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TITLE: Prove that I -xx* is singular if and only x*x = 1
QUESTION [0 upvotes]: (=>) Suppose I - xx* is singular if and only there is a y such that (I−xx*)y=0, i.e. xx* y=y. Now set λ=x* y. Then y=λx, i.e. xx* λx=λx
Thus λx(x* x) = λx => x* x = 1
(<=) Suppose x*x = 1
Please help
REPLY [1 votes]: Just start from the basics. $I-xx^*$ is singular if and only there is a $y$ such that $(I-xx^*)y=0$, i.e. such that $xx^*y=y$. Now set $\lambda=x^*y$. Then $y=\lambda x$, i.e. $xx^*\lambda x=\lambda x$, and you have almost proved one direction.
Make sure that you are aware at each step which expressions are scalars. | 0.117905 |
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5 comments ↓
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frompo.com | 0.225772 |
\begin{document}
\begin{frontmatter}
\author{Mahouton Norbert Hounkonnou\corref{cor1}${}^1$}
\cortext[cor1]{norbert.hounkonnou@cipma.uac.bj}
\author{Sama Arjika\corref{cor2}${}^1$}
\cortext[cor2]{rjksama2008@gmail.com}
\author{ Won Sang Chung\corref{cor3}${}^2$ }
\cortext[cor3]{mimip4444@hanmail.net}
\title{\bf New families of
$q$ and $(q;p)-$Hermite polynomials
}
\address{${}^1$International Chair of Mathematical Physics
and Applications \\
(ICMPA-UNESCO Chair), University of
Abomey-Calavi,\\
072 B. P.: 50 Cotonou, Republic of Benin,\\
${}^2$Department of Physics and Research Institute of Natural Science, \\
College of Natural Science, \\
Gyeongsang National University, Jinju 660-701, Korea
}
\begin{abstract}
In this paper, we construct
a new family of $q-$Hermite polynomials denoted by
$H_n(x,s|q).$ Main properties and relations
are
established and proved.
In addition, is deduced a sequence of novel polynomials,
$\mathcal{L}_n(\cdot,\cdot |q),$ which appear to be connected with well known $(q,n)-$exponential functions $E_{q,n}(\cdot)$\, introduced by Ernst in his work entitled: {\it A New Method for $q-$calculus,} (Uppsala Dissertations in
Mathematics, Vol. {\bf 25}, 2002). Relevant results spread in the literature
are retrieved as particular cases. Fourier integral transforms are explicitly computed and discussed. A $(q;p)-$extension of the $H_n(x,s|q)$ is also provided.
\end{abstract}
\begin{keyword}
Hermite polynomials, $q-$Hermite polynomials, generating function, $q-$derivative, inversion formula, Fourier integral transform
\end{keyword}
\end{frontmatter}
\section{Introduction}
\label{int1}
The classical orthogonal polynomials and the quantum orthogonal polynomials,
also called $q-$orthogonal polynomials, constitute an interesting set of
special functions.
Each family of these polynomials occupies different levels
within the so-called Askey-Wilson scheme (Askey and Wilson, 1985; Koekoek and Swarttouw, 1998;
Lesky, 2005; Koekoek et {\it al}, 2010).
In this scheme, the Hermite polynomials
$H_n(x)$ are the ground level and are characterized
by a set of properties:
(i) they are solutions of a hypergeometric second order differential equation,
(ii) they are generated by a recursion relation,
(iii) they are orthogonal with respect to a weight function and
(iv) they obey the Rodrigues-type formula.
Therefore, there are many ways to construct the Hermite
polynomials. However, they are more commonly
deduced from their
generating function, i.e.,
\be
\sum_{n=0}^\infty\frac{{\bf H}_n(x)}{n!}t^n=e^{2xt-t^2}
\ee
giving rise to the so-called {\it physicists} Hermite polynomials \cite{Johann Cigler}.
Another family of Hermite polynomials, called
the {\it probabilists} Hermite polynomials, is defined as \cite{Johann Cigler}
\be
\sum_{n=0}^\infty\frac{H_n(x)}{n!}t^n=e^{xt-\frac{t^2}{2}}.
\ee
The Hermite polynomials are at the bottom of a
large class of hypergeometric polynomials
to which most of their properties can be
generalized \cite{Bukweli&Hounkonnou12a}, \cite{Dattoli}-\cite{Habibullah}.
In \cite{Johann Cigler}, Cigler introduced
another family of Hermite polynomials $H_n(x,s)$ generalizing
the {\it physicists} and {\it probabilists} Hermite polynomials as
\be
\label{samamasaaa}
\sum_{n=0}^\infty\frac{H_n(x,s)}{n!}t^n=e^{xt-s\frac{t^2}{2}}
\ee
with $H_n(x,1)=H_n(x)$ and $H_n(2x,2)={\bf H}_n(x).$
In this work, we deal with a construction of two new families of $q$ and $(q;p)-$Hermite
polynomials.
The paper is organized as follows. In Section \ref{sction1}, we
give a quick overview on
the Hermite polynomials $H_n(x,s)$ introduced in \cite{Johann Cigler}.
Section \ref{sction2} is devoted to the construction of
a new family of $q-$Hermite polynomials $H_n(x,s|q)$
generalizing the
discrete $q-$Hermite polynomials. The
inversion formula and relevant properties of these polynomials are computed and discussed. Their Fourier integral
transforms are performed in the Section \ref{sction3}.
Doubly indexed Hermite polynomials and some concluding remarks are introduced in Section \ref{sction4}.
\section{On the Hermite polynomials $H_n(x,s)$}
\label{sction1}
In \cite{Johann Cigler}, Cigler showed that the Hermite polynomials $H_n(x,s)$ satisfy
\be
\label{4Hermdiff}
DH_n(x,s)= n\, H_{n-1}(x,s)
\ee
and
the
three term recursion relation
\be\label{4Hermttr}
H_{n+1}(x,s)=x\, H_n(x,s)-s\,n \,H_{n-1}(x,s),\quad n\geq 1
\ee
with $H_0(x,s):=1.\;D:=d/dx$ is the usual differential operator. Immediatly, one can see that
\be
\label{sama:initial}
H_{2n}(0,s)=(-s)^{n}\prod_{k=1}^n(2k-1),\quad H_{2n+1}(0,s)=0.
\ee
The computation of the first fourth polynomials gives:
\be
H_{1}(x,s)= x,\;
H_{2}(x,s)=x^2-s,\;
H_{3}(x,s)=x^3-3\,s\,x,\;
H_{4}(x,s)=x^4-6\,s\,x^2+3\,s^2.
\ee
More generally, the explicit formula of $H_{n}(x,s)$ is written as
\cite{Johann Cigler}
\be
\label{sasama}
H_{n}(x,s)=n!\sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{ (-1)^k\,s^k }{ (2k)!! } \frac{ x^{n-2k} }{(n-2k)! }
=x^n\,{}_2F_0\Bigg(\begin{array}{c}-\frac{n}{2},\frac{1-n}{2}
\\-\end{array}\Big|\;-\frac{2s}{x^2}\Bigg),
\ee
where $({}^n_k)=n!/k!(n-k)!$ is a binomial coefficient,
$n!:=n(n-1)\cdots 2\cdot1,\; (2n)!!:=2n(2n-2)\cdots 2.$\\\\
The symbol $\lfloor\,x\,\rfloor$ denotes the greatest integer
in $x$ and ${}_2F_0$ is called the hypergeometric
series \cite{ASK}.
From
(\ref{4Hermdiff}) and (\ref{4Hermttr}), we have
\be
H_{n }(x,s)= (x-sD)\,H_{n-1 }(x,s),
\ee
where the operator $x-sD $ can be expressed as
\cite{Johann Cigler}
\be
x-s D = e^{\frac{x^2}{2s}} (-sD)\,e^{-\frac{x^2}{2s}}.
\ee
The
Rodrigues formula takes the form
\be
\label{sama:abelmarie}
e^{-\frac{x^2}{2s}}\,H_n (x,s)= (-sD)^n\, e^{-\frac{x^2}{2s}}
\ee
while the second
order
differential equation satisfied by $H_n(x,s)$ is
\be
\label{second}
\big(sD^2- x D+ n\big)\,H_n(x,s)=0.
\ee
Furthermore, from the relation (\ref{sasama}) we derive the result
\be
\label{labelsaj}
H_n(x+sD,s)\cdot(1)=x^n,
\ee
and the inverse formula for $H_n(x,s)$
\be
\label{labelsaaf}
x^n=n!\,\sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{s^k }{ (2k)!! } \frac{ H_{n-2k}(x,s) }{(n-2k)! }.
\ee
We then obtain
\be
\label{props}
\sum_{k,\,n \,(even)}
\frac{ 1 }{(n-k)!\,k! }=\sum_{k,\,n \,(odd)}
\frac{ 1 }{(n-k)!\,k! },\quad 0\leq k\leq n,\quad n\geq 0.
\ee
From (\ref{sasama}), it is also straighforward to note that the polynomials
$H_n(x,s)$ have an alternative expression given by
\be
\label{mirer}
H_n(x,s)=\exp\left( -s\frac{D^2}{2}\right)\cdot (x^n).
\ee
For any integer $k=0, 1, ..., \lfloor\,n/2\,\rfloor,$ we have the following result
\be
\label{e7j}
D^{2k}\,H_n(x,s)=\frac{n!}{(n-2k)!} H_{n-2k}(x,s).
\ee
\begin{corollary}
The Hermite polynomials $H_n(x,s)$ obey
\be
\label{nversion}
\mathcal{T}_n(s, D)\,H_n(x,s)=x^n
\ee
where the polynomial
\be
\label{samapolynom}
\mathcal{T}_n(\alpha,\beta)= \sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{ 1 }{ (2k)!! } \alpha^k\beta^{2k}.
\ee
\end{corollary}
We are now in a position to formulate and prove the following.
\begin{lemma}
\label{malem}
\be
\mathcal{T}_{2n}(\alpha,\beta)=\frac{(\alpha\beta^2)^n}{(2n)!!}\,{}_2F_0\Bigg(
\begin{array}{c}-n,1\\
-\end{array}\Big|-\frac{2}{\alpha\beta^2}\Bigg)
\ee
and
\be
\mathcal{T}_\infty(\alpha,\beta)=e^{\frac{\alpha\beta^2}{2}}.
\ee
\end{lemma}
{\bf Proof.} From (\ref{samapolynom}), we have
\bea
\label{calcal}
\mathcal{T}_{2n}(\alpha,\beta)&=&\sum_{k=0}^{n}\frac{1}{(2k)!!}(\alpha\beta^2)^{k}\cr
&=&\frac{(\alpha\beta^2)^n}{(2n)!!}\sum_{k=n}^{\infty}\frac{(2n)!!}{(2k)!!}(\alpha\beta^2)^{k-n}.
\eea
By substituting $m=n-k$ in the latter expression and using various identities, we arrive at
\be
\label{calcale}
\mathcal{T}_{2n}(\alpha,\beta)
=\frac{(\alpha\beta^2)^n}{(2n)!!}\sum_{m=0}^{\infty}(-n)_m\;\left(\frac{-2}{\alpha\beta^2}\right)^{m},
\ee
where $(a)_j:=a(a+1)\cdots(a+j-1),\; j \geq 1$ and $(a)_0:=1.$
When $n$ goes to $\infty,$ the polynomial (\ref{samapolynom}) takes the form
\be
\mathcal{T}_\infty(\alpha,\beta)= \sum_{k=0}^{\infty}
\frac{\alpha^k\beta^{2k} }{ (2k)!! } = \sum_{k=0}^{\infty}
\frac{ 1 }{ k!}\left(\frac{\alpha\beta^2}{2}\right)^k
\ee
where $(2k)!!=2^k\, k!$ is used. $\square$
To end this section, let us investigate the Fourier transform of the function $e^{-x^2/2s}H_n(x,s)$. In
\cite{Johann Cigler}, Cigler has proven that
\be
\label{samafouri}
\frac{1}{\sqrt{2\pi s}}\int_{\mathbb{R}}e^{i x y-\frac{x^2}{2s}} d x=
e^{-s\frac{y^2}{2}}.
\ee
Hence,
\be
\label{samaM}
\frac{1}{\sqrt{2\pi\, s}}\int_{\mathbb{R}}e^{i x y+ i(n- 2k)
\kappa x- \frac{x^2}{2s}} d x
=e^{-s\frac{y^2}{2}-(n- 2k)s\,y\,\kappa},
\ee
where $e^{-2s\kappa^2}=1$.
By differentiating the relation (\ref{samafouri}) $2n-2k$ times with respect to $y$, one obtains
\be
\frac{1}{\sqrt{2\pi \,s}}\int_{\mathbb{R}}(-1)^{n-k}x^{2n-2k}e^{ix y-\frac{x^2}{2s}} d x=
D^{2n-2k} e^{-s\frac{y^2}{2}}.
\ee
Evaluating the latter expression at $y=0$ and by making use of (\ref{sama:abelmarie}), one gets
\be
\label{tate}
\frac{(-1)^{n-k}}{\sqrt{2\pi \,s}}\int_{\mathbb{R}} x^{2n-2k}e^{-\frac{x^2}{2s}} d x=
{D^{2n-2k} e^{-s\frac{y^2}{2}}}_{\big|y=0}={(-s)^{2n-2k}H_{2n-2k}(y,s^{-1})e^{-s\frac{y^2}{2}}}_{\big|y=0}.
\ee
\\
\begin{theorem}
\label{sama:propo1}
The Fourier transform of the function $e^{-x^2/2 s}H_{n}(x,s )$ is given by
\be
\label{samafourier}
\frac{1 }{\sqrt{2\pi \,s}}\int_{\mathbb{R}}H_{n} (a\, e^{i \kappa x},s)e
^{ix y-\frac{x^2}{2s}}d x= H_{n}
(a\, e^{-s\,\kappa\, y}, s ) e^{-s\frac{y^2}{2}}
\ee
where $a$ is an arbitrary constant factor. For $y=0,$ we have
\be
\label{sama:fourier}
\frac{1 }{\sqrt{2\pi\, s}}\int_{\mathbb{R}}H_{n} (x,s)e
^{-\frac{x^2}{2s}}d x= 0.
\ee
\end{theorem}
{\bf Proof.}
Using (\ref{sasama}) and (\ref{samaM}), we obtain
\bea
\frac{1}{\sqrt{2\pi\, s}}\int_{\mathbb{R}}H_{n}(a \,e^{i\kappa x},s)
e^{ix y-\frac{x^2}{2\,s}}d x &=&\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^{k}n!\,s^k\,a^{n-2k}}{(n-2k)!\,(2k)!!}
\frac{1}{\sqrt{2\pi\, s}}\int_{\mathbb{R}}e^{i x y+ i(n- 2k)
\kappa x- \frac{x^2}{2s}} d x \cr
& =& \sum_{k=0}^{\lfloor
n/2\,\rfloor}\frac{(-1)^{k}\,n!\;s^k\,a^{n-2k}}{(n-2k)!\,(2k)!!}
e^{-\frac{s}{2}[\kappa (n-2k)+y]^2} \cr
& =&e^{-s\frac{y^2}{2}} H_{n} (a\, e^{-s\,\kappa\, y}
, s ).
\eea
Combining (\ref{sasama}) and (\ref{tate}) for $n=2n$, we have
\bea
\frac{1}{\sqrt{2\pi \,s}}\int_{\mathbb{R}}H_{2n}(x,s)
e^{-\frac{x^2}{2\,s}}d x &=&\sum_{k=0}^n
\frac{(-1)^k\,(2n)!\,s^k}{(2n-2k)!\,(2k)!!}\,
\frac{1}{\sqrt{2\pi\, s}}\int_{\mathbb{R}}x^{2n-2k}\,e^{ixy- \frac{x^2}{2\,s}} d x_{\big|y=0} \cr
& =&(-1)^n\sum_{k=0}^n
\frac{ (2n)!\,s^k}{(2n-2k)!\,(2k)!!}{ D^{2n-2k} e^{-s\frac{y^2}{2}}}_{\big|y=0} \cr
& =&(-1)^{n}\,s^{2n}\,e^{-s\frac{y^2}{2}}\sum_{k=0}^n
\frac{(2n)!\,s^{-k}}{(2n-2k)!\,(2k)!!}{H_{2n-2k}(y,s^{-1})}_{\big|y=0}\cr
& =&s^{2n}\,(2n)! \,\sum_{k=0}^n
\frac{(-1)^k}{(2n-2k)!!\,(2k)!!} \cr
&=&0
\eea
where (\ref{props}) is used. $\square$
\section{New $q-$Hermite polynomials $H_n(x,s|q)$}
\label{sction2}
In this section, we construct
through the $q-$chain rule
a new family of $q-$Hermite polynomials denoted by
$H_n(x,s|q).$
We first introduce some standard $q-$notations.
For $n\geq 1,\;q\in\mathbb{C}$,
we denote the $q-$deformed number \cite{Ernst} by
\be
\{n\}_{q}:=\sum_{k=0}^{n-1}q^k.
\ee
In the same way, we define the $q-$factorials
\be
\{n\}_{q}!:=\prod_{k=1}^n \{k\}_{q}, \quad \{2n\}_{q}!!:=\prod_{k=1}^n \{2k\}_{q},\quad \{2n-1\}_{q}!!:=\prod_{k=1}^n \{2k-1\}_{q}
\ee
and, by convention,
\be
\{0\}_{q}!:=1=: \{0\}_{q}!! \quad \mbox{ and } \quad \{-1\}_{q}!!=1.
\ee
For any positive number $c,$ the $q-$Pochhammer symbol $\{c\}_{n,q}$
is defined as follows:
\be
\{c\}_{n,q}:=\prod_{k=0}^{n-1} \{c+k\}_{q}
\ee
while the $q-$binomial coefficients are defined by
\bea
\label{samasama:sa}
{n\atopwithdelims\{\} k}_q:=\frac{ \{n\}_{q}!}{ \{n-k\}_{q}! \{k\}_{q}!}=\frac{(q;q)_n}{(q;q)_{n-k}(q;q)_k}\quad \mbox{ for }\quad
0\leq k\leq n,
\eea
and zero otherwise, where $(a;q)_n:=\prod_{k=0}^{n-1}(1-a\,q^k),\;(a;q)_0:=1.$
\begin{definition}\cite{HahnW, sama}
The Hahn $q-$addition $\op$ is the function: $\mathbb{C}^3\rightarrow \mathbb{C}^2$
given by:
\be
(x,y,q)\mapsto (x,y)\equiv x\op y,
\ee
where
\bea
\label{addition}
(x \op y)^n:&=&(x+y)(x+q \,y)\ldots (x+q^{n-1}\,y)\cr
&=&\sum_{k=0}^n{n
\atopwithdelims\{\} k}_q\, q^{({}^k_2)}\,x^{n-k}\,y^k,\quad n\geq 1,\quad (x\op y)^0:=1,
\eea
while the $q-$subtraction $\om$ is defined as follows:
\be
\label{additionm}
x \om y:= x\op(-y).
\ee
\end{definition}
Consider a function $F$
\bea
F:D_R \longrightarrow\mathbb{C},\quad z \longmapsto \sum_{n=0}^\infty c_n\, z^n,
\eea
where $D_R$ is a disc of radius $R.$ We define $F(x\op y)$ to mean the formal series
\bea
\sum_{n=0}^\infty c_n(x\op y)^n\equiv\sum_{n=0}^\infty \sum_{k=0}^n\,c_n\,{n
\atopwithdelims\{\} k}_q\, q^{({}^k_2)}\,x^{n-k}\,y^k.
\eea
Let $e_q,\;E_q,\; \cos_q$ and $\sin_q$ be the fonctions defined as follows:
\bea
e_q (x) :&=& \sum_{n=0}^{\infty} \frac{1}{ \{n\}_{q}!}x^n\\
E_q (x) :&=& \sum_{n=0}^{\infty} \frac{q^{n(n-1)/2}}{ \{n\}_{q}!}x^n\cr
\cos_q(x):&=&\frac{e_q (i\,x)+e_q (-i\,x)}{2}= \sum_{n=0}^{\infty} \frac{(-1)^n}{ \{2n\}_{q}!}x^{2n}\\
\sin_q(x):&=&\frac{e_q (i\,x)-e_q (-i\,x)}{2\,i}= \sum_{n=0}^{\infty} \frac{(-1)^n}{ \{2n+1\}_{q}!}x^{2n+1}.
\eea
We immediately obtain the following rules for the product of two exponential functions
\be
e_q (x) E_q (y) = e_q ( x \op y ).
\ee
The new family of $q-$Hermite polynomials $H_n(x,s|q) $ can be determined
by the generating function
\be
\label{functiongeneratrice}
e_q\big( tx \ominus_{q,q^2} st^2/ \{2\}_{q} \big) =
e_q( tx)E_{q^2} (- st^2/ \{2\}_{q} ) := \sum_{n=0}^{\infty} \frac{H_n(x,s|q) }{ \{n\}_{q}!}t^n,\quad |t| <1,
\ee
where \cite{sama}
\be
(a\ominus _{q, q^{2}}b )^n := \sum_{k=0}^{n} \frac{ \{n\}_{q}!}{ \{n-k\}_{q}! \, \{k\}_{q^{2} }!}
(-1)^k q^{k(k-1)} a^{n-k} \,b^{k},\quad (a\ominus_{q, q^{2}} b )^0 := 1.
\ee
Performing the $q-$derivative $D_x^q$ of both sides
of (\ref{functiongeneratrice}) with respect to $x$, one obtains
\bea
\label{lowering}
D_x^q\, H_n(x,s|q)= \{n\}_{q}\, H_{n-1}(x,s|q),
\eea
where
\be
\label{lowm}
D_x^q\, f(x)=\frac{f(x)-f(qx)}{(1-q)x}
\ee
satisfying
\bea
D_x^q (a\, x \op b)^n= \{n\}_{q}\,(a\, x \op b)^{n-1}.
\eea
Recall \cite{Ismails} that the Al-Salam-Chihara polynomials $P_n(x;a,b,c)$ satisfy the following recursion relation:
\be
\label{sama:al}
P_{n+1}(x;a,b,c)=(x-a\,q^n)\,P_n(x;a,b,c)-(c+b\,q^{n-1})\, \{n\}_{q}\,P_{n-1}(x;a,b,c)
\ee
with $P_{-1}(x;a,b,c)=0$ and $P_0(x;a,b,c)=1$.\\
Performing the $q-$derivative
of both sides of
(\ref{functiongeneratrice}) with respect to $t$, we have
\bea
\label{raising}
H_{n+1}(x,s|q)= x\, H_{n}(x,s|q)-s\, \{n\}_{q} \,q^{n-1}\,H_{n-1}(x,s|q),\quad n\geq 1
\eea
with $H_0(x,s|q):=1.$ \\
By setting
$a=0=c$ and $b=s$ in (\ref{sama:al}), one obtains the recursion relation (\ref{raising}).
From the latter equation,
one can see that
\be
\label{sama:qinitial}
H_{2n}(0,s|q)=(-s)^n\,q^{n(n-1)}\, \{2n-1\}_{q}!!,\quad H_{2n+1}(0,s|q)=0.
\ee
The first fourth new polynomials are given by
\bea
H_1(x,s|q) &=& x,\\
H_2(x,s|q)& =& x^2 -s,\\
H_3(x,s|q)&=& x^3 - \{3\}_{q}s x,\\
H_4(x,s|q)&=& x^4 -(1+q^2) \{3\}_{q}sx^2+ q^2\,\{3\}_{q}s^2.
\eea
More generally, we have the following.
\begin{theorem}
The explicit formula for the new Hermite polynomials $H_n(x,s|q)$ is given by
\bea
\label{sama:qhermite}
H_n(x,s|q) &=& \sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{ (-1)^k q^{k(k-1)} \{n\}_{q}! }{ \{n-2k\}_{q}! \, \{2k\}_{q}!! }s^k x^{n-2k}\\
&=&x^n\,{}_2\phi_0\Bigg(\begin{array}{c}q^{-n},q^{1-n}\\
-\end{array}\Big|\;q^2;\;\frac{sq^{2n-1}}{(1-q)x^2}\Bigg),
\eea
where ${}_2\phi_0$ is the $q-$hypergeometric series \cite{ASK}.
\end{theorem}
{\bf Proof.} Expanding the generation function
given in (\ref{functiongeneratrice}) in Maclaurin series, we have
\bea
\label{sammas}
e_q( t\,x)E_{q^2} (- s\,t^2 /\{2\}_{q}) &=& \sum_{k=0}^{\infty} \frac{( x\,t)^k }{\{k\}_{q}!}
\sum_{m=0}^{\infty} \frac{(-1)^m q^{m(m-1)} }{\{m\}_{q^2} ! } \left(\frac{s\,t^2}{\{2\}_{q}}\right)^m \cr
&=& \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m q^{m(m-1)}
x^k }{\{k\}_{q}! \,\{m\}_{q^2} ! } \left(\frac{s}{\{2\}_{q}}\right)^m t^{k+ 2m}.
\eea
By substituting
\be
k +2m =n ~\Rightarrow~ m \le\lfloor\,n/2\,\rfloor,
\ee
and
\be
\{2\}_{q}\,\{m\}_{q^2}=\{2m\}_{q}
\ee
in (\ref{sammas}), we have
\be
e_q( t\,x)E_{q^2} (- s\,t^2/\{2\}_{q} ) = \sum_{n=0}^{\infty} \lb
\sum_{m=0}^{\lfloor\,n/2\,\rfloor} \frac{(-1)^m q^{m(m-1)}
s^m\,x^{n-2m} }{\{n-2m\}_{q}!\, \{2m\}_{q} !! }\rb t^{n},
\ee
which achieves the proof. $\square$
In the limit case when $x\to \{2\}_{q}\,x,\;s\to (1-q)\,\{2\}_{q},$ the
polynomials $H_n(x,s|q)$ are reduced to $H_n^q(x)$ investigated by Chung et {\it al} \cite{sama}. When $s\to 1-q$, they are
reduced to the discrete $q-$Hermite I polynomials \cite{ASK}.\\
The relation (\ref{raising}) allows us to write
\be
H_{n}(x,s|q)=(x-sq^N\circ D_x^q)\,H_{n-1}(x,s|q),
\ee
where
the operator $N$ acts on the polynomials $H_n(x,s|q)$ as follows:
\be
NH_n(x,s|q):= n\,H_n(x,s|q),\quad q^N\circ D_x^q=D_x^q\circ q^{N-1}.
\ee
It is straightforward to show that the polynomials (\ref{sama:qhermite}) satisfy the
following $q-$difference equation
\be
\label{sama:hounk}
\big(s\, (D_x^q)^2 -x\,q^{2-n}\,D_x^q+q^{2-n}\,\{n\}_{q}\big)\,H_n(x,s|q)=0.
\ee
In the limit case when $q$ goes to $1$, the $q-$difference
equation (\ref{sama:hounk}) reduces to the well-known differential equation (\ref{second}).
For $n$ even or odd, the polynomials $H_{n}(x,s|q)$ obey the following generating functions
\be
\sum_{n=0}^\infty\frac{H_{2n}(x,s|q)}{\{2n\}_{q}!}(-t)^n=\cos_q(x\sqrt{t})\,E_{q^2}(s\,t/\{2\}_{q}),\; |t| <1
\ee
or
\be
\sum_{n=0}^\infty\frac{H_{2n+1}(x,s|q)}{\{2n+1\}_{q}!}(-t)^n=
\frac{1}{\sqrt{t}}\sin_q(x\sqrt{t})\,E_{q^2}(s\,t/\{2\}_{q}), \; |t| <1,
\ee
respectively.
\begin{theorem}
\label{theo:poly}
The polynomials $H_n(x,s|q)$ can be expressed as
\be
H_n(x,s|q) =\prod_{k=1}^{n} \big( x -s\,q^{n-1-k}\, D_x^q \big)\cdot(1).
\ee
We also have
\bea
\label{sama:reso}
H_n \left(x+sq^N\circ D_x^q,s|q\right)\cdot(1)=x^n.
\eea
\end{theorem}
{\bf Proof.}
Since (\ref{lowering}) and (\ref{raising}) are satisfied, we have
\bea
H_{n}(x,s|q)&=& x\ H_{n-1}(x,s|q)-s\, q^{n-2}\, \{n-1\}_{q}\, H_{n-2}(x,s|q)\cr
&=& x \,H_{n-1}(x,s|q)-s q^{n-2}\,D_x^q\,H_{n-1}(x,s|q).
\eea
The rest holds by induction on $n$.\\
To prove the relation (\ref{sama:reso}) we replace $x^{n-2k}$ in (\ref{sama:qhermite}) by
$(x+sq^N\circ D_x^q)^{n-2k}$ and apply the corresponding linear operator
to $1$. The relation (\ref{sama:reso}) is true for $n=0$ and $n=1.$
For $n=2$, we have
\bea
H_2\left(x+sq^N\circ D_x^q,s|q\right)\cdot(1)&=&\left(x+sq^N\circ D_x^q\right)^2\cdot(1)-s\cr
&=&\left(x+sq^N\circ D_x^q\right)\cdot (x)-s\cr
&=&x^2.
\eea
Assume
that (\ref{sama:reso}) is true for $n-1,\;n\geq 3.$
Then we must prove that
\be
H_{n} \left(x+sq^N\circ D_x^q,s|q\right) \cdot(1)=x^{n}.
\ee
From
(\ref{raising}), we have
\bea
H_{n} \left(x+sq^N\circ D_x^q,s|q\right) \cdot1&=& \left(x+sq^N\circ D_x^q\right)H_{n-1}
\left(x+sq^N\circ D_x^q,s|q\right) \cdot(1)\cr
&-&s\{n-1\}_{q}\,q^{n-2}H_{n-2} \left(x+sq^N\circ D_x^q,s|q\right)\cdot(1)\cr
&=&\left(x+sq^N\circ D_x^q\right)\cdot x^{n-1}
-s\{n-1\}_{q}\,q^{n-2} x^{n-2}\cr
&=&x^{n}
\eea
which achieves the proof. $\square$\\
From the {\bf Theorem} \ref{theo:poly}, we obtain the following.
\begin{corollary}
\label{sama:theorem}
The polynomials (\ref{sama:qhermite}) have the following inversion formula
\be
\label{inversion}
x^n = \{n\}_{q}!\sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{ q^{k(k-1)} \;s^k }{ \{2k\}_{q}!! }\frac{ H_{n-2k}(x,s|q) }{\{n-2k\}_{q}! }.
\ee
\end{corollary}
{\bf Proof.} Let $h_n^q(x,s)$ be the polynomial defined by
\be
h_n^q(x,s)=\left(x+sq^N\circ D_x^q\right)^n\cdot(1).
\ee
Note that
$
h_n^q(x,-s)=H_n(x,s|q).
$
From (\ref{sama:reso}), we have
\bea
x^n&=&\sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{(-1)^k q^{k(k-1)}\{n\}_{q}! }{\{n-2k\}_{q}! \{2k\}_{q}!! }s^k
\left(x+sq^N\circ D_x^q\right)^{n-2k}\cdot(1)\cr
&=&\sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{ q^{k(k-1)}\{n\}_{q}! \,s^k }{\{n-2k\}_{q}!\{2k\}_{q}!! } h_{n-2k}^q(x,-s)
\eea
which achives the proof.
$\square$
From (\ref{lowering}), one readily deduces that, for integer
powers $k=0, 1, ..., \lfloor\,n/2\,\rfloor$ of the operator
$D_x^q,$
\be
\label{e7}
(D_x^q)^{2k}H_n(x,s|q)=\gamma_{n,k}(q)H_{n-2k}(x,s|q),\quad
\gamma_{n,k}(q)=\frac{\{n\}_{q}!}{\{n-2k\}_{q}!}.
\ee
Therefore, we have the following decomposition of unity
\be
\sum_{k=0}^{\lfloor\,n/2\,\rfloor}\frac{ (-1)^kq^{k(k-1)} s^k }{ \{2k\}_{q}!!
}(D_x^q)^{2k}\sum_{m=0}^{\lfloor\,n/2\rfloor}
\frac{ q^{m(m-1)} s^m }{ \{2m\}_{q}!! }(D_x^q)^{2m}={\bf 1}
\ee
and the new
$q-$Hermite polynomials $H_n(x,s|q)$
obey
\bea
\label{rinversion}
\mathcal{L}_n(s, D_x^q|q)H_n(x,s|q)=x^n
\eea
where the polynomial $\mathcal{L}_n(\alpha,\beta|q)$ is defined as follows:
\be
\label{sama:polynom}
\mathcal{L}_n(\alpha,\beta|q)= \sum_{k=0}^{\lfloor\,n/2\,\rfloor}
\frac{ q^{k(k-1)} }{ \{2k\}_{q}!! } \alpha^k\beta^{2k}.
\ee
This polynomial is essentially the $(q,n)-$exponential
function $E_{q,n}(x)$ investigated by Ernst \cite{Ernst}, i.e.,
$\mathcal{L}_{n-1}(\alpha,\beta|q)=E_{q^{-2},\lfloor\,n/2\,\rfloor}(\alpha\beta^2/\{2\}_{q}).$
We are now in a position to formulate and prove the following.
\begin{lemma}
From the polynomial (\ref{sama:polynom}) we have
\be
\mathcal{L}_{2n}(\alpha,\beta|q)=\frac{(\alpha\beta^2)^nq^{n(n-1)}}{\{2n\}_{q}!!}{}_3\phi_2\Bigg(
\begin{array}{c}q^{-n},-q^{-n},q\\
0,0\end{array}\Big|q;-\frac{q^2}{(1-q)\alpha\beta^2}\Bigg)
\ee
and
\be
\mathcal{L}_\infty(\alpha,\beta|q)=E_{q^2}(\alpha\beta^2/\{2\}_{q}).
\ee
\end{lemma}
{\bf Proof.} As it is defined in (\ref{sama:polynom}), we have
\begin{eqnarray}
\mathcal{L}_{2n}(\alpha,\beta|q)&=&\sum_{k=0}^{n}\frac{q^{ k(k-1)}}{\{2k\}_{q}!!}(\alpha\beta^2)^{k}\cr
&=&\frac{(\alpha\beta^2)^n}{\{2n\}_{q}!!}\sum_{k=n}^{\infty}\frac{q^{ k(k-1)}\{2n\}_{q}!!}{\{2k\}_{q}!!}(\alpha\beta^2)^{k-n}.
\end{eqnarray}
By substituting
$m=n-k$ in the latter expression, we arrive at
\begin{eqnarray}
\mathcal{L}_{2n}(\alpha,\beta|q)
&=&\frac{(\alpha\beta^2)^n}{\{2n\}_{q}!!}\sum_{m=0}^{\infty}
\frac{q^{ (n -m)(n -m-1)}\{2n\}_{q}!!}{\{2n-2m\}_{q}!!}(\alpha\beta^2)^{-m}\cr
&=&\frac{(\alpha\beta^2)^nq^{n(n-1)}}{\{2n\}_{q}!!}\sum_{m=0}^{\infty}
(q^{-2n};q^2)_m\left(-\frac{q^2}{(1-q)\alpha\beta^2}\right)^m.
\end{eqnarray}
When $n\to\infty$, (\ref{sama:polynom}) takes the form
\be
\mathcal{L}_\infty(\alpha,\beta|q)= \sum_{k=0}^{\infty}
\frac{ q^{k(k-1)} }{ \{2n\}_{q}!! } (\alpha\beta^2)^k = \sum_{k=0}^{\infty}
\frac{ q^{k(k-1)} }{ \{k\}_{q^2 }!}\left(\frac{\alpha\beta^2}{\{2\}_{q}}\right)^k
\ee
which achieves the proof. $\square$
In the limit, when $q\to1$, the polynomial $\mathcal{L}_n(\alpha,\beta|q)$ is reduced
to the classical one's $\mathcal{T}_n(\alpha,\beta)$, i.e., $\lim_{q\to 1}
\mathcal{L}_n(\alpha,\beta|q)=\mathcal{T}_n(\alpha,\beta),\;\forall\,n.$
\section{Fourier transforms of the new $q-$Hermite
polynomials $H_n(x,s|q)$ }
\label{sction3}
In this section,
we compute the Fourier integral
transforms associated to the new $q-$Hermite
polynomials $H_n(x,s|q)$.
\subsection{ $q^{-1}-$Hermite polynomials $H_n(x,s|q^{-1})$ }
Let us rewrite the new $q-$Hermite polynomials
(\ref{sama:qhermite}) in the following form
\be
\label{sama:pipp}
H_{n}(x,s|q)= \sum_{k=0}^{\lfloor\,n/2\,\rfloor}
c_{n,k}(q)\,s^{k}x^{n-2k},
\ee
where the associated coefficients
$c_{n,\,k}(q)$ are given by
\be
\label{sama:coef}
c_{n,\,k}(q):= \frac{ (-1)^k q^{k(k-1)}\,\{n\}_{q}!
}{\{n-2k\}_{q}! \,\{2k\}_{q}!! }.
\ee
By a direct computation, one can easily check that these coefficients
satisfy the following recursion relation
\be
\label{rett}
c_{n+1,\,k}(q)=c_{n,\,k}(q)-q^{n-1}\{n\}_{q}\,c_{n-1,\,k-1}(q),
\ee
with $
c_{0,\,k}(q)=\delta_{0,k}, \;\; c_{n,\,0}(q)=1.$\\
From the definition of the $q-$binomial coefficients
in (\ref{samasama:sa}), it is not hard to derive an
inversion formula
\bea
\label{samasamasa}
{n\atopwithdelims\{\} 2k}_{q^{-1}}=
q^{2k(2k-n)}\,{n\atopwithdelims\{\} 2k}_q,\qquad
0\leq k\leq \lfloor n/2\rfloor.
\eea
Then, one readily deduces that
\be
\label{sama:labas}
c_{n,\,k}(q^{-1})= q^{k(k+3-2n)}c_{n,\,k}(q),
\ee
allowing to define the $q^{-1}-$Hermite
polynomials $H_n(x,s|q^{-1})$ in the following form
\be
\label{sama:inm}
H_n(x,s|q^{-1}):= \sum_{k=0}^{\lfloor\,n/2\,\rfloor}\,c_{n,\,k}(q^{-1})\,s^{k}x^{n-2k}. \ee
The recursion relation
\be
\label{sama:lab}
c_{n+1,\,k}(q^{-1})= q^{-2k}\,c_{n,\,k}(q^{-1})- q^{3-n-2k}\,\{n\}_{q}\,c_{n-1,\,k-1}(q^{-1}),\quad n\geq 1
\ee
is valid for the coefficients (\ref{sama:labas})
with $c_{0,\,k}(q^{-1})=q^{k(k+3)}\,\delta_{0,k}, \;\; c_{n,\,0}(q^{-1})=1.$ \\
Since (\ref{sama:lab}) is satisfied, the $q^{-1}-$Hermite polynomials $H_n(x,s|q^{-1} )$ obey
the relation
\be
\label{samanm}
H_{n+1}(x,s|q^{-1})=xH_n(x,sq^{-2}|q^{-1})-sq^{1-n}\{n\}_{q}\,H_{n-1}(x,sq^{-2}|q^{-1}),\quad n\geq 1,
\ee
with $H_0(x,sq^{-2}|q^{-1}):=1.$\\
The action of the operator $D_x^q$ on
the polynomials (\ref{sama:inm}) is given by
\bea
\label{samanma}
D_x^q\,H_n(x,s|q^{-1})=\{n\}_{q}\,H_{n-1}(x,sq^{-2}|q^{-1}).
\eea
Let $\epsilon$
denote the operator which maps $f(s)$ to $f(qs)$. Then, from (\ref{samanm}) and (\ref{samanma}) one
can establish that
\be
H_{n}(x,s|q^{-1}) =\prod_{k=1}^{n} \big(x\,\epsilon^{-2}-sq^{k+1-n}D_x^q\big)\cdot(1).
\ee
\subsection{Fourier transforms of the new $q-$Hermite polynomials $H_n(x,s|q)$ }
Considering
the well-known Fourier transforms (\ref{samafouri})
for the Gauss exponential function $e^{-x^2/2s},$ the
Fourier integral transforms
for the exponential function $\exp(i(n-2k)\kappa x-x^2/2s)$ is computed as follows:
\be
\frac{1}{\sqrt{2\pi s}}\int_{\mathbb{R}}e^{i x y+ i(n- 2k)
\kappa x- \frac{x^2}{2s}} d x
=q^{\frac{n^2}{4}+k(k-n)}e^{-s\frac{y^2}{2}-(n- 2k)sy\kappa},
\ee
where $q=e^{-2s\kappa^2} \leq1$ and $0\leq \kappa < \infty$.
\begin{theorem}
\label{sama:propo1}
The new $q-$Hermite polynomials $H_n(x,s|q)$ and
$H_n(x,s|q^{-1})$ defined in (\ref{sama:pipp}) and (\ref{sama:inm}),
respectively, are connected by the integral Fourier transform of the
following form
\be
\label{sama:fourier}
\frac{1 }{\sqrt{2\pi s}}\int_{\mathbb{R}}H_{n} (b e^{i \kappa x},s|q)e
^{ix y-\frac{x^2}{2s}}d x= q^{\frac{n^2}{4}}H_{n}
(b e^{-s\kappa y}, q^{n-3} s| q^{-1})\, e^{-s\frac{y^2}{2}}
\ee
where $b$ is an arbitrary constant factor.
\end{theorem}
{\bf Proof.} To prove this theorem, let us make use of (\ref{sama:pipp}) and
evaluate the left hand side of (\ref{sama:fourier}). Then,
\bea
\frac{1}{\sqrt{2\pi s}}\int_{\mathbb{R}}H_{n}(b e^{i\kappa x},s|q)
e^{ix y-\frac{x^2}{2s}}d x &=&\sum_{k=0}^{\lfloor n/2\rfloor}c_{n,k} (q)s^{k}
b^{n-2k}\,\frac{1}{\sqrt{2\pi s}}\int_{\mathbb{R}}e^{i x y+ i(n- 2k)
\kappa x- \frac{x^2}{2s}} d x \cr
& =&\sum_{k=0}^{\lfloor
n/2\,\rfloor}c_{n,k}(q)s^k b^{n-2k}e^{-\frac{s}{2}[\kappa (n-2k)+y]^2} \cr
& = &q^{\frac{n^2}{4}}\sum_{k=0}^{\lfloor
n/2\,\rfloor}c_{n,k}(q)q^{-k(n-k)}s^k b^{n-2k}e^{-s\frac{y^2}{2}-(n- 2k)sy\kappa} \cr
& = &q^{\frac{n^2}{4}}\sum_{k=0}^{\lfloor
n/2\,\rfloor}c_{n,k}(q^{-1})(q^{n-3}s)^k (be^{-sy\kappa})^{n-2k} e^{-s\frac{y^2}{2}}\cr
& =& q^{\frac{n^2}{4}}H_{n}
(b e^{-s\kappa y}, q^{n-3} s |q^{-1})\, e^{-s\frac{y^2}{2}}.
\eea
$\square$
\section{Doubly indexed Hermite polynomials $\mathcal{H}_{n,p}(x,s|q)$}
\label{sction4}
In this section, we construct a novel family of Hermite polynomials
called {\it doubly indexed Hermite polynomials}, $\mathcal{H}_{n,p}(x,s|q).$
First, let us
defined the
$(q;p)-$shifted factorials $ (a;q)_{pk}$
and the $(q;p)-$number as follows:
\be
\label{sama:newdef}
(a;q)_0:=1,\quad (a;q)_{pk}:=(a,aq,\cdots,aq^{p-1};q^p)_k, \quad p\geq 1,\;k=1,2,3,\cdots
\ee
and
\be
\{pk\}_{q}:=\frac{1-q^{pk}}{1-q},\quad \{pk\}_{q}!!:=\prod_{l=1}^k \{pl\}_{q},\quad \{0\}_{q}!!:=1,
\ee
respectively.
\begin{definition}
For a positive integer $p$, a class of doubly indexed
Hermite polynomials $\big\{\mathcal{H}_{n,p}\big\}_{n,p}$ is defined
such that
\be
\label{sama:doublyH}
\mathcal{H}_{n,p}(x,s|q) := E_{q^p} \lb - s\frac{(D_x^q )^p}{ \{p\}_{q} } \rb\cdot (x ^n).
\ee
\end{definition}
If $p=2,$ a subclass of the polynomials
(\ref{sama:doublyH}) is reduced to the class of polynomials
(\ref{sama:qhermite}).
More generally,
their explicit formula is given by
\bea
\label{sama:qdoublyH}
\mathcal{H}_{n,p}(x,s|q)&=& \{n\}_{q}!\sum_{k=0}^{\lfloor\,n/p\,\rfloor}
\frac{ (-1)^k q^{p({}^k_2)} \,s^k}{ \{pk\}_{q}!!}\frac{ x^{n-pk} }{ \{n-pk\}_{q}! }\\
&=&x^n{}_p\phi_0\left(\begin{array}{c}q^{-n},q^{-n+1},\cdots,q^{-n+p-1}\\
-\end{array}\Big|\;q^p;\;\frac{sq^{p(n+(1-p)/2)}}{(1-q)^{p-1}x^p}\right),
\eea
where ${}_p\phi_0$ is the $q-$hypergeometric series \cite{ASK}.
Since $D_x^qe_q(\omega\,x)=\omega\,e_q(\omega\,x),$ we derive
the generating function of the
polynomials (\ref{sama:doublyH})
as
\be
\label{sama:functionqdoublygeneratrice}
f_q(x,s;p):=e_q( tx)E_{q^p} (- st^p/ \{p\}_{q} ) =
\sum_{n=0}^{\infty} \frac{\mathcal{H}_{n,p}(x,s|q) }{ \{n\}_{q}!}t^n, \quad |t| < 1.
\ee
These polynomials are the solutions of the $q-$analogue of the generalized heat equation \cite{Dattoli}
\be
(D_x^q)^p f_q(x,s;p)= -\{p\}_qD_s^qf_q(x,s;p),\quad f_q(x,0;p)=x^n.
\ee
For any real number $c$ and a positive integer $p$, $|q| <1,$ we have
\be
\sum_{n=0}^{\infty} \frac{\{c\}_{n,q}\mathcal{H}_{n,p}(x,s|q) }{\{n\}_{q}!}t^n=
\frac{1}{(xt;q)_c}{}_p\phi_p\Bigg(\begin{array}{c}q^{c},q^{c+1},\cdots,q^{c+p-1}\\
xtq^{c},xtq^{c+1},\cdots,xtq^{c+p-1}\end{array}\Big|\;q^p;\; \frac{s\,t^p}{(1-q)^{p-1}}\Bigg), \;|xt| <1.
\ee
Performing the $q-$derivative of both sides
of (\ref{sama:functionqdoublygeneratrice}) with respect to $x$ and $t$, one obtains
\be
\label{sa:lowering}
D_x^q \,\mathcal{H}_{n,p}(x,s|q)= \{n\}_{q} \,\mathcal{H}_{n-1,p}(x,s|q)
\ee
and
\be
\label{sa:raising}
\mathcal{H}_{n+1,p}(x,s|q)= x \mathcal{H}_{n,p}(x,s|q)-s q^{n-p+1}
\{n\}_{q} \{n-1\}_{q} \cdots \{n-p+2\}_{q} \mathcal{H}_{n-p+1}(x,s|q),\; n\geq 1,
\ee
with $\mathcal{H}_{0,p}(x,s|q):=1.$
The polynomials (\ref{sama:doublyH}) obey the following $p-$th order difference equation
\be
\Big( s\,(D_x^q)^p-q^{p-n}\,x\,D_x^q+q^{p-n}\, \{n\}_{q}\Big)\,\mathcal{H}_{n,p}(x,s|q)=0.
\ee
\section{Concluding remarks}
In this paper, we have constructed a family of new $q-$Hermite polynomials
$H_n(x,s|q)$.
Several properties related to these
polynomials have been computed and discussed.
Finally, we have constructed
a novel family of Hermite polynomials $\mathcal{H}_{n,p}(x,s|q)$
called doubly indexed Hermite polynomials.
In the limit cases, when $q$ goes to $1$ and $s$ goes to $-py,$ the
polynomials $\mathcal{H}_{n,p}(x,s|q)$ are reduced to
the higher-order Hermite polynomials, sometimes
called the Kamp\'e de F\'eriet or the Gould Hopper polynomials \cite{Dattoli}-\cite{GOULDHOPPER}, i.e.,
\be
\mathcal{H}_{n,p}(x,-py|1)\equiv g_{n}^p(x,y):= n!\sum_{k=0}^{\lfloor\,n/p\,\rfloor}
\frac{ y^kx^{n-pk}}{ k! \, (n-pk)! }.
\ee
When $q$ goes to $1,\;x\to px$ and $s\to p,$ the polynomials $\mathcal{H}_{n,p}(x,s|q)$
become the Hermite
polynomials investigated by Habibullah
and Shakoor \cite{Habibullah}, i.e.,
\be
\mathcal{H}_{n,p}(px,p|1)\equiv S_{p,n}(x):= n!\sum_{k=0}^{\lfloor\,n/p\,\rfloor}
\frac{ (-1)^k(px)^{n-pk}}{ k! \, (n-pk)! }.
\ee
For $p=2$, the doubly indexed polynomials $\mathcal{H}_{n,p}(x,s|q)$ are reduced to the new $q-$Hermite polynomials $H_{n}(x,s|q),$ i.e.,
$\mathcal{H}_{n,2}(x,s|q)\equiv H_{n}(x,s|q)$.
\section*{Acknowledgements}
This work is partially supported by the Abdus Salam International
Centre for Theoretical Physics (ICTP, Trieste, Italy) through the
Office of External Activities (OEA)-\mbox{Prj-15}. The ICMPA
is in partnership with
the Daniel Iagolnitzer Foundation (DIF), France. | 0.002062 |
\begin{document}
\hbox{}
{\parindent=0pt
{\large \bf Lagrangian and Hamiltonian dynamics of
submanifolds}
\bigskip
{\bf G.Giachetta}, {\bf L.Mangiarotti}$^1$, {\bf G.Sardanashvily}$^2$
\bigskip
\begin{small}
$^1$ Department of Mathematics and Informatics, University of
Camerino, 62032 Camerino (MC), Italy
\medskip
$^2$ Department of Theoretical Physics, Moscow State University,
117234 Moscow, Russia
\bigskip
{\bf Abstract}
Submanifolds of a manifold are described as sections of a certain fiber
bundle that enables one to consider their Lagrangian and (polysymplectic)
Hamiltonian dynamics as that of a particular classical field theory.
In particular, their Lagrangians and Hamiltonians must satisfy rather
restrictive Noether identities. For
instance, this is the case of relativistic mechanics and classical string
theory.
\end{small}
\bigskip
}
\section{Introduction}
As is well known, fiber bundles and jet manifolds of their sections
provide an adequate mathematical formulation of classical field theory.
In particular, field Lagrangians and their Euler--Lagrange operators are
algebraically described as elements of the variational bicomplex
\cite{tak2,ander,jmp01}. This description is extended to Lagrangian
theory of odd fields \cite{barn,cmp04,jmp05}.
The Hamiltonian counterpart of first-order Lagrangian theory on fiber
bundles is covariant Hamiltonian formalism developed in
multisymplectic, polysymplectic, and Hamilton -- De Donder)
variants (see, e.g., \cite{book,jpa99,krup,ech,leon}).
Jets of sections of fiber bundles are particular jets of submanifolds.
Namely, a space of jets of submanifolds admits a cover by charts of jets
of sections \cite{book,kras,modu}. Three-velocities in relativistic
mechanics exemplify first order jets of submanifolds
\cite{sard98,book98}. A problem is that differential forms on jets of
submanifolds do not constitute a variational bicomplex because horizontal
forms (e.g., Lagrangians) are not preserved under coordinate
transformations.
We consider $n$-dimensional submanifolds of an $m$-dimensional smooth
real manifold $Z$, and associate to them sections of a trivial fiber
bundle $Z_Q=Q\times Z\to Q$, where $Q$ is some $n$-dimensional manifold.
Here, we restrict our consideration to first order jets of submanifolds,
and state their relation to jets of sections of the fiber bundle $Z_Q\to
Q$ (the formulas (\ref{s17}), (\ref{s31}), and Proposition 1).
This
relation fails to be one-to-one correspondence. The ambiguity contains,
e.g., diffeomorphisms of $Q$.
Then Lagrangian and (polysymplectic)
Hamiltonian formalism on a fiber bundle $Z_Q\to Q$ is developed in a
standard way, but Lagrangians and Hamiltonians are required to be
variationally invariant under the above mentioned diffeomorphisms of $Q$.
This invariance however leads to rather restrictive Noether identities
(\ref{s60}) and (\ref{s110}) which these Lagrangians and Hamiltonians
must satisfy, unless other fields are introduced.
In a different way, one can choose some subbundle of the fiber bundle
$Z_Q\to Q$ in order to avoid the above mentioned ambiguity between jets
of subbundles of $Z$ and jets of sections of $Z_Q\to Q$. Since such a
subbundle itself need not be a jet manifold of some fiber bundle, it is
a nonholonomic constraint. For
instance, this is the case of relativistic mechanics, phrased in terms of
four-velocities.
\section{Jets of submanifolds}
Given an $m$-dimensional smooth real manifold $Z$, a
$k$-order jet of $n$-dimensional submanifolds of $Z$ at a point
$z\in Z$ is defined as the equivalence class $j^k_zS$
of
$n$-dimensional imbedded submanifolds of $Z$ through $z$ which
are tangent to each other at $z$ with order $k\geq 0$.
Namely, two submanifolds
$i_S: S\hookrightarrow Z$, $i_{S'}:
S'\hookrightarrow Z$
through a point $z\in Z$ belong to the same equivalence class $j^k_zS$
iff the images of the $k$-tangent morphisms
\be
T^ki_S: T^kS\hookrightarrow T^kZ, \qquad T^ki_{S'}:
T^kS'\hookrightarrow T^kZ
\ee
coincide with each other.
The set $J^k_nZ=\op\bigcup_{z\in Z} j^k_zS$ of $k$-order jets is a
finite-dimensional real smooth manifold.
One puts $J^0_nZ =Z$. If $k>0$,
let $Y\to X$ be an $m$-dimensional
fiber bundle over an
$n$-dimensional base $X$ and $J^kY$ the $k$-order jet manifold of
sections of $Y\to X$ (or, shortly, the jet manifold of $Y\to X$). Given
an imbedding
$\Phi:Y\to Z$, there is the natural injection
\be
J^k\Phi: J^kY\to J^k_nZ, \qquad j^k_xs \mapsto [\Phi\circ
s]^k_{\Phi(s(x))},
\ee
where $s$ are sections of $Y\to X$. This injection defines a
chart on
$J^k_nZ$. These charts provide a manifold atlas of $J^k_nZ$.
Here, we restrict our consideration to first order jets of submanifolds.
There is obvious one-to-one correspondence
\mar{s10}\beq
\zeta: j^1_zS \mapsto V_{j^1S}\subset T_zZ \label{s10}
\eeq
between the jets $j^1_zS$ at a point $z\in Z$ and the $n$-dimensional
vector subspaces of the tangent space $T_zZ$ of $Z$ at $z$. It follows
that
$J^1_nZ$ is a fiber bundle
\mar{s3}\beq
\rho:J^1_nZ\to Z \label{s3}
\eeq
in Grassmann manifolds. It possesses the following coordinate atlas.
Let $\{(U;z^\m)\}$ be a coordinate atlas of $Z$.
Though $J^0_nZ=Z$, let us provide
$J^0_mZ$ with the atlas obtained by replacing every chart $(U,z^A)$ of
$Z$ with the
\be
{m\choose n}=\frac{m!}{n!(m-n)!}
\ee
charts on $U$ which
correspond to different partitions of $(z^A)$ in
collections of $n$ and $m-n$ coordinates
\mar{5.8}\beq
(U; x^a,y^i), \qquad a=1,\ldots,n, \qquad i=1,\ldots,m-n.\label{5.8}
\eeq
The transition functions between the coordinate charts (\ref{5.8}) of
$J^0_nZ$ associated with a coordinate chart
$(U,z^A)$ of $Z$ reduce to an exchange between
coordinates $x^a$ and $y^i$.
Transition functions between arbitrary coordinate charts of the
manifold $J^0_nZ$ take the form
\mar{5.26} \beq
x'^a = x'^a (x^b, y^k), \qquad y'^i = y'^i (x^b, y^k).
\label{5.26}
\eeq
Given an atlas of coordinate charts (\ref{5.8}) -- (\ref{5.26}) of the
manifold
$J^0_nZ$, the first order jet manifold $J^1_nZ$ is endowed with the
coordinate charts
\mar{5.31}\beq
(\rho^{-1}(U)=U\times\Bbb R^{(m-n)n}; x^a,y^i,y^i_a), \label{5.31}
\eeq
possessing the following transition functions.
With respect to the coordinates (\ref{5.31}) on the jet manifold $J^1_nZ$
and the induced fiber coordinates $(\dot x^a, \dot y^i)$ on the
tangent bundle $TZ$, the above mentioned correspondence $\zeta$
(\ref{s10}) reads
\be
\zeta: (y^i_a) \mapsto \dot x^a(\dr_a +y^i_a(j^1_zS)\dr_i).
\ee
It implies the relations
\mar{s0,1}\ben
&& y'^j_a= (\frac{\dr y'^j}{\dr y^k} y^k_b + \frac{\dr y'^j}{\dr x^b})
(\frac{\dr x^b}{\dr y'^i}y'^i_a + \frac{\dr x^b}{\dr x'^a}), \label{s0}\\
&& (\frac{\dr x^b}{\dr y'^i}y'^i_a + \frac{\dr x^b}{\dr x'^a})
(\frac{\dr x'^c}{\dr
y^k} y^k_b + \frac{\dr x'^c}{\dr x^b})=\dl^c_a,\label{s1}
\een
which jet coordinates $y^i_a$ must satisfy under coordinate
transformations (\ref{5.26}). Let consider a nondegenerate $n\times n$
matrix $M$ with the entries
\be
M^c_b=(\frac{\dr x'^c}{\dr
y^k}y^k_b + \frac{\dr x'^c}{\dr x^b}).
\ee
Then the relations (\ref{s1}) lead to the equalities
\be
(\frac{\dr x^b}{\dr y'^i} y'^i_a + \frac{\dr x^b}{\dr x'^a})=
(M^{-1})^b_a.
\ee
Hence, we obtain the
transformation law of first order jet coordinates
\mar{s2}\beq
y'^j_a=
( \frac{\dr y'^j}{\dr y^k} y^k_b+ \frac{\dr y'^j}{\dr x^b})
(M^{-1})^b_a.
\label{s2}
\eeq
For instance, these are the Lorentz transformation of three-velocities
in relativistic mechanics. In particular, if coordinate
transition functions $x'^a$ (\ref{5.26}) are independent of
coordinates $y^k$, the transformation law (\ref{s2}) comes to the familiar
transformations of jets of sections.
A glance at the transformations (\ref{s2}) shows that, in contrast
with a fiber bundle of jets of sections, the fiber bundle
(\ref{s3}) is not affine. In particular, one generalizes the notion of
a connection on fiber bundles and treat global sections of the jet
bundle (\ref{s3}) as preconnections
\cite{modu}. However, a global section of this bundle need not
exist (\cite{ste}, Theorem 27.18).
Given a coordinate chart (\ref{5.31}) of $J^1_nZ$, one can regard
$\rho^{-1}(U)\subset J^1_nZ$ as the first order jet manifold $J^1U$ of
sections of the fiber bundle
\mar{s11}\beq
U\ni (x^a,y^i)\to (x^a)\in U_X. \label{s11}
\eeq
The graded differential algebra $\cO^*(\rho^{-1}(U))$ of exterior forms
on
$\rho^{-1}(U)$ is generated by horizontal forms $dx^a$ and contact forms
$dy^i-y^i_adx^a$. Coordinate transformations (\ref{5.26}) and
(\ref{s2}) preserve the ideal of contact forms, but horizontal forms
are not transformed into horizontal forms, unless coordinate
transition functions $x'^a$ (\ref{5.26}) are independent of
coordinates $y^k$. Therefore, one can
develop first order Lagrangian formalism with a Lagrangian $L=\cL d^nx$
on a coordinate chart
$\rho^{-1}(U)$, but this Lagrangian fails to be globally defined on
$J^1_nZ$.
In order to overcome this difficulty, let us consider an above mentioned
product
$Z_Q=Q\times Z$ of $Z$ and an
$n$-dimensional real smooth manifold $Q$. We have a trivial fiber bundle
\mar{s12}\beq
\pi: Z_Q=Q\times Z\to Q, \label{s12}
\eeq
whose trivialization throughout holds fixed. This fiber bundle is
provided with an atlas of coordinate charts
\mar{s20}\beq
(U_Q\times U; q^\m,x^a,y^i), \label{s20}
\eeq
where $(U; x^a,y^i)$ are the above mentioned coordinate charts
$(\ref{5.8})$ of the manifold
$J^0_n Z$. The coordinate charts (\ref{s20}) possess
transition functions
\mar{s21}\beq
q'^\m=q^\m(q^\nu), \qquad x'^a = x'^a (x^b, y^k), \qquad y'^i = y'^i
(x^b, y^k). \label{s21}
\eeq
Let $J^1Z_Q$ be the first order jet manifold of the fiber bundle
(\ref{s12}). Since the trivialization (\ref{s12}) is fixed, it is a
vector bundle $\pi^1:J^1Z_Q\to Z_Q$ isomorphic to the tensor
product
\mar{s30}\beq
J^1Z_Q= T^*Q\op\ot_{Q\times Z} TZ \label{s30}
\eeq
of the cotangent bundle $T^*Q$ of $Q$ and the
tangent bundle $TZ$ of $Z$ over $Z_Q$.
Given a coordinate atlas
(\ref{s20}) - (\ref{s21}) of
$Z_Q$, the jet manifold $J^1Z_Q$ is endowed with the
coordinate charts
\mar{s14}\beq
((\pi^1)^{-1}(U_Q\times U)=U_Q\times U\times\Bbb R^{mn};
q^\m,x^a,y^i,x^a_\m, y^i_\m), \label{s14}
\eeq
possessing transition functions
\mar{s16}\beq
x'^a_\m=(\frac{\dr x'^a}{\dr
y^k}y^k_\nu + \frac{\dr x'^a}{\dr x^b}x^b_\nu )\frac{\dr q^\nu}{\dr
q'^\m}, \qquad y'^i_\m=(\frac{\dr
y'^i}{\dr y^k}y^k_\nu + \frac{\dr y'^i}{\dr x^b}x^b_\nu)\frac{\dr
q^\nu}{\dr q'^\m}. \label{s16}
\eeq
Relative to coordinates (\ref{s14}), the isomorphism (\ref{s30}) takes
the form
\be
(x^a_\m, y^i_\m) \to dq^\m\ot(x^a_\m \dr_a + y^i_\m \dr_i).
\ee
Obviously, a jet $(q^\m,x^a,y^i,x^a_\m, y^i_\m)$ of
sections of the fiber bundle (\ref{s12}) defines some jet of
$n$-dimensional subbundles of the manifold
$\{q\}\times Z$ through a point $(x^a,y^i)\in Z$ if an $m\times n$
matrix with the entries $x^a_\m, y^i_\m$ is of maximal rank $n$.
This property is
preserved under the coordinate transformations (\ref{s16}).
An element of $J^1Z_Q$ is called regular if it possesses this property.
Regular elements constitute an open subbundle of the jet bundle
$J^1Z_Q\to Z_Q$.
Since regular elements of $J^1Z_Q$ characterize jets of submanifolds of
$Z$, one hopes to describe the dynamics of
submanifolds of a manifold $Z$ as that of sections of the fiber bundle
(\ref{s12}). For this purpose, let us refine the relation
between elements of the jet manifolds $J^1_nZ$ and $J^1Z_Q$.
Let us consider the manifold product $Q\times J^1_nZ$. Of course, it is
a bundle over $Z_Q$. Given a coordinate atlas
(\ref{s20}) - (\ref{s21}) of
$Z_Q$, this product is endowed
with the coordinate charts
\mar{s13}\beq
(U_Q\times \rho^{-1}(U)=U_Q\times U\times\Bbb R^{(m-n)n};
q^\m,x^a,y^i, y^i_a), \label{s13}
\eeq
possessing transition functions (\ref{s2}).
Let us assign to an element $(q^\m,x^a,y^i, y^i_a)$ of the chart
(\ref{s13}) the elements $(q^\m,x^a,y^i,x^a_\m, y^i_\m)$ of the
chart (\ref{s14}) whose coordinates obey the relations
\mar{s17}\beq
\bx{y^i_a x^a_\m = y^i_\m.} \label{s17}
\eeq
These elements make up an $n^2$-dimensional vector space. The relations
(\ref{s17}) are maintained under coordinate transformations (\ref{s21})
and the induced transformations of the charts (\ref{s14}) and (\ref{s13})
as follows:
\be
&& y'^i_a x'^a_\m =
( \frac{\dr y'^i}{\dr y^k} y^k_c+
\frac{\dr y'^i}{\dr x^c}) (M^{-1})^c_a
(\frac{\dr x'^a}{\dr
y^k}y^k_\nu + \frac{\dr x'^a}{\dr x^b}x^b_\nu)\frac{\dr q^\nu}{\dr q'^\m}
= \\
&& \qquad ( \frac{\dr y'^i}{\dr y^k} y^k_c+
\frac{\dr y'^i}{\dr x^c}) (M^{-1})^c_a
(\frac{\dr x'^a}{\dr
y^k}y^k_b + \frac{\dr x'^a}{\dr x^b} )x^b_\nu\frac{\dr q^\nu}{\dr q'^\m}
=\\
&& \qquad (\frac{\dr y'^i}{\dr
y^k}y^k_b + \frac{\dr y'^i}{\dr x^b})x^b_\nu\frac{\dr q^\nu}{\dr q'^\m}=
(\frac{\dr y'^i}{\dr
y^k}y^k_\nu + \frac{\dr y'^i}{\dr x^b}x^b_\nu)\frac{\dr q^\nu}{\dr
q'^\m}= y'^i_\m.
\ee
Thus, one can associate
\mar{s25}\beq
\zeta': (q^\m,x^a,y^i, y^i_a) \mapsto \{(q^\m,x^a,y^i,x^a_\m, y^i_\m) \,
| \, y^i_a x^a_\m = y^i_\m\} \label{s25}
\eeq
to each element of the manifold $Q\times J^1_nZ$ an
$n^2$-dimensional vector space in the jet manifold $J^1Z_Q$. This
is a subspace of elements $x^a_\m dq^\m\ot(\dr_a + y^i_a\dr_i)$ of a
fiber of the tensor bundle (\ref{s30}) at a point
$(q^\m,x^a,y^i)$. This subspace always contains regular elements, e.g.,
whose coordinates $x^a_\m$ form a nondegenerate $n\times n$ matrix.
Conversely, given a regular element $j^1_zs$ of $J^1Z_Q$, there is a
coordinate chart (\ref{s14}) such that coordinates $x^a_\m$ of
$j^1_zs$ constitute a nondegenerate matrix, and $j^1_zs$ defines a unique
element of
$Q\times J^1_nZ$ by the relations
\mar{s31}\beq
\bx{y^i_a=y^i_\m(x^{-1})^\m_a.} \label{s31}
\eeq
For instance, this is the well-known relation between three- and
four-velocities in relativistic mechanics.
Thus, we have shown the following. Let $(q^\m,z^A)$ further be arbitrary
coordinates on the product
$Z_Q$ (\ref{s12}) and $(q^\m,z^A,z^A_\m)$ the corresponding coordinates
on the jet manifold $J^1Z_Q$. In these coordinates, an element of
$J^1Z_Q$ is regular if an $m\times n$ matrix with the entries $z^A_\m$ is
of maximal rank $n$.
\begin{prop} \label{s50} \mar{s50}
(i) Any jet of submanifolds through a point $z\in Z$ defines some (but
not unique) jet of sections of the fiber bundle $Z_Q$ (\ref{s12}) through
a point $q\times z$ for any $q\in Q$ in accordance with the relations
(\ref{s17}).
(ii) Any regular element of
$J^1Z_Q$ defines a unique element of the jet manifold $J^1_nZ$ by means
of the relations (\ref{s31}). However, nonregular elements of $J^1Z_Q$
can correspond to different jets of submanifolds.
(iii) Two elements $(q^\m,z^A,z^A_\m)$ and $(q^\m,z^A,z'^A_\m)$ of
$J^1Z_Q$ correspond to the same jet of submanifolds if $z'^A_\m= M^\nu_\mu
z^A_\nu$, where $M$ is some matrix, e.g., it
comes from a diffeomorphism of $Q$.
\end{prop}
Basing on this result, we can describe the
dynamics of $n$-dimensional submanifolds of a manifold $Z$ as that of
sections of the fiber bundle $Q\times Z\to Q$ for some $n$-dimensional
manifold $Q$.
\section{Lagrangian dynamics of submanifolds}
Let $Z_Q$ be a fiber bundle (\ref{s12}) coordinated by $(q^\m, z^A)$ with
transition functions $q'^\m(q^\nu)$ and $z'^A(z^B)$. Then the first
order jet manifold $J^1Z_Q$ of this fiber bundle is provided with
coordinates $(q^\m, z^A, z^A_\m)$ possessing transition functions
\be
z'^A_\m=\frac{\dr z'^A}{\dr z^B} \frac{\dr q^\nu}{\dr q'^\m} z^B_\nu.
\ee
Let us recall the notation of contact forms $\th^A=dz^A-z^A_\m dq^\m$,
operators of total derivatives
\be
d_\m=\dr_\m + z^A_\m\dr_A + z^A_{\m\nu}\dr_A^\nu,
\ee
the total differential $d_H (\f)= dq^\m\w d_\m(\f)$ acting on exterior
forms $\f$ on $J^1Z_Q$, and the horizontal projection $h_0(\th^A)= 0$.
A first order Lagrangian in Lagrangian formalism on a fiber bundle
$Z_Q\to Q$ is defined as a horizontal density
\mar{s40}\beq
L=\cL(z^A, z^A_\m) \om, \qquad \om=dq^1\w\cdots \w dq^n, \label{s40}
\eeq
on the jet manifold $J^1Z_Q$. The corresponding Euler--Lagrange operator
reads
\mar{s41}\beq
\dl L= \cE_A dz^A\w \om, \qquad \cE_A= \dr_A\cL - d_\m \dr_A^\m\cL.
\label{s41}
\eeq
It yields the Euler--Lagrange equations
\mar{s90}\beq
\cE_A= \dr_A\cL - d_\m \dr_A^\m\cL =0. \label{s90}
\eeq
Let $u=u^\m\dr_\m + u^A\dr_A$ be a vector field on $Z_Q$. Its jet
prolongation onto $J^1Z_Q$ reads
\mar{s52}\beq
u= u^\m\dr_\m + u^A\dr_A + (d_\m u^A -z^A_\nu d_\m u^\nu)\dr_A^\m.
\label{s52}
\eeq
It admits the vertical splitting
\mar{s53}\beq
u= u_H + u_V= u^\m d_\m +
[(u^A-u^\nu z^A_\nu)\dr_A + d_\m(u^A -z^A_\nu u^\nu)\dr_A^\m].
\label{s53}
\eeq
The Lie derivative $\bL_uL$ of a Lagrangian $L$ along a vector field
$u$ obeys the first variational formula
\mar{s42}\beq
\bL_uL = u_V\rfloor \dl L + d_H(h_0(u\rfloor H_L))=((u^A-u^\m
z^A_\m) \cE_A + d_\m
\gJ^\m)\om,
\label{s42}
\eeq
where
\mar{s43}\beq
H_L=L +\dr_A^\m\cL\th^A\w \om_\m,
\qquad \om_\m=\dr_\m\rfloor \om, \label{s43}
\eeq
is the Poincar\'e--Cartan form and
\mar{s44}\beq
\gJ=\gJ^\m\om_\m=( \dr^\m_A\cL(u^A- u^\nu z^A_\nu) + u^\m\cL)\om_\m
\label{s44}
\eeq
is the Noether current. A vector field $u$ is called a variational
symmetry of a Lagrangian $L$ if the Lie derivative (\ref{s42}) is
$d_H$-exact, i.e., $\bL_u L=d_H\si$. In this case, there is the weak
conservation law $0\ap d_H(\gJ-\si)$
on the shell (\ref{s90}).
One can show that a vector field
$u$ (\ref{s52}) is a variational symmetry only if it is projected onto
$Q$, i.e., its components $u^\m$ are functions on $Q$, and iff its
vertical part $u_V$ (\ref{s53}) is a variational symmetry.
In a general setting, one deals with generalized vector fields
$u$ depending on parameter functions $\xi^r(q^\nu)$, their derivatives
$\dr_{\la_1\ldots\la_k}\xi^r$, and higher order jets
$z^A_{\la_1\ldots\la_k}$ \cite{cmp04,jpa05}. A vertical variational
symmetry depending on parameters is called a gauge symmetry. Here we
restrict our consideration to gauge symmetries
$u$ which are linear in parameters and their first derivatives, i.e.,
\mar{s55}\beq
u= u^A\dr_A + d_\m u_A\dr^\m_A, \qquad u^A= u^A_r\xi^r +
u^{A,\m}_r\dr_\m\xi^r, \label{s55}
\eeq
where $u^A$ are functions of $q^\m$, $z^B$ and the jets
$z^B_{\m_1\ldots\m_k}$ of bounded jet order $k<N$. By virtue of the
Noether second theorem
\cite{jmp05,jpa05,jmp05a},
$u$ (\ref{s55}) is a gauge
symmetry of a Lagrangian $L$ (\ref{s40}) iff the variational derivatives
$\cE_A$ (\ref{s41}) of
$L$ obey the Noether identities
\mar{s56}\beq
(u^A_r-d_\m u^{A,\m}_r)\cE_A - u^{A,\m}_r d_\m\cE_A=0. \label{s56}
\eeq
For instance, let us consider an arbitrary vector field
$u=u^\m(q^\nu)\dr_\m$ on
$Q$. It is an infinitesimal generator of a one-parameter group of local
diffeomorphisms of $Q$. Since $Z_Q\to Q$ is a trivial bundle, this
vector field gives rise to a vector field $u=u^\m\dr_\m$ on $Z_Q$, and its
jet prolongation (\ref{s52}) onto
$J^1Z_Q$ reads
\mar{s59}\beq
u= u^\m \dr_\m - z^A_\nu\dr_\m u^\nu \dr_A^\m= u^\m d_\m +[-
u^\nu z^A_\nu\dr_A - d_\m (u^\nu z^A_\nu)\dr_A^\m].
\label{s59}
\eeq
One can regard it as a generalized vector field depending on parameter
functions $u^\m(q^\nu)$. In accordance with Proposition 1,
it seems reasonable to require that, in order to describe jets of
submanifolds of $Z$, a Lagrangian $L$ on $J^1Z_Q$ is independent on
coordinates of $Q$ and must be variationally invariant under $u$
(\ref{s59}) or, equivalently, its vertical part
\be
u_V= -
u^\nu z^A_\nu\dr_A - d_\m(u^\nu z^A_\nu)\dr_A^\m.
\ee
Then the variational derivatives of this Lagrangian obey
irreducible Noether identities (\ref{s56}) which read
\mar{s60}\beq
\bx{z^A_\nu\cE_A=0.} \label{s60}
\eeq
These Noether identities are rather restrictive, unless other fields are
introduced.
In order to extend Lagrangian formalism on $Z_Q$ to other
fields, one can use the following two constructions:
(i) a bundle product
$Z_Q\times Z'$ of $Z_Q$ and some bundle $Z'\to Q$ bundle over
$Q$ (e.g., a tensor bundle $\op\ot^kTQ\op\ot^rT^*Q$), (ii) a bundle
$E\to Z$ and its pull-back $E_Q$ onto $Q\times Z$ which is a composite
bundle
\mar{s61}\beq
E_Q=Q\times E\to Z_Q \to Q. \label{s61}
\eeq
Let $E\to Z$ be provided with bundle coordinates $(z^A, s^i)$. Its
pull-back $E_Q$ onto $Z_Q$ possesses coordinates $(q^\m,z^A,s^i)$.
Accordingly, the pull-back
$Q\times J^1E$ of the first order jet manifold
$J^1E$ of
$E\to Z$ onto
$Z_Q$ is endowed with coordinates $(q^\m,z^A, s^i, s^i_A)$. It is a
subbundle $Q\times J^1E\subset J^1E_Q$ of the first order jet manifold
$J^1E_Q\to Z_Q$ of the fiber bundle $E_Q\to Z_Q$. This subbundle consists
of jets of sections of $E_Q\to Z_Q$ which are the pull-back of sections of
$E\to Z$. Given a composite fiber bundle (\ref{s61}), there is the
canonical bundle morphism
\mar{s62}\beq
\g: J^1Z_Q\op\times_{Z_Q} J^1E_Q \op\to_{Z_Q} J^1_QE_Q \label{s62}
\eeq
of the bundle product of jet manifolds $J^1Z_Q$, $J^1E_Q$ of the bundles
$Z_Q\to Q$, $E_Q\to Z_Q$ to the first order jet manifold $J^1_QE_Q$ of
the fiber bundle $E_Q\to Q$ \cite{book,sau,book00}. The jet
manifold $J^1_QE_Q$ is coordinated by $(q^\m,z^A,s^i,z^A_\m,s^i_\m)$.
Restricted to
$Q\times J^1E\subset J^1E_Q$, the morphism (\ref{s62}) takes the
coordinate form
\mar{s63}\beq
(z^A_\m,s_\m^i)=\g(z^A_\m,s^i_A)=(z^A_\m,s_A^iz^A_\m). \label{s63}
\eeq
Due to the morphism (\ref{s63}), any connection
\mar{s64}\beq
\G=dz^A\ot(\dr_A + \G^i_A(z^B,s^j)\dr_i) \label{s64}
\eeq
on a fiber bundle $E\to Z$ yields the covariant derivative
\mar{s65}\beq
D_\m s^i=s^i_\m - \G^i_A(z^B,s^j)z^A_\m \label{s65}
\eeq
on the composite bundle $E_Q\to Q$.
Given a fiber bundle $Z_Q\times Z'$ or a composite fiber bundle
$E_Q$ (\ref{s61}), an extended Lagrangian is defined on the
jet manifolds $J^1Z_Q\times J^1Z'$ or $J^1_QE_Q$, respectively. For
instance, any horizontal $n$-form
\be
\frac1{n!}\f_{A_1\ldots A_n}dz^{A_1}\w\cdots\w dz^{A_n}
\ee
on $E\to Z$ yields a horizontal density
\be
\frac1{n!}\f_{A_1\ldots A_n}z^{A_1}_{\m_1}\cdots
z^{A_n}_{\m_n}dq^{\m_1}\w\cdots\w dq^{\m_n}
\ee
on $J^1_QE_Q\to Q$ which may contribute to a Lagrangian (see,
e.g., a relativistic particle in
the presence of an electromagnetic field).
\section{Hamiltonian dynamics of submanifolds}
Here, we follow
polysymplectic Hamiltonian formalism which aims to describe
field systems with nonregular Lagrangians
\cite{book,jpa99,sard94,book95}. Lagrangian and polysymplectic
Hamiltonian formalisms are equivalent in the case of hyperregular
Lagrangians, but a nonregular Lagrangian admits different associated
Hamiltonians, if any. At the same time, there is a comprehensive relation
between these formalisms in the case of almost-regular Lagrangians.
Given a fiber bundle $Z_Q$ (\ref{s12}) and its vertical cotangent bundle
$V^*Z_Q$, let us consider the fiber bundle
\mar{s70}\beq
\Pi=V^*Z_Q\w (\op\w^{n-1} T^*Q). \label{s70}
\eeq
It plays the role of a momentum phase space
of covariant Hamiltonian field theory. Given coordinates
$(q^\m,z^A)$ on
$Z_Q$, this fiber bundle is coordinated by $(q^\m, z^A, p^\m_A)$, where
$p^\m_A$ are treated as coordinates of momenta. It is provided with the
canonical polysymplectic form
\mar{s73}\beq
\Om_\Pi= dp_A^\m\w dz^A\ot \dr_\m. \label{s73}
\eeq
Every Lagrangian $L$ on the jet manifold
$J^1Z_Q$ yields the Legendre map
\mar{s71}\beq
\wh L: J^1Z_Q\op\to_{Z_Q} \Pi, \qquad p^\m_A\circ\wh L =\dr^\m_A\cL,
\label{s71}
\eeq
whose range $N_L=\wh L(J^1Z_Q)$ is called the Lagrangian constraint
space. A Lagrangian $L$ is called hyperregular (resp. regular), if the
Legendre map (\ref{s71}) is a diffeomorphism (resp. local diffeomorphism,
i.e., of maximal rank). A Lagrangian
$L$ is said to be almost-regular if the Lagrangian constraint space is a
closed imbedded subbundle
$i_N:N_L\to\Pi$ of the Legendre bundle $\Pi\to Z_Q$ and the surjection
$\wh L:J^1Z_Q\to N_L$ is a fibered manifold possessing connected fibers.
A multisymplectic momentum phase space is the homogeneous Legendre
bundle
\be
\ol \Pi=T^*Z_Q\w (\op\w^{n-1} T^*Q),
\ee
coordinated by $(q^\m, z^A, p^\m_A, p)$. It is endowed with the canonical
multisymplectic form
\be
\Xi=p\om + p^\m_A dz^A\w\om_\m.
\ee
There is a trivial one-dimensional bundle $\ol\Pi\to \Pi$.
Then a Hamiltonian $\cH$ on the momentum phase space $\Pi$ (\ref{s70}) is
defined as a section
$p=-\cH$ of this fiber bundle. The pull-back of the multisymplectic form
$\Xi$ onto
$\Pi$ by a Hamiltonian $\cH$ is a Hamiltonian form
\mar{s75}\beq
H= p^\m_Adz^A\w\om_\m -\cH\om \label{s75}
\eeq
on $\Pi$. The corresponding Hamilton equations with respect to the
polysymplectic form $\Om_\Pi$ (\ref{s73}) read
\be
z^A_\m-\dr^A_\m\cH=0, \qquad -p^\m_{\m A}- \dr_A\cH=0.
\ee
A key point is that these Hamilton equations coincide with
the Euler--Lagrange equations of the first order Lagrangian
\mar{s77}\beq
L_H= (p^\m_Az^A_\m - \cH) \label{s77}
\eeq
on the jet manifold $J^1\Pi$ of $\Pi\to Q$. Indeed, its variational
derivatives are
\mar{s101}\beq
\cE^A_\m=z^A_\m-\dr^A_\m\cH, \qquad \cE_A=-p^\m_{\m A}- \dr_A\cH.
\label{s101}
\eeq
Any Hamiltonian form $H$ (\ref{s75}) on $\Pi$ yields the Hamiltonian map
\mar{s80}\beq
\wh H: \Pi\op\to_{Z_Q} J^1Z_Q, \qquad z^A_\m\circ \wh H= \dr^A_\m\cH.
\label{s80}
\eeq
A Hamiltonian $\cH$ on $\Pi$ is said to be associated to a Lagrangian $L$
on $J^1Z_Q$ if it satisfies the relations
\mar{s81,2}\ben
&& p^\m_A=\dr^\m_A\cL(q^\nu, z^B, z^B_\la=\dr^B_\la\cH), \label{s81}\\
&& p^\m_A\dr_\m^A\cH -\cH=\cL(q^\nu, z^B, z^B_\la=\dr^B_\la\cH).
\label{s82}
\een
If an associated Hamiltonian $\cH$ exists, the Lagrangian constraint space
$N_L$ is given by the coordinate equalities (\ref{s81}).
The relation between Lagrangian and polysymplectic Hamiltonian
formalisms is based on the following facts.
(i) Let a Lagrangian $L$ be
almost regular, and let us assume that it admits an associated
Hamiltonian $\cH$, which however need not be unique, unless $L$ is
hyperregular. In
this case, the Poincar\'e--Cartan form
$H_L$ (\ref{s43}) is the pull-back $H_L=\wh L^*H$ of the Hamiltonian form
$H$ (\ref{s75}) for any associated Hamiltonian $\cH$. Note that a local
associated Hamiltonian always exists. The Poincar\'e--Cartan form $H$ is
a Lepagean equivalent both of the original Lagrangian $L$ on $J^1Z_Q$ and
the Lagrangian
\be
\ol L=(\cL + (\wh z^A_\m - z^A_\m)\dr^\m_A\cL)\om
\ee
on the repeated jet manifold $J^1J^1Z_Q$. Its Euler--Lagrange equations
are the Cartan equations for $L$. Any solution of the Euler--Lagrange
equations (\ref{s90}) for $L$ is also a solution of the Cartan equations.
Furthermore, Euler--Lagrange equations and the Cartan equations are
equivalent in the case of a regular Lagrangian.
(ii) If a Lagrangian $L$ is almost regular, all
associated Hamiltonian forms $H$ coincide with each other on the
Lagrangian constraint space $N_L$, and define the constrained Lagrangian
$L_N= h_0(i_N^*H)$ on the jet manifold $J^1N_L$ of the fiber bundle
$N_L\to Q$. The Euler--Lagrange equations for this Lagrangian are called
the constrained Hamilton equations. In fact, the Lagrangian $L_H$
(\ref{s77}) is defined on the bundle product
\mar{s100}\beq
\Pi\op\times_{Z_Q} J^1Z_Q, \label{s100}
\eeq
and the constrained Lagrangian $L_N$ is the restriction of
$L_H$ to $N_L\times J^1Z_Q$.
As a result, one can show that a section $\ol
S$ of the jet bundle $J^1Z_Q\to Q$ is a solution of the Cartan equations
for $L$ iff $\wh L\circ \ol S$ is a solution of the constrained Hamilton
equations. In particular, any solution $r$ of the constrained Hamilton
equations provides the solution $\ol S=\wh H\circ r$ of the Cartan
equations.
Turn now to symmetries of a Lagrangian $L_H$ (\ref{s77}). Any vector
field $u$ on $Z_Q$ gives rise to the vector field
\be
u_\Pi= u^\m\dr_\m + u^A\dr_A +(-\dr_A u^B p^\m_B -\dr_\la u^\la p^\m_A
+\dr_\la u^\m p^\la_A)\dr^A_\m
\ee
onto the Legendre bundle $\Pi$. Then we obtain its prolongation
\be
u_\Pi= u^\m\dr_\m + u^A\dr_A + (d_\m u^A -z^A_\nu d_\m
u^\nu)\dr_A^\m + (-\dr_A u^B p^\m_B -\dr_\la u^\la p^\m_A +\dr_\la u^\m
p^\la_A)\dr^A_\m
\ee
onto the product (\ref{s100}). It is a variational symmetry of the
Lagrangian $L_H$ if the Lie derivative $\bL_{u_\Pi}L_H$ is $d_H$-exact.
For instance, let $u=u^\m\dr_\m$ be an arbitrary vector field on $Q$.
Since $Z_Q\to Q$ is a trivial bundle, this
vector field gives rise to a vector field $u=u^\m\dr_\m$ on $Z_Q$ whose
lift onto the Legendre bundle $\Pi$ is
\be
u_\Pi= u^\m\dr_\m + (-\dr_\la u^\la p^\m_A +\dr_\la u^\m
p^\la_A)\dr^A_\m.
\ee
Then we obtain its prolongation
\be
u_\Pi=u^\m\dr_\m - z^A_\nu\dr_\m u^\nu\dr^\m_A + (-\dr_\la u^\la p^\m_A
+\dr_\la u^\m p^\la_A)\dr^A_\m
\ee
onto the product (\ref{s100}), and take its vertical part
\mar{s105}\beq
u_V= -u^\nu z^A_\nu\dr_A - d_\m(u^\nu z^A_\nu)\dr_A^\m + (-\dr_\la u^\la
p^\m_A +\dr_\la u^\m p^\la_A -u^\nu p_{\nu A}^\m)\dr^A_\m. \label{s105}
\eeq
Let us regard it as a generalized vector field dependent on parameter
functions $u^\m(q)$. In accordance with Proposition 1, let us require
that a Lagrangian $L_H$ is independent on coordinates on $Q$ and
possesses the gauge symmetry $u_V$ (\ref{s105}). Then its
variational derivatives (\ref{s101}) of $L_H$ obey
the Noether identities
\be
z^A_\nu \cE_A+ p^\m_{\m A} \cE^A_\nu + p^\m_A(d_\m\cE_\nu^A -
d_\nu\cE^A_\m) =0,
\ee
which reduce to rather restrictive conditions
\mar{s110}\beq
\bx{\dl^\m_\nu \cH =(n-1) p^\m_A\dr^A_\nu\cH}
\label{s110}
\eeq
which a Hamiltonian $\cH$ must satisfy. For instance, $\cH=0$ if $n=1$.
In this case, momenta are scalars relative to transformations of $q$
and, therefore, no function of them is a density with respect to these
transformations.
\section{Example. $n=1,2$}
Given an $m$-dimensional manifold $Z$ coordinated by $(z^A)$, let us
consider the jet manifold $J^1_1Z$ of its one-dimensional submanifolds.
Let us provide $Z=Z^0_1$ with coordinates $(x^0=z^0, y^i=z^i)$
(\ref{5.8}). Then the jet manifold $J^1_1Z$ is endowed with coordinates
$(z^0,z^i,z^i_0)$ possessing transition functions (\ref{5.26}),
(\ref{s2}) which read
\mar{s120}\beq
z'^0=z'^0(z^0,z^k), \quad z'^0=z'^0(z^0,z^k), \quad z'^i_0=
(\frac{\dr z'^i}{\dr z^j} z^j_0 + \frac{\dr z'^i}{\dr z^0} )
(\frac{\dr z'^0}{\dr z^j} z^j_0 + \frac{\dr z'^0}{\dr z^0} )^{-1}.
\label{s120}
\eeq
A glance at the transformation law (\ref{s120}) shows that $J^1_1Z\to Z$
is a fiber bundle in projective spaces.
For instance, put $Z=\Bbb R^4$ whose Cartesian coordinates are subject
to the Lorentz transformations
\mar{s122}\beq
z'^0= z^0{\rm ch}\al - z^1{\rm sh}\al, \qquad z'^'= -z^0{\rm sh}\al +
z^1{\rm ch}\al, \qquad z'^{2,3} = z^{2,3}. \label{s122}
\eeq
Then $z'^i$ (\ref{s120}) are exactly the Lorentz transformations
\be
z'^1_0=\frac{ z^1_0{\rm ch}\al -{\rm sh}\al}{ - z^1_0{\rm
sh}\al+ {\rm ch}\al} \qquad
z'^{2,3}_0=\frac{z^{2,3}_0}{ - z^1_0{\rm
sh}\al + {\rm ch}\al}
\ee
of three-velocities in relativistic mechanics \cite{sard98,book98}.
Let us consider a one-dimensional manifold $Q=\Bbb R$
and the product $Z_Q=\Bbb R\times Z$. Let $\Bbb R$ be provided with a
Cartesian coordinate $\tau$ possessing transition function $\tau'=\tau +
{\rm const}$, unless otherwise stated.
Then the jet manifold
$J^1Z_Q$ of the fiber bundle $\Bbb R\times Z\to Z$ is endowed with the
coordinates $(\tau, z^0, z^i, z^0_\tau, z^i_\tau)$ with the transition
functions
\mar{s123}\beq
z'^0_\tau=\frac{\dr z'^0}{\dr z^k} z^k_\tau + \frac{\dr z'^0}{\dr z^0}
z^0_\tau, \qquad
z'^i_\tau=\frac{\dr z'^i}{\dr z^k} z^k_\tau + \frac{\dr z'^i}{\dr z^0}
z^0_\tau. \label{s123}
\eeq
A glance at this transformation law shows that, unless
nonadditive transformations of
$\tau$ are considered, there is
an isomorphism
\mar{s121}\beq
J^1_1Z_Q =VZ_Q=\Bbb R\times TZ \label{s121}
\eeq
of the jet manifold $J^1_1Z_Q$ to the vertical tangent bundle $VZ_Q$ of
$Z_Q\to \Bbb R$ which, in turn, is a product of
$\Bbb R$ and the tangent bundle $TZ$ of $Z$.
Returning to the example of $Z=\Bbb R^4$ and Lorentz transformations
(\ref{s122}), one easily observed that transformations (\ref{s123}) are
transformations of four-velocities in relativistic mechanics where $\tau$
is a proper time.
Let us consider coordinate charts $(U';\tau, z^0,z^i,z^i_0)$ and
$(U'';\tau, z^0,z^i,z^0_\tau, z^i_\tau)$ of the manifolds $\Bbb R\times
J^1_1Z$ and
$J^1Z_Q$ over the same coordinate chart $(U;\tau, z^0,z^i)$ of $Z_Q$.
Then one can associate to each element $(\tau, z^0,z^i,z^i_0)$ of
$U'\subset \Bbb R\times
J^1_1Z$ the elements of $U''\subset J^1Z_Q$ which obey the relations
\mar{s125}\beq
z^i_0 z^0_\tau= z^i_\tau \label{s125}
\eeq
and, in particular, the relations
\mar{s126}\beq
z^i_0= \frac{z^i_\tau}{z^0_\tau}, \qquad z^0_\tau\neq 0. \label{s126}
\eeq
Given a point $(\tau,z)\in \Bbb R\times
Z$, the relations (\ref{s125}) -- (\ref{s126}) are exactly the
correspondence between elements of a one-dimensional vector subspace of
the tangent space $T_zZ$ and the corresponding element of the projective
space of these subspaces.
In the above mentioned example of relativistic
mechanics, the relations (\ref{s125}) -- (\ref{s126}) are familiar
equalities between three- and four-velocities. It should be emphasized
that, in relativistic mechanics, one avoids the ambiguity between
three- and four-velocities by means of the nonholonomic constraint
\mar{s131}\beq
(z^0_\tau)^2- \op\sum_i(z^i_\tau)^2 =1. \label{s131}
\eeq
In a general setting, Lagrangian formalism on the jet manifold $J^1Z_Q$
can be developed if a Lagrangian $L$ is independent of $\tau$, and it is
variationally invariant under transformations reparametrizations
$\tau'(\tau)$, i.e., its Euler--Lagrange operator obeys the Noether
identity
\mar{s130}\beq
z^A_\tau\cE_A=0. \label{s130}
\eeq
For instance, let $Z$ be a locally affine manifold, i.e., a toroidal
cylinder $\Bbb R^{m-k}\times T^k$. Its tangent bundle can be provided
with a constant nondegenerate fiber metric $\eta_{AB}$. Then
\mar{s132}\beq
L=(\eta_{AB}z^A_\tau z^B_\tau)^{1/2}d\tau \label{s132}
\eeq
is a
Lagrangian on $J^1Z_Q$. It is easily justified that this Lagrangian
satisfies the Noether identity (\ref{s130}). Furthermore, given a one-form
$\cA_Bdz^B$ on $Z$, one can consider the Lagrangian
\label{s133}\beq
L'=[(\eta_{AB}z^A_\tau z^B_\tau)^{1/2} - \cA_Bz^B_\tau]d\tau, \label{s133}
\eeq
which also obeys the Noether identity (\ref{s130}).
In relativistic mechanics, the Euler--Lagrange equations of the
Lagrangians $L$ (\ref{s132}) and $L'$ (\ref{s133}) restricted to the
constraint space (\ref{s131}) restart the familiar equations of motion of
a free relativistic particle particle and a relativistic particle in the
presence of an electromagnetic field $\cA$.
As was mentioned above, no Hamiltonian obeys the Noether identities
(\ref{s110}) if $n=1$. However, Hamiltonian relativistic
mechanics can be developed in the framework of Hamiltonian theory of
mechanical systems with nonholonomic constraints
\cite{book98,sard99,sard03}. A key is that the constraint condition
(\ref{s131}) is not preserved under transformations of $\tau$, and a
Hamiltonian of a mechanical system with this constraint need not
satisfy the Noether identities (\ref{s110}).
In comparison with the case of one-dimensional submanifolds, a
description of the Lagrangian and Hamiltonian dynamics of two-dimensional
submanifolds follows general theory of $n$-dimensional submanifolds. This
is the case of classical string theory
\cite{scherk,hatf,polch}.
For instance, let $Z$ be again an $m$-dimensional locally affine manifold,
i.e., a toroidal cylinder $\Bbb R^{m-k}\times T^k$,
and let $Q$ be a
two-dimensional manifold. As was mentioned above, the tangent bundle
of $Z$ can be provided with a
constant nondegenerate fiber metric
$\eta_{AB}$. Let us consider the $2\times 2$ matrix with the
entries
\be
h_{\m\nu}=\eta_{AB} z^A_\m z^B_\nu.
\ee
Then its determinant provides a
Lagrangian
\mar{s140}\beq
L=(\det h)^{1/2} d^2q =([\eta_{AB} z^A_1 z^B_1] [\eta_{AB} z^A_2
z^B_2]- [\eta_{AB} z^A_1 z^B_2]^2 )^{1/2} d^2q \label{s140}
\eeq
on the jet manifold $J^1Z_Q$ (\ref{s30}). This is the well known
Nambu--Goto Lagrangian of string theory. It satisfies the Noether
identities (\ref{s60}). Let
\be
F=\frac12 F_{AB} dz^A\w dz^B
\ee
be a two-form on a manifold $Z$.
Then
\be
F=\frac12 F_{AB} z^A_\m z^B_\nu dq^\m\w dq^\nu
\ee
is a horizontal density on $J^1Z_Q$ which can be treated as an
interaction term of submanifolds and an external classical field $F$ in
a Lagrangian.
Turn now to Hamiltonian theory of two-dimensional submanifolds on the
momentum phase space $\Pi$ (\ref{s70}). In this case, the Noether
identities (\ref{s110}) take the form
\mar{s145}\beq
\dl^\m_\nu\cH=p^\m_A\dr^A_\nu\cH. \label{s145}
\eeq
For instance, let $Z$ be the above mentioned toroidal cylinder whose
cotangent bundle is provided with a constant nondegenerate fiber metric
$\eta_{AB}$. Let us consider the $2\times 2$ matrix with the
entries
\be
H^{\m\nu}=\eta^{AB} p_A^\m p_B^\nu.
\ee
Then its determinant provides a Hamiltonian
\be
\cH=(\det H)^{1/2} d^2q =([\eta^{AB} p_A^1 p_B^1] [\eta^{AB} p_A^2
p_B^2]- [\eta^{AB} p_A^1 p_B^2]^2 )^{1/2} d^2q
\ee
on the momentum phase space $\Pi$ which satisfies the Noether identities
(\ref{s145}. This Hamiltonian is associated to the Lagrangian
(\ref{s140}). | 0.001571 |
\begin{document}
\title[On Foreman's maximality principle]{On Foreman's maximality principle}
\author[Mohammad Golshani and Yair Hayut]{Mohammad
Golshani and Yair Hayut}
\thanks{The first author's research was in part supported by a grant from IPM (No. 91030417).}
\begin{abstract}
In this paper we consider Foreman's maximality principle, which says that any non-trivial forcing notion either adds a new real or collapses some cardinals. We prove the consistency of some of its consequences. We observe that it is consistent that every $c.c.c.$ forcing adds a real and that for every uncountable regular cardinal $\kappa$, every $\kappa$-closed forcing of size $2^{<\kappa}$ collapses some cardinal.
\end{abstract}
\maketitle
\section{introduction}
Foreman's maximality principle \cite{foreman-magidor-shelah} says that any non-trivial forcing notion either adds a new real or collapses some cardinal. The consistency of this principle was asked by Foreman-Magidor-Shelah \cite{foreman-magidor-shelah}, who showed that if $0^\sharp$ exists, then any non-trivial constructible forcing notion adds a new real over $V$ (see also \cite{stanley}, where a generalization of this result is proved).
In this paper we consider the following two consequences of Foreman's maximality principle, and prove some consistency results related to them:
\begin{enumerate}
\item Any non-trivial $c.c.c.$ forcing notion adds a new real.
\item For every uncountable regular cardinal $\kappa$, every $\kappa$-closed forcing of size $2^{<\kappa}$ collapses some cardinal.
\end{enumerate}
We show that statement $(1)$ is equivalent to the assertion ``there are no Souslin trees'', and hence by Solovay-Tennenbaum \cite{solovay-tennenbaum}, it is consistent that all non-trivial $c.c.c.$ forcing notion add a new real.
We also consider statement $(2)$, and prove that it is consistent, relative to the existence of a strong cardinal, that for all uncountable cardinals $\kappa,$ all $\k$-closed (and in fact all $\k$-strategically closed) forcing notions of size $\leq 2^{<\kappa}$ collapse $2^{<\k}.$ In such a model $GCH$ must fail everywhere, and hence we need some large cardinals to get the result. In the opposite direction, we build, assuming some large cardinals, a model in which $GCH$ fails everywhere, but for each infinite cardinal $\k,$ there exists a $\k^+$-closed forcing notion of size $2^\k$ which preserves all cardinals. Our work extends an earlier work of Foreman and Woodin \cite{foreman-woodin} by reducing their use of a supercompact cardinal to a strong cardinal.
In order to avoid trivialities, in this paper, the phrase ``forcing notion" is used only for separative non-atomic forcing notions.
\section{Consistency of any $c.c.c.$ forcing notion adding a new real}
In this section we consider statement $(1)$, and prove its consistency.
\begin{theorem}
Souslin hypothesis (SH) holds iff any non-trivial $c.c.c.$ forcing notion adds a new real.
\end{theorem}
\begin{proof}
One direction is trivial, since a Souslin tree, considered as a forcing notion, is $c.c.c.$ and adds no new reals.
For the other direction suppose there is a non-trivial $c.c.c.$ forcing notion $\mathbb{P}$ which adds no new reals. Let $\mathbb{B}=R.O(\mathbb{P})$ be the boolean completion of $\mathbb{P}$. $\mathbb{B}$ is a $c.c.c.$ complete Boolean algebra which is $(\omega, \omega)$-distributive, hence it is in fact $(\omega, \infty)$-distributive, thus it is a Souslin algebra, which implies the existence of a Souslin tree (see \cite{jech}). Hence SH fails.
\end{proof}
\begin{remark}
The above theorem is also proved by Shelah \cite{shelah2}, by completely different methods.
\end{remark}
As a corollary of the above theorem, and results of Solovay-Tennenbaum \cite{solovay-tennenbaum} (see also \cite{shelah1}), we have the following.
\begin{corollary}
It is consistent that any non-trivial $c.c.c.$ forcing notion adds a new real.
\end{corollary}
\section{Consistently, for every uncountable $\kappa,$ every $\kappa$-closed forcing of size $2^{<\kappa}$ collapses some cardinal}
In this section, we consider statement $(2),$ and prove the following consistency result.
\begin{theorem}
Assuming the existence of an $\aleph_{\kappa^{++}}$-strong cardinal $\kappa$, it is consistent that for all uncountable cardinals $\lambda,$ any non-trivial $\lambda$-closed forcing notion of size $\leq 2^{<\lambda}$ collapses some cardinal.
\end{theorem}
\begin{remark}
The conclusion of the theorem implies $GCH$ fails everywhere, so some very large cardinals are needed for the theorem.
\end{remark}
\begin{proof}
To prove the theorem, we need two lemmas, which are of some independent interest.
\begin{lemma}\label{lem: singular continuum collapsing} if $2^{\kappa}$ is singular, then every non-trivial $\kappa^+$-closed forcing of size $2^{\kappa}$ collapses $2^{\kappa}$.
\end{lemma}
\begin{proof}
Let $\mathbb{P}$ be a non-trivial $\kappa^+$-closed forcing notion of size $2^{\kappa}$. Then forcing with $\mathbb{P}$ adds a new sequence $\tau$ of ordinals of size $\lambda < 2^{\kappa}$ (since the minimal such $\lambda$ must be regular). We will encode $2^\kappa$ into $\tau$.
If we choose $\lambda$ to be minimal, every initial segment of $\tau$ is in $V$. Now, we can define a function $F\in V$ from all possible initial segments of $\tau$ onto $^{\kappa}2$, such that for every $p\in \mathbb{P}$ and every $x\in$~$^{\kappa}2$ there is $q\leq p$ and $\beta < \lambda$ such that $q\Vdash \tau \restriction \beta = \check{a}$ and $F(a)=x$.
Let $\langle (p_i, x_i) : i < 2^{\kappa}\rangle$ enumerate $\mathbb{P}\times$~$ ^{\kappa}2$. For every $\alpha < 2^{\kappa}$, we use the $\kappa^+$-closure of $\mathbb{P}$ and the fact that $\tau \notin V$ in order to find $q \leq p_\alpha$ such that $q \neq p_i$ for every $i < \alpha$, $q\Vdash \tau \restriction \beta = \check{a}$ (for some $\beta$) and $F(a)$ is not determined yet, and set $F(a)=x_\alpha$.
Let $p = p_\alpha \in \mathbb{P}$. We start by building a tree of $2^{<\kappa}$ incompatible conditions $q_s,\, s\in$~$^{<\kappa}2$ such that $q_\emptyset = p$ and for every $s\in$~$^{<\kappa}2$, $q_{s^\smallfrown (0)}, q_{s^\smallfrown(1)} \leq q_s$, there is $\beta_s < \lambda$ such that $q_{s^\smallfrown(i)} \Vdash \tau \restriction \check\beta_s = \check{a}_{s^\smallfrown(i)}$ for $i\in\{0,1\}$, $a_{s^\smallfrown(0)} \neq a_{s ^\smallfrown(1)}$. This is possible since $\tau \notin V$ but every initial segment of it is in $V$.
For every $f\in$~$^{\kappa}2$, let us pick a condition $q_f \in \mathbb{P}$ such that for all $\alpha<\kappa, q_f \leq q_{f\restriction \alpha}$ (this is possible by the closure of $\mathbb{P}$). Let $\beta_f = \sup_{\alpha < \kappa} \beta_{f\restriction \alpha}$. Then $\forall f \in$~$^{\kappa}2,$ $q_f\Vdash \tau \restriction \check\beta_f = \check{a_f}$, where $a_f = \bigcup_{\alpha < \kappa} a_{f\restriction \alpha}$, and for every $f\neq f^\prime$, $a_f \neq a_{f^\prime}$. Since we chose already only $|\alpha |<2^{\kappa}$ values for $F$, there must be some $f\in$~$^{\kappa}2$ such that $q_f \neq p_i$ for every $i < \alpha$ and $F(a_f )$ was not already defined.
At the end of this process, there might be still possible initial segment of $\tau$ such that $F$ is undefined on, so we define $F(x)$ to be arbitrary on those values.
By density arguments, in $V[G]$, $\{F(\tau \restriction \beta) : \beta < \lambda\} = (2^\kappa)^V$.
\end{proof}
We can slightly generalize the lemma, and show that every $\kappa^+$-strategically closed forcing of size $2^\kappa$ collapses a cardinal. The argument is the same, and the only difference is in the construction of the function $F$. There, in the $\alpha$-th step, when we build the tree of extensions of $p_\a,$ we use the strategy in order to ensure that in limit stages of the tree we can always find $q_s$, stronger than $q_{s\restriction \alpha}$ for every $\alpha < \dom(s)$. This means that when picking $q_{s^\smallfrown(\epsilon)}$ for $\epsilon \in \{0,1\}$ we first extend $q_s$ into two incompatible conditions that force different information about $\tau$ as we did in the former case, and then extend those conditions to $q_{s^\smallfrown(0)}, q_{s^\smallfrown(1)}$ according to the strategy, assuming that the last step was made by the bad player. The limit stages are completely defined by the strategy, since the good player plays those steps.
The lemma also holds in a slightly more general setting. For every regular cardinal $\kappa$, if $2^{<\kappa}$ is singular then any $\kappa$-closed forcing of cardinality $2^{<\kappa}$ collapses some cardinal. The proof is essentially the same.
We also need the following.
\begin{lemma}\label{lem: singular continuum everywhere}
Assume $GCH$ holds and $\kappa$ is $\aleph_{\kappa^{++}}$-strong. Then there is a generic extension of the universe in which $\kappa$ remains inaccessible and for all infinite cardinals $\lambda<\kappa, 2^{\l}$ is a singular cardinal.
\end{lemma}
\begin{proof}
We use the extender based Radin forcing as developed in \cite{merimovich}, and continued in \cite{friedman-golshani}. Our presentation follows \cite{friedman-golshani}. We assume the reader is familiar with these papers and use the definitions and results from them without any mention.
Let $V^*$ denote the ground model. Let $j\colon V^* \to M^*$ be an elementary embedding witnessing the $\aleph_{\k^{++}}$-strongness of $\k$ and let $\bar{E}\in V^*$ be an extender sequence system derived from $j$, $\bar{E}=\langle \bar{E}_\a: \a\in \dom(\bar{E}) \rangle$, where $\dom(\bar{E})=[\k, \aleph_{\k^{+}})$ and $\len(\bar{E})=\k^+.$ Then the ultrapower $j_{\bar{E}}\colon V^* \to M^*_{\bar{E}}\simeq Ult(V^*,\bar{E})$ has critical point $\k$ and $M^*_{\bar{E}}$ contains $V^*_{\aleph_{\k^{+}}}$. Consider the following elementary embeddings $\forall \tau' < \tau < \len(\bar{E})$
\begin{align*} \label{E-system}
& j_\gt\func \VS \to \MSt \simeq \Ult(\VS, E(\gt))=
\{j_\tau(f)(\bar E_\alpha \restricted \tau)\mid f\in V^*, \alpha\in [\kappa, \aleph_{\kappa^+})\},
\notag \\
& k_\gt(j_\gt(f)(\Es_\ga \restricted \gt))=
j(f)(\Es_\ga \restricted \gt),
\\
\notag & i_{\gt', \gt}(j_{\gt'}(f)(\Es_\ga \restricted \gt')) =
j_\gt(f)(\Es_\ga \restricted \gt'),
\\
\notag & \ordered{\MSE,i_{\gt, \Es}} = \limdir \ordered {
\ordof{\MSt} {\gt < \len(\Es)},
\ordof{i_{\gt',\gt}} {\gt' \leq \gt < \len(\Es)}
}.
\end{align*}
We demand that
$\Es \restricted \gt \in \MSt$ for all $\tau<\len(\bar E)$.
Thus we get the following commutative diagram.
\[
\xymatrix{
\VS \ar[rrrrr]^j \ar[rrrrd]^{j_{\Es}} \ar[rrrdd]_{j_\gt} \ar[rdd]_{j_{\gt'}} & & & & & \MS \\
& & & & \MSE\ar[ur]^{k_\Es} & \\
& M^*_{\gt'}\ar@/_6pc/[uurrrr]_{k_{\gt'}} \ar[urrr]^{i_{\gt', \Es}} \ar[rr]_{i_{\gt', \gt}} & & \MSt = \Ult(\VS, E(\gt)) \ar[ru]_{i_{\gt, \Es}}\ar@/_5pc/[rruu]_{k_\gt} & & \\
}
\]
Also factor through the normal ultrafilter to get the following commutative diagram
\begin{align*}
\begin{aligned}
\begin{diagram}
\node{\VS}
\arrow{e,t}{j_\Es}
\arrow{se,t}{j_\gt}
\arrow{s,l}{i_U}
\node{\MSE}
\\
\node{\NS \simeq \Ult(\VS, U)}
\arrow{e,b}{i_{U, \gt}}
\arrow{ne,b}{i_{U, \Es}}
\node{\MSt}
\arrow{n,b}{i_{\gt, \Es}}
\end{diagram}
\end{aligned}
\begin{aligned}
\qquad
\begin{split}
& U = E_\gk(0),
\\
& i_U \func \VS \to \NS \simeq \Ult(\VS, U),
\\
& i_{U, \gt}(i_U(f)(\gk)) = j_\gt(f)(\gk),
\\
& i_{U, \Es}(i_U(f)(\gk)) = j_\Es(f)(\gk).
\end{split}
\end{aligned}
\end{align*}
Force with
\begin{center}
$\mathbb{R}=\Add(\k^+, (\aleph_{\k^{++}})^{M^*_{\bar{E}}})\times \Add(\k^{++}, (\aleph_{\k^{+3}})^{M^*_{\bar{E}}})\times \Add(\k^{+3}, (\aleph_{j_{\bar{E}}(\k)^+})^{M^*_{\bar{E}}}).$
\end{center}
Let $G$ be $\mathbb{R}-$generic over $V$.
By essentially the same arguments as those given in \cite{friedman-golshani}, \cite{merimovich},
we can find filters $G_{\bar{E}}, G_U$ and $G_\tau, \tau<\len(\bar{E}),$ such that $G_{\bar{E}}$
is $j_{\bar{E}}(\PP)$-generic over $M^*_{\bar{E}},$ $G_U$ is $i_U(\PP)$-generic
over $N^*$ and $G_\tau$ is $j_\tau(\PP)$-generic over $M^*_\tau$
and such that
the following diagram is commutative:
\begin{align*}
\begin{diagram}
\node{V = \VS[G]}
\arrow[2]{e,t}{j_\Es}
\arrow{s,l}{i_{U}}
\arrow{se,b}{j_{\gt'}}
\arrow{see,b}{j_\gt}
\node{}
\node{\ME = \MSE[G_\Es]}
\\
\node{N = \NS[G_U]}
\arrow{e,b}{i_{U, \gt'}}
\node{M_{\gt'} = M^*_{\gt'}[G_{\gt'}]}
\arrow{ne,t,3}{i_{\gt', \Es}}
\arrow{e,b}{i_{\gt', \gt}}
\node{\Mt = \MSt[G_\gt]}
\arrow[1]{n,b}{i_{\gt, \Es}}
\end{diagram}
\end{align*}
Set
\begin{center}
$\mathbb{R}_U=\Big(\Add(\k^+, \aleph_{\k^{++}})\times \Add(\k^{++}, \aleph_{\k^{+3}})\times \Add(\k^{+3}, \aleph_{i_{U}(\k)^+})\Big)^{N^*}.$
$\mathbb{R}_\tau=\Big(\Add(\k^+, \aleph_{\k^{++}})\times \Add(\k^{++}, \aleph_{\k^{+3}})\times \Add(\k^{+3}, \aleph_{j_{\tau}(\k)^+})\Big)^{M_\tau}.$
$\mathbb{R}_{\bar{E}}=\Big(\Add(\k^+, \aleph_{\k^{++}})\times \Add(\k^{++}, \aleph_{\k^{+3}})\times \Add(\k^{+3}, \aleph_{j_{\bar{E}}(\k)^+})\Big)^{M^*_{\bar{E}}}.$
\end{center}
Then we can find $I_{U}, I_{\tau}$ and $I_{\bar{E}}$ in $V=V^*[G]$ such that:
\begin{enumerate}
\item $I_{U}$ is $\mathbb{R}_{U}$-generic over $N^{*}[G_{U}]$,
\item $I_{\tau}$ is $\mathbb{R}_{\tau}$-generic over $M_{\tau}^{*}[G_{\tau}]$,
\item $I_{\bar{E}}$ is $\mathbb{R}_{\bar{E}}$-generic over $M_{\bar{E}}^{*}[G_{\bar{E}}]$,
\item The generics are so that we have the following lifting diagram
\begin{align*}
\begin{diagram}
\node{}
\node{}
\node{\ME[I_\Es]}
\\
\node{N[I_U]}
\arrow{e,b}{i^*_{U, \gt'}}
\node{M_{\gt'}[I_{\gt'}]}
\arrow{ne,t,3}{i^*_{\gt', \Es}}
\arrow{e,b}{i^*_{\gt', \gt}}
\node{\Mt[I_\gt]}
\arrow[1]{n,b}{i^*_{\gt, \Es}}
\end{diagram}
\end{align*}
\end{enumerate}
Iterate $j_\Es$ and consider the following diagram:
\begin{align*}
\begin{diagram}
\node{V}
\arrow[2]{e,t}{j_\Es = j^{0,1}_\Es}
\arrow{se,t,1}{j_{\gt_1}}
\arrow{s,l}{i_{U}}
\node{}
\node{\ME}
\arrow[2]{e,t}{j^{1,2}_\Es}
\arrow{se,t,1}{j^2_{\gt_2}}
\arrow{s,l}{i^2_{U}}
\node{}
\node{M_\Es^2}
\arrow[2]{e,t}{j^{2,3}_\Es}
\arrow{se,t,1}{j^3_{\gt_3}}
\arrow{s,l}{i^3_{U}}
\node{}
\node{M_\Es^3}
\arrow[1]{e,..}
\\
\node{N}
\arrow[1]{e,b}{i_{U, \gt_1}}
\arrow[1]{nee,t,3}{i_{U, \Es}}
\node{M_{\gt_1}}
\arrow[1]{ne,b,1}{i_{\gt_1, \Es}}
\node{N^2}
\arrow[1]{e,b}{i^2_{U, \gt_2}}
\arrow[1]{nee,t,3}{i^2_{U, \Es}}
\node{M^{2}_{\gt_2}}
\arrow[1]{ne,b,1}{i^2_{\gt_2, \Es}}
\node{N^3}
\arrow[1]{e,b}{i^3_{U, \gt_3}}
\arrow[1]{nee,t,3}{i^3_{U, \Es}}
\node{M^{3}_{\gt_3}}
\arrow[1]{ne,b,1}{i^3_{\gt_3, \Es}}
\end{diagram}
\end{align*}
where
\begin{align*}
& j^0_\Es = \id,
\\
& j^{n}_\Es = j^{0, n}_\Es,
\\
& j^{m, n}_\Es = j^{n-1, n}_\Es \circ \dotsb \circ j^{m+1, m+2}_\Es \circ j^{m ,m+1}_\Es.
\end{align*}
Let $R(-,-)$ be a function such that
\begin{center}
$i_{U}^{2}(R)(\kappa, i_{U}(\kappa))=\mathbb{R}_U,$
\end{center}
where $i_{U}^{2}$ is the second iterate of $i_{U}.$ It is also clear that
\begin{center}
$j_{\bar{E}}^{2}(R)(\kappa, j_{\bar{E}}(\kappa))=\mathbb{R}_{\bar{E}}.$
\end{center}
As in \cite{merimovich} (see also \cite{friedman-golshani}), in the prepared model $V=V^*[G]$, we define a new extender sequence system $\bar{F}= \langle \bar{F}_{\alpha}: \alpha \in \dom(\bar{F})\rangle$ by:
\begin{itemize}
\item $\dom(\bar{F})=\dom(\bar{E}),$ \item $\len(\bar{F})=\len(\bar{E})$ \item $\leq_{\bar{F}}=\leq_{\bar{E}},$ \item $F(0)=E(0),$ \item $I(\tau)=I_{\tau},$ \item $\forall 0< \tau < \len(\bar{F}), F(\tau)= \langle \langle F_{\alpha}(\tau): \alpha \in \dom(\bar{F}) \rangle, \langle \pi_{\beta, \alpha}: \beta, \alpha \in \dom(\bar{F}), \beta \geq_{\bar{F}} \alpha \rangle \rangle$ is such that
\begin{center}
$X \in F_{\alpha}(\tau) \Leftrightarrow \langle \alpha, F(0), I(0), ..., F(\tau^{'}), I(\tau^{'}), ...: \tau^{'} < \tau \rangle \in j_{\bar{E}}(X),$
\end{center}
and
\begin{center}
$\pi_{\beta, \alpha}(\langle \xi, d \rangle)= \langle \pi_{\beta, \alpha}(\xi), d \rangle, $
\end{center}
\item $\forall \alpha \in \dom(\bar{F}), \bar{F}_{\alpha}= \langle \alpha, F(\tau),I(\tau): \tau < \len(\bar{F}) \rangle.$
\end{itemize}
Also let $I(\bar{F})$ be the filter generated by $\bigcup_{\tau < \len(\bar{F})} i_{\tau, \bar{E}}^{''}I(\tau).$ Then $I(\bar{F})$ is $\mathbb{R}_{\bar{F}}$-generic over $M_{\bar{F}}.$
Working in $V$, let $\mathbb{P}_{\bar{F}}$ be the corresponding extender based Radin forcing, as defined in \cite{merimovich} (see also \cite[Definition 5.1]{friedman-golshani}). Note that the definition of $\mathbb{P}_{\bar{F}}$ depends on the function $R$, and this is were the iterability of the extender sequence plays a role. Let $H$ be $\mathbb{P}_{\bar{F}}$-generic over $V$. By reflection, we may assume that each $\bar{\mu}$ which appears in some condition in $\mathbb{P}_{\bar{F}}$ has $\dom(\bar{\mu})=[\k^0(\bar{\mu}), \aleph_{\k^0(\bar{\mu})^{+}}).$ For $\alpha \in \dom(\bar{F})$ set
\begin{center}
$C_{H}^{\alpha} = \{ \max\kappa(p_{0}^{\bar{F}_{\alpha}}): p \in H \}.$
\end{center}
The following clauses can be proved as in \cite{friedman-golshani}, \cite{merimovich}.
\begin{enumerate}
\item $V[H]$ and $V$ have the same cardinals.
\item $\kappa$ remains strongly inaccessible in $V[H]$.
\item $C_{H}^{\kappa}$ is a club in $\kappa.$
\item Let $\k_0=\min(C_{H}^{\kappa}),$ and let $K$ be $\Add(\omega, \aleph_{\k_0^{+}})_{V[H]}$-generic over $V[H].$ Then the following hold in $V[H][K]:$
\begin{itemize}
\item $\forall \l <\k_0, 2^\l=\aleph_{\k_0^+}$.
\item If $\l < \l_*$ are two successive points in $C_{H}^{\kappa}$, then $2^{\l^{+n}}=\aleph_{\l^{+n+1}},$ for $n=0,1,2$ and $2^{\l^{+3}}=\aleph_{\l_*^+}$.
\end{itemize}
\end{enumerate}
In particular $V[H][K] \models $``$\k$ is strongly inaccessible and for all $\l<\k,$ $2^\l$ is a singular cardinal''. The lemma follows.
\end{proof}
We are now ready to complete the proof of Theorem 3.1.
Let $V$ be a model of $GCH+\k$ is $\aleph_{\kappa^{++}}$ strong. By Lemma~\ref{lem: singular continuum everywhere}, there exists a cardinal preserving generic extension $V$ of $V^*$ in which $\k$ remains inaccessible, and for all infinite cardinals $\l<\k, 2^\l$ is a singular cardinal. Let $\k_*\leq \k$ be the least inaccessible cardinal of $V$, and consider the model $V_{\k_*}.$ It is a model of $ZFC$. We show that $V_{\k_*}$ is as required. So let $\l<\k_*$ be an uncountable cardinal, and let $\mathbb{P}\in V_{\k_*}$ be a non-trivial $\l$-closed forcing notion of size $2^{<\l}.$ As for a singular cardinal $\l,$ being $\l$-closed implies $\l^+$-closed, we can assume without loss of generality that $\l$ is a regular cardinal, and hence by our choice of $\k_*, \l=\mu^+,$ for some cardinal $\mu.$ So it follows from Lemma~\ref{lem: singular continuum collapsing} that forcing with $\mathbb{P}$ collapses $2^\mu$. The theorem follows.
\end{proof}
\section{consistency strength of statement $(2)$ for forcing notions of arbitrary size}
In this section we discuss the consistency of statement $(2)$ for forcing notions of arbitrary large size. We show that in this case the problem is much more difficult, and it requires some very large cardinals. For the sake of simplicity, let's just consider the case $\kappa=\aleph_0.$
The following shows that non-triviality of the forcing is essential. Suppose $\lambda>\aleph_1$ is regular and let $\mathbb{P}=\lambda$, as set of ordinals. Order $\mathbb{P}$ with the reversed ordinal order separated at $\omega_1$, i.e. $p \leq_\mathbb{P} q$ iff $p \geq q$ as ordinals and $p,q < \omega_1$ or $\omega_1 \leq p,q$.
It is clear that $\mathbb{P}$ is $\aleph_1$-closed (but not $\aleph_2$-closed), and it preserves all cardinals.
\begin{theorem}
Let $\l$ be a regular cardinal such that $2^{<\l} = \l$. For every regular cardinal $\mu < \l$ there is a separative forcing notion $\mathbb{P}$ which is $\mu$-closed, not $\mu^+$-closed and $\l$-distributive. Moreover, $|\mathbb{P}| = \l$, so $\mathbb{P}$ does not collapse cardinals.
\end{theorem}
\begin{proof}
The forcing $\mathbb{P}$ will be the forcing that adds a non-reflecting stationary subset of $S^{\l}_{<\mu} = \{\alpha < \l \mid \cf\alpha < \mu\}$, by bounded conditions. Let us describe $\mathbb{P}$ precisely:
The conditions in $\mathbb{P}$ are functions $p \colon S^{\l}_{<\mu} \cap \alpha \to 2$ such that $\alpha < \l$ and $p^{-1}(1)$ is non-stationary at every $\beta \leq \alpha$ such that $\cf\beta\geq\mu$.
The order of $\mathbb{P}$ is end extension. Let us show that $\mathbb{P}$ has the required properties.
First, since $2^{<\l} = \l$, there are exactly $\l$ bounded subsets of $\l$. In particular, $|\mathbb{P}| \leq \l$. On the other hand, for every $\alpha \in S^{\l}_{<\mu}$, the function $p\colon S^{\l}_\mu \cap \alpha \to 2$ defined by $p(\b) = 0$ for every $\b$ is a condition in $\mathbb{P}$, so $|\mathbb{P}| = \l$.
Next, $\mathbb{P}$ is $\mu$-closed since if $\langle p_\alpha \mid \alpha < \eta\rangle$, $\eta < \mu$, is a decreasing sequence of conditions, then $p = \bigcup_{\alpha<\eta} p_\alpha$ is a condition in $\mathbb{P}$. The only thing that we need to verify is that $p^{-1}(1)$ is non-stationary, or that $\cf(\dom(p)) < \mu$, and clearly if the sequence is not eventually constant the second option occurs.
Let us show that $\mathbb{P}$ is not $\mu^+$-closed. For each $\alpha< \mu$ let $p_\alpha \in \mathbb{P}$ be the condition with $\dom(p_\alpha) = \alpha$ and $p_\alpha (\b) = 1$ for every $\b$. Then the sequence of all $p_\alpha$, $\alpha < \mu$ has no lower bound, since for any condition $q$ such that $q\leq p_\alpha$ for every $\alpha$ we would have $\alpha \subseteq q^{-1}(1)$ for every $\alpha < \mu$ and therefore $\mu \subseteq q^{-1} (1)$, so $q^{-1}(1)$ is stationary at $\mu$. Similar argument shows that we can construct such sequences below any condition $p \in \mathbb{P}$, so $\mathbb{P}$ is nowhere $\mu^+$-closed.
The main property of $\mathbb{P}$ is its $\l$-distributiveness. Let us show that $\mathbb{P}$ is $\l$-strategically closed, and in particular $\l$-distributive.
Let $\langle p_\alpha \mid \alpha < \eta\rangle$ be the play until step $\eta$. If $\eta$ is a successor ordinal, the strategy of the good player is to pick some ordinal in $S^{\l}_{<\mu}$ above the supremum of the domain of $p_{\eta - 1}$ and extend $p_{\eta - 1}$ up to this ordinal by appending zeros. If $\eta$ is limit less than $\l$ and $\bigcup_{\alpha < \eta} p_\alpha$ is a condition (so the good player did not lose already), the strategy will be to set $p_\eta = \bigcup_{\alpha < \eta} p_\alpha \cup \{\langle\eta,0 \rangle\}$, if $\eta \in S^{\l}_{<\mu}$, and otherwise just $p_\eta = \bigcup_{\alpha < \eta} p_\alpha$.
Let us prove that this is indeed a winning strategy, namely that at every limit stage of the play below $\l$, the union of the conditions until this step is a condition. Let $\eta$ be a limit ordinal of cofinality at least $\mu$. Then $\{\supp(p_\alpha) : \alpha < \eta,\,\text{limit}\}$ is a club at $\sup_{\alpha < \eta}\supp (p_\alpha)$ that witnesses that $\bigcup p_\alpha ^{-1} (1)$ is non-stationary, as wanted.
Since $\mathbb{P}$ is $\l$-distributive, it doesn't collapse cardinals $\leq \l$. Moreover, since $|\mathbb{P}|=\l$, $\mathbb{P}$ is $\l^{+}$-$c.c.$ and therefore it does not collapse cardinals above $\l$, so $\mathbb{P}$ does not collapse cardinals at all.
\end{proof}
\begin{corollary}
Assume there is an uncountable regular cardinal $\l$ such that $2^{<\l}=\l.$ Then there is a non-trivial $\aleph_1$-closed (but not $\aleph_2$-closed) forcing notion which preserves all cardinals.
\end{corollary}
\begin{corollary}
Suppose there is no inner model with a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}.$ Then there is a non-trivial $\aleph_1$-closed but not $\aleph_2$-closed forcing notion which preserves all cardinals.
\end{corollary}
\begin{proof}
By results of Gitik and Mitchell \cite{mitchell}, there is a strong limit singular cardinal $\kappa$ such that $2^\kappa=\kappa^+$. Applying the previous lemma with $\mu=\aleph_1$ we get the desired result.
\end{proof}
It follows that if we want a model in which all non-trivial $\aleph_1$-closed forcing notions collapse some cardinals, then $GCH$ should fail everywhere in that model.
Let us close this section by showing that there exists a non-trivial $\aleph_1$-closed cardinal preserving forcing notion in the model $V[H][K]$ of section 3. We assume that $V^*$ (and hence $V=V^*[G]$) has no inner model with a Woodin cardinal. It then follows that the combinatoric principle $\Box_\mu$ holds for every singular $\mu$ (see \cite{schimmerling-zeman}).
Let $\mu=\aleph_{\k_0^+}$ and $\l=\mu^+.$ Note that $V[H][K]=V[K][H],$ and that
\begin{center}
$V[K]\models$``$2^{\aleph_0}=\mu + \Box_\mu + SCH$ holds at $\mu$''.
\end{center}
It follows from \cite{rinot} that there exists an $\aleph_1$-closed $\l$-Souslin tree $T$ in $V[K]$. We show that $T$ remains an $\aleph_1$-closed $\l$-Souslin tree in $V[K][H].$
To do this, we need a finer analysis of the model $V[K][H].$ For inaccessible cardinals $\a<\b$ set
\begin{center}
$\mathbb{P}(\a,\b)=(\Add(\a^+, \aleph_{\a^{++}})\times \Add(\a^{++}, \aleph_{\a^{+3}})\times Add(\a^{+3}, \aleph_{\b^+}))^V.$
\end{center}
Then it is easily seen that
\begin{center}
$V[K][H]=V[K\times H_1][H_2]=V[K][H_1][H_2],$
\end{center}
where
\begin{enumerate}
\item $K\times H_1$ is $\Add(\aleph_0, \mu)\times\mathbb{P}(\k_0, \k_1)$-generic over $V,$ where $\k_0 < \k_1$ are the first two elements of the club $C_H^\k.$
\item $V_{\k_1}^{V[K][H]}=V_{\k_1}^{V[K][H_1]}.$
\end{enumerate}
Now $V\models$``$\Add(\aleph_0, \mu)\times\mathbb{P}(\k_0, \k_1)$ is $\k_0^{+4}$-Knaster'', hence $V[K]\models$``$\mathbb{P}(\k_0, \k_1)$ is $\k_0^{+4}$-Knaster, and $\k_0^+$-distributive''.
So after forcing with $\mathbb{P}(\k_0, \k_1)$ over $V[K]$, $T$ remains an $\aleph_1$-closed $\l$-Souslin tree. So by clause $(2)$ above
\begin{center}
$V[K][H]\models$`` $T$ is an $\aleph_1$-closed $\l$-Souslin tree''.
\end{center}
Thus $T$, considered as a forcing notion, is a non-trivial $\aleph_1$-closed forcing notion of size $\l$ which preserves all cardinals.
\section{Adding club sets in the absence of $GCH$}
In this section, we continue our study of the last section, and present a different method for producing cardinal preserving forcing notions in the absence of instances of $GCH$. Our forcing notions will be of size of a regular cardinal $\l$ such that $\l^{<\l} = \l$. Given any regular cardinal $\kappa$ with $\kappa^+ < \lambda,$ note that there is a natural forcing (i.e., $\Add(\l,1)$) which is $\kappa^+$-closed and $\l$-distributive. However the forcing $\Add(\l, 1)$
is also $\kappa^{++}$-closed and it preserves stationary subsets of $\lambda.$ The next theorem shows that it is consistent that there are
$\kappa^+$-closed but not $\kappa^{++}$-closed forcing notions of size $\lambda$ which are $\l$-distributive and destroy a stationary subset of $\lambda$.
The results of this section are joint work with Moti Gitik, and are presented here with his kind permission.
\begin{theorem} Assume $GCH$. Let $\kappa$ and $\lambda>\kappa^+$ be regular cardinals. Then there is a cofinality preserving generic extension in which the following hold: $2^\kappa=\lambda$ and there exists a non-trivial $\kappa^+$-closed, but not $\kappa^{++}$-closed, forcing notion of size $\lambda$ which preserves all cardinals (in fact the forcing is $\lambda$-distributive, $\lambda^+$-$c.c.$).
