text:
\begin{document}

\maketitle

\begin{abstract}
A profinite group is index-stable
if any two isomorphic open subgroups have the same index.
Let $p$ be a prime, and let $G$ be a compact $p$-adic analytic group with
associated $\mathbb{Q}_p$-Lie algebra $\cl{L}(G)$.
We prove that  $G$ is index-stable
whenever $\cl{L}(G)$ is semisimple.
In particular, a just-infinite  compact $p$-adic analytic group is index-stable if and only if it is not virtually abelian.
Within the category of compact $p$-adic analytic groups, this gives a positive answer to a question of C. Reid.

In the Appendix, J-P. Serre proves that $G$ is index-stable if and only if
the determinant of any automorphism of $\cl{L}(G)$ has $p$-adic norm 1.
\end{abstract}
{
\let\thefootnote\relax
\footnotetext{\textit{Mathematics Subject Classification (2020): }
Primary 20E18, 22E20; Secondary 22E60.}
\footnotetext{\textit{Key words:} index-stable group,
just-infinite profinite group,
$p$-adic analytic group, pro-$p$ group, $p$-adic Lie lattice,
commensurator.}
}

\section*{Introduction}

Throughout, let $p$ be a prime. A pro-$p$ groug $G$ is said to be
\emph{powerful} if $p \ges 3$ and
$[G,G]\les G^p$, or $p=2$ and $[G,G]\les G^4$.
A finitely generated torsion-free powerful pro-$p$ group is called \emph{uniform}.
In his seminal paper \emph{Groupes analytiques $p$-adiques} \cite{Laz65}, M. Lazard obtained the following algebraic characterization of $p$-adic analytic groups:
a topological group is $p$-adic analytic if and only if it contains an open uniform pro-$p$ subgroup (see \cite[Theorems 8.1 and 8.18]{DixAnaProP}).
With every uniform pro-$p$ group $U$ one can naturally associate a $\mathbb{Z}_p$-Lie lattice $L_U$.
The $ \mathbb{Q}_p$-Lie algebra associated with a compact $p$-adic analytic group $G$ is defined as $\mathcal{L}(G): = L_U \otimes_{\bb{Z}_p} \mathbb{Q}_p$, where $U$ is an open uniform pro-$p$ subgroup of $G$.

\medskip

The following theorem is the main result of this paper.

\begin{theorem}\label{index-group}
Let $G$ be a  compact $p$-adic analytic group, let $\mathcal{L}(G)$ be the
$\mathbb{Q}_p$-Lie algebra associated with $G$, and assume that $\mathcal{L}(G)$
is semisimple.
If  $H$ and $K$ are two isomorphic closed subgroups  of $G$ then $|G:H|= |G:K|$.
\end{theorem}

\noindent
\textbf{Remark}\hspace{3pt}
Soon after a preliminary version of this paper was published on the arXiv, the
authors received a letter from Jean-Pierre Serre with a different proof
of a more general version of Theorem \ref{index-group};
see Theorem \ref{thserre}.
We thank Professor Serre for kindly agreeing to include his letter as an appendix to this paper.
\\

A profinite group is said to be \emph{index-unstable}
if it contains a pair of isomorphic open subgroups of different indices;
otherwise, it is said to be \emph{index-stable}.
This definition was introduced by C. Reid in \cite{ReiJust-Infinite},
where he also raised the following question, which is still open.

\begin{question}\label{question1}
Let $G$ be a (hereditarily) just-infinite profinite group which is index-unstable.
Is $G$ necessarily virtually abelian?
\end{question}

\smallskip

Recall that a profinite group $G$ is said to be \emph{just-infinite} if it is infinite, and every non-trivial closed normal subgroup of $G$ is of finite index.
A just-infinite profinite group $G$ is  \emph{hereditarily just-infinite} if every open subgroup of $G$ is just-infinite.

