TITLE: How to calculate combinations with duplicates?
QUESTION [11 upvotes]: This site http://webviewworks.com/math/index.html explains that the word BABY has 7 combinations(AB, AY, BA, BB, BY, YA, YB).
Can you explain me how is it beeing calculated?
It's not:

REPLY [1 votes]: What you are doing is, making permutation. $^nC_r$ is used when making combination. The permutation you are making has B repeated twice so you cannot use $^nP_r$ either. How to find the total number of possible permutations is already explained in other answers.


TITLE: solution to non-linear difference equation
QUESTION [1 upvotes]: Did anyone ever come across the global solution for a non-linear difference equation that looks like this:
$y(t+1)=y(t)+a+b \sqrt{c y(t)+d}$.
The initial condition is $y(0)=y_0$, and a,b,c and d are real numbers.
Any help is more than welcome!
Thanks a lot in advance.
PS: Mathematica is able to find a solution this that looks like
$
\left\{\left\{y[t]\to a t^2+\text{y0}-\frac{2 t \sqrt{a (d+c \text{y0})}}{\sqrt{c}}\right\},\left\{y[t]\to a t^2+\text{y0}+\frac{2 t \sqrt{a (d+c \text{y0})}}{\sqrt{c}}\right\}\right\}.
$
I could use this of course but I'd like to know whether it is possible to solve this analytically by hand. Or, basically, how to get Mathematica's solution. There should be some way of transforming the problem by substitution. But - how?
PPS:  Harald Hanche-Olsen just noticed that Mathematica's solution only holds under a special case for $t=1$.

REPLY [1 votes]: You can try a solution on the form $y(t)=At^2+Bt+C$. Substituting that in the recurrence relation, you get $2At+A+B-a=b\sqrt{cAt^2+cBt+cC+d}$. Square this, collect similar powers of $t$, and compare coefficients to get three equations to determine $A$, $B$, $C$.
But then you need a fourth equation to satisfy the initial condition, which indicates that you cannot find a solution on this form in general. For special choices of the data it will be possible, however. Trying higher degree polynomials seems to lead nowhere.


TITLE: Is there anything special about ebonite and fur?
QUESTION [2 upvotes]: I'm from Czech Republic, born 1980.  From elementary school, we all remember this mantra:

When ebonite rod is rubbed with fox fur, electrostatic charge is created.
  Electrostatic charge is created by rubbing ebonite rod with fox fur.
  Rubbing ebonite fur with fox fur creates electrostatic charge.

Etc. ad nauseam.
So...
Is there anything special about the combination of ebonite and fox fur that makes it especially useful for teaching kids about electricity?
Does there even exist a clear distinction between things that do and things that don't create electrostatic charge by rubbing?

The irony: I can't remember ever hearing the word 'ebonite' in any other context than this particular strange example.  (I never even knew what ebonite was until about 15 minutes ago when I googled it.)

REPLY [1 votes]: The electrons in fur are much less tightly bound than electrons in ebonite (very strong relative bond, ebonite is at the bottom of the negative Triboelectric series, see [1]) and hence ebonite gets a strong relative negative charge [1].
"A material towards the bottom of the Triboelectric series table, when touched to a material near the top of the series, will acquire a more negative charge."
So to answer the question, this particular combination of materials will cause a very strong electrostatic effect. Ebonite is very strong negatively static charged.


TITLE: Compute integral of $e^{i\sqrt{x^2+a^2}}/(a^2+x^2)$ over $\mathbb{R}$, why is it wrong?
QUESTION [3 upvotes]: I'd like to compute this integral:
$$\int_{-\infty}^{+\infty} \frac{e^{i\sqrt{x^2+a^2}}}{a^2+x^2}\mathrm{d}x$$
But I'm a little bit rusty with complex analysis ... What I've been trying to do so far is the following using cauchy lemma ($\gamma$ goes straight from $-\infty$ to $+\infty$ and do half-circle from $+\infty$ to $-\infty$ on the upper part of the complex plan):
$$\int_{-\infty}^{+\infty} \frac{e^{i\sqrt{x^2+a^2}}}{a^2+x^2}\mathrm{d}x=\int_{\gamma} \frac{e^{i\sqrt{z^2+a^2}}}{a^2+z^2}\mathrm{d}z = 2\pi i \;\mathrm{Res}(ai)$$
And then computing the residue from the pole on the imaginary axis:
$$\mathrm{Res}(ai) = \frac{e^{i\sqrt{a^2-a^2}}}{2ia} = \frac{1}{2ia}$$
and then the result is obvisously wrong, as this integral cannot have imaginary part equal zero ...
So I wonder:

Why is it wrong to do that ?
How can I compute this integral ? Numerically I guess the result is gonna have something to do with Hankel function but I'd like a mathematical proof.

Thank you.

REPLY [0 votes]: Thanks to Maxim I was able to compute this integral, here is a more detailed solution:
We start with the substitution $u^2a^2=x^2+a^2$, that leads to $\mathrm{d}x = ua^2\mathrm{d}u/\sqrt{u^2-1}$ thus:
$$\int_{-\infty}^{+\infty} \frac{e^{i\sqrt{x^2+a^2}}}{a^2+x^2}\mathrm{d}x =
2\int_{0}^{+\infty} \frac{e^{i\sqrt{x^2+a^2}}}{a^2+x^2}\mathrm{d}x =
\frac{2}{a}\int_{1}^{+\infty} \frac{e^{iau}}{u\sqrt{u^2-1}}\mathrm{d}u$$
We can now write it as a fourier tranform for $u>1$:
$$\frac{2}{a}\int_{1}^{+\infty} \frac{e^{iau}}{u\sqrt{u^2-1}}\mathrm{d}u = \frac{2}{a}\mathcal{F}\left(u\mapsto\frac{1}{u\sqrt{u^2-1}}\right)(a) =  \frac{1}{ai}\int_a^{+\infty}\mathcal{F}\left(u\mapsto\frac{2}{\sqrt{u^2-1}}\right)(t)\mathrm{d}t$$
we can conclude introducing the Hankel function:
$$\frac{1}{ai}\int_a^{+\infty}\mathcal{F}\left(u\mapsto\frac{2}{\sqrt{u^2-1}}\right)(t)\mathrm{d}t = \frac{\pi}{a}\int_a^{+\infty}H^{(1)}_0(t)\mathrm{d}t $$


TITLE: Fine-Tuning, the Hiearchy Problem, and Mass in the Standard Model
QUESTION [2 upvotes]: In Chapter 1 of his book String Theory in a Nutshell, Kiritsis states the following.

The [Standard M]odel is unstable as we increase the energy (hierarchy problem of mass scales) and the theory loses predictivity as one starts moving far from current accelerator energies and closer to the Planck scale.  Gauge bosons are protected from destabilizing corrections because of gauge invariance.  The fermions are equally protected due to chiral symmetries.  The real culprit is the Higgs boson.

Let $\phi$ be a quantum field and define $G(p,q):=\langle \widehat{\phi}(p)\widehat{\phi}(q)\rangle$ (this is of course a vacuum expectation value).  Then, it turns out that, $G(p,q)=\delta (p+q)H(p^2)$.  Under the assumption $H$ has a unique pole which is a non-negative real number (anybody have a proof of why this would be true in general?), define the mass of $\phi$ to be the square-root of this pole.
My understanding was that this mass is given to us by experiment; instead, in calculations, we use this condition to calculate the counter-terms, not the mass.  Is my understanding of this in-correct?
For some reason I was interpreting his words to mean something along the lines of "In quantum field theory, you can calculate particle masses via perturbation theory, and sometimes certain symmetries tells us that anomalies cannot arise, but in the case of the Higgs boson, there is no such symmetry, and so for the result to come out correct, some crazy cancellation has to occur." (now that I think about it, my assumption regarding his meaning here must have been heavily influenced by what I have been told about the Hierarchy Problem from other sources).  This of course, doesn't jive with my aforementioned understanding of masses in quantum field theory (i.e. that you don't calculate them).
Can somebody explain to me the Hierarchy Problem given my understanding of mass in quantum field theory?  What precisely are the theorems that "protect" gauge bosons and fermions?  I've seen time and again people talk about some how finely-tuned cancellation needs to happen regarding the Higgs.  Could somebody tell me very explicitly just what this cancellation is, and where and how it comes up?  How does this imply the instability of the Standard Model?
Thanks much in advance.

REPLY [3 votes]: The gauge bosons such as the photon and chiral fermions may be massless due to the symmetry – gauge symmetry and chiral symmetry, respectively. We may consistently require the theory to respect these symmetries at the exact level, including all quantum corrections.
For the gauge symmetry, it means that $1+1=2$ polarizations of the photon or another particle become unphysical. Only 2 transverse polarizations are left which is only possible if the little group is just $SO(2)$ i.e. if the particle stays massless. So the gauge symmetry protects the masslessness of the photon to all orders. The degree of freedom for the longitudinal photon is just not there – so the mass can't arise.
The same holds for gluons i.e. confined non-Abelian gauge fields although the mass is only meaningful at short distances.
For W-bosons and Z-bosons, the mass is nonzero but at energies much higher than the Higgs scale from the Higgs that gives them the mass, all these masses may be neglected and the longitudinal W-boson or Z-boson may be interpreted as a field from the Higgs multiplet again while the field from the gauge field decouples just like for the photon again. So for spontaneously broken gauge bosons, the mass is protected against large corrections and divergences that would go beyond the tree-level Higgs mass which is however nonzero.
For fermions, the mass term mixes two 2-component spinors, $m\chi_\alpha \psi^\alpha$ etc. However, one may impose a $U(1)$ symmetry rotating the phase of $\chi$ separately (or just $\psi$ separately) which bans the mass term. It is OK for gauge fields and fermions to postulate that Nature likes the symmetries above, at least in an approximate form, which simply explains that the mass is zero (or small if the symmetries are approximate).
For a scalar field, there is no symmetry that could ban the mass terms. And indeed, all corrections to the mass that could occur – finite or divergent ones – do occur. So the loop corrections to $m_H^2$ are of order $c\Lambda^2$, here $\Lambda$ is a cutoff, and they have to cancel with the accuracy $10^{30}$ (Planck scale cutoff is 15 orders of magnitude above the Higgs mass and it's being squared). Cancellation of such terms with the relative accuracy $10^{-30}$ which is needed is "unlikely" for a generic value of all the parameters – the probability is of order $10^{-30}$ by itself. If you reject the anthropic selection etc. and assume that the calculation of the probability above is basically right, then the light Higgs mass is extremely unlikely and an explanation – one that modifies the calculation of the probability that the cancellation is this precise – is needed.
SUSY, for example, may cancel the quadratic corrections (one can see it in two ways: either the Higgs has an approximately equally heavy higgsino that may be protected by the chiral symmetry as a fermion; or from SUSY cancellation to the Higgs scalar mass between particles and their superpartners). The expected order-of-magnitude Higgs mass is of order the superpartners' masses (mass differences) and not $\Lambda$. One may still ask why the tree-level Higgs mass is much smaller than the Planck mass but this is an easier problem to explain by various mechanisms.