\end{theorem}
\begin{proof}
Let $\mathbb{P}_1=Add(\lambda, 1)$ and let $G_1$ be $\mathbb{P}_1$-generic over $V$. Let $F=\bigcup G_1.$ Then $F: \lambda \rightarrow 2.$ Let $\mathbb{P}_2=Add(\kappa, \lambda)^{V[G_1]}$ and let $G_2$ be $\mathbb{P}_2$-generic over $V[G_1].$ In $V[G_1*G_2], 2^\kappa=\lambda.$ We show that in $V[G_1*G_2],$ there is a non-trivial $\kappa^+$-closed, but not $\kappa^{++}$-closed, forcing notion of size $\lambda$ which preserves all cardinals. Let
\begin{center}
$S=\{ \alpha <\lambda : cf(\alpha)\leq \kappa$ or $(cf(\alpha)>\kappa, F(\alpha)=1) \}.$
\end{center}
$S$ is easily seen to be a stationary co-stationary subset of $\lambda$ in models $V[G_1]$ and $V[G_1*G_2]$. Let $\mathbb{R} \in V[G_1*G_2]$ be the forcing notion for adding a club into $S$ by approximations of cardinality $<\lambda.$
\begin{claim}
$(a)$ $\mathbb{R}$ is $\lambda^+$-$c.c.$
$(b)$ $\mathbb{R}$ is $\kappa^+$-closed but not $\kappa^{++}$-closed.
$(c)$ $\mathbb{R}$ is $\lambda$-distributive.
\end{claim}
\begin{proof}
$(a)$ and $(b)$ are trivial, so let's prove $(c).$ Let $K$ be $\mathbb{R}$-generic over $V[G_1*G_2]$,
and let $h\in V[G_1*G_2*K], h:\mu \rightarrow On,$ where $\mu<\l$ is regular. We show that $h\in V[G_1*G_2].$
Work in $V[G_1 * G_2]$. Let $F=\bigcup G_1: \lambda \rightarrow 2$ be as above and let $\theta$ be large enough regular. By density arguments, we can find a continuous chain $\langle M_\a: \a<\mu \rangle$ of elementary submodels of $H(\theta)$ such that:
\begin{enumerate}
\item Every initial segment of $\langle M_\a: \a<\mu \rangle$ is in $V$,
\item $|M_\a| <\l,$
\item $\a<\b \Rightarrow M_\a \subseteq M_\beta,$
\item $\langle M_\a: \a<\beta \rangle\in M_{\beta+1},$
\item $\forall \a<\mu, M_\a\cap \l\in \l,$
\item $\forall \a<\mu, F(M_\a\cap \l)=1,$
\item $F(M\cap \l)=1,$ where $M=\bigcup_{\a<\mu}M_\a.$
\end{enumerate}
We may define the sequence $\langle M_\a: \a<\mu \rangle$ in such a way that
$M$ contains all the relevant information. Let $\delta=M\cap\l$. Using the $\lambda$-closure of $Add(\lambda,1)$ and the chain condition of $Add(\kappa,\lambda)$, we can easily arrange $G_1\upharpoonright \delta=G_1\cap \Add(\delta, 1)^M$ is $\Add(\delta, 1)^M$-generic over $M$ and $\{\a<\delta: F(\a)=1 \}$ contains a club $C$ in $\delta$ (e.g. $C = \{ M_\a \cap \l \mid \alpha < \delta\}$). By our construction we may also assume that all the initial segments of $C$ are in $M[G_1\upharpoonright\delta].$ Let $G_2\upharpoonright \delta=G_2\cap \Add(\k,\delta)^{M[G_1\upharpoonright\delta]}.$ Then $G_2\upharpoonright\delta$ is
$\Add(\k,\delta)^{M[G_1\upharpoonright\delta]}$-generic over $M[G_1\upharpoonright\delta].$
Consider the model $N=M[G_1\upharpoonright\delta *G_2\upharpoonright\delta].$ Now we can decide values of $h$ inside $N$ and use $C$ to ensure that the conditions of $\mathbb{Q}$
used in the process go up to $\delta$. This will allow us to extend them finally to a single
condition deciding all the values of $h$. It follows that every initial segment of $h$ is in $N$ and hence $h\in V[G_1*G_2],$ as required.
\end{proof}
It follows that $\mathbb{R}$ preserves all the cardinals, and the theorem follows.
\end{proof}
Although we shoot a club through the fat stationary set $S = \{ \alpha < \kappa \mid F(\alpha) = 1\}$, we can't use Abraham-Shelah's general theorem from \cite{abraham-shelah}, since their cardinal arithmetic assumptions do not hold.
We now present a generalization of the above theorem, whose proof is essentially the same.
\begin{theorem}
Let $\k$ and $\l>\k^+$ be regular cardinals, and let $V[G_1*G_2]$ be a generic extension of $V$ by $\Add(\l,1)*\lusim{\Add}(\k, \l).$ Suppose $\mathbb{Q}\in V[G_1*G_2]$ is a cardinal preserving forcing notion of size $<\l,$ and let $H$ be $\mathbb{Q}$-generic over $V[G_1*G_2]$. Let $S=\{ \alpha <\lambda : cf(\alpha)\leq \kappa$ or $(cf(\alpha)>\kappa, \bigcup G_1(\alpha)=1) \}$. Then $S$ is a stationary subset of $\l$ in $V[G_1*G_2*H],$ and if $\mathbb{R}$ is the forcing notion for adding a club subset of $S$, using approximations of size $<\l,$ then
\begin{center}
$V[G_1*G_2*H]\models$``$\mathbb{R}$ is a $\k^+$-closed cardinal preserving forcing notion''.
\end{center}
\end{theorem}
As an application of the above theorem, let us prove an analogue of Theorem 5.1 for $\k$ singular.
\begin{theorem}
Suppose $\kappa$ is a strong cardinal and $\lambda>\kappa^+$ is regular. Then there is a forcing extension in which the following hold: $\kappa$ is a strong limit singular cardinal, $2^\kappa=\lambda$ and there exists a non-trivial $\kappa^+$-closed, but not $\kappa^{++}$-closed, cardinal preserving forcing notion of size $\lambda$.
\end{theorem}
\begin{proof}
By results of Gitik-Shelah and Woodin \cite{gitik-shelah}, we can assume that $\k$ is indestructible under the following kind of forcing notions:
\begin{itemize}
\item $\Add(\k, \delta),$ for any ordinal $\delta,$
\item Any $\k^+$-weakly closed forcing notion which satisfies the Prikry condition.
\end{itemize}
Fix any regular cardinal $\delta<\k.$ Force with $\Add(\l,1)*\lusim{\Add}(\k, \l)$ and let $G_1*G_2$ be $\Add(\l,1)*\lusim{\Add}(\k, \l)$-generic over $V$. Then $\k$ remains strong in $V[G_1*G_2].$ Let $\mathbb{Q}$ be the Prikry or Magidor forcing for changing the cofinality of $\k$ into $\delta$ and let $H$ be $\mathbb{Q}$-generic over $V[G_1*G_2].$ Then by Theorem 5.3, we can find, in $V[G_1*G_2*H],$ a $\k^+$-closed, but not $\k^{++}$-closed cardinal preserving forcing notion.
\end{proof}
\section{Consistently, for all $\k, 2^\kappa>\k^+$ and there is a $\kappa^+$-closed cardinal preserving forcing notion}
In this section, we again consider statement $(2),$ and prove a global consistency result, which is, in some sense, in the opposite direction of Theorem 3.1.
\begin{theorem}
Assuming the existence of a strong cardinal and infinitely many inaccessible cardinals above it, it is consistent that $GCH$ fails everywhere, and for each regular uncountable cardinal $\k,$ there exists a non-trivial $\k$-closed forcing notion which preserves all cardinals.
\end{theorem}
\begin{proof}
To prove the theorem, we need two auxiliary lemmas.
\begin{lemma}
Assume $2^\k$ is weakly inaccessible, and for all $\k<\l<2^\k, 2^\l=2^\k.$ Then there exists a $\k^+$-closed forcing notion which preserves all cardinals.
\end{lemma}
\begin{proof}
By Lemma 4.1.
\end{proof}
The next lemma is a generalization of the main theorem of Foreman-Woodin \cite{foreman-woodin}, where the use of a supercompact cardinal with infinitely many inaccessible
cardinals above it, is replaced by the much weaker assumption of the existence of a strong cardinal with an inaccessible above it.
\begin{lemma}
Assume $GCH$ holds, $\k$ is a strong cardinal and there exists an inaccessible cardinal above $\k$. Then there is a generic extension in which $\k$ remains inaccessible, for all $\l<\k, 2^\l$ is weakly inaccessible and $\l<\mu<2^\l$ implies $2^\mu=2^\l.$
\end{lemma}
\begin{proof}
The proof is similar to the proof of Lemma 3.4, so we follow that proof and just mention the changes which are required.
Let $V^*$ denote the ground model.
Also, for an ordinal $\alpha$, let us denote the next inaccessible cardinal above $\alpha$ by $\inacc(\alpha)$.
Let $j: V^* \rightarrow M^*$ be an elementary embedding witnessing the $\inacc(\kappa)^+$-strongness of $\kappa$ and let $\bar{E} \in V^*$
be an extender sequence system derived from $j$, $\bar{E}=\langle \bar{E}_\a: \a\in \dom(\bar{E}) \rangle$, where $\dom(\bar{E})=[\k, \inacc(\kappa))$ and $\len(\bar{E})=\k^+.$ Then the ultrapower $j_{\bar{E}}\colon V^* \to M^*_{\bar{E}}\simeq Ult(V^*,\bar{E})$ has critical point $\k$ and $M^*_{\bar{E}}$ contains $V^*_{\inacc(\kappa)}$. As before, consider the resulting elementary embeddings $j_{\bar{E}}, j_\tau, k_\tau, i_{\tau', \tau}$ and $i_{\tau, \bar{E}}$, where
$\tau' < \tau < \kappa^+.$
Also factor through the normal ultrafilter to get the normal measure $U$ and embeddings $i_U, i_{U, \tau}$
and $i_{U, \bar{E}}$.
Force with
\begin{center}
$\MPB=\Add(\inacc(\kappa), \inacc(\kappa)^{+3})$.
\end{center}
Let $G$ be $\MPB$-generic over $V^*$ and let $V=V^*[G]$. As before, we can find suitable
filters $G_{\bar{E}}, G_U$ and $G_\tau, \tau<\len(\bar{E}),$ such that $G_{\bar{E}}$
is $j_{\bar{E}}(\PP)$-generic over $M^*_{\bar{E}},$ $G_U$ is $i_U(\PP)$-generic
over $N^*$ and $G_\tau$ is $j_\tau(\PP)$-generic over $M^*_\tau$ and such that the resulting
diagram commutes.
Set
$\MPB_U= \Add(\inacc(\kappa), i_U(\kappa))^{N^*},$
and define the forcing notions $\MPB_\tau$
and $\MPB_{\bar{E}}$ similarly, where $i_U, N^*$ are replaced with
$j_\tau, M^*_\tau$ and $j_{\bar{E}}, M^*_{\bar{E}}$ respectively.
We show that there are
$I_{U}, I_{\tau}$ and $I_{\bar{E}}$ in $V=V^*[G]$ such that
$I_{U}$ is $\mathbb{P}_{U}$-generic over $N^{*}[G_{U}]$,
$I_{\tau}$ is $\mathbb{P}_{\tau}$-generic over $M_{\tau}^{*}[G_{\tau}]$, $I_{\bar{E}}$ is $\mathbb{P}_{\bar{E}}$-generic over $M_{\bar{E}}^{*}[G_{\bar{E}}]$, and the generics are so that the corresponding diagram lifts.
Set $I_{\bar{E}}=G \cap \MPB_{\bar{E}}.$ We show that it is $\MPB_{\bar{E}}$-generic over $M_{\bar{E}}^{*}[G_{\bar{E}}]$. Using Easton's lemma, it suffices to
show genericity over $M_{\bar{E}}^{*}.$ Let $A \subseteq \mathbb{P}_{\bar{E}}$ be a maximal antichain in $M_{\bar{E}}^{*}$
and $X= \bigcup \{\dom(p): p \in A \}.$ Then $|X| \leq \inacc(\kappa),$ and $A$ is a maximal antichain of $\Add(\inacc(\kappa), X)^{M_{\bar{E}}^{*}}$.
As $\Add(\inacc(\kappa), X)^{M_{\bar{E}}^{*}}=\Add(\inacc(\kappa), X)^{V^*}$, $A$ is a maximal antichain of $\Add(\inacc(\kappa), X)^{V^*},$ hence a maximal antichain
of $\Add(\inacc(\kappa), \inacc(\kappa)^{+3})^{V^*}$. Let $p \in A \cap G(\kappa).$ Then $p \in A \cap I_{\bar{E}}.$
Now set
\begin{center}
$I_U = \langle i_{U, \bar{E}}^{-1''}[I_{\bar{E}}] \rangle,$ ~the filter generated by ~ $i_{U, \bar{E}}^{-1''}[I_{\bar{E}}]$
\end{center}
and
\begin{center}
$I_\tau = \langle i_{\tau, \bar{E}}^{-1''}[I_{\bar{E}}] \rangle,$ ~the filter generated by ~ $i_{\tau, \bar{E}}^{-1''}[I_{\bar{E}}]$
\end{center}
It is easily seen that $I_U$ and $I_\tau$ are respectively $\MPB_U$-generic over $N^*$ and $\MPB_\tau$-generic over $M^*_\tau$. Now by applying Easton's lemma, we can conclude the desired result.
Let $R(-,-)$ be a function such that
\begin{center}
$i_{U}^{2}(\kappa, i_{U}(\kappa))=\mathbb{P}_U,$
\end{center}
where $i_{U}^{2}$ is the second iterate of $i_{U}.$ Working in $V=V^*[G],$ define the new extender sequence system
$\bar{F}= \langle \bar{F}_{\alpha}: \alpha \in \dom(\bar{F})\rangle$ as before.
Working in $V$, let $\mathbb{P}_{\bar{F}}$ be the corresponding extender based Radin forcing, and let $H$ be $\mathbb{P}_{\bar{F}}$-generic over $V$. By reflection, we may assume that each $\bar{\mu}$ which appears in some condition in $\mathbb{P}_{\bar{F}}$ has $\dom(\bar{\mu})=[\k^0(\bar{\mu}), \inacc(\k^0(\bar{\mu}))).$ For $\alpha \in \dom(\bar{F})$ set
\begin{center}
$C_{H}^{\alpha} = \{ \max\kappa(p_{0}^{\bar{F}_{\alpha}}): p \in H \}.$
\end{center}
The following clauses can be proved as before:
\begin{enumerate}
\item $V[H]$ and $V$ have the same cardinals.
\item $\kappa$ remains strongly inaccessible in $V[H]$
\item $C_{H}^{\kappa}$ is a club in $\kappa,$
\item If $\l < \l_* $ are two successive points in $C_{H}^{\kappa}$, then $2^\l = \inacc(\lambda)$
and $2^{\inacc(\l)}=\l_*,$
\item Let $\gamma_0=\min(C_{H}^{\kappa}).$ If $\gamma_0\leq \l< \mu<2^\l<\k$, then $2^\mu=2^\l.$
\end{enumerate}
It follows immediately that
\begin{center}
$V[H]\models$`` If $\gamma_0\leq \l<\k,$ then $2^\l$ is weakly inaccessible and $\l< \mu<2^\l \Rightarrow 2^\mu=2^\l$''.
\end{center}
Force over $V[H]$ with $Add(\aleph_0, \gamma_0),$ and let $K$ be $Add(\aleph_0, \gamma_0)$-generic over $V[H].$ Then
\begin{center}
$V[H][K]\models$``If $\l<\k,$ then $2^\l$ is weakly inaccessible and $\l< \mu<2^\l \Rightarrow 2^\mu=2^\l.$
\end{center}
The lemma follows.
\end{proof}
We are now ready to complete the proof of Theorem 6.1.
Let $V^*$ be a model of $GCH+\k$ is a strong cardinal $+$ there are infinitely many inaccessible cardinals above $\k.$ By Lemma 6.3 there exists a generic extension $V$ of $V^*$ in which $\k$ remains measurable and for all infinite cardinals $\l<\k, 2^\l$ is weakly inaccessible and if $\l<\mu<2^\l,$ then $2^\mu=2^\l.$ Let $\k_*\leq \k$ be the least inaccessible of $V$, and consider the $ZFC$ model $V_{\k_*}.$ By Lemma 6.2, $V_{\k_*}$ is as required and the theorem follows.
\end{proof}
\begin{remark}
The use of an inaccessible cardinal above the strong cardinal is essentially to help us in constructing a generic extension in which the power function is such that for all infinite cardinals $\kappa, 2^\kappa$ is weakly inaccessible and if $\kappa <\lambda<2^\kappa$ then $2^\lambda=2^\kappa.$ It seems that we need such a behavior of the power function if we want to produce cardinal preserving forcing notions using Theorem 4.1, as otherwise the corresponding forcing will have a larger size and we will face some troubles in checking the chain condition.
\end{remark}
Note that in the above arguments, we always have $\k<\l<2^\k \Rightarrow 2^\l=2^\k,$ so it is natural to ask if it is consistent that $\k^+ < 2^\k < 2^{<2^\k}$ and there exists a $\k^+$-closed
( but not $\k^{++}$-closed) forcing notion which preserves all cardinal.
The next lemma gives a positive answer to this question.
\begin{lemma}
Suppose $GCH$ holds and $\k < \mu < \delta < \l$ are regular cardinals. Let $\mathbb{P}=\Add(\k, \delta),$ $\mathbb{Q}=\Add(\mu, \l)$ and let $G\times H$ be $\mathbb{P}\times \mathbb{Q}$-generic over $V$. Let $\mathbb{R}=\Add(\delta, 1)_{V[G]}.$ Then the following hold in $V[G \times H]:$
$(a)$ $\mathbb{R}$ is $\mu$-closed but not $\mu^+$-closed.
$(b)$ $\mathbb{R}$ is $\delta$-distributive and has size $\delta$.
In particular $V[G\times H]\models$`` $2^\k=\delta, 2^\mu=\l$ and there exists a $\mu$-closed (but not $\mu^+$-closed ) forcing notion of size $\delta=2^{<\mu}$ which preserves all cardinals''.
\end{lemma}
\begin{remark}
In fact our proof shows that it suffices to have:
\begin{itemize}
\item $\mathbb{P}$ is $\delta$-Knaster and it forces $2^{<\delta}=\delta,$
\item $\mathbb{Q}$ is $\mu$-closed and $\delta$-$c.c.$
\end{itemize}
\end{remark}
\begin{proof}
Recall that since $\mathbb{P}$ is $\kappa^+$-$c.c.$ and $\mathbb{Q}$ is $\mu$-closed, $\mu \geq \kappa^+$, by Easton's lemma,
\begin{center}
$\Vdash_{\mathbb{P} } \mathbb{Q}$ is
$\mu$-distributive, and
$\Vdash_{\mathbb{Q} } \mathbb{P}$ is $\kappa^+$-c.c.
\end{center}
Note that $V[G]\models \mathbb{R}$ is $\delta$-closed, and in particular $\mu$-closed. $\mathbb{Q}$ does not add any sequence of elements of $\mathbb{R}$ of length shorter than $\mu$, so $\mathbb{R}$ is still $\mu$-closed in $V[G][H]$.
Let us show now that $\mathbb{R}$ is $\delta$-distributive in $V[G][H]$ (and in particular it does not collapse cardinals).
For this end, we first show that $\mathbb{Q}$ is $\delta$-$c.c.$ after forcing with $\mathbb{P}\ast \lusim{\mathbb{R}}$. Assume otherwise, and let $\mathcal{A}\subseteq \mathbb{Q}$ be an antichain of cardinality $\delta$. Since $\mathbb{R}$ is $\delta$-closed in $V[G]$, one can decide the values of $\mathcal{A}$ by induction on $\alpha < \delta$ and obtain an antichain of cardinality $\delta$ in $V[G]$.
Let $\langle a_i \mid i < \delta \rangle$ enumerates $\mathcal{A}$ in $V[G]$ and let $p_i \Vdash$`` $\lusim{a}_i = \check{q_i}$'' by some condition $p_i$ in $\mathbb{P}$, where $\lusim{a}_i$ is a $\mathbb{P}$-name for $a_i$. Since $\mathbb{P}$ is $\delta$-Knaster there is $I\subset \delta$ such that for every $i, j \in I$, $p_i$ is compatible with $p_j$. Therefore, $q_i \perp q_j$ for every $i \neq j$ in $I$, and this implies that already in $V$ there is an antichain in $\mathbb{Q}$ of cardinality $\delta$, which is a contradiction.
Let $K$ be a $V[G][H]$-generic for $\mathbb{R}$ and let $x$ be a sequence of ordinals of length $<\delta$ in $V[G][H][K]$. Since $V[G][H][K] = V[G][K][H]$, and $\mathbb{Q}$ is $\delta$-$c.c.$ in $V[G][K]$, there is a set $\lusim{x}$ of cardinality $<\delta$ which is a $\mathbb{Q}$-name for $x$ (so $\lusim{x} \subseteq \mathbb{Q}\times On$).
Since $\mathbb{R}$ is $\delta$-distributive in $V[G]$, $\lusim{x}\in V[G]$. Therefore, $x\in V[G][H]$, as wanted.
\end{proof} | 0.00691 |
TITLE: All-pairs shortest paths in trees?
QUESTION [12 upvotes]: This is a reference request, since I'm sure what follows isn't new, but I can't seem to find it.
Suppose that we have a finite tree $T$ with non-negative weights on the edges. Naively, computing the path lengths (i.e., sum of the weights along the unique path) between every pair requires $O(n^3)$ steps: there are $\binom{n}{2}$ pairs of vertices and we can always bound the number of edges on any path by $n-1$.
We can, however, do a great deal better with the following trick. Pick a root $r$ for $T$ arbitrarily. Define the least common ancestor of $i$ and $j$ as the vertex $a$ where the path from $i$ to $r$ meets the path from $j$ to $r$. Then if $d(\cdot,\cdot)$ denotes the distance in $T$, we get $d(i,j) = d(i,r) + d(j,r) - 2d(a,r)$.
It's easy to see that all the $d(i,r)$ can be computed in $O(n)$ steps with BFS. There's also a data structure of Harel and Tarjan that, after $O(n)$ preprocessing will answer least common ancestor queries in $O(1)$ time. So the whole thing becomes $O(n^2)$.
REPLY [1 votes]: Another variant would be to start with an arbitrary vertex and then to update the all pairs shortest paths table each time a new vertex is discovered, that is adjacent to one of the previously discovered ones.
When discovering vertex $v$ via edge $(u,v)$, the entries of row $u$ and of column $v$ of the distance table $T$ are first duplicated into row $v$ and colum $v$ respectively; then to every newly generated non-zero entry in the distance table the weight $w(u,v)$ of edge $(u,v)$ is added and finally the distance from $u$ to $v$ and from $v$ to $u$ are set to $w(u,v)$
The sequential complexity is $O(n^2)$, but the algorithm is much simpler and in contrast to the other variants, this algorithm also works for dynamically growing trees, i.e. initially unknown trees and/or trees with no upper bound on final size.
That algorithm can also exploit parallel computing capability, e.g. of GPUs; the complexity would then be $O(n)$ under the assumption that an unlimited number of processors is available. | 0.141792 |
Most.
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TITLE: Complex normal coordinates in Kähler manifolds
QUESTION [6 upvotes]: Let $(M, g, J, \omega)$ be a Kähler manifold. That is, $(M, J)$ is a complex manifold, $g$ is a Hermitian metric on $M$ and
$$\omega (X, Y) = g(JX, Y)$$
is a closed two form.
As a Riemannian manifold $(M, g)$, for each $x\in M$, one can find a geodesic normal coordinates around each $x$. In the case of Kähler metric, one actually have more:
Proposition: (Complex Normal Coordinates on Kähler manifolds) For each $x\in M$, there is a local holomorphic coordinates around $x$ so that the metric $g = g_{i\bar j}$ satisfies
$$g_{i\bar j}(x) = \delta_{ij}, \ \ d g_{i\bar j} (x) = 0, \ \ \ \frac{\partial^2 g_{i\bar j}}{\partial z_k \partial z_l} (x) = 0.$$
While the first two conditions are similar to what we have for geodesic normal coordinates in Riemannian geometry, there is no corresponding analogue for the last condition. Also, even in a Kähler manifold, the geodesic normal coordinates might not be holomorphic.
I am looking for a proof of this proposition.
REPLY [4 votes]: The result is local, we may assume that $p =0 \in U$, where $U \subset \mathbb C^n$ is an open set. By an complex linear transformation, we may also assume $h_{i\bar j}(0) = \delta_{ij}$. Consider the holomorphic mapping $\phi : B_\epsilon \to U$, given by
$$z_i=\phi (w)_i = w_i + A^i_{mn} w_mw_n + B^i_{pqr} w_pw_qw_r,$$
here $A^i_{mn}, B^i_{pqr}$ are constants to be chosen, and repeated indices means sum over. We also assume that $A^i_{mn}, B^i_{pqr}$ are fully symmetric with respect to the lower indices. Since $d\phi_0 = Id$, $\phi$ is an biholomorphism onto its image (restricting to smaller sets when necessary). Let $g = \phi^* h$ be the pullback metric. Then using the new cordinates $(w_1, \cdots, w_n)$,
\begin{align}\tag{1} g_{\alpha \bar \beta} &= \frac{\partial z_i}{\partial w^\alpha}\overline{\frac{\partial z_j}{\partial w^\beta}}h_{i\bar j}
\end{align}
and
$$\tag{2} \frac{\partial z_i}{\partial w_\alpha} = \delta_{i\alpha} + 2A^i_{\alpha m} w_m + 3B^i_{\alpha pq } w_pw_q$$
Then we have
\begin{align}g_{\alpha\bar\beta} &= h_{\alpha\bar\beta} +2A^i_{\alpha m} h_{i\bar\beta}w_n + 2\overline{A^j_{\beta n}} h_{\alpha \bar j} \bar w_n \\
&\ \ \ + 3B^i_{\alpha pq } w_pw_q h_{i\bar\beta} + 3\overline{B^j_{\beta rs} } \bar w_r \bar w_s h_{\alpha \bar j} + 4 A^i_{\alpha m} \overline{A^j_{\beta n} } h_{i\bar j}w_m \bar w_n\\
&\ \ \ + O(|w|^3).
\end{align}
From here it's obvious that $g_{\alpha\bar\beta}(0) = \delta_{\alpha\beta}$. Also, we have
\begin{align} \partial_\gamma g_{\alpha\bar\beta} (0) &= \partial_\gamma h_{\alpha\bar\beta} (0) + 2A^\beta_{\alpha\gamma}, \\
\bar\partial_\gamma g_{\alpha\bar\beta} (0) &= \bar\partial_\gamma h_{\alpha\bar\beta} (0) + 2\overline{A^\alpha_{\beta\gamma}}, \\
\partial_\eta \partial _\gamma g_{\alpha\bar\beta} (0) &= \partial_\eta \partial _\gamma h_{\alpha\bar\beta} (0) + 6B^\beta_{\alpha\gamma\eta}.
\end{align}
Now we choose $A^\beta_{\alpha\gamma} = -\frac 12 \partial_\gamma h_{\alpha\bar\beta}$. First of all, $A$ so chosen is really symmetric in the lower indice, since
$$\tag{3}\partial_\gamma h_{\alpha\bar\beta} =\partial_\alpha h_{\gamma \bar\beta}$$
when $h$ is a Kähler metric (see here). Next, since $h$ is Hermitian,
$$\bar \partial_\gamma h_{\alpha\bar\beta} (0) = \overline{\partial_\gamma \overline{h_{\alpha\bar\beta}(0)}} = \overline{\partial_\gamma h_{\beta\bar\alpha}(0)} = -\frac{1}{2} \overline{A^\alpha_{\beta\gamma}},
$$
where in the last equality we used the definition of $A$. Thus we have $dg_{i\bar j}(0) = 0$.
Lastly, we choose $B^\beta_{\alpha\gamma\delta} = - \partial_\eta \partial _\gamma h_{\alpha\bar\eta} (0)$. Again by (3), $\partial_\eta \partial _\gamma h_{\alpha\bar\eta}$ is symmetric in $\alpha, \gamma, \eta$. Thus $B$ is again well defined, and this finishes the proof.
In this book, they state the following proposition:
Proposition 1.6: (Normal coordinates in Kähler case) Let $M$ be a Kähler manifold with a real analytic Kähler metric. Given $x\in M$, there exist local complex coordinates $(z_1, \cdots, z_n)$ unique modulo unitary linear transformations such that $g_{i\bar j}(x) = \delta_{ij}$, $dg_{i\bar j}(x)= 0$ and
$$ \frac{\partial ^l g_{i\bar j}}{\partial z_{i_1} \cdots \partial z_{i_k}} (x) = 0$$
for all $l\ge 0$ and $i_1 + \cdots + i_k = l$, and this also holds for its conjugate.
They also suggest a reference, p.286, claiming that there's an elegant proof. | 0.011017 |
\begin{document}
\begin{center}
{\large\bf A $p$-adic analogue of Chan and Verrill's formula for $1/\pi$}
\end{center}
\vskip 2mm \centerline{Ji-Cai Liu}
\begin{center}
{\footnotesize Department of Mathematics, Wenzhou University, Wenzhou 325035, PR China\\
{\tt jcliu2016@gmail.com } \\[10pt]
}
\end{center}
\vskip 0.7cm \noindent{\bf Abstract.}
We prove three supercongruences for sums of Almkvist--Zudilin numbers, which confirm some conjectures of Zudilin and Z.-H. Sun. A typical example is the Ramanujan-type supercongruence:
\begin{align*}
\sum_{k=0}^{p-1}\frac{4k+1}{81^k}\gamma_k \equiv \left(\frac{-3}{p}\right)p\pmod{p^3},
\end{align*}
which is corresponding to Chan and Verrill's formula for $1/\pi$:
\begin{align*}
\sum_{k=0}^{\infty}\frac{4k+1}{81^k}\gamma_k=\frac{3\sqrt{3}}{2\pi}.
\end{align*}
Here $\gamma_n$ are the Almkvist--Zudilin numbers.
\vskip 3mm \noindent {\it Keywords}: Supercongruences; Almkvist--Zudilin numbers; Harmonic numbers
\vskip 2mm
\noindent{\it MR Subject Classifications}: 11A07, 11B65, 11Y55, 05A19
\section{Introduction}
For $n\ge 0$, the following sequence:
\begin{align*}
\gamma_n=\sum_{j=0}^{n}(-1)^{n-j}\frac{3^{n-3j}(3j)!}{(j!)^3}{n\choose 3j}{n+j\choose j}
\end{align*}
are known as Almkvist--Zudilin numbers (see \cite{az-b-2007} and A125143 in \cite{sloane-2020}).
This sequence appears to be first recorded by Zagier \cite{zagier-b-2009} as integral solutions to Ap\'ery-like recurrence equations.
These numbers also appear as coefficients of modular forms.
Let $q=e^{2\pi i \tau}$ and
\begin{align*}
\eta(\tau)=q^{1/24}\prod_{n=1}^{\infty}(1-q^n)
\end{align*}
be the Dedekind eta function. Chan and Verrill \cite{cv-mrl-2009} showed that if
\begin{align*}
t_3(\tau)=\left(\frac{\eta(3\tau)\eta(6\tau)}{\eta(\tau)\eta(2\tau)}\right)^4\quad
\text{and}\quad F_3(\tau)=\frac{\left(\eta(\tau)\eta(2\tau)\right)^3}{\eta(3\tau)\eta(6\tau)},
\end{align*}
and $|t_3(\tau)|$ is sufficiently small, then
\begin{align*}
F_3(\tau)=\sum_{n=0}^{\infty}\gamma_n t_3^n(\tau).
\end{align*}
They also constructed some new series for $1/\pi$ in terms of the numbers $\gamma_n$, one of the typical examples is the following formula \cite[Theorem 3.14]{cv-mrl-2009}:
\begin{align}
\sum_{k=0}^{\infty}\frac{4k+1}{81^k}\gamma_k=\frac{3\sqrt{3}}{2\pi}.\label{aa-1}
\end{align}
The above interesting example motivates us to prove the following supercongruence, which was originally conjectured by Zudilin \cite[(33)]{zudilin-jnt-2009}.
\begin{thm}\label{t-1}
For any prime $p\ge 5$, we have
\begin{align}
\sum_{k=0}^{p-1}\frac{4k+1}{81^k}\gamma_k \equiv \left(\frac{-3}{p}\right)p\pmod{p^3},\label{aa-2}
\end{align}
where $\left(\frac{\cdot}{p}\right)$ denotes the Legendre symbol.
\end{thm}
The supercongruence \eqref{aa-2} may be regarded as a $p$-adic analogue of \eqref{aa-1}. In the past two decades, Ramanujan-type series for $1/\pi$ as well as related supercongruences and $q$-supercongruences have attracted many experts' attention (see, for instance, \cite{ccl-am-2004,cv-mrl-2009,cwz-ijm-2013,guo-aam-2020,gl-itsf-2020,gs-rm-2020,
gz-am-2019,liu-jsc-2019,liu-jmaa-2020,lp-a-2020,sunzw-scm-2011,sunzw-era-2020,van-b-1997,zudilin-jnt-2009}).
The second result of this paper consists of the following two related supercongruences involving
the numbers $\gamma_n$, which were originally conjectured by Z.-H. Sun \cite[Conjecture 6.8]{sunzh-itsf-2015}.
\begin{thm}\label{t-2}
For any prime $p\ge 5$, we have
\begin{align}
\sum_{k=0}^{p-1}(4k+3)\gamma_k \equiv 3\left(\frac{-3}{p}\right)p\pmod{p^3}.\label{aa-3}
\end{align}
\end{thm}
\begin{thm}\label{t-3}
For any prime $p\ge 5$, we have
\begin{align}
\sum_{k=0}^{p-1}\frac{2k+1}{(-9)^k}\gamma_k\equiv \left(\frac{-3}{p}\right)p\pmod{p^3}.\label{aa-4}
\end{align}
\end{thm}
We remark that congruence properties for the Almkvist--Zudilin numbers have been widely investigated by Amdeberhan and Tauraso \cite{at-aa-2016}, Chan, Cooper and Sica \cite{ccs-ijnt-2010}, and Z.-H. Sun \cite{sunzh-itsf-2015,sunzh-a-20-2,sunzh-a-20-4}.
The rest of the paper is organized as follows. In Section 2, we recall some necessary combinatorial identities involving harmonic numbers and prove a preliminary congruence. The proofs of Theorems \ref{t-1}--\ref{t-3} are presented in Sections 3--5, respectively.
\section{Preliminary results}
Let
\begin{align*}
H_n=\sum_{j=1}^n\frac{1}{j}
\end{align*}
denote the $n$th harmonic number. The Fermat quotient of an integer $a$ with respect to an odd prime $p$ is given by $q_p(a)=(a^{p-1}-1)/p$.
In order to prove Theorems \ref{t-1} and \ref{t-2}, we need the following two lemmas.
\begin{lem}
For any non-negative integer $n$, we have
\begin{align}
&\sum_{i=0}^{n}(-1)^i{n\choose i}{n+i\choose i}=(-1)^n,\label{bb-1}\\
&\sum_{i=0}^{n}(-1)^i{n\choose i}{n+i\choose i}H_i=2(-1)^nH_n,\label{bb-2}\\
&\sum_{i=0}^{n}(-1)^i{n\choose i}{n+i\choose i}H_{n+i}=2(-1)^nH_n.\label{bb-3}
\end{align}
\end{lem}
In fact, such identities can be discovered and proved by the symbolic summation package {\tt Sigma} developed by Schneider \cite{schneider-slc-2007}. One can also refer to \cite{liu-jsc-2019} for the same approach to finding and proving identities of this type. For human proofs of \eqref{bb-1}--\eqref{bb-3}, one refers to \cite{prodinger-integers-2008}.
\begin{lem}
For any prime $p\ge 5$, we have
\begin{align}
\sum_{k=0}^{p-1}
\frac{(3k)!}{3^{3k} k!^3}\left(H_{3k}-H_k\right)\equiv \left(\frac{-3}{p}\right)q_p(3)\pmod{p}.
\label{bb-4}
\end{align}
\end{lem}
{\noindent\it Proof.}
Note that
\begin{align}
\sum_{k=0}^{p-1}
\frac{(3k)!}{3^{3k} k!^3}\left(3H_{3k}-H_k\right)
=\sum_{k=0}^{p-1}\frac{(1/3)_k(2/3)_k}{(1)_k^2}\sum_{j=0}^{k-1}
\left(\frac{1}{1/3+j}+\frac{1}{2/3+j}\right).
\label{bb-5}
\end{align}
Recall the following identity due to Tauraso \cite[Theorem 1]{tauraso-integers-2012}:
\begin{align}
\frac{(1/3)_k(2/3)_k}{(1)_k^2}\sum_{j=0}^{k-1}\left(\frac{1}{1/3+j}+\frac{1}{2/3+j}\right)
=\sum_{j=0}^{k-1}\frac{(1/3)_j(2/3)_j}{(1)_j^2}\cdot \frac{1}{k-j}.
\label{bb-6}
\end{align}
Substituting \eqref{bb-6} into \eqref{bb-5} and exchanging the summation order gives
\begin{align}
\sum_{k=0}^{p-1}
\frac{(3k)!}{3^{3k} k!^3}\left(3H_{3k}-H_k\right)
&=\sum_{k=0}^{p-1}\sum_{j=0}^{k-1}\frac{(1/3)_j(2/3)_j}{(1)_j^2}\cdot \frac{1}{k-j}\notag\\[10pt]
&=\sum_{j=0}^{p-2}\frac{(1/3)_j(2/3)_j}{(1)_j^2}H_{p-1-j}\notag\\[10pt]
&\equiv \sum_{j=0}^{p-2}\frac{(3j)!}{3^{3j} j!^3}H_{j}\pmod{p},\label{bb-7}
\end{align}
where we have utilized the fact that $H_{p-1-j}\equiv H_j\pmod{p}$.
By \eqref{bb-7}, we obtain
\begin{align}
\sum_{k=0}^{p-1}\frac{(3k)!}{3^{3k} k!^3}\left(H_{3k}-H_k\right)
&\equiv \frac{1}{3}\left(\sum_{k=0}^{p-2}\frac{(3k)!}{3^{3k} k!^3}H_{k}-2\sum_{k=0}^{p-1}\frac{(3k)!}{3^{3k} k!^3}H_k\right)\notag\\[10pt]
&\equiv -\frac{1}{3}\sum_{k=0}^{\lfloor p/3\rfloor}\frac{(3k)!}{3^{3k} k!^3}H_{k} \pmod{p},
\label{bb-8}
\end{align}
because $(3k)!\equiv 0\pmod{p}$ for $k>\lfloor p/3\rfloor$.
Let $m=\lfloor p/3\rfloor$. From \cite[Lemma 2.3]{at-aa-2016}, we see that for $0\le k\le m$,
\begin{align}
\frac{(3k)!}{3^{3k} k!^3}\equiv (-1)^k{m\choose k}{m+k\choose k}\pmod{p}.
\label{bb-9}
\end{align}
It follows from \eqref{bb-2}, \eqref{bb-8} and \eqref{bb-9} that
\begin{align*}
\sum_{k=0}^{p-1}\frac{(3k)!}{3^{3k} k!^3}\left(H_{3k}-H_k\right)
&\equiv -\frac{1}{3}\sum_{k=0}^m(-1)^k{m\choose k}{m+k\choose k}H_k\pmod{p}\\[10pt]
&=-\frac{2(-1)^m}{3}H_m.
\end{align*}
Finally, noting
\begin{align}
(-1)^{\lfloor p/3\rfloor}=\left(\frac{-3}{p}\right),\label{bb-10}
\end{align}
and the following congruence \cite[page 359]{lehmer-am-1938}:
\begin{align}
H_{\lfloor p/3\rfloor}\equiv -\frac{3}{2}q_p(3) \pmod{p^2},\label{bb-11}
\end{align}
we complete the proof of \eqref{bb-4}.
\qed
\section{Proof of Theorem \ref{t-1}}
We begin with the transformation formula due to Chan and Zudilin \cite[Corollary 4.3]{cz-m-2010}:
\begin{align}
\gamma_n=\sum_{i=0}^n{2i\choose i}^2{4i\choose 2i}{n+3i\choose 4i}(-3)^{3(n-i)}.\label{cc-1}
\end{align}
Using \eqref{cc-1} and exchanging the summation order, we obtain
\begin{align}
\sum_{k=0}^{p-1}\frac{4k+1}{81^k}\gamma_k
&=\sum_{k=0}^{p-1}\frac{4k+1}{81^k}\sum_{i=0}^k{2i\choose i}^2{4i\choose 2i}{k+3i\choose 4i}(-3)^{3(k-i)}\notag\\[10pt]
&=\sum_{i=0}^{p-1}\frac{1}{(-3)^{3i}}{2i\choose i}^2{4i\choose 2i}\sum_{k=i}^{p-1}\frac{4k+1}{(-3)^k}{k+3i\choose 4i}.\label{cc-2}
\end{align}
Note that
\begin{align}
\sum_{k=i}^{n-1}\frac{4k+1}{(-3)^k}{k+3i\choose 4i}=(n-i){n+3i\choose 4i}(-3)^{1-n},\label{cc-3}
\end{align}
which can be easily proved by induction on $n$. Combining \eqref{cc-2} and \eqref{cc-3} gives
\begin{align}
\sum_{k=0}^{p-1}\frac{4k+1}{81^k}\gamma_k
=3^{1-p}\sum_{i=0}^{p-1}\frac{p-i}{(-3)^{3i}}{2i\choose i}^2{4i\choose 2i}{p+3i\choose 4i}.\label{cc-4}
\end{align}
Furthermore, we have
\begin{align*}
&(-1)^i(p-i){2i\choose i}^2{4i\choose 2i}{p+3i\choose 4i}\\[10pt]
&=\frac{(-1)^ip(p+3i)\cdots (p+1)(p-1)\cdots (p-i)}{i!^4}\\[10pt]
&\equiv \frac{p(3i)!}{i!^3}\left(1+p\left(H_{3i}-H_i\right)\right)\pmod{p^3}.
\end{align*}
Thus,
\begin{align*}
\sum_{k=0}^{p-1}\frac{4k+1}{81^k}\gamma_k
&\equiv 3^{1-p}p\sum_{i=0}^{p-1}
\frac{(3i)!}{3^{3i} i!^3}\left(1+p\left(H_{3i}-H_i\right)\right)\pmod{p^3}.
\end{align*}
Finally, noting \eqref{bb-4} and Mortenson's supercongruence \cite[(1.2)]{mortenson-tams-2003}:
\begin{align*}
\sum_{i=0}^{p-1}
\frac{(3i)!}{3^{3i} i!^3}
\equiv \left(\frac{-3}{p}\right) \pmod{p^2},
\end{align*}
we arrive at
\begin{align*}
\sum_{k=0}^{p-1}\frac{4k+1}{81^k}\gamma_k
&\equiv p\left(\frac{-3}{p}\right)\left(3^{1-p}+3^{1-p}pq_p(3)\right)\pmod{p^3}\\[10pt]
&=p\left(\frac{-3}{p}\right),
\end{align*}
as desired.
\section{Proof of Theorem \ref{t-2}}
Recall the following transformation formula \cite[(5.1)]{sunzh-itsf-2015}:
\begin{align}
\gamma_n=\sum_{i=0}^{\lfloor n/3 \rfloor}{2i\choose i}^2{4i\choose 2i}{n+i\choose 4i}(-3)^{n-3i}.\label{dd-1}
\end{align}
By \eqref{dd-1}, we have
\begin{align}
\sum_{k=0}^{p-1}(4k+3)\gamma_k
&=\sum_{k=0}^{p-1}(4k+3)\sum_{i=0}^{\lfloor k/3 \rfloor}{2i\choose i}^2{4i\choose 2i}{k+i\choose 4i}(-3)^{k-3i}\notag\\[10pt]
&=\sum_{i=0}^{p-1}\frac{1}{(-3)^{3i}}{2i\choose i}^2{4i\choose 2i}\sum_{k=i}^{p-1}(-3)^k(4k+3){k+i\choose 4i}.\label{dd-2}
\end{align}
It can be easily proved by induction on $n$ that
\begin{align}
\sum_{k=i}^{n-1}(-3)^k(4k+3){k+i\choose 4i}=3(n-3i){n+i\choose 4i}(-3)^{n-1}.\label{dd-3}
\end{align}
It follows from \eqref{dd-2} and \eqref{dd-3} that
\begin{align*}
\sum_{k=0}^{p-1}(4k+3)\gamma_k
=3^{p}\sum_{i=0}^{\lfloor p/3\rfloor}\frac{p-3i}{(-3)^{3i}}{2i\choose i}^2{4i\choose 2i}{p+i\choose 4i}.
\end{align*}
Note that
\begin{align*}
&(-1)^i(p-3i){2i\choose i}^2{4i\choose 2i}{p+i\choose 4i}\\[10pt]
&=\frac{(-1)^ip(p+i)\cdots(p+1)(p-1)\cdots(p-3i)}{i!^4}\\[10pt]
&\equiv \frac{p(3i)!}{i!^3}\left(1-p\left(H_{3i}-H_i\right)\right)\pmod{p^3}.
\end{align*}
Thus,
\begin{align}
\sum_{k=0}^{p-1}(4k+3)\gamma_k
&\equiv 3^{p}p
\sum_{i=0}^{\lfloor p/3\rfloor}\frac{(3i)!}{3^{3i} i!^3}\left(1-p\left(H_{3i}-H_i\right)\right)\pmod{p^3}.\label{dd-4}
\end{align}
Let $m=\lfloor p/3\rfloor$. Since
\begin{align*}
\frac{(3i)!}{3^{3i} i!^3}=(-1)^i{-1/3\choose i}{-1/3+i\choose i}
=(-1)^i{-2/3\choose i}{-2/3+i\choose i},
\end{align*}
we have
\begin{align}
\frac{(3i)!}{3^{3i} i!^3}
&=(-1)^i{m-p/3\choose i}{m-p/3+i\choose i}\notag\\[10pt]
&=\frac{(-1)^i(m+i-p/3)\cdots (m-i+1-p/3)}{i!^2}\notag\\[10pt]
&\equiv (-1)^i{m\choose i}{m+i\choose i}\left(1-\frac{p}{3}\left(H_{m+i}-H_{m-i}\right)\right)\pmod{p^2}.
\label{dd-5}
\end{align}
Substituting \eqref{dd-5} into the right-hand side of \eqref{dd-4} gives
\begin{align}
&\sum_{k=0}^{p-1}(4k+3)\gamma_k\notag\\
&\equiv 3^{p}p
\sum_{i=0}^{m}(-1)^i{m\choose i}{m+i\choose i}\notag\\
&\times \left(1-\frac{p}{3}\left(H_{m+i}-H_{m-i}+3H_{3i}-3H_i\right)\right)
\pmod{p^3}.\label{dd-6}
\end{align}
Furthermore, we have
\begin{align}
H_{3i}&=\frac{1}{3}\left(H_i+\sum_{j=1}^i\frac{1}{j-1/3}+\sum_{j=1}^i\frac{1}{j-2/3}\right)\notag\\[10pt]
&\equiv \frac{1}{3}\left(H_i+\sum_{j=1}^i\frac{1}{m+j}-\sum_{j=1}^i\frac{1}{m+1-j}\right)\pmod{p}\notag\\[10pt]
&=\frac{1}{3}\left(H_i+H_{m+i}+H_{m-i}-2H_m\right).\label{dd-7}
\end{align}
It follows from \eqref{bb-1}--\eqref{bb-3}, \eqref{dd-6} and \eqref{dd-7} that
\begin{align*}
&\sum_{k=0}^{p-1}(4k+3)\gamma_k\\
&\equiv 3^{p}p
\sum_{i=0}^{m}(-1)^i{m\choose i}{m+i\choose i}\left(1-\frac{2p}{3}\left(H_{m+i}-H_m-H_i\right)\right)
\pmod{p^3}\\
&=3^{p}p(-1)^m\left(1+\frac{2p}{3}H_m\right).
\end{align*}
Finally, using \eqref{bb-10} and \eqref{bb-11}, we obtain
\begin{align*}
\sum_{k=0}^{p-1}(4k+3)\gamma_k
&\equiv 3p\left(\frac{-3}{p}\right)3^{p-1}\left(2-3^{p-1}\right)\\[10pt]
&=3p\left(\frac{-3}{p}\right)\left(1-\left(3^{p-1}-1\right)^2\right)\\[10pt]
&\equiv 3p\left(\frac{-3}{p}\right)\pmod{p^3},
\end{align*}
where we have used the Fermat's little theorem in the last step.
\section{Proof of Theorem \ref{t-3}}
Recall the following transformation formula \cite[Lemma 4.1]{sunzh-itsf-2015}:
\begin{align}
\gamma_n=\sum_{i=0}^{n}(-9)^{n-i}{2i\choose i}{n+i\choose 2i}\sum_{j=0}^i{i\choose j}^2{2j\choose j}.\label{ee-1}
\end{align}
Let
\begin{align*}
g_n=\sum_{k=0}^n{n\choose k}^2{2k\choose k}.
\end{align*}
By \eqref{ee-1}, we have
\begin{align}
\sum_{k=0}^{p-1}\frac{2k+1}{(-9)^k}\gamma_k
&=\sum_{k=0}^{p-1}\frac{2k+1}{(-9)^k}\sum_{i=0}^{k}(-9)^{k-i}{2i\choose i}{k+i\choose 2i}g_i\notag\\[10pt]
&=\sum_{i=0}^{p-1}\frac{g_i}{(-9)^i}\sum_{k=i}^{p-1}(2k+1){k+i\choose 2i}{2i\choose i}.\label{ee-2}
\end{align}
Note that
\begin{align}
\sum_{k=i}^{n-1}(2k+1){k+i\choose 2i}{2i \choose i}
=\frac{n^2}{i+1}{n-1\choose i}{n+i\choose i},\label{ee-3}
\end{align}
which can be proved by induction on $n$. It follows from \eqref{ee-2} and \eqref{ee-3} that
\begin{align*}
\sum_{k=0}^{p-1}\frac{2k+1}{(-9)^k}\gamma_k
=p^2\sum_{i=0}^{p-1}\frac{g_i}{(-9)^i(i+1)} {p-1\choose i}{p+i\choose i}.
\end{align*}
Since
\begin{align*}
{p-1\choose i}{p+i\choose i}\equiv (-1)^i\pmod{p^2},
\end{align*}
we have
\begin{align}
\sum_{k=0}^{p-1}\frac{2k+1}{(-9)^k}\gamma_k\equiv p^2\sum_{i=0}^{p-1}\frac{g_i}{9^i(i+1)}\pmod{p^3}.
\label{ee-4}
\end{align}
From \cite[Lemma 2.7]{jv-rj-2010}, we see that for $0\le i\le p-1$,
\begin{align*}
\frac{g_i}{9^i}\equiv \left(\frac{-3}{p}\right)g_{p-1-i}\pmod{p},
\end{align*}
and so
\begin{align*}
\sum_{i=0}^{p-2}\frac{g_i}{9^i(i+1)}
&\equiv \left(\frac{-3}{p}\right)\sum_{i=0}^{p-2}\frac{g_{p-1-i}}{i+1}\\[10pt]
&=\left(\frac{-3}{p}\right)\sum_{i=1}^{p-1}\frac{g_{i}}{p-i}\\[10pt]
&\equiv -\left(\frac{-3}{p}\right)\sum_{i=1}^{p-1}\frac{g_{i}}{i}\pmod{p}.
\end{align*}
Using the congruence \cite[(1.8)]{sunzw-rj-2016}:
\begin{align*}
\sum_{i=1}^{p-1}\frac{g_{i}}{i}\equiv 0\pmod{p},
\end{align*}
we obtain
\begin{align}
\sum_{i=0}^{p-2}\frac{g_i}{9^i(i+1)}\equiv 0\pmod{p}.\label{ee-5}
\end{align}
Furthermore, combining \eqref{ee-4} and \eqref{ee-5} gives
\begin{align*}
\sum_{k=0}^{p-1}\frac{2k+1}{(-9)^k}\gamma_k\equiv \frac{pg_{p-1}}{9^{p-1}}\pmod{p^3}.
\end{align*}
By \cite[Lemma 3.2]{sunzw-rj-2016}, we have
\begin{align*}
g_{p-1}\equiv \left(\frac{-3}{p}\right)\left(2\cdot3^{p-1}-1\right)\pmod{p^2},
\end{align*}
and so
\begin{align*}
\sum_{k=0}^{p-1}\frac{2k+1}{(-9)^k}\gamma_k
&\equiv p\left(\frac{-3}{p}\right)
\left(1-\frac{\left(3^{p-1}-1\right)^2}{9^{p-1}}\right)\\[10pt]
&\equiv p\left(\frac{-3}{p}\right)\pmod{p^3},
\end{align*}
where we have utilized the Fermat's little theorem.
{\noindent\bf Remark.}
Z.-H. Sun \cite[Conjecture 6.8]{sunzh-itsf-2015} also conjectured a companion supercongruence of
\eqref{aa-4}:
\begin{align}
\sum_{k=0}^{p-1}\frac{2k+1}{9^k}\gamma_k \equiv \left(\frac{-3}{p}\right)p\pmod{p^3}.
\label{ee-6}
\end{align}
In a similar way, by using \eqref{ee-1} and the following identity:
\begin{align*}
\sum_{k=i}^{n-1}(-1)^k(2k+1){k+i\choose 2i}{2i \choose i}
=(-1)^{n-1}n{n-1\choose i}{n+i\choose i},
\end{align*}
we can show that
\begin{align*}
\sum_{k=0}^{p-1}\frac{2k+1}{9^k}\gamma_k
\equiv p\sum_{i=0}^{p-1}\frac{g_i}{9^i}\pmod{p^3}.
\end{align*}
Thus, the conjectural supercongruence \eqref{ee-6} is equivalent to
\begin{align*}
\sum_{i=0}^{p-1}\frac{g_i}{9^i}\equiv \left(\frac{-3}{p}\right)\pmod{p^2},
\end{align*}
which was originally conjectured by Z.-W. Sun \cite[Remark 1.1]{sunzw-rj-2016}.
\vskip 5mm \noindent{\bf Acknowledgments.}
This work was supported by the National Natural Science Foundation of China (grant 11801417). | 0.01293 |
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Local in the June rioting in Maungdaw and Buthidaung townships.
A group of Rakhine victims after their homes are burned down by the Bangladeshi immigrants
According to a police officer from Buthidaung, all legal proceedings including investigation have already done and the accused are only waiting for the decision when they receive the release order. It is said that Saw Myint (a) Mamud, Zaw Myint and Than Naing (a) Adu High of UNHCR from Buthidaung and Win Naing and Kyaw Hla Aung of Doctors Without Borders (MSF) from Sittwe prison are among the 10 terrorist suspects who were suddenly released by the Government without any official explanation. "The release order is signed by an officer named Pyone Cho from the Home Ministry affairs" the officer added. A Buthidaung resident who is also a lawyer told RMG that he didn't believe the suspects can be released easily like this without a special diplomatic pressure from one of the world's super power country like United States. Since they belong to illegally migrated Bangladeshi Muslim community and not the US citizens, the UN and INGO operatives must have paid a large amount of money to the officials. I am sure that they (UN and INGO officials) exploited our country's unstable law and order situation, and bad governance in helping terrorists."
Illegal Bangladeshi immigrants and terrorist suspects Kyaw Hla Aung and Win Naing in front of Sthe ittwe prison gate. (photo-Mizzima)
It is reported that the accused Bengali Muslim migrants used their organizational computers and internet facilities to communicate with the Muslim extremist groups from Pakistan, UK and Middle Eastern countries. Their irresponsible propagation of false Muslim massacre by the local Buddhist Rakhine and Myanmar armed security forces led an uncontrollable community riot between local Buddhist Rakhine and Bengali Muslim immigrants. The said violence, according to Myanmar government report, have resulted 31 Rakhines and 46 Bengalis dead and more than four thousand houses burned down.
The UNHCR and INGO organizations that are working in Rakhine state have a long history of discriminating against the local people as they are only giving a special benefits to the Bengali immigrants while totally neglecting local Rakhine and other Myanmar ethnic people. As a result of this, the Rakhine and other ethnic victims from all over Rakhine state denied the assistance offered to them by the UN and INGOs after many Rakhine died and thousands of their houses are burned down by the Bengali migrants in the riot. "The government has a weakness in implementing law and order of the country. We are glad to hear that a man is freed from the prison.", the Rakhine Nationalities Development Party (RNDP) general secretary U U Hla Saw told to Narinjara News in an interview. Those accused are somehow involved in the riot. No one should be above the law but the release of the accused without the punishment is a sad news for our Rakhine people", he continued.
UN and INGOs are being accused of racial discrimination and demanded by local ethnic people to get out of their land
Many Myanmar citizens including Myanmar people living around the world are also concerned about government's decision to release the terrorist suspects, as this move will encourage Bangladeshi illegal immigrants who want to create a new Islamic country, to create more problems against the local ethnic people. | 0.011828 |
In this section, we take a closer look at the geometry of $\mathfrak X$, and compare it with related moduli spaces.
\subsection{Comparison of $\mathfrak{X}$ with the spaces of weighted admissible covers $\overline{\mathcal{H}}_4^3(1/6+\epsilon)$}
Recall that $\overline{\mathcal H}_4^3(1/6 + \epsilon)$ is the moduli space of weighted admissible covers where up to 5 branch points are allowed to coincide.
Let $U \subset \overline{\mathcal H}_4^3(1/6 + \epsilon)$ be the open substack parametrizing $\phi \from C \to P$ where $P \cong \PP^1$, the curve $C$ is smooth, and the Tschirnhausen bundle $E_\phi$ of $\phi$ is $\O(3) \oplus \O(3)$.
We have a morphism $\Phi \from U \to \mathfrak X$ given by the transformation
\[ (\phi \from C \to P) \mapsto (\PP E_\phi, C).\]
\begin{theorem}\label{thm:phi_extends}
The map $\Phi$ extends to a morphism of stacks $\Phi \from \overline{\mathcal{H}}_4^3(1/6+\epsilon) \to \mathfrak X$.
\end{theorem}
Since $\overline{\mathcal{H}}_4^3(1/6+\epsilon)$ is proper and $\mathfrak X$ is separated, the map $\Phi$ is also proper.
For the proof, we need extension lemmas for morphisms of stacks, extending some well-known results for schemes.
Let $\mathcal X$ and $\mathcal Y$ be separated Deligne--Mumford stacks of finite type over a field $\k$.
Let $\mathcal U \subset \mathcal X$ be a dense open substack.
\begin{lemma}\label{lem:unique_extension}
Assume that $\mathcal X$ is normal.
If $\Phi_1, \Phi_2 \from \mathcal X \to \mathcal Y$ are two morphisms whose restrictions to $\mathcal U$ are equal (2-isomorphic), then $\Phi_1$ and $\Phi_2$ are equal (2-isomorphic).
\end{lemma}
\begin{proof}
We have the following diagram where the square is a pull-back
\[
\begin{tikzpicture}
\node (Y) {$\mathcal Y$};
\node[below of=Y] (YY) {$\mathcal Y \times \mathcal Y$};
\draw (YY) ++ (-2cm,0) node (X) {$\mathcal X$};
\node (U) [left of=X] {$\mathcal U$};
\node[above of=X] (Z) {$\mathcal Z$};
\draw[->] (Z) edge (Y) (Z) edge (X) (U) edge (X) (X) edge node[above] {\tiny $(\Phi_1, \Phi_2)$} (YY) (Y) edge node[right] {\tiny $\Delta$} (YY);
\end{tikzpicture}
\]
Since $\mathcal Y$ is a separated Deligne--Mumford stack, the diagonal map $\Delta$ is representable, proper, and unramified.
Therefore, so is the pullback $\mathcal Z \to \mathcal X$.
Since $\Phi_1$ and $\Phi_2$ agree on $\mathcal U$, the inclusion $\mathcal U \to \mathcal X$ lifts to $\mathcal U \to \mathcal Z$.
Since $\mathcal Z \to \mathcal X$ is unramified and $\mathcal U \to \mathcal X$ is an open immersion, so is the lift $\mathcal U \to \mathcal Z$.
Let $\overline {\mathcal U} \subset \mathcal Z$ be the closure of $\mathcal U$ and $\overline{\mathcal U}^\nu \to \overline{\mathcal U}$ its normalization.
Since $\mathcal X$ is normal, Zariski's main theorem implies that $\overline {\mathcal U}^\nu \to \mathcal X$ is an isomorphism.
Hence the map $\mathcal Z \to \mathcal X$ admits a section $\mathcal X \to \mathcal Z$.
In other words, the map $(\Phi_1, \Phi_2) \from \mathcal X \to \mathcal Y \times \mathcal Y$ factors through the diagonal $\mathcal Y \to \mathcal Y \times \mathcal Y$.
\end{proof}
\begin{example}
In \autoref{lem:unique_extension}, we can drop the normality assumption on $\mathcal X$ if $\mathcal Y$ is an algebraic space, but not otherwise.
An example of distinct maps that agree on a dense open substack can be constructed using twisted curves (see \cite[Proposition~7.1.1]{abr.cor.vis:03}).
Let $\mathcal X$ be the stack
\[ \mathcal X = \left[ \spec \C[x,y]/xy / \mu_n \right],\]
where $\zeta \in \mu_n$ acts by $(x, y) \mapsto (\zeta x, \zeta^{-1} y)$.
Every $\zeta \in \mu_n$ defines an automorphism of $t_\zeta \from \mathcal X \to \mathcal X$ given by $(x,y) \mapsto (x, \zeta y)$.
The map $t_\zeta$ is the identity map on the complement of the node of $\mathcal X$, but not the identity map on $\mathcal X$ if $\zeta \neq 1$.
\end{example}
The map $\Phi \from \mathcal U \to \mathcal Y$ induces a map $|\Phi| \from |\mathcal U| \to |\mathcal Y|$ on the set of points.
Let $\phi \from |\mathcal X| \to |\mathcal Y|$ be an extension of $|\Phi|$.
We say that $\phi$ is \emph{continuous in one-parameter families} if for every DVR $\Delta$ and a map $i \from \Delta \to \mathcal X$ that sends the generic point $\eta$ of $\Delta$ to $\mathcal U$, the map $\Phi \circ i \from \eta \to \mathcal Y$ extends to a map $\Delta \to \mathcal Y$ and agrees with the map $\phi$ on the special point.
\begin{lemma}\label{lem:exists_extension}
Suppose $\mathcal X$ is smooth, and $\Phi \from \mathcal U \to \mathcal Y$ is a morphism.
Let $\phi \from |\mathcal X| \to |\mathcal Y|$ be an extension of $|\Phi| \from |\mathcal U| \to |\mathcal Y|$.
If $\phi$ is continuous in one-parameter families, then it is induced by a morphism $\Phi \from \mathcal X \to \mathcal Y$ that extends $\Phi \from \mathcal U \to \mathcal Y$.
\end{lemma}
By \autoref{lem:unique_extension}, the extension is unique.
\begin{proof}
Consider the map $(\id, \Phi) \from \mathcal U \to \mathcal X \times \mathcal Y$.
Let $\mathcal Z \subset \mathcal X \times \mathcal Y$ be the scheme theoretic image of $\mathcal U$ (see \cite[Tag 0CMH]{sta:20}), and let let $\mathcal Z^\nu \to \mathcal Z$ be the normalization.
By construction, the map $\mathcal Z \to \mathcal X$ is an isomorphism over $\mathcal U$ \cite[Tag 0CPW]{sta:20}.
Since $\mathcal U$ is smooth (and hence normal), the map $\mathcal Z^\nu \to \mathcal X$ is also an isomorphism over $\mathcal U$.
Our aim is to show that $\mathcal Z^\nu \to \mathcal X$ is in fact an isomorphism.
Let $Z^\nu \to X$ be the morphism on coarse spaces induced by $\mathcal Z^\nu \to \mathcal X$.
Let $(x, y) \in |\mathcal X| \times |\mathcal Y|$ be a point from $Z^\nu$.
Since the image of $\mathcal U$ is dense in $\mathcal Z^\nu$, there exists a DVR $\Delta$ with a map $\Delta \to \mathcal Z^\nu$ whose generic point maps into the image of $\mathcal U$ and whose special point maps to $(x,y)$.
The continuity of $\phi$ in one-parameter families implies that $y = \phi(x)$.
As a result, $Z^\nu \to X$ is a bijection on points.
As $Z^\nu$ and $X$ are normal spaces, $Z^\nu \to X$ must be an isomorphism.
By hypothesis, for every DVR $\Delta$, a map $\Delta \to \mathcal X$ that sends the generic point to $\mathcal U$ lifts to a map $\Delta \to \mathcal Y$, and hence to a map $\Delta \to \mathcal Z^\nu$.
This implies that $\mathcal Z^\nu \to \mathcal X$ is unramified in codimension 1.
It follows by the same arguments as in \cite[Corollary~6]{ger.sat:17} that $\mathcal Z^\nu \to \mathcal X$ is an isomorphism.
Since \cite[Corollary~6]{ger.sat:17} is stated with slightly stronger hypotheses, we recall the proof.
Let $V$ be a scheme and $V \to \mathcal X$ an \'etale morphism.
Set $\mathcal W = \mathcal Z^\nu \times_{\mathcal X} V$ and $U = \mathcal U \times_{\mathcal X} V$.
Let $\mathcal W \to W$ be the coarse space.
The map $W \to V$ is an isomorphism over the dense open subset $U \subset V$, and is a quasi-finite map between normal spaces.
By Zariski's main theorem, it is an isomorphism.
Furthermore, as $\mathcal W \to V$ is unramified in codimension 1, so is $\mathcal W \to W$.
Since $\mathcal W$ is normal and $W$ is smooth, purity of the branch locus \cite[Tag 0BMB]{sta:20} implies that $\mathcal W \to W$ is \'etale.
As $\mathcal W \to V$ is an isomorphism over $U$, we see that $\mathcal W$ contains a copy of $U$ as a dense open substack.
In particular, $\mathcal W$ has trivial generic stabilizers.
By \cite[Lemma~4]{ger.sat:17}, we conclude that $\mathcal W \to W$ is an isomorphism.
Since both $\mathcal W \to W$ and $W \to V$ are isomorphisms, their composite $\mathcal W \to V$ is an isomorphism.
We have proved that $\mathcal Z^\nu \to \mathcal X$ is an isomorphism \'etale locally on $\mathcal X$.
We conclude that $\mathcal Z^\nu \to \mathcal X$ is an isomorphism.
The composite of the inverse of $\mathcal Z^\nu \to \mathcal X$, the map $\mathcal Z^\nu \to \mathcal X \times \mathcal Y$, and the projection onto $\mathcal Y$ gives the required extension $\Phi \from \mathcal X \to \mathcal Y$.
\end{proof}
\begin{proof}[Proof of \autoref{thm:phi_extends}]
Define a map $\phi \from |\overline{\mathcal {H}}_4^3(1/6+\epsilon)| \to |\mathcal X|$, consistent with the map induced by $\Phi$ on $U$ as follows.
Let $k$ be an algebraically closed field and $a \from \spec k \to \overline{\mathcal{H}}_4^3(1/6 + \epsilon)$ a map.
Let $\Delta$ be a DVR with residue field $k$ and $\alpha \from \Delta \to \overline{\mathcal{H}}_4^3(1/6 + \epsilon)$ a map that restricts to $a$ at the special point and maps the generic point $\eta$ to $U$.
By \autoref{prop:stab}, there exists an extension $\beta \from \Delta \to \mathfrak X$ of $\Phi \circ \alpha |_\eta$.
Let $b \from \spec k \to \mathfrak X$ be the central point of the extension.
Set $\phi(a) = b$.
\autoref{prop:stab} guarantees that $b$ depends only on $a$, and not on $\alpha$; so the map $\phi$ is well-defined.
\autoref{prop:stab} also guarantees that $\phi$ is continuous in one-parameter families.
Since $\overline{\mathcal H}^3_4(1/6 + \epsilon)$ is smooth, \autoref{lem:exists_extension} applies and yields the desired extension.
\end{proof} | 0.005571 |
PUBLIC RECORD
312 Illinois St Lawrence, KS 66044 (Pinckney)See your commute times
- Single-Family Home
- 3 Bedrooms
- 2 full Bathrooms
- 900 sqft
- Lot size: 6,098 sqft
- Built in 1950
- Edit Home Facts
Property Details for 312 Illinois St
Description provided by Trulia
This is a Single-Family Home located at 312 Illinois Street, Lawrence, KS. 312 Illinois St has 3 beds, 2 baths, and approximately 900 square feet. The property has a lot size of 1 and was built in 1950. The average list price for similar homes for sale is $188,170. 312 Illinois St is in the Pinckney neighborhood in Lawrence, KS. The average list price for Pinckney is $98,981.
Public Records for 312 Illinois St
Official property, sales, and tax information from county (public) records:
- 3 Bedrooms
- Lot Size: 6,098 sqft
- 2 Bathrooms
- Built In 1950
- 900 sqft
- County: Douglas | 0.001024 |
TITLE: Proof of Cancellation Law for Multiplication in N (Induction)
QUESTION [2 upvotes]: I'm struggling a bit to prove the Cancellation Law for multiplication for the Natural numbers, using induction.
So far I have this:
Having $a, b \in N$ e $S = \lbrace c \in N | a \cdot c = b \cdot c \longrightarrow a = b \rbrace$
We know that $1 \in S$, since $a \cdot 1 = a = b = b \cdot 1$, 0 is not the base case because even if $a \neq b$, $a \cdot 0 = 0 = b \cdot 0$.
Using $k \in S$, by definition $a \cdot k = b \cdot k \longrightarrow a=b$, we have to show that $k' \in S$
Assuming that $a \cdot k' = b \cdot k'$, we have $a \cdot k + a = b \cdot k + b$, and I don't know where to go from here.
I tried using something like this, however I'm still stuck.
Thanks for any help
REPLY [0 votes]: There is another approach which requires defining a map:-
Let $x,y\in\mathbb{N}$. Define set $T=\{z\in\mathbb{N}\setminus\{0\}:xz=yz \implies x=y\}$.
Instead of showing that $T=\mathbb{N}\setminus\{0\}$, we define $S=T\cup\{0\}$ and show that $S=N$ by peano's inductive axiom.
Let $s:\mathbb{N}\to \mathbb{N}$ be the successor function with properties as defined in Peano's $5$ axioms characterizing $\mathbb{N}$.
$0\in S$ (by construction).
Assume for $z\in\mathbb{N}$, we have $z\in S$, i.e., $xz=yz \implies x=y$.
Define map $f:\mathbb{N}\to\mathbb{N}$ as $f(n)=n\times s(z)$, where $z$ is a fixed number as assumed above.
It is easy to show that $f$ is injective.
(Inductive step): $x\cdot s(z) = y\cdot s(z)$
$\implies f(x)=f(y)$
$\implies x=y$ ($\because f$ is injective).
Thus, we have shown: $z\in S \implies s(z)\in S$. Hence, $S=\mathbb{N}$. | 0.018726 |
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MAZ 5432
Tested: 1.8.2.5s
- 2014-01-18 10:36
- Author and chassis tuning: Bruieser
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- Author envelope in ETS2: Dalnoboishik ™ (ie I)
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- Reteksturing salon vneshke: Versetti (Gray Tabby) | 0.0036 |
The.
[RELATED: Mark Ragan and Jim Ylisela present advanced writing and editing tips for corporate communicators. Join them in Chicago, Washington D.C., or Denver!]
So you're telling me I spent all that time memorizing state abbreviations only to throw them out now?
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So you're telling me I spent all that time memorizing state abbreviations only to throw them out now?
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Thank you to our members and guests for attending today’s Educational Short Course on the Valuation Process in Ag Lending.
Thank you to Paige Gilligan and Stan Xavier for their time and for sharing their knowledge and experiences. The presentations were very insightful and tied together very well.
Also, thank you to MetLife Agricultural for allowing Ag Lenders' use of the beautiful conference room.
Past Short CoursesEducational Short Course - August 1, 2019, presented by Paige Gilligan and Stan XavierLegal Short Course - February 26, 2019, presented by Baker Manock & JensenLegal Short Course - October 26, 2018, presented by Steven Turner, Michael Gomez, and Frank OliverAccounting Short Course - September 19, 2018, presented by Baker Peterson Franklin, CPAEducational Short Course - July 12, 2018 - presented by First American TitleLegal Short Course - February 22, 2018 - presented by Bolen Fransen Sawyers Legal Short Course - February 28, 2017 - presented by FrandzelEducational Short Course - July 14, 2017 - presented by Old Republic Title Legal Short Course - 2016 - presented by Bolen Fransen Sawyers LLPAccounting Short Course - 2015 - presented by Baker Peterson Franklin, CPA, LLP
Copyright © 2020 Ag Lenders Society of California | Contact Us | 0.365674 |
Day one of Investinblue, (15-16thJune) will include an interactive, industry led, panel debate, highlighting the environmental challenges faced by the maritime industry.
“This House believes that the UK and other countries are meeting the economic resilience requirements to address increasing storms” will be the topical debate closing the first day of the Investinblue conference, taking place during IFB2016.
Peter Green, CEO of National Maritime (NMDG), organisers of Investinblue, said; “The debate will essentially focus on the increasing ratio and force of storms, and whether the UK and other countries may have to re-address the economic resilience factor and necessary infrastructure and innovations that may be required.” The complex pattern of storm damage affecting economic marine activity, inland urban and rural areas and key infrastructure is questioning whether existing resources and capabilities are sufficient. It is likely that if economic activities and growth plans are to be realised, the scale and approach towards such resilience may have to be more transformational.
This presents opportunities for new industry growth and refined flood management, within a climate challenging economy. Recent events have proven the case that despite investment in capital projects, the customer (house owners and businesses), are being negatively impacted by flood.
The cost of mopping up is huge and the process is time consuming. One example of a capital project is that Environment Ministers are going ahead with a five year ‘River Catchment Scheme’, designed to re-direct river flows into natural holding plains before reaching towns or cities.
The cost of repairing the damage spans many sectors and amounts to a significant investment opportunity. Given the complexity of these funding sources, can the case be made for switching the current costs of inundation damage into innovation and improving our use of science and engineering to reduce future inundation risk?
Chaired by Bruce Thomson, Managing Director, Accelerate Corporation, the debate’s participants will include Professor Peter Guthrie, Professor of Engineering for Sustainable Development, University of Cambridge, Professor Sean Smith BSc, PhD, FIOA, FRSA, Professor of Construction Innovation & Director of ISC, Institute for Sustainable Construction, School of Engineering & the Built Environment, Edinburgh Napier University, and also aims to include the day’s key speakers.
For further information and how to book for the event visit:
Further information on sponsorship:
Twitter: @maritimedevgrp @Jobsinmaritime
- Investinblue, a two day event during at IFB2016, a biennial marketplace which brings together thousands of businesses from around the world for three weeks of events, networking and deal-making.. Investinblue will promote UK maritime infrastructure development, high technology projects, products, systems and services to the world, bringing together the UK maritime sector with the goal of working together to showcase, draw interest and inbound investment from UK and overseas stakeholders, dedicated to stimulating the growth of the UK and maritime economy.
-: | 0.996265 |
Enroll in Centennial Schools Today!
Ellie Jensen, a seventh grade student at Centennial Middle School, is a semifinalist eligibile to compete in the 2014 Minnesota State Geographic Bee, Friday, April 4.
The North Suburban Cougars finished third in the state tournament. Congrats to team and coaches on a superb 13-1-1 season!
Congratulations to the CHS girls basketball team, Section 5AAAA champions and MSHSL State Girls Basketball Tournament participant.
Odyssey, a school-within-a school, provides full-time instruction delivered in self-contained classrooms for Level Four gifted students in grades three and four. Odyssey begins in the fall of the 2014-2015 school year.
Centennial Schools to Add Full-Time Gifted Program (Star Tribune, 4/1/14)
Congratulations to 2014 All-State Band Members Emily Green and Mackenzie Smith; soloists Lexie Glaser and Emily Green for "Best @ Site" commendations at the Region 5AA Instrumental Solo & Ensemble Contest; and Centennial students receiving 27 "excellent" ratings and 28 "superior" ratings at the contest.
MPR's Music with Minnesotans: Emily Green | 0.026335 |
This Man Ate 2 Tbsp. Of Coconut Oil For 60 Days…See What Happened To His Brain
Dr..’
According to Dr. Mary Newport, traditional medicines they were using weren’t working at all and in fact, and were only producing terrible side effects for Steve. She decided that she wanted him to try more natural healing methods and when she did some research, she found out something surprising. | 0.894895 |
Monday, June 25, 2018
The new OneIT Strategic Plan details strategic goals OneIT will pursue in support of the institutional mission, and how OneIT will evolve to meet technology needs now and in the future. Drafting the plan was a highly collaborative effort that involved significant engagement of campus stakeholders, IT leaders, and the broader IT community. | 0.483533 |
Item #: AB-GR
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Response from WorkBoots.com | 0.259346 |
: Business News, National News, News Tags: Companies, Fortune 500, Walmart Send to a friend! Share on Facebook Share on Twitter Pin it! Reddit This! Share on Tumblr! | 0.01882 |
5thRound.com will once again be covering UFC 108 tomorrow night. Spike TV will be kicking off the day’s festivities with a live broadcast of two preliminary bouts beginning at 9PM ET/6PM. Unfortunately, you’re gonna have to fork over $44.95 (or 54.95 to have it in HD) for the good stuff starting at 10PPM ET/7PM PT.
The card will be headlined by a light heavyweight scrap between former champ Rashad Evans and the electrifying Thiago Silva. A win is imperative for both if they intend to avenge their respective losses to reigning title holder Lyoto Machida.
Don’t forget that you can stream the UFC 108 post-fight press conference live right here on 5thRound.com at approximately 1AM ET/10PM PT.
Here is a complete breakdown of today’s schedule.
Preliminary Bouts:
Rafaello Oliveira vs. John Gunderson
Jake Ellenberger vs. Mike Pyle
Mark Munoz vs. Ryan Jensen
| 0.000884 |
Mayhem 45
The hybrid top and panel opening of this pack makes it ideal for accessibility in tight environments like the local crag, hostels, or backcountry huts. Built with Spectra fiber, the Mayhem delivers maximum strength and abrasion resistance with a light and comfortable carry.
Features/Specifications:
- Front Panel Expansion Pocket With Daisy Chain Attachment Loops And Built In Haul Handle
- Hybrid Panel/ Top Loader System, Ideal For Trail And Travel
- Hex Ridged Eva Backpanel Padding For Omni-Directional Airflow
- Dual Density Eva/Pu Shoulder Straps
- High Density Foam Framesheet With Wireframe Load Dispersion
- Forward Pull Waitbelt With Side Adjustment Webbing
- Ventilated Waistbelt Padding
- Stretch Mesh Waistbelt Pocket
- Top Lid Zippered Compartment With Secondary Map Pocket Underneath
- Side Panel Mesh Bottle Pockets With Over/Under Strap Pass Thru
- Dual Tool/Trekking Pole Carry With Elastic Pole Tip Loops
- Ice And Mixed Tool Lower Panel Storage Sleeves
Specifications:
- Dimensions: 22X10.5X7 (Hxwxd)
- Torso Size: 15-18
- Waistbelt: 28-48
- Weight: 2 Lbs 9 Oz / 1.2 Kg
- Volume: 29 L
- Capacity: Up To 40 Lbs | 0.006919 |
\begin{document}
\title[Eigenfunctions of perturbed Laplacians]{Arbitrarily small perturbations of Dirichlet Laplacians are quantum unique ergodic}
\author{Sourav Chatterjee}
\address{\newline Department of Statistics \newline Stanford University\newline Sequoia Hall, 390 Serra Mall \newline Stanford, CA 94305\newline \newline \textup{\tt souravc@stanford.edu}}
\author{Jeffrey Galkowski}
\address{\newline Department of Mathematics \newline Stanford University\newline 380 Serra Mall \newline Stanford, CA 94305\newline \newline \textup{\tt jeffrey.galkowski@stanford.edu}}
\thanks{Sourav Chatterjee's research was partially supported by NSF grant DMS-1441513}
\thanks{Jeffrey Galkowski's research was partially supported by the Mathematical Sciences
Postdoctoral Research Fellowship DMS-1502661.}
\keywords{Quantum unique ergodicity, quantum chaos, Laplacian eigenfunction}
\subjclass[2010]{58J51, 81Q50, 35P20, 60J45}
\begin{abstract}
Given an Euclidean domain with very mild regularity properties, we prove that there exist perturbations of the Dirichlet Laplacian of the form $-(I+S_\e)\Delta$ with $\|S_\e\|_{L^2\to L^2}\leq \e$ whose high energy eigenfunctions are quantum uniquely ergodic (QUE). Moreover, if we impose stronger regularity on the domain, the same result holds with $\|S_\e\|_{L^2\to H^\gamma}\leq \e$ for $\gamma>0$ depending on the domain. We also give a proof of a local Weyl law for domains with rough boundaries.
\end{abstract}
\maketitle
\section{Introduction}\label{intro0}
In quantum mechanics, the Laplace operator on a manifold describes the behavior of a free quantum mechanical particle confined to the manifold. The eigenvalues of the Laplacian (under suitable boundary conditions) are the possible values of the energy of the particle and the eigenfunctions are the energy eigenstates. The square of an energy eigenstate gives the probability density function for the location of a particle with the given energy.
The subject of quantum chaos connects the properties of high energy eigenstates with the chaotic properties of the geodesic flow. One important result is the quantum ergodicity theorem due to \citet{snirelman74}, \citet{colindeverdiere85}, and \citet{zelditch87} on manifolds without boundary and generalized to manifolds with boundary by \citet{gerardleichtnam93} and \citet{zelditchzworski96}. The theorem states that if the geodesic flow on a manifold is ergodic, then almost all high energy eigenfunctions (in any orthonormal basis of eigenfunctions) equidistribute over the manifold in the sense that $|u|^2\to 1$ as a distribution. This phenomenon, or more precisely, its analog for equidistribution in both position and momentum, is known as \emph{quantum ergodicity}.
The question of whether all (rather than almost all) high energy eigenfunctions equidistribute in phase space has remained open. This property was christened \emph{quantum unique ergodicity} by \citet{rudnicksarnak94}, who conjectured that the Laplacian on any compact negatively curved manifold is quantum unique ergodic (QUE).
Although the Rudnick--Sarnak conjecture is still open, it is now known that quantum unique ergodicity is not always valid, even if classical particles are chaotic; see \citet{faurenonnenmacher04}, \citet{faureetal03} and \citet{hassell10}. QUE has been verified in only a handful of cases; in particular for the Hecke orthonormal basis on an arithmetic surface by \citet{lindenstrauss06}, \citet{silbermanvenkatesh07} as well as for modular cusp forms on the modular surface \citet{holowinskysoundararajan10} and \citet{soundararajan10}. \citet{anantharaman08} made partial progress towards the general Rudnick--Sarnak conjecture by showing that high energy Laplace eigenfunctions on compact negatively curved manifolds cannot concentrate very strongly. For example, they cannot concentrate on a single closed geodesic. For a more comprehensive survey of results on quantum unique ergodicity, see~\citet{sarnak11}. For more on quantum ergodicity and semiclassical chaos, see \citet{zelditch10}.
In spite of the availability of counterexamples to QUE, it is believed that QUE is generically valid for domains with ergodic billiard ball flow (see \citet{sarnak11}). In other words, QUE is expected to be true for almost all ergodic domains. There are at present no results like this.
The main result of this paper (Theorem~\ref{quethm}) says that for any Euclidean domain satisfying some very regularity conditions, there exists $S_\e:L^2\to L^2$ with $\|S_\e\|_{L^2\to L^2}\leq \e$ such that the perturbation of the Laplacian (with Dirichlet boundary condition) $-(I+S_\e)\Delta$ is self adjoint and has QUE eigenfunctions. In other words, Dirichlet Laplacians lie in the closure (in the $H^2\to L^2$ norm topology) of the set of operators with QUE eigenfunctions. If we impose more regularity on the domain, then we can improve the regularity of $S_\e$. The required operator is constructed using a probabilistic method (described briefly in Section \ref{sec:outline}) and it is then shown that this random operator satisfies the required property with probability one. Notice that, although we show that Laplacians are close in the operator norm to QUE operators, this is very far from showing that one can perturb the domain to obtain a QUE Laplacian. Indeed, one should probably not expect such a result to hold for arbitrary domains.
Our result is closely related to those in \citet{zelditch92,zelditch96,zelditch14}, \citet{maples13} and \citet{chang15} where it is shown that certain unitary randomizations of eigenfunctions are quantum ergodic. In effect, this shows that $-U_k\Delta U_k^*$ is quantum ergodic for $U_k$ random unitary operator which mixes blocks of eigenfunctions. See Section \ref{sec:compare} for a more detailed comparison of the results.
\section{Results}\label{intro}
\subsection{Definitions}
We start by defining the class of domains to which our results apply. These domains may have boundaries which are quite rough and in particular include all domains where the solution of the Dirichlet problem has the property $u(x)\to 0$ as $x\to \partial\Omega$.
Take any $d\ge 2$ and let $\om$ be a Borel subset of $\rr^d$. Let $B_t$ be a standard $d$-dimensional Brownian motion, started at some point $x\in \rr^d$. The exit time of $B_t$ from $\om$ is defined as
\begin{align}\label{exit}
\tau_\om := \inf\{t>0: B_t\not\in \om\}\,.
\end{align}
In this paper we will say that $\om$ is a \emph{regular domain} if it is nonempty, bounded, open, connected, and satisfies the following boundary regularity conditions:
\begin{enumerate}[(i)]
\item $\vol(\partial \om) = 0$, where $\partial \om$ is the boundary of $\om$ and $\vol$ denotes $d$ dimensional Lebesgue measure.
\item For any $x\in \partial \om$, $\pp^x(\tau_\om = 0) = 1$,
where $\pp^x$ denotes the law of Brownian motion started at $x$ and $\tau_\om$ is the exit time from $\om$.
\end{enumerate}
Condition (ii) may look strange to someone who unfamiliar with probabilistic potential theory, but it is actually the well-known sharp condition for the existence of solutions to Dirichlet problems on $\om$~\cite[p.~225]{mortersperes10}. A useful sufficient condition for (ii) is that every point on the boundary satisfies the so-called `Poincar\'e cone condition'~\cite[p.~68]{mortersperes10}. The cone condition stipulates that for every point $x\in \partial \Omega$, there is a cone based at $x$ whose interior lies outside $\Omega$ in a small neighborhood of $x$. Using the Poincar\'e cone condition, it is not difficult to verify that domains with $W^{2,\infty}$ boundaries, considered in \citet{gerardleichtnam93}, satisfy the condition~(ii). However, (i) and (ii) allow more general domains than those with $W^{2,\infty}$ boundary. For example, any convex open set satisfies the cone condition, irrespective of the smoothness of the boundary. Various kinds of regions with corners, such as polygons, also satisfy the cone condition.
An example of a domain that {\it does not} satisfy (ii) is the open unit disk in $\rr^2$ minus the interval $(0,1)$. More generally, domains with very sharp cusps at the boundary may not satisfy condition (ii) (see `Lebesgue's thorn' in~\cite[Section 8.4]{mortersperes10}).
Henceforth, we will assume that $\om$ is a regular domain and $\dbar$ will denote the closure of~$\om$.
Given any measurable function $f: \dbar\ra \mathbb{C}$, we denote by $\|f\|$ the $L^2(\om)$ norm of $f$. For such $f$ there is a natural probability measure associated with $f$ that has density $|f(x)|^2$ with respect to Lebesgue measure on $\dbar$. We will denote this measure as $\nu_f$. Note that in the definition of $\|f\|$ it does not matter whether we integrate over $\om$ or $\dbar$ since $\vol(\partial \om) = 0$. We will denote the $L^2$ inner product of two functions $f$ and $g$ by $\langle f,g\rangle$.
Recall that a sequence of probability measures $\{\mu_n\}_{n\ge 1}$ on $\dbar$ is said to converge weakly to a probability measure $\mu$ if
\[
\lim_{n\ra\infty}\int_{\dbar} fd\mu_n = \int_{\dbar} fd\mu
\]
for every bounded continuous function $f:\dbar \ra\rr$. A probability measure that will be of particular importance in this paper is the uniform probability measure on $\dbar$. This is simply the restriction of Lebesgue measure to $\dbar$, normalized to have total mass one.
\subsection{Defect Measures}
For every bounded sequence of functions $\{f_n\}\in L^2(\re^d)$ with $f_n\underset{L^2}{\rightharpoonup} 0$, we can also associate a family of measures in phase space, $S^*\re^d$ (the cosphere bundle of $\re^d$), called \emph{defect measures}, defined as follows. Recall the notation $\Psi^m(\re^d)$ for the pseudodifferential operators of order $m$ on $\re^d$, and $S^m_{\text{hom}}(T^*\re^d)$ for smooth functions on $T^*\re^d\setminus\{0\}$ homogeneous of degree $m$ in the fiber variable. Let $S_{\text{phg}}^m(T^*\re^d)$ denote the associated polyhomogeneous symbol classes. That is, $a\in S^m_{phg}$ if there exists $a_j\in S^{m-j}_{\text{hom}}$ so that
\begin{equation}
\label{e:expand}\left|\partial_x^\alpha\partial_{\xi}^\beta\left(a(x,\xi)-\sum_{j=0}^{N-1}a_j(x,\xi)\right)\right|\leq C_{\alpha\beta}(1+|\xi|^2)^{m-N-|\beta|},\quad |\xi|\geq 1.
\end{equation}
(See \cite{HOV3} for more details.)
We sometimes write $\Psi(\re^d)$, $S_{\text{hom}}(T^*\re^d)$, and $S_{phg}(T^*\re^d)$ for $\Psi^0(\re^d)$, $S^0_{\text{hom}}(T^*\re^d)$ and $S^0_{phg}(T^*\re^d)$ respectively. We also sometimes omit the $\re^d$ or $T^*\re^d$ when the relevant space is clear from context.
Let
\[
\sigma:\Psi^m\to S^m_{\text{hom}}
\]
be the principal symbol map on $\Psi^m(\re^d)$. For $b\in S^m_{phg}$, we write $b(x,D)\in \Psi^m$ for a quantization of $b$ and observe that
$$\sigma(b(x,D))=b_0(x,\xi)$$
where $b_0\in S^m_{\text{hom}}$ is the first term in the expansion \eqref{e:expand} for $b$.
Let $\chi \in \Cc(\re^d)$ have $\chi\equiv 1$ in a neighborhood of 0. For $a\in \Cc(S^*\re^d)$, let
\[
\tilde{a}(x,\xi)=a(x,\xi/|\xi|)(1-\chi(\xi))\,.
\]
Then $\tilde{a}\in S_{phg}$.
Define the distribution $\mu_n\in \mc{D}'(S^*\re^d)$ by
\[
\mu_n(a)=\la \tilde{a}(x,D)f_n,f_n\ral
\]
where $\la \cdot,\cdot\ral$ denotes the inner product in $L^2(\re^d)$ and $D:=-i\partial$ is $-i$ times the gradient operator. Not that the weak convergence of $f_n$ to zero implies that for every subsequence of $\{\mu_n\}_{n\ge 1}$ there is a further subsequence that converges in the $\mc{D}'(S^*\re^d)$ topology. Moreover, it can be shown that every limit point $\mu$ of $\{\mu_n\}_{n\ge 1}$ in the $\mc{D}'(S^*\re^d)$ topology is a positive radon measure, with the property that there exists a subsequence $\{f_{n_k}\}_{k\ge 1}$ so that for all $A\in \Psi(\re^d)$
\[
\la Af_{n_{k}},f_{n_{k}}\ral\to \int_{S^*\re^d}\sigma(A)d\mu\,.
\]
(See for example \cite{Burq} or \cite{gerard91}.) The set of such limit points $\mu$ is denoted by $\mc{M}(\{f_n\}_{n\ge 1})$ and is called the \emph{set of defect measures associated to the family $\{f_n\}_{n\ge 1}$}. We will write $\mc{M}(f_n)$ instead of $\mc{M}(\{f_n\}_{n\ge 1})$ to simplify notation. Note that while $\mu_n$ depends on the choice of quantization procedure used to define $\tilde{a}(x,D)$ and the function $\chi$, the set $\mc{M}(f_n)$ is independent of such choices.
\subsection{QUE operators}
If $H$ is a linear operator from a dense subspace of $L^2(\dbar)$ into $L^2(\dbar)$, we will say that a function $f$ belonging to the domain of $H$ is an eigenfunction of $H$ with eigenvalue $\lambda\in \mathbb{C}$ if $f\ne 0$ and $Hf = \lambda f$. We will say that an eigenfunction $f$ is normalized if $\|f\|=1$.
\begin{defn}\label{quedef}
Let $H$ be a linear operator from some dense subspace of $L^2(\dbar)$ into $L^2(\dbar)$ having compact resolvent. We say that $H$ is QUE if for any sequence of normalized eigenfunctions $\{f_n\}_{n\ge 1}$ of $H$,
\begin{equation}
\label{queeqn}
\mc{M}(1_{\dbar}f_n)= \left\{\frac{1}{\vol(\om)}1_{\dbar}dxdS(\xi)\right\}
\end{equation} where $S$ is the normalized surface measure on $S^{d-1}$.
\end{defn}
\noindent In particular, notice that if \eqref{queeqn} holds then for all $A\in \Psi^0(\re^d)$,
$$\la A1_{\dbar}f_n,1_{\dbar}f_n\ral\to \frac{1}{\vol(\om)}\int_{S^*\re^d}\sigma(A)1_{\dbar}dxdS(\xi)$$
and hence that $\nu_{f_n}$ converges weakly as a measure to the uniform probability distribution on $\om$.
With this in mind, we define the weaker notion of equidistribution as follows.
\begin{defn}\label{quedef2}
Let $H$ be a linear operator from some dense subspace of $L^2(\dbar)$ into $L^2(\dbar)$ having compact resolvent. We say that $H$ is uniquely equidistributed if for $\{f_n\}_{n\ge 1}$ any sequence normalized eigenfunctions of $H$, $\nu_{f_n}$ converges weakly to the uniform probability distribution on $\om$.
\end{defn}
\subsection{The main result}
Let $-\Delta $ be the Dirichlet Laplacian on $\om$ with domain $\mfd$ (defined in Section \ref{prelim}). The following theorem is the main result of this paper.
\begin{thm}\label{quethm}
Let $\om$ be a regular domain. Then for any $\ep>0$, there exists a linear operator $S_\ep:L^2(\dbar)\ra L^2(\dbar)$ such that:
\begin{enumerate}
\item[\textup{(i)}] $\|S_\ep\|_{L^2\to L^2}\le \ep$.
\item[\textup{(ii)}] $-(I+S_\ep)\Delta$ is a positive operator on $L^2(\dbar)$ with domain $\mfd$
\item[\textup{(iii)}] $-(I+S_\ep)\Delta$ is QUE in the sense of Definition \ref{quedef}.
\end{enumerate}
If $\om$ has $C^\infty$ boundary, then for all $\gamma<1$, there exist such an $S_\ep:L^2(\dbar)\to H^\gamma(\dbar)$ with $\|S_\e\|_{L^2\to H^\gamma}\le \ep$. Moreover, if $\om$ has smooth boundary and the set of periodic billiards trajectories has measure zero (see Section \ref{billiards}), then this holds for $\gamma\leq 1$.
\end{thm}
It would be interesting to see if a different version of this theorem can be proved, where instead of perturbing the Laplacian, it is the domain $\om$ that is perturbed. Alternatively, one can try to perturb the Laplacian by some explicit kernel rather than saying that `there exists $S_\ep$'. Yet another possible improvement would be to show that a generic perturbation, rather than a specific one, results in an operator with QUE eigenfunctions. Indeed, the proof of Theorem \ref{quethm} gets quite close to this goal.
\subsection{Additional results}
The techniques of this paper yield the following version of the local Weyl law for regular domains.
\begin{thm}
\label{localWeyl}
Suppose that $\om\subset \re^d$ is a regular domain, where regularity is defined at the beginning of this section. Let $\{(u_j,\lambda_j^2)\}_{j\ge1}$ be a complete orthonormal basis of eigenfunctions of the Dirichlet Laplacian on $\om$. Then for $A\in \Psi(\re^d)$ with $\sigma(A)$ supported in a compact subset of $\om$ and any $E>1$,
$$\sum_{\lambda_j\in [\lambda,\lambda E]}\la A1_{\dbar}u_j,1_{\dbar}u_j\ral =\frac{\lambda^d}{(2\pi )^d}\iint_{1\leq |\xi|\leq E}\sigma(A)1_{\dbar}dxd\xi+\o{}(\lambda^{d}).$$
\end{thm}
In order to state the next theorem, we need the following definition.
\begin{defn}\label{assume}
Let $C_0(S^*\om)$ be the set of continuous functions on $S^*\re^d$ that vanish on $(\re^d\setminus \om)\times S^{d-1}$ with the sup-norm topology.
Let $\alpha:\re_+\to \re_+$ be nonincreasing. Let $\{(u_j,\lambda_j^2)\}_{j\ge 1}$ be a complete orthonormal basis of eigenfunctions of the Dirichlet Laplacian on $\om$. Suppose that there exists $\mc{A}\subset \Psi(\re^d)$ so that for all $A\in \mc{A}$, $\sigma(A)$ is supported compactly inside $S^*\om$, the set
\[
\sigma(\mc{A}):=\{\sigma(A)|_{S^*\re^d}: A\in \mc{A}\}
\]
is dense in $C_0(S^*\om)$, and for each $A\in\mc{A}\subset \Psi(\re^d)$,
\begin{align*}
\sum_{\lambda_j\in [\lambda,\lambda(1+\alpha(\lambda))]}&\la A1_{\dbar}u_j, 1_{\dbar}u_j\ral \\
&=\frac{\lambda^d}{(2\pi )^d}\iint_{1\leq |\xi|\leq 1+\alpha(\lambda)}\sigma(A)1_{\dbar}dxd\xi+\o{}(\alpha(\lambda)\lambda^{d}).
\end{align*}
Then we say that the domain $\om$ is \emph{average quantum ergodic (AQE) at scale~$\alpha$}.
\end{defn}
Theorem \ref{localWeyl} implies that regular domains $\om$ are AQE at scale $E$ for any $E>0$. In Section~\ref{sec:localWeyl}, we recall Weyl laws holding on domains with $C^\infty$ boundaries which imply that these domains are AQE at scale $\alpha(\lambda)=\lambda^{-\gamma}$ for some $\gamma>0$. For $\gamma\in [0,2]$, let $\mfd^\gamma$ denote the complex interpolation space $(L^2(\om),\mfd)_{\gamma/2}$. Then the following theorem implies Theorem \ref{quethm}.
\begin{thm}\label{quethm2}
Suppose that $\om$ is a regular domain that is AQE at scale $\alpha(\lambda)=\lambda ^{-\gamma}$ for some $1\geq \gamma\geq 0$. Then for any $\ep>0$, there exists a linear operator $S_\ep:L^2(\dbar)\ra \mfd^\gamma$ such that:
\begin{enumerate}
\item[\textup{(i)}] $\|S_\ep\|_{L^2\to \mfd^\gamma}\le \ep$.
\item[\textup{(ii)}] $-(I+S_\ep)\Delta$ is a positive operator on $\mfd$ with compact resolvent
\item[\textup{(iii)}] $-(I+S_\ep)\Delta$ is QUE in the sense of Definition \ref{quedef}.
\end{enumerate}
\end{thm}
A consequence of Theorem \ref{quethm} is that $-\Delta$ has a sequence of `quasimodes' that are equidistributed in the limit. Moreover, when $\om$ is AQE at some scale $\alpha(\lambda)=\o{}(1)$, then there is a full orthonormal basis of (slightly weaker) quasimodes that are QUE. This is the content of the following corollary.
\begin{cor}\label{almostcor}
Let all notation be as in Theorem \ref{quethm}. Suppose that $\om$ is AQE at scale $\alpha(\lambda)=\lambda^{-\gamma}$ for some $\gamma\geq 0$. Then
\begin{itemize}
\item[(i)] there is a sequence of functions $\{f_n\}_{n\ge 1}$ belonging to $\mfd$ and a sequences of positive real numbers $\{\alpha_n\}_{n\ge 1}$, such that $\|f_n\|=1$, $\alpha_n \ra\infty$ and
$
(-\alpha_n^{-2} \Delta -1) f_n=\o{L^2}(\alpha_n^{-\gamma})
$
and
$$\mc{M}(f_n)=\left\{\frac{1}{\vol(\om)}1_{\dbar}dxd\sigma(\xi)\right\}.$$
\item[(ii)] there is an orthonormal basis of $L^2(\dbar)$, $\{g_n\}_{n\ge 1}$ belonging to $\mfd$ and a sequences of positive real numbers $\{\beta_n\}_{n\ge 1}$, such that $\beta_n \ra\infty$ and
$
(-\beta_n^{-2} \Delta -1) g_n=\O{L^2}(\beta_n^{-\gamma}).
$
and $$\mc{M}(g_n)=\left\{\frac{1}{\vol(\om)}1_{\dbar}dxd\sigma(\xi)\right\}.$$
\end{itemize}
\end{cor}
\begin{remark}
Note that up to this point, all results apply equally well to compact manifolds with or without boundary, but we chose to present them for the case of $\om\Subset \re^d$ for concreteness.
\end{remark}
\subsection{Improvements on compact manifolds without boundary}
Together with the analog of Theorem \ref{quethm}, a stronger version of the Weyl law valid on compact manifolds without boundary (see Section \ref{sec:localWeyl}), implies the following corollary.
\begin{cor}\label{almostcor2}
Let $(M,g)$ be a compact Riemannian manifold without boundary so that the set of closed geodesics has measure 0. Then there is an orthonormal basis of $L^2(\dbar)$, $\{f_n\}_{n\ge 1}$, belonging to $C^\infty(M)$ and a sequence of positive real numbers $\{\alpha_n\}_{n\ge 1}$ such that $\alpha_n \ra\infty$,
\[
(-\alpha_n^{-2} \Delta_g -1)f_n =\o{L^2}(\alpha_n^{-1})\, ,
\]
and $\nu_{f_n}\to \frac{1}{\vol(M)}dx$. That is, $f_n$ are uniquely equidistributed in the sense of Definition \ref{quedef2}.
\end{cor}
Unfortunately, the authors were unable to prove a version of Corollary \ref{almostcor2} where the basis of quasimodes is QUE rather than uniquely distributed. This is because the remainder in the strong version of the local Weyl law (see Theorem \ref{thm:weylStrong}) may depend on derivatives of the symbol in $\xi$.
\begin{remark}
Notice that if $M$ has ergodic geodesic flow, then the set of periodic geodesics has measure zero and hence Corollary \ref{almostcor2} applies and there is an orthonormal basis of $\o{L^2}(\alpha^{-1})$ quasimodes that are equidistributed. In particular, being $\o{L^2}(\alpha^{-1})$ quasimodes implies that these functions respect the dynamics at the level of defect measures, that is, defect measures associated to the family of quasimodes are invariant under the geodesic flow. See \cite{Burq} or \cite[Chapter 5]{EZB} for details.
\end{remark}
\subsection{Comparison with previous results}
\label{sec:compare}
One can view the results here as a companion to those in \cite{zelditch92,zelditch96,zelditch14} and \cite{maples13}. In these papers, the authors work on a compact manifold $M$ and fix a basis of eigenfunctions of the Laplacian, $\{u_n\}_{n=1}^\infty$. Their results then show that for almost every block diagonal (in the orthonormal basis $u_n$) unitary operator
$$U=\oplus_{k=1}^\infty U_k$$
(with respect to the product Haar measure)
such that for all $k$, $ \dim \ran U_k<\infty$ and $\dim \ran U_k\to \infty$ at least polynomially in $k$, the basis $\{Uu_n\}_{n=1}^\infty$ has
$$\mc{M}(Uu_n)=\left\{\frac{1}{\vol(M)}dxdS(\xi)\right\}.$$
One reformulation considers a certain basis of eigenfunctions for the operator $-U\Delta U^*$. By taking $U_k$ close to the identity, we may write
$$\tilde{P}:=-U\Delta U^*=-(I+\tilde{S})\Delta$$
where $\tilde{S}$ is small in $L^2\to L^2$ norm. However, $\tilde{P}$ may not be QUE if there is high multiplicity in the spectrum of $-\Delta$.
One can think of the results in the present paper as replacing the $U_k$ by some nearly unitary operator. By choosing these operators carefully, and employing the Hanson--Wright inequality in place of the law of large numbers, we are able to use smaller windows than those in previous work. This allows us to prove that the perturbation is regularizing under various conditions, and to show that the resulting operator is QUE.
\subsection{Outline of the proof and organization of the paper}
\label{sec:outline}
In order to prove Theorem \ref{quethm}, we show that a local Weyl law with a certain window implies the existence of the desired perturbation $S_\e$. The local Weyl law essentially says that when averaged over a certain size window, say $\lambda^{-\gamma}$, the matrix elements $\langle Au_k,u_k\rangle$ behave as though the eigenfunctions were uniquely ergodic. In Section \ref{rotationsec} we give a rigorous meaning to this statement. In particular, we use a modern version of the Hanson--Wright inequality from \cite{rudelsonvershynin13} (see \cite{hansonwright71} for the original) to show that random rotations (with respect to Haar measure) of small groups of eigenfunctions are uniquely ergodic. Here, the size of the group allowed depends on the remainder in the local Weyl law. Thus, the smaller the remainder, the smaller the required group of eigenfunctions.
In Section \ref{last}, we obtain the perturbation, $S_\e$. In order to do this, we make a two scale partition of the eigenvalues, $\lambda_i^2$, of the Laplacian. In particular, we divide the eigenvalues of the Laplacian into
\begin{align*} L_{n,j}&:=\left\{\lambda_i: \left(1+\frac{\e j}{\lceil(1+\e)^{n\gamma}\rceil}\right) \leq \frac{\lambda_{i}}{(1+\e)^n}<\left(1+\frac{\e (j+1)}{\lceil(1+\e)^{n\gamma}\rceil}\right)\right\},\\
&\qquad \qquad 0\leq n,\quad 0\leq j\leq \lceil(1+\e)^{n\gamma}\rceil-1
\end{align*}
where $\gamma$ is determined by the remainder in the local Weyl law.
For each $L_{n,j}$ we then make a random rotation of the corresponding eigenfunctions and perturb the eigenvalues, $\lambda_i\to \lambda'_i$ so that each new eigenvalue, $\lambda_i'$ is simple and lies in $L_{n,j}$.
Because of the fact that random rotations of eigenfunctions on the scale $\lambda^{-\gamma}$ are QUE, this results in an operator that is almost surely QUE. The regularizing nature of the perturbation results from the second scale in $L_{n,j}$. That is, the fact the eigenfunctions with eigenvalue similar to $(1+\e)^n$ are mixed only with those whose eigenvalues are at a distance $(1+\e)^{n(1-\gamma)}\e$. In particular, the larger $\gamma$, the more regularizing the perturbation.
In order to prove Theorem \ref{quethm}, we need to prove the local Weyl law for regular domains (Theorem \ref{localWeyl}), but we postpone this proof until Appendix \ref{ptwisesec}. The key ingredient here is to compare the heat trace for the Dirichlet Laplacian on $\Omega$ with the heat trace for the Laplacian on $\re^d$ as in \cite{gerardleichtnam93, dodziuk81}. Let $k(t,x,y)$ and $k_D(t,x,y)$ be respectively the kernels of $e^{t\Delta}$ and $e^{t\Delta_D}$, where $\Delta$ is the free Laplacian and $\Delta_D$ the Dirichlet Laplacian. The key estimate in proving Theorem \ref{localWeyl} is
$$|\partial_x^\alpha(k(t,x,y)-k_D(t,x,y))|\leq C_\delta t^{-N_\alpha}e^{-c_\delta/t},\quad\quad d(x,\partial\Omega)>\delta.$$
\begin{remark}
The work of \citet{LiStroh} extends this type of estimate to the heat kernel for general self-adjoint extensions of the Laplacian and gives explicit constants independently of the extension.
\end{remark}
We prove this estimate using the relationship between killed Brownian motion on $\Omega$ with the Dirichlet heat Laplacian together with the fact that Brownian motion has independent increments. Because of this approach, we are able to complete the proof on domains which are only regular.
The paper is organized as follows. Section \ref{sec:prelim} recalls local Weyl laws valid for domains with smooth boundary and manifolds without boundary, the functional analytic definition of the Dirichlet Laplacian, and some geometric preliminaries. Section \ref{rotationsec} presents the results on random rotations of eigenfunctions. Section \ref{last} finishes the proof of Theorems \ref{quethm}, \ref{quethm2} and Corollary \ref{almostcor}. Section~\ref{last2} contains the adjustments necessary to obtain the improvements on manifolds without boundary, in particular proving Corollary~\ref{almostcor2}. Finally, Appendix \ref{ptwisesec} contains the proof of Theorem \ref{localWeyl}.
\section{Preliminaries}
\label{sec:prelim}
Throughout the paper, we will adopt the notation that $C$ denotes any positive constant that may depend only on the set $\om$, the dimension $d$, and nothing else. The value of $C$ may change from line to line. In case we need to deal with multiple constants, they will be denoted by $C_1,C_2,\ldots$. From this point forward we will assume that
\[
\vol(\om) = 1\,.
\]
This does not result in any loss of generality since we may always rescale $\om$ with positive volume to have unit volume.
\subsection{Local Weyl Laws}
\label{sec:localWeyl}
We first recall some now classical local Weyl laws for domains $\om$ more regular than those in Theorem \ref{localWeyl}. In this setting, we have the following version of the local Weyl law~\cite{DuiGui, safarovVassiliev}.
\begin{thm}
Suppose that $\om$ has $C^\infty$ boundary. Let $\{(u_j,\lambda_j^2)\}_{j\ge 1}$ be a complete orthonormal basis of eigenfunctions of the (Dirichlet) Laplacian on $\om$. Then for $A\in \Psi(\re^d)$ with $A$ having kernel supported in a compact subset of $\om\times \om$,
$$\sum_{\lambda_j\in [\lambda,\lambda E]}\la A1_{\dbar}u_j,1_{\dbar}u_j\ral =\frac{\lambda^d}{(2\pi )^d}\iint_{1\leq |\xi|\leq E}\sigma(A)1_{\dbar}dxd\xi+\O{}(\lambda^{d-1}).$$
In particular, $\om$ is AQE at scale $\lambda^{-\gamma}$ for any $\gamma<1$.
Moreover if the set of closed trajectories for the billiard flow (see Section~\ref{billiards}) has measure zero, then
\begin{align*}
\sum_{\lambda_j\in [\lambda,\lambda(1+\lambda^{-1})]}&\la A1_{\dbar}u_j,1_{\dbar}u_j\ral \\
&=\frac{\lambda^d}{(2\pi )^d}\iint_{1\leq |\xi|\leq 1+\lambda^{-1}}\sigma(A)1_{\dbar}dxd\xi+\o{}(\lambda^{d-1}).
\end{align*}
In particular, $\om$ is AQE at scale $\lambda^{-1}$.
\end{thm}
\subsection{Manifolds without boundary}
Let $(M,g)$ be a smooth compact Riemannian manifold without boundary (i.e. a smooth manifold with smooth metric). Then the Laplace operator is given in local coordinates by
$$-\Delta_g:=\frac{1}{\sqrt{|g|}}\partial_i(\sqrt{|g|}g^{ij}\partial j)$$
where $|g|=\det g_{ij}$ and $g(\partial_{x_i},\partial_{x_j})=g_{ij}$ with inverse $g^{ij}$. The operator $-\Delta_g$ has domain $H^2(M)$ and is invertible as an operator $L^2_m(M)\to H^2_m(M)$ where $B_m(M)$ is the set of functions in $B$ with $0$ mean.
In this setting, we have the following version of the pointwise Weyl law~\cite{safarovVassiliev}.
\begin{thm}
\label{thm:weylStrong}
Let $\{(u_j,\lambda_j^2)\}_{j\ge 1}$ be the eigenfunctions of $-\Delta_g$. Then
$$\sum_{\lambda_j\leq \lambda} |u_j(x)|^2=\frac{\lambda^d}{(2\pi)^d\vol(M)}\vol(B_d)+\O{}(\lambda^{d-1})$$
where $B_d$ denotes the unit ball in $\re^d$.
If the set of closed geodesics has zero measure, then $\O{}(\lambda^{d-1})$ can be replaced by $\o{}(\lambda^{d-1}).$ Moreover, the asymptotics are uniform for $x\in M$.
\end{thm}
Theorem \ref{thm:weylStrong} provides estimates uniform in $x$ that are used to prove Corollary \ref{almostcor2}.
\subsection{Functional Analysis}
\label{prelim}
Recall our convention that $\|f\|$ denotes the $L^2$ norm of a function $f$ and $\la f,g\ral$ denotes the $L^2$ inner product of $f$ and $g$.
Let $\Omega\subset \re^d$ a bounded open set. We now recall the definition of the Dirichlet Laplacian as a self adjoint unbounded operator on $L^2(\om)$. Let $H_0^1(\om)$ denote the closure of $\Cc(\om)$ with respect to the $H^1$ norm where for $k\in \mathbb{N}$,
$$\|u\|^2_{H^k(\om)}:=\sum_{|\alpha|\leq k}\|\partial^\alpha u\|^2.$$
Here for a multiindex $\alpha\in \mathbb{N}^d$,
$$\partial^\alpha=\partial_{x_1}^{\alpha_1}\partial_{x_2}^{\alpha_2}\dots\partial_{x_d}^{\alpha_d},\quad\quad |\alpha|=\alpha_1+\alpha_2+\dots \alpha_d.$$
Then $H_0^1(\om)$ is a Hilbert space with inner product
$$(u,v)=\la u,v\ral+\la \nabla u,\nabla v\ral.$$
Define the quadratic form $Q:H_0^1(\om)\times H_0^1(\om)\to \mathbb{C}$ by
$$Q(u,v)=\la \nabla u, \nabla v\ral.$$
Then $Q$ is a symmetric, densely defined quadratic form and for $u,v\in H_0^1(\om)$,
$$|Q(u,v)|\leq C\|u\|_{H^1(\om)}\|v\|_{H^1(\om)},\quad\quad c\|u\|_{H^1(\om)}^2\leq Q(u,u)+C\|u\|^2.$$
Therefore by \cite[Theorem VIII.15]{Reed}, $Q$ defines a unique self-adjoint operator $-\Delta$ with domain
\[
\mfd := \{u\in H_0^1: Q(u,w)\leq C_u\|w\| \text{ for all } w\in H_0^1(\om)\}\,.
\]
This operator is called the \emph{Dirichlet Laplacian}. Let $E_\mu$ denote the resolution of the identity for $-\Delta$, i.e. $E_\mu=1_{(-\infty, \mu]}(-\Delta)$. Then the complex interpolation space between $L^2$ and $\mfd$ is given by
$$(L^2,\mfd)_\theta:=\Big\{f\in L^2\mid \int \langle \mu\rangle ^\theta dE_\mu f\in L^2\Big\},\qquad\langle \mu\rangle :=(1+|\mu|^2)^{1/2}.$$
We recall that
\begin{lmm}
\label{lmmdomain}
Suppose that $\om$ has $C^2$ boundary. Then $\mfd=H_0^1(\om)\cap H^2(\om)$ and in particular $(L^2,\mfd)_\theta\subset H^{2\theta}(\om)$.
\end{lmm}
\subsection{The billiard flow}
\label{billiards}
Let $\om$ be a domain with $C^\infty$ boundary. We now define the billiard flow. Let $S^*\re^d$ be the unit sphere bundle of $\re^d$. We write
$$ S^*\re^d|_{\partial \om}=\partial \om_+\sqcup\partial \om_-\sqcup\partial \om_0\,\,$$
where $(x,\xi)\in \partial \om_+$ if $\xi$ is pointing out of $ \om$, $(x,\xi)\in\partial \om_-$ if it points inward, and $(x,\xi)\in\partial \om_0$ if $(x,\xi)\in S^*\partial \om$. The points $(x,\xi)\in \partial \om_0$ are called \emph{glancing} points. Let $B^*\partial \om$ be the unit coball bundle of $\partial \om$, i.e.
$$B^*\partial\om =\{(x,\xi)\in T^*\partial \om \mid |\xi|_g<1\}$$
and denote by $\pi_{\pm}:\partial \om_{\pm}\to B^*\partial \om$ and $\pi:S^*\re^d|_{\partial \om}\to \overline{B^*\partial \om}$ the canonical projections onto $\overline{B^*\partial \om}$. Then the maps $\pi_{\pm}$ are invertible. Finally, write
$$ t_0(x,\xi)=\inf\{t>0:\exp_t(x,\xi)\in T^*\re^d|_{\partial \om}\}\,\,$$
where $\exp_t(x,\xi)$ denotes the lift of the geodesic flow to the cotangent bundle. That is, $t_0$ is the first positive time at which the geodesic starting at $(x,\xi)$ intersects $\partial \om$.
We define the billiard flow as in \cite[Appendix A]{DyZw}. Fix $(x,\xi)\in S^*\re^d\setminus \partial \om_0$ and denote $t_0=t_0(x,\xi)$. Then since $\partial\om$ is $C^\infty$ and $(x,\xi)\notin \partial\Omega_0$, $t_0\in (0,\infty]$. We assume now that $t_0<\infty$. If $\exp_{t_0}(x,\xi)\in \partial \om_0$, then the billiard flow cannot be continued past $t_0$. Otherwise there are two cases: $\exp_{t_0}(x,\xi)\in \partial \om_+$ or $\exp_{t_0}(x,\xi)\in \partial \om_-$. We let
$$(x_0,\xi_0)=\begin{cases}
\pi_-^{-1}(\pi_+(\exp_{t_0}(x,\xi)))\in \partial \om_-\,,&\text{if }\exp_{t_0}(x,\xi)\in \partial \om_+\\
\pi_+^{-1}(\pi_-(\exp_{t_0}(x,\xi)))\in \partial \om_+\,,&\text{if }\exp_{t_0}(x,\xi)\in \partial \om_-.
\end{cases}$$
That is, $(x_0,\xi_0)$ is the reflection of $\exp_{t_0}(x,\xi)$ along the normal bundle of $\partial\om$ through $T_x^*\partial\om$.
We then define $\varphi_t(x,\xi)$, the \emph{billiard flow}, inductively by putting
\[
\varphi_t(x,\xi)=\begin{cases}\exp_t(x,\xi)&0\leq t<t_0,\\
\varphi_{t-t_0}(x_0,\xi_0)&t\geq t_0.
\end{cases}
\]
We say that the trajectory starting at $(x,\xi)\in S^*\re^d$ is \emph{periodic} if there exists $t>0$ such that $\varphi_t(x,\xi)=(x,\xi).$
\subsection{Probabilistic Notation}
We now introduce a few notations from probability. Recall that $\pp(A)$ denotes the probability of the event $A$ and $\ee(X)$ denotes the expected value of the random variable $X$. Finally, $\ee(X;A)$ denotes the expectation of the random variable $X$ conditioned on~$A$.
\section{Concentration of random rotations}\label{rotationsec}
Let $u_1,\ldots, u_n$ be an orthonormal set of real valued functions belonging to $L^2(\dbar)$. Let $Q$ be an $n\times n$ Haar-distributed random orthogonal matrix. Let $q_{ij}$ denote the $(i,j)^{\textup{th}}$ entry of $Q$. Define a new set of functions $v_1,\ldots, v_n$ as
\[
v_i(x) := \sum_{j=1}^n q_{ij} u_j(x)\,.
\]
Then $v_1,\ldots, v_n$ are also orthonormal, since
\begin{align*}
\la v_i,v_j\ral &= \la \sum_{k,l=1}^n q_{ik}q_{jl} u_k,u_l\ral= \sum_{k,l=1}^n q_{ik} q_{jl} \la u_k,u_l\ral\\
&= \sum_{k=1}^n q_{ik}q_{jk} =
\begin{cases}
1 &\text{ if } i=j\, ,\\
0 &\text{ otherwise.}
\end{cases}
\end{align*}
We will refer to $v_1,\ldots, v_n$ as a \emph{random rotation} of $u_1,\ldots, u_n$. The goal of this section is to prove the following concentration inequality for random rotations.
\begin{thm}\label{rotthm}
Let $u_i$ and $v_i$ be as above. Let $A:L^2(\om )\to L^2(\om )$ be a bounded operator.
Then for any $1\le i\le n$ and any $t>0$,
\begin{align*}
\pp\biggl(\biggl|\la Av_i,v_i\ral - \frac{1}{n}\sum_{i=1}^n\la Au_i,u_i\ral\biggr| \ge t\biggr) &\le C_1 \exp\bigl(-C_2(\|A\|) \min\{t^2, \, t\} n\bigr)\,,
\end{align*}
where $C_1$ depends only of $d$ and $\om $, and $C_2(\|A\|)$ depends on $d$, $\om $ and the operator norm, $\|A\|$.
\end{thm}
The key ingredient in the proof of Theorem \ref{rotthm} is the Hanson--Wright inequality~\cite{hansonwright71} for quadratic forms of sub-Gaussian random variables. The original form of the Hanson--Wright inequality does not suffice for our objective. Instead, the following modern version of the inequality, proved recently by \citet{rudelsonvershynin13}, is the one that we will use.
The reason why $\la Av_i, v_i\ral$ is concentrated around its mean is that it can be expressed approximately as a quadratic form of i.i.d.~Gaussian random variables, and the eigenvalues of the matrix defining this quadratic form are roughly of equal size. The spectral decomposition then implies that this quadratic form can be written as a linear combination of squares of i.i.d.~Gaussian random variables, where the coefficients are roughly of equal size. The details are worked out below.
Define the $\psi_2$ norm of a random variable $X$ as
\[
\|X\|_{\psi_2} := \sup_{p\ge 1} p^{-1/2}(\ee|X|^p)^{1/p}\,.
\]
The random variable $X$ is called sub-Gaussian if its $\psi_2$ norm is finite. In particular, Gaussian random variables have this property.
Let $M= (m_{ij})_{1\le i,j\le n}$ be a square matrix with real entries. The Hilbert--Schmidt norm of $M$ is defined as
\[
\|M\|_{\textup{HS}} := \biggl(\sum_{i,j=1}^n m_{ij}^2\biggr)^{1/2}\,,
\]
and the operator norm of $M$ is defined as
\[
\|M\| := \sup_{x\in \rr^n,\, \|x\|=1}\|Mx\|\,,
\]
where the norm on the right side is the Euclidean norm on $\rr^n$. Rudelson and Vershynin's version of the Hanson--Wright inequality states that if $X_1,\ldots, X_n$ are independent random variables with mean zero and $\psi_2$ norms bounded by some constant $K$, and
\[
R := \sum_{i,j=1}^n m_{ij} X_iX_j\,,
\]
then for any $t\ge 0$,
\begin{equation}\label{hs}
\pp(|R - \ee(R)|\ge t) \le 2\,\exp\biggl(-C \min\biggl\{\frac{t^2}{K^4\|M\|_{\textup{HS}}^2}, \frac{t}{K^2\|M\|}\biggr\}\biggr)\,,
\end{equation}
where $C$ is a positive universal constant.
\begin{proof}[Proof of Theorem \ref{rotthm}]
Fix $1\le i\le n$. Define
\[
A_i := \la Av_i,v_i\ral,\quad\quad
B := \frac{1}{n}\sum_{i=1}^n\la Au_i,u_i\ral.
\]
Notice that for each $j$,
\[
\sum_{i=1}^n q_{ij}^2 = 1
\]
and for each $j\ne k$,
\[
\sum_{i=1}^n q_{ij} q_{jk}=0.
\]
Note that the distribution of $Q$ remains invariant under arbitrary permutations of rows. Therefore, for any $j$ and $k$, $\ee(q_{ij}q_{ik})$ is the same for each $i$. Thus, the above identities imply that
\[
\ee(q_{ij} q_{ik}) =
\begin{cases}
1/n &\text{ if } j=k\, ,\\
0 &\text{ otherwise.}
\end{cases}
\]
Therefore
\begin{equation}\label{hs1}\begin{aligned}\mathbb{E}(\la Av_i,v_i\ral)&=\sum_{jk}\mathbb{E}(\la A q_{ij}u_j,q_{ik}u_k\ral)=\sum_{jk}\mathbb{E}(q_{ik}q_{ij})\la Au_j,u_k\ral \\
&=\frac{1}{n}\sum_{j=1}^n\la Au_j,u_j\ral=B.\end{aligned}\end{equation}
Let $q_i$ be the vector whose $j^{\textup{th}}$ component is $q_{ij}$. Since $Q$ is a Haar-distributed random orthogonal matrix, symmetry considerations imply that $q_i$ is uniformly distributed on the unit sphere $S^{n-1}$. Now recall that if $z$ is an $n$-dimensional standard Gaussian random vector, then $z/\|z\|$ is uniformly distributed on $S^{n-1}$, and is independent of $\|z\|$. Therefore if $r_i$ is a random variable that has the same distribution as $\|z\|$ and is independent of $q_i$, then the vector $r_i q_i$ is a standard Gaussian random vector. Let $w_{ij} := r_iq_{ij}$, so that $w_{i1},\ldots, w_{in}$ are i.i.d.~standard Gaussian random variables.
Let $H$ be the matrix with $(j,k)^{\text{th}}$ entry
$$h_{jk}=\la Au_j,u_k\ral$$
so that
$$A_i=\sum_{j,k} q_{ij}q_{ik}h_{jk}.$$
Then define
\[
A_i' := r_i^2 A_i =\sum_{j,k=1}^n w_{ij} w_{ik} h_{jk}\,.
\]
Note that $H$ can also be written as $H=\Pi A\Pi$, where $\Pi$ denotes orthogonal projection onto $\text{span}\{u_j: 1\le j\le n\}$. Therefore
$$\|H\|\leq \|A\|$$
and
\begin{align*}
\|H\|_{\text{HS}}&=\sqrt{\tr(H^*H)}= \sqrt{\tr(\Pi^*A^*A\Pi)}\\
&= \sqrt{\tr(A^*A \Pi \Pi^*)} = \sqrt{\tr(A^*A)} = \|A\|_{\text{HS}}\leq \|A\|\sqrt{n}.
\end{align*}
Therefore by the Hanson--Wright inequality \eqref{hs}, with $X_j=w_{ij}$ and $m_{ij}=h_{ij}$ gives
\begin{align}\label{hs2}
\pp(|A_i'-\ee(A_i')|\ge t) \le 2\exp\biggl(-C(\|A\|)\min\biggl\{\frac{t^2}{n},\, t\biggr\}\biggr)\,,
\end{align}
where $C(\|A\|)=\min (\|A\|^{-2},\|A\|)$. Again note that by the Hanson--Wright inequality and the fact that $\ee(r_i^2)=n$,
\begin{align}\label{hs3}
\pp(|r_i^2-n|\ge t)\le 2\exp\biggl(-C\min\biggl\{\frac{t^2}{n},\, t\biggr\}\biggr)\,.
\end{align}
Next, note that
\begin{align}
|A_i| &\le \|A\|_{L^2\to L^2}\|v_i\|^2= \|A\|_{L^2\to L^2} \sum_{j,k=1}^n q_{ij} q_{ik} \la u_j, u_k\ral\\
&= \|A\|_{L^2\to L^2}\sum_{j=1}^n q_{ij}^2 = \|A\|_{L^2\to L^2}\,.\label{hs4}
\end{align}
Finally, observe that since $r_i^2$ is the square norm of a Gaussian random varianble in $\mathbb{C}^n$, $\mathbb{E}r_i^2=n$ and
\begin{equation}\label{hs5}
\ee(A_i') = n\ee(A_i)\,.
\end{equation}
Combining \eqref{hs1}, \eqref{hs2}, \eqref{hs3}, \eqref{hs4} and \eqref{hs5} we get
\begin{align*}
\pp(|A_i-B|\ge t) &\le \pp(|nA_i - A_i'|\ge nt/2) + \pp(|A_i' - \ee(A_i')|\ge nt/2)\\
&\le \pp(|(r_i^2-n)A_i|\ge nt/2) + \pp(|A_i' - \ee(A_i')|\ge nt/2)\\
&\le \pp(|r_i^2-n|\ge nt/(2\|A\|_{L^2\to L^2})) + \pp(|A_i' - \ee(A_i')|\ge nt/2)\\
&\le C_1 \exp\bigl(-C_2(\|A\|) \min\{t^2, \, t\} n\bigr)\,,
\end{align*}
which concludes the proof of the theorem.
\end{proof}
\section{Construction of the perturbed Laplacian}\label{last}
As described in Section \ref{sec:outline}, our strategy will be to break up the spectrum of $-\Delta$ into blocks and to `mix' the eigenfunctions in each block to produce a new operator that is QUE. To do this, it is convenient to work on the spectral side. We will need a few linear algebra lemmas.
Let $\Psi = (\psi_i)_{i\ge 1}$ be a complete orthonormal basis of $L^2(\dbar)$. Let $\Lambda = (\mu_i)_{i\ge 1}$ be a sequence of real numbers. For $s\geq 0$, let $\mf^s(\Psi, \Lambda)$ be the Hilbert space (with complex scalars) consisting of all $f\in L^2(\dbar)$ such that the norm
\[
\|f\|^2_{\mf^s(\Psi,\Lambda)}:=\sum_{i=1}^\infty \la\mu_i\ral^{2s} |\la f,\psi_i\ral|^2 <\infty\,.
\]
Here $\la \mu\ral:=(1+|\mu|^2)^{1/2}.$ For $s<0$, $\mf^s(\Psi,\Lambda):=(\mf^{-s}(\Psi,\Lambda))^*$ is the completion of $L^2(\dbar)$ with respect to $\|\cdot \|_{\mf^s(\Psi,\Lambda)}.$
For any $f\in \mf(\Psi, \Lambda):=\mf^1(\Psi,\Lambda)$, the series
\[
T_{\Psi,\Lambda}f:=\sum_{i=1}^\infty \mu_i \la f,\psi_i\ral\psi_i
\]
converges in $L^2(\dbar)=\mf^0(\Psi,\Lambda)$. When $\Psi$ and $\Lambda$ are clear from context, we will sometimes write $\mf^s$ instead of $\mf^s(\Psi,\Lambda)$.
\begin{remark}
In our applications, $T_{\Psi,\Lambda}$ will be the Laplacian and $\mf^s(\Psi,\Lambda)$ will be $H^{2s}$.
\end{remark}
\begin{lmm}\label{compare1}
Let $T_{\Psi, \Lambda}$ be as above. Let $\Lambda' = (\mu_i')_{i\ge 1}$ be another sequence of real numbers. Let $\ep\in (0,1)$ and $\gamma\geq 0$ be numbers such that for all $i$,
$$|\mu_i' - \mu_i| \le \ep \la\mu_i\ral^{1-\gamma}.$$
Then $\|\cdot\|_{\mf^s(\Psi, \Lambda')}$ is equivalent to $\|\cdot\|_{\mf^s(\Psi, \Lambda)}$, and for all $s\in \re$, $T_{\Psi,\Lambda'}-T_{\Psi,\Lambda}:\mf^s(\Psi,\Lambda)\to \mf^{s-1+\gamma}(\Psi,\Lambda)$ with
$$\|T_{\Psi,\Lambda'} - T_{\Psi,\Lambda} \|_{\mf^s(\Psi,\Lambda)\to\mf^{s-1+\gamma}(\Psi,\Lambda)} \le \ep.$$
\end{lmm}
\begin{proof}
Since $\la\mu_i'\ral\le (1+\ep)\la\mu_i\ral$, we have $\|\cdot\|_{\mf^s(\Psi, \Lambda')}\leq C\|\cdot\|_{ \mf^s(\Psi, \Lambda)}$. On the other hand since $\la\mu_i\ral\le \la \mu_i'\ral/(1-\ep)$, so $\|\cdot\|_{\mf^s(\Psi, \Lambda)} \leq C\|\cdot\|_{\mf^s(\Psi, \Lambda')}$.
Next, let $f\in \mf^s(\Psi,\Lambda)$ with $s\geq 1$.
Then
$$(T_{\Psi,\Lambda}-T_{\Psi,\Lambda'})f=\sum_i (\mu_i-\mu_i')\la f,\psi_i\ral\psi_i.$$
Therefore,
\begin{align*} \|(T_{\Psi,\Lambda}-T_{\Psi,\Lambda'})f\|^2_{\mf^{s-1+\gamma}(\Psi,\Lambda)}&=\sum_i \la \mu_i\ral^{2s-2+2\gamma}|\mu_i-\mu_i'|^2|\la f,\psi_i\ral|^2\\
&\leq \sum_i\la \mu_i\ral^{2s-2+2\gamma}\e^2\la\mu_i\ral^{2(1-\gamma)}|\la f,\psi_i\ral|^2\\
&\leq \e^2\sum_i\la\mu_i\ral^{2s}|\la f,\psi_i\ral|^2\leq \e^2\|f\|_{\mf^{s}(\Psi,\Lambda)}^2\,.
\end{align*}
The density of $\mf(\Psi,\Lambda)$ in $\mf^{s}(\Psi,\Lambda)$ for $s\leq 1$ implies that the result extends to $s\in \re$.
