text:
\begin{document}

\title{Integrability of Invariant Geodesic Flows on $n$-Symmetric Spaces}

\author{ Bo\v zidar Jovanovi\'c \\ \small Mathematical Institute, Serbian Academy of Sciences and
Arts
\\\small Kneza Mihaila 36, 11000 Belgrade, Serbia}
\date{}

\maketitle

\begin{abstract} In this paper, by modifying the argument shift
method,
we prove Liouville integrability of geodesic flows of normal
metrics (invariant Einstein metrics) on the Ledger-Obata
$n$-symmetric spaces $K^n/\diag(K)$, where $K$ is a semisimple
(respectively, simple) compact Lie group.
\\

{\bf Keywords:} noncommutative and commutative integrability,
invariant polynomials, translation of argument, homogeneous
spaces, Einstein metrics
\\

{\bf MSC:} 70H06, 37J35, 53D25
\end{abstract}

\section{Introduction}

\paragraph{Invariant Geodesic Flows.}
We study integrability of $G$-invariant geodesic flows on a class
of homogeneous spaces
\begin{equation}
 \label{L-O} Q=G/H, \quad
G=\underbrace{K\times \cdots \times K}_n, \quad
H=\diag(K)=\{(g,\dots,g)\,\vert\, g\in K\},
\end{equation}
where $K$ is a compact connected semisimple Lie group. The
homogeneous space $Q$ is diffeomorphic to the direct product
$K^{n-1}$, however as a $G$-homogeneous space it is a basic
example of a $n$-symmetric Riemannian space, see Ledger and Obata
\cite{LO}.

Let $\mathfrak g=\mathfrak k^n=\mathfrak k_1\oplus\mathfrak
k_2\oplus \dots\oplus\mathfrak k_n$, $\mathfrak
h=\{(x,\dots,x)\,\vert \, x\in \mathfrak k\}$, $\mathfrak k$ be
the Lie algebras of $G$, $H$ and $K$, respectively ($\mathfrak
k_i\cong\mathfrak k$ is the $i$-th factor). For simplicity, both
negative Killing forms on $\mathfrak g$ and $\mathfrak k$ will be
denoted by $\langle\cdot,\cdot\rangle$. Let
\begin{equation}
\mathfrak g=\mathfrak h\oplus \mathfrak v, \qquad \mathfrak
v=\{x=(x_1,\dots,x_n)\in\mathfrak k^n\, \vert \, x_1+\dots+x_n=0\}
\label{v}
\end{equation}
be the orthogonal decomposition with respect to the Killing form.

The linear subspace $\mathfrak v$ can be naturally identified with
$T_{\rho(e)}Q$, where $\rho: G\to Q=G/H$ is the canonical
projection. Then $G$-invariant metrics on $Q$, via restrictions to
$T_{\rho(e)}Q \cong \mathfrak v$, are in one-to-one correspondence
with $\Ad_H$-invariant scalar  products (e.,g., see \cite{Be})
\begin{equation}\label{I-metric}
(\,\cdot\,,\,\cdot\,)_{\mathfrak v}=\langle I(\,\cdot\,),\,\cdot\,
\rangle, \quad I: \mathfrak v\to \mathfrak v, \quad
I\circ\Ad_h=I\circ\Ad_h, \quad h\in H.
\end{equation}

The negative Killing form itself defines {\it normal (or standard)
metric} $ds^2_0$ \cite{Be}. Note that, in the case when $K$ is a
simple group, the normal $G$-invariant metric on $Q$ is Einstein
(see Wang and Ziller \cite{WZ}). Besides, Nikonorov proved that,
up to the isometry and homothety, the homogeneous space $Q$ for
$n=3$ ($n\ge 4$) admits exactly (respectively, at least) two
$G$-invariant Einstein metrics \cite{Nik}.

\paragraph{General Setting.}
Let $\mathcal F$ be a collection of functions closed under the
Poisson bracket on a Poisson manifold $(M,\{\cdot,\cdot\})$ and
let $\Lambda$ be the Poisson bivector related to
$\{\cdot,\cdot\}$. Consider the linear space $\mathrm F_x \subset
T_x^* M$ spanned by differentials of functions in $\mathcal F$.
Suppose that the numbers $\dim\mathrm F_x$ and $ \dim \ker \Lambda
\vert_{\mathrm F_x}$ are constant  almost everywhere on $M$ and
denote them by $\ddim\mathcal F$ and $\dind\mathcal F$,
respectively ({\it differential dimension} and {\it differential
index} of $\mathcal F$). The set $\mathcal F$ is called {\it
complete} if: $ \ddim{\mathcal F}+\dind{\mathcal F}=\dim
M+\corank\{\cdot,\cdot\}. $ It is {\it complete at} $x\in M$ if $
\dim\mathrm F_x + \dim \ker \Lambda \vert_{\mathrm F_x}= \dim M +
\ker\Lambda$, i.e., $\mathrm F_x$ is isotropic: $ \mathrm
F^\Lambda_x \subset \mathrm F_x$, $\mathrm F_x^\Lambda=\{\xi\in
T^*_xM \,\vert\, \Lambda_x(\xi,\mathrm F_x)=0\}.$

The Hamiltonian system $\dot f=\{f,h\}$ is {\it completely
integrable in the noncommutative sense} if it possesses a complete
set of first integrals $\mathcal F$. Then (under compactness
condition) $M$ is almost everywhere foliated by $(\dind{\mathcal
F}-\corank\{\cdot,\cdot\}$)-dimensional invariant tori. As in the
Liouville theorem, the Hamiltonian flow restricted to regular
invariant tori is quasi-periodic (see \cite{N, MF2, BJ2, Zu}).
Mishchenko and Fomenko stated the conjecture that non-commutative
integrable systems are integrable in the usual commutative sense
by means of integrals that belong to the same functional class as
the original non-commutative integrals \cite{MF2}. In the analytic
case, when $\mathcal F$ is a {finite-dimensional Lie algebra}, the
conjecture has been proved by Sadetov \cite{Sa}. The conjecture is
also proved in $C^\infty$-smooth case for infinite-dimensional
algebras (see \cite{BJ2}).

Now, let $Q=G/H$ be a homogeneous space of a compact Lie group
$G$, $\Phi: T^*Q\to \g^* $ be the momentum mapping of the natural
$G$-action on $T^*Q$, $\mathcal F_1=\Phi^*(\R[\mathfrak g])$ be
the set of Noether's functions and $\mathcal F_2$ be the set of
$G$-invariant functions, polynomial in momenta. Both $\mathcal
F_1$ and $\mathcal F_2$ are Lie subalgebras of
$(C^\infty(T^*Q),\{\cdot,\cdot\})$, where $\{\cdot,\cdot\}$ is the
canonical Poisson bracket. From Noether's theorem we have
$\{\mathcal F_1,\mathcal F_2\}=0$. Also $\mathcal F_1+\mathcal
F_2$ is a {\it complete set} of functions on $T^*Q$ (see Bolsinov
and Jovanovi\' c \cite{BJ1, BJ3}).

The Hamiltonian function $H_0$ of the normal metric $ds^2_0$ is a
Casimir  function within $\mathcal F_2$, so it Poisson commute
both with $\mathcal{F}_1$ and $\mathcal{F}_2$. Thus the geodesic
flow of the normal metric  is completely integrable in the
non-commutative sense by means of analytic functions, polynomial
in momenta.

\paragraph{Integrable Pairs.}
Within the class of Noether's integrals $\mathcal F_1$, for
example by using the argument translation method \cite{MF1}, one
can always construct a complete commutative subset of function
$\mathcal F^0\subset \mathcal F_1$ ($ \ddim {\mathcal F}^0 =
\frac12\left(\ddim \mathcal F_1 + \dind \mathcal F_1\right)$).
Thus, for the case of the geodesic flow of the normal metric, the
Mishchenko-Fomenko conjecture reduces to the construction of a
complete commutative subset $\mathcal F \subset \mathcal F_2$:
\begin{equation}
\ddim {\mathcal F} = \frac12\left(\ddim \mathcal F_2 + \dind
\mathcal F_2\right). \label{C}
\end{equation}
Indeed, from the completeness of $\mathcal F_1+\mathcal F_2$ it
follows that $\mathcal F^0+\mathcal F$ is a complete commutative
set on $T^*Q$ (see \cite{BJ1, BJ3}).

If the required subset $\mathcal F\subset \mathcal F_2$ exist, we
say that $(G,H)$ is an {\it integrable pair}. In \cite{BJ3} the
conjecture is stated that all pairs $(G,H)$ are integrable. If
$(G,H)$ is a spherical pair, in particular if $G/H$ is a symmetric
space, the algebra $\mathcal F_2$ is already commutative. In this
case we need only Noether's integrals $\mathcal F_1$ to integrate
the geodesic flow (see Mishchenko \cite{Mis}, Brailov \cite{Br}
and Mikityuk \cite{Mik}).

There are several known classes of integrable pairs (see
\cite{BJ1, BJ3, MP, DGJ}) but the general problem rest still
unsolved. For a related problem on the integrability of geodesic
flows on homogeneous spaces of noncompact Lie groups see, e.g.,
\cite{Bu, MS}.

\paragraph{Results and Outline of the Paper.}
Let $Q=G/H$ be the  Ledger-Obata $n$-symmetric space \eqref{L-O}.
By using the flag of subalgebras
\begin{equation}\label{flag}
\g_1=\mathfrak k_1\subset \g_2=\mathfrak k_1 \oplus \mathfrak k_2
\subset\dots\subset \g_n=\g=\mathfrak
k_1\oplus\dots\oplus\mathfrak k_n,
\end{equation}
we modify the argument shift method to construct a complete set of
polynomials on $\g$ with respect to the usual Lie-Poisson bracket
(Theorem \ref{prva}, Section 2). It allows us to  find a complete
commutative subset of polynomials within $\mathcal F_2$ (Theorem
\ref{n=3}, Corollary \ref{potpun}, Section 2) implying:

\begin{theorem}
The geodesic flow of the normal metric on the Ledger-Obata
$n$-symmetric space \eqref{L-O} is Liouville integrable by means
of analytic integrals, polynomial in momenta.
\end{theorem}

As a corollary, the complete commutative integrability of the
geodesic flows of invariant Einstein metrics constructed by
Nikonorov \cite{Nik} (Corollary 3, Section 3) is obtained.

\section{Liouville Integrability of Geodesic Flows}

\paragraph{$H$-invariant Euler Equations.}
Consider the left trivialization $T^*G \cong \mathfrak g \times
G$, where the identification $\g^*\cong\mathfrak g$ is given by
$\langle \cdot,\cdot\rangle$. Let $\hat I: \mathfrak g\to
\mathfrak g$ be a positive definite operator which defines
left-invariant metric $ds^2_{\hat I}$ on $G$.

The left $G$-reduction of the geodesic flow of the metric
$ds^2_{\hat I}$ is described by the Euler equations on $\mathfrak
g^* \cong\mathfrak g$:
\begin{equation}
\dot x_i=[x_i,\xi_i], \quad \xi_i=\nabla_{x_i} \hat
h(x_1,\dots,x_n)=\pr_{\mathfrak k_i} \hat A(x_1,\dots,x_n), \quad
i=1,\dots,n, \label{euler}\end{equation} where $\hat
h=\frac12\langle \hat A(x),x\rangle$ is the Hamiltonian, $\hat
A=\hat I^{-1}$ and $\pr_{\mathfrak k_i}$ is the projection to
$i$-th factor: $\pr_{\mathfrak k_i}(x_1,\dots,x_n)=x_i.$

The Euler equations are Hamiltonian with respect to the
Lie-Poisson bracket (the product of the Lie-Poisson brackets on
factors $\mathfrak k_i$):
\begin{equation}
\{f,g\}(x_1,\dots,x_n)=-\sum_{i=1}^n \langle x_i, [\nabla_{x_i}
f,\nabla_{x_i} g]\rangle. \label{LP}
\end{equation}

The  right $H$-action on $T^*G$ is Hamiltonian with momentum
mapping, in the left-trivialization, given by
\begin{equation}
\mu(x)=x_1+\dots+x_n. \label{mu}\end{equation}

The geodesic flow is invariant with respect to the right
$H$-action if and only if the Hamiltonian $\hat h$ is
$\Ad_H$-invariant, i.e,
\begin{equation*}
\langle [x,\mathfrak h], \hat A(x) \rangle=0
\,\,\Leftrightarrow\,\, \pr_{\mathfrak h} [\hat A(x),x]=0
\,\,\Leftrightarrow\,\, \sum_{i=1}^n {[\pr_{\mathfrak k_i} \hat A
(x),x_i]=0}, \label{i1}
\end{equation*}
where we used
\begin{equation}\label{proj-h}
 \pr_{\mathfrak
h}(x_1,\dots,x_n)=\frac1n(x_1+\dots+x_n,\dots,x_1+\dots+x_n).
\end{equation}

If the Hamiltonian $\hat h$ is $\Ad_H$-invariant, then the
momentum $\mu$ is preserved by geodesic flow and we can perform
the symplectic reduction of the flow to $\mu^{-1}(0)/H\cong T^*Q$.
The reduced flow is the geodesic flow of a $G$-invariant
submersion metric on $Q$.