TITLE: Are profinite groups of cardinality $|\mathbb{R}|$ determined by their finite quotients?
QUESTION [8 upvotes]: Question: Let $G,H$ be profinite groups of cardinality $|\mathbb{R}|$, with the same finite quotients (here I only consider quotients by normal, open subgroups). Then are $G$ and $H$ isomorphic?
Background: First, note that one needs some sort of cardinality assumption, as otherwise one could take $G=\mathbb{F}_2^I$ and $H=\mathbb{F}_2^J$ where $I$ and $J$ are infinite of distinct cardinality.  Moreover, it is NOT sufficient to just require that $G$ and $H$ have the same cardinality, as one could take $G=\mathbb{F}_2^I\times \mathbb{F}_3^J$ and $H=\mathbb{F}_2^J\times\mathbb{F}_3^I$.
As for motivation, if we take $C$ to be an affine hyperbolic curve over the algebraic closure of a finite field, then it is known which subgroups occur as finite quotients of the etale fundamental group. Given this phrasing, I assume that this is not enough information to ecover the actual etale fundamental group, and so I wanted an explicit counterexample.
EDIT: Due to the comments, I think a better question is as follows: Rather than assuming $G$ and $H$ have cardinality $\mathbb{R}$, let me assume instead that they have at most countably many homomorphisms to any finite group.
This avoids (at least) some counterexamples assuming Luzin' hypothesis.
Let me add that the answer is yes if $G$ and $H$ are topologically finitely generated. This is well know, but as its quick I include a proof:
Note that for each positive integer $n$, $G$ has only finitely many homomorphisms to fintie groups of order at most $n$. Let $G(n)$ be the kernel of all of these, and $G_n$ the corresponding finite quotient. Then $G$ is the inverse limit of the $G_n$. To see this note that the natural map $G\to\varprojlim_n G_n$ is clearly injective, and as it is surjective on every quotient it must also be surjective by compactness.
So it is enough to see that $G_n\cong H_n$. But by assumption, $G$ surjects onto $H_n$, and such a surjection must factor through $G_n$, so $G_n$ surjects onto $H_n$. Likewise $H_n$ surjects onto $G_n$ and since these are finite groups they must be isomorphic.

REPLY [1 votes]: A more geometric example is as follows: let $k$ be a countable algebraically closed field of characteristic $p$, and pick a positive integer $n$. Let $G_n = \pi_1(\mathbb{A}^n_k)$, the etale fundamental group of the $n$-dimensional affine space over $k$. By Raynaud's solution of the Abhyankar conjecture, $G_1$ admits as finite quotients (precisely) all finite groups with no prime-to-$p$ quotient. Therefore, the same holds for all $G_n$. However, the cohomological dimension of $G_n$ equals $n$, and hence $G_n\not\cong G_m$ for $n\neq m$.


TITLE: probability density function of n independant random variables
QUESTION [0 upvotes]: $X$,$Y$,$Z$,$T$ are independant random variables with probability density functions $f_{X}(x)$,$f_{Y}(y)$,$f_{Z}(z)$,$f_{T}(t)$.
What is the pdf of $U$ if $U=F(X,Y,Z,T)$ knowing the pdf of $X$,$Y$,$Z$,$T$ ?
The answer should be a function of $U$: $f_{U}(u)$
Maybe this general question has no general answer so here is a practical example:
let $U=\sqrt{(X-Y)^2+(Z-T)^2}$ and $X$,$Y$,$Z$,$T$ are uniformly distributed function between $-a$ and $a$.
With Monte-Carlo simulation, I find that the histogram (the pdf) of U is some sort of bell function. How to have its analytical or numerical value without running a Monte-Carlo spraying?

REPLY [0 votes]: You have to first find the cdf, differentiate it w.r.t $u$ to get the pdf. Let's start with the cdf computation. For $0\leq u\leq 2a$,
\begin{align}
F_U(u) &= \mathbb{P}(U\leq u)=\mathbb{P}(U^2\leq u^2)\\
&= \int_{ \left\{\substack{(x,y,z,t):-a\leq x,y,z,t \leq a\\
(x-y)^2+(z-t)^2\leq u^2}\right\}}f_X(x)f_Y(y)f_Z(z)f_T(t)dxdydzdt.
\end{align}
Here, we first integrate with respect to $y$. The range of $y$ is obtained by rearranging the constraints on $y$:
\begin{align}
(x-y)^2+(z-t)^2\leq u^2 & \implies (x-y)^2+\leq u^2-(z-t)^2\\
&  \implies \begin{cases}
x-\sqrt{u^2-(z-t)^2}\leq y\leq x+\sqrt{u^2-(z-t)^2} \\ (z-t)^2\leq u^2
\end{cases}\\
-a\leq y \leq a & \implies \begin{cases}
\max\{x-\sqrt{u^2-(z-t)^2},-a\}\leq y\leq \min\{ x+\sqrt{u^2-(z-t)^2},a\}\\
(z-t)^2\leq u^2
\end{cases}\\
&\implies \begin{cases}
\begin{cases}
x-f(u,z,t)\leq y\leq  x+f(u,z,t)\\
(z-t)^2\leq u^2
\end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\geq -a\\
x+f(u,z,t)\leq a
\end{cases}\\
\begin{cases}
-a\leq y\leq  x+f(u,z,t)\\
(z-t)^2\leq u^2
\end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\leq -a\\
x+f(u,z,t)\leq a
\end{cases}\\
\begin{cases}
x-f(u,z,t)\leq y\leq  a\\
(z-t)^2\leq u^2
\end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\geq -a\\
x+f(u,z,t)\geq a
\end{cases}\\
\begin{cases}
-a\leq y\leq  a\\
(z-t)^2\leq u^2
\end{cases} &\text{ if } \begin{cases} x-f(u,z,t)\leq -a\\
x+f(u,z,t)\geq a
\end{cases}
\end{cases},
\end{align}
where $f(u,z,t)=\sqrt{u^2-(z-t)^2}$.
Thus, we get
\begin{align}
F_U(u) &= \int_{ \left\{\substack{(x,y,z,t):-a\leq x,y,z,t \leq a\\
x-f(u,z,t)\leq y\leq x+f(u,z,t) \\ (z-t)^2\leq u^2 }\right\}}\frac{1}{16a^4}dxdydzdt\\
&=\frac{1}{16a^4}\underset{I_1}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\
x-f(u,z,t)\geq -a\\x+f(u,z,t)\leq a \\ (z-t)^2\leq u^2}\right\}}2f(u,z,t)dxdzdt }}\\
&\hspace{1cm}+ \frac{1}{16a^4} \underset{I_2}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\
x-f(u,z,t)\leq -a\\x+f(u,z,t)\leq a \\ (z-t)^2\leq u^2}\right\}}\left(x+f(u,z,t)+a \right)dxdzdt }} \\
&\hspace{1cm}+ \frac{1}{16a^4}\underset{I_3}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\
x-f(u,z,t)\geq -a\\x+f(u,z,t)\geq a \\ (z-t)^2\leq u^2}\right\}}\left(a-x+f(u,z,t) \right)dxdzdt}}  \\
&\hspace{1cm}+ \frac{1}{16a^4}\underset{I_4}{\underbrace{\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\
x-f(u,z,t)\leq -a\\x+f(u,z,t)\geq a \\ (z-t)^2\leq u^2}\right\}}2a dxdzdt}} .
\end{align}
Simplifying each integral,
\begin{align}
I_1&=2\int_{ \left\{\substack{(x,z,t):-a\leq z,t \leq a\\
-a+\sqrt{u^2-(z-t)^2}\leq x\leq a-\sqrt{u^2-(z-t)^2}\\ (z-t)^2\leq u^2\leq a}\right\}}\sqrt{u^2-(z-t)^2}dxdzdt\\
&=2\int_{ \left\{\substack{(z,t):-a\leq z,t \leq a\\
(z-t)^2\leq u^2}\right\}}2\sqrt{u^2-(z-t)^2}(a-\sqrt{u^2-(z-t)^2})dzdt\\
&=4\int_{ \left\{\substack{(s,w):-a\leq \frac{w+s}{2},\frac{w-s}{2} \leq a\\
-u\leq s\leq u}\right\}}\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})\frac{1}{2}dsdw\\
&=2\int_{ \left\{\substack{(s,w):-2a+s\leq w \leq 2a-s\\
0\leq s\leq u}\right\}}\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})dsdw \\
&\hspace{2cm}+ \frac{1}{8a^4}\int_{ \left\{\substack{(s,w):-2a-s\leq w\leq 2a+s\\
-u \leq s\leq 0}\right\}}\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})dsdw\\
&=2\int_0^u\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})2(2a-s)ds \\
&\hspace{1cm}+ \frac{1}{8a^4}\int_{-u}^0\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})2(2a+s)ds\\
&=4\int_0^u(2a-s)\sqrt{u^2-s^2}(a-\sqrt{u^2-s^2})ds\\
&=4\int_0^u  2a^2\sqrt{u^2-s^2} ds+4\int_0^u a \sqrt{u^2-s^2} \frac{1}{2}d(u^2-s^2) - 4\int_0^u (2a-s)(u^2-s^2)ds\\
&= 4\left(\frac{\pi}{2}a^2u^2-\frac{5}{3}au^3+\frac{1}{4}u^4\right).
\end{align}
\begin{align}
I_2=\int_{ \left\{\substack{(x,z,t):-a\leq x,z,t \leq a\\
x-\sqrt{u^2-(z-t)^2}\leq -a\\x+\sqrt{u^2-(z-t)^2}\leq a \\ (z-t)^2\leq u^2}\right\}}\left(x+\sqrt{u^2-(z-t)^2}+a \right)dxdzdt
\end{align}
Similarly, simplify the other integrals!


