text:
\begin{document}

\title[Infinite characters]{Infinite characters on $GL_n(\QQ)$,
on  $\SL_n(\ZZ),$  and  on groups acting on trees}

\address{Bachir Bekka \\ Univ Rennes \\ CNRS, IRMAR--UMR 6625\\
Campus Beaulieu\\ F-35042  Rennes Cedex\\
 France}
\email{bachir.bekka@univ-rennes1.fr}
\author{Bachir Bekka}

\thanks{The author acknowledges the support  by the ANR (French Agence Nationale de la Recherche)
through the projects Labex Lebesgue (ANR-11-LABX-0020-01) and GAMME (ANR-14-CE25-0004)}
\begin{abstract}
Answering  a question of J. Rosenberg from \cite{Rose--89}, we construct the first examples of infinite characters on $\GL_n(\KK)$ for a global field $\KK$ and  $n\geq 2.$
The case $n=2$ is deduced from the following more general result. Let $G$ a   non amenable countable subgroup acting on locally finite tree $X$. Assume   either  that  the stabilizer in $G$ of every vertex of $X$ is finite or that
the closure of  the image of $G$ in $\Aut(X)$ is not amenable.
We show  that  $G$  has  uncountably many infinite dimensional irreducible unitary representations $(\pi, \H)$
of $G$ which are  traceable, that is, such that  the $C^*$-subalgebra of $\B(\H)$ generated by $\pi(G)$ contains the algebra  of the compact operators on $\H.$
In the case $n\geq 3,$  we prove the existence of infinitely many characters for  $G=\GL_n(R)$, where $n\geq 3$ and $R$  is  an integral domain  such that $G$ is not amenable. In particular, the group $\SL_n(\ZZ)$ has infinitely many such characters
for $n\geq 2.$

\end{abstract}

\maketitle
\section{Introduction}
\label{S0}
Let $G$ be a  countable discrete group and $\widehat G$  the unitary dual of $G$, that is, the set of
equivalence classes of irreducible unitary representations of $G$.
The space  $\widehat G$, equipped with
a natural Borel  structure, is  a standard Borel space exactly when  $G$ is   virtually abelian,
by results of Glimm and Thoma (see \cite{Glim--61} and \cite{Thom--68}).
So, unless $G$ is   virtually abelian (in which case
the representation theory of $G$ is well understood),  a description of $\widehat G$ is hopeless or useless.
There are at least two other dual objects of $G$, which seem to be more accessible  than $\widehat G$:
\begin{itemize}

\item \textbf{Thoma's dual space} $E(G)$, that is, the set  of indecomposable
positive definite central functions on $G;$
\item the space $\Char(G)$ of \textbf{characters} of $G,$ that is, the space of
 lower semi-continuous semi-finite (not necessarily finite)  traces $t$ on the  \textbf{maximal $C^*$-algebra}  $C^*(G)$
 of $G$  (see Subsection~\ref{SS1}) which satisfies the following extremality condition:
every lower semi-continuous semi-finite trace on $C^*(G)$ dominated by $t$  on the ideal of definition of $t$ is proportional to $t$.
\end{itemize}
\par
The space $\Char(G)$  parametrizes the quasi-equivalence classes of factorial  representations  of $C^*(G)$
which are \textbf{traceable} ; recall that a unitary representation $\pi$ is factorial if the von Neumann algebra $\M$ generated by $\pi(G)$ is  a factor and that a factorial representation $\pi$ is traceable if there exists a faithful normal  (not necessarily finite) trace $\tau$ on $\M$
and a positive element $x\in C^*(G)$  such that $0<\tau(\pi(x))<+\infty.$
If this is the case, then $t=\tau\circ \pi$ belongs to  $\Char(G)$.
Conversely, every  element of $\Char(G)$ is obtained in this way.
Traceable representations are also called \emph{normal} representations.

Two traceable factorial representations  $\pi_1$  and $\pi_2$
are quasi-equivalent if there exists an isomorphism $\Phi: \M_1 \to \M_2$
such that $\Phi(\pi_1(g)) = \pi_2(g)$ for all $g \in G$,  where  $\M_i$ is the factor   generated by $\pi_i(G).$

Observe that an \emph{irreducible} unitary representation $(\pi, \H)$ of $G$ is traceable
 if and only if $\pi(C^*(G))$ contains the algebra of compact operators on $\H$.
 The character associated to such a representation is given by the usual trace on $\B(\H)$
 and so does not belong to $E(G)$ whenever $\H$ is infinite dimensional; in this case,
 the  character is said to be of type $I_\infty,$ in accordance with the type classification
 of von Neumann algebras.
 Observe also that two  irreducible traceable representations of a group $G$ are quasi-equivalent if and only if they are unitarily equivalent.

Thoma's dual space $E(G)$  is a subspace of $\Char(G)$ and classifies the quasi-equivalence classes of the factorial representations $\pi$ of $C^*(G)$ for which  the factor  $\M$ generated by $\pi(G)$ is \emph{finite}, that is, such that the trace $\tau$ on $\M$  takes only finite values (for more detail on all of this, see Chapters 6 and 17 in \cite{Dixm--C*}).
 \par

Thoma's dual space  $E(G)$ was determined for several examples of countable groups $G$,
among them  $G=GL_n(\KK)$ or $G=SL_n(\KK)$ for an infinite field $\KK$ and $n\geq 2$ (\cite{Kirillov}; see also  \cite{PeTh--16}),
and $G=SL_n(\ZZ)$ for $n\geq 3$ (\cite{Bekk--07});
a procedure  is given in \cite[Proposition 3]{Howe--77} to compute $E(G)$ when $G$ is a nilpotent  finitely generated group.

The space $\Char(G)$ has been described for some \emph{amenable} groups $G$:
\begin{itemize}
\item  when $G$ is nilpotent, we have $E(G)=\Char(G)$  (see \cite[Theorem 2.1]{CaMo--84});
\item the space $\Char(G)$ is determined in \cite{Guic--63}
 for the Baumslag-Solitar  group $BS(1,2)$ and in \cite{Vershik} for the infinite symmetric group;
 \item for $G=GL_n(\KK)$ and $n\geq 2$,
it is shown   in  \cite{Rose--89}  that $E(G)=\Char(G)$  in the case where $\KK$ an algebraic  extension of a finite field.
(Observe that $GL_n(\KK)$ is amenable  if and only if  $\KK$ an algebraic extension of a finite field; see Proposition 9 in \cite{HoRo--89} or Proposition~\ref{Pro-AmenableSLn} below.)
  \end{itemize}

  J. Rosenberg asked in  \cite[Remark after Th\'eor\`eme 1]{Rose--89} whether there exists
 an \emph{infinite} character on $G=\GL_n(\KK)$, that is, whether $\Char(G)\neq E(G)$,
  for a field $\KK$   which is not an algebraic extension of a finite field.
 We will show below that   the answer to this question is positive, by exhibiting  as far we know
  the first examples of such characters.
 The case where $n=2$ and $\KK$ is a global field (see below) will  be deduced from a general result concerning groups acting on trees,  which we now state.

\medskip

Recall that   a graph $X$ is locally finite if every vertex on $X$  has only finitely many neighbours.
In this case, the group $\Aut(X)$  of automorphisms of $X,$ equipped with the topology of pointwise convergence,
is a locally compact group for which  the vertex stabilizers are compact.
Concerning the notion of weakly equivalent representations, see Chapters 3 and 18 in \cite{Dixm--C*}  (see also Section~\ref{SS1}).

\begin{theorem}
 \label{Theorem1}
Let $X$ be a  tree and $G$ a countable subgroup acting on $X$. Assume that
\begin{itemize}
\item[(a)] either $G$ is not amenable and the stabilizer in $G$ of every vertex of $X$ is finite, or
\item[(b)] $X$ is locally finite and the closure of  the image of $G$ in $\Aut(X)$ is not amenable.
\end{itemize}
 There exists an uncountable  family  $(\pi_t)_t$ of irreducible unitary representations of $G$ with the following properties:
  $\pi_t$ is infinite dimensional, is traceable and  is not weakly equivalent to $\pi_{t'}$ for $t'\neq t.$
  \end{theorem}

   Recall that a global field is a finite  extension of either the field $\QQ$ of rational numbers or of the field $\FF_p(T)$ of rational
 functions in $T$ over the finite field $\FF_p$ (see Chapter III in \cite{Weil}).

 \begin{corollary}
 \label{Cor-Theorem1}
 Let $G$ be either
\begin{itemize}
\item[(i)]  $\GL_2(\KK)$ or $\SL_2(\KK)$ for  a global field $\KK$, or
\item[(ii)]  $\SL_2(\ZZ)$, or
\item[(iii)] $F_n$, the  free  non abelian group  over $n\in \{2,\dots, +\infty\}$ generators.
\end{itemize}
There exists an uncountable family   $(\pi_t)_t$ of unitary representations of $G$ with the properties from Theorem~\ref{Theorem1}; moreover, in case $G=F_n,$ the representations $\pi_t$ are all faithful.
 \end{corollary}

\medskip
 Turning to the case $n\geq 3,$  we prove a result for $G=\GL_n(R)$ or $G=\SL_n(R)$,
valid for every integral domain $R$ such  that $G$ is not amenable.
   \begin{theorem}
 \label{Theorem2}
 Let $R$ be a countable unital commutative ring which is an integral domain; in case the characteristic of $R$ is positive, assume that  the field of fractions of $R$ is not   an algebraic extension of  its prime field.
 For $n\geq 3,$ let  $G=\GL_n(R)$ or $G=\SL_n(R)$.
 There exists  an  infinite dimensional irreducible unitary representation  of $G$  which is  traceable.
 \end{theorem}

In the  case where $R$ is a field or  the ring of integers, we can even produce infinitely  many non equivalent  representations as in  Theorem~\ref{Theorem2}.
   \begin{corollary}
 \label{Cor-Theorem2}
 \begin{itemize}
 \item[(i)] For $n\geq 3,$ let  $G=GL_n(\KK)$ for a countable field $\KK$ which is not an algebraic extension of a finite field.
 There exists an uncountable family  $(\pi_t)_t$ of    pairwise non equivalent infinite dimensional irreducible unitary representations of $G$   which are  traceable.
 Moreover, the representations $\pi_t$ all have a trivial central character, that is, the $\pi_t$'s are
 representations of $\PGL_n(\KK).$
 \item[(ii)] Let $G=SL_n(\ZZ)$ for $n\geq 3$.   There exists  an infinite family of   pairwise non equivalent infinite dimensional irreducible unitary representations of $G$   which are  traceable.
  \end{itemize}
  \end{corollary}

  The methods of proofs of  Theorem~\ref{Theorem1} and Theorem~\ref{Theorem2} are quite different
  in nature:
  \begin{itemize}
  \item the proof of Theorem~\ref{Theorem1} is based on properties  of a remarkable family of unitary representations
  of groups acting on trees  constructed in \cite{JuVa--84} and used  to show their $K$-theoretic amenability,   a notion
  which originated from  \cite{Cuntz}\   in the case of free  groups;
   \item the traceable representations we construct in Theorem~\ref{Theorem2} are induced representations from suitable subgroups. The case $n\geq 4$ uses the existence of appropriate subgroups of $\GL_n(R)$  with Kazhdan's Property (T).
  \end{itemize}

 \begin{remark}
 \label{Rem-Theo2}
For a group $G$ as in Theorem~\ref{Theorem1} or  Theorem~\ref{Theorem1}, our results show that
 the set $\Char(G)$ contains characters of type $I_\infty$.

 For, say, $G=\GL_n(\QQ)$, we do not know whether  $\Char(G)$ contains  characters of type $II_\infty$,
 that is, characters for which the corresponding factorial representation generates a factor of type $II_\infty$.

 \end{remark}

  This paper is organized as follows. In Section~\ref{S1}, we establish some  preliminary facts
 which are necessary to the proofs of our results. Section~\ref{S2} is devoted to the proofs  of Theorem~\ref{Theorem1}
 and Corollary~\ref{Cor-Theorem1};  Theorem~\ref{Theorem2} and Corollary~\ref{Cor-Theorem2}
 are proved in Section~\ref{S3}.

\section{Some preliminary results}.
\label{S1}
\subsection{$C^*$-algebras}
\label{SS1}

Let $G$ be a countable group. Recall that a unitary representation of $G$  is  a homomorphism
  $\pi:G\to U(\H)$ from $G$ to the unitary group  of a complex separable Hilbert space $\H.$
From now on, we will simply write \textbf{representation} of $G$ instead of  ``unitary representation of $G$''.

 Every representation  $(\pi,\H)$  of $G$ extends naturally to  a $*$-representation, denoted again by $\pi,$  of the group algebra $\CCC[G]$ by bounded operators on $\H.$

Recall that the maximal $C^*$-algebra $C^*(G)$ of $G$ is the completion of $\CCC[G]$  of $G$ with respect to the norm
$$
f\mapsto  \sup_{\pi\in \Rep(G)} \Vert \pi(f) \Vert,
$$
where $\Rep(G)$ denotes the set of  representations $(\pi, \H)$ of $G$ in a separable Hilbert space $\H.$

We can view $G$ as subset of $\CCC[G]$ and hence as a subset of $C^*(G)$.
The $C^*$-algebra  $C^*(G)$ has the following universal property:
every  representation $(\pi, \H)$  of $G$ extends to a unique  representation (that is, $*$-homomorphism) $\pi: C^*(G)\to \B(\H)$.
The correspondence $G\to C^*(G)$ is functorial: every   homomorphism $\varphi: G_1\to G_2$
between two countable groups $G_1$ and $G_2$  extends to a unique morphism
$$\varphi_*: C^*(G_1)\to C^*(G_2)$$ of $C^*$-algebras. In particular, given a subgroup $H$
of a group $G,$ the injection map $i:H\to G$ extends to a morphism  $i_*:C^*(H)\to C^*(G);$
the map $i_*$ is injective and so $C^*(H)$ can be viewed naturally as a subalgebra of  $C^*(G)$:
indeed, this follows from  the fact that every  representation $\sigma$ of $H$ occurs as subrepresentation of
the restriction to $H$ of some  representation $\pi$ of $G$ (one may take as $\pi$
the induced representation $\Ind_H ^G\sigma$, as shown  below in Proposition~\ref{Prop-Induced}).

The following simple lemma will be one of our  tools in order to show that
$\pi(C^*(G))$ contains a non-zero compact operator for a  representation $\pi$ of $G$.