Contrary, for a given $\Ad_H$-invariant positive definite operator
$I:\mathfrak v\to \mathfrak v$, let $ds^2_I$ be a $G$-invariant
metric defined by \eqref{I-metric}.  It can be seen as a
submersion metric of an appropriate left $G$-invariant and right
$H$-invariant metric $ds^2_{\hat I}$, simply by taking
$$
\hat I(x)=s\cdot \pr_\mathfrak v(x)+I\pr_\mathfrak h(x), \qquad
x\in\g,
$$
where $s>0$.

\paragraph{Algebra of $G$-invariant functions on $T^*Q$.}
The algebra $\mathcal F_2$ of $G$-invariant functions on $T^*Q$,
polynomial in momenta, can be identified with
$\mathbb{R}[\mathfrak v]^{H}$ ($\Ad_{H}$-invariant polynomials on
$\mathfrak v$). Within this identification, the Hamiltonian of the
metric $ds^2_{I}$ is given by $h(x)=\frac12\langle A x,x\rangle$,
$A=I^{-1}$, while the Hamiltonian of the normal metric $ds^2_0$ is
simply $h_0(x)=\frac12\langle x,x\rangle$. Further, the canonical
Poisson bracket on $T^* Q$ corresponds to the restriction of the
Lie-Poisson bracket (\ref{LP}) to $\mathbb{R}[\mathfrak v]^{H}$
(see Thimm \cite{Th}):
\begin{equation}
\{f,g\}_\mathfrak v(x)=-\langle x,[\nabla f(x),\nabla
g(x)]\rangle, \qquad f,g: \mathfrak v\to \R.\label{rpz}
\end{equation}

Let $\g_x$, $\mathfrak h_x$ and $\mathfrak k_{x_i}$ be isotropy
algebras of $x$ and $x_i$ in $\g$, $\mathfrak h$ and $\mathfrak
k$.

Consider the space $\mathfrak j_x \subset \mathfrak v$ spanned by
gradients of all polynomials in $\mathbb{R}[\mathfrak v]^{H}$. For
a generic point $x\in \mathfrak v$ we have (see \cite{BJ3, MP}):
\begin{eqnarray*}
&&\mathfrak j_x=([x,\mathfrak h]^\perp)\cap \mathfrak v=\{ \eta\in
\mathfrak v \, \vert\,  \langle \eta,[x,\mathfrak h]\rangle=0\}
=\{\eta\in \mathfrak v\, \vert\, [x,\eta]\subset \mathfrak v\}.\\
&&\mathfrak j_x=\{(\xi_1,\dots,\xi_n)\in\g\,\vert\, \sum_{i=1}^n
\xi_i=0, \quad \sum_{i=1}^n [x_i,\xi_i]=0\}.
\end{eqnarray*}

The Poisson bracket (\ref{rpz}) on the algebra
$\mathbb{R}[\mathfrak v]^{H}$ corresponds to the restriction of
the Lie-Poisson bivector
\begin{eqnarray}
\label{LPB} && \Lambda_x=\lambda_{x_1} \times \cdots \times
\lambda_{x_n}: \g \times \g \to \R, \\
&& \nonumber \lambda_\xi: \mathfrak k\times \mathfrak k \to \R,
\quad \lambda_\xi(\eta_1,\eta_2)=-\langle
\xi,[\eta_1,\eta_2]\rangle, \quad  \xi,\eta_1,\eta_2 \in \mathfrak
k
\end{eqnarray}
to $\mathfrak j_x$. Denote this restriction by $\bar\Lambda$. Note
that the kernel of $\bar\Lambda_x$ is \cite{BJ3}:
$$
\ker\bar\Lambda_x=\pr_\mathfrak v \ker\Lambda_x=\pr_\mathfrak
v\g_x= \pr_\mathfrak v (\mathfrak k_{x_1},\dots,\mathfrak
k_{x_n})\subset \mathfrak j_x.
$$

Here, for simplicity, the gradient operator with respect to the
restriction of $\langle\cdot,\cdot\rangle$ to $\mathfrak v$ is
also denoted by $\nabla$.

Note that \eqref{rpz} is a Poisson bracket within
$\mathbb{R}[\mathfrak v]^{H}$, while it is an almost-Poisson
bracket within the algebra of polynomials on $\mathfrak v$ (it
does not satisfies the Jacobi identity).

We have the following simple basic statement.

\begin{lemma} The differential dimension and index of
$\mathbb{R}[\mathfrak v]^{H}$ are given by:
\begin{equation}\label{DIM}
\ddim\mathbb{R}[\mathfrak v]^{H}=(n-2)\dim K, \qquad
\dind\mathbb{R}[\mathfrak v]^{H}=n\rank K.
\end{equation}
\end{lemma}

\noindent{\it Proof.} The differential dimension  of
$\mathbb{R}[\mathfrak v]^{H}$ is equal to the codimension of a
generic orbit $\Ad_H(x)$ within $\mathfrak v$, that is
$\ddim\mathbb{R}[\mathfrak v]^{H}=\dim\mathfrak j_x=\dim\mathfrak
v-\dim H+\dim\mathfrak h_x$, for a generic $x\in\mathfrak v$.
Since
\begin{equation}\label{HX}
\dim\mathfrak h_x=\dim(\mathfrak k_{x_1} \cap \dots\cap \mathfrak
k_{x_n})=0,
\end{equation}
for a generic $x\in\mathfrak v$, we obtain the first relation in
\eqref{DIM}. On the other side, from \eqref{HX} we get
\begin{eqnarray*}
\dind\mathbb{R}[\mathfrak v]^{H} &=& \dim\ker\bar\Lambda_x=\dim
\pr_\mathfrak v \g_x=\dim\g_x-\dim\mathfrak h_x\\
&=& \dim\g_x=\dim \mathfrak k_{x_1}+\dots+\dim\mathfrak k_{x_n}
=n\rank K,
\end{eqnarray*}
for a generic $x\in\mathfrak v$. $\Box$

\

Let $\R[\mathfrak k]^K$ be the algebra of $\Ad_{K}$-invariant
polynomials on $\mathfrak k$. It is generated by $r=\rank
K$-invariant homogeneous polynomials $f^1,\dots,f^r$. The algebra
of $\Ad_G$-invariant polynomials on $\mathfrak g$ is then
generated by $n\rank K$ polynomials
$$
\mathcal Z=\{f^\alpha_i=f^\alpha\circ\pr_{\mathfrak k_i}\, \vert\,
i=1,\dots,n, \, \alpha=1,\dots,r\}
$$
and the restrictions of invariants $\mathcal Z$ to $\mathfrak v$
give $n\rank K$ independent Casimir functions of
$\mathbb{R}[\mathfrak v]^{H}$.

\paragraph{Translation of Argument and Flag of Subalgebras.}
Mishchenko and Fomenko showed that the set of polynomials induced
from the invariants by shifting the argument ${\mathcal
A}=\{f^\alpha_{a,k}(x)\,\vert\, k=1,\dots,\deg f^\alpha, \,
\alpha=1,\dots,r\}$,
$$
f^\alpha_{a,t}(x)=f^\alpha(x+ta)=\sum_{k=0}^{\deg f^\alpha}
f^\alpha_{a,k}(x)t^k,
$$
is a complete commutative  set on $\mathfrak k$:
\begin{equation}\label{A}
\ddim\mathcal A=\frac12(\dim K+\rank K),
\end{equation}
for a generic $a\in \mathfrak k$ (see \cite{MF1, Bo, PY}). As was
already mentioned, the argument shift method allows us to
construct a complete commutative subalgebra in $\mathcal F_1$. Now
we shall modify the method, by using the flag of subalgebras
\eqref{flag} to use it to construct such a subalgebra within
$\mathcal F_2$.

Let
\begin{eqnarray*}
&&\mathcal B=\mathcal B_1+\mathcal B_2+\dots+\mathcal
B_{n-1}+\mathcal Z, \\
&&\mathcal B_i=\{f^\alpha_{i,k}(x)\,\vert\,k=1,\dots,\deg
f^\alpha, \,\alpha=1,\dots,r\},
\end{eqnarray*}
where polynomials $f^\alpha_{i,k}(x)$ are defined by:
\begin{equation}\label{translacija}
f^\alpha_{i,t}(x_1,\dots,x_n)=f^\alpha(x_1+\dots+x_i+tx_{i+1})=\sum_{k=0}^{\deg
f^\alpha} f^\alpha_{i,k}(x_1,\dots,x_n)t^k.
\end{equation}

\begin{theorem}\label{prva}
\item[(i)] The set $\mathcal B$ is a commutative set of
$\Ad_H$-invariant polynomials on $\g$.

\item[(ii)] The set $\mathcal B+\mu^*(\R[\mathfrak k])$ is a
complete set of polynomials on $\g$. In particular, if $\mathcal
A$ is any complete commutative set on $\mathfrak k$, then
$\mathcal B+\mu^*(\mathcal A)$ will be a complete commutative set
on $\g$.
\end{theorem}

\noindent{\it Proof.} Step 1. The polynomial in $\mathcal B$ are
$\Ad_H$-invariant. Indeed, let $h=(k,\dots,k)\in H$. Then
\begin{eqnarray*}
f^\alpha_{i,t}(\Ad_h(x)) &=&
f^\alpha(\Ad_k(x_1)+\dots+\Ad_k(x_i)+t\Ad_k(x_{i+1}))\\
&=& f^\alpha(\Ad_k(x_1+\dots+x_i+t x_{i+1}))\\
&=& f^\alpha(x_1+\dots+x_i+t x_{i+1})=f^\alpha_{i,t}(x).
\end{eqnarray*}

Step 2. The set $\mathcal B$ is involutive. Take  polynomials
$f^\alpha_{i,t_1}(x)$ and $f^\beta_{j,t_2}(x)$ given by
\eqref{translacija}.