This concludes the proof of the lemma.
\end{proof}
\begin{lmm}\label{compare2}
Let $\Psi$ and $\Lambda$ be as above. Let $L$ be the set of distinct elements of $\Lambda$. For each $\ell \in L$, let $I_\ell$ be the set of all $i$ such that $\mu_i = \ell$. Assume that $|I_\ell|$ is finite for each $\ell$. Let $\Psi' = (\psi_i')_{i\ge 1}$ be another complete orthonormal basis, such that for each $\ell\in L$, the span of $(\psi'_i)_{i\in I_\ell}$ equals the span of $(\psi_i)_{i\in I_{\ell}}$. Then for all $s$, $\mf^s(\Psi', \Lambda)=\mf^s(\Psi, \Lambda)$ and $T_{\Psi',\Lambda} = T_{\Psi, \Lambda}$.
\end{lmm}
\begin{proof}
The lemma follows from the fact that for any $\varphi\in C^\infty(\re)$,
$$\sum_i \varphi(\mu_i)\langle f, \psi_i\rangle \psi_i=\sum_{\ell\in L}\Pi_{\ell}f$$
where $\Pi_{\ell}$ denotes the orthognonal projection onto the span of $\{\phi_i\mid \mu_i=\ell\}$ and the fact that this span is clearly invariant under choices of bases.
\begin{comment}
Take some $\ell\in L$. Let $n = |I_\ell|$. Rename the elements of $(\psi_i)_{i\in I_\ell}$ as $\xi_1,\ldots, \xi_n$ and the elements of $(\psi_i')_{i\in I_\ell}$ as $\xi_1',\ldots, \xi_n'$. By assumption, $n$ is finite. Since the span of $(\xi'_i)_{1\le i\le n}$ equals the span of $(\xi_i)_{1\le i\le n}$, there is a matrix $Q = (q_{ij})_{1\le i,j\le n}$ such that for each $i$,
\[
\xi_i' = \sum_{j=1}^n q_{ij} \xi_j\,.
\]
By orthonormality of $\xi_1,\ldots, \xi_n$ and $\xi_1',\ldots, \xi_n'$,
\begin{align*}
\sum_{k=1}^n q_{ik} q_{jk} &= \la \xi_i', \xi_j'\ral =
\begin{cases}
1 &\text{ if } i=j\, ,\\
0 &\text{ otherwise.}
\end{cases}
\end{align*}
Therefore $Q$ is an orthogonal matrix. Thus, for any $f\in \mf(\Psi, \Lambda)$,
\begin{equation}
\label{eqn:rotate}
\begin{aligned}
\sum_{i=1}^n \la f,\xi_i'\ral\xi_i' &= \sum_{i=1}^n \biggl(\sum_{j=1}^n q_{ij} \la f,\xi_j\ral\biggr) \biggl(\sum_{k=1}^n q_{ik} \xi_k\biggr)\\
&= \sum_{j,k=1}^n \biggl(\sum_{i=1}^n q_{ij} q_{ik}\biggr) \la f,\xi_j\ral \xi_k = \sum_{j=1}^n \la f,\xi_j\ral\xi_j\,.
\end{aligned}
\end{equation}
For any function $g:\re\to \re$, \eqref{eqn:rotate} gives
\begin{align*}
\sum_{i\in I_\ell}g( \mu_i )\la \psi_i', f\ral\psi_i' &= g(\ell)\sum_{i=1}^n \la \xi'_i, f\ral\xi'_i\\
&= g(\ell) \sum_{i=1}^n \la\xi_i,f\ral\xi_i= \sum_{i\in I_\ell} g(\mu_i) \la\psi_i, f\ral\psi_i\,.
\end{align*}
Taking $L^2$ norms of both sides with $g(x)=\la x\ral^s$ and $g(x)=x$ respectively we see that $\mf^s(\Psi', \Lambda)=\mf^s(\Psi,\Lambda)$ and $T_{\Psi', \Lambda} = T_{\Psi, \Lambda}$.
\end{comment}
\end{proof}
\begin{lmm}\label{compare3}
Suppose that $\Lambda$ has $|\mu_i|>c>0$ with $|\mu_i|\to \infty$.
Let $\gamma_i := 1/\mu_i$ and $\Gamma := (\gamma_i)_{i\ge 1}$. Then $\mf(\Phi, \Gamma)= L^2(\dbar)$ and the range of $T_{\Phi, \Gamma}$ on $L^2$ is contained in $\mf(\Phi, \Lambda)$. Moreover, $T_{\Phi, \Lambda}T_{\Phi, \Gamma} = I$.
\end{lmm}
\begin{proof}
If $f\in L^2(\dbar)$, then clearly $f\in \mf(\Phi,\Gamma)$ since $\gamma_i \ra 0$ as $i\ra \infty$. The remainder of the proof follows from elementary computations together with the definition of $\mf(\Phi,\Lambda).$
\end{proof}
Now let $(u_i,\lambda_i^2)_{i\ge 1}$ be a complete orthonormal basis of eigenfunctions of the Dirichlet Laplacian. The let $\Phi=\{u_i\}$ and $\Lambda =\lambda_i^2 $.
\begin{lmm}
\label{inverse}
Let $T_{\Phi,\Gamma}$ be as in Lemma \ref{compare3}. Then $\mf(\Phi,\Lambda)=\mfd$ and $T_{\Phi,\Gamma}\Delta f=-f$.
\end{lmm}
\begin{proof}
Lemma \ref{inverse} is also an easy consequence of the spectral theorem applied to the Dirichlet Laplacian.
\end{proof}
\begin{comment}
\begin{lmm}\label{greenlmm}
Let $T_{\Phi, \Gamma}$ be as in Lemma \ref{compare3}, and let $B_t$, $\ee^x$ and $\tau_\om $ be as in Section \ref{ptwisesec}. Then for any bounded Borel measurable $f:\om \ra\rr$,
\[
T_{\Phi, \Gamma} f(x) = \frac{1}{2}\, \ee^x\biggl( \int_0^{\tau_\om } f(B_t) \, dt\biggr)
\]
for almost every $x\in \om $.
\end{lmm}
\begin{proof}
Define
\[
h(x) := \ee^x\biggl( \int_0^{\tau_\om } f(B_t) \, dt\biggr)\,.
\]
It is not difficult to see from standard properties of Brownian motion that $\pp^x(t<\tau_\om )$ decreases exponentially in $t$. By the boundedness of $f$, this implies that the integral on the right side is finite, and moreover, that the expectation and integral may be exchanged to give
\[
h(x) = \int_0^\infty \ee^x(f(B_t); t<\tau_\om )\, dt\,.
\]
Thus by part (i) and part (iii) of Theorem \ref{summarythm},
\begin{align}\label{heq}
h(x) &= \int_0^\infty \int_\om p(t,x,y) f(y) \, dy\, dt\,.
\end{align}
Note that $f$ is bounded, $p(t,x,y)$ is nonnegative and
\[
\int_0^\infty \int_\om p(t,x,y) \, dy\, dt = \int_0^\infty \pp^x(t<\tau_\om ) \, dt = \ee^x(\tau_\om )\,,
\]
and $\ee^x(\tau_\om )$ is easily shown to be uniformly bounded in $x$. From this it follows by Fubini's theorem and~\eqref{heq} that for any $g\in L^2(\dbar)$,
\begin{equation}\label{gh1}
(g,h) = \int_0^\infty \int_\om \int_\om p(t,x,y) g(x)f(y)\, dy\, dx\,dt\,.
\end{equation}
Part (iii) of Theorem \ref{summarythm} implies that for any $t>0$,
\[
\int_\om \int_\om p(t,x,y) g(x)f(y)\, dy\, dx = \int_\om \int_\om \sum_{i=1}^\infty e^{-\mu_i t} \phi_i(x)\phi_i(y) g(x)f(y)\, dy\, dx\,.
\]
Equation (4.28) on page 133 of \citet{bass95} gives the estimate $e^{-\mu_i t} |\phi_i(x)\phi_i(y)|\le Ce^{-\mu_i t/2} t^{-d}$, which, together with the second assertion of Theorem \ref{ptwisethm} and the square-integrability of $f$ and $g$, allows us to apply Fubini's theorem to interchange the sum and the integrals on the right side of the above display and get
\begin{equation}\label{gh2}
\int_\om \int_\om p(t,x,y) g(x)f(y)\, dy\, dx = \sum_{i=1}^\infty e^{-\mu_i t} (\phi_i, g)(\phi_i, f)\,.
\end{equation}
Since $f$ and $g$ both belong to $L^2(\dbar)$, therefore
\[
\sum_{i=1}^\infty \frac{|(\phi_i, g)(\phi_i, f)|}{\mu_i} \le C\sum_{i=1}^\infty|(\phi_i, g)(\phi_i, f)| \le C\|g\|\|f\|<\infty\,.
\]
Thus, another application of Fubini's theorem gives that
\begin{align*}
\int_0^\infty \sum_{i=1}^\infty e^{-\mu_i t} (\phi_i, g)(\phi_i, f)\, dt &= \sum_{i=1}^\infty (\phi_i,g)(\phi_i, f)\int_0^\infty e^{-\mu_i t}\, dt \\
&= \sum_{i=1}^\infty \frac{(\phi_i, g)(\phi_i, f)}{\mu_i} = 2\,(g, T_{\Phi, \Gamma} f)\,.
\end{align*}
On the other hand by \eqref{gh1} and \eqref{gh2},
\[
\int_0^\infty \sum_{i=1}^\infty e^{-\mu_i t} (\phi_i, g)(\phi_i, f)\, dt = (g,h)\,.
\]
Thus, $2\,(g, T_{\Phi, \Gamma} f) = (g,h)$.
Since this holds for every $g\in L^2(\dbar)$, this shows that $2\,T_{\Phi, \Gamma} f(x)=h(x)$ for almost all $x\in \om $.
\end{proof}
\begin{lmm}\label{greenlmm2}
For any $f\in \mf$, $T_{\Phi, \Gamma} \Delta f = -f$ in $L^2(\dbar)$.
\end{lmm}
\begin{proof}
Take any $f\in \mf$. Then $\Delta f$ is a bounded continuous function on $\om $. Therefore by Lemma~\ref{greenlmm},
\[
T_{\Phi, \Gamma} \Delta f(x) =\frac{1}{2}\, \ee^x\biggl(\int_0^{\tau_\om } \Delta f(B_t)\, dt\biggr)\,
\]
for almost every $x\in \om $. This is the same as saying that the two sides, as functions of $x$, are equal in $L^2(\dbar)$.
Using the boundedness and continuity of $\Delta f$, and the fact that $f$ vanishes continuously on $\partial \om $, a simple argument using It\^{o}'s formula shows that
\[
f(x) = - \frac{1}{2}\,\ee^x\biggl(\int_0^{\tau_\om } \Delta f(B_t)\, dt\biggr)\,.
\]
The details of this argument can be found in the solution to Exercise 8.3 on page 378 of \cite{mortersperes10}.
\end{proof}
\end{comment}
We are now ready to construct the perturbed Laplacian and finish the proof of Theorem \ref{quethm2} (and hence, also of Theorem \ref{quethm}).
\begin{proof}[Proof of Theorem \ref{quethm2}]
Let $\{\lambda_i^2\}_{i\ge 1}$ be the eigenvalues of $-\Delta$
and let $\Lambda=\{\lambda_i^2\}_{i\ge1}$. Recall that we assume $\om$ is AQE at scale $\lambda^{-\gamma}$ for some $0\leq \gamma\leq 2$.
Fix $\e\in(0,1)$. Our strategy will be to split the eigenvalues between $(1+\e)^n$ and $(1+\e)^{n+1}$ into $N_n$ intervals where $N_n\sim \e(1+\e)^{n(1-\frac{\gamma}{2})}$. We will then reassign all of the eigenvalues in each subinterval to the left boundary of that interval, randomly rotate the corresponding eigenfunctions, and reassign eigenvalues so that the spectrum is simple. This will produce an almost surely QUE operator.
Observe that for all $i\geq 1$, either $\lambda_i< 1+\e$ or there exist positive integers $n$, $0 \leq j\leq N_n-1$ where
\begin{equation}
\label{eqn:N}N_n:=\lceil (1+\e)^{n\gamma}\rceil\end{equation} such that
$$(1+\e)^n\left(1+\frac{j\e}{N_n}\right)\leq \lambda_i< (1+\e)^{n}\left(1+\frac{(j+1)\e}{N_n}\right).$$
In the first case, let $\lambda_i'=\lambda_i$. In the second, let
$$\lambda_i'=(1+\e)^n\left(1+\frac{j\e}{N_n}\right).$$
Note that for $\e>0$ small enough (independent of $\lambda_i$,
$$|\lambda_i^2-(\lambda_i')^2|\leq \e(1+\e)^{-n\gamma}(1+\e)^n\leq 3 \e|\lambda_i^2|^{1-\frac{\gamma}{2}}.$$
Therefore, by Lemma \ref{compare1}, for $s\geq 0$
$$\mf^s(\Phi,\Lambda')=\mf^s(\Phi,\Lambda)$$
and for $s\geq 1-\frac{\gamma}{2}$ and $\ep$ small enough,
\begin{equation}
\label{e:12}\|T_{\Phi,\Lambda'}-T_{\Phi,\Lambda}\|_{\mf^s\to \mf^{s-1+\frac{\gamma}{2}}}\leq 3\e.
\end{equation}
Let $L$ be the set of distinct eigenvalues in $\Lambda'$. For each $l\in L$, let $I_l$ be the set of $i$ such that $\lambda_i'=l$. Then since $\lambda_i\to\infty$, $|I_l|<\infty$ for all $l$. For each $l$, let $(u_i')_{i\in I_l}$ be a random rotation of $(u_i)_{i\in I_l}.$ Then, by Lemma \ref{compare2},
$$T_{\Phi,\Lambda'}=T_{\Phi',\Lambda'},\quad\quad \mf^s(\Phi,\Lambda')=\mf^s(\Phi',\Lambda').$$
Now, for each $ l\in L$,
\[
l=(1+\e)^n\left(1+\frac{j\e}{N_n}\right)
\]
for some $n,j$ or $0<l<(1+\e)$. Denote the set of $l$ with $0<l<1+\e$ by $L_<$ and let $I_{<} := \cup_{l\in L_<}I_l$. Let
$(\lambda_i'')_{i\in I_<}$ be an arbitrary set of distinct real numbers with
$$(1-\e)\lambda_i'\leq \lambda_i''<\lambda_i'.$$
For $l\notin L_<$, let $(\lambda_i'')_{i\in I_l}$ be an arbitrary set of distinct real numbers with
$$(1+\e)^n\left(1+\frac{j\e}{N_n}\right)\leq \lambda_i''<(1+\e)^n\left(1+\frac{(j+1)\e}{N_n}\right).$$
Then for any $i$, and $\e>0$ small enough (independently of $i$)
$$|(\lambda_i')^2-(\lambda_i'')^2|\leq 3\e |(\lambda_i')^2|^{1-\frac{\gamma}{2}}$$
and hence
$$\mf^s(\Phi',\Lambda'')=\mf^s(\Phi',\Lambda')=\mf^s(\Phi,\Lambda')=\mf^s(\Phi,\Lambda)$$
and
$$\|T_{\Phi',\Lambda''}-T_{\Phi,\Lambda'}\|_{\mf^s\to\mf^{s-1+\frac{\gamma}{2}}}\leq 3\e.$$
Combining this with \eqref{e:12} gives
$$\|T_{\Phi',\Lambda''}-T_{\Phi,\Lambda}\|_{\mf^s\to \mf^{s-1+\frac{\gamma}{2}}}\leq 6\e.$$
Now, let
$$\Gamma=
\{\lambda_i^{-1}\}_{i\ge 1}.$$
and $G:=T_{\Phi,\Gamma}.$ For convenience, write $T=T_{\Phi,\Lambda}$ and $T''=T_{\Phi',\Lambda''}.$
Then by Lemma \ref{compare3}, $G$ is bounded on $L^2(\dbar)$, has range in $\mf(\Phi,\Lambda)$, and satisfies $TG=I$. Therefore, the operator
$$S:=(T''-T)G$$ maps $L^2$ into $\mf^{\frac{\gamma}{2}}(\Phi,\Lambda)$.
We will show that $S$ satisfies the three assertions of the theorem. Note that the construction of $S$ involves random rotations and what we will actually show is that $S$ satisfies the conditions with probability one. This will suffice to demonstrate the existence of an $S$ that satisfies the requirements.
First, notice that
$$\|Sf\|_{\mf^{\frac{\gamma}{2}}}=\|(T''-T)Gf\|_{\mf^{\frac{\gamma}{2}}}\leq 10 \e\|Gf\|_{\mf}\leq C\e\|f\|.$$
Now, by Lemma \ref{inverse}, $\mfd=\mf(\Phi,\Lambda)$, therefore
\[
\mf^{\frac{\gamma}{2}}(\Phi,\Lambda)=\mfd^{\gamma}=(L^2(\om ),\mfd)_{\frac{\gamma}{2}},
\]
the complex interpolation space of $L^2$ and $\mfd$. Hence (i) holds.
Next, note that by Lemma \ref{inverse} for $f\in \mfd$, $-G\Delta f=f. $ Therefore, for $f\in \mfd$,
$$(I+S)\Delta f=(I+(T''-T)G)\Delta f=T''G\Delta f=-T''f.$$
That is, $-(I+S)\Delta=T''$ on $\mfd$. This proves part (ii) of the theorem. Part (iii) of the theorem follows from the fact that $\{u_i'\}$ is an orthonormal basis for $L^2(\dbar)$ and each $u_i'$ is a linear combination of finitely many $u_i$ which have $u_i\in \mfd^s$ for all $s$.
It remains to show that the eigenfunctions of $T''$ are equidistributed. For this, recall that
\[
l=(1+\e)^n\left(1+\frac{j\e}{N_n}\right)
\]
for $l$ large enough and hence
\begin{gather*}
I_l=\left\{i:\lambda_-\leq \lambda_i<\lambda_+\right\},\\
\lambda_-:=(1+\e)^n\left(1+\frac{j\e}{N_n}\right),\qquad \lambda_+:=(1+\e)^n\left(1+\frac{(j+1)\e}{N_n}\right).
\end{gather*}
Now,
$$r_+:=\frac{\lambda_+}{\lambda_-}=\frac{1+\frac{(j+1)\e}{N_n}}{1+\frac{j\e}{N_n}}=1+\frac{\e}{N_n}+\O{}(\e^2N_n^{-1}).$$
Then since $\om $ is AQE at scale $\alpha(\lambda)=\O{}(\lambda^{-\gamma})$ and $N_n^{-1}\geq c\lambda^{-\gamma}$,
\begin{equation}\label{eqn:locWeyl}\lim_{l\in L,\,l\to \infty}\frac{1}{|I_l|}\left|\sum_{i\in I_l}\la (A -\overline{\sigma(A)})1_{\dbar}u_i,1_{\dbar}u_i\ral\right| =0\end{equation}
for $A\in\mc{A}\subset\Psi(\re^d)$, where
\begin{align*}
\overline{\sigma(A)}&=\frac{1}{\vol(1\leq |\xi|\leq 1+r_+)}\iint_{1\leq |\xi|\leq 1+r_+}\sigma(A)(x,\xi)1_{\dbar}dxd\xi\\
&=\int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi).
\end{align*}
Note that we have used that $\sigma(A)$ is homogeneous of degree $0$.
Now, by Theorem \ref{rotthm}, for any $A\in \mc{A}$ and $t\in (0,1)$,
\begin{align*}
&\mathbb{P}\left(\left|\la A1_{\dbar}u_i',1_{\dbar}u_i'\ral-\frac{1}{|I_l|}\sum_{i\in I_l}\la A1_{\dbar}u_i,1_{\dbar}u_i\ral\right|\geq t\right)\\
&\leq C_1 \exp(-C_2(\|A\|)\min(t^2,t)|I_l|).
\end{align*}
\begin{remark}
Note that we may assume that $u_i$ are real valued without loss of generality.
\end{remark}
So,
\begin{align*}
&\mathbb{P}\left(\max_{i\in I_l}\left|\la A1_{\dbar}u_i',1_{\dbar}u_i'\ral-\frac{1}{|I_l|}\sum_{i\in I_l}\la A1_{\dbar}u_i,1_{\dbar}u_i\ral\right|\geq t\right)\\
&\leq C_1 |I_l|\exp(-C_2(\|A\|)\min(t^2,t)|I_l|).
\end{align*}
The Weyl law (or more precisely the fact that $\om$ is AQE at scale $\lambda^{-\gamma}$) implies that $|I_l|\sim C\e l^{\frac{d-\gamma}{2}}$ and hence that
$$\sum_{l\in L}|I_l|\exp(-C_2(\|A\|)\min(t^2,t)|I_l|)<\infty.$$
Using the Borel--Cantelli lemma we have that
\begin{align*}
\mathbb{P}&\biggl(\biggl|\la A1_{\dbar}u_i',1_{\dbar}u_i'\ral-\frac{1}{|I_l|}\sum_{i\in I_l}\la A1_{\dbar}u_i,1_{\dbar}u_i\ral\biggr|\geq t\\
&\qquad \qquad \text{ for infinitely many $i$ and $l$ with $i\in I_l$}\biggr)=0.
\end{align*}
Thus, by \eqref{eqn:locWeyl} for all $\delta>0$,
$$\mathbb{P}\left(\limsup_{i\to \infty}\left|\la A1_{\dbar}u_i',1_{\dbar}u_i'\ral-\overline{\sigma(A)}\right|\geq \delta\right)=0.$$
The fact that $\mc{A}$ is dense in $C_0(S^*\om )$ and $C_0(S^*\om)$ is separable then implies that $\mc{M}(u_i')=\{1_{\dbar}dxd\sigma(\xi)\}.$
Now, suppose that $f\in \mfd$ is an $L^2$ normalized eigenfunction of $T''$. Then
$$0=\|T''f-\lambda^2 f\|^2=\sum_i((\lambda_i'')^2-\lambda^2)^2|\la f,u_i'\ral|^2.$$
Hence, since $f\neq 0$, $\lambda=\lambda_i''$ for some $i$. Thus, for any $j$
\begin{align*}
\la\phi_j',f\ral&=\frac{1}{(\lambda_i'')^2}\la u_j',T''f\ral\\
&=\frac{1}{(\lambda_i'')^2}\sum_{k}\la u_j',u_k'\ral(\lambda_k'')^2\la u_k',f\ral=\frac{(\lambda_j'')^2}{(\lambda_i'')^2}\la u_j',f\ral.
\end{align*}
Hence $\langle u_j',f\rangle=0$ or $\lambda_j''=\lambda_i''$. But for $j$ large enough, $\lambda_i''\neq \lambda_j''$ for $i\neq j$ and hence $f=u_i'$ and $T''$ has equidistributed eigenfunctions.
Notice also that this implies that for $\{f_n\}_{n=1}^\infty$ the eigenfunctions of $-(I+S)\Delta $ with $-(I+S)\Delta f_n=\alpha_n^2f_n$, and $n$ large enough, $f_n=\phi_{n_j}'$ and hence
$$-(I+S)\Delta f_n=T''f_n=\alpha_n^2f_n.$$
Consequently,
\begin{equation}
\label{perturbEst}\|Sf_n\|=\|(T''-T)G\Delta \phi_{n_j}'\|=\|(T''-T)\phi_{n_j}'\|\leq C\e\la \alpha_n\ral^{-\gamma}.\end{equation}
This completes the proof of both Theorem \ref{quethm} and part (ii) of Corollary \ref{almostcor}.
\end{proof}
\begin{proof}[Proof of Corollary \ref{almostcor}]
By Theorem \ref{quethm}, (taking for example, $\e=n^{-1}$) there exists a sequence of linear operators $\{S_n\}_{n\ge 1}$ such that
\[
\|S_n\|_{L^2\to \mfd^{\gamma}}\ra0
\]
and $-(I+S_n)\Delta$ is positive and has QUE eigenfunctions for each $n$. This implies the existence of an orthonormal basis of $L^2(\dbar)$, $\{f_{n,k}\}_{k=1}^\infty$ and $\alpha_{n,k}$ such that $\|f_{n,k}\|=1$ for each $n$ and $k$, $\alpha^2_{n,k} \ra \infty$ as $k\to \infty$, and
\[
(I+S_n)\Delta f_{n,k} = -\alpha^2_{n,k} f_n\,.
\]
Without loss of generality, $\|S_n\|< 1$. Then the series
\[(I+S_n)^{-1}= \sum_{k=0}^\infty (-1)^k S_n^k
\]
converges in the space of bounded linear operators on $L^2(\dbar)$. Moreover,
\[
(I+S_n)^{-1}-I=-(I+S_n)^{-1}S_n
\]
Therefore, by \eqref{perturbEst}
\begin{align*}
\|-\Delta f_{n,k} - \alpha^2_{n,k} f_{n,k}\|&= \|\alpha^2_{n,k}(I+S_n)^{-1}S_nf_{n,k}\|\\
&\le \alpha^2_{n,k}\|(I+S_n)^{-1}\|_{L^2\to L^2}\|S_nf_n\|\\\
&\le C\frac{\la \alpha_{n,k}\ral ^{2-\gamma}}{1-\|S_n\|_{L^2\to L^2}}\|S_n\|_{L^2\to \mfd^\gamma}\,.
\end{align*}
Dividing both sides by $\alpha^2_{n,k}$ completes the proof since $\|S_n\|\to 0$.
\end{proof}
\section{Improvements on closed manifolds}
\label{last2}
In order to prove Theorem \ref{quethm} on a manifold $M$ with $\vol(M)=1$, we work with $L^2_0(M)$, the set of $0$ mean functions in $L^2$ to remove the 0 eigenvalue of the Laplacian. Let $\{(u_i, \lambda_i^2)\}_{i=1}^\infty$ be the eigenvalues and eigenfunctions of $-\Delta_g$. Then with $T_{\Phi,\Lambda}$ and $T_{\Phi,\Gamma}$ as above, the proof of Theorem~\ref{quethm} for $M$ proceeds as above.
We now prove Corollary \ref{almostcor2}. For this, we need to use the full strength of Theorem \ref{thm:weylStrong}.
\begin{proof}[Proof of Corollary \ref{almostcor2}]
Recall that the set of closed geodesics is assumed to have measure zero in $S^*M$. Let $\gamma=1$ and return to \eqref{eqn:N}, where
we replace $N_n$ with
$$N_n:=\lceil (1+\e)^{n}\rceil\beta_n$$
where $\beta_n\in \mathbb{N}$ has $\beta_n\to\infty$ slowly enough. We then proceed as in the proof of Theorem \ref{quethm} until \eqref{eqn:locWeyl}. At this point we need to show that there exists $\beta_n\to \infty$ slowly enough so that for $\|f\|_{L^\infty(M)}\leq 1$,
$$\lim_{l\in L,i\to \infty}\frac{1}{|I_l|}\Big|\sum_{i\in I_l}\la (f-\overline{f}) u_i,u_i\ral\Big|=0$$
where
$$\overline{f}=\int_Mfd\vol.$$
First, observe that
\begin{gather*} \lambda_-:=(1+\e)^{n}\left(1+\frac{j\e}{N_n}\right),\quad\quad\lambda_+:=(1+\e)^{n}\left(1+\frac{(j+1)\e}{N_n}\right),\\
I_l=\left\{i\left|\,\lambda_-\leq \lambda_i<\lambda_+\right.\right\}.\end{gather*}
Note also that by Theorem \ref{thm:weylStrong},
$$\sum_{\lambda_1\leq \lambda_j\leq \lambda_2}|u_j(x)|^2=\frac{(\lambda_2-\lambda_1)\lambda_2^{d-1}}{(2\pi )^d}\vol(S^{d-1})+g(\lambda_2,\lambda_1,x)$$
where
$$\lim_{\lambda_2\to \infty}\sup_{\lambda_1\leq \lambda_2}\|g(\lambda_2,\lambda_1,x)\|_{L^\infty_{x}}\lambda_2^{-d+1}=0.$$
Therefore, integrating, we have
$$\#\{\lambda_1\leq \lambda_j\leq \lambda_2\}=\frac{(\lambda_2-\lambda_1)\lambda_2^{d-1}}{(2\pi )^d}\vol(S^{d-1})+\int g(\lambda_2,\lambda_1,x)dx$$
and, provided that
$$(\lambda_2-\lambda_1)\lambda_2^{d-1}\gg \sup_{\lambda_1\leq \lambda_2}\|g(\lambda_2,\lambda_1,x)\|_{L^\infty},$$
$$\left|\frac{\sum_{\lambda_1\leq \lambda_j\leq \lambda_2}|u_j(x)|^2}{\#\{\lambda_1\leq \lambda_j\leq \lambda_2\}}-1\right|\leq C\|g(\lambda_2,\lambda_1,x)\|_{L^\infty_{x}}\lambda_2^{-d+1}(\lambda_2-\lambda_1)^{-1} $$
Thus, taking $\lambda_1=\lambda_-$ and $\lambda_2=\lambda_+$, we have
$$\lambda_2\geq (1+\e)^{n},\quad\quad \lambda_2-\lambda_1\sim \frac{\e}{\beta_n}.$$
Therefore, taking $\beta_n\to \infty$ slowly enough so that
$$\lim_{n\to \infty}\sup_{\lambda_-\leq \lambda_+}\|g(\lambda_+,\lambda_-,x)\|_{L^\infty_{x}}\lambda_+^{-d+1}(\lambda_+-\lambda_-)^{-1}=0$$
gives that uniformly for $\|f\|_{L^\infty}\leq 1$,
$$\lim_{l\in L,i\to \infty}\frac{1}{|I_l|}\Big|\sum_{i\in I_l}\la (f-\overline{f}) u_i,u_i\ral\Big|=0.$$
\begin{remark}
Note that the uniformity in $f$ is crucial here and is precisely the reason that we have been unable to prove a version of Corollary \ref{almostcor2} giving an orthonormal basis of QUE eigenfunctions. More precisely, the remainder in the Weyl law involving matrix elements $\langle Au_j,u_j\rangle$ depends on more than just $\sup |\sigma(A)|$. In particular, it involves derivatives $\sigma(A)$.
\end{remark}
Then, using the fact that $f\in C^\infty(M)$ with $\|f\|_{L^\infty(M)}\leq 1$ is dense in the unit ball of the dual space to finite radon measures, that this space is separable, and following the proof of Theorem \ref{quethm} from \eqref{eqn:locWeyl} shows that for all $\e>0$, there exists $S:L^2(M)\to H^1(M)$ so that $\|S\|_{L^2\to H^1}\leq \e$,
$-(I+S)\Delta_g$ has equidistributed eigenfunctions, $\{(f_n,\alpha_n)\}_{n=1}^\infty,$ and by \eqref{perturbEst} $\|S f_n\|=\o{}(\alpha_n^{-1})\|f_n\|.$
Therefore,
$$-(I+S)\Delta f_n=\alpha_n^2f_n.$$
Now,
$$(I+S)^{-1}=\sum_{k=0}^\infty (-1)^kS^k,\quad\quad (I+S)^{-1}-I=-(I+S)^{-1}S.$$
Therefore,
$$(-\Delta -\alpha_n)f_n=-\alpha_n^2(I+S)^{-1}Sf_n$$
and hence,
\begin{align*}
\|(-\Delta-\alpha_n^2)f_n\|&\leq |\alpha_n^2|\|(I+S)^{-1}\o{}(\alpha_n^{-1})\|f_n\|\\
&=\o{}(\alpha_n)\|f_n\|
\end{align*}
Dividing by $\alpha_n^2$ completes the proof of the corollary.
\end{proof}
\vskip.2in
\noindent {\bf Acknowledgements.} The authors would like to thank Persi Diaconis, Peter Sarnak, Andr\'as Vasy, Steve Zelditch and Maciej Zworski for various helpful discussions as well as Dima Jakobson for his careful reading of a previous version. The authors also thank the anonymous referee for many useful comments suggestions.
\appendix
\section{A local Weyl law on regular domains}\label{ptwisesec}
Throughout this section, we assume that $\Omega\subset \re^d$ is a regular domain. Let $B_t$ be a standard $d$-dimensional Brownian motion (in $\mathbb{R}^d$), starting at some point $x\in \om$. Recall the definition \eqref{exit} of the exit time $\tau_\om$ from the domain $\om$.
We will need a few well-known facts about this exit time, summarized in the following theorem.
\begin{thm}[Compiled from Proposition 4.7 and Theorems 4.12 and 4.13 of Chapter II in \citet{bass95} and Section 4 of Chapter 2 in \citet{portstone78}]\label{summarythm}
For any regular domain $\om$ (as defined in Section \ref{intro}), there exists a unique function $p:(0,\infty)\times \dbar\times \dbar \ra[0,\infty)$ such that:
\begin{enumerate}
\item[\textup{(i)}] For any bounded Borel measurable $f:\om\ra\rr$, $x\in \om$, and $t\geq 0$,
\[
\ee^x(f(B_t); t< \tau_\om) = \int_\om p(t,x,y) f(y)\, dy\,,
\]
where $\ee^x$ denotes expectation with respect to the law of Brownian motion started at $x$.
\item[\textup{(ii)}] There is a complete orthonormal basis $(u_i)_{i\ge 1}$ of $L^2(\dbar)$ such that each $u_i$ is $C^\infty$ in $\om$, vanishes continuously at the boundary, and there are numbers $0<\lambda_1^2\le \lambda_2^2\le\cdots$ tending to infinity such that
\[
p(t,x,y) = \sum_{i=1}^\infty e^{-\frac{1}{2}\lambda_i^2 t} u_i(x)u_i(y)\,,
\]
where the right side converges absolutely and uniformly on $\dbar\times \dbar$. Moreover, $-\Delta_D u_i = \lambda_i^2 u_i$ for each $i$ where $-\Delta_D$ is the Dirichlet Laplacian on $\Omega$.
\end{enumerate}
\end{thm}
\begin{remark}
Notice that $p(t,x,y)$ is the Heat kernel of $\Omega$. That is, the kernel of $e^{t\Delta_D/2}.$
\end{remark}
Let $\lambda_i$ be as in the above theorem. For each $\ep>0$ and $\lambda >0$ define a set of indices $J_{\ep,\lambda}$ as
\[
J_{\ep,\lambda} := \{i: \lambda \le \lambda_i< \lambda (1+\ep)\}\,.
\]
Let $|J_{\ep,\lambda}|$ denote the size of the set $J_{\ep,\lambda}$. The following theorem is the main result of this section.
\begin{thm}\label{ptwisethm}
For any fixed $\ep>0$, $J_{\ep, \lambda}$ is nonempty for all large enough $\lambda$ and for $A\in \Psi(\re^d)$, with symbol $\sigma(A)(x,\xi)$ supported in $K_x\times \re^d$ with $K_x\subset \om$ compact,
\begin{align*}
\lim_{\lambda \ra \infty} \biggl|\frac{1}{|J_{\ep,\lambda}|}\sum_{i\in J_{\ep,\lambda}} \la (A-\bar{A})1_{\dbar}\phi_ j,1_{\dbar}\phi_j\ral \ \biggr|\, =0\,.
\end{align*}
where
$$\bar{A}=\int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi)$$
where $S$ is the normalized surface measure on $S^{d-1}$.
Moreover, $$\lim_{\lambda\ra\infty} \lambda^{-d}|J_{\ep, \lambda}| = \frac{(1+\e)^{d}-1}{(4\pi)^{d/2}\Gamma(d/2+1)}.$$
\end{thm}
This theorem implies Theorem \ref{localWeyl} since $\sigma(A)$ is homogeneous of degree $0$ and is a variant of results that are sometimes called `local Weyl laws', as in \cite{zelditch10}. However, we are not aware of a local Weyl law in the literature that applies for a domain as general as the one considered here. Our proof follows closely that in \cite{gerardleichtnam93}, but by using probabilistic methods to obtain estimates on the kernel of $e^{t\Delta_D}$, we are able to weaken the regularity assumptions on the domain.
Since we will have occasion to refer to both the Laplace operator on $L^2(\re^d)$ and the Dirichlet Laplacian in this section, we will denote them respectively by $-\Delta_{\re^d}$ and $-\Delta_D$. Theorem \ref{ptwisethm} will follow from the following lemma
\begin{lmm}
\label{traceLem}
Take $A \in \Psi^0(\re^d)$ with symbol $\sigma(A)(x,\xi)$ supported in $K_x\times \re^d$ where $K_x\subset \om$ is compact. Then for all $t>0$, $1_{\dbar}A1_{\dbar}e^{t\Delta_{D}}$ is trace class as an operator on $L^2(\dbar)$ and
$$\lim_{t\to 0^+}\frac{\tr(1_{\dbar}A1_{\dbar}e^{t\Delta_D})}{\tr (e^{t\Delta_D})}=\int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi)$$
where $\lambda=1_{\dbar}dxd\sigma(\xi)$ and $\sigma$ is the normalized surface measure on $S^{d-1}$.
\end{lmm}
We first show how Theorem \ref{ptwisethm} follows from Lemma \ref{traceLem}. We will need the following classical Tauberian theorem (see for example \cite{Ta}).
\begin{lmm}
\label{taubLem}
Suppose that $F:[0,\infty)\to \re$ is nondecreasing and for some $A,\gamma>0$,
$$\int_0^\infty e^{-t\alpha}dF(\alpha)\sim At^{-\gamma} \ \ \text{ as } t\to 0^+\,.$$
Then
$$F(\tau)\sim \frac{A\tau^{\gamma}}{\Gamma(\gamma+1)}\ \ \text{ as } \tau \to \infty\,.$$
\end{lmm}
The rest of this section is devoted to the proof of Lemma \ref{traceLem} and Theorem \ref{ptwisethm}. We will freely use the notation introduced in the statements of Theorem \ref{summarythm} and Theorem \ref{ptwisethm} without explicit reference.
First, note that the following corollary of Theorem \ref{summarythm} is immediate from the continuity of $p$.
\begin{lmm}\label{immcor}
Take any $x,y\in \om$ and let $A_{y,r}$ be the closed ball of radius $r$ centered at $x$. Then
\[
p(t,x,y) = \lim_{r\ra0} \frac{\pp^x(B_t\in A_{y,r},\, t<\tau_\om)}{\vol(A_{y,r})}\,.
\]
\end{lmm}
\begin{proof}
By assertion (i) of Theorem \ref{summarythm},
\[
\pp^x(B_t\in A_{y,r},\, t<\tau_\om) = \int_{A_{y,r}} p(t,x,z)\, dz\,.
\]
By assertion (ii) of Theorem \ref{summarythm},
\[
\lim_{r\ra0}\frac{1}{\vol(A_{y,r})} \int_{A_{y,r}} p(t,x,z)\, dz = p(t,x,y)\,.
\]
The proof is completed by combining the two displays.
\end{proof}
The following lemma compares the transition density of killed Brownian motion with the transition density of unrestricted Brownian motion when $t$ is small.
\begin{lmm}\label{pdifflmm}
Let
\[
\rho(t,x,y) := \frac{1}{(2\pi t)^{d/2}} e^{-\|x-y\|^2/2t}
\]
be the transition density of Brownian motion. Take any $x,y\in \om$ and let $\delta_y$, $\delta_x$ be respectively the distance of $y$ and $x$ from $\partial \om$.
Then
\begin{gather*}
|\partial_y^\alpha(\rho(t,x,y) - p(t,x,y)) | \le \frac{Ce^{-\delta_y^2/2t}}{t^{d/2+|\alpha|}}\,,\quad \quad 0< t<\delta_y^2/(d+2|\alpha|), \\
|\partial_x^\alpha(\rho(t,x,y) - p(t,x,y)) | \le \frac{Ce^{-\delta_x^2/2t}}{t^{d/2+|\alpha|}}\,,\quad\quad 0< t<\delta_x^2/(d+2|\alpha|),
\end{gather*}
where $C$ is a finite constant that depends only on $d$, $|\alpha|$ and the diameter of the domain $\om$.
\end{lmm}
\begin{proof}
Since $\tau_\om$ is a stopping time, the strong Markov property of Brownian motion implies that $X_s := B_{s+\tau_\om}$ is a standard Brownian motion started from $B_{\tau_\om}$ that is independent of the stopped sigma algebra of $\tau_\om$, which we will denote by $\mf_{\tau_\om}$. Consequently, if $A_{y,r}$ is the closed ball of radius $r<\delta_y/2$ centered at $y$, then for any $s\ge 0$,
\begin{align*}
\pp^x(X_{s} \in A_{y,r}\mid\mf_{\tau_\om}) &= \frac{1}{(2\pi s)^{d/2}}\int_{A(y,r)} e^{-\|z-B_{\tau_\om}\|^2/2s}\, dz\,.
\end{align*}
Consequently,
\begin{align*}
&\pp^x(B_t \in A_{y,r}, \, t\ge \tau_\om) = \pp^x(X_{t-\tau_\om} \in A_{y,r},\, t\ge \tau_\om)\\
&= \ee^x(\pp^x(X_{t-\tau_\om} \in A_{y,r}\mid\mf_{\tau_\om})\,;\, t\ge \tau_\om)\\
&= \ee^x\biggl(\frac{1}{(2\pi (t-\tau_\om))^{d/2}}\int_{A(y,r)} e^{-\|z-B_{\tau_\om}\|^2/2(t-\tau_\om)}\, dz\,;\, t\ge \tau_\om\biggr)\,,
\end{align*}
where the term inside the expectation is interpreted as zero if $t=\tau_\om$.
Dividing both sides by $\vol(A_{y,r})$, sending $r$ to zero, and observing that the term inside the above expectation after division by $\vol(A_{y,r})$ is uniformly bounded by a deterministic constant, we get
\begin{align*}
&\lim_{r\ra 0} \frac{\pp^x(B_t \in A_{y,r}, \, t\ge \tau_\om)}{\vol(A_{y,r})}\\
&= \ee^x\biggl(\frac{1}{(2\pi (t-\tau_\om))^{d/2}} e^{-\|y-B_{\tau_\om}\|^2/2(t-\tau_\om)}\,;\, t\ge \tau_\om\biggr)\,.
\end{align*}
Now note that
\begin{align*}
\pp^x(B_t \in A_{y,r}) - \pp^x(B_t\in A_{y,r}, \, t <\tau_\om)&= \pp^x(B_t\in A_{y,r}, \, t \ge \tau_\om)
\end{align*}
and
\begin{align*}
\lim_{r\ra0} \frac{\pp^x(B_t\in A_{y,r})}{\vol(A_{y,r})} =\rho(t,x,y)\,,
\end{align*}
and by Lemma \ref{immcor},
\begin{align*}
\lim_{r\ra0} \frac{\pp^x(B_t\in A_{y,r},\, t <\tau_\om)}{\vol(A_{y,r})} =p(t,x,y)\,.
\end{align*}
Combining all of the above observations, we get
\begin{align*}
\rho(t,x,y) - p(t,x,y) &= \ee^x\biggl(\frac{1}{(2\pi (t-\tau_\om))^{d/2}} e^{-\|y-B_{\tau_\om}\|^2/2(t-\tau_\om)}\,;\, t\ge \tau_\om\biggr)\,.
\end{align*}
Now note that any derivative of the term inside the expectation (with respect to $y$) is uniformly bounded by a deterministic constant that does not depend on $y$ or $t$. Therefore derivatives with respect to $y$ can be carried inside the expectation. Consequently,
\begin{align*}
&\partial_y^{|\alpha|}\rho(t,x,y) - \partial_{y}^{|\alpha|} p(t,x,y) \\
&= \ee^x\biggl(\frac{1}{(2\pi (t-\tau_\om))^{d/2}} \partial_y^{|\alpha|}(e^{-\|y-B_{\tau_\om}\|^2/2(t-\tau_\om)})\,;\, t\ge \tau_\om\biggr)\,.
\end{align*}
If $t\le \delta_y^2/(d+2|\alpha|)$, an easy verification shows that
\begin{align*}
&\biggl|\frac{1}{(2\pi (t-\tau_\om))^{d/2}} \partial_y^{|\alpha|}(e^{-\|y-B_{\tau_\om}\|^2/2(t-\tau_\om)})\biggr|\\
&\le \frac{C}{(t-\tau_\om)^{d/2+|\alpha|}} e^{-\|y-B_{\tau_\om}\|^2/2(t-\tau_\om)}\,,
\end{align*}
where $C$ depends only on $d$, $|\alpha|$ and the diameter of the domain $\om$.
Another easy calculation shows that the map $u\mapsto (2\pi u)^{-d/2-|\alpha|} e^{-\beta^2/2u}$ is increasing in $u$ when $0< u\le \beta^2/(d+2|\alpha|)$. Therefore if $\tau_\om<t\le \delta_y^2/(d+2|\alpha|)$, then
\[
\frac{1}{(t-\tau_\om)^{d/2+|\alpha|}} e^{-\|y-B_{\tau_\om}\|^2/2(t-\tau_\om)}\le \frac{ e^{-\delta_y^2/2t}}{t^{d/2+|\alpha|}}\,.
\]
Noticing that $p(t,x,y)=p(t,y,x)$ (for example, by part (ii) of Theorem \ref{summarythm}) and $\rho(t,x,y)=\rho(t,y,x)$, this completes the proof of the lemma.
\end{proof}
\begin{proof}[Proof of Theorem \ref{ptwisethm} from Lemmas \ref{traceLem} and \ref{pdifflmm}]
By Lemma \ref{traceLem}, we have that
\begin{equation}\label{eqn:tr}\frac{\tr(1_{\dbar}A1_{\dbar}e^{t\Delta_D})}{\tr (e^{t\Delta_D})}\to\int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi),\quad\quad t\to 0^+.\end{equation}
Since $\{u_j\}_{j\ge 1}$ is an orthonormal basis of $L^2(\dbar)$,
\begin{equation}\label{eqn:trexp}\tr(1_{\dbar}A1_{\dbar}e^{t\Delta_D})=\sum_{j}e^{-t\lambda_j^2}\la A1_{\dbar}u_j,1_{\dbar}u_j\ral.\end{equation}
By Lemma \ref{pdifflmm} and the assumption that $\vol(\dbar)=1$,
\begin{equation}\label{eqn:trnoA}\tr(e^{t\Delta_D})=\sum e^{-t\lambda_j^2}=\int_\om p(2t,x,x)\sim (4\pi t)^{-d/2}\to \infty ,\quad t\to 0^+.\end{equation}
Putting \eqref{eqn:tr}, \eqref{eqn:trexp}, and \eqref{eqn:trnoA} together we have that
$$\sum_{j}e^{-t\lambda_j^2}\la A1_{\dbar}u_j,1_{\dbar}u_j\ral\sim (4\pi t)^{-d/2}\int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi).$$
Now, assuming that $\sigma(A)\geq 0$, and adding a regularizing perturbation $C\in \Psi^{-1}$ if necessary, so that $(A+C)\geq 0$, we may apply Lemma \ref{taubLem} with
$$F_{A}(\tau)=\sum_j 1_{\lambda_j\leq \tau}\la (A+C)1_{\dbar}u_j,1_{\dbar}u_j\ral.$$
More precisely, we apply it with
$$\tilde{F}_{A}(\tau)=\sum_j 1_{\lambda_j\leq \sqrt{\tau}}\la (A+C)1_{\dbar}u_j,1_{\dbar}u_j\ral$$
and rescale so that
$$F_A(\tau)\sim \frac{\tau^{d}}{(4\pi )^{d/2}\Gamma(d/2+1)}\int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi).$$
Now, $C:L^2(\re^d)\to L^2(\re^d)$ is compact. Therefore, $\lim_{j\to \infty}\la C1_{\dbar}u_j,1_{\dbar}u_j\ral =0$, and hence
$$\sum1_{\lambda_j\leq \tau}\la C1_{\dbar}u_j,1_{\dbar}u_j\ral=\o{}(\tau^{d}).$$
Therefore
\begin{align}
&\sum_j1_{\lambda_j\leq \tau}\la A1_{\dbar}u_j,1_{\dbar}u_j\ral \nonumber \\
&\sim \frac{ \tau^{d}}{(4\pi)^{d/2}\Gamma(d/2+1)}\int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi)+o(\tau^d).\label{eqn1}
\end{align}
Using \eqref{eqn:trnoA} together with Lemma \ref{taubLem} also gives
\begin{equation}\label{eqn2}\#\{\lambda_j:\lambda_j\leq \tau\}\sim \frac{\tau^{d}}{(4\pi)^{d/2}\Gamma(d/2+1)}.\end{equation}
Subtraction of two formulae like \eqref{eqn1} and \eqref{eqn2} yields the desired asymptotics.
\end{proof}
The proof of Lemma \ref{traceLem} requires one further lemma.
\begin{lmm}\label{traceLemLem}
Let $A\in \Psi(\re^d)$ with symbol compactly supported in $\Omega$ and $\psi\in C_c^\infty(\Omega)$ with $\psi\equiv 1$ on $\supp \sigma(A)$. Then we have
$$(4\pi t)^{d/2}\tr (A\psi e^{t\Delta_{\re^d}}\psi)\to \int_{S^*\re^d}\sigma(A)(x,\xi)1_{\dbar}dxdS(\xi) \ \ \text{ as } t\to 0^+$$
and there exists $\e>0$, $t_0>0$, so that for $0<t<t_0$,
$$|\tr A\psi(e^{t\Delta_D}-e^{t\Delta_{\re^d}})\psi|\leq \e^{-1}e^{-\e/t}\,.$$
\end{lmm}
\begin{proof}
The kernel $K(t,x,y)$ of $A\psi e^{t\Delta_{\re^d}}\psi$ is given by
\begin{align*}
K(t,x,y)&=(2\pi )^{-d}\int a(x,\xi)\int e^{i\la x-w,\xi\ral -|w-y|^2/4t}(4\pi t)^{-d/2}\psi(w)\psi(y)dwd\xi\\
&=(2\pi )^{-2d}\int a(x,\xi)\int e^{i\la x,\xi\ral}e^{-|\xi-\eta|^2t}e^{-i\la y,\xi-\eta\ral}\hat{\psi}(\eta)\psi(y)d\eta d\xi.
\end{align*}
where
\begin{align*}
&\left|\partial_x^\alpha\partial_\xi^\beta\left(a-\sum_{j=0}^{N-1}a_j(x,\xi)\right)\right|\leq C_{\alpha\beta}\langle \xi\rangle^{-N},\\
&\qquad \qquad a_j\in S^j_{hom}(T^*\re^d),\quad a_0(x,\xi)=\sigma(A)(x,\xi)
\end{align*}
So, changing variables so that $\xi \sqrt{t}=\zeta$,
\begin{align*}
&(4\pi t)^{d/2}\tr \psi e^{it\Delta_{\re^d}}\psi\\
& =\pi^{-d/2}t^{d/2}(2\pi )^{-d}\int a(x,\xi)\int e^{-|\xi-\eta|^2t}e^{i\la x,\eta\ral}\hat{\psi}(\eta)\psi(x)d\eta d\xi dx\\
&=\pi^{-d/2}(2\pi )^{-d}\int a(x,\zeta t^{-1/2})\int e^{-|\zeta-\eta\sqrt{t}|^2}e^{i\la x,\eta\ral}\hat{\psi}(\eta)\psi(x)d\eta d\zeta dx
\end{align*}
Now, since $\psi \in \mc{S}$, we can use the dominated convergence theorem and let $t\to 0^+$ to obtain
\begin{align*}
&\lim_{t\to 0^+}(4\pi t)^{d/2}\tr A\psi e^{t\Delta_{\re^d}}\psi\\
&=\pi^{-d/2}(2\pi )^{-d}\int \sigma(A)(x,\zeta )\int e^{-|\zeta|^2}e^{i\la x,\eta\ral}\hat{\psi}(\eta)\psi(x)d\eta d\zeta dx\\
&=\pi^{-d/2}\int \sigma(A)(x,\zeta )\int e^{-|\zeta|^2}\psi(x)\psi(x)d\zeta dx\\
&=\pi^{-d/2}\int \sigma(A)(x,\zeta )\int e^{-|\zeta|^2}d\zeta dx
\end{align*}
where we have used that $\psi\equiv 1$ on $\supp \sigma(A)$. Now, since $\sigma(A)$ is homogeneous of degree 0 in $\zeta$, this is equal to
\begin{align*}
&\pi^{-d/2}\vol(S^{d-1})\int_{S^*\re^d} \sigma(A)(x,\zeta )1_{\dbar}dxdS(\zeta) \int_0^\infty e^{-r^2}r^{d-1} dr\\
&=\int_{S^*\re^d} \sigma(A)(x,\zeta )1_{\dbar}dxdS(\zeta).
\end{align*}
For the second claim, we use Lemma \ref{pdifflmm}. Let $g(t,x,y)$ denote the kernel of $\psi (e^{t\Delta_D}-e^{t\Delta})\psi$ and $g_A(t,x,y)$ the kernel of $A\psi (e^{t\Delta_D}-e^{t\Delta})\psi$. Then
$$\tr (A\psi (e^{t\Delta_D}-e^{t\Delta_{\re^d}})\psi)=\int_{\dbar} g_A(t,x,x)dx.$$
Using the Sobolev embedding, for $m>\frac{d}{2}$,
$$|g_A(t,x,x)|\leq \|g_A\|_{H^m}\leq \|A\|_{H^m\to H^m}\|g(t,\cdot,y)\|_{H^m}.$$
Letting $\delta_x=d(\supp \psi, \partial \om )$, Lemma \ref{pdifflmm} implies for $t<\frac{\delta_x^2}{2(d+2|\alpha|)}$,
$$|\partial_x^{\alpha} g(t,x,y)|\leq C\frac{e^{-\delta_x^2/4t}}{t^{d/2+|\alpha|}}$$
and hence since $\|A\|_{H^m\to H^m}<\infty$,
$$|g_A(t,x,x)|\leq C\frac{e^{-\delta_x^2/4t}}{t^{d/2+m}}$$
for each $t<\frac{\delta_x^2}{2(d+2m)}$.
\end{proof}
We are now ready to prove Lemma \ref{traceLem}.
\begin{proof}[Proof of Lemma \ref{traceLem}]
Since $1_{\dbar}A1_{\dbar}:L^2(\om )\to L^2(\om )$, and $e^{t\Delta_D}$ is trace class, $1_{\dbar}A1_{\dbar}e^{t\Delta_D}$ is trace class. Let $\psi\in C_c^\infty(\om )$ with $\psi \equiv 1$ on $\supp \sigma(A)$. Then,
$$\psi A= A-(1-\psi)A,\quad\quad A\psi=\psi A+[A,\psi].$$
But, $(1-\psi)A,$ $[A,\psi]\in \Psi^{-1}$ and hence $1_{\dbar}(1-\psi)A$, $1_{\dbar}[A,\psi]$ are compact on $L^2(\re^d)$ and have
$$\|1_{\dbar}(1-\psi)A1_{\dbar}\phi_k\|+\|1_{\dbar}[A,\psi]1_{\dbar}\phi_k\|\to 0,\quad\quad k\to \infty.$$
In particular,
$$\frac{\tr(1_{\dbar}A1_{\dbar}e^{t\Delta_D})}{\tr{e^{t\Delta_D}}}\sim\frac{\tr (\psi A\psi e^{t\Delta_D})}{\tr e^{t\Delta_D}}=\frac{\tr (A\psi e^{t\Delta_D}\psi)}{\tr e^{t\Delta_D}},\quad\quad t\to 0^+.$$
By Lemma \ref{traceLemLem}, the proof of Lemma \ref{traceLem} is now complete.
\end{proof}
\bibliographystyle{plainnat} | 0.002676 |
Psystar, the controversial seller of Mac clones that has drawn a lot of press scrutiny over the past week, seems to be in the process of getting its new office up and running.
Psystar is apparently operating out of an office with an attached warehouse space in the city of Doral in Miami-Dade County.
The front door of the locale, at 10475 NW 28th. St. in Doral, is clearly labeled with the Psystar logo and the building features warehouse-type loading docks. It is in an area of similar warehouse-type buildings that is close to major highways, like the Florida Turnpike, and to Miami International Airport.
During a quick visit by IDG News Service on Monday at around 9 a.m., the front office door was unlocked and a quick peek inside showed cans of paint on the floor, indicating that Psystar is likely in the process of conditioning its new digs.
However, a young man quickly emerged and said the company isn't yet open to the walk-in public. After a few minutes, another young man came out of the locale and said that reporters have to make an appointment to be seen. He said that Rudy Pedraza [cq], Psystar's president, wasn't yet in and politely indicated he would relay notice of IDG News Service's visit to him.
IDG News Service had requested an interview with Pedraza [cq] on Friday both via phone and e-mail, but those queries haven't been answered.
Psystar has attracted a lot of attention in the past week for several reasons, starting with its claim that it makes and sells a Mac clone, something Apple hasn't allowed for years.
In addition to the possibility that it might be in violation of Apple's licensing terms for its Mac OS, Psystar last week had its Web store go down, parted ways in unfriendly terms with its credit-card payment processor and endured accusations from bloggers that its operation is a scam.
Via its Web site and through comments some of its staffers have made to various news outlets, Psystar has defended itself, saying it runs a legitimate business and that it has been overwhelmed by buyers' response to its Mac clone.
The company has said it had to scramble last week to move to a larger location and find a new payments processor. As of Monday, Psystar claims on its Web site that it is again accepting new orders for its Mac clone, called Open Computer, and that it has started to ship computers to buyers.
"It is our pleasure to inform you that our store is up and running thanks to our new high volume payment processor. To all, we challenge you, let's see if we can max this one out," reads a note on the company's home page.
The note also says the company has started shipping computers to its customers. "Orders are being shipped in the order that they were received, don't worry, you'll get yours soon. Upon shipment an e-mail notification including tracking information will be sent to you automatically."
While it's clear that Psystar has put itself in danger of getting sued by Apple, it certainly doesn't look like the company is running a scam operation, as has been suggested by a number of bloggers. The company has a working Web site, staffers who answer its phones and an easy-to-find physical location anyone can drive up to, since there is no gate limiting access to the facility. At this point, it looks more like a case of a startup company that did indeed get overwhelmed by response to its products and is struggling to meet the demand.
A base configuration of the Open Computer, without an operating system, costs US$399, but for an extra $155 Psystar will install Mac OS X 10.5 Leopard with it.
Apple declined to comment.
On Friday, PowerPay, based in Portland, Maine, told Computerworld that it had terminated Psystar's payment processing account earlier in the week because the Miami company had misrepresented what it would sell and failed to properly verify credit cards.
"In its application to PowerPay, PsyStar stated it expected to process a specific amount of credit card transactions per month, and that the product or service being sold was 'consulting for information and communication solutions,' " Stephen Goodrich, CEO of Portland, Maine-based PowerPay, said via e-mail to Computerworld. PowerPay also felt uncomfortable that Psystar changed its address several times last week.
A Psystar employee who identified herself as Maria told Computerworld that the only reason PowerPay gave for canceling the account was that Psystar exceeded its transaction allowance. She denied that PsyStar had misrepresented what it was going to sell when it struck the agreement with Power | 0.036403 |
Stop gassing about gas and produce it!
The UK section of the North Sea used to be sufficient to supply all of the country’s gas requirements, but now some has to be imported from countries such as Norway. With the cold weather, the usage of gas has increased to record levels and there are now concerns for future supplies, especially if the cold weather returns.
However, the National Grid has said that there isn’t a problem, despite a glitch with a Norwegian gas supply. Gas supplies from various sources have been increased to deal with this record demand. There have been calls for Britain to build more gas storage facilities and the National Grid did issue ‘gas balancing alerts’, asking power firms and other large industries to cut back on their gas consumption. There are suggestions that even if supplies of gas aren’t a problem at the moment, we could see serious shortages in a few years.
Following growing demand for gas supplies, wholesale prices rose, but they did fall again when supplies were increased. Prices of household bills could be affected in the future, but for now, it’s too soon to tell. However, rising prices could spell further trouble for ours and other economies suffering from extreme weather on top of a financial crisis. Economic recovery could be put in jeopardy.
This fear of gas shortages and security of supply has led environmental and business groups to argue that Britain needs to diversify its energy supplies and become less dependent on foreign exports. This issue fits in with the latest developments in new investment in wind turbines.
Who knew that something as beautiful as snow could cause so much trouble and provide so much economic analysis!
National Grid warns of UK gas shortage Guardian, David Teather (5/1/10)
Is the United Kingdom facing a natural gas shortage The Oil Drum (9/1/10)
Wind farms: Generating power and jobs? BBC News (8/1/10)
Gas rationing in -22C Britain increases fears of energy crisis Mail Online, Martina Lees (8/1/10)
Gas usage hits new high in UK cold snap BBC News (8/1/10)
Energy fears over gas and kerosene shortages Scotsman (6/1/10)
Gas shortages highlights firms’ exposure to energy security risks Business Green, Tom Young (8/1/10)
Uh-oh: the return of $3 gas CNN Money, Paul R La Monica (7/1/10)
Natural gas prices seen rising with winter shortages Global Times, Chen Xiaomin (4/1/10)
Gas demand hits record on Thursday Reuters (8/1/10)
Gas demand in UK hits another highBBC News, Hugh Pym (7/1/10)
Questions
- Illustrate the effects in the gas market of increasing demand and the resulting shortages. Then show the effects of increasing the supplies of gas. How is equilibrium achieved when there is a shortage in the market?
- Why did energy prices increase and then fall?
- To what extent should the government have been able to forecast this higher demand? Should better contingency plans have been in place?
- The article from CNN Money looks at the effect of rising prices of oil and energy and how this is likely to affect consumer spending. Why could rising prices of these commodities adversely affect economic recovery?
- What is an ‘interruptible contract’ and how useful have they been in dealing with these gas shortages?
- Why has this gas shortage presented environmental groups with an opportunity to promote renewable energy supplies? Think about economic interdependence.
- What alternatives are there to our current gas sources? Are they realistic alternatives? | 0.994438 |
\section{I-MMSE formula} \label{app:immse}
We give for completeness a short calculation to derive the I-MMSE formula \eqref{y-immse} for our (structured) vector setting.
Detailed proofs can be found in \cite{GuoShamaiVerdu_IMMSE} and here we do not go through the technical justifications required to exchange integrals and differentiate under the integral sign.
Thanks to \eqref{eq:true_mutual_info} the MI is represented as follows
\begin{align}
i^{\rm cs} & = -\frac{\alpha B}{2} - \frac{1}{L} \mathbb{E}_{\bm\Phi}\Big[\int d\by {\cal Z}^{\rm cs}(\by)(2\pi\Delta)^{-M/2}\ln ({\cal Z}^{\rm cs}(\by)) \Big]
\nonumber \\ &
= -\frac{\alpha B}{2} - \frac{1}{L} \mathbb{E}_{\bm\Phi}\Big[
\int d\by \int \prod_{l=1}^L d\bs_l P_0(\bs_l)\, \frac{e^{-\frac{1}{2\Delta}\Vert \bm\Phi\bs - \by\Vert^2}}{(2\pi\Delta)^{M/2}}
\ln\Big\{ \int \prod_{l=1}^L d\bx_l P_0(\bx_l)\, e^{-\frac{1}{2\Delta}\Vert \bm\Phi\bx - \by\Vert^2} \Big\} \Big].
\end{align}
We exchange the $\by$ and $\bs$ integrals. Then
for fixed $\bs$ we perform the change of variables $\by \to \bz\sqrt\Delta + \bm\Phi\bs$. This yields
\begin{align}
i^{\rm cs} = - \frac{\alpha B}{2} - \frac{1}{L} \mathbb{E}_{\bm\Phi, \bS, \bZ}\Big[
\ln\Big\{ \int \prod_{l=1}^L d\bx_l P_0(\bx_l)\, e^{-\frac{1}{2}\Vert \frac{\bm\Phi\bx}{\sqrt\Delta} - \frac{\bm\Phi\bS}{\sqrt{\Delta}} - \bZ\Vert^2} \Big\} \Big],
\end{align}
where $\bZ\sim{\cal N}(0,\mathbf{I}_M)$. We now perform the derivative $d/d\Delta^{-1} = (\sqrt{\Delta}/2) d/d \Delta^{-1/2}$.
We use the statistical mechanical notation for the ``posterior average'', namely for any quantity $A(\bx)$,
\begin{align}
\langle A(\bX) \rangle \defeq \frac{\int \prod_{l=1}^Ld\bx_l P_0(\bx_l) A(\bx)
e^{-\frac{1}{2}\Vert \frac{\bm\Phi\bx}{\sqrt\Delta} - \frac{\bm\Phi\bs}{\sqrt{\Delta}} - \bz\Vert^2}}
{\int \prod_{l=1}^Ld\bx_l P_0(\bx_l)
e^{-\frac{1}{2}\Vert \frac{\bm\Phi\bx}{\sqrt\Delta} - \frac{\bm\Phi\bs}{\sqrt\Delta} - \bz\Vert^2}}.
\end{align}
Note that if $A$ does not depend on $\bx$ then $\langle A \rangle = A$.
We get
\begin{align}
\frac{d i^{\rm cs}}{d\Delta^{-1}} & = \frac{\sqrt\Delta}{2L}
\mathbb{E}_{\bm\Phi, \bS, \bZ}\Big[ \Big\langle \Big(\frac{\bm\Phi\bX}{\sqrt\Delta} - \frac{\bm\Phi\bS}{\sqrt \Delta} - \bZ\Big)\cdot
(\bm\Phi\bX - \bm\Phi\bS)\Big\rangle \Big]
\nonumber \\ &
= \frac{1}{2L}
\mathbb{E}_{\bm\Phi, \bS, \bZ}[ \langle \Vert \bm\Phi\bX - \bm\Phi\bS\Vert^2 \rangle ]
-
\frac{\sqrt\Delta}{2L}
\mathbb{E}_{\bm\Phi, \bS, \bZ}[ \bZ \cdot \langle
\bm\Phi\bX - \bm\Phi\bS \rangle ].
\end{align}
Integrating the second term by parts w.r.t the Gaussian noise $\bz$ we get that this term is equal to
\begin{align}
\frac{\sqrt\Delta}{2L}
\mathbb{E}_{\bm\Phi, \bS, \bZ}[ \bZ \cdot \langle
\bm\Phi\bX - \bm\Phi\bS \rangle ] &= \frac{\sqrt\Delta}{2L}\mathbb{E}_{\bm\Phi, \bS, \bZ}\Big[
\Big\langle (\bm\Phi\bX - \bm\Phi\bS)\cdot \Big(\frac{\bm\Phi\bX}{\sqrt\Delta} - \frac{\bm\Phi\bS}{\sqrt\Delta} - \bZ\Big)\Big\rangle
-
\big\langle \bm\Phi\bX - \bm\Phi\bS\big\rangle \cdot \Big\langle \frac{\bm\Phi\bX}{\sqrt\Delta} - \frac{\bm\Phi\bS}{\sqrt\Delta} - \bZ\Big\rangle
\Big]
\nonumber \\
&=
\frac{1}{2L}\mathbb{E}_{\bm\Phi, \bS, \bZ}[
\langle \Vert \bm\Phi\bX - \bm\Phi\bS\Vert^2\rangle
-
\Vert\langle \bm\Phi\bX - \bm\Phi\bS\rangle\Vert^2] .
\end{align}
We therefore obtain
\begin{align}
\frac{d i^{\rm cs}}{d\Delta^{-1}} & = \frac{1}{2L} \mathbb{E}_{\bm\Phi, \bS, \bZ}[\Vert \langle \bm\Phi\bX - \bm\Phi\bS\rangle\Vert^2]
=
\frac{\alpha B}{2} \frac{1}{M} \mathbb{E}_{\bm\Phi, \bS, \bZ}[\Vert \bm\Phi\langle \bX\rangle - \bm\Phi\bS\Vert^2] = \frac{\alpha B}{2} {\rm ymmse}.
\end{align}
In the last step we recognized that
$\langle \bX \rangle$ is nothing else than the MMSE estimator $\mathbb{E}[\bX\vert \bm\phi \bs +\bz\sqrt\Delta]$
entering in the definition of the measurement MMSE \eqref{defymmse}. | 0.018242 |
TITLE: conjectured general continued fraction for the quotient of gamma functions
QUESTION [18 upvotes]: Given complex numbers $a=x+iy$, $b=m+in$ and a gamma function $\Gamma(z)$ with $x\gt0$ and $m\gt0$, it is conjectured that the following general continued fraction which is symmetric on $a$ and $b$ is true
$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}\tag{1}$$
And can be further generalised to
$$\frac{1}{\sqrt{c}}\tan\left(\frac{b-a}{b+a}\tan^{-1}\left(\frac{\sqrt{c}}{d}\right)\right)=\cfrac{(b-a)}{d(a+b)+\cfrac{c(2a)(2b)} {3d(a+b)+\cfrac{c(3a+b)(a+3b)}{5d(a+b)+\cfrac{c(4a+2b)(2a+4b)}{7d(a+b)+\ddots}}}}\tag{2}$$
Corollaries:
(i) Specializing to $a=1/2$ and $b=2z/n+1/2$,we obtain
the continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$ found in this post after applying the functional equation of the gamma function
(ii) and specializing further to $a=-1/2$ and $b=2z/n+3/2$,we obtain immediately the continued fraction for $\cot\left(\frac{z\pi}{4z+2n}\right)$ found here after applying the functional equation of the gamma function.This continued fraction was proved by @GEdgar.
(iii) letting $2a=m-n$ and $2b=m+n$,we find $$\displaystyle\tan\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2-n^2} {3m+\cfrac{4m^2-n^2}{5m+\cfrac{9m^2-n^2}{7m+\cfrac{16m^2-n^2}{9m+\ddots}}}}}$$
From which we obtain its hyperbolic companion $$\displaystyle\tanh\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2+n^2} {3m+\cfrac{4m^2+n^2}{5m+\cfrac{9m^2+n^2}{7m+\cfrac{16m^2+n^2}{9m+\ddots}}}}}$$
(iv)and if $2a=-n$ and $2b=2m+n$,then it follows that
$$\displaystyle\tan\left(\frac{\pi(m+n)}{4m}\right)=\frac{1+\tan\Big(\frac{\pi n}{4m}\Big)}{1-\tan\Big(\frac{\pi n}{4m}\Big)}=\cfrac{m+n}{m+\cfrac{(-n)(2m+n)} {3m+\cfrac{(m-n)(3m+n)}{5m+\cfrac{(2m-n)(4m+n)}{7m+\cfrac{(3m-n)(5m+n)}{9m+\ddots}}}}}$$
Q: Is the conjectured general continued fraction true (for all complex numbers $a$,$b$ with $x\gt0$ and $m\gt0$)?
REPLY [1 votes]: If we consider the following symmetric q-continued fraction,where $q=ab$,$|q|\lt1$ see here and here
$$\cfrac{1}{1-ab+\cfrac{(ab)(1-b^2)(1-a^2)}{1-(ab)^3+\cfrac{(ab)^2(1-ab^3)(1-a^3b)}{1-(ab)^5+\cfrac{(ab)^3(1-a^2b^4)(1-a^4b^2)}{1-(ab)^7+\cfrac{(ab)^4(1-a^3b^5)(1-a^5b^3)}{1-(ab)^9+\ddots }}}}}=\frac{(a^5b^3;(ab)^4)_\infty\,(a^3b^5;(ab)^4)_\infty}
{(ab^3;(ab)^4)_\infty\,(a^3b;(ab)^4)_\infty}$$
and let $a\rightarrow q^a$ ,$b\rightarrow q^b$ then we have
$$\cfrac{1}{1-q^{(a+b)}+\cfrac{q^{(a+b)}(1-q^{2b})(1-q^{2a})}{1-q^{3(a+b)}+\cfrac{q^{2(a+b)}(1-q^{(a+3b)})(1-q^{(b+3a)})}{1-q^{5(a+b)}+\cfrac{q^{3(a+b)}(1-q^{(2a+4b)})(1-q^{(2b+4a)})}{1-q^{7(a+b)}+\cfrac{q^{4(a+b)}(1-q^{(3a+5b)})(1-q^{(5a+3b)})}{1-q^{9(a+b)}+\ddots }}}}}=\frac{(q^{(5a+3b)};q^{4(a+b)})_\infty\,(q^{3a+5b};q^{4(a+b)})_\infty}
{(q^{(a+3b)};q^{4(a+b)})_\infty\,(q^{(3a+b)};q^{4(a+b)})_\infty}$$
now after multiplying both sides by $1-q$ ,and performing equivalence transformation
$$\cfrac{1}{\frac{1-q^{(a+b)}}{1-q}+\cfrac{q^{(a+b)}\Big(\frac{1-q^{2b}}{1-q}\Big)\Big(\frac{1-q^{2a}}{1-q}\Big)}{\frac{1-q^{3(a+b)}}{1-q}+\cfrac{q^{2(a+b)}\Big(\frac{1-q^{(a+3b)}}{1-q}\Big)\Big(\frac{1-q^{(b+3a)}}{1-q}\Big)}{\frac{1-q^{5(a+b)}}{1-q}+\cfrac{q^{3(a+b)}\Big(\frac{1-q^{(2a+4b)}}{1-q}\Big)\Big(\frac{1-q^{(2b+4a)}}{1-q}\Big)}{\frac{1-q^{7(a+b)}}{1-q}+\cfrac{q^{4(a+b)}\Big(\frac{1-q^{(3a+5b)}}{1-q}\Big)\Big(\frac{1-q^{(5a+3b)}}{1-q}\Big)}{\frac{1-q^{9(a+b)}}{1-q}+\ddots }}}}}=(1-q)\frac{(q^{(5a+3b)};q^{4(a+b)})_\infty\,(q^{3a+5b};q^{4(a+b)})_\infty}
{(q^{(a+3b)};q^{4(a+b)})_\infty\,(q^{(3a+b)};q^{4(a+b)})_\infty}$$
then apply the equality $\lim_{q\rightarrow 1}\frac{1-q^x}{1-q}=x$,see q-analog and the q-gamma function $\Gamma_q(x)=(1-q)^{1-x}\frac{(q;q)_{\infty}}{(q^x;q)_{\infty}}$
which concludes the proof of continued fraction (1) | 0.070994 |
I have just walked the wonderful South Downs Way - a 100-mile journey from Winchester to Eastbourne - to raise money for Brighton's World Sacred Music Festival, which is the only event of its kind in the UK. The Festival brings together artists and audiences from many different faiths and cultures with the aim of promoting understanding and tolerance through beautiful and inspiring music.
In these challenging times the Festival, which is a registered charity, relies on donations and volunteers to keep going.
It's not too late to sponsor me.
Do please give what you can and help me reach my target of £2000. Every penny received will go straight to the Festival, which takes place 10-18 October 2009.
Donating through JustGiving is simple, fast and totally secure. Once you donate, they’ll send your money directly to the Festival and make sure Gift Aid is reclaimed on every eligible donation by a UK taxpayer.
Thank you so much for your support. | 0.363552 |
USA Today (March 29)
“Investors are looking for the number of infections to slow before markets can find a bottom, analysts say. Last week, the U.S. topped China as the global leader in virus cases.”
Tags: Virus cases
Chicago Tribune (March 26)
“This isn’t a drill. It’s real life. The illness is hitting all age groups and in Illinois, cropping up in at least 35 counties so far. Grateful to health care workers? Want to offer a sincere thanks? Then stay home.”
The Guardian (March 26)
“How did Spain get its coronavirus response so wrong? Spain saw what happened in Iran and Italy – and yet it just overtook China’s death toll in one of the darkest moments in recent Spanish history.”
The Times of India (March 26)
“On day one of lockdown, supply of fruits and vegetables took a hit, despite the government having marked it out as an ‘essential service.’ Wholesale suppliers…say there are multiple logistics issues.” The largest “is the closure of state entry points and tolls across India. Some 1.2 crore trucks are said to be stranded across India” with some drivers “getting no food or water as dhabas remain closed for miles along highways, even as essentials rot inside the trucks.”
Tags: Closure, Essential service, Food, Fruits, Government, Highways, Lockdown, Logistics, Stranded, Supply, Tolls, Trucks, Vegetables, Water, Wholesale
CNN (March 25)
“Scientists have been more concerned about West Antarctica, where the ice has been melting faster in recent years.” They are now realizing East Antarctica brings perils as well, especially the Denman Glacier. This “giant glacier” has already “retreated almost three miles” and if it “fully thaws, sea levels would rise almost 5 feet.”
Tags: Denman Glacier, East Antarctica, Ice, Melting, Perils, Retreated, Scientists, Thaws, West Antarctica
Institutional Investor (March 25)
Due to the “historic buying opportunity,” a few “hedge funds legends” are “quietly contacting investors” These “superstar managers” are “making an exception” and reopening their funds to new investors citing “the massive drop in asset prices catalyzed by the novel coronavirus pandemic.”
Tags: Asset prices, Buying opportunity, Coronavirus, Hedge funds, Historic, Investors, Legends, Reopening
Bloomberg (March 25)
“It’s the worst epidemic of our times, a health emergency that has now left more than 420,000 infected, 18,800 dead and paralyzed the global economy. The scale has been clear for weeks.”Yet the same “baffling” decisions are .”
Tags: Baffling, Dead, Decisions, Emergency, Epidemic, Global economy, Health, Infected, Italy, Paralyzed, Repeated, U.S.
Reuters (March 23)
“China is consciously uncoupling from Western peers on rates. Its central bank has held lending benchmarks steady as global peers slash…. The People’s Bank of China’s relative immobility has surprised many economists…. The spread between 10-year Chinese government bonds and U.S. Treasuries is nearly two percentage points, its widest since 2015.”
Tags: Benchmarks, Central bank, China, Economists, PBOC, Rates, Spread, Steady, U.S., Uncoupling, Western
The Guardian (March 23)
“Canada said it will not send athletes to Tokyo Olympics, New Zealand said it would consider boycotting Tokyo 2020 and Australia told its olympic athletes to prepare for the games to be held next year, in 2021 – all in the wake of Japan’s prime minister, Shinzo Abe, saying postponement could be an option.”
Tags: 2021, Abe, Athletes, Australia, Boycott, Canada, Japan, New Zealand, Olympics, Tokyo 2020 | 0.9962 |
\begin{document}
\maketitle
\begin{itemize}
\item[] \textbf{Abstract.}
The parameterization process used in the symbolic computation systems
Kenzo and EAT is studied here as a general construction in a categorical framework.
This parameterization process starts from a given specification
and builds a parameterized specification
by transforming some operations into parameterized operations,
which depend on one additional variable called the parameter.
Given a model of the parameterized specification,
each interpretation of the parameter, called an argument,
provides a model of the given specification.
Moreover, under some relevant terminality assumption,
this correspondence between the arguments and the models of the
given specification is a bijection.
It is proved in this paper that the parameterization process
is provided by a free functor
and the subsequent parameter passing process by a natural transformation.
Various categorical notions are used, mainly
adjoint functors, pushouts and lax colimits.
\end{itemize}
\section{Introduction}
Kenzo \cite{Kenzo} and its predecessor EAT \cite{EAT} are software
systems developed by F. Sergeraert. They are devoted to
Symbolic Computation in Algebraic Topology. In particular, they
carry out calculations of homology groups of complex topological
spaces, namely iterated loop spaces. By means of EAT and Kenzo,
some homology groups that had never been obtained with any other
method, neither theoretical nor automatic, have been computed. In
view of the obtained results, some years ago, the first author
of this paper began the formal study of the programs, in order to
reach a good understanding on the internal calculation processes
of these software systems.
In particular, our study of the data types used in EAT and
Kenzo \cite{LPR03,DLR07,DRS06}
shows that there are two different layers of data structures in the systems.
In the first layer, one finds the usual abstract data types,
like the type of integers.
In the second layer, one deals with algebraic structures,
like the structure of groups,
which are implemented thanks to the abstract data types belonging to the first layer.
In addition, we realized that in a system such as EAT,
we do not simply implement one group,
but more generally \emph{parameterized} families of groups.
In \cite{LPR03} an operation is defined,
which is called the \emph{imp} construction
because of its role in the implementation process in the system EAT.
Starting from a specification $\Ss$
in which some operations are labelled as ``pure'' \cite{DRS06},
the \emph{imp} construction builds a new specification $\Ss_A$
with a distinguished sort $A$
which is added to the domain of each non-pure operation.
It follows that each implementation of $\Ss_A$
defines a family of implementations of $\Ss$
depending on the choice of a value in the interpretation of $A$.
Besides, working with the \emph{imp} construction in \cite{LPR03} we were able
to prove that the implementations of EAT algebraic structures are as general as
possible, in the sense that they are ingredients of terminal objects in
certain categories of models;
this result is called the \emph{exact parameterization property}.
Later on, led by this characterization of EAT algebraic structures,
in \cite{LPR03} we reinterpreted our results in terms of object-oriented technologies
like hidden algebras \cite{GM00} or coalgebras \cite{Ru00}.
This paper deals with generalization by parameterization in the sense of Kenzo and EAT,
so that our \emph{parameters} are symbolic constants of a given type,
that will be replaced by \emph{arguments} which are elements in a given set.
The notion of parameterization in programming and specification languages
bears several meanings, where the parameter may be a type or a specification.
For instance, in object-oriented programming,
parametric polymorphism is called generic programming,
in C++ it is characterized by the use of template parameters
to represent abstract data types.
On the other hand, in algebraic specifications,
a parameterized specification is defined as a morphism of specifications
where the parameter is the source
and the parameter passing is defined as a pushout \cite{ADJ80}.
The framework for this paper is provided by \emph{equational logic},
considered from a categorical point of view.
An equational theory, or simply a theory, is a category with chosen finite products.
A model $M$ of a theory $\Tt$ is a functor $M\colon \Tt\to\Set$ which maps
the chosen products to cartesian products.
A theory $\Tt$ can be presented by a specification $\Ss$,
this means that $\Ss$ generates $\Tt$.
In this paper, we are not interested in specifications for themselves,
but as presentations of theories.
So, specifications are used mainly in the examples,
and we feel free to modify a specification whenever needed
as long as the presented theory is not changed.
The \emph{parameterization process} studied in this paper
is essentially the ``\emph{imp} construction'' of \cite{LPR03}.
Starting from a theory $\Tt$ it provides a \emph{parameterized theory}~$\Tt_A$
by adding a \emph{type of parameters} $A$
and by transforming each term $f\colon X\to Y$ in $\Tt$ into
a parameterized term $f'\colon A\times X\to Y$ in $\Tt_A$.
Then clearly $\Tt_A$ generalizes $\Tt$:
the models of $\Tt$ can be identified to the models of $\Tt_A$
which interpret the type of parameters $A$ as a singleton.
There is another way to relate $\Tt$ and $\Tt_A$,
called the \emph{parameter passing process}, which runs as follows.
By adding to $\Tt_A$ a constant $a$ (called the \emph{parameter}) of type $A$ we get
a \emph{theory with parameter}~$\Tt_a$,
such that for each parameterized term $f'\colon A\times X\to Y$ in $\Tt_A$
there is a term $f'(a,-)\colon X\to Y$ in $\Tt_a$.
Then the \emph{parameter passing morphism} $j\colon \Tt \to \Tt_a$
maps each term $f\colon X\to Y$ in $\Tt$ to $f'(a,-)\colon X\to Y$ in $\Tt_a$.
Given a model $M_A$ of $\Tt_A$
an \emph{argument} $\alpha$ is an element of the set $M_A(A)$,
it provides a model $M_{A,\alpha}$ of $\Tt_a$
which extends $M_A$ and satisfies $M_{A,\alpha}(a)=\alpha$.
Thanks to the parameter passing morphism,
the model $M_{A,\alpha}$ of $\Tt_a$ gives rise to a model $M$ of $\Tt$
such that $M(f)=M_A(f')(\alpha,-)$ for each term $f$ in $\Tt$.
Moreover, under some relevant terminality assumption on $M_A$,
this correspondence between the arguments $\alpha\in M_A(A)$
and the models of $\Tt$ is a bijection:
this is the \emph{exact parameterization property}.
The parameterization process and its associated parameter passing process
have been described for each given theory $\Tt$,
but in fact they have the property of preserving the theory structure,
which can be stated precisely in a categorical framework:
this is the aim of this paper.
The parameterization process is defined as a \emph{functor}:
the construction of the parameterized theory~$\Tt_A$ from the given theory~$\Tt$ is
a functor left adjoint to the construction of a coKleisli category,
and more precisely it is a free functor in the sense of section~\ref{sec:free}.
The parameter passing process is defined as a \emph{natural transformation},
along the following lines.
First, the construction of the theory with parameter~$\Tt_a$
from the parameterized theory~$\Tt_A$ is simply a pushout construction,
such that the construction of~$\Tt_a$ from~$\Tt$ is a functor.
Then, each parameter passing morphism $j:\Tt\to\Tt_A$ is defined
from a lax colimit of theories,
in such a way that the parameter passing morphisms are (essentially) the components
of a natural transformation from the identity functor to this functor.
A first version of this approach can be found in \cite{DDLR05},
it relies on \emph{diagrammatic logic} \cite{Du03,Du07}.
In this paper, the explicit use of diagrammatic logic is postponed to the appendix.
With respect to the previous papers like \cite{LPR03},
we provide a new interpretation of the parameterization process
and in addition an interpretation of the parameter passing process.
Moreover, we take into account the fact that there is a pure part in the given theory,
and we derive the exact parameterization property
from a more general result which does not rely on the existence of a terminal model.
In section~\ref{sec:defi} equational theories are defined
and several examples are presented.
The parameterization process and the parameter passing process
are defined categorically in section~\ref{sec:const}.
In section~\ref{sec:free} free functors are defined
as left adjoint functors associated to morphisms of limit sketches,
and it is proved that the parameterization functor is free.
The diagrammatic point of view on equational logic is presented in appendix~\ref{sec:dia}.
Most of the categorical notions used in this paper can be found in \cite{MacLane98}
or in \cite{BW99}.
We omit the size issues: for instance most colimits should be small.
A \emph{graph} always means a directed multigraph,
and in order to distinguish between various kinds of structures
with an underlying graph,
we speak about the \emph{objects} and \emph{morphisms} of a category,
the \emph{types} and \emph{terms} of a theory (or a specification)
and the \emph{points} and \emph{arrows} of a limit sketch.
\section{Examples and definitions}
\label{sec:defi}
\subsection{Equational theories and specifications}
\label{subsec:equa}
In this paper, equational logic is seen from a categorical point of view,
as for instance in \cite{Pitts}.
\begin{defi}
\label{defi:equa-thry}
The category $\catT_{\eq}$ of \emph{equational theories}
is made of the categories with chosen finite products
together with the functors which preserve the chosen finite products.
In addition, $\catT_{\eq}$ can be seen as a 2-category
with the natural transformations as 2-cells.
\end{defi}
Equational theories are called simply \emph{theories}.
For instance, the theory $\Set$ is made of the category of sets
with the cartesian products as chosen products.
\begin{rema}
\label{rema:equa-thry}
The correspondence between
equational theories in the universal algebra style (as in \cite{LEW96})
and equational theories in the categorical style (as defined here)
can be found in \cite{Pitts}.
Basically, the \emph{sorts} and products of sorts become objects, still called types,
the \emph{operations} and \emph{terms} become morphisms, still called terms
(the \emph{variables} correspond to projections, as in example~\ref{exam:sgp})
and the \emph{equations} become equalities:
for instance a commutative square $g_1\circ f_1=g_2\circ f_2$
means that there is a term $h$ such that $g_1\circ f_1=h$ and $g_2\circ f_2=h$.
However a more subtle point of view on equations is presented in appendix~\ref{sec:dia}.
\end{rema}
\begin{defi}
\label{defi:equa-mod}
A \emph{(strict) model} $M$ of a theory $\Tt$ is
a morphism of theories $M\colon \Tt\to\Set$
and a \emph{morphism $m\colon M\to M'$ of models} of $\Tt$ is a natural transformation.
This forms the category $\Mod(\Tt)$ of models of $\Tt$.
\end{defi}
For every morphism of equational theories $\tt\colon \Tt_1\to\Tt$,
we denote by $\tt^\md\colon \Mod(\Tt)\to\Mod(\Tt_1)$
the functor which maps each model $M$ of $\Tt$
to the model $\tt^\md(M)=M\circ \tt$ of $\Tt_1$
and each morphism $m\colon M\to M'$ to $m\circ \tt$.
In addition, for each model $M_1$ of $\Tt_1$,
the category of \emph{models of $\Tt$ over $M_1$} is denoted $\Mod(\Tt)|_{M_1}$,
it is the subcategory of $\Mod(\Tt)$
made of the models $M$ such that $\tt^\md(M)=M_1$
and the morphisms $m$ such that $\tt^\md(m)=\id_{M_1}$.
Whenever $\tt$ is surjective on types, the category $\Mod(\Tt)|_{M_1}$ is discrete.
A theory $\Tt$ can be described by some presentation:
a \emph{presentation} of an equational theory $\Tt$
is an equational specification $\Ss$ which generates $\Tt$;
this is denoted $\Tt\dashv\Ss$.
Two specifications are called \emph{equivalent} when they present the same theory.
An equational specification can be defined either
in the universal algebra style as a signature (made of sorts and operations)
together with equational axioms, or equivalently, in a more categorical style,
as a finite product sketch, see \cite{Lellahi89}, \cite{BW99},
and also section~\ref{subsec:sketch} and appendix~\ref{subsec:dia-equ}.
The correspondence between the universal algebra and the categorical
points of view runs as in remark~\ref{rema:equa-thry}.
\begin{defi}
\label{defi:equa-spec}
The category $\catS_{\eq}$ of \emph{equational specifications}
is the category of finite product sketches.
With (generalized) natural transformations as 2-cells, $\catS_{\eq}$ can be seen as a 2-category.
\end{defi}
Equational specifications are called simply \emph{specifications}.
The category $\catT_{\eq}$ can be identified to a subcategory of $\catS_{\eq}$
(more precisely, to a reflective subcategory of $\catS_{\eq}$).
When $\Ss$ is a presentation of $\Tt$, a model of $\Tt$
is determined by its restriction to $\Ss$, which is called a \emph{model} of $\Ss$,
and in fact $\Mod(\Tt)$
can be identified to the category $\Mod(\Ss)$ of models of $\Ss$.
\begin{figure}[!t]
$$ \begin{array}{|l|l|c|}
\hline
& \textrm{subscript} \; \ttE & \Ss_{\ttE} \\
\hline
\hline
\textrm{type (or sort)} & \Type & X \\
\hline
\textrm{term (or operation)} & \Term &
\xymatrix@C=3pc{X\ar[r]^{f} & Y} \\
\hline
\textrm{selection of identity} & \Selid &
\xymatrix@C=3pc{X\ar[r]^{\id_X} & X} \\
\hline
\textrm{composition} & \Comp &
\xymatrix@C=3pc{X\ar[r]^{f} \ar@/_3ex/[rr]_{g\circ f}^{=} & Y\ar[r]^{g} & Z} \\
\hline
\textrm{binary product} & \bProd &
\xymatrix@C=3pc@R=.7pc{ X & \\
& X\stimes Y \ar[lu]_{p_X} \ar[ld]^{p_Y} \\ Y & \\ } \\
\hline
\textrm{pairing (or binary tuple)} & \bTuple &
\xymatrix@C=3pc@R=1pc{ & X & \\
Z \ar[ru]^{f} \ar[rd]_{g} \ar[rr]|{\,\tuple{f,g}\,} &
\ar@{}[u]|{=}\ar@{}[d]|{=} &
X\stimes Y \ar[lu]_{p_X} \ar[ld]^{p_Y} \\ & Y & \\ } \\
\hline
\end{array}$$
\caption{\label{fig:elem} Elementary specifications}
\end{figure}
We will repeatedly use the fact that $\catT_{\eq}$ and $\catS_{\eq}$,
as well as other categories of theories and of specifications, have colimits,
and that left adjoint functors preserve colimits.
In addition every specification is the colimit of a diagram of elementary specifications.
The \emph{elementary specifications} are the specifications respectively made of:
a type,
a term,
an identity term,
a composed term,
a $n$-ary product
and a $n$-ary tuple, for all $n\geq0$,
as in figure~\ref{fig:elem} (where only $n=2$ is represented).
Let us consider a theory $\Tt$ presented by a specification $\Ss$,
then $\Ss$ is the colimit of a diagram $\Delta$ of elementary specifications,
and $\Tt$ is the colimit of the diagram of theories generated by $\Delta$.
\subsection{Examples}
\label{subsec:exam}
\begin{exam}
\label{exam:term}
Let us consider the theory $\Tt_{\oper,0}$ presented by two types $X,Y$,
and the three following theories extending $\Tt_{\oper,0}$
(the subscript $\oper$ stands for ``operation'',
since $\Tt_{\oper}$ is presented by the elementary specification
for terms or operations $\Ss_{\Term}$).
The unit type is denoted $\uno$ and the projections are not given any name.
$$ \begin{array}{c|c|c|c|}
\cline{2-2}
\Tt_{\oper,A} \dashv &
\xymatrix@R=1pc{A & A\stimes X\ar[l]\ar[d]\ar[rd]^{f'} & \\ & X & Y \\ } &
\multicolumn{2}{c}{} \\
\cline{2-2} \cline{4-4}
\multicolumn{2}{c}{} &
\qquad \Tt_{\oper,a} \dashv &
\xymatrix@R=1pc{A & A\stimes X\ar[l]\ar[d]\ar[rd]^{f'} & \\ \uno \ar[u]^{a} &X & Y \\ } \\
\cline{2-2} \cline{4-4}
\Tt_{\oper} \dashv &
\xymatrix{ & X\ar[r]^{f} &Y \\ } &
\multicolumn{2}{c}{} \\
\cline{2-2}
\end{array}$$
These theories are related by various morphisms (all of them preserving $\Tt_{\oper,0}$):
$\tt_{\oper,A}\colon \Tt_{\oper,A}\to\Tt_{\oper}$ maps $A$ to $\uno$
and $\tt_{\oper,a}\colon \Tt_{\oper,a}\to\Tt_{\oper}$ extends $\tt_{\oper,A}$
by mapping $a$ to $\id_{\uno}$,
while $j_{\oper,A}\colon \Tt_{\oper,A}\to\Tt_{\oper,a}$ is the inclusion.
In addition, here are two other presentations of the theory $\Tt_{\oper,a}$
(the projections are omitted and $ \uno\times X$ is identified to $X$):
$$
\begin{array}{|c|}
\cline{1-1}
\xymatrix@R=1pc{A & A\stimes X\ar[rd]^{f'} & \\
\uno \ar[u]^{a} & X \ar@{}[ru]|(.3){=} \ar[u]^{a\stimes\id_X} \ar[r]_{f''} & Y \\ } \\
\cline{1-1}
\end{array}
\qquad\qquad
\begin{array}{|c|}
\cline{1-1}
\xymatrix@R=1pc{A & A\stimes X\ar[r]^{f'} \ar@{}[rd]|{=} &Y \\
\uno \ar[u]^{a} & X \ar[u]^{a\stimes\id_X} \ar[r]_{f''} & Y \ar[u]_{\id_Y} \\ } \\
\cline{1-1}
\end{array} $$
It is clear from anyone of these new presentations of $\Tt_{\oper,a}$
that there is a morphism $j_{\oper}\colon \Tt_{\oper}\to\Tt_{\oper,a}$
which maps $f$ to $f''$.
In addition, $\tt_{\oper,a}\circ j_{\oper,A} = \tt_{\oper,A}$
and there is a natural transformation
$t_{\oper} \colon j_{\oper} \circ \tt_{\oper,A} \To j_{\oper,A}$
defined by $(t_{\oper})_X=\id_X$, $(t_{\oper})_Y=\id_Y$
and $(t_{\oper})_A=a\colon \uno\to A$.
$$ \xymatrix@R=.8pc{
\Tt_{\oper,A} \ar[dd]_{\tt_{\oper,A}} \\
\mbox{ } \\
\Tt_{\oper} \\
}
\qquad \qquad
\xymatrix@R=.8pc{
\Tt_{\oper,A} \ar[dr]^{j_{\oper,A}} \ar[dd]_{\tt_{\oper,A}} \\
\ar@{}[r]|(.4){=} & \Tt_{\oper,a} \ar[dl]^{\tt_{\oper,a}} \\
\Tt_{\oper} & \\
}
\qquad \qquad
\xymatrix@R=.8pc{
\Tt_{\oper,A} \ar[dr]^{j_{\oper,A}} \ar[dd]_{\tt_{\oper,A}} & \\
\ar@{}[r]|(.4){\trnat}|(.25){t_{\oper}} & \Tt_{\oper,a} \\
\Tt_{\oper} \ar[ur]_{j_{\oper}} \\
}
$$
\textbf{Parameterization process}
(construction of $\Tt_{\oper,A}$ from $\Tt_{\oper}$).
The theory $\Tt_{\oper,A}$ is obtained from $\Tt_{\oper}$
by adding a type $A$, called the \emph{type of parameters},
to the domain of the unique term in $\Tt_{\oper}$.
Then $\Tt_{\oper,A}$ can be seen as a \emph{generalization} of $\Tt_{\oper}$,
since each model $M$ of $\Tt_{\oper}$ can be identified to a model of $\Tt_{\oper,A}$
where $M(A)$ is a singleton.
We will also say that $\Tt_{\oper,A}$ is the \emph{expansion} of $\Tt_{\oper}$.
\textbf{Parameter passing process}
(construction of $\Tt_{\oper,a}$ from $\Tt_{\oper,A}$
and of a morphism from $\Tt_{\oper}$ to $\Tt_{\oper,a}$).
The theory $\Tt_{\oper,a}$ is obtained from $\Tt_{\oper,A}$
by adding a constant term $a\colon \uno\to A$, called the \emph{parameter}.
A model $M_a$ of $\Tt_{\oper,a}$ is made of a model $M_A$ of $\Tt_{\oper,A}$
together with an element $\alpha=M_a(a)\in M_A(A)$,
so that we can denote $M_a=(M_A,\alpha)$.
Now, let $M_A$ be some fixed model of $\Tt_{\oper,A}$,
then the models $M_a$ of $\Tt_{\oper,a}$ over $M_A$ correspond bijectively
to the elements of $M_A(A)$ by $M_a \mapsto M_a(a)$,
so that we get the \emph{parameter adding} bijection
(the category $\Mod(\Tt_{\oper,a})|_{M_A}$ is discrete):
$$ \Mod(\Tt_{\oper,a})|_{M_A} \isoto M_A(A) \hsp \mbox{by} \hsp
M_a = (M_A,\alpha) \mapsto M_a(a)=\alpha \;. $$
On the other hand,
each model $M_a=(M_A,\alpha)$ of $\Tt_{\oper,a}$ gives rise to a model
${j_{\oper}}^\md(M_a)$ of $\Tt_{\oper}$ such that
${j_{\oper}}^\md(M_a)(X)=M_a(X)=M_A(X)$, ${j_{\oper}}^\md(M_a)(Y)=M_a(Y)=M_A(Y)$
and ${j_{\oper}}^\md(M_a)(f)=M_a(f'')=M_A(f')(\alpha,-) $.
Now, let $M_A$ be some fixed model of $\Tt_{\oper,A}$
and $M_0$ its restriction to $\Tt_{\oper,0}$,
then for each model $M_a=(M_A,\alpha)$ of $\Tt_{\oper,a}$ over $M_A$
the model ${j_{\oper}}^\md(M_a)$ of $\Tt_{\oper}$ is over $M_0$.
This yields the \emph{parameter passing} function
(the categories $\Mod(\Tt_{\oper,a})|_{M_A}$ and $\Mod(\Tt_{\oper})|_{M_0}$
are discrete):
$$ \Mod(\Tt_{\oper,a})|_{M_A} \to \Mod(\Tt_{\oper})|_{M_0} \hsp \mbox{by} \hsp
M_a \mapsto {j_{\oper}}^\md(M_a) \;. $$
\textbf{Exact parameterization.}
Let $M_0$ be any fixed model of $\Tt_{\oper,0}$,
it is made of two sets $\setX=M_0(X)$ and $\setY=M_0(Y)$.
Let $M_A$ be the model of $\Tt_{\oper,A}$ over $M_0$ such that
$M_A(A)=\setY^{\setX}$ and
$M_A(f')\colon \setY^{\setX} \times \setX \to \setY$
is the application.
It can be noted that $M_A$ is the terminal model of $\Tt_{\oper,A}$ over $M_0$.
Then the parameter passing function is a bijection,
and composing it with the parameter adding bijection we get
(where $\curry{M(f)}\in\setY^{\setX}$ corresponds by currying to
$M(f)\colon \setX\to\setY$):
$$ \Mod(\Tt_{\oper})|_{M_0} \iso M_A(A) \hsp \mbox{by} \hsp
M_{A,\alpha} \leftrightarrow \alpha \hsp \mbox{i.e., by} \hsp
M \leftrightarrow \curry{M(f)} \;. $$
\end{exam}
\begin{exam}
\label{exam:sgp}
Let $\Tt_{\sgp}$ be the theory for semigroups
presented by one type $G$,
one term $\prd\colon G^2 \to G$
and one equation $\prd(x,\prd(y,z))=\prd(\prd(x,y),z)$
where $x$, $y$, $z$ are variables of type $G$.
As usual with the categorical point of view,
in fact the \emph{variables} are projections;
here, $x,y,z\colon G^3\to G$ are the three projections
and for instance $\prd(x,y)$ is $\prd\circ\tuple{x,y}\colon G^3\to G$,
composed of the pair $\tuple{x,y}\colon G^3\to G^2$ and of $\prd\colon G^2 \to G$
(more details are given in the appendix, example~\ref{exam:dia-sg}).
\textbf{Parameterization process}.
In order to get parameterized families of semigroups,
we consider the theory $\Tt_{\sgp,A}$
presented by two types $A$ and $G$,
one term $\prd'\colon A\times G^2 \to G$
and one equation $\prd'(p,x,\prd'(p,y,z))= \prd'(p,\prd'(p,x,y),z)$
where $x$, $y$, $z$ are variables of sort $G$
and $p$ is a variable of sort $A$.
\textbf{Parameter passing process}.
The theory $\Tt_{\sgp,a}$ is $\Tt_{\sgp,A}$ together with a parameter $a\colon \uno\to A$,
hence with $\prd''=\prd'\circ(a\times\id_{G^2}) \colon G^2 \to G$
(where $\uno \times G^2$ is identified to $G^2$).
Each model $M_A$ of $\Tt_{\sgp,A}$ gives rise to a family of models of $\Tt_{\sgp,a}$,
all of them with the same underlying set $M_A(G)$
but with different interpretations of $a$ in $M_A(A)$.
Mapping $\prd$ to $\prd''$ defines a morphism from $\Tt_{\sgp}$ to $\Tt_{\sgp,a}$.
So, each model $M_a$ of $\Tt_{\sgp,a}$ gives rise to a model $M$ of $\Tt_{\sgp}$ such that
$M(G)=M_a(G)$ and $M(\prd)(x,y)=M_a(\prd')(\alpha,x,y) $ for each $x,y\in M_a(G)$,
where $\alpha=M_a(a)$ is the \emph{argument}.
\end{exam}
\begin{exam}
\label{exam:list}
This example motivates the existence of pure terms in the given theory.
Let us consider the theory $\Tt_{\nat}$ ``of naturals''
presented by a type $N$ and two terms $z\colon \uno\to N$ and $s\colon N\to N$,
and let us say that $z$ is pure.
Let $\Tt_{\nat,0}$ be the subtheory presented by $N$ and $z$,
it is called the pure subtheory of $\Tt_{\nat}$.
We define the theory $\Tt_{\nat,A}$ as made of
two types $A$ and $N$ and two terms
$z\colon \uno\to N$ and $s'\colon A\times N\to N$.
It should be noted that $\Tt_{\nat,A}$ contains $\eps_\uno\colon A\times \uno\to \uno$
and $z'=z\circ \eps_\uno\colon A\times \uno\to N$.
Then $\Tt_{\nat,A}$ is a theory ``of lists of $A$'', with $z$ for the empty list
and $s'$ for concatenating an element to a list.
In this way, the theory of lists of $A$ is built as a generalization
of the theory of naturals;
indeed the naturals can be identified to the lists over a singleton.
\end{exam}
\begin{exam}
\label{exam:dm}
Here is another example where pure terms are required,
this is a simplified version of many structures in Kenzo/EAT.
Let $\Tt_{\mon}$ be the theory for monoids presented by
one type $G$,
two terms $\prd\colon G^2 \to G$ and $\unt\colon \to G$,
and the equations $\prd(x,\prd(y,z)) = \prd(\prd(x,y),z)$,
$\prd(x,\unt) = x$, $\prd(\unt,x) = x$ where $x$, $y$, $z$ are variables of type $G$.
Let $\Tt_{\dm}$ be the theory for \emph{differential monoids},
presented by $\Tt_{\mon}$ together with one term $\dif \colon G \to G$
and the equations $\dif(\prd(x,y)) = \prd(\dif(x),\dif(y))$, $\dif(\unt) = \unt$, $\dif(\dif(x)) = \unt$,
and with the terms in $\Tt_{\mon}$ as its pure terms.
In order to get parameterized families of differential structures on one monoid,
we define the theory $\Tt_{\dm,A}$
presented by two types $G$, $A$ and three terms
$\prd\colon G^2 \to G$, $\unt\colon \uno\to G$ and $\dif' \colon A\times G \to G$
and the equations $\prd(x,\prd(y,z)) = \prd(\prd(x,y),z)$,
$\prd(x,\unt) = x$, $\prd(\unt,x) = x$,
$\dif'(p,(\prd(x,y))) = \prd(\dif'(p,x), \dif'(p,y))$, $\dif'(p, \unt) = \unt$,
$\dif'(p,\dif'(p,x)) = \unt$.
Each model $M_A$ of $\Tt_{\dm,A}$ gives rise to a family
of models of $\Tt_{\dm}$,
all of them with the same underlying monoid $(M_A(G),M_A(\prd),M_A(\unt))$:
there is a model $M_a$ of $\Tt_{\dm}$ over $M_A$ for each element $\alpha$ in $M_A(A)$,
with its differential structure defined by
$M_a(\dif)= M_A(\dif')(\alpha,-)$.
\end{exam}
\begin{exam}
\label{exam:pi}
In the next sections we will use the theories with the following presentations:
$$ \begin{array}{c|c|c|c|}
\cline{2-2}
\Pi_A \dashv & \xymatrix{A} & \multicolumn{2}{c}{} \\
\cline{2-2} \cline{4-4}
\multicolumn{2}{c}{} & \qquad \Pi_a \dashv & \xymatrix@R=1pc{ A \\ \uno \ar[u]^{a} \\ } \\
\cline{2-2} \cline{4-4}
\Pi \dashv & \xymatrix{\uno} & \multicolumn{2}{c}{} \\
\cline{2-2}
\end{array}$$
These theories are related by several morphisms:
$\pi_A\colon \Pi_A\to \Pi$ maps $A$ to $\uno$,
both $i\colon \Pi\to \Pi_a$ and $i_A\colon \Pi_A\to \Pi_a$ are the inclusions,
and $\pi_a\colon \Pi_a\to \Pi$ extends $\pi_A$ by mapping $a$ to $\id_{\uno}$,
so that $\pi_A$ and $\pi_a$ are epimorphisms.
In addition, $\pi_a\circ i_A=\pi_A$ and there is a natural transformation
$p\colon i \circ \pi_A \To i_A$ defined by $p_A=a\colon \uno\to A$.
The diagram below on the right is the \emph{lax colimit of $\pi_A$},
which means that it enjoys the following universal property:
for each $\Pi'_a$ with $i'_A\colon \Pi_A\to \Pi'_a$, $i'\colon \Pi\to \Pi'_a$
and $p'\colon i' \circ \pi_A \To i'_A$,
there is a unique $h\colon \Pi_a\to \Pi'_a$ such that
$h\circ i_A = i'_A$, $h\circ i = i'$ and $h\circ p=p'$.
For instance, given $\Pi$, $\pi_A\colon \Pi_A\to \Pi$, $\id_{\Pi}\colon \Pi\to \Pi$
and $\id_{\pi_A}\colon \pi_A \To \pi_A$,
then $\pi_a\colon \Pi_a\to \Pi$ is the unique morphism such that
$\pi_a\circ i_A = \pi_A$, $\pi_a\circ i = \id_{\Pi}$ and $\pi_a\circ p=\id_{\pi_A}$.
$$ \xymatrix@R=.8pc{
\Pi_A \ar[dd]_{\pi_A} \\
\\
\Pi \\
}
\qquad \qquad
\xymatrix@R=.8pc{
\Pi_A \ar[dr]^{i_A} \ar[dd]_{\pi_A} \\
\ar@{}[r]|(.4){=} & \Pi_a \ar[dl]^{\pi_a} \\
\Pi & \\
}
\qquad \qquad
\xymatrix@R=.8pc{
\Pi_A \ar[dr]^{i_A} \ar[dd]_{\pi_A} & \\
\ar@{}[r]|(.4){\trnat}|(.25){p} & \Pi_a \\
\Pi \ar[ur]_{i} \\
}
$$
\end{exam}
\subsection{Some other kinds of theories}
\label{subsec:thry}
For every theory $\Tt$,
the coslice category of \emph{theories under $\Tt$} is denoted $\Tt\cosl\catT_{\eq}$.
It can be seen as a 2-category, with
the natural transformations which extend the identity on $\Tt$ as 2-cells.
\begin{defi}
\label{defi:A-th}
A \emph{parameterized theory} $\Tt_A$
is a theory $\Tt$ with a distinguished type,
called the \emph{type of parameters} and usually denoted $A$.
The 2-category of parameterized theories
is the coslice 2-category $\catT_A=\Pi_A\cosl\catT_{\eq}$ of theories under $\Pi_A$.
A \emph{theory with a parameter} $\Tt_a$
is a parameterized theory with a distinguished constant of type $A$,
called the \emph{parameter} and usually denoted $a\colon \uno\to A$.
The 2-category of theories with a parameter
is the coslice 2-category $\catT_a=\Pi_a\cosl\catT_{\eq}$ of theories under $\Pi_a$.
\end{defi}
According to the context,
$\Tt_A$ usually denotes the parameterized theory $\gamma_A \colon \Pi_A\to\Tt_A$,
and sometimes it denotes the equational theory $\Tt_A$ itself.
Similarly for $\Tt_a$, which usually denotes $\gamma_a\colon \Pi_a\to\Tt_a$
and sometimes $\Tt_a$ itself.
In addition, it can be noted that $\Pi$ is the initial theory
(which may also be presented by the empty specification)
so that $\Pi\cosl\catT_{\eq}$ is isomorphic to $\catT_{\eq}$.
The 2-categories $\catS_A$ and $\catS_a$ of \emph{parameterized specifications}
and \emph{specifications with a parameter}, respectively,
are defined in a similar way.
On the other hand, the input of the parameterization process
is a theory $\Tt$ together with a wide subtheory $\Tt_0$
(\emph{wide} means: with the same types),
such a structure is called a decorated theory.
\begin{defi}
\label{defi:dec-th}
A \emph{decorated theory} is made of a theory $\Tt$ with a wide subtheory $\Tt_0$
called the \emph{pure} subtheory of $\Tt$.
A morphism of decorated theories is a morphism of theories $\tt\colon \Tt\to\Tt'$
which maps the pure part of $\Tt$ to the pure part of $\Tt'$.
This forms the category $\catT_{\dec}$ of decorated theories.
\end{defi}
So, a decorated theory $\Tt$ is endowed with a distinguished family of terms,
called the \emph{pure} terms,
such that all the identities and projections are pure
and every composition or tuple of pure terms is pure.
Pure terms are denoted with ``$\rpto$''.
When there is no ambiguity we often use the same notation $\Tt$
for the theory $\Tt$ itself
and for the decorated theory made of $\Tt$ and $\Tt_0$.
The decorated specifications are defined in a straightforward way.
For instance, we may consider the decorated specification made of
a type $N$, a pure term $z\colon \uno\rpto N$ and a term $s\colon N\to N$
(see example~\ref{exam:list}).
\section{Constructions}
\label{sec:const}
\subsection{The parameterization process is a functor}
\label{subsec:gene}
In this section we prove that the parameterization process is functorial,
by defining a functor $F_{\expan}\colon \catT_{\dec}\to\catT_A$
(``$\expan$'' for ``expansion'')
which adds the type of parameters to the domain of every non-pure term.
In addition, theorem~\ref{theo:gene} states that $F_{\expan}$ is left adjoint to
the functor $G_{\expan}\colon \catT_A\to\catT_{\dec}$,
which builds the coKleisli category of the comonad $A\times-$.
Moreover, we will see in section~\ref{sec:free} that $F_{\expan}$ is a free functor
associated to a morphism of limit sketches,
and in appendix~\ref{sec:dia} that this morphism of limit sketches
underlies a morphism of diagrammatic logics.
\begin{figure}[!t]
$$ \begin{array}{|l|l|c|c|}
\hline
& \textrm{index} \; \ttE\deco & \Ss_{\ttE\deco} & F_{\expan}\Ss_{\ttE\deco} \\
\hline
\hline
\textrm{type} & \Type\pur & X & X \\
\hline
\textrm{pure term} & \Term\pur &
\xymatrix@C=3pc{X\ar@{~>}[r]^{f} & Y} &
\xymatrix@C=3pc{X\ar[r]^{f} & Y} \\
\hline
\textrm{term} & \Term\gen &
\xymatrix@C=3pc{X\ar[r]^{f} & Y} &
\xymatrix@C=3pc{A\stimes X\ar[r]^{f'} & Y} \\
\hline
\textrm{sel. of identity} & \Selid\pur &
\xymatrix@C=3pc{X\ar@{~>}[r]^{\id_X} & X} &
\xymatrix@C=3pc{X\ar[r]^{\id_X} & X} \\
\hline
\textrm{pure composition} & \Comp\pur &
\xymatrix@C=3pc{X\ar@{~>}[r]^{f} \ar@/_3ex/@{~>}[rr]_{g\circ f}^{=} & Y\ar@{~>}[r]^{g} & Z} &
\xymatrix@C=3pc{X\ar[r]^{f} \ar@/_3ex/[rr]_{g\circ f}^{=} & Y\ar[r]^{g} & Z} \\
\hline
\textrm{composition} & \Comp\gen &
\xymatrix@C=3pc{X\ar[r]^{f} \ar@/_3ex/[rr]_{g\circ f}^{=} & Y\ar[r]^{g} & Z} &
\xymatrix@C=3pc{A\stimes X \ar[r]^{\tuple{\proj_X,f'}}
\ar@/_3ex/[rr]_{g'\circ\tuple{\proj_X,f'}}^{=} & A\stimes Y\ar[r]^{g'} & Z \\} \\
\hline
\textrm{binary product} & \bProd\pur &
\xymatrix@C=3pc@R=.7pc{ X & \\
& X\stimes Y \ar@{~>}[lu]_{p_X} \ar@{~>}[ld]^{p_Y} \\ Y & \\ } &
\xymatrix@C=3pc@R=.7pc{ X & \\
& X\stimes Y \ar[lu]_{p_X} \ar[ld]^{p_Y} \\ Y & \\ } \\
\hline
\textrm{pure pairing} & \bTuple\pur &
\xymatrix@C=3pc@R=1pc{ & X & \\
Z \ar@{~>}[ru]^{f} \ar@{~>}[rd]_{g} \ar@{~>}[rr]|{\tuple{f,g}} & \ar@{}[u]|{=}\ar@{}[d]|{=} &
X\stimes Y \ar@{~>}[lu]_{p_X} \ar@{~>}[ld]^{p_Y} \\ & Y & \\ } &
\xymatrix@C=3pc@R=1pc{ & X & \\
Z \ar[ru]^{f} \ar[rd]_{g} \ar[rr]|{\tuple{f,g}} &
\ar@{}[u]|{=}\ar@{}[d]|{=} &
X\stimes Y \ar[lu]_{p_X} \ar[ld]^{p_Y} \\ & Y & \\ } \\
\hline
\textrm{pairing} & \bTuple\gen &
\xymatrix@C=3pc@R=1pc{ & X & \\
Z \ar[ru]^{f} \ar[rd]_{g} \ar[rr]|{\tuple{f,g}} & \ar@{}[u]|{=}\ar@{}[d]|{=} &
X\stimes Y \ar@{~>}[lu]_{p_X} \ar@{~>}[ld]^{p_Y} \\ & Y & \\ } &
\xymatrix@C=3pc@R=1pc{ & X & \\
A\stimes Z \ar[ru]^{f'} \ar[rd]_{g'} \ar[rr]|{\tuple{f',g'}} &
\ar@{}[u]|{=}\ar@{}[d]|{=} &
X\stimes Y \ar[lu]_{p_X} \ar[ld]^{p_Y} \\ & Y & \\ } \\
\hline
\end{array}$$
\caption{\label{fig:F-elem} The functor $F_{\expan}$
on elementary decorated specifications}
\end{figure}
In order to define the functor $F_{\expan}$
we use the fact that it should preserve colimits.
It has been seen in section~\ref{subsec:equa} that
every specification is the colimit of a diagram of elementary specifications.
Similarly, every decorated specification is the colimit of
a diagram of elementary decorated specifications,
denoted $\Ss_{\ttE\deco}$ where
$\mathtt{x}=\mathtt{p}$ for ``pure'' or $\mathtt{x}=\mathtt{g}$ for ``general''.
Informally, the functor $F_{\expan}$ explicits the fact that every
general feature in a decorated specification gets parameterized,
while every pure feature remains unparameterized.
Figure~\ref{fig:F-elem} defines the parameterized specification
$F_{\expan}\Ss_{\ttE\deco}$
for each elementary decorated specification $\Ss_{\ttE\deco}$
(many projection arrows are omitted, and
when needed the projections from $A\times X$ are denoted
$\proj_X\colon A\times X\to A$ and $\eps_X\colon A\times X\to X$).
The morphisms of parameterized specifications $F_{\expan}\ss$,
for $\ss$ between elementary decorated specifications,
are straightforward.
For instance, let $c\colon \Ss_{\Term\gen}\to\Ss_{\Term\pur}$
be the conversion morphism,
which corresponds to the fact that every pure term can be seen as a general term,
then $F_{\expan}c$ maps $f'\colon A\times X\to Y$ in $F_{\expan}\Ss_{\Term\gen}$
to $f\circ \eps_X\colon A\times X\to Y$ in $F_{\expan}\Ss_{\Term\pur}$.
Now, given a decorated theory $\Tt$ presented by
the colimit of a diagram $\Delta$ of elementary decorated specifications,
we define $F_{\expan}\Tt$ as the parameterized theory presented by
the colimit of the diagram $F_{\expan}\Delta$ of parameterized specifications.
\begin{defi}
\label{defi:gene}
The functor $F_{\expan}:\catT_{\dec}\to\catT_A$ defined above
is called the \emph{parameterization functor}.
\end{defi}
Clearly the parameterization functor preserves colimits.
In addition, let $\Tt_A$ be the parameterized theory $F_{\expan}\Tt$,
it follows from the definition of $F_{\expan}$ that the equational theory $\Tt_A$
is a theory under $\Tt_0$.
Now the functor $G_{\expan}$ is defined independently from $F_{\expan}$.
Let $\Tt_A$ be a parameterized theory.
The endofunctor of product with $A$ forms a comonad on $\Tt_A$
with the counit $\eps$ made of the projections $\eps_X\colon A\times X\to X$
and the comultiplication made of the terms
$\delta_X\colon A\times X\to A\times A\times X$ induced by the diagonal on $A$.
Let $\Tt$ be the coKleisli category of this comonad:
it has the same types as $\Tt_A$
and a term $\kl{f}\colon X\to Y$ for each term $f\colon A\times X\to Y$ in $\Tt_A$.
There is a functor from $\Tt_A$ to $\Tt$ which is the identity on types
and maps every $g\colon X\to Y$ in $\Tt_A$
to $\kl{g\circ \eps_X}\colon X\to Y$ in $\Tt$.
Then every finite product in $\Tt_A$ is mapped to a finite product in $\Tt$,
which makes $\Tt$ a theory.
Let $\Tt_0$ denote the image of $\Tt_A$ in $\Tt$, it is a wide subtheory of $\Tt$.
In this way, any parameterized theory yields a decorated theory.
The definition of $G_{\expan}$ on morphisms is straightforward.
The next result can be derived directly,
or as a consequence of theorem~\ref{theo:free}.
\begin{theo}
\label{theo:gene}
The parameterization functor $F_{\expan}$ and the functor $G_{\expan}$
form an adjunction $F_{\expan}\dashv G_{\expan}$:
$$ \xymatrix@C=4pc{ \catT_{\dec} \ar@/^/[r]^{F_{\expan}} \ar@{}[r]|{\bot} &
\catT_A \ar@/^/[l]^{G_{\expan}} } $$
\end{theo}
The next result states that $\Tt$ can be easily recovered from $\Tt_A$,
by mapping $A$ to $\uno$.
\begin{prop}
\label{prop:general}
Let $\Tt$ be a decorated theory with pure subtheory $\Tt_0$
and $\gamma_A\colon \Pi_A\to\Tt_A$ the parameterized theory $F_{\expan}\Tt$.
Let $\gamma\colon \Pi\to\Tt$ be the unique morphism from the initial theory $\Pi$
to the theory $\Tt$.
Then there is a morphism $\tt_A\colon \Tt_A\to\Tt$ under $\Tt_0$
such that the following square is a pushout:
$$ \xymatrix{
\ar@{}[rd]|{\PO}
\Pi_A \ar[d]_{\pi_A} \ar[r]^{\gamma_A} & \Tt_A \ar[d]^{\tt_A} \\
\Pi \ar[r]^{\gamma} &\Tt \\
} $$
\end{prop}
\begin{proof}
It can easily be checked that this property is satisfied by each elementary specification.
Then the result follows by commuting two colimits:
on the one hand the colimit that defines the given theory from its elementary components,
and on the other hand the pushout.
\end{proof}
When there is an epimorphism of theories $\tt\colon \Tt_1\to\Tt_2$,
we say that $\Tt_1$ is \emph{the generalization of $\Tt_2$ along $\tt$}.
Indeed, since $\tt$ is an epimorphism,
the functor $\tt^\md\colon \Mod(\Tt_2)\to\Mod(\Tt_1)$ is a monomorphism,
which can be used for identifying $\Mod(\Tt_2)$ to a subcategory of $\Mod(\Tt_1)$.
\begin{coro}
\label{coro:general}
With notations as in proposition~\ref{prop:general},
$\Tt_A$ is the generalization of $\Tt$ along $\tt_A$.
\end{coro}
\begin{proof}
Clearly $\pi_A\colon \Pi_A\to\Pi$ is an epimorphism.
Since epimorphisms are stable under pushouts,
proposition~\ref{prop:general} proves that $\tt_A\colon \Tt_A\to\Tt$
is also an epimorphism.
\end{proof}
Let $F_{\expan}:\catT_{\dec}\to\catT_A$ be the parameterization functor
and let $U\colon \catT_A\to\catT_{\eq}$ be the functor
which simply forgets that the type $A$ is distinguished,
so that $U \circ F_{\expan}\colon \catT_{\dec}\to\catT_{\eq}$
maps the decorated theory $\Tt$ to the equational theory $\Tt_A$.
$$ \xymatrix@C=4pc{ \catT_{\dec} \ar[r]^{F_{\expan}} &
\catT_A \ar[r]^{U} & \catT_{\eq} \\ } $$
Every theory $\Tt$ can be seen as a decorated theory where
the pure terms are defined inductively as the identities, the projections,
and the compositions and tuples of pure terms.
Let $I\colon \catT_{\eq}\to\catT_{\dec}$ denote the corresponding inclusion functor.
Then the endofunctor $U \circ F_{\expan}\circ I\colon \catT_{\eq}\to\catT_{\eq}$
corresponds to the ``\emph{imp} construction'' of \cite{LPR03},
which transforms each term $f\colon X\to Y$ in $\Tt$ into
$f'\colon A\times X\to Y$ for a new type $A$.
\subsection{The parameter passing process is a natural transformation}
\label{subsec:passing}
A theory $\Tt_a$ with a parameter is built simply by adding
a constant $a$ of type $A$ to a parameterized theory $\Tt_A$.
Obviously, this can be seen as a pushout.
\begin{defi}
\label{defi:adding}
Let $\gamma_A\colon \Pi_A\to\Tt_A$ be a parameterized theory.
The theory with parameter \emph{extending} $\gamma_A$
is $\gamma_a\colon \Pi_a\to\Tt_a$ given by the pushout of $\gamma_A$ and $i_A$:
$$ \xymatrix{
\ar@{}[rd]|{\PO}
\Pi_A \ar[d]_{i_A} \ar[r]^{\gamma_A} & \Tt_A \ar[d]^{j_A} \\
\Pi_a \ar[r]^{\gamma_a} &\Tt_a \\
} $$
\end{defi}
The pushout of theories in definition~\ref{defi:adding}
gives rise to a pullback of categories of models,
hence for each model $M_A$ of $\Tt_A$
the function which maps each model $M_a$ of $\Tt_a$ over $M_A$
to the element $M_a(a)\in M_A(A)$ defines a bijection:
\begin{equation}
\label{eq:adding}
\Mod(\Tt_a)|_{M_A} \isoto M_A(A) \;.
\end{equation}
Let us assume that the parameterized theory $\gamma_A\colon \Pi_A\to\Tt_A$
is $F_{\expan}\Tt$ for some decorated theory $\Tt$
with pure subtheory $\Tt_0$.
Then the pushout property in definition~\ref{defi:adding}
ensures the existence of a unique $\tt_a\colon \Tt_a\to \Tt$
such that $\tt_a \circ \gamma_a = \gamma\circ\pi_a$
(which means that $\tt_a$ maps $A$ to $\uno$ and $a$ to $\id_{\uno}$)
and $\tt_a \circ j_A = \tt_A$.
Then $\Tt_A$ is a theory under $\Tt_0$
and the composition by $j_A$ makes $\Tt_a$ a theory under $\Tt_0$
with $j_A$ preserving~$\Tt_0$.
$$ \xymatrix@R=.8pc{
\Tt_A \ar[dr]^{j_A} \ar[dd]_{\tt_A} & \\
\ar@{}[r]|(.4){=} & \Tt_a \ar[dl]^{\tt_a} \\
\Tt & \\
} $$
Lax cocones and lax colimits in 2-categories generalize
cocones and colimits in categories.
For each decorated theory $\Tt$ with pure subcategory $\Tt_0$,
let $\Tt_A=F_{\expan}\Tt$ and $\tt_A\colon \Tt_A\to\Tt$
be as in section~\ref{subsec:gene},
and let $\Tt_a$ and $j_A\colon \Tt_A\to\Tt_a$ be as above.
Let $j\colon \Tt\to\Tt_a$ be the morphism under $\Tt_0$
which maps each type $X$ to $X$
and each term $f\colon X\to Y$ to $f'\circ(a\times\id_X)\colon X\to Y$.
Let $t\colon j\circ\tt_A \To j_A$ be the natural transformation under $\Tt_0$
such that $t_A=a\colon \uno\to A$.
Then the following diagram is a \emph{lax cocone with base $\tt_A$}
in the 2-category $\Tt_0\cosl\catT_{\eq}$, for short it is denoted $(\Tt_a,j_A,j,t)$,
and it is called \emph{the lax colimit associated to}~$\Tt$ because of lemma~\ref{lemm:passing}.
$$ \xymatrix@R=.8pc{
\Tt_A \ar[dr]^{j_A} \ar[dd]_{\tt_A} & \\
\ar@{}[r]|(.4){\trnat}|(.25){t} & \Tt_a \\
\Tt \ar[ur]_{j} & \\
} $$
\begin{lemm}
\label{lemm:passing}
Let $\Tt$ be a decorated theory with pure subcategory $\Tt_0$.
The lax cocone $(\Tt_a,j_A,j,t)$ with base $\tt_A$ defined above
is a lax colimit in the 2-category of theories under $\Tt_0$.
\end{lemm}
\begin{proof}
This means that the given lax cocone
is initial among the lax cocones with base $\tt_A$ in $\Tt_0\cosl\Tt$,
in the following sense.
For every lax cocone $(\Tt'_a,j'_A,j',t')$ with base $\tt_A$ under $\Tt_0$
there is a unique morphism $ h\colon \Tt_a\to\Tt'_a$ such that
$ h\circ j_A=j'_A$, $ h\circ j=j'$ and $ h\circ t = t'$,
it is defined from the pushout in definition~\ref{defi:adding}
by $h\circ j_A=j'_A$, so that $h(A)=A$,
and $h\circ \gamma_a(a)=t'_A:\uno\to A$.
\end{proof}
For instance, given $\Tt$, $\tt_A\colon \Tt_A\to \Tt$, $\id_{\Tt}\colon \Tt\to \Tt$
and $\id_{\tt_A}\colon \tt_A \To \tt_A$,
then $\tt_a$ is the unique morphism such that
$\tt_a\circ j_A = \tt_A$, $\tt_a\circ j = \id_{\Tt}$ and $\tt_a\circ t=\id_{\tt_A}$.
Let $\Tt$ be a decorated theory with pure subtheory $\Tt_0$
and let $(\Tt_a,j_A,j,t)$ be its associated lax colimit, with base $\tt_A\colon \Tt_A\to \Tt$.
Let $M_A$ be a model of $\Tt_A$
and $M_0$ its restriction to $\Tt_0$,
and let $\{ (M,m) \mid m\colon {\tt_A}^\md M \to M_A \}|_{M_0}$
(where as before ${\tt_A}^\md M=M\circ\tt_A$)
denote the set of pairs $(M,m)$ with $M$ a model of $\Tt$ over $M_0$
and $m$ a morphism of models of $\Tt_A$ over $M_0$.
A consequence of the lax colimit property
is that the function which maps each model $M_a$ of $\Tt_a$ over $M_A$
to the pair $(j^\md M_a, t^\md M_a)=(M_a\circ j,M_a\circ t)$
defines a bijection:
\begin{equation}
\label{eq:passing}
\Mod(\Tt_a)|_{M_A} \iso
\{ (M,m) \mid m\colon {\tt_A}^\md M \to M_A \}|_{M_0} \;.
\end{equation}
The bijections~\ref{eq:adding} and~\ref{eq:passing} provide the next result,
which does not involve $\Tt_a$.
\begin{prop}
\label{prop:passing}
Let $\Tt$ be a decorated theory with pure subtheory $\Tt_0$
and let $\Tt_A=F_{\expan}\Tt$ and $\tt_A\colon \Tt_A\to \Tt$.
Then for each model $M_A$ of $\Tt_A$,
with $M_0$ denoting the restriction of $M_A$ to $\Tt_0$,
the function which maps each element $\alpha\in M_A(A)$
to the pair $(M,m)$,
where $M$ is the model of $\Tt$ such that $M(f)=M_A(f')(\alpha,-)$
and where $m:{\tt_A}^\md M \to M_A$ is the morphism of models of $\Tt_A$
such that $m_A:M(\uno)\to M_A(A)$ is the constant function $\star\to\alpha$,
defines a bijection:
\begin{equation}
\label{eq:adding-passing}
M_A(A) \iso
\{ (M,m) \mid m\colon {\tt_A}^\md M \to M_A \}|_{M_0} \;.
\end{equation}
\end{prop}
As an immediate consequence,
we get the \emph{exact parameterization} property from \cite{LPR03}.
\begin{coro}
\label{coro:exact}
Let $\Tt$ be a decorated theory with pure subcategory $\Tt_0$,
and let $\Tt_A=F_{\expan}\Tt$.
Let $M_0$ be a model of $\Tt_0$
and $M_A$ a terminal model of $\Tt_A$ over $M_0$.
Then there is a bijection:
\begin{equation}
\label{eq:exact}
M_A(A) \iso \Mod(\Tt)|_{M_0}
\end{equation}
which maps each $\alpha\in M_A(A)$ to the model
$M_{A,\alpha}$ of $\Tt$ defined by
$M_{A,\alpha}(X)=M_0(X)$ for each type $X$
and $M_{A,\alpha}(f)=M_A(f')(\alpha,-)$ for each term $f$,
so that $M_{A,\alpha}(f)=M_A(f)$ for each pure term $f$.
\end{coro}
The existence of a terminal model of $\Tt_A$ over $M_0$
is a consequence of \cite{Ru00} and \cite{HR95}.
Corollary~\ref{coro:exact} corresponds to the way algebraic structures are implemented
in the systems Kenzo/EAT.
In these systems
the parameter set is encoded by means of a record of Common Lisp functions,
which has a field for each operation in the algebraic structure to be implemented.
The pure terms correspond to functions which can be obtained
from the fixed data and do not require an explicit storage.
Then, each particular instance of the record gives rise to an algebraic structure.
Clearly the construction of $\gamma_a$ from $\gamma_A$
is a functor, which is left adjoint to the functor
which simply forgets that the constant $a$ is distinguished.
So, by composing this adjunction
with the adjunction $F_{\expan}\dashv G_{\expan}$ from theorem~\ref{theo:gene}
we get an adjunction $F'_{\expan}\dashv G'_{\expan}$
where $F'_{\expan}$ maps each decorated theory $\Tt$ to $\Tt_a$, as defined above:
$$ \xymatrix@C=4pc{ \catT_{\dec} \ar@/^/[r]^{F'_{\expan}} \ar@{}[r]|{\bot} &
\catT_a \ar@/^/[l]^{G'_{\expan}} } $$
Let $U'\colon \catT_a\to\catT_{\eq}$ be the functor
which simply forgets that the type $A$ and the constant $a$ are distinguished.
