = r(0) - tK$
$V(2) = \frac 12 V(0)\\
\frac 43 \pi (r(2))^3 = \frac 23 \pi (r(0))^3\\
(r(2))^3 = \frac 12 (r(0))^3\\
(r(0) -2K)^3 = \frac 12(r(0))^3\\
r(0) - 2K = \frac {1}{2^{\frac 13}} r(0)\\
K = \frac 12(1 - 2^{-\frac 13}) r(0)$
Find $t$ such that
$r(0) - Kt = 0\\
r(0) - \frac 12(1 - 2^{-\frac 13}) r(0)t = 0\\
1 = \frac 12(1 - 2^{-\frac 13})t\\
t = \frac  {2}{2^{-\frac 13} - 1}\\
t = \frac  {4}{2 - 2^\frac 23}$
This $t$ is total time.  As the question is "how much longer" we are not starting the clock until the 2 hour point.
$\frac  {4}{2 - 2^\frac 23}-2$


text:
\section{Detailed Experiment Results}\label{sec:tables}

In Tables~\ref{tab:easytomatchconsumers}-\ref{tab:generalcard}, we present the average and standard deviation values for the expected number of matchings obtained for each set of instances. The columns ``Avg UB'' display the average value of the conic relaxation for Problem \eqref{eq:aux_problem} which is detailed in Appendix \ref{sec:conic_relaxation}.

\begin{table}[!h]
    \centering
    \caption{Summary of results. }
    \label{tab:easytomatchconsumers}
    {\footnotesize
\begin{tabular}{|c|HHccccccc|}\hline
\multicolumn{10}{|c|}{{\easyinst} instances}\\\hline
  $n$ &  Avg cpu & Avg& Avg & Avg & Avg  & Avg & Avg& Avg  & Avg   \\
 &  time (s) & bi & UB~\eqref{eq:aux_problem}& UB~\eqref{eq:mnlmnl_concave_unconstrained} & Algorithm~\ref{alg:easy_to_match} & Algorithm~\ref{alg:mnl_concave_unconstrained}& $V^N$ & $R_1$ & $R_2$   \\ \hline
50 & 0.64 & 36.23 & 32.77 & 31.25 & 18.00 $\pm$3.39 & 21.34 $\pm$3.26 & 13.88 $\pm$2.97 & 13.85 $\pm$2.98 & 4.12 $\pm$0.97\\
75 & 1.18 & 54.40 & 42.37 & 40.60 & 26.94 $\pm$4.15 & 29.67 $\pm$3.77 & 19.78 $\pm$3.51 & 19.75 $\pm$3.48 & 4.73 $\pm$0.99\\
100 & 3.85 & 72.29 & 49.59 & 47.73 & 35.38 $\pm$4.76 & 36.80 $\pm$4.10 & 25.05 $\pm$3.83 & 24.97 $\pm$3.83 & 5.28 $\pm$1.04\\
125 & 9.57 & 79.17 & 55.18 & 53.34 & 40.61 $\pm$4.85 & 42.94 $\pm$4.30 & 29.87 $\pm$4.08 & 29.77 $\pm$4.09 & 5.33 $\pm$0.98\\
150 & 16.34 & 86.20 & 59.65 & 57.85 & 45.25 $\pm$4.89 & 48.19 $\pm$4.42 & 34.19 $\pm$4.29 & 34.02 $\pm$4.30 & 5.61 $\pm$1.00\\
200 & 43.84 & 99.72 & 66.39 & 64.71 & 54.44 $\pm$4.97 & 56.70 $\pm$4.53 & 41.64 $\pm$4.53 & 41.64 $\pm$4.53 & 6.02 $\pm$1.01\\
\hline
\end{tabular}
}
\end{table}

\begin{table}[!h]
    \centering
    \caption{Summary of results. }
    \label{tab:etaleasy}
    {\footnotesize
\begin{tabular}{|c|ccHHcHccccccH|}\hline
\multicolumn{14}{|c|}{{\etaleasy} instances}\\\hline
  $n$ & $\lambda_v$ & $\lambda_0$ & Avg cpu & Avg& Avg & Avg & Avg  & Avg & Avg& Avg & Avg & Avg & Avg  \\
 &  & & time (s) & LP & UB~\eqref{eq:mnlmnl_concave_unconstrained} & bi & UB~\eqref{eq:aux_problem}& Algorithm~\ref{alg:lp_unconstrained}  & Algorithm~\ref{alg:mnl_concave_unconstrained}& $V^N$ & $R_1$ & $R_2$ &  \\ \hline
50 & 1 & 1 & 0.15 & 49.17 & 46.90 & 49.17 & 47.75 & 25.67 $\pm$3.14 & 25.90 $\pm$3.15 & 25.67 $\pm$3.12 & 25.36 $\pm$3.12 & 25.33 $\pm$3.14 & \\
 & 1 & 10 & 0.15 & 49.17 & 34.66 & 49.17 & 35.41 & 21.53 $\pm$3.17 & 21.70 $\pm$3.20 & 21.51 $\pm$3.18 & 21.23 $\pm$3.16 & 21.23 $\pm$3.17 & \\
 & 10 & 1 & 0.14 & 49.46 & 47.27 & 49.46 & 47.58 & 26.26 $\pm$3.16 & 26.27 $\pm$3.16 & 26.28 $\pm$3.14 & 26.02 $\pm$3.16 & 26.04 $\pm$3.16 & \\
 & 10 & 10 & 0.15 & 49.46 & 35.11 & 49.46 & 35.39 & 22.00 $\pm$3.21 & 22.00 $\pm$3.21 & 21.98 $\pm$3.21 & 21.78 $\pm$3.21 & 21.79 $\pm$3.20 & \\
75 & 1 & 1 & 0.52 & 73.68 & 56.72 & 73.62 & 57.56 & 35.34 $\pm$3.61 & 35.53 $\pm$3.66 & 35.11 $\pm$3.60 & 34.68 $\pm$3.61 & 34.69 $\pm$3.61 & \\
 & 1 & 10 & 0.36 & 73.71 & 44.24 & 73.65 & 45.09 & 29.85 $\pm$3.66 & 30.09 $\pm$3.71 & 29.66 $\pm$3.64 & 29.31 $\pm$3.62 & 29.35 $\pm$3.65 & \\
 & 10 & 1 & 0.29 & 74.19 & 56.98 & 74.19 & 57.29 & 36.09 $\pm$3.64 & 35.91 $\pm$3.65 & 36.09 $\pm$3.66 & 35.79 $\pm$3.63 & 35.81 $\pm$3.64 & \\
 & 10 & 10 & 0.26 & 74.19 & 44.87 & 74.19 & 45.19 & 30.59 $\pm$3.71 & 30.61 $\pm$3.70 & 30.59 $\pm$3.68 & 30.33 $\pm$3.69 & 30.33 $\pm$3.70 & \\
100 & 1 & 1 & 2.18 & 97.09 & 63.20 & 96.82 & 64.04 & 43.78 $\pm$3.94 & 43.39 $\pm$3.96 & 42.82 $\pm$3.93 & 42.33 $\pm$3.90 & 42.35 $\pm$3.90 & \\
 & 1 & 10 & 1.69 & 97.06 & 51.37 & 96.79 & 52.25 & 37.37 $\pm$4.02 & 37.28 $\pm$4.02 & 36.65 $\pm$3.96 & 36.20 $\pm$3.99 & 36.23 $\pm$3.97 & \\
 & 10 & 1 & 2.64 & 98.82 & 63.40 & 98.67 & 63.70 & 44.14 $\pm$3.95 & 43.82 $\pm$3.96 & 44.09 $\pm$3.94 & 43.85 $\pm$3.93 & 43.77 $\pm$3.95 & \\
 & 10 & 10 & 2.05 & 98.82 & 51.96 & 98.66 & 52.28 & 37.86 $\pm$4.03 & 37.86 $\pm$4.03 & 37.83 $\pm$4.07 & 37.58 $\pm$4.01 & 37.56 $\pm$4.02 & \\
125 & 1 & 1 & 0.60 & 100.00 & 68.05 & 99.22 & 68.87 & 46.65 $\pm$4.15 & 50.02 $\pm$4.13 & 49.17 $\pm$4.04 & 48.64 $\pm$4.06 & 35.01 $\pm$3.13 & \\
 & 1 & 10 & 0.59 & 100.00 & 56.96 & 99.24 & 57.85 & 40.17 $\pm$4.21 & 43.53 $\pm$4.23 & 42.50 $\pm$4.12 & 42.07 $\pm$4.16 & 31.59 $\pm$3.31 & \\
 & 10 & 1 & 1.58 & 100.00 & 68.66 & 99.31 & 68.94 & 46.88 $\pm$4.13 & 50.76 $\pm$4.13 & 51.17 $\pm$4.12 & 50.80 $\pm$4.12 & 38.03 $\pm$3.32 & \\
 & 10 & 10 & 1.57 & 100.00 & 57.50 & 99.30 & 57.82 & 40.32 $\pm$4.20 & 44.12 $\pm$4.24 & 44.13 $\pm$4.22 & 43.77 $\pm$4.21 & 33.24 $\pm$3.41 & \\
150 & 1 & 1 & 0.96 & 100.00 & 71.86 & 99.12 & 72.64 & 46.93 $\pm$4.29 & 55.71 $\pm$4.23 & 54.49 $\pm$4.16 & 54.08 $\pm$4.18 & 31.57 $\pm$2.71 & \\
 & 1 & 10 & 0.84 & 100.00 & 61.29 & 99.09 & 62.16 & 40.36 $\pm$4.32 & 48.79 $\pm$4.37 & 47.56 $\pm$4.31 & 47.13 $\pm$4.30 & 27.96 $\pm$2.81 & \\
 & 10 & 1 & 0.58 & 100.00 & 72.28 & 98.67 & 72.55 & 46.96 $\pm$4.32 & 56.40 $\pm$4.23 & 56.79 $\pm$4.22 & 56.39 $\pm$4.18 & 23.54 $\pm$2.02 & \\
 & 10 & 10 & 0.56 & 100.00 & 61.85 & 98.69 & 62.15 & 40.50 $\pm$4.35 & 49.47 $\pm$4.36 & 49.39 $\pm$4.36 & 49.11 $\pm$4.35 & 22.15 $\pm$2.21 & \\
200 & 1 & 1 & 0.81 & 100.00 & 77.52 & 98.69 & 78.19 & 47.48 $\pm$4.47 & 64.83 $\pm$4.27 & 63.21 $\pm$4.20 & 62.75 $\pm$4.18 & 24.28 $\pm$1.98 & \\
 & 1 & 10 & 0.81 & 100.00 & 67.83 & 98.70 & 68.68 & 40.77 $\pm$4.48 & 57.38 $\pm$4.46 & 55.59 $\pm$4.39 & 55.16 $\pm$4.39 & 22.47 $\pm$2.13 & \\
 & 10 & 1 & 0.76 & 100.00 & 77.59 & 98.18 & 77.83 & 47.50 $\pm$4.50 & 65.25 $\pm$4.25 & 65.62 $\pm$4.19 & 65.33 $\pm$4.21 & 19.95 $\pm$1.69 & \\
 & 10 & 10 & 0.78 & 100.00 & 68.38 & 98.18 & 68.66 & 40.93 $\pm$4.49 & 58.10 $\pm$4.46 & 58.10 $\pm$4.45 & 57.74 $\pm$4.42 & 18.30 $\pm$1.79 & \\
\hline
\end{tabular}
}
\end{table}