Let $\nabla f^\alpha=\nabla f^\alpha\vert_{x_1+\dots+x_i+t_1
x_{i+1}}$ and $\nabla f^\beta=\nabla
f^\beta\vert_{x_1+\dots+x_j+t_2x_j}$. Then $[\nabla
f^\alpha,x_1+\dots+x_i+t_1x_{i+1}]=0$, $[\nabla
f^\beta,x_1+\dots+x_j+t_2x_{j+1}]=0$ and
\begin{eqnarray}
&&\nabla f^\alpha_{i,t_1}(x)=(\underbrace{\nabla
f^\alpha,\dots,\nabla f^\alpha}_i,t_1\nabla f^\alpha,0,\dots,0),
\\
&&\nabla f^\beta_{j,t_2}(x)=(\underbrace{\nabla
f^\beta,\dots,\nabla f^\beta}_j,t_2\nabla f^\beta,0,\dots,0).
\end{eqnarray}

First, consider the case $i<j$. We have
\begin{eqnarray*}
\{f^\alpha_{i,t_1},f^\beta_{j,t_2}\}(x)&=&-\langle x_1,[\nabla
f^\alpha,\nabla f^\beta]\rangle-\dots-\langle x_i,
[\nabla f^\alpha,\nabla f^\beta]\rangle\\
&&-t_1\langle x_{i+1},[\nabla f^\alpha,\nabla f^\beta]\rangle\\
&=&-\langle x_1+\dots+x_i+t_1x_{i+1},[\nabla f^\alpha,\nabla f^\beta]\rangle\\
&=& \langle [\nabla f^\alpha,x_1+\dots+x_i+t_1x_{i+1}],\nabla
f^\beta\rangle=0.
\end{eqnarray*}

Now, let $i=j$. Then we have
\begin{eqnarray*}
\{f^\alpha_{i,t_1},f^\beta_{i,t_2}\}(x)&=&-\langle
x_1+\dots+x_i,[\nabla f^\alpha,\nabla
f^\beta]\rangle \\
&&-t_1 t_2\langle x_{i+1},[\nabla f^\alpha,\nabla f^\beta]\rangle\\
&=& -\langle x_1+\dots+x_i+ t_1 x_{i+1},[\nabla f^\alpha,\nabla
f^\beta] \rangle\\
&&+t_1 \langle x_{i+1},[\nabla f^\alpha,\nabla
f^\beta]\rangle \\
&&-t_1\langle x_1+\dots+x_i+t_2x_{i+1},[\nabla f^\alpha,\nabla
f^\beta]\rangle\\
&&+
t_1 \langle x_1+\dots+x_i,[\nabla f^\alpha,\nabla f^\beta]\rangle\\
&=& t_1\langle x_1+\dots+x_{i}+x_{i+1},[\nabla f^\alpha,\nabla
f^\beta]\rangle.
\end{eqnarray*}

In the same way:
$$
\{f^\alpha_{i,t_1},f^\beta_{i,t_2}\}(x)=
 t_2\langle x_1+\dots+x_{i}+x_{i+1},[\nabla f^\alpha,\nabla
f^\beta]\rangle.
$$
Therefore $\{f^\alpha_{i,t_1},f^\beta_{i,t_2}\}=0$ for $t_1\ne
t_2$ and taking the limit $t_1 \mapsto t_2$, we get
$$
\{f^\alpha_{i,t_1},f^\beta_{i,t_2}\}=0
$$
for all $t_1, t_2$. It follows that $\{\mathcal B,\mathcal B\}=0$.
Item (i) is proved.

\

Step 3.  For a generic $x_1,\dots,x_i$, due to the
Mishchenko-Fomenko shifting of argument method, the set of
polynomials $\mathcal B_i$, considered as polynomials in variable
$x_{i+1}$, form a complete set on $\mathfrak k_{i+1}$ with respect
to the corresponding Lie-Poisson bracket. Therefore
\begin{equation}
\dim \pr_{\mathfrak k_{i+1}}\mathrm B_{i,x} x= \frac12(\dim
K+\rank K), \quad \mathrm B_{i,x}=\Span\{\nabla
f^\alpha_{i,k}(x)\}, \label{dimBi}
\end{equation}
for a generic $x\in\g$.

Let $\mathrm B_x$ be the linear space spanned by gradients of
polynomial in $\mathcal B$ at $x\in\g$. From \eqref{dimBi} we get
\begin{eqnarray}
\dim\mathrm B_x &\ge& \rank K+
\dim\pr_{\mathfrak k_2}\mathrm B_{1,x}+\dots+\dim\pr_{\mathfrak k_n}\mathrm B_{n-1,x}\nonumber\\
&\ge&  \rank K+\frac{n-1}2(\dim K+\rank K),
\label{dimB}
\end{eqnarray} where we used that $\mathcal B$
contains invariants in variable $x_1$. Thus $\ddim\mathcal
B\ge\frac12((n-1)\dim K+(n+1)\rank K)$.

\

Step 4. Fix a generic ($n\dim K-n\rank K$)-dimensional adjoint
orbit
$$
\mathcal O=\Ad_G(x_1,\dots,x_n)=\mathcal
O_1(x_1)\times\dots\times\mathcal O_n(x_n), \quad \mathcal
O_i(x_i)=\Ad_K(x_i),
$$
such that \eqref{HX} holds. This mean that the action of
$H=\diag(K)$ is locally free.

The orbit $\mathcal O$ with the Konstant-Kirillov symplectic form
$\omega$ is a symplectic leaf in $(\g,\{\cdot,\cdot\})$. The
$\Ad_H$-action, restricted to $\mathcal O$, is Hamiltonian with
the momentum mapping \eqref{mu} (e.g., see \cite{P}).

The algebra of $H$-invariant and Noether's functions
$C^\infty_H(\mathcal O)+\mu^*(C^\infty(\mathfrak k))$ is a
complete algebra on $(\mathcal O,\omega)$ and
\begin{eqnarray*}
&&\ddim C^\infty_H(\mathcal O)=(n-1)\dim K-n\rank K, \\
&&\ddim \mu^*(C^\infty(\mathfrak k))=\dim K \\
&&\ddim\left(C^\infty_H(\mathcal O)+\mu^*(C^\infty(\mathfrak k))\right)=n\dim K-(n+1)\rank K\\
 &&\dind C^\infty_H(\mathcal O)=\dind \mu^*(C^\infty(\mathfrak
k))
=\rank K\\
&&\dind\left(C^\infty_H(\mathcal O)+\mu^*(C^\infty(\mathfrak
k))\right)=\rank K
\end{eqnarray*}
 (see Theorem 2.1 and Remark 2.1 in \cite{BJ2}, where we used that a generic
$\Ad_K$-orbit in $\mu(\mathcal O)$ is regular and that the
$\Ad_H$-action is locally free at a generic point $x\in\mathcal
O$). In particular, a commutative set $\mathcal C\subset
C^\infty_H(\mathcal O)$ is a complete subset if
\begin{equation}\label{uslov}
\ddim\mathcal C=\frac12\left(\ddim C^\infty_H(\mathcal O)+\dind
C^\infty_H(\mathcal O)\right)=\frac{n-1}2(\dim K-\rank K).
\end{equation}

Let $\mathcal C=\{f\vert_\mathcal O\, \vert\, f\in \mathcal B\}$.
The invariants $\mathcal Z$ restricted to $\mathcal O$ are
constants, so we have
\begin{equation}\label{uslov*}
\ddim \mathcal C=\ddim \mathcal B-n\rank K \ge \frac12((n-1)\dim
K-(n-1)\rank K).
\end{equation}
From \eqref{uslov}, we get that $\mathcal C$ is a complete
commutative subset of $C^\infty_H(\mathcal O)$. In particular,
inequalities in \eqref{dimB} and \eqref{uslov*} are equalities.

Since the set of $\mathcal B+\mu^*(\R[\mathfrak k])$ is a complete
set restricted to a generic symplectic leaf $(\mathcal O,\omega)$,
it is a complete set on $(\g,\{\cdot,\cdot\})$. This completeness
the proof. $\Box$

\

By using Theorem \ref{prva}, we obtain the following integrable
model. Consider a left-invariant metric on $G$ defined by the
Hamiltonian function
\begin{equation}\label{novi}
\hat h_{s,t}=\sum_{i=1}^{n-1} \frac12  \langle
s_i(x_1+\dots+x_i)+t_i x_{i+1},s_i(x_1+\dots+x_i)+t_i
x_{i+1}\rangle,
\end{equation}
where parameters $s_i,t_i$ are chosen such that $\hat h$ is a
positive definite Hamiltonian of the left-invariant metric.

\begin{corollary}
The Euler equations on $\g$ determined with Hamiltonian
\eqref{novi}
\begin{eqnarray*}
&& \dot x_1=[x_1,\sum_{i=1}^{n-1}\left(s_i^2(x_1+\dots+x_i)+t_is_i
x_{i+1}\right)],\\
 &&\dot
x_k=[x_k,s_{k-1}t_{k-1}(x_1+\dots+x_{k-1})+\sum_{i=k}^{n-1}\left(s_i^2(x_1+\dots+x_i)+t_is_i
x_{i+1}\right)],\\
&&\dot x_n=[x_n,s_{n-1}t_{n-1}(x_1+\dots+x_{n-1})], \qquad \qquad
\qquad k=2,\dots,n-1
\end{eqnarray*}
are completely integrable.
\end{corollary}

\begin{lemma}{\rm \cite{BJ1}}\label{restrikcija}
\label{komutativnost} If $f$ and $g$ are $\Ad_H$-invariant
polynomials on $\g$ and $\{f,g\}=0$, then $\{f\vert_\mathfrak
v,g\vert_\mathfrak v\}_\mathfrak v=0$, where
$\{\cdot,\cdot\}_\mathfrak v$ is the bracket given by \eqref{rpz}.
\end{lemma}

Let $\mathcal F$ be the set of polynomials, obtained by
restriction of polynomials in $\mathcal B$ to $\mathfrak v$.

\begin{theorem}\label{n=3}
The set $\mathcal F$ is a complete commutative subset of
$\R[\mathfrak v]^H$.
\end{theorem}

\noindent{\it Proof.} According to Theorem \ref{prva} and Lemma
\ref{restrikcija}, the set $\mathcal F$ is commutative. Further,
from \eqref{C} and Lemma 1, it is complete if and only if
\begin{equation}
\ddim\mathcal F=\frac12((n-2)\dim K+n\rank K).
\label{dimF}\end{equation}

Since
\begin{equation*}
\mathrm F_x=\Span \{\nabla_x f(x)\, \vert \, f\in\mathcal F\}
=\Span\{\pr_\mathfrak v \nabla f(x)\,\vert\, f\in\mathcal
B\}=\pr_\mathfrak v\mathrm B_x,
\end{equation*}
we have
\begin{equation}\label{F-B}
\dim\mathrm F_x=\dim \mathrm B_x-\dim(\mathrm B_x\cap\mathfrak h).
\end{equation}

The relation \eqref{dimBi} is satisfied for a generic
$x\in\mathfrak v$ and $i<n-1$, while for  $i=n-1$ it does not
hold. Indeed, from $x_1+\dots+x_n=0$, we get that
$f^\alpha_{n-1,t}(x)=f^\alpha((1-t)x_n)=(1-t)^{\deg
f^\alpha}f^\alpha(x_n)$. Thus
\begin{eqnarray}
\dim\mathrm B_x &\ge& \rank K+ \dim\pr_{\mathfrak k_2}\mathrm
B_{1,x}+\dots+\dim\pr_{\mathfrak
k_{n-1}}\mathrm B_{n-2,x}+\rank K\nonumber\\
&=& 2\rank K+\frac{n-2}2(\dim K+\rank K),
\label{dimBv}
\end{eqnarray}
for a generic $x\in \mathfrak v$.

On the other hand, it is obvious that
\begin{equation}\label{presek}
\dim(\mathrm B_x \cap \mathfrak h) \le \dim(\Span\{\nabla
f^\alpha(x_n) \,\vert\, \alpha=1,\dots,\rank K\})=\rank K.
\end{equation}

Combining (\ref{dimBv}), \eqref{F-B} and \eqref{presek} we get
\begin{equation}
\dim \mathrm F_x \ge \frac12((n-2)\dim K +n\rank K), \label{dimF2}
\end{equation}
for a generic $x\in\mathfrak v$. According to (\ref{dimF}) we have
$\ddim\mathcal F \le\frac12((n-2)\dim K +n\rank K)$,  i.e., the
relation (\ref{dimF2}) is an equality.  $\Box$

\

Let $h_{s,t}$ be the restriction of the Hamiltonian \eqref{novi}
to $\mathfrak v$ and $ds^2_{s,t}$ be the corresponding
$G$-invariant submersion metric on $Q=G/H$.

\begin{corollary}\label{potpun}
The geodesic flow of the metric $ds^2_{s,t}$ is completely
integrable. The complete commutative set of analytic functions,
polynomial in momenta is
$$
\{\tau(f^\alpha_{i,k}\vert_\mathfrak v),
\tau(f^\alpha_i\vert_\mathfrak
v),\Phi^*(f^\alpha_{a_i,k})\,\vert\, i=1,\dots,n-1, k=1,\dots,\deg
f^\alpha, \alpha=1,\dots,r\}.
$$
\end{corollary}

Here $\tau$ denotes the bijection $\R[\mathfrak v]^H\to \mathcal
F_2$, $\Phi: T^*Q\to \g^* \cong \g$ is the momentum mapping of the
canonical $G$-action,
$$
f^\alpha(x_i+ta_i)=\sum_{k=0}^{\deg f^\alpha}
f^\alpha_{a_i,k}(x_1,\dots,x_n)t^k
$$
and $a_i\in\mathfrak k$, $i=1,\dots,n$ are in generic position.
That is, $\{f^\alpha_{a_i,k}\}$ is a complete commutative set on
$\g$ induced from the invariants by the argument translation with
$a=(a_1,\dots,a_n)$.