Then the functor $U' \circ F'_{\expan} \colon \catT_{\dec}\to\catT_{\eq}$
maps the decorated theory $\Tt$ to the equational theory $\Tt_a$.
$$ \xymatrix@C=3pc{
\catT_{\dec} \ar[r]^{F'_{\expan}} & \catT_a \ar[r]^{U'} & \catT_{\eq} \\ } $$
The morphism of theories $j\colon \Tt\to\Tt_a$ depends on the decorated theory $\Tt$,
let us denote it $j=J_{\Tt}$.
Let $H\colon \catT_{\dec}\to\catT_{\eq}$ be the functor which maps
each decorated theory $\Tt$ to the equational theory $\Tt$.
The next result is easy to check.
\begin{theo}
\label{theo:passing}
The morphisms of theories $J_{\Tt}\colon \Tt\to\Tt_a$ form the components of
a natural transformation $J\colon H \To U'\circ F'_{\expan}\colon \catT_{\dec}\to\catT_{\eq}$.
$$ \xymatrix@C=3pc{
\catT_{\dec} \ar[r]^{F'_{\expan}} \ar@/_4ex/[rr]_{H}^(.4){\Uparrow}^(.45){J} &
\catT_a \ar[r]^{U'} & \catT_{\eq} \\ } $$
\end{theo}
\begin{defi}
\label{defi:passing}
The natural transformation $J\colon H \To U'\circ F'_{\expan}\colon \catT_{\dec}\to\catT_{\eq}$
in theorem~\ref{theo:passing} is called the \emph{parameter passing natural transformation}.
\end{defi}
\begin{exam}
\label{exam:adding}
Starting from $\Tt_{\oper}$ and $\Tt_{\oper,0}$ as in example~\ref{exam:term},
the pushouts of theories from proposition~\ref{prop:general}
and definition~\ref{defi:adding} are respectively:
$$ \begin{array}{|c|c|c|c|c|c|c|}
\cline{1-1} \cline{3-3} \cline{5-5} \cline{7-7}
\xymatrix{A \\} &
\longrightarrow &
\xymatrix@R=1pc{A & A\stimes X \ar[l]\ar[d]\ar[rd]^{f'} & \\ & X & Y \\} &
\qquad \qquad &
\xymatrix@R=1pc{A \\} &
\longrightarrow &
\xymatrix@R=1pc{A & A\stimes X \ar[l]\ar[d]\ar[rd]^{f'} & \\ & X & Y \\} \\
\cline{1-1} \cline{3-3} \cline{5-5} \cline{7-7}
\col{\downarrow} &
\col{} &
\col{\downarrow} &
\col{} &
\col{\downarrow} &
\col{} &
\col{\downarrow} \\
\cline{1-1} \cline{3-3} \cline{5-5} \cline{7-7}
\xymatrix@R=1pc{ \\ \uno \\ } &
\longrightarrow &
\xymatrix@R=1pc{ \\ & X\ar[r]^{f} &Y \\ } & &
\xymatrix@R=1pc{A \\ \uno \ar[u]^{a} \\ } &
\longrightarrow &
\xymatrix@R=1pc{ A & A\stimes X \ar[l]\ar[d]\ar[rd]^{f'} & \\ \uno \ar[u]^{a} & X & Y \\ } \\
\cline{1-1} \cline{3-3} \cline{5-5} \cline{7-7}
\end{array} $$
We have seen in example~\ref{exam:term}
two other presentations of the vertex $\Tt_{\oper,a}$ of the second pushout,
with $f''=f'\circ(a\times\id_X):X\to Y$.
For each decorated theory $\Tt$,
the morphism of equational theories $j_{\oper}=J_{\Tt_{\oper}}:\Tt\to\Tt_a$
maps $f$ to $f''$, as in example~\ref{exam:term}.
A model $M_0$ of $\Tt_{\oper,0}$ is simply made of two sets
$\setX=M_0(X)$ and $\setY=M_0(Y)$.
On the one hand, a model of $\Tt$ over $M_0$ is characterized
by a function $\varphi\colon \setX\to\setY$.
On the other hand,
the terminal model $M_A$ of $\Tt_{\oper,A}$ over $M_0$
is such that $M_A(A)=\setY^\setX$ and
$M_A(f')\colon \setY^{\setX} \times \setX \to \setY$ is the application.
The bijection $\Mod(\Tt)|_{M_0}\iso M_A(A)$ then
corresponds to the currying bijection $\varphi\mapsto\curry{\varphi}$.
\end{exam}
\begin{exam}
\label{exam:dm-terminal}
Let $\Tt_{\dm}$ be the theory for differential monoids from example~\ref{exam:dm},
with the pure subtheory $\Tt_{\dm,0}=\Tt_{\mon}$ of monoids.
They generate the parameterized theory $\Tt_{\dm,A}$ as in example~\ref{exam:dm}.
Let $M_0$ be some fixed monoid
and $M_A$ any model of $\Tt_{\dm,A}$ over $M_0$,
then each element of $M_A(A)$ corresponds to a
differential structure on the monoid $M_0$.
If in addition $M_A$ is the terminal model of $\Tt_{\dm,A}$ over $M_0$,
then this correspondence is bijective.
\end{exam}
\begin{exam}
\label{exam:state}
When dealing with an imperative language,
the states for the memory are endowed with an operation $\lup$
for observing the state and an operation $\upd$ for modifying it.
There are two points of view on this situation: either the state is hidden,
or it is explicit.
Let us check that the parameterization process allows to generate
the theory with explicit state from the theory with hidden state.
First, let us focus on observation:
the theory $\Tt_{\st}$ is made of two types $L$ and $Z$
(for locations and integers, respectively) and a term $v\colon L\to Z$
for observing the values of the variables.
The pure subtheory $\Tt_{\st,0}$ is made of $L$ and $Z$.
We choose a model $M_0$ of $\Tt_{\st,0}$ made of a countable set
of locations (or addresses, or ``variables'') $\bL=M_0(L)$
and of the set of integers $\bZ=M_0(Z)$.
Let $\bA=\bZ^\bL$, then as in example~\ref{exam:adding}
the terminal model $M_A$ of $\Tt_{\st,A}$ over $M_0$
is such that $M_A(A)=\bA$ and $M_{\st,A}(v')\colon \bA\times\bL\to\bZ$
is the application, denoted $\lup$.
The terminal model $M_A$ does correspond to an ``optimal'' implementation
of the state.
Now, let us look at another model $N_A$ of $\Tt_{\st,A}$ over $M_0$,
defined as follows:
$N_A(A)=\bA\times \bL\times \bZ$ and
$N_A(v')\colon \bA\times \bL\times \bZ\times \bL\to\bZ$
maps $(p,x,n,y)$ to $n$ if $x=y$ and to $\lup(p,y)$ otherwise.
The terminality property of $M_A$ ensures that there is a unique
function $\upd\colon \bA\times \bL\times \bZ \to \bA$ such that
$\lup(\upd(p,x,n),y)$ is $n$ if $x=y$ and $\lup(p,y)$ otherwise.
So, the updating operation $\upd$ is defined coinductively from the observation operation $\lup$.
\end{exam}
\section{Free functors}
\label{sec:free}
In this section some basic facts about limit sketches and their associated adjunction
are mentioned,
and it is proved that the parameterization functor $F_{\expan}$
from section~\ref{subsec:gene} is a free functor,
in the sense that it is the left adjoint
associated to a morphism of limits sketches.
\subsection{Limit sketches}
\label{subsec:sketch}
It is quite usual to define a \emph{free} functor
as the left adjoint of a forgetful functor,
but there is no unique definition of a forgetful functor.
In this section forgetful functors are defined
from morphisms of limit sketches, they are not always faithful.
There are several definitions of limit sketches
(also called projective sketches) in the litterature,
see for instance \cite{CL84} or \cite{BW99}.
These definitions are different but all of them serve the same purpose:
each limit sketch generates a category with limits,
so that limit sketches generalize equational specifications
in allowing some interdependence between the variables.
In this paper, limit sketches are used at the meta level,
in order to describe each category of theories or specifications
as the category of realizations (or models) of a limit sketch.
While a category with limits is a graph with identities, composition,
limit cones and tuples, satisfying a bunch of axioms,
a limit sketch is a graph with \emph{potential} identities, composition,
limit cones and tuples, which are not required to satisfy any axiom.
Potential limit cones, or simply \emph{potential limits},
may also be called \emph{specified limits} or \emph{distinguished cones}.
\begin{defi}
\label{defi:sketch}
A \emph{limit sketch} is a graph where
some points $X$
have an associated potential identity arrow $\id_X\colon X\to X$,
some pairs of consecutive arrows $f\colon X\to Y$, $g\colon Y\to Z$
have an associated potential composed arrow $g\circ f\colon X\to Z$,
some diagrams $\Delta$
have an associated potential limit, which is a cone with base $\Delta$,
and when there is a potential limit with base $\Delta$ then some cones with base $\Delta$
have an associated potential tuple, which is a morphism of cones with base $\Delta$
from the given cone to the potential limit cone.
A morphism of limit sketches is a morphism of graphs which preserves the
potential features. This yields the category of limit sketches.
\end{defi}
Whenever this definition is restricted to potential limits with a finite discrete base
(called potential finite products),
we get the category of \emph{finite product sketches}:
this is the category $\catS_{\eq}$ of equational specifications, from section~\ref{subsec:equa}.
\begin{defi}
\label{defi:realization}
Given a limit sketch $\skE$ and a category $\catC$,
a \emph{realization} (or \emph{loose model}) of $\skE$ with values in $\catC$
is a graph homomorphism which maps the potential features of $\skE$
to real features of $\catC$.
A morphism of realizations is (an obvious generalization of) a natural transformation.
This gives rise to the category $\Rea(\skE,\catC)$
of realizations of $\skE$ with values in $\catC$.
By default, $\catC$ is the category of sets.
\end{defi}
By default, $\catC$ is the category of sets.
A category is called \emph{locally presentable}
if it is equivalent to the category of set-valued realizations
of a limit sketch $\skE$; then $\skE$ is called a limit sketch \emph{for} this category.
Let $\olskE$ denote the category generated by $\skE$
such that every potential potential feature of $\skE$ becomes a real feature of $\olskE$.
The \emph{Yoneda contravariant realization} $\funY_{\skE}$ of $\skE$
is the contravariant realization of $\skE$ with values in $\Rea(\skE)$
such that $\funY_{\skE}(E)=\Hom_{\olskE}(E,-)$
for every point or arrow $E$ in $\skE$.
Then for each theory $\Tt$ and each point $E$ in $\skE$,
the set $\Tt(E)$ is in bijection with $\Hom_{\Rea(\skE)}(\funY_{\skE}(E),\Tt)$.
The Yoneda contravariant realization is injective on objects and faithful.
In addition it is \emph{dense}:
although $\Rea(\skE)$ may be ``much larger'' than $\skE$,
every realization of $\skE$ is the vertex of a colimit
with its base in the image of $\funY_{\skE}$.
Let $\ske\colon \skE_1\to\skE_2$
be a morphism of limit sketches and
$G_{\ske}\colon \Rea(\skE_2)\to\Rea(\skE_1)$ the precomposition with $\ske$.
A fundamental result due to Ehresmann states that there is an adjunction,
that will be called the adjunction \emph{associated with} $\ske$:
$$ \xymatrix@C=4pc{ \Rea(\skE_1) \ar@/^/[r]^{F_{\ske}} \ar@{}[r]|{\bot} &
\Rea(\skE_2) \ar@/^/[l]^{G_{\ske}} } $$
Moreover, the functor $F_{\ske}$ \emph{(contravariantly) extends} $\ske$
via the Yoneda contravariant realizations,
in the sense that there is an isomorphism:
$$F_{\ske} \circ \funY_{\skE_1} \iso \funY_{\skE_2} \circ \ske \;.$$
Our definition of forgetful and free functors relies on this adjunction.
\begin{defi}
\label{defi:free}
A \emph{forgetful} functor is a functor of the form
$G=-\circ\ske\colon \Rea(\skE_2)\to\Rea(\skE_1)$
for a morphism of limit sketches $\ske\colon \skE_1\to\skE_2$.
A \emph{free} functor is a left adjoint to a forgetful functor
(as every adjoint functor, it is unique up to a natural isomorphism).
\end{defi}
\begin{rema}
\label{rema:free}
It is easy to describe the forgetful functor $G_{\ske}$, using its definition:
for each realization $R_2$ of $\skE_2$,
the realization $R_1=G_{\ske}(R_2)$ of $\skE_1$ is such that $R_1(E)=R_2(\ske(E))$
for every point or arrow $E$ in $\skE_1$.
It is also quite easy to describe the left adjoint functor $F_{\ske}$,
using the fact that $F_{\ske}$ extends $\ske$:
let $R_1$ be a realization of $\skE_1$ and $R_2=F_{\ske}(R_1)$,
if $R_1=\funY_{\skE_1}(E)$ for some point $E$ in $\skE_1$
then $R_2=\funY_{\skE_2}(\ske(E))$,
and the general case follows thanks to the density of $\funY_{\skE_1}$
and to the fact that $F_{\ske}$ preserves colimits (since it is a left adjoint).
\end{rema}
\subsection{A limit sketch for equational theories}
\label{subsec:sk-equa}
The construction of various ``sketches of categories'' and ``sketches of sketches''
is a classical exercise about sketches \cite{CL84,CL88,BW99}.
Here we build (a significant part of) a limit sketch $\skE_{\eq}$
for the category $\catT_{\eq}$ of equational theories,
i.e., for the category of categories with chosen products.
\subsubsection*{$\bullet$ Graphs}
Let us start from the following limit sketch $\skE_{\gr}$ for the category of graphs,
simply made of two points $\Type$ and $\Term$ (for types and terms)
and two arrows $\dom$ and $\codom$ (for domain and codomain):
$$
\xymatrix@C=4pc{
\;\Type\; & \;\Term\; \ar[l]_{\dom} \ar@<1ex>[l]^{\codom} \\
}
$$
The image of $\skE_{\gr}$ by its Yoneda contravariant realization
is the following diagram of graphs:
$$ \begin{array}{|c|c|c|}
\cline{1-1}\cline{3-3}
\xymatrix{
X \\} &
\xymatrix@C=3pc{
\mbox{} \ar[r]^{X\mapsto X} \ar@<-2ex>[r]_{X\mapsto Y} & \mbox{} \\} &
\xymatrix{
X \ar[r]^{f} & Y \\} \\
\cline{1-1}\cline{3-3}
\end{array}$$
\subsubsection*{$\bullet$ Categories}
First, let us build a limit sketch $\skE'_{\gr}$
by adding to $\skE_{\gr}$ a point $\Cons$ for consecutive terms,
as the vertex of the following potential limit, where the projections
$\fst$ and $\snd$ stand for the first and second component
of a pair of consecutive terms and $\midd$ stands for its ``middle type''
(codomain of the first component and domain of the second one):
$$
\xymatrix@=1.5pc{
& \Cons \ar[dl]_{\fst} \ar[dr]^{\snd} \ar[d]|{\midd} & \\
\Term \ar[r]_{\codom} & \Type & \Term \ar[l]^{\dom} \\
} $$
hence the equations
$ \codom\circ\fst = \midd \,,\; \dom\circ\snd = \midd $ hold,
so that $\midd$ may be omitted.
Adding such a potential limit, with new vertex and projections over a known base,
is an equivalence of limit sketches:
the realizations of $\skE'_{\gr}$ are still the graphs.
Now, a limit sketch $\skE_{\cat}$ for categories is obtained by adding to $\skE'_{\gr}$
two arrows $\selid$ for the selection of identities and $\comp$ for the composition
and several equations:
$$
\xymatrix@C=4pc{
\;\Type\; \ar@<1ex>@/^/[r]^{\selid} &
\;\Term\; \ar[l]_{\dom} \ar@<1ex>[l]^{\codom} &
\;\Cons\; \ar[l]_{\fst} \ar@<1ex>[l]^{\snd} \ar@<-1ex>@/_/[l]_{\comp} \\
} $$
$$
\dom\circ\selid = \ttid_{\Type} \,,\;
\codom\circ\selid = \ttid_{\Type} \,,\;
\dom\circ\comp = \dom\circ\fst \,,\;
\codom\circ\comp = \codom\circ\snd \,,
$$
and with the equations which ensure that the three axioms of categories are satisfied.
The image of this part of $\skE_{\cat}$ by its Yoneda contravariant realization
is the following diagram of categories:
$$ \begin{array}{|c|c|c|c|c|}
\cline{1-1}\cline{3-3} \cline{5-5}
\xymatrix{
X \ar@(lu,ru)^(.6){\id_X} \\} &
\xymatrix@C=3pc{
\mbox{} \ar[r]^{X\mapsto X} \ar@<-2ex>[r]_{X\mapsto Y} &
\mbox{} \ar@<-3ex>@/_/[l]_{f\mapsto\id_X} \\ } &
\xymatrix{
X \ar@(lu,ru)^(.6){\id_X} \ar[r]^{f} &
Y \ar@(lu,ru)^(.6){\id_Y} \\} &
\xymatrix@C=3pc{
\mbox{} \ar[r]^{f\mapsto f} \ar@<-2ex>[r]_{f\mapsto g}
\ar@<3ex>@/^/[r]^{f\mapsto g\circ f} &
\mbox{} \\} &
\xymatrix{
X \ar@(lu,ru)^(.6){\id_X} \ar[r]^{f} \ar@/_4ex/[rr]_{g\circ f} &
Y \ar@(lu,ru)^(.6){\id_Y} \ar[r]^{g} &
Z \ar@(lu,ru)^(.6){\id_Z} \\} \\
\cline{1-1}\cline{3-3} \cline{5-5}
\end{array}$$
\subsubsection*{$\bullet$ Theories}
We build a limit sketch $\skE'_{\cat}$
by adding to $\skE_{\cat}$ for each $n\in\bN$ the following potential limits,
with vertex $\nType$ for $n$-tuples of types and $\nCone$ for $n$-ary discrete cones:
$$
\xymatrix@=1pc{
& \nType \ar[dl]_{\prbone} \ar[dr]^{\prbn} & \\
\Type & \dots \qquad \dots & \Type \\
} \qquad\qquad
\xymatrix@=1pc{
& \nCone \ar[dl]_{\prcone} \ar[dr]^{\prcn} \ar[dd]|(.4){\vertex} & \\
\Term \ar[rd]_{\dom} & \dots \qquad \dots & \Term \ar[ld]^{\dom} \\
& \Type & \\
}
$$
The arrow $\vertex$ may be omitted when $n>0$.
When $n=0$, the potential limits mean that
$\zType$ is a unit type (also denoted $\Unit$)
and $\zCone$ is isomorphic to $\Type$.
We also add the tuple
$\nbase=\tuple{\codom\circ\prcone, \dots, \codom\circ\prcn}$
which maps each cone to its base:
$$
\xymatrix@C=4pc{
\nType & \nCone \ar[l]_{\nbase}
}
$$
The realizations of $\skE'_{\cat}$ are still the categories.
Now, a limit sketch $\skE_{\thry}$ for equational theories is obtained by adding to $\skE'_{\cat}$
the following features, for each $n\in\bN$.
First an arrow $\nprod\colon \nType\to\nCone$
together with the equation $\nbase\circ\nprod=\ttid_{\nType}$,
for building the product cone of each family of $n$ types.
Then for building tuples, an arrow $\ntuple\colon \nCone\to\Term$
together with the equations $\dom\circ\ntuple=\vertex$
and $\codom\circ\ntuple=\vertex\circ\nprod\circ\nbase$
and with several additional equations for ensuring that the universal property of a product
is satisfied.
So, here is a relevant part of this limit sketch $\skE_{\thry}$ for theories
(equations are omitted, and only one arity $n$ is represented):
$$
\xymatrix@C=4pc@R=3pc{
\;\Type\; \ar@<1ex>@/^/[r]^{\selid} &
\;\Term\; \ar[l]_{\dom} \ar@<1ex>[l]^{\codom} &
\;\Cons\; \ar[l]_{\fst} \ar@<1ex>[l]^{\snd} \ar@<-1ex>@/_/[l]_{\comp} \\
\;\nType\; \ar@<1ex>[u]^{\prbone} \ar@{}[u]|{\dots} \ar@<-1ex>[u]_{\prbn}
\ar@<-1ex>@/_/[r]_{\nprod} &
\;\nCone\; \ar@<1ex>[u]^{\prcone} \ar@{}[u]|{\dots} \ar@<-1ex>[u]_{\prcn}
\ar@<-2ex>@/_/[u]_{\ntuple} \ar[l]_{\nbase} \ar[ul]|{\vertex} &
} $$
Let us focus on the following part of $\skE_{\thry}$:
$$
\xymatrix@C=4pc{
\;\bType\; \ar@<-1ex>@/_/[r]_{\bprod} &
\;\bCone\; \ar[l]_{\bbase} \ar@/_/[r]_{\btuple} &
\;\Term\; \\
}
$$
and its image by the Yoneda contravariant realization (only presentations are given):
$$ \begin{array}{|c|c|c|c|c|}
\cline{1-1}\cline{3-3}\cline{5-5}
\xymatrix@C=3pc@R=1pc{ X & \\
& X\stimes Y \ar[lu]_{p} \ar[ld]^{q} \\
Y & \\ } &
\xymatrix@C=3pc@R=1pc{ \\
\mbox{} \ar@<1ex>[r]^{\subseteq} &
\mbox{} \ar@<1ex>@/^/[l]^{Z\mapsto X\stimes Y} \\ } &
\xymatrix@C=3pc@R=1pc{ & X & \\
Z \ar[ru]^{f} \ar[rd]_{g} \ar[rr]|{\,\tuple{f,g}\,} & \ar@{}[u]|{=}\ar@{}[d]|{=} &
X\stimes Y \ar[lu]_{p} \ar[ld]^{q} \\ & Y & \\ } &
\xymatrix@C=3pc@R=1pc{ \\
\mbox{} &
\mbox{} \ar@/^/[l]^{h\mapsto \tuple{f,g}} \\ } &
\xymatrix@C=3pc@R=1pc{ \mbox{} \\ W \ar[r]^{h} & T \\ } \\
\cline{1-1}\cline{3-3}\cline{5-5}
\end{array}$$
\subsection{The parameterization process is a free functor}
\label{subsec:sk-free}
\subsubsection*{$\bullet$ Parameterized theories}
A limit sketch $\skE_A$ for parameterized theories is obtained
by adding to $\skE_{\thry}$ an arrow $\ttA \colon \Unit \to \Type$.
\subsubsection*{$\bullet$ Decorated theories}
A limit sketch $\skE_{\dec}$ for decorated theories comes with
a morphism $\ske_{\undec}\colon \skE_{\dec}\to\skE_{\thry}$
which forgets about the decorations
(``$\mathit{undec}$'' for ``undecoration'').
Here are two slightly different choices,
the first one is simpler but the second one better reflects the idea of decoration.
A limit sketch $\skE_{\dec}$ for decorated theories is made of
two related copies of $\skE_{\thry}$:
one copy $\skE\pur$ for the pure features
and another copy $\skE\gen$ for the general features,
together with a monomorphic transition arrow $\ttt_{\ttE} \colon \ttE\pur \to \ttE\gen$
for each point $\ttE$ in $\skE_{\thry}$ with $\ttt_{\Type}$ an identity
and with the transition equations
$\ttt_{\ttE'}\circ\tte\pur=\tte\gen\circ\ttt_{\ttE}$
for each arrow $\tte\colon \ttE\to\ttE'$ in $\skE_{\thry}$.
The morphism $\ske_{\undec}$ maps both copies $\skE\pur$ and $\skE\gen$ to $\skE$.
Another limit sketch for decorated theories, still denoted $\skE_{\dec}$,
is the \emph{sketch of elements} (similar to the more usual \emph{category of elements})
of a model $\catD$ of $\skE_{\thry}$ with values in $\catT_{\eq}$,
then the morphism $\ske_{\undec}$ is provided by the construction.
This model $\catD$ formalizes the fact that
identities and projections are always pure
while the composition or pairing of pure terms is pure.
Precisely, the theory $\catD(\Type)$ is generated by one type $D$
and the theory $\catD(\Term)$ by two types $p$, $g$
(for ``pure'' and ``general'', respectively) and a monomorphism $p\to g$
(for ``every pure term can be seen as a general term'').
As for the functors, $\catD(\selid)$ maps $D$ to $p$,
$\catD(\nprod)$ maps $\tuple{D,\dots,D}$ to $p$,
while $\catD(\comp)$ maps
$\tuple{p,p}$ to $p$ and $\tuple{p,g},\tuple{g,p},\tuple{g,g}$ to $g$
and $\catD(\ntuple)$ maps
$\tuple{p,\dots,p}$ to $p$ and everything else to $g$.
The resulting sketch of elements $\skE_{\dec}$ is made of
one point $\Type\mathtt{.D}$ over the point $\Type$ of $\skE_{\thry}$,
two points $\Term\pur$ and $\Term\gen$ over the point $\Term$ of $\skE_{\thry}$,
four points over $\Cons$, $2^n$ over $\nType$ and $\nCone$,
a monomorphic arrow $\Term\pur \to \Term\gen$, and so on.
\subsubsection*{$\bullet$ From decorated theories to parameterized theories}
Let us consider the functors $F_{\expan}\colon \catT_{\dec}\to\catT_A$
and $G_{\expan}\colon \catT_A\to\catT_{\dec}$ from section~\ref{subsec:gene}.
We can now prove that $G_{\expan}$ is a forgetful functor
and $F_{\expan}$ is its associated free functor,
in the sense of definition~\ref{defi:free}.
\begin{theo}
\label{theo:free}
There is a morphism of limit sketches $\ske_{\expan}\colon \skE_{\dec} \to \skE_A$
such that the associated adjunction is $F_{\expan}\dashv G_{\expan}$
from section~\ref{subsec:gene}.
\end{theo}
\begin{proof}
In section~\ref{subsec:gene} the functor $F_{\expan}$ has been defined
on $\funY(\skE_{\dec})$ (see figure~\ref{fig:F-elem}).
Since $F_{\expan}$ extends $\ske_{\expan}$ via the Yoneda contravariant realizations,
this provides the definition of a unique morphism $\ske_{\expan}$
with associated left adjoint $F_{\expan}$.
For instance, the point $\Term\pur$ is mapped to $\Term$
and the point $\Term\gen$ to the point of $\skE_A$
characterized by the fact
that its image by Yoneda is presented by $X$, $Y$ and $f'\colon A\times X\to Y$.
Then it is easy to check that the precomposition with $\ske_{\expan}$
is the functor $G_{\expan}$.
\end{proof}
\subsubsection*{$\bullet$ A span of limit sketches}
Altogether, the following span of limit sketches provides a framework
for the process that starts from an equational theory,
choose the pure terms, and forms the corresponding parameterized theory:
$$ \xymatrix@C=4pc {
\skE_{\eq} & \skE_{\dec} \ar[l]_{\ske_{\undec}} \ar[r]^{\ske_{\expan}} & \skE_A \\
\\ } $$
\subsection{A limit sketch for equational specifications}
\label{subsec:sk-spec}
In this section we build a limit sketch $\skE_{\spec}$ for equational specifications
from the limit sketch $\skE_{\thry}$ for theories,
thus providing another point of view on the elementary specifications
in section~\ref{subsec:equa}.
This construction can be seen as an illustration of the
factorization theorem in \cite{Du03}.
A direct detailed construction of a limit sketch for equational specifications
can be found in appendix~\ref{subsec:dia-spec}.
In the part of $\skE_{\thry}$ shown in section~\ref{subsec:sk-equa}
there are four arrows that are neither in $\skE_{\gr}$ nor
projections in a potential limit:
$\selid$, $\comp$, $\nprod$, $\ntuple$.
These arrows stand for features that are always defined in a theory
but only partially defined in a specification.
So, $\skE_{\spec}$ is obtained by replacing each of these arrows
$\tte\colon \ttE_1\to\ttE_2$ by a span:
$$\xymatrix@C=3pc{
\ttE_1 & \;\;\ttE'_1\;\; \ar@{>->}[l]_{\tte'_1} \ar[r]^{\tte'} & \ttE_2 \\
}$$
where the arrow ``$\rightarrowtail$'' stands for a potential monomorphism
(which can be expressed as a potential limit).
So, here is (a relevant part of) $\skE_{\spec}$:
$$
\xymatrix@C=4pc@R=3pc{
\;\Selid\;\; \ar@{>->}[r] \ar@/^4ex/[rr]^{\selid} &
\;\Type\; &
\;\Term\; \ar[l]_{\dom} \ar@<1ex>[l]^{\codom} &
\;\Cons\; \ar[l]_{\fst} \ar@<1ex>[l]^{\snd} &
\;\;\Comp\; \ar@{>->}[l] \ar@/_4ex/[ll]_{\comp} \\
\;\nProd\;\; \ar@{>->}[r] \ar@/_4ex/[rr]_{\nprod} &
\;\nType\; \ar@<1ex>[u]^{\prbone} \ar@{}[u]|{\dots} \ar@<-1ex>[u]_{\prbn} &
\;\nCone\; \ar@<1ex>[u]^{\prcone} \ar@{}[u]|{\dots} \ar@<-1ex>[u]_{\prcn}
\ar[l]_{\nbase} \ar[ul]|{\vertex} &
\;\;\nTuple\; \ar@{>->}[l] \ar@/^/[ul]_(.4){\ntuple} & \\
} $$
The elementary specifications from section~\ref{subsec:equa}
are the images by the Yoneda contravariant realization
of the points in $\skE_{\spec}$ which are not vertices of potential limits, namely:
$\Type$, $\Term$, $\Selid$, $\Comp$, $\nProd$, $\nTuple$:
our notations are such that $\Ss_{\ttE}=\funY(\ttE)$ for each of these points $\ttE$.
\section{Conclusion}
This paper provides a neat categorical formalization for
the parameterization process in Kenzo and EAT.
Future work includes the generalization of this approach
from equational theories to other families of theories, like distributive categories,
and to more general kinds of parameters, like data types. | 0.002952 |
Abstract. Bioabsorbable and functionally graded apatites (fg-HAp) ceramics were designed using calcined bovine bone (b-HAp) by the partial dissolution-precipitation methods. The fg-HAp ceramics with micro-pores of 10-160 nm had larger specific surface areas (30-40 m2･g-1) than the b-HAp ceramics, although the two HAp ceramics exhibited same macro-pore sizes of 100-600 µm and porosities of 60-80 %. Surface structure of these ceramics was modified by soaking at 309.5 K for 1-90 days in a simulated body fluid (SBF). At 8 days after the soaking, microstructure of the fg-HAp changed from small grains to dense cocoon-like ones by rapid precipitation of HAp microcrystals, while at 14 days, that of the b-HAp was porous urchin-like grains, suggesting that the fg-HAp had higher bone-bonding ability than the b-HAp. | 0.029084 |
\begin{document}
\title{On U-Statistics and Compressed Sensing I: Non-Asymptotic Average-Case Analysis}
\author{
\authorblockN{Fabian Lim${}^*$ and Vladimir Marko Stojanovic}\\
\ifthenelse{\boolean{dcol}}{
\thanks{F. Lim and V. M. Stojanovic are with the Research Laboratory of Electronics,
Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139.
Email: \{flim,vlada\}@mit.edu.
This work was supported by NSF grant ECCS-1128226. \newline
\indent Part of this work will be presented at the 2012 IEEE International Conference on Communications (ICC), Ottawa, Canada.}
}{
\authorblockA{Research Laboratory of Electronics\\ Massachusetts Institute of Technology,
77 Massachusetts Avenue, Cambridge, MA 02139\\
\{flim,vlada\}@mit.edu
\thanks{This work was supported by NSF grant ECCS-1128226.
Part of this work will be presented at the 2012 IEEE International Conference on Communications (ICC), Ottawa, Canada.}
}}
}
\maketitle
\begin{abstract}
Hoeffding's U-statistics model combinatorial-type matrix parameters (appearing in CS theory) in a natural way.
This paper proposes using these statistics for analyzing random compressed sensing matrices, in the non-asymptotic regime (relevant to practice).
The aim is to address certain pessimisms of ``worst-case'' \emph{restricted isometry} analyses, as observed by both Blanchard \& Dossal,~et.~al.
We show how U-statistics can obtain ``average-case'' analyses, by relating to \emph{statistical restricted isometry property} (StRIP) type recovery guarantees.
However unlike standard StRIP, random signal models are not required; the analysis here holds in the \emph{almost sure} (probabilistic) sense.
For Gaussian/bounded entry matrices,
we show that both $\ell_1$-minimization and LASSO essentially require on the order of $k \cdot [\log((n-k)/u) + \sqrt{2(k/n) \log(n/k)}]$ measurements
to respectively recover at least $1-5u$ fraction, and $1-4u$ fraction, of the signals.
Noisy conditions are considered.
Empirical evidence suggests our analysis to compare well to Donoho \& Tanner's recent large deviation bounds for $\ell_0/\ell_1$-equivalence, in the regime of block lengths $1000 \sim 3000$ with high undersampling ($50\sim 150$ measurements); similar system sizes are found in recent CS implementation.
In this work, it is assumed throughout that matrix columns are independently sampled.
\end{abstract}
\begin{IEEEkeywords}
approximation, compressed sensing, satistics, random matrices
\end{IEEEkeywords}
\ifthenelse{\boolean{dcol}}{
\markboth{Lim and Stojanovic: On U-Statistics and Compressed Sensing I: Non-Asymptotic Average-Case Analysis}{}
}{
\markboth{IEEE Transactions on Signal Processing, Lim and Stojanovic, Prepared using \LaTeX}{}
}
\section{Introduction}
Compressed sensing (CS) analysis involves relatively recent results from random matrix theory~\cite{Candes}, whereby recovery guarantees are framed in the context of matrix parameters known as \emph{restricted isometry constants}.
Other matrix parameters are also often studied in CS.
Earlier work on sparse approximation considered a matrix parameter known as \emph{mutual coherence}~\cite{Grib,Elad,Tropp}.
Fuchs' work on \emph{Karush-Kuhn-Tucker (KKT)} conditions for sparsity pattern recovery considered a parameter involving a matrix \emph{pseudoinverse}~\cite{Fuchs}, re-occurring in recent work~\cite{Tropp,Cand2008,Barga}.
Finally, the \emph{null-space property}~\cite{Zhang,Cohen2008,DAspremont2009} is gaining recent popularity - being the parameter closest related to the fundamental compression limit dictated by \emph{Gel'fand widths}.
All above parameters share a similar feature, that is they are defined over subsets of a certain fixed size $k$.
This combinatorial nature makes them difficult to evaluate, even for moderate block lengths $n$. Most CS work therefore involve some form of randomization to help the analysis.
While the celebrated $k \log (n/k)$ result was initially approached via asymptotics,
\textit{e.g.},~\cite{Candes,Donoho,RIPsimp,Blanchard}, implementations require finite block sizes.
Hence, non-asymptotic analyses are more application relevant.
In the same practical aspect, recent work deals with non-asymptotic analysis of \emph{deterministic} CS matrices, see~\cite{Calder,Gan,Tropp,Barga}.
On the other hand certain situations may not allow control over the sampling process, whereby the sampling may be inherently random, \textit{e.g.}, prediction of clinical outcomes of various tumors based on gene expressions~\cite{Cand2008}.
Random sampling has certain desirable simplicity/efficiency features -
see~\cite{Sartipi2011} on data acquisition in the distributed sensor setting.
Also recent hardware implementations point out energy/complexity-cost benefits of implementing \emph{pseudo-random binary sequences}~\cite{Fred,Mishali,EPFL}; these sequences mimic statistical behavior.
Non-asymptotic analysis is particularly valuable, when random samples are costly to acquire. For example, each clinical trial could be expensive to conduct an excessive number of times.
In the systems setting, the application could be running on a tight energy budget - whereby processing/communication costs depend on the number of samples acquired.
This work is inspired by the \emph{statistical} notion of the restricted isometry property (StRIP), initially developed for deterministic CS analysis~\cite{Calder,Gan}. The idea is to relax the analysis, by allowing sampling matrix parameters (that guarantee signal recovery) to be satisfied for a \emph{fraction} of subsets.
Our interest is in ``average-case'' notions in the context of randomized sampling, reason being that certain pessimisms of
``worst-case'' restricted isometry analyses have been observed in past works~\cite{Blanchard,Dossal2009,Bah}.
On the other hand in~\cite{DT}, Donoho \& Tanner remarked on potential benefits of the above ``average-case'' notion, recently pursued in an adaptation of a previous asymptotic result~\cite{DTExp}.
In the multichannel setting, ``average-case'' notions are employed to make analysis more tractable~\cite{F2007,Eldar2009}.
In~\cite{Epfl2008}~a simple ``thresholding'' algorithm is analyzed via an ``average'' coherence parameter.
However the works in this respect are few, most random analyses are of the ``worst-case'' type, see~\cite{Rudel,RIPsimp,Blanchard,Bah}.
We investigate the unexplored, with the aim of providing new insights and obtaining new/improved results for the ``average-case''.
Here we consider a random analysis tool that is well-suited to the CS context, yet seemingly left untouched in the literature.
Our approach differs from that of deterministic matrices, where ``average-case'' analysis is typically made accessible via mutual coherence, see~\cite{Calder,Gan,Mishali}.
For random matrices, we propose an alternative approach via \emph{U-statistics}, which do not require random signal models typically introduced in StRIP analysis, see~\cite{Calder,Epfl2008,Eldar2009}; here, the results are stated in the \emph{almost sure} sense.
U-statistics apply naturally to various kinds of non-asymptotic CS analyses, since they are designed for combinatorial-type parameters.
Also, they have a natural ``average-case'' interpretation, which we apply to recent recovery guarantees
that share the same ``average-case'' characteristic.
Finally thanks to the wealth of U-statistical literature, the theory developed here is open to other extensions, \textit{e.g.}, in related work~\cite{Lim2} we demonstrate how U-statistics may also perform ``worst-case'' analysis.
\textbf{Contributions}:
``Average-case'' analyses are developed based on U-statistics, which are i) empirically observed to have good potential for predicting CS recovery in non-asymptotic regimes, and ii) theoretically obtain measurement rates that incorporate a non-zero failure rate (similar to the $k\log (n/k)$ rate from ``worst-case'' analyses).
We utilize a
U-statistical large deviation concentration theorem,
under the assumption that the matrix columns are independently sampled.
The large deviation error bound holds \emph{almost surely} (Theorem \ref{thm:Ustat}). No random signal model is needed, and the error is of the order $(n/k)^{-1} \log (n/k)$, whereby $k$ is the U-statistic kernel size (and $k$ also equals sparsity level).
Gaussian/bounded entry matrices are considered.
For concreteness, we connect with StRIP-type guarantees (from~\cite{Barga,Cand2008}) to study the fraction of recoverable signals (\textit{i.e.}, ``average-case'' recovery) of: i) \emph{$\ell_1$-minimization} and ii) \emph{least absolute shrinkage and selection operator (LASSO)}, under noisy conditions.
For both these algorithms we show $\const\cdot k [\log((n-k)/u) + \sqrt{2(k/n) \log(n/k)}]$
measurements are essentially required, to respectively recover at least $1-5u$ fraction (Theorem \ref{thm:L1}), and $1-4u$ fraction (Theorem \ref{thm:Lasso}), of possible signals.
This is improved to $1-3u$ fraction for the noiseless case.
Here $\const = \max(4/(a_1 a_2)^2, 2c_1/(0.29-a_1)^2)$ for to be specified constants $a_1,a_2,c_1$, where $c_1$ depends on the distribution of matrix entries.
Note that the term $\sqrt{2(k/n) \log(n/k)}$ is at most 1 and vanishes with small $k/n$.
Empirical evidence suggests that our approach compares well with recent results from Donoho \& Tanner~\cite{DTExp} - improvement is suggested for system sizes found in implementations~\cite{Fred}, with large undersampling (\textit{i.e.}, $m = 50 \sim 100$ and $n = 1000 \sim 3000$).
The large deviation analysis here does show some pessimism in the size of $\const$ above, whereby $\const \geq 4$ (we conjecture possible improvement).
For Gaussian/Bernoulli matrices, we find $\const \approx 1.8$ to be inherently smaller, \textit{e.g.}, for $k=4$ this predicts recovery of $1\times 10^{-6}$ fraction with $153$ measurements - empirically $m= 150$.
\textbf{Note}:
StRIP-type guarantees~\cite{Barga,Cand2008} seem to work well, by simply \emph{not} placing restrictive conditions on the maximum eigenvalues of the size-$k$ submatrices.
Our theory applies fairly well for various considered system sizes $k,m,n$ (\textit{e.g.}, Figure \ref{fig:CompL1}), however in \emph{noisy} situations, a (relatively small) factor of $\sqrt{k}$ losses is seen without making certain maximum eigenvalue assumptions.
For $\ell_1$-recovery, the estimation error is now bounded by a $\sqrt{k}$ factor of its best $k$-term approximation error (both errors measured using the $\ell_1$-norm).
For LASSO, the
the non-zero signal magnitudes must now be bounded below by a factor $\sqrt{2 k \log n}$ (with respect to noise standard deviation), as opposed to $\sqrt{2 \log n}$ in~\cite{Cand2008}.
These losses occur not because of StRIP analyses, but because of the estimation techniques employed here.
\textbf{Organization}:
We begin with relevant background on CS in Section \ref{sect:params}. In Section \ref{sect:Dist} we present a general U-statistical theorem for large-deviation (``average-case'') behavior.
In Section \ref{sect:KKT} the U-statistical machinery is applied to StRIP-type ``average-case'' recovery.
We conclude in Section \ref{sect:conc}.
\newcommand{\Real}{\mathbb{R}}
\newcommand{\eigm}{\sigma^2_{\scriptsize \mbox{\upshape min}}}
\newcommand{\eigM}{\sigma^2_{\scriptsize \mbox{\upshape max}}}
\newcommand{\sigm}{\sigma_{\scriptsize \mbox{\upshape min}}}
\newcommand{\sigM}{\sigma_{\scriptsize \mbox{\upshape max}}}
\newcommand{\Tr}{ \mbox{\upshape Tr}}
\newcommand{\Eigm}{\varsigma_{\scriptsize \mbox{\upshape min}}}
\newcommand{\EigM}{\varsigma_{\scriptsize \mbox{\upshape max}}}
\textbf{Notation}:
The set of real numbers is denoted $\Real$.
Deterministic quantities are denoted using $a,\mat{a}$, or $\mat{A}$, where bold fonts denote vectors (\textit{i.e.}, $\mat{a}$) or matrices (\textit{i.e.}, $\mat{A}$).
Random quantities are denoted using upper-case \emph{italics}, where $A$ is a random variable (RV), and $\pmb{A}$ a random vector/matrix.
Let $\Pr\{A\leq a\}$ denote the probability that event $\{A\leq a\}$ occurs.
Sets are denoted using braces, \textit{e.g.}, $\{1,2,\cdots\}$.
The notation $\mathbb{E}$ denotes expectation.
The notation $i,j,\ell,\omega$ is used for indexing.
We let $\Norm{\cdot}{p}$ denote the $\ell_p$-norm for $p=1$ and $2$.
\newcommand{\x}{\alpha}
\newcommand{\xbk}{\overline{\pmb{\alpha}}_{k}}
\section{Preliminaries} \label{sect:params}
\subsection{Compressed Sensing (CS) Theory} \label{ssect:params}
\newcommand{\Sens}{\pmb{\Phi}}
\renewcommand{\S}{\mathcal{S}}
\newcommand{\Fm}{F_{\sigma^2_{\scriptsize \mbox{\upshape min}}}}
\newcommand{\FM}{F_{\sigma^2_{\scriptsize \mbox{\upshape max}}}}
\renewcommand{\a}{a}
\newcommand{\matt}[1]{\pmb{#1}}
\newcommand{\Bas}{\mat{D}}
A vector $\mat{a}$ is said to be $k$-sparse, if at most $k$ vector coefficients are non-zero (\emph{i.e.}, its $\ell_0$-distance satisfies $\Norm{\mat{\a}}{0} \leq k$).
Let $n$ be a positive integer that denotes {block length}, and let $\matt{\x}=[\x_1,\x_2,\cdots, \x_n]^T$ denote a length-$n$ signal vector with signal coefficients $\x_i$.
The \emph{best $k$-term approximation} $\xbk $ of $\matt{\x}$, is obtained by finding the $k$-sparse vector $\xbk $ that has minimal approximation error $\Norm{\xbk - \matt{\x}}{2}$.
\newcommand{\col}{\pmb{\phi}}
\newcommand{\y}{b}
\newcommand{\eps}{\epsilon}
Let $\Sens$ denote an $m\times n$ CS sampling matrix, where $m < n$.
The length-$m$ \textbf{measurement vector} denoted $\mat{\y}=[\y_1,\y_2,\cdots, \y_m]^T $ of some length-$n$ signal $\matt{\x}$, is formed as $\mat{\y} =\Sens \matt{\x}$.
Recovering $\matt{\x}$ from $\mat{\y}$ is challenging as $\Sens$ possesses a \emph{non-trivial null-space}.
We typically recover $\matt{\x}$ by solving the (convex) $\ell_1$-\textbf{minimization} problem
\bea
\min_{\tilde{\matt{\x}} \in \Real^n} \Norm{\tilde{\matt{\x}}}{1}~~~\mbox{ s. t. } \Norm{\tilde{\mat{\y}} - \Sens \tilde{\matt{\x}}}{2} \leq \eps. \label{eqn:L1}
\eea
The vector $\tilde{\mat{\y}}$ is a \emph{noisy} version of the original measurements $\mat{\y}$, and here $\eps$ bounds the noise error, \emph{i.e.}, $\eps \geq \Norm{\tilde{\mat{\y}}-\mat{\y}}{2}$.
Recovery conditions have been considered in many flavors~\cite{Donoho,DTExp,Grib,Elad,DT}, and mostly rely on studying parameters of the sampling matrix $\Sens$.
\newcommand{\RIC}{\delta}
For $k \leq n$, the $k$-th \textbf{restricted isometry constant} $\RIC_k$ of an $m\times n$ matrix $\Sens$, equals the smallest constant that satisfies
\bea
(1-\RIC_k) \Norm{\matt{\x}}{2}^2 \leq \Norm{\Sens\matt{\x}}{2}^2 \leq (1+\RIC_k) \Norm{\matt{\x}}{2}^2, \label{eqn:RIC}
\eea
for any $k$-sparse $\matt{\x} \mbox{ in } \Real^n$.
The following well-known recovery guarantee is stated w.r.t. $\RIC_k$ in (\ref{eqn:RIC}).
\newcommand{\ConstZero}{c_1}
\newcommand{\ConstOne}{c_2}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\RIPthm}{A}
\begin{thmN}{\RIPthm, \textit{c.f.},~\cite{RIP}}
Let $\Sens$ be the sensing matrix.
Let $\matt{\x}$ denote the signal vector.
Let $\mat{\y}$ be the measurements, \emph{i.e.}, $\mat{\y} = \Sens\matt{\x}$.
Assume that the $(2k)$-th restricted isometry constant $\RIC_{2k}$ of $\Sens$ satisfies $\RIC_{2k} < \sqrt{2} -1$, and further assume that the noisy version $\tilde{\mat{\y}}$ of $\mat{\y}$ satisfies $ \Norm{\tilde{\mat{\y}}-\mat{\y}}{2} \leq \epsilon$. Let $\xbk$ denote the best-$k$ approximation to $\matt{\x}$. Then the $\ell_1$-minimum solution $\matt{\x}^*$ to (\ref{eqn:L1}) satisfies
\[
\Norm{\matt{\x}^* - \matt{\x}}{1} \leq \ConstZero \Norm{\matt{\x} - \xbk}{1} + \ConstOne \epsilon,
\]
for small constants $\ConstZero = 4\sqrt{1+\RIC_{2k}}/(1-\RIC_{2k}(1+\sqrt{2}))$ and $\ConstOne = 2 (\RIC_{2k} (1-\sqrt{2}) - 1)/(\RIC_{2k} (1+\sqrt{2}) - 1) $.
\end{thmN}
\newcommand{\Gram}{\Sens^T\Sens}
\newcommand{\ord}[1]{{(#1)}}
\newcommand{\GramS}{\Sens^T_\S\Sens_\S}
\newcommand{\SensS}{\Sens_\S}
\newcommand{\SensSi}{\Sens_{\S_i}}
\newcommand{\GramSi}{\Sens^T_{\S_i}\Sens_{\S_i}}
\newcommand{\GramSoi}[1]{\Sens_{\S_\ord{#1}}}
\newcommand{\GrampSoi}[1]{\Sens'_{\S_\ord{#1}}}
\newcommand{\N}{N}
\newcommand{\Bin}[2]{{#1 \choose #2}}
\newcommand{\define}{\stackrel{\Delta}{=}}
\renewcommand{\fn}{\footnote{We aim to relax this fairly restrictive assumption in future work.}}
\renewcommand{\fn}{\footnote{For simplicity, we omitted small deviation constants in Theorem B, see~\cite{Candes2004} p. 18 for details.}}
\newcommand{\A}{\pmb{A}}
Theorem \RIPthm~is very powerful, on condition that we know the constants $\RIC_k$. But because of their combinatoric nature, computing the restricted isometry constants $\RIC_k$ is NP-Hard~\cite{Blanchard}.
Let $\S$ denote a size-$k$ subset of indices.
Let $\SensS$ denote the size $m\times k$ submatrix of $\Sens $, indexed on (column indices) in $\S$.
Let $\eigM(\SensS)$ and $\eigm(\SensS)$ respectively denote the minimum and maximum, \emph{squared-singular values} of $\SensS$. Then from (\ref{eqn:RIC}) if the columns $\col_i$ of $\Sens$ are properly normalized, \emph{i.e.}, if $\Norm{\col_i}{2}=1$, we deduce that $\RIC_{k}$ is the smallest constant in $\Real$ that satisfies
\bea
\RIC_k &\geq& \max(\eigM(\Sens_\S) - 1, 1-\eigm(\Sens_\S)),
\label{eqn:RIC2}
\eea
for all $\Bin{n}{k}$ size-$k$ subsets $\S$.
For large $n$, the number $\Bin{n}{k}$ is huge.
Fortunately $\RIC_k$ need not be explicitly computed, if we can estimate it after incorporating \emph{randomization}~\cite{Candes,Donoho}.
\newcommand{\func}{\zeta}
\newcommand{\1}{\mathbb{1}}
\newcommand{\LedThm}{B}
Recovery guarantee Theorem \RIPthm~involves ``worst-case'' analysis.
If the inequality (\ref{eqn:RIC2}) is violated for \emph{any} one submatrix $\Sens_\S$, then the \emph{whole} matrix $\Sens$ is deemed to have restricted isometry constant larger than $\RIC_k$.
A common complaint of such ``worst-case'' analyses is pessimism, \textit{e.g.}, in~\cite{Dossal2009} it is found that for $n=4000$ and $m=1000$, the restricted isometry property is not even satisfied for sparsity $k=5$.
This motivates the ``average-case'' analysis investigated here, where the recovery guarantee is relaxed to hold for a large ``fraction'' of signals (useful in applications that do not demand all possible signals to be completely recovered).
We draw ideas from the statistical StRIP notion used in deterministic CS, which only require ``most'' of the submatrices $\Sens_\S$ to satisfy some properties.
In statistics, a well-known notion of a U-statistic (introduced in the next subsection) is very similar to StRIP.
We will show how U-statistics naturally lead to ``average-case'' analysis.
\subsection{U-statistics \& StRIP} \label{ssect:UstatStr}
\newcommand{\E}{\mathbb{E}}
\newcommand{\g}{g}
\newcommand{\ZO}{\Real_{[0,1]}}
A function $\func : \Real^{m \times k} \rightarrow \Real$ is said to be a \textbf{kernel},
if for any $\mat{A},\mat{A}' \in \Real^{m \times k}$, we have $\func(\mat{A}) = \func(\mat{A}')$ if matrix ${\mat{A}}'$ can be obtained from $\mat{A}$ by \emph{column reordering}.
Let $\ZO$ be the set of real numbers bounded below by $0$ and above by $1$, i.e., $\ZO = \{a \in \Real : 0 \leq a \leq 1\}$.
U-statistics are associated with functions $g : \Real^{m \times k} \times \Real \rightarrow \ZO$ known as \textbf{bounded kernels}.
To obtain bounded kernels $g$ from indicator functions, simply use some kernel $\func$ and set
$g(\mat{A},a) = \Ind{\func(\mat{A})\leq a}$ or $g(\mat{A},a) = \Ind{\func(\mat{A}) > a}$, \textit{e.g.} $\1\{\sigM^2(\mat{A}) \leq a\}$.
\newcommand{\V}{U}
\begin{defn}[Bounded Kernel U-Statistics] \label{def:Ustat}
Let $\A$ be a random matrix with $n$ columns. Let $\Sens$ be sampled as $\Sens=\A$.
Let $g : \Real^{m \times k} \times \Real \mapsto \ZO$ be a bounded kernel. For any $a\in \Real$, the following quantity
\bea
\U_n(a) \define \frac{1}{\Bin{n}{k}} \sum_{\S} g(\Sens_{\S}, a) \label{eqn:Ustat}
\eea
is a U-statistic of the sampled realization $\Sens=\A$, corresponding to the kernel $g$.
In
(\ref{eqn:Ustat}), the matrix $\SensS$ is the submatrix of $\Sens$ indexed on column indices in $\S$, and the sum takes place over all subsets $\S$ in $\{1,2,\cdots, n\}$.
Note, $0 \leq U_n(a) \leq 1$.
\end{defn}
For $k\leq n$ and positive $u$ where $u\leq 1$, a matrix $\Sens$ has $u$-\textbf{StRIP constant} $\RIC_k$, if $\RIC_k$ is the smallest constant s.t.
\bea
(1-\RIC_k) \Norm{\matt{\x}}{2}^2 \leq \Norm{\Sens_\S\matt{\x}}{2}^2 \leq (1+\RIC_k) \Norm{\matt{\x}}{2}^2, \label{eqn:SRIC}
\eea
for any $\matt{\x}\in\Real^k $ and fraction $u$ of size-$k$ subsets $\S$.
The difference between (\ref{eqn:SRIC}) and (\ref{eqn:RIC}) is that $\Sens_S$ is in place of $\Sens$.
This StRIP notion coincides with~\cite{Barga}.
Consider $\func(\mat{A}) = \max(\eigM(\mat{A}) -1,1 -\eigm(\mat{A}) )$ where here $\func$ is a kernel.
Obtain a bounded kernel $g$ by setting $g(\mat{A},a) = \1\{\func(\mat{A}) > a \}$.
Construct a U-statistic $\U_n(\RIC)$ of $\Sens$ the form $\U_n(\RIC)= \Bin{n}{k}^{-1}\sum_\S \1\{\func(\SensS) > \RIC \}$.
Then if this U-statistic satisfies $U_n(\RIC)=1-u$, the $u$-StRIP constant $\RIC_k$ of $\Sens$ is at most $\RIC$, \emph{i.e.}, $\RIC_k \leq \RIC$.
\renewcommand{\fn}{\footnote{If $\mat{A}$ has full column rank, then $\mat{A}^\dagger = (\mat{A}^T\mat{A})^{-1} \mat{A}^T$,}}
To exploit apparent similarities between U-statistics and StRIP, we turn to two ``average-case'' guarantees found in the StRIP literature.
In the sequel, the conditions required by these two guarantees,
will be analyzed in detail via U-statistics - for now let us recap these guarantees.
First, an $\ell_1$-minimization recovery guarantee recently given in~\cite{Barga}, is a StRIP-adapted version of the ``worst-case'' guarantee Theorem \RIPthm.
For any non-square matrix $\mat{A}$, let $\mat{A}^\dagger$ denote the \textbf{Moore-Penrose pseudoinverse}\fn.
A vector $ \matt{\beta}$ with entries in $\{-1,1\}$ is termed a \textbf{sign vector}.
For $\matt{\x}\in \Real^n$, we write $\matt{\x}_\S$ for the length-$k$ vector supported on $\S$.
Let $\S_c$ denote the complementary set of $\S$, \emph{i.e.}, $\S_c = \{1,2,\cdots, n\}\setminus \S$.
The ``average-case'' guarantees require us to check conditions on $\Sens$ for fractions of subsets $\S$, or \textbf{sign-subset} pairs $(\matt{\beta},\S)$.
\newcommand{\BargThm}{B}
\begin{thmN}{\BargThm,~\textit{c.f.},~Lemma 3,~\cite{Barga}}
Let $\Sens$ be an $m\times n$ sensing matrix. Let $\S$ be a size-$k$ subset, and let $ \matt{\beta} \in \{-1,1\}^k$. Assume that $\Sens$ satisfies
\bitm
\item invertibility: for at least a fraction $1-u_1$ of subsets $\S$, the condition $\sigm(\SensS) > 0$ holds.
\item small projections: for at least a fraction $1-u_2$ of sign-subset pairs $(\matt{\beta},\S)$, the condition
\bea
\left|(\Sens_\S^\dagger \col_i)^T \matt{\beta} \right| \leq a_2 \mbox{ for every } i \notin \S\nonumber
\eea
holds where we assume the constant $a_2 < 1$.
\item worst-case projections: for at least a fraction $1-u_3$ of subsets $\S$, the following condition holds
\bea
\Norm{\Sens_\S^\dagger \col_i}{1} \leq a_3 \mbox{ for every } i \notin \S. \nonumber
\eea
\eitm
Then for a fraction $1-u_1-u_2-u_3$ of sign-subset pairs $(\matt{\beta},\S)$, the following error bounds are satisfied
\bea
\Norm{\matt{\x}^*_\S - \matt{\x}_\S}{1} &\leq& \frac{2a_3}{1-a_2} \Norm{\matt{\x} - \xbk}{1}, \nn
\Norm{\matt{\x}^*_{\S_c} - \matt{\x}_{\S_c}}{1} &\leq& \frac{2 }{1-a_2} \Norm{\matt{\x} - \xbk}{1}, \nonumber
\eea
where $\matt{\x}$ is a signal vector that satisfies $\sgn(\matt{\x}_\S)= \matt{\beta}$, and $\xbk$ is the best-$k$ approximation of $\matt{\x}$ and $\xbk$ is supported on $\S$, and finally $\matt{\x}^*$ is the solution to (\ref{eqn:L1}) where the measurements $\mat{\y}$ satisfy $\mat{\y} = \Sens\matt{\x}$.
\end{thmN}
\newcommand{\Reg}{\theta_n}
\newcommand{\noisesd}{c_Z}
For convenience, the proof is provided in Supplementary Material \ref{app:recov}.
The second guarantee is a StRIP-type recovery guarantee for the \emph{LASSO} estimate, based on~\cite{Cand2008} (also see~\cite{Barga}).
Consider recovery from noisy measurements
\[
\tilde{\mat{\y}} = \Sens \matt{\x}+ \mat{z},
\]
here $\mat{z}$ is a length-$m$ noise realization vector. We assume that the entries $z_i$ of $\mat{z}$, are sampled from a zero-mean Gaussian distribution with variance $\noisesd^2$.
The LASSO estimate considered in~\cite{Cand2008}, is the optimal solution $\matt{\x}^*$ of the optimization problem
\bea
\min_{\tilde{\matt{\x}} \in \Real^n} \frac{1}{2} \Norm{\tilde{\mat{\y}} - \Sens \tilde{\matt{\x}}}{2} + 2 \noisesd \cdot\Reg \Norm{\tilde{\matt{\x}}}{1}. \label{eqn:Lasso}
\eea
The $\ell_1$-regularization parameter is chosen as a \emph{product} of two terms $\noisesd$ and $ \Reg$, where we specify $\Reg=(1+a)\sqrt{2 \log n }$ for some positive $a$.
What differs from convention is that the regularization depends on the noise standard deviation $\noisesd$.
We assume $\noisesd > 0$, otherwise there will be no $\ell_1$-regularization.
\newcommand{\CandThm}{C}
\begin{thmN}{\CandThm, \textit{c.f.},~\cite{Cand2008}}
Let $\Sens$ be the $m\times n$ sensing matrix. Let $\S$ be a size-$k$ subset, and let $ \matt{\beta} \in \{-1,1\}^k$.
\bitm
\item invertability: for at least a fraction $1-u_1$ of subsets $\S$, the condition $\sigm(\SensS) > {a_1}$ holds.
\item small projections: for at least a fraction $1-u_2$ of subsets $\S$, same as Theorem \BargThm.
\item invertability projections: for at least a fraction $1-u_3$ of sign-subset pairs $(\matt{\beta},\S)$, the following condition holds
\bea
\Norm{(\SensS^T\SensS)^{-1}\matt{\beta}}{\infty} \leq a_3. \nonumber
\eea
\eitm
Let $\noisesd$ denote noise standard deviation.
Assume Gaussian noise realization $\mat{z}$ in measurements $\mat{\tilde{\y}}$, satisfy
\bitm
\item[i)] $\Norm{(\Sens_\S^T \Sens_\S)^{-1}\Sens_\S^T \mat{z}}{\infty} \leq (\noisesd \sqrt{2\log n})/a_1$, for the constant $a_1$ in the invertability condition.
\item[ii)] $\Norm{\Sens_{\S_c}^T (\mat{I} - \Sens_\S \Sens_\S^\dagger)\mat{z}}{\infty} \leq \noisesd 2\sqrt{\log n}$, where $\S_c$ is the complementary set of $\S$.
\eitm
For some positive $a$, assume that constant $a_2$ in the small projections condition, satisfies
\bea
(\sqrt{2}(1+a))^{-1} + a_2 < 1.\label{eqn:Cand1}
\eea
Then for a fraction $1-u_1-u_2-u_3$ of sign-subset pairs $(\matt{\beta},\S)$, the LASSO estimate $\matt{\x}^*$ from (\ref{eqn:Lasso}) with regularization $\Reg=(1+a)\sqrt{2 \log n }$ for the same $a$ above, will successfully recover both signs and supports of $\matt{\x}$, if
\bea
|\alpha_i| \geq \left[a_1^{-1} + 2 a_3 (1+a) \right] \cdot \noisesd\sqrt{2 \log n}~\mbox{ for all }~i \in \S \label{eqn:Cand2}
\eea
\end{thmN}
Because of some differences from~\cite{Cand2008}, we also provide the proof in Supplementary Material \ref{app:recov}.
In~\cite{Cand2008} it is shown that the noise conditions i) and ii) are satisfied with large probability at least $1 - n^{-1} (2\pi \log n)^{-\half}$ (see Proposition \ref{pro:noise} in Supplementary Material \ref{app:recov}).
Theorem \CandThm~is often referred to as a \emph{sparsity pattern recovery} result, in the sense that it guarantees recovery of the sign-subset pairs $(\matt{\beta}, \S)$ belonging to a $k$-sparse signal $\matt{\x}$.
Fuchs established some of the earlier important results, see~\cite{Fuchs,Tropp2,Fuchs2005}.
In Theorems \BargThm~and \CandThm,
observe that the \emph{invertability} condition can be easily checked using an U-statistic; simply set the bounded kernel $g$ as $g(\mat{A},a_1) = \I{\sigm(\mat{A}) \leq a_1}$ for some positive $a_1$ and measure the fraction $U_n(a_1) = u_1$.
Other conditions require slightly different kernels, to be addressed in upcoming Section \ref{sect:KKT}.
But first we first introduce the main U-statistical large deviations theorem (central to our analyses) in the next section.
\section{Large deviation theorem: ``average-case'' behavior} \label{sect:Dist}
Consider two bounded kernels $g$ defined for $\mat{A}\in \Real^{m\times k}$, corresponding to maximum and minimum squared singular values
\bea
g(\mat{A}, a) &=& \I{\eigM(\mat{A}) \leq a}, \mbox{ and } \label{eqn:geig_max} \\
g(\mat{A}, a) &=& \I{\eigm(\mat{A}) \leq a}. \label{eqn:geig_min}
\eea
Note that restricted isometry conditions (\ref{eqn:RIC}) and (\ref{eqn:SRIC}) depend on both $\eigm$ and $\eigM$ behaviors, although
the conditions in the previous StRIP-recovery guarantees Theorem \BargThm~are explicitly imposed only on $\eigm$.
See~\cite{Blanchard,Edelman} for the different behaviors and implications of these two extremal eigenvalues.
In this section we consider two U-statistics, corresponding separately to (\ref{eqn:geig_max}) and (\ref{eqn:geig_min}).
\newcommand{\gauss}{\int_{-\infty}^a (1/\sqrt{2\pi}) e^{-\t^2/2}d\t}
\renewcommand{\t}{t}
\ifthenelse{\boolean{dcol}}{
\begin{figure}[!t]
\centering
\epsfig{file={Intro.eps},width=.6\linewidth}
\caption{Gaussian measure. Concentration of U-statistic $\U_n(a)$ for squared singular value $\eigm$ and $\eigM$ kernels $g$, see (\ref{eqn:geig_max}) and (\ref{eqn:geig_min}). Shown for $m=25, k=2$ and two values of $n = 25$ and $100$.}
\label{fig:Intro}
\vspace*{-10pt}
\end{figure}
}{
\begin{figure}[!t]
\centering
\epsfig{file={Intro.eps},width=.5\linewidth}
\caption{Gaussian measure. Concentration of U-statistic $\U_n(a)$ for squared singular value $\eigm$ and $\eigM$ kernels $g$, see (\ref{eqn:geig_max}) and (\ref{eqn:geig_min}). Shown for $m=25, k=2$ and two values of $n = 25$ and $100$.}
\label{fig:Intro}
\vspace*{-10pt}
\end{figure}
}
Let $\A_i$ denote the $i$-th column of $\A$, and assume $\A_i$ to be IID.
For an bounded kernel $g$, let $p(a)$ denote the expectation $\E g(\A_\S,a)$, \textit{i.e.}, $p(a) = \E g(\A_\S,a)$ for any size-$k$ subset $\S$.
Since $p(a) = \E \U_n(a)$, thus the U-statistic mean $\E \U_n(a)$ does not depend on block length $n$.
\begin{thm} \label{thm:Ustat}
Let $\A$ be an $m\times n$ random matrix, whereby the columns $\A_i$ are IID. Let $g$ be a bounded bounded kernel that maps $\Real^{m\times k} \times \Real \rightarrow \ZO$ and let $p(a)=\E g(\A_\S,a) = \E \V_n(a)$. Let $\V_n(a)$ be a U-statistic of the sampled realization $\Sens=\A$ corresponding to the bounded kernel $g$. Then almost surely when $n$ is sufficiently large, the deviation $|\V_n(a) - p(a)| \leq \eps_n(a)$ is bounded by an error term $\eps_n(a)$ that satisfies
\bea
\eps_n^2(a) = 2 p(a) (1-p(a)) \cdot (n/k)^{-1} \log (n/k) . \label{eqn:UstatThm}
\eea
\end{thm}
Theorem \ref{thm:Ustat} is shown by piecing together (5.5) in~\cite{Hoef} and Lemma 2.1 in~\cite{Sen}. The proof is given in Appendix \ref{app:proofDev}. Figure \ref{fig:Intro} empirically illustrates this concentration result for $g$ in (\ref{eqn:geig_max}) and (\ref{eqn:geig_min}),
corresponding to $p(a) = \E g(\A_\S,a) = \Pr\{\eigM(\A_\S) \leq a\}$ and $p(a) = \Pr\{\eigm(\A_\S) \leq a\}$. Empirical simulation of restricted isometries is very difficult, thus we chose small values $k=2$, $m=25$ and block lengths $n=25$ and $n=100$. For $n=25$ the deviation $|U_{25}(a) - p(a)|$ is very noticeable for all values of $a$ and both $\eigM$ and $\eigm$. However for larger $n=100$, the deviation $|U_{100}(a) - p(a)|$ clearly becomes much smaller.
This is predicted by vanishing error $\eps_n(a)$ given in Theorem \ref{thm:Ustat}, which drops as the ratio $n/k$ increases.
In fact if $k$ is kept constant then the error behaves as $\mathcal{O}(n^{-1}\log n )$.
\ifthenelse{\boolean{dcol}}{
\begin{figure}[!t]
\centering
\epsfig{file={GaussMeans2.eps},width=1\linewidth}
\caption{Means $p(a) = \E \U_n(a)$ for predicting the concentration of $\U_n(a)$. Shown for the Gaussian case, $(a)$ $m=50$ and $(b)$ $m=150$.}
\label{fig:Gauss}
\end{figure}
}{
\begin{figure}[!t]
\centering
\epsfig{file={GaussMeans.eps},width=.8\linewidth}
\caption{Means $p(a) = \E \U_n(a)$ for predicting the concentration of $\U_n(a)$. Shown for the Gaussian case, $(a)$ $m=50$ and $(b)$ $m=150$.}
\label{fig:Gauss}
\end{figure}
}
\renewcommand{\fn}{\footnote{We point out that Bah actually defined two separate restricted isometry constants, each corresponding to $\eigm$ and $\eigM$ in~\cite{Bah}. In this paper to coincide the presentation with our discussion on squared singular values, their results will be discussed in the domain of $\eigm$ and $\eigM$.}}
\newcommand{\fnT}{\footnote{The analysis in~\cite{Bah} was performed for the large limit of $k,m$ and $n$, where both $k/m$ and $m/n$ approach fixed constants.}}
Table \ref{tab:1} reproduces\fn~a sample of (asymptotic) estimates for both $\eigM$ and $\eigm$ cases, taken from~\cite{Bah}.
These estimates are derived for ``worst-case'' analysis,
under assumption that every entry $A_{ij}$ of $\pmb{A}$ is IID and Gaussian distributed (\textit{i.e.}, $A_{ij}$ is Gaussian with variance $1/m$).
Table \ref{tab:1} presents the estimates according\fnT~to fixed ratios $k/m$ and $m/n$.
To compare, Figure \ref{fig:Gauss} shows the expectations $p(a)=\E \U_n(a)$.
The values $p(a)$ are interpreted as fractions, and as $n/k$ becomes large $p(a)$ is approached by $\U_n(a)$ within a stipulated error $\eps_n$.
Figure \ref{fig:Gauss} is empirically obtained, though note that in Gaussian case
for $p(a)$ we also have exact expressions~\cite{Edelman,Koev}, and the \emph{Bartlett decomposition}~\cite{Edelman2005}, available.
Again $p(a)$ is a marginal quantity (\textit{i.e.} does not depend on $n$)
and simulation is reasonably feasible.
In the spirit of non-asymptotics, we consider relatively small $k,m$ values as compared to other works~\cite{Bah,Dossal2009}; these adopted values are nevertheless ``practical'', in the sense they come an implementation paper~\cite{Fred}.
\begin{table}[t]
\centering
\caption{Asymptotic Lower and Upper Bounds on ``Worst-Case'' Eigenvalues,~\cite{Bah}}
\begin{tabular}{|@{\hspace{.3ex}}c@{\hspace{.3ex}}|c|c|c|c||c|c|c|}
\cline{1-8}
& & \multicolumn{3}{|c||}{Minimum: $\sigma^2_{\mbox{\scriptsize min}}$}
& \multicolumn{3}{|c|}{Maximum: $\sigma^2_{\mbox{\scriptsize max}}$} \\\cline{3-8}
& & \multicolumn{3}{|c||}{$m/n$} & \multicolumn{3}{|c|}{$m/n$} \\\cline{3-8}
& & \emph{0.1} & \emph{0.3} & \emph{0.5} & \emph{0.1} & \emph{0.3} & \emph{0.5}\\\hline
\multirow{3}{*}{\begin{sideways}$k/m$ \end{sideways}}
& \emph{0.1} & 0.095 & 0.118 & 0.130 & 3.952 & 3.610 & 3.459 \\\cline{2-8}
& \emph{0.2} & 0.015 & 0.026 & 0.034 & 5.587 & 4.892 & 4.535 \\\cline{2-8}
& \emph{0.3} & 0.003 & 0.006 & 0.010 & 6.939 & 5.806 & 5.361 \\\hline
\end{tabular}
\label{tab:1}
\end{table}
Differences are apparent from comparing ``average-case'' (Figure \ref{fig:Gauss}) and ``worst-case'' (Table \ref{tab:1}) behavior. Consider $k/m = 0.3$ where Table \ref{tab:1} shows for all undersampling ratios $m/n$, the worst-case estimate of $\eigm$ is very small, approximately $0.01$.
But for fixed $m=50$ and $m=150$, Figures \ref{fig:Gauss}$(a)$ and $(b)$ show that for respectively $k= 0.3\cdot (150)=15$ and $k=45$, a large fraction of subsets $\S$ seem to have $\eigm(\Sens_\S)$ lying above $0.1$.
From Table \ref{tab:1}, the estimates for $\eigm$ gets worse (\textit{i.e.}, gets smaller) as $m/n$ decreases.
But the error $\eps_n(a)$ in Theorem \ref{thm:Ustat} vanishes with larger $n/k$.
For the other $\eigM$ case, we similarly observe that the values in Table \ref{tab:1} also appear more ``pessimistic''.
We emphasize that Theorem \ref{thm:Ustat} holds regardless of distribution. Figure \ref{fig:BernUnif} is the counterpart figure for Bernoulli and Uniform cases (\textit{i.e.}, each entry $A_{ij}$ is respectively drawn uniformly from $\{-1/\sqrt{m},1/\sqrt{m}\}$, or $\{a \in \Real : |a| \leq \sqrt{3/m}\}$), shown for $m=50$. Minute differences are seen when comparing with previous Figure \ref{fig:Gauss}.
For $k=3$, we observe the fraction $p(a)$ corresponding to $\eigM$ to be roughly 0.95 in the latter case, whereas in the former we have roughly $0.9$ in Figure \ref{fig:BernUnif}$(a)$, and $0.88$ in Figure \ref{fig:BernUnif}$(b)$.
\begin{rem} \label{rem:1}
Exponential bounds on $\Pr\{\min_\S \sigm^2(\pmb{A}_\S) < 1-\RIC\}$ and $\Pr\{\max_\S \sigM^2(\pmb{A}_\S) > 1+\RIC\}$ for $\max(\RIC,\sqrt{k/m}) < \sqrt{2}-1$, see (\ref{eqn:RIC2}), employed in ``worst-case'' analyses, give the optimal $m = \mathcal{O}(k\log(n/k))$ rate, see~\cite{Candes,Rudelson2010,RIPsimp}.
However the implicit constants are inherently not too small (\textit{i.e.}, these constants cannot be improved).
\end{rem}
These comparisons motivate ``average-case'' analysis. Marked out on Figures \ref{fig:Gauss} and \ref{fig:BernUnif} are the ranges for which $\eigM$ and $\eigm$ must lie to apply Theorem \RIPthm~(``worst-case'' analysis). In the cases shown above, the observations are somewhat disappointing - even for small $k$ values, a substantial fraction of eigenvalues lie outside of the required range.
Thankfully, there exist ``average-case'' guarantees, \textit{e.g.}, previous Theorems~\BargThm~and~\CandThm, addressed in the next section.
\ifthenelse{\boolean{dcol}}{
\begin{figure}[!t]
\centering
\epsfig{file={BernUnifMeans.eps},width=1\linewidth}
\caption{Means $p(a) = \E \U_n(a)$ for $m=50$ and the $(a)$ Bernoulli and $(b)$ Uniform cases.}
\label{fig:BernUnif}
\vspace*{-15pt}
\end{figure}
}{
\begin{figure}[!t]
\centering
\epsfig{file={BernUnifMeans.eps},width=.8\linewidth}
\caption{Means $p(a) = \E \U_n(a)$ for $m=50$ and the $(a)$ Bernoulli and $(b)$ Uniform cases.}
\label{fig:BernUnif}
\vspace*{-15pt}
\end{figure}
}
\newcommand{\mtail}{p}
\newcommand{\jtail}{q}
\newcommand{\z}{a}
\newcommand{\Lam}{\lambda_n}
\newcommand{\kernel}{\zeta}
\renewcommand{\define}{=}
\newcommand{\PoiThm}{2.N}
\newcommand{\Iv}{^{-1}}
\newcommand{\R}{\mathcal{R}}
\section{U-statistics \& ``Average-case'' Recovery Guarantees} \label{sect:KKT}
\newcommand{\s}{\pmb{\beta}}
\newcommand{\jp}{\ell}
\newcommand{\Rj}{{\R\setminus \{j\}}}
\subsection{Counting argument using U-statistics} \label{ssect:count}
Previously we had explained how the \emph{invertability} conditions required by Theorems \BargThm~and \CandThm~naturally relate to U-statistics.
We now go on to discuss the other conditions, whereby the relationship may not be immediate.
We begin with the \emph{projections} conditions, in particular
the \emph{worst-case projections} condition.
For given $\Sens$, we need to \emph{upper bound} the fraction of subsets $\S$, for which there exists \emph{at least one} column $\col_j$ where $j\notin\S$, such that $\Norm{\Sens_{\S}^\dagger \col_j }{\infty}$ exceeds some value $a$.
To this end, let $\R$ denote a size-$(k+1)$ subset, and $\Rj$ is the size-$k$ subset excluding the index $j$.
Consider the bounded kernel $g : \Real^{m \times (k+1)} \times \Real \mapsto \ZO$ set as
\bea
g(\mat{A},a) = \frac{1}{k+1} \sum_{j=1}^{k+1} \I{ \Norm{\mat{A}_{\Rj}^\dagger \mat{a}_j}{\infty}> a},
\label{eqn:g2}
\eea
where here $\R = \{1,2,\cdots, k+1\}$, and $\mat{a}_j$ denotes the $j$-th column of $\mat{A}$.
Consider the U-statistic with bounded kernel (\ref{eqn:g2}). We claim that
\ifthenelse{\boolean{dcol}}{
\begin{align}
(n&-k)\cdot U_n(a) \nn
&= \frac{n-k}{(k+1)\Bin{n}{k+1}} \sum_{\R} \! \sum_{j\in \R} \I{ \Norm{\Sens_\Rj^\dagger \col_j}{\infty} > a}, \nn
&= \frac{1}{\Bin{n}{k}} \sum_{\S} \! \sum_{j\notin \S} \I{ \Norm{\Sens_\S^\dagger \col_j}{\infty} > a}, \nonumber
\end{align}}{
\bea
(n-k)\cdot U_n(a)
&=& \frac{n-k}{(k+1)\Bin{n}{k+1}} \sum_{\R} \! \sum_{j\in \R} \I{ \Norm{\Sens_\Rj^\dagger \col_j}{\infty} > a}, \nn
&=& \frac{1}{\Bin{n}{k}} \sum_{\S} \! \sum_{j\notin \S} \I{ \Norm{\Sens_\S^\dagger \col_j}{\infty} > a}, \nonumber
\eea}
where the summations over $\R$ and $\S$ are over all size-$(k+1)$ subsets, and all size-$k$ subsets, respectively.
The first equality follows from Definition \ref{def:Ustat} and (\ref{eqn:g2}).
The second equality requires some manipulation.
First the coefficient $\Bin{n}{k}\Iv$ follows from the binomial identity $ \Bin{n}{k+1}\cdot (k+1) = \Bin{n}{k}\cdot (n-k)$.
Next for some subset $\S$ and index $j$, write the indicator $\I{ \Norm{\Sens_\S^\dagger \col_j}{\infty} > a}$ as $\1_{\S,j}$ for brevity's sake.
By similar counting that proves the previous binomial identity, we argue $\sum_\R \sum_{j\in \R} \1_{\Rj,j} = \sum_\S \sum_{j\notin \S}\1_{\S,j}$, which then proves the claim.
Imagine a grid of ``pigeon-holes'', indexed by pairs $(\S,j)$, where $j \notin\S$.
For each size-$(k+1)$ subset $\R$, we assign $k+1$ indicators $\1_{\Rj,j}$ to $k+1$ pairs $(\S,j)$.
No ``pigeon-hole'' gets assigned more than once.
In fact we infer from the binomial identity, that every ``pigeon-hole'' is in fact assigned exactly once, and argument is complete.
Similarly for the \emph{small projections} condition, we define a different bounded kernel $g : \Real^{m \times (k+1)} \times \Real \mapsto \ZO$ as
\bea
g(\mat{A},a) = \frac{1}{2^k(k+1)} \sum_{\jp=1}^{2^k} \sum_{j=1}^{k+1} \I{ \left|(\mat{A}_{\Rj}^\dagger \mat{a}_j)^T\s_\jp \right|> a},
\label{eqn:g3}
\eea
where $\R = \{1,2,\cdots, k+1\}$, and $\mat{a}_j$ denotes the $j$-th column of $\mat{A}$, and $\s_1,\s_2,\cdots,\s_{2^k}$ enumerate all $2^k$ unique sign-vectors in the set $\{-1,1\}^k$.
By similar arguments as before, we can show for the U-statistic $U_n(a)$ of $\Sens$ corresponding to the bounded kernel (\ref{eqn:g3}) satisfies
\ifthenelse{\boolean{dcol}}{
\bea
(n-k)\cdot U_n(a)
\!\!\! &=&
\!\!\!\frac{1}{2^k \Bin{n}{k}} \sum_{\jp=1}^{2^k} \sum_{\S} \! \sum_{j\notin \S} \I{\left|(\Sens_\S^\dagger \col_j)^T \s_\jp\right| > a}, \nonumber
\eea}{
\bea
(n-k)\cdot U_n(a)
&=& \frac{1}{2^k \Bin{n}{k}} \sum_{\jp=1}^{2^k} \sum_{\S} \! \sum_{j\notin \S} \I{\left|(\Sens_\S^\dagger \col_j)^T \s_\jp\right| > a}, \nonumber
\eea}
For indicators $\1_{\S,j}$, note that $\sum_{j\notin \S} \1_{\S,j} \geq 1$ if \emph{at least one} indicator satisfying $\1_{\S,j} =1$, and we proved the following.
\begin{pro} \label{cor:count}
Let $\V_n(a_3)$ be the U-statistic of $\Sens$, corresponding to the bounded kernel $g(\mat{A},a_3)$ in (\ref{eqn:g2}).
Then the fraction of subsets $\S$ of size-$k$, for which the worst-case projections condition is violated for some $a_3\in \Real$, is at most $(n-k)\cdot U_n(a_3)$.
Similarly if $\V_n(a_2)$ corresponds to $g(\mat{A},a_2)$ in (\ref{eqn:g3}), the fraction sign-subset pairs $(\s,\S)$, for which the small projections condition is violated for some $a_2\in \Real$, is at most $(n-k)\cdot U_n(a_2)$.
\end{pro}
Referring back to Theorem \BargThm, we point out that the \emph{small projections} condition is more stringent than the \emph{worst-case projections} condition. We mean the following:
in the former case, the value $a_2$ must be chosen such that $a_2 < 1$;
in the latter case, the value $a_3$ is allowed to be larger than $1$, its size only affects the constant $2 a_3/(1-a_2)$ appearing in the error estimate $\Norm{\matt{\x}_\S^*-\matt{\x}_\S}{1}$.
In fact if the signal $\matt{\x}$ is $k$-sparse, then $\Norm{\matt{\x} - \xbk}{1} = 0$ and the size of $a_3$ is inconsequential, \textit{i.e.}, the \emph{worst-case projections} condition is not required in this special case.
In this special case, it is best to set $a_2 = 1-\eps$ for some arbitrarily small $\eps$.
Theorem \BargThm~is in fact a stronger version of Fuchs' early work on $\ell_0/\ell_1$-\emph{equivalence}~\cite{Fuchs}.
In the same respect, Donoho \& Tanner also produced early seminal results from counting faces of random polytopes~\cite{DTExp,DT}.
Figure \ref{fig:CompL1} shows empirical evidence, where the $k,m,n$ values are inspired by practical system sizes taken from an implementation paper~\cite{Fred}.
These experiments consider $\Sens$ sampled from Gaussian matrices $\pmb{A}$, \emph{exactly} $k$-sparse signals with non-zero $\alpha_i$ sampled from $\{-1,1\}$, and uses $\ell_1$-minimization recovery (\ref{eqn:L1}).
Figure \ref{fig:CompL1}$(a)$ plots simulated (sparsity pattern recovery) results for 3 measurement sizes $m=50,100$ and $150$ and block sizes $n \geq 200$ and $n \leq 3000$.
For example the contour marked ``0.1'', delineates the $k,n$ values for which recovery fails for a 0.1 fraction of (random) sparsity patterns (sign-subset pairs $(\s,\S)$).
We examine the U-statistic $U_n(a_2)$ with kernel (\ref{eqn:g3}), related to the small projections condition.
Since $\pmb{A}$ has Gaussian distribution, we set $a_2=1$ in the kernel $g(\mat{A},a_2)$, as $\Pr\{(\pmb{A}_\S^\dagger\pmb{A}_i)^T\s = 1\}=0$ for any $(\s,\S)$ and $j\notin\S$.
Figure \ref{fig:CompL1}$(b)$ plots the expectation $(n-k)\cdot p(1)$, where $p(1) = \E U_n(1)= \E g(\pmb{A}_\R,1)$ for any size-$(k+1)$ subset $\R$.
Again the contour marked ``0.1'', delineates the $k,n$ values for which $(n-k)\cdot p(1)= 0.1$.
Here the values $p(1)$ are empirical.
We observe that both Figures \ref{fig:CompL1}$(a)$ and $(b)$ are remarkably close for fractions $0.5$ and smaller.
Figures \ref{fig:CompL1}$(c)$ incorporates the large deviation error $\eps_n$ given in Theorem \ref{thm:Ustat} (in doing so, we assume $n$ sufficiently large).
The bound is still reasonably tight for fractions $\leq 0.5$.
Comparing with recent Donoho \& Tanners' (also ``average-case'') results for $\ell_1$-recovery (for only the noiseless case), taken from~\cite{DTExp}.
For fractions $0.5$ and $0.01$, we observe that for system parameters $m=50$ and $n\leq 1000$ (chosen in hardware implementation~\cite{Fred}), we do not obtain reasonable predictions. For $m=100$, the bounds~\cite{DTExp} work only for very small block lengths $n\leq 300$. The only reasonable case here is $m=150$, where the bounds~\cite{DTExp} perform better than ours only for lengths $n\leq 400$ (\textit{i.e.}, Figure \ref{fig:CompL1}$(c)$ shows that for $n=300$, the large deviation bounds predict a 0.01 fraction of size $k=5$ unrecoverable sparsity patterns, but~\cite{DTExp} predict a 0.01 fraction of size $k=11$ unrecoverable sparsity patterns).
\begin{figure*}[!t]
\centering
\epsfig{file={CompL1.eps},width=.9\linewidth}
\caption{Gaussian case. Comparing $(a)$ empirical results for $\ell_1$-minimization recovery, $(b)$ mean parameter $(n-k)\cdot p(1)$ (empirically obtained), and $(c)$ after accounting for large deviations (Thm. \ref{thm:Ustat}). We show cases $m=50, 100$ and $150$. We also compare with Donoho \& Tanners' (DT) large deviation bounds~\cite{DTExp}.}
\label{fig:CompL1}
\vspace*{-15pt}
\end{figure*}
The above experiments suggest the deviation error $\eps_n(a)$ in Theorem \ref{thm:Ustat} to be over-conservative.
Fortunately in the next two subsections (pertaining to U-statistics treastise of $\ell_1$-recovery Theorem \BargThm~(Section \ref{ssect:ell1}), and LASSO recovery Theorem \CandThm~(Subsection \ref{ssect:Lasso})), this conservative-ness does not show up from a rate standpoint (it only shows up in implicit constants).
In fact by empirically ``adjusting'' these constants, we find good measurement rate predictions (akin to moving from Figure \ref{fig:CompL1}$(c)$ to $(b)$).
\subsection{Rate analysis for $\ell_1$-recovery (Theorem \BargThm)} \label{ssect:ell1}
In ``worst-case'' analysis, it is well-known that it is sufficient to have measurements $m$ on the order of $k\log(n/k)$, in order to have the restricted isometry constants $\delta_k$ defined by (\ref{eqn:RIC}), satisfy the conditions in Theorem \RIPthm.
We now go on to show that for ``average-case'', a similar expression for this rate can be obtained. To this end we require tail bounds on salient quantities. Such bounds have been obtained for the \emph{small projections} condition, see~\cite{Cand2008,Barga,Eldar2009}, where typically an equiprobable distribution is assumed over the sign-vectors $\matt{\beta}_\jp$.
To our knowledge these techniques were born from considering deterministic matrices.
Since $\Sens$ is randomly sampled here, we proceed slightly differently (though essentially using similar ideas) without requiring this random signal model.
For simplicity, the bound assumes zero mean matrix entries, either i) Gaussian or ii) bounded.
\begin{pro} \label{pro:proj}
Let $\pmb{A}$ be an $m \times n$ random matrix, whereby its columns $\A_i$ are identically distributed.
Assume every entry $A_{ij}$ of $\A$ has zero mean, i.e., $\E A_{ij} = 0$.
Let every $A_{ij}$ be either i) Gaussian with variance $1/m$, or ii) bounded RVs satisfying $|A_{ij}| \leq 1/\sqrt{m}$.
Let the rows $[A_{i1},A_{i2},\cdots, A_{in}]$ of $\A$ be IID.
Let $\S$ be a size-$k$ subset, and let index $\omega$ be outside of $\S$, i.e., $\omega\notin\S$. Then for any sign vector $\s$ in $\{-1,1\}^k$, we have
\ifthenelse{\boolean{dcol}}{
\begin{align}
\Pr\left\{ \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right\}
\leq &~2\exp\left( - \frac{m a^2 \delta}{2 k} \right) \nn
&+ \Pr\{ \eigm(\A_\S) \leq \delta\} \label{eqn:proj0}
\end{align}}{
\bea
\Pr\left\{ \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right\}
\leq 2\exp\left( - \frac{m a^2 \delta}{2 k} \right)
+ \Pr\{ \eigm(\A_\S) \leq \delta\} \label{eqn:proj0}
\eea}
for any positive $\delta \in \Real$.
\end{pro}
{
\renewcommand{\A}{\pmb{A}}
\renewcommand{\b}{\pmb{B}}
\newcommand{\w}{\omega}
\newcommand{\Ev}{\mathcal{E}}
\renewcommand{\eigM}{\varsigma_{\scriptsize \mbox{\upshape max}}}
\begin{proof}
For $\tau \in \Real$, let $\Ev(\tau) = \{\s^T (\A_\S^T \A_\S)^\dagger \s \leq \tau \}$ where $\Ev(\tau)$ is an probabilistic event. Let $\Ev_c(\tau)$ denote the complementary event.
Bound the probability as
\ifthenelse{\boolean{dcol}}{
\begin{align}
\Pr\left\{ \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right\}
\leq& \Pr\left\{\left. \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right| \Ev(\tau) \right\} \nn
&+ \Pr\{\Ev_c(\tau)\}.
\label{eqn:proj1}
\end{align}
}{
\bea
\Pr\left\{ \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right\} \leq
\Pr\left\{\left. \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right| \Ev(\tau) \right\} + \Pr\{\Ev_c(\tau)\}.
\label{eqn:proj1}
\eea}
We upper bound the first term as follows.
Denote constants $c_1,c_2, \cdots, c_m$.
For entries $(\A_\w)_i$ of $\A_\w$, consider the sum $ \sum_{i=1}^m c_i \cdot ( m^{-\half}\A_\w)_i = \frac{1}{m}\sum_{i=1}^m c_i X_i$, where RVs $X_i$ satisfy $X_i = (\sqrt{m} \A_\w)_i$.
By standard arguments (see Supplementary Material \ref{sup:stdbound}) we have the double-sided bound $\Pr\left\{\left|\sum_{i=1}^m c_i X_i \right| > m t \right\} \leq 2\exp\left(-(mt)^2/( 2 \cdot \Norm{\mat{c}}{2}^2) \right)$, where vector $\mat{c}$ equals $[c_1,c_2,\cdots, c_m]^T$.
Next write $(\A_\S^\dagger \A_\w)^T \s = ( \sqrt{m} \cdot \s^T \A_\S^\dagger ) (m^{-\half} \A_\w)$.
When conditioning on $ \s^T \A_\S^\dagger $, then $ \sqrt{m} \cdot \s^T \A_\S^\dagger $ is fixed, say equals some vector $\mat{c}$. Put $X_i = (\sqrt{m} \A_\w)_i$ and $X_i$'s are independent (by assumed independence of the rows of $\pmb{A}$).
Then use the above bound for $\Pr\left\{\sum_{i=1}^m c_i X_i > t \right\}$, set $t=a$ and conclude
\ifthenelse{\boolean{dcol}}{
\begin{align}
&\Pr\left\{\left. \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right| \s^T \A_\S^\dagger \right\} \nn
&\leq 2\exp\left(- \frac{(m a)^2}{ 2 m \Norm{\s^T \A_\S^\dagger}{2}^2 } \right) = 2\exp\left(- \frac{m a^2}{ 2 \cdot \s^T (\A_\S^T \A_\S)^\dagger \s } \right),
\end{align}}{
\[
\Pr\left\{\left. \left|(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s \right| > a \right| \s^T \A_\S^\dagger \right\}
\leq 2\exp\left(- \frac{(m a)^2}{ 2 m \Norm{\s^T \A_\S^\dagger}{2}^2 } \right)
= 2\exp\left(- \frac{m a^2}{ 2 \cdot \s^T (\A_\S^T \A_\S)^\dagger \s } \right),
\]}
where the last equality follows from the identity $\A_\S^\dagger (\A_\S^\dagger)^T = (\A_\S^T \A_\S)^\dagger $.
Further conclude that the first term in (\ref{eqn:proj1}) is bounded by $2\exp( - m a^2/(2 \tau))$, due to further conditioning on the event $\Ev(\tau) = \{\s^T (\A_\S^T \A_\S)^\dagger \s \leq \tau \}$.
To bound the second term, let $\eigM(\mat{A})$ denote the maximum eigenvalue of matrix $\mat{A}$.
Since $\A_\S^T \A_\S$ is positive semidefinite, note that $\s^T (\A_\S^T \A_\S)^\dagger \s $ is upper bounded by $\Norm{\s}{2}^2 \cdot \eigM((\A_\S^T \A_\S)^\dagger) $, which equals $ k \cdot \eigM((\A_\S^T \A_\S)^\dagger) $.
Furthermore $\eigM((\A_\S^T \A_\S)^\dagger) \leq 1/\eigm(\A_\S)$, where here $\sigm(\mat{A})$ is the minimum singular value of $\mat{A}$.
Thus $\Pr\{\Ev_c(\tau) \} \leq \Pr\{k/\eigm(\A_\S) > \tau\}$.
Finally put $\tau = \delta k$
to get $\Pr\{\Ev_c(\tau) \} \leq \Pr\{ \eigm(\A_\S) \leq \delta\Iv\}$.
\end{proof}
}
\renewcommand{\fn}{\footnote{For $m \geq 2k$, we have $\Pr\{\sigm(\pmb{A}) < c \cdot 0.29 - t \} \leq \Pr\{\sigm(\pmb{A}) < 1 - c \cdot \sqrt{k/m} - t \}\leq \exp(-m t^2/c_1)$ for some constants $c,c_1$, where $\pmb{A}$ has size $m\times k$ and with proper column normalization. For simplicity we drop the constant $c$ in this paper; one simply needs to add $c$ in appropriate places in the exposition. In particular for the Gaussian and Bernoulli cases $c=1$, and $c_1=2$ and $c_1=16$, respectively, see Theorem B,~\cite{Lim2}.}}
Proposition \ref{pro:proj} is used as follows. First recall that previous Proposition \ref{cor:count} allows us to upper bound the fraction $u_2$ of sign-subset pairs $(\s, \S)$ \emph{failing} the \emph{small projections} condition, with the (scaled) U-statistic $(n-k)\cdot U_n(a_2)$ with kernel $g$ in (\ref{eqn:g3}) and $|\S|=k$.
By Theorem \ref{thm:Ustat} the quantity $(n-k)\cdot U_n(a_2)$ concentrates around $(n-k)\cdot p(a_2)$, where $p(a_2)=\E g(\A_\R,a_2)$, where $g$ in (\ref{eqn:g3}) is defined for size-$(k+1)$ subsets $\R$.
We use Proposition \ref{pro:proj} to upper estimate $p(a_2)$ using the RHS of (\ref{eqn:proj0}).
Indeed verify that $p(a_2) = 2^{-k} \sum_{\jp} \Pr\{ |(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s_\jp | > a_2 \}$ for any $S$ and $\omega \notin S$, and the bound (\ref{eqn:proj0}) holds for any $\s=\s_\ell$. Now $p(a_2)$ is bounded by two terms.
By $u_2 \leq (n-k) \cdot U_n(a_2)$, thus to have $u_2$ small, we should have the (scaled) first term $2(n-k)\cdot\exp(-ma_2^2\RIC/(2 k))$ of (\ref{eqn:proj0}) to be at most some small fraction $u$. This requires
\bea
m \geq \const \cdot k \log \left(\frac{n-k}{u} \right) \label{eqn:rate}
\eea
with $\const = 2/(a_2^2\RIC)$ (and we dropped an insignificant $\log 2$ term).
Next, for $m \geq 2k$ and $\RIC < (0.29)^2$, we can bound\fn~the second term $\Pr\{ \eigm(\A_\S) \leq \delta\}$ of (\ref{eqn:proj0}) by $\exp(-m\cdot (0.29-\sqrt{\RIC})^2/c_1)$ where $c_1$ is some constant, see~\cite{Rudel}, Theorem 5.39.
Roughly speaking, $\eigm(\A_\S) \geq 0.29$ with ``high probability''.
Figures \ref{fig:Gauss} and \ref{fig:BernUnif} (in the previous Section \ref{sect:Dist}) empirically support this fact.
Again to have $u_2$ small the second term of (\ref{eqn:proj0}) must be small. This requires $ (n-k) \cdot \exp(-m \cdot (0.29-\sqrt{\RIC})^2/c_1) \leq u $ for some small fraction $u$, in which it suffices to have $m$ satisfy (\ref{eqn:rate})
with $\const = c_1/(0.29-\sqrt{\RIC})^2$.
\newcommand{\w}{\omega}
For the \emph{invertability} condition in Theorem \BargThm, we also need to upper bound the corresponding fraction $u_1$ of size-$k$ subsets $\S$. We simply use an U-statistic $U_n(a_1)$ with kernel $g(\mat{A},a_1) = \Ind{\sigm(\mat{A}) > a_1}$ for some positive $a_1$ (see also Theorem \CandThm).
Here Proposition \ref{cor:count} is not needed.
To make $p(a_1)$ small, where $p(a_1) = \E g(\pmb{A}_\S,a_1)$, use the previous bound $p(a_1) \leq \exp(-m \cdot (0.29-a_1)^2/c_1) $, where we set $a_1 = \sqrt{\RIC}$ with $a_1 \leq 0.29$.
Clearly $p(a_1)$ cannot exceed some fraction $u$, if $m$ satisfies (\ref{eqn:rate}) with $\const = c_1/(0.29-a_1)^2$.
For the time being consider \emph{exactly} $k$-sparse signals $\matt{\x}$.
In this special case the \emph{worst-case projections} condition in Theorem \BargThm~is superfluous (\textit{i.e.}, with no consequence $a_3$ can be arbitrarily big) -
only \emph{invertability} and \emph{small projections} conditions are needed.
While we have yet to consider the large deviation error $\eps_n(a)$ from Theorem \ref{thm:Ustat}, doing so will not drastically change the rate. For $U_n(a)$ with kernel $g$ and $p(a)$, where $p(a)=\E g(\pmb{A},a)$, almost surely
\ifthenelse{\boolean{dcol}}{
\bea
\!\!\!\!\!\!\!\!\!U_n(a) \leq p(a) + \eps_n(a)
\!\!\!&\leq& \!\!\!(p(a))^\half + \sqrt{2 p(a) \w^{-1} \log \w } \nn
\!\!\! &\leq& \!\!\! (p(a))^\half \left(1 + \sqrt{2 \w^{-1} \log \w }\right) \label{eqn:Ubound}
\eea}{
\bea
U_n(a) \leq p(a) + \eps_n(a) \leq (p(a))^\half + \sqrt{2 p(a) \w^{-1} \log \w } \leq (p(a))^\half \left(1 + \sqrt{2 \w^{-1} \log \w }\right) \label{eqn:Ubound}
\eea}
where the second inequality follows because $p(a) \leq 1$, and by setting $\w = n/k$.
Taking $\log$ of the RHS, we obtain $(1/2) \log p(a) + \log (1 + \sqrt{2 \w^{-1} \log \w }) $.
Note $\log (1 + \sqrt{2 \w^{-1} \log \w }) \leq \sqrt{2 \w^{-1} \log \w }$, since $\log(1+\alpha) \leq \alpha$ holds for all positive $\alpha$.
For the \emph{small projections} condition, bound $(p(a))^\half$ by the sum of the square-roots of each term in (\ref{eqn:proj0}).
Then to have $u_2 \leq (n-k) \cdot U_n(a_2) \leq 2 u$, it follows similarly as before that it suffices that (see Supplementary Material \ref{sup:rate})
\bea
m \geq \const \cdot k \left[ \log \left(\frac{n-k}{u} \right) + \sqrt{2 \cdot (k/n) \log (n/k) } \right] \label{eqn:rate2}
\eea
with $\const = \max(4/(a_2^2\RIC), 2c_1/(0.29-\sqrt{\RIC})^2)$ where we had set $\sqrt{\RIC} = a_1$ (we dropped an insignificant $\log 2$ term).
For \emph{invertability} condition do the same.
To have $u_1 = U_n(a_1) \leq u$ it suffices that $m$ satisfies (\ref{eqn:rate2}) with the same $\const$.
Observe that the term $\sqrt{2 \cdot (k/n) \log (n/k) }$ is at most 1, and vanishes with high undersampling (small $k/n$).
Hence (\ref{eqn:rate}) and (\ref{eqn:rate2}) are similar from a rate standpoint.
\renewcommand{\fn}{\footnote{Comparing (\ref{eqn:rate2}) and (\ref{eqn:rate}) and the respective expressions for $\const$, dropping $\const$ from 4 to 1.8 is akin to ignoring the deviation error $\eps_n(a)$.
This, and as Figure \ref{fig:CompL1} suggests, the U-statistic ``means'' $(n-k) \cdot p(1)$ seem to predict recovery remarkably well, with similar rates to (\ref{eqn:rate2}), and inherent $\const$ smaller than that derived here.}}
We conclude the following: for exactly $k$-sparse signals
the rate (\ref{eqn:rate2}) suffices to recover at least $1-3u$ fraction of sign-subset $(\s,\S)$ pairs. While $\const$ in (\ref{eqn:rate2}) must be at least $4$ (recall that Figure \ref{fig:CompL1}$(c)$ was somewhat pessimistic),
for matrices with Gaussian entries
we empirically find that $\const$ is inherently smaller, whereby $\const \approx 1.8$.
This is illustrated in Figure \ref{fig:MeasRates}, for two fractions $0.1$ and $0.01$ of unrecoverable sign-subset pairs.
We observe good match with simulation results shown in the previous Figure \ref{fig:CompL1}$(a)$, and quantities\fn~$(n-k) \cdot p(1)$ plotted in Figure \ref{fig:CompL1}$(b)$.
For example, $m=150$ suffices for a 0.01 fractional recovery failure, for $n=300 \sim 1000$ and $k = 6\sim 7$, and for 0.1 fraction then $k = 7 \sim 10$.
We conjecture possible improvment for $\const$.
\ifthenelse{\boolean{dcol}}{
\begin{figure}[!t]
\centering
\epsfig{file={MeasRates.eps},width=.7\linewidth}
\caption{Measurement rates predicted by equation (\ref{eqn:rate2}), with $\const$ taken to equal $1.8$, required to recover at least $1- 3u = 0.9$ and $0.99$ fractions of sign-subset pairs $(\s,\S)$ (when the signal is exactly $k$-sparse), shown respectively in $(a)$ and $(b)$.}
\label{fig:MeasRates}
\vspace*{-10pt}
\end{figure}
}{
\begin{figure}[!t]
\centering
\epsfig{file={MeasRates.eps},width=.4\linewidth}
\caption{Measurement rates predicted by equation (\ref{eqn:rate2}), with $\const$ taken to equal $1.8$, required to recover at least $1- 3u = 0.9$ and $0.99$ fractions of sign-subset pairs $(\s,\S)$ (when the signal is $k$-sparse), shown respectively in $(a)$ and $(b)$.}
\label{fig:MeasRates}
\vspace*{-10pt}
\end{figure}
}
\renewcommand{\fn}{\footnote{We used an assumption that $(n-k)/u$ is suitably larger than $2$, see Supplementary Material \ref{sup:rate}.}}
In the more general setting for \emph{approximately} $k$-sparse signals, we can also have rate (\ref{eqn:rate2}).
To see this, observe that Proposition \ref{pro:proj} also delivers an exponential bound for the \emph{worst-case projections} condition, see (\ref{eqn:g2}).
This is because $\Norm{\pmb{A}_\S^\dagger \pmb{A}_\omega }{1} = \max_{\jp : \; 1 \leq \jp \leq 2^k} |(\pmb{A}_\S^\dagger \pmb{A}_\omega)^T \s_\jp |$, and we take a
union bound over $2^k$ terms.
Set $a_3 = a_2 \sqrt{k} $, where $a_2$ and $a_3$ respectively correspond to \emph{small projections} and \emph{invertability} conditions.
Then we proceed similarly as before (see Supplementary Material \ref{sup:rate}) to show\fn~that
the rate for recovering at least $1-5u$ fraction of $(\s,\S)$ pairs
suffices to be (\ref{eqn:rate2}). The following is the main result summarizing the exposition so far.
\begin{thm} \label{thm:L1}
Let $\Sens$ be an $m \times n$ matrix, where assume $n$ sufficiently large for Theorem \ref{thm:Ustat} to hold. Sample $\Sens = \pmb{A}$ whereby the entries $A_{ij}$ are IID, and are Gaussian or bounded (as stated in Proposition \ref{pro:proj}).
Then all three conditions in $\ell_1$-recovery guarantee Theorem \BargThm~for $(\s,\S)$ with $|\S|=k$, with the invertability condition taken as $\sigm(\SensS) \geq a_1$ with $a_1 \leq 0.29$. and with $a_3 = a_1 \sqrt{k}$, are satisfied for $u_1 + u_2 + u_3 = 5u$ for some small fraction $u$, if $m$ is on the order of (\ref{eqn:rate2}) with
$\const = \max(4/(a_1a_2)^2, 2c_1/(0.29-a_1)^2)$, and $c_1$ depends on the distribution of $A_{ij}$'s.
Note $\const \geq 4$.
In the exactly $k$-sparse case where only the first 2 conditions are required, this improves to $u_1 + u_2 = 3u$.
\end{thm}
We end this subsection with two comments on the rate (\ref{eqn:rate2}) derived here for ``average-case'' analysis.
Firstly (\ref{eqn:rate2}) is
very similar to that of $k \log (n/k)$ for ``worst-case'' analysis.
This justifies the counting employed in previous Subsection \ref{ssect:count}, Proposition \ref{cor:count}, and is reassuring since we know that ``worst-case'' analysis provides the optimal rate~\cite{Candes,Donoho}.
Secondly to have (\ref{eqn:rate2}) hold for the approximately $k$-sparse case, we lose a factor of $\sqrt{k}$ in the error estimate $\Norm{\matt{\x}^*_\S - \matt{\x}_\S}{1}$, as compared to ``worst-case'' Theorem \RIPthm.
This is because we need to set $a_3 = a_2 \sqrt{k} $, as mentioned in the previous paragraph.
However, the ``average-case'' analysis here achieves our primary goal, that is to predict well for system sizes $k,m,n$ when ``worst-case'' analysis becomes too pessimistic.
\renewcommand{\Bas}{\mat{C}}
\newcommand{\am}{\delta_{\scriptsize \mbox{\upshape min}}}
\newcommand{\aM}{\delta_{\scriptsize \mbox{\upshape max}}}
\newcommand{\eeigm}{\varsigma_{\scriptsize \mbox{\upshape min}}}
\newcommand{\eeigM}{\varsigma_{\scriptsize \mbox{\upshape max}}}
\ifthenelse{\boolean{dcol}}{
\begin{figure}[!t]
\centering
\epsfig{file=Lasso2.eps,width=.7\linewidth}
\caption{Empirical LASSO recovery performance, Bernoullli case. In $(a)$ the non-zero signal magnitudes $|\x_i|$ equal 1, and in $(b)$ they are in $\ZO$. Noise variances denoted $\noisesd^2$.}
\label{fig:Lasso}
\vspace*{-10pt}
\end{figure}
}{
\begin{figure}[!t]
\centering
\epsfig{file={lasso.eps},width=.8\linewidth}
\caption{Empirical LASSO recovery performance, Bernoullli case. In $(a)$ the non-zero signal magnitudes $|\x_i|$ equal 1, and in $(b)$ they are in $\ZO$. Noise variances denoted $\noisesd^2$.}
\label{fig:Lasso}
\vspace*{-10pt}
\end{figure}
}
\subsection{Rate analysis for LASSO (Theorem \CandThm)} \label{ssect:Lasso}
Next we move on to the LASSO estimate of~\cite{Cand2008}.
Recall from (\ref{eqn:Lasso}) that the regularizer depends on the noise standard deviation $\noisesd$, and the term $\Reg=(1+a)\sqrt{2 \log n }$ that depends on block length $n$ and some non-negative constant $a$ that we set.
This constant $a$ impacts performance~\cite{Cand2008}.
For matrices with Bernoulli entries, Figure \ref{fig:Lasso} shows recovery failure rates for two data sets $m=50, n=1000$ and $m=150, n=1000$; the sparsity patterns (sign-subset pairs $(\s,\S)$) were chosen at random, and failure rates are shown for various sparsity values $k$, and noises $\noisesd$.
In Figure \ref{fig:Lasso}$(a)$ we set $a=0$, and in $(b)$ we set $a=1$.
Also, in $(a)$ the non-zero signal magnitudes $|\x_i|$ are in $\{1,-1\}$, and in $(b)$ they are in $\ZO$.
The performances are clearly different.
``Threshold-like'' behavior is seen in $(a)$ for both data sets, whereby the performances stay the same for $\noisesd $ in the range $ 5 \times 10^{-2} \sim 1 \times 10^{-4}$, and then catastrophically failing for $\noisesd = 1 \times 10^{-1}$.
However in $(b)$, for various $\noisesd$ the performances seem to be limited by a ``noise-floor''.
We see that in the noiseless limit (more specifically when $\noisesd \rightarrow 0$), the performances become the same.
In this subsection, we apply U-statistics on the various conditions of Theorem \CandThm, in particular
the \emph{invertability} and \emph{small projections} conditions have already been discussed in the previous subsection.
We account for the observations in Figure \ref{fig:Lasso}.
In the noiseless limit, the previously derived rate (\ref{eqn:rate2}) holds.
Here, the regularizer in (\ref{eqn:Lasso}) becomes so small that $a$ (equivalently $\Reg$) does not matter. As mentioned in~\cite{Fuchs}, LASSO then becomes equivalent to $\ell_1$-minimization (\ref{eqn:L1}), hence the (noiseless) performances in Figures \ref{fig:Lasso}$(a)$ and $(b)$ are the same.
That is, in this special case the rate (\ref{eqn:rate2}) suffices to recover at least $1-3u$ fraction of $(\s,\S)$.
To test, take $k = 4$, $n=3000$, and fraction $1-3u = 1- 6 \times 10^{-6}$, and with $\const = 1.8$ gives $153$, close to $m$ here which is set to $150$.
In the noisy case, we are additionally concerned with the noise conditions i) and ii), conditions (\ref{eqn:Cand1}) and (\ref{eqn:Cand2}), and \emph{invertability projections}.
Recall that the noise conditions are satisfied with probability $1- n^{-1} (2\pi \log n)^{-\half}$, that goes to 1 superlinearly~\cite{Cand2008} (Proposition \ref{pro:noise}, Supplementary Material \ref{app:recov}).
The remaining conditions are influenced by the value $a$ set in the $\Reg$ regularization term in (\ref{eqn:Lasso}).
In condition (\ref{eqn:Cand1}), the value $a$ sets the maximal value for $a_2$ (when $a=0$ then $a_2< 0.2929$, and when $a=1$ then $a_2< 0.6464$).
This affects the \emph{small projections} condition, to which constant $a_2$ belongs, which in turn affects performance.
However from a rate standpoint (\ref{eqn:rate2}) still holds, only now
the value of $\const$ (which has the term $4/(a_2^2\RIC)$) becomes larger.
In condition (\ref{eqn:Cand2}), the value $a$ affects the size of the term $a_1^{-1} + 2 a_3(1+a)$.
The larger $a$ is,
the more often (\ref{eqn:Cand2}) fails to satisfy.
Here there are two constants $a_1$ and $a_3$.
Recall $a_1$ belongs to the \emph{invertability} condition discussed in the previous subsection, which holds with rate (\ref{eqn:rate2}) with $\const = 2c_1/(0.29-a_1)^2$ and $a_1 \leq 0.29$.
Consider the case where the non-zero signal magnitudes $|\x_i|$ are independently drawn from $\ZO$. Then we observe $(\min_{i \in \S } |\x_i|) < t$ with probability $1 - (1-t)^k$ where $t \in \ZO$ and $|\S|=k$.
For $t$ set equal to the RHS of (\ref{eqn:Cand2}), this gives the probability that condition (\ref{eqn:Cand2}) fails.
Figure \ref{fig:Lasso}$(b)$ shows good empirical match when setting $a_1 = 0.29$ and $a_3 = 1$, where the dotted curves predict the ``error-floors'' for various $k$, measurements $m=50$ and $m=150$, and noise $\noisesd$.
In the other case where $|\x_i| =1$ (as in Figure \ref{fig:Lasso}$(a)$), condition (\ref{eqn:Cand2}) remains un-violated as long as $\noisesd$ (and $a_1,a_3,n$) allow the RHS to be smaller than 1.
Figure \ref{fig:Lasso}$(a)$ suggests that for the appropriate choices for $a_1,a_3$, condition (\ref{eqn:Cand2}) is always un-violated when $\noisesd \leq 5 \times 10^{-2}$, and violated when $\noisesd \geq 1 \times 1^{-1}$.
For more discussion on noise effects see Supplementary Material \ref{sup:Lasso}.
\renewcommand{\fn}{\footnote{For $m \geq 2k$ we have $\Pr\{\sigM(\pmb{A}) > 1.71 + t \} \leq \Pr\{\sigM(\pmb{A}) > 1 + \sqrt{k/m} + t \}\leq \exp(-m t^2/c_1)$ for some $c_1$, see~\cite{Rudel}, Theorem 5.39.}}
The constant $a_3$ belongs to the remaining \emph{invertability projections} condition.
The fraction $u_3$ of size-$k$ subsets \emph{failing} the \emph{invertability projections} condition for some $a_3$, can be addressed using U-statistics.
Consider the bounded kernel $g : \Real^{m\times k} \times \Real \rightarrow \ZO$, set as
\bea
g(\mat{A},a) = \frac{1}{2^k}\sum_{\jp=1}^{2^k} \Ind{ (\mat{A}^T\mat{A})^\dagger \s_\jp > a} \label{eqn:g4}
\eea
where $\s_\jp\in \{-1,1\}^k$ and $(\mat{A}^T\mat{A})^\dagger$ is the pseudoinverse of $\mat{A}^T\mat{A}$.
Then $u_3 = U_n(a_3)$, and as before Theorem \ref{thm:Ustat} guarantees the upper bound (\ref{eqn:Ubound}), which depends on
$p(a_3)$ where $p(a_3) = \E g(\pmb{A}_\S,a_3)$.
We go on to discuss a bound on $p(a_3)$ under some general conditions.
In~\cite{Cand2008}, analysis on $p(a_3)$ (see Lemma 3.5) requires $\eigM(\pmb{A}_\S) \leq 1.5$, a condition not explicitly required in Theorem \CandThm.
Also, empirical evidence suggests not to assume that $\eigM(\pmb{A}_\S) \leq 1.5$.
For $m=150$ and $k=5$ we see from Figure \ref{fig:Lasso} that (in the noiseless limit) the \emph{failure} rate is on the order of $1 \times 10^{-4}$, but in Figure \ref{fig:Gauss}$(b)$ we see $\eigM(\pmb{A}_\S) > 1.5$ occurs with much larger fraction 0.1.
Hence we take a different approach.
Using ideas behind Bauer's generalization of \emph{Wielandt's inequality}~\cite{Householder1965},
the following proposition allows $\eigM(\pmb{A}_\S)$ to arbitrarily exceed 1.5.
Also, it does not assume any particular distribution on entries of $\pmb{A}$.
{\newcommand{\Ev}{\mathcal{E}}
\begin{pro} \label{pro:invproj}
Let $\S$ be a size-$k$ subset. Assume $k\geq 2$.
Let $\pmb{A}_\S$ be an $k \times n$ random matrix.
Let $\am,\aM$ be some positive constants.
For any sign vector $\s$ in $\{-1,1\}^k$, we have
\ifthenelse{\boolean{dcol}}{
\begin{align}
&\Pr\left\{ \Norm{(\pmb{A}_\S^T\pmb{A}_\S)^\dagger\s}{\infty} > \frac{(\sqrt{k} +1 ) \cdot| \tau_k -1 |}{\am^2\cdot(\tau_k + 1)} \cdot \right\} \nn
&~~~~~~~\leq~
\Pr\{ \Ev_c(\am,\aM) \} \label{eqn:invproj0}
\end{align}}
{\bea
\Pr\left\{ \Norm{(\pmb{A}_\S^T\pmb{A}_\S)^\dagger\s}{\infty} > \frac{(\sqrt{k} +1 ) \cdot| \tau_k -1 |}{\am^2\cdot(\tau_k + 1)} \cdot \right\}
\leq
\Pr\{ \Ev_c(\am,\aM) \} \label{eqn:invproj0}
\eea}
where $\Ev(\am,\aM) = \{ \am \leq \sigm(\pmb{A}_\S) \leq \sigM(\pmb{A}_\S) \leq \aM \}$, and $\Ev_c(\am,\aM)$ is the complementary event of $\Ev(\am,\aM)$, and the constant $\tau_k$ satisfies
\bea
\tau_k = \tau_k(\aM,\am) = \left(\frac{\aM }{\am}\right)^2 \cdot \frac{ 1 + k^{-\half}}{1 - k^{-\half}} . \label{eqn:invprojtau}
\eea
\end{pro}
}
\renewcommand{\fn}{\footnote{For $m \geq 2k$ we have $\Pr\{\sigM(\pmb{A}) > 1.71 + t \} \leq \Pr\{\sigM(\pmb{A}) > 1 + \sqrt{k/m} + t \}\leq \exp(-m t^2/c_1)$ for some $c_1$, see~\cite{Rudel}, Theorem 5.39.}}
We defer the proof for now.
If $\pmb{A}_\S^T\pmb{A}_\S$ is ``almost'' an identity matrix, then we expect $\Norm{(\pmb{A}_\S^T\pmb{A}_\S)^{-1}\matt{\beta}}{\infty} \approx 1$ for any sign vector $\matt{\beta}$ (hence our above hueristic whereby we set $a_3=1$).
Proposition \ref{pro:invproj} makes a slightly weaker (but relatively general) statement.
Now for some appropriately fixed $\aM$ and $\am$, we expect $\Pr\{\mathcal{E}_c(\am,\aM)\}$ in (\ref{eqn:invproj0}) to drop exponentially in $m$.
Just as the term $\Pr\{ \sigm(\A_\S) \leq \am\}$ in Proposition \ref{pro:proj} can be bounded by $\exp(-m\cdot (0.29-\am)^2/c_1)$,
we can bound\fn~$\Pr\{\sigM(\pmb{A}) > \aM \} \leq \exp(-m ( \aM - 1.71 )^2/c_1)$ for some $\aM \geq 1.71$.
Roughly speaking, $\sigM(\A_\S) \leq 1.71$ (or $\eigM(\A_\S) \leq 2.92$) with ``high probability''.
We fix $\am = a_1$, where $a_1$ belongs to the \emph{invertability} condition.
So to bound $p(a_3)$, both (\ref{eqn:g4}) and Proposition \ref{pro:invproj} imply $p(a_3) \leq \Pr\{\mathcal{E}_c(\am,\aM)\}$ for $a_3 = (\sqrt{k}+1) \cdot |\tau_k - 1| / (\am^2\cdot (\tau_k+1)) $.
Now $\Pr\{\mathcal{E}_c(\am,\aM)\} \leq 2 \exp(-m \cdot t^2/c_1) $, where we set $t = \aM - 1.71 = 0.29-a_1 $ and $\am = a_1$.
By (\ref{eqn:Ubound}), the rate (\ref{eqn:rate2}) suffices to ensure $u_3 = U_n(a_3) \leq u$ for some fraction $u$,
with the same $\const$.
Thus we proved the other main theorem, similar to Theorem \ref{thm:L1}.
\begin{thm} \label{thm:Lasso}
Let $\Sens$ be an $m \times n$ matrix, , where assume $n$ sufficiently large for Theorem \ref{thm:Ustat} to hold. Sample $\Sens = \pmb{A}$ whereby the entries $A_{ij}$ are IID, and are Gaussian or bounded (as stated in Proposition \ref{pro:proj}).
Then all three \textit{invertability}, \textit{small projections}, and \textit{invertability projections} conditions in LASSO Theorem \CandThm~for $(\s,\S)$ with $|\S|=k \geq 2$, with $a_1 \leq 0.29$, with $a_2$ satisfying (\ref{eqn:Cand1}) for some $a$ set in the regularizer $\Reg$, and with $a_3 = (\sqrt{k}+1) \cdot |\tau_k - 1| / (a_1^2\cdot (\tau_k+1)) $ for $\tau_k=\tau_k(1.42-a_1,a_1)$ in (\ref{eqn:invprojtau}),
are satisfied for $u_1 + u_2 + u_3 = 4u$ for some small fraction $u$, if $m$ is on the order of (\ref{eqn:rate2}) with
$\const = \max(4/(a_1a_2)^2, 2c_1/(0.29-a_1)^2)$, and $c_1$ depends on the distribution of $A_{ij}$'s. Note $\const \geq 4$.
In the noiseless limit where only the first 2 conditions are required, this improves to $u_1 + u_2 = 3u$.
\end{thm}
\begin{rem} \label{rem:LassoNoise}
We emphasize again that the rate (\ref{eqn:rate2}) is measured w.r.t. to the three conditions in Theorem \ref{thm:Lasso}.
The probability for which both noise conditions i) and ii) are satisfied, and for which condition (\ref{eqn:Cand2}) imposed on $\min_{i \in \S}|\x_i|$ is satisfied, require additional consideration. For the former the probability is at least $1 - n^{-1} (2\pi \log n)^{-\half}$, see~\cite{Cand2008}.
For the latter, it has to be derived based on signal statistics, \textit{e.g.}, for $|\x_i| \in \ZO$ then $(\min_{i \in \S}|\x_i|) > t$ is observed with probability $(1-t)^k$ with $|\S|=k$.
\end{rem}
Note that the choice for $a_3$ in Theorem \ref{thm:Lasso} implies $\Norm{(\pmb{A}_\S^T\pmb{A}_\S)^\dagger\s}{\infty}$ is roughly on the order $\sqrt{k}$.
Indeed this is true since $\tau_k \geq 1$, and we note $\tau_k = (\aM/\am)^2 + 2k^{-\half} + o(k^{-\half})$, thus $\tau_k \approx (\aM/\am)^2 $ for moderate $k$.
Now LASSO recovery also depends on the probability that condition (\ref{eqn:Cand2}) holds.
Our choice for $a_3$ causes the RHS of (\ref{eqn:Cand2}) to be roughly of the order $\noisesd \sqrt{2 k \log n}$.
Compare this to~\cite{Cand2008} (see Theorem 1.3) where it was assumed that
$\sigm(\pmb{A}_\S) \leq 1.5$,
they only require $a_3 = 3$,
\textit{i.e.}, a factor of $\sqrt{k}$ is lost without this assumption (which was previously argued to be fairly restrictive).
To improve Proposition \ref{pro:invproj}, one might additionally assume some specific distributions on $\pmb{A}$. We leave further improvements to future work.
{
\newcommand{\Ev}{\mathcal{E}}
\newcommand{\X}{\pmb{X}}
\newcommand{\Xe}{X}
\newcommand{\ang}{\omega}
\renewcommand{\Bas}{\mat{B}}
\newcommand{\D}{\mat{D}}
\renewcommand{\eigM}{\varsigma_{\scriptsize \mbox{\upshape max}}}
\renewcommand{\eigm}{\varsigma_{\scriptsize \mbox{\upshape min}}}
\begin{proof}[Proof of Proposition \ref{pro:invproj}]
For notational convenience, put $\X = (\pmb{A}_\S^T\pmb{A}_\S)^\dagger$.
Bound the probability
\ifthenelse{\boolean{dcol}}{
\begin{align}
\Pr\left\{ \Norm{\X\s}{\infty}> a\sqrt{k} \right\} \leq &
\Pr\left\{\left. \Norm{\X\s}{\infty} > a\sqrt{k} \right| \Ev(\am,\aM) \right\} \nn&+ \Pr\{\Ev_c(\am,\aM)\}.
\label{eqn:invproj1}
\end{align}}{
\bea
\Pr\left\{ \Norm{\X\s}{\infty}> a\sqrt{k} \right\} \leq
\Pr\left\{\left. \Norm{\X\s}{\infty} > a\sqrt{k} \right| \Ev(\am,\aM) \right\} + \Pr\{\Ev_c(\am,\aM)\}.
\label{eqn:invproj1}
\eea}
where we take $a$ to mean
\bea
a = \frac{|\tau_k-1|}{\tau_k+1} \cdot \frac{1+k^{-\half}}{\sigm^2(\pmb{A}_\S)} \label{eqn:invproja}
\eea
for $\tau_k$ chosen as in (\ref{eqn:invprojtau}).
We claim that every entry $(\X\s)_i$ of $\X\s$ is upper bounded by $a \sqrt{k}$, for $a$ as in (\ref{eqn:invproja}).
Then by definition of $\Ev(\am,\aM)$, the first term in (\ref{eqn:invproj1}) equals $0$ and
we would have proven the bound (\ref{eqn:invproj0}).
Let $\mat{C}$ denote a $k \times 2$ matrix.
The first column $\mat{C}$ is be a normalized version of $\s$, more specifically it equals $ k^{-\half}\s_i$.
The second column equals the canonical basis vector $\mat{c}_i$, where $\mat{c}_i$ is a 0-1 vector whereby $(\mat{c}_i)_j = 1$ if and only if $j=i$.
Consider the $2 \times 2$ matrix $\X'$ that satisfies $\X' = \mat{C}^T \X \mat{C}$.
This matrix $\X'$ is symmetric (from symmetry of $\X$) and $k^{-\half} (\X\s)_i = X'_{1,2} = X'_{2,1}$ (from our construction of $\mat{C}$).
That is the entry $X'_{1,2}$ (and $X'_{2,1}$) of $\pmb{X}'$, correspond to the (scaled) quantity $k^{-\half} (\X\s)_i$ that we want to bound.
Condition on the event $\Ev_c(\am,\aM)$, then $\pmb{A}_\S$ has rank $k$ and therefore $\X = (\pmb{A}_\S^T\pmb{A}_\S)^{\dagger} = (\pmb{A}_\S^T\pmb{A}_\S)^{-1}$.
Let $\det(\cdot)$ and $\Tr(\cdot)$ denote determinant and trace.
As in~\cite{Householder1965} equation (11), we have
\ifthenelse{\boolean{dcol}}{
\bea
\!\!\!\!\!\!\!\!\!
1 - \frac{X'_{1,2}X'_{1,2}}{X'_{1,1} X'_{2,2}} &=& \frac{4 \det (\X') }{(\Tr(\X'))^2 - (X'_{1,1} - X'_{2,2})^2 } \nn
&\geq& \frac{4 \eigM(\X') \cdot \eigm(\X') }{(\Tr(\X'))^2} = \frac{4t}{(1+t)^2}
\label{eqn:invproj2}
\eea}{
\bea
1 - \frac{X'_{1,2}X'_{1,2}}{X'_{1,1} X'_{2,2}} = \frac{4 \det (\X') }{(\Tr(\X'))^2 - (X'_{1,1} - X'_{2,2})^2 }
\geq \frac{4 \eigM(\X') \cdot \eigm(\X') }{(\Tr(\X'))^2} = \frac{4t}{(1+t)^2}
\label{eqn:invproj2}
\eea}
where $t = \eigM(\X')/\eigm(\X')$ and $\eigM$ and $\eigm$ respectively denote the maximum and minimum eigenvalues.
Now $t = \eigM(\X')/\eigm(\X') \geq 1$.
If $t=1$ then $ 4 t/(1+t)^2 = 1$, and for $t \geq 1$ the function $ 4 t/(1+t)^2$ decreases monotonically.
We claim that $\tau_k$ in (\ref{eqn:invprojtau}) upper bounds $\eigM(\X')/\eigm(\X')$, and (\ref{eqn:invproj2}) then allows us to produce the following upper bound
\ifthenelse{\boolean{dcol}}{
\begin{align}
|X'_{1,2}| &\leq \sqrt{X_{1,1}' X_{2,2}' \cdot \left( 1 - \frac{4 \tau_k}{(1+\tau_k)^2}\right)} \nn
&= \sqrt{X_{1,1}' X_{2,2}'} \cdot \frac{|\tau_k-1|}{1 + \tau_k}.
\end{align}}{
\[
|X'_{1,2}| \leq \sqrt{X_{1,1}' X_{2,2}' \cdot \left( 1 - \frac{4 \tau_k}{(1+\tau_k)^2}\right)}
= \sqrt{X_{1,1}' X_{2,2}'} \cdot \frac{|\tau_k-1|}{1 + \tau_k}.
\]}
Bound $(X_{1,1}' X_{2,2}')^\half$ by the maximum eigenvalue $\eigM(\X')$ of $\X'$.
Then, further bound $\eigM(\X')$ by $(1+k^{-\half})/\sigm^2(\pmb{A}_\S)$, which gives the form (\ref{eqn:invproja}).
This bound is argued as follows.
For $k\geq 2$, we have the columns in $\mat{C}$ to be \emph{linearly independent}.
Since $\X' = \mat{C}^T \X \mat{C}$ and $\X$ is \emph{positive definite}, it is then clear that $\eigM(\X') \leq \eigM(\mat{C}^T\mat{C}) \cdot \eigM(\X)$.
Now $\mat{C}^T \mat{C}$ is a $2 \times 2$ matrix with diagonal elements 1, and off-diagonal elements $\pm 1/\sqrt{k}$.
Hence $\eigM(\mat{C}^T \mat{C}) = 1 + k^{-\half}$.
Also $\eigM(\X) \leq 1/\sigm^2(\pmb{A}_\S)$, and the bound follows.
To finish, we show the claim $\tau_k \geq \eigM(\X')/\eigm(\X')$. By similar arguments as above, it follows that
\[
\frac{\eigM(\X')}{\eigm(\X')} \leq \frac{\eigM(\mat{C}^T \mat{C})}{\eigm(\mat{C}^T \mat{C})}\cdot \frac{\eigM(\X)}{\eigm(\X)} = \frac{1 + k^{-\half}}{1 - k^{-\half}}\cdot \frac{\sigM^2(\pmb{A}_\S)}{\sigm^2(\pmb{A}_\S)} \leq \tau_k
\]
since $\eigm(\X') \geq \eigm(\mat{C}^T\mat{C}) \cdot \eigm(\X)$, and $\eigm(\X')= 1- k^{-\half}$,
and $\X = (\pmb{A}_\S^T\pmb{A}_\S)^{-1}$. We are done.
\end{proof}
}
\section{Conclusion} \label{sect:conc}
We take a first look at U-statistical theory for predicting the ``average-case'' behavior of salient CS matrix parameters.
Leveraging on the generality of this theory, we consider two different recovery algorithms i) $\ell_1$-minimization and ii) LASSO.
The developed analysis is observed to have good potential for predicting CS recovery, and compares well (empirically) with Donoho \& Tanner~\cite{DTExp} recent ``average-case'' analysis for system sizes found in implementations.
Measurement rates that incorporate fractional $u$ failure rates, are derived to be on the order of $ k [\log((n-k)/u) + \sqrt{2(k/n) \log(n/k)}]$, similar to the known optimal $k\log(n/k)$ rate.
Empirical observations suggest possible improvement for $\const$ (as opposed to typical ``worst-case'' analyses whereby implicit constants are known to be inherently large).
There are multiple directions for future work.
Firstly while restrictive maximum eigenvalue assumptions are avoided (as StRIP-recovery does not require them), the applied techniques could be fine-tuned.
It is desirable to overcome the $\sqrt{k}$ losses observed here for noisy conditions.
Secondly, it is interesting to further leverage the general U-statistical techniques to other different recovery algorithms, to try and obtain their good ``average-case'' analyses.
Finally, one might consider similar U-statistical ``average-case'' analyses for the case where the sampling matrix columns are dependent, which requires appropriate extensions of Theorem \ref{thm:Ustat}.
\section*{Acknowledgment}
The first author is indebted to A. Mazumdar for discussions, and for suggesting to perform the rate analysis.