\begin{table}[!h]
    \centering
    \caption{Summary of results.}
    \label{tab:general}
    {\footnotesize
\begin{tabular}{|c|ccHHcHccccccH|}\hline
\multicolumn{14}{|c|}{{\general} instances}\\\hline
  $n$ & $\lambda_v$ & $\lambda_0$ & Avg cpu & Avg& Avg & Avg & Avg  & Avg & Avg& Avg & Avg & Avg & Avg  \\
 &  & & time (s) & LP & UB~\eqref{eq:mnlmnl_concave_unconstrained} & bi & UB~\eqref{eq:aux_problem}& Algorithm~\ref{alg:lp_unconstrained}  & Algorithm~\ref{alg:mnl_concave_unconstrained}& $V^N$ & $R_1$ & $R_2$ &  \\ \hline
 50 & 1 & 1 & 0.17 & 49.18 & 47.12 & 49.18 & 47.55 & 26.15 $\pm$3.16 & 26.16 $\pm$3.16 & 26.19 $\pm$3.15 & 25.80 $\pm$3.16 & 25.79 $\pm$3.16 & \\
& 1 & 10 & 0.18 & 49.17 & 35.03 & 49.17 & 35.42 & 21.94 $\pm$3.21 & 21.92 $\pm$3.21 & 21.91 $\pm$3.21 & 21.59 $\pm$3.21 & 21.60 $\pm$3.20 & \\
 & 10 & 1 & 0.17 & 49.46 & 47.06 & 49.46 & 47.33 & 26.21 $\pm$3.16 & 26.22 $\pm$3.15 & 26.21 $\pm$3.18 & 25.96 $\pm$3.15 & 25.98 $\pm$3.16 & \\
& 10 & 10 & 0.17 & 49.46 & 35.19 & 49.46 & 35.45 & 22.04 $\pm$3.20 & 22.04 $\pm$3.21 & 22.01 $\pm$3.22 & 21.81 $\pm$3.20 & 21.84 $\pm$3.21 & \\
75 & 1 & 1 & 0.32 & 73.76 & 57.03 & 73.76 & 57.45 & 36.06 $\pm$3.64 & 35.88 $\pm$3.65 & 36.05 $\pm$3.61 & 35.61 $\pm$3.63 & 35.60 $\pm$3.65 & \\
 & 1 & 10 & 0.32 & 73.76 & 44.68 & 73.76 & 45.11 & 30.46 $\pm$3.71 & 30.47 $\pm$3.71 & 30.43 $\pm$3.73 & 30.06 $\pm$3.69 & 30.06 $\pm$3.72 & \\
 & 10 & 1 & 0.29 & 74.19 & 56.71 & 74.19 & 56.98 & 36.01 $\pm$3.65 & 35.83 $\pm$3.66 & 36.02 $\pm$3.65 & 35.70 $\pm$3.68 & 35.71 $\pm$3.64 & \\
 & 10 & 10 & 0.26 & 74.19 & 44.90 & 74.19 & 45.17 & 30.61 $\pm$3.70 & 30.62 $\pm$3.70 & 30.65 $\pm$3.71 & 30.35 $\pm$3.69 & 30.37 $\pm$3.70 & \\
100 & 1 & 1 & 5.84 & 98.33 & 63.89 & 98.22 & 64.30 & 44.33 $\pm$3.95 & 44.04 $\pm$3.95 & 44.31 $\pm$3.93 & 43.81 $\pm$3.96 & 43.83 $\pm$3.95 & \\
 & 1 & 10 & 6.01 & 98.33 & 51.82 & 98.22 & 52.24 & 37.73 $\pm$4.03 & 37.73 $\pm$4.03 & 37.74 $\pm$4.06 & 37.28 $\pm$4.03 & 37.27 $\pm$4.02 & \\
 & 10 & 1 & 5.59 & 98.92 & 63.77 & 98.90 & 64.03 & 44.37 $\pm$3.94 & 44.07 $\pm$3.95 & 44.37 $\pm$3.93 & 44.03 $\pm$3.94 & 44.03 $\pm$3.94 & \\
 & 10 & 10 & 5.44 & 98.92 & 51.98 & 98.91 & 52.26 & 37.88 $\pm$4.04 & 37.88 $\pm$4.03 & 37.87 $\pm$4.05 & 37.62 $\pm$4.04 & 37.60 $\pm$4.02 & \\
125 & 1 & 1 & 0.64 & 100.00 & 68.61 & 98.33 & 69.00 & 46.58 $\pm$4.20 & 50.71 $\pm$4.13 & 51.07 $\pm$4.13 & 50.53 $\pm$4.13 & 21.08 $\pm$2.13 & \\
 & 1 & 10 & 0.63 & 100.00 & 57.29 & 98.35 & 57.71 & 39.74 $\pm$4.27 & 43.94 $\pm$4.23 & 43.93 $\pm$4.23 & 43.44 $\pm$4.26 & 19.30 $\pm$2.27 & \\
 & 10 & 1 & 0.66 & 100.00 & 69.00 & 98.45 & 69.25 & 46.80 $\pm$4.18 & 51.05 $\pm$4.14 & 51.43 $\pm$4.08 & 51.10 $\pm$4.13 & 22.82 $\pm$2.20 & \\
 & 10 & 10 & 0.55 & 100.00 & 57.50 & 98.43 & 57.76 & 39.87 $\pm$4.25 & 44.12 $\pm$4.24 & 44.10 $\pm$4.22 & 43.79 $\pm$4.26 & 20.65 $\pm$2.33 & \\
150 & 1 & 1 & 0.73 & 100.00 & 72.12 & 98.76 & 72.49 & 50.41 $\pm$4.30 & 56.25 $\pm$4.23 & 56.67 $\pm$4.21 & 56.10 $\pm$4.21 & 20.53 $\pm$2.01 & \\
 & 1 & 10 & 0.75 & 100.00 & 61.76 & 98.81 & 62.16 & 43.47 $\pm$4.37 & 49.34 $\pm$4.38 & 49.29 $\pm$4.37 & 48.87 $\pm$4.35 & 19.13 $\pm$2.13 & \\
 & 10 & 1 & 0.83 & 100.00 & 72.51 & 98.82 & 72.75 & 50.65 $\pm$4.28 & 56.59 $\pm$4.22 & 56.99 $\pm$4.23 & 56.63 $\pm$4.20 & 22.21 $\pm$2.06 & \\
 & 10 & 10 & 0.72 & 100.00 & 61.93 & 98.83 & 62.18 & 43.52 $\pm$4.37 & 49.53 $\pm$4.36 & 49.50 $\pm$4.36 & 49.20 $\pm$4.37 & 20.45 $\pm$2.19 & \\
200 & 1 & 1 & 1.04 & 100.00 & 77.51 & 99.54 & 77.84 & 58.78 $\pm$4.34 & 65.15 $\pm$4.24 & 65.60 $\pm$4.20 & 65.09 $\pm$4.22 & 22.61 $\pm$1.98 & \\
 & 1 & 10 & 0.96 & 100.00 & 68.22 & 99.47 & 68.59 & 51.27 $\pm$4.49 & 57.94 $\pm$4.46 & 57.91 $\pm$4.48 & 57.46 $\pm$4.48 & 21.06 $\pm$2.11 & \\
 & 10 & 1 & 1.13 & 100.00 & 77.77 & 99.54 & 77.98 & 59.15 $\pm$4.32 & 65.44 $\pm$4.24 & 65.93 $\pm$4.23 & 65.52 $\pm$4.24 & 24.99 $\pm$2.07 & \\
 & 10 & 10 & 0.87 & 100.00 & 68.40 & 99.55 & 68.63 & 51.69 $\pm$4.48 & 58.14 $\pm$4.45 & 58.15 $\pm$4.44 & 57.85 $\pm$4.46 & 23.24 $\pm$2.21 & \\
 \hline
\end{tabular}
}
\end{table}

\begin{table}[!h]
    \centering
    \caption{Summary of results.}
    \label{tab:generalcard}
    {\footnotesize
\begin{tabular}{|c|cHHHHcHHccHcHH|}\hline
\multicolumn{15}{|c|}{{\generalcard} instances}\\\hline
  $n$ & $K_i$ & $\lambda_v$& $\lambda_0$ & Avg cpu & Avg& Avg& Avg& Avg & Avg & Avg  & & Avg & Avg &  \\
 &  && &  time (s) & LP & UB~\eqref{eq:mnl_concave_card} & bi & UB~\eqref{eq:aux_problem} & Algorithm~\ref{alg:cardinality} & Algorithm~\ref{alg:mnl_mnl_cardinality}  & &$R_1$ & $R_2$ &  \\ \hline
 50 & 5 & 1 & 1 & 0.29 & 41.46 & 43.08 & 41.45 & 47.55 & 22.67 $\pm$3.17 & 23.83 $\pm$3.26 &  & 10.32 $\pm$1.67 & 0.00 $\pm$0.00 & \\
75 & 5 & 1 & 1 & 0.59 & 62.18 & 52.94 & 62.11 & 57.45 & 32.04 $\pm$3.67 & 33.02 $\pm$3.75 &  & 11.47 $\pm$1.56 & 0.00 $\pm$0.00 & \\
100 & 5 & 1 & 1 & 3.84 & 82.89 & 59.98 & 82.52 & 64.30 & 40.33 $\pm$4.00 & 40.70 $\pm$4.07 &  & 12.43 $\pm$1.54 & 0.00 $\pm$0.00 & \\
125 & 5 & 1 & 1 & 0.94 & 100.00 & 64.92 & 97.04 & 69.00 & 46.82 $\pm$4.24 & 47.20 $\pm$4.25 &  & 13.07 $\pm$1.47 & 0.00 $\pm$0.00 & \\
150 & 5 & 1 & 1 & 0.74 & 100.00 & 68.65 & 97.49 & 72.49 & 48.74 $\pm$4.36 & 52.63 $\pm$4.36 &  & 13.75 $\pm$1.50 & 0.00 $\pm$0.00 & \\
200 & 5 & 1 & 1 & 0.87 & 100.00 & 74.47 & 98.82 & 77.84 & 56.33 $\pm$4.41 & 61.58 $\pm$4.39 &  & 14.58 $\pm$1.46 & 0.00 $\pm$0.00 & \\
50 & 10 & 1 & 1 & 0.28 & 45.23 & 45.14 & 45.22 & 47.55 & 24.37 $\pm$3.18 & 25.05 $\pm$3.19 &  & 15.07 $\pm$2.17 & 0.00 $\pm$0.00 & \\
75 & 10 & 1 & 1 & 0.57 & 67.84 & 55.04 & 67.77 & 57.45 & 34.01 $\pm$3.65 & 34.47 $\pm$3.70 &  & 17.95 $\pm$2.14 & 0.00 $\pm$0.00 & \\
100 & 10 & 1 & 1 & 4.45 & 90.43 & 62.01 & 89.96 & 64.30 & 42.45 $\pm$3.97 & 42.45 $\pm$4.00 &  & 19.82 $\pm$2.05 & 0.00 $\pm$0.00 & \\
125 & 10 & 1 & 1 & 0.62 & 100.00 & 66.84 & 98.28 & 69.00 & 46.48 $\pm$4.20 & 49.00 $\pm$4.19 &  & 21.24 $\pm$2.01 & 0.00 $\pm$0.00 & \\
150 & 10 & 1 & 1 & 0.69 & 100.00 & 70.47 & 98.65 & 72.49 & 50.05 $\pm$4.31 & 54.51 $\pm$4.28 &  & 22.03 $\pm$1.96 & 0.00 $\pm$0.00 & \\
200 & 10 & 1 & 1 & 0.85 & 100.00 & 76.08 & 99.42 & 77.84 & 58.33 $\pm$4.35 & 63.45 $\pm$4.32 &  & 23.56 $\pm$1.89 & 0.00 $\pm$0.00 & \\
50 & 25 & 1 & 1 & 0.26 & 47.83 & 46.48 & 47.83 & 47.55 & 25.57 $\pm$3.16 & 25.82 $\pm$3.16 &  & 22.06 $\pm$2.87 & 0.00 $\pm$0.00 & \\
75 & 25 & 1 & 1 & 0.59 & 71.74 & 56.39 & 71.73 & 57.45 & 35.34 $\pm$3.65 & 35.43 $\pm$3.67 &  & 28.81 $\pm$3.15 & 0.00 $\pm$0.00 & \\
100 & 25 & 1 & 1 & 5.29 & 95.65 & 63.29 & 95.24 & 64.30 & 43.73 $\pm$3.96 & 43.51 $\pm$3.98 &  & 33.89 $\pm$3.20 & 0.00 $\pm$0.00 & \\
125 & 25 & 1 & 1 & 0.74 & 100.00 & 68.05 & 99.03 & 69.00 & 47.70 $\pm$4.16 & 50.13 $\pm$4.16 &  & 37.73 $\pm$3.23 & 0.00 $\pm$0.00 & \\
150 & 25 & 1 & 1 & 0.88 & 100.00 & 71.60 & 99.26 & 72.49 & 51.95 $\pm$4.25 & 55.65 $\pm$4.26 &  & 40.61 $\pm$3.21 & 0.00 $\pm$0.00 & \\
200 & 25 & 1 & 1 & 0.93 & 100.00 & 77.07 & 99.69 & 77.84 & 60.57 $\pm$4.27 & 64.58 $\pm$4.28 &  & 44.93 $\pm$3.08 & 0.00 $\pm$0.00 & \\
 \hline
\end{tabular}
}
\end{table}