\smallskip

The following corollary gives a positive answer to Question \ref{question1} within the category of $p$-adic analytic groups.

\begin{corollary}\label{index-stable-Reid}
Let $G$ be a  just-infinite compact $p$-adic analytic group.
Then $G$ is index-stable if and only if it is not virtually abelian.
\end{corollary}

\smallskip

Question \ref{question1} was motivated by the study of commensurators of profinite groups,
in the spirit of Y. Barnea, M. Ershov and T. Weigel \cite{BEWAbs-Com}.  Let us recall briefly some relevant definitions.
A virtual automorphism of a profinite group $G$ is  an isomorphism between two open subgroups of $G$. Two virtual automorphisms of $G$ are said to be equivalent if they agree on some open subgroup of $G$.
Given two virtual automorphisms, one can define their composition up to equivalence by composing suitable equivalence-class representatives.
 With respect to this composition, the  equivalence classes of virtual automorphisms of $G$ form a group $\textrm{Comm}(G)$ which is called the \emph{commensurator} of $G$.
 In \cite{BEWAbs-Com}, a topology is defined on $\textrm{Comm}(G)$, called the strong topology, so that it becomes a
  topological group $\textrm{Comm}(G)_S$.
The \emph{virtual center} of $G$ is defined to be
$V\!Z(G) = \{ x \in G :\, \textrm{Cent}_G(x) \textrm{ is open in }  G \}$.
If $V\!Z(G) =\{1\}$ then $\textrm{Comm}(G)_S$ is locally compact;
in this case, $\textrm{Comm}(G)_S$ is said to be \emph{unimodular}
if its modular function is trivial, or equivalently, if the Haar measure is both left and right invariant.

Suppose that $U$ and $V$ are isomorphic open subgroups of $G$.
 Given an isomorphism $\theta: U \to V$, let $\iota(\theta) = \frac{\lvert G:U \rvert}{\lvert G:V \rvert}$; note that this is invariant under equivalence.
  Hence, given $\phi \in \textrm{Comm}(G)$, one can define $\iota(\phi)$ as $\iota(\theta)$ for any  $\theta$ that represents $\phi$.
  This yields a function $\iota : \textrm{Comm}(G) \to \mathbb{Q}_{+}$, where $\mathbb{Q}_{+}$ is the multiplicative group of positive rationals; this function is called the \emph{index ratio}.
If the virtual center of $G$ is trivial then the index ratio turns out to be the
modular function for the strong topology.
Moreover, it is not difficult to see that if $G$ is a just-infinite profinite group then $V\!Z(G) =\{1\}$ if and only if $G$ is not virtually abelian.

\bigskip

The following result is a direct consequence of Corollary \ref{index-stable-Reid}
and the above discussion.

 \begin{corollary}
 Let $G$ be a just-infinite compact $p$-adic analytic group.
  If $G$ is not virtually abelian then $\mathrm{Comm}(G)_S$ is unimodular.
 \end{corollary}

Solvable just-infinite $p$-adic analytic pro-$p$ groups are irreducible
$p$-adic space groups; in particular, they are virtually abelian (cf.  \cite{KLP-Linear}).
Hence, they contain many pairs of open subgroups of different indices that are isomorphic to each other. In contrast,
by \cite[Lemma III.11]{KLP-Linear},  an \emph{unsolvable} just-infinite $p$-adic analytic pro-$p$ group $G$
has the remarkable property of not being isomorphic to any of its proper closed subgroups.
C. Reid generalized this result to all profinite groups by proving that a just-infinte profinite group $G$  that contains an open proper subgroup  isomorphic to $G$ is virtually abelian (see \cite[Theorem~E]{ReiJust-Infinite}).
The following corollary shows that within the category of compact $p$-adic analytic groups a much stronger result holds.

\begin{corollary}\label{just-infinite-pro-p}
Let $G$ be an  unsolvable just-infinite $p$-adic analytic pro-$p$ group.
Then $G$ is index-stable.
\end{corollary}

Open pro-$p$ subgroups  of  $p$-adic Chevalley groups form a rich source  of unsolvable (hereditarily) just-infinite $p$-adic analytic pro-$p$ groups; for $n\ges 2$,
the first congruence subgroup $\textrm{SL}_n^1(\mathbb{Z}_p)$ of  $\mathrm{SL}_n(\mathbb{Z}_p)$ and its open subgroups are typical examples of such groups.
More generally, given a simple finite dimensional $\mathbb{Q}_p$-Lie algebra $\mathcal{L}$, any open pro-$p$
 subgroup of  $\textrm{Aut}(\mathcal{L})$ is an unsolvable (hereditarily) just-infinite $p$-adic analytic pro-$p$ group (see  \cite[Proposition III.9]{KLP-Linear}). Moreover,
 given  an unsolvable just-infinite $p$-adic analytic pro-$p$ group $G$, there is a semisimple $\mathbb{Q}_p$-Lie algebra $\mathcal{L}$ such that
 $G$ is an open subgroup of  $\textrm{Aut}(\mathcal{L})$ (cf.  \cite[Section III.9]{KLP-Linear}).