\paragraph{Gaudin Type Systems on $G=K^n$.}
Consider the Hamiltonian
\begin{equation*}
\hat h_a(x)=\frac12\left\langle \frac{1}{a_1} x_1 + \dots +
\frac{1}a_n x_n, \frac{1}{a_1} x_1 + \dots + \frac{1}a_n x_n
\right\rangle. \label{GH}\end{equation*}

 The
corresponding Euler equations on $\g=\mathfrak k^n$ are
\begin{equation}
\dot x_i=\sum_{j=1}^n \frac 1{a_ia_j} [x_i,x_j], \qquad
i=1,\dots,n. \label{gaudin}\end{equation}

Following \cite{P}, we refer to system (\ref{gaudin}) as a Gaudin
type system on $\mathfrak k^n$ (the Gaudin system is originally
defined for $\mathfrak k=su(2)$).

The system is $H$-invariant, so the momentum mapping (\ref{mu}) is
conserved along the flow. By using the pencil of compatible
Poisson brackets (e.g., see Bolsinov \cite{Bo}) defined by the
Lie-Poisson bivector (\ref{LPB}) and the bivector
$$
\hat{\Lambda}_x=a_1\lambda_{x_1} \times \cdots \times a_n
\lambda_{x_n}: \g \times \g \to \R,
$$
Panasyuk proved the integrability of equations (\ref{gaudin})
restricted to {\it admissible adjoint orbits} $\mathcal
O=\Ad_G(x_1,\dots,x_n)$ for a generic value of parameters
$a=(a_1,\dots,a_n)$ \cite{P}. The complete algebra of integrals is
$ \mathcal P+\mu^*(\R[\mathfrak k])$, where
\begin{equation*}
\mathcal P=\left\{f\left(\frac{x_1}{t_1 + a_1 t_2} + \dots
+\frac{x_n}{t_1 + a_n t_2}\right)\, \Big\vert\, t_1,t_2 \in \R,
t_1^2+t_2^2\ne 0, \, f\in \R[\mathfrak k]^K \right\}.
\end{equation*}

The set $\mathcal P$ is commutative. It could be proved that the
set of polynomials, obtained by the restriction of polynomials in
$\mathcal P$ to $\mathfrak v$ is a complete commutative subset of
$\R[\mathfrak v]^H$.

\section{Einstein Metrics}

Recall that the Riemannian manifold $(Q,g)$ is called {\it
Einstein} if the Ricci curvature $\mathrm{Ric}(g)$ satisfies the
equation $\mathrm{Ric}(g)=C\cdot g$, for some constant $C$
\cite{Be}.

From now on we assume  that $K$ is a simple Lie group. The normal
$G$-invariant metric $ds^2_0$ on \eqref{L-O} is Einstein (see
Proposition 5.5, \cite{WZ}). Up to the isometry and homothety, the
homogeneous space $Q$ for $n=3$ ($n\ge 4$) admits exactly
(respectively, at least) two $G$-invariant Einstein metrics that
we shall describe below.

Let $(\cdot,\cdot)_{\mathfrak v}$ be an $\ad_\mathfrak
h$-invariant scalar product on $\mathfrak v$. We can diagonalize
$(\cdot,\cdot)_{\mathfrak v}$ and $\langle
\cdot,\cdot\rangle\vert_\mathfrak v$ simultaneously (see
\cite{Nik}). Namely, let $\nu=(\nu_1,\dots,\nu_n)\in \R^n$ be an
unit vector and $\mathfrak k_\nu \subset \g$ be a linear subspace
defined by
\begin{equation}\label{modul}
\mathfrak k_\nu=\{(\nu_1 \xi,\dots,\nu_n \xi) \, \vert
\, \xi\in\mathfrak k\}.
\end{equation}
There exist $n-1$ orthogonal $\ad_\mathfrak h$-invariant
irreducible submodules $\mathfrak v_1,\cdots,\mathfrak
v_{n-1}\subset \mathfrak v$ and $n-1$ positive numbers
$s_1,\dots,s_{n-1}$ such that
$$
\mathfrak v=\mathfrak v_1 \oplus \mathfrak v_2 \oplus \dots \oplus
\mathfrak v_{n-1}, \qquad \mathfrak v_1=\mathfrak
k_{\nu^1},\dots,\mathfrak v_{n-1}=\mathfrak k_{\nu^{n-1}}
$$
and
\begin{equation}
(\cdot,\cdot)_{\mathfrak v}=s_1 \langle \cdot,\cdot
\rangle\vert_{\mathfrak v_1} \oplus s_2 \langle \cdot,\cdot
\rangle\vert_{\mathfrak v_2}\oplus \cdots \oplus s_{n-1} \langle
\cdot,\cdot \rangle\vert_{\mathfrak v_{n-1}},
\label{metric}\end{equation} where $\nu^1,\dots,\nu^{n-1}$ is the
orthonormal base of the hyperplane orthogonal to
$(1,\dots,1)\in\R^n$. The diagonalization is unique if all $s_i$
are different.

Now, let $ds^2_{p,q}$ be a $G$-invariant metric defined by the
scalar product (\ref{metric}), where
\begin{eqnarray}
&&\nu^j=\frac{1}{\sqrt{j^2+j}}(\underbrace{1,\dots,1}_j,-j,0,\dots,0),\qquad  j=1,\dots,n-1, \nonumber\\
&&s_1=\dots=s_{n-2}=1/p, \qquad s_{n-1}=1/q.
\label{s}\end{eqnarray} It is Einstein for $ p={n}^{1/(n-1)}$,
$p^{n-2}q=1$. Moreover, for $n=3$, up to isometry and homothety,
this is the only $G$-invariant Einstein metric different from the
normal one $p=q=1$ (see \cite{Nik}).

Together with the scalar product (\ref{metric}), (\ref{s}) it is
natural to consider its extension to an $\Ad_H$-invariant scalar
product on $\g$
\begin{equation}
(\cdot,\cdot)_{\mathfrak g}=\frac{1}{s}\langle \cdot,\cdot
\rangle\vert_{\mathfrak h}\oplus \frac{1}{p} \langle \cdot,\cdot
\rangle\vert_{\mathfrak v_1} \oplus \dots \oplus \frac{1}{p}
\langle \cdot,\cdot \rangle\vert_{\mathfrak v_{n-2}}\oplus
\frac{1}{q} \langle \cdot,\cdot \rangle\vert_{\mathfrak v_{n-1}}
\label{metric2}
\end{equation}
and the corresponding left-invariant metric $ds^2_{p,q,s}$ on $G$.
Then $ds^2_{p,q}$ can be seen as a submersion metric, induced by
$ds^2_{p,q,s}$.

The Hamiltonian of the metric $ds^2_{p,q,s}$, in the
left-trivialization, read
\begin{eqnarray}
\hat h &=&\frac{s}{2}  \langle \pr_\mathfrak h x,\pr_\mathfrak h x
\rangle+\frac{p}{2} \langle x-\pr_\mathfrak h x-\pr_{\mathfrak
v_{n-1}} x,x-\pr_\mathfrak h x- \pr_{\mathfrak v_{n-1}} x
\rangle\nonumber\\  &&+\frac{q}{2} \langle \pr_{\mathfrak v_{n-1}}
x,\pr_{\mathfrak v_{n-1}} x \rangle. \label{EH}
\end{eqnarray}

Note that the orthogonal projection to \eqref{modul} with respect
to the Killing form is given by
\begin{equation}\label{L3}
\pr_{\mathfrak k_\nu}(x_1,\dots,x_n)=(\nu_1(\nu_1 x_1+\dots+\nu_n
x_n),\dots,\nu_n(\nu_1 x_1+\dots+\nu_n x_n)).
\end{equation}

By using \eqref{proj-h} and \eqref{L3}, we easily get:

\begin{lemma}
The Hamiltonian \eqref{EH} has the form
\begin{eqnarray}
\hat h&=&\frac{p}{2}\sum_{k=1}^{n-1}\langle x_k,x_k\rangle +
\frac12\left(\frac{q n}{n-1}-\frac{p}{n-1}\right)\langle
x_n,x_n\rangle\nonumber \\
&&+\frac12\left(\frac{s}{n}-\frac{p}{n-1}+\frac{q}{n^2-n}\right)
\langle\mu,\mu\rangle+ \left(\frac{p}{n-1}-\frac{q}{n-1}\right)
\langle\mu,x_n\rangle \label{EH2}\end{eqnarray} where $\mu$ is the
momentum mapping \eqref{mu}.
\end{lemma}

The Euler equations with Hamiltonian (\ref{EH2}) are
\begin{eqnarray}
\nonumber && \dot x_k=[x_k, u (x_1+\dots+x_{n-1})+ vx_n  ], \qquad k=1,\dots,n-1,\\
\label{EN} && \dot x_n=[x_n, v(x_1+\dots+x_{n-1}) ],
\end{eqnarray}
where $u={s}/{n}-{p}/{(n-1)}+{q}/{(n^2-n)}$ and
$v={s}/{n}-{q}/{n}$. In particular, the set $\mathcal B$ is a set
of integral of the system (\ref{EN}). Whence, the functions
$\mathcal F$ commute with the Hamiltonian $h=\hat h\vert_\mathfrak
v$ of the metric $ds^2_{p,q}$. Applying Corollary \ref{potpun} we
obtain:

\begin{corollary}
The geodesic flow of the $G$-invariant Nikonorov's Einstein metric
on \eqref{L-O} is completely commutatively integrable by means of
analytic integrals, polynomial in momenta.
\end{corollary}

Note that, restricted to the invariant subspace $\mathfrak
v=\mu^{-1}(0)$, the equations (\ref{EN}) take the form
\begin{eqnarray}
&& \dot x_k=[x_k, (v-u)x_n  ], \qquad   k=1,\dots,n-1,\nonumber\\
&& \dot x_n=0, \label{ENR}\\
&& x_1+\dots+x_n=0.\nonumber
\end{eqnarray}
The generic solution of (\ref{ENR}) is given by
\begin{eqnarray*}
&&x_k(t)=\Ad_{\exp(t\xi)} x_k^0, \qquad x_k^0=x_k(0), \\
&&\xi=(v-u)(x_1^0+\dots +x^0_{n-1}), \quad k=1,\dots,n-1,
\end{eqnarray*}
where $\exp:\mathfrak k\to K$ is the exponential mapping.

\subsection*{Acknowledgments.} I am greatly thankful to Alexey
Bolsinov on useful discussions. This research was supported by the
Serbian Ministry of Science Project 144014 Geometry and Topology
of Manifolds and Integrable Dynamical Systems.


text:
\section{Proof of Proposition~\ref{prop:replicas_product_structure}}\label{sec:app_equivalence_replicas}

In this section, we prove Proposition~\ref{prop:replicas_product_structure}: we start from Conjecture~\ref{conjecture:replicas_general} and derive eq.~\eqref{eq:Phi_noniid}.
Note that by \ref{hyp:phi_product} we have $\langle\lambda\rangle_\nu = \alpha \EE_{\nu_B}[X] / \delta$.
We begin by recalling some sup-inf formulas, before turning to the actual proof.