TITLE: Continuity between measure and "'cumulative" function
QUESTION [1 upvotes]: Let $\mu$, a finite measure and $F$ increasing,right continuous function which is defined as $F(x)=\mu((-\infty,x])$.
We also know that $\mu$ is continuous at a single point $x$ if $\mu(\left\{x\right\})=0$.
I would like to show that $F(x)$ is continuous at x if and only if $\mu(\left\{x\right\})=0$.
Any idea on how to start it ??

REPLY [2 votes]: Observe that $\{x\}=\bigcap_{n=1}^{\infty}(x-\frac1n,x]$ and prove that: $$\mu(\{x\})=\lim_{n\to\infty}\mu((x-\frac1n,x])$$
Further note that $\mu((x-\frac1n,x]=F(x)-F(x-\frac1n$) so the RHS of $(1)$ takes value $0$ iff (the non-decreasing function) $F$ is left continuous at $x$.
We allready know that $F$ is right continuous at $x$ so the condition of being left continuous at $x$ is the same as the condition of being continuous at $x$.


text:
\begin{document}
\title[The $p$-adic Solid-on-Solid model]
{On $p$-adic Solid-on-Solid model on a Cayley tree}

\author{O. N. Khakimov}

\address{O.N. Khakimov\\ Institute of mathematics,
29, Do'rmon Yo'li str., 100125, Tashkent, Uzbekistan.}
\email {hakimovo@mail.ru}

\maketitle

\begin{abstract} We consider a nearest-neighbor $p$-adic Solid-on-Solid
(SOS) model
 with $m+1$ spin values and coupling constant $J\in\mathbb Q_p$ on a
Cayley tree. It is found conditions under which a phase
transition does not occur for this model. It is shown that under condition
$p\mid m+1$ for some $J$ a phase transition occurs. Moreover, we give
criterion of boundedness of
$p$-adic Gibbs measures for the $m+1$-state SOS model.
\end{abstract}
\maketitle

{\bf{Key words.}} $p$-adic number, $p$-adic SOS model, Cayley tree,
$p$-adic Gibbs measure.

\section{introduction}

The $p$-adic numbers were first introduced by the German
mathematician K.Hensel. They form an integral part of number
theory, algebraic geometry,
representation theory and other branches of modern mathematics.
However, numerous applications of these numbers to
theoretical physics have been proposed \cite{14}
to quantum mechanics and to $p$-adic valued physical
observable \cite{k91}. A number of $p$-adic models in physics
cannot be described using ordinary probability theory based on the
Kolmogorov axioms.

In \cite{33} a theory of stochastic processes with values in
$p$-adic and more general non-Archimedean fields was developed,
having probability distributions with non-Archimedean values.

One of the basic branches of mathematics lying at the base of the
theory of statistical mechanics is the theory of probability and
stochastic processes. Since the theories of probability and
stochastic processes in a non-Archimedean setting have been
introduced, it is natural to study problems of statistical
mechanics in the context of the $p$-adic theory of probability.

We note that $p$-adic Gibbs measures were studied for several
$p$-adic
models of statistical mechanics \cite{GRR,16,TMP2},
\cite{MRasos,MNGM,rh,bookroz}.

The SOS model can be considered as a generalization of the Ising model
(which arises when $m=1$). It is known that a phase transition does not occur
for the $p$-adic Ising model on a Cayley tree. We prove that there is no phase transition
for the $m+1$-state $p$-adic SOS model if $p\nmid m+1$. Moreover, we show that if
$p\mid m+1$ then a phase transition may occur for this model.

The organization of this paper as follows. Section 2 is
a mathematically preliminary. In section 3 we give a construction
of $p$-adic Gibbs measures for the $m+1$-state $p$-adic SOS model
on a Cayley tree of order $k\geq1$. Moreover, we study
a problem of boundedness of such measures. In section 4 we study
the set of all translation-invariant $p$-adic Gibbs measures for
$3$-state $p$-adic SOS model. In section 5 we give a criterion of uniqueness
of the $p$-adic Gibbs measure.

\section{definitions and preliminary results}

\subsection{\bf $p$-adic numbers and measures.}

Let $\mathbb Q$ be the field of rational numbers.
For a fixed prime number $p$, every rational number $x\ne0$ can
be represented in the form $x = p^r{n\over m}$, where $r,n\in\mathbb Z$,
$m$ is a positive integer, and $n$ and $m$ are relatively
prime with $p$: $(p, n) = 1$, $(p, m) = 1$. The $p$-adic norm of
$x$ is given by
$$
|x|_p= \left\{\begin{array}{ll}
p^{-r},& \text{ if }  x\neq 0,\\
0,& \text{ if } x=0.
 \end{array}\right.
$$

This norm is non-Archimedean  and satisfies the so called strong
triangle inequality
$$|x+y|_p\leq \max\{|x|_p,|y|_p\}.$$

From this property immediately follow the following facts ({\it non-Archimedean norm's property}):

1) if  $|x|_p\neq |y|_p$, then $|x-y|_p=\max\{|x|_p,|y|_p\}$;

2) if  $|x|_p=|y|_p$, then  $|x-y|_p\leq |x|_p$;

The completion of $\mathbb Q$ with respect to the $p$-adic norm defines
the $p$-adic field $\mathbb Q_p$ (see \cite{29}).

The completion of the field of rational numbers $\mathbb Q$ is either
the field of real numbers $\mathbb R$ or one of the fields of
$p$-adic numbers $\mathbb Q_p$ (Ostrowski's theorem).

Any $p$-adic number $x\ne 0$ can be uniquely represented
in the canonical form
\begin{equation}\label{ek}
x = p^{\gamma(x)}(x_0+x_1p+x_2p^2+\dots),
\end{equation}
where $\gamma=\gamma(x)\in \mathbb Z$ and the integers $x_j$ satisfy: $x_0 > 0$,
$0\leq x_j \leq p - 1$ (see \cite{29,sc,48}). In this case $|x|_p =
p^{-\gamma(x)}$.
\begin{thm}\label{tx2}\cite{48}
The equation
$x^2 = a$, $0\ne a =p^{\gamma(a)}(a_0 + a_1p + ...), 0\leq a_j
\leq p - 1$, $a_0 > 0$ has a solution $x\in \mathbb Q_p$ iff
hold true the following:\\
$(a)$ $\gamma(a)$ is even;\\
$(b)$ $y^2\equiv a_0(\operatorname{mod} p)$ is solvable for $p\ne 2$;
the equality $a_1=a_2=0$ holds if $p=2$.
\end{thm}

We respectively denote the sets of all {\it $p$-adic integers}
and {\it units} of $\mathbb Q_p$ by
$$
\mathbb Z_p=\left\{x\in\mathbb Q_p: |x|_p\leq1\right\},\quad
\mathbb Z^*_p=\left\{x\in\mathbb Q_p: |x|_p=1\right\}.
$$
\begin{lemma}\label{lem1}(Hensel's lemma, \cite{48}). Let $f(x)$ be a polynomial
whose the coefficients are $p$-adic integers. Let $a_0$ be a $p$-adic
integer such that for some $i\geq0$ we have
$$\begin{array}{ll}
f(a_0)\equiv0(\operatorname{mod }p^{2i+1}),\\
f^\prime(a_0)\equiv0(\operatorname{mod }p^{i}),\quad f^\prime(a_0)\not\equiv0(\operatorname{mod }p^{i+1}).
\end{array}
$$
Then $f(x)$ has a unique $p$-adic integer root $x_0$ which satisfies
$x_0\equiv a_0(\operatorname{mod }p^{i+1})$.
\end{lemma}

For $a\in \mathbb Q_p$ and $r> 0$ we denote
$$B(a, r) = \{x\in \mathbb Q_p : |x-a|_p < r\}.$$

$p$-adic {\it logarithm} is defined by the series
$$\log_p(x) =\log_p(1 + (x-1)) =
\sum_{n=1}^\infty (-1)^{n+1}{(x-1)^n\over n},$$ which converges
for  $x\in B(1, 1)$;  $p$-adic {\it exponential} is defined by
$$\exp_p(x) =\sum^\infty_{n=0}{x^n\over n!},$$
which converges
for $x \in B(0, p^{-1/(p-1)})$.
\begin{lemma}\label{lem2} Let $x\in B(0, p^{-1/(p-1)})$. Then
$$|\exp_p(x)|_p = 1,\ \ |\exp_p(x)-1|_p = |x|_p, \ \
|\log_p(1 + x)|_p = |x|_p,$$
$$\log_p(\exp_p(x)) = x,\ \ \exp_p(\log_p(1 + x)) = 1 + x.$$
\end{lemma}
Denote
$$
\mathcal E_p=\left\{x\in\mathbb Q_p: |x|_p=1, |x-1|_p<p^{-1/(p-1)}\right\}.
$$

A more detailed description of $p$-adic calculus and $p$-adic
mathematical physics can be found in \cite{29,sc,48}.

Let $(X,{\mathcal B})$ be a measurable space, where ${\mathcal B}$
is an algebra of subsets of $X$. A function $\mu: {\mathcal B}\to
\mathbb Q_p$ is said to be a $p$-adic measure if for any $A_1, . . .
,A_n\in {\mathcal B}$ such that $A_i\cap A_j = \emptyset$, $i\ne
j$, the following holds:
$$\mu(\bigcup^n_{j=1}A_j)=\sum^n_{j=1}\mu(A_j).$$
A $p$-adic measure is called a probability measure if $\mu(X) =
1$. A $p$-adic probability measure $\mu$ is called {\it bounded}
if $\sup\{|\mu(A)|_p:A\in\mathcal B\}<\infty$ (see \cite{k91}).

We call a $p$-adic measure a probability measure \cite{16} if $\mu(X)=1$.