\appendix
\subsection{Proof of Theorem \ref{thm:Ustat}} \label{app:proofDev}
\newcommand{\floor}[1]{\left\lfloor#1\right\rfloor }
\newcommand{\perm}{\pi}
\renewcommand{\t}{\eps_n}
\newcommand{\M}{\omega_n}
\newcommand{\Sm}{S_\perm}
For notational simplicity we shall henceforth drop explicit dependence on $a$ from all three quantities $U_n(a), p(a)$ and $\g(\mat{A},a)$ in this appendix subsection.
While $U_n$ is made explicit in Definition \ref{def:Ustat} as a statistic corresponding to the realization $\Sens = \pmb{A}$, this proof considers $U_n$ consisting of random terms $g(\pmb{A}_\S)$ for purposes of making probabalistic estimates.
Theorem \ref{thm:Ustat} is really a \emph{law of large numbers} result.
However even when the columns $\pmb{A}_i$ are assumed to be IID, the terms $g(\pmb{A}_\S)$ in $U_n$ depend on each other.
As such, the usual techniques for IID sequences do not apply.
Aside from large deviation results such as Thm. \ref{thm:Ustat}, there exist \emph{strong law} results, see~\cite{Berk}. The following proof is obtained by combining ideas taken from~\cite{Hoef} and~\cite{Sen}. We use the following new notation just in this subsection of the appendix. Partition the index set $ \{1,2,\cdots,n\}$ into $\M = \floor{n/k}$ subsets denoted $\S_i$ each of size $k$, and a single subset $\R$ of size at most $k$.
More specifically, let $\S_i = \{(i-1)\cdot k+1,(i-1)\cdot k+2,\cdots, i \cdot k\}$ and let $\R = \{\floor{n/k}\cdot k+1,\floor{n/k}\cdot k+2,\cdots, n \}$. Let $\perm $ denote a \emph{permutation} (bijective) mapping $ \{1,2,\cdots, n\} \rightarrow \{1,2,\cdots, n\}$. The notation $\perm(\S)$ denotes the set of all \emph{images} of each element in $\S$, under the mapping $\perm$. Following Section 5c in~\cite{Hoef} we express the U-statistic $\V_n$ of $\pmb{A}$ in the form
\bea
\V_n &=& \frac{1}{n!} \sum_{\perm} \left( \frac{1}{\M} \sum_{i=1}^{\M} g(\pmb{A}_{\perm(\S_i)}) \right), \label{eqn:U}
\eea
the first summation taken over all $n!$ possible permutations $\perm$ of $ \{1,2,\cdots,n\}$. To verify, observe that any subset $\S$ is counted exactly $\M \cdot k! (n-k)!$ times in the RHS of (\ref{eqn:U}).
Recall $p=\E g(\pmb{A}_\S) = \E \V_n$. From the theorem statement let the term $\eps_n^2$ equal $c p (1-p) \cdot \M^{-1} \log \M$ where $c>2$. We show that the probabilities $\Pr\{|\V_n - p| > \t\}$ for each $n$ are small. For brevity, we shall only explicitly treat the upper tail probability $\Pr\{\V_n - p > \t\}$, where standard modifications of the below arguments will address the lower tail probability $\Pr\{-\V_n + p > \t\}$ (see comment in p. 1,~\cite{Hoef}). Using the expression (\ref{eqn:U}) for $\V_n$, write the probability $\Pr\{\V_n - p > \t\}$ for any $h> 0$ as
\ifthenelse{\boolean{dcol}}{
\bea
\Pr\{\V_n - p > \t\} &\leq& \E \exp (h( \V_n - p + \t)) \nn
&=& \E \exp\left( \frac{1}{n!} \left( \sum_{\pi} h(\Sm - p + \t ) \right)\right), \nonumber
\eea}{
\bea
\Pr\{\V_n - p > \t\} &\leq& \E \exp (h( \V_n - p + \t))
= \E \exp\left( \frac{1}{n!} \left( \sum_{\pi} h(\Sm - p + \t ) \right)\right), \nonumber
\eea}
where here $\Sm$ is a RV that equals the inner summation in (\ref{eqn:U}), i.e. $\Sm = \frac{1}{\M}\sum_{i=1}^{\M} g(\pmb{A}_{\perm(\S_i)})$. Using convexity of the function $\exp(\cdot)$ we express
\bea
\Pr\{\V_n - p > \t\} &\leq& \frac{1}{n!} \sum_{\perm} \E \exp(h(\Sm- p + \t) ). \nonumber
\eea
Now observe that the RV $\Sm$ is an average of $\M$ IID terms $g(\pmb{A}_{\perm(\S_i)})$. This is due to the assumption that the columns $\pmb{A}_i$ of $\pmb{A}$ are IID, and also due to the fact that the sets $\perm(\S_i)$ are disjoint (recall sets $\S_i$ are disjoint).
Hence for any permutation $\perm$, by this independence we have $\E \exp(h\Sm) = (\E \exp( h' \cdot g(\pmb{A}_{\perm(\S_1)})))^{\M} $, where the normalization $h' = h/\M$ bears no consequence.
The RV $g(\pmb{A}_{\perm(\S_1)}) $ is bounded, i.e. $0 \leq g(\pmb{A}_{\perm(\S_i)}) \leq 1$, and its expectation $\E g(\pmb{A}_{\perm(\S_1)}) $ equals $p$. By convexity of $\exp(\cdot)$ again and for all $h >0$, the inequality
$e^{h\alpha} \leq e^{h }\alpha + 1 - \alpha $
holds for all $0 \leq \alpha \leq 1$. Therefore putting $\alpha = g(\pmb{A}_{\perm(\S_1)})$ we get the inequality $\exp( h \cdot g(\pmb{A}_{\perm(\S_1)})) \leq 1 + (e^h-1)\cdot g(\pmb{A}_{\perm(\S_1)}) $.
By the irrelevance of $\perm$ in previous arguments, by putting $\E g(\pmb{A}_{\perm(\S_1)}) = p$
\[
\Pr\{\V_n - p > \t\} \leq e^{-h (\t+p)} \left(1 - p + p e^h\right)^{\M}.
\]
We optimize the bound by putting $pe^h = (1-p)(p+\t)/(1-p-\t)$, see (4.7) in~\cite{Hoef}, to get
\ifthenelse{\boolean{dcol}}{
\begin{align}
&\Pr\{\V_n - p > \t\} \nn
&\leq \left((1+ \t p\Iv)^{p+\t} (1-\t(1-p)\Iv)^{1-p-\t} \right)^{-\M}. \label{eqn:Uproof1}
\end{align}}{
\bea
\Pr\{\V_n - p > \t\} \leq \left((1+ \t p\Iv)^{p+\t} (1-\t(1-p)\Iv)^{1-p-\t} \right)^{-\M}. \label{eqn:Uproof1}
\eea}
Following (2.20) in~\cite{Sen} we use the relation $\log(1+\alpha) = \alpha - \frac{1}{2}\alpha^2 + o(\alpha^2)$ as $\alpha\rightarrow 0$, to express the logarithmic exponent on the RHS of (\ref{eqn:Uproof1}) as
\[
\frac{- \M \t^2\cdot(1+ o(1))}{2p(1-p)}.
\]
Therefore by the form $\t^2 = c p (1-p) \cdot \M^{-1} \log \M$ where $c>2$, for sufficiently large $n$ we have
\[
\Pr\{\V_n - p > \t\} \leq \M^{-c/2} < \M^{-1}
\]
which in turn implies $\sum_{n = k}^\infty \Pr\{\V_n - p > \t\} < \infty$. Repeating similar arguments for the lower tail probability $\Pr\{-\V_n + p > \t\}$, we eventually prove $\sum_{n = k}^\infty \Pr\{|\V_n - p| >\t\} < \infty$ which implies the claim. | 0.001042 |
\begin{document}
\begin{abstract}
We study the energy decay rate of the Kelvin-Voigt damped wave equation with piecewise smooth damping on the multi-dimensional domain. Under suitable geometric assumptions on the support of the damping, we obtain the optimal polynomial decay rate which turns out to be different from the one-dimensional case studied in \cite{LR05}. This optimal decay rate is saturated by high energy quasi-modes localised on geometric optics rays which hit the interface along non orthogonal neither tangential directions. The proof uses semi-classical analysis of boundary value problems.
\end{abstract}
\maketitle
\setlength{\parskip}{0.3em}
\section{Introduction}
\label{in}
\subsection{Kelvin-Voigt damped wave equation}
In this article, we study the decay rate of the Kelvin-Voigt damped wave equation on the multi-dimensional bounded domain $\Omega\subset \R^d$, $d\geq 2$:
\begin{equation}\label{KVDW-main}
\left\{
\begin{aligned}
&\big(\partial_t^2-\Delta-\mathrm{div}a(x)\nabla\partial_t\big)u=0,\quad (t,x)\in\R_t\times\Omega,\\
&u(t,\cdot)|_{\partial\Omega}=0,\\
& (u,\partial_tu)|_{t=0}=(u_0,u_1)
\end{aligned}
\right.
\end{equation}
The damping $a(x)\geq 0$ is assumed to be piecewise smooth.
Denote by $\mathcal{H}^1=H_0^1\times L^2$. The solution of \eqref{KVDW-main} can be written as
$$ U(t)=\binom{u(t)}{\partial_tu(t)}=e^{t\mathcal{A}}\binom{u_0}{u_1},
$$
where the generator
\begin{equation}\label{A}
\mathcal{A}=\left(\begin{matrix}
0 &1 \\
\Delta &\mathrm{div}a\nabla
\end{matrix}\right)
\end{equation}
with domain
$$ \mathcal{D}(\mathcal{A})=\{(u_0,u_1)\in H_0^1\times L^2: \Delta u_0+\mathrm{div} a\nabla u_1\in L^2, u_1\in H_0^1 \}.
$$
Note that the energy
$$ E[u](t)=\frac{1}{2}\|e^{t\mathcal{A}}(u_0,u_1)\|_{\mathcal{H}^1}^2 =\frac{1}{2}\int_{\Omega}\big(|\partial_tu|^2+|\nabla u|^2 \big)dx
$$
satisfies
$$ E[u] (t) - E[u] (0) = - \int_{0}^t\int_{\Omega} a(x) |\nabla_x \partial_t u |^2 (s,x) ds$$
It was proved in \cite{BC15} and \cite{BS} (see also \cite{LR06},\cite{Te16} for related results) that if $a$ is smooth, vanishing nicely and the region $\{x\in\Omega: a(x)>0\}$ controls geometrically $\Omega$, then the rate of decay of the energy is exponential:
$$ E[u](t)\leq C\mathrm{e}^{-ct}E[u](0).
$$
In this article, we investigate the different case where the damping $a(x)$ is piecewise smooth and has a jump across some hypersurface $\Sigma\subset\Omega$. Unlike the smooth damping vanishing nicely, the problem with piecewise damping can be seen as an elliptic-hyperbolic transmission system on the two sides of the interface $\Sigma$ connected by some transmission condition. The interface becomes a wall to reduce the energy transmission from the hyperbolic region to the damped region. This phenomenon is known as {\em overdamping}. It turns out that this discontinuous Kelvin-Voigt damping $\nabla\cdot(a(x)\nabla\partial_t u)$ does not follow the principle that the ``geometric control condition" implies the exponential stabilization, which holds for the wave equation with localized viscous damping $a(x)\partial_tu$ (see~\cite{LR05, Zh18} for results using the multiplier methods)
\subsection{The main result}
To state our main result, we first make some geometric assumptions.
Let $\Omega\subset \R^d$ with $d\geq 2$. We consider the piecewise smooth damping $a\in C^{\infty}(\ov{\Omega}_1)$, $a|_{\Omega\setminus\Omega_1}=0$, such that there exists $\alpha_0>0$,
$$ \inf_{x\in\partial\Omega_1}a(x)\geq \alpha_0,
$$
where $\Omega_1\subset\Omega.$ We assume that $\partial\Omega_1$ consists of $\partial\Omega$ and $\Sigma=\partial\Omega_1\setminus\partial\Omega$ where $\Sigma\subset \Omega$. Denote by $\Omega_2=\Omega\setminus (\Omega_1\cup \Sigma)$, then $\partial\Omega_2=\Sigma$ is the interface. We will fix this geometry in this article and assume that $\Omega_1, \Omega_2$ and $\Sigma$ are smooth ($C^\infty$, though this assumption could be relaxed to a finite number of derivatives).
\begin{center}
\begin{tikzpicture}
\draw[fill=blue!40] (0,0) ellipse[x radius=3.4cm, y radius =2.4 cm];
\draw[black] (0,1.8) node {$\Omega_1:a(x)\geq \alpha_0$};
\draw[fill=green!40] (-0.15,0) ellipse [x radius = 1.4cm, y radius =1.6cm];
\draw[black] (0,-0.4) node[above] {$\Omega_2:a(x)=0$};
\draw [->] (-0.15,-1.6) -- (-0.15,-0.8);
\draw[black] (0.2,-0.9) node{$\nu$};
\draw[black] (-0.6,-1.7) node{$\Sigma$};
\draw[black] (0,-3) node {
\text{Geometry of the damped region}};
\end{tikzpicture}
\end{center}
\begin{definition}[Geometric control condition]\label{GCC}
We say that $\Omega_1$ satisfies the geometric control condition, if all generalized rays (geometric optics reflecting on the boundary $\partial\Omega$ according to the laws of geometric optics) of $\Omega$ eventually reach the set $\Omega_1$ in finite time.
\end{definition}
An alternative (equivalent in this context) property is the following
\begin{itemize}
\item[(H)] All the bicharacteristics of $\Omega_2$ will reach a non-diffractive point (with respect to the domain $\Omega_2$) at the boundary $\Sigma$.
\end{itemize}
\begin{thm}\label{thm:decay}
Assume that $\Omega,\Omega_1,\Omega_2$ and $a(x)$ satisfy the above geometric conditions. Then under the hypothesis (H), there exists a uniform constant $C>0$, such that for every $(u_0,u_1)\in\mathcal{D}(\mathcal{A})$ and $t\geq 0$,
\begin{equation}\label{eq:decayrate}
\|e^{t\mathcal{A}}(u_0,u_1)\|\leq \frac{C}{1+t}\|(u_0,u_1)\|_{\mathcal{D}(\mathcal{A})}.
\end{equation}
Moreover, the decay rate is optimal in the following sense: when $\Omega\subset \R^d$, $d\geq 2$ and $\Omega_2=\mathbb{D}\subset \Omega$ is a unit ball, $\Omega_1=\Omega\setminus\Omega_2$, the semi-group $e^{t\mathcal{A}}$ associated with the damping $a(x)=\mathbf{1}_{\ov{\Omega}_1(x)}$ satisfies
\begin{align}\label{lowerbound}
\sup_{0\neq (u_0,u_1)\in \mathcal{D}(\mathcal{A}) }\frac{ \|e^{t\mathcal{A}}(u_0,u_1)\|_{\mathcal{H}^1}
} {\|(u_0,u_1)\|_{\mathcal{D}(\mathcal{A})} }\geq \frac{C'}{1+t},
\end{align}
for all $t\geq 0$, where $C'>0$ is a uniform constant.
\end{thm}
\begin{rem} In~\cite{B19}, under the geometric control condition, a weaker decay rate, namely $\frac 1 {\sqrt{1+t}}$ was achieved with a simpler and very robust general proof requiring much less rigidity on the geometric setting. Notice also that in dimension $1$, a stronger decay rate, namely $\frac 1 {{(1+t)^2}}$ is known to hold~\cite[Section 3, Example 1]{LR05}. It is hence remarquable that in higher dimensions we can construct examples of geometries where the $\frac{ 1} {(1+ t)}$ decay rate is saturated. This phenomenon is linked to the fact that in higher dimensions there exists sequences of eigenfunctions of the Laplace operator in $\Omega _2$ with Dirichlet boundary condidtions (or at least high order quasi-modes), with mass concentrated along rays which do not encounter the boundary at normal incidence (a fact which is clearly false in dimension $1$, seeing that in this case the incidence is always normal).
\begin{center}
\begin{tikzpicture}
\draw[fill=blue!40] (0,0) ellipse[x radius=3.4cm, y radius =2.4 cm];
\draw[black] (0,1.8) node {$\Omega_1$};
\draw[fill=green!40] (-0.15,0) ellipse [x radius = 1.4cm, y radius =1.6cm];
\draw[black] (0,-0.4) node[above] {$\Omega_2$};
\draw[->,blue](-1.3,0.7) -- (-0.15,1.6);
\draw[->,blue] (-0.15,1.6)--(1.1,0.6);
\draw[->,blue] (1.1,0.6)--(-0.2,-1.58);
\draw[->,blue] (-0.2,-1.58)--
(-1.2,0);
\draw[->,blue] (-1.1,0.7)--(0,1.57);
\draw[->,blue] (0,1.57)--(1,0.8);
\draw[->,blue] (1,0.8)--(-0.4,-1.56);
\draw[->,blue]
(-0.4,-1.56)--(-1.4,0);
\draw[black] (-0.6,-1.7) node{$\Sigma$};
\draw[black] (0,-3) node {
\text{Angle of incidence is acute}};
\end{tikzpicture}
\end{center}
\end{rem}
\begin{rem}
Let us mention that the non-exponential stability for \eqref{KVDW-main} and a more general (theromo)viscoelastic system were studied in \cite{MRa}, where the authors obtained a rougher polynomial decay rate $O(t^{-\frac{1}{3}})$. Moreover, in our result, the damped region ($\Omega_1$) only needs to satisfy the geometric control condition, so the geometric configuration in Munoz Rivera-Racke is contained in our assumption.
\end{rem}
\begin{rem} The choice of Dirichlet boundary conditions on $\partial \Omega$ plays no particular role, and we could have taken any type of boundary conditions for which the system is well posed and we have propagation of singularities (e.g. Neumann boundary conditions)
\end{rem}
\begin{rem}
The picture for Kelvin Voigt damping is now quite complete for smooth (essentially $C^2$) dampings \cite{BC15} and \cite{BS} (and also \cite{LR06},\cite{Te16}), or discontinuous dampings, see in dimension $1$ ~\cite[Section 3, Example 1]{LR05}, and the present paper. It would be interesting to understand the intermediate situation ($C^\alpha$, $\alpha \in (0, 2)$, dampings). We refer to~\cite{HZZ} for resuts in this direction in dimension $1$.
\end{rem}
\begin{rem}
In this article, we do not treat the case where $\Sigma\cap\partial\Omega\neq \emptyset$. In that case, $\partial\Omega_2$ can be only Lipchitz, and more technical treatments for the propagation of singularities are needed near the points $\Sigma\cap\partial\Omega$.
\end{rem}
\begin{thm}\label{thm:resolvent}
We have Spec$(\mathcal{A})\cap i\R=\emptyset$. Moreover, there exists $C>0$, such that for all $\lambda\in\R, |\lambda|\geq 1$,
\begin{equation}\label{eq:resolvent}
\big\|(i\lambda-\mathcal{A})^{-1}\big\|_{\mathcal{L}(\mathcal{H})}\leq C|\lambda|.
\end{equation}
Moreover, when $\Omega\subset \R^d$, $d\geq 2$ and $\Omega_2=\mathbb{D}\subset \Omega$ is a unit ball, $\Omega_1=\Omega\setminus\Omega_2$, we actually have a lower bound:
$$\limsup_{\lambda\rightarrow + \infty} \lambda^{-1} \big\|(i\lambda-\mathcal{A})^{-1}\big\|_{\mathcal{L}(\mathcal{H})}=c >0.$$ In other words, there exist sequences $(U_n)\subset \mathcal{H}^1$ and $\lambda_n\rightarrow +\infty $ such that
\begin{equation}\label{lowerboundbis} \|U_n\|_{\mathcal{H}}=1,\; \|(i\lambda_n-\mathcal{A})U_n\|_{\mathcal{H}}=O(\lambda_n^{-1}).
\end{equation}
\end{thm}
\noi
$\bullet${ Theorem \ref{thm:decay} and Theorem \ref{thm:resolvent}} are essentially equivalent. Indeed, the equivalence between the resolvent estimate \eqref{eq:resolvent} and the decay rate \eqref{eq:decayrate} is covered by Theorem 2.4 of \cite{BoTo}. It is very likelly that~\eqref{lowerbound} and~\eqref{eq:resolvent} are also equivalent. However, we prove here only the fact that ~\eqref{lowerboundbis} implies \eqref{lowerbound}. We argue as follows:
Let $U_n$ be a sequence of quasi-modes associated with $\lambda_n$ ($\lambda_n\rightarrow+\infty$) that saturates \eqref{eq:resolvent}. Denote by $F_n=(i\lambda_n-\mathcal{A})U_n$. We have
$$ \|U_n\|_{\mathcal{H}}=1,\;\|F_n\|_{\mathcal{H}}=O(\lambda_n^{-1}),\quad \|U_n\|_{\mathcal{D}(\mathcal{A})}\sim \lambda_n.
$$
Define $U_n(t)=\mathrm{e}^{t\mathcal{A}}U_n$ and we write
$$
U_n(t)=\mathrm{e}^{i\lambda_n t}U_n+R_n(t),
$$
then
$$ (\partial_t-\mathcal{A})R_n=-(i\lambda_n-\mathcal{A})\mathrm{e}^{it\lambda_n}U_n=O_{\mathcal{H}}(\lambda_n^{-1}),\; R_n(0)=0.
$$
Since $$R_n(t)=-\int_0^t\mathrm{e}^{(t-s)\mathcal{A}}(i\lambda_n-\mathcal{A})\mathrm{e}^{is\lambda_n}U_nds,$$
we deduce that $\|R_n(t)\|_{\mathcal{H}}=O(\lambda_n^{-1}t)$ for $t>0$. Assume that $\kappa(t)$ is the optimal decay rate of the energy, then by $E[U_n(t)]^{\frac{1}{2}}=\|U_n(t)\|_{\mathcal{H}}\leq \kappa(t)^{\frac{1}{2}}\|U_n\|_{\mathcal{D}(\mathcal{A})}$ we have
$$ C_1\kappa(t)^{\frac{1}{2}} \lambda_n\geq 1-\|R_n(t)\|_{\mathcal{H}}=1-C_2\lambda_n^{-1}t.
$$
For fixed $t>0$, we choose $n$ large enough such that $C_2\lambda_n^{-1}t=\frac{1}{2}$, thus we obtain that
$$ \kappa(t)^{\frac{1}{2}}\geq\frac{1}{2C_1\lambda_n}=\frac{1}{C_1C_2 t}.
$$
This proves \eqref{lowerbound}. As a consequence, we shall in the sequel reduce the analysis to the proof of Theorem~\ref{thm:resolvent}.
This article is organized as follows. We present the proof of \eqref{eq:resolvent} of Theorem \ref{thm:resolvent} in Section 2, Section 3 and Section 4. The proof follows from a contradiction argument which reduces the matter to study the associated high energy quasi-modes. In Section 2, we reduce the equation of quasi-modes to a transmission problem, consisting of an elliptic system in $\Omega_1$ and a hyperbolic system in $\Omega_2$, coupled at the interface $\Sigma$. Both systems are semi-classical but with different scales $h, \hbar= h^{1/2}$. Next in Section 3, we study the elliptic system and obtain the information of the quasi-modes restricted to the interface by transmission conditions. Then in Section 4, we prove the propagation theorem for the hyperbolic problem in $\Omega_2$ which will lead to a contradiction. We need to analyze two semi-classical scales corresponding to the elliptic and hyperbolic region, connected by the transmission condition on the interface.
Finally in Section 5, we construct a sequence of quasi-modes saturating the inequality \eqref{eq:resolvent} in a simple geometry. In particular this proves the optimality of the resolvent estimate. We collect various toolboxes in the final section of the appendix.
Throughout this article, we adopt the standard notations in semi-classical analysis (see for example \cite{EZB}). We will use the standard quantization for classical and semi-classical pseudo-differential operators $\mathrm{Op},\mathrm{Op}_h,\mathrm{Op}_{\hbar}$. We will also adopt the usual asymptotic notations, such as $O(h^{\alpha}),O(\hbar^{\alpha})$ and $o(h^{\alpha}),o(\hbar^{\alpha})$, as $h\rightarrow 0$. Moreover, for a Banach space $X$ and $h$-dependent families of functions $f_h,g_h$, we mean $f_h=O_X(h^{\alpha})$, $g_h=o_X(h^{\alpha})$, if
$$ \|f_h\|_X=O(h^{\alpha}),\quad \|g_h\|_X=o(h^{\alpha}),
$$
as $h\rightarrow 0$.
\subsection*{Acknowledgment}
The first author is supported by Institut Universitaire de France and
ANR grant ISDEEC, ANR-16-CE40-0013. The second author is supported by the postdoc programe: ``Initiative d'Excellence Paris Seine" of CY Cergy-Paris Universit\'e and ANR grant
ODA (ANR-18-CE40- 0020-01).
\section{Reduction to a transmission problem }
It was proved by the first author in \cite{B19} that
$$ \|(\mathcal{A}-i\lambda)^{-1}\|_{\mathcal{L}(\mathcal{H})}\leq Ce^{c|\lambda|}
$$
under more general conditions for the damping. Therefore, the proof of the first part of Theorem \ref{thm:resolvent} (i.e. \eqref{eq:resolvent}) is reduced to the high energy regime $|\lambda|\rightarrow+\infty$.
For this, we argue by contradiction. Assume that \eqref{eq:resolvent} is not true, then there exist $h$-dependent functions $U=\binom{u}{v}, F=\binom{f}{g}$, such that
\begin{align}\label{seq:contradiction}
&\|U_j\|_{H^1\times L^2}=O(1),\;\|F_j\|_{H^1\times L^2}=o(h)\\
& (\mathcal{A}-ih^{-1})U=F.
\end{align}
Let $\nu$ be the unit normal vector pointing to the undamped region $\Omega$. Denote by $a_1(x)=a(x)\mathbf{1}_{\ov{\Omega}_1}$
Let $U=\binom{u}{v}$ and $F=\binom{f}{g}$. Then for $U\in D(\mathcal{A})$ and $F\in\mathcal{H}$, the equation
$$ (\mathcal{A}-i\lambda)U=F
$$
is equivalent to ($h=\lambda^{-1}$) the following system for $u_j=u\mathbf{1}_{\Omega_j}, f_j=f\mathbf{1}_{\Omega_j},$ and $g_j=g\mathbf{1}_{\Omega_j}$, $j=1,2$:
\begin{align}\label{system1}
\begin{cases}
& u_1=ih(f_1-v_1), \text{ in }\Omega_1\\
& h\Delta u_1+h\nabla_x\cdot(a_1(x)\nabla v_1)-i v_1=hg_1,\text{ in } \Omega_1 \\
& u_2=ih(f_2-v_2), \text{ in } \Omega_2\\
& h\Delta u_2-iv_2=hg_2,\text{ in }\Omega_2
\end{cases}
\end{align}
with boundary condition on the interface
\begin{align}\label{BC:interface}
u_1|_{\Sigma}=u_2|_{\Sigma},\quad \partial_{\nu}u_2|_{\Sigma}=(\partial_{\nu}u_1+a_1\partial_{\nu}v_1 )|_{\Sigma},
\end{align}
Indeed, the equations inside $\Omega_1,\Omega_2$ can be verified directly. The first boundary condition is just the fact that the function $u$ equal to $u_j$ in $\Omega_j$ must have no jump at the interface to enssure taht ity belongs to $H^1( \Omega)$. To check the second boundary condition, we take an arbitrary test function $\varphi\in C_c^{\infty}(\Omega)$ and multiply the equation $h\Delta u-iv+h\Div a\nabla v=0$ by $\varphi$. We obtain that
\begin{align*}
0=&-h\int_{\Omega}\nabla u\cdot\nabla\ov{\varphi}-h\int_{\Omega}a\nabla v\cdot\nabla\ov{\varphi}-i\int_{\Omega}v\ov{\varphi}\\
=&-\sum_{j=1}^2\int_{\Omega_j}\big(h\nabla u_j\cdot\nabla\ov{\varphi}-iv_j\ov{\varphi}\big)-h\int_{\Omega_1}a_1(x)\nabla v_1\cdot\nabla\ov{\varphi}\\
=&\sum_{j=1}^2\int_{\Omega_j}\big(h\Delta u_j\ov{\varphi}-iv_j\ov{\varphi}\big)+\int_{\Omega_1}h\nabla_x\cdot(a_1(x)\nabla v_1)\cdot \ov{\varphi}+h\int_{\Sigma}(\partial_{\nu}u_2-\partial_{\nu}u_1-a_1\partial_{\nu}v_1)|_{\Sigma}\cdot\ov{\varphi}.
\end{align*}
Using the differential equations in $\Omega_1,\Omega_2$, the last term on the right side is equal to $$h\int_{\Sigma}(\partial_{\nu}u_2-\partial_{\nu}u_1-a_1\partial_{\nu}v_1)|_{\Sigma}\cdot\ov{\varphi}|_{\Sigma},$$
hence it must vanish for all $\varphi$. This verifies \eqref{BC:interface}.
First we prove an a priori estimate for these functions:
\begin{lem}[A priori estimate]\label{apriori}
Denote by $U_j=\binom{u_j}{v_j}, F_j=\binom{f_j}{g_j}$, for $j=1,2$.
Assume that $\|U_j\|_{H^1\times L^2}=O(1)$ and $\|F_j\|_{H^1\times L^2}=o(h)$, then we have
$$ \|\nabla v_1\|_{L^2}=o(h^{\frac{1}{2}}), \quad \|v_1\|_{L^2}=o(h)
$$
and
$$ \|\nabla u_1\|_{L^2}=o(h^{\frac{3}{2}}),\quad \|u_1\|_{L^2}=o(h^2).
$$
Consequently, by the trace theorem, we have
$$ \|u_1\|_{H^{\frac{1}{2}}(\Sigma)}=o(h^{\frac{3}{2}}),\quad \|v\|_{H^{\frac{1}{2}}(\Sigma)}=o(h^{\frac{1}{2}}).
$$
\end{lem}
\begin{proof}
First we observe that, from the relation between $u$ and $v$, we deduce that $\nabla v\in L^2(\Omega)$ and
\begin{equation}\label{eq1}
\|\nabla v_j\|_{L^2(\Omega_j)}=O(h^{-1}),\quad j=1,2.
\end{equation}
Moreover, by the trace theorem, $v_1|_{\Sigma}=v_2|_{\Sigma}$ as functions in $H^{\frac{1}{2}}(\Sigma)$.
From the system \eqref{system1}, we have
\begin{align}
& (\nabla u_1, \nabla v_1)_{L^2(\Omega_1)}\notag \\
&=ih(\nabla f_1,\nabla v_1)_{L^2(\Omega_1)}-ih\|\nabla v_1\|_{L^2(\Omega_1)}^2-(\nabla u_1,\nabla v_1)_{L^2(\Omega_1)}\notag \\
& \qquad -\|a_1^{1/2}\nabla v_1\|_{L^2(\Omega_1)}^2+(\partial_{\nu}u_1+a_1\partial_{\nu}v_1,v_1)_{L^2(\Sigma)} \label{eq2}\\
&=ih^{-1}\|v_1\|_{L^2(\Omega_1)}^2+(g_1,v_1)_{L^2(\Omega_1)}-(\nabla u_2, \nabla v_2)_{L^2(\Omega_2)}\label{eq3}\\
&=ih(\nabla f_2,\nabla v_2)_{L^2(\Omega_2)}-ih\|\nabla v_2\|_{L^2(\Omega_2)}^2-(\nabla u_2,\nabla v_2)_{L^2(\Omega_2)}-(\partial_{\nu}u_2,v_2)_{L^2(\Sigma)}\label{eq4}\\
&=ih^{-1}\|v_2\|_{L^2(\Omega_2)}^2+(g_2,v_2)_{L^2(\Omega_2)} \label{eq5}.
\end{align}
Taking the real part of \eqref{eq2}+\eqref{eq3}-\eqref{eq4}+\eqref{eq5}, we deduce that
$ \|\nabla v_1\|_{L^2(\Omega_1)}^2=o(h)
$, thanks to the boundary condition \eqref{BC:interface} and $v_1|_{\Sigma}=v_2|_{\Sigma}$. Therefore, from the first equation of \eqref{system1}, we have $\|\nabla u_1\|_{L^2(\Omega_1)}^2=o(h^3)$. Then, using this fact and the second equation of \eqref{system1}, we deduce that
$$ iv_1=h\Delta u_1+h\nabla\cdot(a_1\nabla v_1)-hg_1=O_{H^{-1}(\Omega_1)}(h^{\frac{3}{2}}).
$$
By interpolation, we have $v_1=o_{L^2(\Omega_1)}(h)$, and from $u_1=ih(f_1-v_1)$, $u_1=o_{L^2(\Omega_1)}(h^2)$. This completes the proof of Lemma \ref{apriori}.
\end{proof}
\section{Estimates of the elliptic system}\label{sectionelliptic}
\subsection{Standard theory}
We briefly recall the semiclassical elliptic boundary value problem near the interface $\Sigma$.
In what follows, we will sketch the parametrix construction for \eqref{elliptic1}, following \cite{BL03}. Near a point $x_0\in\Sigma$, we use the coordinate system $(y,x')$ where $\Omega_1=\{(y,x'):y>0\}$ near $x_0$.
\begin{equation}\label{elliptic1} L_{\hbar}w=\kappa=o_{L^2}(\hbar^2),\quad w|_{\Omega_1}=o_{H^1}(\hbar),\; w|_{\Sigma}=o_{H^{\frac{1}{2}}}(\hbar)
\end{equation}
where in the local coordinate chart,
$$ L_{\hbar}:=\hbar^2D_y^2-R(y,x',\hbar D_{x'})+\sum_{j=1}^{d-1}\hbar M_j(y,x')\hbar\partial_{x_j'}+\hbar H(y,x')\hbar\partial_y.
$$
Here $R(y,x',\hbar D_{x'})$ is a second order semiclassical differential operator in $x'$ with the principal symbol $r(y,x',\xi')$. The principal symbol of
$L_{\hbar}$ is
$$ l(y,x',\eta,\xi')=\eta^2-r(y,x',\xi'),
$$
and we denote by
$$ m(y,x',\eta,\xi')=\sum_{j=1}^{d-1}M_j(y,x')\xi_j'+H(y,x')\eta.
$$
The set of elliptic points in $T^*\partial\Omega$ is given by
$$ \mathcal{E}:=\{(y=0,x',\xi'): r(0,x',\xi')<0 \}
$$
By homogeneity, near a point $\rho_0\in\mathcal{E}$
\begin{equation}\label{localization-elliptic}
-r(y,x',\xi')\geq c(\rho_0)|\xi'|^2.
\end{equation}
Denote by $\ud{w}:=w\mathbf{1}_{y\geq 0}$ the extension by zero of $w$, and the same for $\ud{\kappa},$ etc. Then $\ud{w}$ satisfies the equation
\begin{equation}\label{elliptic2}
L_{\hbar}\ud{w}=-\hbar(\hbar\partial_yw)|_{y=0}\otimes\delta_{y=0}+\hbar^2w|_{y=0}\otimes\delta'_{y=0}+\hbar^2H(0,x')w|_{y=0}\otimes\delta_{y=0}+\ud{\kappa}.
\end{equation}
Let $\varphi(y,x')$ be a cut-off to the local chart. Let $\psi\in C^{\infty}(\R^{d-1})$, be a Fourier multiplier in $S_{\xi'}^0$ such that on the support of $\varphi(y,x')\psi(\xi')$, \eqref{localization-elliptic} holds and $\varphi(y,x')\psi(\xi')=1$ near $\rho_0$. We define
\begin{equation}\label{symbol:elliptic}
e^0(y,x',\eta,\xi'):=\frac{\varphi(y,x')\psi(\xi')}{l(y,x',\eta,\xi')}
\end{equation}
and $e^j,j\geq 1$ inductively by
\begin{equation*}
\begin{split}
e^1\cdot l=&-\sum_{|\alpha|=1}\frac{1}{i}\partial_{\xi',\eta}^{\alpha}e^0\cdot\partial_{x',y}^{\alpha}l-e^0\cdot m,\\
e^j\cdot l=-&\sum_{|\alpha|+k=n,k\neq n}\frac{1}{i^{|\alpha|}}\partial_{\xi',\eta}^{\alpha}e^k\cdot\partial_{x',y}^{\alpha}l-\sum_{|\alpha|+k=n-1}\frac{1}{i^{|\alpha|}}\partial_{\xi',\eta}^{\alpha}e^k\cdot\partial_{x',y}^{\alpha}m.
\end{split}
\end{equation*}
For any $N\in\N$, we define
$$ e_N=\sum_{j=0}^N\hbar^je^j,\quad E_N=\mathrm{Op}_{\hbar}(e_N),
$$
and then
$$ E_NL_{\hbar}=\varphi(y,x')\psi(\xi')\mathrm{Id}+R_N,
$$
where
$$ R_N=\mathcal{O}(\hbar^{N+1}): L_{x',y}^2\rightarrow L_{x',y}^2,\quad R_N=\mathcal{O}(\hbar^{N+1-2M}): H_{x',y}^s\rightarrow H_{x',y}^{s+2M},
$$
and
$$ E_N=\mathcal{O}(1):L_{x',y}^2\rightarrow L_{x',y}^2,\quad E_N=\mathcal{O}(\hbar^{-2}): H_{x',y}^s\rightarrow H_{x',y}^{s+2},
$$
thanks to Lemma \ref{technical1}.
Applying $E_N$ to the equation \eqref{elliptic2}, we obtain that
\begin{equation*}
\begin{split}
\varphi(y,x')\psi(\hbar D_{x'})\ud{w}=&-\hbar^2E_N((\partial_{y}w)|_{y=0}\otimes\delta_{y=0})+\hbar^2E_N(w|_{y=0}\otimes\delta'_{y=0} )+\hbar^2E_N(Hw|_{y=0}\otimes\delta_{y=0} )\\+&E_N\ud{\kappa}-R_N\ud{w}.
\end{split}
\end{equation*}
Note that $e_N(y,x',\eta,\xi')$ is meromorphic in $\eta$ with poles $\eta_{\pm}=\pm i\sqrt{-r(y,x',\xi')}$.
Denote by $G(x')=\partial_yw(0,x')+H(0,x')w(0,x')$, we calculate for $y>0, x'\in\R^{d-1}$ that
\begin{equation*}
\begin{split}
&\hbar^2E_N((\partial_yw+Hw)|_{y=0}\otimes\delta_{y=0})(y,x')\\=&\frac{\hbar^2}{(2\pi\hbar)^d}\int G(\widetilde{x}')\mathrm{e}^{\frac{i(x'-\widetilde{x}')\cdot\xi'}{\hbar}}d\widetilde{x}'d\xi'\int e_N(y,x',\eta,\xi')\mathrm{e}^{\frac{iy\eta}{\hbar}}d\eta\\
=&\frac{i\hbar}{(2\pi\hbar)^{d-1}}\int \mathrm{e}^{\frac{iy\eta_+}{\hbar}}n_N(y,x',\xi')\mathrm{e}^{\frac{i(x'-\widetilde{x'})\cdot\xi'}{\hbar}}G(\widetilde{x}')d\widetilde{x}'d\xi',
\end{split}
\end{equation*}
where $n_N(y,x',\xi')=\mathrm{Res}(e_N(y,x',\eta,\xi');\eta=\eta_+)$. Similarly, for $y>0, x'\in\R^{d-1}$,
\begin{equation*}
\begin{split}
\hbar^2E_N(w|_{y=0}\otimes\delta'_{y=0})(y,x')=&\frac{i\hbar}{(2\pi\hbar)^d}\int w(0,\widetilde{x}')\mathrm{e}^{\frac{i(x'-\widetilde{x}')\cdot\xi'}{\hbar}}d\widetilde{x}'d\xi'\int \eta\mathrm{e}^{\frac{iy\eta}{\hbar}}
e_N(y,x',\eta,\xi')d\eta\\
=&-\frac{1}{(2\pi\hbar)^{d-1}}\int \mathrm{e}^{\frac{iy\eta_+}{\hbar}}d_N(y,x',\xi')\mathrm{e}^{\frac{i(x'-\widetilde{x}')\cdot\xi'}{\hbar}}w(0,\widetilde{x}')d\widetilde{x}'d\xi',
\end{split}
\end{equation*}
where $d_N=\eta_+n_N$. Therefore,
\begin{multline}\label{pseudo-relation}
\varphi(y,x')\psi(\hbar D_{x'})\ud{w}\\
=i\mathrm{Op}_{\hbar}(\mathrm{e}^{iy\eta_+/\hbar}n_N(y,\cdot))\big(-(\hbar\partial_yw)|_{y=0}+\hbar (Hw)|_{y=0}\big)-\mathrm{Op}_{\hbar}(\mathrm{e}^{iy\eta_+/\hbar}d_N(y,\cdot))(w|_{y=0})\\+E_N\ud{\kappa}-R_N\ud{w},
\end{multline}
where the two operators in the expression above are tangential. Note that by Lemma \ref{technical1}
$$ R_N\ud{w},\quad E_N\ud{\kappa}=o_{L_{x',y}^2}(\hbar^2)=o_{H_{x',y}^2}(1),
$$
hence from the interpolation and the trace theorem, we have
$$ (R_N\ud{w})|_{y=0}=o_{H_{x'}^{1/2}}(\hbar),\quad (E_N\ud{\kappa})|_{y=0}=o_{H_{x'}^{1/2}}(\hbar).
$$
Taking the trace $y=0$ for \eqref{pseudo-relation}, we obtain that
\begin{equation}\label{trace-relation} \mathrm{Op}_{\hbar}(\varphi(0,x')\psi(\xi')+d_N(0))(w|_{y=0})=-\mathrm{Op}_{\hbar}(in_N(0))((\hbar\partial_yw)|_{y=0}+\hbar(Hw)|_{y=0} )+o_{H_{x'}^{1/2}}(\hbar).
\end{equation}
Note that the principal symbols of $n_N(0), d_N(0)$ are
$$ \sigma(in_N(0))=\frac{\varphi(0,x')\psi(\xi')}{2\sqrt{-r(0,x',\xi')}},\quad \sigma(d_N(0))=\frac{\varphi(0,x')\psi(\xi')}{2}.
$$
In summary, there exists (near $\rho_0$) a $\hbar$-P.d.O $\mathcal{N}_{\hbar}$, elliptic and of order 1 classic and of order 0 semi-classic, in the sense that
$$ \mathcal{N}_{\hbar}=\mathcal{O}(\hbar): H_{x'}^s\rightarrow H_{x'}^{s-1},
$$
such that
$$ (\hbar\partial_yw)|_{y=0}=\mathcal{N}_{\hbar}(w|_{y=0}+ O_{H^{1/2}}(\hbar)).
$$
\subsection{Control of the semi-classical wave-front set of the trace}
For the further need, we should also control the wave front set of the precise elliptic equation (with $\hbar=h^{\frac{1}{2}}$) $$\hbar^2\Delta w-\frac{i}{a_1}w+\hbar\frac{\nabla a_1}{a_1}\cdot\hbar\nabla w=\kappa,
$$
where the $h$-semiclassical wave front set of the Neumann data
$ \mathrm{WF}_h(\partial_{\nu}w|_{\Sigma}).
$
Here we need to pay attention to two different semi-classical scales.
\begin{prop}\label{control:frontdonde}
Assume that $w$ satisfies the $\hbar$-semiclassical elliptic equation (with $\hbar=h^{\frac{1}{2}}$)
$$ \hbar^2\Delta w-\frac{i}{a_1}w+\hbar\frac{\nabla a_1}{a_1}\cdot\hbar\nabla w=\kappa
$$
with Neumann trace $\partial_{\nu}w|_{\Sigma}$ and $ \mathrm{WF}_h(\partial_{\nu}w|_{\Sigma})
$ is contained in a compact subset of $T^*\Sigma\setminus\{0\}$. Assume moreover that $w=O_{H^1}(h^{\frac{1}{2}})$ and $\kappa=O_{L^2}(h)$, then we have
$$ \mathrm{WF}_h(w|_{\Sigma})\subset \mathrm{WF}_h(\partial_{\nu}w|_{\Sigma})\cup \pi\big(\mathrm{WF}_h(\kappa)\big),
$$
where $\pi: T^*\Omega_1\rightarrow T^*\Sigma$ is the projection defined for points near $T^*\Sigma$, and
$$ \pi\big(\mathrm{WF}_h(\kappa)\big)=\big\{\rho_0\in T^*\Sigma: \exists \rho\in T^*\Omega_1, \text{ near } T^*\Sigma, \text{ such that } \rho\in \mathrm{WF}_h(\kappa) \text{ and } \pi(\rho)=\rho_0 \big\}.
$$
\end{prop}
\begin{proof}
Let $(x_0,\xi_0)\notin \mathrm{WF}_h(\partial_{\nu}w|_{\Sigma})\cup\pi\big(\mathrm{WF}_h(\kappa)\big)$. Locally near $x_0\in\Sigma$, we can choose local coordinate system as in the previous subsection. Here the cutoff $\psi(\xi')$ can be chosen as $1$, since the operator $\hbar^2\Delta-i$ is always elliptic. Consider the tangential $h$-P.d.O $A_h$ which is elliptic near $(x_0,\xi_0)$ and its principal symbol is supported away from $\mathrm{WF}_h(\partial_{\nu}w|_{\Sigma})\cup \pi\big(\mathrm{WF}_h(\kappa)\big)$. We need to show that $(A_hw)|_{y=0}=O_{L^2(\Sigma)}(h^{\infty})$.
From \eqref{pseudo-relation} we have
\begin{align*}
\varphi(y,x')\ud{w}=&i\mathrm{Op}_{\hbar}\big(\mathrm{e}^{\frac{iy\eta_+}{\hbar}}n_N(y) \big)\big(-(\hbar\partial_yw)|_{y=0}+\hbar(Hw)|_{y=0} \big)-\mathrm{Op}_{\hbar}\big(\mathrm{e}^{\frac{iy\eta_+}{\hbar}}d_N(y) \big)(w|_{y=0})\\+&E_N\ud{\kappa}+O_{H^{1}}(h^{\frac{N}{2}}),
\end{align*}
where we gain $\hbar^N$ for $R_N\ud{w}$. By taking the trace $y=0$ and using the fact that $d_N(0)=\frac{1}{2}\varphi(0,x'),$ we obtain that
\begin{align*}
&(A_h\varphi(y,x')\ud{w})|_{y=0}+\big(A_h\mathrm{Op}_h(d_N(0))w\big)|_{y=0}-i\hbar (A_h\mathrm{Op}_{\hbar}(n_N)(Hw )|_{y=0}\\=&-iA_h\mathrm{Op}_{\hbar}(n_N(0))(\hbar\partial_{y}w)|_{y=0}+ (A_hE_N\ud{\kappa})|_{y=0}
+O_{H_{y=0}^{\frac{1}{2}}}(h^{N/2}).
\end{align*}
We claim that it suffices to show that
\begin{align}\label{WFh1} A_h\mathrm{Op}_{\hbar}(n_N(0))(\hbar\partial_{y}w)|_{y=0}=O_{L_{y=0}^2}(h^{\infty}) \text{ and } (A_hE_N\ud{\kappa})|_{y=0}=O_{L_{y=0}^2}(h^{\infty}).
\end{align}
Indeed, once this is done, we obtain that, at least $(A_h\ud{w})|_{y=0}=O_{L^2}(\hbar)$\footnote{The operator $A_h$ is of the form $\chi A_h\chi$ for some cutoff $\chi$ such that $\chi\equiv 1$ on supp$(\varphi)$. }.
Now we can replace $A_h$ by another tangential operator $\widetilde{A}_h$ with principal symbol $\widetilde{a}$ such that $\widetilde{a}$ is supported in a slightly larger region containing supp$(a)$ and $\widetilde{a}=1$ on supp$(a)$. We still have $(\widetilde{A}_h\ud{w})|_{y=0}=O_{L^2}(\hbar)$. Now we write
$$ \hbar A_h\mathrm{Op}_{\hbar}(n_N)Hw=\hbar A_h\mathrm{Op}_{\hbar}(n_N)H\widetilde{A}_hw+
\hbar A_h\mathrm{Op}_{\hbar}(n_N)H(1-\widetilde{A}_h)w.
$$
From Lemma \ref{commutator:2echelles}, the trace of the second term on the right side is $\mathcal{O}_{L^2}(h^{\infty})$. Therefore, the trace of the first term on the right side is $O_{L^2}(\hbar^2)$, hence $(A_h\ud{w})|_{y=0}=O_{L^2}(\hbar^2)$. Then we can continuously apply this argument to conclude.
It remains to prove \eqref{WFh1}. For this, we just need to interchange the operator $A_h$ with $E_N$ and $\mathrm{Op}_{\hbar}(n_N(0))$.
Here additional attentions are needed, since $\mathrm{Op}_{\hbar}(n_N(0)), E_N$ are $\hbar$-P.d.O. This can be verified from the following lemma:
\begin{lem}\label{commutator:2echelles}
Assume that $a,b,q\in S^0(\R_x^n\times\R_{\xi}^n)$, compactly supported in the $x$ variable such that
$$ \mathrm{dist}\big(\mathrm{supp}(a),\mathrm{supp}(b) \big)\geq c_0>0.
$$
Then for any $s\in\R, N\in\N, N\geq 2n$, we have
$$ a(x,hD_x)q(x,h^{\frac{1}{2}}D_x) b(x,hD_x)=\mathcal{O}_{L^2\rightarrow L^2}(h^{N}).
$$
\end{lem}
\begin{proof}
Denote by
$$ A(x,y,\xi,\eta)=a(x,h(\xi+\eta))q(x+y,h^{\frac{1}{2}}\xi).
$$
Then from Lemma \ref{symbolic},
\begin{align*}
a(x,hD_x)q(x,h^{\frac{1}{2}}D_x)=&\sum_{|\beta|\leq N}\mathrm{Op}\big(\frac{h^{|\beta|}}{i^{|\beta|}\beta!}(\partial_{\xi}^{\beta}a)(x,h\xi)(\partial_x^{\beta}q)(x,h^{\frac{1}{2}}\xi) \big)\\
+&O_{\mathcal{L}(L^2)}(h^{N+1-n}),
\end{align*}
since for any $\beta\in \N^{2n}$,
$$ \sup_{|\alpha|=N+1}\sup_{(x,\xi)}\iint_{\R^{2n}}|\partial_{x,\xi}^{\beta}\partial_z^{\alpha}\partial_{\zeta}^{\alpha}(a(x,h(\xi+\zeta)) q(x+z,h^{\frac{1}{2}}\xi) ) |dz d\zeta =O(h^{N+1-n}).
$$
Using the fact that $\frac{h^{|\beta|}}{i^{|\beta}|\beta!}\partial_{\xi}^{\beta}a\cdot \partial_x^{\beta}q\cdot b=0$, thanks to the support property, we have, using again Lemma \ref{symbolic},
$$ a(x,hD_x)q(x,h^{\frac{1}{2}}D_x)b(x,hD_x)=O_{\mathcal{L}(L^2)}(h^N)
$$
for any $N$ large enough. This completes the proof.
\end{proof}
Therefore the proof of Proposition \ref{control:frontdonde} is now complete.
\end{proof}
\subsection{Estimate of the traces}
Let $u_1,v_1$ be solutions of the first two equations of \eqref{system1}. Consider $w=u_1+a_1v_1$, then under the assumption of Lemma \ref{apriori},
$$ w=o_{H^1}(h^{\frac{1}{2}}),\quad w=o_{L^2}(h),\quad w|_{\Sigma}=o_{H^{\frac{1}{2}}}(h^{\frac{1}{2}}).
$$
Note that $w$ satisfies the elliptic equation (with $\hbar=h^{\frac{1}{2}}$)
\begin{equation}\label{w:elliptic}
\begin{split}
\hbar^2\Delta w+\hbar\frac{\nabla a_1}{a_1}\cdot \hbar\nabla w-\frac{i}{a_1}w=\hbar^2g_1-\hbar^2\Delta a_1\cdot v_1+\hbar^2\frac{|\nabla a_1|^2}{a_1}v_1-\frac{\hbar \nabla a_1}{a_1}\cdot\hbar\nabla u_1+\frac{i}{a_1}u_1
\end{split}
\end{equation}
In particular,
$$ \hbar^2\Delta w-\frac{i}{a_1}w+\hbar\frac{\nabla a_1}{a_1}\cdot \hbar\nabla w=o_{L^2}(\hbar^2).
$$
In this case, $\mathcal{N}_{\hbar}$ defined in the last subsection is the usual $\hbar$-semiclassical Dirichlet-Neumann operator:
$$ \mathcal{N}_{\hbar}(w|_{\Sigma}+ o_{H^{1/2}} ( \hbar^2)):=(\hbar\partial_{\nu}w)|_{\Sigma}.
$$ We can apply the standard theory (to $h^{-1} w$)with the particular choice $\psi(\xi')\equiv 1$ in \eqref{symbol:elliptic} and obtain the following:
\begin{prop}\label{estimatetrace1}
Let $\chi\in C_c^{\infty}(\R)$. Then under the hypothesis of Lemma \ref{apriori} and in the local chart near $\Sigma$, we have $\varphi\chi(hD_{x'})\varphi (\partial_{y}w)|_{y=0}=o_{L_{x'}^2}(1)$, where $h=\hbar^2$. Consequently,
$$ u_2|_{\Sigma}=o_{H^{1/2}}(h^{\frac{3}{2}}),\quad \varphi\chi(hD_{x'})\varphi h\partial_yu_2|_{y=0}=o_{L^2(\Sigma)}(h).
$$
\end{prop}
\begin{proof}
Assume that $\varphi,\varphi_1$ are supported in a local chart and satisfy $\varphi_1|_{\text{supp}(\varphi)}=1$. In apriori, we have $\varphi\partial_{y}(\varphi_1w)|_{y=0}=\varphi\hbar^{-1}\mathcal{N}_{\hbar}((\varphi_1w)|_{y=0})=o_{H_{x'}^{-1/2}}(\hbar)$. Thus by Lemma \ref{technical1} we have
$\varphi\hbar^{-1}\chi(\hbar^2D_{x'})\varphi\mathcal{N}_{\hbar}((\varphi_1w)|_{y=0})=o_{L_{x'}^2}(1)$.
\end{proof}
\section{Propagation estimate}
In this section, we will deal with the propagation estimate for $u_2$ in $H^1$, satisfying
\begin{equation}\label{eq:u2}
\begin{split}
& (h^2\Delta+1)u_2=ihf_2+h^2g_2=o_{H^1}(h^2)+o_{L^2}(h^3),\text{ in } \Omega_2,\\
&\|u_2\|_{H^1(\Omega_2)}=O(1),\; \|u_2|_{\Sigma}\|_{H^{1/2}(\Sigma)}=o(h^{3/2}),\\ &\|h\partial_{\nu}u_2|_{\Sigma}\|_{H^{-1/2}(\Sigma)}=o(h^{3/2}),\;\|\varphi\psi(hD_{x'})\varphi h\partial_{\nu} u_2|_{\Sigma}\|_{L^2(\Sigma)}=o(h).
\end{split}
\end{equation}
Set $w_2=h^{-1}u_2$. From \eqref{eq:u2},
$$ -\|\nabla u_2\|_{L^2(\Omega_2)}^2+\|h^{-1}u_2\|_{L^2(\Omega_2)}^2=\big\langle(\partial_{\nu}u_2)|_{\Sigma}\cdot u_2|_{\Sigma}\big\rangle^{H^{1/2}(\Sigma)}_{H^{-1/2}(\Sigma)}+o(1)=o(1).
$$
Hence we could equivalently deal with the propagation estimate for $w_2$ in $L^2$, satisfying
\begin{equation}\label{eq:w2}
\begin{split}
& (h^2\Delta+1)w_2=if_2+hg_2=o_{H^1}(h)+o_{L^2}(h^2),\text{ in } \Omega_2,\\
&\|w_2\|_{H^1(\Omega_2)}=O(h^{-1}),\|w_2\|_{L^2(\Omega_2)}=O(1),\; \|w_2|_{\Sigma}\|_{H^{1/2}(\Sigma)}=o(h^{1/2}),\\ &\|h\partial_{\nu}w_2|_{\Sigma}\|_{H^{-1/2}(\Sigma)}=o(h^{1/2}),\; \|\varphi \psi(hD_{x'})\varphi h\partial_{\nu}w_2|_{\Sigma}\|_{L^2(\Sigma)}=o(1).
\end{split}
\end{equation}
The goal of this section is to prove the invariance of the semiclassical measure $\mu$ associated with (a subsequence of) $w_2$ and finally prove that $\mu=0$ from the boundary conditions in \eqref{eq:w2} on the interface $\Sigma$. This will end the contradiction argument.
\subsection{Propagation away from \texorpdfstring{$\Sigma$}{Sigma}}
The defect measure in the interior of $\Omega_2$ for $u_2$ is defined via the following quadratic form:
$$ \phi(Q_h,w_2)=(Q_hw_2,w_2)_{L^2(\Omega_2)}:=\int_{\Omega_2}Q_hw_2\cdot \ov{w}_2dx.
$$
\begin{prop}[Interior propagation]\label{propagation:interior}
Let $Q_h=\widetilde{\chi}Q_h\widetilde{\chi}$ be a $h$-pseudodifferential operator of order 0, where $\widetilde{\chi}\in C_c^{\infty}(\Omega_2)$, then we have
$$ \frac{1}{ih}\big([h^2\Delta+1,Q_h] w_2, w_2 \big)_{L^2}=o(1).
$$
\end{prop}
\begin{proof}
By developing the commutator and using the equation \eqref{eq:u2}, we have
\begin{align*}
\big(\frac{1}{ih}[h^2\Delta+1,Q_h]w_2, w_2\big)=& \frac{1}{ih}\big( (h^2\Delta+1)Q_hw_2,w_2 \big)-\frac{1}{ih}\big( Q_h(if_2+hg_2), w_2 \big)\\
=& \frac{1}{ih}\big(Q_hw_2,if_2+hg_2 \big)+\frac{1}{ih}\big( Q_h(if_2+hg_2), w_2 \big)\\
=&o(1),
\end{align*}
where we used the integration by part without boundary terms, since the kernel of $Q_h$ is supported away from the boundary $\Sigma=\partial\Omega_2$.
This completes the proof of Proposition \ref{propagation:interior}.
\end{proof}
\subsection{Geometry near the interface}\label{geometry}
Near $\Sigma=\partial\Omega_2$, we adopt the local coordinate system $x=(y,x')$ in $U:=(-\epsilon_0,\epsilon_0)_y\times X_{x'}$ for the tubular neighborhood of $\Sigma$, similar as in the previous section but with the new convention
$\Omega_2\cap U=(0,\epsilon_0)_y\times X_{x'}$. In this coordinate system, the Euclidean metric $dx^2$ is identified as the matrix
$$
\ov{g}=\left(\begin{matrix}
1 &0\\
0 &g(y,x')
\end{matrix}\right),\text{ or } \ov{g}^{-1}:=\left(\begin{matrix}
1 &0\\
0 &g^{-1}(y,x')
\end{matrix}
\right).
$$
Near $\Sigma$, the defect measure $\mu$ for $w_2$ is defined via the quadratic form for tangential operators:
$$ \phi(Q_h,w_2):=\int_{U}Q_hw_2\cdot \ov{w}_2 \sqrt{|\ov{g}|}dydx',
$$
where $|\ov{g}|:=\mathrm{det}(\ov{g})$.
The principal symbol of the operator $P_{h,0}=-(h^2\Delta+1)$ is $$p(y,x',\eta,\xi')=\eta^2+|\xi'|_{\ov{g}}^2-1:=\eta^2+\langle\xi',\ov{g}^{-1}\xi'\rangle_{\R^{d-1}}-1.$$
By Char($P$) we denote the characteristic variety of $p$:
$$\textrm{Char}(P):=\{(x,\xi)\in T^*\mathbb{R}^{d}|_{\ov{\Omega}}:p(x,\xi)=0\}.
$$
Denote by $^bT\ov{\Omega}_2$ the vector bundle whose sections are the vector fields $X(p)$ on $\ov{\Omega}_2$ with $X(p)\in T_p\partial\Omega_2$ if $p\in\partial\Omega_2$. Moreover, denote by $^bT^*\ov{\Omega}_2$ the Melrose's compressed cotangent bundle which is the dual bundle of $^bT\ov{\Omega}_2$. Let
$$ j:T^*\ov{\Omega}_2\rightarrow ^bT^*\ov{\Omega}_2
$$
be the canonical map. In our geodesic coordinate system near $\partial\Omega_2$, $^bT\ov{\Omega}_2$ is generated by the vector fields $\frac{\partial}{\partial x'_1},\cdot\cdot\cdot,
\frac{\partial}{\partial x'_{d-1}},y\frac{\partial}{\partial y}$ and thus $j$ is defined by
$$ j(y,x';\eta,\xi')=(y,x';v=y\eta,\xi').
$$
Let $Z:=j(\textrm{Char}(P))$.
By writing in another way
$$p=\eta^2-r(y,x',\xi'),\; r(y,x',\xi')=1-|\xi'|_{\ov{g}}^2,
$$
we have the standard decomposition
$$ T^*\partial\Omega_2=\mathcal{E}\cup\mathcal{H}\cup\mathcal{G},
$$
according to the value of $r_0:=r|_{y=0}$ where
$$\mathcal{E}=\{r_0<0\},
\mathcal{H}=\{r_0>0\},
\mathcal{G}=\{r_0=0\}.
$$
The sets $\mathcal{E},\mathcal{H},
\mathcal{G}$ are called elliptic, hyperbolic and glancing, with respectively. We define also the set
$$ \mathcal{H}_{\delta}:=\{ \delta<r_0<1-\delta \}
$$
with $0<\delta<\frac{1}{2}$ for the non-tangential and non incident points.
Note that here the elliptic points $\mathcal{E}$ is different from those defined in Section \ref{sectionelliptic}.
To classify different situations as a ray approaching the boundary, we need more accurate decomposition of the glancing set $\mathcal{G}$. Let $r_1=\partial_yr|_{y=0}$ and define
$$ \mathcal{G}^{k+3}=\{(x',\xi'):r_0(x',\xi')=0,H_{r_0}^j(r_1)=0,\forall j\leq k;H_{r_0}^{k+1}(r_1)\neq 0\}, k\geq 0\}
$$
$$
\mathcal{G}^{2,\pm}:=\{(x',\xi'):r_0(x',\xi')=0,\pm r_1(x',\xi')>0\},\mathcal{G}^2:=\mathcal{G}^{2,+}\cup\mathcal{G}^{2,-}.
$$
Next we recall the definition of the generalized bicharacteristic:
\begin{definition}\label{generalbicha}
A generalized bicharacteristic of $\Omega_2$ is a piecewise continuous map from $\R$ to $^bT^*\ov{\Omega}_2$ such that at any discontinuity point $s_0$, the left and right limits $\gamma(s_0\mp)$ exist and are the two points above the same hyperbolic point on the boundary (this property translates the specular reflection of geometric optics) and except at these isolated points the curve is $C^1$ and satisfies
\begin{itemize}
\item $\frac{d\gamma}{ds}(s)=H_p(\gamma(s))$ if $\gamma(s)\in T^*\Omega_2$ or $\gamma(s)\in \mathcal{G}^{2,+}$
\item $\frac{d\gamma}{ds}(s)=H_p(\gamma(s))-\frac{H_p^2y}{H_y^2p}H_y $ if $\gamma(s)\in \mathcal{G}\setminus\mathcal{G}^{2,+}$ where $y$ is the boundary defining function.
\end{itemize}
\end{definition}
\begin{rem}
The first property in the definition above is the fact that the curve is a geodesic in the interior or passing though a non diffractive point. The second one is that passing through a non diffractive gliding point it is curved to be forced to remain in the interior of $T^*\partial\Omega_2$ for a while. When the domain is smooth and does not have infinite order of contact with its tangents, then (see \cite{MS-1}) through each point passes a unique generalized bicharacteristic. In general only existence is known.
\end{rem}
\begin{rem}
In the statement of the geometric control condition \ref{GCC}, the generalized rays are the projection of the generalized bicharacteristics of $\Omega$ onto $\ov{\Omega}$.
\end{rem}
\subsection{Elliptic regularity}
\begin{lem}\label{PoissonItegral}
Denote by $\lambda(y,x',\xi')=\sqrt{|\xi'|_{\ov{g}}^2-1}$. Let $\psi\in C^{\infty}(\R^{d-1}), \varphi_1,\varphi_2\in C_c^{\infty}(\R^{d})$, such that on the support of $\psi(\xi')\varphi_1(y,x')$ and $\psi(\xi')\varphi_2(y,x')$, $|\xi'|_g>1+\delta$ for some $\delta>0$. Then we have
$$ \mathrm{Op}_h\big(\mathbf{1}_{y\geq 0}\varphi_2\mathrm{e}^{\frac{-y\lambda}{h}}\psi(\xi')\big)\varphi_1=\mathcal{O}(1): H^{-\frac{1}{2}}(\R_{x'}^{d-1})\rightarrow L^2(\R_+^d)
$$
\end{lem}
\begin{proof}
Denote by $T_y:=\mathrm{Op}_h\big(\mathbf{1}_{y\geq 0}\varphi_2\mathrm{e}^{\frac{-y\lambda}{h}}\psi(\xi')\big)$. By definition, we have for $f_0\in H_{x'}^{-1/2}$ and $y>0$ that
\begin{equation*}
\begin{split}
(T_yf_0)(x'):=\frac{1}{(2\pi h)^{d-1}}\iint \mathrm{e}^{-\frac{y\lambda(y,x',\xi')}{h}}\psi(\xi')\varphi_2(y,x')\mathrm{e}^{\frac{i(x'-z')\cdot\xi'}{h}}f_0(z')dz'd\xi'.
\end{split}
\end{equation*}
Denote by $F_0:=\langle D_x'\rangle^{-1/2}f_0$, then this term can be written as
$$ \mathrm{Op}\big(e^{-\frac{y\lambda(y,x',h\xi')}{h}}\psi(h\xi')\langle\xi'\rangle^{\frac{1}{2}}\varphi_2(y,x') \big)F_0.
$$
For fixed $y>0$, from the Calder\'on-Vaillancourt theorem and the support property of $\psi$, we have, for any $M>0$ that
$$\big\|\mathrm{Op}\big(e^{-\frac{y\lambda(y,x',h\xi')}{h}}\psi(h\xi')\langle\xi'\rangle^{\frac{1}{2}}\varphi_2(y,x') \big)F_0\big\|_{L_{x'}^2}\leq C_Mh^{-\frac{1}{2}}e^{-\frac{cy}{h}}\big(1+\frac{y^M}{h^M}\big)\|F_0\|_{L_{x'}^2}.
$$
and the constants $C_M, c$ are independent of $y$. Squaring the inequality above and integrating in $y$ yields the bound $O(1)\|F_0\|_{L_{x'}^2}^2=O(1)\|f_0\|_{H_{x'}^{-\frac{1}{2}}}^2.$
This completes the proof of Lemma \ref{PoissonItegral}.
\end{proof}
\begin{prop}\label{elliptic}
$$ \mu\mathbf{1}_{\mathcal{E}}=0.
$$
\end{prop}
\begin{proof}
Applying \eqref{pseudo-relation} to $\kappa=o_{H^1}(h)+o_{L^2}(h^2)$ and $\hbar=h$, we obtain that
\begin{equation}\label{eqelliptic}
\begin{split}
\varphi(y,x')\psi(hD_{x'})\ud{w_2}=&-\mathrm{Op}_h\big(\mathrm{e}^{\frac{iy\eta_+}{h}}in_N(y,\cdot) \big)(h\partial_yw_2|_{y=0})-\mathrm{Op}_h\big(\mathrm{e}^{\frac{iy\eta_+}{h}}d_N(y,\cdot)\big)(w_2|_{y=0})\\
+&\mathrm{Op}_h\big(\mathrm{e}^{\frac{iy\eta_+}{h}}in_N(y,\cdot) \big)(hH(0,x')w_2|_{y=0})+o_{L_{y,x'}^2}(h^2).
\end{split}
\end{equation}
Applying Lemma \ref{PoissonItegral}, we have
$ \mathrm{Op}_h\big(\mathrm{e}^{\frac{iy\eta_+}{h}}in_N(y,\cdot) \big)(h\partial_yw_2|_{y=0})
=o_{L_{y,x'}^2}(h^{\frac{1}{2}}).
$
By the same way, the other terms on the right side of \eqref{eqelliptic} are at most $o_{L_{y,x'}^2}(h^{\frac{1}{2}})$. Hence $\varphi(y,x')\psi(hD_{x'})\ud{w_2}=o_{L_{y,x'}^2}(h^{\frac{1}{2}})$, and this completes the proof of Proposition \ref{elliptic}.
\end{proof}
\subsection{Propagation formula near the interface}
Consider the operator
$$ B_h=B_{0,h}+B_{1,h}h\partial_y
$$
where $B_{j,h}=\widetilde{\chi}_1\op_h(b_j)\widetilde{\chi}_1$, $j=0,1$ are two tangential operators and $\widetilde{\chi}_1$ has compact support near a point $z_0\in \Sigma$. The symbols $b_j$ are compactly supported in $(x',\xi')$ variables. Note that in the local coordinate system,
$$ P_{h,0}=-h^2\Delta-1=-\frac{1}{\sqrt{|\ov{g}|}}h\partial_y\sqrt{|\ov{g}|}h\partial_y-R_h,
$$
where $R_h$ is a self-adjoint tangential differential operator of order $2$ classic and of order $0$ semiclassic.
\begin{lem}[Boundary propagation]\label{propagation:boundary}
Let $(\widetilde{w}_h)$ be a $h$-dependent family of functions
satisfying $\widetilde{w}_h=O_{L^2(\Omega_2)}=O(1)$ and $\widetilde{w}_h=O_{H^{1}(\Omega_2)}(h^{-1})$.
Assume moreover that $\widetilde{w}_h$ satisfies the equation
$$ P_{h,0}\widetilde{w}_h=o_{H^1(\Omega_2)}(h)+o_{L^2(\Omega_2)}(h^2)
$$
and the boundary condition: $\widetilde{w}_h|_{\Sigma}=o_{H^{\frac{1}{2}}}(h^{\frac{1}{2}})$ and $h\partial_{\nu}\widetilde{w}_h=O_{H^{-\frac{1}{2}}}(h^{\frac{1}{2}})$. Then we have
\begin{equation}\label{eq:propagationinterface} \frac{1}{ih}\big([P_{h,0},B_h]\widetilde{w}_h,\widetilde{w}_h\big)_{L^2(\Omega_2)}=i\big(B_{1,h}|_{y=0}(h\partial_y\widetilde{w}_h)|_{y=0},(h\partial_y\widetilde{w}_h)|_{y=0} \big)_{L^2(\Sigma)}+o(1).
\end{equation}
\end{lem}
\begin{proof}
First we remark that the right hand side of \eqref{eq:propagationinterface} makes sense, since $B_{1,h}|_{y=0}$ is a classical smoothing operator (but of semi-classical order $0$).
We denote by $\widetilde{w}=\widetilde{w}_h$ for simplicity. Without loss of generality, we may assume that $B_{0,h}=0$, since the treatment for the term $\frac{1}{ih}\big([P_{h,0},B_{0,h}]\widetilde{w},\widetilde{w}\big)_{L^2}$ is the same as in the proof of Proposition \ref{propagation:interior}, which contributes only $o(1)$ terms. By expanding the commutator, we have
\begin{equation*}
\begin{split}
&\frac{1}{ih}\big([P_{0,h},B_h]\widetilde{w},\widetilde{w}\big)_{L^2}\\=&\frac{1}{ih}\big(P_{0,h}B_{1,h}h\partial_y\widetilde{w},\widetilde{w}\big)-\frac{1}{ih}\big(B_{1,h}h\partial_yP_{0,h}\widetilde{w},\widetilde{w} \big)_{L^2}\\
=&\frac{1}{ih}\big(B_{1,h}h\partial_y\widetilde{w},P_{0,h}\widetilde{w}\big)_{L^2}-\frac{1}{ih}\big(B_{1,h}h\partial_yP_{0,h}\widetilde{w},\widetilde{w}\big)_{L^2}\\
+&i\big(B_{1,h}|_{y=0}(h\partial_y\widetilde{w})|_{y=0},(h\partial_y\widetilde{w})|_{y=0}\big)_{L^2(\Sigma)}-i\big((h\partial_yB_{1,h}h\partial_y\widetilde{w})|_{y=0},\widetilde{w}|_{y=0}\big)_{L^2(\Sigma)}
\end{split}
\end{equation*}
Observe that $B_{1,h}h\partial_y\widetilde{w}=O_{L^2(\Omega_2)}(1)$, $P_{0,h}\widetilde{w}=o_{H^1(\Omega_2)}(h)+o_{L^2(\Omega_2)}(h^2)$, and $B_{1,h}h\partial_yP_{0,h}\widetilde{w}=o_{L^2(\Omega_2)}(h)+o_{H^{-1}(\Omega_2}(h^2)$, thus
$$ \frac{1}{ih}\big(B_{1,h}h\partial_yw_2,P_{0,h}\widetilde{w}\big)_{L^2}-\frac{1}{ih}\big(B_{1,h}h\partial_yP_{0,h}\widetilde{w},\widetilde{w}\big)_{L^2}=o(1)
$$
as $h\rightarrow 0$. Write
$h\partial_yB_{1,h}h\partial_y\widetilde{w}=h(\partial_yB_{1,h})h\partial_y\widetilde{w}+B_{1,h}h^2\partial_y^2\widetilde{w}$ and using the equation satisfied by $\widetilde{w},$ we obtain that
$$ h\partial_yB_{1,h}h\partial_y\widetilde{w}=A_hh\partial_y\widetilde{w}-B_{1,h}R_h\widetilde{w}-B_{1,h}P_{h,0}\widetilde{w},
$$
where $A_h$ is a tangential operator of order $0$ semi-classic. Thanks to Lemma \ref{technical1}, $B_{1,h}=\mathcal{O}_{L^2\rightarrow H^1}(h^{-1})$, thus
$ B_{1,h}P_{h,0}\widetilde{w}=o_{H^1(\Omega_2)}(h)
$ and by the trace theorem $(B_{1,h}P_{h,0}\widetilde{w})|_{\Sigma}=o_{H^{\frac{1}{2}}(\Sigma)}(h)$. Next since $R_h\widetilde{w}|_{\Sigma}=o_{H^{\frac{1}{2}}(\Sigma) }(h^{\frac{1}{2}})+o_{H^{-\frac{3}{2}}(\Sigma) }(h^{\frac{5}{2}})$, we have $B_{1,h}R_h\widetilde{w}|_{\Sigma}=o_{H^{\frac{1}{2}}(\Sigma)}(h^{\frac{1}{2}})$.
We then deduce that $(h\partial_yB_{1,h}h\partial_y\widetilde{w})|_{y=0}=O_{H^{-\frac{1}{2}}(\Sigma)}(h^{\frac{1}{2}})$, which implies that $$\big((h\partial_yB_{1,h}h\partial_y\widetilde{w})|_{y=0},\widetilde{w}|_{y=0}\big)_{L^2(\Sigma)}=o(1).$$ This completes the proof of Lemma \ref{propagation:boundary}.
\end{proof}
To derive the propagation formula for the semiclassical measure, we consider a family of functions $(\widetilde{w}_h)$ satisfying the equation
$$P_{h,0}\widetilde{w}_h=o_{H^1(\Omega_2)}(h)+o_{L^2(\Omega_2)}(h^2)
$$
with a weaker boundary conditions, compared with \eqref{eq:w2}.
$$ \|\widetilde{w}_h\|_{L^2(\Omega_2)}=O(1),\;\|h\nabla \widetilde{w}_h\|_{L^2(\Omega)}=O(1),\; \|\widetilde{w}_h|_{\Sigma}\|_{H^{\frac{1}{2}}(\Sigma)}=o(h^{\frac{1}{2}}),\; \|(h\partial_{\nu}\widetilde{w}_h)|_{\Sigma}\|_{H^{-\frac{1}{2}}(\Sigma)}=O(h^{\frac{1}{2}}).
$$
Denote by $\mu$ is the semiclassical measure associated with $(\widetilde{w}_h)$.
\begin{prop}\label{muH=0}
\begin{align}
&(1) \quad \mu\mathbf{1}_{\mathcal{H}}=0;\\
&(2)\quad \limsup_{h\rightarrow 0}\big|\big(\op_h(b_0)h\partial_y \widetilde{w}_h,\widetilde{w}_h \big)_{L^2}\big|\leq \sup_{\rho\in\text{supp}(b_0)}|r(\rho)|^{\frac{1}{2}}|b_0(\rho)|,
\end{align}
for any tangential symbol $b_0(y,x',\xi')$ of order 0.
\end{prop}
\begin{proof}
(1) follows from the transversality of the rays reaching $\mathcal{H}$, and the proof is the same as in \cite{BL03} (see also the proof of Proposition 2.14 in \cite{BS} by taking $M_h=0$ there). The proof of (2) is also similar as in \cite{BS}, with an additional attention when doing the integration by part. Indeed, by Cauchy-Schwartz,
$$ \big|\big(\mathrm{Op}_h(b_0)h\partial_y\widetilde{w}_h,\widetilde{w}_h\big)_{L^2}\big|\leq \big|\big(\mathrm{Op}_h(b_0)h\partial_y\widetilde{w}_h,\mathrm{Op}_h(b_0)h\partial_y\widetilde{w}_h \big)\big|^{\frac{1}{2}}\|\widetilde{w}_h\|_{L^2}.
$$
Doing the integration by part,
$$ \big(\mathrm{Op}_h(b_0)h\partial_y\widetilde{w}_h,\mathrm{Op}_h(b_0)h\partial_y\widetilde{w}_h \big)_{L^2}=O(h)-\big(\mathrm{Op}_h(b_0)h^2\partial_y^2\widetilde{w}_h,\mathrm{Op}_h(b_0)\widetilde{w}_h\big)_{L^2},
$$
where $O(h)$ comes from the commutators and the boundary term, since by the assumption on $\widetilde{w}_h$, $\big(\mathrm{Op}_h(b_0)(h\partial_y\widetilde{w}_h)|_{y=0},h\mathrm{Op}_h(b_0)(\widetilde{w}_h|_{y=0}) \big)_{L_{x'}^2}=o(h^2)$. For the rest argument, we just replace $-h^2\partial_y^2\widetilde{w}_h$ by $-R_h\widetilde{w}_h$ plus errors in $O_{L^2}(h)$. From the symbolic calculus, the contribution $\sup_{\rho}|r|^{\frac{1}{2}}|b_0(\rho)|$ comes from the principal term $\big|\big(\mathrm{Op}_h(b_0)R_h\widetilde{w}_h,\mathrm{Op}_h(b_0)\widetilde{w}_h \big)_{L^2}\big|^{\frac{1}{2}},$ after taking limsup in $h$. This completes the proof of Proposition \ref{muH=0}.
\end{proof}
\begin{lem}\label{H1measure}
Let $B_{0,h}, B_{1,h}$ are tangential semiclassical operators of order $0$, with principal symbols $b_0, b_1$ with respectively, supported near a point $\rho_0$ of $T^*\Sigma$. Then
\begin{equation}\label{measureidentity}
\big(\big(B_{0,h}+B_{1,h}\frac{h}{i}\partial_y\big)\widetilde{w}_h,\widetilde{w}_h \big)= \langle\mu,b_0+b_1\eta\rangle+o(1),
\end{equation}
as $h\rightarrow 0$.
\end{lem}
\begin{proof}
First we remark that the expression $\langle\mu,b_0+b_1\eta \rangle$ is well-defined, since $\mu$ belongs to the dual of $C^0(Z)$ and $\mu ( \mathcal{H})=0$, and in particular, by elliptic regularity, $\mu\mathbf{1}_{|\eta|>1}=0$. The convergence of the quadratic form $(B_{0,h}\widetilde{w}_h,\widetilde{w}_h)$ to $\langle\mu,b_0\rangle$ is just the definition of the semiclassical measure $\mu$. If $\rho_0\in\mathcal{E}$, the contributions of both sides of \eqref{measureidentity} is $o(1)$, thanks to the elliptic regularity (see the proof of Proposition \ref{elliptic}). Next we assume that $\rho_0\in\mathcal{H}\cup\mathcal{G}$. Take $\varphi\in C_c^{\infty}(-1,1)$, $\varphi$ is equal to $1$ in a neighborhood of $(-1/2,1/2)$. For $\epsilon>0$, we write
$$ B_{1,h,\epsilon}:=\big(1-\varphi\big(\frac{y}{\epsilon}\big)\big)B_{1,h}, \quad B_{1,h}^{\epsilon}:=B_{1,h}-B_{1,h,\epsilon}.
$$
Taking $h\rightarrow 0$ first we obtain that
$$ \big(B_{1,h,\epsilon}\frac{h}{i}\partial_y\widetilde{w}_h,\widetilde{w}_h\big)_{L^2(\Omega_2)}\rightarrow \langle\mu,\big(1-\varphi\big(\frac{y}{\epsilon}\big)\big)b_1\eta \rangle.
$$
If $\rho_0\in\mathcal{H}$, then taking $\epsilon\rightarrow 0$, we obtain that
$$ \lim_{\epsilon\rightarrow 0}\langle\mu,\big(1-\varphi\big(\frac{y}{\epsilon}\big)\big)b_1\eta \rangle=\langle\mu,\mathbf{1}_{y>0}b_1\eta \rangle=\langle\mu, b_1\eta\rangle,
$$
since $\mu\mathbf{1}_{\mathcal{H}\cup\mathcal{E}}=0$. It remains to estimate the contribution of $\big(B_{1,h}^{\epsilon}\frac{h}{i}\partial_y\widetilde{w}_h,\widetilde{w}_h\big)$. For fixed $\epsilon>0$, we have
\begin{align*}
&\Big|\limsup_{h\rightarrow 0}\big(\big(B_{1,h}^{\epsilon}h\partial_y\widetilde{w}_h,\widetilde{w}_h\big) \big)_{L^2(\Omega_2)} \Big| \leq \limsup_{h\rightarrow 0}\Big(\|\varphi(y/\epsilon)B_{1,h}^*\widetilde{w}_h\|_{L^2(\Omega_2)}\|h\partial_y\widetilde{w}_h\|_{L^2(\Omega_2)} \Big).
\end{align*}
Since on supp$(\mu\mathbf{1}_{y>0})$, $|\eta|\leq 1$, together with the fact that $\rho_0\in\mathcal{H}\cap\mathcal{E}$, we deduce that the right side converges to $0$ as $\epsilon\rightarrow 0$.
Now suppose that $\rho_0\in\mathcal{G}$. For any $\epsilon>0, \delta>0$, we decompose $B_{1,h}=B_{1,h}^{\epsilon}+B_{1,h}^{\epsilon,\delta}+B_{1,h,\delta}^{\epsilon}$, with
\begin{align*}
& B_{1,h,\epsilon}=\Big(1-\varphi\big(\frac{y}{\epsilon}\big)\Big)B_{1,h},\\
& B_{1,h}^{\epsilon,\delta}=\mathrm{Op}_h\Big(\varphi\big(\frac{y}{\epsilon}\big)\varphi\big(\frac{r}{\delta}\big)\Big)B_{1,h},\\
&B_{1,h,\delta}^{\epsilon}=\mathrm{Op}_h\Big(\varphi\big(\frac{y}{\epsilon}\big)\Big(1-\varphi\big(\frac{r}{\delta}\big)\Big)\Big)B_{1,h}.
\end{align*}
By the same argument, we have
$$ \lim_{\epsilon\rightarrow 0}\lim_{h\rightarrow 0}\big(B_{1,h,\epsilon}hD_y\widetilde{w}_h,\widetilde{w}_h\big)_{L^2(\Omega_2)}=\langle\mu,b_1\eta\mathbf{1}_{y>0}\rangle=\langle\mu,b_1\eta\mathbf{1}_{\rho\notin\mathcal{H}}\rangle,
$$
since $\mu\mathbf{1}_{\mathcal{H}\cup\mathcal{E}}=0$. Next, from (2) of Proposition \ref{muH=0}, we have
$$ \limsup_{\epsilon\rightarrow 0}\limsup_{h\rightarrow 0}\big(B_{1,h}^{\epsilon,\delta}hD_y\widetilde{w}_h,\widetilde{w}_h\big)_{L^2(\Omega_2)}\leq C\delta,
$$
which converges to $0$ if we let $\delta\rightarrow 0$. Finally, by Cauchy-Schwartz,
$$\big|\big(B_{1,h,\delta}^{\epsilon}hD_y\widetilde{w}_h,\widetilde{w}_h \big)_{L^2(\Omega_2)}\big|\leq \|hD_y\widetilde{w}_h\|_{L^2(\Omega_2)}\Big\|B_{1,h}^*\mathrm{Op}_h\Big(\varphi\big(\frac{y}{\epsilon}\big)\Big(1-\varphi\big(\frac{r}{\delta}\big)\Big) \Big)^*\widetilde{w}_h \Big\|_{L^2}.
$$
Taking the triple limit, we have
$$ \limsup_{\delta\rightarrow 0}\limsup_{\epsilon\rightarrow 0}\limsup_{h\rightarrow 0}\big|\big(B_{1,h,\delta}^{\epsilon}hD_y\widetilde{w}_h,\widetilde{w}_h\big)_{L^2(\Omega_2)}\big|\leq \langle\mu,|b_1|^2\mathbf{1}_{y=0}\mathbf{1}_{r\neq 0}\rangle=0,
$$
since $\mu\mathbf{1}_{\mathcal{E}\cup\mathcal{H}}=0$. This completes the proof of Lemma \ref{H1measure}.
\end{proof}
As in \cite{BL03}, we define the function
\begin{align*}
\theta(y,x';\eta,\xi')=\frac{\eta}{|\xi'|} \text{ if } y>0; \quad \theta(y,x',\eta,\xi')=i\frac{\sqrt{-r_0(x',\xi')}}{|\xi'|} \text{ on } \mathcal{E}.
\end{align*}
Since $\mu\mathbf{1}_{\mathcal{H}}=0$, $\theta$ is $\mu$ almost everywhere defined as a function on $Z$. Formally,
$$ \sigma\big(\frac{i}{h}[P_{h,0},B_h]\big)=\{\eta^2-r,b_0+b_1\eta \}=a_0+a_1\eta+a_2\eta^2,
$$
where
\begin{align}\label{relation}
a_0=b_1\partial_yr-\{r,b_0\}',\quad a_1=2\partial_yb_0-\{r,b_1\}',\quad a_2=2\partial_yb_1,
\end{align}
and $\{\cdot,\cdot\}'$ is the Poisson bracket for $(x',\xi')$ variables. By expanding the commutator, we find
\begin{align}\label{Malagrange}
\frac{i}{h}[P_{h,0},B_h]=A_0+A_1hD_y+A_2h^2D_y^2+h\op_h(S_{\partial}^{0}+S_{\partial}^0\eta),
\end{align}
where $A_0,A_1,A_2$ are tangential operators with symbols $a_0,a_1,a_2$, with respectively, and $S_{\partial}^0$ stands for the tangential symbol class of order $0$.
We now have all the ingredients to present the propagation formula for the defect measure in the spirit of \cite{BL03}:
\begin{prop}\label{propagationformula}
Assume that $B_h=B_{h,0}+B_{h,1}hD_y$, where $B_{h,0}, B_{h,1}$ are tangential operators of order 0 with symbols $b_0, b_1$, with respectively. Assume that $b=b_0+b_1\eta$. Define the formal Poisson bracket
$$ \{p,b\}=(a_0+a_2r)+a_1\theta|\xi'|\mathbf{1}_{\rho\notin\mathcal{H}},
$$
where $a_0,a_1,a_2$ are given by \eqref{relation}.
Then any defect measures $\mu,\nu_0$ of $(\widetilde{w}_h)$, $(h\partial_{\nu}\widetilde{w}_h)|_{\Sigma}$ satisfy the relation
$$ \langle\mu,\{p,b\}\rangle=-\langle\nu_0,b_1 \rangle.
$$
Moreover, if $b\in C^0(Z)$, we have
\begin{equation}\label{propag-1}
\langle\mu,\{p,b\} \rangle=0.
\end{equation}
\end{prop}
\begin{proof}
See \cite{BL03}.
\end{proof}
Moreover, we have
\begin{prop}\label{zerodiffractive}
$\mu\big(\mathcal{G}^{2,+}\big)=0$.
\end{prop}
As showed in \cite{BL03}, we obtain that the measure $\mu$ is invariant by the flow of Melrose-Sj\"ostrand. More precisely, we have
\begin{thm}[\cite{BL03}]\label{propagationtheorem}
Assume that $\mu$ is a semi-classical measure on $^bT^*\ov{\Omega}$ associated with the sequence $(\widetilde{w}_h)$ satisfying \eqref{propag-1} and Proposition~\ref{zerodiffractive}. Then $\mu$ is invariant under the Melrose-Sj\"ostrand flow $\phi_s$.
\end{thm}
\begin{rem}
This is a consequence of Theorem 1 in \cite{BL03} which asserts the equivalence between the measure invariance and the propagation formula $H_p(\mu)=0$ together with $\mu(\mathcal{G}^{2,+})=0$. Though Theorem 1 in \cite{BL03} is stated and proved in the context of micro-local defect measure, it also holds true in the context of the semiclassical measure from the word-by-word translation.
\end{rem}
\subsection{The last step to the proof of the resolvent estimate in Theorem \ref{thm:resolvent}}
In this subsection, we take $\widetilde{w}_h=w_2$. To finish the contradiction argument in the proof of \eqref{eq:resolvent}, it suffices to show that $\mu=0$. Let $\mu$ be the corresponding semiclassical measure and $\nu_0$ be the semiclassical measure of $h\partial_{\nu}u_2|_{\Sigma}$. Since $h\partial_{\nu}u_2|_{\Sigma}=o_{H^{-\frac{1}{2}}(\Sigma}(h^{\frac{1}{2}})$, we have that $\langle\nu_0,b_1\rangle=0$ for any compactly supported symbol $b_1(x',\xi')$. Thanks to (H), along the Melrose-Sj\"ostrand flow of $^bT^*\ov{\Omega}_2$ issued from points in $^bT^*\ov{\Omega}_2$, there must be some points reaching $\mathcal{H}(\Sigma)\cup\mathcal{G}^{2,-}(\Sigma)$. By the property of the Melrose-Sj\"ostrand flow on $^bT^*\ov{\Omega}_2$, to show that $\mu=0$, we need to verify that
$$ \mu\big(\mathcal{G}^{2,-}(\Sigma)\big)=0
$$
and $\mu=0$ near a neighborhood of $\rho_0\in\mathcal{H}(\Sigma)$.
\begin{prop}\label{zerogliding}
$\mu\big(\mathcal{G}^{2,-}(\Sigma)\big)=0.$
\end{prop}
\begin{proof}
The proof is exactly the same as the proof of Proposition \ref{zerodiffractive}. We will make use of the formula
$$ \langle\mu,\{p,b\}\rangle=\langle\nu_0,b_1\rangle
$$
by choosing $b=b_{1,\epsilon}\eta$ with
$$ b_{1,\epsilon}(y,x',\xi')=\psi\big(\frac{y}{\epsilon^{\frac{1}{2}}}\big)\psi\big(\frac{r(y,x',\xi')}{\epsilon}\big)\kappa(y,x',\xi'),
$$
where $\psi\in C_c^{\infty}(\R)$ equals to $1$ near the origin and $\kappa(y,x',\xi')\geq 0$ near a point $\rho_0\in\mathcal{G}^{2,-}$. Note that $\{p,b_{\epsilon} \}=(a_{0}+a_2r)+a_1\eta\mathbf{1}_{\rho\notin\mathcal{H}}$, and $a_0,a_1,a_2$ are given by the relation \eqref{relation}. In particular for our choice, by direct calculation we have
$$ a_0=b_{1,\epsilon}\partial_y r, \quad a_1=-\{r,\kappa\}'\psi\big(\frac{y}{\epsilon^{\frac{1}{2}}}\big)\psi\big(\frac{r}{\epsilon}\big),$$
and
$$ a_2=2\partial_yb_{1,\epsilon}=2\epsilon^{-\frac{1}{2}}\psi'\big(\frac{y}{\epsilon^{\frac{1}{2}}}\big)\psi\big(\frac{r}{\epsilon}\big)\kappa+2\frac{\partial_yr}{\epsilon}\psi\big(\frac{y}{\epsilon^{\frac{1}{2}}}\big)\psi'\big(\frac{r}{\epsilon}\big)\kappa+2\psi\big(\frac{y}{\epsilon^{\frac{1}{2}}}\big)\psi\big(\frac{r}{\epsilon}\big)\partial_y\kappa.
$$
Observe that $a_2$ is uniformly bounded in $\epsilon$ and for any fixed $(y,x',\xi')$, $ra_2\rightarrow 0$ as $\epsilon\rightarrow 0$. Thus from the dominating convergence, we have
$$ \lim_{\epsilon\rightarrow 0}\langle\mu,\{p,b_{\epsilon} \}\rangle=\langle\mu,\kappa|_{y=0}\partial_yr\mathbf{1}_{r=0}\rangle.
$$
Since $\partial_yr<0$ on $\mathcal{G}^{2,-}$, while $-\langle\nu_0,b_{\epsilon}\rangle=0$, we deduce that $\mu\mathbf{1}_{\mathcal{G}^{2,-}}=0$. This completes the proof of Proposition \ref{zerogliding}.
\end{proof}
\begin{prop}\label{zerohyperbolic}
Let $\rho_0\in\mathcal{H}(\Sigma)$. Let $b(y,x',\xi')$ be a tangential symbol, supported near $\rho_0$. Then
$$ \|\mathrm{Op}_h(b)w_2\|_{L^2(\Omega_2)}+\|h\partial_y\mathrm{Op}_h(b)w_2\|_{L^2(\Omega_2)}=o(1),
$$
as $h\rightarrow 0$.
\end{prop}
\begin{proof}
Since the Melrose-Sj\"ostrand flow is transverse to $\mathcal{H}$, by localizing the symbol $b$, it suffices to prove the same estimate by replacing $b$ to $q^{\pm}$, where $q^{\pm}$ is the solutions of
$$ \partial_yq^{\pm}\mp H_{\sqrt{r}(y,x',\xi')}q^{\pm}=0,\quad q^{\pm}|_{y=0}=q_0,
$$
and $q_0$ is supported in a sufficiently small neighborhood of $\rho_0$. Near $\rho_0$, it follows from \cite{BL03} that we can factorize $P_{h,0}$ as $\big(hD_y-\Lambda_h^+(y,x',hD_{x'})\big)\big(hD_y+\Lambda_h^-(y,x',hD_{x'}) \big)+O_{H^{\infty}}(h^{\infty})$, and also
$ \big(hD_y-\widetilde{\Lambda}_h^+(y,x',hD_{x'})\big)\big(hD_y+\widetilde{\Lambda}_h^-(y,x',hD_{x'}) \big)+O_{H^{\infty}}(h^{\infty}),
$
where $\Lambda_h^{\pm}$ and $\widetilde{\Lambda}_h^{\pm}$ have principal symbols $\pm\sqrt{r(y,x',\xi')}$. Denote by $Q_h^{\pm}=\mathrm{Op}_h(q^{\pm})$ and set
$$ w_2^{+}:=\varphi(y)Q_h^{+}(hD_y-\Lambda_h^{-})w_2,\quad w_2^-:=\varphi(y)Q_h^{-}(hD_y-\widetilde{\Lambda}_h^+)w_2,
$$
where the cutoff $\varphi(y)$ is supported on $0\leq y\leq \epsilon_0\ll 1$ and is equal to $1$ for $0\leq y\leq \epsilon_0/2$. From the equation of $w_2$, we have
\begin{align*}
(hD_y-\Lambda_h^+)w_2^{+}=&\varphi(y)[hD_y-\Lambda_h^+,Q_h^+](hD_y-\Lambda_h^-)w_2-ih\varphi'(y)Q_h^+(hD_y-\Lambda_h^-)w_2+o_{L_{y,x'}^2}(h)\\
=&-ih\varphi'(y)Q_h^+(hD_y-\Lambda_h^-)w_2+o_{L_{y,x'}^2}(h),
\end{align*}
since the principal symbol of $\frac{1}{ih}[hD_y-\Lambda_h^+,Q_h^+]$ is zero, thanks to the choice of symbols $q^{\pm}$. Multiplying by $\ov{w}_2^+$ to both sides and integrating, we have for $y\leq \epsilon_0/2$ (thus $\varphi'(y)=0$) that
\begin{align}\label{energyestimate}
h\|w_2^+(y,\cdot)\|_{L_{x'}^2}^2\leq h\|w_2^{+}(0,\cdot)\|_{L_{x'}^2}^2+o(h).
\end{align}
Since $\mathrm{Op}_h(q_0)(h\partial_yw_2)|_{y=0}=o_{L_{x'}^2}(1)$, we deduce by definition that $w_2^+(0)=o_{L_{x'}^2}(1)$. This together with \eqref{energyestimate} yields $w_2^+(y)=o_{L_{x'}^2}(1)$, uniformly for all $0\leq y\leq \epsilon_0/2$. Thus $w_2^+=o_{L_{y,x'}^2}(1)$. Similar argument for $w_2^-$ yields $w_2^-=o_{L_{y,x'}^2}(1)$. Note that $hD_y-\Lambda_h^-$ is elliptic on the support of $q^+$, we deduce that $Q_h^+w_2=o_{L_{y,x'}^2}(1)$. This means that $\mu$ is zero near the support of $q^+$, hence the proof of Proposition \ref{zerohyperbolic} is complete.
\end{proof}
Consequently, we have shown that the measure $\mu$ is invariant along the bicharacteristic flow on $\Omega _2$, it vanishes near every hyperbolic point of $\Sigma$, and $\mu( \mathcal{G}^{2,-}) =0$. Thus $\mu$ is supported on bicharacteristics which encounter $\Sigma$ only at points of
$$\mathcal{G}^{2<}= \cup_{k\geq 3 }\mathcal{G}^k.$$
These bicharacteristics are consequently near $\Sigma$ integral curves of $H_p$ (because in Definition~\ref{generalbicha}, the two vector fields $H_p$ and $H_p - \frac{ H_p^2(y}{H^2 _y p } H_y$ coincide on $\mathcal{G}^{2<}$). However, according to the geometric condition assumption, all such bicharacteristics must leave $\Omega_2$. As a consequence, $\mu$ is supported on the emptyset, and hence $\mu =0$. This gives a contradiction. The proof of \eqref{eq:resolvent} in Theorem~\ref{thm:resolvent} is now complete.
\section{Optimality of the resolvent estimate}
In this section we prove the second part of Theorem \ref{thm:resolvent}. For simplicity, we consider the model case $\Omega_2=\mathbb{D}:=\{x\in\R^2:|x|<1 \}$ and $a(x)=\mathbf{1}_{\Omega_1}$ and $\Sigma=\mathbb{S}^1$.
To prove the second part in Theorem~\ref{thm:resolvent} we need to construct functions $u_1,v_1,u_2,v_2$, such that
$$ \|(u_j,v_j)\|_{H^1\times L^2(\Omega_j)}\sim 1,\; \|(f_j,g_j)\|_{H^1\times L^2(\Omega_j)}=O(h), \quad j=1,2
$$
\begin{align*}
\begin{cases}
& u_1=ih(f_1-v_1), \text{ in }\Omega_1\\
& h\Delta u_1+h\Delta v_1-i v_1=hg_1,\text{ in } \Omega_1 \\
& u_2=ih(f_2-v_2), \text{ in } \Omega_2\\
& h\Delta u_2-iv_2=hg_2,\text{ in }\Omega_2
\end{cases}
\end{align*}
together with the boundary condition on the interface
\begin{align*}
u_1|_{\Sigma}=u_2|_{\Sigma},\quad \partial_{\nu}u_2|_{\Sigma}=(\partial_{\nu}u_1+\partial_{\nu}v_1 )|_{\Sigma},
\end{align*}
The key point in the construction is that in $\Omega_1$, we construct quasi-modes concentrated at the scale $|D_x|\sim \hbar^{-1}=h^{-\frac{1}{2}}$ while in $\Omega_2$, the quasi-modes are concentrated at the scale $|D_{x'}|\sim |D_y|\sim |D_x|\sim h^{-1}$ near the interface $\Sigma$, where $x'$ is the tangential variable near $\Sigma$ and $y$ is the normal variable.
Now we describe the construction.
\noi
$\bullet$ {\bf Step 1: Construction at the zero order:}
We first choose $u_2^{(0)}$, such that
$$ h^2\Delta u_2^{(0)}+u_2^{(0)}=0, \quad u_2^{(0)}|_{\Sigma}=0;\quad \|\nabla u_2^{(0)}\|_{L^2(\Omega_2)}\sim h^{-1}\|u_2^{(0)}\|_{L^2(\Omega_2)}\sim 1.
$$
Moreover, we require that $u_2^{(0)}$ such that they are hyperbolically localized, in the sense that
\begin{align}\label{localizationstep1}
\|\partial_{\nu}u_2^{(0)}|_{\Sigma}\|_{H^s(\Sigma)}\sim h^{-s} ,
\quad
\mathrm{WF}_h(\partial_{
\nu}u_2^{(0)}|_{\Sigma})\subset \mathcal{H}_{\delta}(\Sigma):=\{(x',\xi'):\delta<r_0(x',\xi')<1-\delta \}
\end{align}
for some $0<\delta<\frac{1}{2}$.
The existence of such sequence of eigenfunctions is not difficult to prove in the case of a disc or an ellipse, we postpone this fact in Lemma \ref{energynorm} of the Appendix. This will actually be the only point where in Theorem~\ref{thm:decay} we use the particular choice $\Omega_2 = \mathbb{D}$.
Next we define
$$ v_2^{(0)}=ih^{-1}u_2^{(0)},\quad f_2^{(0)}=g_2^{(0)}=0.
$$
From \eqref{localizationstep1}, we have
\begin{equation}
\partial_{\nu}u_2^{(0)}|_{\Sigma}= \begin{cases} O_{L^2(\Sigma)}(1)\\ O_{H^{-\frac{1}{2}}(\Sigma)}(h^{\frac{1}{2}})\\ O_{H^{\frac{1}{2}}(\Sigma)}(h^{-\frac{1}{2}}).\end{cases}
\end{equation}
We remark that here we use the fact that the dimension $d\geq 2$.
Next we solve the elliptic equation with the mixed Dirichlet-Neumann data (with $\hbar=h^{\frac{1}{2}}$):
$$ (\hbar^2\Delta-i)w^{(0)}=0,\quad \partial_{\nu}w^{(0)}|_{\Sigma}=\partial_{\nu}u_2^{(0)}|_{\Sigma},\quad w^{(0)}|_{\partial\Omega_1\setminus\Sigma}=0.
$$
From Proposition \ref{pb:mixed}, there exists a unique solution $w^{(0)}$ of this system, which satisfies
\begin{equation}
w^{(0)}=\begin{cases}
O_{H^2(\Omega_1)}(\hbar^{-1})\\
O_{H^1(\Omega_1)}(\hbar) \\
O_{L^2(\Omega_1)}(\hbar^2),
\end{cases}
\end{equation}
and hence by interpolation
$$ w^{(0)}=O_{H^{\frac 3 2}(\Omega_1)}(1)
$$
and by trace theorems
\begin{equation}
w^{(0)}\mid_{\Sigma} = \begin{cases}
O_{H^{\frac 1 2} ( \Sigma)} (\hbar)\\
O_{H^{1} ( \Sigma)} (1)
\end{cases}
\end{equation}
Moreover, from the information of $\mathrm{WF}_h(\partial_{\nu}u_2^{(0)}|_{\Sigma})$ and Proposition \ref{control:frontdonde}, we have
$$\mathrm{WF}_h(w^{(0)}|_{\Sigma})\subset \mathrm{WF}_h(\partial_{\nu}w^0|_{\Sigma})\subset \mathcal{H}_{\delta}(\Sigma).$$ Hence
$$ \| w^{(0)}\mid_{\Sigma}\|_{H^1( \Sigma)} \sim h^{- \frac 1 2} \| w^{(0)}\mid_{\Sigma}\|_{H^{\frac 1 2}( \Sigma)} = O(1)
$$
Next we define $u_1^{(0)}, v_1^{(0)}$ such that
$$ v_1^{(0)}=ih^{-1}u_1^{(0)}, \quad w^{(0)}=u_1^{(0)}+v_1^{(0)} = (1 + i h^{-1}) u_1^{(0)};\quad f_1^{(0)}=0,\quad g_1^{(0)}=ih^{-1}u_1^{(0)}= v^{(0)}_1.
$$
This implies
\begin{equation}
u_1^{(0)}=\begin{cases}
O_{H^1(\Omega_1)}(h^{\frac{3}{2}})\\
O_{L^2(\Omega_1)}(h^{2}),
\end{cases}
\end{equation}
and consequently $g_1^{(0)}=O_{L^2(\Omega_1)}(h)$.
In summary, as the first step, we have constructed quasi-modes $(u_1^{(0)},v_1^{(0)};u_2^{(0)},v_2^{(0)})$ and $(f_1^{(0)}=0,g_1^{(0)};f_2^{(0)}=0,g_2^{(0)}=0)$ such that
\begin{equation}\label{0approximation}
\begin{cases}
& h\Delta(u_1^{(0)}+v_1^{(0)})-iv_1^{(0)}=hg_1^{(0)}, \qquad g_1 = O_{L^2( \Omega_1)}(h)\\
&u_1^{(0)}=-ihv_1^{(0)}\\
& h\Delta u_2^{(0)}-iv_2^{(0)}=0\\
& u_2^{(0)}=-ih v_2^{(0)}\\
&\partial_{\nu}u_2^{(0)}|_{\Sigma}=\big(\partial_{\nu}u_1^{(0)}+\partial_{\nu}v_1^{(0)} \big)|_{\Sigma},\\
& \big(u_2^{(0)}-u_1^{(0)}\big)|_{\Sigma}=O_{H^{\frac{1}{2}}(\Sigma)}(h^{\frac{3}{2}}),\quad \big(v_2^{(0)}-v_1^{(0)}\big)|_{\Sigma}=O_{H^{\frac{1}{2}}(\Sigma)}(h^{\frac{1}{2}}),
\end{cases}
\end{equation}
and to conclude the proof of Theorem~\ref{thm:resolvent}, it remains to eliminate the error term in the last boundary condition in~\eqref{0approximation}. An important point is that both $u^{(0)}_2$ and $u^{(0)}_1$ (and hence also $uv{(0)}_2$ and $v^{(0)}_1$) have their wave front included in $\mathcal{H}_\delta ( \Sigma)$.
\noi
$\bullet$ {\bf Step 2: Construction at the first order:}
We now introduce correction terms to eliminate the error term in the last boundary condition of \eqref{0approximation}.
We are looking for a correction term $e_2 ^{(1)}$,
$$ u_2^{(1)}=u_2^{(0)}+e_2^{(1)},\quad v_2^{(1)}=i h^{-1} u_2^{(1)}=v_2^{(0)}+i h^{-1} e_2^{(1)},
$$
while keeping all other terms identical
$$ u_1^{(1)}= u_1^{(0)}, \quad v_1^{(1)}= v_1^{(0)},
$$
First, using the geometric optics construction (see Appendix), we construct $\widetilde{e}_2^{(1)}$, solving near $\Sigma$, solving for $N$ large enough to be fixed later
$$ (h^2\Delta+1)\widetilde{e}_2^{(1)}=O_{L^2}(h^N)$$
near $\Sigma$, and the boundary conditions
\begin{equation}\label{boundary-cond}
\widetilde{e}_2^{(1)}|_{\Sigma}=(u_1^{(0)}-u_2^{(0)})|_{\Sigma}+O_{L^2(\Sigma)}(h^N), \qquad \partial_\nu \widetilde{e}^{(1)}_2 \mid _{\Sigma} = O_{L^2( \Sigma)} ( h^N).
\end{equation}
with $h$-semiclassical wave front sets of all the functions are localized near $\mathcal{H}_\delta (\Sigma)$.
\begin{align}\label{go}
(h^2\Delta+1)\widetilde{e}_2^{(1)}=O_{L^2(\Omega_2)}(h^N),\quad \widetilde{e}_2^{(1)}=(u_1^{(0)}-u_2^{(0)})|_{\Sigma}+O_{H^N(\Sigma)}(h^N),\quad h\partial_{\nu}\widetilde{e}_2^{(1)}|_{\Sigma}=O_{H^N(\Sigma)}(h^N)
\end{align}
locally near $x_0\in\Sigma$. We then take a cutoff $\chi$, such that $\chi\equiv 1$ on $\Sigma$, and with support sufficiently close to $\Sigma$ so that $\widetilde{e}$ is defined on the support of $\chi$ (i.e. $\chi$ vanishes along the bicharacteristics, before the formation of the caustics). Let $e_2^{(1)}:=\chi \widetilde{e}_2^{(1)}.$ Hence
\begin{equation}
\begin{gathered}
(h^2\Delta+1)e_2^{(1)}=[h^2\Delta,\chi]\widetilde{e}_2^{(1)}+O_{L^2(\Omega_2)}(h^4)=O_{L^2(\Omega_2)}(h^3),\\
e_2^{(1)}|_{\Sigma}=\widetilde{e}_2^{(1)}|_{\Sigma},\quad h\partial_{\nu}e_2^{(1)}|_{\Sigma}=h\partial_{\nu}\widetilde{e}_2^{(1)}|_{\Sigma}.
\end{gathered}
\end{equation}
Again, all the functions and the errors are microlocalized near $(x_0,\xi_0)\in \mathcal{H}_\delta(\Sigma)$. Moreover, from the boundary conditions~\eqref{boundary-cond} which determine the values of the symbols $b^{\pm}$ in the geometric optics construction, we have
$$ \|e_2^{(1)}\|_{H^1(\Omega_2)}=O(h),\quad \|e_2^{(1)}\|_{L^2(\Omega_2)}=O(h^2),\quad \partial_{\nu}e_2^{(1)}|_{\Sigma}=O_{L^2(\Sigma)}(h^{N-1}).
$$
The geometric optics constructions in the appendix are local, but using a partition of unity of $\Sigma$, we choose a finite cutoff functions $(\chi_j)_{j=1}^M$ to replace $\chi$ and modify the function $e_2^{(1)}$ by
$$ e_2^{(1)}:=\sum_{j=1}^M \chi_j\widetilde{e}_{2,j}^{(1)},
$$
where $\widetilde{e}_{2,j}^{(1)}$ is the corresponding geometric optics near supp$(\chi_j)$.
Next we define $g_2^{(1)}=h^{-2}\cdot(h^2\Delta+1)e_2^{(1)}$.
We now have
\begin{equation}\label{1approximation}
\begin{cases}
& h\Delta(u_1^{(1)}+v_1^{(1)})-iv_1^{(1)}=hg_1^{(1)}, \qquad g_1 = O_{L^2( \Omega_1)}(h)\\
&u_1^{(1)}+ihv_1^{(1)}= 0\\
& h\Delta u_2^{(1)}-iv_2^{(1)}=h g_2^{(1)}, \qquad g_2^{(1)} = O_{L^2( \Omega_1)} (h)\\
& u_2^{(1)}=-ih v_2^{(1)}\\
&\partial_{\nu}u_2^{(1)}|_{\Sigma}=\big(\partial_{\nu}u_1^{(1)}+\partial_{\nu}v_1^{(1)} \big)|_{\Sigma} +O_{H^{N}(\Sigma)}(h^N)\\
& \big(u_2^{(1)}-u_1^{(1)}\big)|_{\Sigma}=O_{H^N(\Sigma)}(h^N),\quad \big(v_2^{(1)}-v_1^{(1)}\big)|_{\Sigma}=O_{H^{N}(\Sigma)}(h^{N-1}),
\end{cases}
\end{equation}
It now remains to eliminate completely the errors in the last boundary condition in~\eqref{1approximation}. For this we just use the trace operators. Recall that if $s> \frac 3 2$, the map
$$ \Gamma: u \in H^s( \Omega_1) \mapsto ( u \mid_\Sigma, \partial_\nu u \mid_{\Sigma}) \in H^{s- 1/2}( \Sigma)\times H^{s- 3/2}(\Sigma)
$$
is continuous surjective and admits a bounded right inverse.
As a consequence, if $N$ is large enough, there exists $e_2^{(2)}\in H^{N- \frac 3 2}$ (supported near $\Sigma$) such that
$$ \|e_2^{(2)} \|_{ H^{N- \frac 3 2}(\Omega_2)} = O(h^N), \quad e_2^{(2)}\mid_\Sigma = (u_1^{(1)}-u_2^{(1)}) \mid_{\Sigma}, \quad \partial_{\nu}e_2^{(2)}|_{\Sigma}=\big(\partial_{\nu}u_1^{(1)}-\partial_{\nu}v_1^{(1)}\big)\mid_{\Sigma} - \partial_\nu u_2^{(1)} \mid_{\Sigma}
$$
Choosing now
$$ u_2^{(2)}=u_2^{(1)}+ e_2^{(2)}, \qquad v_2^{(2)}=v_2^{(1)}+ i h^{-1} e_2^{(2)}, \qquad g_2^{(2)} = g_2^{(1)} +h^{-1} ( h^2 \Delta + 1) e_2^{(2)}
$$
and keeping the other terms identical
$$ u_1^{(2)}= u_1^{(0)}, \quad v_1^{(2)}= v_1^{(0)},\quad g_1^{(2)} = g_1^{(1)},
$$
we get (if $N$ is large enough)
\begin{equation}\label{2approximation}
\begin{cases}
& h\Delta(u_1^{(2)}+v_1^{(2)})-iv_1^{(2)}=hg_1^{(2)}, \qquad g_1 = O_{L^2( \Omega_1)}(h)\\
&u_1^{(2)}+ihv_1^{(2)}=0\\
& h\Delta u_2^{(2)}-iv_2^{(2)}= h g_2^{(2)}, \qquad g_2^{(2)} = O_{L^2( \Omega_1)} (h)\\
& u_2^{(2)}=-ih v_2^{(2)}\\
&\partial_{\nu}u_2^{(2)}|_{\Sigma}=\big(\partial_{\nu}u_1^{(2)}+\partial_{\nu}v_1^{(2)} \big)|_{\Sigma} \\
& \big(u_2^{(2)}-u_1^{(2)}\big)|_{\Sigma}=0,\quad \big(v_2^{(2)}-v_1^{(2)}\big)|_{\Sigma}=0
\end{cases}
\end{equation}
This ends the proof of the construction of quasi-modes in Theorem~\ref{thm:resolvent}. \qed
\section{Appendix: Technical ingredients }
\subsection{Elliptic problem with mixed Dirichlet Neumann data}
Let $U\subset \R^d$ be a bounded domain with smooth boundary. For $F\in C^{\infty}(\ov{U})$, we denote by
$$ \gamma^0(F)=F|_{\partial U},\; \gamma^1(F)=(\partial_{\nu}F)|_{\partial U}
$$
the Dirichlet and Neumann trace, with respectively. From the trace theorem, we know that
$$ \gamma^0: H^s(U)\rightarrow H^{s-\frac{1}{2}}(U)
$$
is bounded and surjective.
Let
$$ \mathcal{H}^1_{0} ( \Omega_1) = \{ v \in H^1( \Omega); v\mid_{\partial\Omega_1 \setminus \Sigma} =0\},
$$
We prove the following existence result of the mixed Dirichlet-Neumann boundary value problem:
\begin{prop}\label{pb:mixed}
For any $F \in H^{-\frac 1 2} ( \Sigma)$, the boundary value problem (note that $\partial\Omega_1=\Sigma\cup\partial\Omega $ and $\Sigma,\partial\Omega$ are separated)
\begin{align}
(\hbar ^2 \Delta -i) w&=0, \label{syst-1}\\
\partial_\nu w \mid_\Sigma = F, \qquad & w \mid_{\partial \Omega_1 \setminus \Sigma} =0 \label{syst-2}
\end{align}
admits a unique solution $w\in \mathcal{H}^1_0(\Omega_1)$ satisfying
$$ \Bigl( \hbar \| \nabla_x w\|_{L^2( \Omega_1)} + \| w\|_{L^2( \Omega_1)} \Bigr) \leq C\hbar \| F\|_{H^{-\frac 1 2} ( \Sigma)}. $$
Furthermore, if $F\in H^{\frac 1 2} ( \Sigma)$, then $w\in {H}^2(\Omega_1)$ and
$$ \| \nabla^2_x w\|_{L^2( \Omega_1)} \leq C\Bigl( \| F\|_{H^{\frac 1 2} ( \Sigma)} + \hbar^{-1}\| F\|_{H^{- \frac 1 2} ( \Omega_1)}\Bigr). $$
\end{prop}
\begin{proof}
We just sketch the proof which is a variation around very classical ideas. Multiplying~\eqref{syst-1} by $\overline{\varphi}$ vanishing on $\partial\Omega_1 \setminus \Sigma$ and integrating by parts using Greens formula, we get
$$ 0 = \int_{\Omega_1 } (\hbar ^2 \Delta -i) w\overline{\varphi} (x) dx = \int_{\Omega_1 } -\hbar ^2 \nabla_x w \nabla_x \overline{\varphi} -iw \overline{\varphi} (x) dx + \int_\Sigma \hbar^2 \partial_\nu w \overline{\varphi} (x) d \sigma
$$
As a consequence, if
the function $w$ satisifes~\eqref{syst-1} ~\eqref{syst-2} if an only if
\begin{equation}\label{Lax-Mil}
\forall v \in \mathcal{H}^1_{0} ( \Omega_1), \qquad Q(w, v ) := \int_{\Omega_1 } \hbar ^2 \nabla_x w \nabla_x \overline{v}+ iw \overline{v} (x) dx = T_F (v) := \int_{\Sigma} \hbar^2 F \overline{v} (x) d \sigma.
\end{equation}
From the trace theorem, the map
$$ v \in \mathcal{H}^1_{0}( \Omega_1) \mapsto v \mid_{\Sigma} \in H^{\frac 1 2} ( \Sigma)$$
is continuous and hence for any $F\in H^{- \frac 1 2 } ( \Sigma)$, the map
$$ v \mapsto T_F(v) \in \mathbb{C}$$ is a continuous antilinear form on $\mathcal{H}^1_{0}(\Omega_1).$
The existence of a unique solution to~\eqref{Lax-Mil} (and consequently the solution to~\eqref{syst-1}, ~\eqref{syst-2}) now follows from Lax-Milgram Theorem.
Applying~\eqref{Lax-Mil} to $v = w$, we get
$$ \| h \nabla_x w\|_{L^2( \Omega_1)} ^2 +\| w\|_{L^2( \Omega_1)} ^2 \leq 2 |T_F(w)| \leq C \hbar^2 \| F\|_{H^{- \frac 1 2}( \Sigma)} \| w\|_{H^1(\Omega_1)},$$
which implies
$$ \| w\| _{H^1( \Omega_1)} \leq C \| F\|_{H^{- \frac 1 2} ( \Sigma)}, $$
and using again~\eqref{Lax-Mil} with $v=w$,
$$ \| w\| ^2_{L^2( \Omega_1)} \leq \hbar^2 \| \nabla_x w\| _{L^2( \Omega_1)} +\hbar^2 | T_F (w)| \leq C \hbar^2 \| F\|^2_{H^{- \frac 1 2} ( \Sigma)}.
$$
This proves the first part in Proposition~\ref{pb:mixed}. The proof of the second part is standard elliptic regularity results. Indeed,
we have
$$ \Delta w = i \hbar^{-2} w, \qquad \partial_\nu w \mid_\Sigma = F \in H^{\frac 1 2} ( \Sigma), \qquad w \mid_{\partial \Omega_1 \setminus \Sigma} =0,$$
and we deduce by standard elliptic regularity results,
$$ \| w \|_{H^2( \Omega_1)} \leq C \Bigl(\hbar^{-2} \| w\|_{L^2( \Omega_1)} + \| F\|_{H^{\frac 1 2 } ( \Sigma)}\Bigr) \leq C \Bigl( \hbar^{-1} \| F\|_{H^{- \frac 1 2} ( \Sigma)} + \| F\|_{H^{\frac 1 2 } ( \Sigma)}\Bigr)
$$
This completes the proof of Proposition \ref{pb:mixed}.
\end{proof}
\subsection{Estimates for some operators}
\begin{lem}\label{technical1}
If $b(x,\xi)\in S^{-m}$ ($m\geq 0$) is compactly supported in $x\in\R^n$, then for any $s\in\R$,
$$\mathrm{Op}_h(b)=\mathcal{O}(h^{-\theta}): H^{s}(\R^n)\rightarrow H^{s+\theta}(\R^n),\quad \forall \theta\in[0,m].
$$
\end{lem}
\begin{proof}
First we show that $\mathrm{Op}_h(b)$ is bounded from $H^s$ to $H^s$. It is equivalent to show that the operator $T_h:=\langle D_x\rangle^s\mathrm{Op}_h(b)\langle D_x\rangle^{-s}$ is bounded (independent of $h$) from $L^2$ to $L^2$. By definition, we have
\begin{align*}
\widehat{(T_hf)}(\xi)=\frac{1}{(2\pi)^d}\int_{\R^n}\langle\xi\rangle^s\widehat{b}(\xi-\eta,h\eta)\langle\eta\rangle^{-s}\widehat{f}(\eta)d\eta,
\end{align*}
where $\widehat{b}(\zeta,\eta)=(\mathcal{F}_{x\rightarrow\zeta}a)(\zeta,\eta)$ is a well-defined function. Thus $\widehat{T_hf}$ can be viewed as an operator acting on $\widehat{f}\in L^2(\R_{\xi}^d)$ with Schwartz kernel
$$ K_h(\xi,\eta):=\frac{1}{(2\pi)^d}\langle\xi\rangle^s\langle\eta\rangle^{-s}\widehat{b}(\xi-\eta,h\eta).
$$
By Schur's test, to check the boundedness of this operator, it suffices to check that
$$ \sup_{\xi,h}\int_{\R^n}|K_h(\xi,\eta)|d\eta<\infty,\quad \sup_{\eta,h}\int_{\R^n}|K_h(\xi,\eta)|d\xi<\infty.
$$
Since $K_h(\xi,\eta)$ is rapidly decaying in $\langle\xi-\eta\rangle$, these conditions can be simply verified by the elementary convolution inequalities:
\begin{align}\label{convolution1}
\int_{\R^n} \frac{1}{\langle\eta\rangle^s\langle\xi-\eta\rangle^M}d\eta\leq C_M\langle\xi\rangle^{-s},\quad \forall M>d , s\geq 0,
\end{align}
and
\begin{align}\label{convolution2} \int_{\R^n}\frac{\langle\eta\rangle^{\sigma}}{\langle\xi-\eta\rangle^M}d\eta \leq C_{M,\sigma}\langle\xi\rangle^{\sigma},\quad \forall M>d+\sigma,\sigma\geq 0.
\end{align}
By interpolation, to finish the proof, it suffices to estimate the operator bound of $\mathrm{Op}_h(b)$ from $H^s$ to $H^{s+m}$. Similarly, we need to check that the kernel
$$ G_h(\xi,\eta)=h^{m}\langle\xi\rangle^{s+m}\widehat{b}(\xi-\eta,h\eta)\langle\eta\rangle^{-s}
$$
satisfies the conditions for Schur's test.
First note that for any $\alpha\in\N^n$,
$$ (i(\xi-\eta))^{\alpha}\widehat{b}(\xi-\eta,\eta)=\frac{1}{(2\pi)^d}\int_{\R^n}(\partial_x^{\alpha}b)(x,\eta)e^{-ix\cdot(\xi-\eta)}dx,
$$
thus $\widehat{b}(\xi-\eta,h\eta)=O\big(\langle\xi-\eta\rangle^{-M}\langle h\eta\rangle^{-m}\big)$ for any $M\in\N$. Note that
$$\langle hm\rangle^{-m}\sim (1+h|\eta|)^{-m}\leq h^{-m}\langle\eta\rangle^{-m}.$$
This implies that
$$ |G_h(\xi,\eta)|\leq C_M \langle\xi\rangle^{s+m}\langle\eta\rangle^{-(s+m)}\langle\xi-\eta\rangle^{-M}.
$$
Now the boundeness of the integration $\int G_h(\xi,\eta)d\eta$ or $\int G_h(\xi,\eta)d\xi$ follows from the same convolution inequalities \eqref{convolution1} and \eqref{convolution2}. This completes the proof of Lemma \ref{technical1}.
\end{proof}
\begin{lem}\label{symbolic}
Let $a\in S^0(\R^{2n}), b\in S^0(\R^{2n})$ be two symbols with compact support in the $x$ variable. Then for any $N\in\N, N\geq 2n$,
\begin{align*}
&\Big\|\mathrm{Op}(a)\mathrm{Op}(b)-\sum_{|\alpha|\leq N}\frac{1}{i^{|\alpha|}\alpha !}\mathrm{Op}\big(\partial_{\xi}^{\alpha}a\partial_x^{\alpha}b \big) \Big\|_{\mathcal{L}(H^s\rightarrow H^s)}\\ \leq &C_{N}\sum_{|\beta|\leq K(n)}\sup_{|\alpha|=N+1}\sup_{(x,\xi)\in\R^{2n}}\iint_{\R^{2n}}\big|\partial_{x,\xi}^{\beta}\partial_z^{\alpha}\partial_{\zeta}^{\alpha}A(x,z,\xi,\zeta)\big|dz d\zeta,
\end{align*}
where
$$ A(x,x,\xi,\zeta)=a(x,\xi+\zeta)b(x+z,\xi).
$$
\end{lem}
\begin{proof}
The symbol of the operator $$\mathrm{Op}(a)\mathrm{Op}(b)-\sum_{|\alpha|\leq N}\frac{1}{i^{|\alpha|}\alpha !}\mathrm{Op}\big(\partial_{\xi}^{\alpha}a\partial_x^{\alpha}b \big)$$
is given by
$$ r_N(x,\xi):=\frac{1}{N!}\iint _{\R^{2n}}\int_0^1(1-t)^N\sum_{|\alpha_1|+|\alpha_2|=N+1}(\partial_y^{\alpha_1}\partial_{\eta}^{\alpha_2}A)(x,tz,\xi,t\zeta)z^{\alpha_1}\zeta^{\alpha_2}\mathrm{e}^{-iz\cdot\zeta}dzd\zeta dt,
$$
with
$$ A(x,z,\xi,\zeta)=a(x,\xi+\zeta)b(x+z,\xi).
$$
Using the identity
$$ z^{\alpha_1}\zeta^{\alpha_2}\mathrm{e}^{-iz\cdot\zeta}=i^{N+1}\partial_{z}^{\alpha_2}\partial_{\zeta}^{\alpha_1}(\mathrm{e}^{-iz\cdot\zeta})
$$
and doing the integration by part, we have
\begin{align*} r_N(x,\xi)=&\sum_{|\alpha|=N+1}\frac{i^{N+1}}{N!}\int_0^1(1-t)^Nt^{N+1}dt\iint_{\R^{2n}}(\partial_z^{\alpha}\partial_{\zeta}^{\alpha}A)(x,tz,\xi,t\zeta)\mathrm{e}^{-iz\cdot\zeta}dzd\zeta\\
=&\sum_{|\alpha|=N+1}\frac{i^{N+1}}{N!}\int_0^1(1-t)^Nt^{N+1-2n}dt\iint_{\R^{2n}}(\partial_z^{\alpha}\partial_{\zeta}^{\alpha}A)(x,z,\xi,\zeta)\mathrm{e}^{-it^{-2}z\cdot\zeta}dzd\zeta
\end{align*}
Hence the integral converges absolutely.
Viewing $r_N(x,\xi)$ as a symbol of order $0$, we obtain the desired bound, thanks to the Caldr\'on-Vaillancourt theorem.
\end{proof}
\subsection{Special sequence of eigenfunctions of a disc}
First we recall that
$$ J_m(z)=\Big(\frac{z}{2}\Big)^m\sum_{k=0}^{\infty}\frac{(-1)^k\big(\frac{z}{2}\big)^{2k}}{k!(m+k)!}
$$
are the Bessel functions satisfying the Bessel differential equation:
$$ z^2J_m''(z)+zJ_m'(z)+(z^2-m^2)J_m(z)=0.
$$
By definition, one has
\begin{equation}\label{recurrence}
J_{m+1}(z)+J_{m-1}(z)=\frac{2m}{z}J_m(z),\quad J_{m-1}(z)-J_{m+1}(z)=2J_m'(z).
\end{equation}
Denote by $\lambda_{m,n}$ the $n$-th zero of $J_m(z)$. It is well known that
$$ \lambda_{m,1}<\lambda_{m,2}<\cdots< \lambda_{m,n}<\cdots
$$
and the functions
$$ \varphi_{m,n}(r,\theta)=J_m(\lambda_{m,n}r)\mathrm{e}^{im\theta}
$$
form an orthogonal sequence of eigenfunctions of $\Delta_{\mathbb{D}}$, associated with eigenvalues $\{\lambda_{m,n}^2:m\in\Z, n\in\N \}$. We will chose a special sequence
$$ J_{\alpha n}(\lambda_{\alpha n,n}r)\mathrm{e}^{i\alpha n\theta}
$$
for some $\alpha\in\N$, to be fixed later. Let us recall some facts about the zeros of Bessel functions:
\begin{prop}[\cite{E84}]\label{zeros}
There exists a continuous function $\iota:[-1,\infty)$, such that $$\lambda_{\alpha n,n}<n\iota(\alpha),\text{and } \lim_{n\rightarrow\infty}\frac{\lambda_{\alpha n,n}}{n}=\iota(\alpha).
$$
Moreover, there exists $0<\beta_1<\beta_2$, such that for all $\alpha\geq 1$,
$$ 1+\beta_1\alpha^{-\frac{2}{3}}<\frac{\iota(\alpha)}{\alpha}\leq 1+\beta_2\alpha^{-\frac{2}{3}}.
$$
\end{prop}
Thanks to this proposition, we have:
\begin{lem}\label{energynorm}
Fix $\alpha\in\N$, large enough and let
$$ w_n:=\frac{\varphi_{\alpha n,n}}{\lambda_{\alpha n,n}\|\varphi_{\alpha n,n}\|_{L^2(\mathbb{D})}}.
$$
Then we have
$$ \|(\partial_{\nu}w_n)|_{\partial\mathbb{D}}\|_{L^2(\partial\D)}=O(1),\quad \mathrm{WF}_h(\partial_{\nu}w_h|_{\Sigma})\subset\mathcal{H}_{\delta}(\partial\mathbb{D}):=\{\delta<r_0<1-\delta \}
$$
where $h=(h_n)_{n\in\N}, h_n=\lambda_{\alpha n,n}^{-1}\sim (\iota(\alpha)n)^{-1}$ and the semiclassical wave-front set is taken for the sequence $(w_n)_{n\in\N}$, with a little abuse of the notation.
\end{lem}
\begin{proof}
To simplify the notation, we write $m=\alpha n$ and $\iota:=\iota(\alpha).$ From Proposition \ref{zeros}, we have
$$ 1+\beta_1\alpha^{-\frac{2}{3}}-o(1)<\frac{\iota}{\alpha}-o(1)=\frac{\lambda_{m,n}}{m}<\frac{\iota}{\alpha}\leq 1+\beta_2\alpha^{-\frac{2}{3}},\text{ as }n\rightarrow\infty.
$$
Note that at $r=1$, $\partial_{\nu}=\partial_r$ and $|\nabla w|^2=|\partial_r w|^2+\frac{1}{r^2}|\partial_{\theta}w|^2$. The hyperbolicity at the boundary is essentially due to the fact that
$$ \partial_{\theta}w_n=imw_n
$$
and
\begin{equation}\label{i}
\frac{|m|}{\lambda_{m,n}}=\frac{\alpha}{\iota}+o(1)\leq 1-\delta(\alpha)
\end{equation} for $n\gg 1$. Let $0<\epsilon_0<\delta(\alpha)$, $\chi\in C^{\infty}(\R)$ such that $\chi(s)\equiv 0$ if $|s|>1-\epsilon_0$. From \eqref{i} we have $w_n=\chi(h_n\partial_{\theta})w_n.$ Since $\partial_{\theta}^2$ is just the Laplace operator on $L^2(\partial\mathbb{D})$, we have, near $\partial\mathbb{D}$, $\mathrm{WF}_h(w_n)$ is contained in $r>\epsilon_0>0$, thus $w_n$ is microlocalized near $\mathcal{H}(\Sigma)$. The estimate $\|\partial_{r}w_n|_{r=1}\|_{L^2(\partial\mathbb{D})}=O(1)$ then follows from the hyperbolicity and the fact that $\|\nabla w_n\|_{H^1(\mathbb{D})}=1$. This completes the proof of Lemma \ref{energynorm}.