\medskip

By  \cite[Proposition 6.1]{GSKpsdimJGT}, every solvable
just-infinite pro-$p$ group other than $ \mathbb{Z}_p$ has torsion.
Thus, a non-procyclic  torsion-free just-infinite $p$-adic analytic pro-$p$ group
must be unsolvable. Hence, we can deduce the following result, which
is the correct formulation of \cite[Conjecture~2.10]{NSSelf}.

\begin{corollary}\label{just-torsion-free}
Let $G$ be a non-procyclic torsion-free just-infinite $p$-adic analytic pro-$p$ group.
Then $G$ is index-stable.
\end{corollary}

Next,
we observe that Theorem \ref{index-group} relies on its Lie-algebra counterpart,
which is an interesting result on its own.
By definition, a $\bb{Z}_p$-Lie lattice is a $\bb{Z}_p$-Lie algebra the
underlying module of which is finitely generated and free.
A $\bb{Z}_p$-Lie lattice $L$ is said to be \emph{index-stable} if for any pair
$M$ and $N$ of
isomorphic finite-index subalgebras of $L$ we have $|L:M|=|L:N|$.

 \begin{theorem}\label{index-Lie}
Let $L$ be a $\bb{Z}_p$-Lie lattice, and assume that $L\otimes_{\bb{Z}_p} \bb{Q}_p$
is semisimple as $\bb{Q}_p$-Lie algebra. Then $L$ is index-stable.
\end{theorem}

An $n$-dimensional $\mathbb{Z}_p$-Lie lattice  $L$ is said to be \emph{just infinite} if
every non-zero ideal of $L$ has dimension $n$. The following corollary proves
(the correct formulation of) \cite[Conjecture~2.9]{NSSelf}.

\begin{corollary}\label{index-stable-Lattice}
Let $L$ be a just-infinite $\bb{Z}_p$-Lie lattice, and assume
that $\mathrm{dim}\,L >1$.
Then $L$ is index-stable.
\end{corollary}

\noindent
We remark that Corollary \ref{just-infinite-pro-p} and Corollary \ref{index-stable-Lattice}
have applications to the study of self-similar actions
of hereditarily just-infinite $p$-adic analytic pro-$p$ groups using the language of virtual endomorphisms
(cf.  \cite{NSSelf}).

\medskip

We close the introduction by stating the theorem proved by Serre in the Appendix.

\begin{theorem} {\normalfont\textbf{(Serre)}}\label{thserre}
Let $G$ be a compact $p$-adic analytic group, and let $\cl{L}(G)$ be the $\bb{Q}_p$-Lie algebra
associated with $G$. Then $G$ is index-stable if and only if
all automorphisms $s$ of $\cl{L}(G)$ satisfy $|\mr{det}(s)| = 1$, where
$|\cdot |$ is the $p$-adic norm of $\bb{Q}_p$.
\end{theorem}

\begin{remark}
The proof of Theorem \ref{thserre} (see the Appendix) relies on $p$-adic integration,
for which the reader may consult, for instance,
\cite[Section 7.4]{Igu00}.
Now,  denote by (*) the condition on $\cl{L}(G)$ in the statement of Theorem \ref{thserre}.
We note that there exist $\bb{Q}_p$-Lie algebras that satisfy condition (*) but are not semisimple.
J. L. Dyer
constructed
a nilpotent Lie algebra of dimension 9 and nilpotency class 6
with the property that its group of automorphisms is unipotent \cite{Dyer70}.
In general, by a result of G. Leger and E. Luks \cite[Theorem (*)]{Leger72},
if the automorphism group $\mr{Aut}(\cl{L})$ of a nilpotent Lie algebra $\cl{L}$ of dim $>1$
is nilpotent then it is unipotent.
J. Dixmier and W. G. Lister constructed an
example of a nilpotent Lie algebra $\cl{M}$ of dimension 8
and nilpotency class 3 such that $\mr{Aut}(\cl{M})$ is not nilpotent but the derivation algebra
$\mr{Der}(\cl{M})$ is nilpotent \cite{Dixmier57}.
The latter condition implies that any automorphism of $\cl{M}$
has eigenvalues that are roots of unity.
Clearly, all of these Lie algebras satisfy condition (*).
\end{remark}

\begin{corollary}
Let $G$ be a compact $p$-adic analytic group of dimension $>1$, let $\cl{L}$ be the $\bb{Q}_p$-Lie algebra
associated with $G$, and assume that $\cl{L}$ is nilpotent.
If $\mr{Aut}(\cl{L})$ is nilpotent  or $\mr{Der}(\cl{L})$
is nilpotent then $G$ is index-stable.
\end{corollary}