\subsection{\bf Cayley tree.}

The Cayley tree $\Gamma^k$
of order $ k\geq 1 $ is an infinite tree, i.e., a graph without
cycles, such that exactly $k+1$ edges originate from each vertex.
Let $\Gamma^k=(V, L)$ where $V$ is the set of vertices and  $L$
the set of edges.
Two vertices $x$ and $y$ are called {\it nearest neighbors} if
there exists an
edge $l \in L$ connecting them.
We shall use the notation $l=\langle x,y\rangle$.
A collection of nearest neighbor pairs $\langle x,x_1\rangle,
\langle x_1,x_2\rangle,...,\langle x_{d-1},y\rangle$ is called a
{\it
path} from $x$ to $y$. The distance $d(x,y)$ on the Cayley tree
is the number of edges of the shortest path from $x$ to $y$.

For a fixed $x^0\in V$, called the root, we set
\begin{equation*}
W_n=\{x\in V\,| \, d(x,x^0)=n\}, \qquad V_n=\bigcup_{m=0}^n W_m
\end{equation*}
and denote
$$
S(x)=\{y\in W_{n+1} :  d(x,y)=1 \}, \ \ x\in W_n, $$ the set  of
{\it direct successors} of $x$.

Let $G_k$ be a free product of $k + 1$ cyclic groups of the second
order with generators $a_1, a_2,\dots, a_{k+1}$,
respectively.
It is known that there exists a one-to-one correspondence between
the set of vertices $V$ of the Cayley tree $\Gamma^k$ and the group $G_k$.

\subsection{\bf $p$-adic SOS model}

Let $\mathbb Q_p$ be the field of $p$-adic numbers and
$\Phi$ be a finite set. A configuration $\sigma$ on $V$ is then defined
as a function $x\in V\mapsto\sigma(x)\in\Phi$; in a similar fashion
one defines a configuration $\sigma_n$ and $\sigma^{(n)}$ on $V_n$
and $W_n$ respectively. The set of all configurations on $V$
(resp. $V_n,\ W_n$) coincides with $\Omega=\Phi^V$
(resp.$\Omega_{V_n}=\Phi^{V_n},\ \Omega_{W_n}=\Phi^{W_n}$). Using
this, for given configurations $\sigma_{n-1}\in\Omega_{V_{n-1}}$
and $\omega\in\Omega_{W_n}$ we define their concatenations
by
$$
(\sigma_{n-1}\vee\omega)(x)=\left\{\begin{array}{ll}
\sigma_{n-1}(x),& \text{if}\  x\in V_{n-1},\\
\omega(x),& \text{if}\  x\in W_n.
 \end{array}\right.
$$
It is clear that $\sigma_{n-1}\vee\omega\in\Omega_{V_n}.$\\

Let $G^*_k$ be a subgroup of the group $G_k$. A function $h_x$ (for example,
a configuration $\sigma(x)$) of $x\in G_k$ is called $G^*_k$-periodic if
$h_{yx}=h_x$ (resp. $\sigma(yx)=\sigma(x)$) for any $x\in G_k$ and
$y\in G^*_k.$

A $G_k$-periodic function is called {\it translation-invariant}.

We consider {\it $p$-adic SOS model} on a Cayley tree,
where the spin takes values in the set
$\Phi:=\{1,2,\dots,m\}$, and is assigned to the vertices
of the tree.

The (formal) Hamiltonian is of a $p$-adic SOS form:
\begin{equation}\label{ham}
H(\sigma)=J\sum_{\langle x,y\rangle\in L}
\left|\sigma(x)-\sigma(y)\right|_\infty,
\end{equation}
where $J\in B(0, p^{-1/(p-1)})$ is a coupling constant,
$\langle x,y\rangle$ stands for nearest neighbor vertices
and $|\cdot|_\infty$ stands for usual absolute value.

The $p$-adic SOS model of this type can be considered as a generalization
of the $p$-adic Ising model (which arises when $m=1$).

\section{The system of $p$-adic vector-valued functional equations}
We use standard definition of a $p$-adic Gibbs measure, a translation-invariant
(TI) measure. Also, call measure $\mu$ {\it symmetric} if it is preserved
under simultaneous change $j\mapsto m-j$ at each vertex $x\in V$.

Let $z: x\mapsto z_x=(z_{0,x}, z_{1,x},\dots, z_{m,x})\in\mathcal E^{m+1}_p$
be a $p$-adic vector-valued function of $x\in V\setminus\{x_0\}$. Given
$n=1,2,\dots,$ consider the $p$-adic probability distribution $\mu^{(n)}$ on $\Omega_{V_n}$
defined by
\begin{equation}\label{mu_n}
\mu_{\tilde z}^{(n)}(\sigma_n)=Z_{n,\tilde z}^{-1}\exp_p\left\{H_n(\sigma_n)\right\}
\prod_{x\in W_n}{\tilde z}_{\sigma(x),x}.
\end{equation}
Here, $\sigma_n: x\in V_n\mapsto\sigma_n(x)$ and $Z_{n,\tilde z}$ is the corresponding
partition function:

\begin{equation}\label{Z_n}
Z_{n,\tilde z}=\sum_{\sigma\in\Omega_{V_n}}\exp_p\left\{H_n(\sigma_n)\right\}
\prod_{x\in W_n}{\tilde z}_{\sigma(x),x}.
\end{equation}

We say that the $p$-adic probability distributions (\ref{mu_n}) are
compatible if for all
$n\geq 1$ and $\sigma_{n-1}\in \Phi^{V_{n-1}}$:
\begin{equation}\label{cc}
\sum_{\omega_n\in\Omega_{W_n}}\mu_{\tilde z}^{(n)}(\sigma_{n-1}\vee \omega_n)
=\mu_{\tilde z}^{(n-1)}(\sigma_{n-1}).
\end{equation}
Here $\sigma_{n-1}\vee \omega_n$ is the concatenation of the configurations.\\
We note that an analog of the Kolmogorov extension theorem for distributions
can be proved for $p$-adic distributions given by (\ref{mu_n}) (see \cite{16}).
According to this theorem there exists a unique $p$-adic measure
$\mu_{\tilde z}$ on $\Omega$ such that,
for all $n$ and $\sigma_n\in \Omega_{V_n}$,
$$\mu_{\tilde z}(\{\sigma|_{V_n}=\sigma_n\})=\mu_{\tilde z}^{(n)}(\sigma_n).$$

Such a measure is called a {\it $p$-adic Gibbs measure} (pGM)
corresponding to the Hamiltonian (\ref{ham}) and vector-valued
function ${\tilde z}_x, x\in V$. We denote by $G(H)$ the set of all $p$-adic Gibbs
measures for the hamiltonian $H$. If $|G(H)|\geq2$ then we say that for this model
there is {\it a phase transition}.

The following statement describes conditions on ${\tilde z}_x$
guaranteeing compatibility of $\mu_{\tilde z}^{(n)}(\sigma_n)$.

\begin{pro}\label{pro1}
The $p$-adic probability distributions
$\mu^{(n)}_{\tilde z}(\sigma_n)$, $n=1,2,\dots$ in
(\ref{mu_n}) are compatible for $p$-adic SOS model iff for any $x\in V\setminus\{x^0\}$
the following system of equations holds:
\begin{equation}\label{p***}
 z_{i,x}=\prod_{y\in S(x)}\frac{\sum_{j=0}^{m-1}\theta^{|i-j|_\infty}z_{j,y}+\theta^{m-i}}
 {\sum_{j=0}^{m-1}\theta^{m-j}z_{j,y}+1}\ ,\quad i=0,1,\dots,m-1.
\end{equation}
Here $\theta=\exp_p(J)$ and $z_{i,x}=\frac{\tilde{z}_{i,x}}{\tilde{z}_{m,x}},\ i=0,1,\dots,m-1.$
\end{pro}
\begin{proof}
{\it Necessity}. Suppose that (\ref{cc}) holds; we want to prove (\ref{p***}).
Substituting (\ref{mu_n}) in (\ref{cc}), obtain, for any configurations
$\sigma_{n-1}\in\Omega_{V_n}$:
$$
\frac{Z_{n-1,\tilde z}}{Z_{n,\tilde z}}\sum_{\omega\in\Omega_{W_n}}\prod_{x\in W_{n-1}}
\prod_{y\in S(x)}\exp_p\left(J|\sigma_{n-1}(x)-\omega(y)|_\infty\right)\tilde{z}_{\omega(y),y}=
\prod_{x\in W_{n-1}}\tilde{z}_{\sigma_{n-1}(x),x}.
$$
From this for any $i\in\{0,1,\dots,m\}$ we get
\begin{equation}\label{p**}
\prod_{y\in S(x)}\frac{\sum_{j=0}^{m}\exp_p\left(J|i-j|_\infty\right)\tilde{z}_{j,y}}
 {\sum_{j=0}^{m}\exp_p\left(J(m-j)\right)\tilde{z}_{j,y}}=\frac{\tilde{z}_{i,x}}{\tilde{z}_{m,x}}.
\end{equation}
Denoting $\theta=\exp_p(J)$ and $z_{i,x}=\frac{\tilde{z}_{i,x}}{\tilde{z}_{m,x}}$, we get
(\ref{p***}) from (\ref{p**}).

{\it Sufficiency}. Suppose that (\ref{p***}) holds. It is equivalent to the
representations
\begin{equation}\label{p1*}
\prod_{y\in S(x)}\sum_{j=0}^m\theta^{|i-j|_\infty}\tilde{z}_{j,y}=a(x)\tilde{z}_{i,x},
\quad i=0,1,\dots,m-1
\end{equation}
for some function $a(x)$. We have

\begin{equation}\label{p1**}
\mbox{LHS of }(\ref{cc})=\frac{1}{Z_{n,\tilde{z}}}\exp_p\left(H(\sigma_{n-1})\right)
\prod_{x\in W_{n-1}}\prod_{y\in S(x)}\sum_{j=0}^m\theta^{|i-j|_\infty}\tilde{z}_{i,y}.
\end{equation}

Substituting (\ref{p1*}) into (\ref{p1**}) and denoting $A_n(x)=\prod_{y\in W_{n-1}}a(y)$,
we get
\begin{equation}\label{p1***}
\mbox{RHS of }(\ref{p1**})=\frac{A_{n-1}}{Z_{n,\tilde{z}}}\exp_p\left(H(\sigma_{n-1})\right)
\prod_{x\in W_{n-1}}\tilde{z}_{\sigma_{n-1}(x),x}.
\end{equation}

Since $\mu^{(n)}_{\tilde{z}},\ n\geq1$ is a probability, we should have
$$
\sum_{\sigma_{n-1}\in\Omega_{V_{n-1}}}\sum_{\omega\in\Omega_{W_n}}\mu^{(n)}_
{\tilde{z}}(\sigma_{n-1},\omega)=1.
$$

Hence from (\ref{p1***}) we get $Z_{n,\tilde{z}}=A_{n-1}Z_{n-1,\tilde{z}}$,
and (\ref{cc}) holds.
\end{proof}

The following proposition is straightforward.
\begin{pro}\label{pro2} Let $H$ be a hamiltonian of $m+1$-state $p$-adic
SOS model on a Cayley tree $\Gamma^k$. Then it hold the following
statements:

$1)$ Any measure $\mu$ with local distributions $\mu^{(n)}$ satisfying
(\ref{mu_n}),(\ref{cc}) belongs to $G(H)$.