\end{proof}
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\end{center}
\subsection{Geometric optics construction}
In this part we recall the geometric optics construction for the hyperbolic boundary value problem. In the tubular neighborhood of the interface $\Sigma$, we use the geodesic normal coordinate $x=(y,x')$, such
$$ \Delta=\frac{1}{\kappa}\partial_y(\kappa\partial_y)+\frac{1}{\kappa}\partial_i(g_0^{ij}\kappa\partial_j),
$$
where $\kappa=\sqrt{\det(g_0)}$ and $\partial_j=\partial_{x_j'}$. The semiclassical operator
$$ P_{h}=h^2\Delta_{g_0}+1=h^2\partial_y^2+h^2g_0^{ij}\partial_i\partial_j+1+\frac{h}{\kappa}(\partial_y\kappa) h\partial_y+\frac{h}{\kappa}\partial_i(g_0^{ij}\kappa)h\partial_j.
$$
Let $f_0^{\pm}\in L^2(\R_{x'}^{d-1})$ such that $\mathrm{WF}_h(f_0^{\pm})$ lies in a neighborhood of $(y=0,x_0';\eta=0,\xi_0')$, such that
$$ r_0(0,x_0',\xi_0')\geq c_0>0.
$$ Denote by $\theta^{\pm}(\xi)=\mathcal{F}_h(\chi f_0^{\pm})(\xi)$, where $\chi\in C_c^{\infty}(\R_{x'}^{d-1})$, supported near $x_0'$.
Consider the semi-classical Fourier integral operators $U^{\pm}$, represented by
$$ U^{\pm}(\chi f_0^{\pm})(y,x')=\frac{1}{(2\pi h)^{d-1}}\int_{\R^{d-1}} \mathrm{e}^{\frac{i}{h}\varphi^{\pm}(y,x',\xi')}b^{\pm}(y,x',\xi')\theta^{\pm}(\xi')d\xi'.
$$
We have
\begin{align*}
P_h(U^{\pm}(\chi f_0^{\pm}))=\frac{1}{(2\pi h)^{d-1}}\int_{\R^{d-1}} (h^2\Delta_g+1)(\mathrm{e}^{\frac{i\varphi^{\pm}}{h}} b^{\pm})\theta^{\pm}(\xi')d\xi'.
\end{align*}
Observing that
\begin{align*}
(h^2\Delta_{g_0}+1)(\mathrm{e}^{\frac{i\varphi^{\pm}}{h}} b^{\pm})=&(1-|\nabla_{g_0}\varphi^{\pm}|^2)b^{\pm}\mathrm{e}^{\frac{i\varphi^{\pm}}{h}}
+ih(2\nabla_{g_0}\varphi^{\pm}\cdot\nabla_{g_0}b^{\pm}+\Delta_{g_0}\varphi^{\pm}b^{\pm} )\mathrm{e}^{\frac{i\varphi^{\pm}}{h}}\\
+& h^2(\Delta_{g_0}b^{\pm})\mathrm{e}^{\frac{i\varphi^{\pm}}{h}}.
\end{align*}
Near $\mathrm{WF}_h(f_0)$ and for small $y$, we can solve the eikonal equation
\begin{equation}\label{eikonal} 1-|\nabla_{g_0}\varphi^{\pm}|^2=0,\quad \varphi^{\pm}|_{y=0}(x',\xi')=x'\cdot\xi'.
\end{equation}
Note that
$ |\nabla_{g_0}\varphi^{\pm}|^2=|\partial_y\varphi^{\pm}|^2+g_0^{jk}\partial_j\varphi^{\pm}\partial_k\varphi^{\pm}.
$ Near $(x_0',\xi_0')\in\mathcal{H}(\Sigma)$, for each fixed $\xi'$, we find a Lagrangian submanifold of $T^*\Sigma$, locally of the form
$$ \mathcal{L}_{0,\xi'}:=\{(x',\xi=\partial_{x'}\varphi_0(x,\xi')): \varphi_0(x',\xi')=x'\cdot\xi' \}.
$$
At each point $(x',\xi=\xi')\in\mathcal{L}_{0,\xi'}$, there are two distinct roots $\eta^{\pm}$ of the equation
$$ \eta^2+g_0^{jk}\xi_j'\xi_k'=1,
$$
and each root determines a flow $\Phi_y^{\pm}$ of the bicharacteristics $p=\eta^2-r(y,x',\xi')$ on $\{p=0\}$. Then we can define the Lagrangian
$ \mathcal{L}_{y,\xi'}^{\pm}:=\exp(\Phi_y^{\pm})(\mathcal{L}_{0,\xi'})
$ locally, which is again a Lagrangian of $T^*\Sigma$ (viewing $y$ as a parameter) and can be written locally as $\mathcal{L}_{y,\xi'}^{\pm}=\{(x',\partial_{x'}\varphi^{\pm})\}$. Then $\varphi^{\pm}$ is the desired solutions of \eqref{eikonal} with the property
$$ \partial_y\varphi^{+}+\partial_y\varphi^{-}=0,\text{ at }y=0.
$$
Next we set
$$ b^{\pm}(y,x',\xi')=\sum_{j=0}^Nh^jb_j^{\pm}(y,x',\xi'),
$$
with coefficients $b_j$
solving the transport equations
\begin{equation}\label{transport}
\begin{split}
&2\partial_y\varphi^{\pm}\partial_y b_0^{\pm}+g_0^{jk}\partial_j\varphi^{\pm}\partial_kb_0^{\pm}+(\Delta_{g_0}\varphi^{\pm})b_0^{\pm}=0,\\
&2i\partial_y\varphi^{\pm}\partial_y b_j^{\pm}+ig_0^{jk}\partial_j\varphi^{\pm}\partial_kb_j^{\pm}+i(\Delta_{g_0}\varphi^{\pm})b_{j}^{\pm}+\Delta_{g_0}b_{j-1}^{\pm}=0,\; 1\leq j\leq N.
\end{split}
\end{equation}
Then
$$ P_h(\chi f_0^{\pm})=\frac{h^{N+2}}{(2\pi h)^{d-1}}\int_{\R^{d-1}}\mathrm{e}^{\frac{i\varphi^{\pm}}{h}}\Delta_{g_0}b_N^{\pm}(y,x',\xi')\theta^{\pm}(\xi')d\xi'=O_{L^2}(h^{N+2}).
$$
To determine the datum $b_j^{\pm}|_{y=0}$, we need the boundary conditions. Note that
the approximate quasi-mode is given by
$$ u_h=\sum_{\pm} U^{\pm}(\chi f_0^{\pm})
$$
and we want to determine $f_0^{\pm}$.
Denote by $B_h^{\pm}=\mathrm{Op}_h(b^{\pm})$ and $B_h^{0,\pm}=\mathrm{Op}_h(b^{\pm}|_{y=0})$,
then the Dirichlet trace is given by
$$ B_h^{0,+}(\chi f_0^{+})+B_h^{0,-}(\chi f_0^{-}),
$$
and the Neumann trace is given by
$$ \sum_{\pm}\pm\mathrm{Op}_h(\sqrt{r_0}b^{\pm}|_{y=0} )(\chi f_0^{\pm})+h\sum_{\pm}\mathrm{Op}_h(\partial_yb^{\pm}|_{y=0} )|_{y=0}(\chi f_0^{\pm}).
$$
Now we choose $b_0^{+}|_{y=0}=b_0^-|_{y=0}=\chi(x')\psi(\xi')$ localized near $(x_0',\xi_0')$ and $b_j^{\pm}|_{y=0}=0$ for all $1\leq j\leq N$. Then the symbol (matrix-valued)
\begin{align*}
\Theta=\Theta_0+h
\left(\begin{matrix}
0 &0\\
\partial_yb_0^+ &\partial_yb_0^-
\end{matrix}
\right)|_{y=0}
\end{align*}
with
$$\Theta_0:=\left(
\begin{matrix}
b_0^+ &b_0^-\\
\sqrt{r_0}b_0^+ &-\sqrt{r_0}b_0^-
\end{matrix}\right)|_{y=0}
$$
is invertible. For such an elliptic system, we can construct a symbol (matrix-valued) $\Upsilon$, such that
$$ \mathrm{Op}_h(\Theta)\mathrm{Op}_h(\Upsilon)=\mathrm{Id}+\mathcal{O}_{H^s\rightarrow H^{s+m}}(h^{N+1-m}).
$$
In particular, for a given Dirichlet trace $\sigma_{Dir}$ and Neumann trace $\sigma_{Neu}$ with wave front sets located near $(x_0',\xi_0')$, we find
$$\binom{\chi f_0^+}{\chi f_0^-}=\chi\mathrm{Op}_h(\Upsilon)\chi\binom{\sigma_{Dir}}{\sigma_{Neu}}.
$$
Then microlocally near $(x_0',\xi_0')\in\mathcal{H}(\Sigma)$, $u_h$ satisfies
$$ (h^2\Delta_{g_0}+1)u_h=O_{L^2}(h^N),\quad u|_{h=0}=\sigma_{Dir}+O_{H^{\frac{1}{2}}}(h^{N}),\; h\partial_yu_h|_{y=0}=\sigma_{Neu}+O_{H^{-\frac{1}{2}}}(h^N),
$$
and microlocally near $(x_0',\xi_0')$, $u_h=O_{L^2}(1)$,$\mathrm{WF}_h(u_h)$ lies in a small neighborhood of $(x_0',\xi_0')$.
Finally, due to the microlocalisation in the hyperbolic region, we can exchange in the error terms powers of $h$ against derivatives, leading to
$$ (h^2\Delta_{g_0}+1)u_h=O_{H^{k}}(h^{N-k}),\; u|_{h=0}=\sigma_{Dir}+O_{H^{\frac{1}{2} +k }}(h^{N-k }),\; h\partial_yu_h|_{y=0}=\sigma_{Neu}+O_{H^{-\frac{1}{2} + k }}(h^{N-k}).
$$ | 0.00264 |
Opinion
DeMarco’s Chairshot Power Rankings: Week Ending 11/7/19
It’s the 11/7/19 edition of The Chairshot Power Rankings! Did the NXT Invasion overcome all this week? Who topped the list, and why was it not The Fiend? Was a strong promo enough to get Cody in?
It’s the 11/7/19 edition of The Chairshot Power Rankings! Did the NXT Invasion overcome all this week? Who topped the list, and why was it not The Fiend? Was a strong promo enough to get Cody in?.
Greg DeMarco’s Power Rankings: week of 10/31/2019 – 11/6/2019
Missing The Cut
- Airplane Mechanics – C’mon, who else defeated 175 people and altered a show watched by 2.5 million viewers this week?
- SHO & YOH – The Roppongi 3K duo picked up their third straight New Japan Junior Tag League win this week.
- Tommaso Ciampa – Daddy has proven to be back in a big way, with a win over The Miz on Smackdown and leading the War Games charge against The Undisputed Era.
Missing The Cut is listed in no particular order.
5. Tetsuya Naito
Tetsuya Naito snagged a win over Taichi, and did so in a more serious manner. Being in the IWGP Intercontinental Championship is important for Naito, as New Japan has confirmed that the Intercontinental Champion will face the IWGP Heavyweight Champion at night two of WrestleKingdom on January 5, with both championships on the line. This win puts Naito in position to leave the Tokyo Dome as a double champion.
4. Cody Rhodes
Headed into AEW Full Gear, Cody has all the momentum in the world behind him. Many are calling his promo from Wednesday night the best of his career, and I have even seen people call it the best of any career. I don’t know of all this leads to Cody winning the AEW World Championship, or if another plan is in place. I would assume the latter, and here’s hoping that the plan is one that will do the promo justice.
3. Colt Cabana
Thanks to his win on NWA Power, Colt Cabana is (once again) the new NWA National Champion. The match was obviously taped several weeks ago, but I count the airing and am giving the nod to Cabana here. Outside of the televised product, NWA Power is one of the hottest streaming commodities on today. Cabana’s win might be the biggest of his career in terms of overall eyeballs and attention.
2. Bray Wyatt/The Fiend
After failing in his bid to capture the WWE Universal Championship from Seth Rollins at Hell In A Cell, Bray Wyatt’s The Fiend was successful in that same conquest at Crown Jewel. The win also moved the Universal Championship to Smackdown, as The Fiend will remain on wrestling’s most watched program on a weekly basis.
1. Adam Cole
I mean I can’t think of a person with more momentum than ol’ Adam Cole BAY BAY. Adam Cole was at the forefront of the hottest angle in wrestling this week, NXT’s “Invasion” of Smackdown. He got a career highlight clean victory over Daniel Bryan with Triple H and Shawn Michaels looking on from ringside–a win that was the impetus for Triple H’s brand defining promo. From there he followed it up with a great performance with Seth Rollins on Raw. To me, this was enough to edge out Bray Wyatt’s Universal Championship win at Crown Jewel, even if not by much.<< | 0.001069 |
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Buddy Barrett wowed us with this beautifully kept '59 Corvette convertible. Though an enthusiast for sentimental reasons, Buddy was a novice when it came to the stringent rules and regulations of Corvette judging. It was a good thing he had a dead ringer on his hands. His Corvette is a gorgeous survivor of over 46 years. We let him tell his story.
"Marilyn, my wife, and I had been looking for a C1 for several years, starting in the late '90s, but we couldn't reach a meeting of the minds. She had her sights set on a '62 Fawn Beige fuelie and I, being the caring husband, was desperately trying to track down and purchase that same color and model; but, as fate would have it, this would prove unsuccessful.
"The '59 was decided upon, and ended up in our garage-not only because of the merits of the vehicle, but because it was the same color and horsepower as my first ride in a friend's Corvette back in 1959. In addition, the VIN, J59S100351 (351 of 9,670 produced for that year), correlates with everything, making it an all numbers-matching car, and most components are believed to be original. The paint code is 506A/518B. It was also from the same year I received my driver's license, finished high school, joined the U.S. Marine Corps and, most importantly, my 59th birthday was only a month away from the purchase date.
"I brought the '59 to the NCRS Regionals at the Wildwood, New Jersey, Convention Center, thanks to many of my friends who pushed me in that direction. I was thrilled beyond words when the little red Corvette won its Top Flight Award, but the highlight of the show was talking with many NCRS judges and learning so much more about what makes these classics what they are.
"The dual carbs, according to NCRS, are believed to have never been rebuilt and appear almost factory fresh. This corresponds with the mileage on the odometer, 16,400, which is also believed to be original. I am the second owner of record, and I purchased the '59 in June 2002 from an individual in California who had the car in storage for many years. It had 12,009 miles. The NCRS also stated that the engine stamp pad and number F923CT was one of the finest examples they had ever seen. Another national judge said the body and body panels were rare because they were exceptionally straight and fitting for that year. The judge also said this particular NCRS meet was unique because they had never before seen the original hood and deck ornaments on the '59 in such as-new condition.
"As we are still making corrections and doing restoration work, our future plans include the NCRS National Convention in Park City, Utah, in July, then the Duntov Award judging for operations and performance in 2006. Owning this '59 Corvette is seemingly a never-ending, nerve-shattering, but well-rewarding hobby and ongoing experience."
Entering a Corvette in the gauntlet of judging may seem a bit daunting, but Buddy went into the game with a stacked deck. His '59 convertible is a prime example of a survivor, having endured over 40 years of life on the road without a blemish, dent, or scratch. Even for those who might not like the idea of garaging these vehicles, you have to admit it's gorgeous. | 0.134713 |
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tel: 250.812.8877 | 0.885932 |
\begin{document}
\def\smfbyname{}
\begin{abstract}
In this article, we study Newton-Okounkov bodies on projective vector bundles over curves. Inspired by Wolfe's estimates used to compute the volume function on these varieties, we compute all Newton-Okounkov bodies with respect to linear flags. Moreover, we characterize semi-stable vector bundles over curves via Newton-Okounkov bodies.
\end{abstract}
\maketitle
\tableofcontents
\section{Introduction}
Let $C$ be a smooth projective curve over an algebraically closed field of characteristic zero and let $E$ be vector bundle over $C$ of rank $r\geq 2$. It is well-known since Hartshorne's work \cite{Ha71} that numerical information coming from semi-stability properties of $E$ can be translated into positivity conditions. Namely, the shape of the nef and pseudo-effective cone of the projective vector bundle $\mathbb{P}(E)$ were determined by Hartshorne \cite{Ha71} (cf. \cite[\S 6.4.B]{Laz04}) and Nakayama \cite[Chapter IV]{Nak04} (cf. \cite{MDS15}) in terms of the Harder-Narasimhan filtration of $E$. More precisely, if we denote by $\xi$ the class of the tautological line bundle $\mathcal{O}_{\mathbb{P}(E)}(1)$ and by $f$ the class of a fiber of $\pi:\mathbb{P}(E)\to C$ then we have that for $t\in \mathbb{R}$ the class $\xi - tf$ is nef (resp. pseudo-effective) if and only if $t\leq \mu_{\min}(E)$ (resp. $t\leq \mu_{\max}(E)$), where $\mu_{\min}(E)$ (resp. $\mu_{\max}(E)$) is the minimal (resp. maximal) slope of $E$. In particular, the nef and pseudo-effective cone of $\mathbb{P}(E)$ coincide if and only if $E$ is semi-stable (cf. \cite[Theo. 3.1]{Miy87} and \cite{Ful11}).
Indirectly, the pseudo-effective cone can be also deduced from the work of Wolfe \cite{Wol05} and Chen \cite{Che08}, who explicitly computed the volume function on $\mathbb{P}(E)$. In fact, they showed that for every $t\in \mathbb{R}$ the volume of the class $\xi - tf$ on $\mathbb{P}(E)$ can be expressed in terms of numerical information coming from the Harder-Narasimhan filtration of $E$,
$$\operatorname{HN}_\bullet(E) : 0=E_\ell \subseteq E_{\ell-1} \subseteq \cdots \subseteq E_1 \subseteq E_0=E $$
with successive semi-stable quotients $Q_i=E_{i-1}/E_i$ of rank $r_i$ and slope $\mu_i$. More precisely, if we consider $\sigma_1 \geq \ldots \geq \sigma_r$ to be the ordered slopes of $E$ counted with multiplicities equal to the rank of the corresponding semi-stable quotient\footnote{In other words, $\mu_{\max}(E)=\sigma_1 \geq \ldots \geq \sigma_r=\mu_{\min}(E)$ can be viewed as the components of the vector $\boldsymbol{\sigma}=(\sigma_1,\ldots,\sigma_r)=(\underbrace{\mu_\ell,\ldots,\mu_\ell}_{r_\ell\text{ times}},\underbrace{\mu_{\ell-1},\ldots,\mu_{\ell-1}}_{r_{\ell-1} \text{ times}},\ldots,\underbrace{\mu_1,\ldots,\mu_1}_{r_1 \text{ times}})\in \mathbb{Q}^r$.}, their results can be summarized as follows.
\begin{theo}\label{theo: volume result, wolfe and chen}
Let $C$ and $E$ as above. Then,
$$\vol_{\mathbb{P}(E)}(\xi-tf)=r! \cdot \int_{\widehat{\Delta}_{r-1}} \max \left\{\sum_{j=1}^r \sigma_{r+1-j} \lambda_j - t,0 \right\}\; d\lambda, $$
\noindent where ${\widehat{\Delta}}_{r-1}\subseteq \mathbb{R}^r$ is the standard $(r-1)$-simplex with coordinates $\lambda_1,\ldots,\lambda_r$ and $d\lambda$ is the standard induced Lebesgue measure for which $\widehat{\Delta}_{r-1}$ has volume $\frac{1}{(r-1)!}$.
\end{theo}
Following the idea that numerical information encoded by the Harder-Narasimhan filtration of $E$ should be related to asymptotic numerical invariants of $\mathbb{P}(E)$, we study the geometry of Newton-Okounkov bodies on $\mathbb{P}(E)$. These compact convex bodies were introduced by Okounkov in his original article \cite{Ok96} and they were studied later on by Kaveh and Khovanskii \cite{KK12} and Lazarsfeld and Musta\c{t}\u{a} \cite{LM09}, who associated to any big divisor $D$ on a normal projective variety $X$ of dimension $r$, and any complete flag of subvarieties $Y_\bullet$ on $X$ satisfying suitable conditions, a convex body $\Delta_{Y_\bullet}(D)\subseteq \mathbb{R}^r$ depending only on the numerical equivalence class of $D$. Moreover, there exists a {\it global Newton-Okounkov body} $\Delta_{Y_\bullet}(X)\subseteq \mathbb{R}^r\times \N^1(X)_{\mathbb{R}}$ such that the slice of $\Delta_{Y_\bullet}(X)$ over any big rational class $\eta\in \N^1(X)_{\mathbb{Q}}$ is given by $\Delta_{Y_\bullet}(\eta)\subseteq \mathbb{R}^r\times \{\eta\}$.
Newton-Okounkov bodies of big divisors on geometrically ruled surfaces with respect to {\it linear flags} (see Definition \ref{defi: natural flag}) can be computed via Zariski decomposition (see Example \ref{ex:ruled surfaces}). In higher dimension, we will use methods similar to those used by Wolfe to compute the volume function in \cite{Wol05}. More precisely, we will first reduce ourselves to the case of the nef and big class $\xi-\mu_{\min}(E)f$ (see Lemma \ref{lemma:reduction to big nef}). Afterwards, we will need to understand the Harder-Narasimhan filtration of the symmetric products $S^m E$ for $m\geq 1$ and then to consider suitable refinements of these filtrations.
Let $Y_\bullet$ be a complete {\it linear flag which is compatible with the Harder-Narasimhan filtration of $E$} (see Definition \ref{defi: compatible natural flags}).
With the notation of Theorem \ref{theo: volume result, wolfe and chen} above, define for each real number $t\in \mathbb{R}$ the following polytope inside the full dimensional standard simplex $\Delta_{r-1}$ in $\mathbb{R}^{r-1}$ (see Notation \ref{notation: Wolfe's square})
$$\square_t=\left\{(\nu_2,\ldots,\nu_r)\in \Delta_{r-1}\;\left|\; \sum_{i=2}^r \sigma_{i-1}\nu_i+\sigma_{r}\left(1-\sum_{i=2}^r \nu_i\right) \geq t \right.\right\}.$$
\noindent Then, we prove the following result (see Corollary \ref{cor: main result}).
\begin{thma}\label{thm: main theorem}
Let $C$ be a smooth projective curve and let $E$ be a vector bundle over $C$ of rank $r\geq 2$. Then, for every real number $t<\mu_{\max}(E)=\mu_\ell$ we have that
$$\Delta_{Y_\bullet}(\xi-t f)=\left\{(\nu_1,\ldots,\nu_r)\in \mathbb{R}^r_{\geq 0}\;|\;0 \leq \nu_1 \leq \mu_\ell - t,\;(\nu_2,\ldots,\nu_r)\in \square_{t+\nu_1} \right\}. $$
In particular, the global Newton-Okounkov body $\Delta_{Y_\bullet}(\mathbb{P}(E))$ is a rational polyhedral cone and it depends only on $\operatorname{gr}(\operatorname{HN}_\bullet(E))$, the graded vector bundle associated to the Harder-Narasimhan filtration of $E$.
\end{thma}
Moreover, we obtain the following characterization of semi-stability in terms of Newton-Okounkov bodies.
\begin{propb}\label{propo:semi-stable case}
Let $C$ be a smooth projective curve and let $E$ be a vector bundle over $C$ of rank $r\geq 2$. The following conditions are then equivalent:
\begin{itemize}
\item[(1)] $E$ is semi-stable.
\item[(2)] For every big rational class $\eta=a(\xi-\mu_\ell f)+bf$ on $\mathbb{P}(E)$ and every linear flag $Y_\bullet$ on $\mathbb{P}(E)$ we have that $\Delta_{Y_\bullet}(\eta)=[0,b]\times a\Delta_{r-1} \subseteq \mathbb{R}^r$.
\end{itemize}
Here, $a\Delta_{r-1} =\{(\nu_2,\ldots,\nu_r)\in \mathbb{R}_{\geq 0}^{r-1}\;|\;\sum_{i=2}^r \nu_i \leq a \}$ is the full dimensional standard $(r-1)$-simplex with side length $a$.
\end{propb}
\subsubsection*{{\bf Outline of the article}} First of all, we establish some notation and recall some basic facts in $\S 2$. Secondly, we recall in $\S 3$ definitions and well-known results about Newton-Okounkov bodies and semi-stability of vector bundles over curves. Next, $\S 4$ is devoted to the different cones of divisors on $\mathbb{P}(E)$ as well as results concerning their volume and restricted volume. Finally, we prove both Theorem A and Proposition B in $\S 5$.
\subsubsection*{{\bf Acknowledgements}} I would like to express my gratitude to my thesis supervisors, St\'ephane \textsc{Druel} and Catriona \textsc{Maclean}, for their advice, helpful discussions and encouragement throughout the preparation of this article. I also thank Bruno \textsc{Laurent}, Laurent \textsc{Manivel} and Bonala \textsc{Narasimha Chary} for fruitful discussions. Finally, I would like to thank the anonymous referee for a very helpful and detailed report.
\section{{Preliminaries}}
Throughout this article all varieties will be assumed to be reduced and {irreducible} schemes of finite type over a fixed algebraically closed field of characteristic zero\footnote{In positive characteristic one should take into account iterates of the absolute Frobenius as in \cite{BP14,BHP14}. It is worth mentioning that in positive characteristic the semistability of a vector bundle over a curve does not imply the semi-stability of its symmetric powers.} $k$.
\subsection{Numerical classes and positivity} We denote by $\N^1(X)$ the group of numerical equivalence classes of Cartier divisors on $X$, and we define $\N^1(X)_{\Bbbk}=\N^1(X)\otimes_{\mathbb{Z}}\Bbbk$ for $\Bbbk=\mathbb{Q}$ or $\mathbb{R}$. All the $\mathbb{R}$-divisors that we consider are $\mathbb{R}$-Cartier. Dually, we denote by $\N_1(X)$ the group of numerical equivalence classes of 1-cycles on $X$. Inside $\N_1(X)_{\mathbb{R}}=\N_1(X)\otimes_\mathbb{Z}\mathbb{R}$ we distinguish the {\it Mori cone} $\overline{\NE}(X)\subseteq \N_1(X)_\mathbb{R}$, which is the closed and convex cone generated by numerical classes of 1-cycles with non-negative real coefficients.
Let $E$ be a locally free sheaf on a variety $X$. We follow Grothendieck's convention and we define the {\it projectivization $\mathbb{P}_X(E)=\mathbb{P}(E)$ of $E$} to be ${\mathbf{Proj}_{\mathcal{O}_X} \oplus_{m\geq 0} S^m E}$, where $S^m E$ denotes de $m$th symmetric power of $E$. This variety is endowed with a natural projection $\pi:\mathbb{P}(E)\to X$ and a tautological line bundle $\mathcal{O}_{\mathbb{P}(E)}(1)$.
Let $X$ be a normal projective variety. Following \cite{Laz04}, we say that a numerical class $\eta\in \N^1(X)_{\mathbb{R}}$ is {\it big} if there exists an effective $\mathbb{R}$-divisor $E$ such that $\eta-E$ is ample. We denote by $\Bg(X)\subseteq \N^1(X)_{\mathbb{R}}$ the open convex cone of big numerical classes. A numerical class $\eta\in \N^1(X)_{\mathbb{R}}$ is called {\it pseudo-effective} if it can be written as the limit of classes of effective $\mathbb{R}$-divisors. The pseudo-effective cone is the closure of the big cone: $\overline{\Bg}(X)=\Psef(X)$ (see \cite[Theo. 2.2.26]{Laz04}, for instance). Moreover, a numerical class $\eta\in \N^1(X)_{\mathbb{R}}$ is {\it nef} if $\eta\cdot [C]\geq 0$ for every $[C]\in \overline{\NE}(X)$, and is {\it ample} if it is the numerical class of an $\mathbb{R}$-divisor that can be written as a finite sum of ample Cartier divisors with positive real coefficients. The cone $\Nef(X)\subseteq \N^1(X)_{\mathbb{R}}$ of nef classes is closed convex, and its interior $\Amp(X)$ is the cone of ample classes, by Kleiman's ampleness criterion. A line bundle $L$ is big (resp. pseudo-effective, ample, nef) if and only if its numerical class $c_1(L)\in \N^1(X)_{\mathbb{R}}$ is big (resp. pseudo-effective, ample, nef).
Let us recall that the {\it stable base locus} of a $\mathbb{Q}$-divisor $D$ on $X$ is the closed set
$$\mathbf{B}(D)=\bigcap_{m>0}\bs(mD) $$
where $\bs(mD)$ is the base locus of the complete linear series $|mD|$. Following \cite{ELMNP06}, we define the {\it augmented base locus} of $D$ to be the Zariski closed set
$$\mathbf{B}_+(D)=\bigcap_{A}\mathbf{B}(D-A), $$
where the intersection runs over all ample $\mathbb{Q}$-divisors $A$. Similarly, the {\it restricted base locus} of $D$ is defined by
$$\mathbf{B}_-(D)=\bigcup_{A}\mathbf{B}(D+A), $$
where the union runs over all ample $\mathbb{Q}$-divisors $A$. The restricted base locus $\mathbf{B}_-(D)$ consist of at most a countable union of subvarieties whose Zariski closure is contained in $\mathbf{B}_+(D)$ (see \cite[Rema. 1.13]{ELMNP06} and \cite{Les14}). By \cite[Prop. 1.4, Exam. 1.8, Prop. 1.15, Exam. 1.16]{ELMNP06}, both $\mathbf{B}_{-}(D)$ and $\mathbf{B}_+(D)$ depend only on the numerical class of $D$, there is an inclusion $\mathbf{B}_{-}(D) \subseteq \mathbf{B}_+(D)$, and for any rational number $c>0$ we have ${\mathbf{B}_{-}(cD)=\mathbf{B}_{-}(D)}$ and ${\mathbf{B}_+(cD)=\mathbf{B}_+(D)}$. Moreover by \cite[Exam. 1.7, Exam. 1.18]{ELMNP06} we have that ${\mathbf{B}_+(D)=\emptyset}$ if and only if $D$ is ample, and that $\mathbf{B}_-(D)=\emptyset$ if and only if $D$ is nef.
\subsection{Flag varieties and Schubert cells} Let us denote by $\mathbb{F}_r$ the full flag variety parametrizing all complete linear flags on $\mathbb{P}^{r-1}$. Recall that if we fix a reference complete linear flag $Y_\bullet$ on $\mathbb{P}^{r-1}$, then there is a decomposition of $\mathbb{F}_r$ into {\it Schubert cells}
$$\mathbb{F}_r=\coprod_{w\in \mathfrak{S}_r}\Omega_w. $$
More explicitly, if we consider homogeneous coordinates $[x_1:\ldots:x_r]$ on $\mathbb{P}^{r-1}$, we assume that for every $i=1,\ldots,r-1$ we have
$$Y_i=\{x_1=\ldots=x_i=0\}\subseteq \mathbb{P}^{r-1}$$
and we regard the permutation group $\mathfrak{S}_r$ as a subgroup of $\operatorname{PGL}_r(k)$ via its natural action on the standard basis points $e_1,\ldots,e_r\in \mathbb{P}^{r-1}$ then we have that $\Omega_w$ is the orbit $B\cdot Y_\bullet^w$, where $B$ denotes the (Borel) subgroup of $\operatorname{PGL}_r(k)$ that fixes the flag $Y_\bullet$ and $Y_\bullet^w$ is the complete linear flag such that for every $i=1,\ldots,r-1$ we have
$$Y_i^w = \{x_{w(1)}=\ldots=x_{w(i)}=0 \} \subseteq \mathbb{P}^{r-1}.$$
We refer the reader to \cite[$\S 1.2$]{Bri05} and \cite[$\S 3.6$]{Man98} for further details.
\section{{Newton-Okounkov bodies and Semi-stability}}
In this section, we review the construction of Newton-Okounkov bodies and semi-stability of vector bundles over smooth projective curves.
\subsection{Newton-Okounkov bodies}
Let $X$ be a smooth projective variety of dimension $n$ and let $L$ be a big line bundle on $X$. A full flag of closed subvarieties of $X$ centered at the point $p\in X$
$$Y_\bullet: X = Y_0 \supseteq Y_1 \supseteq Y_2 \supseteq \cdots \supseteq Y_{n-1} \supseteq Y_n = \{p\} $$
\noindent is an {\it admissible flag} if $\codim_X(Y_i)=i$, and each $Y_i$ is smooth at the point $p$. In particular, $Y_{i+1}$ defines a Cartier divisor on $Y_i$ in a neighborhood of the point $p$. Following the work of Okounkov \cite{Ok96,Ok00}, Kaveh and Khovanskii \cite{KK12} and Lazarsfeld and Musta\c{t}\u{a} \cite{LM09} independently associated to $L$ and $Y_\bullet$ a convex body $\Delta_{Y_\bullet}(X,L)\subseteq \mathbb{R}^n$ encoding the asymptotic properties of the complete linear series $|L^{\otimes m}|$. We will follow the presentation of \cite{LM09} and we refer the interested reader to the survey \cite{Bo12} for a comparison of both points of view.
Let $D$ be any divisor on $X$ and let $s=s_1\in H^0(X,\mathcal{O}_X(D))$ be a non-zero section. We shall compute successive vanishing orders of global sections in the following manner: let $D_1=D+\operatorname{div}(s_1)$ be the effective divisor in the linear series $|D|$ defined by $s_1$ and set $\nu_1(s)=\ord_{Y_1}(D_1)$ the coefficient of $Y_1$ in $D_1$. Then $D_1 - \nu_1(s)Y_1$ is an effective divisor in the linear series $|D-\nu_1(s)Y_1|$, and does not contain $Y_1$ in its support, so we can define $D_2=(D_1-\nu_1(s)Y_1)|_{Y_1}$ and set $\nu_2(s)=\ord_{Y_2}(D_2)$. We proceed inductively in order to get $\nu_{Y_\bullet}(s)=(\nu_1(s),\ldots,\nu_n(s))\in \mathbb{N}^d$. This construction leads to a valuation-like function
$$\nu_{Y_\bullet}: \h^0(X,\mathcal{O}_X(D))\setminus \{0\} \to \mathbb{Z}^n,\;s\mapsto (\nu_1(s),\ldots,\nu_n(s)). $$
We then define the {\it graded semigroup} of $D$ to be the sub-semigroup of $\mathbb{N}^n\times \mathbb{N}$ defined by
$$\Gamma_{Y_\bullet}(D)=\left\{(\nu_{Y_\bullet}(s),m)\in \mathbb{N}^n\times \mathbb{N}\;|\;0\neq s \in \h^0(X,\mathcal{O}_X(mD) \right\}. $$
Finally, we define the {\it Newton-Okounkov body of $D$ with respect to $Y_\bullet$} to be
$$\Delta_{Y_\bullet}(D)=\operatorname{cone}(\Gamma_{Y_\bullet}(D))\cap (\mathbb{R}^n\times \{1\}), $$
where $\operatorname{cone}(\Gamma_{Y_\bullet}(D))$ denotes the closed convex cone in $\mathbb{R}^n \times \mathbb{R}$ spanned by $\Gamma_{Y_\bullet}(D)$.
These convex sets $\Delta_{Y_\bullet}(D)$ are compact and they have non-empty interior whenever $D$ is big. Moreover, by \cite[Theo. A]{LM09}, we have the following identity
$$\vol_{\mathbb{R}^n}(\Delta_{Y_\bullet}(D))=\dfrac{1}{n!}\cdot \vol_X(D), $$
where $\vol_X(D)=\lim_{m\to \infty}\frac{h^0(X,\mathcal{O}_X(mD))}{m^n/n!}$. In particular, if $D$ is big and nef, then $\vol_{\mathbb{R}^n}(\Delta_{Y_\bullet}(D))=\frac{1}{n!}D^n$, by the Asymptotic Riemann-Roch theorem.
The Newton-Okounkov bodies of big divisors depend only on numerical classes: if $D\equiv_{\text{num}} D'$ are big divisors then $\Delta_{Y_\bullet}(D)=\Delta_{Y_\bullet}(D')$ for every admissible flag $Y_\bullet$ on $X$, by \cite[Prop. 4.1]{LM09} (see \cite[Theo. A]{Jow10} for the converse). This fact, along with the identity $\Delta_{Y_\bullet}(pD)=p\cdot \Delta_{Y_\bullet}(D)$ for every positive integer $p$, enables us to define an Newton-Okounkov body $\Delta_{Y_\bullet}(\eta)\subseteq \mathbb{R}^n$ for every big rational class $\eta\in \Bg(X)\cap \N^1(X)_\mathbb{Q}$. Moreover, by \cite[Theo. B]{LM09}, there exists a {\it global Newton-Okounkov body}: a closed convex cone
$$\Delta_{Y_\bullet}(X)\subseteq \mathbb{R}^n\times \N^1(X)_\mathbb{R} $$
such that for each big rational class $\eta \in \Bg(X)_\mathbb{Q}=\Bg(X)\cap \N^1(X)_\mathbb{Q}$ the fiber of the second projection over $\eta$ is $\Delta_{Y_\bullet}(\eta)$. This enables us to define Newton-Okounkov bodies for big real classes by continuity.
The above construction works for {\it graded linear series} $A_\bullet$ associated to a big divisor $D$ on $X$. A graded linear series is a collection of subspaces $A_m\subseteq \h^0(X,\mathcal{O}_X(mD))$ such that $A_{\bullet}=\oplus_{m\geq 0}A_m$ is a graded subalgebra of the section ring ${R(D)=\oplus_{m\geq 0} \h^0(X,\mathcal{O}_X(mD))}$. The construction enables us to attach to any graded linear series $A_\bullet$ a closed and convex set $\Delta_{Y_\bullet}(A_\bullet)\subseteq \mathbb{R}^n$. This set $\Delta_{Y_\bullet}(A_\bullet)$ will be compact and will compute the volume of the linear series under some mild conditions listed in \cite[$\S$ 2.3]{LM09}. We will be specially interested on {\it restricted complete linear series} of a big divisor $D$, namely graded linear series of the form
$$A_m = \h^0(X|F,\mathcal{O}_X(mD))=\text{Im}\left(\h^0(X,\mathcal{O}_X(mD))\xrightarrow{\operatorname{rest}} \h^0(F,\mathcal{O}_F(mD)) \right) $$
where $F\subseteq X$ is an irreducible subvariety of dimension $d\geq 1$. Under the hypothesis that $F\not\subseteq \mathbf{B}_+(D)$, the conditions listed in \cite[$\S$ 2.3]{LM09} are satisfied by \cite[Lemm. 2.16]{LM09}. Therefore, the Newton-Okounkov body associated to $A_\bullet$ above, the {\it restricted Newton-Okounkov body} (with respect to a fixed admissible flag)
$$\Delta_{X|F}(D)\subseteq \mathbb{R}^d, $$
is compact and
$$\vol_{\mathbb{R}^d}(\Delta_{X|F}(D)) = \frac{1}{d!}\vol_{X|F}(D),$$
\noindent where
$$\vol_{X|F}(D)= \lim_{m\to \infty} \frac{\dim_k A_m}{m^d/d!}$$
\noindent is the {\it restricted volume} on $F$ of the divisor $D$. In particular, if $D$ is big and nef, then ${\vol_{X|F}(D)=(D^d\cdot F)}$, by \cite[Cor. 2.17]{ELMNP09}. Restricted Newton-Okounkov bodies depend only on numerical classes (see \cite[Rema. 4.25]{LM09}), so it is meaningful to consider $\Delta_{X|F}(\eta)$ for every big rational class $\eta$ such that $F\not\subseteq \mathbf{B}_+(\eta)$.
As before, there exists a global Newton-Okounkov body $\Delta_{Y_\bullet}(X|F)$ that enables us to define, by continuity, $\Delta_{X|F}(\eta)$ for any big real numerical class $\eta$ such that $F\not\subseteq \mathbf{B}_+(\eta)$. See \cite[Exam. 4.24]{LM09} for details.
Restricted Newton-Okounkov bodies can be used to describe slices of Newton-Okounkov bodies.
\begin{thm}[{\cite[Theo. 4.26, Cor. 4.27]{LM09}}]\label{slices} \noindent Let $X$ be a normal projective variety of dimension $n$, and let $F\subseteq X$ be an irreducible and reduced Cartier divisor on $X$. Fix an admissible flag
$$Y_\bullet: X = Y_0 \supseteq Y_1 \supseteq Y_2 \supseteq \cdots \supseteq Y_{n-1} \supseteq Y_n = \{p\} $$
\noindent with divisorial component $Y_1=F$. Let $\eta\in \Bg(X)_\mathbb{Q}$ be a rational big class, and consider the Newton-Okounkov body $\Delta_{Y_\bullet}(\eta)\subseteq \mathbb{R}^n$. Write $\operatorname{pr}_1:\Delta_{Y_\bullet}(\eta) \to \mathbb{R}$ for the projection onto the first coordinate, and set
\begin{equation*}
\begin{aligned}
\Delta_{Y_\bullet}(\eta)_{\nu_1=t}&=\operatorname{pr}_1^{-1}(t)\subseteq \{t\}\times \mathbb{R}^{n-1}\\
\Delta_{Y_\bullet}(\eta)_{\nu_1\geq t}&=\operatorname{pr}_1^{-1}\left([t,+\infty) \right)\subseteq \mathbb{R}^n
\end{aligned}
\end{equation*}
Assume that $F\not\subseteq \mathbf{B}_+(\eta)$ and let
$$\tau_F(\eta)=\sup\{s>0\;|\;\eta - s\cdot f \in \Bg(X)\}, $$
\noindent where $f\in \N^1(X)$ is the numerical class of $F$. Then, for any $t\in \mathbb{R}$ with $0\leq t < \tau_F(\eta)$ we have
\begin{enumerate}
\item $\Delta_{Y_\bullet}(\eta)_{\nu_1\geq t}=\Delta_{Y_\bullet}(\eta-tf)+t\cdot \vec{e}_1$, where $\vec{e}_1 = (1,0,\ldots,0)\in \mathbb{N}^n$ is the first standard unit vector. \footnote{In fact, this statement remains true even if we do not assume that $E\not\subseteq \mathbf{B}_+(\eta)$. See \cite[Prop. 1.6]{KL15b}.}
\item $\Delta_{Y_\bullet}(\eta)_{\nu_1=t}=\Delta_{X|F}(\eta-tf)$.
\item The function $t\mapsto \vol_X(\eta+tf)$ is differentiable at $t=0$, and
$$\dfrac{d}{dt}\left(\vol_X(\eta+tf) \right)|_{t=0}= n\cdot \vol_{X|F}(\eta). $$
\end{enumerate}
\end{thm}
We will need the following observation by K\"{u}ronya, Lozovanu and Maclean.
\begin{propo}[{\cite[Prop. 3.1]{KLM12}}]\label{propo:psef=nef} Let $X$ be a normal projective variety together with an admissible flag $Y_\bullet$. Suppose that $D$ is a big $\mathbb{Q}$-divisor such that $Y_1\not\subseteq \mathbf{B}_+(D)$ and that $D-tY_1$ is ample for some $0\leq t < \tau_{Y_1}(D)$, where $ \tau_{Y_1}(D)=\sup\{s>0\;|\;D-sY_1 \text{ is big} \}$. Then
$$\Delta_{Y_\bullet}(X,D)_{\nu_1=t}=\Delta_{Y_\bullet|Y_1}(Y_1,(D-tY_1)|_{Y_1}), $$
where $Y_\bullet|Y_1 : Y_1\supseteq Y_2 \supseteq \cdots \supseteq Y_n$ is the induced admissible flag on $Y_1$.
In particular, if $\Psef(X)=\Nef(X)$ then we have that the Newton-Okounkov body $\Delta_{Y_\bullet}(D)$ is the closure in $\mathbb{R}^n$ of the following set
$$\{(t,\nu_2,\ldots,\nu_n)\in \mathbb{R}^n\;|\;0\leq t < \tau_{Y_1}(D), \;(\nu_2,\ldots,\nu_n)\in \Delta_{Y_\bullet|Y_1}(Y_1,(D-tY_1)|_{Y_1}) \}. $$
\end{propo}
Let us finish this section with the case of Newton-Okounkov bodies on surfaces (see \cite[\S 6.2]{LM09} for details). We will use this description in Example \ref{ex:ruled surfaces} in order to illustrate the shape of Newton-Okounkov bodies on ruled surfaces.
\begin{exmp}[Surfaces]\label{ex:surfaces}
Let $S$ be a smooth projective surface together with a flag $Y_\bullet: Y_0=S \supseteq Y_1= C \supseteq Y_2=\{p\}$, where $C\subseteq S$ is a smooth curve and $p\in C$.
Let $D$ be a big $\mathbb{Q}$-divisor on $S$. Any such divisor admits a {\it Zariski decomposition}, that is we can uniquely write $D$ as a sum
$$D=P(D)+N(D) $$
\noindent of $\mathbb{Q}$-divisors, with $P(D)$ nef and $N(D)$ either zero or effective with negative definite intersection matrix. Moreover, $P(D)\cdot \Gamma = 0$ for every irreducible component $\Gamma$ of $N(D)$ and for all $m\geq 0$ there is an isomorphism $\h^0(S,\mathcal{O}_S(\lfloor mP(D) \rfloor)) \cong \h^0(S,\mathcal{O}_S(\lfloor mD \rfloor))$. In this decomposition $P(D)$ is called the {\it positive part} and $N(D)$ the {\it negative part}. See \cite[\S 2.3.E]{Laz04} and references therein for proofs and applications.
With the above notation, we have that
$$\Delta_{Y_\bullet}(D)=\left\{(t,y)\in \mathbb{R}^2\;|\; \nu \leq t \leq \tau_C(D),\; \alpha(t)\leq t \leq \beta(t) \right\} $$
where
\begin{enumerate}
\item $\nu \in \mathbb{Q}$ the coefficient of $C$ in $N(D)$,
\item $\tau_C(D)=\sup\{t>0\;|\;D-tC \text{ is big} \}$,
\item $\alpha(t)=\ord_p(N_t\cdot C)$,
\item $\beta(t)=\ord_p(N_t\cdot C)+P_t\cdot C$,
\end{enumerate}
where $D-tC=P_t+N_t$ is a Zariski decomposition, $P_t$ being the positive and $N_t$ the negative part. Moreover, these bodies are finite polygons, by \cite[Theo. B]{KLM12}.
\end{exmp}
\subsection{Semi-stability and Harder-Narasimhan filtrations}
Throughout this section, $C$ is a smooth projective curve and $E$ is a locally free sheaf on $C$ of rank $r>0$ and degree $d=\deg(E)=\deg (c_1(E))$. Given such a bundle we call the rational number
$$\mu(E)=\dfrac{d}{r} $$
the {\it slope} of $E$.
\begin{defn}[Semi-stability]
Let $E$ be a vector bundle on $C$ of slope $\mu$. We say that $E$ is {\it semi-stable} if for every non-zero sub-bundle $ S \subseteq E$, we have $\mu(S)\leq \mu$. Equivalently, $E$ is semi-stable if for every locally-free quotient $E \twoheadrightarrow Q$ of non-zero rank, we have $\mu \leq \mu(Q)$. A non semi-stable vector bundle will be called unstable.
\end{defn}
Following \cite[Prop. 5.4.2]{LP97}, there is a canonical filtration of $E$ with semi-stable quotients.
\begin{propo}\label{propo: HN filtration}
Let $E$ be a vector bundle on $C$. Then $E$ has an increasing filtration by sub-bundles
$$ \operatorname{HN}_\bullet(E): 0=E_\ell \subseteq E_{\ell-1} \subseteq \cdots \subseteq E_1 \subseteq E_0 = E $$
where each of the quotients $E_{i-1}/E_i$ satisfies the following conditions:
\begin{enumerate}
\item Each quotient $E_{i-1}/E_i$ is a semi-stable vector bundle;
\item $\mu(E_{i-1}/E_i)<\mu(E_{i}/E_{i+1})$ for $i=1,\ldots,\ell-1$.
\end{enumerate}
This filtration is unique.
\end{propo}
The above filtration is called the {\it Harder-Narasimhan filtration} of $E$.
\begin{notation}\label{notation: HN quotients}
Let $E$ be a vector bundle on a smooth projective curve $C$. We will denote by $Q_i=E_{i-1}/E_i$ the semi-stable quotients of the Harder-Narasimhan filtration of $E$, each one of rank $r_i=\rk(Q_i)$, degree $d_i=\deg(c_1(Q_i))$ and slope $\mu_i=\mu(Q_i)=d_i/r_i$. With this notation, $\mu_1$ and $\mu_\ell$ correspond to the minimal and maximal slopes, $\mu_{\min}(E)$ and $\mu_{\max}(E)$, respectively.
\end{notation}
From a cohomological point of view, semi-stable vector bundles can be seen as the good higher-rank analogue of line bundles. For instance, we have the following classical properties (see \cite{RR84} or \cite[Lemm. 1.12, Lemm. 2.5]{But94}).
\begin{lemma}\label{lemma: properties slope}
Let $E$ and $F$ be vector bundles on $C$ and $m\in \mathbb{N}$. Then,
\begin{enumerate}
\item $\mu_{\max}(E\otimes F)=\mu_{\max}(E)+\mu_{\max}(F)$.
\item $\mu_{\min}(E\otimes F)=\mu_{\min}(E)+\mu_{\min}(F)$.
\item $\mu_{\max}(S^m E)=m \mu_{\max}(E)$.
\item $\mu_{\min}(S^m E)=m \mu_{\min}(E)$.
\item If $\mu_{\max}(E)<0$, then $\dim_k \h^0(C,E)=0$.
\item If $\mu_{\min}(E)>2g-2$, then $\dim_k \h^1(C,E)=0$.
\end{enumerate}
In particular, if $E$ and $F$ are semi-stable then $S^m E$ and $E\otimes F$ are semi-stable.
\end{lemma}
If $E_1,\ldots, E_\ell$ are vector bundles on $C$ and $m_1,\ldots,m_\ell$ be non-negative integers. By the splitting principle \cite[Rema. 3.2.3]{Ful84} we can prove the following formula:
$$ \mu(S^{m_1}E_1 \otimes \cdots \otimes S^{m_\ell} E_\ell)=\sum_{i=1}^\ell m_i \mu(E_i).$$
Moreover, we have that for every $m\geq 1$ the Harder-Narasimhan filtration of the symmetric product $S^m E$ can be computed in terms of the one for $E$ (see \cite[Prop. 3.4]{Che08} and \cite[Prop. 5.10]{Wol05}, for instance).
\begin{propo}\label{propo: HN filtration of S^mE}
Let $E$ be a vector bundle on $C$ with Harder-Narasimhan filtration
$$\operatorname{HN}_\bullet(E): 0=E_\ell \subseteq E_{\ell-1} \subseteq \cdots \subseteq E_1 \subseteq E_0 = E $$
and semi-stable quotients $Q_i=E_{i-1}/E_i$ with slopes $\mu_i=\mu(Q_i)$, for $i=1,\ldots,\ell$.
For every positive integer $m\geq 1$, let us consider the vector bundle $S^m E$ with Harder-Narasimhan filtration
$$\operatorname{HN}_\bullet(S^mE): 0=W_M \subseteq W_{M-1} \subseteq \cdots \subseteq W_1 \subseteq W_0 = S^m E $$
and semi-stable quotients $W_{j-1}/W_j$ with slopes $\nu_j=\mu(W_{j-1}/W_j)$, for ${j=1,\ldots,M}$. Then, for every $j=1,\ldots,M$ we have that
$$W_j=\sum_{{\sum_i m_i\mu_i \geq \nu_{j+1}}}S^{m_1}E_0\otimes \cdots\otimes S^{m_\ell}E_{\ell-1} $$
and
$$W_{j-1}/W_j \cong \bigoplus_{\sum_i m_i \mu_i = \nu_j} S^{m_1}Q_1\otimes \cdots \otimes S^{m_\ell} Q_\ell, $$
where the sums are taken over all partitions $\mathbf{m}=(m_1,\ldots,m_\ell)\in \mathbb{N}^\ell$ of $m$, and
$S^{m_1}E_0\otimes \cdots \otimes S^{m_\ell}E_{\ell-1}$ denotes the image of the composite
$$E_0^{\otimes m_1}\otimes \cdots \otimes E_{\ell-1}^{\otimes m_\ell}\to E^{\otimes m} \to S^m E. $$
In particular, there is a refinement $F_\bullet$ of $\operatorname{HN}_\bullet(S^m E)$ of length $L=L(m)$ and whose respective successive quotients are of the form
$$F_{i-1}/F_i\cong Q_{\mathbf{m}(i)}=S^{m_1}Q_1\otimes \cdots \otimes S^{m_\ell}Q_\ell $$
for some partition $\mathbf{m}(i)=(m_1,\ldots,m_\ell)\in \mathbb{N}^\ell$ of $m$, and such that for every $i\in \{1,\ldots L\}$ we have $\mu(Q_{\mathbf{m}(i)})\leq \mu( Q_{\mathbf{m}(i+1)})$. Moreover, given any partition $\mathbf{m}\in \mathbb{N}^\ell$ of $m$, there is one and only one $i\in \{1,\ldots,L\}$ such that $\mathbf{m}(i)=\mathbf{m}$.
\end{propo}
\section{Divisors on projective bundles over curves}
Let $E$ be a vector bundle on a smooth projective curve $C$, of rank $r\geq 2$ and degree $d$. In this section we study divisors on the projective bundle $\pi:\mathbb{P}(E)\to C$ of one-dimensional quotients. Let us recall that in this case the N\'eron-Severi group of $\mathbb{P}(E)$ is of the form
$$\N^1(\mathbb{P}(E))=\mathbb{Z}\cdot f \oplus \mathbb{Z}\cdot \xi, $$
\noindent where $f$ is the numerical class of a fiber of $\pi$ and $\xi=\xi_E$ is the numerical class of a divisor representing the tautological line bundle $\mathcal{O}_{\mathbb{P}(E)}(1)$. Moreover, if $[\text{pt}]$ denotes the class of a point in the ring $\N^*(\mathbb{P}(E))$ then we have the following relations:
$$f^2 = 0,\; \xi^{r-1}f = [\text{pt}], \; \xi^r = d\cdot [\text{pt}]. $$
The cone of nef divisors can be described via Hartshorne's characterization of ample vector bundles over curves \cite[Theo. 2.4]{Ha71} (cf. \cite[Lemm. 2.1]{Ful11}).
\begin{lemma}\label{lemma: nef cone} $\Nef(\mathbb{P}(E))=\langle \xi - \mu_{\min} f,f \rangle. $
\end{lemma}
The cone of pseudo-effective divisors was obtained by Nakayama in \cite[Cor. IV.3.8]{Nak04}, and it was indirectly computed by Wolfe \cite{Wol05} and Chen \cite{Che08} who independently obtained the volume function $\vol_{\mathbb{P}(E)}$ on $\N^1(\mathbb{P}(E))_{\mathbb{R}}$. A more general result on the cone of effective cycles of arbitrary codimension can be found in \cite[Theo. 1.1]{Ful11}.
\begin{lemma}\label{lemma: psef cone}
$\Psef(\mathbb{P}(E))=\langle \xi - \mu_{\max} f,f \rangle. $
\end{lemma}
In particular, we recover a result of Miyaoka \cite[Theo. 3.1]{Miy87} on semi-stable vector bundles over curves that was generalized by Fulger in \cite[Prop. 1.5]{Ful11}.
\begin{cor}\label{Miyaoka} A vector bundle $E$ on a smooth projective curve $C$ is semi-stable if and only if $\Nef(\mathbb{P}(E))=\Psef(\mathbb{P}(E))$.
\end{cor}
We finish this section by recalling Wolfe's computation of the volume function on $\N^1(\mathbb{P}(E))$. See also \cite[Theo. 1.2]{Che08}.
\begin{notation}\label{notation: simplex}
Let $d\geq 1$ be an integer. We define the {\it standard $d$-simplex} $\widehat{\Delta}_d$ with $d+1$ vertices in $\mathbb{R}^{d+1}$ to be
$$\widehat{\Delta}_d=\Big\{(x_1,\ldots,x_{d+1})\in \mathbb{R}^{d+1}\;\Big|\; \textstyle{\sum_{i=1}^{d+1} x_i=1} \text{ and }x_i\geq 0 \text{ for all }i \Big\}. $$
By projecting $\widehat{\Delta}_d$ onto the hyperplane $x_1=0$, we can identify $\widehat{\Delta}_d$ with the {\it full dimensional standard $d$-simplex} (or just {\it $d$-simplex}) in $\mathbb{R}^d$ given by
$$\Delta_d=\Big\{(x_2,\ldots,x_{d+1})\in \mathbb{R}^{d}\;\Big|\; \textstyle{\sum_{i=2}^{d+1} x_i\leq 1} \text{ and }x_i\geq 0 \text{ for all }i \Big\}. $$
Via the previous identification, we will denote by $\lambda$ the Lebesgue measure on $\widehat{\Delta}_d$ induced by the standard Lebesgue measure on $\Delta_d\subseteq \mathbb{R}^d$. In particular, we will have $\lambda(\widehat{\Delta}_d)=\frac{1}{d!}$.
Given a positive real number $a>0$, we define the {\it $d$-simplex with side length $a$} by $a\Delta_d=\big\{(x_2,\ldots,x_{d+1})\in \mathbb{R}^{d}\;\big|\; \sum_{i=2}^{d+1} x_i\leq a \text{ and }x_i\geq 0 \text{ for all }i \big\}$. Similar for $a \widehat{\Delta}_d \subseteq \mathbb{R}^{d+1}$, the {\it standard $d$-simplex with side length $a$}.
\end{notation}
\begin{thm}[{\cite[Theo. 5.14]{Wol05}}]\label{theo: Wolfe volume} Let $E$ be a vector bundle with Harder-Narasimhan filtration of length $\ell$ and semi-stable quotients $Q_i$ of ranks $r_i$ and slopes $\mu_i$. Then, for any $t\in \mathbb{R}$
$$\vol_{\mathbb{P}(E)}(\xi-tf)=r!\cdot \int_{\widehat{\Delta}_{\ell-1}}\max\left\{\sum_{i=1}^\ell \mu_i \beta_i - t,0\right\}\dfrac{\beta_1^{r_1-1}\cdots\beta_\ell^{r_\ell-1}}{(r_1-1)!\cdots (r_\ell-1)!} d\beta $$
\noindent where $\widehat{\Delta}_{\ell-1}\subseteq \mathbb{R}^\ell$ is the standard $(\ell-1)$-simplex with coordinates $\beta_1,\ldots,\beta_\ell$, and $\beta$ be the standard induced Lebesgue measure.
\end{thm}
\begin{remark}\label{remark: Huayi volume}
Alternatively, Chen computed in \cite[Theo. 1.2]{Che08} a similar volume formula, but slightly simplified by integrating in $\mathbb{R}^r$ instead of $\mathbb{R}^\ell$ (cf. \cite[Prop. 3.5]{Che08}). More precisely, with the same notation as above
$$\vol_{\mathbb{P}(E)}(\xi-tf)= r! \cdot \int_{\widehat{\Delta}_{r-1}}\max\left\{\sum_{j=1}^r s_j \lambda_j - t,0\right\} d\lambda $$
\noindent where $\widehat{\Delta}_{r-1}\subseteq \mathbb{R}^r$ is the standard $(r-1)$-simplex with coordinates $\lambda_1,\ldots,\lambda_r$, $d\lambda$ is the standard induced Lebesgue measure\footnote{Unlike Chen, we do not normalize the measure in order to have $\lambda(\Delta_{r-1})=1$.}, and ${\mathbf{s}=(s_1,\ldots,s_r)}$ is a vector in $\mathbb{R}^r$ such that the value $\mu_i$ appears exactly $r_i$ times in the coordinates of $\mathbf{s}$ as in Notation \ref{notation: Wolfe's square} below.
\end{remark}
\begin{notation}\label{notation: Wolfe's square}
Fix $\ell\geq 1$ and $r\geq 1$ two integers, $(r_1,\ldots,r_\ell)\in \mathbb{N}^\ell$ a partition of $r$ and $t\in \mathbb{R}$. We define for $(\mu_1,\ldots,\mu_\ell)\in \mathbb{Q}^\ell$ the following polytopes:
$$\widehat{\square}_{t}=\left\{(\beta_1,\ldots,\beta_{\ell})\in\widehat{\Delta}_{\ell-1}\subseteq \mathbb{R}^{\ell}\;\left|\;\sum_{i=1}^\ell \mu_i\beta_i\geq t \right.\right\} $$
\noindent and
$$\square_t=\left\{(\lambda_1,\ldots,\lambda_r)\in \widehat{\Delta}_{r-1}\subseteq \mathbb{R}^r\;\left|\; \sum_{i=1}^r s_i\lambda_i \geq t \right.\right\}, $$
\noindent where
$$\mathbf{s}=(\underbrace{\mu_1,\ldots,\mu_1}_{r_1\text{ times}},\underbrace{\mu_2,\ldots,\mu_2}_{r_2 \text{ times}},\ldots,\underbrace{\mu_\ell,\ldots,\mu_\ell}_{r_\ell \text{ times}})\in \mathbb{Q}^r.$$
Similarly, for every permutation $w\in \mathfrak{S}_r$ we define
$$\square_t^{w}=\left\{(\lambda_1,\ldots,\lambda_r)\in \widehat{\Delta}_{r-1}\subseteq \mathbb{R}^r\;\left|\; \sum_{i=2}^r \sigma_{w(i-1)}\lambda_i + \sigma_{w(r)}\lambda_1 \geq t \right.\right\}, $$
\noindent where
$$\boldsymbol{\sigma}=(\underbrace{\mu_\ell,\ldots,\mu_\ell}_{r_\ell\text{ times}},\underbrace{\mu_{\ell-1},\ldots,\mu_{\ell-1}}_{r_{\ell-1} \text{ times}},\ldots,\underbrace{\mu_1,\ldots,\mu_1}_{r_1 \text{ times}})\in \mathbb{Q}^r.$$
By abuse of notation, we will also denote by $\square_t$ and $\square_t^{w}$ the full dimensional polytopes in $\mathbb{R}^{r-1}$ obtained via the projection of $\widehat{\Delta}_{r-1}\subseteq \mathbb{R}^r$ onto $\Delta_{r-1}\subseteq \mathbb{R}^{r-1}$. Explicitly, if $(\nu_2,\ldots,\nu_r)$ are coordinates in $\mathbb{R}^{r-1}$ then
$$\square_t=\left\{(\nu_2,\ldots,\nu_r)\in \Delta_{r-1}\subseteq \mathbb{R}^{r-1}\;\left|\;s_1\left(1-\sum_{i=2}^r \nu_i\right)+\sum_{i=2}^r s_i\nu_i \geq t \right.\right\} $$
\noindent and
\begin{equation*} \square_t^{w}=\left\{(\nu_2,\ldots,\nu_r)\in \Delta_{r-1}\;\left|\; \sum_{i=2}^r \sigma_{w(i-1)}\nu_i+\sigma_{w(r)}\left(1-\sum_{i=2}^r \nu_i\right) \geq t \right.\right\}. \tag{$\star$} \label{eq:star} \end{equation*}
\end{notation}
\begin{cor}\label{cor: volume Okounkov bodies}
For any admissible flag $Y_\bullet$ on $\mathbb{P}(E)$ and any big rational class $\xi-tf$ on $\mathbb{P}(E)$ we have that
$$\vol_{\mathbb{R}^r}(\Delta_{Y_\bullet}({\xi-tf}))=\int_{\widehat{\square}_t} \left(\sum_{i=1}^\ell \mu_i \beta_i - t\right)\dfrac{\beta_1^{r_1-1}\cdots\beta_\ell^{r_\ell-1}}{(r_1-1)!\cdots (r_\ell-1)!}d\beta,$$
\noindent for $\widehat{\square}_t \subseteq \mathbb{R}^\ell$ as in Notation \ref{notation: Wolfe's square}.
\end{cor}
\begin{remark}\label{remark: augmented base locus}
Let $F=\pi^{-1}(q)\cong \mathbb{P}^{r-1}$ be any fiber of $\pi:\mathbb{P}(E)\to C$. Then, for any big $\mathbb{R}$-divisor $D$ on $\mathbb{P}(E)$ we have that $F\not\subseteq \mathbf{B}_{+}(D)$. In fact, if $D\sim_{\mathbb{R}}A+E$, with $A$ ample $\mathbb{R}$-divisor and $E$ effective $\mathbb{R}$-divisor, and if $F\subseteq \operatorname{Supp}(E)$, then we can write
$E=aF+E'$ with $a>0$ and $F\not\subseteq \operatorname{Supp}(E')$, which implies that $F\not\subseteq \mathbf{B}_{+}(D)$ since $A+aF$ is ample.
\end{remark}
As a direct consequence, we compute the restricted volume function on a fiber $F=\pi^{-1}(q)\cong \mathbb{P}^{r-1}$.
\begin{cor}\label{cor: restricted volume}
Let $F$ be a fiber of $\pi:\mathbb{P}(E)\to C$ and let $\xi-tf$ be a big rational class. Then,
$$\vol_{\mathbb{P}(E)|F}(\xi-tf)=(r-1)!\cdot\int_{\widehat{\square}_t}\dfrac{\beta_1^{r_1-1}\cdots\beta_\ell^{r_\ell-1}}{(r_1-1)!\cdots (r_\ell-1)!}d\beta, $$
\noindent for $\widehat{\square}_t \subseteq \mathbb{R}^\ell$ as in Notation \ref{notation: Wolfe's square}. In particular, if $0\leq \tau \leq \mu_\ell-t$ then we have that
$$\vol_{\mathbb{R}^{r-1}}(\Delta_{Y_\bullet}({\xi-tf})_{\nu_1=\tau})=\int_{\widehat{\square}_{t+\tau}}\dfrac{\beta_1^{r_1-1}\cdots\beta_\ell^{r_\ell-1}}{(r_1-1)!\cdots (r_\ell-1)!}d\beta, $$
\noindent where $Y_\bullet$ is any admissible flag on $\mathbb{P}(E)$ with divisorial component $Y_1=F$. In particular, these volumes depend only on $\operatorname{gr}(\operatorname{HN}_\bullet(E))$, the graded vector bundle associated to the Harder-Narasimhan filtration of $E$.
\end{cor}
\begin{proof}
We consider
$$v(\tau)=\vol_{\mathbb{P}(E)}(\xi-tf+\tau f)=r!\cdot \int_{\widehat{\square}_{t-\tau}} \left(\sum_{i=1}^\ell \mu_i \beta_i - t+\tau\right)\dfrac{\beta_1^{r_1-1}\cdots\beta_\ell^{r_\ell-1}}{(r_1-1)!\cdots (r_\ell-1)!}d\beta.$$
Since $F \not\subseteq \mathbf{B}_{+}(\xi-tf)$ by Remark \ref{remark: augmented base locus}, Theorem \ref{slices} and differentiation under the integral sign give
$$\vol_{\mathbb{P}(E)|F}(\xi-tf)=\dfrac{1}{r}\cdot\left.\dfrac{d}{d\tau}v(\tau)\right|_{\tau=0}=(r-1)!\cdot \displaystyle\int_{\widehat{\square}_{t}} \dfrac{\beta_1^{r_1-1}\cdots\beta_\ell^{r_\ell-1}}{(r_1-1)!\cdots (r_\ell-1)!}d\beta.$$
\end{proof}
\begin{lemma}\label{lemma: Huayi volume slices}
Following Notation \ref{notation: Wolfe's square} \eqref{eq:star}, we have that
$$\vol_{\mathbb{R}^r}(\Delta_{Y_\bullet}(\xi-tf))=\int_{\square_t}\left(\sum_{j=1}^r s_j \lambda_j - t\right)d\lambda $$
\noindent and
$$\vol_{\mathbb{R}^{r-1}}(\Delta_{Y_\bullet}({\xi-tf})|_{\nu_1=\tau})=\int_{\square_{t+\tau}}d\lambda=\vol_{\mathbb{R}^{r-1}}(\square_{t+\tau}). $$
Moreover, $\vol_{\mathbb{R}^{r-1}}(\square_{t+\tau})=\vol_{\mathbb{R}^{r-1}}(\square_{t+\tau}^{w})$ for every $w\in \mathfrak{S}_r$.
\end{lemma}
\begin{proof}
The first two equalities follow from Remark \ref{remark: Huayi volume}. For the last assertion consider $\square_{t}\subseteq \mathbb{R}^r$ and $\square_{t}^{w}\subseteq \mathbb{R}^r$ as in Notation \ref{notation: Wolfe's square}. Then, there is a linear transformation $T:\mathbb{R}^r \to \mathbb{R}^r$, whose associated matrix in the canonical basis of $\mathbb{R}^r$ is given by a permutation matrix, such that $T(\square_t)=\square_t^{w}$ and $|\det(T)|=1$.
\end{proof}
\section{Newton-Okounkov bodies on projective bundles over curves}
Let $E$ be a vector bundle on a smooth projective curve $C$, of rank $r\geq 2$ and degree $d$. In this section we study the geometry of Newton-Okounkov bodies of rational big classes in $\N^1(\mathbb{P}(E))$ in terms of the numerical information of the Harder-Narasimhan filtration of $E$. In particular, Theorem A will be a consequence of Lemma \ref{lemma:reduction to big nef}, Theorem \ref{theo: okounkov bodies} and Corollary \ref{cor: main result}. We follow Notation \ref{notation: HN quotients}.
\begin{defn}[Linear flag]\label{defi: natural flag} A complete flag of subvarieties $Y_\bullet$ on the projective vector bundle $\mathbb{P}(E)\xrightarrow{\pi} C$ is called a {\it linear flag centered at $p\in \mathbb{P}(E)$, over the point $q\in C$} (or simply a {\it linear flag}) if $Y_0=\mathbb{P}(E)$, $Y_1=\pi^{-1}(q)\cong \mathbb{P}^{r-1}$ and $Y_i \cong \mathbb{P}^{r-i}$ is a linear subspace of $Y_{i-1}$ for $i=2,\ldots,r$, with $Y_r=\{p\}$.
\end{defn}
\subsection{Ruled surfaces}
Let us begin with the following example that illustrates the general case. Namely, that the shape of Newton-Okounkov bodies on $\mathbb{P}(E)$ will depend on the semi-stability of $E$.
\begin{exmp}\label{ex:ruled surfaces} Suppose that $\rk(E)= 2$ and let $\eta=a(\xi-\mu_\ell f)+bf \in \N^1(\mathbb{P}(E))_\mathbb{Q}$ be a big class. In other words, $a,b\in \mathbb{Q}_{>0}$. Let
$$Y_\bullet: \mathbb{P}(E) \supseteq F=\pi^{-1}(q) \supseteq \{p\}$$
\noindent be the linear flag centered at $p\in \mathbb{P}(E)$, over $q\in C$. The Newton-Okounkov body of $\eta$ can be computed by applying \cite[Theo. 6.4]{LM09} (see Example \ref{ex:surfaces}).
\medskip
\noindent\textit{Semi-stable case.} If $E$ is semi-stable then Corollary \ref{Miyaoka} implies that every big class is ample. In particular, for every $\eta \in \Bg(X)_{\mathbb{Q}}$ and every $\mathbb{Q}$-divisor $D_\eta$ with numerical class $\eta$, we have that $N(D_\eta)=0$ and $P(D_\eta)=D_\eta$. It follows that, with the notation as in Example \ref{ex:surfaces}, $\nu=0$, $\tau_F(\eta)=b$, $\alpha(t)=0$ for every $t\in [0,b]$ and $\beta(t)=a$ for every $t\in [0,b]$. The Newton-Okounkov bodies are given by rectangles in this case.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\draw [->] (0,0) -- (4.5,0);
\node [right] at (4.5,0) {$t$};
\draw [->] (0,0) -- (0,3.5);
\node [above] at (0,3.5) {$y$};
\draw [fill=gray, ultra thick] (0,0) -- (0,3) -- (4,3) -- (4,0) -- (0,0);
\node [below left] at (0,0) {$(0,0)$};
\node [left] at (0,3) {$(0,a)$};
\node [below] at (4,0) {$(b,0)$};
\end{tikzpicture}
\end{center}
\caption{Newton-Okounkov body $\Delta_{Y_\bullet}(\eta)$ for $E$ semi-stable}\label{fig:1}
\end{figure}
\medskip
\noindent \textit{Unstable case.} If $E$ is unstable we consider its Harder-Narasimhan filtration
$$0 \to E_1 \to E \to Q_1 \to 0, $$
and we note that in this case $E_1 = Q_\ell$. The quotient $E\to Q_1 \to 0$ corresponds to a section $s:C\to \mathbb{P}(E)$ with $[s(C)]=\xi-\mu_\ell f$ in $\N^1(\mathbb{P}(E))$. The curve $s(C)$ is the only irreducible curve on $\mathbb{P}(E)$ with negative self-intersection. On the other hand, if $D_{\eta}$ is any $\mathbb{Q}$-divisor with numerical class $\eta$ then, either $\eta$ is inside the nef cone of $\mathbb{P}(E)$ and thus $P(D_\eta)=D_\eta$ and $N(D_\eta)=0$, or $\eta$ is big and not nef in which case we compute that
$$[P(D_\eta)]=\frac{b}{\mu_\ell - \mu_1}(\xi-\mu_1 f) \;\; \text{and}\;\; [N(D_\eta)]=\left(\frac{a(\mu_\ell - \mu_1)-b}{\mu_\ell - \mu_1} \right)(\xi - \mu_\ell f) $$
in $\N^1(\mathbb{P}(E))_\mathbb{Q}$. We notice that $N(D_\eta)=\left(\frac{a(\mu_\ell - \mu_1)-b}{\mu_\ell - \mu_1} \right)\cdot s(C)$ as $\mathbb{Q}$-divisor, by minimality of the negative part and by the fact that the negative part is unique in its numerical equivalence class. See \cite[Lemm. 14.10, Cor. 14.13]{Bad01} for details.
Let $t^*=b-a(\mu_\ell-\mu_1)$. We have that $\eta$ is big and nef if and only if $t^*\geq 0$, in which case the class $\eta - tf$ is big and nef for $0\leq t \leq t^*$. For $t^* \leq t \leq b$, the same computation above enables us to find the Zariski decomposition of $D_\eta - t F$, which is big and not nef.
We notice that, with the notation as in Example \ref{ex:surfaces}, $\nu=0$. On the other hand, the functions $\alpha(t)$ and $\beta(t)$ will depend on whether or not we have $\{p\}=s(C)\cap F$. A straightforward computation shows that the Newton-Okounkov bodies are given by the following finite polygons in $\mathbb{R}^2$.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\draw [->] (0,0) -- (4.5,0);
\node [right] at (4.5,0) {$t$};
\draw [->] (0,0) -- (0,3.5);
\node [above] at (0,3.5) {$y$};
\draw [fill=gray, ultra thick] (0,0) -- (0,3) -- (2,3) -- (4,0) -- (0,0);
\draw (2,0) -- (2,3);
\node [below left] at (0,0) {$(0,0)$};
\node [below] at (2,0) {$(t^*,0)$};
\node [left] at (0,3) {$(0,a)$};
\node [below] at (4,0) {$(b,0)$};
\end{tikzpicture}
\hspace{7mm}
\begin{tikzpicture}
\draw [->] (0,0) -- (4.5,0);
\node [right] at (4.5,0) {$t$};
\draw [->] (0,0) -- (0,3.5);
\node [above] at (0,3.5) {$y$};
\draw [fill=gray, ultra thick] (0,0) -- (0,3) -- (4,3) -- (2,0) -- (0,0);
\draw (2,0) -- (2,3);
\draw [dashed] (4,0) -- (4,3);
\node [below left] at (0,0) {$(0,0)$};
\node [below] at (2,0) {$(t^*,0)$};
\node [left] at (0,3) {$(0,a)$};
\node [below] at (4,0) {$(b,0)$};
\end{tikzpicture}
\end{center}
\caption{$\Delta_{Y_\bullet}(\eta)$ for $E$ unstable and $\eta$ big and nef \\ {(a) if $\{p\}\neq s(C)\cap F $} \hspace{3.5cm} (b) if $\{p\}=s(C)\cap F$}\label{fig:2}
\end{figure}
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\draw [->] (0,0) -- (2.5,0);
\node [right] at (2.5,0) {$t$};
\draw [->] (0,0) -- (0,3.5);
\node [above] at (0,3.5) {$y$};
\draw [fill=gray, ultra thick] (0,0) -- (0,3) -- (2,0) -- (0,0);
\node [below left] at (0,0) {$(0,0)$};
\node [left] at (0,3) {$(0,\frac{b}{\mu_\ell-\mu_1})$};
\node [below] at (2,0) {$(b,0)$};
\end{tikzpicture}
\hspace{1cm}
\begin{tikzpicture}
\draw [->] (0,0) -- (2.5,0);
\node [right] at (2.5,0) {$t$};
\draw [->] (0,0) -- (0,3.5);
\node [above] at (0,3.5) {$y$};
\draw [fill=gray, ultra thick] (0,0.5) -- (0,3) -- (2,3) -- (0,0.5);
\draw [dashed] (2,0) -- (2,3);
\node [below left] at (0,0) {$(0,0)$};
\node [left] at (0,0.5) {$(0,\frac{-t^*}{\mu_\ell-\mu_1})$};
\node [left] at (0,3) {$(0,a)$};
\node [below] at (2,0) {$(b,0)$};
\end{tikzpicture}
\end{center}
\caption{$\Delta_{Y_\bullet}(\eta)$ for $E$ unstable and $\eta$ big and not nef \\ {(a) if $\{p\}\neq s(C)\cap F $} \hspace{3.5cm} (b) if $\{p\}=s(C)\cap F$}\label{fig:3}
\end{figure}
We notice that Figure \ref{fig:3} provides examples of big and not nef divisors classes $\eta$ such that the origin $\vec{0}\in \Delta_{Y_\bullet}(\eta)$ for almost every linear flag $Y_\bullet$ except for one. This shows in particular that condition (2) in the characterization of nefness given in \cite[Cor. 2.2]{KL15b} has to be checked for {\it all} linear flags.
\end{exmp}
\subsection{Some reductions}
We first observe that Proposition B in $\S 1$ states that the shape of Newton-Okounkov bodies on projective semi-stable vector bundles will be similar as in Example \ref{ex:ruled surfaces} above. Moreover, this provides a characterization of semi-stability in terms of Newton-Okounkov bodies.
\begin{proof}[Proof of Proposition B]
\noindent $(1)\Rightarrow (2)$. Let $\eta=a(\xi-\mu_\ell f)+bf$ be a big rational class on $\mathbb{P}(E)$ and $Y_\bullet: X = Y_0 \supseteq Y_1 \supseteq Y_2 \supseteq \cdots \supseteq Y_{r-1} \supseteq Y_r = \{p\}$ be a linear flag centered at $p\in \mathbb{P}(E)$, over the point $q\in C$. Since $E$ is semi-stable we have that $\Bg(\mathbb{P}(E))=\Amp(\mathbb{P}(E))$ by Corollary \ref{Miyaoka}. Equivalently, we have that $\mathbf{B}_+(\eta)= \emptyset$ for every big rational class $\eta$.
We notice that $\tau_F(\eta)=\sup\{s>0\;|\;\eta - s f \in \Bg(\mathbb{P}(E)) \}=b$. The implication follows from Proposition \ref{propo:psef=nef} and \cite[Cor. 4.11]{Bo12}, by noting that if $D_\eta$ is a $\mathbb{Q}$-divisor with numerical class $\eta$ then
$$\Delta_{Y_\bullet}(\eta)_{\nu_1=t}=\Delta_{Y_\bullet|F}(F,(D_\eta-tF)|_F)=a \Delta_{Y_\bullet|F}(F,H) = a\Delta_{r-1} \subseteq \mathbb{R}^{r-1},$$
where $H\subseteq F\cong \mathbb{P}^{r-1}$ is an hyperplane section.
\medskip
\noindent $(2)\Rightarrow (1)$. We notice that if for all linear flags $Y_\bullet$ the Newton-Okounkov body of $\eta=a(\xi-\mu_\ell f)+bf$ is given by $\Delta_{Y_\bullet}(\eta)=[0,b]\times a\Delta_{r-1} \subseteq \mathbb{R}^r$ then $\eta$ is a big and nef class, by \cite[Cor. 2.2]{KL15b}. We can therefore compute the volume of $\eta$ via the top self-intersection $\vol_{\mathbb{P}(E)}(\eta)=\eta^r = r a^{r-1}\left(b-a(\mu_\ell-\mu(E))\right)$. On the other hand, we have that
\begin{equation*}
\begin{aligned}
\vol_{\mathbb{R}^n}\left(\Delta_{Y_\bullet}(\eta) \right) = \vol_{\mathbb{R}^r}\left([0,b]\times a\Delta_{r-1} \right) = \frac{a^{r-1}}{(r-1)!}b,
\end{aligned}
\end{equation*}
and hence \cite[Theo. A]{LM09} leads to $\mu_\ell = \mu(E)$, implying the result.
\end{proof}
We will first reduce our problem to computing the Newton-Okounkov body of the big and nef divisor class $\xi-\mu_1 f$.
\begin{lemma}\label{lemma:reduction to big nef}
Let $E$ be a unstable vector bundle on a smooth projective curve $C$, with $\rk(E)=r\geq 2$. Then for every big class $\eta = a(\xi-\mu_\ell f)+bf$ and every linear flag $Y_\bullet$ we have that
\begin{itemize}
\item[(1)] $\Delta_{Y_\bullet}(\eta)=\left([0,t^*]\times a\Delta_{r-1} \right) \cup \left(a \Delta_{Y_\bullet}(\xi-\mu_1 f)+t^*\vec{e}_1\right)$ if $\eta$ is big and nef;
\item[(2)] $\Delta_{Y_\bullet}(\eta)=a \Delta_{Y_\bullet}(\xi-\mu_1 f)_{\nu_1 \geq - t^*}+t^*\vec{e}_1$ if $\eta$ is big and not nef.
\end{itemize}
Here $t^*=b-a(\mu_\ell-\mu_1)$ and $a\Delta_{r-1}\subseteq \mathbb{R}^{r-1}$ is the $(r-1)$-simplex with side length $a$.
\end{lemma}
\begin{proof}
If $\eta$ is an ample rational class then ${\Delta_{Y_\bullet}(\eta)_{\nu_1=t}=a\Delta_{r-1}}$ for $0\leq t \leq t^*$, by Proposition \ref{propo:psef=nef} and \cite[Exam. 1.1]{LM09}. On the other hand, we have that ${\Delta_{Y_\bullet}(\eta)_{\nu_1 \geq t^*}= a \Delta_{Y_\bullet}(\xi - \mu_1 f)+t^*\vec{e}_1}$ by Theorem \ref{slices}.
If $\eta$ is big and not nef then $t^*<0$. Theorem \ref{slices} implies therefore that ${a \Delta_{Y_\bullet}(\xi - \mu_1 f)_{\nu_1\geq -t^*}= \Delta_{Y_\bullet}(\eta)-t^*\vec{e}_1}$, which leads to (2).
\end{proof}
\subsection{A toric computation}
We will need the following result concerning the Newton-Okounkov bodies of some toric graded algebras.
\begin{lemma}\label{lemm: NO of special algebras}
Fix $A\in \mathbb{R}$, an integer $r\geq 2$, $\boldsymbol{\sigma}=(\sigma_1,\ldots,\sigma_r)\in \mathbb{R}^r$ and homogeneous coordinates $[x_1:\ldots:x_r]$ on $\mathbb{P}^{r-1}$. Put $B_0=k$ and consider for every integer $m\geq 1$ the vector subspace $B_m\subseteq \h^0(\mathbb{P}^{r-1},\mathcal{O}_{\mathbb{P}^{r-1}}(m))$ generated by monomials ${x^\alpha=x_1^{\alpha_1}\cdots x_r^{\alpha_r}}$ of total degree $|\alpha|=m$ such that $\sum_{i=1}^r \sigma_i \alpha_i > A$. Suppose that $A\geq 0$ and $\sigma_1\geq \sigma_2 \geq \cdots \geq \sigma_r$. Then
\begin{enumerate}
\item $B_\bullet=\oplus_{m\geq 0}B_m$ is a graded subalgebra of the coordinate ring $k[x_1,\ldots,x_r]$.
\item Let us denote by $V_\bullet$ the flag of linear subspaces
$$V_\bullet: V_0=\mathbb{P}^{r-1}\supseteq V_1 \supseteq \cdots \supseteq V_{r-1},$$
where $V_i=\{x_1=\ldots=x_i=0\}$ for $i=1,\ldots,r-1$ and consider the Schubert cell decomposition of the full flag variety parametrizing complete linear flag on $\mathbb{P}^{r-1}$
$$\mathbb{F}_r = \coprod_{w\in \mathfrak{S}_r} \Omega_w, $$
with respect to $V_\bullet$. Fix a permutation $w\in \mathfrak{S}_r$ and let $Y_\bullet \in \Omega_w$. Then $\Delta_{Y_\bullet}(B_\bullet)$ is given by the projection of the polytope
$$\left\{(\nu_1,\ldots,\nu_r)\in \mathbb{R}^r_{\geq 0}\;\left|\;\sum_{i=1}^r \nu_i=1 \text{ and } \sum_{i=1}^r \sigma_{w(i)} \nu_i \geq 0 \right.\right\} $$
onto the hyperplane $\{\nu_r=0\}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let us suppose that $x^\alpha \in B_m$ and $x^{\alpha'}\in B_{m'}$, then $x^{\alpha+\alpha'}$ belongs to $b_{m+m'}$ since we have $\sum_{i=1}^r \sigma_i (\alpha_i+\alpha_i') > 2A \geq A$ as long as $A\geq 0$. This proves (1).
In order to prove (2) we will first show that every automorphism $\varphi$ of $\mathbb{P}^{r-1}$ that fixes the flag $V_\bullet$ induces an automorphism of the graded algebra $B_\bullet$. For this, we remark that automorphisms $\varphi$ of $\mathbb{P}^{r-1}$ fixing the flag $V_\bullet$ correspond to lower triangular matrices in $\operatorname{PGL}_r(k)$ and hence, given $\varphi=(\varphi_{i,j})_{1\leq i,j \leq r}$ such a matrix, the image of the monomial $x^\alpha \in B_m$ via the induced action on $k[x_1,\ldots,x_r]$ is given by
$$\varphi(x^\alpha)=\prod_{i=1}^r (\varphi_{i,1}x_1+\ldots+\varphi_{i,i}x_i)^{\alpha_i}.$$
The above product can be written as a linear combination of monomials of the form $x^{\alpha'}=x^{k_1+\ldots+k_r}$ where $k_i=(k_{i,1},\ldots,k_{i,i},0,\ldots,0)\in \mathbb{N}^r$ is such that $|k_i|=\alpha_i$. Let us prove that all these monomials belong to $B_m$ as well. In fact, we have that $\alpha_i'=\sum_{j=i}^r k_{j,i}$ and hence
$$
\begin{array}{ccll}
\sum_{i=1}^r \sigma_i \alpha_i' & = & \sum_{i=1}^r \sum_{j=i}^r \sigma_i k_{j,i}=\sum_{i=1}^r \sum_{j=1}^i \sigma_j k_{i,j} \\
& \geq & \sum_{i=1}^r \sum_{j=1}^i \sigma_i k_{i,j} & \text{ since } \sigma_1\geq\cdots \geq \sigma_r \\
& = & \sum_{i=1}^r \sigma_i |k_i| = \sum_{i=1}^r \sigma_i \alpha_i > A.
\end{array}
$$
It follows that $\varphi$ induces an automorphism of the graded algebra $B_\bullet$.
In order to compute Newton-Okounkov bodies with respect to linear flags on a given Schubert cell, we note that \cite[Prop. 1.2.1]{Bri05} implies that given a permutation $w\in \mathfrak{S}_r$ and a linear flag $Y_\bullet \in \Omega_w$ there exists an automorphism $\varphi$ of $\mathbb{P}^{r-1}$ that fixes the reference flag $V_\bullet$ and such that the image of the flag $Y_\bullet^w$ via the induced action of $\varphi$ on $\mathbb{F}_r$ is $Y_\bullet$, where $Y_\bullet^w$ is the linear flag such that for every $i=1,\ldots,r-1$ we have
$$Y_i^w = \{x_{w(1)}=\ldots=x_{w(i)}=0 \} \subseteq \mathbb{P}^{r-1}.$$
It follows from the previous paragraph that $\varphi$ induces an automorphism of the graded algebra $B_\bullet$ and then for every $m\geq 1$ and every $P\in B_m$ we have that
$$\nu_{Y_\bullet}(\varphi(P))=\nu_{\varphi(Y_\bullet^w)}(\varphi(P))=\nu_{Y_\bullet^w}(P).$$
In particular, we have $\{ \nu_{Y_\bullet}(P) \}_{P\in B_{m}} = \{ \nu_{Y_\bullet^w}(P) \}_{P\in B_{m}}\subseteq \mathbb{N}^{r-1}$ and consequently ${\Delta_{Y_\bullet}(B_{\bullet})=\Delta_{Y_\bullet^w}(B_{\bullet})}$. Therefore, we can suppose that $Y_\bullet=Y_\bullet^w\in \Omega_w$ in order to prove (2).
Since $B_m$ is generated by monomials we have that
$$\Delta_{Y_\bullet^w}(B_{\bullet}) = \overline{\bigcup_{m\geq 1}\left\{\left.\frac{\nu_{Y_\bullet^w}(x^\alpha)}{m}\;\right|\;x^\alpha\in B_{m} \right\}}.$$
We compute for every monomial $x^\alpha=x_1^{\alpha_1}\cdots x_r^{\alpha_r}\in B_m$ that
$$\nu_{Y_\bullet^w}(x^\alpha)=(\alpha_{w(1)},\alpha_{w(2)},\ldots,\alpha_{w(r-1)})\in \mathbb{N}^{r-1} $$
and hence for every $m\geq 1$ the set $\left\{\left.\frac{\nu_{Y_\bullet^w}(x^\alpha)}{m}\;\right|\;x^\alpha\in B_{m} \right\}$ is given by the set of points of the form $\left(\frac{\alpha_{w(1)}}{m},\ldots,\frac{\alpha_{w(r-1)}}{m} \right) \in \mathbb{Q}^{r-1}_{\geq 0}$ where $\alpha=(\alpha_1,\ldots,\alpha_r)\in \mathbb{N}^r$ is such that $|\alpha|=\alpha_1+\ldots+\alpha_r=m$ and $\sum_{i=1}^r \sigma_i \alpha_i > A$. Equivalently, is given by the set of points $\left(\frac{\alpha_{w(1)}}{m},\ldots,\frac{\alpha_{w(r-1)}}{m} \right) \in \mathbb{Q}^{r-1}_{\geq 0}$ such that
\begin{itemize}
\item[$\circ$] $\frac{\alpha_{w(1)}}{m}+\ldots+\frac{\alpha_{w(r-1)}}{m}\leq 1$ and
\item[$\circ$] ${\textstyle \begin{array}{ll} \sum_{i=1}^r \sigma_{i} \frac{\alpha_{i}}{m} &=\sum_{i=1}^r \sigma_{w(i)} \frac{\alpha_{w(i)}}{m}\\
& = \sum_{i=1}^{r-1} \sigma_{w(i)} \frac{\alpha_{w(i)}}{m} + \sigma_{w(r)}\left(1-\sum_{i=1}^{r-1} \frac{\alpha_{w(i)}}{m} \right) > \frac{A}{m}. \end{array}}$
\end{itemize}
We conclude therefore that
$$\Delta_{Y_\bullet^w}(B_{\bullet}) = \left\{(\nu_1,\ldots,\nu_{r-1})\in \Delta_{r-1}\;\left|\; \sum_{i=1}^{r-1} \sigma_{w(i)} \nu_i + \sigma_{w(r)}\left(1-\sum_{i=1}^{r-1} \nu_i \right) \geq 0 \right.\right\},$$
from which (2) follows.
\end{proof}
\begin{remark}
The assumption $\sigma_1\geq \sigma_2 \geq \cdots \geq \sigma_r$ in Lemma \ref{lemm: NO of special algebras} can always be fulfilled by considering an automorphism of $\mathbb{P}^{r-1}$ permuting the chosen homogeneous coordinates.
\end{remark}
\subsection{Proofs of main results}
Let us note that given $q\in C$, the Harder-Narasimhan filtration
$$\operatorname{HN}_\bullet(E) : 0=E_\ell \subseteq E_{\ell-1} \subseteq \cdots \subseteq E_1 \subseteq E_0=E$$
induces a (not necessarily complete) flag on $\mathbb{P}(E)$ in the following way: let $Y_0=\mathbb{P}(E)$ and for every $i=2,\ldots,\ell$ the exact sequence
$$0\to Q_i\to E/E_i\to E/E_{i-1}\to 0 $$
\noindent induces an inclusion $\mathbb{P}((E/E_{i-1})|_q)\hookrightarrow \mathbb{P}((E/E_i)|_q) $. We obtain therefore the following (possibly partial) flag of linear subvarieties
$$\mathbb{P}((E/E_1)|_q)\subseteq \mathbb{P}((E/E_2)|_q)\subseteq \cdots \subseteq \mathbb{P}((E/E_{\ell-1})|_q)\subseteq \mathbb{P}(E|_q)=\pi^{-1}(q)\subseteq \mathbb{P}(E) $$
\noindent with $\operatorname{codim}_{\mathbb{P}(E)}\mathbb{P}((E/E_i)|_q)=\operatorname{rank}(E_i)+1$. We also note that this flag is complete if and only if all the semi-stable quotients $Q_i$ are line bundles. In general, it will be necessary to choose a complete linear flag on each $\mathbb{P}((E_{i-1}/E_i)|_q)=\mathbb{P}(Q_i|_q)$ in order to complete the flag above.
We shall consider linear flags that are compatible with the Harder-Narasimhan filtration of $E$ in the sense that they complete the previous flag.
\begin{defn}[Compatible linear flag]\label{defi: compatible natural flags} A linear flag $Y_\bullet$ on $\mathbb{P}(E)$ over $q\in C$ is said to be {\it compatible with the Harder-Narasimhan filtration of $E$} if
$$Y_{\operatorname{rank}E_i+1}=\mathbb{P}((E/E_i)|_q)\cong \mathbb{P}^{r-\operatorname{rank}E_i-1} \subseteq \mathbb{P}(E)$$
\noindent for every $i=1,\ldots,\ell$.
\end{defn}
We will adopt the following convention.
\begin{convention}\label{convention:schubert cells}
Let $V_\bullet$ be a fixed linear flag on $\mathbb{P}(E)$ over $q\in C$ and consider the corresponding Schubert cell decomposition
$$\mathbb{F}_r = \coprod_{w\in \mathfrak{S}_r} \Omega_w, $$
where $\mathbb{F}_r$ is the full flag variety parametrizing complete flags of linear subspaces of $V_1 = \pi^{-1}(q)\cong \mathbb{P}^{r-1}$. We say that a linear flag $Y_\bullet$ on $\mathbb{P}(E)$ over $q\in C$ {\it belongs} to a Schubert cell $\Omega_w$ if the induced linear flag $Y_\bullet|Y_1$ belongs to $\Omega_w$.
\end{convention}
We can prove now the main reduction step. Namely, compute the Newton-Okounkov body of the big and nef class $\xi-\mu_1f$ with respect to any linear flag.
\begin{thm}\label{theo: okounkov bodies}
Let $C$ be a smooth projective curve and let $E$ be a vector bundle over $C$ of rank $r\geq 2$. Fix a linear flag $Y_\bullet^{\HN}$ on $\mathbb{P}(E)$ over $q\in C$ which is compatible with the Harder-Narasimhan filtration of $E$ and let
$$\mathbb{F}_r=\coprod_{w\in \mathfrak{S}_r}\Omega_w $$
be the corresponding Schubert cell decomposition of the full flag variety $\mathbb{F}_r$ parametrizing linear flags on $\pi^{-1}(q)\cong \mathbb{P}^{r-1}$. Then, for every linear flag $Y_\bullet$ on $\mathbb{P}(E)$ over $q\in C$ that belongs to $\Omega_w$ we have that
$$\Delta_{Y_\bullet}(\xi-\mu_1f)=\left\{(\nu_1,\ldots,\nu_r)\in \mathbb{R}_{\geq 0}^r \;|\;0\leq \nu_1\leq \mu_\ell-\mu_1,\;(\nu_2,\ldots,\nu_r)\in \square_{\mu_1+\nu_1}^{w} \right\}, $$
where $\square_{\mu_1+\nu_1}^{w}\subseteq \mathbb{R}^{r-1}$ is the full dimensional polytope defined in Notation \mbox{\ref{notation: Wolfe's square} \eqref{eq:star}}, with $(\mu_1,\ldots,\mu_\ell)$ and $(r_1,\ldots,r_\ell)$ given by the Harder-Narasimhan \mbox{filtration} of $E$ as in Notation \ref{notation: HN quotients}.
\end{thm}
\begin{proof}
We follow Notation \ref{notation: HN quotients}. We first note that if $E$ is semi-stable then $\mu_1=\mu_\ell=\mu(E)$ and hence the Theorem follows from Proposition B. Let us suppose from now on that $E$ is unstable.
\vspace{2mm}
For the reader's convenience, the proof is subdivided into several steps. We first observe that it is enough to compute rational slices of $\Delta_{Y_\bullet}(\xi-\mu_1f)$. In order to do so, we consider the restricted algebra $A_\bullet$ whose Newton-Okounkov body computes the desired slice. After performing a suitable Veronese embedding $A_{n\bullet}\subseteq A_\bullet$, we define a graded subalgebra $B_{n\bullet}\subseteq A_{n\bullet}$ which turns out to be a toric graded algebra as in Lemma \ref{lemm: NO of special algebras}. A comparison of volumes leads to the result.
\vspace{2mm}
\noindent {\it Step 1. Reduction to rational slices.} It follows from Theorem \ref{slices} and Lemma \ref{lemma: psef cone} that the projection of the Newton-Okounkov body of $\xi-\mu_1f$ onto the first coordinate is given by
$$\operatorname{pr}_1(\Delta_{Y_\bullet}(\xi-\mu_1f))=[0,\mu_\ell-\mu_1].$$
By continuity of slices of Newton-Okounkov bodies (cf. \cite[Lemm. 1.7]{KL15b}), it suffices to consider a fixed $t\in ]0,\mu_\ell-\mu_1[\cap \mathbb{Q}$ and show that the slice of $\Delta_{Y_\bullet}(\xi-\mu_1f)$ at $t\in ]0,\mu_\ell-\mu_1[\cap \mathbb{Q}$ is given by
$$\Delta_{Y_\bullet}(\xi-\mu_1f)|_{\nu_1=t}=\square_{\mu_1+t}^{w}\subseteq \mathbb{R}^{r-1} $$
\noindent for linear flags $Y_\bullet$ on $\mathbb{P}(E)$ that belong to the Schubert cell $\Omega_w$ (see Convention \ref{convention:schubert cells}).
Let us fix from now on a permutation $w\in \mathfrak{S}_r$ and a linear flag $Y_\bullet$ on $\mathbb{P}(E)$, over $q\in C$, and let us denote by $Y_\bullet|F$ the induced flag on $F$. Suppose that $Y_\bullet|F$ belongs to the Schubert cell $\Omega_w$ with respect to the reference flag $Y_\bullet^{\HN}|F$.
\vspace{2mm}
\noindent {\it Step 2. Restricted algebra $A_\bullet$.} Consider a $\mathbb{Q}$-divisor $D$ such that $[D]=\xi-\mu_1f$ in $\N^1(\mathbb{P}(E))_\mathbb{Q}$. For every integer $m\geq 1$, let us define the subspace
$$\begin{array}{ll}
A_m=A_{m,t}&=\h^0(\mathbb{P}(E)|F,\mathcal{O}_{\mathbb{P}(E)}(\lfloor m(D-tF)\rfloor)) \\
&=\text{Im}(\h^0(\mathbb{P}(E),\mathcal{O}_{\mathbb{P}(E)}(\lfloor m(D-tF)\rfloor)) \xrightarrow{\operatorname{rest}} \h^0(F,\mathcal{O}_F(m)))\\
&\subseteq \h^0(F,\mathcal{O}_F(m)).
\end{array} $$
If follows from \cite[Prop. 4.1, Rem. 4.25]{LM09} that the restricted algebra $A_\bullet$ above computes the desired slice
$$\Delta_{Y_\bullet|F}(A_\bullet)=\Delta_{Y_\bullet}(\xi-\mu_1f)|_{\nu_1=t},$$
and that for every integer $n\geq 1$ we have
$$\Delta_{Y_\bullet}(nD)|_{\nu_1=nt}=\Delta_{Y_\bullet|F}(n(D-tF))=n\Delta_{Y_\bullet|F}(D-tF)=n\Delta_{Y_\bullet}(D)|_{\nu_1=t}.$$
\vspace{2mm}
\noindent {\it Step 3. Veronese embedding $A_{n\bullet}\subseteq A_\bullet$.} We will consider $A_{n\bullet}=\{a_{nm}\}_{m\geq 0}$ instead of $A_\bullet=\{a_m \}_{m\geq 0}$, for $n$ fixed and divisible enough such that $n\mu_1\in \mathbb{Z}$ and $nt\in \mathbb{Z}$. We also note that for $0<t<\mu_\ell-\mu_1$ we have that
$$\mu_{\max}(S^m E\otimes \mathcal{O}_C(-m(\mu_1+t)\cdot q))=m(\mu_\ell-\mu_1-t)>0$$
and
$$\mu_{\min}(S^m E\otimes \mathcal{O}_C(-m(\mu_1+t)\cdot q))=-mt<0. $$
Therefore, by considering $n$ above large enough we may also assume that ${\mu_{\max}(S^{nm} E\otimes \mathcal{O}_C(-nm(\mu_1+t)\cdot q))>2g-1}$ for every $m\geq 1$.
Let $[x_1:\cdots:x_r]$ be homogeneous coordinates on $F\cong \mathbb{P}^{r-1}$. Since $Y_\bullet$ is a linear flag on $\mathbb{P}(E)$, there is an isomorphism of graded algebras
$$\phi:\bigoplus_{m\geq 0}H^0(F,\mathcal{O}_F(m))\to k[x_1,\ldots,x_r] $$
such that $A_m$ can be regarded as a subspace of $k[x_1,\ldots,x_r]_m$, the $k$-vector space of homogeneous polynomials of degree $m$ in the variables $x_1,\ldots,x_r$, for all $m\geq 0$. Via this identification, $A_\bullet$ can be seen as a graded subalgebra of $k[x_1,\ldots,x_r]$. Moreover, the projection formula implies that we can identify $A_m$ with
$$\text{Im}\Big(\h^0(C,S^mE\otimes \mathcal{O}_C(-m(\mu_1+t) \cdot q)) \xrightarrow{\operatorname{rest}} \h^0(C,(S^mE\otimes \mathcal{O}_C(-m(\mu_1+t) \cdot q))|_q)\Big). $$
\vspace{2mm}
\noindent {\it Step 4. Toric graded subalgebra $B_{n\bullet}\subseteq A_{n\bullet}$.} We shall define a graded subalgebra $B_{n\bullet} \subseteq A_{n\bullet}$ for which we can explicitly compute that
$$\Delta_{Y_\bullet|F}(B_{n\bullet}) = n\square_{\mu_1+t}^{w},$$
and we will prove that $ \Delta_{Y_\bullet|F}(B_{n\bullet}) = \Delta_{Y_\bullet|F}(A_{n\bullet})=n\Delta_{Y_\bullet|F}(A_{\bullet})$.
\medskip
In order to construct $B_{n\bullet}$ let us note that Proposition \ref{propo: HN filtration of S^mE} implies that for every $m\geq 1$ there is a filtration
$$F_\bullet : 0=F_L\subseteq F_{L-1} \subseteq \cdots \subseteq F_1 \subseteq F_0=S^{nm} E \otimes \mathcal{O}_C(-nm(\mu_1+t)\cdot q) $$
whose successive quotients have the form
$$F_{j-1}/F_j \cong Q_{\mathbf{m}(j),\mu_1+t}=S^{m_1}Q_1 \otimes \cdots \otimes S^{m_\ell}Q_\ell\otimes \mathcal{O}_C(-nm(\mu_1+t)\cdot q) $$
\noindent for some partition $\mathbf{m}(j)\in \mathbb{N}^\ell$ of $nm$, and $\mu(Q_{\mathbf{m}(j),\mu_1+t})\leq \mu(Q_{\mathbf{m}(j+1),\mu_1+t})$ for every $j\in \{1,\ldots,L\}$.
Let us define $J=J(m)\in\{1,\ldots,L\}$ to be the largest index such that ${\mu(Q_{\mathbf{m}(J),\mu_1+t})\leq 2g-1}$. We have that for every $j\in\{J,\ldots,L-1\}$, the short exact sequence
$$0\to F_{j+1}\otimes \mathcal{O}_C(-q)\to F_{j+1}\to F_{j+1}|_q\to 0 $$
\noindent gives an exact sequence in cohomology
$$0\to \h^0(C,F_{j+1}\otimes \mathcal{O}_C(-q))\to \h^0(C,F_{j+1})\to \h^0(C, F_{j+1}|_q)\to 0, $$
\noindent since we have that $h^1(C,F_{j+1}\otimes \mathcal{O}_C(-q))=0$, by Lemma \ref{lemma: properties slope}. In particular, we get for every $j\in \{J,\ldots,L-1\}$ a surjection $\h^0(C,F_{j+1})\to \h^0(C,F_{j+1}|_q)$. Therefore, let us consider the subspaces
$$B_{nm}=\text{Im}(\h^0(C,F_{J+1}) \xrightarrow{\operatorname{rest}} \h^0(C,F_{J+1}|_q) )=\h^0(C,F_{J+1}|_q)\subseteq A_{nm}. $$
Let us choose homogeneous coordinates $[x_1:\ldots:x_r]$ on $F$ such that
$$Y_{i+1}^{\HN}=\{x_1=\ldots = x_{i}=0 \}\subseteq \mathbb{P}(E)$$
for $i=1,\ldots,r-1$. In particular, we have that
$$Y_{\rk E_i+1}^{\HN}=\mathbb{P}((E/E_i)|_q)=\{x_1=\ldots=x_{\rk E_i}=0\}\subseteq \mathbb{P}(E)$$
for $i=1,\ldots,\ell$ and therefore the degree 1 part of the isomorphism $\phi$,
$$\phi_1:\h^0(F,\mathcal{O}_F(1))\cong \h^0(C,(E \otimes \mathcal{O}_C(-(\mu_1+t)\cdot q))|_q) \to k[x_1,\ldots,x_r]_1, $$
is such that for every $i=0,\ldots, \ell-1$ the image of the subspace
$$\h^0(C,(E_i\otimes \mathcal{O}_C(-(\mu_1+t)\cdot q))|_q)\subseteq \h^0(C,(E \otimes \mathcal{O}_C(-(\mu_1+t)\cdot q))|_q) $$
via $\phi_1$ coincide with the subspace generated by the variables $x_1,\ldots,x_{\rk E_i}$. By taking symmetric powers it follows from Proposition \ref{propo: HN filtration of S^mE} that for each $m\geq 1$ we have that ${B_{nm}\subseteq k[x_1,\ldots,x_r]_{nm}}$, the image of ${\h^0(C,F_{J+1}|_q)\subseteq \h^0(C,F_0|_q)}$, corresponds to the subspace of homogeneous polynomials of degree $nm$ generated by polynomials of the form
$$P(\mathbf{x})=P_1(\mathbf{x}_1)\cdots P_\ell(\mathbf{x}_\ell) $$
\noindent where $P_i$ is an homogeneous polynomial of degree $m_i\geq 0$ in the variables $\mathbf{x}_i=(x_{i,1},\ldots,x_{i,{r_i}})$, where $(x_1,\ldots,x_r)=(\mathbf{x}_\ell,\ldots,\mathbf{x}_1)$, and the $m_i$ are such that $m_1+\ldots+m_\ell=nm$ and
$$\mu_1m_1+\ldots+\mu_\ell m_\ell > nm(\mu_1+t)+2g-1. $$
In other words, $B_{nm}$ is the subspace generated by monomials $x^\alpha=x_1^{\alpha_1}\cdots x_r^{\alpha_r}$ of total degree $|\alpha|=nm$ such that $\sum_{i=1}^r (\sigma_i-\mu_1-t) \alpha_i > 2g-1$, where $\boldsymbol{\sigma}=(\underbrace{\mu_\ell,\ldots,\mu_\ell}_{r_\ell\text{ times}},\underbrace{\mu_{\ell-1},\ldots,\mu_{\ell-1}}_{r_{\ell-1} \text{ times}},\ldots,\underbrace{\mu_1,\ldots,\mu_1}_{r_1 \text{ times}})\in \mathbb{Q}^r$.
\vspace{2mm}
\noindent {\it Step 5. Volume comparison and conclusion.} It follows from Lemma \ref{lemm: NO of special algebras} applied\footnote{We note that if $C\cong \mathbb{P}^1$ then all the slopes $\mu_i=\mu(Q_i)$ are integer numbers and hence the inequality ``$>2g-1$'' becomes ``$\geq 0$''.} to the collection of subspaces $\{B_{nm} \}_{m\geq 1}$ that $B_{n\bullet}$ is a graded subalgebra of $k[x_1,\ldots,x_r]$ whose Newton-Okounkov body, with respect to a linear flag $Y_\bullet$ that belongs to the Schubert cell $\Omega_w$ (see Convention \ref{convention:schubert cells}) is given by
$$ \Delta_{Y_\bullet|F}(B_{n\bullet})=n\square_{\mu_1+t}^{w}, $$
\noindent where
$$\square_{\mu_1+t}^{w}={\textstyle \Big\{(\nu_2,\ldots,\nu_r)\in \Delta_{r-1}\;\left|\; \sum_{i=2}^r \sigma_{w(i-1)}\nu_i+\sigma_{w(r)}\left(1-\sum_{i=2}^r \nu_i\right) \geq \mu_1+t \right.\Big\}}$$
and $\boldsymbol{\sigma}=(\underbrace{\mu_\ell,\ldots,\mu_\ell}_{r_\ell\text{ times}},\underbrace{\mu_{\ell-1},\ldots,\mu_{\ell-1}}_{r_{\ell-1} \text{ times}},\ldots,\underbrace{\mu_1,\ldots,\mu_1}_{r_1 \text{ times}})\in \mathbb{Q}^r$. Finally, we have that
$$
\begin{array}{ll}
\vol_{\mathbb{R}^{r-1}}(\Delta_{Y_\bullet|F}(B_{n\bullet}))&=\vol_{\mathbb{R}^{r-1}}(n\square_{\mu_1+t}^{w}) \\
&=\vol_{\mathbb{R}^{r-1}}(n\Delta_{Y_\bullet}(\xi-\mu_1f)|_{\nu_1=t}) \text{ by Lemma \ref{lemma: Huayi volume slices}}\\
&= \vol_{\mathbb{R}^{r-1}}(\Delta_{Y_\bullet|F}(A_{n\bullet}))
\end{array} $$
and hence the inclusion $\Delta_{Y_\bullet|F}(B_{n\bullet})\subseteq \Delta_{Y_\bullet|F}(A_{n\bullet})$ leads to the equality $\Delta_{Y_\bullet|F}(B_{n\bullet}) = \Delta_{Y_\bullet|F}(A_{n\bullet})$, as the two convex bodies have equal volume. From this we conclude that
$$\Delta_{Y_\bullet}(\xi-\mu_1f)|_{\nu_1=t}=\square_{\mu_1+t}^{w}\subseteq \mathbb{R}^{r-1}. $$ \end{proof}
The following result (from which Theorem A is easily deduced) is an immediate consequence. We keep the same notation as in Theorem \ref{theo: okounkov bodies}.
\begin{cor}\label{cor: main result}
For every linear flag $Y_\bullet$ on $\mathbb{P}(E)$ that belongs to the Schubert cell $\Omega_w$ and every big rational class $\eta=a(\xi-\mu_\ell f)+bf$ we have that
$$\Delta_{Y_\bullet}(\eta)=\left\{(\nu_1,\ldots,\nu_r)\in \mathbb{R}^r_{\geq 0}\;|\;0 \leq \nu_1 \leq b,\;(\nu_2,\ldots,\nu_r)\in a\square_{\mu_\ell - \frac{1}{a}(b-\nu_1)}^{w} \right\}, $$
and hence the global Newton-Okounkov body of $\mathbb{P}(E)$ with respect to $Y_\bullet$ is given by
$$\begin{array}{ll}
\Delta_{Y_\bullet}(\mathbb{P}(E))=&\Big\{((a(\xi-\mu_\ell f)+bf),(\nu_1,\ldots,\nu_r))\in \N^1(\mathbb{P}(E))_\mathbb{R}\times \mathbb{R}^{r} \text{ such that } \\
& \;\; 0 \leq \nu_1 \leq b \text{ and } (\nu_2,\ldots,\nu_r)\in a\square_{\mu_\ell - \frac{1}{a}(b-\nu_1)}^{w} \Big\}.
\end{array} $$
In particular, the global Newton-Okounkov body $\Delta_{Y_\bullet}(\mathbb{P}(E))$ is a rational polyhedral cone and it depends only on $\operatorname{gr}(\operatorname{HN}_\bullet(E))$, the graded vector bundle associated to the Harder-Narasimhan filtration of $E$.
\end{cor}
\begin{proof}
We note that if $\eta=a(\xi-\mu_\ell f)+bf$ is a big rational class on $\mathbb{P}(E)$ and the induced linear flag $Y_\bullet$ belongs to the Schubert cell $\Omega_w$ with respect to a reference flag $Y_\bullet^{\HN}$, then Theorem \ref{theo: okounkov bodies} gives
$$\Delta_{Y_\bullet}(\xi-\mu_1f)=\left\{(\nu_1,\ldots,\nu_r)\in \mathbb{R}_{\geq 0}^r \;|\;0\leq \nu_1\leq \mu_\ell-\mu_1,\;(\nu_2,\ldots,\nu_r)\in \square_{\mu_1+\nu_1}^{w} \right\} $$
and hence
$$a\Delta_{Y_\bullet}(\xi-\mu_1f)=\left\{(\nu_1,\ldots,\nu_r)\in \mathbb{R}_{\geq 0}^r \;|\;0\leq \nu_1\leq a(\mu_\ell-\mu_1),\;(\nu_2,\ldots,\nu_r)\in a\square_{\mu_1+\frac{1}{a}\nu_1}^{w} \right\}.$$
We compute that for $t^*=b-a(\mu_\ell-\mu_1)$ we have
$$a\Delta_{Y_\bullet}(\xi-\mu_1f)+t^*\vec{e}_1=\left\{(\nu_1,\ldots,\nu_r)\in \mathbb{R}_{\geq 0}^r \;|\;0\leq \nu_1\leq b,\;(\nu_2,\ldots,\nu_r)\in a\square_{\mu_1+\frac{1}{a}(\nu_1-t^*)}^{w} \right\}$$
with $\mu_1+\frac{1}{a}(\nu_1-t^*)=\mu_\ell - \frac{1}{a}(b-\nu_1)$. The result follows from Lemma \ref{lemma:reduction to big nef}. \end{proof}
{\bibliography{bibliokounkov}{}}
\end{document} | 0.001343 |
TITLE: Colimit in the category of (all) simply transitive group actions
QUESTION [3 upvotes]: Let $\mathcal{C}$ be the category of all group actions, i.e. :
the objects are the pairs $(G,F)$ where $G$ is a group and $F$ is a functor $F\colon G\to\mathbf{Sets}$
a morphism between $(G_1,F_1)$ and $(G_2,F_2)$ is a pair $(L,\lambda)$, where $L$ is a functor $L \colon G_1 \to G_2$ and $\lambda$ is a natural transformation from $F_1$ to $F_2 \circ L$.
Consider the subcategory $\mathcal{C}_r$, where the functor $F$ of any object $(G,F) \in \mathcal{C}_r$ is representable.
The general questions are :
Does $\mathcal{C}$ have all colimits ? If so, how are they constructed ?
Does $\mathcal{C}_r$ have all colimits ?
For some specific examples: consider $G_1=G_2=\mathbb{Z}_2$, acting on two elements sets $\{p_1,p_2\}$ and $\{q_1,q_2\}$ (where the action of the non-trivial element sends $p_1$ to $p_2$, etc.). What are the colimits in $\mathcal{C}$ and in $\mathcal{C}_r$ if they exist ?
Edit: I've edited the question so that the questions are explicitly stated.
REPLY [2 votes]: The category $\mathcal{C}$ of all group actions is complete and cocomplete. First, observe that there is an evident forgetful functor $U : \mathcal{C} \to \mathbf{Set} \times \mathbf{Set}$ that preserves and creates limits. You can also check that $U : \mathcal{C} \to \mathbf{Set} \times \mathbf{Set}$ preserves and creates filtered colimits. With further work, you can eventually show that $\mathcal{C}$ is finitely accessible and hence locally finitely presentable, so cocomplete a fortiori.
On the other hand, the full subcategory $\mathcal{C}_r$ of simply transitive group actions is neither complete nor cocomplete. Indeed, it is easy to check that $\mathcal{C}_r$ is closed under products and filtered colimits in $\mathcal{C}$. It is even a finitely accessible category. On the other hand, as Jeremy Rickard observed, $\mathcal{C}_r$ does not even have an initial object. (Notice that, for $(G, X)$ and $(H, Y)$ in $\mathcal{C}_r$, there is a (non-canonical) bijection between the set of morphisms $(G, X) \to (H, Y)$ and the set $\mathrm{Hom} (G, H) \times Y$, so there is no possible initial object.) It follows that $\mathcal{C}_r$ cannot be complete either – if it did, then it would be locally finitely presentable and hence cocomplete. | 0.064303 |
\begin{document}
\author{Cody Holdaway}
\address{Department of Mathematics, Box 354350, Univ.
Washington, Seattle, WA 98195}
\email{codyh3@math.washington.edu}
\keywords{quotient category; representations of quivers; path algebras.}
\subjclass[2010]{14A22, 16B50, 16G20, 16W50}
\begin{abstract}
Let $k$ be a field, $Q$ a finite directed graph, and $kQ$ its path algebra. Make $kQ$ an $\NN$-graded algebra by assigning each arrow a positive degree. Let $I$ be a homogeneous ideal in $kQ$ and write $A=kQ/I$. Let $\QGr A$ denote the quotient of the category of graded right $A$-modules modulo the Serre subcategory consisting of those graded modules that are the sum of their finite dimensional submodules. This paper shows there is a finite directed graph $Q'$ with all its arrows placed in degree 1 and a homogeneous ideal $I'\subset kQ'$ such that $\QGr A \equiv \QGr kQ'/I'$. This is an extension of a result obtained by the author and Gautam Sisodia in \cite{HG}.
\end{abstract}
\maketitle
\pagenumbering{arabic}
\setcounter{section}{0}
\section{Introduction}
\subsection{} In noncommutative projective geometry, there seems to be a consensus that being generated in degree $1$ is ``good.''
For example, consider Serre's Theorem: If $A$ is a locally finite commutative graded $k$-algebra generated in degree $1$, then $\QGr A \equiv \Qcoh(\Proj A)$. Serre's Theorem can fail if the algebra is not generated in degree $1$, a counterexample being the polynomial algebra $k[x,y]$ with $\deg x=1$ and $\deg y=2$.
Another nice theorem that uses generation in degree $1$ is Verevkin's result about the equivalence
$$
\QGr A \equiv \QGr A^{(d)}
$$
where $A^{(d)}$ is the $d$-th Veronese subalgebra of $A$ \cite{ABV}.
Given a graded algebra $A$, is it possible to find a graded algebra $A'$ generated in degree one such that
$$
\QGr A \equiv \QGr A'?
$$
In \cite{HG} it was shown that the answer is yes when $A$ is a path algebra or a monomial algebra. This article extends these results to include the case where $A$ is any quotient of a path algebra by a finitely generated homogeneous ideal.
Lets consider the example with the commutative polynomial algebra $A=k[x,y]$ where $\deg x=1$ and $\deg y=2$. $A$ is the quotient of the path algebra $kQ$ modulo the ideal $I=(xy-yx)$ where $Q$ is the quiver
$$
\UseComputerModernTips
\xymatrix{ {\bullet} \ar@(ul,dl)_{x} \ar@(ur,dr)^{y} & {.} }
$$
Let $Q'$ be the quiver:
$$
\UseComputerModernTips
\xymatrix{ {\bullet} \ar@(ul,dl)_{x} \ar@/^1pc/[r]^{y'} & {\bullet} \ar@/^1pc/[l]^{y''}}
$$
and give $kQ'$ the grading where all arrows have degree $1$. It is shown in \cite{HG} that $\QGr kQ \equiv \QGr kQ'$. That is, the noncommutative projective schemes $\Proj_{nc}kQ$ and $\Proj_{nc}kQ'$ are isomorphic.
The scheme $\Proj_{nc}k[x,y]$ is a ``closed subsecheme'' of $\Proj_{nc}kQ$ defined by the ideal $I=(xy-yx)$. Since $\Proj_{nc}kQ \cong \Proj_{nc}kQ'$, the space $\Proj_{nc}k[x,y]$ should correspond to some ``closed subscheme'' of $\Proj_{nc}kQ'$.
One guess might be that $\Proj_{nc}k[x,y]$ corresponds to the closed subscheme of $\Proj_{nc}kQ'$ cut out by the ideal $I'=(xy'y''-y'y''x)$. The methods of this paper show this is true. More explicitly, the main result shows
$$
\QGr k[x,y]\equiv \QGr kQ'/I'.
$$
This equivalence is rather interesting. The algebra $k[x,y]$ is a connected Noetherian domain while $kQ'/I'$ is none of these. However, $kQ'/I'$ is generated in degree $1$. Thus, in trying to understand $\QGr k[x,y]$, one can use whichever algebra is most suited to the question at hand.
The principal result of this paper is:
\begin{Thm} \label{thm.main}
Let $Q$ be a weighted quiver and $I$ a finitely generated homogeneous ideal in $kQ$. There is a quiver $Q'$ with all arrows having degree $1$, a finitely generated homogeneous ideal $I'\subset kQ'$, and an equivalence of categories
$$
F:\QGr kQ/I \equiv \QGr kQ'/I'
$$
which respects shifting. That is, $F(\sM(1))\cong F(\sM)(1)$ for all $\sM \in \QGr kQ/I$.
\end{Thm}
\subsection{Notation and definitions} \label{sec.def}
Throughout, $Q=(Q_0,Q_1,s,t)$ will always denote a finite quiver, i.e., a finite directed graph. The set $Q_0$ is called the vertex set, $Q_1$ the arrow set and $s,t:Q_1\to Q_0$ will be the source and target maps respectively. Given a field $k$, the path algebra $kQ$ is the algebra with basis consisting of all paths in $Q$, including a trivial path $e_v$ at each vertex $v$.
Given two paths $p=a_1\cdots a_n$ and $q=b_1\cdots b_m$, the product $pq$ is the path $a_1\cdots a_nb_1\cdots b_m$ if $t(a_n)=s(b_1)$ and is zero otherwise.
Call the pair $(Q,\deg)$ a {\it weighted quiver} if $Q$ is a finite quiver and $\deg:Q_1 \to \NN_{>0}$. Usually, the $\deg$ part of the notation $(Q,\deg)$ will be dropped.
A weighted quiver determines an $\NN$-graded path algebra $kQ$ where the degree of the arrow $a$ is $\deg(a)$ and the trivial paths have degree zero. The term {\it weighted path algebra} will mean the path algebra of a weighted quiver. The term path algebra will always mean the arrows have degree $1$.
Given an $\NN$-graded $k$-algebra $A$, $\Gr A$ will denote the category of $\ZZ$-graded right $A$ modules with degree preserving homomorphisms. $\Fdim A$ will denote the localizing subcategory of $\Gr A$ consisting of all graded modules which are the sum of their finite-dimensional submodules. The quotient of $\Gr A$ by $\Fdim A$ is denoted $\QGr A$ and the canonical quotient functor will be denoted
$$
\pi^*: \Gr A \to \QGr A.
$$
The functor $\pi^*$ is exact and the subcategory $\Fdim A$ is localizing, that is, $\pi^*$ has a right adjoint which will be denoted $\pi_*$.
\section{The category of graded representations with relations.}
Associated to a weighted quiver $Q$ is the category of graded representations $\GrRep Q$. A graded representation is the data $M=(M_v,M_a)$ where for each vertex $v$, $M_v$ is a $\ZZ$-graded vector space over $k$ ($k$ is in degree zero) and for each arrow $a$, $M_a:M_{s(a)}\to M_{t(a)}$ is a degree $\deg(a)$ linear map.
A morphism $\phi:M \to N$ is a collection of degree $0$ linear maps $\phi_v:M_v\to N_v$ for each vertex $v$ such that for each arrow $a\in Q_1$, the diagram
$$
\UseComputerModernTips
\xymatrix{ {M_{s(a)}} \ar[r]^{M_a} \ar[d]_{\phi_{s(a)}} & {M_{t(a)}} \ar[d]^{\phi_{t(a)}}\\
{N_{s(a)}} \ar[r]_{N_a} & {N_{t(a)}} }
$$
commutes.
The categories $\Gr kQ$ and $\GrRep Q$ are equivalent. An explicit equivalence is given by sending a graded module $M$ to the data $(Me_v,M_a)$ where $M_a:Me_{s(a)}\to Me_{t(a)}$ is the degree $\deg(a)$ linear map induced by the action of $a$.
If $p=a_1\cdots a_m$ is a path in $Q$, then given any graded representation $(M_v,M_a)$, $p$ determines a degree $\deg(p)$ linear map $M_p:M_{s(a_1)}\to M_{t(a_m)}$ which is the composition
$$
M_p=M_{a_m}\circ \cdots \circ M_{a_1}.
$$
Given a linear combination $\rho=\sum \alpha_ip_i$, where $\alpha_i\in k$ and the $p_i$ are paths in $Q$ with the same source and target, we get a linear map
$$
M_{\rho}=\sum \alpha_iM_{p_i}.
$$
Let $A=kQ/I$ be a weighted path algebra modulo an ideal $I$ generated by a finite number of homogeneous elements. Because of the idempotents $e_v$, we can write
$$
I=(\rho_1,\ldots,\rho_n)
$$
where $\rho_i$ is a linear combination of paths of the same degree all of which have the same source and target.
Let $\GrRep (Q,\rho_1,\ldots,\rho_n)$ denote the full subcategory of $\GrRep Q$ consisting of all the graded representations $(M_v,M_a)$ such that $M_{\rho_i}=0$ for all $i=1,\ldots,n$. The equivalence $\Gr kQ \equiv \GrRep Q$ induces an equivalence $\Gr kQ/I \equiv \GrRep (Q,\rho_1,\ldots,\rho_n)$. From now on, the categories $\Gr kQ/I$ and $\GrRep (Q,\rho_1,\ldots,\rho_n)$ will be identified.
\section{Proof of Theorem \ref{thm.main}}
\subsection{} \label{sec.secmain}
The proof of Theorem \ref{thm.main} follows section 3 in \cite{HG} very closely. The details, with the appropriate modifications for the more general case, are reproduced here for convenience of the reader.
Given a weighted quiver $Q$, define the {\it weight discrepancy} to be the non negative integer
$$
D(Q):=\left(\sum_{a\in Q_1}\deg(a)\right)-|Q_1|.
$$
Note that $D(Q)=0$ if and only if each arrow in $Q$ has degree $1$. The proof of Theorem \ref{thm.main} will be based on induction on $D(Q)$.
Let $Q$ be a weighted quiver and suppose $b$ is an arrow with $\deg(b)>1$. Define a new quiver $Q'$ from $Q$ by declaring
\begin{eqnarray*}
Q'_0&:=& Q_0 \sqcup \{z\} \\
Q'_1&:=& (Q_1\-\{b\}) \sqcup \{b':s(b) \to z, b'':z\to t(b)\}.
\end{eqnarray*}
Make $Q'$ a weighted quiver by letting each arrow in $Q'_1\-\{b',b''\}$ have the same degree as it had in $Q_1$ and letting $\deg(b')=1$ and $\deg(b'')=\deg(b)-1$. From the construction of $Q'$ it follows that
$$
D(Q')=D(Q)-1.
$$
\begin{example} \label{ex.ex1}
Let $Q$ be the quiver
$$
\UseComputerModernTips
\xymatrix{ {\bullet} \ar@(ul,dl)_{a} \ar@/^1pc/[r]^{b} \ar@/_1pc/[r]_{c} & {\bullet} \ar@(ur,dr)^{d} }
$$
with $\deg(b)>1$. The associated quiver $Q'$ is
$$
\UseComputerModernTips
\xymatrix{ {\bullet}\ar@(ul,dl)_{a} \ar@/_0.5pc/[rr]_{c} \ar@/^1pc/[r]^{b'} & {z}\ar@/^1pc/[r]^{b''} & {\bullet} \ar@(ur,dr)^{d} }
$$
with $\deg(b')=1$ and $\deg(b'')=\deg(b)-1$.
\end{example}
Let $Q$ be a weighted quiver and $Q'$ the associated quiver constructed above. Given a path $p=a_1\cdots a_m$ in $Q$, let $f(p)$ be the path in $Q'$ which is obtained by replacing every occurrence of $b$ with $b'b''$ while leaving the path unchanged if there is no occurrence of $b$. For the quiver in example \ref{ex.ex1},
$$
f(a^2bd)=a^2b'b''d
$$
while
$$
f(acd)=acd.
$$
As $\deg(b'b'')=\deg(b)$, the map $f$ preserves the degree of paths. Hence, $f$ determines a graded $k$-linear map $f:kQ \to kQ'$ which can be seen to respect multiplication.
\subsection{} Let $Q$ be a weighted quiver and $Q'$ the associated quiver as in section \ref{sec.secmain}. Given a graded representation $M \in \Gr kQ$, let $F(M)$ be the following graded representation in $\Gr kQ'$:
For the vertices;
\begin{itemize}
\item $F(M)_v:=M_v$ for all $v\in Q'_1\-\{z\}$,
\item $F(M)_z:=M_{s(b)}(-1)$,
\end{itemize}
while for the arrows;
\begin{itemize}
\item $F(M)_a:=M_a$ for all $a\in Q'_1\-\{b',b''\}$,
\item $F(M)_{b'}:=\id:M_{s(b)} \to M_{s(b)}(-1)$ considered a linear map of degree $1$,
\item $F(M)_{b''}:=M_b:M_{s(b)}(-1) \to M_{t(b)}$ considered a linear map of degree $\deg(b)-1$.
\end{itemize}
Given a morphism $\phi:M \to M'$ in $\Gr kQ$, define $F(\phi):F(M) \to F(M')$ by
\begin{itemize}
\item $F(\phi)_v:=\phi_v$ for all $v\in Q'_0\-\{z\}=Q_0$, and
\item $F(\phi)_z:=\phi_{s(b)}(-1):M_{s(b)}(-1) \to M'_{s(b)}(-1)$.
\end{itemize}
It is shown in \cite{HG} that $F:\Gr kQ \to \Gr kQ'$ is an exact functor for which
$$
F(M(1))\cong F(M)(1).
$$
Let $p=a_1\cdots a_m$ be a path in $Q$ and $f(p)$ the associated path in $Q'$. From the definition of the functor $F$,
$$
F(M)_{f(p)}=M_p.
$$
To see this, note $f(p)=f(a_1)\cdots f(a_m)$ so
$$
F(M)_{f(p)}=F(M)_{f(a_m)}\cdots F(M)_{f(a_1)}.
$$
If $a_i\neq b$, then $f(a_i)=a_i$ and thus $F(M)_{f(a_i)}=F(M)_{a_i}=M_{a_i}$. If $a_i=b$, then $f(a_i)=b'b''$ and thus $F(M)_{f(a_i)}=F(M)_{b'b''}=F(M)_{b''}F(M)_{b'}=M_b\circ \id=M_b$. Hence, if $\rho=\sum \alpha_ip_i$ is a linear combination of paths with the same source and target, then
$$
F(M)_{f(\rho)}=\sum \alpha_iF(M)_{f(p_i)}=\sum \alpha_iM_{p_i}=M_{\rho}.
$$
Let $I=(\rho_1,\ldots,\rho_n)\subset kQ$ be a homogeneous ideal. As before,
$$
\rho_i=\sum_{j=1}^m \alpha_jp_j
$$
is a linear combination of paths of the same degree such that $s(p_j)=s(p_{j'})$ and $t(p_j)=t(p_{j'})$ for all pairs $(j,j')$.
Suppose $M\in \Gr kQ/I$. For all $\rho_i\in I$, $M_{\rho_i}=0$. Hence, for the representation $F(M)$, $F(M)_{f(\rho_i)}=M_{\rho_i}=0$ which implies $F(M) \in \Gr kQ'/I'$ where $I'$ is the ideal
$$
I'=(f(\rho_1),\ldots,f(\rho_n)).
$$
Therefore, the functor $F:\Gr kQ \to \Gr kQ'$ induces a functor $F:\Gr kQ/I \to \Gr kQ'/I'$.
Let $N$ be a representation of $kQ'$. Define $G(N)$ to be the following representation of $kQ$:
For the vertices,
\begin{itemize}
\item $G(N)_v:=N_v$ for all vertices $v\in Q_0=Q'_0\-\{z\}$,
\end{itemize}
while for the arrows
\begin{itemize}
\item $G(N)_a:=N_a$ for all $a\in Q_1\-\{b\}$, and
\item $G(N)_b:=N_{b''}\circ N_{b'}$ which is a linear map of degree $\deg(b''b')=\deg(b)$.