Let $\A$ be a $C^*$-algebra.
Recall that  a representation $\pi: \A\to \B(\H)$ weakly contains another
representation  $\rho: \A\to \B(\K)$  if
$$
\Vert \rho(a) \Vert \leq \Vert \pi(a) \Vert \tous a\in \A,
$$
or, equivalently, $\ker \pi\subset \ker \rho$ (see Chapter 3 in \cite{Dixm--C*}).
Two representations $\pi$ and $\rho$ are weakly equivalent if $\pi$ weakly contains $\rho$ and
$\rho$ weakly contains $\pi,$ that is, if $\ker \pi=\ker \rho.$
 \begin{lemma}
\label{lem-CompactOp}
Let $\A$ be a $C^*$-algebra and $\pi: \A\to \B(\H)$  a representation of $\A$.
Assume that $\H$ contains a non-zero finite dimensional $\pi(\A)$-invariant subspace
$\K$ and that the restriction $\pi_1$ of $\pi$ to $\K$ is not weakly contained in the restriction $\pi_0$ of $\pi$ to
the orthogonal complement $\K^\perp$.
Then $\pi(\A)$ contains a non-zero compact operator.
\end{lemma}
\begin{proof}
 The ideal $\ker \pi_0$ is not contained in $\ker \pi_1,$  since $\pi_1$ is not weakly contained in $\pi_0$.
 Hence, there exists $a\in \A$ with  $\pi_0(a)=0$ and $\pi_1(a)\neq 0.$
 Then  $\pi(a)=\pi_1(a)$ has a finite dimensional range and is non-zero.
 \end{proof}

 Knowing that a representation of $\A$ contains in its image a non-zero compact operator,
 the following lemma enables us to construct an \emph{irreducible} representation of $\A$ with the same property.

 \begin{lemma}
\label{lem-IrredCompactOp}
Let $\A$ be a $C^*$-algebra and $\pi: \A\to \B(\H)$  a representation of $\A$  in a separable Hilbert space $\H$
Let $a\in \A$ be such that  $\pi(a)$ is a non-zero compact operator.
Then  there exists an irreducible subrepresentation $\sigma$ of $\pi$ such that $\sigma(a)$  is a compact operator
and such that  $\Vert \sigma(a)\Vert=\Vert \pi(a)\Vert.$
\end{lemma}

 \begin{proof}
 We can decompose $\pi$ as a direct integral $ \int^{\oplus}_{\Omega} \pi_{\omega} d\mu (\omega)$ of irreducible representations $\pi_\omega;$ thus, we can find
 a probabilility measure $\mu$ on a standard Borel space $\Omega$,
a  measurable field  $\omega \to \pi_{\omega}$ of irreducible  representations of $\A$
in a measurable field $\omega \to \H_{\omega}$
   of separable Hilbert spaces on $\Omega$,
and a Hilbert space isomorphism $U:\H \to \int^{\oplus}_{\Omega} \H_{\omega} d \mu (\omega)$
such that
$$
U\pi(x)U^{-1} = \int^{\oplus}_{\Omega} \pi_{\omega}(x) d\mu (\omega).
\tous x\in \A
$$
(see \cite[\S 8.5]{Dixm--C*}). Without loss of generality, we will
identify $\pi$ with  $\int^{\oplus}_{\Omega} \pi_{\omega} d\mu (\omega)$.

 Let $a\in \A$ be such that $\pi(a)$ is a non-zero compact operator.
Since $\Vert \sigma(a^*a)\Vert =\Vert \sigma(a)\Vert^2$ for every
representation $\sigma$ of $\A,$
upon replacing $a$ by $a^*a,$
we can assume  that $a$ is a positive element of $\A$.
So  $\pi(a)$ is  a positive selfadjoint compact operator on $\H$ with $\pi(a)\neq 0.$

There exists an orthonormal basis $(F_n)_{n\geq 1}$ of $\H=\int^{\oplus}_{\Omega} \H_{\omega} d \mu (\omega)$
consisting of eigenvectors of $\pi(a)$,
with corresponding eigenvalues $(\lambda_n)_{n\geq 1}$, counted with multiplicities.
For every $\omega\in \Omega$  and  every $n\geq 1,$ we have
$$
\pi_\omega(a) (F_n(\omega))=\lambda_n F_n(\omega).
\leqno{(*)}
$$

Let $n_0\geq 1$ be such that $\lambda_{n_0}=\max\{\lambda_{n}\mid n\geq 1\}$. Then
$\Vert \pi(a)\Vert=\lambda_{n_0}.$
Set
$$
\Omega_0=\{\omega\in \Omega\mid F_{n_0}(\omega)\neq 0.\}.
$$
Since $F_{n_0}\neq 0,$ we have $\mu(\Omega_0)>0.$
We claim that $\Omega_0$ is a finite subset of $\Omega$
consisting  of atoms of $\mu.$
Indeed, assume by contradiction that is not the case. Then there exists
an infinite  sequence $(A_k)_k$ of pairwise disjoint Borel subsets of $\Omega_0$ with $\mu(A_k)>0$.
Observe  that  $\Un_{A_k} F_{n_0}$ is a  non-zero vector in $\H$
and that $\langle \Un_{A_k} F_{n_0},\mid \Un_{A_l} F_{n_0}\rangle=0$ for
every $k\neq l.$ Moreover,
we have
$$
\begin{aligned}
\pi(a)(\Un_{A_k} F_{n_0})&= \int^{\oplus}_{A_k} \pi_{\omega}(a) (F_{n_0}(\omega)) d\mu (\omega)\\
&= \lambda_{n_0}\int^{\oplus}_{A_k} F_{n_0}(\omega) d\mu (\omega)\\
&=\lambda_{n_0} \Un_{A_k} F_{n_0}.
\end{aligned}
$$
Since $\pi(a)$ is a compact operator and $\lambda_{n_0}\neq 0,$ this is a contradiction.

Let $\omega_0\in \Omega_0$ be such that $\mu(\{\omega_0\})>0.$
We claim that the linear span of $\{F_n(\omega_0)\mid n\geq 1\}$ is dense in $\H_{\omega_0}.$
Indeed, let $v\in \H_{\omega_0}$ be such that
$$\langle v \mid F_n(\omega_0)\rangle=0 \tous n\geq 1.$$
 Let $F=\Un_{\omega_0}\otimes v\in \H$ be defined
by $F(\omega_0)=v$ and $F(\omega)=0$ for $\omega\neq \omega_0.$
Then $\langle F\mid F_n \rangle=0$
for all $n\geq 1.$ Hence, $F=0,$ that is, $v=0$, since $(F_n)_{n\geq 1}$ is a
basis of $\H.$

By $(*)$, $F_n(\omega_0)$ is an eigenvector of $\pi_{\omega_0}(a)$
with eigenvalue $\lambda_n$ for every $n\geq 1$ such that $F_n(\omega_0)\neq 0.$
Since   $\{F_n(\omega_0)\mid   n\geq 1 \}$ is a
 total subset of $\H_{\omega_0}$, it follows that there exists a basis of $\H_{\omega_0}$
 consisting of eigenvectors of $\pi_{\omega_0}(a)$.
As
$$\lim_{n\to \infty} \lambda_n=0$$
 (in case the sequence $(\lambda_n)_{n\geq 1}$ is infinite),  it follows that $\pi_{\omega_0}(a)$ is a
  compact operator on $\H_{\omega_0}.$
 Moreover,   we have
 $$\Vert \pi_{\omega_0}(a)\Vert =\max\{\lambda_{n}\mid n\geq 1\}= \lambda_{n_0} =\Vert \pi(a)\Vert.$$

Finally,   an equivalence  between $\pi_{\omega_0}$  and a subrepresentation of $\pi$
is provided by the unitary linear map
$$\H_{\omega_0}\to \H,\,  v\mapsto \Un_{\omega_0}\otimes v.$$
 \end{proof}

\subsection{Induced representations of groups}
\label{SS2}
In the sequel, we will often  consider   group representations  which are induced representations.
 Let $G$ be a countable group, $H$ a subgroup of $G$ and $(\sigma, \K)$ a  representation of $H$.
  Recall that  the induced representation $ \Ind_H^G \sigma$ of $G$  may be realized as follows.
 Let $\H$ be the Hilbert space of maps $f : G \to \K$ with the following properties
\begin{itemize}
\item[(i)]
$f(hx) = \sigma(h) f(x)$ for all $x \in G, h \in H$;
\item[(ii)]
$\sum_{x \in H \backslash G} \Vert f(x) \Vert^2 < \infty.$
(Observe that
 $\Vert f(x) \Vert$ only depends on the coset of $x$ in $H\backslash G$.)
\end{itemize}

The induced representation
$\pi = \Ind_H^G \sigma$
is given on $\H$ by right translation:
$$
(\pi(g) f)(x) \, = \, f(xg)
\hskip.5cm \text{for all} \hskip.2cm
g \in G, \hskip.1cm f \in \H
\hskip.2cm \text{and} \hskip.2cm
x \in G.
$$

Recall that the \textbf{commensurator} of $H$ in $G$
is the subgroup, denoted by $\Comm_G (H)$, of the elements $g \in G$ such that
$ gH g^{-1}\cap H$ is of finite index in both $H$ and $g^{-1} Hg$.

The following result  appeared in \cite{Mack--51} in the case where $\sigma$ is of dimension 1
and was extended to its present form in \cite{Klep--61} and \cite{Corw--75}.
\begin{theorem}
\label{Theo-Induced}
Let $G$ be a countable group and $H$ a subgroup of $G$
  such that  $\Comm_G (H)=H.$
\begin{itemize}
 \item [(i)] For every   \emph{finite dimensional} irreducible  representation $\sigma$ of $H,$
 the induced representation $\Ind_H^G \sigma$ is   irreducible.

 \item [(ii)] Let  $\sigma_1$ and $\sigma_2$ be non equivalent \emph{finite dimensional} irreducible  representations of
 $H$. The representations $\Ind_H^G \sigma_1$ and $\Ind_H^G \sigma_2$ are non equivalent.

\end{itemize}
\end{theorem}
We will need to decompose the restriction to a subgroup  of an induced representation
$\Ind_H^G \sigma$ as in Theorem~\ref{Theo-Induced}.
For $g\in G,$ we denote by  $\sigma^g$
the representation of $g^{-1} Hg$ defined by $\sigma^g(x)= \sigma(gxg^{-1})$ for $x\in g^{-1} Hg.$

For the convenience of the reader, we give a short and elementary proof
of the following  special case of the far more general result  \cite[Theorem 12.1]{Mack--52}.

 \begin{proposition}
\label{Prop-Induced}
Let $G$ be a countable group, $H,L$  subgroups of $G$ and $(\sigma, \K)$ a  representation of
$H.$  Let  $S$ be a system of representatives for the double coset space $H\backslash G/L$.
 The restriction  $\pi|_L$ to $L$ of the induced representation $\pi=\Ind_H^G \sigma$  is equivalent to the direct sum
$$
\bigoplus_{s\in S} \Ind_{s^{-1}H s\cap L}^L (\sigma^s|_{s^{-1}H s\cap L})
$$
\end{proposition}

\begin{proof}
Let $\H$ be the Hilbert space  of $\pi$, as  described above.
For every $s\in S,$ let $\H_s$ be the space of maps $f\in \H$ such that $f=0$ outside
the double coset $HsL.$
We have an orthogonal $L$-invariant decomposition
$$
\H=\bigoplus_{s\in S} \H_s.
$$

Fix $s\in S.$ The Hilbert space $\H'_s$ of $\Ind_{s^{-1}H s\cap L}^L (\sigma^s|_{s^{-1}H s\cap L})$
consists of the maps $f: L\to \K$ such that
\begin{itemize}
\item
$f(t x) = \sigma(sts^{-1}) f(x)$ for all $t \in s^{-1}H s\cap L, x \in L$;
\item
$\sum_{x \in s^{-1}H s\cap L\backslash L} \Vert f(x) \Vert^2 < \infty$.
\end{itemize}
Define a linear map $U: \H_s\to \H_s'$  by
$$
Uf(x)= f(sx) \tous f\in \H_s, x\in L.
$$
Observe that, for $t\in s^{-1}H s\cap L$, $x\in L$ and $f\in \H_s,$ we have
$$
Uf(tx)=f(stx)= f((sts^{-1}) sx)= \sigma(sts^{-1}) f(sx)= \sigma^s(t) Uf(x)
$$
and that
$$
\sum_{x \in s^{-1}H s\cap L\backslash L} \Vert Uf(x) \Vert^2 =
\sum_{x \in s^{-1}H s\cap L\backslash L} \Vert f(sx) \Vert^2=\sum_{y\in H\backslash G} \Vert f(y) \Vert^2 <\infty,
$$
so that $Uf\in \H_s'$  and $U$ is an isometry. It is easy to check that the map $U$ is invertible, with inverse given by
$$
U^{-1}f(y)=
\begin{cases}
  \sigma(h)f(x)   &\text{if } y=hsx\in HsL \\
   0    & \text{otherwise },
  \end{cases}
$$
for $f\in  \H_s'.$
Moreover, $U$ intertwines  the restriction of $\pi|_L$ to $\H_s$ and $\Ind_{s^{-1}H s\cap L}^L (\sigma^s|_{s^{-1}H s\cap L}):$
for $g, x\in L$ and $ f\in \H_s'$, we have
$$
\begin{aligned}
\left(U \pi(g)U^{-1}f\right)(x)&= \left(\pi(g)U^{-1}f\right)( sx)\\
&= (U^{-1}f( sxg) \\
&= f(xg) \\
&= \left(\Ind_{s^{-1}H s\cap L}^L (\sigma^s|_{s^{-1}H s\cap L})(g) f\right)(x)
\end{aligned}
$$
\end{proof}

We will need  the following elementary lemma about induced representations containing a
finite dimensional representation.
Recall that a representation $\pi$ of a group $G$ contains another representation $\sigma$ of $G$ if
 $\sigma$ is equivalent to a subrepresentation of $\pi.$
Recall also that,  if $\pi$ is finite dimensional representation of a group $G$, then
$\pi\otimes \bar{\pi}$ contains the trivial representation $1_G$,
where $\bar\pi$ is the conjugate
representation of $\pi$ and $\pi\otimes \rho$ denotes the (inner) tensor product
of the  representations $\pi$ and $\rho$ (see \cite[Proposition A. 1.12]{BHV}).

\begin{proposition}
\label{Prop-Induced-SubRep}
Let $G$ be a countable group, $H$ a subgroup of $G$,
 and $\sigma$ a  representation of $H.$
Assume that  the induced representation $\Ind_H^G \sigma$ contains
a finite dimensional representation of $G.$ Then $H$ has finite index in $G.$
\end{proposition}
\begin{proof}
By assumption, $\pi:=\Ind_H^G \sigma$ contains a  finite dimensional representation $\sigma.$
Hence, $\pi\otimes \bar{\pi}$ contains $1_G.$
On the other hand,
$$\pi\otimes \bar{\pi}= (\Ind_H^G \sigma)\otimes \bar{\pi}$$
is equivalent to $\Ind_H^G (\rho)$, where $\rho= \sigma \otimes (\bar{\pi}|_H)$;
see \cite[Proposition E. 2.5]{BHV}. So, there exists  a non-zero map $f : G \to \K$
in the Hilbert space of $\Ind_H^G (\rho)$ which is $G$-invariant, that is,
such that $f(xg)=f(x)$ for all $g,x \in G$. This implies that the
$L^2$-function $x\mapsto \Vert f(x) \Vert^2$ is constant on $H \backslash G.$
This is only possible if $H \backslash G$ is finite.