$2)$ An $\mu\in G(H)$ is TI iff $z_{i,x}$ does not depend on $x: z_{i,x}\equiv z_i,\ x\in V,
\ i\in\Phi$,
and symmetric TI if and only if $z_i=z_{m-i},\ i\in\Phi$.
\end{pro}

\subsection{Boundedness of $p$-adic Gibbs measure}
\begin{thm}
Let $H$ be a $m+1$-state $p$-adic SOS model on a Cayley tree of order $k$.
Then a measure $\mu\in G(H)$ is bounded if and only if $m+1$ is not
divisible by $p$.
\end{thm}
\begin{proof}
Let $z_x=(z_{0,x},z_{1,x},\dots,z_{m-1,x})$ is a solution to (\ref{p***}).
Then from (\ref{p1*}) for all $x\in V\setminus\{x_0\}$
we find
$$
a(x)=\prod_{y\in S(x)}\left(\sum_{j=0}^{m-1}\theta^{m-j}z_{j,y}+1\right)=
\prod_{y\in S(x)}\left(\sum_{j=0}^{m-1}(\theta^{m-j}z_{j,y}-1)+m+1\right).
$$
Since $\theta\in\mathcal E_p$ and $z_x\in\mathcal E^{m}_p$, by
non-Archimedean norm's property we get
$$
|a(x)|_p=\left\{\begin{array}{ll}
1, & \mbox{if }\ p\nmid m+1;\\
\leq p^{-k}, & \mbox{if }\ p\mid m+1.
\end{array}\right.
$$
From this we obtain
\begin{equation}\label{A_n}
\left|A_{n}(x)\right|_p=\prod_{y\in W_{n-1}}|a(y)|_p=\left\{\begin{array}{ll}
1, & \mbox{if }\ p\nmid m+1;\\
\leq p^{-k|W_{n-1}|}, & \mbox{if }\ p\mid m+1.
\end{array}\right.
\end{equation}
We use the following recurrence formula
$$
Z_{n,z}=A_{n-1}Z_{n-1,z}.
$$
From (\ref{A_n}) we get
\begin{equation}\label{Z_nA}
\left|Z_{n,z}\right|_p=\prod_{x\in V_{n-1}}\left|A_{n-1}(x)\right|_p=
\left\{\begin{array}{ll}
1, & \mbox{if }\ p\nmid m+1;\\
\leq p^{-k|V_{n-1}|}, & \mbox{if }\ p\mid m+1.
\end{array}\right.
\end{equation}
For any configuration $\sigma\in\Omega_{V_n}$ by (\ref{Z_nA}) we have
$$
\left|\mu^{(n)}_z(\sigma)\right|_p=\frac{\left|\exp_p\{H_n(\sigma)\}\prod_{x\in W_n}
z_{\sigma(x),x}\right|_p}{\left|Z_{n,z}\right|_p}=$$
$$
\frac{1}{\left|Z_{n,z}\right|_p}=\left\{\begin{array}{ll}
1, & \mbox{if }\ p\nmid m+1;\\
\geq p^{k|V_{n-1}|}, & \mbox{if }\ p\mid m+1.
\end{array}\right.
$$
which means the measure $\mu_z$ is bounded if and only if $p\nmid m+1$.
\end{proof}

\section{Three-state SOS model}

From Proposition \ref{pro2} it follows that for any $z=\{z_x,\ x\in V\}$
satisfying (\ref{p***}) there exists a unique $p$-adic Gibbs measure $\mu$.
Denote by $\mbox{TIpGM}$ the set of all translation-invariant $p$-adic Gibbs measures
for the model (\ref{ham}). Note that $\mbox{TIpGM}\subset G(H)$. However,
description of the set $\mbox{TIpGM}$ for an arbitrary $m$ is not easy.
We now suppose that $m=2$, i.e.
$\Phi=\{0,1,2\}$. We assume that $z_{2,x}\equiv1$
($z_{m,x}\equiv1$ for general $m$).

\subsection{Translation-invariant solutions}

It is natural begin with TI solutions where $z_{x}=z\in\mathcal E^m_p$
is constant. Unless otherwise stated, we concentrate on the simplest case
where $m=2$, i.e., spin values are $0,1$ and $2$. In this case we obtain
from (\ref{p***}):
\begin{equation}\label{z_0}
z_0=\left(\frac{z_0+\theta z_1+\theta^2}{\theta^2 z_0+\theta z_1+1}\right)^k,
\end{equation}
\begin{equation}\label{z_1}
z_1=\left(\frac{\theta z_0+z_1+\theta}{\theta^2 z_0+\theta z_1+1}\right)^k.
\end{equation}

\begin{pro}\label{pro3}
If $p\neq3$ then the system of equations (\ref{z_0}),(\ref{z_1}) has no
solution in $\mathcal A_p=\{(z_0,z_1)\in\mathcal E^2_p: z_0\neq1\}$.
\end{pro}
\begin{proof}
Let $p\neq3$. Denote $A=z_0+\theta z_1+\theta^2$, $B=\theta^2 z_0+\theta z_1+1$.
Then from (\ref{z_0}) we get:
\begin{equation}\label{z_0-1}
(z_0-1)\left(B^k+(\theta^2-1)\sum_{i=0}^{k-1}A^{k-1-i}B^i\right)=0.
\end{equation}
Since $\theta,z_0,z_1\in\mathcal E_p$ we have
$$A\equiv3(\operatorname{mod }p)\quad\mbox{and}\quad B\equiv3(\operatorname{mod }p).$$
Consequently,
\begin{equation}\label{Bk}
\left|B^k+(\theta^2-1)\sum_{i=0}^{k-1}A^{k-1-i}B^i\right|_p=\left|3^k\right|_p=1.
\end{equation}
From (\ref{z_0-1}) and (\ref{Bk}) we obtain $z_0=1$.
\end{proof}
Observe that $z_0=1$ satisfies equation (\ref{z_0}) independently of
$k, \theta$ and $z_1$.
Substituting
$z_0=1$ into (\ref{z_1}), we have
\begin{equation}\label{z_1*}
z_1=\left(\frac{2\theta+z_1}{\theta^2+\theta z_1+1}\right)^k.
\end{equation}
We consider the function
$$
f(x)=\left(\frac{2a+x}{a^2+ax+1}\right)^k.
$$
Let $p\neq3$ and $a\in\mathcal E_p$. Then for any $x\in\mathcal E_p$ from non-Archimedean
norm's property we get
$$
\left|f(x)\right|_p=\left(\frac{\left|2a+x\right|_p}{\left|a^2+ax+1\right|_p}\right)^k=1,
$$
and
$$
\left|f(x)-1\right|_p=\left|\frac{(a-1)(1-x-a)\sum_{i=0}^{k-1}(2a+x)^{k-1-i}(a^2+ax+1)^i}{(a^2+ax+1)^k}\right|_p<p^{-1/(p-1)}.
$$
Thus, we have shown that $f:\mathcal E_p\mapsto \mathcal E_p$ if $p\neq3$ and $a\in\mathcal E_p$.

Now we shall show that $|f'(x)|_p<1$ for any $x\in\mathcal E_p$.
$$
f'(x)=-\frac{k(a^2-1)}{\left(a^2+ax+1\right)^2}\left(\frac{2a+x}{a^2+ax+1}\right)^{k-1}.
$$
Since $p\neq3$ and $a\in\mathcal E_p$ we obtain
$$
|f'(x)|_p=\left|k\left(a^2-1\right)\right|_p\leq\frac{1}{p}.
$$
Hence, we get
$$
\left|f(x)-f(y)\right|_p\leq\frac{1}{p}|x-y|_p\qquad\mbox{for all }x,y\in\mathcal E_p.
$$

Consequently, the function $f$ has a unique fixed point $x^*$ as $\mathcal E_p$ is compact.
Thus, we have proved the following proposition
\begin{pro}\label{pro4}
If $p\neq3$ then the system of equations (\ref{z_0}),(\ref{z_1}) has a unique
solution in $\mathcal E^2_p\setminus\mathcal A_p$.
\end{pro}
From Proposition \ref{pro3} and Proposition \ref{pro4} we get the following
\begin{thm}
Let $p\neq3$. Then there exists a unique translation-invariant $p$-adic
Gibbs measure for the three-state $p$-adic SOS model on a Cayley tree of order $k$.
\end{thm}

\begin{pro}\label{pro5} Let $p=3$. If
$\theta\in\{x\in\mathcal E_3: |x-1|_3<3^{-2}\}$ then the system of equations
(\ref{z_0}),(\ref{z_1}) has a solution in $\mathcal E_3^2$.
\end{pro}
\begin{proof} Let $p=3$. We consider the following polynomial with
$p$-adic integers coefficients:
\begin{equation}\label{g(x)}
g(x)=\theta x^{k+1}-x^k+(\theta^2+1)x-2\theta.
\end{equation}
We will check conditions of Hensel's lemma.
\begin{equation}\label{hensel}
g(1)=\theta(\theta-1)\quad\mbox{and}\quad g^\prime(1)=(\theta-1)(\theta+k+2)+3.
\end{equation}
Since $\theta\in\{x\in\mathcal E_3: |x-1|_3<3^{-2}\}$ we get
\begin{equation}\label{theta}
\theta\equiv1(\operatorname{mod }27).
\end{equation}
Using this, from (\ref{hensel}) we have
$$\begin{array}{ll}
g(1)\equiv0(\operatorname{mod }27),\\[2mm]
g^\prime(1)\equiv0(\operatorname{mod }3),\quad g^\prime(1)\not\equiv0(\operatorname{mod }9).
\end{array}
$$
Thus, we have shown that for the function $g(x)$ all conditions of Hensel's lemma are satisfied.
Then the function $g(x)$ has a unique $p$-adic integer root $x_*$ which satisfies
$x_*\equiv1(\operatorname{mod }9)$.