\subsection{Some sup-inf formulas}

We recall Corollary~8 of \cite{barbier2019optimal}, stated here as a lemma:
\begin{lemma}[\cite{barbier2019optimal}]\label{lemma:supinf_barbier}
    Let $f : \bbR_+ \to \bbR$ be a $\mathcal{C}^1$ convex, non-decreasing, Lipschitz function. Define $\rho \equiv ||f'||_\infty$. Let $g : [0,\rho] \to \bbR$ be a convex, non-decreasing, Lipschitz function. For $(q_1,q_2) \in \bbR_+ \times [0,\rho]$ we define $\psi(q_1,q_2) \equiv f(q_1) + g(q_2) - q_1 q_2$. Then:
    \begin{align*}
        \sup_{q_1 \geq 0} \inf_{q_2 \in [0,\rho]} \psi(q_1,q_2) &= \sup_{q_2 \in [0,\rho]} \inf_{q_1 \geq 0} \psi(q_1,q_2).
    \end{align*}
\end{lemma}
We can state a corollary for functions of two variables.
\begin{corollary}\label{corollary:supinf_two_variables}
    Let $f : \bbR_+^2 \to \bbR$ be a $\mathcal{C}^1$ convex, Lipschitz function which is nondecreasing in each of its variables. Define $\rho_1 \equiv ||\partial_1 f||_\infty,\rho_2 \equiv ||\partial_2 f||_\infty$. Let $g : [0,\rho_1] \to \bbR$, $g_2 : [0,\rho_2] \to \bbR$ be two convex, non-decreasing, Lipschitz functions. For $(x_1,x_2,y_1,y_2) \in \bbR_+^2 \times [0,\rho_1] \times [0,\rho_2]$ we define $\psi(x_1,x_2,y_1,y_2) \equiv f(x_1,x_2) + g_1(y_1) + g_2(y_2) - x_1 y_1 - x_2 y_2$. Then:
    \begin{align*}
        \sup_{x_1,x_2 \geq 0} \inf_{y_1,y_2 \in [0,\rho_1]\times[0,\rho_2]} \psi(x_1,x_2,y_1,y_2) &=
       \sup_{y_1,y_2 \in [0,\rho_1]\times[0,\rho_2]} \inf_{x_1,x_2 \geq 0} \psi(x_1,x_2,y_1,y_2).
    \end{align*}
\end{corollary}

\begin{proof}[Proof of Corollary~\ref{corollary:supinf_two_variables}]
The proof is a verbatim of the proof of Corollary~8 in \cite{barbier2019optimal}, using that at fixed $y$, $x \mapsto f(x,y)$ is $\rho_1$-Lipschitz, while at fixed $x$, $y \mapsto f(x,y)$ is $\rho_2$-Lipschitz.
\end{proof}

\subsection{Core of the proof}

We now turn to the proof of Proposition~\ref{prop:replicas_product_structure}.
We begin by simplifying the free entropy potential using the Gaussian prior. We start from Conjecture~\ref{conjecture:replicas_general}.
Since $P_0$ is Gaussian by \ref{hyp:prior}, we can easily simplify the prior term $I_0$ as:
\begin{align*}
   I_0(q_x) &= \inf_{\hat{q}_x \geq 0} \Big[\frac{\beta \hat{q}_x (1-q_x)}{2} - \frac{\beta}{2} \ln(1+\hat{q}_x)\Big] = \frac{\beta q_x}{2} + \frac{\beta}{2} \ln(1-q_x).
\end{align*}
We now turn to the term $I_\mathrm{int}(q_x,q_z)$.
We can write it as:
\begin{align}\label{eq:Iint_gaussian_prior}
    I_\mathrm{int}(q_x,q_z) &= \inf_{\gamma_x,\gamma_z \geq 0} \Big[\frac{\beta}{2} (1 - q_x) \gamma_x + \frac{\alpha \beta}{2} (Q_z - q_z) \gamma_z - \frac{\beta}{2} \langle \ln (1 + \gamma_x + \lambda \gamma_z) \rangle_\nu \Big] \\
    & \hspace{1cm} - \frac{\beta}{2} \ln(1-q_x) - \frac{\beta q_x}{2} - \frac{\alpha \beta}{2} \ln(Q_z - q_z) - \frac{\alpha \beta q_z}{2 Q_z}. \nonumber
\end{align}
So we have, using Corollary~\ref{corollary:supinf_two_variables}, that if $f \equiv \sup_{q_x \in [0,1]} \sup_{q_z \in [0,Q_z]} [I_0(q_x) + \alpha I_\mathrm{out}(q_z) + I_\mathrm{int}(q_x,q_z)]$ is the conjectured limit of the free entropy:
\begin{align}
f &= \sup_{q_x \in [0,1]} \sup_{q_z \in [0,Q_z]} \inf_{\gamma_x,\gamma_z \geq 0} \Big[\alpha I_\mathrm{out}(q_z) +
\frac{\beta}{2} (1 - q_x) \gamma_x + \frac{\alpha \beta}{2} (Q_z - q_z) \gamma_z \nonumber\\
&\hspace{2cm}- \frac{\beta}{2} \langle \ln (1 + \gamma_x + \lambda \gamma_z) \rangle_\nu - \frac{\alpha \beta}{2} \ln(Q_z - q_z) - \frac{\alpha \beta q_z}{2 Q_z}\Big], \nonumber \\
\label{eq:f_supinf}
&=\sup_{\gamma_x,\gamma_z \geq 0} \inf_{q_z \in [0,Q_z]} \inf_{q_x \in [0,1]}  \Big[\alpha I_\mathrm{out}(q_z) +
\frac{\beta}{2} (1 - q_x) \gamma_x + \frac{\alpha \beta}{2} (Q_z - q_z) \gamma_z \\
&\hspace{2cm}- \frac{\beta}{2} \langle \ln (1 + \gamma_x + \lambda \gamma_z) \rangle_\nu - \frac{\alpha \beta}{2} \ln(Q_z - q_z) - \frac{\alpha \beta q_z}{2 Q_z}\Big]. \nonumber
\end{align}
The infimum on $q_x$ is very easily solved, as we have $\inf_{q_x \in [0,1]}[-\beta q_x \gamma_x / 2] = - \beta \gamma_x / 2$. Note that at fixed $\gamma_z \geq 0$, the variables $\gamma_x,q_z$ are completely decoupled in eq.~\eqref{eq:f_supinf}, so we have $\sup_{\gamma_x} \inf_{q_z} = \inf_{q_z} \sup_{\gamma_x}$. This yields:
\begin{align*}
f &= \sup_{\gamma_z \geq 0} \inf_{q_z \in [0,Q_z]} \sup_{\gamma_x \geq 0}  \Big[\alpha I_\mathrm{out}(q_z) +
\frac{\alpha \beta}{2} (Q_z - q_z) \gamma_z \\
&\hspace{2cm}- \frac{\beta}{2} \langle \ln (1 + \gamma_x + \lambda \gamma_z) \rangle_\nu - \frac{\alpha \beta}{2} \ln(Q_z - q_z) - \frac{\alpha \beta q_z}{2 Q_z}\Big], \\
&= \sup_{\gamma_z \geq 0} \inf_{q_z \in [0,Q_z]} \Big[ \frac{\beta}{2} \big[
\alpha (Q_z - q_z) \gamma_z - \alpha \frac{q_z}{Q_z} - \langle \ln (1 + \lambda \gamma_z) \rangle_\nu -\alpha \ln(Q_z - q_z) \big] +\alpha I_\mathrm{out}(q_z)\Big].
\end{align*}
Recall the form of $I_\mathrm{out}$ in Conjecture~\ref{conjecture:replicas_general} and that $\hat{Q}_z = 1/Q_z$. Using the form of $I_\mathrm{out}$, we have with the notations of Proposition~\ref{prop:replicas_product_structure}:
\begin{align*}
f &= \sup_{\gamma_z \geq 0} \inf_{q_z \in [0,Q_z]} \inf_{\hat{q}_z \geq 0} \Big[ \frac{\beta}{2} \big[
\alpha (Q_z - q_z) \gamma_z - \alpha \frac{q_z}{Q_z} - \langle \ln (1 + \lambda \gamma_z) \rangle_\nu -\alpha \ln(Q_z - q_z) \\
&- \alpha q_z \hat{q}_z - \alpha \ln(\hat{q}_z+1/Q_z) + \alpha Q_z \hat{q}_z\big]+  \alpha \Psi_\mathrm{out}(\sqrt{Q_z^2 \hat{q}_z / (1+Q_z \hat{q}_z)})
\Big].
\end{align*}
Again, we use that at fixed $q_z$, the variables $\hat{q}_z,\gamma_z$ are decoupled. So using again Lemma~\ref{lemma:supinf_barbier}, we have schematically $\sup_{\gamma_z} \inf_{q_z} \inf_{\hat{q}_z} = \sup_{q_z} \inf_{\hat{q}_z} \inf_{\gamma_z} = \sup_{\hat{q}_z}  \inf_{\gamma_z} \inf_{q_z}$. We can then explicitly solve the infimum on $q_z$, which yields:
\begin{align*}
f &= \sup_{\hat{q}_z \geq 0} \inf_{\gamma_z \geq 0} \Big[ \frac{\beta}{2} \big[
- \langle \ln (1 + \lambda \gamma_z) \rangle_\nu + \alpha \ln(1+\gamma_z(Q_z-q(\hat{q}_z))) \big]
+  \alpha \Psi_\mathrm{out}(q(\hat{q}_z))
\Big],
\end{align*}
with
\begin{align}
   q(\hat{q}_z) &\equiv \frac{Q_z^2 \hat{q}_z}{1 + Q_z \hat{q}_z}.
\end{align}
Note that $q$ is a strictly increasing smooth function of $\hat{q}_z$, with $q(0) = 0$ and $q(+\infty) = Q_z$.
So we have:
\begin{align}\label{eq:f_gammaz}
f &= \sup_{q \in [0,Q_z]} \inf_{\gamma_z \geq 0} \Big[ \frac{\beta}{2} \big[
- \langle \ln (1 + \lambda \gamma_z) \rangle_\nu + \alpha \ln(1+\gamma_z(Q_z-q)) \big]
+  \alpha \Psi_\mathrm{out}(q)
\Big],
\end{align}
We then state a technical lemma:
\begin{lemma}\label{lemma:technical_phi_structure}
Under hypothesis~\ref{hyp:phi_product}, one has for every $q \in [0,Q_z]$:
   \begin{align*}
        \inf_{\gamma_z \geq 0} [\alpha \ln(1+\gamma_z(Q_z-q))- \langle \ln (1 + \lambda \gamma_z) \rangle_\nu] &= \inf_{\hat{q} \geq 0} [\delta\hat{q}(Q_z - q)- \EE_{\nu_B} \ln(1+\hat{q}X)].
   \end{align*}
\end{lemma}
Using Lemma~\ref{lemma:technical_phi_structure} in eq.~\eqref{eq:f_gammaz}, and inverting the sup-inf by Lemma~\ref{lemma:supinf_barbier} finishes the proof of Proposition~\ref{prop:replicas_product_structure}.
In the remaining of the section we prove Lemma~\ref{lemma:technical_phi_structure}