\end{proof}

\subsection{Amenability }
\label{SS3}
Let $\GG$ be a topological group and $UCB(\GG)$ the  Banach space  of the
 left uniformly continuous bounded functions on $\GG,$
 equipped with the uniform norm. Recall  that $\GG$ is amenable
 if there exists a $\GG$-invariant mean on $UCB(\GG)$ (see Appendix G in \cite{BHV}).

The following proposition characterizes the integral domains $R$ for which $\GL_n(R)$ or $\SL_n(R)$ is amenable;
the proof is an easy extension of the proof given in Proposition 9 in \cite{HoRo--89} for the case where $R$ is a field.

\begin{proposition}
\label{Pro-AmenableSLn}
Let $R$ be a countable unital commutative ring which is an integral domain.
Let $\KK$ be the field of fractions of $R$
and $G=\GL_n(R)$ or $G=\SL_n(R)$ for an integer $n\geq 2$.
The following properties are equivalent:
\begin{itemize}
\item[(i)]   $G$ is not amenable.
\item[(ii)] $\KK$  is not an  algebraic extension of a finite field.
\item[(iii)]  $R$ contains $\ZZ$ if the characteristic of $\KK$ is $0$
or the polynomial ring $\FF_p[T]$  if the characteristic of $\KK$ is $p>0.$
\end{itemize}
\end{proposition}

\begin{proof}
Assume  that $\KK$ is an algebraic extension of a finite field $\FF_q$.
Then $\KK = \bigcup_{m} \KK_m$
for an increasing family of finite extensions $\KK_m$ of $\FF_q$;
hence, $\GL_n(\KK) = \bigcup_m \GL_n(\KK_m)$ is the
inductive limit of the finite and hence amenable groups $\GL_n(\KK_m)$;
it follows that $\GL_n(\KK)$ is amenable and therefore $\GL_n(R)$ and $\SL_n(R)$
are amenable. This shows that (i) implies (ii).
\par

Assume that (ii) holds.
If the characteristic of $\KK$ is $0$, then $\KK$ contains $\QQ$ and hence $R$ contains
$\ZZ.$  So, we can assume that the characteristic of $\KK$ is $p>0.$ We claim
 that $R$ contains  an element which is not algebraic over the prime field $\FF_p.$
Indeed, otherwise, every element in $R$ is algebraic over $\FF_p.$
As the set of  elements in   $\KK$  which are algebraic over  $\FF_p$ is a field,
 it would follow that  the field fraction field $\KK$ is algebraic over  $\FF_p.$
 This  contradiction shows that (ii) implies (iii).

Assume that (iii) holds.
Then $\SL_n(R)$ contains a copy of $\SL_2(\ZZ)$ or a copy of $\SL_2(\FF_p[T])$.
It  is well-known that both $\SL_2(\ZZ)$ and $\SL_2(\FF_p[T])$  contain a subgroup which is isomorphic
to the free group on two generators.
Therefore, $G$ is not amenable and so  (iii) implies (i).
\end{proof}

Let   $\GG$ be a  \emph{locally compact} group, with Haar measure $m.$
Recall that  the amenability of $\GG$
is characterized by the Hulanicki-Reiter theorem (see \cite[Theorem G.3.2]{BHV}):
$\GG$ is amenable if and only if the regular representation $(\lambda_{\GG}, L^2(\GG, m))$
weakly contains the trivial representation $1_{\GG}$, where $m$ is Haar measure on $\GG$;
when this is the  case, $\lambda_{\GG}$ weakly contains every  representation of $\GG$

The following result shows the amenability of  $\GG$ can be detected by the restriction
of $\lambda_{\GG}$ to a dense subgroup; for a more general result, see
\cite[Proposition~1]{GuivAsym} or \cite[Theorem 5.5]{Bekk--16}.

\begin{proposition}
\label{Pro-RestRegRep}
Let $\GG$ be a locally compact group and $G$ a countable dense subgroup of $\GG.$
Assume that the restriction to $G$  of the regular representation
$\lambda_{\GG}$ of $\GG$  weakly contains the trivial representation $1_{G}$.
Then $\GG$ is amenable.
\end{proposition}
\begin{proof}
By assumption, there exists a sequence $(f_n)_n$ in $ L^2(\GG, m)$ with $\Vert f_n\Vert=1$ such that
$$
\lim_{n} \Vert  \lambda_{\GG}(g) f_n-f_n\Vert =0 \tous g\in G.
$$
Then, since $\left| |f_n(g^{-1}x)|- |f_n(x)|\right|\leq  \left|f_n(g^{-1}x)-f_n(x)\right|$ for $g,x\in \GG,$
we have
$$
\lim_{n} \Vert  \lambda_{\GG}(g) |f_n|-|f_n|\Vert =0 \tous g\in G.\leqno{(*)}
$$
Set $\varphi_n:= \sqrt{|f_n|}.$ Then $\varphi_n\geq 0$ and $\int_{\GG} \varphi_n dm= 1.$
Every $\varphi_n$ defines a mean $M_n:f\mapsto \int_{\GG} f\varphi_n dm$
on $UCB(\GG)$.
 Let $M$ be  a limit of  $(M_n)_n$ for  the weak-*-topology on the dual space of $UBC(\GG).$
 It follows from $(*)$ that $M$ is invariant under $G.$ Since,
 for every $f \in UCB(\GG),$   the map
 $$\GG\to UCB(\GG), \quad g\mapsto _gf$$
 is continuous (where $_gf$ denotes left translation by $g\in \GG$),
 it follows that $M$ is   invariant under $\GG.$
 Hence, $\GG$ is amenable.

 \subsection{Special linear groups over a subring of a field}
 \label{SS4}
We will  use  the following elementary lemma
about subgroups of $\SL_n(\KK)$ which stabilize a line in $\KK^n.$
\begin{lemma}
\label{Lem-InfiniteIndexSubring}
  For an infinite field $\KK$ and $n\geq 2,$ let $L$ be a subgroup of $\SL_n(\KK)$ which stabilizes a line
 $\ell$ in $\KK^n.$  Then $L\cap \SL_n(R)$  has infinite index in  $\SL_n(R)$
 for every infinite unital subring $R$ of $\KK.$
 \end{lemma}

 \begin{proof}
 Let $\{v_1, \dots, v_{n}\}$ be a basis of $\KK^{n}$ with $\ell= \KK{v_1}.$
Fix $i,j\in \{1,\dots, n\}$ with $i\neq j$ and, for $\lambda\in \KK,$ let
$E_{ij}(\lambda)$ be the corresponding elementary matrix
in $\SL_{n}(\KK),$ that is,
$$E_{ij}(\lambda)=  I_{n} + \lambda \Delta_{ij},$$
 where $\Delta_{ij}$ denotes the matrix with $1$ at the position $(i,j)$ and $0$ otherwise.

For every $l=1,\dots, n,$ let $\varphi_l:\KK\to \KK$  be  defined by
$$
E_{ij}(\lambda)(v_1)= \sum_{i=1}^{n} \varphi_l(\lambda) v_i \quad \text{for} \quad \lambda\in \KK.
$$
Every  $\varphi_l$ is a polynomial function (in fact, an affine function)
on $\KK$ and, for  $l=2, \dots, n,$  we have
$\varphi_l(\lambda)=0$ for every $\lambda\in \KK$ such that $E_{ij}(\lambda)\in L.$

Assume, by contradiction, that $L\cap \SL_n(R)$ has finite index in $\SL_{n}(R)$
for an infinite subring $R$ of $\KK.$
Then the subgroup
  $$L_{i,j}(R):=L\cap \{E_{ij}(\lambda)\mid \lambda\in R\}$$ has finite index in
the subgroup $\{E_{ij}(\lambda)\mid \lambda\in R\}$ of $SL_{n}(R).$
 In particular,  $L_{i,j}(R)$ is infinite. It follows that
 $\varphi_l$ has infinitely many roots in $\KK$ and hence that $\varphi_l=0$, for every $l=2, \dots, n.$
 Therefore, every elementary
matrix $E_{ij}(\lambda)$ fixes the line $\ell$, for
 $i,j\in \{1,\dots, n\}$ and $\lambda\in \KK$.
Since $\SL_{n}(\KK)$ is generated by elementary matrices,
it follows that every matrix in $\SL_{n}(\KK)$ fixes the line $\ell$; this of course is impossible.

 \end{proof}
\section{Proofs of Theorem~\ref{Theorem1} and Corollary~\ref{Cor-Theorem1}}
\label{S2}

\subsection{Proof of Theorem~\ref{Theorem1}}
\label{S2-SS1}
Let $X$ be a  tree, with $X^0$ the set of vertices   and $X^1$ the set of edges of $X.$
Let $\GG$ be  a locally compact group acting on $X$.

Julg and Valette constructed in \cite{JuVa--84} (see also \cite{Szwa--91} and \cite{Julg--15})
a remarkable family of  representations $(\pi_t)_{t\in [0, 1]}$ of $\GG$, all defined  on $\ell^2(X^0)$, with the following properties:
\begin{itemize}
\item[(i)]  $\pi_0$ is the natural  representation of $\GG$ on $\ell^2(X^0)$ and $\pi_1$ is equivalent to
$1_\GG\oplus \rho_1,$ where $\rho_1$ is the natural  representation of $\GG$ on $\ell^2(X^1);$
\item[(ii)]  for every $t\in [0, 1],$ there exists a bounded operator $T_t$ on $\ell^2(X^0)$
with inverse $T_t^{-1}$ defined on the subspace of functions of $X^0$ with finite support
such that $\pi_t(g):=T_t^{-1} \pi_0(g) T_t$ extends to a unitary operator on $\ell^2(X^0)$
for every $g\in \GG$; so, a unitary representation $g\mapsto \pi_t(g)$  of $\GG$ is defined on  $\ell^2(X^0)$;
\item[(iii)]  $\pi_t(g)-\pi_0(g)$ is a finite-rank operator on $\ell^2(X^0),$ for every $t\in[0,1]$ and $g\in \GG;$
  \item[(iv)] we have
  $$\langle \pi_t(g) T_t^{-1}\delta_x\mid T_t^{-1}\delta_y\rangle= t^{d(gx,y)},$$
   for every $t\in (0,1)$, $g\in \GG$ and $x,y\in X^0,$ where $d$ denotes the natural distance on $X^0;$
 \item[(v)] the map
 $$[0,1]\to \RRR^+, \ t\mapsto  \Vert\pi_t(g)-\pi_0(g)\Vert $$  is continuous   for every $g\in \GG.$
\end{itemize}
(Our representation $\pi_t$ is  $g\mapsto U_{\lambda} \rho_\lambda(g) U_{t}^{-1}$
with $\lambda=-\log t,$  for the representation $\rho_\lambda$ and the operator $U_\lambda$ appearing in  \S 2 of  \cite{JuVa--84}.)

Let  $G$ be a countable group acting on $X$. Assume that
\begin{itemize}
\item[(a)] either $G$ is not amenable and the stabilizer in $G$ of every vertex of $X$ is finite or
\item[(b)] $X$ is locally finite and the closure of  the image of $G$ in $\Aut(X)$ is not amenable.
\end{itemize}
Set $\GG=G$ in case (a)  and  let $\GG$ be the closure of $G$ in $\Aut(X)$ in case (b).
 Let $(\pi_t)_{t\in [0, 1]}$ be the family of  representations of $\GG$ as above.

\vskip.2cm
$\bullet$ {\it First step.} For every $a\in C^*(G)$ and every $t\in[ 0,1],$ the operator $ \pi_t(a)-\pi_0(a)$
is compact and the map
 $$[0,1]\to \RRR^+, \ t\mapsto  \Vert\pi_t(a)-\pi_0(a)\Vert $$  is continuous.

 Indeed, this follows from  Properties (iii) and (v) of the family $(\pi_t)_{t}$ and from the fact that
 $\CCC[G]$ is dense in $C^*(G).$

\vskip.2cm
$\bullet$ {\it Second step.} The restriction $\pi_0|_G$ of  $\pi_0$ to $G$
does not weakly contain the trivial representation  $1_G.$

Indeed, the  representation $\pi_0$ of $\GG$  is  equivalent to the direct sum  $\oplus_{s\in T}  \lambda_{\GG/ \GG_s},$ where $S$ is a system
of representatives for the $\GG$-orbits in $X^0$
and $\GG_s$ is the stabilizer in $\GG$ of $s\in S$.
Since $\GG_s$ is compact (and even finite in case (a)) and hence
amenable, $\lambda_{\GG/ \GG_s}=\Ind_{\GG_s} ^\GG 1_{\GG_s}$ is weakly contained in
the regular representation $\lambda_{\GG}$ of $\GG$ and so  $\pi_0$
is weakly contained in $\lambda_{\GG}.$ Hence, $\pi_0$ does not weakly contain the trivial representation  $1_G$
in case (a). In case (b),  the claim follows from
Proposition~\ref{Pro-RestRegRep}, since  $\GG$ is not amenable and $G$ is dense in $\GG.$

\vskip.2cm
$\bullet$ {\it Third step.} There exists an element
$a\in C^*(G)$ and $0\leq t_0<1$ with the following properties:
$\pi_{t_0}(a)=0$,  $\pi_t(a)$ is a  non zero compact operator
for every $t\in(t_0, 1]$,  and the map
$$[t_0,1] \to \RRR+, \ t\mapsto \Vert \pi_t(a)\Vert$$
is continuous.

Indeed,  by the second step, there exists $a\in C^*(G)$ such that $\pi_0(a)=0$
and $1_G(a)\neq 0$. Therefore, $\pi_1(a)\neq 0$ and  $ \pi_t(a)=\pi_t(a)-\pi_0(a)$
for every $t\in[0,1]$ and  so the claim follows from the first step.