It is easy to see $x_*^k\in\mathcal E_3$ for any $k\geq1$ and $z_1^*=x_*^k$ is a
solution to (\ref{z_1*}). Consequently, $(1,z_1^*)\in\mathcal E_3^2$ is a
solution to (\ref{z_0}),(\ref{z_1}).
\end{proof}

\begin{pro}\label{pro6} Let $p=3$. If
$k$ is an even and it is not divisible by $p$ then a system of equations
(\ref{z_0}),(\ref{z_1}) has a solution in $\mathcal E_3^2$.
\end{pro}
\begin{proof} Let $k$ is an even and it is not divisible by $3$. Then we get
\begin{equation}\label{2^k}
2^k\equiv1(\operatorname{mod }3)\quad\mbox{and}\quad
k\not\equiv0(\operatorname{mod }3).
\end{equation}
Consider the function $g(x)$ (see (\ref{g(x)})). We have
$$
g(2)=2^{k+1}\theta-2^k+2(\theta^2-\theta+1)
$$
and
$$
g^\prime(2)=(k+1)2^k\theta-k2^{k-1}+\theta^2+1.
$$
Since $\theta\in\mathcal E_3$ and using (\ref{2^k}),
we obtain
$$
g(2)=2^{k+1}\theta-2^k+2(\theta^2-\theta+1)\equiv0(\operatorname{mod }3),
$$
and
$$
g^\prime(2)=(k+1)2^k\theta-k2^{k-1}+\theta^2+1\not\equiv0(\operatorname{mod }3).
$$
Then by Hensel's lemma there exists a unique $p$-adic integer $\tilde{x}_{*}$
such that $g(\tilde{x}_{*})=0$
and $\tilde{x}_{*}\equiv2(\operatorname{mod }3)$.
It is easy to check that $\tilde{x}_*^k\in\mathcal E_3$. Denote $\tilde{z}_*=\tilde{x}_*^k$.
Then $\tilde{z}_*$ is a solution to (\ref{z_1}). Consequently,
$(1,\tilde{z}_1^*)\in\mathcal E_3^2$ is a
solution to (\ref{z_0}),(\ref{z_1}).
\end{proof}

\begin{cor}\label{cor} Let $p=3$ and
$\theta\in\{x\in\mathcal E_3: |x-1|_3<3^{-2}\}$. If
$k$ is an even and it is not divisible by $p$ then a system of equations
(\ref{z_0}),(\ref{z_1}) has at least two solutions in $\mathcal E_3^2$.
\end{cor}
\begin{thm}
Let $p=3$ and $\theta\in\{x\in\mathcal E_3: |x-1|_3<3^{-2}\}$.
If
$k$ is an even and it is not divisible by $p$ then a phase transition occurs for the
three state $p$-adic SOS model on a Cayley tree of order $k$.
\end{thm}
\begin{pro}\label{pro7}
Let $p=3$ and $k=2$.\\
1) If
$\theta\not\in\{x\in\mathcal E_3: |x-10|_3<3^{-2}\}$ then the system of equations
(\ref{z_0}),(\ref{z_1}) has no solutions in
$\mathcal A_3=\left\{(z_0,z_1)\in\mathcal E_3^2: z_0\neq1\right\}$.\\
2) If
$\theta\in\{x\in\mathcal E_3: |x-37|_3<3^{-3}\}$ then the system of equations
(\ref{z_0}),(\ref{z_1}) has two solutions in
$\mathcal A_3=\left\{(z_0,z_1)\in\mathcal E_3^2: z_0\neq1\right\}$.
\end{pro}
\begin{proof} Let $p=3$ and $k=2$.
Since $z_i\in\mathcal E_p,\ i=1,2$ by Theorem \ref{tx2} there exist $\sqrt{z_i}$
in $\mathbb Q_p$. Denote
$x=\sqrt{z_0}$ and $y=\sqrt{z_1}$. Rewrite (\ref{z_0}) and (\ref{z_1}) as
\begin{equation}\label{x}
x=\frac{x^2+\theta y^2+\theta^2}{\theta^2 x^2+\theta y^2+1},
\end{equation}
\begin{equation}\label{y}
y=\frac{\theta x^2+y^2+\theta}{\theta^2 x^2+\theta y^2+1}.
\end{equation}
From (\ref{x}) we get
\begin{equation}\label{x-1}
(x-1)\left(\theta^2x^2+\theta y^2+1+(\theta^2-1)(x+1)\right)=0.
\end{equation}
Let $x\neq1$. Then from (\ref{x-1}) we have
\begin{equation}\label{x,y1}
\theta^2x^2+\theta y^2+1=(1-\theta^2)(x+1).
\end{equation}
Put this to (\ref{y}) and obtain
\begin{equation}\label{x,y2}
\theta y=\frac{x}{x+1}.
\end{equation}
Hence we conclude that
$$
x\equiv1(\operatorname{mod }3)\quad\mbox{and}\quad y\equiv2(\operatorname{mod }3).
$$
From (\ref{x,y1}) and (\ref{x,y2}) we get
\begin{equation}\label{x4}
\theta^3x^4+\theta(3\theta^2-1)x^3+(4\theta^3-2\theta+1)x^2+\theta(3\theta^2-1)x+\theta^3=0.
\end{equation}
Note that $x=1$ is a solution to (\ref{x4}) if and only if $12\theta^3-4\theta+1=0$.

Let $12\theta^3-4\theta+1=0$. Then from (\ref{x4}) we get
$$
(x-1)^2\left(\theta^3x^2+\theta\left(5\theta^2-1\right)x+\theta^3\right)=0.
$$
Consider the following equation
\begin{equation}\label{kv}
\theta^3x^2+\theta\left(5\theta^2-1\right)x+\theta^3=0.
\end{equation}
This equation is solvable in $\mathbb Q_3$ if and only if $\sqrt{1-7\theta^2}$
exists in
$\mathbb Q_3$. We will show that the quadratic equation (\ref{kv}) is not
solvable in $\mathbb Q_3$.
From $12\theta^3-4\theta+1=0$ and $\theta\in\mathcal E_3$ we get
$\theta\equiv10(\operatorname{mod }27)$. Then
we have
$$
1-7\theta^2\equiv3(\operatorname{mod }27).
$$
Hence by Theorem \ref{tx2} there does not exist $\sqrt{1-7\theta^2}$
in
$\mathbb Q_3$. Consequently, the equation (\ref{kv}) is not solvable
in $\mathbb Q_3$.

Let $12\theta^3-4\theta+1\neq0$. Denote $t=x+x^{-1}$. Then from
(\ref{x4}) we get
\begin{equation}\label{kv2}
\theta^3t^2+\theta\left(3\theta^2-1\right)t+2\theta^3-2\theta+1=0.
\end{equation}
Denote $D=\theta^4+2\theta^2-4\theta+1$. The equation (\ref{kv2})
has two distinct solutions if there exists $\sqrt{D}$ in $\mathbb Q_3$.
Moreover,
we have $t_i\equiv2(\operatorname{mod }3),\ i=1,2$.

The solutions of the equation (\ref{x4}) are
$$
x^\pm=\frac{1-3\theta^2+\sqrt{D}\pm\sqrt{\left(1-7\theta^2+\sqrt{D}\right)
\left(1+\theta^2+\sqrt{D}\right)}}{4\theta^2},
$$
$$
x_\pm=\frac{1-3\theta^2-\sqrt{D}\pm\sqrt{\left(1-7\theta^2-\sqrt{D}\right)
\left(1+\theta^2-\sqrt{D}\right)}}{4\theta^2}.
$$
Note that the existence of solutions $x^\pm$ equivalent to the existence
of $\sqrt{D}$ and the existence of
$\sqrt{2\left(1-7\theta^2-\sqrt{D}\right)}$ in $\mathbb Q_3$. Moreover,
if the solutions $x^\pm$ exists then
$x^\pm\equiv1(\operatorname{mod }3)$.

We will show that the number $\sqrt{2\left(1-7\theta^2+\sqrt{D}\right)}$
does not exist in $\mathbb Q_3$.
At first we will check the existence of $\sqrt{D}$. We have
$$
D=\theta^4+2\theta^2-4\theta+1=(\theta-1)\left(4\theta+(\theta-1)(\theta+1)^2\right).
$$
Since $\theta\in\mathcal E_3$ by Theorem \ref{tx2} we conclude that
the $\sqrt{D}$ exists if and only if
\begin{equation}\label{theta3}
\theta=1+3^{2n}(1+\varepsilon),\quad n\in\mathbb N,\ |\varepsilon|_3<1.
\end{equation}
Then we get
$$
\sqrt{D}=3^n(1+\varepsilon'),\qquad |\varepsilon'|_3<1.
$$
Hence, for all $n\in\mathbb N$ we have
$$
\left|1-7\theta^2+\sqrt{D}\right|_3=\left|1-7\left(1+3^{2n}
(1+\varepsilon)\right)+3^n(1+\varepsilon')\right|_3=
$$
\begin{equation}\label{+}
\left|-6+3^n+3^n\varepsilon'-14(1+\varepsilon)3^{2n}+7(1+\varepsilon)^2
3^{4n}\right|_3=\frac{1}{3}.
\end{equation}
By Theorem \ref{tx2} the number $\sqrt{2\left(1-7\theta^2+\sqrt{D}\right)}$
does not exist in $\mathbb Q_3$.

Now we will check the existence of $\sqrt{2\left(1-7\theta^2-\sqrt{D}\right)}$.
Using (\ref{theta3}) we obtain
$$
\left|1-7\theta^2+\sqrt{D}\right|_3=\left\{\begin{array}{cc}
\frac{1}{3},\quad\mbox{if }n>1;\\[2mm]
<\frac{1}{3},\quad\mbox{if }n=1.
\end{array}\right.
$$
Hence, by Theorem \ref{tx2} there does not exist
$\sqrt{2\left(1-7\theta^2-\sqrt{D}\right)}$ in $\mathbb Q_3$ if $n>1$.
So, we must check the case $n=1$.