  \section{Proofs of the main results}

  In this section we prove  Theorem \ref{index-group}, Theorem  \ref{index-Lie}, Corollary \ref{index-stable-Reid} and Corollary \ref{index-stable-Lattice}.

   \vspace{5mm}

\noindent
\textbf{Proof of Theorem  \ref{index-Lie}.}
Let $\mathcal{L}: = L \otimes_{\bb{Z}_p} \mathbb{Q}_p$
and fix a $\mathbb{Z}_p$-basis $\{a_1, ..., a_d \}$ of $L$.
Then, clearly, $\{a_1, ..., a_d \}$ is a $\mathbb{Q}_p$-basis of $\mathcal{L}.$
Let $\kappa: \mathcal{L} \times \mathcal{L} \to \mathbb{Q}_p$, defined by $\kappa(x, y) = \textrm{tr}(\textrm{ad}(x)\textrm{ad}(y))$,
  be the Killing form of $\mathcal{L}$.
  Denote by $A$ the matrix representing
  $\kappa$ with respect to the given basis; in other words,
  $A = (A_{ij})$, where $A_{ij} = \kappa (a_i, a_j)$ for $1 \les i, j \les d.$  Since $\mathcal{L}$ is a semisimple Lie algebra over a field of characteristic 0,
   by Cartan's criterion \cite[Page 69]{JacLieA}
    we have that $\kappa$ is a non-degenerate bilinear form; in particular, $\det(A) \neq 0$.
Let  $\{m_1, ..., m_d \}$ be a $\mathbb{Z}_p$-basis of $M$, and let $\varphi: M \to N$ be an isomorphism of $\bb{Z}_p$-Lie lattices.
Then $\{\varphi(m_1), ..., \varphi(m_d) \}$ is a $\mathbb{Z}_p$-basis of $N$.
Let $B= (B_{ij})$ and $C= (C_{ij})$ be the change-of-basis matrices
defined by  $m_j = \sum_i B_{ij} a_i$ and
$\varphi(m_j) = \sum_i C_{ij} a_i$.
Note that $[L:M]= p^{v_p(\textrm{det}(B))}$ and
$[L:N] = p^{v_p(\textrm{det}(C))}$,
where $v_p$ is the $p$-adic valuation
(one way to prove this claim is to recall that
there exist a basis $\{b_i\}$ of $L$ and non-negative integers
$\{k_i\}$ such that $\{p^{k_i}b_i\}$ is a basis of $M$;
similarly for $N$; cf. \cite[Lemma 10.7.2]{Grillet99}).
The matrices of $\kappa$ with respect to the bases $\{m_i\}$ and $\{\varphi(m_i)\}$
are $B^TAB$ and $C^TAC$, respectively.
Since the automorphism of $\cl{L}$ induced by $\varphi$
preserves the Killing form, $B^TAB= C^TAC$.
Taking the determinant on both sides, and recalling that $\mr{det}(A)\neq 0$,
we see that $v_p(\mr{det}(B))=v_p(\mr{det}(C))$, and the theorem follows.
\ep

\vspace{5mm}

\noindent
\textbf{Proof of Corollary \ref{index-stable-Lattice}.}
Since $L$ is a just-infinite $\mathbb{Z}_p$-Lie lattice of dimension greater than 1,
it is not difficult to see that $L\otimes_{\bb{Z}_p} \bb{Q}_p$
is a simple $\mathbb{Q}_p$-Lie algebra. Hence, the corollary follows from Theorem \ref{index-Lie}.
\ep

\vspace{5mm}

A $\mathbb{Z}_p$-Lie lattice $L$ is called \emph{powerful} if
$[L,L] \subseteq pL$ for $p$ odd, or $[L,L] \subseteq 4L$ for $p=2$.