\vskip.2cm
$\bullet$ {\it Fourth step.} Let $a\in C^*(G)$ and  $0\leq t_0<1$ be as in the third step.
 There exists an irreducible infinite dimensional subrepresentation
$\sigma_t$ of $\pi_t$ such that  $\sigma_t(a)$ is  a compact operator
and such that  $\Vert \sigma_t(a)\Vert=\Vert \pi_t(a)\Vert$ for every $t\in (t_0,1).$

Indeed, it  follows from the third step and Lemma~\ref{lem-IrredCompactOp}
that $\pi_t|_G$  contains an irreducible subrepresentation $\sigma_t$
 such that  $\sigma_t(a)$ is  a compact operator
with $\Vert \sigma_t(a)\Vert=\Vert \pi_t(a)\Vert$.
 It remains to show that $\sigma_t$ is infinite dimensional
 for every $t\in(t_0,1).$

 Assume, by contradiction, $\sigma_t$ is finite dimensional for some $t\in(t_0,1).$
 Since $G$ is dense in $\GG,$ the closed  subspace $\K_t$ of $\ell^2(X^0)$
 defining $\sigma_t$ is invariant under $\GG$ and so $\sigma_t$ is the restriction to $G$
of a dimensional subrepresentation of $\pi_t$, again denoted by  $\sigma_t$.
On the one hand, $\GG$ acts properly on $X^0$, since  the stabilizers of vertices
are compact (and even finite in case (a)). So, we have
$$\lim_{g\to +\infty: g\in \GG} d(gx,x)=0 \tous x\in X^0.$$  It follows
from Property (iv) of the family $(\pi_t)_{t}$ that
$\pi_t$ (and hence $\sigma_t$) is a $C_0$-representation, that is,
$$
\lim_{g\to +\infty: g\in \GG}\langle \pi_t(g)v\mid w\rangle= 0
$$
for every $v,w\in \ell^2(X^0)$.
On the other hand, since $\sigma_t$ is  finite dimensional,
$\sigma_t\otimes\overline{\sigma_t}$ contains $1_{\GG}$.
As $\GG$ is not compact, this is a contradiction to the fact that
$\sigma_t$ is a $C_0$-representation.

\vskip.1cm

\vskip.2cm
$\bullet$ {\it Fifth step.} There exists uncountably many real numbers
$t\in(t_0,1)$
such that the  subrepresentations
$\sigma_t$ of $\pi_t|_G$ as in the fourth step are pairwise non weakly equivalent.

 Indeed, by the third step, the function $f:t\mapsto \Vert \pi_t(a)\Vert$
 is continuous on $[t_0,1]$, with $f(t_0)=0$ and $f(1)>0.$
 So, the range of $f$ contains a whole interval.
 Let  $t,s\in (t_0,1)$ be such that  $f(t)\neq f(s).$ Then
 $$\Vert \sigma_t(a)\Vert=\Vert \pi_t(a)\Vert =f(t)\neq f(s) =  \Vert \pi_s(a)\Vert=\Vert \sigma_s(a)\Vert,$$
 and so $\sigma_t$ and $\sigma_s$ are not weakly equivalent.

\subsection{Proof of Corollary~\ref{Cor-Theorem1}}
\label{Section-Proofs-Cor-Theo1-2}
The following remarks show how Corollary~\ref{Cor-Theorem1}
follows from Theorem~\ref{Theorem1}.

\medskip
\n
(i) Let $\KK$ be  global field $\KK$.
Choose a non trivial  discrete valuation  $v:\KK^*\to \ZZ$.
The completion  of $\KK$ at $v$ is a non archimedean local field $\KK_v$.
The  tree  $X_v$ associated to $v$  (see Chapter II in \cite{Serr--80}) is a locally finite regular graph.
The group $G=\GL_2(\KK)$  acts as a group of automorphisms of $X_v,$
with vertex stabilizers conjugate to $\GL_2(\O_v\cap \KK),$ where $\O_v$ is the compact subring of the integers  in $\KK_v.$
The closure of the image of $G$ in $\Aut(X_v)$ coincides with $\PGL_2(\KK_v)$
and is therefore non amenable. A similar remark applies to $G=\SL_2(\KK).$

\medskip
\n
(ii)  As is well-known, the group  $G=\SL_2(\ZZ)$ is an amalgamated product $\ZZ/4\ZZ\ast_{\ZZ/2\ZZ}\ZZ/6\ZZ.$
It follows that $G$ acts on a tree with vertices of valence 2 or 3
with vertex stabilizers of order $4$ or $6$ (see Chapter I, Examples 4.2. in \cite{Serr--80})

\medskip
\noindent
(iii) The  free  non abelian group $F_2$ acts freely on its Cayley graph $X,$ which is
a $4$-regular tree. It follows that $F_n$ acts freely on $X$ for  every $n\in \{2,\dots, +\infty\}.$
Observe that, in this case, the representations $\pi_t$ and $\sigma_t$ as in the proof of Theorem~\ref{Theorem1}
are faithful for $t\neq 1$  (since there are even $C_0$-representations).
\end{proof}

\section{Proofs of Theorem~\ref{Theorem2} and Corollary~\ref{Cor-Theorem2}}
\label{S3}
\subsection{Proof of Theorem~\ref{Theorem2}}
\label{SS-Proofs-Theo1}
Let $R$ be a countable unital commutative ring which is an integral domain and
 $\KK$ its  field of fractions.
 In case the characteristic of $\KK$ is positive, assume that $\KK$ is not  an algebraic extension of  its prime field.

Let $n\geq 3$ and $G= \GL_n(R)$.
We consider the   natural action
of $G$ on the projective space $\PP(\KK^n).$
Let $\ell_0=\KK e_1\in \PP(\KK^n)$ be the line defined by the first unit vector $e_1$ in $\KK^n.$
The stabilizer of $\ell_0$ in $G$ is
$$H= \begin{pmatrix} R^\times & R^{n-1}\\ 0 & GL_{n-1}(R) \end{pmatrix}.$$
Let $\sigma$ be a finite dimensional  representation of $H$
and $\pi:= \Ind_H ^G \sigma$.
We claim that $\pi$ is irreducible and that $\pi(C^*(G))$ contains a non zero compact operator.
For the proof of this claim, we have to treat separately the cases $n=3$ and $n\geq 4.$

\subsubsection{\textbf{Case $n=3$}}
$\bullet$ {\it First  step.} We claim that $gHg^{-1}\cap H$ is amenable,  for every $g\in G\setminus H.$

Indeed, let $g\in G\setminus H.$
 Then $\ell_0$ and $g\ell_0$  are distinct lines in $\KK^n$  and are both stabilized by $gHg^{-1}\cap H$.
 Hence,  $gHg^{-1}\cap H$  is isomorphic to a subgroup of the solvable group
$$ \begin{pmatrix} \KK^* & 0& \KK \\ 0& \KK^*& \KK \\0 & 0&\KK^* \end{pmatrix}$$
and is therefore amenable.

\vskip.2cm
$\bullet$ {\it Second  step.} We claim that the representation $\pi$ is irreducible.

Indeed, in view of  Theorem~\ref{Theo-Induced}, we have to show that
$\Comm_G(H)=H.$  Let $g\in G\setminus H.$
On the one hand, $gHg^{-1}\cap H$  is amenable, by the first step.
 On the other hand,  $H$ is non amenable, by Proposition~\ref{Pro-AmenableSLn}.
 This  implies that $gHg^{-1}\cap H$ is not of finite index in $H$ and so  $g$ is not in
the commensurator of $H$ in $G.$

\vskip.2cm
$\bullet$ {\it Third step.}  We claim that the $C^*$-algebra $\pi(C^*(G))$  contains a non-zero compact operator.

Indeed, let $S$ be a system of representatives for the double cosets space $H\backslash G/H$
with $e\in S.$
 By Proposition~\ref{Prop-Induced}, the restriction  $\pi|_H$ of $\pi$  to $H$  is equivalent to the direct sum
$$
\bigoplus_{s\in S} \Ind_{s^{-1}H s\cap H}^H (\sigma^s|_{s^{-1}H s\cap H}) = \sigma\oplus \bigoplus_{s\in S\setminus\{e\}} \Ind_{s^{-1}H s\cap H}^H (\sigma^s|_{s^{-1}H s\cap H})
$$
Let $s\in S\setminus\{e\}.$ By the first step, $s^{-1}H s\cap H$ is amenable
and hence $\sigma^s|_{s^{-1}H s\cap H} $ is weakly contained in the regular representation $\lambda_{s^{-1}H s\cap H}$ of ${s^{-1}H s\cap H}$, by the Hulanicki-Reiter theorem. By continuity of induction (see \cite[Theorem F.3.5]{BHV}), it follows that
$\Ind_{s^{-1}H s\cap H}^H (\sigma^s|_{s^{-1}H s\cap H})$ is weakly contained in
 the regular representation $\lambda_H$  of $H$.  Therefore,
$$\pi_0:=\bigoplus_{s\in S\setminus\{e\}} \Ind_{s^{-1}H s\cap H}^H (\sigma^s|_{s^{-1}H s\cap H})$$
is weakly contained in $\lambda_H$. It follows that
$\pi_0$ does not weakly contain $\sigma;$ indeed, assume by contradiction that
$\sigma$ is weakly contained in $\pi_0.$ Then
$\lambda_H\otimes \overline{\lambda_H},$ which is a multiple of $ \lambda_H,$
 weakly contains $\sigma\otimes \overline{\sigma}$. However, since  $\sigma$ is finite dimensional,
 $\sigma\otimes \overline{\sigma}$ contains $1_H$. Hence, $1_H$  is weakly contained in $ \lambda_H$
 and this is a contradiction to the non amenability of $H.$

 It follows from Lemma~\ref{lem-CompactOp} that $\pi(C^* (H))$ contains a non-zero compact
 operator. Since $C^* (H)$ can be viewed a subalgebra of $C^* (G)$, the claim is proved
 for $G=\GL_3(R)$.

\subsubsection{\textbf{Case $n\geq 4$}}

For every unital  subring $R'$ of $R,$ set
$$
L(R'):=\begin{pmatrix} 1& 0\\ 0 & \SL_{n-1}(R') \end{pmatrix},
$$
which is a subgroup of $H$  isomorphic to $SL_{n-1}(R').$

\vskip.2cm
$\bullet$ {\it First step.} Let $g_0\in G\setminus H$ and $R'$ an infinite unital subring of $R.$
We claim that $g_0Hg_0^{-1}\cap L(R')$ has infinite index in $L(R')$.

Indeed, the group $L:=g_0Hg_0^{-1}\cap L(R')$ stabilizes the two lines $\ell_0$ and $g_0\ell_0.$ Let
 $V$ be the linear span of the $n-1$ unit vectors $e_2,\dots, e_{n}$. Denote
 by $\ell$ the projection on $V$ of the line $g_0\ell_0$, parallel to $\ell_0.$
 As $g\ell_0\neq \ell_0$, we have $\ell\neq\{0\}$. Moreover, $L$
 stabilizes $\ell$, since $L$ stabilizes $\ell_0$ and $V$.
 So,  identifying $L(R')$ with the  group $ \SL_{n-1}(R'),$
 we see can view $L$ as a subgroup of   $ \SL_{n-1}(\KK)$
 which stabilizes a line in $\KK^{n-1}.$
 Lemma~\ref{Lem-InfiniteIndexSubring} shows that $L$ has infinite
index in $\SL_{n-1}(R')$, as claimed.

\vskip.2cm
$\bullet$ {\it Second step.}  We claim that the representation $\pi$ is irreducible.
In view of Theorem~\ref{Theo-Induced}, it suffices to show that $\Comm_G(H)=H.$

 Let $g_0\in G\setminus H$.  By the first step, $g_0Hg_0^{-1}\cap L(R)$ has infinite index in $L(R)$; hence,
  $g_0Hg_0^{-1}\cap H$  has infinite index in $H$, since $L(R)$ is a subgroup of $H.$

\vskip.2cm
$\bullet$ {\it Third step.} We claim that $\pi(C^*(G))$  contains a non-zero compact operator.

Indeed,  since $\KK$ is not  an algebraic extension over its prime field,
$R$ contains a subring $R'$ which is a copy $\ZZ$ or a copy of the polynomial ring $\FF_p[T]$, by Proposition~\ref{Pro-AmenableSLn}. The corresponding subgroup
$$L:=L(R')$$
 of $G$ is  isomorphic to $\SL_{n-1}(\ZZ)$ or $\SL_{n-1}(\FF_p[T])$.
Observe that $L$ is a lattice in the locally group $\GG=SL_{n-1}(\RRR)$ or
 $\GG=SL_{n-1}(\FF_p((T^{-1})))$,
where $\FF_p((T^{-1}))$ is the local field of Laurent series over $\FF_p$.
Since $n-1\geq 3,$  the group $\GG$ and hence $L$  has Kazhdan's Property (T); see \cite[\S. 1.4, 1.7]{BHV}.

Let $S$ be a system of representatives for the double cosets space $H\backslash G/H$
with $e\in S.$
 By Proposition~\ref{Prop-Induced}, the restriction  $\pi|_{L}$ to $L$ of $\pi$  is equivalent to the direct sum
 $\sigma|_L\oplus \pi_0,$ where
 $$\pi_0:=\bigoplus_{s\in S\setminus\{e\}} \Ind_{s^{-1}H s\cap L}^L (\sigma^s|_{s^{-1}H s\cap L}).$$
 We claim that  $\pi_0$ does not weakly contain $\sigma|_L.$
Indeed, assume by contradiction that $\pi_0$  weakly contains $\sigma|_L.$
Since $\sigma$ is finite dimensional and $L$ has Property (T),  it follows that
$\pi_0$ contains   $\sigma|_L$ (see \cite[Theorem 1.2.5]{BHV}).
Therefore, $\Ind_{s^{-1}H s\cap L}^L  (\sigma^s|_{s^{-1}H s\cap L})$ contains a  subrepresentation
of $\sigma|_L$ for some $s\in S\setminus\{e\}.$
Hence,  $s^{-1}H s\cap L$ has finite index in $L$, by Proposition~\ref{Prop-Induced-SubRep}.
Since $L=L(R')$ for an infinite unital subring $R'$ of $R,$
this is a contradiction to the first step.

As in the proof for the case $n=3,$ we conclude that $\pi(C^*(G))$  contains a non-zero compact operator.

This proves Theorem~\ref{Theorem2} for $G= \GL_n(R)$ when $n\geq 3.$
 The case $G=\SL_n(R)$ is proved in exactly the same way.

 \subsection{Proof of Corollary~\ref{Cor-Theorem2}}
 \label{SS-Proofs-Cor-Theo2}

 For $n\geq 3,$ let  $G=GL_n(R)$ for a ring $R$ as above.
 The irreducible  traceable representations of $G$ constructed in the proof of  Theorem~\ref{Theorem2}
 are of the form $\pi= \Ind_H ^G \sigma$ for a  finite dimensional  representation of
 the subgroup   $H= \begin{pmatrix} R^\times & R^{n-1} \\ 0 & GL_{n-1}(R) \end{pmatrix}$.
 Observe that, $\pi= \Ind_H ^G \sigma$ is trivial on the center $Z$ of $G,$ since $H$ contains $Z.$

By Theorem~\ref{Theo-Induced},   there are infinitely (respectively, uncountably)  many non equivalent such representations $\pi$, provided there exists infinitely (respectively, uncountably)  non equivalent  finite dimensional irreducible representations of $H.$
This will be the case if $GL_{n-1}(R) \ltimes R^{n-1},$ which  is a quotient of $H$,  has infinitely (respectively, uncountably)  many non equivalent   finite dimensional  irreducible representations.