\end{itemize}
Given a morphism $\psi:N \to N'$ in $\Gr kQ'$, define $G(\psi):G(N) \to G(N')$ by
\begin{itemize}
\item $G(\psi)_v:=\psi_v$ for all $v\in Q_0=Q'_0\-\{z\}$.
\end{itemize}
$G$ is a functor $\Gr kQ' \to \Gr kQ$.
Let $N$ be a representation in $\Gr kQ'$ and $p=a_1\cdots a_m$ a path in $Q$. Since $G(N)_b=N_{b''}N_{b'}$ and $G(N)_a=N_a$ for $a\in Q_1\-\{b\}$, it follows that
$$
G(N)_p=N_{f(p)}
$$
and more generally,
$$
G(N)_{\rho}=N_{f(\rho)}
$$
for any linear combination of paths with the same source and target. Hence, if $N$ is a representation in $\Gr kQ'/I'$, then for all $\rho_i\in I$,
$$
G(N)_{\rho_i}=N_{f(\rho_i)}=0.
$$
Hence, the functor $G:\Gr kQ' \to \Gr kQ$ induces a functor $G:\Gr kQ'/I' \to \Gr kQ/I$.
From the definitions of $F$ and $G$, it can be seen that $GF=\id_{\Gr kQ/I}$.
Let $N\in \Gr kQ'/I'$, then the module $FG(N)$ is given by the data
\begin{itemize}
\item $FG(N)_v=N_v$ for $v\in Q'_0\-\{z\}$,
\item $FG(N)_z=N_{s(b)}(-1)$,
\item $FG(N)_a=N_a$ for all $a\in Q'_1\-\{b',b''\}$,
\item $FG(N)_{b'}=\id:N_{s(b)} \to N_{s(b)}(-1)$ considered a degree one linear map,
\item $FG(N)_{b''}=N_{b''}\circ N_{b'}:N_{s(b)}(-1) \to N_{t(b)}$.
\end{itemize}
For each $N\in \Gr kQ'/I'$, define $\epsilon_N:FG(N) \to N$ by $(\epsilon_N)_v=\id$ for $v\neq z$ and $(\epsilon_N)_z=N_{b'}$ considered as a degree zero map from $FG(N)_z=N_{s(b)}(-1) \to N_z$.
\begin{Prop}
The assignment $N\mapsto \epsilon_N$ is a natural transformation $\epsilon:FG \to \id_{\Gr kQ'/I'}$. Let $\eta:\id_{\Gr kQ/I} \to GF$ be the identity natural transformation. Then $F$ is left adjoint to $G$ with unit $\eta$ and counit $\epsilon$.
\end{Prop}
\begin{proof}
See Propositions $3.3$ and $3.4$ in \cite{HG}.
\end{proof}
\subsection{}
Let $\pi^*:\Gr kQ'/I' \to \QGr kQ'/I'$ be the canonical quotient functor and $\pi_*$ it's right adjoint. Let $\sigma:\id_{\Gr kQ'/I'} \to \pi_*\pi^*$ be the unit and $\tau:\pi^*\pi_* \to \id_{\QGr kQ'/I'}$ the counit of the adjoint pair $(\pi^*,\pi_*)$. Using the adjoint pair $(F,G)$, we get the adjoint pair $(\pi^*F,G\pi_*)$ where
\begin{itemize}
\item $G\sigma F\cdot \eta:\id_{\Gr kQ/I}\to G\pi_*\circ \pi^*F$ is the unit and
\item $\tau \cdot \pi^*\epsilon \pi_*:\pi^*F \circ G\pi_* \to \id_{\QGr kQ'/I'}$ is the counit.
\end{itemize}
As $\pi^*$ and $F$ are exact so is $\pi^*F$.
\begin{lemma}
The kernel of $\pi^*F:\Gr kQ/I \to \QGr kQ'/I'$ is
$$
\Ker \pi^*F=\Fdim kQ/I.
$$
\end{lemma}
\begin{proof}
Same as the proof of Lemma $3.5$ in \cite{HG}.
\end{proof}
\begin{Prop} \label{prop.prop2}
For every module $N\in \Gr kQ'/I'$, $\pi^*(\epsilon_N)$ is an isomorphism.
\end{Prop}
\begin{proof}
For each vertex $v\in Q'_0\-\{z\}$, $\epsilon_N=\id_{N_v}$. Hence, $(\Ker \epsilon_N)_v$ and $(\Coker \epsilon_N)_v$ are zero for all vertices $v\in Q'_0\-\{z\}$. Hence, the modules $\Ker \epsilon_N$ and $\Coker \epsilon_N$ are supported only on the vertex $z$. Thus, every arrow acts trivially on $\Ker \epsilon_N$ and $\Coker \epsilon_N$ showing they are both in $\Fdim kQ'/I'$. Hence, the map $\pi^*(\epsilon_N)$ is an isomorphism.
\end{proof}
\begin{Thm} \label{thm.main2}
The functor $\pi^*F:\Gr kQ/I \to \QGr kQ'/I'$ induces an equivalence of categories
$$
\QGr kQ/I \equiv \QGr kQ'/I'.
$$
\end{Thm}
\begin{proof}
As $F$ and $\pi^*$ preserve shifting, $\pi^*F$ preserves shifting.
The functor $\pi^*F$ is an exact functor with a right adjoint $G\pi_*$. For every object $\sN \in \QGr kQ'/I'$, the map $\pi^*(\epsilon_{\pi_*\sN})$ is an isomorphism by Proposition \ref{prop.prop2}. By \cite[Prop. 4.3, pg. 176]{Pop}, the counit $\tau$ of the adjoint pair $(\pi^*,\pi_*)$ is a natural isomorphism. Hence, the counit $\tau \cdot \pi^*\epsilon \pi_*$ is a natural isomorphism as
$$
(\tau \cdot \pi^*\epsilon \pi_*)_{\sN}=\tau_{\sN}\circ \pi^*(\epsilon_{\pi_*\sN})
$$
is a composition of isomorphisms for all $\sN \in \QGr kQ'/I'$.
Thus, the right adjoint $G\pi_*$ is fully faithful. By \cite[Theorem 4.9, pg. 180]{Pop}, $\pi^*F$ induces an equivalence
$$
\frac{\Gr kQ/I}{\Ker \pi^*F} \equiv \QGr kQ'/I'
$$
which preserves shifting. As $\Ker \pi^*F=\Fdim kQ/I$, the Theorem is proved.
\end{proof}
\subsection{Proof of Theorem \ref{thm.main}.}
The proof of Theorem \ref{thm.main} now follows by induction on the weight discrepancy. If $kQ/I$ is a quotient of a weighted path algebra for which $D(Q)=0$, then every arrow in $Q$ has degree $1$ and there is nothing to prove. Suppose $D(Q)>1$ and let $b$ be an arrow in $Q$ of degree greater than $1$. Let $Q'$ be the quiver obtained from $Q$ by replacing the arrow $b$ with two arrows as in Section \ref{sec.secmain} and $I'$ the ideal obtained from the ideal $I$. By Theorem \ref{thm.main2} there is an equivalence
$$
\QGr kQ/I \equiv \QGr kQ'/I'.
$$
which respects shifting. Since $D(Q')=D(Q)-1$, we can find, by induction, a quiver $Q''$ with all arrows in degree $1$ and a homogeneous ideal $I''\subset kQ''$ such that
$$
\QGr kQ'/I' \equiv \QGr kQ''/I''
$$
where the equivalence respects shifting. Hence, $\QGr kQ/I \equiv \QGr kQ''/I''$ via an equivalence which respects shifting. | 0.001263 |
Energy Initiatives and Engineering
Energy Initiatives: Washington and Lee is committed to reducing its carbon footprint and achieving carbon neutrality as outlined in the 2010 Climate Action Plan and as demonstrated in the "Five for Five" program. The Director of Energy Initiatives coordinates the efforts of energy consultants and contractors to implement short, medium, and long-term goals:
- 2011 - Reduce annual utilities cost by $1 million
- 2013 - Reduce BTUs per square foot by 25%
- 2020 - Reduce greenhouse production 20%
- 2050 - Achieve carbon neutrality
Current Energy Conservation Measures (ECMs)
- Building retro-commissioning of selected buildings
- Lighting retro-fits
- Controls upgrades
- Sub-metering of selected buildings
- Alternative energy learning lab
ENERGY DASHBOARD: The Energy Intelligence System (E.I.S.) allows users to track the amount of energy consumed to heat , cool and provide electricity to the buildings on campus.
Environmental Management: The Director works closely with the University Sustainability Committee to create, organize, and provide operational support for sustainability efforts across campus departments. These include recycling, composting, the Campus Garden, energy conservation, and related educational opportunities. The Director maintains the University Sustainability Website and Calendar, and records and reports campus sustainability efforts.
Engineering: The Engineer provides support to all University Facilities departments including:
- Analyzing operational problems with University equipment and systems such as steam and chilled water production and distribution, power and water distribution and sanitary facilities.
- Monitoring campus energy consumption and identify opportunities for improved energy efficiency.
- Participating in the design, planning, evaluation, procurement, construction, operation, and maintenance of major capital projects.
- Supporting project management activities throughout construction document review and through periodic inspection of project progress and assuring conformance with the established specifications and standards. | 0.976867 |
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Video games are beneficial to teens
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Posted by cheyennebodie 2 years ago
Takes way too much of their time sitting around. Limited use is much more to be desired than being addicted to them.One way to find out an addiction is turn it off and see the. | 0.005439 |
Release Date ~ April 17, 2012
Little, Brown Books for Young Readers ~ Hachette Book Group
ISBN13: 9780316182881
ARC received from HBG Canada for review.
Also, make sure you check out the giveaway at the end of this review because one commenter will win a copy of this book!
-.
P.S. You have to go check out this cool website, Hachette created for the book. It's just like the one in the book!
AND more importantly, you must go check out the YouTube they have for the book! They have some SUPER CREEPY video clips to go with it, although they some are a teensy bit spoiler-y because they hint at what happens later on. But it really isn't any HUGE secret.
So this isn't the technical book trailer, but it's the ad they made based off the one from the book. Personally, I love the one showing Antoine's transmission from the moon - but it's the one I'm warning you gives you a pretty good hint of what's to come in the book :) So don't click on this link if you don't want ANY spoilers.
Because I loved this book so much, I'm going to buy this book for one lucky commenter. I'll be ordering from The Book Depository, so it's open to all the countries TBD ships to :) Standard contest rules apply, just comment with an answer this question and leave me a way to contact you:
Would you want to go on a trip to the moon?
31 comments:
I don't know if I'm the only one not seeing the review (just a blank space)? But I'm looking forward to your review of this book.
I've read the premise on BookDepository & I hope it'll be a good read!
angkarina_angkasuwan at hotmail dot com
I so wouldn't go to the moon at all. I'm afraid of heights and just thinking of being in OUTER SPACE is making my feet and palms sweat!!
itlnsilver (at) yahoo (dot) ca
This one sounds INCREDIBLE!!
I love science fiction, and the fact that it's teens in space is such a pleasant twist:) And it's a huge feat when, in a plot-based sciency story, you still connect with the characters.
I'm with you on loving horror that's more subtle and creepy other than the blood and guts kind of scary.
Thanks for providing the links
It's hugely important that the translator is so good you forget it's originally in a different language. That's so important and such a difficult thing to do, to translate a whole novel and keep the character and personality of the story itself in a second language without jarring the reader out of the story with awkward phrasing.
Thanks for all the extra info you gave us:)
ccfioriole at gmail dot com
Got so caught up I forgot to answer the question!
I would NEVER go on a trip to the moon. I'd say yes to spending a night in a haunted house (I'd love to do that actually;) or anything else that's creepy, but in space, who do you call for help??
You're just floating there and if something goes wrong, then what? There's no Space AAA if the spaceship decides to stop working.
ccfioriole at gmail dot com
Thanks for the review. This reminds me of Willy Wonka & The Chocolate Factory, but it's more modern. =D
AWESOME review. While I've heard mixed things about this one, I'm determined to get my hands on a copy, because any book that can genuinely terrify readers is a book that I want to read. Plus, however improbable, the premise is also really freaking awesome.
As far as the question . . . I'm not certain. On the one hand, I don't think I'd have the guts or the fitness for it. But on the other hand, I'm overwhelmingly curious, so one never knows.
nerdfighterwriter@gmail.com
Hadn't heard of this one before, but now I'm really intrigued. Thanks for the giveaway. Btw, videos were great.
curlypow3 AT gmail DOT com
I totally agree this book has been underrated so far. It's so amazingly creepy and good. If you liked it, you should check out the movie "Moon" it has a similar vibe to it.
This looks like a great book. Thanks for the giveaway!!
I would like to go to the moon, especially the dark side. I'm curious to see what's there. Mind you I'm also really afraid that something would happen.
Jessica
jbronderblogs at aol dot com
I have to admit I don't think I could take the wait time it actually takes to get to the moon, however I would love to be on the moon and explore it!
I haven't read a lot of science fiction...yet, but this one sounds like such an intriguing, fun and creepy read! I really like the cover! It is really creepy and grabs your attention must-read-me status. Well, for me that is.
Awesome review!
Thanks for sharing and the chance to win it!! :D
DeAnna Schultz
sacredmoon1(at)gmail(dot)com
AaaahhH! This book has been sitting on my ARC pile for a while now. While it's not usually the type of book I would get excited about, after reading your review I must admit I am kind of sold :D Definitely intriguing!!
If given the chance to go to the moon I would take it. To view Earth from space would be amazing.
Fabulous giveaway thank you.
marypres(AT)gmail(DOT)com
I don't think I would want to go to the moon, i'm not a very big fan of heights. Thanks for the chance to win!
natasha_donohoo_8 at hotmail dot com
Wow. Great review! Of course. I've always wondered how it is to be out there! I'd love to go to the MOON! :)
franchie15_nina@yahoo.com
Oh my gosh, even that cover made me shudder a little bit. And I would love to go to the moon! With plenty of other people and certified moon expert types along too. :)
liforda(at)yahoo(dot)com
I think it'd be pretty awesome to go to the moon.Granted, it wouldn't be with the Doctor on the TARDIS as I dreamed it would be, but still, to see the moon! My goodness, it'd just be so surreal. I love observing different sceneries, and the moon would definitely prove to be a wonderful sight to take in! Soooo, to some it up, yes, yes I would like to go to the moon.
melissa.riley.95@gmail.com
Sounds like a must read!
As for the trip to the moon - I would totally go! I love traveling, I love exploring and it's just so different than anything we know. Plus, since I was a child, I had a fascination for anything space related (it blew up to be a mini obsession in the 5th grade, but settled down to healthy interest since then).
advakramer at gmail dot com
Going to the moon would be fantastic! I love new places and being lost in them. Haha! Only, if it would be like that in 172 hours, I'm not sure I would really want to go. Haha!
tomakunisnumberone at gmail dot com
Yeah, I would love to go to the moon! And thanks for this giveaway! The book sure looks like a must read! :D
dianne.acoba at gmail dot com
As long as it's free! I love free things so thanks for this giveaway! dianne_dld at yahoo dot com
Yes I would definitely love to go to the moon! It would be amazing to go :)
Thanks for the giveaway!
laceyblossom@yahoo.com
How awesome!! I've heard how creepy this book is! I don't think I'd want to go to the moon... maybe more interested in the actual spaceship ride and how that is.
Thanks! :) you know my email ;)
So in my earlier comment I didn't say if I would want to go to the moon!
As a die hard Star Wars/Trek fan I would definitely want to go to the moon. Anywhere in space really. It would be so neat so go where so few people have gone before.
Thank you for the giveaway! I would love to go to the moon! It sounds and looks awesome to defy gravity :)
I think the moon, although explored, always appeared mysteious to me so I would totally take a trip to the moon.
Just to start out I never plan on going to space
I just finishes reading this book...
yeah i am not ever going to see the moon the same way ever again. never ever ever. Just thinking about the moon makes me shudder and want to dig a hole in the ground and stay there forever! This book is amazing but actually really scary! when i got this book from the library i didn't know it was like this! | 0.000876 |
text Anelia Alexandrova, photos Ilian Rujin.
Back to the RootsThe three boys on the picture are only the core of Root Souljah, our new favourite BG gang. After their first performance ever – on December 6th last year – they managed to produce a small album with 6 tracks. If it is not obvious, they sing about love, the good and the bad, the system and about all important things in general. These true rasta people turned our hearts red-hot.
He and SheSantra is keeping us under steam since that track with Spens and the tiger, I don’t want to stand in your way, which was in her 2002 debut solo album. After a short break, last year came the next assault – this time with Kristo’s help – for two, hmm, pulse-quickening singles. Now Santra and Kristo are ready with a new club remix of Strength – Make Me Stronger, arranged by Maga and music by Bobo. And there is no force capable of keeping us away from the dance floor.
The Magical ViolinVesko Eshkenazi’s violin is not truly magic – he just plays on it with all his heart. He is young and talented, one of our greatest classical musicians famous beyond the borders of Bulgaria, and since 1999 he is first violinist of the Royal Konzertgebau Orchestra in Amsterdam. The news is, that maestro Eshkenazi is coming to Sofia to take part in a BNR Symphony Orchestra Concert. We were more than excited and called him in order to understand what is going on in classical music. Or, in short –to hear a genuine musician
I Swear!You've probably heard the news – the Switched On! project gathered a dozen British DJs and trusted them with remixing classic songs from modern Bulgarian music. The new versions of Chernata Ovtsa and Edna Bulgarska Roza would be performed exclusively on February 24 at 4km party centre with a host of DJS and legendary acts Revu and Nova Generazia performing. We learned that our favourite Bulgarian-British dame Mira Aroyo of Ladytron fame has remixed Sturcite's Kletva, so we quickly called her at her London appartment for details.
Underground HouseWell… John Digweed has just released his new compilation and is already coming to Bulgaria for a big party of the Viva Sound Factory series. It is organized by Metropolis, and the location is fantastic – second level of the Sofia City Centre underground parking lot. The party is on February 17th, so start warming up with Transitions Vol. 2 or dig out the first Renaissance, recorded together with Sasha and remember the beginning. Is it necessary to explain who the hell is John Digweed? He is simply big – in 2001 DJ?Magazine declared him #1, and today, despite his heavy schedule as a producer promoter and a DJ, is also a resident in London’s Heaven and Brighton’s The?Beach.
Music for the MassesAs they themselves point out, Laibach isn't an ordinary band of humble pop musicians. This innovative Slovenian collective is an art institution in itself with its own virtual state (NSK), organizes exhibitions, lectures and seminares, although its main activity remains musical. Their latest ambitious project is entitled Volk and contains 14 national hymns in new symphonic, martial industrial arrangements. Mere days before their second performance in Sofia, Laibach's spoekesman Ivan Novak had this to say about some very important issues.
Napalm Burns
Simply DivineThey call him the kaval player of the world. His style is unique, a mixture of folklore and jazz. He is considered one of the most talented musicians from Eastern Europe. And he always has something up his sleeve – a day before the New Year Music Festival’s closing Teodosii Spasov will present the premiere of his concert Into the Infinity. | 0.069407 |
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\begin{document}
\title{On Finitely Stable Additive Bases}
\author{Lucas Y. Obata, Luan A. Ferreira, Giuliano G. La Guardia
\thanks{Lucas Obata and Giuliano La Guardia (corresponding author) are with the Department of Mathematics and Statistics,
State University of Ponta Grossa (UEPG), 84030-900, Ponta Grossa,
PR, Brazil, e-mails: (obatalucas@gmail.com;gguardia@uepg.br). Luan Ferreira is with Institute of Mathematics and
Statistics of University of S\~ao Paulo, Rua do Mat\~ao, 1010, Cidade Universit\'aria, S\~ao Paulo, SP 05508-090, Brazil,
e-mail: (luan@ime.usp.br)}}
\maketitle
\begin{abstract}
The concept of additive basis has been investigated in the literature for several
mathematicians which works with number theorem. Recently, the concept of
finitely stable additive basis was introduced. In this note we provide a counterexample of the reciprocal of
Theorem 2.2 shown in [Ferreira, L.A.: Finite Stable Additive Basis; Bull. Aust. Math. Soc.].
The idea of the construction of such a counterexample can possibly help the process of
finding additive bases of some specific order.
\end{abstract}
\section{Introduction}\label{sec1}
The concept of additive basis as well as finitely stable (additive) basis are well-known
and much investigated for number theory theorists. For more details about such a theory,
we refer the reader to the textbook \cite{Nathanson:1996}; see also the interesting papers
\cite{Nathanson:1972,Nathanson:1974,Nathanson:1996A,Nathanson:2005,Nathanson:2012,Nathanson:2014,Tafula:2019}.
An additive basis $A$ is a subset $A\subseteq {\mathbb N}=\{0, 1, 2, 3, \ldots \}$ having the property
that there exists a positive integer $h$ such that every $n \in {\mathbb N}$ can
be written as the sum of $h$ elements of $A$. The order $\operatorname{o}(A)$ of $A$ is the minimum
among all $h$ satisfying such condition.
From Lagrange's Theorem it follows
that the set of squares
$${\mathbb N}^2 =\{ 0, 1, 4, 9, \ldots, n^2, \ldots\}$$
is an additive basis of order $4$.
Applying Wieferich's Theorem it follows that the set of cubes
$${\mathbb N}^3 = \{0, 1, 8, \ldots, n^3 , \ldots \}$$
has order $9$.
Let us recall the concept of finitely stable basis.
\begin{definition}\label{fsbasis}
Let $A$ be an additive basis. Then $A$ is said to be finitely stable if
$\operatorname{o}(A)=\operatorname{o}(A\cup F)$ for every finite set
$F \subset {\mathbb N}$.
\end{definition}
\section{The Counterexample}\label{sec2}
We maintain the notation adopted in \cite{Ferreira:2018}. Let $A, B \subseteq
{\mathbb N}$. We define
$$ A + B = \{a + b : a \in A, b \in B\}.$$
If $t$ is a positive integer and $A \subseteq {\mathbb N}$ we write
$$tA := \underbrace{A + A + \cdots + A}_{ t \ \mbox{times}}$$ and $0A = \{0\}$.
If $n \in {\mathbb N}$, we denote $A(n)=| \{a \in A : 1 \leq a\leq n \} |$.
One of us provided in Ref.~\cite{Ferreira:2018} conditions in order to guarantee
finitely stability of a given (additive) basis.
\begin{theorem}\cite[Theorem 2.2.]{Ferreira:2018}\label{LF}
Let $A$ be an additive basis whose order satisfies $\operatorname{o}(A)=h \geq 3$. If
$$\displaystyle\lim_{n\rightarrow \infty} \frac{((h-2)A)(n)}{n}=0$$ and
$$\displaystyle\limsup_{n\rightarrow \infty} \frac{((h-1)A)(n)}{n}< 1,$$ then $A$ is finitely stable.
\end{theorem}
We here present a counterexample for the reciprocal of Theorem~\ref{LF}.
\vspace{0.3cm}
\textbf{Counterexample}
\vspace{0.3cm}
Let
$A = \{0, 1, \ldots , 10\}\cup \{22, 23, \ldots, 100\}\cup \ldots \cup
\{ 2\cdot 10^{n-1}+2, 2 \cdot 10^{n-1}+3, \ldots , 10^{n}\}\cup \ldots =
A_1 \cup A_2 \cup \ldots \cup A_n \cup \ldots$.
\vspace{0.5cm}
It is easy to see that $A$ is an additive basis of order 3, i.e., all nonnegative integers can be written
as sum of three elements of $A$. Moreover, $A$ is finitely stable since, for
every finite set $F \subset {\mathbb N}$, the set $A\cup F$ has not order 2 because
the set $\{ 2 10^{n} + 1 : n \in {\mathbb Z}^{+}\}$ has infinite many numbers.
We will now construct a subsequence of $\frac{(h-2)A(n)}{n} = \frac{A(n)}{n}$ in the following way:
\vspace{0.3cm}
$\{\frac{A(21)}{21}, \frac{A(201)}{201}, \frac{A(2001)}{2001}, \ldots, \frac{A(2 10^{n}+1)}{2 10^{n}+1}, \ldots \}$.
\vspace{0.3cm}
Let $A_n = \{ 2 \cdot 10^{n-1}+2, \ldots , 10^{n}\}$, $n \geq 2$. We then have
\begin{eqnarray*}
\displaystyle\sum_{i=2}^{n} |A_i | = \displaystyle\sum_{i=2}^{n} [10^{i}-(2 \cdot 10^{i-1}+2)]\\
=\displaystyle\sum_{i=2}^{n} (8 \cdot 10^{i-1}-2) = \displaystyle\sum_{i=1}^{n-1} (8 \cdot 10^{i}-2);
\end{eqnarray*}
hence,
\begin{eqnarray*}
\displaystyle\liminf_{n\rightarrow\infty} \frac{A(n)}{n}\leq
\displaystyle\lim_{n\rightarrow\infty}\frac{10 + \displaystyle\sum_{i=1}^{n-1}
(8 \cdot 10^{i}-2)}{2 \cdot 10^{n}+1}=\frac{4}{9}.
\end{eqnarray*}
We next construct another subsequence of $\frac{A(n)}{n}$ given by:
\vspace{0.3cm}
$\{\frac{A(10)}{10}, \frac{A(100)}{100}, \ldots, \frac{A(10^n)}{10^n}, \ldots\}$.
\vspace{0.3cm}
It is easy to see that
\begin{eqnarray*}
\displaystyle\limsup_{n\rightarrow\infty} \frac{A(n)}{n}\geq \displaystyle\lim_{n\rightarrow\infty}\frac{10 +
\displaystyle\sum_{i=1}^{n-1} (8 \cdot 10^{i}-2)}{10^{n}}=\frac{8}{9}.
\end{eqnarray*}
Therefore, $\displaystyle\lim_{n\rightarrow\infty}\frac{A(n)}{n}$ does not exist. Then, $A$ is a
finitely stable additive basis of order three
such that $\displaystyle\lim_{n\rightarrow\infty}\frac{(h-2)A(n)}{n}$ does not exist.
\section{Final Remarks}\label{sec3}
In this paper we have presented a constructive counterexample for Theorem 2.2 in
[Ferreira, L.A.: Finite Stable Additive Basis; Bull. Aust. Math. Soc.]. It seems
that the techniques proposed here can be used in the construction of additive basis of some order.
\begin{center}
\textbf{Acknowledgements}
\end{center}
This research has been partially supported by the Brazilian Agencies
CAPES and CNPq. | 0.001111 |
Which gemstones are copper-bearing?
The most famous cuprian (copper-bearing) gemstone is Paraiba tourmaline. There are a number of other gemstones that contain copper, including malachite, azurite, turquoise, chrysocolla, Larimar and oregon sunstone. It would be misleading to think that a gemstone is valuable simply because it is copper-bearing, but many valuable gems do get their distinctive color from traces of copper. For more information see our article on copper-bearing gemstones. | 0.797239 |
Geometric proofs taught me about failure.
Up until encountering them, I had certainly failed before.
Failed to dunk.
Failed to learn a legitimate skateboard trick.
You know, things like that.
However, the freshness of this type of failure was that success was more of a mirage and less of a clear impossibility.
When riding skateboards it was obvious that in order to get better, I needed to ride more hours and take more risks. I simply didn’t do these things, so it wasn’t a surprise when I wiped out.
When writing proofs in Geometry class, no matter how much I studied and how much extra help I attended, the correct answers evaded me like sand between my fingers.
A few months ago I felt this feeling all over again when in an Escape Room in D.C.
Having experienced one before, I was confident of my critical thinking skills and even found myself sizing up the other people in the room before we began. I had visions of being the Rosetta Stone, the one calling out clear directions, the one untying the Gordian Knot.
Instead, I found myself right back in sophomore year – frustrated, dejected, defeated.
Just to be clear, I’ve failed countless times, but these two seemed to fit together perfectly.
For most of my life I’ve looked at solutions as the aha moment, when statements and reasons are clear and acceptable, when the door finally opens.
I love, but have always struggled with Matthew 7:7-8,
Ask, and it will be given you; search, and you will find; knock, and the door will be opened for you. For everyone who asks receives, and everyone who searches finds, and for everyone who knocks, the door will be opened.
Sure, problems can be solved, doors can be opened, it might just take a while I guess. Whether it’s in this life is not for me to know, and I’m okay with that.
I’m comfortable knowing there are those out there with superior intellects, more money, or exotic opportunities. Comfortable because it’s true, the only false belief is me thinking I have a handle on things.
The reality is, there’s no way out.
Now, instead of searching for proof or a way out, I’m getting more and more comfortable with being right here.
Our society is obsessed with the next best_________
thing
person
job
opportunity.
If I’ve put in the effort, there is nothing wrong with problems remaining unsolved.
My life is not something to be figured out.
My life is not something to be escaped. | 0.717222 |
Action on Sustainability
Sustainability is an integral part of Kendrion’s overall business strategy. Kendrion’s Corporate Social Responsibility (CSR) programme is aligned with Kendrion’s strategic pillars ‘Simplify, Focus and Grow’ and supportsthe development of Kendrion’s business in a responsible and sustainable way.
The three pillars of value creation are: Natural Capital, Social and Human Capital and Responsible Business Conduct. They form the basis of Kendrion’s CSR programme. The CSR programme reflects Kendrion’s mission and commitment to conduct business in a responsible and sustainable way and aspires to strike the right balance between long-term value creation and profitability on the one hand and the objective of playing a meaningful role in addressing key societal and environmental matters on the other. This means strategic decisions are taken based on a range of financial and non- financial considerations. Key principles and measurable targets support the Kendrion CSR programme.
Natural Capital
The Natural Capital pillar focuses on improving Kendrion’s environmental performance and its ambition to reduce its environmental footprint. Material themes for the Natural Capital pillar include energy consumption, CO2 emissions and waste management.
Energy consumption & CO2 emissions
Kendrion aims to reduce the amount of energy used during its production process and its CO2 emission. Kendrion applies an environmental reporting system that tracks the CO2 emissions and energy consumption of all the production plants. Year-on-year, Kendrion focuses on improving the production processes with the overall objective of reducing the environmental footprint of the production plants. The global certification ISO 50001 Energy Management System supports the production plants in their efforts to use energy more efficiently by developing and maintaining an energy management system. Kendrion’s environmental management systems are in accordance with ISO 14001. ISO 14001 Environmental Management Systems specifies requirements for an environmental management system in order to enhance environmental performance. Nine Kendrion production plants are ISO 50001 certified and all plants except for Mishawaka are ISO 14001 certified. Kendrion’s drive to use energy efficiently and reduce CO2 emissions is reflected in a range of activities and investments. Targeted actions to reduce energy consumption include the application of technologies such as high-pressure air, electric devices and other efficient production equipment for various production lines. Other efforts aimed at reducing production-line energy consumption include installing air-conditioning and ventilation that is more efficient and extending LED lighting to certain additional production areas. Activities aimed at reducing CO2 emissions include using recycled paper, concluding or extending additional CO2 neutral energy contracts.
Waste management
Kendrion is committed to continuously improving its management of all waste throughout its lifecycle and to helping reduce waste in order to minimise its adverse impact on the environment. This involves the minimisation and responsible disposal of waste related to production. Kendrion’s environment management systems are set up in accordance with the global certification ISO 14001 (all production plants except for Mishawaka are ISO 14001 certified) and the certified plants maintain effective records of their production and processing of all waste and work exclusively with certified waste processing companies when this is required pursuant to local regulations. New waste reduction measures must be implemented each year as part of the ISO 14001 certification process.
Social and Human Capital
The Social and Human Capital pillar concerns people’s competences, capabilities and motivations. Material themes for the Social and Human Capital pillar include community connection, health & safety, diversity and inclusion and company culture.
Community connection
Local employer and a good neighbour
Kendrion maintains strong ties to the communities in which it operates by encouraging an open dialogue with local management and through Kendrion’s long-standing commitment to being a local employer and a good neighbour.
Together@Kendrion
Kendrion values the social good that can be achieved by demonstrating initiative and taking a long-term perspective. Kendrion supports the economic and social well-being of the local communities in which it operates through Together@Kendrion and other efforts initiated by the local management.
Volunteering
In addition to sponsoring and financially supporting various good causes, Kendrion’s employees are encouraged to invest personal time in the local communities by taking part in fundraising activities and volunteering at events.
Health and Safety
Safety is given the highest priority with respect to every aspect of Kendrion’s operations. Kendrion is committed to providing a safe and healthy workplace for all employees by implementing the most stringent quality and safety standards to avoid any potential risks to people, communities and the environment. Kendrion’s employees are periodically trained to implement the best sustainability practices. The health and safety of employees are essential to the successful conduct of Kendrion’s business and are in the best interest of Kendrion’s other stakeholders.
In addition to certain centrally coordinated health and safety- related initiatives, day-to-day responsibility for health and safety is concentrated within the business units in which health and safety are managed systematically and in a standardised manner with clear rules and procedures based on recognised industry standards and best practices that are laid down in Health, Safety & Environmental (HSE) policies. Each production plant further implements tailored initiatives to enhance their HSE standards depending on plant-specific needs, production lines and technologies. With HSE audits, the implementation and compliance with HSE policies are assessed at regular intervals.
Diversity and inclusion
Kendrion believes in the strength of a diverse workforce. Kendrion is committed to attracting and retaining a diverse global workforce through its employment strategy. Kendrion’s workforce in 2018 comprised 37 nationalities (2017: 38) employed in 10 countries. 51% of our workforce is female. Kendrion continues to undertake action to further improve diversity in technical and non-technical roles and at all levels of the organisation.
Company culture and ethical behaviour
Kendrion believes it is important that all activities are conducted with integrity and in a transparent manner. To this end, Kendrion fosters a culture in which shared norms, universal ethical values and behaviours are the standard. Shared norms, ethical values and expected behaviours are laid down in a set of internal policies and procedures that form part of the Global Legal Compliance and Governance Framework (GLC&GF). In addition to setting norms, values and expected behaviours, the GLC&GF is aimed at ensuring compliance with applicable laws and regulations.
The GLC&GF includes the following key policies and procedures: Code of Conduct, Anti-Bribery and Anti- Corruption Policy, Speak-up Procedure, Competition Compliance Manual, Insider Trading Code, Data Protection Governance Guidelines, Personal Data Breach Reporting Procedure, Supplier Code of Conduct and related internal policies and procedures.
Kendrion considers it essential that every employee understands, complies with and conveys the shared norms and universal ethical values and behaviours as laid down in the policies and procedures of the GLC&GF. The policies and procedures of the GLC&GF are fundamental to ensuring responsible business conduct. It is the responsibility of senior management to lead by example and to ensure that all Kendrion employees are aware of the GLC&GF and behave in accordance with the spirit and the letter of Kendrion’s policies and procedures.
Responsible Business Conduct
The Responsible Business Conduct pillar focuses on business conduct and integrity, accountability and transparency. Material themes for the Responsible Business Conduct pillar include supply chain management and anti-bribery and anti- corruption.
Supply chain management
Kendrion operates as part of a supply chain with a central focus on manufacturing and production processes..
Kendrion intends to play a meaningful role in the supply chain in which it is active. In many instances, Kendrion is a relatively minor link in the supply chain. In order to achieve realistic results, it is of great importance that Kendrion continues the dialogue with its suppliers and, in the selection and assessment of suppliers, continues to consider their performance with respect to sustainability.
Supplier Code of Conduct and implementation audits
The Kendrion Supplier Code of Conduct requires suppliers to accept their responsibility for matters concerning the environment, human rights, working conditions and fair trade. Kendrion regularly conducts implementation audits to review whether suppliers comply with the standards and principles of the Supplier Code of Conduct. Audits that reveal that the relevant supplier does not meet the requirements of the Supplier Code of Conduct are followed by a meeting to prepare a remediation plan. Failure to adequately follow up the remediation plan may result in the termination of the relationship with the relevant supplier.
Anti-Bribery and Anti-Corruption
Kendrion does not tolerate bribery or any form of corruption. Bribery can include the offering, promising or giving of payments or other advantages to any person (including government officials or public officials) to influence a business outcome improperly, but it also means accepting payment or advantages given to influence a business outcome improperly. Integrity of financial reporting is also a key principle. The Kendrion Anti-Bribery and Anti-Corruption Policy specifically addresses these matters.
Three Pillars of Value Creation:
More info about Sustainability within Kendrion
If you would like to read more on Sustainability, please see our 2018 Annual Integrated Report.
Read the Kendrion Code of Conduct
Read the Supplier Code of Conduct
If you like to read the Supplier Code of Conduct please click below.
Kendrion Supplier Code of Conduct
Read Kendrion's Anti-Bribery and Anti-Corruption Policy
Kendrion Anti-Bribery and Anti-Corruption Policy | 0.996237 |
Jason Smith Newswire (Page 7)
Results 121 - 140 of 1,070 in Jason Smith
Fantasy Basketball: Where Will Amare Stoudemire End Up?
w/Photo
... at least a more consistent fantasy impact than he did in New York. The tanking Knicks are using Cole Aldrich and Jason Smith at center, which severely limited Stoudemire's minutes. If you have owned him at all this year, you know how frustrating ... Comment?
Amar'e Stoudemire, Knicks reach buyout agreement
w/Photo
... forward Serge Ibaka (9) dives for a loose ball against Knicks power forward Lou Amundson (21) and power forward Jason Smith (14). Jan 27, 2015: Raptors guard DeMar DeRozan (10) drives to the basket against Pacers forward Solomon Hill at Bankers Life ... Comment?
ESPN: Carmelo likely done after ASG
... e Knicks in their quest to finish with the worst record in the league, as Melo is their second-best player after Jason Smith. Comment?
Magic 89, Knicks 83: "Jason Smith's bucket chart is a thing of beauty"
.... ... Comment?
Knicks lose again, enter break with NBA's worst record
... owner James Dolan fired off an inappropriate email to a disgruntled fan who begged him to sell the team. Jason Smith scored 21 of his 25 points in the first half. Stoudemire, contemplating a buyout this weekend, might have made his Knicks swan song ... Comment?
Melo sits as Knicks enter All-Star break with loss to Magic
w/Photo
... just two of the Knicks' first 51 games due to hamstring and calf injuries. He scored 11 points in 17 minutes and Jason Smith netted 21 of his season-high 25 in the first half, but the Knicks (10-43) dropped their fifth straight game heading into the ... Comment?
Final: Magic 89, Knicks 83
... between the rim and the backboard. Fisher immediately removed him from the game, because nothing gold can stay. Jason Smith carried the Knicks early with a shocking 21 first-half points, while Pablo Prigioni made sweet Pablo magic in the second, ... Comment?
FantasyAlarm.com: Daily fantasy basketball playbook for Feb. 11
w/Photo
... best place to save money at tonight as Josh Smith has been consistent and still has a low price tag. Meanwhile, Jason Smith is a great play and doesn't cost a lot. Mitch McGary finally faces a good defensive team, so we could see a huge decline from ... Comment?
Previewing What New York Knicks Fans Should Expect Without Carmelo Anthony
w/Photo
... perimeter more before reaching their preferred point of initiation, but whether it's Quincy Acy, Lou Amundson, Jason Smith or Anthony headlining the high post, the offensive structure doesn't change. Notice how radically the Knicks' three-point ... Comment?
Irving Bierman stands up for Knicks fans everywhere
w/Photo
Long time Knicks fan Irving Bierman joins The Jason Smith Show to talk about his letter to Knicks Owner James Dolan regarding the state of the franchise. Their much publicized feud erupted when James Dolan responded with a personal attack and a suggestion that Bierman go root for the Brooklyn Nets. Comment?
Greg Cote: Knicks good for what is ailing Miami Heat
w/Photo
... Arena in Miami, Florida, February 9, 2015. Miami Heat's Hassan Whiteside is defended by New York Knicks' Jason Smith and Carmelo Anthony in the first quarter at the AmericanAirlines Arena in Miami, Florida, February 9, 2015. Miami Heat's Luol Deng ... Comment?
Green, Warriors rebound with win over Knicks
... g his troublesome knee in the second game of a back-to- back, Langston Galloway paced the Knicks with 15 points. Jason Smith totaled 14 points and 13 rebounds in New York's third straight loss. Comment?
Hyde: Bosh plays like the All-Star the Heat need
Miami Heat forward Chris Bosh scores over New York Knicks center Jason Smith during the second half. Before Monday's game, with dramatic pomp and ceremony, Heat coach Erik Spoelstra stood at center-court, reached into individual boxes and awarded Chris Bosh and Dwyane Wade, their All-Star jerseys for Sunday's game. Comment?
Heat, Knicks tangle in South Beach
... Carmelo Anthony did not play to rest his sore knee and Langston Galloway paced the Knicks with 15 points. Jason Smith totaled 14 points and 13 rebounds for New York, which shot 40.2 percent from the field and trailed 32-19 after the first quarter. ... Comment?
Knicks soon could have a very different look
w/Photo
... rookie Langston Galloway, second-year guards Tim Hardaway Jr. and Shane Larkin and well-traveled young veterans Jason Smith, Quincy Acy and Lance Thomas played in the fourth quarter Saturday. They could be the nucleus of the team that ends this ... Comment?
Green, Golden State Warriors rebound with win over New York Knicks
w/Photo
... g his troublesome knee in the second game of a back-to- back, Langston Galloway paced the Knicks with 15 points. Jason Smith totaled 14 points and 13 rebounds in New York's third straight loss. Comment?
Warriors have to work to beat lowly Knicks 106-92
w/Photo
... Galloway, who scored seven of the first 10 points for New York in the 16-0 run, led the Knicks with 15 points. Jason Smith added 14 points and 13 rebounds. "This was a game we should have blown out, and we let go of the rope," said Green, who had 20 ... Comment?
Nets rewind: Garnett shows he's not done yet amid buyout rumors
w/Photo
... rebounds, two assists and three blocked shots in 21 minutes. Now it has to be said that the Knicks were playing Jason Smith, Lou Amundson, Amar'e Stoudemire and Lance Thomas as their bigs for most of this game - not exactly a murderer's row of NBA ... Comment?
Jarrett Jack hits clutch 3 as Nets hold off Knicks, 92-88
w/Photo
... win over the Knicks. Brook Lopez made an uncontested lay-in with 1:38 to give Brooklyn (21-28) an 87-84 lead. Jason Smith's two free throws cut the lead to one with 1:21 left. Langston Galloway stole the ball along the baseline with 56 seconds left ... Comment?
Battle for New York heads to Brooklyn
... contributed 21 points, while Jose Calderon and Tim Hardaway Jr. had 17 and 14 points, respectively, in the loss. Jason Smith chipped in 12 and Lance Thomas scored 11 off the bench. The Celtics jumped on the Knicks early, taking a 26-19 lead after ... Comment? | 0.000837 |
.
A welcoming environment where anyone of any ability level can come to play a game of chess.
Location: Library
Time: Wednesday 2:45-3:45 p.m.
There are several levels of choir at Dakota. These classes meet during the school day and have concerts at DRHS as well at festivals around the state. For more information please visit our website.
Location: HR 108
Time: Please email Traci Nemechek for more information on time/dates of meetings.
The Student Voice
ESA is the broader CTSO (Career and Technical Student Organization) for technology students. It will include the broadcasting and computer programming students.
Location: ST 112
Time: Cardinal Mondays at 7:00 a.m.
Our club focuses on coordinating outreach initiatives in the community for kids to attend in order to find their passion in the environment and explore issues we can address once we begin to contact our state representatives.
Location: HR 141
Time: Tuesdays 2:45-3:30 p.m.
Fellowship of Christian Athletes club focuses on understanding how to mix being an athlete and Christian in the high school environment.
Location: HR 132
Time: Every Cardinal Thursday or Friday at 7:15 a.m.
Family, Career and Community Leaders of America club promotes personal growth and leadership development through Family and Consumer Sciences education. Focusing on the multiple roles of family member, wage earner and community leader, members develop skills for life through character development, creative and critical thinking, interpersonal communications, practical knowledge, and career preparation. For more information please visit our website.
Location: HR 110
Time: Cardinal Fridays from 7-8 a.m. as well as other days as needed
The Gamer's Club provides a place where students can go to be social and play games.
Location: B 217
Time: Every Wednesday from 2:45-4:45 pm
Have you been looking for a place to hang out with your Christian peers and strengthen your relationship with Jesus? Come to God Squad and join us for fellowship, Bible lessons, snacks, worship, and fun! You're welcome to come every week or just once in a while. We'd love to see you there!
Location: B 218
Time: Thursdays 2:45-4:00 purpose of the Women's Hiking Club is to share the common interest of hiking and the outdoors, to get to know new people and build connections among females at DRHS and to promote physical and mental health and self-confidence.
We plan to go on a hike approximately once every three weeks. Most hikes will be scheduled on weekdays after school and others may take place on weekend days. Check out the DRHS Women's Hiking Club webpage and/or see Ms. Wilson in the counseling office for more details and to be added to the Group Me account.
The Dakota Ridge Historical Society strives to create a strong community related around learning and understanding history. This club is for students who are interested in learning outside their regular history class and who strive for a better understanding of how the world once was.
Location: B 237
Time: Every first Cardinal day of the week at 7:15 a.m.
The instrumental music ensembles at Dakota Ridge High School! These classes meet during the school day and have concerts at DRHS as well as at festivals around the state. Please visit our website for more information.
We provide a safe and inclusive environment for anyone who wishes to learn the Latin language and its benefits.
Location: AR 210
Time: Cardinal Tuesdays 2:45!
MOD/SWOD is a senior group of leaders nominated by the staff, and their job is to fill the leadership role and mentor role within the halls of Dakota Ridge. Please contact the sponsors for more details if you are interested in MOD/SWOD.
Awareness leads to knowledge which is subtle but critical to let go of negative and stressful thoughts that restrict you from achieving your full potential. Our own negative thoughts work detrimentally against ourselves, however, you can practice the technique of meditation to let go of such emotions and focus on wise choices, that lead to a happier and more meaningful life! Join us on this journey every second cardinal day morning at 7:30!
Location: B 235 (Ms. Weikel's classroom)
Time: 7:30-8:00 a our sponsors..
Student Government is a class students are elected to or appointed by executives/sponsors which represents the student body. In this class students work collectively on producing activities for all students that leave a positive memory of high school.
Student Government Auxiliary - please speak with Keary Sullivan if you are interested in Student Government Auxiliary.
Student Voice is a district-wide initiative that empowers high school students from diverse backgrounds with leadership, cultural awareness, teamwork and advocacy skills. Jeffco Student Voice is a forum for students to collaborate with the Superintendent and other school, district and community leaders to create a sustainable movement fro social justice in our community through compassionate leadership.
Location: ST 222
Time: Every other Tuesday 3-4 pm. | 0.473319 |
Alabama
Week 1 Preview
The Crimson Tide men’s basketball team entered Southeastern Conference play with a 6-7 record. Although Alabama’s losses came against potential tournament teams such as UCLA, Oklahoma, and Wichita State, Anthony Grant’s team must have a stellar conference record to have a chance at making the NCAA tournament. Alabama picked up a 68-63 win over Vanderbilt to open the league schedule and travel to Athens to take on Georgia on Saturday (Jan. 11).
Auburn Week 1 Preview
After suffering double-digit losses to Northwestern State, Iowa State, and Illinois, the Auburn men’s basketball team finally started clicking on both ends of the court. Tony Barbee’s squad won huge home games over two ACC teams in Boston College and Clemson. The Tigers enter SEC play at 8-3, just win away from matching last season’s win total. Auburn opened the SEC season at Ole Miss on Thursday (Jan. 9) and return home on Saturday to take on Missouri.
SEC Men’s Basketball Rankings
1. Florida (12-2, 1-0)
2. Kentucky (11-3, 1-0)
3. Tennessee (10-4, 1-0)
4. Missouri (12-2, 0-1)
5. Ole Miss (9-4, 0-0)
6. LSU (9-4, 0-1)
7. Arkansas (11-3, 0-1)
8. Alabama (7-7, 1-0)
9. Texas A&M (10-4, 1-0)
10. Georgia (7-6, 1-0)
11. Auburn (8-3, 0-0)
12. Vanderbilt (8-5, 0-1)
13. Mississippi State (10-4, 0-1)
14. South Carolina (7-7, 0-1)
Week 1 Previews and Predictions
Saturday, Jan. 11
Missouri at Auburn. Missouri was 12-1 entering SEC play with wins over UCLA, West Virginia and N.C. State. However, the Tigers lost their conference home opener to Georgia. Missouri has a chance to get back on track against Auburn, who was picked to finish at the bottom of the SEC this season. Prediction: Missouri 73, Auburn 65.
LSU at South Carolina. Frank Martin’s Gamecocks are off to a rough start yet again. South Carolina struggled throughout much of last season, and this year hasn’t been any better as the Gamecocks were blown out by Florida to start league play. LSU lost its conference opener to Tennessee, and this game is a must-win for the Tigers if they want a chance to make the NCAA tournament. Prediction: LSU 63, South Carolina 58.
Florida at Arkansas. Right now, Billy Donovan’s Gators are the best team in the SEC. Florida is poised to make another deep run in the NCAA tournament. Arkansas is still having trouble winning on the road under Mike Anderson, but the Hogs are one of the best in the league at home. Prediction: Florida 77, Arkansas 75.
Kentucky at Vanderbilt. Kentucky might have the most talent in the country. However, the Wildcats are playing like an inexperienced team, although they made improvements since the beginning of the season. Vanderbilt had high expectations entering the season, but a plethora of injuries and off-the-court issues have hurt the Commodores tremendously thus far this year. Prediction: Kentucky 75, Vanderbilt 67.
Alabama at Georgia. At 7-7, Alabama cannot afford to lose many conference games this year. A loss to Georgia would be devastating for the Crimson Tide. Georgia upset Missouri on the road earlier in the week, just days after head coach Mark Fox’s father passed away. The Bulldogs have improved as the season has pro-gressed and are looking to move to 2-0 in league play on Saturday. Prediction: Alabama 56, Georgia 52.
Ole Miss at Mississippi State. Despite a couple of disappointing home losses, the Ole Miss Rebels still have a chance to be a tournament team by the end of the season. The defending conference champions have an opportunity to knock off their rival on the road, and the Rebels are seeking revenge after losing to the Bulldogs in Starkville a season ago. Prediction: Ole Miss 72, Mississippi State 66.
Texas A&M at Tennessee. The Aggies started conference play by handily defeating Arkansas, while the Volunteers knocked off LSU on the road. The Volunteers have a chance to contend with Florida and Kentucky for the top spot in the SEC but cannot afford to lose at home to Texas A&M. Prediction: Tennessee 70, Texas A&M 57. | 0.000868 |
PRX - Pieces for Format: Interview
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Open Orchard Productions
Series: The Harvest by Open Orchard Productions
Series: The Harvest by Open Orchard Productions
Did you know there are currently more cell phones in the world than toothbrushes? Meet Joselyn Miller, co-founder of Global Grins, a philanthropy t...
- Added: Oct 17, 2014
- Length: 04:12
Humankind
Series: Humankind Weekly
Series: Humankind Weekly
An American pediatric brain surgeon describes his remarkable recent journey to Africa and David Scheffer recalls his years of behind-the-scenes neg...
- Added: Sep 08, 2014
- Length: 58:59
Pod Academy
The term ‘ethnomusicology’ was coined in 1959 by Dutch academic, Jaap Kunst. Put simply, it is the social and cultural study of music. Jo Barratt ...
- Added: Jul 28, 2014
- Length: 21:33
Gareth Stack
Series: Mad Scientists of Music
Series: Mad Scientists of Music
This episode looks at how innovative new ways of making and distributing music are coming into conflict with our legal system. Some argue that copy...
- Added: Jul 08, 2014
- Length: 30:00
- Purchases: 1
The Bahamas is composed of more than 700 islands and is the nearest resort destination to the U.S. mainland after Cuba. It has a varied history.
- Added: Jun 07, 2014
- Length: 01:03:57
During World War II, some 400,000 captured German soldiers were shipped across the Atlantic to prison camps dotted across the U.S. Suddenly the ene...
Bought by WCPN, Spokane Public Radio, KQED, WSHU, KWGS and more
- Added: May 19, 2014
- Length: 54:00
- Purchases: 28
EU'sCapitalCitiesofCulture, 2014, include Riga,Latvia & Umea, Sweden. Both are known for Art,Music&Dance, Festivals,Historic Sites.
- Added: Apr 28, 2014
- Length: 01:00:05
Let's Travel! Radio
Cruising will feature a number of prominent executives who best know about this exploding market.
- Added: Mar 24, 2014
- Length: 01:01:57
Let's Travel! Radio explores how living, studying, teaching, praying in other cultures can expand not only one's own horizons but may also lead to ...
- Added: Jan 27, 2014
- Length: 02:04:14
VOCO sings folk music - but not the usual coffeehouse fare. It’s music that's steeped in generations of tradition, from far-off, tucked-away corner...
Bought by American Voices
- Added: Dec 02, 2013
- Length: 07:45
- Purchases: 1
Constance A. Dunn
Series: On the Couch...
Series: On the Couch...
Singer Djixx of the dance hall band Sevdah Baby talks to Constance A. Dunn about life, music and almonds.
- Added: Aug 13, 2013
- Length: 38:02
An enactment of the anonymous collective's words. Smetnjak wrote the answers, Constance A. Dunn interprets them. Smetnjak is an anonymous collecti...
- Added: Jul 08, 2013
- Length: 20:26
Carl Conradi, of the Romeo Dallaire Child Soldiers Initiative, considers the notion of 'bearing witness' as it relates to the Initiative's efforts ...
- Added: Jul 03, 2013
- Length: 10:05
Interview with Serbian musician Nemanja Kojic (Hornsman Coyote) by Constance A. Dunn. Interview from Coyote's living room in Belgrade.
- Added: Jul 03, 2013
- Length: 44:12
As Burma transitions from dictatorship to democracy, hundreds of political prisoners have been freed after decades behind bars. On this edition, we...
- Added: May 02, 2013
- Length: 29:00
A short piece about a long anticipated Cinco de Mayo party. As it turned out, almost everybody at the party, including the hosts thought they were ...
- Added: May 01, 2013
- Length: 04:12
In this edition of WTIP's ongoing historical series, Moments in Time, producer Carah Thomas learns about an ancient style of timber framing from so...
- Added: Apr 02, 2013
- Length: 07:03
- Purchases: 1
Open Orchard Productions
Series: The Elements by Open Orchard Productions
Series: The Elements by Open Orchard Productions
One amazing woman, Anila Ali, is transforming the stereotypes of Muslims in America through her work as a teacher, journalist, author, and activist...
- Added: Feb 26, 2013
- Length: 08:04
The music of and an in-studio interview with Celtic Thunder songwriter and vocalist Keith Harkin as he releases his first solo CD.
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Many American expats gathered in Brussels on 4th July this year to celebrate American Independence Day. Reporter, Alison Turner, joined the celebr...
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Guided tour through the creation and application of Belgium's communal administrative sanctions.
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Series: Peace Talks Radio - Series of Half-Hours
Series: Peace Talks Radio - Series of Half-Hours
Peace Talks Radio rebroadcasts excerpts from the 2011 program produced by Amy Mayer. Produced in recognition of the 50th anniversary of the Peace ...
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How a documentary about Rwanda's genocide is being used as an educational tool for the youth around the world.
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Bashora Galgalo is a 47 year old member of the Watha Tribe a minority indigenous hunter gather tribe from Kenya. He was born a hunterat Ndololo in ...
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For several of my posts, there’s been a very vigorous exchange about American foreign policy. Just recently, one of my readers, who espouses what I would call a conservative viewpoint, commented about “fundamental truths” common to all people. Another reader who, on foreign policy, is I think hostile to current Bush initiatives wrote this “I accept none of your ‘fundamental truths’, and disagree with the main point of every single one of your paragraphs. I don’t think there’s much point even arguing about them because we’d be at it for years and wouldn’t make any progress.”
I think that’s a great point. That is, buried under every single argument, no matter the layers of facts, all of us have animating beliefs that function as our starting points for argument. Regarding events in the Middle East (and, indeed, around the world), these are some of my fundamental beliefs, which affect everything I say and think:
I believe that, at this precise moment in time, American culture is more civilized and beneficent than Islamist culture. (You’ll notice I was careful not to say Muslim, Islamic, Persian or Arab. I’m comparing us to the radical Islamic movement cropping up in the new all over the world.) I also believe that American values, which are grounded in Judeo-Christian doctrine, are better than the multiculturalist values that currently dominate Europe and that allow, I believe, fertile soil for the most hate-filled Islamists and the recent resurgence of aggressive anti-Semitism.
I believe that Islamists have declared war on us, dozens of times in the past 20 years, with the loudest declaration occuring on 9/11. I believe we’d be suicidal idiots not to listen. I believe that there is no meaningful negotiation available with someone whose end goal is your death or total subjugation — which is why I think the peace movement naive and misguided.
I believe that Israel absolutely, historically, morally, legally, whatever, has the right to exist unmolested by her neighbors. I believe that her closest neighbors — the Palestinians — did not exist as a nation before 1948, and that the Arab nations in 1948 (Egypt, Jordan, Syria, etc), created the notion of a Palestianian nation state (when, before, there were merely Arab fellahins barely working land held by distant Arab overlords) to justify to the world the Arabs’ continued hostility to Israel; to play into the Marxist/Cold War desire for an imperialist enemy (Israel); and to detract their own citizens’ eyes from these nations’ overwhelming corruption and inequities.
Finally, I believe that I’ve got to get my kids up to go to the dentist, and I’ll follow with more later — if any more occurs tome. | 0.797789 |
TITLE: Mermin-Wagner and superconductivity
QUESTION [6 upvotes]: Why can superconductivity exist in 2D since Mermin-Wagner should forbit it? This question was asked here before, but I don't think anyone gave a satisfactory answer, so let me revisit it.
I have read both the rigorous proof of BCS/Hartree-Fock (1) and that of the quantum Mermin-Wagner (2). Of course, the two rigorous statements don't exactly exclude each other. However, on a conceptual level, it seems hard to reconcile. Indeed, superconductivity breaks $U(1)\cong SO(2)$ gauge symmetry at some finite temperature $T>0$ for all spatial dimension $d\ge 1$, but this is in principle forbidden by Mermin-Wagner since it doesn't allow any continuous symmetry breaking at finite temperatures for $d\le 2$.
There's also this explanation, but I don't quite understand its answer. Mathematically speaking, a 3D system that is translationally invariant in 1 dimension is equivalent to a 2D system so I don't quite understand why it matters. Although anyons are the correct way to describe statistics in a 2D system, the mathematical formalism of fermions/bosons do exist (not to mention that fermions/bosons are also types of anyons). Finally, the only way for the $U(1)$-symmetry to not have long range order (the phase of the complex order parameter $\psi$) seems to be that the amplitude $|\psi|=0$, so superconductivity indeed seems to be a long-range order phase transition (correct me if I'm wrong).
Bach, V., Lieb, E.H. & Solovej, J.P. Generalized Hartree-Fock theory and the Hubbard model. J Stat Phys 76, 3–89 (1994). https://doi.org/10.1007/BF02188656
http://www.scholarpedia.org/article/Mermin-Wagner_Theorem
REPLY [0 votes]: Leggett does not mention superconductivity, where the situation is even worse than for crystal order. The sample would have to be larger than the observable universe for long-wavelength fluctuations to have an effect on the SC Tc in 2D. See our paper Physical limitations of the Hohenberg–Mermin–Wagner theorem, also available on https://arxiv.org/abs/2107.09714. | 0.018794 |
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FLOW STATE
is an optimal state of consciousness where you Feel and Perform your BEST.
It has many names – the Zone, the Groove, etc – but the feeling and benefits are always the same.
When you’re in Flow, whether it’s work-related or play related:
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- You get flashes of insight and innovation that seem to come from out of the blue that afterwards seem to be obvious and you often wonder why you hadn’t thought of them earlier.
Many people have experienced Flow State but they can’t control when it Happens
Playing a musical instrument, creating art, participating in high-performance sports, practicing yoga or mediation or even just out walking alone in nature – these are all ways that people typically tap into Flow State.
Most people have a very difficult time getting into Flow State at will – they just tend to fall into it ‘when the stars align perfectly’.
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When you use your Personal Colors to tap into Flow, you get clarity about any areas in your life where you you may be feeling blocked or stuck. And you start to get unstuck. Fast.
Getting into Creative Flow on a regular basis has also been scientifically shown to dramatically improve your physical and mental health and reduce stress, anxiety and depression.
Learn Leanne’s 8 Simple Steps for Tapping into Flow State
One of Leanne Venier’s international award-winning, healing Flow State paintings.
SCIENTIFIC RESEARCH
shows that being able to still your mind and get into a right-brain, meditative FLOW state on a regular basis:
- Allows your natural creativity to surface and flow which can be then channeled into all projects, activities or relationships in your life
- Increases work productivity by 500% McKinsey did a 10 year study on top executives and found that when they’re operating in Flow State, they’re 500% more productive. Work on Monday and get a full week of work completed!
- Dramatically reduces stress and depression
- Allows you to tap into your natural healing abilities (See Dr. Deepak Chopra quote below)
- Slows down the aging process by reducing stress levels and triggering healing responses within the body
- Improves concentration, self esteem and overall health (See Dr. Mihaly Cziksentmihalyi’s quote below)
- Improves your quality of life, sleep and overall health by restoring the body-mind natural balance (See Dr. Deepak Chopra quote below)
- Helps you achieve Peak Performance and Ultimate Happiness
More Scientific Research & Quotes about Flow & Meditative Right Brain States
How they Impact Creativity, Peak Productivity, Innovation and Optimal Health & Healing …
“It has been found that people who experience a lot of flow in their daily lives also develop other positive traits, such as high concentration, high self-esteem, and even greater health
– Mihaly Cziksentmihalyi, Psychologist & originator of the term ‘Flow’:
“Almost every day, new scientific research is confirming what the ancient sages knew thousands of years ago: The regular practice of meditation and meditative activities (like Flow State) lowers blood pressure, reduces cholesterol, strengthens immune function, relieves insomnia, decreases the production of stress hormones such as cortisol and glucagon (making it easier to lose weight), and restores the body and mind’s natural balance.”
– Dr. Deepak Chopra, M.D.
On the Medical Science & Healing benefits of Meditative/Flow states:
“When studying the brainwaves of meditating monks, Dr. Richard Davidson, director of the Laboratory for Affective Neuroscience at the University of Wisconsin, found that brain circuitry is different in long-time meditators (or people who practice meditative activities).”
– Dr. Andrew Weil, M.D.
On the Neuroscience of rewiring the brain for Self-healing using Meditative/Flow states
How Does Color Tie in with Triggering Flow States?
Color has been heavily researched and is now known to have dramatic beneficial effects on your body and psyche.
When you use your Personal Colors in a particular way, one of the many positive benefits is that they can catapult you into this heightened state of consciousness, Flow State, where healing begins, clarity abounds and your creativity and productivity soars.
Discover your Personal Colors and What they Say about You:
Read more about the Scientific Research about the Healing Effects of Color and Light HERE.
It all boils down to how our bodies take in and use color and light.
Harvard Medical School Research Study – Flow State IMPROVES your Brain
Mindfulness practice leads to increases in regional brain gray matter density
A new study has found that participating in an eight-week meditation or mindfulness (aka Flow State) training program for just 27 minutes per day, can have measurable effects on how the brain functions even when someone is not actively meditating.
.”
“This is the first time that meditation training has been shown to affect emotional processing in the brain outside of a meditative state.”.”
Requirements for Team Flow – McKinsey 20 year study on Flow State
.”
McKinsey Quarterly Article – On the Psychological Science of Using Flow for Peak Performance, Increased Productivity & Greater Job & Life Satisfaction
‘flow’. Athletes describe the same feeling as ‘being in the zone’; musicians talk about ‘being in the groove’.
More important, Csìkszentmihàlyi found that individuals who frequently experienced it (flow).Flow sounds great in theory, but few business leaders have mastered the skill of generating it reliably in the workplace…
McKINSEY’S QUESTION: … Can employees in the workplace experience similar performance peaks and, if so, what can top management do to encourage the mental state that brings them about?”
ANSWER:
YES, they can – and so can you!
When you attend Leanne Venier’s Creative Flow Training you’ll learn to quickly and reliably enter Flow whenever you’d like ……
for Peak Productivity, Optimal Health, Enhanced Creativity, Heightened Performance, and increased Job and Life satisfaction … in addition to physical & emotional healing and mind-body wellness benefits that being in regular Flow brings. You’ll learn to use your Personal Colors for triggering Flow State whenever you’d like.
In Leanne’s Creative Flow Training you’ll also learn to use your Personal Colors to clearly understand what is going on in your unconscious mind and psyche that may be holding you back from attaining true Joy, Fulfillment, Satisfaction and Abundance in all areas of your personal and work life as well as in your relationships.
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Good points: hot water for bath and shower, bed comfortable, lovely view from deck, great tasting morning coffee. Not so good points: thin towels and face cloths, dated inside and out, very thin walls (can hear everything in adjacent rooms), small toilet area with an even smaller shower, musty smell from old carpet. The owners were friendly and accommodating.
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\subsection{The case when $|X|\geq 3$ and $X$ is an independent set}\label{sec:indset}
We start handling the general case of $|X| \geq 3$. In this section, we
prove the upper bound for the case when $X$ is an independent set.
\begin{theorem}\label{thm:independent}
Let $G(V,E)$ be a graph and $X\subseteq V$ be a set of vertices such that
$|X|=d\geq 3$ and $G[V\setminus X]$ is a disjoint union of cliques.
If $X$ is an independent set, then $\chicf{} \leq d+1$.
\end{theorem}
In order to show the above theorem, we first prove
Lemma \ref{lem:D2C_mod_indep_case1}, where we handle the case when every vertex
in $V \setminus X$ has at most one neighbor in $X$. After this, we
prove Lemma \ref{lem:D2C_mod_indep_case2}, where there is a vertex $v \in V\setminus X$
that has at least two neighbors in $X$. The proof of Lemma \ref{lem:D2C_mod_indep_case2}
uses Lemma \ref{lem:isolated_in_X}, which also serves as the completion phase for all the remaining cases (including those where $X$ is not an independent set).
\begin{lemma}\label{lem:D2C_mod_indep_case1}
Let $G=(V,E)$ be a graph and $X\subseteq V$ be a set of vertices such that
$|X|=d\geq 3$ and $G[V\setminus X]$ is a disjoint union of cliques.
If $X$ is an independent set and
every vertex
in $V\setminus X$ has at most one neighbor in $X$,
then $\chicf{} \leq d+1$.
\end{lemma}
\begin{proof}
We explain how to assign $C: V \rightarrow [d+1]$ such that
$C$ is a CFON coloring of $G$.
Let $X = \{v_1, v_2, \dots, v_d\}$.
\noindent\textbf{Initial phase:} For each $v_i\in X$, assign $C(v_i)=d+1$.
For each $v_i\in X$, choose an arbitrary neighbor $w_i\in V \setminus X$
and assign $C(w_i)=i$. We get that $\ucn(v_i)=w_i$.
Now, each vertex in $X$ is colored and has a uniquely colored neighbor.
\noindent\textbf{Completion phase:} Each uncolored singleton clique in $G[V\setminus X]$ is assigned the color
$d+1$. Note that all the singleton cliques have exactly one neighbor in $X$,
and this neighbor is the uniquely colored neighbor.
What remains to be addressed are cliques of size at least 2.
\begin{itemize}
\item \textbf{Clique $K \subseteq G[V\setminus X]$ with at least two colored vertices.}
Color the uncolored vertices with the color $d+1$.
Let $v,v' \in K$ be two of the vertices that were colored prior
to this step. Hence it follows that $C(v), C(v') \in [d]$ and
$C(v) \neq C(v')$.
Since $\deg_X(y)\leq 1$ for all $y \in G[V \setminus X]$, one of
$v$ and $v'$ will be the uniquely colored neighbor
of all vertices in $K$.
\item \textbf{Clique $K \subseteq G[V\setminus X]$ with exactly 1 colored vertex $v$}. Let $C(v)=j$ and hence $v_j\in N(v)$.
\begin{itemize}
\item If $|K|=2$. Let $K=\{v,v'\}$.
\begin{itemize}
\item If
$v_j\notin N(v')$,
we assign $C(v')=j$.
We get that $\ucn(v)=v'$ and $\ucn(v')=v$.
\item Else, we have $N(v)\cap X = N(v')\cap X = \{v_j\}$.
We assign $C(v')$ arbitrarily from $[d]\setminus \{j\}$.
We get that $\ucn(v)=v'$ and $\ucn(v')=v$.
\end{itemize}
\item Else if $|K|\geq 3$.
Let $v'$ be arbitrarily chosen from $K \setminus \{v\}$.
Since $|X| \geq 3$ and $\deg_X(v')\leq 1$, there exists a vertex $v_{\ell} \in X$, $v_{\ell} \neq v_j$ such that $v_{\ell} \notin
N(v') \cap X$.
Assign $C(v') = \ell$ and
color the rest of the vertices in $K\setminus \{v,v'\}$
with the color $d+1$.
We get that for all vertices $w\in K$, either $v$ or $v'$ is a uniquely colored neighbor.
\end{itemize}
\item \textbf{Clique $K \subseteq G[V\setminus X]$ with no colored vertices}. We first select two vertices $v, v' \in K$.
Since $\deg_X(v)\leq 1$, we can choose $j \in [d]$ such that
$v_j \notin N(v) \cap X$. Since $deg_X(v')\leq 1$ and $|X| \geq 3$,
we can choose $\ell \in [d]$ such that $\ell \neq j$ and $v_{\ell}
\notin N(v') \cap X$.
Assign $C(v) = j$, $C(v') = \ell$ and the rest of the vertices in
$K\setminus \{v,v'\}$ the color $d+1$. For all vertices $w\in K$, either $v$ or $v'$ is a uniquely colored neighbor.
\end{itemize}
\end{proof}
We first state Lemma \ref{lem:isolated_in_X}, which will serve as the completion phase
for almost all the remaining cases (even for those where $X$ is not an independent set).
This lemma states that the graph can be \cf{} colored provided it has been
partially colored satisfying certain rules.
\begin{lemma}\label{lem:isolated_in_X}
Let $G = (V,E)$ be a graph with $X = \{v_1, v_2, \ldots, v_d\} \subseteq V$ such that $d\geq 3$
and $G[V \setminus X]$ is a disjoint union of cliques. Further,
$Y = \{v_i \in X: \deg_X(v_i) \geq 1\}$ and $C: V \rightarrow [d+1]$
be a partial coloring that satisfies the below rules.
Then, $C$ can be extended to
a CFON coloring $\widehat C: V \rightarrow [d+1]$ of all the vertices in $V$.
\end{lemma}
\begin{mybox}{Rules}
\begin{enumerate}[topsep=0pt,itemsep=-0.5ex,partopsep=1ex,parsep=1ex, label=(\roman*)]
\item For all $v_i\in X$, $C(v_i)=i$.
\item For some number of cliques $K$ in $G[V \setminus X]$,
all the vertices in $K$ are colored using colors from $[d+1]$. The remaining cliques are uncolored.
\item All the vertices in
$Y$ and all the colored vertices in $V\setminus X$ have a uniquely colored neighbor.
Moreover, some vertices in $X \setminus Y$ have a uniquely colored neighbor.
\item The uniquely colored neighbor is identified for all the vertices in $X$ whose
entire neighborhood is colored.
\item There exists $1\leq f \leq d$, such that (a) $v_f \in X$ has a uniquely colored
neighbor and (b) for each vertex in $X$, the color $f$ is not the unique color in its neighborhood.
We refer to $f$ as the \emph{free color}.
\item If a vertex $v_i\in X\setminus Y$ does not have a uniquely colored neighbor,
then the color $i$ is not assigned to
any vertex in $V\setminus X$ yet.
\end{enumerate}
\end{mybox}
Before proving Lemma \ref{lem:isolated_in_X}, we prove the upper bound when
$X$ is an independent set and there is a vertex $v \in V \setminus X$ that has at
least two neighbors in $X$.
\begin{lemma}\label{lem:D2C_mod_indep_case2}
Let $G(V,E)$ be a graph and $X\subseteq V$ be a set of vertices such that
$|X|=d\geq 3$ and $G[V\setminus X]$ is a disjoint union of cliques.
If $X$ is an independent set and
there exists a vertex $v\in V\setminus X$, such that
$\deg_X(v)\geq 2$, then $\chicf{} \leq d+1$.
\end{lemma}
\begin{proof}
The goal here is to partially color some vertices of $G$
so that the rules of Lemma \ref{lem:isolated_in_X} are satisfied.
We then use Lemma \ref{lem:isolated_in_X} to extend the partial coloring and obtain a CFON coloring of $G$.
Let $X = \{v_1, v_2, \dots, v_d\}$.
We explain how to assign $C: V \rightarrow [d+1]$ such that
$C$ is a partial coloring that satisfies the rules of Lemma \ref{lem:isolated_in_X}.
For each vertex $v_i\in X$, we assign a distinct color $C(v_i)=i$.
There are two cases depending on the neighborhood of
vertices in $V\setminus X$.
\begin{itemize}
\item \textbf{There exists a singleton clique $K=\{v\}$ such that
$\deg_X(v)\geq 2$}.
Let $N(v)\cap X = \{ v_{i_1}, v_{i_2}, \dots , v_{i_m} \}$, with $m\geq 2$.
We assign $C(v)=i_1$.
We get that $\ucn(v)=v_{i_1}$ and for all $1 \leq j \leq m$, $\ucn(v_{i_j})=v$.
The color $i_2$ will not be the unique color of any
vertex in $X$ and will be the free color.
\item \textbf{All singleton cliques have degree equal to 1.}
By assumption, there is
a vertex $v\in V\setminus X$, such that
$\deg_X(v)\geq 2$. It follows that $v \in K$ where
$K$ is a clique in $G[V \setminus X]$ and
$|K|\geq 2$.
Let $N(v)\cap X = \{v_{i_1}, v_{i_2}, \dots , v_{i_m}\}$, with $m\geq 2$.
Now we have two cases depending on whether there exists vertices in $X$
whose neighbors belong to only $K$.
We refer to these vertices as $S_K$. Formally, $S_K = \{
v_i \in X : N(v_i) \subseteq K\} \setminus N(v)$.
The vertices in $N(v) \cap X$ rely on $v$ for their uniquely colored
neighbor and hence does not require special attention.
\begin{itemize}
\item $S_K \neq \emptyset$.
First, we assign $C(v)=i_1$ and choose $i_2$ as the free color.
For each vertex $v_j\in S_K$, we choose a vertex $w_j \in N(v_j)$.
Note that by the definition of $S_K$, it follows that $w_j \in K$.
We assign $C(w_j)=j$ if it is not already assigned ($w_j$ could have been the
chosen neighbor for some other vertex $v_{j'}$ as well). Now all the
vertices in $S_K$ have a uniquely colored neighbor.
\begin{itemize}
\item If all vertices in $K$ are colored because of the above coloring,
every vertex in
$K$ has a uniquely colored neighbor.
We get that $\ucn(v) = v_{i_1}$.
Each vertex $w\in K\setminus \{v\}$ is assigned a distinct color, say $j$, because it is adjacent to
$v_j \in S_K$, which serves as its uniquely colored neighbor.
\item There exists at least one uncolored vertex $K$.
\begin{itemize}
\item If there exists a uniquely colored neighbor for each uncolored vertex
in $K$, assign the color $d+1$ to all the uncolored vertices.
We get that $\ucn(v) = v_{i_1}$. The vertices $w_j \in K \setminus \{v\}$
rely on the corresponding $v_j$'s as mentioned above.
\item Else, let $v' \in K$ be an uncolored vertex that does not
see a uniquely colored neighbor.
This implies that
$N(v') \cap \{v_{i_1}, v_{i_2}, \cdots, v_{i_m}\} = \{v_{i_1}\}$.
We reassign $C(v)=i_2$, assign $C(v')=d+1$, and
designate $i_1$ as the free color instead of $i_2$.
We assign the color $i_1$ to the remaining uncolored vertices in $K$.
We have $\ucn(v')=v$ and $\ucn(w)=v'$, for all
$w\in K\setminus \{v'\}$.
\end{itemize}
\end{itemize}
\item $S_K= \emptyset$.
This implies that,
there is no vertex in $X\setminus \{v_{i_1}, v_{i_2}, \cdots, v_{i_m}\}$ that was relying on $K$ for its uniquely colored neighbor.
We first assign $C(v)=i_1$ and choose $i_2$ as the free color.
We have the following cases.
\begin{itemize}
\item There exists a vertex $v'\in K\setminus \{v\}$
such that
$v_{i_1}\notin N(v')$.
Assign $C(v')=d+1$ and assign the remaining vertices of $K\setminus \{v,v'\}$
the color $i_2$.
For every vertex $w\in K\setminus \{v'\}$, $\ucn(x)=v'$.
Finally, we have $\ucn(v')=v$.
\item
Else, for every vertex $w \in K$, we have $v_{i_1} \in N(w) \cap X$.
Reassign $C(v) = i_2$ and assign the color $d+1$ to all the
vertices in $K \setminus \{v\}$.
The color $i_1$ is the redesignated free color.
And for each $w\in K$, $\ucn(w)=v_{i_1}$.
\end{itemize}
\end{itemize}
\end{itemize}
Now $C$ is a partial color assignment satisfying all the conditions in the
rules of Lemma \ref{lem:isolated_in_X}.
Hence by Lemma \ref{lem:isolated_in_X}, we can extend $C$ to a full CFON coloring
of $G$ that uses at most $d+1$ colors.
\end{proof}
We conclude this section with the proof of Lemma \ref{lem:isolated_in_X}.
\begin{proof}[Proof of Lemma \ref{lem:isolated_in_X}]
For each colored vertex $w\in G$, $\widehat C (w) =C(w)$.
We explain how to extend $\widehat C: V \rightarrow [d+1]$
to all vertices
such that
$\widehat C$ is a CFON coloring of $G$.
Let $X = \{v_1, v_2, \dots, v_d\}$ and $Y = \{v_i \in X : \mbox{deg}_X (v_i)\geq 1\}$.
\noindent \textbf{Process to identify uniquely colored neighbors for $X\setminus Y$:} For every $v_j\in X\setminus Y $,
that does not have a uniquely colored neighbor,
choose an
uncolored neighbor of $v_j$, say $w_j\in V\setminus X$
and assign $\widehat C(w_j)=j$.
Rules (iv) and (vi) of Lemma \ref{lem:isolated_in_X} allow us to do this.
Since $v_j$ does not have a uniquely colored neighbor, rule (iv) implies that
$v_j$ has an uncolored neighbor, and as per rule (vi), no vertex in
$V\setminus X$ is assigned the color $j$.
\noindent \textbf{Observation:} It is possible that all the neighbors of $v_j\in X\setminus Y$ may be colored
by the above coloring process on other vertices $v_i \in X \setminus Y$
even before applying the process on $v_j$.
In such a case, we choose an arbitrary neighbor of $v_j$
that was already colored by this process and
assign it as the uniquely colored neighbor for $v_j$.
This neighbor
acts as the uniquely colored neighbor
for at least 2 vertices in $X\setminus Y$.
This fact will be useful later.
Now, every vertex in $X\setminus Y$ has a uniquely colored
neighbor. We now look at the previously uncolored cliques $K$ in $G[V\setminus X]$.
For each such clique $K$, we color $K$ as per the applicable case below.
\begin{comment}
\begin{enumerate}
\item As a result of the above rule, all the neighbors of $v_j\in X\setminus Y$ may be colored even before applying the rule on $v_j$.
We choose a neighbor that was colored by this rule and
treat it as the uniquely colored neighbor for $v_j$.
This means that a vertex in $K$ acts as the uniquely colored neighbor
for at least 2 vertices in $X\setminus Y$. We keep in mind this fact when
$|K|\geq 2$ and has at least 2 colored vertices.
Now, every vertex in $X\setminus Y$ will have a uniquely colored
neighbor.
\item While coloring a clique $K$ which has at least 2 colored vertices,
(because of the above rule),
we keep in mind the vertices that were relying on
this clique for their
uniquely colored neighbor.
Note that this is only for cliques $K$ of size at least 2.
\end{enumerate}
\end{comment}
\noindent
\textbf{Case 1: $K$ has no colored vertices}
\begin{itemize}[topsep=-0.25pt,itemsep=-0.5ex,partopsep=1ex,parsep=1ex]
\item $|K|=1$. Let $K = \{w\}$.
We assign $\widehat C(w) = d+1$. As all the
neighbors of $w$ are distinctly colored, we assign one of the neighbors
as $\ucn(w)$.
\item $|K|\geq 2$. We have two subcases here.
\begin{itemize}
\item There exists a vertex $w\in K$, such that
$N(w)\cap X= \emptyset$.
Choose another vertex $w'\in K\setminus \{w\}$
such that $N(w') \cap X \neq \emptyset$.
We have two subcases.
\begin{itemize}
\item $N(w') \cap X = \{ v_f\}$, where $v_f\in X$ is the vertex that corresponds to the free color $f$.
Assign $\widehat C(w') = d+1$ and $\widehat C(w) = c$, where
$c \in [d] \setminus \{f\}$, chosen arbitrarily.
For all the vertices (if any) $x \in K \setminus \{w, w'\}$,
assign $\widehat C(x) = f$.
We have $\ucn(w') = w$ and for all vertices $x\in K\setminus \{w'\}$, we have $\ucn(x)=w'$.
\item There exists a vertex $v_i \in N(w') \cap X$, where $v_i \neq v_f$.
Assign $\widehat C(w')=d+1$.
For all the vertices $x \in K \setminus \{w'\}$, assign $\widehat C(x) = f$.
We have $\ucn(w') = v_i$ and for all vertices $x\in K\setminus \{w'\}$, we have $\ucn(x)=w'$.
\end{itemize}
\item For all $w \in K$, $N(w) \cap X \neq \emptyset$.
Assign all the vertices in $K$ the color $d+1$.
For all the vertices in $K$, we assign one of the
neighbors in $X$ as the respective uniquely colored neighbor.
\end{itemize}
\end{itemize}
\noindent
\textbf{Case 2: $K$ has exactly one colored vertex}
WLOG, let $v\in K$ be such that $\widehat C(v)=j$. This implies that
$v_j\in N(v)\cap X$.
\begin{itemize}[topsep=-0.25pt,itemsep=-0.5ex,partopsep=1ex,parsep=1ex]
\item $|K|=1$.
We have $\ucn(v)=v_j$ and $\ucn(v_j) = v$ (as was already assigned).
\item $|K|=2$. Let $K=\{v,v'\}$.
\begin{itemize}
\item $N(v')\cap X=\emptyset$.
We assign
$\widehat C(v')=d+1$.
We get that $\ucn(v)=v'$, $\ucn(v')=v$ and $\ucn(v_j)=v$.
\item $v'$ has a neighbor other than $v_j$ in $X$.
That is, $\exists v_k\in N(v')\cap X$, with $v_k \neq v_j$.
We assign $\widehat C(v')=d+1$.
We get that $\ucn(v)=v'$, $\ucn(v')=v_k$ and $\ucn(v_j)=v$.
\item $N(v') \cap X = \{v_j\}$.
\begin{itemize}
\item There exists a vertex
$v_\ell\in X\setminus Y$, $v_{\ell} \neq v_j$
such that $\ucn(v_{\ell}) = v$.
We reassign $\widehat C(v)=\ell$ and $\widehat C(v')=d+1$.
We have that $\ucn(v)=v'$ and $\ucn(v')=v$.
Note that $\ucn(v_j)= \ucn(v_\ell)=v$, as before.
\item No vertex in $X$ other than $v_j$ sees $v$ as its uniquely colored neighbor and
$v$ has another neighbor in $X$ besides $v_j$, say $v_k$.
We reassign $\widehat C(v)=d+1$ and assign $\widehat C(v')=j$.
We get that $\ucn(v')=v$, $\ucn(v)=v_k$ and we reassign $\ucn(v_j)=v'$.
\item $N(v) \cap X = N(v') \cap X = \{v_j\}$.
We reassign $\widehat C(v)$ to an arbitrarily chosen value from $[d]\setminus\{j, f\}$.
Note that such a value exists since $d \geq 3$.
We assign $\widehat C(v')=d+1$. We get that $\ucn(v)=v'$,
$\ucn(v')=v$ and $\ucn(v_j)=v$, as before.
\end{itemize}
\end{itemize}
\item $|K|\geq3$.
\begin{itemize}
\item There exists a vertex $v'\in K\setminus \{v\}$ such that
$v_j\notin N(v')$.
We assign
$\widehat C(v')=d+1$ and the vertices in $K \setminus \{v, v'\}$ are colored with the free color $f$.
We get that $\ucn(v')=v$ and for all $w\in K\setminus \{v'\}$, $\ucn(w)=v'$.
As before, $\ucn(v_j)=v$.
\item Every vertex in $K$ is adjacent to $v_j$.
\begin{itemize}
\item There exists a vertex $v'\in K\setminus \{v\}$ such that
$(N(v') \cap X) \setminus~\{v_j, v_f\} \neq \emptyset$.
Let $v_k \in N(v') \cap X$, where $v_k \neq v_j$ and $v_k \neq v_f$.
Assign
$\widehat C(v')=d+1$ and rest of the vertices in $K\setminus \{v,v'\}$ are
colored with the free color $f$.
We get that $\ucn(v')=v_k$ and for all $w\in K\setminus \{v'\}$, $\ucn(w)=v'$. As before, $\ucn(v_j)=v$.
\item For all $w \in K\setminus \{v\}$, $N(w) \cap X \subseteq \{v_j, v_f\}$.
We choose $k \in [d]\setminus\{j, f\}$ arbitrarily.
Note that such a value exists since $d \geq 3$.
Choose two vertices $v', v''$ arbitrarily from $K\setminus \{v\}$ and assign $\widehat C(v')=d+1$, $\widehat C(v'')=k$
and the remaining vertices in $K\setminus \{v,v', v''\}$ are colored with $f$.
We get that $\ucn(v')=v''$, and for all $w\in K\setminus \{v'\}$,
$\ucn(w)=v'$. As before $\ucn(v_j)=v$.
\end{itemize}
\end{itemize}
\end{itemize}
\noindent
\textbf{Case 3: $K$ has at least two colored vertices and there exists a
vertex in $K$ that is a uniquely colored neighbor for
at least two vertices in $X\setminus Y$}
Let $v'\in K$ be such that $\ucn(v_j) = \ucn(v_k) = v'$, where $v_j, v_k \in X \setminus Y$. The vertex $v'$ was colored in the process to identify
uniquely colored neighbors for vertices in $X\setminus Y$. WLOG, we may assume that
$v'$ was colored when assigning a unique colored neighbor for $v_j$. That is,
$\widehat C(v') = j$. This also implies that no vertices in $K$ are colored $k$.
There are two cases here.
\begin{itemize}[topsep=-0.25pt,itemsep=-0.5ex,partopsep=1ex,parsep=1ex]
\item All vertices in $K$ are colored.
In this case, every vertex in $K$ will have a uniquely colored neighbor.
This is because every vertex in $K$ would have been assigned a distinct color.
If $w \in K$ is such that $\widehat C(w) = \ell$, then $\ucn(w) = v_{\ell} \in X \setminus Y$.
\item There exists an uncolored vertex $v\in K$. There are two subcases.
\begin{itemize}
\item $v$ is adjacent to $v_k$.
We assign $\widehat C(v)=d+1$ and the remaining uncolored vertices
in $K$ (if any) are assigned the free color $f$.
We get that $\ucn(v)=v_k$, and for all $w\in K\setminus \{v\}$, $\ucn(w)=v$.
Note that $\ucn(v_j)= \ucn(v_k) = v'$, as before.
\item $v$ is not adjacent to $v_k$.
Reassign $\widehat C(v')=k$, assign $\widehat C(v)=d+1$ and the remaining uncolored
vertices in $K$ (if any) are assigned the free color $f$.
We get that $\ucn(v)=v'$, and for all $w\in K\setminus \{v\}$,
$\ucn(w)=v$. Note that $\ucn(v_j)= \ucn(v_k) = v'$, as before.
\end{itemize}
\end{itemize}
\noindent \textbf{Case 4: $K$ has at least two colored vertices and every colored vertex in $K$ is the uniquely colored neighbor for exactly one vertex in $X\setminus Y$}
Let
$v,v'\in K$ be two colored vertices
such that $\widehat C(v)=j$ and $\widehat C(v')=k$.
The colors $j$ and $k$ are assigned because they are adjacent to $v_j$ and $v_k$ respectively, where $v_j, v_k \in X\setminus Y$.
We have cases depending on the neighborhood of $K$.
\begin{itemize}[topsep=-0.25pt,itemsep=-0.5ex,partopsep=1ex,parsep=1ex]
\item There exists a colored vertex in $K$ that is adjacent to both $v_j$
and $v_k$.
\begin{itemize}
\item At least one of $v$ or $v'$ is adjacent to
both $v_j$ and $v_k$.
WLOG let that vertex be $v$.
Reassign $\widehat C(v')=d+1$ and assign the color $f$ to
the remaining uncolored vertices (if any).
We have that $\ucn(v')=v_k$, and for all vertices $w\in K\setminus \{v'\}$, $\ucn(w)=v'$. We reassign $\ucn(v_k) = v$, while $\ucn(v_j) = v$ as before.
\item There exists a colored vertex $v'' \in K\setminus \{v,v'\}$ such that
$\{v_j,v_k\} \subseteq N(v'')$.
Let $\widehat C(v'') = \ell$ because it was the chosen neighbor for
$v_{\ell} \in X \setminus Y$ in the coloring process stated in the beginning
of this proof.
We reassign $\widehat C(v)=d+1$ and $\widehat C(v')=f$.
It is important to note that because of the case definition, this reassignment
does not affect the uniquely colored neighbors of vertices in $X \setminus (Y \cup \{v_j, v_k\})$.
The remaining uncolored vertices in $K$ (if any) are assigned the color $f$.
We have $\ucn(v)=v_j$, and for every vertex $w\in K\setminus \{v\}$, $\ucn(w)=v$.
We reassign $\ucn(v_j)=\ucn(v_k) = v''$, while $\ucn(v_{\ell}) = v''$ as before.
\end{itemize}
\item There exists an uncolored vertex $v''\in K$ such that $\{v_j,v_k\}
\subseteq N(v'')$.
We reassign $\widehat C(v)=d+1$,
$\widehat C(v')=f$ and
assign $\widehat C(v'')=k$.
The remaining uncolored vertices in $K$ (if any) are assigned the color $f$.
We have $\ucn(v)=v_j$, and for every vertex $w\in K\setminus \{v\}$, $\ucn(w)=v$.
We reassign $\ucn(v_j)=\ucn(v_k) = v''$.
\item No vertex in $K$ is adjacent to both $v_j$ and $v_k$.
Assign the color $f$ to the remaining uncolored vertices (if any).
Every vertex in $K\setminus\{v,v'\}$ will see either $v$ or $v'$ as its uniquely colored neighbor.
Also $\ucn(v)=v_j$ and $\ucn(v')=v_k$, while $\ucn(v_j)=v$ and $\ucn(v_k)=v'$, as before.
\end{itemize}
\end{proof} | 0.004299 |
Become a Developer/9.2
(Sorry for the inconvenience)
If you like programming, and especially coding on FreeBSD, we would love to see you join the txt=Missing Link[1] as a PC-BSD® committer. Developers who want to help improve the PC-BSD® codebase are always welcome! If you would like to participate in core development, subscribe to the developers mailing list[2]. Once you have signed up, feel free to browse the active TODO list, or search for bugs that need fixing in the PC-BSD® Trac database[3]. | 0.009441 |
I used to be able to obtain the last 5 tle's for an object from OIG by telnet'ing to "oig1.gsfc.nasa.gov", login as OIG with password of "goddard". Does the OIG web site support obtaining the last most recent tles by a single identifier? If not, where can I obtain them easily/readily? Also, does anyone have current logon info to get on to the OIG telnet site? Thanks.....in advance. Clear skies...Bill e-mail: bkrosney@richgreen.com bkrosney@mbnet.mb.ca | 0.032973 |
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Strasburg's debut up there with the biggest
Strasburg's debut up there with the biggest
In fact, it was such a big deal, bunting lined the railings at the Metrodome, and there were pregame ceremonies, introductions on the base lines and everything.
Oh, sure, the fact it was Opening Day -- not just the ascension of one of the game's best prospects -- had something to do with the pomp and circumstance surrounding Mauer's debut on April 5, 2004.
Still, Mauer knows a little bit about what Stephen Strasburg is going through heading into his first Major League game on Tuesday.
"It was a little different for me because Opening Day's kind of crazy whether you're a rookie or a veteran," Mauer said. "I'm sure it's all going to seem kind of crazy for him. It's an exciting time, and you know a lot of people are looking forward to it."
No doubt, when the Nationals' 21-year-old pitching phenom steps to the mound in front of a packed house at Nationals Park to make one of the most heralded debuts in years, it will be a big deal, and then some. It's a debut that in terms of attention-wrenching anticipation ranks with those of Ken Griffey Jr. in 1989, Mark Prior in 2002 ... and Mauer, whose promise was recognized well beyond the Twin Cities, just as Strasburg's is outside the Beltway.
In a way, Mauer might have had it easier in that all the festivities surrounding Opening Day kept the attention from being focused directly on his performance. Still, it was a big day.
"I was nervous, that's for sure," Mauer said. "Obviously, for me growing up there, it's like hearing the same song, everything was so familiar for me. And I realized that I wasn't there sitting and watching, I was actually playing."
That's a storyline that played out again this year with Jason Heyward, the Braves rookie whose three-run homer in his first at-bat on Opening Day showed he'd arrived on his hometown team's roster just in time to shine.
The much-ballyhooed premiere is a storyline that has played out for decades in baseball. It's part of the fabric of the game -- the big prospect taking his first steps into the Majors.
There was the ultimate debut, by Jackie Robinson on April 15, 1947, and others that resonated through time like Joe Nuxhall's with the Reds at age 15 -- or 60 years before he'd retire from broadcasting Reds games -- and those of sure-fire greats like Willie Mays and Mickey Mantle. More recently, future stars like Dwight Gooden and Pudge Rodriguez came in young and with a lot of fanfare.
Indeed, many of today's superstars' arrivals were celebrated, but to different degrees.
Starting pitchers like Justin Verlander, Tim Lincecum and David Price have entered the Major Leagues within a short time after being drafted, arriving with some serious hype of their own, albeit none quite rivaling Strasburg's on a national level. Then there was the case of Daisuke Matsuzaka, the Japanese right-hander whose arrival in Boston was the biggest in that city since Nomar Garciaparra in 1996 or perhaps Roger Clemens in 1984, and was much more of an international phenomenon.
"I think that the expectations placed on [Strasburg] are even higher than what was placed on me," Matsuzaka said through a translator to The Associated Press last week. "But the fact that expectation exists means that there's talent there."
Whether that talent translates to long-term success is another matter entirely, especially with starting pitchers.
Perhaps one of the splashiest debuts ever was that of David Clyde of the Rangers, a No. 1 overall pick from a Texas high school whose debut ... well, it couldn't be as well described as it was by the late Ron Fimrite in Sports Illustrated:
"It remained for the mayor of Arlington, Texas, himself to put the Big Event in historical perspective. 'From now on,' said the Hon. Tom Vandergriff, 'time here shall be marked from June 27, 1973.'
"It is a date that shall live in infancy, for on this night in Arlington Stadium, David Clyde, a stripling of 18, began his Major League baseball career by pitching the Texas Rangers to a 4-3 win over the Minnesota Twins."
Clyde threw 112 pitches, struck out eight and walked seven in five innings, allowing just one hit, a "startling performance for a youngster only 19 days out of Westchester High School in Houston," as Fimrite wrote. But it wound up being Clyde's greatest highlight. Clyde finished his Major League career with an 18-33 record and 4.63 ERA over parts of five seasons.
Other top picks like Ben McDonald of the Orioles rose quickly but didn't quite live up to the hype over the long haul.
And then there's Prior, who lived up to his billing in a big way from his very first start -- striking out 10 and setting Chicago into a tizzy. Even in comparison to the Kerry Wood debut four years before Prior's, this was a big deal in Wrigleyville.
"We get to see the No. 1 guy who everybody's been waiting for," then-slugger Sammy Sosa said before Prior's first outing.
Prior delivered on that promise until arm problems derailed his career, but his debut stands out as a superb start to a career.
Maybe it's the nature of the beast, but not every position player came in with the attention showered on Mauer, even if their debuts were eagerly awaited. In recent years, debuts from the likes of Evan Longoria with the Rays in 2008 and Matt Wieters of the Orioles last year turned some heads, but didn't shake up the game like, say, Dice-K.
Albert Pujols' debut obviously was anticipated as he'd rocketed up the Cardinals chain by age 21, but he was on the bubble to make the 2001 Opening Day roster for the Cards. He started as more of a role player -- knocking that notion out of the park quickly, mind you. Ryan Howard made a pretty big splash in 2004 as a September call-up, but his first two homers were as a pinch-hitter.
Before that, Seattle was home to two debuts that were among the most memorable of their era, with Junior becoming a household name at age 19 on Opening Day 1989 and Alex Rodriguez making his debut at age 18 as a skinny shortstop barely a year out of Westminster Christian High in Miami.
"I was young. It was a long time ago," Rodriguez recalled Sunday before he took the field for the Yankees. "I was nervous. I was 18 years old, right out of prom, right after my high school graduation. It was definitely a wet-behind-the-ears thing. Yeah, real nervous."
And then there was Mauer, back in 2004. You'd say little did they know what was to come from that day, but everybody knew what would come: Gold Glove, Silver Slugger and MVP hardware.
Naturally, he was 2-for-3 with two walks in his debut, while batting (ahem) eighth. He also tagged out the potential tying run for the final out in the 11th inning. Unfortunately, he suffered a knee injury the next day and was limited to 35 games in his rookie season, but the rest has been a history of success in Minnesota.
And like Strasburg, he had only one debut, and it'll be with him the rest of his career.
"There was a lot of excitement," Mauer says in an understated tone, "and it's something I'll never forget."
A-Rod's advice to Strasburg? "Enjoy it," he said.
John Schlegel is a reporter for MLB.com. Tim Britton contributed. This story was not subject to the approval of Major League Baseball or its clubs.
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Red Hook, Brooklyn, residents Carly Yates and Mark Chin started a group called No Toxic Red Hook. They along with other residents of the neighborhood are saying "No" to the EPA's proposal to clean up the toxic Gowanus Canal, which runs through their community as well as Carroll Gardens and Boerum Hill.
But folks who live in Red Hook, a lower income community with a large minority population, feel that the EPA's plan puts them more at risk.
Part of the cleanup plan involves turning the toxic sludge in to a concrete substance and dump it in to a steel-enclosed container in New York Harbor. There would also be a treatment plant to squeeze the water out of the contaminated sludge. Both of these structures would be built right next to ball fields, a swimming pool, and a massive complex of low-income housing.
Fox 5 reached out to the EPA for comment, but the agency did not get back to us. In recent months, EPA has said the plan is "completely safe or we wouldn't have proposed it."
Residents in Red Hook will have to wait until the end of the summer to see if the EPA listened to their concerns. That is when the EPA is scheduled to announce its final plans for the project. | 0.89869 |
Music Composers Unite!
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I am thrilled you listened to my song. I think it would fun to develop this song. I listened with earphones and discovered it's sound might need a bass boost and it needs some effects to be added to improve the sound.
Yes, the techno parts are loud. I agree.
Happy composing! :)
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From January onwards, you won’t have to sleep with one eye open while travelling in a train. The Western Railway authorities have decided that either the August Kranti Rajdhani Express or the Mumbai Rajdhani will get CCTVs installed in their coaches next year.
According to The Indian Express, the advanced surveillance cameras will be installed for monitoring and surveillance purposes.
“All coaches of either August Kranti or Rajdhani Express will be fitted with a camera within three months. This could assure better security to passengers in the train,” AK Gupta, general manager, WR, told the publication.
This comes after several cases of theft were reported on both the trains, on the Mumbai-Delhi route this year.
Udai Shukla, chief security commissioner, Railway Protection Force, WR, told The Indian Express, “We will install them in each coach of the train. There is a need for more than one camera in the coach as surveillance can be enhanced.”
Installing these CCTV cameras will also eradicate the need for personal monitoring by the RPF staff, who have been manning each trip of the Rajdhani trains.
So far, they’ve had to question travellers and railway officials in the train and rely on their answers to solve theft cases. The idea is to get rid of the interrogation process altogether so that no railway employee is affected by it. | 0.367082 |
Author, television producer and breast cancer survivor Geralyn Lucas will be the keynote speaker at Emanuel Medical Center’s 14th annual Women’s Cancer Awareness event..
American Cancer Society statistics show an estimated 513,000 cancer survivors are between the ages of 20 to 39 years. The mortality rate for patients in that age range has dropped 19 percent for men and 15 percent for women since 1999.
Lucas’ memoir delves into her daily dealings with the diagnosis, both the triumphs and defeats, while mixing in witty observations and healthy doses of inspiration.
“When I was diagnosed with breast cancer, I promised that I would share my story with other women to help take away the fear of breast cancer and encourage them to get checked,” Lucas wrote when the movie based on her book came out. “Although it is ‘my’ story, it is all of ours. Please share your story with your family and friends. The more we talk about it, the more we can take away the fear. And remember, live up to your lipstick!”
The Women’s Cancer Awareness Event will be held Oct. procession of cancer survivors will begin at 7 p.m. with the performance events to follow.
Tickets are free, but must be reserved in advance. Reserve tickets by e-mailing your name, mailing address and number of tickets needed to cancercenter@emanuelmed.org, by phoning Emanuel Cancer Center at 664-2434, or by texting “TICKETS” to 88788.
To contact Sabra Stafford, e-mail sstafford@turlockjournal.com or call 634-9141 ext. 2002.
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This is getting tougher and tougher each week. Writing about a defense that really has no hope of performing better unless they get a lot better really quickly. Unfortunately, it looks like this unit is stuck here and the improvements we see from here on will be marginal.
The Tampa Bay Buccaneers are coming off of a bye week after the Chicago Bears beat them 48-10, and they hope to get back to their 27+ points per game magic this week. Lucky for them, they play the Falcons defense which is giving up just over 30 points a game.
Can Atlanta keep the Bucs offense down for one more week? Let’s take a look.
In the trenches
There is very little desirable play along either units’ line of scrimmage. For Atlanta, struggles rushing the quarterback and getting running backs to the ground have plagued this group all season long. In Tampa, the league’s worst rushing defense and questions at quarterback linger over an incredibly hot start to the season.
Atlanta is incredibly depleted and has struggled mightily thus far this season, but I am going to put some faith in Takkarist McKinley and Vic Beasley getting to a familiar face in Jameis Winston. I don’t think it happens routinely but I do think it happens enough to make him uncomfortable. Donovan Smith and Demar Dotson have been solid in pass protection so far but I think Takk comes out a little pissed off after the performance last week.
In terms of running the ball, this is truly the battle of the below-average. Tampa will be able to have more success than they are used to on the ground because, well—Falcons. With Grady Jarrett still out due to his sprained ankle, the middle of the defense is vulnerable and while the Bucs’ interior offensive line has kind of sucked this season, Atlanta has a knack for helping teams get back on track.
Bottom line, each team will perform better than they expect in one area or another, but neither will have an all-around solid performance.
Advantage: Push
Skill positions
Typically, I feel better about the Falcons matchups in the back end of the defense than I do up front, however, I think this group of wideouts and tight ends may be trouble for the Atlanta secondary. Mike Evans always gives Desmond Trufant all that he can handle and I think Desean Jackson is explosive enough to hurt Robert Alford deep a time or two.
This is really going to come down to whether or not the Falcons can force the Bucs into some third-and-longs and force Winston to throw under some semblance of pressure. If Atlanta gets beat deep or gets caught in third-and-short situations, they will struggle against the skill players.
The Bucs don’t have any game-breakers at running back or tight end but it doesn’t take that much to impact a game against the Falcons defense. Peyton Barber and Cameron Brate are plenty capable of giving the Falcons linebackers and safeties issues—especially if they are on an island with little help.
Advantage: Tampa Bay
Conclusion
I wish I had more faith in this group and that they were playing better, however it is pretty clear that we are going to have to watch this group struggle week after week this season. There is potential for the Falcons to show flashes—like I expect Takk and Vic to do this week—but there is very little hope in the Atlanta defense truly stopping Jameis and the rest of the Bucs’ offense.
Advantage: Tampa Bay
Let’s hope the offense can do their job. | 0.000844 |
NetID: How to Log onto the HCC Network
You will need to use the default password when registering your password with the NetID Password Portal (). The default password is: Uppercase first
letter of their first name + Lowercase first letter of their last name + the
entire seven digit student number. For
example, John Doe (0123456) would have an initial password of Jd0123456. Your password will most likely NOT be the same as any of your other HCC passwords. Once you have registered with the NetID Password
Portal, change your password to something you will remember that others can't figure out.
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\begin{document}
\maketitle
\begin{abstract}
Consider a smooth manifold $M$. Let $G$ be a compact Lie group which acts on $M$ with cohomogeneity one. Let $Q$ be a singular orbit for this action. We study the gradient Ricci soliton equation $\Hess(u)+\Ric(g)+\frac{\epsilon}{2}g=0$ around $Q$. We show that there always exists a solution on a tubular neighbourhood of $Q$ for any prescribed $G$-invariant metric $g_Q$ and shape operator $L_Q$, provided that the following technical assumption is satisfied: if $P=G/K$ is the principal orbit for this action, the $K$-representations on the normal and tangent spaces to $Q$ have no common sub-representations. We also show that the initial data are not enough to ensure uniqueness of the solution, providing examples to explain this indeterminacy. This work generalises the papaer "The initial value problem for cohomogeneity one Einstein metrics" of 2000 by J.-H. Eschenburg and McKenzie Y. Wang to the gradient Ricci solitons case.
\end{abstract}
\small \textbf{Keywords}: Ricci solitons, cohomogeneity one manifolds
\small \textbf{MSC}: Differential geometry
\section{Introduction}
A \emph{Ricci soliton} is a triple $(M^n,\widehat{g},\widehat{X})$, where $(M,\widehat{g})$ is a complete Riemannian manifold of dimension $n$ and $\widehat{X}$ is a vector field on $M$, which satisfies the equation
\begin{equation}\label{eqn:0.2}
\Ric(\widehat{g})+\frac{1}{2}\mathcal L_{\widehat{X}}\widehat{g}+\frac{\epsilon}{2}\widehat{g}=0,
\end{equation}
where $\epsilon$ is some real constant and $\Ric(\widehat{g})$ is the Ricci tensor of the metric $\widehat{g}$.
Equation \eqref{eqn:0.2} may be written also as
\begin{equation}\label{eqn:0.3}
\Ric(\widehat{g})+\delta^*\widehat{\omega}+\frac{\epsilon}{2}\widehat{g}=0,
\end{equation}
where $\widehat{\omega}=\widehat{X}^\flat$ and $\delta^*$ is the symmetrized covariant derivative. If the dual form of $X$ is exact, i.e. if there exists a smooth function $u$ on $M$ such that $\widehat{X}^\flat=du$, the Ricci soliton is called a \emph{gradient Ricci soliton} and equation \eqref{eqn:0.3} takes the following form
\begin{equation}\label{eqn:0.4}
\Ric(\widehat{g})+\Hess(u)+\frac{\epsilon}{2}\widehat{g}=0,
\end{equation}
where $\Hess(u)$ is the Hessian of $u$.
Ricci solitons are important for many reasons. Firstly, they generate particular solutions to the Ricci flow equation. In fact, considering the one-parameter family of vector fields on $M$ given by
\begin{equation*}
X(t)=\frac{\widehat{X}}{1+\epsilon t},
\end{equation*}
and integrating it to a one-parameter family $\varphi(t)$ of diffeomorphisms of $M$, we have that the one-parameter family of Riemannian metrics
\begin{equation*}
g(t)=(1+\epsilon t)\varphi(t)^*\widehat{g}
\end{equation*}
evolves under the Ricci flow equation
\begin{equation*}
\begin{cases}
&\frac{\partial g(t)}{\partial t}=-2\Ric(g(t)),\\
&g(0)=\widehat{g},
\end{cases}
\end{equation*}
for $t\in[0,T)$, where $T\in(0,+\infty)$. Secondly, the notion of gradient Ricci soliton has motivated the discovery of monotonicity formulas for the Ricci flow (see \cite{Perelman:1}), which have many geometric applications. Thirdly, Ricci solitons often appear as limits of dilations of singularities in the Ricci flow (see \cite{Hamilton:1}). Finally, they are generalisations of Einstein metrics. In fact, if we take $\widehat{X}$ to be the zero vector field in \eqref{eqn:0.2} or if we take $u$ to be constant in \eqref{eqn:0.4}, we obtain the Einstein condition for the metric $\widehat{g}$, which is
\begin{equation*}
\Ric(\widehat{g})+\frac{\epsilon}{2}\widehat{g}=0.
\end{equation*}
Therefore, it is natural to ask whether techniques used to produce Einstein metrics can also produce Ricci solitons. For example, we may consider spaces with a certain amount of symmetry. By \cite{Petersen-Wylie:1}, we have that the maximal amount of symmetry on a nontrivial (i.e. non Einstein) gradient Ricci soliton is given by a cohomogeneity one action. This means that the symmetry group of the manifold acts on it in such a way that the orbits with maximal dimension, which are called \emph{principal orbits}, have codimension one. In this context, orbits with lower dimension are called \emph{singular orbits} or \emph{special orbits}. With this kind of action, the orbit space is one-dimensional and can be an interval (open, closed or semi-open), $\mathbb R$ itself or the circle $S^1$, depending on the number of singular orbits (cf. \cite{Bergery:1}).
Many examples of cohomogeneity one gradient solitons have been constructed; e.g., the Bryant soliton (see \cite{Ivey:1}, \cite{Chow:1}), the cigar soliton (\cite{Hamilton:3}), the $U(n)$-symmetric soliton on $\mathbb C^n$ discovered by Cao (\cite{Cao:2}, \cite{Cao:3}) and the K\"{a}hler generalisations \cite{Koiso:1}, \cite{Yang:1}, \cite{Dancer-Wang:2}, \cite{Eminenti-Mantegazza-LaNave:1} and \cite{Feldman-Ilmanen-Knopf:1}.
In \cite{Wang-Eschenburg:1}, the authors consider local existence and uniqueness of a smooth $G$-invariant Einstein metric around a singular orbit in the cohomogeneity one setting. They prove, under a technical assumption, that, given any $G$-invariant Riemannian metric and any shape operator in a neighbourhood of a singular orbit $Q$, there always exists an invariant Einstein metric around $Q$ with any prescribed sign of the Einstein constant. We generalise the Einstein case to the case of gradient Ricci solitons. In particular, we study the existence and uniqueness of $G$-invariant gradient Ricci solitons around the singular orbit $Q$. Our main theorem is the following.
\begin{theorem}
Let $G$ be a compact Lie group acting on a connected Riemannian manifold $(M,\widehat{g})$ with cohomogeneity one and by isometries of $\widehat{g}$. Let $Q=G/H$ be a singular orbit of codimension $k+1$, $k\geq1$. So, $H$ acts linearly with cohomogeneity one on $V=\mathbb R^{k+1}$ and the Lie algebra of $G$ splits as $\mathfrak g=\mathfrak h\oplus\mathfrak p_-$, where $\mathfrak h$ is the Lie algebra of $H$. Let $v_0\in S^k$ have isotropy group $K$ with respect to the $H$-action. Then, $G/K$ is a principal orbit for the action.
Assume that $V$ and $\mathfrak p_-$ have no irreducible common factors as $K$- representations. Then, given any $G$-invariant metric $g_Q$ on $Q$ and any shape operator $L_1:NQ\to\Sym^2(T^*Q)$, where $NQ=G\times_H V$ is the normal bundle over $Q$, there exists a $G$-invariant gradient Ricci soliton on some open disk bundle of $NQ$.
\end{theorem}
We write the Ricci soliton condition around $Q$ as an initial value problem with initial data given by a $G$-invariant metric and shape operator $L_1$ around $Q$. We note that the smoothness condition of the metric implies that $\tr{(L_1)}=0$, so $Q$ must be a minimal submanifold in $M$. As we are working around a singular orbit, we only need to solve the gradient Ricci soliton equation in the directions tangent and orthogonal to the orbits. Moreover, we can write the initial value problem as a system of ordinary non-linear differential equations with a singular point at the origin. We solve this system using the same technique as in \cite{Wang-Eschenburg:1}, which consists of applying the method of asymptotic series to find a solution and then showing that this solution is in fact a smooth $G$-invariant gradient Ricci soliton. We can always show existence, but the initial data given are not sufficient to ensure uniqueness and the indeterminacy of the problem, which is the same as in the Einstein case, is always finite and can be computed using representation theory. Finally, the technical assumption about the irreducible summands of $V$ and $\mathfrak p_-$ is motivated by \cite{Wang-Eschenburg:1} and, as the authors explain (cf. Remark 2.7 of \cite{Wang-Eschenburg:1}), it appears quite natural in the context of the Kaluza-Klein construction.
\section{The cohomogeneity one Ricci soliton equation}
In this section, following the approach and notation of \cite{Wang-Eschenburg:1} and \cite{Dancer-Wang:2}, we recall the Ricci soliton equation in the cohomogeneity one setting.
Let $(M,\widehat{g})$ be a connected Riemannian manifold of dimension $n+1$ and let $G$ be a compact Lie group which acts on $M$ by isometries and with cohomogeneity one.
Now, let us choose a unit speed geodesic
\begin{equation*}
\gamma:I\longrightarrow M,
\end{equation*}
where $I\subset\mathbb R$ is an open interval, and take $\gamma$ such that it intersects all the principal orbits orthogonally. Then, it is possible to define an equivariant diffeomorphism
\begin{equation*}
\begin{aligned}
\Phi:\,&I\times G/K\longrightarrow M_0\subset M,\\
&(t,g\cdot K)\longmapsto g\cdot\gamma(t),
\end{aligned}
\end{equation*}
where $K$ is the isotropy group of $\gamma(t)$ with respect to the $G$-action. Hence, $\Phi(t,G/K)$ is the principal orbit $P_t$ passing through $\gamma(t)$ and $M_0$ is an open dense subset in $M$ which is the union of all the principal orbits.
Every orbit $P_t$ is naturally equipped with a $G$-invariant Riemannian metric, which depends on $t$. Hence, through $\Phi$ we obtain a family $g(t)$ of $G$-invariant metrics on the homogeneous space $P$, where $P$ denotes an abstract copy of the principal orbit $G/K$. Moreover, the map $\Phi$ sends $I\times\{p\}$, with $p\in P$, to the geodesic through $p$ orthogonal to the principal orbits, and we have that the canonical parametrisation of $I$ corresponds to the arclength parametrisation of the geodesic. Then, if we pullback the metric $\widehat{g}$ on $M$ through $\Phi$ we get
\begin{equation*}
\Phi^*(\widehat{g})=d t^2+g_t,
\end{equation*}
where $g_t$ is a family of $G$-invariant Riemannian metrics on $P$.
Let $\widehat{\nabla}$ and $\widehat{\Ric}$ denote the Levi-Civita connection and the Ricci tensor of the manifold $(M,\widehat{g})$, respectively. Let $\nabla_t$ and $\Ric_t$ denote the Levi-Civita connection and the Ricci tensor of $(P_t,g_t)$, respectively. Let $L_t$ be the shape operator on $P_t$ defined by
\begin{equation*}
L_t(X)=\widehat{\nabla}_XN,
\end{equation*}
where $X$ is a vector field on $P_t$ and $N=\Phi_*(\frac{\partial}{\partial t})$ is a unit normal $G$-invariant vector field along $P_t$ such that $\widehat{\nabla}_NN=0$. We then have a family $(L_t)_{t\in I}$ of $G$-invariant, $g_t$-symmetric endomorphisms of the tangent space of $P$. In particular, the trace of $L_t$ is constant along $P_t$. Moreover, the following equality holds
\begin{equation}\label{eqn:2.1}
\dot{g}_t(X,Y)=2g_t(L_t(X),Y),
\end{equation}
for every pair of vector fields $X,Y$ on $P_t$ and for every $t\in I$.
Consider now the Ricci soliton equation for $(M,\widehat{g},\widehat{\omega})$
\begin{equation}\label{eqn:2.6}
\widehat{\Ric}(\widehat{g})+\widehat{\delta}^*\widehat{\omega}+\frac{\epsilon}{2}\widehat{g}=0.
\end{equation}
First of all, we can take the form $\widehat{\omega}$ to be $G$-invariant, or, if we are dealing with gradient Ricci solitons, we can take the potential function to be $G$-invariant (cf. \cite{Dancer-Wang:2}, p. 4). Hence, if we consider the pull-back of $\widehat{\omega}$ through $\Phi$, we obtain
\begin{equation}\label{omega}
\Phi^*\widehat{\omega}=\xi(t)d t+\omega_t,
\end{equation}
where $\xi=\xi(t)$ is a function on $I$ and $\omega_t$ is a one-parameter family of $G$-invariant one-forms on $P$.
Let $\mathcal X(P_t)$ denote all the vector fields on $P_t$. We then have the following proposition.
\begin{proposition}[\cite{Dancer-Wang:2}]
Let $(M^{n+1},\widehat{g})$ be a connected Riemannian manifold which admits a cohomogeneity one action by a compact Lie group $G$ of isometries of $\widehat{g}$. Let $\widehat{\omega}$ be a $G$-invariant one-form on $M$. Under the parametrisation induced by a unit speed geodesic orthogonal to the principal orbits, the Ricci soliton equation for $\widehat{g}$ and the vector field dual to $\widehat{\omega}$ is given by
\begin{align*}
&-(\delta^{\nabla_t}L_t)^\flat-d(\tr(L_t))+\frac{1}{2}\dot{\omega}_t-\omega_t\circ L_t=0,\\
&-\tr(\dot{L}_t)-\tr(L^2_t)+\dot{\xi}(t)+\frac{\epsilon}{2}=0,\\
\begin{split}
&\Ric_t(X,Y)-\tr(L_t)g_t(L_t(X),Y)-g_t(\dot{L}_t(X),Y)\\
&+\xi(t)g_t(L_t(X),Y)+\delta^*_t\omega_t(X,Y)+\frac{\epsilon}{2}g_t(X,Y)=0,
\end{split}
\end{align*}
for all $X,Y\in\mathcal X(P_t)$ and $t\in I$, where, viewing $L_t$ as an endomorphism of $\mathcal X(P_t)$, the operator $\delta^{\nabla_t}:\mathcal X^*(P_t)\otimes\mathcal X(P_t)\to\mathcal X(P_t)$ is the codifferential.
Conversely, if $g_t$ and $\omega_t$ are one-parameter families of metrics and one-forms on $P_t$, respectively, and $\xi=\xi(t)$ is a smooth function on $I$ such that the above system is satisfied with $L_t$ defined by $\dot{g}_t(X,Y)=2g_t(L_t(X),Y)$ for all $X,Y\in\mathcal X(P_t)$, then $\widehat{g}=d t^2+g_t$ and $\widehat{\omega}=\xi(t)d t+\omega_t$ give a local Ricci soliton on $M_0$.
\end{proposition}
If we are looking for gradient Ricci solitons, that is when there exists a $G$-invariant smooth function $u$ such that $\widehat{\omega}=d u$, equation \eqref{omega} becomes
\begin{equation*}
\Phi^*\widehat{\omega}=\dot{u}(t)d t,
\end{equation*}
and the Ricci soliton equation in the cohomogeneity one setting is equivalent to the following system
\begin{align}
\label{eqn:2.11} &-(\delta^{\nabla_t}L_t)^\flat-d(\tr(L_t))=0,\\
\label{eqn:2.12} &-\tr(\dot{L}_t)-\tr(L^2_t)+\ddot{u}(t)+\frac{\epsilon}{2}=0,\\
\label{eqn:2.13}\begin{split}
&\Ric_t(X,Y)-\tr(L_t)g_t(L_t(X),Y)-g_t(\dot{L}_t(X),Y)+\dot{u}(t)g_t(L_t(X),Y)\\
&+\frac{\epsilon}{2}g_t(X,Y)=0,\end{split}
\end{align}
for all $X,Y\in \mathcal X(P_t)$ and $t\in I$, where $u(t)(p)=u\circ\Phi(t,p)$, for all $p\in P_t$.
\section{Smoothness of tensors around a singular orbit}\label{section 3}
In this section, following \cite{Wang-Eschenburg:1}, we will discuss briefly the smoothness criterion for the metric $\widehat{g}$ and the one-form $\widehat{\omega}$, in the case when there is a special orbit.
Let $(M,\widehat{g})$ be a connected $(n+1)$-dimensional Riemannian manifold and $G$ a compact Lie group which acts on $M$ by isometries of $\widehat{g}$ and with cohomogeneity one. Let $Q=G\cdot q$ be a singular orbit of codimension $k+1$, with $k\geq1$, with isotropy group $H=G_q$. As $Q=G/H$ is a homogeneous space, the Lie algebra $\mathfrak g$ of $G$ decomposes in the following way:
\begin{equation}\label{eqn:3.1}
\mathfrak g=\mathfrak h\oplus\mathfrak p_-,
\end{equation}
where $\mathfrak h$ is the Lie algebra of $H$ and $\mathfrak p_-$ is the $\Ad(H)$-invariant complement of $\mathfrak h$ in $\mathfrak g$, which can be identified with the tangent space $T_qQ$.
Let $V=T_qM/T_qQ\backsimeq\mathbb R^{k+1}$ be the normal space at $q$ of $Q$, on which $H$ acts linearly with cohomogeneity one, i.e. it acts transitively on the sphere $S^k=H/K$, where $K\subset H$. We can identify a tubular neighbourhood of $Q$ with the total space of the normal bundle $NQ=G\times_HV$ of $Q$. We have that
\begin{equation*}
T(NQ)_{|_V}=V\times(V\oplus\mathfrak p_-).
\end{equation*}
In fact, the tangent space to $NQ$ splits into orthogonal and vertical parts:
\begin{equation*}
T(NQ)=\pi^*NQ\oplus\pi^*TQ,
\end{equation*}
where $\pi:NQ\rightarrow Q$ is the bundle map, and the two pull-back bundles are trivial on $V$, which can be viewed as the fibre of $NQ$ over $q\in Q$. Hence, a smooth $G$-invariant symmetric bilinear form $a$ is determined by an $H$-equivariant smooth map
\begin{equation*}
a:V\longrightarrow \Sym^2(V\oplus\mathfrak p_-).
\end{equation*}
Let $W$ be the vector space of all smooth $H$-equivariant maps $L:S^k\to\Sym^2(V\oplus\mathfrak p_-)$. We then have that, for $v_0\in S^k$, the evaluation map
\begin{equation*}
\begin{aligned}
\text{ev}:\,&W\longrightarrow \Sym^2(V\oplus\mathfrak p_-)^K,\\
&L\longmapsto\text{ev}(L)=L(v_0),
\end{aligned}
\end{equation*}
is a linear isomorphism. Here, $\Sym^2(V\oplus\mathfrak p_-)^K$ denotes the elements of $\Sym^2(V\oplus\mathfrak p_-)$ which are $K$-invariant, where $K=H_{v_0}$. Let $W_m$ be the subspace of $W$ consisting of all maps which are restrictions to $S^k$ of $H$-equivariant homogeneous polynomials of degree $m$. We then have a necessary and sufficient condition for $a$ to be smooth.
\begin{lemma}[\cite{Wang-Eschenburg:1}]\label{Lemma:3.1}
Let $t\mapsto a_t$, where $a_t: S^k\to\Sym^2(V\oplus\mathfrak p_-)^K$ for all $t\in[0,\infty)$, be a smooth curve, i.e. at zero the right-hand derivatives of all orders exist and are continuous from the right. Let $\sum_p{a_pt^p}$ be its Taylor expansion at zero. Then the map $a$ defined by
\begin{equation*}
\begin{aligned}
a:V&\backslash\{0\}\longrightarrow\Sym^2(V\oplus\mathfrak p_-),\\
&v\longmapsto a(v)=a_{|v|}\left(\frac{v}{|v|}\right)
\end{aligned}
\end{equation*}
can be extended smoothly at zero if and only if $a_p\in\text{ev}(W_p)$ for all $p\geq0$.
\end{lemma}
Motivated by \cite{Wang-Eschenburg:1}, we now assume that the representations of $K$ on $\mathfrak p_-$ and $V$ have no irreducible common factors. As a consequence of this, we have that
\begin{equation}\label{assumption}
\Sym^2(V\oplus\mathfrak p_-)^K=\Sym^2(V)^K\oplus \Sym^2(\mathfrak p_-)^K,
\end{equation}
and each $W_m$ splits as $W_m^+\oplus W_m^-$, where the polynomials in $W_m^+$ take values in $\Sym^2(V)$ and the ones in $W_m^-$ take values in $\Sym^2(\mathfrak p_-)$.
The smoothness criterion for $\widehat{\omega}$ is obtained essentially in the same way as for the metric $\widehat{g}$. Under the above assumption, we have that on a tubular neighbourhood around $Q$, $\widehat{\omega}$ is determined by an $H$-equivariant map
\begin{equation*}
\widehat{\omega}:V\longrightarrow V^*\oplus\mathfrak p_-^*.
\end{equation*}
In this case, $W_m$ is defined as the space of $H$-equivariant maps $L:V\to V^*\oplus\mathfrak p_-^*$ which are restrictions to the unit sphere $S^k$ of homogeneous polynomials of degree $m$. The necessary and sufficient condition for $\widehat{\omega}$ of the form \eqref{omega} to be smooth is that its pth Taylor coefficient, viewing $\widehat{\omega}$ as function of $t$, lives in $\text{ev}(W_p)$, for all $p\geq0$.
Finally, if we consider gradient Ricci solitons with potential function $u$, in the case of a special orbit the smoothness criterion for $\widehat{\omega}=d u$ implies that $u(t)$ must be even in $t$. In fact, around zero, $u$ is given by
\begin{equation}\label{eqn:3.3}
u(t)=\sum_{t=0}^\infty{\frac{u_p}{p!}t^p},
\end{equation}
where
\begin{equation*}
u_p=\frac{d^p}{dt^p}u(t)\Big\vert_{t=0}.
\end{equation*}
The smoothness condition implies that $u_p$ must be a homogeneous polynomial of degree $p$ on the sphere $S^k$, on which $H$ acts transitively. Moreover, we can take $u$ to be $H$-invariant. We now show that $u_p=0$ if $p$ is odd. Given $x\in S^k$, there exists $h\in H$ such that $h\cdot x=-x\in S^k$. If $p$ is odd we have that
\begin{equation*}
-u_p(x)=u_p(-x)=u_p(h\cdot x)=u_p(x)\Longrightarrow u_p(x)=0.
\end{equation*}
Hence, if $p$ is odd, $u_p(x)=0$, for all $x\in S^k$. This implies that $u(t)$ given by \eqref{eqn:3.3} is even in t.
\section{Initial value problem for gradient Ricci solitons around a singular orbit}\label{sec:4}
First of all note that by Proposition 2.17 of \cite{Dancer-Wang:2}, if we are looking for gradient Ricci solitons in the case when there is a singular orbit, instead of considering the system \eqref{eqn:2.11}-\eqref{eqn:2.13}, we can consider the following system
\begin{align}
\label{eqn:4.1} &\frac{d^3}{d t^3}u(t)+\tr(L_t)\ddot{u}(t)+\tr(\dot{L}_t)\dot{u}(t)-2\ddot{u}(t)\dot{u}(t)-\epsilon\dot{u}(t)=0\\
\label{eqn:4.2} \begin{split}
&\Ric_t(X,Y)-\tr(L_t)g_t(L_t(X),Y)-g_t(\dot{L}_t(X),Y)+\dot{u}(t)g_t(L_t(X),Y)\\
&+\frac{\epsilon}{2}g_t(X,Y)=0,
\end{split}
\end{align}
together with \eqref{eqn:2.1}. Note that equation \eqref{eqn:4.1}, which can be viewed as an equation in $\dot{u}(t)$, is the first integral which arises from the contracted second Bianchi identity and was observed in \cite{Ivey:1} (p. 242) and more generally in \cite{Cao:1} (p. 123) and \cite{Hamilton:1} (pp. 84-85). Moreover, the smoothness condition on the function $u$ implies that
\begin{equation*}
\dot{u}(0)=0.
\end{equation*}
Using the Ricci endomorphisms $r_t$ on $P_t$, defined by
\begin{equation*}
\Ric_t(X,Y)=g_t(r_t(X),Y),
\end{equation*}
for all $X,Y\in\mathcal X(P_t)$, equation \eqref{eqn:4.2} becomes
\begin{equation}\label{eqn:4.3}
r_t-\tr(L_t)L_t-\dot{L}_t+\dot{u}(t)L_t+\frac{\epsilon}{2}\MI=0,
\end{equation}
where $\MI$ is the identity matrix.
Let $\mathfrak h$ in \eqref{eqn:3.1} decompose in the following way
\begin{equation*}
\mathfrak h=\mathfrak k\oplus\mathfrak p_+,
\end{equation*}
and let $\mathfrak p=\mathfrak p_+\oplus\mathfrak p_-$, so that
\begin{equation*}
\mathfrak g=\mathfrak k\oplus\mathfrak p.
\end{equation*}
Hence, $\mathfrak p$ is the tangent space at a point to the principal orbit $G/K$, while $\mathfrak p_+$ and $\mathfrak p_-$ can be identified with the tangent spaces to $H/K$ and $G/H=Q$, respectively.
As we are working around the singular orbit $Q$, by assumption \eqref{assumption}, we have that $g_t$ and $L_t$ split in $+$ and $-$ parts. Hence, following \cite{Wang-Eschenburg:1}, let us choose $x(t),\eta(t)\in\End(\mathfrak p)^K$ preserving the splitting of $\mathfrak p$ and such that
\begin{align*}
&g_t=t^2x_+(t)\oplus x_-(t),\\
&L_t=\left(\frac{1}{t}\MI_++\eta_+(t)\right)\oplus\eta_-(t),
\end{align*}
with initial conditions given by
\begin{align*}
&x(0)=\MI,\\
&\eta_+(0)=0\quad\text{and}\quad\eta_-(0)=L_1(v_0),
\end{align*}
where $L_1$ is the shape operator of the singular orbit $Q$, which is an $H$-equivariant linear map from $V\longrightarrow\Sym^2(\mathfrak p_-)$. We will now drop the $t$-dependence in order to simplify the notation. In these new variables, equations \eqref{eqn:2.1}, \eqref{eqn:4.1} and \eqref{eqn:4.3} become
\begin{align*}
&\dot{x}=2x\eta,\\
&\frac{d^3}{d t^3}u+\frac{k}{t}\ddot{u}+\tr{(\eta)}\ddot{u}+\tr(\dot{\eta})\dot{u}-\frac{k}{t^2}\dot{u}-2\ddot{u}\dot{u}-\epsilon\dot{u}=0,\\
&\dot{\eta}=-\frac{k}{t^2}\MI_++\frac{1}{t^2}\MI_+-\frac{k}{t}\eta-\frac{1}{t}\tr{(\eta)}\MI_++\frac{1}{t}\dot{u}\MI_++r-\tr{(\eta)}\eta+\dot{u}\eta+\frac{\epsilon}{2}\MI.
\end{align*}
It is convenient to change again variables using
\begin{equation*}
y=x\eta,
\end{equation*}
so that we do not have to deal with the quadratic term $x\eta$. We then obtain
\begin{align}
\label{eqn:4.4}
&\dot{x}=2y,\\ \notag
&\frac{d^3}{d t^3}u+\frac{k}{t}\ddot{u}+\tr{(x^{-1}y)}\ddot{u}-\frac{k}{t^2}\dot{u}-2\tr{(x^{-1}yx^{-1}y)}\dot{u}+\tr{(x^{-1}\dot{y})}\dot{u}\\
\label{eqn:4.5}
&-2\ddot{u}\dot{u}-\epsilon\dot{u}=0,\\
\notag
&\dot{y}=(1-k)\frac{1}{t^2}x_+-\frac{k}{t}y-\frac{1}{t}\tr{(x^{-1}y)}x_++\frac{1}{t}\dot{u}x_++2yx^{-1}y+xr\\
\label{eqn:4.6}
&-\tr{(x^{-1}y)}y+\dot{u}y+\frac{\epsilon}{2}x,
\end{align}
with initial conditions on $y$ given by
\begin{equation*}
y_+(0)=0\quad\text{and}\quad y_-(0)=L_1(v_0).
\end{equation*}
At this point, we need the formula for the Ricci tensor of the homogeneous metric $g$ on the homogeneous space $P=G/K$, where $G$ is a compact Lie group. It has the following expression (see \cite{Besse:1} p. 185 for a derivation of this formula):
\begin{equation*}
\begin{aligned}
\Ric(X,Y)=&-\frac{1}{2}\tr_{\mathfrak g}(\ad(X)\ad(Y))-\frac{1}{2}\sum_{ij}{g([X,X_i]_{\mathfrak p},[Y,X_j]_{\mathfrak p})g^{ij}}\\
&+\frac{1}{4}\sum_{ijpq}{g(X,[X_i,X_p])g(Y,[X_j,X_q])g^{ij}g^{pq}},
\end{aligned}
\end{equation*}
for any basis $\{X_i\}_{i=1}^n$ of $\mathfrak p$ and for all $X,Y\in\mathfrak p$. Note that our expression for the Ricci tensor is simpler than $(7.38)$ in \cite{Besse:1}. This is due to the fact that $G$ is compact and hence unimodular.