Let $\left|\theta-37\right|_3<3^{-3}$, i.e., $\theta=37+\beta$, $|\beta|_3<3^{-3}$.
Then we obtain
$$
\left(1-7\theta^2+\sqrt{D}\right)\left(1-7\theta^2-\sqrt{D}\right)=
4\theta\left(12\theta^3-4\theta+1\right)=
$$
\begin{equation}\label{+-}
4(37+\beta)\left(12(37+\beta)^3-4(37+\beta)+1\right)=27(1+2\cdot3+3^2+\cdots)
\end{equation}
On the other hand by (\ref{+}) one can find
$$
1-7\theta^2+\sqrt{D}=6\left(1+\alpha_1\cdot3+\alpha_2\cdot3^2+\cdots\right),
\quad\alpha_i\in\{0,1,2\},\ i=1,2,\dots
$$
From this and by (\ref{+-}) we get
$$
2\left(1-7\theta^2-\sqrt{D}\right)=9\left(1+\gamma_1\cdot3+\gamma_2\cdot3^2+
\cdots\right),\quad\gamma_i\in\{0,1,2\},\ i=1,2,\dots
$$
Then by Theorem \ref{tx2} there exists $\sqrt{2\left(1-7\theta^2-\sqrt{D}\right)}$
in $\mathbb Q_3$.
Hence, $(x_+,y_+)$ and $(x_-,y_-)$ are the solutions to (\ref{x}),(\ref{y}),
where $y_\pm=\frac{x_\pm}{\theta(x_\pm+1)}$.
Redenote $z^\pm_0=x_\pm$ and $z^\pm_1=y_\pm$. It is clear that
$z^\pm_i\in\mathcal E_3,\ i=0,1$ and $z^\pm_0\neq1$.

Thus, we have shown that
the system of equations (\ref{z_0}),(\ref{z_1}) has two solutions in
$\mathcal A_3$ if $|\theta-37|_3<3^{-3}$ and
it has no solution in $\mathcal A_3$ if
$|\theta-1|_3<3^{-2}$.
\end{proof}

Thus by Proposition \ref{pro7} we have
\begin{thm} A phase
transition occurs for the three state $3$-adic
SOS model on a Cayley tree of order two if one of the following statements hold:\\
1) $\theta\in\{x\in\mathcal E_3: |x-1|_3<3^{-2}\}$;\\
2) $\theta\in\{x\in\mathcal E_3: |x-37|_3<3^{-3}\}$.
\end{thm}

\section{The uniqueness of $p$-adic Gibbs measures}

In the previous section we have shown that if $p\neq3$ then there is no phase transition
for the three state $p$-adic SOS model. A natural question arises:
what should be the relation between a number $m$ and prime $p$ in order to have
a phase transition for the $m+1$-state $p$-adic SOS model? In this section we
shall find this relation.

Let us first prove some technical results.

\begin{lemma}\label{lem3}\cite{MRasos}
If $a_i\in\mathcal E_p,\ i=1,2,\dots,n$ then $\prod_{i=1}^na_i\in\mathcal E_p$.
\end{lemma}
Recall that the $p$-adic norm of $x\in\mathbb Q^m_p$ defined as
$$
\Vert x\Vert_p=\max_{1\leq i\leq m}\{|x_i|_p\}.
$$

Let the collection of functions $F_i:\mathbb Q^{m+1}_p\mapsto\mathbb Q_p,\ i=0,1,\dots,m$
given by
$$
F_i(z,a_i,b_i,m)=\frac{\sum_{j=0}^{m}a_{ij}z_j}{\sum_{j=0}^{m}b_{ij}z_j},
$$
where
$$
a_{i}=(a_{i0},a_{i1},\dots,a_{im})\in\mathcal E^{m+1}_p\quad\mbox{and}\quad b_{i}=(b_{i0},b_{i1},\dots,b_{im})\in\mathcal E^{m+1}_p,
$$
for any $i=0,1,\dots,m$. For convenience we write $F_i(z)$ instead of $F_i(z,a_i,b_i,m)$.
\begin{lemma}\label{lem4}
Let $m+1$ is not divisible by $p$. Then $F_i$ is a function from
$\mathcal E^{m+1}_p$ to $\mathcal E_p$ for any $i=0,1,\dots,m$.
\end{lemma}
\begin{proof}
Let $z\in\mathcal E^{m+1}_p$. We will show that
$$
\left|F_i(z)-1\right|_p<p^{-1/(p-1)}.
$$

By Lemma \ref{lem3} we have $a_{ij}z_j,\ b_{ij}z_j\in\mathcal E_p$ for all $j=0,1,\dots,m$
as $a_{ij},\ b_{ij},\ z_j\in\mathcal E_p$. Since $m+1\not\equiv0(\operatorname{mod }p)$ and using
non-Archimedean norm's property we get
\begin{equation}\label{l21}
\left|\sum_{j=0}^mb_{ij}z_j\right|_p=\left|\sum_{j=0}^m\left(b_{ij}z_j-1\right)+m+1\right|_p=1.
\end{equation}
From this we obtain
$$
\left|F_i(z)-1\right|_p=\frac{\left|\sum_{j=0}^m(a_{ij}-b_{ij})z_j\right|_p}
{\left|\sum_{j=0}^mb_{ij}z_j\right|_p}\leq\max_{0\leq j\leq m}\{|a_{ij}-b_{ij}|_p\}<p^{-1/(p-1)}.
$$
\end{proof}
\begin{lemma}\label{lem5}
Let $m+1$ is not divisible by $p$. Then for any $z,t\in\mathcal E^{m+1}_p$ it holds
$$
\left|F_i(z)-F_i(t)\right|_p\leq\frac{1}{p}\Vert z-t\Vert_p,\qquad i=0,1,\dots,m.
$$
\end{lemma}
\begin{proof} Let $m+1\not\equiv0(\operatorname{mod }p)$. Then for any
$z,t\in\mathcal E_p$ from (\ref{l21}) we get
$$
\left|F_i(z)-F_i(t)\right|_p=\frac{\left|\sum_{j=0}^m\sum_{l=0}^ma_{ij}b_{il}(z_jt_l-z_lt_j)\right|_p}
{\left|\sum_{j=0}^ma_{ij}z_j\sum_{l=0}^mb_{ij}t_l\right|_p}=
$$
$$
\left|\sum_{j=0}^m\sum_{l=0}^ma_{ij}b_{il}\left(t_l(z_j-t_j)-t_j(z_l-t_l)\right)\right|_p=
$$
\begin{equation}\label{l31}
\left|\sum_{s=0}^m(z_s-t_s)\left(\sum_{l\neq s}a_{is}b_{il}-\sum_{j\neq s}a_{ij}b_{is}\right)\right|_p
\leq\frac{1}{p}\Vert z-t\Vert_p.
\end{equation}
As $a_{ij}, b_{il}\in\mathcal E_p$ then by Lemma \ref{lem3} we get $a_{ij}b_{il}\in\mathcal E_p$.
Hence, for any $i,j,l\in\{0,1,\dots,m\}$ we have
$$
\left|a_{ij}-b_{il}\right|_p=\left|a_{ij}-1+1-b_{il}\right|_p\leq
\max\{\left|a_{ij}-1\right|_p,\left|b_{il}-1\right|_p\}<p^{-1/(p-1)}\leq\frac{1}{p}.
$$
From this and by (\ref{l31}) we obtain
$$
\left|F_i(z)-F_i(t)\right|_p\leq\frac{1}{p}\Vert z-t\Vert_p.
$$
\end{proof}
\begin{lemma}\label{lem6}
Let $\{z^{(r)}\}_{r\in\mathbb N}$ and $\{t^{(r)}\}_{r\in\mathbb N}$ are the sequences
in $\mathcal E^{m+1}_p$. Then for any $n\in\mathbb N$ it holds
\begin{equation}\label{l41}
\left|\prod_{s=1}^nF_i(z^{(s)})-\prod_{s=1}^nF_i(t^{(s)})\right|_p\leq\frac{1}{p}
\max_{1\leq s\leq n}\left\Vert z^{(s)}-t^{(s)}\right\Vert_p,\qquad i=0,1,\dots,m.
\end{equation}
\end{lemma}
\begin{proof}
We prove this by induction on $n$. By Lemma \ref{lem5} condition (\ref{l41})
is satisfied for $n=1$. Suppose that (\ref{l41}) is valid for $n$.
We will prove that it is satisfied for $n+1$. We have
$$
\left|\prod_{s=1}^{n+1}F_i(z^{(s)})-\prod_{s=1}^{n+1}F_i(t^{(s)})\right|_p=
\left|\prod_{s=1}^{n}F_i(z^{(s)})F_i(z^{(n+1)})-\prod_{s=1}^{n}F_i(t^{(s)})F_i(t^{(n+1)})\right|_p=
$$
\begin{equation}\label{l42}
\left|\left(\prod_{s=1}^{n}F_i(z^{(s)})-\prod_{s=1}^{n}F_i(t^{(s)})\right)F_i(z^{(n+1)})+
\left(F_i(z^{(n+1)})-F_i(t^{(n+1)})\right)\prod_{s=1}^nF_i(t^{(s)})\right|_p
\end{equation}
By Lemma \ref{lem3} and Lemma \ref{lem4} we get
$$
\left|F_i(z^{(n+1)})\right|_p=1\qquad\mbox{and}\qquad\left|\prod_{s=1}^nF_i(t^{(s)})\right|_p=1.
$$
Using non-Archimedean norm's property from (\ref{l42}) we obtain
$$
\left|\prod_{s=1}^{n+1}F_i(z^{(s)})-\prod_{s=1}^{n+1}F_i(t^{(s)})\right|_p\leq
\max\left\{\left|\prod_{s=1}^{n}F_i(z^{(s)})-\prod_{s=1}^{n}F_i(t^{(s)})\right|_p,
\left|F_i(z^{(n+1)})-F_i(t^{(n+1)})\right|_p\right\}
$$
$$
\leq\frac{1}{p}\max\left\{\max_{1\leq s\leq n}\left\Vert z^{(s)}-t^{(s)}\right\Vert_p,
\left\Vert z^{(n+1)}-t^{(n+1)}\right\Vert_p\right\}=\frac{1}{p}\max_{1\leq s\leq n+1}\left\Vert z^{(s)}-t^{(s)}\right\Vert_p.
$$
\end{proof}
Consider the following functional equations
\begin{equation}\label{pro31}
z_{i,x}=\prod_{y\in S(x)}F_i(z),\quad i=0,1,\dots,m,\quad x\in V\setminus\{x_0\}.
\end{equation}
\begin{pro}\label{pro9} Let $m+1$ is not divisible by $p$. Then (\ref{pro31})
has a unique solution.
\end{pro}
\begin{proof}
Write
\begin{equation}\label{pro32}
\mathcal F_i(z_x)=\prod_{y\in S(x)}F_i(z_x).
\end{equation}
Let $z_x,t_x\in\mathcal E^{m+1}_p$ for any $x\in V\setminus\{x_0\}$.
Since $|S(x)|=k$ for all $x\in V\setminus\{x_0\}$ then by Lemma \ref{lem6} we have
\begin{equation}\label{pro33}
\left|\mathcal F_i(z_x)-\mathcal F_i(t_x)\right|_p\leq\frac{1}{p}\max_{y\in S(x)}
\Vert z_y-t_y\Vert_p\qquad\mbox{for all }\ x\in V\setminus\{x_0\}.
\end{equation}
Denote
$$
\mathcal F(z)=\left(\mathcal F_0(z),\mathcal F_1(z),\dots,\mathcal F_m(z)\right).
$$