\medskip
\noindent
 (i)
 Assume that $R=\KK$ is an infinite field.
It is easy to show that the finite dimensional irreducible  representations of $GL_{n-1}(\KK) \ltimes \KK^{n-1}$
are all of the form
$$
 \begin{pmatrix} * & *\\ 0 & A \end{pmatrix} \mapsto \chi(\det A), \qquad A\in \GL_{n-1}(\KK)
 $$
 for some $\chi$ in the unitary dual $\widehat{\KK^*}$ of $\KK^*$;
 as $\KK^*$ is infinite, $\widehat{\KK^*}$ is a compact infinite group and is therefore uncountable.

\medskip
\noindent
 (ii)  Assume that  $R=\ZZ$.

  \noindent
$\bullet$ {\it Case $n=3.$}  The  free group $F_2$
is a subgroup of finite index in  $\GL_2(\ZZ)$. There exists uncountably many
 unitary characters  (that is one-dimensional unitary representations)  of $F_2$.
For every such unitary character   $\chi$, the representation $\Ind_{F_2}^{\GL_2(\ZZ)}\chi$
is finite dimensional and so has a decomposition  $\oplus_{i} \sigma_i^{(\chi)}$
 as a direct sum of  finite  dimensional   irreducible  representations $ \sigma_i^{(\chi)}$ of $\GL_2(\ZZ)$.
One can choose uncountably many   pairwise non equivalent
representations among the $\sigma_i^{(\chi)}$'s and  we obtain in this way
uncountably many non equivalent  finite dimensional irreducible representations of $\GL_2(\ZZ)$
and hence of $\GL_{2}(\ZZ)\ltimes \ZZ^{2}.$

\noindent
$\bullet$ {\it Case $n\geq 4.$}  The   group $\GL_{n-1}(\ZZ)\ltimes \ZZ^{n-1}$
 has Kazhdan's property (T)  and so has at most  countably many
 non equivalent finite dimensional  representations (see \cite[Theorem 2.1]{Wang}).
 There are indeed \emph{infinitely}  many  such representations:
  for every  integer $N\geq 1,$ the finite group
  $$G_N=\GL_{n-1}(\ZZ/N\ZZ))\ltimes (\ZZ/N\ZZ)^{n-1}$$
  is a quotient of $\GL_{n-1}(\ZZ)\ltimes \ZZ^{n-1}$;
infinitely many  representations among the irreducible representations of the $G_N$'s
 are  pairwise non equivalent  when viewed as representations of  $\GL_{n-1}(\ZZ)\ltimes \ZZ^{n-1}$.

config: arxiv


TITLE: Why is that when adding 3+(x+9) i don't need to have the parenthesis surrounding it?
QUESTION [1 upvotes]: Why is that when adding 3+(x+9) i don't need to have the parenthesis surrounding it?
I never understood this?

REPLY [3 votes]: They can be removed because addition is an associative operation. It makes no difference to the result if the parens are where you put them.
Compare this with an operation that isn’t associative, like $3-(x-9)$ and $(3-x)-9$

REPLY [0 votes]: Because addition is associative: that is, $a+(b+c) = (a+b)+c$, for all $a, b, c$.
So you might as well write $a+b+c$, since it doesn't matter where you put the parentheses.


TITLE: "Find the unique x in the interval [0,pi] with..."
QUESTION [1 upvotes]: I don't understand what I should calculate if is said: "find the unique x", e.g. in the following exercise:
"Find the unique x in the interval [0,pi] with cos(x) = 0,5 * sqrt(3)"

REPLY [0 votes]: "find the unique $x$" means
1, Such $x$ must exist;
2, it is the only one.


TITLE: Proving the Range of $2 - (x + 1)^2$
QUESTION [1 upvotes]: What I've been taught is that a good way to prove the range of a function is by using the two statements $\infty < x \leq 0$ and $0 \leq x < \infty$ then working from there to reach the function's form. Doing this here leads to the range being ]-infinity,1] which obviously isn't true as 2 is clearly in the range.
My steps: (I'll use $\vee$ to separate the two statements until they're the same)
$(-\infty < x \leq 0) \vee (0 \leq x < \infty)$
$(-\infty < x+1 \leq 1) \vee (1 \leq x+1 < \infty$
$(1 \leq (x+1)^2 < \infty) \vee (1 \leq (x+1)^2 < \infty)$
$-\infty < -(x+1)^2 \leq -1$
$-\infty < 2-(x+1)^2 \leq 1$
What am I doing wrong ? Is the method I've been taught wrong to begin with ?

REPLY [0 votes]: Let $u=x+1$. Since $x\in(-\infty,\infty)$, we have $$\begin{align}u^2\in[0,\infty)&\implies -u^2\in(-\infty,0]\\&\implies 2-u^2\in(-\infty,2]\\&\implies\color{red}{2-(x+1)^2\in(-\infty,2]}\end{align}$$ as required.


text:
\chapter{Stochastic Volatility Filter for Heston Model}

\noindent With discrete observations, we derive a Bayesian stochastic volatility filter for a Heston model. Let $Y_t$ denote the log-price of the equity, and let $\sqrt{X_t}$ denote volatility. The dynamics of the processes are,
\begin{eqnarray}
\label{eq:heston}
dX_t &=& \kappa(\bar X-X_t)dt+\gamma\sqrt{X_t}dB_t\\
\label{eq:logReturns}
dY_t& =& \left(\mu-\frac{1}{2}X_t\right)dt+\sqrt{X_t}\left(\rho dB_t+\sqrt{1-\rho^2}dW_t\right)
\end{eqnarray}
and the interpretation of the model parameters is as follows:
\begin{eqnarray}
\nonumber
\bar X&=&\hbox{the long-time average of $X_t$}\\
\nonumber
\kappa&=&\hbox{the rate of mean-reversion (on $X_t$)}\\
\nonumber
\gamma&=&\hbox{volatility of volatility}\\
\nonumber
\rho&=&\hbox{models volatility leverage effect when $\rho\in(-1,0)$}\\
\nonumber
\mu&=&\hbox{the mean-rate of returns on the stock}
\end{eqnarray}
Certain restrictions on the parameters need to be put in place, such as the Feller condition: $\gamma\leq \sqrt{2\kappa \bar X}$ to insure the $X_t$ is well-defined (see chapter volatility time scales in \cite{FPS00}). There are times $(t_n)_n$ for which the process is actually observed, and if our model is correct\footnote{`correct' not only means that the processes follow these parametric SDEs, but it also means that $\{W_t\}$ and $\{B_t\}$ are idiosyncratic noises that are endemic to this system and not correlated with other data in the market.} then $\mathcal F_t^m=\mathcal F_t^Y = \sigma\{Y_{t_n}:t_n\leq t\}$.

\section{The Filter}
\label{sec:volFilter}
We showed in previous lectures that $X_t$ is measurable when $Y$ is observed continuously. It is also straight forward to show that $\mathbb E[g(X_t)|\mathcal F_t^Y]\rightarrow g(X_t)$ as the partition of $[0,t]$ shrinks to zero.

\begin{lemma}
\label{lemma:discreteToCont}
For any $N\in\mathbb Z^+$ let there be a partition of $[0,t]$ into $N$-many points, and let $\mathcal F_t^N$ denote the filtration generated by the observations of $Y$ at the partitioned points $(i.e.~\mathcal F_t^N = \sigma\{\{Y_{t_n}\}_{n=0}^N\})$. If the filtrations are increasing with $N$, then  in the context of the model given by (\ref{eq:heston}) and (\ref{eq:logReturns}) we have
\[\mathbb E[g(X_t)|\mathcal F_t^N]\rightarrow g(X_t)\qquad\hbox{a.s.}\]
as $N\nearrow\infty$ for any function $g(x)$.
\end{lemma}
\begin{proof}
Let $\mathcal F_t^Y=\sigma\{Y_s:s\leq t\}$. We assume that the filtrations are increasing with $N$, and they are certainly bounded by $\mathcal F_t^Y$,
\[\mathcal F_t^N\subset\mathcal F_t^{N+1}\subset\dots\dots\subseteq\mathcal F_t^Y.\]
Therefore there exists $\mathcal F_t^*$ such that $\bigvee_{N=1}^\infty\mathcal F_t^N=\mathcal F_t^*$. Now because the path of $Y_t$ is continuous on all sets of non-zero probability, the information contained in $\mathcal F_t^*$ is enough to measure the event $\{Y_t=y\}$ for any $y\in\mathbb R$ even if $t$ is not a partition point for any finite $N$. Therefore, $\mathcal F_t^* = \mathcal F_t^Y$ a.s., and by the L\'evy 0-1 law we have
\[\lim_N\mathbb E[g(X_t)|\mathcal F_t^N]=\mathbb E[g(X_t)|\mathcal F_t^*] = \mathbb E[g(X_t)|\mathcal F_t^Y]= g(X_t)\]
which proves the lemma.

\end{proof}
From lemma \ref{lemma:discreteToCont} we know that our posterior estimates are consistent as we refine the partition. Now we need to determine the filter for a specific partition. From here forward consider a specific finite partition of the time domain, and we only consider the filter at times when data has arrived. For ease in notation we let $X_n=X_{t_n}$, $Y_n=Y_{t_n}$ and $\mathcal F_n^Y=\mathcal F_{t_n}^Y$.

\begin{proposition}
\label{prop:SVMfilter}
Let $(Y_t,X_t)$ be the price and volatility processes in the Heston model from (\ref{eq:heston}) and (\ref{eq:logReturns}), and assume the Feller condition $\gamma^2\leq 2\bar X\kappa$. Then there is a kernel that gives $X$'s transition density, $e^{Q^*\Delta t}(x|v)=\frac{d}{dx}\mathbb P(X_{t+\Delta t}\leq x|X_t=v)$ for any $x,v\in\mathbb R^+$, and the filtering distribution for $X_n$ at observation time $t_n=n\Delta t$ has a density. This density is given recursively as
\begin{align}
\nonumber
&\pi_n(x)\\
\label{eq:piDensity}
&=\frac{1}{c_n}\int \mathbb E\left[\mathbb L(y|(X_u)_{\{t_{n-1}\leq u\leq t_n\}},Y_{n-1})\Big|X_n=x,X_{n-1}=v,Y_{n-1}\right]e^{Q^*\Delta t}(x|v)\pi_{n-1}(dv)\Bigg|_{y=Y_n}\ ,
\end{align}
for almost-everywhere $x\in\mathbb R^+$, where $c_n$ is a normalizing constant, and $\mathbb L$ is the likelihood of the any path $(x_u)_{\{t_{n-1}\leq u\leq t_n\}}$ given observations $Y_n$ and $Y_{n-1}$, and is given by
\[\mathbb L(y|(x_u)_{\{t_{n-1}\leq u\leq t_n\}},Y_{n-1}) =\frac{\exp\left\{ -\frac{1}{2}\left(\frac{ (y-Y_{n-1})-\left(\mu\Delta t-.5\int_{t_{n-1}}^{t_n}x_udu\right)-\rho\xi_n(x)}{\sqrt{(1-\rho^2)\int_{t_{n-1}}^{t_n}x_udu}}\right)^2 \right\}}{ \sqrt{(1-\rho^2)\int_{t_{n-1}}^{t_n}x_udu}}\]
with
\[\xi_n(x) = \frac{1}{\gamma}\left\{ \Delta x_{n-1}-\kappa\left(\bar X\Delta t-\int_{t_{n-1}}^{t_n}x_udu\right)\right\}\ .\]
\end{proposition}

\begin{proof} Given the Feller condition, the CIR process $dX_t=\kappa(\bar X-X_t)dt+\gamma\sqrt{X_t}dB_t$ is well-known to have a transition density that can be written in terms of a modified Bessel function (see \cite{aitSahalia1999}), and so $\Gamma^{\Delta t}(\cdot|v)$ is a smooth density function for all $v\geq 0$. Furthermore, it was shown in \cite{DY2002} that $(Y_t,X_t)$ has a smooth transition density function, that is,
\[\mbox P^{\Delta t}(y,x|s,v)\doteq \frac{\partial^2}{\partial y\partial x}\mathbb P(Y_n\leq y,X_n\leq x|Y_{n-1}=s,X_{n-1}=v)\]
is smooth for $x> 0$, $y> 0$, and $\Delta t>0$, and does not collect mass at $x=0$ or $y=0$. Hence, the filter has a density that can be written using Bayes rule:
\[\pi_n(x) = \frac{\int\mbox P^{\Delta t}(Y_n,x|Y_{n-1},v)\pi_{n-1}(v)dv}{\int\hbox{[numerator]}dx}\ ,\]
where we don't need to assume smoothness of $\pi_{n-1}$ because it is smoothed by its convolution with $\mbox P^{\Delta t}$ in the $dv$-integral.

Now, from equation (\ref{eq:logReturns}) we notice the following:
\begin{align*}
Y_n-Y_{n-1} &=  \mu\Delta t-\frac 12\int_{t_{n-1}}^{t_n}X_udu+\rho\int_{t_{n-1}}^{t_n}\sqrt{X_u}dB_u+\sqrt{1-\rho^2}\int_{t_{n-1}}^{t_n}\sqrt{X_u}d W_u\\
&=_d  \mu\Delta t-\frac 12\int_{t_{n-1}}^{t_n}X_udu+\rho\int_{t_{n-1}}^{t_n}\sqrt{X_u}dB_u+\sqrt{(1-\rho^2)\int_{t_{n-1}}^{t_n}X_udu}~~\mathcal Z
\end{align*}
where ``$=_d$" signifies equivalence in distribution, and $\mathcal Z$ is another independent standard normal random variable. This means that conditional on the path $(X_u)_{t_{n-1}\leq u\leq t_n}$ and $S_{n-1}$,

\[\frac{Y_n-Y_{n-1} -\left(\mu\Delta t-\frac 12\int_{t_{n-1}}^{t_n}X_udu+\rho\int_{t_{n-1}}^{t_n}\sqrt{X_u}dB_u\right)}{\sqrt{(1-\rho^2)\int_{t_{n-1}}^{t_n}X_udu}}=_d\mathcal Z\ .\]
Then noticing $\xi_n$ evaluated at $(X_u)_{t_{n-1}\leq u\leq t_n}$ is the the same as $\xi_n\left(X\right)=\int_{t_{n-1}}^{t_n}\sqrt{X_u}dB_u$, it follows that
\[\frac{Y_n-Y_{n-1} -\left(\mu\Delta t-\frac 12\int_{t_{n-1}}^{t_n}X_udu+\rho\xi_n\left(X\right)\right)}{\sqrt{(1-\rho^2)\int_{t_{n-1}}^{t_n}X_udu}}=_d\mathcal Z\ .\]
This shows the likelihood of the path $(X_u)_{t_{n-1}\leq u\leq t_n}$ given $S_{n-1}$ and $S_n=y$ is in fact the function $\mathbb L$.