\subsection{Proof of Lemma~\ref{lemma:technical_phi_structure}}
If $q=Q_z$, the equality is trivially satisfied, so let us assume $0 \leq q < Q_z$.
Let us denote $h(\gamma_z) \equiv \alpha \ln(1+\gamma_z(Q_z-q))- \langle \ln (1 + \lambda \gamma_z) \rangle_\nu$.
Recall that $Q_z = \EE_{\nu_B}[X]/\delta$.
Since $\alpha \geq 1-\nu(\{0\})$ and $q < Q_z$, one easily checks that $h$ is lower-bounded, so the infimum is always well-defined.
We introduce $\mu$ the asymptotic measure of $\bPhi \bPhi^\dagger / n$, and we denote $g_\mu(z) \equiv \langle (\lambda-z)^{-1}\rangle_\mu$ its Stieltjes transform. For every function $f$, one has $\langle f(\lambda) \rangle_\nu = \alpha \langle f(\lambda) \rangle_\mu + (1-\alpha) f(0)$. This allows to write:
\begin{align*}
    h(\gamma_z)  = \alpha \ln(1+\gamma_z(Q_z-q))- \alpha \langle \ln (1 + \lambda \gamma_z) \rangle_\mu.
\end{align*}
We will use the following equation, valid for every $\gamma_z \geq 0$ and any positively supported measure $\mu$:
\begin{align}\label{eq:log_potential}
    \langle \ln(\gamma_z+\lambda)\rangle_\mu &= \inf_{\tilde{\gamma_z} \geq 0} \Big[\gamma_z \tilde{\gamma_z} + \int_0^{\tilde{\gamma_z}} \mathcal{R}_\mu(-t) \mathrm{d}t - \ln \tilde{\gamma_z} -1\Big],
\end{align}
in which $\mathcal{R}_\mu$ is the so-called ``$R$-transform'' of $\mu$, defined as $\mathcal{R}_\mu(-x) \equiv g_\mu^{-1}(x) + 1/x$. It is a classical result of random matrix theory \cite{tulino2004random} that if $\mu$ is positively supported, $t \mapsto \mathcal{R}_\mu(-t)$ is well-defined on $\bbR_+$. We finish the proof of Lemma~\ref{lemma:technical_phi_structure}, before proving eq.~\eqref{eq:log_potential}.
By a classical result of random matrix theory \cite{marchenko1967distribution}, we know the $R$-transform of $\mu$ as a function of $\nu_B$:
\begin{align}\label{eq:R_transform_product}
   \mathcal{R}_\mu(-t) &= \EE_{\nu_B} \Big[\frac{X}{\delta + \alpha t X}\Big].
\end{align}
Combining eq.~\eqref{eq:log_potential} and eq.~\eqref{eq:R_transform_product}, we reach:
\begin{align*}
   \inf_{\gamma_z \geq 0} h(\gamma_z) &= \inf_{\gamma_z \geq 0} \sup_{\tilde{\gamma_z} \geq 0} \Big[\alpha \ln(1+\gamma_z(Q_z-q))+ \alpha - \alpha\frac{\tilde{\gamma_z}}{\gamma_z} + \alpha\ln \frac{\tilde{\gamma_z}}{\gamma_z} - \EE_{\nu_B} \ln \Big(1 + \frac{\alpha}{\delta} X \tilde{\gamma_z}\Big) \Big].
\end{align*}
Using Lemma~\ref{lemma:supinf_barbier} to invert the inf-sup, we have:
\begin{align*}
   \inf_{\gamma_z \geq 0} h(\gamma_z) &= \inf_{\tilde{\gamma_z} \geq 0}\sup_{\gamma_z \geq 0}  \Big[\alpha \ln(1+\gamma_z(Q_z-q))+ \alpha - \alpha\frac{\tilde{\gamma_z}}{\gamma_z} + \alpha\ln \frac{\tilde{\gamma_z}}{\gamma_z} - \EE_{\nu_B} \ln \Big(1 + \frac{\alpha}{\delta} X \tilde{\gamma_z}\Big) \Big].
\end{align*}
The supremum on $\gamma_z$ is now completely tractable, and we have:
\begin{align*}
   \inf_{\gamma_z \geq 0} h(\gamma_z) &= \inf_{\tilde{\gamma_z} \geq 0} \Big[\alpha (Q_z-q)\tilde{\gamma_z} - \EE_{\nu_B} \ln \Big(1 + \frac{\alpha}{\delta} X \tilde{\gamma_z}\Big) \Big].
\end{align*}
Doing the replacement $\hat{q} \equiv \alpha \tilde{\gamma_z} / \delta$ yields Lemma~\ref{lemma:technical_phi_structure}.
We now prove eq.~\eqref{eq:log_potential}, which will finish the proof. It follows from a classical result used in random matrix theory, see e.g.~\cite{guionnet2005fourier} for an application of these calculations to spherical integrals. Recall that $g_\mu$ is smooth and strictly increasing on $(-\infty,0)$, as $\mu$ is positively supported. It is easy to see by differentiation that the infimum in eq.~\eqref{eq:log_potential} is attained at $\tilde{\gamma_z} = g_\mu(-\gamma_z)$. We then use some manipulations:
\begin{align*}
    \inf_{\tilde{\gamma_z} \geq 0} \Big[\gamma_z \tilde{\gamma_z} + \int_0^{\tilde{\gamma_z}} \mathcal{R}_\mu(-t) \mathrm{d}t - \ln \tilde{\gamma_z}\Big]
    &= \gamma_z g_\mu(-\gamma_z) + \int_0^{g_\mu(-\gamma_z)} \mathcal{R}_\mu(-t) \mathrm{d}t - \ln g_\mu(-\gamma_z), \\
    &\hspace{-1cm}= \gamma_z g_\mu(-\gamma_z) + \int_\epsilon^{g_\mu(-\gamma_z)} g_\mu^{-1}(t) \mathrm{d}t - \ln \epsilon + \int_0^\epsilon \mathcal{R}_\mu(-t) \mathrm{d}t,
\end{align*}
this equation being valid for all $\epsilon > 0$ sufficiently small.
By regularity of the $R$-transform around $0$ \cite{tulino2004random}, $\int_0^\epsilon \mathcal{R}_\mu(-t) \mathrm{d}t = \smallO_\epsilon(1)$. Moreover, we can change variables in the other integral, and we reach:
\begin{align*}
    \inf_{\tilde{\gamma_z} \geq 0} \Big[\gamma_z \tilde{\gamma_z} + \int_0^{\tilde{\gamma_z}} \mathcal{R}_\mu(-t) \mathrm{d}t - \ln \tilde{\gamma_z}\Big]
    &= \gamma_z g_\mu(-\gamma_z) + \int_{-g_\mu^{-1}(\epsilon)}^{\gamma_z} u g_\mu(-u) \mathrm{d}u - \ln \epsilon + \smallO_\epsilon(1), \\
    &\overset{(a)}{=} - \ln \epsilon - \epsilon g_\mu^{-1}(\epsilon) + \int_{-g_\mu^{-1}(\epsilon)}^{\gamma_z} g_\mu(-u) \mathrm{d}u + \smallO_\epsilon(1), \\
    &\overset{(b)}{=} 1 + \langle \ln(\lambda+\gamma_z)\rangle_\mu + \smallO_\epsilon(1),
\end{align*}
in which we used integration by parts in $(a)$ and the definition of the Stieltjes transform in $(b)$.  Since $\epsilon$ was taken arbitrarily small, taking the limit $\epsilon \to 0$ ends the proof.

config: arxiv


TITLE: cohomology of $\mathbb{CP}^n$ and $\mathbb{RP}^m$
QUESTION [0 upvotes]: How to obtain the cohomology of $\mathbb{CP}^n$ and $\mathbb{RP}^m$, and what are the precise answers:
$$
H^i(\mathbb{CP}^n,\mathbb{Z}_2)=\text{?}
$$
$$
H^i(\mathbb{CP}^n,\mathbb{Z} )=\text{?}
$$
$$
H^i(\mathbb{RP}^n,\mathbb{Z}_2)=\text{?}
$$
$$
H^i(\mathbb{RP}^n,\mathbb{Z} )=\text{?}
$$

REPLY [2 votes]: You can find the cohomology of $\mathbb{CP}^{n}$: https://topospaces.subwiki.org/wiki/Cohomology_of_complex_projective_space
$$
H^p(\mathbb{P}^n(\mathbb{C});M) = \left\lbrace\begin{array}{rl} M, & \qquad p \text{ even, } 0 \le p \le 2n\\ 0, & \qquad \text{otherwise} \end{array}\right.
$$
find the cohomology of $\mathbb{RP}^{n}$:
https://topospaces.subwiki.org/wiki/Cohomology_of_real_projective_space
$$
H^p(\mathbb{P}^n(\mathbb{R}); M) = \left\lbrace \begin{array}{rl} M, & \qquad p=0,n\\ M/2M, &\qquad p \text{ even, } 0 < p < n\\ T, & \qquad p \text{ odd, } 0 < p < n \\ 0, & \qquad \text{otherwise}\end{array}\right.
$$
Here $M$ is an Abelian group.
Here $T$ denotes the $2$-torsion subgroup, i.e., the subgroup comprising elements of order dividing $2.$


TITLE: Simplexes in $\mathbb R^n$ have at most $n+1$ points
QUESTION [3 upvotes]: This is an exercise from the book Espaços Métricos (metric spaces) by Elon Lima. I'm translating it (the part of it that I'm having trouble with):

Show that if $X\subset\mathbb R^n$ is such that $\left\|x-y\right\|=1$ whenever $x\neq y$ and $x,y\in X$, then $X$ has at most $n+1$ points.

This is pretty straightforward when $n=1$, easy with some algebra when $n=2$, a little bit exhaustive with lots of algebra when $n=3$, but I'm not able to generalize the argument to $\mathbb R^n$ at all.
I know this looks like induction, but I've been trying for some time, and I can't do it.

REPLY [3 votes]: Let's try induction on $n$.  Say you've proved it for $1,2,3,\ldots,n-1$.  Suppose we have a set of $n$ points at unit distance from each other in $\mathbb R^n$ and we seek more points in that space that we can add to that set of $n$ points.  Those $n$ points must be affinely independent and affinely span an $(n-1)$-dimensional affine subspace of $\mathbb R^n$, since if they spanned a space of lower dimension than that, then our proposition would fail to be true in at least of the cases $1,2,3,\ldots,n-1$.  No generality is lost by assuming that that $(n-1)$-dimensional subspace is $\{(x_1,\ldots,x_{n-1},0) \in\mathbb R^n\}$.  Any point in $\mathbb R^n$ at unit distance from all $n$ of those points in that affine subspace must be in either of two hyperplanes parallel to that subspace.  The two points in those planes at unit distance from all $n$ of the points we started with are not at unit distance from each other, so we must discard one and keep one.


TITLE: Question about limits with variable on exponent
QUESTION [2 upvotes]: So I have to find the following limit $$\lim_{n\to\infty}\left(1+\frac{2}{n}\right)^{1/n}.$$I said that this is $$\lim_{n\to\infty}\left[\left(1+\frac{2}{n}\right)^n\right]^{1/n^2}=\left(e^2\right)^{\lim_{n\to\infty}1/n^2}=1.$$Now I know that the final answer is correct, but my method seems to me to be wrong. Can I seperate a limit in the way I  did - i.e. is my working correct? Is there any other more elegant way to find the limit?
Thanks in advance.

REPLY [3 votes]: You don't actually need to go through this all:
$$\lim_{n\rightarrow\infty} 1+\frac{2}{n}=1$$ and
$$\lim_{n\rightarrow\infty} \frac{1}{n}=0$$
therefore
$$\lim_{n\rightarrow\infty} \left(1+\frac{2}{n}\right)^{1/n}=1^0=1$$


TITLE: Show sum of any element is equal to zero in a ring
QUESTION [1 upvotes]: Given $x$ is in $\mathbb Z_n$, show the sum with $n$ terms $x+x+...+x = 0$
, by using the ring axioms.
My approach so far:
$~~~~x+x+...+x$
$=1\cdot x+1\cdot x +...+1\cdot x~$  (Uniqueness of 1)
$= (1+...+1)\cdot x~~~~~~~~~~~~~~~~$     (Distributivity)
$=  0\cdot x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ i doubt this if i can directly take $1+...+1=n=0~(mod~n)$
$= 0$
I doubt why I can take $1+...+1=0$ but not using $x+x+...+x=n\cdot x =0$ directly

REPLY [1 votes]: You are concerned whether we can assume $1+1+.... + 1 = n*1 = 0$.  And if we can assume it, the why can't we simply assume $x+x + ... + x = n*x = 0$?
Lemma:  $n*1 = 0$.
Pf:  if not, then $n*1 = m\ne 0; m \in \mathbb Z_n$.  So $0 < m < n$.  So $(n-m)*1 = n*1 + m*1 = n-m = 0$.  But $0 < n-m < n$ so $n-m \in \mathbb Z_n$ so $m = n \equiv 0 \mod n$.
i.e.  $n*1 = 0$.
And with this now proven, your proof is just fine.