TITLE: What does Entropy mean, other than just "disorder of the system"? Could it be called "Tendency to move towards Thermal Equilibrium"?
QUESTION [3 upvotes]: What is meant by Entropy? My professor tells me "disorder"(I make no comments on that, except that it really adds no real understanding). The way the concept of entropy kept developing in class, I was forced to conclude that the statement "Entropy of an isolated system always increases" is equivalent to the statement "An isolated system always reaches thermodynamic equilibrium (among its parts, i.e. all parts of the isolated system reach the same temperature)". If that is so, why invent Entropy? Why not just say "Thermal Equilibrium"? Then, I understand that $ΔS=\frac{q_{rev}}{T}$. Meaning, if the reaction is reversible, Entropy change may be interpreted as heat exchanged divided by temperature at which heat flows. Other than this I have no understanding of $S$ or $ΔS$.
So, What is meant by "Entropy"? And why is $ΔS=\frac{q_{rev}}{T}$? And what is up with this thing called "Disorder"?

REPLY [1 votes]: "Measurement of Disorder" is not a particularly good metaphor for entropy. Your title suggestion is closer to a sounder notion: in these terms I would call entropy a measure of "progress towards thermodynamic equilibrium".
What the entropy of a system's macrostate actually measures is the likelihood of that macrostate, i.e. the logarithm of the number of quantum (microscopic) states that are consistent with an observed macrostate, assuming all possible microscopic states are equally probable.
Another way of saying this is that it is the size of the maximally compressed document or codeword, in bits, that you would need to specify the system's exact state given that you already know it is in the observed macrostate. Actually we multiply the bit length by a constant to make this definition consistent with the classical, Clausius definition that you cite, but a document size in bytes is a good thought picture to hold on to.
I say more about these ideas in my anser here.
The reason I think one could think, in your words, of entropy as a "progress" towards equilibrium is that in any thermodynamic system studied in all undergrad statistical mechanics there are so many possible states that the law of large numbers enforces the following statement: in any such thermodynamic system, there are states that look very much like the maximum likelihood state and almost nothing else, as I discuss in my answer here. Thus, if a system beginning in any state undergoes a random walk in state space, then almost certainly it will wind up near the maximum likelihood, equilibrium state.


TITLE: What is the smallest value of the expresion $\sqrt{a^2 +9}+\sqrt{(b-a)^2+4}+\sqrt{(8-b)^2 +16}$
QUESTION [3 upvotes]: How can you find the smallest possible value of $\sqrt{a^2 +9}+\sqrt{(b-a)^2+4}+\sqrt{(8-b)^2 +16}$ using only algebra or analytic geometry.
I guessed that it would be the value when $a=0,b=8$ as then the first and third terms would have smallest values but am unable to prove it.
Any suggestions?

REPLY [10 votes]: Hint: This is the sum of the distances from $(0,0)$ to $(a,3)$ to $(b,5)$ to $(8,9)$. Pick $a$ and $b$ so these are all on a line.


TITLE: Coupon Collector problem with unequal probabilities
QUESTION [1 upvotes]: Suppose there are two different coupons and we want to find the average number of coupons drawn until we get the two coupons.
The probabilities of getting the two coupon types are not equal. Denote the probabilities by $p$ and $1-p$.
I thought about the next solution:
$$1 + p\cdot \frac{1}{1-p} + (1-p) \cdot \frac{1}{p}$$
Explanation: $1$ is for the first coupon. with probability p the fisrt one was of the first type, and now (by geometric distribution) we will have to try $\frac{1}{1-p}$ times in average. the third component $(1-p) \cdot\frac{1}{p}$ is the complementary situation.
Is this correct?

REPLY [1 votes]: Community wiki answer so the question can be closed as answered: As noted in a comment, yes, that's correct.


TITLE: How prove this $ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$
QUESTION [5 upvotes]: let $a,b,c\ge 0$, show that

$$ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$$

my idea use the SOS methods, But I don't work at last. Thank you

REPLY [2 votes]: Since this expression is symmetric in $b$ and $c$, we can make the substitution $x=b+c$, $y=bc$ to obtain the equivalent inequality
$$
y^2+axy+a(x^3-3xy)+a^3x \geq 2a^2(x^2-y) \, ;
$$
the condition $b,c \geq 0$ is equivalent to requiring $x,y \geq 0$.
This inequality is quadratic in $y$. By collecting terms, we see that it suffices to show that
\begin{eqnarray}
f_{x,a}(y):&=&y^2+y(ax-3ax+2a^2)+ax^3+a^3x-2a^2x^2\\
&=&y^2-2ya(x-a)+ax(x-a)^2
\end{eqnarray}
is non-negative whenever $a,x,y$ are.
For fixed $x,a$, the function $f$ is minimized when $y=a(x-a)$. There are two possible cases:

If $a>x$, this minimum value lies outside the domain of consideration. Since $f$ is a quadratic with positive leading coefficient, it is minimized on the domain of consideration at its boundary; that is, when $y=0$. But $a,x \geq 0$, which means $f_{x,a}(0)=ax(x-a)^2 \geq 0$.
If $a \leq x$, $f$ is minimized on the domain of consideration at $y=a(x-a)$. Moreover,
\begin{eqnarray}
f_{x,a}(a(x-a))&=&a^2(x-a)^2-2a^2(x-a)^2+ax(x-a)^2\\
&=& a(x-a)^3
\end{eqnarray}
which is non-negative as $a \geq 0$ and $x \geq a$.

So, for any choices of $x$ and $a$, $f$'s minimum value in $y$ is non-negative. Thus $f$ is non-negative on the entire domain of consideration.


TITLE: Showing that the infinite grid is Eulerian
QUESTION [4 upvotes]: In a post to usenet in 2004, I wrote:

I'm currently remembering learning [sic] some (long forgotten) things about
  Graph Theory via Robin J. Wilson's "Introduction to Graph Theory", 2nd.
  ed., 1972.
Unfortunately, I'm having a hard time with one of the exercises, which
  asks for the reader to show that the infinite square grid is an Eulerian
  graph by showing an explicit two-way Eulerian path (i.e., one path that
  covers every edges of the graph and that extends in both directions).
Where can I find a hint for this excercise?

At that point, I had already seen that the infinite square grid (considering only the vertical or horizontal lines as being the "edges" of the grid) is Hamiltonian by a simple drawing of two "concentric" spirals, as the following figure shows:

One of the replies that I received was from David Eppstein, who told me: "Hint: spiral."
Unfortunately, I have revisited the problem from time to time and I have not found a way to solve it. I asked some colleagues and they were not also able to come up with an answer.
So, how can one systematically traverse all the edges of the unit grid without getting stuck at some point by the two-sided infinite path bumping into itself?

REPLY [3 votes]: Spiral around, leaving one half-row unoccupied, then approach the origin from infinity on the left:


TITLE: mathematical statistics written by Jun Shao P29
QUESTION [1 upvotes]: For a symmetric  $k\times k$ matrix $M$, let $R_M = \{y \in \mathbb{R}^k:y = Mx \text{ with some }x \in R^k\}$ . If a random $k$-vector $X$ has a finite $\operatorname{Var}(X)$, then $P(X-EX \in R_{\operatorname{Var}(X)})=1$.
How to understand this?

REPLY [0 votes]: Idea: let $z \in R_{\operatorname{Var}(X)}^\perp$, i.e. $z\cdot y = 0$ for all $y\in R_{\operatorname{Var}(X)}$. Then $$\operatorname{var}(z\cdot X) = \mathrm E\bigl(z^\top (X-\mathrm E X)(X-\mathrm E X)^\top z\bigr) = z^\top \operatorname{Var}(X) z = z\cdot (\operatorname{Var}(X) z) = 0,$$
whence $\mathrm P(z\cdot X = 0)=1$.


TITLE: Is Witten's proof of the positive mass theorem rigorous?
QUESTION [22 upvotes]: I noticed that the only official reason given for awarding Edward Witten the Fields medal was his 1981 proof of the positive mass theorem with spinors, so I was assuming that the proof was fully rigorous.
However, I came across this paper https://projecteuclid.org/download/pdf_1/euclid.cmp/1103921154 by Taubes and Parker which claims to make Witten's proof 'mathematically rigorous' and to justify assumptions which Witten made about Dirac operators.  Does this mean that the Witten proof is not rigorous, or is it just the case that there were some unjustified lemmas to clear up which do not affect the validity or rigour of the argument (similar to the case of Perelman's proof of the Poincaré conjecture, where some lemmas and slight gaps had to be filled in)?
I am just curious as I have never really heard of the Taubes-Parker paper so I was assuming that the Witten paper was fully rigorous.
Later: I realise now that this is a bit of a misleading question, as 'proofs' can in reality come in many different flavours and/or levels of rigour ie. all the i's dotted and t's crossed, or some things left unsaid or which still remain to be proved.