Finally, given Bayes rule for the density $\pi_n$, the expression in equation (\ref{eq:piDensity}) displays the filter using a probabilistic representation of the transition density:

\begin{align*}
&\mbox P^{\Delta t}(y,x|Y_{n-1},v)\\
&=\frac{\partial}{\partial y}\int_0^y\mbox P^{\Delta t}(z,x|Y_{n-1},v)dz \\
& =\frac{\partial}{\partial y}\mathbb P(Y_n\leq y|X_n=x,X_{n-1}=v,Y_{n-1})\Gamma^{\Delta t}(x|v)\\
& =\frac{\partial}{\partial y}\mathbb E\left\{\mathbf 1_{Y_n\leq y}\Big |X_n=x,X_{n-1}=v,Y_{n-1}\right\}\Gamma^{\Delta t}(x|v)\\
&=\frac{\partial}{\partial y}\mathbb E\left[\mathbb E\left\{\mathbf 1_{Y_n\leq y}\Big|( X_u)_{\{t_{n-1}\leq u\leq t_n\}},Y_{n-1}\right\}\Big| X_n=x, X_{n-1}=v,Y_n,Y_{n-1}\right]\Gamma^{\Delta t}(x|v)\\
&=\mathbb E\left[\frac{\partial}{\partial y}\mathbb E\left\{\mathbf 1_{Y_n\leq y}\Big|( X_u)_{\{t_{n-1}\leq u\leq t_n\}},Y_{n-1}\right\}\Big| X_n=x, X_{n-1}=v,Y_n,Y_{n-1}\right]\Gamma^{\Delta t}(x|v)\\
&\propto\mathbb E\left[\mathbb L(y|(X_u)_{\{t_{n-1}\leq u\leq t_n\}},Y_{n-1})\Big|X_n=x, X_{n-1}=v,Y_{n-1}\right]e^{Q^*\Delta t}(x|v)\ .
\end{align*}
Lastly, when computing the likelihood based on the time-$n$ observation, the last line is evaluated at $y=Y_n$. This completes the proof of the proposition.

\end{proof}

 At this point it seems that the filter is rather complicated, and would be difficult to implement in real-time. Often times, what one might do is consider a discrete scheme that approximates the SDEs:
\begin{eqnarray}
\label{eq:discreteSDEy}
\Delta Y_{n-1}&=& (\mu-.5X_{n-1})\Delta t+\sqrt{X_{n-1}}(\rho \Delta B_{n-1}+\sqrt{1-\rho^2}\Delta W_{k-1})\\
\label{eq:discreteSDEx}
\Delta X_{n-1} &=& \kappa(\bar X-X_{n-1})\Delta t+\gamma\sqrt{X_{n-1}}\Delta B_{n-1}
\end{eqnarray}
where $\Delta Y_{n-1}=Y_n-Y_{n-1}$ and $\Delta X_{n-1}=X_n-X_{n-1}$. Using (\ref{eq:discreteSDEy}) and (\ref{eq:discreteSDEx}), we can impute an approximate filtering density to the density given in Proposition \ref{prop:SVMfilter}:

\begin{equation}
\label{eq:volFilter}
\tilde\pi_n(x) = \frac{1}{c_n}\int \psi_n(x,v)e^{Q^*\Delta t}(x|v)\tilde \pi_{n-1}(dv)
\end{equation}
where $\psi_n(x,v)$ is the likelihood of $\{X_n=x,X_{n-1}=v\}$ given $\{Y_n,Y_{n-1}\}$,
\[\psi_n(x,v) = \frac{1}{\sqrt{v(1-\rho^2)\Delta t}}\exp\left\{-\frac{(\Delta Y_{n-1}-(\mu-.5v)\Delta t-\sqrt{v}\rho\Delta B_{n-1}(x,v))^2}{2v(1-\rho^2)\Delta t}\right\}\]

\[\hbox{with}\qquad \Delta B_{n-1}(x,v) = \frac{1}{\gamma\sqrt v}\left(x-v-\kappa(\bar X-v)\Delta t\right).\]
If the model simplification given by (\ref{eq:discreteSDEy}) and (\ref{eq:discreteSDEx}) can be considered `correct', then there is no need to dispute the validity of the filter given by (\ref{eq:volFilter}). But in general, if one knows apriori that the continuous-time SDEs are the correct model, then there needs to be some analysis to verify that the approximate filter (such as that in(\ref{eq:volFilter})) converges as $\Delta t\searrow 0$, that is
\[\left|\int g(x)\tilde\pi_n(dx)-\int g(x)\pi_n(dx)\right|\rightarrow 0\]
in a strong sense as $\Delta t\rightarrow 0$. As was mentioned earlier, it is well-known (see Kushner \cite{kushner08}) that approximate filters that are sometimes consistent, but the results in \cite{kushner08} do not apply to the Heston model.

\section{Extracting the Risk-Premium}
Under the physical measure, we have,

\[\mathbb EX_t = \mathbb EX_0e^{-\kappa t}+\bar X(1-e^{-\kappa t})\]
and so the expected value of realized variance ($RV_{[0,T]} \doteq \frac 1T\int_0^TX_sds$) is
\[\mathbb E_0RV_{[0,T]} = \bar X-\frac{\bar X-\mathbb E_0X_0}{\kappa T}\left(1-e^{-\kappa T}\right)\]
where we have (without loss of generality) considered the case at time 0, and we have denoted the posterior expectation as $\mathbb E_0^*[~\cdot~]=\mathbb E^*[~\cdot~|\mathcal F_0^Y]$ and $\mathbb E_0[~\cdot~]=\mathbb E[~\cdot~|\mathcal F_0^Y]$. Therefore, there is the following close-formula for the risk-premium

\[\mathcal{RP}_0\doteq \mathbb E_0^*RV_{[0,T]}-\mathbb E_0RV_{[0,T]}\]

\[=\mathbb E_0^*RV_{[0,T]}-\bar X+\frac{\bar X-\mathbb E_0X_0}{\kappa T}\left(1-e^{-\kappa T}\right).\]
This expression for the risk-premium holds whenever volatility-squared is modeled with a mean-reverting SDE with drift term $\kappa(\bar X-X_t)$, not just the Heston model.

Under the risk-neutral measure, the market adds a risk-premium term to $dX_t$:
\[dX_t = \kappa(\bar X-X_t)dt-\Lambda_tX_tdt+\gamma\sqrt{X_t}dB_t^*\]
where $B_t^*$ Brownian motion under the market measure, and $\Lambda_t$ is the market price of volatility risk. We leave the modeling of $\Lambda_t$ open here because we will not delve deeply into its correlation structure. However, $\Lambda_t$ is most likely thought of as a mean-reverting process and could be modeled as such. Under the market's measure there is the following expectation of variance
\[\mathbb E_0^*X_t =\mathbb E_0X_t-\int_0^t\mathbb E_0^*[X_s\Lambda_s]e^{-\kappa(t-s)}ds.\]
From this, we see that $\mathbb E_0^*RV_{[0,T]}$ can be written as follows,

\[\mathbb E_0^*RV_{[0,T]} = \frac{1}{T}\int_0^T\mathbb E_0^*X_tdt = \frac{1}{T}\int_0^T\mathbb E_0X_tds-\frac{1}{T}\int_0^T\int_0^t\mathbb E_0^*[X_s\Lambda_s]e^{-\kappa(t-s)}dsdt\]

\[=\mathbb E_0[RV_{0,T}] - \underbrace{\frac{1}{\kappa T}\int_0^T\left(1-e^{-\kappa(T-s)}\right)\mathbb E_0^*[X_s\Lambda_s]ds}_{\hbox{risk-premium}}\qquad\qquad(**)\ .\]
 In $(**)$, notice that if $X$ and $\Lambda$ are independent, then for $\kappa\gg 1$ the risk-premium simplifies to
 \[\frac{1}{\kappa T}\int_0^T\left(1-e^{-\kappa(T-s)}\right)\mathbb E_0^*[X_s\Lambda_s]ds\sim -\frac{\bar X}{T}\mathbb E_0^*\Lambda_T\ .\]

\section{Filtering Average Volatility in Fast Time-Scales}
The purpose of fast time-scales in volatility modeling is to capture mean reverting effects that occur on the order of 2 to 3 days. In the Heston model, suppose we are in a fast time-scale where $\gamma\sim\sqrt\kappa$ for $\kappa$ large. Then, it can be shown that the distribution of $X_t$ settles into a $\Gamma$ distribution almost instantaneously,
\[X_t\Rightarrow\Gamma\left(2\bar X,\frac{1}{2}\right)\qquad\hbox{as }\kappa\nearrow\infty\]
for all $t>0$. Therefore, there is a fast-averaging of the realized variance,
\[\frac{1}{T}\int_0^TX_sds\rightarrow \bar X\qquad\hbox{in probability as }\kappa\nearrow\infty\]
and so the expected payoff of any contract that is a function of realized variance will be deterministic unless $\bar X$ is random and/or unknown. Thus, building a risk-premium into the dynamics of $X_t$ in the manner that we did in the previous section will not be meaningful in fast time-scales. An alternative idea would be to take $\bar X$ as a hidden regime-process that is governed by another Markov chain, and thus adds another dimension to the HMM. Then, we can take realized variance as our observations and write a filter to estimate the regime $\bar X_t$. We do this as follows:

Take $\bar X_t$ to be a Markov chain with generator $Q$, for which we assume the standard structure for changes; changes in $\bar X_t$ are governed by a Poisson jump process so that over a time interval of length $\Delta t$, the probability of $\bar X_t$ changing states more than once is $o(\Delta t)$. The new dynamics of $X_t$ are then
\[dX_t = \kappa(\bar X_t-X_t)dt+\gamma\sqrt{X_t}dB_t\]
and for such a model the realized variance is a random variable in fast time-scales,
\[\frac{1}{T}\int_0^TX_sds \sim \frac{1}{T} \int_0^T\bar X_sds\qquad\hbox{for $\kappa$ large.}\]

Let $(t_{n,\ell})_{n,\ell}$ be a partition of some finite time interval, say $10$ years, where $n$ denotes the $nth$ week, and $\ell$ denotes the $\ell th$ trade. Then
\[t_{n+1,\ell}-t_{n,\ell}=\Delta t=\hbox{1 week,}\qquad\qquad\hbox{for any $\ell$}\]
 \[t_{n,\ell+1}-t_{n,\ell}=\hbox{ time from $\ell th$ quote until the next trade during week $n$.}\]
We have observations on $Y_t$ at each $t_{n,\ell}$ (i.e. $Y_{n,\ell}=Y_{t_{n,\ell}}$ is the $\ell th$ observations on the $nth$ week). For any $n$ and $\ell$ let $\Delta Y_{n,\ell} = Y_{n,\ell+1} -Y_{n,\ell} $, and for simplicity let $t_n=t_{n,0}$. We then have the following model for weekly observations on realized variance,
\[Z_{n+1}\doteq\frac{1}{\Delta t}\sum_{\ell}(\Delta Y_{n,\ell})^2=\frac{1}{\Delta t}\int_{t_{n}}^{t_{n+1}}\bar X_sds+\epsilon_{n+1}\]
where $\epsilon_{n+1}$ is a noise process with $\mathbb E\epsilon_n\epsilon_{n'} = 0$ if $n\neq n'$. Clearly, $Z_n$ is observable  and $(Z_n,\bar X_n)$ is a Markov process. From here the goal is to estimate the the state-space and transition rates of $\bar X_t$, and then apply the nonlinear filtering results in estimating the variance risk-premium. Letting $\bar\pi_n(x_i) = \mathbb P(\bar X_{t_n}=x_i|Y_{0:n})$, we have  an estimate of the physical measure's expectation of realized variance:

\[\mathbb E_{t_n}\left[\frac{1}{T}\int_{t_n}^{t_n+T} X_sds\right]\approx\mathbb E_{t_n}\left[\frac{1}{T}\int_{t_n}^{t_n+T}\bar X_sds\right]=\frac{1}{T}\int_{t_n}^{t_n+T}\sum_ix_ie^{Q^*(s-t_n)}\bar\pi_n(x_i)ds\ ,\]
for $\kappa\sim \gamma^2\gg 1$.


TITLE: Real Numbers as Well Defined Sets
QUESTION [1 upvotes]: For every construction of the reals, we define a real number to be some kind of set of rational numbers (such as cuts or sequences).
However, the number of symbols we have to formulate the description of a set is finite (analytic functions, infinite sums, etc), and I assume sets can only be defined as finite strings of logical symbols.
Every number I can think of, including transcendentals, and even the Chaitin Constant, can be described as such.
Where is the fallacy in saying, then, that the real numbers are countable (as, for instance, well defined cuts seem to be)?

REPLY [1 votes]: Let me add something that was overlooked by the other answers.
First of all, the collection of Dedekind cuts itself is definable if we allow second-order quantification. That is, we can write a formula $\varphi(A)$ in the language of $\leq$ such that $A$ is a set of rational numbers and $\varphi(A)$ is true if and only if $A$ is a Dedekind cut (this depends on the flavor you chose for the cuts, if those are sets, or partitions of $\Bbb Q$ etc. in either case you can define what it means to be a cut).
But the fact that we can define what it means to be a cut doesn't mean that every cut is definable. Look at $\Bbb Q$ with the language of $\leq$ with first-order logic. Every element is rational, so to define the rational numbers we only need to write the formula $x=x$. But no element is definable, because $\Bbb Q$ has plenty of automorphisms when only considered as an ordered set.
So even if we consider $\Bbb Q$ in the language of $\leq$, and allow second-order variables with full semantics (i.e. we consider all the subsets of $\Bbb Q$ as our second-order sets), we can still only define what it is to be a cut, not each and every cut for itself.
And if you think about it, it makes sense. What does it mean to be a cut? It really is a conjunction of infinitely many formulas all using parameters. $r$ is the cut between $A$ and $B$ if $r$ satisfies $q<r$ for every $q\in A$ and $r\leq q$ for every $q\in B$. That's the conjunction of $\aleph_0$ formulas.
So while the number of formula which are the conjunction of finitely many formulas (with parameters!) is indeed countable, the number of infinite conjunctions like that is no longer countable.

To sum up, your mistakes lie in two crucial points:

The fact that the set of cuts is [second-order] definable, does not mean that every cut is definable. Id est, the elements of a definable set need not be definable.
Cuts are definable from infinitely many parameters, and there are uncountably many such sets of parameters, each defining a different cut.