TITLE: Prove that the integral $\int_0^1\frac{\log x}{x-1}dx$ exists
QUESTION [2 upvotes]: How do you show that \begin{equation*}\int_{0}^{1}\frac{\log x}{x-1}\mathop{}\!\mathrm{d}x\end{equation*} exists? (without using the $\text{Li}_2$ function etc. - I just want to show existence, not calculate the value).

REPLY [1 votes]: Note
\begin{align*}
\int_{0}^{1}\frac{\log x}{x-1}dx
 &=  \int_{0}^{1}\frac{\log x} {x^2-1}dx+ \int_{0}^{1}\frac{x\log x}{x^2-1}dx\\
 &
= \frac12 \int_{0}^{1} \left( \frac{\log x}{x-1}- \frac{\log x}{x+1} \right)dx + \frac14\int_{0}^{1}\frac{\log t}{t-1}dt\\
\end{align*}
which leads to
$$ \int_{0}^{1}\frac{\log x}{x-1}dx=-2 \int_{0}^{1}\frac{\log x}{x+1}dx< -2 \int_{0}^{1}\log x dx=2
$$
Thus, the integral is bounded, hence existing.


TITLE: Integrating $\frac{\sin ^2(x)}{x^2}$
QUESTION [1 upvotes]: Fourier transforming the function:
$$f(t) =
\left\{
 \begin{array}{ll}
  1;  & \mbox{ } |t| \leq 1 \\
  0; & \mbox{otherwise}
 \end{array}
\right.$$
We get:
$$F(y)=2 \frac{\sin y}{y}$$
And now applying Parseval's identity for the Fourier transform,
$$\int_{- \infty}^\infty {4 \frac{\sin ^2 y}{y^2}dy}=\int_{-1}^1 {f(t)^2}dt=2$$
By linearity of the integral we get the result $\frac{1}{2}$, however the result is $\pi$. Where did I go wrong? Thank you.

REPLY [4 votes]: Your result for the Fourier transform of $f$ indicates that you use the definition
$$\mathcal{F}[f](y) = \int_{-\infty}^\infty f(t) e^{-ity}\,dt$$
of the Fourier transform (or maybe even with $e^{ity}$ instead of $e^{-ity}$).
With that definition, the Fourier transform is not an isometry of $L^2(\mathbb{R})$, and Plancherel's theorem becomes
$$\int_{-\infty}^\infty \lvert \mathcal{F}[f](y)\rvert^2\,dy = 2\pi \int_{-\infty}^\infty \lvert f(t)\rvert^2\,dt.$$
Then the result you get is indeed $\pi$.


TITLE: Contraposition of "P if and only if Q"
QUESTION [4 upvotes]: I'm understanding the basic idea of contraposition, when it comes to propositional logic and writing proofs, but I'm having trouble figuring out what the contraposition of "P if and only if Q" would be. It seems simple enough that the contraposition of "If P then Q" would be "If not Q then not P", but that seems too simple to be the case with if and only if. Anyone able to help?

REPLY [5 votes]: Contrapositives are typically understood to be for conditionals. That is, conjunctions and disjunctions don't have any contrapositive. and the biconditinal ... well ... that's unusual too.
BUT it *is * true that $P \leftrightarrow Q \Leftrightarrow \neg Q \leftrightarrow \neg P$, so you could see this as the Contrapositive of the biconditional ... But again, that's pretty unusal. There will probably be a few websites, and maybe even some textbooks that talk about the contrapositive of a biconditional this way, but it is not standard nomenclature.


TITLE: If $u$ is harmonic, then $u\left(\frac{y_1}{|y|^2}, -\frac{y_2}{|y|^2}\right)$ is also harmonic
QUESTION [1 upvotes]: Let $u$ harmonic in $\Omega=\{x\in \mathbb{R}^2:|x|>1\}$. Show that
  $$v(y) = u\left(\frac{y_1}{|y|^2}, -\frac{y_2}{|y|^2}\right)$$ is harmonic in $U=\{y\in\mathbb{R}^2: 0<|y|<1\}$. Suggestion: to simplify the calculations, define $g_1 = y_1|y|^{-2}, g_2 = -y_2|y|^{-2}$ and show that, first, $(g_1)_{y_1} = (g_2)_{y_2}$ and also that $(g_1)_{y_2} = -(g_2)_{y_1}$.

I guess $(g_1)_{y_1}$ is the partial of $g_1$ with respect to $y_1$ and so on.
Recall that
$$p = \frac{1}{|y|^2} = \frac{1}{y_1^2 + y_2^2} \implies \frac{\partial p}{\partial y_1}= -1\frac{1}{(y_1^2 + y_2^2)^2}2y_1 = -1\frac{1}{|y|^4}2y_1,$$
$$\frac{\partial g_1}{\partial y_1} = |y|^{-2} -1y_1\frac{1}{|y|^4}2y_1,$$
$$\frac{\partial g_1}{\partial y_1} = |y|^{-2} -1y_2\frac{1}{|y|^4}2y_1,$$
which are not the same. What is wrong?
Suppose I did carry the calculations right. Then I must show that $\Delta v(y) = 0$ for $y$ with $0<|y|<1$.
$$\frac{\partial v}{\partial y_i} = \partial_1 u\frac{\partial g_1}{\partial y_i} + \partial_2 u\frac{\partial g_2}{\partial y_i},$$
$$\frac{\partial^2 v}{\partial y_i^2} = \partial_1^2 u\frac{\partial g_1}{\partial y_i} + \partial_1 u\frac{\partial^2 g_1}{\partial y_i^2} + \partial_2^2 u\frac{\partial g_2}{\partial y_i} + \partial_2 u\frac{\partial^2 g_2}{\partial y_i^2},$$
so
$$\Delta v(y) = \partial_1^2 u\frac{\partial g_1}{\partial y_1} + \partial_1 u\frac{\partial^2 g_1}{\partial y_1^2} + \partial_2^2 u\frac{\partial g_2}{\partial y_1} + \partial_2 u\frac{\partial^2 g_2}{\partial y_1^2} + \\ \partial_1^2 u\frac{\partial g_1}{\partial y_2} + \partial_1 u\frac{\partial^2 g_1}{\partial y_2^2} + \partial_2^2 u\frac{\partial g_2}{\partial y_2} + \partial_2 u\frac{\partial^2 g_2}{\partial y_2^2}=\\(\partial_1^2 u + \partial_2^2 u)(g_1)_{y_1} + (-\partial_1^2 u + \partial_2^2 u)(g_1)_{y_1} + \\
 \partial_1 u\frac{\partial^2 g_1}{\partial y_1^2} + \partial_2 u\frac{\partial^2 g_2}{\partial y_1^2} + \partial_1 u\frac{\partial^2 g_1}{\partial y_2^2} + \partial_2 u\frac{\partial^2 g_2}{\partial y_2^2},
$$
which must be $0$. I do not think I should take the second derivatives of $g_1$ and $g_2$, because it would be huge. Maybe the conditions $(g_1)_{y_1} = (g_2)_{y_2}$, $(g_1)_{y_2} = -(g_2)_{y_1}$ imply something about the second derivatives that I can use. I do not know. The only thing I can see being $0$ is $(\partial_1^2 u + \partial_2^2 u)(g_1)_{y_1}$, because $u$ is harmonic.

REPLY [0 votes]: Is your new $u$ even well-defined?! Because the division by $|y|^2$ takes a point to the perimeter of the unit circle, where the original $u$ is not assigned a value, at least immediately.
Secondly, consider the mean/average property of harmonic maps.


TITLE: Convert a real range to an integer range.
QUESTION [1 upvotes]: Let's say that we have a set of integers in the range $[1, 4]$. Now, I have a function that will calculate a distance between two vectors, and this function returns a real number in the range $[0, 1]$.
Currently, to calculate the error made by this function, I convert the numbers from the real range $[0, 1]$ to the integer range $[1, 4]$ using the following mapping:
$$
M(x) =
\begin{cases}
    1,&\quad\text{if}~~~0~~~~~ \le x \le 0.25\\
    2,&\quad\text{if}~~~0.25 \lt x \le 0.5\\
    3,&\quad\text{if}~~~0.5~~ \lt x \le 0.75\\
    4,&\quad\text{if}~~~0.75 \lt x \le 1\\
\end{cases}
$$
But I think this is not really precise, because values such as $0.49$ will be converted to $2$ when they are closer to $3$.
I am looking for a better way to map values from the real range $[0, 1]$ to the integer range $[1, 4]$ that will not lose so much precision as the one above.

REPLY [2 votes]: With the information given, you can't do better. If you map an interval of reals to a finite set of discrete values, there will always be points of discontinuity. The smaller the number of those discrete values, the bigger the average size of the pre-image of a range value becomes.
In your example, it's obvious that if you use a 'sensible' mapping (something measurable), then you need to map at least an interval (or union of intervals) with aggregate length $\frac14$ to one of your $4$ values. So the mapping you gave is in some sense best, you have reached, but not exceeded, that limit.
No matter what you do, if you choose just $4$ possible mapping values, it will feel imprecise. Think of them as school grades. With just $4$ different grades available, as a teacher you'd have lots of cases where a small difference in 2 students' work quality would result in a different grade, even when the difference doesn't feel like "one grade".
Now, if you know or suspect something about the actual distribution of your value in the $[0,1]$, that might help you find a better mapping. For example, if values above $0.5$ are much less likey than values below, maybe have a mapping where the cutoff points are $0.1, 0.2, 0.6$ and $1$.
My advise would be to think about what you actually need. Why are you converting $[0,1]$ values to discrete 'grades' when you are concerned about loss of accurracy? If you need discrate 'grades', maybe use more of them such that each 'jump' from one grade to the next doesn't feel as big. Maybe use an odd number of grades, if you think that having a 'middle grade' would be benficial to map values around $0.5$ to.
None of those are generally better than your function. But depending on what else you know about the problem at hand, they might be able to better express what you want to achieve.


TITLE: Prove that if $A \subset B$ then $P(A) \leq P(B)$
QUESTION [8 upvotes]: I'm supposed to prove that if $A \subset B$, then $P(A) \leq P(B)$. The hint it gives is confusing me even more. It says use a venn diagram to convince yourself $ B = A \cup (A^c \cap B)$ and $A$ and $A^c \cap B$ are disjoint.
I know that I can do this:
$A \cup (A^c \cap B) = (A \cup A^c) \cap (A \cup B)$
Since $(A \cup A^c) = U$ that means if it intersects with $A \cup B$ we just have a venn diagram with everything shaded in.
How do I use this to my advantage to prove the original question? And wouldn't the two sets $A$ and $A^c \cap B$ being disjoint hurt me? Since I want to show that everything in $A$ is in $B$?

REPLY [1 votes]: This is my first post on math StackExchange. I was reading up on probabilities when I came across this question. The following is my shot at the problem. Do let me know in the comments if this is a valid proof.
$P(B) = P(A) + P(A^C \cap B)$
The above equation can be formulated by imagining a simple Venn diagram. Now, by removing the second probability from R.H.S., we get,
$P(B) \geq P(A)$


TITLE: Continuity at $x=0$ of this function
QUESTION [2 upvotes]: Not a hard exercise:$$f(x)=\frac{1}{x^3}\cdot \int_{-x}^x \sin(4t^2) \, \text{d}t \quad \text{where} \space x\ne 0\:$$
$$f(x)=5\:;\:x=0\:$$
Checking it's continuity at $x=0$ by using L'Hospital's rule and Newton-Leibniz axiom:
$$\begin{align} \lim _{x\to 0}\frac{1}{x^3}\cdot \int_{-x}^x \sin(4t^2) \, \text{d}t & =\lim_{x\to 0}\frac{2\sin(4x^2)}{3x^2} \\[10pt]
& =\lim_{x\to 0}\frac{2\sin(4x^2)}{8x^2}\frac{8x^2}{3x^2} \\[10pt]
& =\lim_{x\to 0}\frac{\sin(4x^2)}{4x^2}\frac{8x^2}{3x^2} \\[10pt]
& =\frac{8}{3}\end{align} $$

REPLY [0 votes]: In the way it is written here, you have made a minor mistake. I recommend you to pass from the first line to the third line directly and omit the x^2s.