With a uniform pro-$p$ group $G$ one may associate a powerful $\mathbb{Z}_p$-Lie lattice $L_G$ in the following way: $G$ and $L_G$ are identified as sets,
and the Lie operations are defined by
\begin{displaymath}
g+h=\lim_{n \to \infty}(g^{p^n}h^{p^n})^{p^{-n}},  ~ ~ ~  [g,h]_{\mr{Lie}}=\lim_{n \to \infty}[g^{p^n},h^{p^n}]^{p^{-2n}}=\lim_{n\to \infty} (g^{-p^n}h^{-p^n}g^{p^n}h^{p^n})^{p^{-2n}} .
\end{displaymath}
On the other hand, if $L$ is a powerful $\mathbb{Z}_p$-Lie lattice, then the Campbell-Hausdorff formula induces a group structure on $L$; the resulting group is a uniform pro-$p$ group. If this construction is applied to the $\mathbb{Z}_p$-Lie Lattice $L_G$ associated with a uniform group $G$, one recovers the original group.
Indeed, the assignment $G\mapsto L_G$ gives an isomorphism between the category of uniform pro-$p$ groups and the category of powerful $\mathbb{Z}_p$-Lie lattices (see \cite[Theorems 4.30 and 9.10]{DixAnaProP}).
Recall that every compact $p$-adic analytic group contains an open subgroup that is a uniform pro-$p$ group  (\cite[Corollary 8.34]{DixAnaProP}).
As we already mentioned in the introduction, the $ \mathbb{Q}_p$-Lie algebra associated
with a compact $p$-adic analytic group $G$ is defined as
$\mathcal{L}(G): = L_U \otimes_{\bb{Z}_p} \mathbb{Q}_p$, where $U$ is an open uniform pro-$p$ subgroup of $G$ (see \cite[Section 9.5]{DixAnaProP}).
A key invariant of a $p$-adic analytic group $G$ is its dimension as a $p$-adic manifold,  denoted by $\textrm{dim}(G)$.
Algebraically, $\textrm{dim}(G)$ can be described  as $d(U)$,
where $U$ is any uniform open pro-$p$ subgroup of $G$ and $d(U)$ denotes the minimal cardinality of a topological generating set for $U$.

\vspace{5mm}

\noindent
\textbf{Proof of Theorem \ref{index-group}.}
Let $H$ and $K$ be two isomorphic closed subgroups of $G$. If one of these subgroups, say $H$,  has infinite index in $G$,
then $\textrm{dim}(H) < \textrm{dim}(G)$. Hence $\textrm{dim}(K) < \textrm{dim}(G)$, and therefore $| G:K|$ is infinite as well.
Since $G$ is virtually pro-$p$,  it is  not difficult to see that   $| G:K|$ and $| G:H|$  coincide as supernatural numbers.

Now suppose that $H$ and $K$ are open subgroups in $G$. Let $\varphi: H \to K$  be an isomorphism, and let $U$ be an open uniform subgroup of $G$; note that, since $G$ is compact, $U$ is of finite index in $G$. Let $U_H$ be a uniform open subgroup of  $H$. Then $U_K:  = \varphi(U_H)$ is a uniform open subgroup of $K$. Moreover,
$| H:U_H | = | K:U_K |.$
Choose a positive integer $p^m$ such that $V_H : = {(U_H)}^{p^m}$ and  $V_K : = {(U_K)}^{p^m}$ are contained in $U$;
by \cite[Theorem~3.6]{DixAnaProP},  $V_H$ and $V_K$ are uniform. Let $L_U$ be the $\bb{Z}_p$-Lie lattice associated with $U$.
Then  the $\mathbb{Q}_p$-Lie algebra associated with $G$ is given by $\mathcal{L}(G) = L_U\otimes_{\bb{Z}_p} \mathbb{Q}$,
which is semisimple by assumption.
Since $U_H$ and $U_K$ are isomorphic, we have that $V_H$ and $V_K$ are isomorphic open uniform subgroups of $U$.
This implies that $L_{V_H}$ and $L_{V_K}$ are isomorphic $\bb{Z}_p$-Lie lattices of $L_U$ of finite index. By Theorem \ref{index-Lie},
$| L_U:L_{V_H} | = | L_U:L_{V_K} |$. Hence, by \cite[Proposition 4.31]{DixAnaProP}, $| U: V_H | = | U:V_K |$.
Clearly, this implies that $ | G: V_H | = | G:V_K |$. Now, since  $| H:U_H | = | K:U_K |$
and  $ | U_H:V_H | = | U_K:V_K |$, it follows that  $| G:H | = | G:K |$, as desired.
\ep