TITLE: Equality of the transpose of a vector-matrix product
QUESTION [1 upvotes]: This is a rather trivial question, but in my lecture notes for two vectors $x,y\in \mathbb{R}$ and a symmetric matrix $C$ I have given
$$
x^TCy=y^TC^Tx
$$
and I do not understand why this is true. I get that
$$
(x^TCy)^T=y^TC^Tx
$$
but why is $(x^TCy)^T=x^TCy=y^TC^Tx$

REPLY [2 votes]: It is because $C$ is a symmetric matrix (i.e., $C^T = C$)
\begin{align*}
(x^T C y)^T = (y^T C^T (x^T)^T)^T = (y^T C x)^T = x^T C^T (y^T)^T = x^T C y.
\end{align*}


TITLE: Are $x$ and $y$ divisible by $n$, if so how do I prove it?
QUESTION [0 upvotes]: If $y$, $x$,are natural numbers, and $n$ is a prime number, $y = x + n$, $y>x>n$, and $y$ and $x$ are not coprime, is it true that $n$ is a divisor of both $x$ and $y$? If so could you please start off the proof and point me in the general direction of how to prove it, please don't show me the full proof, I want to solve it myself.

REPLY [3 votes]: You can even say that any divisor of both $x$ and $y$ is also a divisor of $n$, hence $\gcd(x,y)=p$.
If $x$ and $y$ are coprime this is not necessarily true. Counter-example: twin primes differ by $2$, a prime number.


TITLE: If $\mathcal{S}=\{\emptyset, X\}$ the only $\mathcal{S}$-measurable functions from $X\to\mathbb{R}$ are the constant functions
QUESTION [1 upvotes]: I have proved the following statement and I would like to know if there are any gaps in my reasoning, thank you:
"Suppose $(X,\mathcal{S})$ is a measurable space (i.e. $X$ is a set and $\mathcal{S}$ is a $\sigma$-algebra on $X$).
We say that a function $f:X\to\mathbb{R}$ is $\mathcal{S}$-measurable if $f^{-1}(B)\in\mathcal{S}$ for every Borel set $B\subset\mathbb{R}$."

If $\mathcal{S}=\{\emptyset, X\}$ then the only $\mathcal{S}$-measurable functions from $X\to\mathbb{R}$ are the constant functions".

My proof:
Let $X$ be a set and $f:X\to\mathbb{R}$ be a non-constant measurable function: then there will be $y_1,y_2\in\mathbb{R}, y_1\neq y_2$ such that $y_1, y_2\in f(X)$ (and there will also be $x_1,x_2\in X, x_1\neq x_2$ such that $f(x_1)=y_1$ and $f(x_2)=y_2$). Now, $f^{-1}( \{y_1\} )=\mathcal{B_1}$ and $f^{-1}(\{y_2\})=\mathcal{B_2}$, where $\mathcal{B_1}$ and $\mathcal{B_2}$ are Borel sets.
There are four possible cases: either $\mathcal{B_1}=\mathcal{B_2}=\emptyset$, only one of them is $\emptyset$ and the other one is $X$ or $\mathcal{B_1}=\mathcal{B_2}=X$.
Suppose that one of $\mathcal{B_1}$ or $\mathcal{B_2}$, say $\mathcal{B_1}$, were the empty set: then $x_1\in\mathcal{B_1}=\emptyset$, a contradiction: thus the first three cases are impossible.
Suppose now that $\mathcal{B_1}=\mathcal{B_2}=\emptyset$: then $f(x_1)=f(x_2)=X$ so $y_1=y_2$, a contradiction.
Thus $f:X\to\mathbb{R}$ cannot be non-constant.
If it were constant instead, $f:X\to\mathbb{R}, f(x)=c$ for every $x\in X$ then $f^{-1}(B)=X\in\mathbb{S}$ for every Borel set containing $c$ and $f^{-1}(B)=\emptyset\in\mathcal{S}$ for every Borel set which does not contain $c$ thus $f^{-1}(B)\in\mathcal{S}$ in every possible case.
We have thus showed that if $\mathcal{S}={\emptyset, X}$ then the only $\mathcal{S}$-measurable functions are the constant functions, as desired.

REPLY [3 votes]: Your proof looks correct, but can be written much shorter. In particular, you don't need to distinguish cases.
Suppose that $f: X \to \mathbb{R}$ is $\mathcal{S}$-measurable and consider $y\in \operatorname{Im}(f)$. Then
$$\{x \in X: f(x) = y\} \in \mathcal{S}$$
but since $y$ is in the image of $f$, this set is non-empty. Hence, this set must be $X$ and thus $f(x) = y$ for all $x \in X$, i.e. $f$ is contant.


TITLE: How to check if a non-negative matrix is primitive (in the stochastic sense)?
QUESTION [4 upvotes]: A square non-negative matrix $A$ is primitive if $A^k$ has all entries positive for some positive integer $k$. Is there an algorithm to do this check in practice? Given a non-negative square matrix $A$ of finite dimension $N$, can we determine if it is primitive?
Note: By non-negative matrix $A$, I mean that all its entries are non-negative.

REPLY [6 votes]: Corollary 8.5.8 in Horn & Johnson, Matrix Analysis.
Let $A$ be an $n \times n$ non-negative matrix. Then $A$ is primitive if and only if $A^{n^2 - 2n + 2}$ has only positive entries.


TITLE: Solving Linear Congruences with Euclidean algorithm
QUESTION [1 upvotes]: I have the following equation: $70x \equiv 2 \pmod {58}$ and this equation: $X = X_i + i \times \frac{n}{d}$.
Using the table method with Euclidean algorithm I do $6$ steps to find out $\rm{gcd}(70,58) = d = 2\quad\quad$ I know that $n = 58$ however what are $X_i$ & $i$.
How do I solve this equation ?
I assumed $X_i$ was denominator at step $i$ and $i$ was the number of step where the Euclidean algorithm ended. Seems I am wrong.

REPLY [1 votes]: $X_i$ is any solution to the linear congruence $70x≡2 (\mod 58)$, $n$ is 58 as you mentioned and $i$ is any integer.
Therefore, what the equation really means is that from any particular solution $X_i$, we can construct an infinite amount of solutions $X = X_i + \frac{i*n}{d}$, where $d = \gcd(a,m)$.
You can find that particular solution $X_i$ either by inspection (if there is a trivial or obvious value of x that would make the equation hold) or by the Extended Euclidean Algorithm.
If you haven't seen the EEA, this post might help: What's the difference between the euclidean algorithm and the extended euclidean algorithm?


TITLE: Taylor expansion of $\sqrt{f(x)}$ at $ x=0.$
QUESTION [8 upvotes]: Let $f$ be a strictly positive function. Then we can compute the Taylor expansion of $\sqrt{f(x)}$ at zero. Wolfram alpha gives the first terms as
$$\sqrt{f(x)}= \sqrt{f(0)} + \frac{f'(0)x}{2\sqrt{f(0)}}+...$$
Now, I was wondering if there is a closed representation of this as a power series? At first glance this does not look at hard, as we just apply chain and quotient-rule, but I could not come up with a closed equation.

REPLY [4 votes]: Making the problem more general, if you consider $\sqrt[n]{f(x)}$ $$\sqrt[n]{f(x)}=f(0)^{\frac{1}{n}}+\frac{ f(0)^{\frac{1}{n}-1} f'(0)}{n}x+\frac{
   f(0)^{\frac{1}{n}-2} \left(nf(0)  f''(0)-n f'(0)^2+f'(0)^2\right)}{2
   n^2}x^2+O\left(x^3\right)$$ which is what Wolfram Alpha gave you for $n=2$.
Otherwise, you could consider $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$ $$\sqrt[n]{f(x)}=\sqrt[n]{f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)}$$ and use the generalized binomial theorem writing $$\sqrt[n]{f(x)}=\sqrt[n]{f(0)}\left(1+\frac{f'(0)}{f(0)}x+\frac12 \frac{f''(0)}{f(0)}x^2+O\left(x^3\right)\right)^{\frac 1n}$$ using $$(1+z)^{\frac 1n}=1+\frac{1}{n}z+\frac{(1-n) }{2 n^2}z^2+O\left(z^3\right)$$ in which $$z=\frac{f'(0)}{f(0)}x+\frac12 \frac{f''(0)}{f(0)}x^2+O\left(x^3\right)$$ For sure, the same applies with more terms.


TITLE: Orthogonal polynomials derive normalization factor
QUESTION [1 upvotes]: I have the following problem, and have literally no idea where to start! Even a hint to get me going would be apreciated. I believe the question is concerning Laguerre polynomials.

Let $\alpha$ > -1. Suppose that a monic orthogonal polynomial sequence
  $\{P_n\}_{n\geq0}$ (for $\mathscr{L}$) satisfies the orthogonality
  relation
$\mathscr{L}[x^m P_n(x)] = \frac{1}{{\Gamma(\alpha +1)}}\int_0^\infty P_n(x)x^me^{-x}x^\alpha dx = N_n\delta_{n,m},$ for $m\leq n$, and $m,n
\geq 0$
with $N_n =\mathscr{L}[x^n P_n(x)] \neq 0$, for $n\geq 0$
Obtain an explicit expression for $N_n$.

REPLY [0 votes]: Hint: $$\int_0^\infty x^{p-1}e^{-x} dx = \Gamma(p),$$
where $\Gamma$ is the Gamma function.


TITLE: Riemann integral of product of sequence of function with a continuous function at a point.
QUESTION [4 upvotes]: Here is the setup to a questions I've been looking at. I mainly just need a hint on how to proceed from where I'm at. Or suggest a theorem/corollaries that would help.
The problem:
Let $\{g_n\}$ be a sequence of nonnegative Riemann integrable function on $[-1,1]$ satisfying

$\int_{-1}^{1} g_n(x) dx = 1$ for $n\in\mathbb{N}^+$
For every $\delta>0$, $g_n \to 0$ uniformly on $[-1,-\delta] \cup [\delta, 1]$ as $n\to \infty$.

Prove that if $\psi:[-1,1] \to \mathbb{R}$ is Riemann integrable and continuous at $x=0$, then
$$
\lim_{n\to \infty} \int_{-1}^{1} \psi(x)g_n(x)dx = \psi(0).
$$
Attempt at solution
So by the properties of the integral and limit, and uniform convergence of $g_n$ we have that
$$
\lim_{n\to \infty} \int_{-1}^{1} \psi(x)g_n(x) dx = \lim_{n\to \infty} \int_{-1}^{-\delta} \psi(x)g_n(x) dx + \lim_{n\to \infty} \int_{-\delta}^{\delta} \psi(x)g_n(x) dx + \lim_{n\to \infty} \int_{\delta}^{1} \psi(x)g_n(x) dx \\
=\lim_{n\to \infty} \int_{-\delta}^{\delta} \psi(x)g_n(x) dx.
$$
Question
Am I correct in the above computations?
If so then it is clear here that I have to use condition "1" along with continuity of $\psi$ at $x=0$. However, this is where I'm entirely lost and not sure how to approach the integral
$$
\lim_{n\to \infty} \int_{-\delta}^{\delta} \psi(x)g_n(x) dx
$$

REPLY [5 votes]: Since $\psi$ is continuous at $x=0$, for any $\epsilon>0$, there is a $\delta>0$ that for any $|x|<\delta$, $\:|\psi(x)-\psi(0)|<\epsilon$. Also, since $\psi$ is Riemann integrable, on $[-1,-\delta] \cup [\delta, 1]$, $|\psi(x)-\psi(0)|<M$. Thus
\begin{align}
\left|\int_{-1}^{1} \psi(x)\,g_n(x)\,dx-\psi(0)\right|&=\left|\int_{-1}^{1} \psi(x)\,g_n(x)\,dx-\int_{-1}^{1} \psi(0)\,g_n(x)\,dx\right|
\\
&\leqslant\int_{-1}^{1} |\psi(x)-\psi(0)|\,g_n(x)\,dx
\\
&=\int_{-1}^{-\delta} |\psi(x)-\psi(0)|\,g_n(x)\,dx+\int_{-\delta}^{\delta} |\psi(x)-\psi(0)|\,g_n(x)\,dx+\int_{\delta}^{1} |\psi(x)-\psi(0)|\,g_n(x)\,dx
\\
&<M\int_{-1}^{-\delta} g_n(x)\,dx+ \epsilon \int_{-\delta}^{\delta} g_n(x)\,dx+M\int_{\delta}^{1} g_n(x)\,dx
\\
&<\epsilon + M\int_{-1}^{-\delta} g_n(x)\,dx+ M\int_{\delta}^{1} g_n(x)\,dx
\end{align}
Since $g_n \to 0$ uniformly on $[-1,-\delta] \cup [\delta, 1]$, there is a $N$ that for any $n>N,$
$$
\int_{-1}^{-\delta} g_n(x)\,dx<\epsilon/M\quad\text{and}\quad \int_{\delta}^{1} g_n(x)\,dx<\epsilon/M
$$
Thus
$$
\left|\int_{-1}^{1} \psi(x)\,g_n(x)\,dx-\psi(0)\right|<3\epsilon\quad\text{and}\quad\lim_{n\to \infty}\,\int_{-1}^{1} \psi(x)\,g_n(x)\,dx=\psi(0)
$$


TITLE: How can I prove this proposition from Peano Axioms: (Cancellation law). Let a, b, c be natural numbers such that a + b = a + c. Then we have b = c.
QUESTION [1 upvotes]: Peano Axioms.
Axiom 2.1
$0$ is a natural number.
Axiom 2.2
If $n$ is a natural number then $n++$ is also a natural number. (Here $n++$
  denotes the successor of $n$ and previously in the book the notational
  implication has been bijected to the familiar $1,2…$).
Axiom 2.3
$0$ is not the successor of natural number; i.e. we have $n++≠0$ for every
  natural number $n$.
Axiom 2.4
Different natural numbers must have different successors; i.e., if $n,m$
  are natural numbers and $n≠m$, then $n++≠m++$.
Axiom 2.5
Let $P(n)$ be any property pertaining to a natural number $n$. Suppose
  that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n++)$ is
  also true. Then $P(n)$ is true for every natural number $n$.
Definition of Addition: Let m be a natural number. We define, $0+m=m$
  and suppose we have inductively defined the addtion $n+m$ then we
  define, $(n++)+m=(n+m)++$. Where $n++$ is the successor of $n$.