TITLE: What did I do wrong when I solved this differential equation
QUESTION [0 upvotes]: If $\frac{dy}{dx}=y\sec^2x$ and $y=5$ when $x=0$, then $y=$?
A) $e^{\tan x} + 4$
B) $e^{\tan x} + 4$
C) $5e^{\tan x}$
D) $\tan x +5$
E) $\tan x + 5e^x$
This is how I solved it:
$\frac{dy}{y}=\sec^2x dx$
$\int \frac{dy}{y}= \int \sec^2x dx$
$\ln|y|= \tan x +C$
$e^{\ln|y|}=e^{\tan x}+C$
$|y|=e^{\tan x} + C$
$y=\pm (e^{\tan x} + C)$
Since $y=5$ we choose $y=e^{\tan x} + C$
But the answer is C)
I know that I can find $dy/dx$ for each answer choice and C) is the only one that fits.. however I don't know what I did wrong in my solution.

REPLY [0 votes]: Here is where you made a mistake

$\ln|y|= \tan x +C$
$e^{\ln|y|}=e^{\tan x}+C$

If $\ln|y|= \tan x +C$, after exponentiation you get $e^{\ln|y|}=e^{\tan x+C}$


TITLE: Find f(32) given f(x) = f(x-1) + 2^(f(x-1)+1) and f(0) = 5
QUESTION [2 upvotes]: I would like to find f(32) given:
f(x) = f(x-1) + 2^(f(x-1)+1)
f(0) = 5
I figured that this function grows tetrationally (if that is a word) but I don't really know
Where I got this equation from:
In the collectable trading card game Magic: The Gathering, there are these cards:
Miirym, sentinel wyrm,
Astral Dragon,
Parallel Lives
If Astral Dragon where to come into play with Miirym and Parallel Lives already on the battlefield and all of the tokens Astral Dragon created where copies of Parallel Lives, you would end up with f(32) Parallel Lives in play.
Edit: I am not looking for a specific answer, just a rough estimate of how large that number is. I know it is larger than 2^^32 bit is it Larger than 2^^33? 2^^34?

REPLY [1 votes]: I used Hypercalc at the suggestion of Gottfried Helms
The answer is 30 PT (3.553934904655 × 10^20)
Which means a power tower of 10s, 30 high with (3.553934904655 × 10^20) on top ie: 10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^(3.553934904655 × 10^20)


text:
\section{Half-Open Real Interval is Closed in some Open Intervals}
Tags: Real Intervals, Closed Sets

\begin{theorem}
Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as an [[Definition:Euclidean Space|Euclidean space]].
Let $\left[{a \,.\,.\, b}\right) \subset \R$ be a [[Definition:Half-Open Real Interval|half-open interval]] of $\R$.
Let $c < a$.
Then $\left[{a \,.\,.\, b}\right)$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $\left({c \,.\,.\, b}\right)$.
Similarly, let $d > b$.
Then the [[Definition:Half-Open Real Interval|half-open interval]] $\left({a \,.\,.\, b}\right]$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $\left({a \,.\,.\, d}\right)$.
\end{theorem}

\begin{proof}
Consider:
: $A := \left({c \,.\,.\, b}\right) \setminus \left[{a \,.\,.\, b}\right) = \left({c \,.\,.\, a}\right)$
Then $A$ is an [[Definition:Open Real Interval|open interval]].
By [[Open Real Interval is Open Set]], $A$ is [[Definition:Open Set (Metric Space)|open]] in $\R$.
By the definition of the [[Definition:Topological Subspace|subspace topology]] it follows that $A$ is [[Definition:Open Set (Metric Space)|open]] in $\left({c \,.\,.\, b}\right)$.
Thus by the definition of [[Definition:Closed Set (Metric Space)|closed set]], $\left({c \,.\,.\, b}\right) \setminus A = \left[{a \,.\,.\, b}\right)$ is [[Definition:Closed Set (Metric Space)|closed]] in $\left({c \,.\,.\, b}\right)$.
[[Definition:Mutatis Mutandis|Mutatis mutandis]], the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $\left({a \,.\,.\, d}\right)$.
{{qed}}
\end{proof}

config: wiki


TITLE: To prove being not isomorphic via connectivity of vertices of certain degrees
QUESTION [1 upvotes]: I'm reading Richard J. Trudeau's book "Introduction to Graph Theory", in ch 2 there's an example

Example. The graphs of Figure 35
 are not isomorphic, even though they share the four
  properties: in each v = 8 and e = 10; each has four vertices of degree
  2 and four of degree 3; and they are both in one piece. The structural
  difference is that the vertices of degree are not related in the same
  way in the two graphs. In the first graph the vertices of degree 2
  come in adjacent pairs: B is adjacent to H and D is adjacent to F.
  But in the second graph they are completely separated from one another by vertices of degree 3: no vertex of degree is adjacent to
  any other vertex of degree 2. To see how this prevents the graphs from
  being isomorphic, we shall systematically try to construct an
  isomorphism in all possible ways. We begin by noticing that B, being
  of degree 2, would have to correspond to 1, 3, 6, or 8.
First case: B corresponds to 1.
Then H, being adjacent to B, would have to correspond to 2 or 7. But H
  must correspond to a vertex having the same degree, so H cannot
  correspond to either 2 or 7. Therefore no isomorphism is possible if B
  is made to correspond with 1.
Second case: ...

after discussed all 4 cases, it announces

We have seen that no isomorphism is possible if B is made to
  correspond to any of 1, 3, 6, or 8. But under any isomorphism, B would
  have to correspond to 1, 3, 6, or 8. Therefore no isomorphism is
  possible, and the graphs of Figure 35 are not isomorphic.

I agree with all the proof process, just, I wonder, is it enough to assert being not isomorphic via connectivity of vertices of certain degrees? Such as, with such differences

In the first graph the vertices of degree 2 come in adjacent pairs... But in the second graph they are completely separated from
  one another by vertices of degree 3

is it ok to announce the two graphs are not isomorphic? Is there a such general theorem?

REPLY [1 votes]: Yes, it is sufficient. For a graph $G$, let $G_d$ be the induced subgraph made by vertices of degree $d$. We have the following
Lemma: suppose $f:G \to H$ is an isomorphism of graphs. Then $f| G_d :G_d \to H_d$ is an isomorphism for every $d$.
Indeed, $f$ maps vertices of degree $d$ to vertices of degree $d$: the neighbors of $f(v)$ are exactly $\{f(w)\}$ where $w$ is a neighbor of $w$. So $f|G_d : G_d \to H_d$ is injective and surjective on vertices: injective because it is the restriction of an injective map, surjective because every vertex of degree $d$ must be taken by a vertex of degree $d$. Moreover, for $v,w \in G_d$:
$$(v,w) \in G_d \ \ \Leftrightarrow \ \  (v,w) \in G \ \ \Leftrightarrow \ \  (f(v),f(w) ) \in H \ \ \Leftrightarrow \ \  (f(v),f(w) ) \in H $$
using the property that $f$ is an iso and that $G_d, H_d$ are induced subgraphs.
In your case, set $G$ the graph on the left, $H$ the graph on the right. You have that $G_2$ has two connected components, while $H_2$ has four connected components, so they cannot be isomorphic.


TITLE: if one of the sets A and B is compact then d(A,B)>0.
QUESTION [1 upvotes]: Let $A$ and $B$ be two nonempty disjoint subsets of $\mathbb{R}^{n}$. Put $d(A,B)=inf\left \{ ||a-b||:a\in A, b\in B  \right \}$.
a) Show that if one of the sets $A$ and $B$ is compact then $d(A,B)>0$.
b) Let $n=2$, $A=\left \{ (x,1/x):x>0 \right \}$ and $B=\left \{ (x,0):x>0 \right \}$. Show that the sets $A$ and $B$ are closed in $\mathbb{R}^{2}$, $A\cap B=\emptyset$ but $d(A,B)=0$.
My thoughts: Let $A$ be compact then it is not only closed but also bounded. Assume $d(A,B)=0$, so $inf||a-b||=0$. So $a=b$ for some $a$ and $b$. (I think  $A$ and $B$ being closed is the reason of that, but I'm not sure how to show this.) but that contradicts the fact that the sets are disjoint.
for part b:$ d(A,B)=\sqrt{((x-x)^{2}+(1/x)^{2})}$, as $x$ increases distance decrases, since $lim_{x\rightarrow \infty}(1/x)^{2}=0$, $d(A,B)=0$. For showing their intersection is an empty set $1/x\neq 0$ if $x\in \mathbb{R}$. I need to show every convergent sequence converges to a point in the sets, for showing $A$ and $B$ are closed, but I couldn't find how to.

REPLY [0 votes]: For part (b): Try showing that the sets are (finite intersections of) pre-images of closed sets under suitable continuous functions $\mathbb R^2\to \mathbb R$.  (And I think it must be "$x\ge 0$" in the definition of $B$.)
For part (a): As the example in part (b) shows, you need more than $A$ and $B$ being closed to move from $d(A,B)=0$ to the existence of $a,b$ such that $a=b$.  Compactness is crucial here.  Hint: Suppose $A$ is compact.  Show that, for each $a\in A$, there exists $\varepsilon_a > 0$ such that $d(\{a\},B)>\varepsilon_a$.  Now what you would like to be able to do is take $\varepsilon = \min_{a\in A} \varepsilon_a$, and conclude $d(A,B)>\varepsilon$.  But you can't do this because there might be infinitely many values of $\varepsilon_a$, so there might be no minimum, and if you take the infimum, well, that might be zero.  If only you could restrict attention to only finitely many $a$!


TITLE: Proof of Luzin's Theorem on Axler's Measure, Integration & Real Analysis
QUESTION [3 upvotes]: I would appreciate some help understanding a step on the proof of Luzin's Theorem on Sheldon Axler's Measure, Integration & Real Analysis (open access here, Theorem 2.91 pg 66).
Basically, there is a finite collection of disjoint Borel sets $D_1, \dots, D_n$, colsed sets $F_k \subset D_k$ and open sets $G_k\supset D_k$
Then, he defines a set
$$F=\left(\bigcup_{k=1}^n F_k\right)\cup \left(\bigcap_{k=1}^n \mathbb{R}\setminus G_k\right)$$
And then claims that $\mathbb{R}\setminus F = \bigcup_{k=1}^n (G_k \setminus F_k)$. I cannot see how this follows, unless  $G_i \cap F_j= \emptyset$ for $i\neq j$, and I don't see why this would be the case.
Many thanks!

REPLY [4 votes]: You are correct; the $=$ in the equation in your question below the displayed equation should be $\subset$. Only the set inclusion is used in the proof, so everything else should be fine. I will correct this typo in the next printing of the book. Thank you for pointing out this typo.


TITLE: Is $R/I$ flat over $R$ when $I$ is the nilradical of $R$ and $I=I^2$?
QUESTION [3 upvotes]: This is a flowup of this question. To be precise, let $R$ is a commutative unitary ring with nilradical $I$. In the above URL, it is proved that if $R/I$ is a flat $R$-module, then $I=I^2$.
My question is whether the converse holds in general. Thanks !

REPLY [4 votes]: Not in general, no.
Let $R$ be a zero dimensional local ring, and suppose that its maximal ideal $I$
satisfies $I^2 = I$.  Then $R/I$ is flat over $R$ only if $I = 0$.  (See this MO answer.)
E.g. if $S$ is a one-dimensional local domain with maximal ideal $\mathfrak m $ such that $\mathfrak m^2 = \mathfrak m$
and $x \in \mathfrak m$ is non-zero, then $R := S/(x)$ is a zero-dimensional local ring whose maximal ideal $I$ satifies $I^2 = I$.
E.g. Let $S = \overline{\mathbb Z}_p$ denote the integral closure of $\mathbb Z_p$ in $\overline{\mathbb Q}_p$, and take $s = p$.  Then $R = \overline{\mathbb Z}_p/(p)$ gives a concrete example.