REPLY [9 votes]: The positive mass theorem is more or less to do with the geometry of a type of positive scalar curvature condition.
Witten's work considers harmonic spinors, which are solutions to a certain linear elliptic system of partial differential equations. In his paper he presents a calculation which proves a rigidity theorem for harmonic spinors under a type of positive scalar curvature, directly comparable to Bochner's famous rigidity theorem for harmonic 1-forms in positive Ricci curvature. It is a little more complicated only since spinors are more complicated than differential forms.
Given the existence of a harmonic spinor with certain asymptotics, an integrated version of Witten's Bochner-type formula proves that the relevant positive scalar curvature condition implies the nonnegativity of the mass, now being expressed as the integral of the sum of squares of expressions built out of the harmonic spinor.
Witten's proof of the existence of such harmonic spinors is openly incomplete; he says "We have shown that the Dirac operator has no zero eigenvalue. Using this fact, we presume that standard methods can be used to yield" a key analytical step. The problem is existence of a Green's function for the relevant elliptic operator over noncompact spaces. Parker and Taubes gave a complete proof. I think it is not completely accurate to say that their proof only consists of "standard methods," since care is needed about weighted Sobolev spaces which can be somewhat delicate.
So I think it is inaccurate/misleading to put Witten's proof of positive energy theorem with some of his other works in terms of "inspiration and insight" for mathematics, or to just cite "origin in supergravity" (as Atiyah's laudatio does). His work here is a pretty direct and rigorous mathematical argument, in the vein of standard differential geometry. The gap is only due to his not being an expert in PDE methods. Even so he makes plausible that the relevant harmonic spinors exist. I think that a PDE expert reading his paper would likely even find it informally convincing.
As far as the Fields medal goes, I think some of his other work must have been more relevant. It's not hard to imagine someone else having discovered the main parts of Witten's proof, having to do with differential identities for spinors, with much less fanfare. The calculation relating the mass to the spinor is maybe the most striking part but I think many people (whether reasonably or not) would not regard it as a high point of mathematics (or whatever Fields medal is supposed to be about).


text:
\begin{document}

\title[M\"obius  rigidity of invariant metrics]{M\"obius  rigidity of invariant metrics in boundaries of symmetric spaces of rank-1}

\author[I.D. Platis \& V. Schroeder]{I.D. Platis \& V. Schroeder}

\email{jplatis@math.uoc.gr, viktor.schroeder@math.uzh.ch}
\address{Department of Mathematics and Applied Mathematics\\
University of Crete\\
University Campus\\
GR 700 13 Voutes\\ Heraklion Crete\\Greece}

\bigskip

\address{Institut f\"ur Mathematik\\
Universit\"at Z\"urich\\
Winterthurerstrasse 190\\
CH 8057 Z\"urich\\Switzerland}

\begin{abstract}
Let $\bH^n_\K$ denote the symmetric space of rank-1 and of non-compact type and let $d_\fH$ be the Kor\'anyi metric defined on its boundary. We prove that if $d$ is a metric on $\partial\bH^n_\K$ such that all Heisenberg similarities are $d$-M\"obius maps, then under a topological condition $d$ is a constant multiple of a power of  $d_\fH$.

\end{abstract}

\date{{17 March 2016}\\
{\it 2010 Mathematics Subject Classifications.} 51B10, 51F99.
\\ \it{Key words.} M\"obius space, invariant metric, boundary of symmetric space.}

\maketitle

\section{Introduction and Statement of Results}
Let $(S, d)$ be a metric space and suppose that there exists a remote point $\infty$ such that $\overline{S}=S\cup\{\infty\}$ is compact. We may extend $d$ to the compactification by agreeing  that $d(p,\infty)=+\infty$ for every $p\in S$ and also  $d(\infty,\infty)=0$. A natural metric cross-ratio $|\X^d|$ is defined for each quadruple of pairwise distinct points $\fp=(p_1,p_2,p_3,p_4)$ by setting
$$
|\X^d|(\fp)=\frac{d(p_4,p_2)}{d(p_4,p_1)}\cdot\frac{d(p_3,p_1)}{d(p_3,p_2)},
$$
with obvious modifications when one of the points is $\infty$. The {\it M\"obius group} $\cM_d=\cM_d(S)$ is the group of homeomorphisms of $\overline{S}$ which leave $|\X^d|$ invariant for each quadruple $\fp$ of pairwise distinct points of $\overline{S}$. The {\it similarity group} (or {\it homothety group}) ${\rm Sim}_d={\rm Sim}_d(S)$ is the subgroup of $\cM_d$ comprising homeomorphisms $\phi$ such that there exists a positive constant  $K(\phi)$ which satisfies $d\left(\phi(p),\phi(q)\right)=K(\phi)\cdot d(p,q),$ for all $p,q\in S$. If $K(\phi)=1$ then $\phi$ is an {\it isometry}.

Given two metrics $d_1$ and $d_2$ on $S$ for which $\infty$ is their remote point, we say that they define the same {\it M\"obius structure} on $S$ if $|\X^{d_1}|=|\X^{d_2}|$. This happens if and only if $d_1$ and $d_2$ are homothetic, that is, there exists a $c>0$ such that $d_1=c\cdot d_2$ (see Lemma 2.1 of \cite{BS}). It is immediate that for such metrics we have $\cM_{d_1}=\cM_{d_2}$.

In this paper we are concerned with the following problem: Suppose that we are given a representative $d_1$ of a M\"obius structure on $\overline{S}$ and suppose also that there exists a metric $d_2$ defined on $S$ which has the same remote point as $d_1$ and also satisfies $\cM_{d_1}\subseteq \cM_{d_2}$. Then, assuming that $\cM_{d_1}$ is sufficiently rich, what can we say about the relation of $d_1$ and $d_2$?

We give a quite precise answer to this question in the case where $\overline{S}$ is the boundary of a symmetric space of rank-1 and of non-compact type and the given M\"obius structure is the canonical one, i.e., the one arising from the Kor\'anyi metric.  We recall below some known facts about the aforementioned notions; for details we refer to Section \ref{sec:prel}.
Let $\K=\R,\C,\mH$ or $\bO$ be the set of real, complex, quaternionic and octonionic numbers, respectively. With $\bH^n_\K$ we shall thereafter denote the $n$-dimensional $\K$-hyperbolic space (in the case $\K=\bO$, $n=2$). Symmetric spaces of rank-1 and of non-compact type are necessarily the $\K$-hyperbolic spaces. There is a natural conjugation $z\mapsto\overline{z}$ in $\K$ (in the case when $\K=\R$ this is of course the identity map); denote by $\Im(\K)$ the subset of $\K$ comprising points $z$ such that $z=-\overline{z}$. The boundary $\partial\bH^n_\K$ is a sphere, isomorphic to the one point compactification of the $\K$-{\it Heisenberg group} $\fH_\K$: That is, the set $\K^{n-1}\times\Im(\K)$ endowed with the group multiplication
$$
(\zeta,v)*(\zeta',v')=\left(\zeta+\zeta',v+v+2\omega(\zeta,\zeta')\right),
$$
for each $(\zeta,v),(\zeta',v')\in\fH_\K$. Here, $\omega$ is the usual symplectic form in $\K^{n-1}$. A natural conjugation $J$ is defined on $\K^{n-1}\times\Im(\K)$ as the restriction of the conjugation of $\bH^n_\K$; that is,  $(\zeta,v)\mapsto (\overline{\zeta},-v)$. There is a gauge $|\cdot|_\K$ on $\fH_\K$, called the Kor\'anyi gauge, which is given for each $(\zeta,v)$ by
$$
|(\zeta,v)|_\K=|-\|\zeta\|^2+v|^{1/2}=\left(\|\zeta\|^4+|v|^2\right)^{1/4}.
$$
The Kor\'anyi (or, Kor\'anyi-Cygan) metric $d_\fH$ is then defined in $\fH_\K$ by
\begin{equation}\label{eq:Kor}
d_\fH\left((\zeta,v)\;,\;(\zeta',v')\right)=\left|(\zeta',v')^{-1}*(\zeta,v)\right|_\K,
\end{equation}
for each $(\zeta,v),(\zeta',v')\in\fH_\K$.
The canonical M\"obius structure of $\partial\bH^n_\K$ is the one arising from $d_\fH$. The M\"obius group $\cM_{d_\fH}$ of the metric $d_\fH$ is generated by the similarities of $d_\fH$ and an inversion $I$ defined on $\partial\bH^n_\K$, see Section \ref{sec:prel} for details.

One of our main results can be stated in the following way:

\medskip

\noindent{\bf Theorem.} {\it Let $d$ be a metric on $\partial\bH^n_\K$, which induces the same topology and has the same remote point as $d_\fH$. Assume further that $\cM_d=\cM_{d_\fH}$. Then there exist $\alpha\in(0,1]$ and $\beta>0$ such that $d=\beta\cdot d^\alpha_\fH$.}

\medskip

This special result is embedded in much more general statements: Suppose that we  are given a metric $d$ in $\partial\bH^n_\K$  such that:

\medskip

\begin{enumerate}
 \item [{{\bf (Sim)}}] ${\rm Sim}_{d_\fH}\subseteq{\rm Sim}_d$.
\end{enumerate}

\medskip

We shall further presuppose the next quite plausible topological condition:

\medskip

\begin{enumerate}
 \item  [{{\bf (Top)}}]
 If a sequence of points $p_\nu\in\partial\bH^n_\K$, $\nu\in\mathbb{N}$, is converging in the $d_\fH$-topology of $\partial\bH^n_\K$, then it converges in the $d$-topology of $\partial\bH^n_\K$.
\end{enumerate}

\medskip

Before stating our results, we  list the following set of conditions:

\medskip

\begin{enumerate}
  \item  [{{\bf (Inv)}}] Inversion $I$ is  in $\cM_d$.
  \item [{{\bf ($\alpha$-H\"ol)}}] There exists a constant $\alpha\in (0,1]$ such that the identity map of $(\partial\bH^n_\K, d_\fH)\to(\partial\bH^n_\K, d)$ is H\"older continuous with exponent $\alpha$. Explicitly, there exists a $\beta>0$ such that for each $p,q\in\partial\bH^n_\K$,
  $$
  d(p,q)\le\beta\cdot d_\fH^\alpha(p,q).
  $$
  \item [{{\bf (G)}}] There exists an $\alpha\in (0,1]$ such that for each $p=(\zeta,v)\in\fH_\K$, $\K\neq\R$,
  \begin{equation*}
   d^{4/\alpha}(o,p)=d^{4/\alpha}\left(o,\Pi_{\K^{n-1}}(p)\right)+d^{4/\alpha}\left(\Pi_{\K^{n-1}}(p),\Pi_{\Im(\K)}(p)\right).
  \end{equation*}
Here, $o=({\bf 0}_{\K^{n-1}},{\bf 0}_{\Im(\K)})$ is the origin of $\fH_\K$, where ${\bf 0}_{\K^{n-1}}$ and ${\bf 0}_{\Im(\K)}$ are the origins of $\K^{n-1}$ and $\Im(\K)$, respectively. Moreover,  $\Pi_{\K^{n-1}}$,  $\Pi_{\Im(\K)}$ are defined respectively by $\Pi_{\K^{n-1}}(p)=\zeta$ and $\Pi_{\Im(\K)}(p)=v$.
\item [{{\bf (Eq)}}] The metric $d$ satisfies $$d\left(o,(1,0)\right)=d\left(o,(0,1)\right).$$ Here, $(1,0)=({\bf e}_1,{\bf 0}_{\Im(\K)})$, where ${\bf e}_1=(1,0,\dots,0)$ is the first coordinate vector of $\K^{n-1}$ and $(0,1)=({\bf 0}_{\K^{n-1}},{\bf f}_1)$ where ${\bf f}_1$ is the first coordinate vector of $\Im(\K)$.
\item [{{\bf (biLip)}}] The metric $d$ is bi-Lipschitz equivalent to $d_\fH^\alpha$ for some $\alpha\in (0,1]$.
\item [{{\bf ($\alpha$-Met)}}]  There exist  constants $\beta>0$ and  $\alpha\in(0,1]$ such that $d=\beta\cdot d^\alpha_\fH$.
\end{enumerate}

\medskip

Our first theorem assumes the least number of conditions for $d$, that is, {\bf (Sim)}  and {\bf (Top)}.