\vspace{5mm}

\noindent
\textbf{Proof of Corollary \ref{index-stable-Reid}.}
Clearly, if $G$ is virtually abelian then $G$ is index-unstable.
Suppose that $G$ is not virtually abelian.
By \cite[Theorem~3.6]{Gri-branch}, a just-infinite profinite group is either a branch profinite group or it contains an open normal subgroup
which is isomorphic to the direct product of a finite number of copies  of some hereditarily just-infinite profinite group. Note that $G$ cannot be a branch profinite group since,  by  \cite[Theorem~8.33]{DixAnaProP},
$G$ is of finite rank (that is, there exists a positive integer $d$ such that every closed subgroup of $G$ can be  generated topologically by at most $d$ elements).
Hence, $G$ contains an open normal subgroup $H$
which is isomorphic to the direct product of a finite number, say $k$, of copies  of some hereditarily just-infinite uniform pro-$p$ group $U$; in particular, $H$ is uniform.  Note that if $U$ is solvable
then $U$ is isomorphic to $\mathbb{Z}_p$, and consequently $G$ is virtually abelian,
which is a contradiction. Hence, $U$ is an unsolvable hereditarily just-infinite uniform pro-$p$ group.
By \cite[Proposition F]{GSKpsdimJGT}, the associated $\mathbb{Q}_p$-Lie algebra
$\mathcal{L}(U) = L_U\otimes_{\bb{Z}_p} \mathbb{Q}_p$ is simple.  Thus, the $\mathbb{Q}_p$-Lie algebra $\mathcal{L}(G) = L_H\otimes_{\bb{Z}_p} \mathbb{Q}_p$ associated with $G$ is semisimple,   since it is isomorphic to the direct sum of $k$ copies of  $\mathcal{L}(U)$.
Now Theorem  \ref{index-group} yields that $G$ is index-stable.
\ep

\section*{Appendix - A letter from Jean-Pierre Serre}

\hspace{\parindent}Dear MM. Noseda and Snopce,\\

I have just seen your arXiv paper on ``Index-stable compact $p$-adic analytic
groups''. Your proof uses Lazard's very nice results, but in fact  these results are not
necessary: $p$-adic integration is enough. Let me explain:
\\

Let $L$ be a finite-dimensional Lie algebra over $\bb{Q}_p$. Denote by $|x|$ the
$p$-adic norm of $\bb{Q}_p$. Consider the following property of $L$:
\\

  (*) For every automorphism  $s$  of $L$, we have $|\mr{det}(s)| = 1$.\\

\noindent
This is true for instance if $L$ is semisimple since $\mr{det}(s)=\pm 1$, which is the case
you consider.
\\

    Assume property (*). Let $n = \dim L.$ Let $u$ be a nonzero element of $\wedge^nL^*$, where $L^*$ is the $\Q_p$-dual of $L$. Let $G$ be a compact $p$-adic analytic group with Lie algebra $L$. Then $u$ defines a right-invariant differential form $\omega_u$ on $G$ of degree $n$. The corresponding measure $\mu_u = |\omega_u|$ is a non-zero right-invariant positive measure on $G$, hence is a Haar
    measure since $G$ is compact. Property (*) implies that $\mu_u$ is
    invariant by every automorphism of $L$. Hence every isomorphism of $G$ onto another group $G'$ carries $\mu_u$ (for $G$) into $\mu_u$ (for $G'$). This implies that, if $G, G'$ are open subgroups of some compact $p$-adic group $G''$, then they have the same index -  as
  wanted.
\\

  Conversely, if (*) is not true for some  $s$, and if $G''$ is compact with Lie algebra $L$,
$s$ defines a local automorphism of  $G''$, and if  $G$  is a small enough open subgroup
of $G''$, it is transformed by  $s$  into another open subgroup  $G'$, and the ratio
$(G'':G)/(G'':G')$ is equal to $|\mr{det} (s)|$, which is  $\neq 1$.
\\

  Best wishes,\\

  J-P. Serre

\begin{footnotesize}

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text:
\subsection{Proof of \prettyref{cor:IBUpperBound}}
Without loss of generality, assume $k_j$ is even for all $j \in [m]$. After the random \IB, there are $mk/2$ independent
pairwise comparisons and let $L$ denote the Laplacian of the comparison graph after the breaking.