By Lemma \ref{lem4} $\mathcal F$ is a function from $\mathcal E^{m+1}_p$ to $\mathcal E^{m+1}_p$.
From (\ref{pro33}) for any $x\in V\setminus\{x_0\}$ we get
$$
\Vert\mathcal F(z_x)-\mathcal F(t_x)\Vert_p\leq\frac{1}{p}\max_{y\in S(x)}\Vert z_y-t_y\Vert_p,
$$
which means that the function $\mathcal F$ is contractive. Then the function has a unique fixed
point in $\mathcal E^{m+1}_p$.
\end{proof}
\begin{thm}\label{uniq}
Let $H$ be a hamiltonian of $m+1$-state $p$-adic SOS model on a Cayley tree $\Gamma^k$. If
$p\nmid m+1$ then $|G(H)|=1$. Moreover, a measure $\mu\in G(H)$ is a
translation-invariant and symmetric.
\end{thm}
\begin{proof}
Let $F_m(z)\equiv1$ and
$$
F_i(z)=\frac{\sum_{j=0}^{m-1}\theta^{|i-j|_\infty}z_{j,y}+\theta^{m-i}}
{\sum_{j=0}^{m-1}\theta^{m-j}z_{j,y}+1}\ ,\quad i=0,1,\dots,m-1.
$$
Then by Proposition \ref{pro1} and Proposition \ref{pro9} there is a unique
$p$-adic Gibbs measure for the $m+1$-state $p$-adic SOS model on a Cayley tree
$\Gamma^k$ if $p\nmid m+1$.

Denote
$$
\mathcal I=\{(z_0,z_1,\dots,z_m): z_j\in\mathcal E_p\mbox{ and } z_j=z_{m-j},\ j=0,\dots,m\}.
$$
It is easy to see that for any $i=1,2,\dots,m+1$ it holds
$F_i(z)=F_{m-i}(z)$ if
$z\in\mathcal I$.
From the proof of the previous Proposition a unique solution of (\ref{pro31}) belongs
to $\mathcal I$. Then by Proposition \ref{pro2} a unique $p$-adic Gibbs measure
is a symmetric TI.
\end{proof}
{\bf Acknowledgments.} The author is grateful to Professor U. A. Rozikov and Professor F. M.
Mukhamedov for the useful advice and
valuable comments.

config: arxiv


TITLE: Examples of matrix groups with centralizer isomorphic to $\Bbb C$ or $\Bbb H$
QUESTION [0 upvotes]: By Schur's Lemma, the centralizer $C(G)$ of an irreducible matrix groups $G\subseteq\mathrm{GL}(\Bbb R^d)$ is an $\Bbb R$-division algebra, and thus, isomorphic to either $\Bbb R$, $\Bbb C$ or $\Bbb H$ (the quaternions). I always had the feeling that $C(G)\cong \Bbb R$ is the generic case. This is because I only know the example $G\cong \mathrm U(1)$ (considered as a real matrix group of dimension two) with $C(G)\cong\Bbb C$. I do not know an example for $C(G)\cong\Bbb H$ at all.

Questions:

What examples are there with $C(G)\cong \Bbb H$?
What other examples are there with $C(G)\cong \Bbb C$?
Have the groups with a given centralizer been "classified" in some sense? That is, is there a result of the form "all $G$ with $C(G)\cong \Bbb C$ are of the following form: ..." or the like?

Note that I want $G$ to be irreducible! But I don't care whether $G$ is finite or a Lie group or not nice at all.

REPLY [3 votes]: Well, that is just a homework exercise I set in my representation theory class a few weeks ago. (It is based on a problem I was given myself in a class I took in Aachen last milennium.) It is giving an example for which the endomorphism ring is $\mathbb{H}$. Since I'm lazy you get the problem text:
Let $i=\sqrt{-1}$ and $G=\langle{
\left(\begin{array}{rr}
0&1\\%
-1&0\\%
\end{array}\right),
\left(\begin{array}{rr}
i&0\\
0&-i\\%
\end{array}\right)%
}\rangle$ be the quaternion group of order 8.
a) Construct an irreducible representation of $G$ over the
real numbers, acting on a $4$-dimensional vectorspace $V\cong \mathbb{R}^4$.
( Hint: Use an $\mathbb{R}$-basis of $\mathbb{C}$ to get an $\mathbb{R}$-basis of $\mathbb{C}^2$.
To show that no 2-dimensional
submodule exists, consider images of a nonzero vector $(a,b,c,d)$ in this subspace under different elements of $G$, and show that they will yield a basis of at least a 3-dimensional subspace.)
b) Determine the endomorphism ring $End_{\mathbb{R} G}(V)$.
(Hint: The elements of $End_{\mathbb{R} G}(V)$ are $4\times 4$ matrices that
commute with the generators of $G$. Use this to deduce conditions on their
entires. Then show that every matrix fulfilling these conditions commutes
with $G$.)
c) By Schur's lemma $End_{\mathbb{R} G}(V)$ must be a division ring. Can you
identify it?


TITLE: Revisit "What is the rank of $Q+Q^T-Q\circ I$ if $Q = qq^T$"
QUESTION [0 upvotes]: Based on my previous problem: What is the rank of $Q+Q^T-Q\circ I$ if $Q = qq^T$

$Q = qq^T$, $q\in \mathbb{R}^n$. Suppose $q^Tq=1$.
$A\circ B:$ Hadmard (elementwise) product
$Q\circ I$ just takes the diagonal entries of $Q$.

My question: Is $Q+Q^T-Q\circ I$ still positive semidefinite?

I guess the answer is not fixed by Gershgorin circle theory since the diagonal terms become small. Not sure if I am correct.

REPLY [1 votes]: It is clearly not always positive semidefinite. E.g. when $q^T=(x,y)$, we have
$$
qq^T+qq^T-(qq^T)\circ I=\pmatrix{x^2&2xy\\ 2xy&y^2},
$$
whose determinant $-3x^2y^2$ is nonnegative only when $x=0$ or $y=0$.


TITLE: Magnetic field current loop
QUESTION [2 upvotes]: Why does a loop of irregular shape carrying current located in a magnetic field change to a circular shape?
Will this always happen assuming the wire to be flexible? Even when the force is inwards?
I think the reason might be to maximise flux but why should flux be maximised?

REPLY [3 votes]: Imagine the coil is totally free to move on a flat surface the plane of which is perpendicular to the external magnetic field.
The torques and forces acting on the coil will make the coil orientate itself so that the magnetic field that is produced inside the coil is in the same direction as the external magnetic field as this is the lowest potential energy state.
The "plane" of the coil is now at right angles to the external magnetic field and so the forces on the coil will be outwards.
These forces will thus make the coil form a circle to reach a stable equilibrium state with the net force and the net torque on the coil both zero.
If the coil found itself with its magnetic field in the opposite direction to the external magnetic field and was unable to flip over then the forces would be inwards and the coil would collapse in on itself as this would be a minimum potential energy state for this configuration.


TITLE: Product measure; support on a graph
QUESTION [1 upvotes]: Let $\mu$ and $\nu$ two probability measures on $\mathbb{R}$. Let $\pi$ a probability measure in $\mathbb{R}^{2}$ such as
$$
\pi (A\times \mathbb{R}) =\mu(A) \text{  and  } \pi(\mathbb{R} \times B)=\nu(B)
$$
and
$$
\forall (x,y),(x',y')\in \Gamma, x<x' \Rightarrow y\le y'
$$
with $\Gamma$ the support of the measure $\pi$.

Theorem : If $\mu$ is atomless then it exists a map $T : \mathbb{R} \rightarrow \mathbb{R}$ defined $\mu$-a.e. such
  that $\pi$ is concentrated on the graph of $T$.

Does somebody know how to prove it? It would be a relief for me because I don't see how.
And just to be sure the sentence "$\pi$ is concentrated on the graph of $T$" means $\Gamma = N \cup G(T)$ with $\pi(N)=0$ and $G(T)$ the graph of $T$ ?
Thank you so much for your help.

REPLY [0 votes]: I made a progress,  let $\mu$ and $\nu$ two probability measures on $\mathbb{R}$. Let $\pi$ a probability measure in $\mathbb{R}^{2}$ such as
$$
\pi (A\times \mathbb{R}) =\mu(A) \text{  and  } \pi(\mathbb{R} \times B)=\nu(B)
$$
Let $\Gamma$ be the support of $\pi$. If
$$
\Gamma \subseteq G_{T}
$$
with $G_{T}$ the graph of the $\mu$ mesurable function $T$ then
$$
T_{\text{#}} \mu = \nu
$$
A question, for the beauty of mathematics, is the controverse true ?
Anyway, let's go back to our problem, this step is not important to answer to my question. But it's a important application for me, I hope you will understand why I'm really interested in this problem now.
Thank you for your help and regards.
PS : Maybe we can use convex range for atomless measure to solve it !