\begin{thm}\label{thm:main1}
Let $d$ be a metric in $\fH_\K$ and denote again by $d$ its extension to $\partial\bH^n_\K=\fH\cup\{\infty\}$.
Then the following hold:
\begin{enumerate}
\item If $\K=\R$, conditions {\bf (Sim)} and {\bf (Top)} together imply {\bf ($\alpha$-Met)}.
\item If $\K\neq\R$, conditions {\bf (Sim)} and {\bf (Top)}  together imply {\bf ($\alpha$-H\"ol)}.
\end{enumerate}
\end{thm}

We note the significant difference between the real case and all the other cases. With the least possible assumptions, the metric $d$ is a power of the Euclidean metric when $\K=\R$. The picture is entirely different in case when  $\K\neq\R$; metrics which satisfy {\bf ($\alpha$-H\"ol)} may be of nature entirely different from the one of a power of $d_\fH$ (e.g., the Carnot-Carath\'eodory metric). It is therefore quite necessary to add more assumptions for $d$ in this case. To that end, the strongest version of our first main result for the case $\K\neq\R$ follows:

\begin{thm}\label{thm:main}
With the assumptions of Theorem \ref{thm:main1}, suppose $\K\neq\R$. Then conditions {\bf (Sim)}, {\bf (Top)} and {\bf (Inv)} together imply {\bf ($\alpha$-Met)}.
\end{thm}

But again, adding {\bf (Inv)} to our basic assumptions seems to be excessive for the proof of {\bf ($\alpha$-Met)}; in the case $\K=\R$ {\bf (Inv)} holds {\it a posteriori}. It is natural therefore to ask if {\bf (Inv)} is really necessary. Theorem \ref{thm:main1} tells us that we obtain a rather weak result if we entirely drop {\bf (Inv)}. It turns out though that {\bf ($\alpha$-Met)} follows by replacing {\bf (Inv)} with {\bf (G)} and {\bf (Eq)}. In fact, we have:

\begin{thm}\label{thm:mainw2}
 With the assumptions of Theorem \ref{thm:main1}, suppose $\K\neq\R$. Then:
\begin{enumerate}
 \item Conditions {\bf (Sim)}, {\bf (Top)} and {\bf (G)} together imply {\bf (biLip)}.
 \item Conditions {\bf (Sim)}, {\bf (Top)}, {\bf (G)} and {\bf (Eq)} together imply {\bf ($\alpha$-Met)}.
\end{enumerate}
\end{thm}

Therefore, {\bf (Inv)} follows as a side result of  Theorem \ref{thm:mainw2} and we may further observe  that {\bf (G)} and {\bf (Eq)} hold vacuously in the real case. Moreover, {\bf (G)} can be replaced with an equivalent statement which is the closest to parallelogram law in the $\K$-Heisenberg group setting, $\K\neq\R$.  For this set for each $p\in\fH$,
$$
|p|=d(o,p).
$$
It turns out that if {\bf (Sim)} holds, then {\bf (G)} is equivalent to the following condition, see Proposition \ref{prop:p-l}:

\medskip

\begin{enumerate}
\item [{{\bf (P-L)}}] For $\alpha\in (0,1]$ and for each $p,q\in\fH$,
\begin{eqnarray*}
\left|p*q\right|^{4/\alpha}+ \left|p^{-1}*q\right|^{4/\alpha}+\left|p*q^{-1}\right|^{4/\alpha}+\left|p^{-1}*q^{-1}\right|^{4/\alpha}&=&\\
2\left(\left|\Pi_{\K^{n-1}}(p*q)\right|^{4/\alpha}+\left|\Pi_{\K^{n-1}}\left(p^{-1}*q\right)\right|^{4/\alpha}\right)&+&\\
\left|\Pi_{\Im(\K)}(p*q)\right|^{4/\alpha}+ \left|\Pi_{\Im(\K)}\left(p^{-1}*q\right)\right|^{4/\alpha}+\left|\Pi_{\Im(\K)}\left(p*q^{-1}\right)\right|^{4/\alpha}+\left|\Pi_{\Im(\K)}\left(p^{-1}*q^{-1}\right)\right|^{4/\alpha},
\end{eqnarray*}
where $\Pi_{\K^{n-1}}$ and $\Pi_{\Im(\K)}$ are the projections of $\fH$ to $\K^{n-1}$ and $\Im(\K)$, respectively.
\end{enumerate}

\medskip

Thus an equivalent to the second statement of Theorem \ref{thm:mainw2} (2) is:

\medskip

\begin{enumerate}
\item [{($2^{'}$)}] Conditions {\bf (Sim)}, {\bf (Top)}, {\bf (P-L)} and {\bf (Eq)} together imply {\bf ($\alpha$-Met)}.
\end{enumerate}

\medskip

Recall now that a metric $d$ defined in a space $S$ is Ptolemaean if for each quadruple of points $\fp=(p,q,r,s)$ of $S$ the following inequality is satisfied for all possible permutations of points in $\fp$:
\begin{equation}\label{eq:Ptolineq}
 d(p,r)\cdot d(q,s)\le d(p,q)\cdot d(r,s)+d(p,s)\cdot d(r,q).
\end{equation}
A Ptolemaean circle is a subset $\sigma$ of $S$ which is topologically equivalent to the unit circle $S^1$ and for each quadruple of points $\fp=(p,q,r,s)$ of $\sigma$ such that $p$ and $r$ separate $q$ and $s$, Inequality \ref{eq:Ptolineq} holds as an equality.
It is well known that $(\partial\bH^n_\K, d_\fH)$ is Ptolemaean, see for instance \cite{P}; therefore it is natural to ask which of the metrics $d$ who satisfy {\bf ($\alpha$-Met)} satisfies also:

\medskip

\begin{enumerate}
  \item [{{\bf (Ptol)}}] The metric $d$ is Ptolemaean.
  \end{enumerate}

  \medskip

  It turns out that we will also need:

  \medskip

  \begin{enumerate}
  \item [{{\bf (Circ)}}] The metric $d$ has a Ptolemaean circle.
 \end{enumerate}

 \medskip

 We have the following corollary to Theorem \ref{thm:mainw2}:

\begin{thm}\label{thm:pt}
Condition {\bf ($\alpha$-Met)} implies {\bf (Ptol)} and {\bf (Ptol)} together with {\bf (Circ)}  are equivalent to  {\bf ($1$-Met)}. Therefore a metric $d$ in $\partial\bH^n_\K$ which satisfies conditions:
\begin{enumerate}
\item {\bf (Sim)}, {\bf (Top)} and {\bf (Circ)} if $\K=\R$ and
\item {\bf (Sim)}, {\bf (Top)}, {\bf (P-L)} or {\bf (G)}, {\bf (Eq)}  and {\bf (Circ)} if $\K\neq\R$,
\end{enumerate}
is necessarily a constant multiple of the Kor\'anyi metric $d_\fH$.
\end{thm}

This is in the spirit of the old result of Schoenberg, see \cite{S}. That particular result was on metrics which were derived from semi-norms which  also share  Ptolemaean property.

\thanks{ Part of this work was carried out while IDP was visiting University of Z\"urich, Switzerland. Hospitality of Instit\"ut f\"ur Mathematik, University of Z\"urich, is gratefully appreciated.

\section{Preliminaries}\label{sec:prel}

This section is divided in two parts. In the first part (Section \ref{sec:hyp}), we state for clarity some known results about the Kor\'anyi metric in $\partial\bH^n_\K$ and its properties. In the second part (Section \ref{sec:Mob}), we state in brief some elementary facts concerning M\"obius geometry, mainly focusing on properties of the similarity group.

\subsection{The $\K$-Heisenberg group $\fH_\K$ and the Kor\'anyi metric $d_\fH$}\label{sec:hyp}
The following are well known; we refer the reader to the classical book of Mostow, \cite{M}, or to \cite{P}. Another useful reference for the case $\K=\C$ is the book of Goldman, \cite{Gol}. For the octonionic case in particular, we refer to \cite{A} and to \cite{M-P}.

\medskip

We shall use hereafter the following notation ($n>1$):
$$
{\rm G}_\K=\left\{\begin{matrix} {\rm SO}(n,1)& & \text{if}\; \;\K=\R,\\
{\rm SU}(n,1)& & \text{if}\;\; \K=\C,\\
{\rm Sp}(n,1)& & \text{if}\; \;\K={\mathbb H},\\
F_{4(-20)} & & \text{if}\; \;\K={\mathbb O}\;(n=2).\end{matrix}\right.
$$
Also,
$$
{\rm F}(n)=\left\{\begin{matrix} {\rm SO}(n)& & \text{if}\; \;\K=\R,\\
{\rm SU}(n)& & \text{if}\;\; \K=\C,\\
{\rm Sp}(n)& & \text{if}\; \;\K={\mathbb H},\\
{\rm Spin}_7(\R) & & \text{if}\; \;\K={\mathbb O}\;(n=2).\end{matrix}\right.
$$
$\K$-hyperbolic space $\bH^n_\K$ is ${\rm G}_\K/{\rm F}(n)$. The  metric $d_\fH$ as given in Equation (\ref{eq:Kor}) is invariant under the following transformations (and their extensions to $\infty$).
\begin{enumerate}
 \item Left translations which come from the action of $\fH_\K$ on itself: For any fixed point $(\zeta',v')\in\fH_\K$ let
$$
T_{(\zeta',v')}(\zeta,v)=(\zeta',v')*(\zeta,v), \quad T_{(\zeta',v')}(\infty)=\infty.
$$

In the particular case where $\K=\R$, $d_\fH$  is also invariant under the right translations.
\item Rotations: For the cases $\K\neq{\mathbb O}$, these come from the action of ${\rm F}(n-1)$ on $\K^{n-1}$. Specifically, given a $U\in{\rm F}(n-1)$, $\K\neq{\mathbb O}$, we define
$$
S_U(\zeta,v)=(U\cdot\zeta,v),\quad S_U(\infty)=\infty.
$$
Only in the case where $\K=\mH$ we have the action of ${\rm F}(1)={\rm Sp}(1)$ given by
$$
(\zeta_1,\dots,\zeta_{n-1},v)\mapsto (\mu\zeta_1\mu^{-1},\dots,\mu\zeta_{n-1}\mu^{-1},\mu v\mu^{-1}),\quad \mu\in{\rm Sp}(1);
$$
observe that in all other cases this action is vacuous.
In the particular case $\K={\mathbb O}$, for given unit imaginary octonion $\mu$, let
\begin{equation*}
S_{\mu}(\zeta,v)=(\zeta\cdot\overline{\mu},\mu v\overline{\mu}),\quad S_{\mu}(\infty)=\infty.
\end{equation*}
We stress at this point that in general $S_\mu\circ S_\nu\neq S_{\mu\nu}$ for $\mu,\nu$ unit imaginary octonions. The group generated by transformations $S_\mu$ is the compact group ${\rm Spin}_7(\R)$.
\end{enumerate}
All these actions generate the group ${\rm Isom}_{d_\fH}(\partial(\bH_\K^n)$ of (orientation-preserving) $d_\fH$-{\it isometries}; this acts transitively on $\fH_\K$.
We also consider two other kinds of transformations of $\partial\bH_\K^n$.
\begin{enumerate}
\item[{(3)}] Dilations: If $\delta\in\R_*^+$ we define
$$
D_\delta(\zeta,v)=(\delta\zeta,\delta^2v),\quad D_\delta(\infty)=\infty.
$$
(In the octonionic case, $\delta^2$ is used in \cite{M-P} instead of $\delta$, but the model for $\fH_{\mathbb{O}}$ is somewhat different).
One verifies that for every $(\zeta,v),(\zeta',v')\in\partial\bH_\K^n$ we have
$$
d_\fH\left(D_\delta(\zeta,v),D_\delta(\zeta',v')\right)=\delta\cdot d_\fH\left((\zeta,v),(\zeta',v')\right).
$$
Thus the metric $d_\fH$ is  scaled up to multiplicative constants by the action of dilations. We mention here that together with $d_\fH$-isometries, dilations generate the $d_\fH$-{\it similarity group} ${\rm Sim}_{d_\fH}(\partial\bH_\K^n)$.
\item[{(4)}] Inversion $I$ is given by
$$
I(\zeta,v)=\left(\zeta(-\|\zeta\|^2+v)^{-1}\;,\;\overline{v}\left|-\|\zeta\|^2+v\right|^{-2}\right),\;\;\text{if}\;(\zeta,v)\neq o,\infty,\;\quad I(o)=\infty,\;I(\infty)=o.
$$
Inversion $I$ is an involution of $\partial\bH^n_\K$. Moreover, for all $p=(\zeta,v),p'=(\zeta',v')\in\fH_\K\setminus\{o\}$ we have
$$
d_\fH(I(p),o)=\frac{1}{d_\fH(p,o)},\quad d_\fH(I(p),I(p'))=\frac{d_\fH(p,p')}{d_\fH(p,o)\;d_\fH(o,p')}.
$$
\end{enumerate}