Terence Tao has a proof in his Analysis I book, but I couldn't understand . I want a proof line by line like this:
Thm: 3 is a natural number
\begin{align*}
   & 0 \text{ is natural } && \text{Axiom 2.1 }\\
   & 0++ = 1 \text{ is natural } && \text{Axiom 2.2}\\
   & 1++ = 2 \text{ is natural } && \text{Axiom 2.2}\\
   & 2++ = 3 \text{ is natural } && \text{Axiom 2.2}
   \end{align*}

REPLY [0 votes]: Here is a formal proof done in Fitch:
(Note: I use $s(x)$ instead of $x$++)


TITLE: Pontryagin Forms and Special Holonomy
QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian manifold. Recall that the $k^{th}$-Pontryagin class is a topological invariant which, by classical Chern-Weil theory, can be represented using the so-called Pontryagin forms. These forms are obtained by appropriate constant linear combinations and wedge multiplications of the closed $4p$-forms $\operatorname{Tr}(R^{p})$, where
$$\operatorname{Tr}(R^{p})(X_1,....,X_{4p}):=\sum_{\sigma}(-1)^{|\sigma|}\operatorname{Tr}(R(X_{\sigma(1)},X_{\sigma(2)})\circ\cdots\circ R(X_{\sigma(4k-1)},X_{\sigma(4p)}))$$
and $R$ is the curvature tensor of $(M,g)$. It is known that the Pontryagin forms depend only on the Weyl part of the curvature tensor (in particular they vanish for conformally flat manifolds).
In concrete cases, they can in principle be explicitly computed.
However I am interested in the case where $M$ has special holonomy (contained in) $SU(5)$, i.e. it is a not necessarily compact Calabi-Yau $5$-fold. My question is:

Are there any conditions under which the Pontryagin forms (or, equivalently, the above trace forms) are proportional to the appropriate powers of the Kahler form $\omega$?
More generally, are there any conditions under which the Pontryagin forms are parallel?

Note that the $k^{th}$-Pontryagin form of a Kahler manifold is always of type $(2k,2k)$ so that, when the value $h^{2k,2k}$ of the Hodge diamond is one, this form is necessarily a multiple of $\omega^{2k}$ up to exact terms.
Any example using algebraic geometry or any argument which might enlighten some properties (if any) of Pontryagin forms for manifolds with special holonomy in the sense of Berger's classification are appreciated.

REPLY [4 votes]: Here are a few remarks to show that your Questions 1 and 2 are almost equivalent.  (What the answers are is another matter, and I expect that to be somewhat difficult, as I'll explain below.)  Of course, Question 1 is a special case of Question 2 in the OP's situation because, if a closed form is a scalar multiple of $\omega^2$ or $\omega^4$ on $M^{10}$, then that scalar multiple must be constant, and hence the form is parallel.  What's interesting is that, in spite of appearances, having the Chern-Weil representatives of the Pontryagin forms be parallel when the holonomy is a subgroup of $\mathrm{SU}(5)$ is nearly the same as having these forms be scalar multiples of $\omega^{2k}$.
First, for Question 2, you can reduce to the case in which the holonomy acts irreducibly:  If $(M,g)$ is a (simply-connected) Riemannian manifold whose holonomy preserves a parallel splitting $TM = V_1\oplus V_2$, then, locally, $g$ is a product metric and the formula for the total Pontryagin class $p(TM) = p(V_1)p(V_2)$ and that fact that the holonomy is the product of the holonomies acting on the two subbundles $V_1$ and $V_2$ shows that the Chern-Weil representatives of the classes in $p(TM)$ are parallel with respect to the holonomy of the product if and only if each of the two factor metrics has the property that its Chern-Weil representatives of the classes $p(V_i)$ are also parallel.  By reduction, then, it suffices to classify the metrics $(M,g)$ for which the Pontryagin forms are parallel and the holonomy acts irreducibly.
Second, although you say you are mainly interested in the case of an $(M^{10},g)$ whose holonomy is (contained in) $\mathrm{SU}(5)$, you might as well generalize your question to considering the case of $(M^{2n},g)$ with holonomy (contained in) $\mathrm{SU}(n)$.  Since, by the first remark, we can assume the holonomy acts irreducibly, that means, by the Berger classification, that the holonomy has to be either $\mathrm{SU}(n)$ or, if $n$ is even, $\mathrm{Sp}(n/2)$.
In the case the holonomy is $\mathrm{SU}(n)$, the only parallel forms of type $(2k,2k)$ are the (even) powers of the Kähler form $\omega$, so, in this case, having the Pontryagin forms be parallel implies having them be (constant) multiples of the  powers of $\omega$.  Meanwhile, for $k<n$ the only multiples of $\omega^k$ that are closed are the constant multiples.  In particular, when $n=5$, if the Chern-Weil representative of $p_1(M)$ (respectively,$p_2(M)$) is a multiple of $\omega^2$ (respectively, $\omega^4$), then it is a constant multiple, and hence parallel.
In the case the holonomy is $\mathrm{Sp}(n/2)$, things are a little more complicated:  In this case, the algebra of parallel forms is generated by $\omega$, $\phi$, and $\bar\phi$, where $\phi$ is a parallel holomorphic $(2,0)$-form.  Thus, the parallel $(2k,2k)$-forms are constant linear combinations of the forms $\omega^{2(k-j)}\wedge\phi^j\wedge\bar\phi^j$ for $0\le j\le k\le n/2$.  Since we only care about the cases $n\le 5$ for the OP's problem, we really only have to deal with $n=2$ and $n=4$ in this situation.  When $n=2$, since $\mathrm{Sp}(1)=\mathrm{SU}(2)$, this case is already handled. When $n=4$, the only case to deal with is $k=1$, where there are two parallel $(2,2)$-forms, $\omega^2$ and $\phi\wedge\bar\phi$.  However, in this case, because there is a $2$-sphere of parallel complex structures and because the Pontryagin form $p_1(M,g)$ will have to be of type $(2,2)$ with respect to all the complex structures of this family, one finds that, if the Chern-Weil Pontryagin form $p_1(M,g)$ is to be parallel, it must be a constant multiple of $\omega^2 + \phi\wedge\bar\phi$.  (Here, I am assuming that $\phi$ and $\omega$ have been suitably normalized.)  This multiple can't be zero because it is essentially the squared norm of the curvature; if it were zero, the metric would be flat, and the holonomy wouldn't be $\mathrm{Sp}(2)$.
Thus, the only case that you care about in which the two Questions essentially differ is the case of an $(M^{10},g)$ that is a product  $(N^8,h)\times (\mathbb{C},g_0)$ where $g_0$ is the flat metric and $(N^8,h)$ has holonomy $\mathrm{Sp}(2)$ with $p_1(N,h)$ being a (constant) multiple of $\omega^2 + \phi\wedge\bar\phi$ and $p_2(N,h)$ being a constant multiple of the volume form.
Finally, there is the question of whether any of these irreducible cases can actually occur.  Here, I don't have much to say.  I do not know of a single example of even a local, non-flat  Calabi-Yau metric for which the square norm of the curvature tensor is constant, even in complex dimension $2$, which is the very first nontrivial case.  The problem gets more and more overdetermined as the dimension goes up, so it seems unlikely that such metrics exist, even locally, but I think that actually writing down a proof might not be easy.


TITLE: Solve for x Given Two Square Roots: Algebra Problem
QUESTION [0 upvotes]: I am trying to solve for all real numbers for $x$ given $5=\sqrt{9-x^2}+\sqrt{16-x^2}$.
The answer is that $x=\pm \frac{12}{5}$. I am looking for a clean way to do it. I am stumped, and I feel like I am missing something about conjugates here.

REPLY [1 votes]: There is no clean way that I know of. We have
$5=\sqrt{9-x^2}+\sqrt{16-x^2} \implies $
$5-\sqrt{9-x^2}=\sqrt{16-x^2} \implies $
$(5-\sqrt{9-x^2})^2=16-x^2 \implies $
$25+9-x^2-10\sqrt{9-x^2} = 16-x^2 \implies $
$34-x^2-10\sqrt{9-x^2}  = 16-x^2 \implies $
$18 =10\sqrt{9-x^2} \implies $
$\frac{9}{5} = \sqrt{9-x^2} \implies $
$\frac{81}{25} = 9-x^2 \implies $
$ \frac{-144}{25} = -x^2  \implies $
$ x = \pm \frac{12}{5} $


TITLE: About the definition of orbifolds
QUESTION [1 upvotes]: I am new to orbifolds. By reading the definition (classical ones, not in terms of stacks or groupoids), I am wondering why only finite group action is allowed in the definition of local charts. I am expecting that a quotient manifold is an orbifold where the action is proper. However, it seems that it is not the case unless the action is, in fact, properly discontinuous. It seems that the crux here is that stabilizers cannot be too big and need to be finite.
My questions are:

is the finite group action absolutely necessary, or is it merely a technical convenience?
What happens if compact groups are allowed in local charts? Is there already such generalization?

Thanks.

REPLY [4 votes]: One purpose of the definition of an orbifold was to be able to describe quotient objects of general properly discontinuous group actions, such as triangle group actions on the 2-sphere, on the euclidean plane, or on the hyperbolic plane.
In the special case where the group action is also free, a very nice theory already existed, namely the theory of regular covering maps of manifolds. In that situation, every point stabilizer is trivial.
In the general case where the group action may not be free, proper discontinuity implies that the point stabilizers are finite. Those point stabilizers of the group action become the finite groups that label points in the singular locus of the quotient orbifold. For a properly discontinuous group action on a surface, for example, its easy to prove that the groups that can arise as point stabilizers are finite cyclic groups and finite dihedral groups. For that reason, the groups that label the singular locus of an orbifold of dimension 2 are all either finite cyclic groups or finite dihedral groups.


TITLE: How do I notate this statement about a state of affairs (similar to a possible world)?
QUESTION [3 upvotes]: I'd like to notate this statement formally:
If any given agent desires that a certain state of affairs obtains, then there is no state of affairs in which she enjoys greater security than that one.
$ \forall{a}\forall s_1 \in S (\mathbf{D_a}Os_{1} \rightarrow \neg \exists s_2 \in S...) $
In the ellipsis, I want to express 'such that the security, $c$, of $a$, in $s_{2}$ is greater than the $c$ of $a$ in $s_1$, although I don't know how to represent that notationally.
How might I notate this? Even speculative answers would help.
Thank you.
-Hal

REPLY [1 votes]: I would go for something like this (assuming we have a set of actors $A$, a set of states $S$, and a set of levels of security $L$):
$$\forall a \in A \forall s \in S( aDs \to \neg \exists s'\in S (c(a,s') > c(a,s))$$
where $D \subseteq A \times S$ is the "desires" relation, $c: A \times S \to L$ is the "level of security function" and $\mathord < \subseteq L \times L$ is the "ordering on levels of security".

The guiding principles in writing these things are:

Represent booleans as relation symbols;
Represent properties and mappings as function symbols.

It is often easier to use many-sorted expressions (in this case, we have the "sorts" $A, S, L$) to represent these things. Otherwise, you will have to tire yourself with unary predicates "$x$ is an actor", "$x$ is a state of affairs", etcetera, which only serve to obscure what you're trying to express. (But do note that conventional model theory usually assumes a single-sorted universe, so it might be useful to make this transition if you want to apply model-theoretic results.)


TITLE: What graph is denoted by I_t?
QUESTION [1 upvotes]: I'm reading a paper (found here paper) and a type of graph is denoted as $I_t$. What kind of graph is meant by this?

REPLY [0 votes]: This is in fact defined in the paper. Midway through the third page, we've got:

What happens if the forbidden graph $H$ is not complete? Let $I_t$ be a graph on $t$ vertices with no edges, and for graphs $G$ and $H$, let $G + H$ be a graph obtained from the disjoint union of $G$ and $H$ by adding an edge between each vertex of $G$ and each vertex of $H$. For positive integers $s$ and $t$, let $K^∗_{s,t}$ be a graph obtained from $K_s + I_t$ by subdividing every edge joining vertices of the subgraph $K_s$ once.

In general, you should expect notation that's not completely standard to be defined somewhere early on in the paper!


TITLE: Marbles Probability question using Bayes Rule
QUESTION [0 upvotes]: Marbles Probability question using Bayes Rule
I have this question from my revision booklet that I cant understand.
A bag initially contained 4 white marbles and 9 black marbles. A marble was drawn and replaced by two marbles of the opposite color. Then another marble was drawn and also replaced by two marbles of the opposite color. A third marble was then drawn.
Find the probability that the first marble drawn was white if:
a) The second marble drawn was white
b) The third marble drawn was white
c) Both the second and third marbles drawn were white
My Attempt:
For a I used Bayes rule for a and computed the various probabilities and found out that a is 2/11. I'm not at all sure about b and c.

REPLY [1 votes]: Let $W_k$ be the event of drawing a white marble on the $k$-th draw, and $B_k$ the complementary event.
So your solution should look somewhat like:
(a) $$\begin{align}\mathsf P(W_1\mid W_2)&=\dfrac{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)}{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)+\mathsf P(B_1)~\mathsf P(W_2\mid B_1)}\\[1ex]&=\dfrac{\tfrac{4}{4+9}\tfrac{3}{3+11}}{\tfrac{4}{4+9}\tfrac{3}{3+11}+\tfrac{9}{4+9}\tfrac{6}{6+8}}&=\dfrac{4(4-1)}{4(4-1)+9(4+2)}\\[1ex]&=\dfrac{2}{11}&\checkmark\end{align}$$
The others use Bayes' Rule too.  Hint: tackle (c) first.
(b) $$\begin{align}\mathsf P(W_1\mid W_3)&=\dfrac{\mathsf P(W_1,W_2,W_3)+\mathsf P(W_1,B_2,W_3)}{\mathsf P(W_1,W_2,W_3)+\mathsf P(W_1,B_2,W_3)+\mathsf P(B_1,W_2,W_3)+\mathsf P(B_1,B_2,W_3)}\\[2ex]&=\tfrac{\mathsf P(W_1)~\big(\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)+\mathsf P(B_2\mid W_1)~\mathsf P(W_3\mid W_1,B_2)\big)}{{\mathsf P(W_1)~\big(\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)} + {\mathsf P(B_2\mid W_1)~\mathsf P(W_3\mid W_1,B_2)\big)+\mathsf P(B_1)~\big(\mathsf P(W_2\mid B_1)~\mathsf P(W_3\mid B_1,W_2)+\mathsf P(B_2\mid B_1)~\mathsf P(W_3\mid B_1,B_2)\big)}}\end{align}$$
(c) $$\begin{align}\mathsf P(W_1\mid W_2,W_3)&=\dfrac{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)}{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)+\mathsf P(B_1)~\mathsf P(W_2\mid B_1)~\mathsf P(W_3\mid B_1,W_2)}\end{align}$$


TITLE: Pigeon hole principle Cartesian product?
QUESTION [0 upvotes]: Let $A = \{0, 1, 2\}^{10}$. Show that every subset of $A$ with more than $1200$ elements contain two elements in which the $0’s$ appear in exactly the same positions.
Ok. so A is just a subset of ordered pairs. Of which there are $9$ possibilities. I guess I could group them by which have the same locations of $0's$? But from there I'm lost

REPLY [0 votes]: We count how many different ways a subset can contain $0$'s, where the placement of $0$ matters.
Every spot in the ordered pair either does not have a $0$ or has a $0$ (it does not matter whether the spot has a $1$ or $2$ in it). Therefore, the number of ways to arrange where the $0$'s are in an ordered pair is $2^{10}=1024$.
By Pigeonhole, we get that there must be at least $\left\lceil\frac{1200}{1024}\right\rceil=2$ ordered pairs with the $0$'s in the same places.