Recall that $L= \sum_{j=1}^m L_j$. With random \IB, we have
$
\mathbb{E}[L_j]= \frac{k_j}{n-1} \left( I- \frac{1}{n} \mathbf{1}\mathbf{1}^\top \right).
$
Define $Z_j=L_j- \mathbb{E}[L_j]$. Then $Z_1, \ldots, Z_m$ are independent symmetric random matrices with zero mean. Moreover,
\begin{align*}
\|Z_j\| \le  \|L_j\| + \|\mathbb{E}[L_j]\| \le  2 + \frac{k_j}{n-1} \le 4,
 \end{align*}
and
\begin{align*}
\mathbb{E}[Z_j^2]= \frac{2k_j}{n-1} \left( I- \frac{1}{n} \mathbf{1}\mathbf{1}^\top \right)- \frac{k_j^2}{(n-1)^2} \left( I- \frac{1}{n} \mathbf{1}\mathbf{1}^\top \right).
\end{align*}
Therefore, $\| \sum_{j=1}^m \mathbb{E}[Z_j^2] \| \le  \frac{2mk}{n-1}$.

Following the same argument for proving \prettyref{cor:MLEUpperBound}, we can show that $ \lambda_2(L_{\IB}) \ge \frac{mk}{2(n-1)}$ and the corollary follows by \prettyref{thm:MLEUpperBound} with $k=2$.

\subsection{Proof of \prettyref{thm:FBUpperBound}}

It follows from the definition of $\calL(\theta)$ given by \prettyref{eq:loglikelihoodbreaking} that
\begin{equation}  \label{eq.gradL_breaking}
\nabla_i \calL(\theta^*) = \sum_{j: i \in S_j} \frac{1}{k_j-1}   \sum_{i' \in S_j: i'\neq i} \left[ \1{\sigma_j^{-1}(i)< \sigma_j^{-1}(i') } - \frac{\exp(\theta^\ast_i)}{\exp(\theta^\ast_i)+\exp(\theta^\ast_{i'} ) } \right] := \sum_{j: i \in S_j} Y_j,
\end{equation}

which is a sum of $d_i$ independent random variables with mean zero and bounded by $1$. By Hoeffding's inequality, $|\nabla_i L(\theta^*)| \le \sqrt{d_i \log n} $ with probability at least $1-2n^{-2}$. By union bound, $\|\nabla L(\theta^\ast)\|_2 \le \sqrt{m k \log n} $ with probability at least $1-2n^{-1}$.
The Hessian matrix is given by
\begin{align*}
H(\theta)=- \sum_{j=1}^m  \frac{1}{2(k_j-1)}\sum_{i,i' \in S_j}  (e_i-e_{i'} ) (e_i - e_{i'} )^\top  \frac{\exp(\theta_i+ \theta_{i'} ) }{ \left[ \exp( \theta_{i}) +  \exp(\theta_{i'} ) \right]^2 }.
\end{align*}
If $|\theta_i | \le b, \forall i \in [n]$,
$
\frac{\exp(\theta_i+ \theta_{i'} ) }{ \left[ \exp( \theta_{i}) +  \exp(\theta_{i'} ) \right]^2 } \ge \frac{e^{2b}}{(1+e^{2b})^2}.
$
It follows that $-H(\theta) \ge \frac{e^{2b}}{(1+e^{2b})^2} L$ for $\theta \in \Theta_b$ and the theorem follows from \prettyref{eq:MLEUpperBound}.


TITLE: Contribution of a change in volume in a liquid to the change in heat capacity
QUESTION [0 upvotes]: I have come upon this question that asks me to estimate the contribution of an expansion in a liquid to its heat capacity. However, I do not understand how a change in its volume can affect it's heat capacity.
Furthermore, after thinking about it for some time and then looking at the solution I have become more confused as it states that what I am looking for is the ratio of the heat absorbed by the sample to the work done by the sample.
I am obviously not understanding something conceptually here so any pointers would be much appreciated.