\medskip

The similarity group ${\rm Sim}_{d_\fH}(\partial\bH_\K^n)$ is the semidirect product $\R\rtimes{\rm Isom}_{d_\fH}(\partial\bH_\K^n)$. The group generated by ${\rm Sim}_{d_\fH}(\partial\bH_\K^n)$ and inversion $I$ is isomorphic to $G_\K$ (with the exception of the case $\K=\R$, $n$ odd: in this case, $I$ reverses orientation).
Given two distinct points on the boundary, we can find an element of $G_\K$  mapping those points to $o$ and $\infty$ respectively; in particular $G_\K$  acts doubly transitively on the boundary. In the exceptional case where $\K=\R$, the action of $G_\R$  is triply transitive; this follows from the fact that we can map three distinct points of the boundary to the points $o$, $\infty$ and ${\bf e}_1$, respectively.

\subsection{M\"obius Group, Similarity Group}\label{sec:Mob}

Recall from the introduction that in the most general setting we start from a mere metric space $(S,d)$ and we suppose that there is a {\it remote point} $\infty$ such that:
\begin{enumerate}
 \item $\overline{S}=S\cup\{\infty\}$ is compact;
 \item $d$ is extended to the compactification $\overline{S}$ by setting
 $
 d(p,\infty)=+\infty$,
for each $p\in S$ and also $d(\infty,\infty)=0$.
\end{enumerate}

The notion of metric cross-ratio is next. The following proposition may be verified straightforwardly, detailed discussions about cross-ratios may be found among others in \cite{F}, \cite{FP}, \cite{KR1} and \cite{P}:

\medskip

\begin{prop}\label{prop:abs-cr}
 Let $(S,d)$ be a metric space with a remote point $\infty$. Denote by $\cC$ the space $S\times S\times S\times S\setminus\{diagonals\}$. Then the map $|\X^d|:\cC\to \R_*^+$ defined for each $\fp=(p_1,p_2,p_3,p_4)\in\cC$ by
 \begin{equation}\label{eq:abs-cr}
  |\X^d|(\fp)=\frac{d(p_4,p_2)}{d(p_4,p_1)}\cdot\frac{d(p_3,p_1)}{d(p_3,p_2)},
 \end{equation}
satisfies the following:
\begin{enumerate}
 \item {\it Symmetries:}
 $$
 |\X^d|(p_1,p_2,p_3,p_4)=|\X^d|(p_2,p_1,p_4,p_3)=|\X^d|(p_3,p_4,p_1,p_2)=|\X^d|(p_4,p_3,p_2,p_1).
 $$
 \item Let
 $$
 |\X^d_1|(\fp)=|\X^d|(p_1,p_2,p_3,p_4),\quad\text{and}\quad |\X^d_2|(\fp)=|\X^d|(p_1,p_3,p_2,p_4).
 $$
 Then
 \begin{eqnarray*}
  &&
  |\X^d|(p_1,p_3,p_4,p_2)=\frac{1}{|\X^d_2|},\quad |\X^d|(p_1,p_4,p_3,p_2)=\frac{|\X^d_1|}{|\X^d_2|},\\
  &&
   |\X^d|(p_1,p_2,p_4,p_3)=\frac{1}{|\X^d_1|},\quad |\X^d|(p_1,p_4,p_2,p_3)=\frac{|\X^d_2|}{|\X^d_1|}.
 \end{eqnarray*}
Thus for any possible permutation $\tilde\fp$ of points of a given quadruple $\fp$, $|\X^d(\tilde\fp)|$ depends only on $|\X^d_1|(\fp)$ and $|\X^d_1|(\fp)$.
\end{enumerate}

\end{prop}

A homeomorphism $\phi:S\to S$ shall be called a {\it $d$-M\"obius map} if $|\X^d(\fp)|=|\X^d(\phi(\fp))|$ for each $\fp=(p_1,p_2,p_3,p_4)\in\cC$, where, by $\phi(\fp)$ we denote the quadruple $\left(\phi(p_1),\phi(p_2),\phi(p_3),\phi(p_4)\right)$. This is equivalent to say that for each given quadruple $\fp$,
$$
|\X^d_1|(\phi(\fp))=|\X^d_1|(\fp) \quad\text{and} \quad |\X^d_2|(\phi(\fp))=|\X^d_2|(\fp).
$$
The set of all M\"obius maps of $S$ form a group $\cM_d=\cM_d(S)$, which we shall call the {\it M\"obius group} of $(S,d)$.
In particular, we consider the subset ${\rm Sim}_d$ of $\cM_d$ consisting of similarities: An element $\phi\in\cM_d$ is a {\it similarity}, if there exists a $K_\phi>0$ such that for every $p,q\in S$
 $$
 d\left(\phi(p),\phi(q)\right)=K_\phi\cdot d(p,q).
 $$
 By definition, $K_\phi=K(\phi)$ is independent of the choice of points.

 \begin{prop}\label{prop:homoth}
The set ${\rm Sim}_d$ is a subgroup of $\cM_d$. There exists a group homomorphism $K:{\rm Sim}_d\to$ $ \R^+_*$ given for each $\phi\in{\rm Sim}_d$ by
\begin{equation*}\label{eq:homoth}
K(\phi)=K_\phi.
\end{equation*}
Here $\R^+_*$ is the multiplicative topological group of the positive reals, inheriting its topology from the real line.
\end{prop}
\begin{proof}
We show that ${\rm Sim}_d$ is a subgroup of $\cM_d$. If $\phi,\psi\in{\rm Sim}_d$, then for every $p,q\in S$,
 $$
 d\left(\phi(p),\phi(q)\right)=K_\phi\cdot d(p,q)\quad\text{and}\quad d\left(\psi(p),\psi(q)\right)=K_\psi\cdot d(p,q).
 $$
 Therefore for each $p,q\in S$,
 \begin{eqnarray*}
  d\left((\phi\circ\psi)(p),(\phi\circ\psi)(q)\right)&=&d\left((\phi\left(\psi(p)\right),(\phi\left(\psi
  (q)\right)\right)\\
  &=&K_\phi\cdot d\left(\psi(p),\psi(q)\right)\\
  &=&K_\phi\cdot K_\psi\cdot d(p,q).
 \end{eqnarray*}
Since this shows that $\phi\circ\psi$ is a similarity, we must also have
$$
d\left((\phi\circ\psi)(p),(\phi\circ\psi)(q)\right)=K_{\phi\circ\psi}\cdot d(p,q).
$$
Hence we also must have $K_{\phi^{-1}}=(K_\phi)^{-1}$.
\end{proof}

We call $K$ the {\it similarity homomorphism}.  Under our assumptions for $(S,d)$ in the beginning of this section, we endow $\cM_d$ with the compact-open topology and accordingly ${\rm Sim}_d$ inherits the relative topology. We then have:

\begin{prop}\label{prop:cont}
The similarity homomorphism $K$ is continuous.
\end{prop}
\begin{proof}
 For any $\phi\in {\rm Sim}_d$ we have
 $$
 K(\phi)=\frac{d(\phi(p),\phi(q))}{d(p,q)},
 $$
 for any fixed $p,q\in S$. Now the map ${\rm Sim}_d\to S$, $\phi\mapsto\phi(p)$, is continuous in the compact-open topology for any fixed $p$, and $d:S\times S\to\R^+$ is continuous by definition.
\end{proof}

We are interested in the stabiliser of $\infty$ in $\cM_d$, that is,
$$
{\rm Stab}_d(\infty):={\rm Stab}_{\cM_d}(\infty)=\{\phi\in\cM_d\;|\;\phi(\infty)=\infty \}.
$$

\begin{prop}\label{prop:K}
${\rm Stab}_d(\infty)={\rm Sim}_d$.
\end{prop}

\begin{proof}
We show first that ${\rm Sim}_d\subseteq {\rm Stab}_d(\infty)$. Supposing the contrary, assume that there exists a $\phi\in {\rm Sim}_d$ such that $\phi(\infty)=p\in S$. Then for any $q\in S$, $q\neq p$, we would have
$$
+\infty=d\left(\phi^{-1}(p),\phi^{-1}(q)\right)=K_{\phi^{-1}}\cdot d(p,q),
$$
a contradiction.

To show that ${\rm Stab}_d(\infty)\subseteq {\rm Sim}_d$ we only have to prove that any element $\phi\in{\rm Stab}_d(\infty)$  is a similarity.
For this, we consider an arbitrary $\phi\in{\rm Stab}_d(\infty)$ and  we fix two arbitrary points $p,q\in S$. We also  consider the quantity
$$
K(\phi,p,q)=\frac{d\left(\phi(p),\phi(q)\right)}{d(p,q)}.
$$
Now for any $r\in S$, let $\fp=(r,q,\infty,p)$. Relations
$$
|\X^d_1|(\fp)=|\X^d_1|\left(\phi(\fp)\right)

\;\;\text{and}\;\;
|\X^d_2|(\fp)=|\X^d_2|\left(\phi(\fp)\right),
$$
yield
$$
K(\phi,p,q)=K(\phi,p,r)\;\;\text{and}\;\;K(\phi,p,q)=K(\phi,q,r).
$$
From the left equality it follows that $K(\phi,p,q)$ does not depend on $q$. But then the right equality shows that it also does not depend on $p$. Since $p,q$ are  arbitrary, $K=K(\phi)$
and the proof is complete.
\end{proof}

In view of Propositions \ref{prop:homoth} and \ref{prop:K} we obtain:

\begin{cor}\label{cor:K} If the similarity homomorphism $K$ is onto, then
\begin{equation*}
{\rm Sim}_d/{\rm Isom}_d\simeq \R^+_*,
\end{equation*}
where ${\rm Isom}_d$ is the subgroup of $\cM_d$ consisting of isometries and $\R^+_*$ is the multiplicative group of positive real numbers. Thus we have the short exact sequence
\begin{equation}\label{eq:short-ex}
1\to {\rm Isom}_d\to {\rm Sim}_d\to \R^+_*\to 1.
\end{equation}
\end{cor}

\begin{rem}
Note that if there exists an inverse $L:\R^+_*\to{\rm Sim}_d$ to  the similarity homomorphism  $K$ of the short exact sequence \ref{eq:short-ex} such that $L\circ K$ is the identity homomorphism when restricted to ${\rm Isom}_d$, then by Splitting Lemma, see \cite{H}, we have
$$
{\rm Sim}_d\simeq \R^+_*\rtimes{\rm Isom}_d,
$$
the semidirect product of $R^+_*$ and ${\rm Isom}_d$. This is for instance the case where $S=\partial\bH^n_\K$ and $d=d_\fH$: $L$ maps each positive $\delta$ to $D_\delta$ and $L\circ K=id.$ when restricted to ${\rm Isom}_{d_\fH}$.
\end{rem}

We close this section with some remarks concerning inversions. In general, an {\it inversion} between two distinct  points $r,r'\in\overline{S}$ is a map $\phi_{r,r'}\in\cM_d$ such that $\phi^2=id.$ and $\phi(r)=r'$. Of course, an inversion may or may not exist in an arbitrary $\cM_d$. But if it does, it satisfies properties listed in the next proposition; for simplicity, we only treat the case where $r=\infty$ and $r'=o$ is an arbitrary point in $S$.

\begin{prop}\label{prop:invol}
 Let $\phi\in\cM_d$ be an inversion between $o$ and $\infty$. Then there exists a $\beta>0$ such that
 \begin{equation*}\label{eq:invol-cond1}
  d(o,p)\cdot d(o, \phi(p))=\beta^2,
 \end{equation*}
for each $p\in S$ other than $o$, $\infty$. Moreover,
\begin{equation*}\label{eq:invol-cond2}
 d(\phi(p),\phi(q))=\frac{d(p,q)}{d(o,p)\cdot d(o,q)}\cdot \beta^2,
\end{equation*}
for each $p,q\in S$ other than $o$, $\infty$.
\end{prop}
\begin{proof}
 We consider two arbitrary points $p,q\in S$ other than $o$ and $\infty$ and the quadruple $\fp=(\infty,o,$ $p,q)$. Since $\phi\in\cM_d$ we have
 $
 |\X^d_1|(\fp)=|\X^d_1|\left(\phi(\fp)\right),
 $
 which yields
 $$
 \frac{d(o,q)}{d(o,p)}=\frac{d(\phi(p),o)}{d(\phi(q),o)}.
 $$
 Thus the quantity $d(o,p)\cdot d(0,\phi(p))$ is constant for each $p$. By setting $p=p_0$ for some arbitrary $p_0$ other that $o$ and $\infty$ and by letting $\beta$ be the positive square root of $d(o,p_0)\cdot d(0,\phi(p_0))$, we obtain the first relation of our proposition.

On the other hand, from
 $
 |\X^d_2|(\fp)=|\X^d_2|\left(\phi(\fp)\right),
 $
 we have
$$
 \frac{d(p,q)}{d(o,p)}=\frac{d(\phi(p),\phi(q))}{d(o,\phi(q))}.
 $$
 Therefore,
 $$
 d(\phi(p),\phi(q))=d(o,\phi(q))\cdot\frac{d(p,q)}{d(o,p)}=\frac{d(p,q)}{d(o,p)\cdot d(o,q)}\cdot \beta^2.
 $$
 \end{proof}

  Note finally that from Proposition \ref{prop:invol} it follows that inversion $\phi$ leaves invariant the metric sphere $$S^\beta_d=\{p\in S\;|\; d(p,o)=\beta\}.$$

\section{M\"obius Rigidity}\label{sec:proof1}
In this section we prove our results. In Section \ref{sec:prep-lem} we prove an elementary lemma which will lead us to the proof of Theorems  \ref{thm:main1}, \ref{thm:main}, \ref{thm:mainw2} and \ref{thm:pt} which are in Sections \ref{sec:main1}, \ref{sec:main}, \ref{sec:mainw2} and \ref{sec:pt}, respectively.

\subsection{The Preparatory Lemma}\label{sec:prep-lem}
 The following holds.
\begin{lem}\label{lem:step1} We refer to the conditions stated in the introduction. Then conditions {\bf (Sim)} and {\bf (Top)} together imply the following.
\begin{enumerate}
\item If $D_\delta$ is a dilation, then
$K(\delta)=\delta^\alpha$,
where $\alpha\in(0,1]$.
\item The rotational group $\rF(n-1)$ is in ${\rm Isom}_d(\partial\bH^n_\K)$.
\item The group of left translations of $\fH_\K$ is in ${\rm Isom}_d(\partial\bH^n_\K)$.
 \end{enumerate}
\end{lem}
\begin{proof}
 To prove (1),  we will show that if $D_d$ is a dilation, then there exists an $\alpha\in(0,1]$ such that for each $p,q$,
 \begin{equation*}\label{eq:dil}
  d\left(D_\delta(p),D_\delta(q)\right)=\delta^\alpha\cdot d(p,q).
 \end{equation*}
 For this, set $L=\log K$. Then $L$ is a continuous function satisfying the functional equation
 $$
 L(\delta_1\cdot\delta_2)=L(\delta_1)+L(\delta_2),\quad \text{for all}\;\;\delta_1,\delta_2>0.
 $$
 Hence $L(\delta)=\alpha\log (\delta)$ and therefore
$K(\delta)=\delta^\alpha$ for some non-zero $\alpha$ as desired.
Now $\alpha$ cannot be negative, for if that was the case then for every $\delta>0$ we would have
$
d\left(o,(\delta,0)\right)=\delta^\alpha\cdot d\left(o,(1,0)\right),$
where we have written  $({\bf e}_1,{\bf 0}_{\Im(\K)})=(1,0)$, $(\delta{\bf e}_1,{\bf 0}_{\Im(\K)})=(\delta,0)$. Since the function
$f(\delta)=$ $d\left(o,(\delta,0)\right)$ is continuous due to the continuity of $d$, by letting $\delta$ tending to $+\infty$ we have a contradiction.
The proof that $\alpha$ is in $(0,1]$ lies after the proof of (3).

To prove (2), we observe that from  condition ({\bf Top}) we have that the group $\rF(n-1)$ is compact in the compact-open topology arising from $d$.  Therefore its $K$-image in the multiplicative group $\R^+_*$ has to be compact, i.e., equal to 1. Thus the continuous $K$ is constant in $\rF(n-1)$ and thus equal to $1$.

 Finally, to prove (3), let $T$ be a translation and set $\lambda=K(T)>0$. We have to show that $\lambda=1$; asserting the contrary, suppose with no loss of generality that $\lambda<1$. For any $p\in\partial\bH^n_\K$, the sequence $T^{\nu}(p)$ tends to $\infty$ as $\nu\to +\infty$ in the $d_\fH$-topology. On the other hand,
 $$
 d(p,T^{\nu}(p))\le d(p,T(p))+\dots+d(T^{\nu-1}(p),T^{\nu}(p))\le(1+\dots+\lambda^{\nu-1}(p)).
 $$
 But this is bounded in the $d$-topology and this contradicts ({\bf Top}).

 We finally prove that $K(\delta)=\delta^\alpha$, $\alpha\in(0,1]$. For this, observe that for every two arbitrary $\delta_1, \delta_2>0$ we have
$$
d\left(o, D_{\delta_1+\delta_2}(1,0)\right)=(\delta_1+\delta_2)^\alpha\cdot d\left(o,(1,0)\right)
$$
and also by invariance of translations and triangle inequality,
\begin{eqnarray*}
 d\left(o, D_{\delta_1+\delta_2}(1,0)\right)&=&d\left(o, (\delta_1+\delta_2,0)\right)\\
 &=&d\left((-\delta_1,0),(\delta_2,0)\right)\\
 &\le&\delta_1^\alpha\cdot d\left((-1,0),o\right)+\delta_2^\alpha\cdot d\left((1,0),o\right)\\
 &=&\delta_1^\alpha\cdot d\left((1,0),o\right)+\delta_2^\alpha\cdot d\left((1,0),o\right)\\
 &=&\left(\delta_1^\alpha+\delta_2^\alpha\right)\cdot d\left(o, (1,0)\right).
\end{eqnarray*}
But also $ d\left(o, D_{\delta_1+\delta_2}(1,0)\right)=\left(\delta_1+\delta_2\right)^\alpha\cdot d\left(o, (1,0)\right)$; therefore, $(\delta_1+\delta_2)^\alpha\le \delta_1^\alpha+\delta_2^\alpha$. By putting $\delta_1=\delta_2=\delta$ we have
$2^\alpha\cdot \delta^\alpha\le 2\cdot\delta^\alpha$, i.e., $2^{1-\alpha}\ge 1$ which can happen only if $\alpha\in(0,1]$.
 This concludes the proof.
\end{proof}

\begin{rem}\label{rem:equid}
 (Notation convention) It is clear from the proof of Lemma \ref{lem:step1} that
 $$
  d\left(o,(\pm{\bf e}_i,{\bf 0}_{\Im(\K)})\right)=d\left(o,({\bf e}_1,{\bf 0}_{\Im(\K)})\right).
 $$
 where ${\bf e}_i$ is the $i$-coordinate vector of $\K^{n-1}$. All these quantities shall be hereafter denoted by $d\left((1,0),o\right)$. Also
 $$
 d\left(o,({\bf 0}_{\K^{n-1}},\pm{\bf f}_i)\right)=d\left(o,({\bf 0}_{\K^{n-1}},{\bf f}_1)\right).
 $$
 where ${\bf f}_i$ is the $i$-unit vector of $\Im(\K)$. This is is true because according to Proposition \ref{prop:K}, $J$ is in ${\rm Sim}_d$; in particular,  $K(J)=1$. All these quantities shall be denoted by $d\left((0,1),o\right)$.
\end{rem}

\subsection{Proof of Theorem \ref{thm:main1}}\label{sec:main1}
For the case $\K=\R$ (case (1) of the theorem), let $\zeta,\zeta'\in\R^{n-1}$. We have
\begin{eqnarray*}
 d\left(\zeta,\zeta'\right)&=& d\left(\zeta-\zeta',0\right)\\
 &=& d\left(\|\zeta-\zeta'\|\cdot\mu,0\right),\quad \mu\in{\rm O}(n-1),\\
 &=& d(0,1)\cdot\|\zeta-\zeta'\|^\alpha=d(0,1)\cdot d_\fH^\alpha(\zeta,\zeta').
\end{eqnarray*}
Here, the penultimate equation follows from the transitive action of ${\rm SO}(n-1)$ on $S^{n-1}$: We may map any $r\in S^{n-1}$ to $1={\bf e}_1$ via an element of ${\rm SO}(n-1)$.