REPLY [0 votes]: I assume they take the heat capacity to be constant from 4 to 100 degrees. This sample of water can take up some heat $\Delta Q = c_p m \Delta T  $. So now the question is: what does the heat do to the sample ? For an isobaric process it can increase its temperature and its volume. Both processes contribute to the uptake of the heat, which is described by the former equation. And for an isobaric process we know $dQ = dU +dW$.


TITLE: Conditional Probability - Red and black marbles
QUESTION [0 upvotes]: I saw this question on one of the app.
A wise man presents you with an urn containing an equal number of red and black marbles.
If you reach in and draw two marbles (uniformly and independently) at random from the urn, the probability that they're of matching color is necessarily less than ½. Here's the tricky part: if the colors of those first two marbles do match, and you reach in to grab two more, the probability of those next two being of matching color is exactly ½.
How can that be? What is the total number of balls in the urn — red plus black — before you take any out?
I am unable to work it out with equal number of initial red and black marbles.

REPLY [1 votes]: Suppose there are $n$ red marbles and $n$ black ones.  We choose a marble from the urn.  There are $n-1$ matching marbles left among the $2n-1$ marbles in the urn, so the probability that the first two match is $${n-1\over2n-1}<\frac12$$
Now suppose the first two match.  We may assume that there are $n-2$ black marbles and $n$ red marbles left.  There are ${n-2\choose2}$ ways to choose two matching black marbles, ${n\choose2}$ ways to choose two matching red marbles, and ${2n-2\choose2}$ ways to choose two marbles. Can you finish it from here?


TITLE: Compact Hausdorff space and its double dual: Gelfand Naimark.
QUESTION [1 upvotes]: I am trying to prove the equivalence  of categories between compact Hausdorff spaces and unital $C^*$ algebras.

The maps
      \begin{align*}
 f: X \mapsto C(X) \\
 g: A \mapsto \hat{A}
 \end{align*}
       is a contravariant category equivalence between the category of compact Hausdorff spaces  and continuous maps and the category of commutative unital $C^*$ algebras and unital $*$-homomoprhisms.

I know one direction $fg \simeq id:A \rightarrow C(\hat{A})$ is Gelfand Naimark.
It then suffices to prove $gf \simeq id$, i.e.
$$ X \mapsto \widehat{C(X)}, \quad x \mapsto \phi_x:f \mapsto f(x)$$
is a homeomorphism?

REPLY [2 votes]: Yes. And the only nontrivial part is that the map $x\longmapsto \phi_x$ is surjective. Below is the argument that I know.
Assume that $\varphi:C(X)\to\mathbb C$ is a character. We prove a few things

if $\varphi(f)=0$, then there exists $x_f\in X$ such that $f(x_f)=0$. Indeed, if $f(x)\ne0$ for all $x$, then by compactness there exists $c>0$ with $|f(x)|\geq c$ for all $x$. Then $1/f\in C(X)$ and $1=\varphi(f/f)=\varphi(f)\varphi(1/f)$ and $\varphi(f)\ne0$.
for each $f\in C(X)$, there exists $x_f\in X$ with $\varphi(f)=f(t_x)$. Apply the above to the function $f-\varphi(f)$.
For each $f\in C(X)$, consider the set $T_f=\{x:\ f(x)=\varphi(f)\}$. By the above, $T_f\ne\varnothing$. As $T_f=f^{-1}(\{\varphi(f)\})$, it is compact.
Given $f_1,\ldots,f_n\in C(X)$, we have $\bigcap_{j=1}^n T_{f_j}\ne\varnothing$. Indeed, let
$$
g=\sum_{j=1}^n|f_j-\varphi(f_j)|^2.
$$
As $\varphi$ is linear and multiplicative,
$$
\varphi(g)=\sum_{j=1}^n \varphi[(f_j-\varphi(f_j))\overline{(f_j-\varphi(f_j))}]
=\sum_{j=1}^n \varphi[f_j-\varphi(f_j)]\varphi[\overline{f_j-\varphi(f_j)}]=0
$$
As we showed above, there exists $x\in X$ with $g(x)=0$. It follows that $f_j(x)=\varphi(f_j)$ for $j=1,\ldots,n$ so $x\in\bigcap_{j=1}^n T_{f_j}$.
The family $\{T_f\}_{f\in C(X)}$ has the finite intersection property. As $X$ is compact, it follows that $\bigcap_{f\in C(X)}T_f\ne\varnothing$. That is, there exists $x\in X$ such that $\varphi(f)=f(x)$ for all $f\in C(X)$.