For $\K\neq\R$ (case (2) of the theorem), let  $$\beta_1=\max\left\{d\left((1,0),o\right), d\left((0,1),o\right)\right\},\;\;
\beta_2=\min\left\{d\left((1,0),o\right), d\left((0,1),o\right)\right\}.$$
Since for each $p=(\zeta,v)\in\fH_\K$:
\begin{eqnarray}
 &&\label{eq:dil1}
 d\left((\zeta,0),o\right)=d\left((1,0),o\right)\cdot\|\zeta\|^\alpha,\\
 &&\label{eq:dil2}
 d\left((0,v),o\right)=d\left((0,1),o\right)\cdot |v|^{\alpha/2},
\end{eqnarray}
from \ref{eq:dil1} and \ref{eq:dil2} we obtain the inequalities:
\begin{eqnarray*}
 &&\label{ineq:dil1}
 \beta_2\cdot \|\zeta\|^\alpha\le d\left((\zeta,0),o\right)\le\beta_1\cdot \|\zeta\|^\alpha,\\
 &&\label{ineq:dil2}
 \beta_2\cdot |v|^{\alpha/2}\le d\left((0,v),o\right)\le\beta_1\cdot |v|^{\alpha/2}.
\end{eqnarray*}
Therefore,
\begin{equation*}
 \beta_2\cdot d^\alpha_\fH(o,p)\le \left(d^{4/\alpha}\left((\zeta,0),o\right)+d^{4/\alpha}\left((0,v),o\right)\right)^{\alpha/4}\le\beta_1\cdot d^\alpha_\fH(o,p).
\end{equation*}
By triangle inequality and $d$-invariance of  translations we have
$$
d(p,o)\le d\left((\zeta,0),o\right)+d\left((0,v),o\right).
$$
We apply H\"older's inequality  with exponent $4/\alpha$   to obtain
\begin{eqnarray*}
 d(p,o)&\le&2^{(4-\alpha)/4}\cdot\left(d^{4/\alpha}\left((\zeta,0),o\right)+d^{4/\alpha}\left((0,v),o\right)\right)^{\alpha/4}\\

 &\le&\beta_1\cdot 2^{(4-\alpha)/4}\cdot\left(d_\fH^4\left((\zeta,0),o\right)+d_\fH^4\left((0,v),o\right)\right)^{\alpha/4}\\
 &=&\beta_1\cdot 2^{(4-\alpha)/4}\cdot d^\alpha_\fH(p,o),
\end{eqnarray*}
and Theorem \ref{thm:main1} follows.\qed

We wish to address  at one important issue at this point. As we have underlined  in the introduction, condition {\bf ($\alpha$-H\"ol)} is rather inadequate to describe in full the nature of metrics which satisfy {\bf (Sim)} and {\bf (Top)}; there might exist metrics which satisfy {\bf ($\alpha$-H\"ol)} on the one hand but on the other, their nature might be entirely different from that of $d_\fH$. A concrete example which illustrates this matter is that  of  the Carnot-Carath\'eodory metric $d_{cc}$ (for details about $d_{cc}$, see for instance \cite{CDPT}); it suffices only to consider the case $\K=\C$, $n=2$. Certainly, since ${\rm Sim}_{d_{cc}}={\rm Sim}_{d_\fH}$, $d_{cc}$ automatically  satisfies  conditions {\bf (Sim)} and {\bf (Top)} of Theorem \ref{thm:main1} (but satisfies neither {\bf (Inv)} nor {\bf (G)}). Since $\alpha=1=\beta_1$,  we have from Theorem \ref{thm:main1} that for each $p$,
$$
d_{cc}(p,o)\le 2^{3/4}\cdot d_{\fH}(p,0).
$$
The reader is invited to compare this inequality to the well known estimate
 $$
 \pi^{-1/2}\cdot d_\fH(p,o)\le d_{cc}(p,o)\le \cdot d_\fH(p,o).
 $$

\subsection{Proof of Theorem \ref{thm:main}}\label{sec:main}
The following lemma is an immediate corollary of Proposition \ref{prop:invol}; it will be needed for the proof of Theorem \ref{thm:main}.

\begin{lem}\label{lem:confinv}
 If inversion  $I$ is in $\cM_d(\fH_\K)$, then
 \begin{equation*}\label{eq:inv-cond1}
  d(o,p)\cdot d(o, I(p))=d(o,p_0)\cdot d\left(o,I(p_0)\right),
 \end{equation*}
 where $p_0$ is any point other than $o,\infty$ and for each $p\in\fH_\K$ other than $o$, $\infty$. Moreover
\begin{equation*}\label{eq:inv-cond2}
 d(I(p),I(q))=\frac{d(p,q)}{d(o,p)\cdot d(o,q)}\cdot d(o,p_0)\cdot d\left(o,I(p_0)\right),
\end{equation*}
for each $p,q\in\fH_\K$ other than $o$, $\infty$.
\end{lem}

We proceed now to the proof of Theorem \ref{thm:main}. Recall that $J$ is in ${\rm Sim}_d$ and $K(J)=1$. By choosing $p_0=(1,0)$ which we shall denote simply with 1, the formulae in the statement of the proposition also read as:
$$
d(o,p)\cdot d(o, I(p))=d^2(o,1),\;\; d(I(p),I(q))=\frac{d(p,q)}{d(o,p)}\cdot d(o,q)\cdot d^2(o,1).
$$
Let now $p=(\zeta,v)\in\fH_\K$. Then by setting $\cA(p)=-\|\zeta\|^2+v$ we have
\begin{eqnarray*}
 d\left(I(p),o)\right)&=&d\left(\left(\zeta\cdot\frac{\cA(\overline{p})}{|\cA(p)|^2},\frac{\overline{v}}{|\cA(p)|^2}\right),o\right)\\
 &=&d\left(D_{|\cA(p)|^{-1}}\left(\zeta\cdot\frac{\cA(\overline{p})}{|\cA(p)|},\overline{v}\right),o\right)\\

&=&|\cA(p)|^{-\alpha}\cdot d\left(\left(\frac{ \cA(p)}{|\cA(p)|}\cdot\overline{\zeta},v\right),o\right)\\
&=&|\cA(p)|^{-\alpha}\cdot d\left(\left(-\overline{\zeta},v\right),o\right)\\

&=&|\cA(p)|^{-\alpha}\cdot d\left(\left(-\zeta,-v\right),o\right)\\
&=&|\cA(p)|^{-\alpha}\cdot d(p^{-1},o)\\
&=&|\cA(p)|^{-\alpha}\cdot d\left(T_p (p^{-1}),o\right)\\
&=&|\cA(p)|^{-\alpha}\cdot d\left(o,p\right).
\end{eqnarray*}
This gives
$$
d\left(I(p),o)\right)=\frac{d^2(o,1)}{d(o,p)}=|\cA(p)|^{-\alpha}\cdot d\left(o,p\right),
$$
and by Lemma \ref{lem:confinv} we conclude
$$
d^2(o,p)=d^2(o,1)\cdot |\cA(p)|^{\alpha}=d^2(o,1)\cdot d_\fH^{2\alpha}(o,p),
$$
which proves Theorem \ref{thm:main}.\qed

Note that implicit in the proof lies the following formula for the inversion $I$: For each $p$,
$$
I(p)=\left(D_{|\cA(p)|^{-1}}\circ J\circ R_{-\frac{\cA(p)}{|\cA(p)|}}\circ J\right)\left(p^{-1}\right).
$$
This formula only dictates the manner that inversion $I$ is defined; it cannot be regarded as formula which expresses $I$ as a compound of similarities, due to the presence of $p^{-1}$.

\subsection{Proof of Theorem \ref{thm:mainw2}}\label{sec:mainw2}

Suppose now that {\bf (G)} holds, that is,
\begin{equation*}\label{eq:pyth1}
d(o,p)=\left(d^{4/\alpha}\left((\zeta,0),o\right)+d^{4/\alpha}\left((0,v),o\right)\right)^{\alpha/4}.
\end{equation*}
Then $\beta_2\cdot d^\alpha_\fH(o,p)\le d(o,p)\le \beta_1\cdot d^\alpha_\fH(o,p)$.
From $d$-invariance and $d_\fH$-invariance of translations we have:
$$
\beta_2\cdot d_\fH^\alpha(p,q)\le d(p,q)\le \beta_1\cdot  d_\fH^\alpha(p,q),
$$
therefore $d$ and $d_\fH^\alpha$ are metrically equivalent. But there exists also a $\beta\ge 1$ depending on $\beta_1,\beta_2$
 such that
$$
\frac{1}{\beta}\cdot d_\fH^\alpha(p,q)\le d(p,q)\le \beta\cdot d_\fH^\alpha(p,q).
$$
If  {\bf (Eq)} holds, then $\beta_1=\beta_2=\beta$ and  $d=\beta\cdot d_\fH^\alpha$. In this manner we have proved Theorem \ref{thm:mainw2}.
\qed

We remark here that condition {\bf (biLip)} is equivalent to that the identity map from $(\partial\bH^n_\K, d)$ to $(\partial\bH^n_\K, d_\fH)$ is bi-Lipschitz.

\medskip

Finally, we prove:

\begin{prop}\label{prop:p-l}
Suppose that condition {\bf (Sim)} holds. Then conditions {\bf (G)} is equivalent to the following:
\begin{enumerate}
\item [{{\bf (P-L)}}] For $\alpha\in (0,1]$ and for each $p,q\in\fH$,
\begin{eqnarray*}
\left|p*q\right|^{4/\alpha}+ \left|p^{-1}*q\right|^{4/\alpha}+\left|p*q^{-1}\right|^{4/\alpha}+\left|p^{-1}*q^{-1}\right|^{4/\alpha}&=&\\
2\left(\left|\Pi_{\K^{n-1}}(p*q)\right|^{4/\alpha}+\left|\Pi_{\K^{n-1}}\left(p^{-1}*q\right)\right|^{4/\alpha}\right)&+&\\
\left|\Pi_{\Im(\K)}(p*q)\right|^{4/\alpha}+ \left|\Pi_{\Im(\K)}\left(p^{-1}*q\right)\right|^{4/\alpha}+\left|\Pi_{\Im(\K)}\left(p*q^{-1}\right)\right|^{4/\alpha}+\left|\Pi_{\Im(\K)}\left(p^{-1}*q^{-1}\right)\right|^{4/\alpha},
\end{eqnarray*}
where $\Pi_{\K^{n-1}}$ and $\Pi_{\Im(\K)}$ are projections of $\fH$ to $\K^{n-1}$ and $\Im(\K)$, respectively.
\end{enumerate}
\end{prop}
\begin{proof}
 To show that {\bf (P-L)} implies {\bf (G)}, just set $q=o$ and use the fact that $\left|p^{-1}\right|=|p|$, due to $d$-invariance of translations which follows from {\bf (Sim)} (cf. Lemma \ref{lem:step1}):
 $$
d\left(o,p^{-1}\right)=d\left(T_p(o),T_p\left(p^{-1}\right)\right)=d(o,p).
$$
To show that {\bf (G)} implies {\bf (P-L)}, just apply {\bf (G)} for the points $p*q$, $p^{-1}*q$, $p*q^{-1}$ and $p^{-1}*q^{-1}$.
\end{proof}

\subsection{Proof of Theorem \ref{thm:pt}}\label{sec:pt}

Assuming condition {\bf ($\alpha$-Met)} holds, we have for each $p,q,r,s$ points in $\partial\bH^n_\K$ that
\begin{eqnarray*}
d^\alpha_\fH(p,r)\cdot d^\alpha_\fH(q,s)&\le&\left(d_\fH(p,q)\cdot d_\fH(r,s)+d_\fH(p,s)\cdot d_\fH(r,q)\right)^{\alpha}\\
&\le& d^\alpha_\fH(p,q)\cdot d^\alpha_\fH(r,s)+d^\alpha_\fH(p,s)\cdot d^\alpha_\fH(r,q),
\end{eqnarray*}
where the first inequality follows from the Ptolemaean property of $d_\fH$ and the second inequality holds because $\alpha\le 1$. Thus $d$ satisfies {\bf (Ptol)}.

Assume now that  {\bf ($\alpha$-Met)} and {\bf (Circ)} hold. Pick any $p,q,r,s$ lying in the Ptolemaean circle $\sigma$ and suppose with no loss of generality that $p$ and $s$ separate $q$ and $r$. We may also normalise so that $q=o$ and $r=\infty$. Then
\begin{eqnarray*}
 d_\fH^\alpha(s,o)+ d_\fH^\alpha(p,0)= d_\fH^\alpha(p,s).
\end{eqnarray*}
Therefore,
$$
d_\fH(p,s)=\left(d_\fH^\alpha(s,o)+ d_\fH^\alpha(p,0)\right)^{1/\alpha}\ge d_\fH(s,o)+ d_\fH(p,0),
$$
since $1/\alpha\ge 1$. But this contradicts triangle inequality unless $\alpha=1$. Moreover, we have that then $\sigma$ is an $\R$-circle and the proof is complete. \qed