The metric $\widehat{g}$ induces a $G$-invariant background metric $\widehat{g}_0$ on $NQ$. In fact, $\widehat{g}$ induces inner products on $\mathfrak p_-$, which can be identified with $T_qQ$, and on $V$, which can be identified with $N_qQ$.
Considering bases $\{U_\alpha\}_{\alpha=1}^k$ of $\mathfrak p_+$ and $\{Z_i\}_{i=k+1}^n$ of $\mathfrak p_-$, which are orthonormal with respect to the background metric $\widehat{g}_0$, the inverse of $g$ splits as follows
\begin{equation*}
g^{\alpha\beta}=\frac{1}{t^2}x_+^{\alpha\beta},\quad g^{ij}=x_-^{ij}.
\end{equation*}
Consequently, the Ricci endomorphism splits into a regular part and a singular part:
\begin{equation*}
r=\frac{1}{t^2}r_{\text{sing}}+r_{\text{reg}},
\end{equation*}
which are given in Lemma 3.1 of \cite{Wang-Eschenburg:1}.
We also have that
\begin{equation*}
\begin{aligned}
x_+r_+&=\frac{1}{t^2}g_+r_+=\frac{1}{t^2}\Ric_+,\\
x_-r_-&=g_-r_-=\Ric_-.
\end{aligned}
\end{equation*}
Hence, equation \eqref{eqn:4.6} becomes
\begin{equation*}
\dot{y}=\frac{1}{t^2}A(x)+\frac{1}{t}B(x,y)+C(x,y,t),
\end{equation*}
where
\begin{equation}\label{eqn:4.7}
\begin{aligned}
&A(x)=(1-k)x_++xr_{\text{sing}},\\
&B(x,y)=-ky-\tr{(x^{-1}y)}x_++\dot{u}x_+,\\
&C(x,y,t)=2yx^{-1}y+xr_{\text{reg}}-\tr{(x^{-1}y)}y+\dot{u}y+\frac{\epsilon}{2}x.
\end{aligned}
\end{equation}
We can now substitute $\dot{y}$ in \eqref{eqn:4.5} with the expression given by \eqref{eqn:4.6}. Then, Equation \eqref{eqn:4.5} becomes
\begin{equation*}
\frac{d^3}{d t^3}u=\frac{1}{t^2}\widetilde{A}(\dot{u})+\frac{1}{t}\widetilde{B}(\dot{u},\ddot{u})+\widetilde{C}(\dot{u},\ddot{u},t),
\end{equation*}
where
\begin{align*}
&\widetilde{A}(\dot{u})=k\dot{u}+(k-1)\tr{(x^{-1}x_+)}\dot{u}-\tr{(r_{\text{sing}})}\dot{u},\\
&\widetilde{B}(\dot{u},\ddot{u})=-k\ddot{u}+k\tr{(x^{-1}y)}\dot{u}+\tr{(x^{-1}y)}\tr{(x^{-1}x_+)}\dot{u}-\tr{(x^{-1}x_+)}\dot{u}^2\\
\begin{split}
&\widetilde{C}(\dot{u},\ddot{u},t)=-\tr{(x^{-1}y)}\ddot{u}-\tr{(r_{\text{reg}})}\dot{u}+\tr{(x^{-1}y)}\tr{(x^{-1}y)}\dot{u}-\tr{(x^{-1}y)}\dot{u}^2\\
&\quad\quad\quad\quad\quad\,-(n+1)\frac{\epsilon}{2}\dot{u}+2\ddot{u}\dot{u}+\epsilon\dot{u},
\end{split}
\end{align*}
are analytic functions.
We then obtain that the system \eqref{eqn:4.4}-\eqref{eqn:4.6}, with the above initial conditions, becomes the initial value problem given by
\begin{align*}
&\dot{x}=2y,\\
&\frac{d^3}{d t^3}u=\frac{1}{t^2}\widetilde{A}(\dot{u})+\frac{1}{t}\widetilde{B}(\dot{u},\ddot{u})+\widetilde{C}(\dot{u},\ddot{u},t),\\
&\dot{y}=\frac{1}{t^2}A(x)+\frac{1}{t}B(x,y)+C(x,y,t),\\
&x(0)=\MI,\\
&y_+(0)=0\quad\text{and}\quad y_-(0)=L_1(v_0),\\
&\dot{u}(0)=0.
\end{align*}
Therefore, the initial value problem for cohomogeneity one Ricci solitons has been reduced to an initial value problem for a system of non linear ordinary differential equations of order one in $x,y$ and of order two in $\dot{u}$ with a singular point at the origin.
By Lemma 2.2 of \cite{Dancer-Wang:2}, Ricci solitons are real analytic. Therefore, we can solve the system by applying the method of asymptotic power series, which is described in \cite{Wasow:1}, Chapter 9. This method consists, first of all, of showing that there always exists a formal power series solution of an appropriate type. Then, after having a formal power series solution, one can apply Theorem 7.1 of \cite{Malgrange:1} to get a genuine solution.
We will see that the reason why there always exists a formal power series solution, is due to the geometric nature of the equations.
\section{Solution to the initial value problem}
The initial value problem considered in Section \ref{sec:4} has the following general form
\begin{align}
&\dot{x}=2y,\label{eqn:5.1}\\
&\frac{d^3}{d t^3}u=\frac{1}{t^2}\widetilde{A}(\dot{u})+\frac{1}{t}\widetilde{B}(\dot{u},\ddot{u})+\widetilde{C}(\dot{u},\ddot{u},t)\label{eqn:5.2}\\
&\dot{y}=\frac{1}{t^2}A(x)+\frac{1}{t}B(x,y)+C(x,y,t),\label{eqn:5.3}\\
&x(0)=a,\label{eqn:5.4}\\
&y(0)=b,\label{eqn:5.5}\\
&\dot{u}(0)=0,\label{eqn:5.6}
\end{align}
where $x(t),y(t),a,b\in \Sym^2(V\oplus\mathfrak p_-)^K$, $u(t)$ is smooth function and $A$, $B$, $C$, $\widetilde{A}$, $\widetilde{B}$ and $\widetilde{C}$ are analytic functions.
As the left-hand sides of \eqref{eqn:5.2} and \eqref{eqn:5.3} do not have $\frac{1}{t}$ or $\frac{1}{t^2}$ terms, $A$ and $\widetilde{A}$ must satisfy the following initial conditions
\begin{align}
&A(x(0))=A(a)=0\quad\text{and}\quad2(d A)_a\cdot b+B(a,b)=0,\label{eqn:5.12}\\
&\widetilde{A}(\dot{u}(0))=\widetilde{A}(0)=0\quad\text{and}\quad(d\widetilde{A})\vert_{t=0}\ddot{u}(0)+\widetilde{B}(0,\ddot{u}(0))=0.\label{eqn:5.13}
\end{align}
We want to show that there always exists a formal power series solution. So let
\begin{equation*}
x(t)=\sum_{m=0}^\infty{\frac{x_m}{m!}t^m},\quad y(t)=\sum_{m=0}^\infty{\frac{y_m}{m!}t^m},
\end{equation*}
with
\begin{equation*}
x_{m+1}=2y_m,\quad\forall m\geq0,
\end{equation*}
and
\begin{equation*}
u(t)=\sum_{m=0}^\infty{\frac{u_m}{m!}t^m}.
\end{equation*}
Then, let
\begin{equation*}
A(x(t))=\sum_{m=0}^\infty{\frac{A_m}{m!}t^m},\quad B(x(t),y(t))=\sum_{m=0}^\infty{\frac{B_m}{m!}t^m},\quad C(x(t),y(t),t)=\sum_{m=0}^\infty{\frac{C_m}{m!}t^m},
\end{equation*}
and
\begin{equation*}
\widetilde{A}(\dot{u}(t))=\sum_{m=0}^\infty{\frac{\widetilde{A}_m}{m!}t^m},\quad\widetilde{B}(\dot{u}(t),\ddot{u}(t))=\sum_{m=0}^\infty{\frac{\widetilde{B}_m}{m!}t^m},\quad\widetilde{C}(\dot{u}(t),\ddot{u}(t),t)=\sum_{m=0}^\infty{\frac{\widetilde{C}_m}{m!}t^m}.
\end{equation*}
Substituting the above expressions in \eqref{eqn:5.2} and \eqref{eqn:5.3}, respectively, we get
\begin{align}
\frac{1}{2}x_{m+2}&=\frac{A_{m+2}}{(m+2)(m+1)}+\frac{B_{m+1}}{m+1}+C_m,\label{eqn:5.8}\\
u_{m+3}&=\frac{\widetilde{A}_{m+2}}{(m+2)(m+1)}+\frac{\widetilde{B}_{m+1}}{m+1}+\widetilde{C}_m.\label{eqn:5.9}
\end{align}
By definition, we have that
\begin{align*}
A_{m+2}&=\frac{d^{m+2}}{d t^{m+2}}(A(x(t))\Big\vert_{t=0}=\frac{d^{m+1}}{d t^{m+1}}\left(\frac{d}{d t}(A(x(t)))\right)\Big\vert_{t=0}\\
&=\frac{d^{m+1}}{d t^{m+1}}(d_{x(t)}A\cdot\dot{x}(t))\Big\vert_{t=0}\\
&\equiv(d_{x(t)} A)_a\cdot x_{m+2}\quad(\text{mod}\,x_1,\dots,x_{m+1}),\\
B_{m+1}&=\frac{d^{m+1}}{d t^{m+1}}(B(x(t),y(t)))\Big\vert_{t=0}=\frac{d^m}{d t^m}\left(\frac{d}{d t}(B(x(t),y(t)))\right)\Big\vert_{t=0}\\
&=\frac{d^m}{d t^m}(\partial_{x(t)}B\cdot\dot{x}(t)+\partial_{y(t)}B\cdot\dot{y}(t)+x_+\ddot{u})\Big\vert_{t=0}\\
&\equiv\frac{1}{2}(\partial_{y(t)}B)_{(a,b)}\cdot x_{m+2}\quad(\text{mod}\,x_1,\dots,x_{m+1}),\\
C_m&\equiv0\quad(\text{mod}\,x_1,\dots,x_{m+1}).
\end{align*}
Using the same strategy, we also have that
\begin{align*}
\widetilde{A}_{m+2}&\equiv(d_{\dot{u}(t)}\widetilde{A})\vert_{t=0}u_{m+3}\quad(\text{mod}\,u_1,\dots,u_{m+2}),\\
\widetilde{B}_{m+1}&\equiv(\partial_{\ddot{u}(t)}\widetilde{B})\vert_{t=0}u_{m+3}\quad(\text{mod}\,u_1,\dots,u_{m+2}),\\
\widetilde{C}_m&\equiv0\quad(\text{mod}\,u_1,\dots,u_{m+2}).
\end{align*}
Hence, equations \eqref{eqn:5.8} and \eqref{eqn:5.9} become
\begin{align}
&x_{m+2}=2\frac{(d_{x(t)} A)_a\cdot x_{m+2}}{(m+2)(m+1)}+\frac{(\partial_{y(t)}B)_{(a,b)}\cdot x_{m+2}}{m+1}+\frac{D_m}{m+1},\label{eqn:5.14}\\
&u_{m+3}=\frac{(d_{\dot{u}(t)}\widetilde{A})\vert_{t=0}u_{m+3}}{(m+2)(m+1)}+\frac{(\partial_{\ddot{u}(t)}\widetilde{B})\vert_{t=0} u_{m+3}}{m+1}+\frac{\widetilde{D}_m}{m+1},\label{eqn:5.15}
\end{align}
for some functions $D_m$ of $x_1,\dots,x_{m+1}$ and $\widetilde{D}_m$ of $u_1,\dots,u_{m+2}$. If we now define the following two operators
\begin{align}
&\mathcal L_{m}=(m+1)\MI-\frac{2}{m+2}(d_{x(t)} A)_a-(\partial_{y(t)}B)_{(a,b)},\label{eqn:5.11}\\
&\mathcal{\widetilde{L}}_m=(m+1)-\frac{1}{m+2}(d_{\dot{u}(t)}\widetilde{A})\vert_{t=0}-(\partial_{\ddot{u}(t)}\widetilde{B})\vert_{t=0},\notag
\end{align}
we need to have that
\begin{equation*}
\mathcal L_{m}\cdot x_{m+2}=D_m\quad\text{and}\quad\mathcal{\widetilde{L}}_mu_{m+3}=\widetilde{D}_m,
\end{equation*}
which give necessary and sufficient conditions to the existence of a formal power series solution:
\begin{equation}\label{eqn:5.10}
D_m\in\image(\mathcal L_{m})\quad\text{and}\quad\widetilde{D}_m\in\image(\mathcal{\widetilde{L}}_m),
\end{equation}
for all $m\geq0$. We have that $\mathcal L_m$ and $\mathcal{\widetilde{L}}_m$ are invertible for all $m\geq m_0$, for some $m_0$. In fact, $\mathcal{\widetilde{L}}_m$ is bounded and $d A$ and $d B$ are bounded as well. For this reason, if $m$ is large, $\mathcal L_m$ is close to a multiple of the identity and hence invertible. This implies that, if \eqref{eqn:5.10} is satisfied for $m<m_0$, we can fix further initial conditions, namely $x_1,\dots,x_{m_0}$ and $u_1,\dots,u_{m_0}$ satisfying equations \eqref{eqn:5.14} and \eqref{eqn:5.15} respectively, such that the formal power series solution is uniquely determined.
As we said before, $a$ and $b$ are two $K$-invariant endomorphisms which preserve the splitting of $\mathfrak p$ in a $+$ part and in a $-$ part. Moreover, as we saw in section \ref{sec:4}, they are given by
\begin{equation*}
a=\MI,\quad b_+=0\quad\text{and}\quad b_-=L_1(v_0),
\end{equation*}
where we have that $L_1\in W_1^-$, which implies that $\tr{(L_1)}=0$.
Now, using expressions given in \eqref{eqn:4.7} and Lemmas 4.2 and 4.4 of \cite{Wang-Eschenburg:1}, we can write the operator $\mathcal L_m$ as follows. First of all, we have that
\begin{align*}
&(d_{x(t)} A)_a\cdot\xi=(d_{x(t)} r_{\text{sing}})_{\MI}\cdot\xi,\\
&B(a,b)=-kb,\\
&(\partial_yB)_{(a,b)}\cdot\xi=-k\xi-\tr{(\xi)}\MI_+,\\
&C(x,y,t)=2yx^{-1}y+xr_{\text{reg}}-\tr{(x^{-1}y)}y+\dot{u}y+\frac{\epsilon}{2}x,
\end{align*}
where $\xi\in \Sym^2(V\oplus\mathfrak p_-)^K$ and
\begin{align*}
(d_{x(t)} r_{\text{sing}})_{\MI}\cdot\xi_+&=(k+1)\xi_+-2\tr{(\xi_+)}\MI_+,\\
(d_{x(t)} r_{\text{sing}})_{\MI}\cdot\xi_-&=\frac{1}{2}\mathcal C\cdot\xi_-,
\end{align*}
where $\mathcal C$ is an operator defined by $\mathcal C=-\sum_{\alpha=1}^k{\ad(U_\alpha)^2}$. Note that, if the $H$-homogeneous standard metric on $S^k$ is normal, we have a bi-invariant metric on $\mathfrak h$ and we can extend the basis $\{U_\alpha\}$ of $\mathfrak p_+\subset\mathfrak h$ to an orthonormal basis $\{V_\alpha\}$ of $\mathfrak h$, equipped with this bi-invariant metric, and we have that $\mathcal C=-\sum_\alpha{\ad(V_\alpha)^2}$ is the Casimir operator for the adjoint representation on $\mathfrak p_-$ and $\End(\mathfrak p_-)$. Note that we obtained, apart from $C$, the same expressions as \cite{Wang-Eschenburg:1} (p. 129).
Substituting these expression in \eqref{eqn:5.11}, we have that
\begin{equation*}
\mathcal L_m\cdot\xi=(m+1)\xi-\frac{2}{m+2}(d r_{\text{sing}})_{\MI}\cdot\xi+k\xi+\tr{(\xi)}\MI_+.
\end{equation*}
Furthermore, by Lemma 4.6 of \cite{Wang-Eschenburg:1}, we have that
\begin{align*}
&(\mathcal L_m\cdot\xi)_+=m\left(1+\frac{k+1}{m+2}\right)\xi_++\left(\frac{4}{m+2}\tr{(\xi_+)}+\tr{(\xi)}\right)\MI_+,\\
&(\mathcal L_m\cdot\xi)_-=(m+1+k)\xi_--\frac{1}{m+2}\mathcal C\cdot\xi_-.
\end{align*}
Again by Lemmas 4.2 and 4.4 of \cite{Wang-Eschenburg:1} and by the above expressions, we also have that
\begin{align*}
(d_{\dot{u}(t)}\widetilde{A})\vert_{t=0}f&=kf+(k-1)\tr{(x^{-1}(0)x_+(0))}f-\tr{(r_{\text{sing}}(0))}f\\
&=kf+(k-1)kf-\tr{((k-1)\MI_+)}f\\
&=kf+(k-1)kf-(k-1)kf\\
&=kf,\\
(\partial_{\ddot{u}(t)}\widetilde{B})\vert_{t=0}f&=-kf,
\end{align*}
so that the operator $\mathcal{\widetilde{L}}_m$ becomes
\begin{equation}\label{eqn:5.16}
\mathcal{\widetilde{L}}_m=(m+1)-\frac{k}{m+2}+k.
\end{equation}
We now have to verify that the initial conditions \eqref{eqn:5.12} and \eqref{eqn:5.13} hold and that \eqref{eqn:5.10} is satisfied. Note that \eqref{eqn:5.12} is satisfied as explained in \cite{Wang-Eschenburg:1} (p. 130). Then, we have that $\widetilde{A}(\dot{u}(0))=0$, because $\dot{u}(0)=0$. Moreover,
\begin{equation*}
\begin{aligned}
(d_{\dot{u}(t)}\widetilde{A})\vert_{t=0}\ddot{u}(0)=&k\ddot{u}(0)+(k-1)\tr(x^{-1}(0)x_+(0))\ddot{u}(0)-\tr{(r_{\text{sing}}(0))}\ddot{u}(0)\\
=&k\ddot{u}(0)
\end{aligned}
\end{equation*}
by Lemmas 4.2 and 4.4 of \cite{Wang-Eschenburg:1} and $\widetilde{B}(0,\ddot{u}(0))=-k\ddot{u}(0)$ by definition. Hence, \eqref{eqn:5.13} holds as well. As the initial conditions hold, we need to verify equation \eqref{eqn:5.10} and to show that
\begin{equation*}
x^l(t):=\sum_{m=0}^l{\frac{x_m}{m!}t^m}
\end{equation*}
defines a smooth $G$-invariant metric around $Q$ for all $l$. By Lemma \ref{Lemma:3.1} this means that we need to prove that $x_m\in\text{ev}(W_m)$ for all $m$.
In \cite{Wang-Eschenburg:1}, the authors show that $\mathcal L_m(\text{ev}(W_{m+2}))\subset\text{ev}(W_m)$, by decomposing $\text{ev}(W_{m+2})$ into eigenspaces of $\mathcal L_m$ and showing that the only eigenspaces corresponding to nonzero eigenvalues lie in $\text{ev}(W_m)\subset\text{ev}(W_{m+2})$, (we can always modify the degree of a homogeneous polynomial by an even factor without changing its value on the sphere). Moreover, $\mathcal L_m$ maps $\text{ev}(W_m)$ bijectively onto itself if $m>0$. Hence, the equation $\mathcal L_m\cdot x_{m+2}=D_m$ has a solution if and only if $D_m\in\text{ev}(W_m)$. Moreover, we also have that the kernel of $\mathcal L_m$ is isomorphic to $W^-_{m+2}/W^-_m$. We need now to show that $x_p\in\text{ev}(W_p)$ for all $p$ and that $D_m\in\text{ev}(W_m)$ for all $m$. In \cite{Wang-Eschenburg:1}, the authors show this by induction over $m$. They show that there exists a solution $x_{m+2}$ of $\mathcal L_m\cdot x_{m+2}=D_m$, but we can add an arbitrary element of the kernel of the operator considered, as it is not trivial.
Considering the multiplicative operator defined by \eqref{eqn:5.16}, we can see that it maps $\text{ev}(W_{m+3})$ to itself. So we need to show that $\widetilde{D}_m\in\text{ev}(W_{m+3})$ and that
\begin{equation*}
u^m(t):=\sum_{p=0}^{m}{\frac{u_p}{p!}t^p}
\end{equation*}
defines a smooth $G$-invariant function around $Q$ for all $m$. This means that we have to show that $u_p\in\text{ev}(W_p)$ for all $p$. We can prove that $\widetilde{D}_m\in\text{ev}(W_{m+3})$ by induction over $m$. We have that $u_1=0\in\text{ev}(W_1)$. Then, suppose that $u_p\in\text{ev}(W_p)$ for $p=2,\dots,m+2$, which implies that $u^{m+2}(t)$ is even in $t$, and consider
\begin{equation*}
\widehat{u}(t)=u^{m+2}(t)=\sum_{p=0}^{m+2}{\frac{u_p}{p!}t^p},
\end{equation*}
which defines a smooth $G$-invariant function around $Q$, because of the discussion in Section \ref{section 3}. By definition, we have that $\widehat{u}$ satisfies equation \eqref{eqn:5.2}. Let $\widehat{A},\widehat{B}$ and $\widehat{C}$ be the the analogues of $\widetilde{A},\widetilde{B},\widetilde{C}$ for $\widehat{u}$. Moreover, let $\widehat{D}_m$ be some function of $\widehat{u}_1,\dots,\widehat{u}_{m+2}$ which satisfies an analogue of equation \eqref{eqn:5.15} for $\widehat{u}$. Now, as $\widehat{u}_{m+3}=0\in\text{ev}(W_{m+3})$ and $\mathcal{\widetilde{L}}_m\widehat{u}_{m+3}=\widehat{D}_m$, we have that $\widehat{D}_m=0$. Then, by equations \eqref{eqn:5.9} and \eqref{eqn:5.15} and by recalling the expression of $\widetilde{C}$, we have that
\begin{equation*}
\frac{\widetilde{D}_m}{m+1}=\frac{\widetilde{D}_m-\widehat{D}_m}{m+1}=\widetilde{C}_m-\widehat{C}_m=0\in\text{ev}(W_{m+3}).
\end{equation*}
Hence, $\widetilde{D}_m\in\text{ev}(W_{m+3})$ and a solution $u_{m+3}$ to $\mathcal{\widetilde{L}}_mu_{m+3}=\widetilde{D}_m$ exists in $\text{ev}(W_{m+3})$.
The indeterminacy is just in the operator $\mathcal L_m$ and it is the same as in the Einstein case. In \cite{Wang-Eschenburg:1}, the authors describe this indeterminacy in the formal power series solution. If $m>0$, after solving the $-$ part of $\mathcal L_m\cdot x_{m+2}=D_m$, the $+$ part is uniquely determined. On the contrary, if $m=0$, the trace free part of $(x_2)_+$ is arbitrary. This is to be expected, as the values of $x(0)$ and $y_+(0)$ are fixed by the geometry of the problem and this implies that the usual freedom in the initial value problem lies in the trace free part of $(x_2)_+$. Furthermore, by section 1 of \cite{Wang-Eschenburg:1}, we see that the spaces $W_m^-$ eventually stabilise. So, suppose that
\begin{equation*}
W^-_{2m}=W^-_{2m_0}\quad\text{and}\quad W^-_{2m+1}=W^-_{2m_1+1},
\end{equation*}
for all $m>m_0$ and for all $m>m_1$, respectively. Hence, as $\ker(\mathcal L_m)\backsimeq W^-_{m+2}/W^-_m$, we have that the indeterminacy of the initial value problem considered is given by
\begin{equation*}
(W^-_{2m_0}/W^-_0)\oplus(W^-_{2m_1+1}/W^-_1),
\end{equation*}
where, in particular,
\begin{equation*}
W_0=\Sym^2(V)^H\oplus \Sym^2(\mathfrak p_-)^H.
\end{equation*}
The formal solution to the initial value problem \eqref{eqn:5.1}-\eqref{eqn:5.6} has the property that if truncated at any order it gives a smooth $G$-invariant metric and a smooth $G$-invariant function on $E$. Now, by Theorem 7.1 in \cite{Malgrange:1}, we obtain a genuine solution to the problem considered which is defined on a small interval $[0,T]$. The genuine solution may also be obtained by carrying out the Picard iteration directly, as shown in \cite{Wang-Eschenburg:1}. From this, one can see that $x_m(t)$ defines a smooth $G$-invariant metric on a tubular neighbourhood of radius $T$ around $Q$. Moreover, by Lemma 1.2 of \cite{Wang-Eschenburg:1}, we can choose $m$ to be at least 3. Hence, the solution $x$ gives a $C^3$ $G$-invariant metric which satisfies the appropriate equation. Finally, by Lemma 2.2 of \cite{Dancer-Wang:2}, the pair $(x,u)$ gives a smooth $G$-invariant Ricci soliton on the tubular neighbourhood of radius $T$ around the singular orbit $Q$.
To conclude this section, as the reader may not be familiar with \cite{Wang-Eschenburg:1}, we describe some examples, which show how to compute the indeterminacy explicitly.
\begin{example}[\cite{Wang-Eschenburg:1}]
Let
\begin{equation*}
G=SO(p+n),\quad H=SO(p)\times SO(n),\quad\text{and}\quad K=SO(p)\times SO(n-1).
\end{equation*}
Then, the manifold we are dealing with has dimension $np+n+1$. $H$ acts effectively and transitively on the unit sphere in $V$. By Lemma 1.2 of \cite{Wang-Eschenburg:1}, we have that $W_m^+\backsimeq\Hom(\Sym^m(V),\Sym^2(V))^H$ is isomorphic to zero if $m$ is odd and that all these spaces are isomorphic if $m$ is even. Hence, we can compute the indeterminacy in $(x_2)_+$, which is given by the dimension of $\Hom(\Sym^2(V),\Sym^2(V))^H$, which is 1, by Lemma 1.2 of \cite{Wang-Eschenburg:1}. Now, we need to compute the dimension of $W_m^-\backsimeq\Hom(\Sym^m(V),\Sym^2(\mathfrak p_-))^H$. As an $H$-representation $\mathfrak p_-$ is the tensor product of the standard representation $\rho_p$ of $SO(p)$ and the standard representation $\rho_n$ of $SO(n)$. On the other hand, $V$ is the tensor product of the trivial representation $\boldsymbol 1$ of $SO(p)$ and $\rho_n$. It is well known (see \cite{Fulton-Harris:1} p. 296) that
\begin{equation*}
\Sym^m(V)=\sigma_m\oplus\sigma_{m-2}\oplus\cdots\oplus\sigma_{m-2[\frac{m}{2}]},
\end{equation*}
where $\sigma_k$ is the irreducible representation of $SO(n)$ with dominant weight $k$ times that of $\rho_n$. We also have that
\begin{equation*}
\Sym^2(\mathfrak p_-)=(\Sym^2(\rho_p)\otimes \Sym^2(\rho_n))\oplus(\Lambda^2(\rho_p)\otimes\Lambda^2(\rho_n)),
\end{equation*}
which decomposes as
\begin{equation*}
(\sigma_2\otimes\sigma_2)\oplus(\boldsymbol 1\otimes\sigma_2)\oplus(\sigma_2\otimes\boldsymbol 1)\oplus(\boldsymbol 1\otimes\boldsymbol 1)\oplus(\ad_p\otimes\ad_n),
\end{equation*}
where we used the fact that the adjoint representation of $SO(n)$ is equivalent to $\Lambda^2V$.
Applying Schur's Lemma, we have that the dimension of $W_m^-$ is 2 when $m\geq2$ is even, it is zero if $m$ is odd. Finally, $\dim(W_0^-)=\dim(\Sym^2(\mathfrak p_-)^H)=1$. Thus, the indeterminacy which occurs with $(x_2)_-$ has dimension 1. We have that the choice of the initial metric is unique up to homothety and the only choice for the initial shape operator is zero, because $V$ is not a summand in $\Sym^2(\mathfrak p_-)$. This means that the singular orbit must be totally geodesic. Anyway, there is a one-dimensional freedom in choosing $(x_2)_+$.
\end{example}
\begin{example}
Let
\begin{equation*}
G=SO(n+2),\quad K=SO(n),\quad H=SO(n+1).
\end{equation*}
The manifold $M$ on which $G$ acts has dimension $2n+2$, the principal orbits for this action are given by the Stiefel manifold, which is the homogeneous manifold $SO(n+2)/SO(n)$ for $n\geq2$, and the singular orbit is given by the sphere $S^{n+1}$, which has codimension $n+1$ in $M$.
We want to compute the indeterminacy in the initial value problem for cohomogeneity one gradient Ricci solitons. We have that $V\backsimeq\mathbb R^{n+1}$, as $H$-representation, is given by the standard orthogonal representation $\rho_{n+1}$. Similarly, $\mathfrak p_-\backsimeq\mathbb R^{n+1}$, as $H$-representation, is given by $\rho_{n+1}$. Hence,
\begin{equation*}
\Sym^2(\mathfrak p_-)=\Sym^2(\rho_{n+1})=\sigma_2\oplus\boldsymbol 1.
\end{equation*}
Note that the assumption \eqref{assumption} is satisfied, because we assume that the metric $g_t$ is diagonal with respect to the following decomposition:
\begin{equation*}
\mathfrak g=\mathfrak k\oplus\underbrace{\mathfrak p_1\oplus\mathfrak p_2\oplus\mathfrak p_3}_{\mathfrak p},
\end{equation*}
where $\mathfrak p_1$ and $\mathfrak p_2$ are n-dimensional $SO(n)$-representations and $\mathfrak p_3$ is the trivial $SO(n)$-representation. With this decomposition we have that $\mathfrak p_-=\mathfrak p_2\oplus\mathfrak p_3$. By Schur's Lemma we then have that the dimension of $W_m^-$ is zero if $m$ is odd and it is one if $m\geq2$ is even. Moreover, $W_0^-$ has dimension one. So, we have that the indeterminacy in $(x_2)_-$ is zero. We can now compute the dimension of $W_m^+$, which is zero if $m$ is odd and 1 if either $m$ is zero or $m\geq2$ is even. So the indeterminacy in $(x_2)_+$, which is given by the dimension of $W_2^+$, is one. So we just have a one-dimensional freedom in choosing $(x_2)_+$.
\newline
Instead of considering $H=SO(n+1)$, we could consider $H=SO(2)\times SO(n)$, so that the singular orbit $Q=G/H$ has codimension 2. We have that $V\backsimeq\mathbb R^2$ and, as an $H$-representation, it is given by $\rho_2\otimes\boldsymbol 1$. On the other hand, $\mathfrak p_-\backsimeq\mathbb R^{2n}$, as an $H$-representation, is given by $\boldsymbol 1\otimes\rho_n\otimes\rho_n$. We then have that
\begin{equation*}
\begin{aligned}
\Sym^2(\mathfrak p_-)&=\Sym^2(\rho_n\otimes\rho_n)=(\Sym^2(\rho_n)\otimes\Sym^2(\rho_n))\oplus(\Lambda^2(\rho_n)\otimes\Lambda^2(\rho_n))\\
&=(\sigma_2\otimes\sigma_2)\oplus(\boldsymbol 1\otimes\sigma_2)\oplus(\sigma_2\otimes\boldsymbol 1)\oplus(\boldsymbol 1\otimes\boldsymbol 1)\oplus(\ad_n\otimes\ad_n).
\end{aligned}
\end{equation*}
Then we have that the dimension of $W_m^-$ is two when $m\geq2$ is even and it is zero when $m$ is odd. Moreover, the dimension of $W_0^-$ is one. So we have that the indeterminacy in $(x_2)_-$ is one. As the dimension of $W_2^+$ is one, the indeterminacy in $(x_2)_+$ is one.
\end{example}
\section*{Acknowledgements}
I would like to thank my supervisor Prof. Andrew Dancer for his useful comments and his supervision in writing the paper and Thomas Madsen for interesting discussions.
\bibliographystyle{amsplain}
\bibliography{citazioni}
\end{document} | 0.019319 |
TITLE: The product of a cofibration with an identity map is a cofibration
QUESTION [6 upvotes]: This is a problem from the book "modern classical homotopy theory" which I can't solve.
Let $i : A \rightarrow X$ be a cofibration and $Y$ any space. Show that $i : A\times Y \rightarrow X\times Y$ is also a cofibration.
I am supposed to use the following result:
$i : A \rightarrow X$ is a cofibration if and only if the canonical
map $T \rightarrow X\times I$ has a retraction, where $T$ is the
pushout of the diagram $A\times I \leftarrow A \rightarrow X$.
I'm not able to construct a map $X\times Y\times I \rightarrow T_2$ without using projections (and I don't think that is the way), and even if that is ok I have not been able to show that the map is the retraction of $T_2 \rightarrow X\times Y \times I$. By $T_2$ I mean the pushout of $A\times Y\times I \leftarrow A\times Y \rightarrow X\times Y$
REPLY [0 votes]: proof that $T\times Y\cong T_2$, or say that $T\times Y$ is the pushout of $A\times Y\times I\leftarrow A\times Y\to X\times Y$
by the universal property, we have two arrow $i_1: A\times Y\times I\to T\times Y$ and $i_2:X\times Y\to T\times Y$. and the pushout diagram in (1) commute
to proof universal arrow, suppose $K$ making the diagram commute, then by adjoint of $\times$, we have $K^Y$ commute in the pushout $A\times I\leftarrow A\to X$, so there is unique $T\to K^Y$ so there is unique $T\times Y\to K$
so the retraction you wanted is $r\times \mathrm{id}_Y: X\times I\times Y\to T\times Y\cong T_2$ | 0.013387 |
This is the first picture of a British tourist and her boyfriend, who died falling off a wall while ‘trying to take beach selfie’.
The pair plunged to their deaths in Portugal from a 30m-tall beach wall, it has been reported.
Louise Benson, 37, and her Australian boyfriend, Michael Kearns, 33, were named at the victims, Mirror Online reports.
The pair lived in Perth, Australia.
They were enjoying a five-month trip when they were discovered in the European country early yesterday morning (Tuesday).
A fisherman recovered their bodies at Pescadores (Fisherman’s) Beach in Ericeira at 6am.
Friend John Keogh paid tribute to Louise, who was a keen diver, on Facebook.
He wrote: “It can be such a cruel work quite, not for long tho, Bristol girl who just loved life.
“Your smile’s infectious, your laugh amazing, your friendship invaluable … in shock and speechless, you will be missed and I will miss you so much.”
Mirror Online reported earlier how the captain of the port of Cascais, 30km west of Lisbon, told Corfina Media he believed the woman may have been trying to take photos when the incident occurred.
“We found a mobile phone [at the bottom of] on the sea wall which indicated they might have been taking a selfie when the phone fell – they leaned over to get it and fell,” he said.
Source : BirminghamMail | 0.087984 |
TITLE: Proof verification: if $A'$ contains all the limit points of $A$ then $A$ is closed
QUESTION [0 upvotes]: Say we have a set $A$ and let us define a set $A'$ which contains all the limit points of $A$. Prove that $A'$ is closed.
Here is my attempt, please tell me if it is correct:
Say $A'$ is not closed. Let x be limit point in A'.
Due to unclosed group definition
We can say there is an $x$ that doesnt belong to $A'$ and this $x$ is limit point in $A'$.l et e1>0.
Due to limit point definition we have infinity number in $A'$ part $(x+e1,x-e1)$
Due to A' definition if X doesnt belong to A',it isnt limit point in A,which means we have finite numbers in the part (x+e1,x-e1)
Let a be limit point A.
Due to A' definition we can say there exist a in the part (x+e1) ,that mean a>x. Let e1= 2a-x.
Due to a definition ,there are infinite numbers(at group A) in the part (a+e2,a-e2).Let e2= a/2,that means we have infinite numbers in the part (3a/2,a/2).
That mean we have infinite numbers in the part (x+e1,x-e1)=(2a,2x-2a) because it containt (3a/2,a/2). And that contradict what i said at the begining that we have finite numbers in this part.
REPLY [1 votes]: Your proof is very confusing, for two reasons:
First of all, the language is horrible. I understand you are not a native speaker, but the english in your post is very unclear at some points. For example, I have no idea what the sentence
Moreover, we have infinity number in $A'$ $(x+e,x-e)$ (as limit point definition)
is supposed to mean.
Second of all, you made a mistake with saying
Assume that $A'$ is not closed. This means there is an $x$ that doesnt belong to $A'$ and this $x$ is limit point in $A'$.
This statement is just false. The second sentence simply does not follow from the first. | 0.037259 |
TITLE: Prove $\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$
QUESTION [0 upvotes]: I want to prove that $\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$. Is this a valid/good/correct proof?
Let $\lim_{x \to a} f(x) = L_1$ and $\lim_{x \to a} g(x) = L_2$. The righthand side can be written as $L_1L_2$.
Then $(\lim_{x \to a} f(x) - L_1) = \lim_{x \to a} f(x) - \lim_{x \to a} L_1 = L_1 - L_1 = 0$ and $(\lim_{x \to a} g(x) - L_2) = \lim_{x \to a} g(x) - \lim_{x \to a} L_2 = L_2 - L_2 = 0$.
Let $\epsilon > 0$. There exists a $\delta_1$ such that $|f(x)-L_1| < \sqrt{\epsilon}$ whenever $0 < |x-a| < \delta_1$, and there exists a $\delta_2$ such that $|g(x)-L_2| < \sqrt{\epsilon}$ whenever $0 < |x-a| < \delta_2$.
Therefore, suppose $0 < |x-a| < \delta$ where $\delta = \min(\delta_1, \delta_2)$ such that $0 < |x-a| < \delta_1$ and $0 < |x-a| < \delta_2$ are both satisfied. Then $|(f(x)-L_1)(g(x)-L_2)| < \sqrt{\epsilon}\sqrt{\epsilon} = \epsilon$.
Consider the expansion $(f(x)-L_1)(g(x)-L_2) = f(x)g(x) - L_2f(x) - L_1g(x) + L_1L_2$. Rearranged, we get $f(x)g(x) = (f(x)-L_1)(g(x)-L_2) + L_2f(x) + L_1g(x) - L_1L_2$.
We now have:
$\lim_{x \to a} f(x)g(x) = \lim_{x \to a}((f(x)-L_1)(g(x)-L_2) + L_2f(x) + L_1g(x) - L_1L_2)$
$\lim_{x \to a} f(x)g(x) = \lim_{x \to a}(f(x)-L_1)(g(x)-L_2) + \lim_{x \to a} L_2f(x) + \lim_{x \to a}L_1g(x) - \lim_{x \to a}L_1L_2$
$\lim_{x \to a} f(x)g(x) = 0 + L_1L_2 + L_1L_2 - L_1L_2$
$\lim_{x \to a} f(x)g(x) = L_1L_2$
REPLY [1 votes]: You simplify $\lim_{x \to a}(f(x)-L_1)(g(x)-L_2)$ to 0, but isn't that using the fact that $f(x)-L_1$ and $g(x)-L_2$ go to zero, thus their product goes to zero, and therefore relies on the very thing you're trying to prove? You also use lim(f(x)+g(x)) = lim(f(x))+lim(g(x)) repeatedly. Have you proved that claim?
In this sort of question, you're being asked to analyze a limit at low, epsilon-delta level, so you should be focusing n that level rather than working a higher, "this limit simplifies to this" level. A general tactic with limit is to define error terms equal to the value minus the limit: error1 = f(x)-L1, error2 = f(x)-L2. Then
f(x)g(x) = (L1-error1)(L2-error2) = L1*L2-L1*error2-L2*error1+error1*error2
and
f(x)g(x)-L1*L2 = -L1*error2-L2*error1+error1*error2
Now you just have to show that this expression can be made arbitrarily small. | 0.02194 |
\begin{document}
\title[Stochastic Control of Tide Equations]{Stochastic Control of Tidal Dynamics Equation with L\'evy Noise}
\author{Pooja Agarwal}
\address{Division of Applied Mathematics\\
Brown University\\
Providence, Rhode Island 02912, USA}
\email{Pooja\_Agarwal@Brown.edu}
\author{Utpal Manna}
\address{Indian Institute of Science Education and Research Thiruvananthapuram\\
Thiruvananthapuram 695016\\
Kerala, India}
\email{manna.utpal@iisertvm.ac.in}
\subjclass{35Q35, 60H15, 76D03, 76D55}
\keywords{Stochastic control, Initial value control, Tide equation, Minty-Browder theory}
\begin{abstract}
In this work we first present the existence, uniqueness and regularity of the strong solution of the tidal dynamics model perturbed by L\'evy noise. Monotonicity arguments have been exploited in the proofs. We then formulate a martingale problem of Stroock and Varadhan associated to an initial value control problem and establish existence of optimal controls.
\end{abstract}
\maketitle
\section{Introduction}
Ocean-tide information has considerably many applications. The data obtained is used to solve vital problems in oceanography and geophysics, and to study earth tides, elastic properties of the Earth's crust and tidal gravity variations. It is also used in space studies to calculate the trajectories of man-made satellites of the Earth and to interpret the results of satellite measurements. The interaction of tides with deep sea ridges and chains of seamounts give rise to deep eddies which transport nutrients from the deep to the surface. The alternate flooding and exposure of the shoreline due to tides is an important factor in the determination of the ecology of the region.
\paragraph*{}One of the first mathematical explanation for tides was given by Newton by describing tide generating forces. The first dynamic theory of tides was proposed by Laplace. Here we consider the tidal dynamics model proposed by Marchuk and Kagan \cite{marchuk}. The existence and uniqueness of weak solutions of the deterministic tide equation and that of strong solutions of the stochastic tide equation with additive trace class gaussian noise have been proved in Manna, Menaldi and Sritharan\cite{manna2008stochastic}. In this work, we consider the stochastic tide equation with L\'evy noise and prove the existence and uniqueness and regularity of solution in bounded domains. Control of fluid flow has numerous applications in control of pollutant transport, oil recovery/transport problems, weather predictions, control of underwater vehicles, etc. Unification of many control problems in the engineering sciences have been done by studying the optimal control problem of Navier-Stokes equations (see \cite{sritharan2000deterministic}, \cite{sritharan}). Here we consider the initial data optimal control of the stochastic tidal dynamics model. We consider the Stroock-Varadhan martingale formulation \cite{var} of the stochastic model to prove the existence of optimal initial value control.
\paragraph*{}The organization of the paper is as following. A brief description of the model has been given in Section \ref{model}. Section \ref{setting} describes the functional setting of the problem and states the monotonicity property of the non-linear operator. In Section \ref{estimate} we consider the a-priori estimates and prove the existence, uniqueness and regularity of strong solution. In Section \ref{stochastic} we consider the stochastic optimal control problem with initial value control.
\paragraph*{} In the framework of Gelfand triple $\h(\mathcal{O})\subset\Ll (\mathcal{O})\subset\mathbb{H}^{-1}(\mathcal{O})$ we consider the following tidal dynamics model with L\'evy noise
\begin{IEEEeqnarray*}{lCl}
du + [Au+B(u)+g\nabla\z]dt
=f(t)dt+\s(t,u(t))dW(t)+\int_Z H(u,z) \n ,\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{in }\mathcal{O}\times [0,T],\\
\dot{\z}+Div(hu)=0\quad\text{in }\mathcal{O}\times [0,T],\\
u(0)=u_0,\qquad \z(0)=\z_0.
\end{IEEEeqnarray*}
The operators $A$ and $B$ are defined in Sections \ref{model} and \ref{setting}. $(W(t)_{t\geq 0})$ is an $\Ll(\mathcal{O})$-valued Wiener process with trace class covariance. $\n=N(dt,dz)-\la dt$ is a compensated Poisson random measure, where $N(dt,dz)$ denotes the Poisson counting measure associated to the point process $p(t)$ on $Z$, a measurable space, and $\la$is a $\sigma$-finite measure on $(Z,\mathcal{B}(Z))$.
\paragraph*{}The following theorem states the main result of Section \ref{estimate}. The functional spaces appearing in the statement of the theorem have been defined in Section \ref{setting}.
\begin{theorem}
\label{thmint1}
Let us consider the above stochastic tide model with $f,u_0$ and $\z_0$ such that
\begin{equation}
f\in L^2(0,T;\mathbb{H}^{-1}(\mathcal{O})),\quad u_0\in \Ll(\mathcal{O}),\quad \z_0\in L^2(\mathcal{O}).
\end{equation}
Assume that $\s$ and $H$ satisfy the following hypotheses:
\begin{enumerate}
\item $\s\in C([0,T]\times\h(\mathcal{O});L_Q(H_0,\Ll)), H\in\mathbb{H}^2_\lambda([0,T]\times Z;\Ll(\mathcal{O}))$,
\item For all $t\in(0,T)$, there exists a positive constant $K$ such that for all $u\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)\|_{L_Q}^2+\int_Z\| H(u,z)\|_{\Ll}^2\la\leq K(1+\|u\|_{\Ll}^2),
\end{equation*}
\item For all $t\in(0,T)$, there exists a positive constant $L$ such that for all $u,v\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)-\s(t,v)\|_{L_Q}^2+\int_Z \|H(u,z)-H(v,z)\|_{\Ll}^2\la
\leq L(\|u-v\|_{\Ll}^2).
\end{equation*}
\end{enumerate}
Then there exist path-wise unique adapted processes $u(t,x,\omega)$ and $\z(t,x,\omega)$ with the regularity
\begin{equation}
\left\{\begin{array}{ll}
&u\in L^2(\Omega, L^{\infty}(0,T;\mathbb{L}^2(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\Ll(\mathcal{O}))),\\
&\z,\dot{\z}\in L^2(\Omega ;L^2(0,T;L^2(\mathcal{O})))
\end{array}\right. ,
\end{equation}
satisfying the above stochastic tide model in the weak sense.
\end{theorem}
In Section \ref{stochastic} we consider the following stochastic optimal control problem with initial value control,
\begin{IEEEeqnarray}{lCl}
\label{scint1}
du + [Au+B(u)+g\nabla\z]dt
=f(t)dt+\s(t,u(t))dW(t)+\int_Z H(u,z) \n\nonumber \\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{in }\mathcal{O}\times [0,T],\\
\label{scint2}
\dot{\z}+Div(hu)=0\quad\text{in }\mathcal{O}\times [0,T],\\
\label{scint3}
u(0)=u_0+U,\qquad \z(0)=\z_0,
\end{IEEEeqnarray}
where $u_0\in\Ll(\mathcal{O})$, $\z_0\in L^2(\mathcal{O})$ and $U\in\Ll(\mathcal{O})$. The regularities on the initial values and the assumptions on $\s$ and $H$ are the same as considered in \ref{thmint1}. The cost functional is given by
\begin{equation}
J(u,U)=\E\left[\int_0^T \int_\mathcal{O} L(t,u,U) dxdt\right],
\end{equation}
where the function $L$ is defined in Section \ref{stochastic}.
\paragraph*{} The main result of Section \ref{stochastic} is
\begin{theorem}
Given $u_0\in\Ll(\mathcal{O})$, there exists a pair
\begin{equation*}
(\hat{u},\hat{U})\in (L^2(0,T;\h(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O})\cap \D(0,T;\mathbb{H}^{-1}(\mathcal{O})))\times L^2(0,T;\Ll(\mathcal{O})),
\end{equation*}
which gives a martingale solution to equations \eqref{scint1}-\eqref{scint3}, and
\begin{IEEEeqnarray*}{rcl}
J(\hat{u},\hat{U})=\min\{&&J(u,U);(u,U)\in (L^2(0,T;\h(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O}))\\
&&\cap \D(0,T;\mathbb{H}^{-1}(\mathcal{O})))\times L^2(0,T;\Ll(\mathcal{O})),\\
&&\text{where the pair }(u,U)\text{ gives a martingale solution}\\
&&\text{to equations }\eqref{scint1}-\eqref{scint3}\}.
\end{IEEEeqnarray*}
\end{theorem}
\section{Tidal Dynamics: The Model}\label{model}
Under the assumptions that: (1) Earth is perfectly solid, (2) ocean tides do not change Earth's gravitational field, and (3) no energy exchange takes place between the mid-ocean and shelf zone, Marchuk and Kagan \cite{marchuk} obtained the following mathematical model
\begin{equation}
\partial_t w + A_1 w -\kappa_h\triangle w +\dfrac{r}{h}|w|w+g\nabla\xi =f,
\end{equation}
\begin{equation}
\partial_t \xi + Div(hw)=0,
\end{equation}
in $\mathcal{O}\times [0,T]$, where $\mathcal{O}$ is a bounded 2-D domain (horizontal ocean basin) with coordinates $x=(x_1,x_2)$ and $t$ represents the time. Here $\partial_t$ denotes the time derivative, $\triangle,\nabla$ and $Div$ are the Laplacian, gradient and the divergence operators respectively.
\paragraph*{}The unknown variables $(w,\xi)$ represent the total transport 2-D vector (i.e., the vertical integral of the velocity from the ocean surface to the ocean floor) and the displacement of the free surface with respect to the ocean floor. For details on the coefficients and the domain description see Manna, Menaldi and Sritharan \cite{manna2008stochastic}.
\paragraph*{} Denote by $A$ the following matrix operator
\begin{equation}
\label{eq3}
A:=\left(\begin{array}{cc}
-\alpha\triangle & -\beta\\
\beta & -\alpha\triangle
\end{array}\right),
\end{equation}
and the nonlinear vector operator
\begin{equation}
v\mapsto\gamma |v|v:=\left(\begin{array}{c}
\gamma(x)v_1\sqrt{v_1^2+v_2^2}\\
\gamma(x)v_2\sqrt{v_1^2+v_2^2}
\end{array}\right),
\end{equation}
where $\alpha:=\kappa_h$ and $\beta:=2\omega\cos (\varphi)$ are positive constants, $\gamma(x):=r/h(x)$ is a strictly positive smooth function. In this model we assume the depth $h(x)$ to be a continuously differentiable function of $x$, nowhere becoming zero, so that
\begin{equation}
\min\limits_{x\in\mathcal{O}} h(x)=\epsilon >0,\qquad\max\limits_{x\in\mathcal{O}}h(x)=\mu,\quad\max\limits_{x\in\mathcal{O}}|\nabla h(x)|\leq M ,
\end{equation}
where M is some positive constant which equals zero at a constant ocean depth.
\paragraph*{} To reduce to homogeneous Dirichlet boundary conditions consider the natural change of unknown functions
\begin{equation}
u(x,t):=w(x,t)-w^0(x,t),
\end{equation}
and
\begin{equation}
\z(x,t):=\xi(x,t)+\int_0^t Div(hw^0(x,s))ds,
\end{equation}
which are referred to the tidal flow and the elevation. The full flow $w^0$ which is given a priori on the boundary $\partial\mathcal{O}$, has been extended to the whole domain $\mathcal{O}\times [0,T]$ as a smooth function and still denoted by $w^0$.\\\\
Then the tidal dynamic equation can be written as
\begin{equation}
\left\{\begin{array}{l}
\partial_t u+Au+\gamma |u+w^0|(u+w^0)+g\nabla\z=f^\prime\quad\text{in}\quad\mathcal{O}\times[0,T],\\
\partial_t \z + Div(hu)=0\quad\text{in}\quad\mathcal{O}\times[0,T],\\
u=0\quad\text{on}\quad\partial\mathcal{O}\times[0,T],\\
u=u_0,\;\;\z=\z_0\quad\text{in}\quad\mathcal{O}\times\{0\},
\end{array}\right.
\end{equation}
where
\begin{IEEEeqnarray}{lCl}
f^\prime =f-\dfrac{\partial w^0}{\partial t}+g\nabla\int_0^t Div(hw^0)dt-Aw^0,\\
u_0(x)=w_0(x)-w^0(x,0),\\
\z_0(x)=\xi_0(x).
\end{IEEEeqnarray}
\section{Functional setting}\label{setting}
We use the (vector-valued) Sobolev spaces $\h(\mathcal{O}):=H^1_0(\mathcal{O},\mathbb{R}^2)$ and $\Ll(\mathcal{O}):=L^2(\mathcal{O},\mathbb{R}^2)$, with:\\
the norm on $\h(\mathcal{O})$ as
\begin{equation}
\|v\|_{\h}:=\left(\int_{\mathcal{O}}|\nabla v|^2 dx\right)
^{1/2},
\end{equation}
and the norm on $\Ll(\mathcal{O})$ as
\begin{equation}
\|v\|_{\Ll}:=\left(\int_{\mathcal{O}}|v|^2 dx\right)^{1/2}.
\end{equation}
Using the Gelfand triple $\h(\mathcal{O})\subset\Ll (\mathcal{O})\subset\mathbb{H}^{-1}(\mathcal{O})$, we may consider $\triangle$ or $\nabla$ as a linear map from $\h(\mathcal{O})$ or $\Ll(\mathcal{O})$ into the dual of $\h(\mathcal{O})$ respectively. The inner product in $\Ll(\mathcal{O})$ or $L^2(\mathcal{O})$ is denoted by $(\cdot,\cdot)$. So
\begin{equation}
(u,v)_{\Ll}=\int_{\mathcal{O}}u(x)\cdot v(x)dx,
\end{equation}
for any $u$ and $v$ in $\Ll(\mathcal{O})$. The duality pairing between $\h$ and $\mathbb{H}^{-1}$ is denoted by $\langle\cdot,\cdot\rangle$.
\paragraph*{}Now we give the definitions and some properties of Hilbert space valued Wiener processes, L\'evy processes and Skorokhod spaces, most of which have been borrowed from the books by Da Prato and Zabczyk \cite{da}, Applebaum \cite{apple} and Metivier \cite{metivier}.
\paragraph*{}For a Hilbert space $U$, let us denote the space of all bounded linear operators from $U$ to $U$ by $L(U)$. We consider two Hilbert spaces $H$ and $U$, and a symmetric non-negative operator $Q\in L(U)$ such that $Tr(Q)<+\infty$. A $U$-valued stochastic process $\{W(t),t\geq 0\}$, is called a $Q$-Wiener process if
\begin{enumerate}
\item[(i)] $W(0)=0$,
\item[(ii)]$W$ has continuous trajectories,
\item[(iii)]$W$ has independent increments,
\item[(iv)]$\mathcal{L}(W(t)-W(s))=N(0,(t-s)Q)),\; t\geq s\geq 0$.
\end{enumerate}
If a process $\{W(t),t\in [0,T]\}$ satisfies $(i)-(iii)$ and $(iv)$ for $t,s\in [0,T]$ the we say that $W$ is a $Q$-Weiner process on $[0,T]$.
\paragraph*{}Let $Q$ be a symmetric non-negative operator $Q\in\Ll$. Define $H_0=Q^{1/2}\Ll$. Then $H_0$ is a Hilbert space equipped with inner product $(\cdot,\cdot)_{0}$,
\begin{equation*}
(u,v)_0=\left(Q^{-1/2}u,Q^{-1/2}v\right)_{\Ll}, \forall u,v\in H_0,
\end{equation*}
where $Q^{-1/2}$ is the pseudo-inverse of $Q^{1/2}$, and $Q$ is the covariance operator of the $\Ll$-valued Wiener process $(W(t)_{t\geq 0})$.
\paragraph*{}Let $L_Q$ denote the space of linear operators $S$ such that $SQ^{1/2}$ is a Hilbert-Schmidt operator from $\Ll$ to $\Ll$. Define the norm on the space $L_Q$ by $\|S\|_{L_Q}^2=Tr(SQS^*)$.
\paragraph*{}Let $(\mathbb{S},\rho)$ be a separable and complete metric space. Let $\D(0,T;\mathbb{S})$ denote the set of $\mathbb{S}$-valued functions on $[0,T]$ which are c\`adl\`ag (if $g\in \D(0,T;\Ll)$ then for all $t\in[0,T]\;\;g$ is right continuous at $t$ and has a left limit at $t$), endowed with the Skorokhod topology. This topology is metrizable by the following metric $\delta_T$
\begin{IEEEeqnarray*}{rcl}
\delta_T(u,v):=\inf_{\lambda\in\Lambda_T}&&\left[\sup_{0\leq t\leq T}\rho (u(t),v\circ\lambda(t))\right. \\
&&\left. +\sup_{0\leq t\leq T} |t-\lambda(t)|+\sup_{s\neq t}|\log\left(\frac{\lambda(t)-\lambda(s)}{t-s}\right)\right],
\end{IEEEeqnarray*}
where $\Lambda_T$ is the set of increasing homeomorphisms of $[0,T]$. More over, $(\D(0,T;\mathbb{S}),\delta_T)$ is a complete metric space.
\begin{remark}
A sequence $(u_n)\subset \D(0,T;\mathbb{S})$ converges to $u\in \D(0,T;\mathbb{S})$ iff there exists a sequence $(\lambda_n)$ of homeomorphisms of $[0,T]$ such that $\lambda_n$ tends to the identity uniformly on $[0,T]$ and $u_n\circ\lambda_n$ tends to $u$ uniformly on $[0,T]$.
\end{remark}
\begin{definition}
Let $(\Omega,\mathcal{F},\mathcal{F}_t,P)$ be a filtered probability space, and $E$ be a Banach space. A process $(X_t)_{t\geq 0}$ with state space $(E,\mathcal{B}(E))$ is called a L\'evy process if
\begin{enumerate}
\item[(i)] $(X_t)_{t\geq 0}$ is adapted to $(\mathcal{F}_t)_{t\geq 0}$,
\item[(ii)] $X_0=0$ a.s.,
\item[(iii)] $X_t - X_s$ is independent of $\mathcal{F}_s$ if $0\leq s< t$,
\item[(iv)]$(X_t)_{t\geq 0}$ is stochastically continuous, i.e., $\forall\epsilon>0,\lim_{s\rightarrow t}P(|X_s-X_t|>\epsilon)=0$,
\item[(iv)]$(X_t)_{t\geq 0}$ is c\`adl\`ag,
\item[(v)]$(X_t)_{t\geq 0}$ has stationary increments, i.e., $\E[X_t-X_s]= \E[X_{t-s}],0\leq s< t$.
\end{enumerate}
\end{definition}
The jump of $X_t$ at $t\geq 0$ is given by $\triangle X_t=X_t-X_{t-}$. Let $Z\in\mathcal{B}(\mathbb{R}^+\times E)$. We define
\begin{equation*}
N(t,Z)=N(t,Z,\omega)=\sum_{s:0<s\leq t} \chi_Z(\triangle X_s).
\end{equation*}
So, $N(t,Z)$ is the number of jumps of size $\triangle X_s\in Z$ which occur before or at time $t$. $N(t,Z)$ is called the Poisson random measure of $(X_t)_{t\geq 0}$. The differential form of this measure is written as $N(dt,dz)(\omega)$.
\paragraph*{}We call $\n=N(dt,dz)-\la dt$ a compensated Poisson random measure, where $\la dt$ is known as the compensator of the L\'evy process $(X_t)_{t\geq 0}$. Here $dt$ denotes the Lebesgue measure on $\mathcal{B}(\mathbb{R}^+)$, and $\la$ is a $\s$-finite measure on $(Z,\mathcal{B}(Z))$.
\begin{lemma}
\label{lemw}For any real-valued smooth function $\varphi$ and $\psi$ with compact support in $\mathbb{R}^2$, the following hold:
\begin{IEEEeqnarray}{lCl}
\|\varphi\psi\|_{L^2}^2\leq 4\|\varphi\partial_1\varphi\|_{L^2}\|\psi\partial_2\psi\|_{L^2},\\
\label{eq13}
\|\varphi\|_{L^4}^4\leq 2\|\varphi\|_{L^2}^2\|\nabla\varphi\|_{L^2}^2.
\end{IEEEeqnarray}
\end{lemma}
For proof see \cite{lady}.\\\\
Notice that by means of the Gelfand triple we may consider $A$, given by \eqref{eq3}, as a mapping of $\h(\mathcal{O})$ into its dual $\mathbb{H}^{-1}(\mathcal{O})$.
\paragraph*{}Define the non-symmetric bilinear form
\begin{equation}
a(u,v):=\alpha[(\partial_1 u_1,\partial_1 v_1)_{\Ll}+(\partial_2 u_2,\partial_2 v_2)_{\Ll}]+\beta[(u_1,v_2)_{\Ll}-(u_2,v_1)_{\Ll}],
\end{equation}
on $\h$. Thus if $u$ has a smooth second derivative then
\begin{equation*}
a(u,v)=(Au,v)_{\Ll},
\end{equation*}
for every $v$ in $\h$. More over, the bilinear form $a(\cdot,\cdot)$ is continuous and coercive in $\h(\mathcal{O})$,i.e.,
\begin{IEEEeqnarray}{lCl}
|a(u,v)|\leq C_1\|u|\|_{\mathbb{H}_0^1}\|v|\|_{\mathbb{H}_0^1}\qquad\forall u,v\in\h,\\
(Au,u)_{\Ll}=a(u,u)=\|u\|_{\mathbb{H}_0^1}^2,
\end{IEEEeqnarray}
for some positive constant $C_1=\alpha + \beta$.
\paragraph*{}Let us denote the nonlinear operator $B(\cdot)$ by
\begin{equation}
\label{eq4}
v\mapsto B(v):=\gamma |v+w^0|[v+w^0].
\end{equation}
Then we have the following lemma:
\begin{lemma}
\label{mon}
Let $u$ and $v$ be in $\mathbb{L}^4(\mathcal{O},\mathbb{R}^2)$. Then the following estimate holds:
\begin{equation}
\langle B(u)-B(v),u-v\rangle\geq 0.
\end{equation}
\end{lemma}
For proof see Lemma 3.3 in \cite{manna2008stochastic}.
\paragraph*{}
The nonlinear operator $B(\cdot)$ is a continuous operator from $\mathbb{L}^4(\mathcal{O})$ to $\Ll(\mathcal{O})$, where
\begin{IEEEeqnarray}{lCr}
\label{e2}
\|B(v)\|_{\Ll}\leq C_2\|v+w^0\|_{\mathbb{L}^4}^2,\\
\label{e1}
\|B(u)-B(v)\|_{\Ll}\leq C_2[\|u+w^0\|_{\mathbb{L}^4}+\|v+w^0\|_{\mathbb{L}^4}]\|u-v\|_{\mathbb{L}^4},
\end{IEEEeqnarray}
where the constant $C_2$ is the sup-norm of the function $\gamma$.
\paragraph*{} With the above functional setting, we write the stochastic tidal dynamics equation with the L\'evy forcing as
\begin{IEEEeqnarray}{lCl}
\label{eq21}
du + [Au+B(u)+g\nabla\z]dt
=f(t)dt+\s(t,u(t))dW(t)+\int_Z H(u,z) \n\nonumber,\\
\\
\label{eq22}
\dot{\z}+Div(hu)=0,\\
\label{eq23}
u(0)=u_0,\qquad \z(0)=\z_0,
\end{IEEEeqnarray}
where $u_0\in\Ll(\mathcal{O})$ and $\z_0\in L^2(\mathcal{O})$. The operators $A$ and $B$ are defined through \eqref{eq3} and \eqref{eq4} respectively. $(W(t)_{t\geq 0})$ is a $\Ll(\mathcal{O})$-valued Wiener process with trace class covariance. $H(u,z)$ is a measurable mapping from $\h\times Z$ into $\Ll$.
\paragraph*{}We define the space $\mathbb{H}^2_\lambda([0,T]\times Z;\Ll(\mathcal{O}))$ as
\begin{equation}
\mathbb{H}^2_\lambda([0,T]\times Z;\Ll(\mathcal{O}))=\{X:\E\left[\int_0^T\int_Z \|X\|_{\Ll}^2 \la dt\right]<\infty\}.
\end{equation}
We assume that $\s$ and $H$ satisfy the following hypotheses:
\begin{enumerate}
\item[H.1] $\s\in C([0,T]\times\h(\mathcal{O});L_Q(H_0,\Ll)), H\in\mathbb{H}^2_\lambda([0,T]\times Z;\Ll(\mathcal{O}))$,
\item[H.2] For all $t\in(0,T)$, there exists a positive constant $K$ such that for all $u\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)\|_{L_Q}^2+\int_Z\|H(u,z)\|_{\Ll}^2\la\leq K(1+\|u\|_{\Ll}^2),
\end{equation*}
\item[H.3] For all $t\in(0,T)$, there exists a positive constant $L$ such that for all $u,v\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)-\s(t,v)\|_{L_Q}^2+\int_Z \|H(u,z)-H(v,z)\|_{\Ll}^2\la
\leq L(\|u-v\|_{\Ll}^2).
\end{equation*}
\end{enumerate}
\section{Energy Estimates and Existence Result}\label{estimate}
Denote by $\Ll_n(\mathcal{O}):=span\{e_1,e_2,\ldots e_n\}$ where $\{e_j\}$ is any fixed orthonormal basis in $\Ll(\mathcal{O})$ with each $e_j\in\h(\mathcal{O})$ . Similarly denote by $L_n^2(\mathcal{O})$ the n-dimensional subspace of $L^2(\mathcal{O})$. Let $P_n$ denote the orthogonal projection of $\Ll(\mathcal{O})$ into $\Ll_n(\mathcal{O})$. Define $u^n=P_n u, W^n=P_n W, \s^n=P_n \s, H^n=P_n H $ and $\int_Z H^n(u^n(t-),z)\n=P_n\int_Z H(u(t-),z)\n$.
\paragraph*{}We define $u^n$ as the solution of the following stochastic equation in variational form:
\begin{IEEEeqnarray}{lrl}
\label{eqn5}
&&(du^n(t),v^n(t))_{\Ll}+(a(u^n(t),v^n(t))+(B(u^n(t)),v^n(t))_{\Ll}+(g\nabla\z^n(t) ,v^n(t))_{\Ll})dt\nonumber\\
&&=(f(t),v^n(t))_{\Ll}dt+(\s^n(t,u^n(t)),v^n(t))_{\Ll}dW^n(t)\nonumber\\
&&+\int_Z (H^n(u^n(t-),z),v^n(t))_{\Ll} \n \qquad\qquad\qquad\qquad \forall v^n\in\mathbb{L}_n^2(\mathcal{O}),\\
\label{eqn6}
&&(\dot{\z}^n(t)+Div(hu^n(t)),\zeta(t))_{L^2}=0\quad\forall \zeta\in L_n^2(\mathcal{O}),\\
&&u^n(0)=u^n_0\qquad\qquad \z^n(0)=\z^n_0.
\end{IEEEeqnarray}
\begin{proposition}
\label{prop}
Under the above mathematical setting let
\begin{equation}
\left\{\begin{array}{l}
w^0\in L^4(0,T;\mathbb{L}^4(\mathcal{O})),\;f\in L^2(0,T;\Ll(\mathcal{O})),\\
\s\in C([0,T]\times\h(\mathcal{O});L_Q(H_0,\Ll)),\;H\in\mathbb{H}^2_\lambda([0,T]\times Z;\Ll(\mathcal{O})),\\
u_0\in\Ll(\mathcal{O}),\z_0\in L^2(\mathcal{O}).
\end{array}\right.
\end{equation}
Let $u^n(t)$ be an adapted process in $\mathcal{D}(0,T,\Ll_n(\mathcal{O}))$ which solves the differential equations \eqref{eqn5} and \eqref{eqn6}. Then we have the following a priori estimates:
\begin{IEEEeqnarray}{l}
\label{eq14}
\mathbb{E}[\|u^n(t)\|_{\Ll}^2+\|\z^n(t)\|_{L^2}^2]+2\alpha\mathbb{E}[\int_0^t\|u^n(t)\|_{\h}^2dt]\leq C_{1(2)} \;\;\forall t\in[0,T],\\
\label{eq15}
\mathbb{E}[\sup\limits_{0\leq t\leq T}(\|u^n(t)\|_{\Ll}^2 + \|\z^n(t)\|_{L^2}^2)]+2\alpha\mathbb{E}[\int_0^T\|u^n(t)\|_{\h}^2dt]\leq C_{2(2)},
\end{IEEEeqnarray}
where the constants $C_{1(2)}$ and $C_{2(2)}$ depend on the coefficients $\alpha, g,M,\mu$ and the norms $\|f\|_{L^2(0,T;\Ll)},\|w^0\|_{L^4(0,T;\mathbb{L}^4)},\|u^n_0\|_{\Ll}$, $\|\z^n_0\|_{L^2}$ and $T$.
\end{proposition}
\begin{proof}
Applying It\^o's formula to $|x|^2$ and the process $u^n(t)$
\begin{IEEEeqnarray*}{lr}
\label{eqn 7}
d(\|u^n(t)\|_{\Ll}^2)+ 2\alpha\|u^n(t)\|^2_{\mathbb{H}_0^1}dt+2(B(u^n(t)),u^n(t))_{\Ll}dt+2(g\nabla\z^n(t) ,u^n(t))_{\Ll}dt\\
=2(f(t),u^n(t))_{\Ll}dt+2(\s^n(t,u^n(t)),u^n(t))_{\Ll}dW^n(t)+\|\s^n(t,u^n(t))\|_{L_Q}^2dt\\
\quad +\int_Z\| H^n(u^n(t-),z)\|_{\Ll}^2 N(dt,dz) +2\int_Z (H^n(u^n(t-),z),u^n(t-))_{\Ll}\n .\\\IEEEyesnumber
\end{IEEEeqnarray*}
Using the definition of the operator $B(\cdot)$ and Lemma \ref{mon}
\begin{IEEEeqnarray*}{rCl}
(B(u^n(t)),u^n(t))_{\Ll}&\geq &\int_\mathcal{O}\gamma(x)|w^0(t)|^2u^n(t)dx\\
&\geq & -\dfrac{r}{\epsilon}\|w^0(t)\|_{\mathbb{L}^4}^2\|u(t)\|_{\Ll}\\
&\geq & -\dfrac{r}{2\epsilon}[\|w^0(t)\|_{\mathbb{L}^4}^4+\|u(t)\|_{\Ll}^2].\IEEEyesnumber
\end{IEEEeqnarray*}
Using the divergence theorem and the inequality
\begin{equation}
2ab\leq \delta a^2 + \dfrac{1}{\delta}b^2,
\end{equation}
we obtain,
\begin{IEEEeqnarray*}{rCl}
|g(\nabla\z^n(t) ,u^n(t))_{\Ll}| &=& |-g(\z^n(t),Div (u^n(t)))_{\Ll}| \\
&\leq &\dfrac{g}{2}[\dfrac{2g}{\alpha}\|\z^n(t)\|_{L^2}^2+\dfrac{\alpha}{2g}\|Div (u^n(t))\|_{L^2}^2].
\end{IEEEeqnarray*}
Since the divergence is bounded by gradient in $L^2$-norm
\begin{equation}
\label{eq26}
|g(\nabla\z^n(t) ,u^n(t))_{\Ll}|\leq \dfrac{g}{2}[\dfrac{2g}{\alpha}\|\z^n(t)\|_{L^2}^2+\dfrac{\alpha}{2g}\| u^n(t)\|_{\mathbb{H}_0^1}^2].
\end{equation}
Using Cauchy-Schwarz inequality
\begin{equation}
|(f(t),u^n(t))_{\Ll}|\leq\dfrac{1}{2}[\|f(t)\|_{\Ll}^2+\|u^n(t)\|_{\Ll}^2].
\end{equation}
Hence the energy equality \eqref{eqn 7} yields
\begin{IEEEeqnarray*}{rcl}
\label{eq8}
d&(&\|u^n(t)\|_{\Ll}^2)+ 2\alpha\|u^n(t)\|^2_{\mathbb{H}_0^1} dt\\
&\leq & (\|f(t)\|_{\Ll}^2+\|u^n(t)\|_{\Ll}^2 +\dfrac{r}{\epsilon}[\|w^0(t)\|_{\mathbb{L}^4}^4+\|u(t)\|_{\Ll}^2]+\dfrac{2g^2}{\alpha}\|\z^n(t)\|_{L^2}^2)dt\\
&&+\dfrac{\alpha}{2}\| u^n(t)\|_{\mathbb{H}_0^1}^2dt +2(\s^n(t,u^n(t)),u^n(t))_{\Ll}dW^n(t)+\|\s^n(t,u^n(t))\|_{L_Q}^2dt\\
&&+\int_Z\| H^n(u^n(t-),z)\|^2_{\Ll} N(dt,dz) +2\int_Z (H^n(u^n(t-),z),u^n(t-))_{\Ll}\n .\\
\IEEEyesnumber
\end{IEEEeqnarray*}
Using equation \eqref{eqn6}
\begin{equation*}
\dfrac{1}{2}\dfrac{d}{dt}\|\z^n(t)\|_{L^2}^2=-(Div(hu^n(t)),\z^n(t))_{L^2}.
\end{equation*}
Now
\begin{IEEEeqnarray*}{rCl}
|(Div(hu^n(t)),\z^n(t))_{L^2}|&=&|(h\,Div(u^n(t)),\z^n(t))_{L^2}+(u^n(t)\cdot\nabla h,\z^n(t))_{L^2}|\\
&\leq & |(h\,Div(u^n(t)),\z^n(t))_{L^2}|+|(u^n(t)\cdot\nabla h,\z^n(t))_{L^2}|\\
&\leq & \| h\|_{L^\infty}\|Div(u^n(t))\|_{L^2}\|\z^n(t)\|_{L^2}\\
&&+\|u^n(t)\|_{\Ll}\|\nabla h\|_{L^\infty}\|\z^n(t)\|_{L^2}.
\end{IEEEeqnarray*}
Using the assumptions on $h$
\begin{IEEEeqnarray*}{rCl}
|(Div(hu^n(t)),\z^n(t))_{L^2}|&\leq &\mu\| u^n(t)\|_{\mathbb{H}_0^1}\|\z^n(t)\|_{L^2}
+M\|u^n(t)\|_{\Ll}\|\z^n(t)\|_{L^2}\\
&\leq &\dfrac{\mu}{2}[\dfrac{\alpha}{2\mu}\|u^n(t)\|_{\mathbb{H}_0^1}^2 +\dfrac{2\mu}{\alpha}\|\z^n(t)\|_{L^2}^2]\\
&&+\dfrac{M}{2}[\|u^n(t)\|_{\Ll}^2+\|\z^n(t)\|_{L^2}^2].
\end{IEEEeqnarray*}
Thus
\begin{equation}
\label{eq9} d\|\z^n(t)\|_{L^2}^2\leq M\|u^n(t)\|_{\Ll}^2 dt+(\dfrac{2\mu^2}{\alpha}+M)\|\z^n(t)\|_{L^2}^2 dt+\dfrac{\alpha}{2}\|u^n(t)\|_{\mathbb{H}_0^1}^2 dt.
\end{equation}
Adding equations \eqref{eq8} and \eqref{eq9}
\begin{IEEEeqnarray*}{lcl}
\label{eq10}
d(&\|&u^n(t)\|_{\Ll}^2+\|\z^n(t)\|_{L^2}^2)+\alpha\|u^n(t)\|^2_{\mathbb{H}_0^1}dt\\
&\leq & (1+M+\dfrac{r}{\epsilon})\|u^n(t)\|_{\Ll}^2 dt
+ (\dfrac{2g^2}{\alpha}+\dfrac{2\mu^2}{\alpha}+M)\|\z^n(t)\|_{L^2}^2 dt\\
&&+\dfrac{r}{\epsilon}\|w^0(t)\|_{\mathbb{L}^4}^4 dt + \|f(t)\|_{\Ll}^2 dt +\|\s^n(t,u^n(t))\|_{L_Q}^2 dt \\
&&+2(\s^n(t,u^n(t)),u^n(t))_{\Ll}dW^n(t)+\int_Z\| H^n(u^n(t-),z)\|^2_{\Ll} N(dt,dz)\\
&&+2\int_Z (H^n(u^n(t-),z),u^n(t-))_{\Ll}\n .\IEEEyesnumber
\end{IEEEeqnarray*}
Define
\begin{equation}
\tau_N=\inf\{t:\|u^n(t)\|_{\Ll}^2+\|\z^n(t)\|_{L^2}^2+\int_0^t\|u^n(s)\|^2_{\mathbb{H}_0^1}ds>N\}.
\end{equation}
Integrating equation \eqref{eq10}
\begin{IEEEeqnarray*}{rcl}
\label{eq11}
\| & u^n & (t\wedge\tau_N)\|_{\Ll}^2+\|\z^n(t\wedge\tau_N)\|_{L^2}^2+\int_0^{t\wedge\tau_N}\alpha\|u^n(s)\|^2_{\mathbb{H}_0^1}ds\\
&\leq & (1+M+\dfrac{r}{\epsilon})\int_0^{t\wedge\tau_N}\|u^n(s)\|_{\Ll}^2 ds
+ (\dfrac{2g^2}{\alpha}+\dfrac{2\mu^2}{\alpha}+M)\int_0^{t\wedge\tau_N}\|\z^n(s)\|_{L^2}^2 ds\\
&&+\dfrac{r}{\epsilon}\int_0^{t\wedge\tau_N}\|w^0(s)\|_{\mathbb{L}^4}^4 ds + \int_0^{t\wedge\tau_N}\|f(s)\|_{\Ll}^2 ds +\int_0^{t\wedge\tau_N}\|\s^n(s,u^n(s))\|_{L_Q}^2 ds \\
&&+2\int_0^{t\wedge\tau_N}(\s^n(s,u^n(s)),u^n(s))_{\Ll}dW^n(s)\\
&&+\int_0^{t\wedge\tau_N}\int_Z\| H^n(u^n(s-),z)\|^2_{\Ll} \ns \\
&&+\int_0^{t\wedge\tau_N}2\int_Z (H^n(u^n(s-),z),u^n(s-))_{\Ll}\ns\\
&&+\int_0^{t\wedge\tau_N}\int_Z\| H^n(u^n(s-),z)\|^2_{\Ll} \la ds +\|u^n_0\|_{\Ll}^2 +\|\z^n_0\|_{L^2}^2 . \IEEEyesnumber
\end{IEEEeqnarray*}
Let $C=\max \{1+M+\dfrac{r}{\epsilon},\dfrac{2g^2}{\alpha}+\dfrac{2\mu^2}{\alpha}+M\}.$ Since $H^n(t)$ is strong 2-integrable w.r.t. $\n$, hence using H\"older's inequality $|H^n(t)|^2$ is strong 1-integrable w.r.t. $\n$. Then by Theorem 4.12 of \cite{rudiger}
\begin{IEEEeqnarray*}{rcl}
\label{eq12}
&&\mathbb{E}\left[\int_0^t\int_Z \| H^n(u^n(s-),z)\|^2_{\Ll} \ns\right] \\
&&\qquad\qquad\qquad \leq 2\int_0^t\int_Z\mathbb{E}\left[\| H^n(u^n(s-),z)\|^2_{\Ll} \la ds\right]. \IEEEyesnumber
\end{IEEEeqnarray*}
Also the stochastic integrals
\begin{equation*}
\int_0^{t\wedge\tau_N}(\s^n(s,u^n(s)),u^n(s))_{\Ll}dW^n(s), \int_0^{t\wedge\tau_N}\int_Z (H^n(u^n(s-),z),u^n(s-))_{\Ll}\ns
\end{equation*}
are martingales with zero averages.\\
Taking expectation of equation \eqref{eq11} and using \eqref{eq12} and the above property, we get
\begin{IEEEeqnarray*}{rcl}
\mathbb{E}&[&\|u^n(t\wedge\tau_N)\|_{\Ll}^2+\|\z^n(t\wedge\tau_N)\|_{L^2}^2]+\mathbb{E}\left[\int_0^{t\wedge\tau_N}\alpha\|u^n(s)\|^2_{\mathbb{H}_0^1}ds\right]\\
&\leq & C\mathbb{E}\left[\int_0^{t\wedge\tau_N}(\|u^n(s)\|_{\Ll}^2 + \|\z^n(s)\|_{L^2}^2) ds\right]
+\dfrac{r}{\epsilon}\mathbb{E}\left[\int_0^{t\wedge\tau_N}\|w^0(s)\|_{\mathbb{L}^4}^4 ds\right] \\&&+\mathbb{E}\left[\int_0^{t\wedge\tau_N}\|f(s)\|_{\Ll}^2 ds\right] +\mathbb{E}\left[\int_0^{t\wedge\tau_N}\|\s^n(s,u^n(s))\|_{L_Q}^2 ds\right] \\
&& +3\mathbb{E}\left[\int_0^{t\wedge\tau_N}\int_Z\| H^n(u^n(s-),z)\|^2_{\Ll} \la ds\right]\\
&&+\mathbb{E}[\|u^n_0\|_{\Ll}^2+\|\z^n_0\|_{L^2}^2].\IEEEyesnumber
\end{IEEEeqnarray*}
Using the assumption H.2
\begin{IEEEeqnarray*}{rcl}
\mathbb{E}&[&\|u^n(t\wedge\tau_N)\|_{\Ll}^2+\|\z^n(t\wedge\tau_N)\|_{L^2}^2]+\mathbb{E}\left[\int_0^{t\wedge\tau_N}\alpha\|u^n(s)\|^2_{\mathbb{H}_0^1}ds\right]\\
&\leq & C\mathbb{E}\left[\int_0^{t\wedge\tau_N}(\|u^n(s)\|_{\Ll}^2 + \|\z^n(s)\|_{L^2}^2) ds\right]
+\dfrac{r}{\epsilon}\mathbb{E}\left[\int_0^{t\wedge\tau_N}\|w^0(s)\|_{\mathbb{L}^4}^4 ds\right]\\
&&+\mathbb{E}\left[\int_0^{t\wedge\tau_N}\|f(s)\|_{\Ll}^2 ds\right]
+3K\mathbb{E}\left[\int_0^{t\wedge\tau_N} (1+ \|u^n(s)\|_{\Ll}^2) ds\right]\\
&& +\mathbb{E}[\|u^n_0\|_{\Ll}^2+\|\z^n_0\|_{L^2}^2].
\end{IEEEeqnarray*}
Hence using Gronwall's inequality and taking limit as $N\rightarrow\infty$ we have the desired a priori estimate \eqref{eq14}.
Proceeding similarly but taking supremum over $[0,T\wedge\tau_N]$ over equation \eqref{eq11} before taking expectation
\begin{IEEEeqnarray*}{rcl}
\label{eq17}
\mathbb{E}&[&\sup_{0\leq t\leq T\wedge\tau_N} (\|u^n(t)\|_{\Ll}^2+\|\z^n(t)\|_{L^2}^2)]+\mathbb{E}\left[\int_0^{T\wedge\tau_N}\alpha\|u^n(s)\|^2_{\mathbb{H}_0^1}ds\right]\\
&\leq & C\mathbb{E}\left[\int_0^{T\wedge\tau_N}(\|u^n(s)\|_{\Ll}^2 + \|\z^n(s)\|_{L^2}^2) ds\right]
+\dfrac{r}{\epsilon}\mathbb{E}\left[\int_0^{T\wedge\tau_N}\|w^0(s)\|_{\mathbb{H}_0^1} ds\right] \\ &&+\mathbb{E}\left[\int_0^{T\wedge\tau_N}\|f(s)\|_{\Ll}^2 ds\right] +3K(T\wedge\tau_N)+3K\mathbb{E}\left[\int_0^{T\wedge\tau_N}\| u^n(s)\|^2_{\Ll} ds\right]\\
&&+2\mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}|\int_0^{t}(\s^n(s,u^n(s)),u^n(s))_{\Ll}dW^n(s)|\right] + \\
&&+\mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}|\int_0^{t}2\int_Z (H^n(u^n(s-),z),u^n(s-))_{\Ll}\ns|\right]\\
&&+\mathbb{E}[\|u^n_0\|_{\Ll}^2+\|\z^n_0\|_{L^2}^2].\IEEEyesnumber
\end{IEEEeqnarray*}
Using Burkholder-Davis-Gundy inequality, Young's inequality and assumption H.2
\begin{IEEEeqnarray*}{lCl}
\label{eq18}
2\mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}|\int_0^{t}(\s^n(s,u^n(s)),u^n(s))_{\Ll}dW^n(s)|\right]\\
\leq C_3\mathbb{E}\left[\left(\int_0^{T\wedge\tau_N}(\s^n(s,u^n(s)),u^n(s))_{\Ll}ds\right)\right]^{1/2}\\
\leq C_3\mathbb{E}\left[\left(\int_0^{T\wedge\tau_N}\|\s^n(s,u^n(s))\|_{L_Q}^2\|u^n(s)\|_{\Ll}^2 ds\right)\right]^{1/2}\\
\leq C_3K\mathbb{E}\left[\left(\int_0^{T\wedge\tau_N}(1+\|u^n(s)\|_{\Ll}^2)\|u^n(s)\|_{\Ll}^2 ds\right)\right]^{1/2}\\
\leq C_3K\mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}\|u^n(s)\|_{\Ll}^2\left(\int_0^{T\wedge\tau_N}(1+\|u^n(s)\|_{\Ll}^2) ds\right)\right]^{1/2}
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{lCl}
\leq \dfrac{1}{4}\mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}\|u^n(s)\|_{\Ll}^2\right]+(C_3K)^2\mathbb{E}\left[\int_0^{T\wedge\tau_N}\|u^n(s)\|_{\Ll}^2 ds\right]\\
\quad+(C_3K)^2(T\wedge\tau_N).\IEEEyesnumber
\end{IEEEeqnarray*}
Again using Burkholder-Davis-Gundy inequality, Young's inequality and assumption H.2
\begin{IEEEeqnarray*}{lCl}
\label{eq19}
2\mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}|\int_0^{t}\int_Z (H^n(u^n(s-),z),u^n(s-))_{\Ll}\ns|\right]\\
\leq C_4 \mathbb{E}\left[\left(\int_0^{T\wedge\tau_N}\int_Z (H^n(u^n(s-),z),u^n(s-))_{\Ll}\la ds\right)\right]^{1/2}\\
\leq C_4 \mathbb{E}\left[\left(\int_0^{T\wedge\tau_N}\int_Z \|H^n(u^n(s-),z)\|_{\Ll}^2\| u^n(s-)\|_{\Ll}^2\la ds\right)\right]^{1/2}\\
\leq C_4K \mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}\|u^n(s)\|_{\Ll}^2\left(\int_0^{T\wedge\tau_N}(1+\|u^n(s)\|_{\Ll}^2) ds\right)\right]^{1/2}\\
\leq \dfrac{1}{4}\mathbb{E}\left[\sup_{0\leq t\leq T\wedge\tau_N}\|u^n(s)\|_{\Ll}^2\right]+(C_4K)^2\mathbb{E}\left[\int_0^{T\wedge\tau_N}\|u^n(s)\|_{\Ll}^2 ds\right]\\
\quad+(C_4K)^2(T\wedge\tau_N).\IEEEyesnumber
\end{IEEEeqnarray*}
Substituting equations \eqref{eq18} and \eqref{eq19} in equation \eqref{eq17} and rearranging the terms we have
\begin{IEEEeqnarray*}{rcl}
\mathbb{E}&[&\sup_{0\leq t\leq T\wedge\tau_N} (\|u^n(t)\|_{\Ll}^2+\|\z^n(t)\|_{L^2}^2)]+2\mathbb{E}\left[\int_0^{T\wedge\tau_N}\alpha\|u^n(s)\|^2_{\mathbb{H}_0^1}ds\right]\\
&\leq & C^\prime\mathbb{E}\left[\int_0^{T\wedge\tau_N}\sup\limits_{0\leq s\leq t}(\|u^n(s)\|_{\Ll}^2 + \|\z^n(s)\|_{L^2}^2) dt\right]
+\dfrac{2r}{\epsilon}\mathbb{E}\left[\int_0^{T\wedge\tau_N}\|w^0(s)\|_{\mathbb{L}^4}^4 ds\right] \\ &&+2\mathbb{E}\left[\int_0^{T\wedge\tau_N}\|f(s)\|_{\Ll}^2 ds\right] +C'' (T\wedge\tau_N)+2\mathbb{E}[\|u^n_0\|_{\Ll}^2+\|\z^n_0\|_{L^2}^2],
\end{IEEEeqnarray*}
where $C^\prime=2[C+(C_3K)^2+(C_4K)^2+3K]$ and $C''=2[(C_3K)^2+(C_4K)^2+3K]$.
Now taking limit as $N\rightarrow\infty$ and Gronwall's inequality we get the desired a priori estimate \eqref{eq15}.
\end{proof}
\subsection{$L^p$ energy estimate}
Let $2<p<\infty$. We assume that $\s$ and $H$ satisfy the following hypotheses:
\begin{enumerate}
\item[Hp.1] $\s\in C([0,T]\times\h(\mathcal{O});L_Q(H_0,\Ll)), H\in\mathbb{H}^p_\lambda([0,T]\times Z;\Ll(\mathcal{O}))$,
\item[Hp.2] For all $t\in(0,T)$, there exists a positive constant $K$ such that for all $u\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)\|_{L_Q}^2+\int_Z\|H(u,z)\|_{\Ll}^2\la\leq K(1+\|u\|_{\Ll}^2),
\end{equation*}
\item[Hp.3] For all $t\in(0,T)$, there exists a positive constant $L$ such that for all $u,v\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)-\s(t,v)\|_{L_Q}^2+\int_Z \|H(u,z)-H(v,z)\|_{\Ll}^2\la
\leq L(\|u-v\|_{\Ll}^2),
\end{equation*}
\item[Hp.4] For all $t\in(0,T)$, there exists a positive constant $M$ such that for all $2<p<\infty$ and $u\in\Ll(\mathcal{O})$
\begin{equation*}
\int_Z\|H(u,z)\|_{\Ll}^p\la\leq M(1+\|u\|_{\Ll}^p).
\end{equation*}
\end{enumerate}
\begin{proposition}
Let $2<p<\infty$ and
\begin{equation}
\left\{\begin{array}{l}
w^0\in L^p(0,T;\h(\mathcal{O})),\;f\in L^p(0,T;\Ll(\mathcal{O})),\\
\s\in C([0,T]\times\h(\mathcal{O});L_Q(H_0,\Ll)),\;H\in\mathbb{H}^p_\lambda([0,T]\times Z;\Ll(\mathcal{O})),\\
u_0\in\Ll(\mathcal{O}),\z_0\in L^2(\mathcal{O}).
\end{array}\right.
\end{equation}
Let $u^n(t)$ be an adapted process in $L^p([0,T];\Ll(\mathcal{O}))\cap L^2([0,T];\h(\mathcal{O}))\cap\mathcal{D}(0,T,\Ll(\mathcal{O}))$ which solves the differential equations \eqref{eqn5} and \eqref{eqn6}. Then we have the following a priori estimate:
\begin{IEEEeqnarray}{l}
\label{eq16}
\mathbb{E}[\sup\limits_{0\leq t\leq T}\|u^n(t)\|_{\Ll}^p+\|\z^n(t)\|_{L^2}^p]+\alpha p\mathbb{E}[\int_0^T \|u^n(t)\|_{\Ll}^{p-2}\|u^n(t)\|_{\h}^2 dt]\leq C_{1(p)},\nonumber\\
\end{IEEEeqnarray}
where the constant $C_{1(p)}$ depends on the coefficients $\alpha, g,M,\mu$ and the norms $\|f\|_{L^p(0,T;\Ll)},$ $\|w^0\|_{L^p(0,T;\h)},\|u^n(0)\|_{\Ll}, \|\z^n(0)\|_{L^2}$ and $T$.
\end{proposition}
The proposition can be proved using the same ideas used in Proposition \ref{prop}.
\subsection{Existence Result}
\begin{definition}
A path-wise strong solution $u$ is defined on a given filtered probability space $(\Omega,\mathcal{F},\mathcal{F}_t,P)$ as a $L^\infty(0,T;\mathbb{L}^2(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2(\mathcal{O}))$ valued function which satisfies the stochastic tide equations \eqref{eq21} and \eqref{eq22} in the weak sense and also the energy inequalities in Proposition \ref{prop}
\end{definition}
\begin{theorem}
\label{thmuniq}
Let $f,u_0$ and $\z_0$ be such that
\begin{equation}
f\in L^2(0,T;\Ll(\mathcal{O})),\quad u_0\in \Ll(\mathcal{O}),\quad \z_0\in L^2(\mathcal{O}).
\end{equation}
Suppose $\s$ and $H$ satisfy the conditions in H.1-H.3, then there exist a unique path-wise strong solution $u(t,x,\omega)$ and $\z(t,x,\omega)$ with the regularity
\begin{equation}
\left\{\begin{array}{ll}
&u\in L^2(\Omega; L^\infty(0,T;\mathbb{L}^2(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2(\mathcal{O}))),\\
&\z,\dot{\z}\in L^2(\Omega ;L^2(0,T;L^2(\mathcal{O}))).
\end{array}\right.
\end{equation}
satisfying the stochastic equations \eqref{eq21}-\eqref{eq22} and the apriori bounds \eqref{eq14}-\eqref{eq15}
\end{theorem}
\begin{proof}
\textbf{Existence}:\\
Define
\begin{equation*}
F(u)=Au+B(u)-f.
\end{equation*}
Then
\begin{equation*}
du^n(t)+F(u^n(t))dt+g\nabla\z^n(t) dt
=\s^n(t,u^n(t))dW^n(t)
+\int_Z H^n(u^n(t-),z) \n .
\end{equation*}
Using the a priori estimates \eqref{eq14}-\eqref{eq15}, it follows from the Banach-Alaoglu theorem that along a subsequence, the Galerkin approximations $\{u^n\}$ have the following limits:
\begin{IEEEeqnarray*}{rCl}
&&u^n\rightarrow u\; \text{weakly star in }L^2(\Omega ;L^\infty(0,T,\mathbb{L}^2(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))),\\
&&\z^n\rightarrow\z\;\text{weakly in }L^2(\Omega ;L^2(0,T;L^2(\mathcal{O}))),\\
&&F(u^n)\rightarrow F_0\;\text{weakly in }L^2(\Omega ;L^2(0,T;\mathbb{H}^{-1}(\mathcal{O}))),\\
&&\s(\cdot ,u^n)\rightarrow\s_0\;\text{weakly in }L^2(\Omega ;L^2(0,T;L_Q)),\\
&&H^n(u^n,\cdot)\rightarrow H_0\;\text{weakly in }\mathbb{H}_\lambda^2 ([0,T]\times Z;\Ll),
\end{IEEEeqnarray*}
where $u$ has the differential form
\begin{equation*}
du(t)+F_0(t)dt+g\nabla\z(t) dt
=\s_0(t)dW(t)
+\int_Z H_0(t,z) \n ,
\end{equation*}
weakly in $L^2(\Omega;L^2(0,T;\mathbb{H}^{-1}(\mathcal{O})))$.\\\\
Applying It\^o's formula to $e^{-Lt}|\sqrt{h}x|^2$ and the process $u^n(t)$
\begin{IEEEeqnarray*}{rCl}
d[e^{-Lt}\|\sqrt{h}u^n(t)\|_{\Ll}^2 &+& e^{-Lt}g\|\z^n(t)\|_{L^2}^2]\\
&&=-Le^{-Lt}\|\sqrt{h}u^n(t)\|_{\Ll}^2 dt -Le^{-Lt}g\|\z^n(t)\|_{L^2}^2 dt\\
&&-e^{-Lt}(2F(u^n(t)),hu^n(t))_{\Ll} dt\\
&&+2e^{-Lt}(\s^n(t,u^n(t)),hu^n(t))_{\Ll}dW^n(t)\\
&&+e^{-Lt}\|\sqrt{h}\s^n(t,u^n(t))\|_{L_Q}^2dt \\
&&+2e^{-Lt}\int_Z (H^n(u^n(t-),z),\sqrt{h}u^n(t-))_{\Ll}\n \\
&&+e^{-Lt}\int_Z\|\sqrt{h} H^n(u^n(t-),z)\|^2_{\Ll} N(dt,dz).
\end{IEEEeqnarray*}
Integrating from $0$ to $T$ and taking expectation
\begin{IEEEeqnarray*}{lCl}
\mathbb{E}&[&e^{-LT}\|\sqrt{h}u^n(T)\|_{\Ll}^2 + e^{-LT}g\|\z^n(T)\|_{L^2}^2 - \|\sqrt{h}u^n(0)\|_{\Ll}^2 - g\|\z^n(0)\|_{L^2}^2]\\
&=&-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u^n(t)\|_{\Ll}^2 dt] -\mathbb{E}[\int_0^T Le^{-Lt}g\|\z^n(t)\|_{L^2}^2 dt]\\
&&-2\mathbb{E}[\int_0^T e^{-Lt}(F(u^n(t)),hu^n(t))_{\Ll} dt]\\
&&+\mathbb{E}[\int_0^T 2e^{-Lt}(\s^n(t,u^n(t)),hu^n(t))_{\Ll}dW^n(t)]\\
&&+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s^n(t,u^n(t))\|_{L_Q}^2dt]\\
&&+\mathbb{E}[\int_0^T 2e^{-Lt}\int_Z (H^n(u^n(t-),z),\sqrt{h}u^n(t-))_{\Ll}\n]
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{lCl}
&&+\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h} H^n(u^n(t-),z)\|^2_{\Ll} N(dt,dz)].
\end{IEEEeqnarray*}
Since
\begin{equation*}
\mathbb{E}[\int_0^T 2e^{-Lt}(\s^n(t,u^n(t)),hu^n(t))_{\Ll}dW^n(t)],
\end{equation*}
and
\begin{equation*}
\mathbb{E}[\int_0^T 2e^{-Lt}\int_Z (H^n(u^n(t-),z),\sqrt{h}u^n(t-))_{\Ll}\n],
\end{equation*}
are martingales with zero averages, and since $\|\sqrt{h} H^n(u^n(t-),z)\|^2$ is strong 1-integrable w.r.t $\n$,
\begin{IEEEeqnarray*}{lrl}
\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h} H^n(u^n(t-),z)\|^2_{\Ll} N(dt,dz)]\\
\qquad\qquad\qquad\qquad=\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h} H^n(u^n(t-),z)\|^2_{\Ll} \la dt].
\end{IEEEeqnarray*}
So
\begin{IEEEeqnarray*}{lCl}
\mathbb{E}[e^{-LT}\|\sqrt{h}u^n(T)\|_{\Ll}^2 + e^{-LT}g\|\z^n(T)\|_{L^2}^2 - \|\sqrt{h}u^n(0)\|_{\Ll}^2 - g\|\z^n(0)\|_{L^2}^2]\\
=-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u^n(t)\|_{\Ll}^2 dt] -\mathbb{E}[\int_0^T Le^{-Lt}g\|\z^n(t)\|_{L^2}^2 dt]\\
-2\mathbb{E}[\int_0^T e^{-Lt}(F(u^n(t)),hu^n(t))_{\Ll} dt]
+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s^n(t,u^n(t))\|_{L_Q}^2dt] \\
+\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h} H^n(u^n(t-),z)\|^2_{\Ll} \la dt].
\end{IEEEeqnarray*}
Using the lower semi-continuity of $\Ll$-norm
\begin{IEEEeqnarray*}{lCl}
\label{eq25}
\liminf_n \{-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u^n(t)\|_{\Ll}^2 dt] -\mathbb{E}[\int_0^T Le^{-Lt}g\|\z^n(t)\|_{L^2}^2 dt]\\
-2\mathbb{E}[\int_0^T e^{-Lt}(F(u^n(t)),hu^n(t))_{\Ll} dt]
+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s^n(t,u^n(t))\|_{L_Q}^2dt] \\
+\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h} H^n(u^n(t-),z)\|^2_{\Ll} \la dt]\}\\
=\liminf_n \{\mathbb{E}[e^{-LT}\|\sqrt{h}u^n(T)\|_{\Ll}^2 + e^{-LT}g\|\z^n(T)\|_{L^2}^2\\
\qquad\qquad\qquad\qquad\qquad - \|\sqrt{h}u^n(0)\|_{\Ll}^2 - g\|\z^n(0)\|_{L^2}^2]\}\\
\geq \mathbb{E}[e^{-LT}\|\sqrt{h}u(T)\|_{\Ll}^2 + e^{-LT}g\|\z(T)\|_{L^2}^2 - \|\sqrt{h}u(0)\|_{\Ll}^2 - g\|\z(0)\|_{L^2}^2]\\
=-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u(t)\|_{\Ll}^2 dt] -\mathbb{E}[\int_0^T Le^{-Lt}g\|\z(t)\|_{L^2}^2 dt]\\
-2\mathbb{E}[\int_0^T e^{-Lt}(F_0(t),hu(t))_{\Ll} dt]
+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s_0(t)\|_{L_Q}^2dt] \\
+\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h} H_0(t,z)\|^2_{\Ll} \la dt].\IEEEyesnumber
\end{IEEEeqnarray*}
Using the monotonicity property of $F$ and assumption H.3, we have for all $v\in L^2(\Omega; L^\infty(0,T;\mathbb{L}^2_{n}(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2_n(\mathcal{O})))$
\begin{IEEEeqnarray*}{lCl}
-2\mathbb{E}[\int_0^T e^{-Lt}(F(u^n(t))-F(v(t)),hu^n(t)-hv(t))_{\Ll} dt]\\
+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s^n(t,u^n(t))-\sqrt{h}\s^n(t,v(t))\|_{L_Q}^2dt] \\
+\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h}H^n(u^n(t-),z)-\sqrt{h}H^n(v(t-),z)\|^2_{\Ll} \la dt]\\
-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u^n(t)-\sqrt{h}v(t)\|_{\Ll}^2 dt]\\
-\mathbb{E}[\int_0^T Le^{-Lt}g\|\z-\z^n\|_{L^2}^2 dt]\\
\leq 0.\IEEEyesnumber
\end{IEEEeqnarray*}
Rearranging the terms
\begin{IEEEeqnarray*}{lCl}
&-2&\mathbb{E}[\int_0^T e^{-Lt}(F(u^n(t)),hu^n(t))_{\Ll} dt]
+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s^n(t,u^n(t))\|_{L_Q}^2dt] \\
&+&\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h}H^n(u^n(t-),z)\|^2_{\Ll} \la dt]
-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u^n(t)\|_{\Ll}^2 dt]\\
&-&\mathbb{E}[\int_0^T Le^{-Lt}g\|\z^n\|_{L^2}^2 dt]\\
&\leq & -2\mathbb{E}[\int_0^T e^{-Lt}(F(v(t)),hu^n(t)-hv(t))_{\Ll} dt]
-\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s^n(t,v(t))\|_{L_Q}^2dt] \\
&&+2\mathbb{E}[\int_0^T e^{-Lt}(\sqrt{h}\s^n(u^n(t)),\sqrt{h}\s^n(v(t)))_{L_Q}dt]\\
&&-\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h}H^n(v(t-),z)\|^2_{\Ll} \la dt]\\
&&+2\mathbb{E}[\int_0^T e^{-Lt}\int_Z\ (\sqrt{h}H^n(u^n(t-),z),\sqrt{h}H^n(v(t-),z))_{\Ll} \la dt]\\
&&+\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}v(t)\|_{\Ll}^2 dt]
-2\mathbb{E}[\int_0^T Le^{-Lt}(\sqrt{h}u^n(t),\sqrt{h}v(t))_{\Ll} dt]\\
&&+\mathbb{E}[\int_0^T Le^{-Lt}g\|\z\|_{L^2}^2 dt]
-2\mathbb{E}[\int_0^T Le^{-Lt}g(\z^n(t),\z(t))_{L^2} dt]\\
&&-2\mathbb{E}[\int_0^T e^{-Lt}(F(u^n(t)),hv(t))_{\Ll} dt].
\end{IEEEeqnarray*}
Taking limit in $n$ and using \eqref{eq25}
\begin{IEEEeqnarray*}{lCl}
&-2&\mathbb{E}[\int_0^T e^{-Lt}(F_0(t),hu(t))_{\Ll} dt]
+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s_0(t)\|_{L_Q}^2dt] \\
&+&\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h}H_0(t,z)\|^2_{\Ll} \la dt]
-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u(t)\|_{\Ll}^2 dt]
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{lCl}
&-&\mathbb{E}[\int_0^T Le^{-Lt}g\|\z\|_{L^2}^2 dt]\\
&\leq &-2\mathbb{E}[\int_0^T e^{-Lt}(F(v(t)),hu(t)-hv(t))_{\Ll} dt]\\
&&-\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s(t,v(t))\|_{L_Q}^2dt] \\
&&+2\mathbb{E}[\int_0^T e^{-Lt}(\sqrt{h}\s_0(t),\sqrt{h}\s(v(t)))_{L_Q}dt]\\
&&-\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h}H(v(t-),z)\|^2_{\Ll} \la dt]\\
&&+2\mathbb{E}[\int_0^T e^{-Lt}\int_Z\ (\sqrt{h}H_0(t,z),\sqrt{h}H(v(t-),z))_{\Ll} \la dt]\\
&&+\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}v(t)\|_{\Ll}^2 dt]
-2\mathbb{E}[\int_0^T Le^{-Lt}(\sqrt{h}u(t),\sqrt{h}v(t))_{\Ll} dt]\\
&&+\mathbb{E}[\int_0^T Le^{-Lt}g\|\z\|_{L^2}^2 dt]
-2\mathbb{E}[\int_0^T Le^{-Lt}g(\z(t),\z(t))_{L^2} dt]\\
&&-2\mathbb{E}[\int_0^T e^{-Lt}(F_0(t),hv(t))_{\Ll} dt].
\end{IEEEeqnarray*}
Rearranging the terms
\begin{IEEEeqnarray*}{lCl}
-2\mathbb{E}[\int_0^T e^{-Lt}(F_0(t)-F(v(t)),hu(t)-hv(t))_{\Ll} dt]\\
+\mathbb{E}[\int_0^T e^{-Lt}\|\sqrt{h}\s_0(t)-\sqrt{h}\s(t,v(t))\|_{L_Q}^2dt] \\
+\mathbb{E}[\int_0^T e^{-Lt}\int_Z\|\sqrt{h}H_0(t,z)-\sqrt{h}H(v(t-),z)\|^2_{\Ll} \la dt]\\
-\mathbb{E}[\int_0^T Le^{-Lt}\|\sqrt{h}u(t)-\sqrt{h}v(t)\|_{\Ll}^2 dt]\\
\leq 0.
\end{IEEEeqnarray*}
This estimate holds for any $v\in L^2(\Omega; L^\infty(0,T;\mathbb{L}^2_{n}(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2_n(\mathcal{O})))$, for any $n\in\mathbb{N}$. It is obvious from the density argument that the above inequality remains the same for any $v\in L^2(\Omega; L^\infty(0,T;\mathbb{L}^2(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2(\mathcal{O})))$. In fact, for any $v\in L^2(\Omega; L^\infty(0,T;\mathbb{L}^2(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2(\mathcal{O})))$ there exists a strongly convergent subsequence $v_m\in L^2(\Omega; L^\infty(0,T;\mathbb{L}^2_{m}(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2_{m}(\mathcal{O})))$ satisfying the above inequality.\\\\
Taking $v=u$, we get $\s_0(t)=\s(t,u(t))$ and $H_0(t,z)=H(u(t),z)$.\\
Now we take $v=u+\lambda w$ where $\lambda>0$ and $w$ is an adapted process in $L^2(\Omega; L^\infty(0,T;\mathbb{L}^2(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O}))\cap \mathcal{D}(0,T;\mathbb{L}^2(\mathcal{O})))$.\\
Then we have
\begin{equation}
\lambda\mathbb{E}[\int_0^T e^{-Lt}(F(u(t)+\lambda w(t)),hw(t))_{\Ll} dt]\geq \lambda\mathbb{E}[\int_0^T e^{-Lt}(F_0(t),hw(t))_{\Ll} dt].
\end{equation}
Dividing by $\lambda$ on both sides on the inequality above and letting $\lambda$ go to 0, we have by the hemicontinuity of $F$
\begin{equation*}
\mathbb{E}[\int_0^T e^{-Lt}(F(u(t))-F_0(t),hw(t))_{\Ll} dt]\geq 0.
\end{equation*}
Since $w$ is arbitrary and $h$ is a positive, bounded, continuously differentiable function, $F_0(t)=F(u(t))$. This proves the existence of a strong solution.\\\\
\textbf{Uniqueness}:\\
If $u$ and $v$ are two solutions then $w=u-v$ solves the stochastic differential equation
\begin{IEEEeqnarray*}{rcl}
dw(t)+Aw(t)dt&+&g\nabla(\z(t)-\tz(t))dt\\
&=&B(v(t))-B(u(t))dt+(\s(t,u(t))-\s(t,v(t)))dW(t)\\
&&+\int_Z (H(u(t-),z)-H(v(t-),z))\n .
\end{IEEEeqnarray*}
Applying It\^o's formula to $|x|^2$ and to the process $w(t)$
\begin{IEEEeqnarray*}{rcl}
d(\|w(t)\|_{\Ll}^2)&+&2\alpha\|w(t)\|_{\h}^2 dt+2g(\nabla(\z(t)-\tz(t)),w(t))_{\Ll} dt\\
&=&2(B(v(t))-B(u(t)),u(t)-v(t))_{\Ll}dt\\
&&+2(\s(t,u(t))-\s(t,v(t)),w(t))_{\Ll} dW(t)\\
&&+\|\s(t,u(t))-\s(t,v(t))\|_{L_Q}^2 dt\\
&&+\int_Z \|H(u(t-),z)-H(v(t-),z)\|^2_{\Ll} N(dt,dz) \\
&&+\int_Z \left((H(u(t-),z)-H(v(t-),z)),w(t-)\right)_{\Ll}\n .
\end{IEEEeqnarray*}
Using the result from Lemma \ref{mon} and equation \eqref{eq26}
\begin{IEEEeqnarray*}{rcl}
\label{eq27}
d&(&\|w(t)\|_{\Ll}^2)+2\alpha\|w(t)\|_{\h}^2 dt\\
&\leq &\dfrac{2g^2}{\alpha}\|\z(t)-\tz(t)\|_{L^2}^2 dt+\dfrac{\alpha}{2}\|w(t)\|_{\h}^2 dt\\
&&+2(\s(t,u(t))-\s(t,v(t)),w(t))_{\Ll} dW(t)+\|\s(t,u(t))-\s(t,v(t))\|_{L_Q}^2 dt\\
&&+\int_Z \|H(u(t-),z)-H(v(t-),z)\|^2_{\Ll} N(dt,dz) \\
&&+\int_Z \left((H(u(t-),z)-H(v(t-),z)),w(t-)\right)_{\Ll}\n .\IEEEyesnumber
\end{IEEEeqnarray*}
Notice that
\begin{equation*}
d(\z(t)-\tz(t))+Div(hw(t)) dt=0.
\end{equation*}
Taking inner product with $\z(t)-\tz(t)$, we have as in equation \eqref{eq9}
\begin{IEEEeqnarray*}{lcl}
\label{eq28}
d(\|\z(t)-\tz(t)\|_{L^2}^2)\\
\leq M\|w(t)\|_{\Ll}^2 dt + \left(\dfrac{2\mu^2}{\alpha}+M\right)\|\z(t)-\tz(t)\|_{L^2}^2 dt + \dfrac{\alpha}{2}\|w(t)\|_{\h}^2 dt.\IEEEyesnumber
\end{IEEEeqnarray*}
Let
\begin{equation*}
C=\dfrac{2g^2}{\alpha}+\dfrac{2\mu^2}{\alpha}+M.
\end{equation*}
Adding equations \eqref{eq27} and \eqref{eq28}
\begin{IEEEeqnarray*}{rcl}
d&(&\|w(t)\|_{\Ll}^2+\|\z(t)-\tz(t)\|_{L^2}^2)+\alpha\|w(t)\|_{\h}^2 dt\\
&\leq &C(\|w(t)\|_{\Ll}^2+\|\z(t)-\tz(t)\|_{L^2}^2) dt\\
&&+2(\s(t,u(t))-\s(t,v(t)),w(t))_{\Ll} dW(t)+\|\s(t,u(t))-\s(t,v(t))\|_{L_Q}^2 dt\\
&&+\int_Z \|H(u(t-),z)-H(v(t-),z)\|^2_{\Ll} N(dt,dz)\\
&& +\int_Z \left((H(u(t-),z)-H(v(t-),z)),w(t-)\right)_{\Ll}\n.
\end{IEEEeqnarray*}
Integrating from 0 to $T$ and taking expectation
\begin{IEEEeqnarray*}{rcl}
\mathbb{E}&[&\|w(T)\|_{\Ll}^2+\|\z(T)-\tz(T)\|_{L^2}^2]+\mathbb{E}[\int_0^T \alpha\|w(t)\|_{\h}^2 dt]\\
&\leq &C\mathbb{E}[\int_0^T(\|w(t)\|_{\Ll}^2+\|\z(t)-\tz(t)\|_{L^2}^2) dt]\\
&&+\mathbb{E}[\int_0^T \|\s(t,u(t))-\s(t,v(t))\|_{L_Q}^2 dt]\\
&&+\mathbb{E}[\int_0^T \int_Z \|H(u(t-),z)-H(v(t-),z)\|^2_{\Ll} \la dt]\\
&&+\mathbb{E}[\|w(0)\|_{\Ll}^2+\|\z(0)-\tz(0)\|_{L^2}^2].
\end{IEEEeqnarray*}
Using assumption H.3
\begin{IEEEeqnarray*}{rcl}
\mathbb{E} &[&\|w(T)\|_{\Ll}^2+\|\z(T)-\tz(T)\|_{L^2}^2]+\mathbb{E}[\int_0^T \alpha\|w(t)\|_{\h}^2 dt]\\
&\leq &C\mathbb{E}[\int_0^T(\|w(t)\|_{\Ll}^2+\|\z(t)-\tz(t)\|_{L^2}^2) dt]
+L\mathbb{E}[\int_0^T \|w(t)\|_{\Ll}^2 dt]\\
&&+\mathbb{E}[\|w(0)\|_{\Ll}^2+\|\z(0)-\tz(0)\|_{L^2}^2].
\end{IEEEeqnarray*}
In particular
\begin{IEEEeqnarray*}{rcl}
\mathbb{E}[\|w(T)\|_{\Ll}^2&+&\|\z(T)-\tz(T)\|_{L^2}^2 ] \\
&\leq &\;\mathbb{E}[\|w(0)\|_{\Ll}^2+\|\z(0)-\tz(0)\|_{L^2}^2]\\
&&+(C+L)\mathbb{E}[\int_0^T(\|w(t)\|_{\Ll}^2+\|\z(t)-\tz(t)\|_{L^2}^2) dt].\\
\end{IEEEeqnarray*}
Hence the uniqueness of pathwise strong solution follows using Gronwall's inequality.
\end{proof}
\subsection{$\h$ regularity}
We assume that $\s$ and $H$ satisfy the following hypotheses:
\begin{enumerate}
\item[A.1] $\s\in C([0,T]\times \h(\mathcal{O});L_Q(H_0,\h)), H\in\mathbb{H}^2_\lambda([0,T]\times Z;\h(\mathcal{O}))$,
\item[A.2] For all $t\in(0,T)$, there exists a positive constant $K$ such that for all $u\in\h(\mathcal{O})$
\begin{equation*}
\|\nabla \s(t,u)\|_{L_Q}^2+\int_Z\|\nabla H(u,z)\|_{\Ll}^2\la\leq K(1+\|u\|_{\h}^2),
\end{equation*}
\item[A.3] For all $t\in(0,T)$, there exists a positive constant $L$ such that for all $u,v\in\h(\mathcal{O})$
\begin{equation*}
\|\nabla\s(t,u)-\nabla\s(t,v)\|_{L_Q}^2+\int_Z \|\nabla H(u,z)-\nabla H(v,z)\|_{\Ll}^2\la
\leq L(\|u-v\|_{\h}^2).
\end{equation*}
\end{enumerate}
\begin{proposition}
Let
\begin{equation}
\left\{\begin{array}{l}
w^0\in L^2(0,T;\h(\mathcal{O})),\;f\in L^2(0,T;\h(\mathcal{O})),\\
\s\in C([0,T]\times \h(\mathcal{O});L_Q(H_0,\h)),\;H\in\mathbb{H}^2\lambda([0,T]\times Z;\h(\mathcal{O})),\\
u_0\in\h(\mathcal{O}),\z_0\in H^1_0(\mathcal{O}).
\end{array}\right.
\end{equation}
Let $u^n(t)$ be an adapted process in $L^2([0,T];W^{2,2}_0(\mathcal{O}))\cap L^\infty([0,T];\h(\mathcal{O}))\cap\mathcal{D}(0,T,\h(\mathcal{O}))$ which solves the differential equations \eqref{eqn5} and \eqref{eqn6}. Then we have
\begin{IEEEeqnarray*}{lr}
\lim\limits_{T\rightarrow 0}\sup\limits_N \mathbb{P}(\sup_{0\leq t\leq T\wedge\tau_N}\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2+ \frac{\alpha}{2}\int_0^{T\wedge\tau_N}\|\triangle u^n(t)\|^2_{\Ll} dt\\
\qquad\qquad\qquad\qquad\qquad > (N-1)+\|u^n_0\|_{\h}^2+\|\z^n_0\|_{H^1_0}^2)=0.\IEEEyesnumber
\end{IEEEeqnarray*}
\end{proposition}
\begin{proof}
Applying It\^o's formula to $|x|^2$ and to the process $\nabla u^n(t)$
\begin{IEEEeqnarray*}{lr}
d(\|u^n(t)\|_{\h}^2)+ 2\alpha\|\triangle u^n(t)\|^2_{\Ll}dt+2(\nabla B(u^n),\nabla u^n)_{\Ll}dt\\
+2(\nabla (g\nabla\z^n) ,\nabla u^n)_{\Ll}dt\\
=2(\nabla f,\nabla u^n(t))_{\Ll}dt+2(\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)\\
\quad+\|\nabla\s^n(t,u^n(t))\|_{L_Q}^2dt +\int_Z\|\nabla H^n(u^n(t-),z)\|_{\Ll}^2 N(dt,dz) \\
\quad+2\int_Z (\nabla H^n(u^n(t-),z),\nabla u^n(t-))_{\Ll}\n .\IEEEyesnumber
\end{IEEEeqnarray*}
Using divergence theorem
\begin{IEEEeqnarray*}{lr}
d(\|u^n(t)\|_{\h}^2)+ 2\alpha\|\triangle u^n(t)\|^2_{\Ll} dt\\
=2( B(u^n),\triangle u^n)_{\Ll}dt+2(g\nabla \z^n ,\triangle u^n)_{\Ll}dt+2(\nabla f,\nabla u^n(t))_{\Ll}dt\\
\quad+2(\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)+\|\nabla\s^n(t,u^n(t))\|_{L_Q}^2dt\\
\quad+\int_Z\|\nabla H^n(u^n(t-),z)\|_{\Ll}^2 N(dt,dz) \\
\quad+2\int_Z (\nabla H^n(u^n(t-),z),\nabla u^n(t-))_{\Ll}\n . \IEEEyesnumber
\end{IEEEeqnarray*}
Using Young's inequality
\begin{equation}
2|g(\nabla\z^n(t) ,\triangle u^n(t))_{\Ll}|\leq \dfrac{2g^2}{\alpha}\|\z^n(t)\|_{H_0^1}^2+\dfrac{\alpha}{2}\|\triangle u^n(t)\|_{\Ll}^2,
\end{equation}
and
\begin{equation}
(\nabla f,\nabla u^n(t))_{\Ll}\leq\dfrac{1}{2}[\|f(t)\|_{\h}^2+\|u^n(t)\|_{\h}^2].
\end{equation}
Using Lemma \ref{lemw}, equation \eqref{e2} and Young's inequality
\begin{IEEEeqnarray*}{lrl}
2(B(u^n),\triangle u^n)_{\Ll}&\leq &\dfrac{2}{\alpha}\|B(u^n)\|_{\Ll}^2+\frac{\alpha}{2}\|\triangle u(t)\|_{\Ll}^2\\
&\leq & C_2(\|u^n(t)\|^4_{\mathbb{L}^4}+\|w^0(t)\|_{\mathbb{L}^4}^4)+\frac{\alpha}{2}\|\triangle u(t)\|_{\Ll}^2\\
&\leq & C_2(2\|u^n(t)\|^2_{\Ll}\|u^n(t)\|_{\h}^2+\|w^0(t)\|_{\mathbb{L}^4}^4)+\frac{\alpha}{2}\|\triangle u(t)\|_{\Ll}^2.
\end{IEEEeqnarray*}
Now using the a-priori estimate given by equation \eqref{eq14} we get
\begin{equation}
2(B(u^n),\triangle u^n)_{\Ll}\leq 2C_2C_{1(2)}\|u^n(t)\|_{\h}^2+C_2\|w^0(t)\|_{\mathbb{L}^4}^4+\frac{\alpha}{2}\|\triangle u(t)\|_{\Ll}^2.
\end{equation}
Hence
\begin{IEEEeqnarray*}{lr}
d(\|u^n(t)\|_{\h}^2)+ \alpha\|\triangle u^n(t)\|^2_{\Ll} dt\\
\leq(2C_2C_{1(2)}+1)\|u^n(t)\|_{\h}^2dt+\dfrac{2g^2}{\alpha}\|\z^n(t)\|_{H_0^1}^2+\|f(t)\|_{\h}^2dt+C_2\|w^0(t)\|_{\mathbb{L}^4}^4\\
+2(\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)+\|\nabla\s^n(t,u^n(t))\|_{L_Q}^2dt\\
+\int_Z\|\nabla H^n(u^n(t-),z)\|_{\Ll}^2 N(dt,dz) \\
+2\int_Z (\nabla H^n(u^n(t-),z),\nabla u^n(t-))_{\Ll}\n . \IEEEyesnumber
\end{IEEEeqnarray*}
Using assumption A.2
\begin{IEEEeqnarray*}{lr}
\label{eeeeq}
d(\|u^n(t)\|_{\h}^2)+ \alpha\|\triangle u^n(t)\|^2_{\Ll} dt\\
\leq(2C_2C_{1(2)}+1)\|u^n(t)\|_{\h}^2dt+\dfrac{2g^2}{\alpha}\|\z^n(t)\|_{H_0^1}^2+\|f(t)\|_{\h}^2dt+C_2\|w^0(t)\|_{\mathbb{L}^4}^4\\
+2(\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)+K(1+\|u^n(t)\|_{\h}^2)dt\\
+\int_Z\|\nabla H^n(u^n(t-),z)\|_{\Ll}^2 N(dt,dz)\\
+2\int_Z (\nabla H^n(u^n(t-),z),\nabla u^n(t-))_{\Ll}\n .\IEEEyesnumber
\end{IEEEeqnarray*}
Using \eqref{eqn6}
\begin{equation}
\dfrac{1}{2}\dfrac{d}{dt}\|\z^n(t)\|_{H_0^1}^2=(Div(hu^n(t)),\triangle\z^n(t))_{L^2}.
\end{equation}
Now
\begin{IEEEeqnarray*}{rCl}
|(Div(hu^n(t)),\triangle\z^n(t))_{L^2}|&\leq &\mu\| u^n(t)\|_{\mathbb{H}_0^1}\|\triangle\z^n(t)\|_{L^2}
+M\|u^n(t)\|_{\Ll}\|\triangle\z^n(t)\|_{L^2}\\
&\leq &\dfrac{\mu}{2}[\dfrac{\alpha}{2\mu}\|u^n(t)\|_{\mathbb{H}_0^1}^2 +\dfrac{2\mu}{\alpha}\|\triangle\z^n(t)\|_{L^2}^2]\\
&&+\dfrac{M}{2}[\|u^n(t)\|_{\Ll}^2+\|\triangle\z^n(t)\|_{L^2}^2].
\end{IEEEeqnarray*}
Using Poincar\'e's inequality
\begin{equation}
\|u^n\|_{\Ll}\leq C_P \|u^n\|_{\h},
\end{equation}
and rearranging
\begin{equation}
\label{eeeeq1}
d(\|\z^n(t)\|_{\h}^2)\leq \left(MC_P^2+\dfrac{\alpha}{2}\right)\|u^n(t)\|_{\h}^2 dt+(\dfrac{2\mu^2}{\alpha}+M)\|\triangle\z^n(t)\|_{L^2}^2 dt.
\end{equation}
Adding equations \eqref{eeeeq} and \eqref{eeeeq1}
\begin{IEEEeqnarray*}{lr}
d(\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2)+ \frac{\alpha}{2}\|\triangle u^n(t)\|^2_{\Ll} dt\\
\leq(2C_2C_{1(2)}+1+MC_P^2+\dfrac{\alpha}{2})\|u^n(t)\|_{\h}^2dt+\dfrac{2g^2}{\alpha}\|\z^n(t)\|_{H_0^1}^2 dt\\
+(\dfrac{2\mu^2}{\alpha}+M)\|\triangle\z^n(t)\|_{L^2}^2 dt+\|f(t)\|_{\h}^2dt+C_2\|w^0(t)\|_{\mathbb{L}^4}^4\\
+2(\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)+K(1+\|u^n(t)\|_{\h}^2)dt\\
+\int_Z\|\nabla H^n(u^n(t-),z)\|_{\Ll}^2 N(dt,dz)\\
+2\int_Z (\nabla H^n(u^n(t-),z),\nabla u^n(t-))_{\Ll}\n . \IEEEyesnumber
\end{IEEEeqnarray*}
Define
\begin{IEEEeqnarray}{lrl}
\tau_N :=\inf\{t:\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2&+&\int_0^t\frac{\alpha}{2}\|\triangle u^n(s)\|^2_{\Ll}ds+\|\triangle\z^n(t)\|_{\Ll}^2\nonumber\\
&&>N+\|u^n_0\|_{\h}^2+\|\z^n_0\|_{\h}^2\}.
\end{IEEEeqnarray}
Integrating and taking supremum over $0\leq t\leq T\wedge\tau_N$
\begin{IEEEeqnarray*}{lr}
\sup_{0\leq t\leq T\wedge\tau_N}\left(\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2\right)+ \frac{\alpha}{2}\int_0^{T\wedge\tau_N}\|\triangle u^n(t)\|^2_{\Ll} dt\\
\leq S\int_0^{T\wedge\tau_N}(\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2 +\|\triangle\z^n(t)\|_{L^2}^2+\|f(t)\|_{\h}^2+\|w^0(t)\|_{\mathbb{L}^4}^4+ 1) dt\\
+2\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t((\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)\\
+\int_0^{T\wedge\tau_N}\int_Z (\nabla H^n(u^n(t-),z),\nabla u^n(t-))_{\Ll}\n \\
+\int_0^{T\wedge\tau_N}\int_Z\|\nabla H^n(u^n(t-),z)\|_{\Ll}^2 N(dt,dz))\\
+\|u^n_0\|_{\h}^2+\|\z^n_0\|_{H^1_0}^2,\IEEEyesnumber
\end{IEEEeqnarray*}
where $S=\max\{2C_2C_{1(2)}+1+MC_P^2+\dfrac{\alpha}{2}+K,\dfrac{2g^2}{\alpha},\dfrac{2\mu^2}{\alpha}+M,1,C_2,K\}$.\\
Hence
\begin{IEEEeqnarray*}{lr}
\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2+ \frac{\alpha}{2}\int_0^{T\wedge\tau_N}\|\triangle u^n(t)\|^2_{\Ll} dt\right.\\
\qquad\qquad\qquad\qquad\qquad\qquad\quad\left.> (N-1)+\|u^n_0\|_{\h}^2+\|\z^n_0\|_{H^1_0}^2\right)\\
\leq \mathbb{P}\left(S\int_0^{T\wedge\tau_N}(\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2+\|\triangle\z^n(t)\|_{L^2}^2\right.\\
\qquad\qquad\qquad\left. +\|f(t)\|_{\h}^2+\|w^0(t)\|_{\mathbb{L}^4}^4+ 1) dt>(N-1)/2\right)\\
+2\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t((\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)\right.\\
\qquad\quad+\int_0^{T\wedge\tau_N}\int_Z (\nabla H^n(u(t-),z),\nabla u^n(t-))_{\Ll}\n \\
\qquad\quad\left.+\int_0^{T\wedge\tau_N}\int_Z\|\nabla H^n(u(t-),z)\|_{\Ll}^2 N(dt,dz)) >(N-1)/2\right).
\end{IEEEeqnarray*}
Now
\begin{IEEEeqnarray*}{lr}
\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t((\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)\right.\\
\qquad+\int_0^{T\wedge\tau_N}\int_Z (\nabla H^n(u(t-),z),\nabla u^n(t-))_{\Ll}\n\\
\qquad\left.+\int_0^{T\wedge\tau_N}\int_Z\|\nabla H^n(u(t-),z)\|_{\Ll}^2 N(dt,dz)) >(N-1)/2\right)
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{lr}
\leq \mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t(\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)>(N-1)/6\right)\\
+\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t \int_Z (\nabla H^n(u(t-),z),\nabla u^n(t-))_{\Ll}\n >(N-1)/6\right)\\
+\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t\int_Z\|\nabla H^n(u(t-),z)\|_{\Ll}^2 N(dt,dz) >(N-1)/6\right).
\end{IEEEeqnarray*}
Using Doob's Inequality
\begin{IEEEeqnarray*}{lr}
\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t(\nabla \s^n(t,u^n(t)),\nabla u^n(t))_{\Ll}dW^n(t)>(N-1)/6\right)\\
\leq \frac{36}{(N-1)^2}\mathbb{E}[\int_0^{T\wedge\tau_N} \|u^n(t)\|_{\h}^2\|\nabla \s^n(t,u^n(t))\|_{L_Q}^2 dt]\\
\leq \frac{36}{(N-1)^2}K\mathbb{E}[\int_0^{T\wedge\tau_N} \|u^n(t)\|_{\h}^2(1+ \|u^n(t)\|_{\h}^2) dt]\\
\leq \frac{36}{(N-1)^2}KN(N+1)(T\wedge\tau_N).
\end{IEEEeqnarray*}
Using Doob's inequality again
\begin{IEEEeqnarray*}{lr}
\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t\int_Z (\nabla H^n(u(t-),z),\nabla u^n(t-))_{\Ll}\n>(N-1)/6\right)\\
\leq \frac{36}{(N-1)^2}\mathbb{E}[\int_0^{T\wedge\tau_N} \|u^n(t)\|_{\h}^2\int_Z\|\nabla H^n(u(t-),z)\|_{\Ll}^2 \la dt]\\
\leq \frac{36}{(N-1)^2}K\mathbb{E}[\int_0^{T\wedge\tau_N} \|u^n(t)\|_{\h}^2(1+ \|u^n(t)\|_{\h}^2) dt]\\
\leq \frac{36}{(N-1)^2}KN(N+1)(T\wedge\tau_N).
\end{IEEEeqnarray*}
Using Doob's inequality and strong 2-integrability of $|H^n(u(t-),z)|$
\begin{IEEEeqnarray*}{lr}
\mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\int_0^t\int_Z\|\nabla H^n(u(t-),z)\|_{\Ll}^2 N(dt,dz) >(N-1)/6\right)\\
\leq \frac{6}{N-1}\mathbb{E}[\int_0^{T\wedge\tau_N}\int_Z\|\nabla H^n(u(t-),z)\|_{\Ll}^2 \la dt]\\
\leq \frac{6}{N-1}K\mathbb{E}[\int_0^{T\wedge\tau_N}(1+\|u(t)\|_{\h}^2) dt]\\
\leq \frac{6}{N-1}K(N+1)(T\wedge\tau_N).
\end{IEEEeqnarray*}
Using Chebychev's inequality
\begin{IEEEeqnarray*}{lr}
\mathbb{P}\left(S\int_0^{T\wedge\tau_N}(\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2+\|\triangle\z^n(t)\|_{L^2}^2\right.\\
\qquad\qquad\qquad\left. +\|f(t)\|_{\h}^2+\|w^0(t)\|_{\mathbb{L}^4}^4+ 1) dt>(N-1)/2\right)\\
\leq \frac{2}{N-1}\mathbb{E}[S\int_0^{T\wedge\tau_N}(\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2+\|\triangle\z^n(t)\|_{L^2}^2\\
\qquad\qquad\qquad\qquad\qquad +\|f(t)\|_{\h}^2+\|w^0(t)\|_{\mathbb{L}^4}^4+ 1) dt]
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{lr}
\leq \frac{2}{N-1}3NS (T\wedge\tau_N)+\frac{2}{N-1}S\mathbb{E}[\int_0^{T\wedge\tau_N}(\|f(t)\|_{\h}^2+\|w^0(t)\|_{\mathbb{L}^4}^4+ 1)dt].
\end{IEEEeqnarray*}
Hence
\begin{IEEEeqnarray*}{lr}
\lim\limits_{T\rightarrow 0}\sup\limits_N \mathbb{P}\left(\sup_{0\leq t\leq T\wedge\tau_N}\|u^n(t)\|_{\h}^2+\|\z^n(t)\|_{H_0^1}^2+ \frac{\alpha}{2}\int_0^{T\wedge\tau_N}\|\triangle u^n(t)\|^2_{\Ll} dt\right.\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \left.> (N-1)+\|u^n_0\|_{\h}^2+\|\z^n_0\|_{H^1_0}^2\right)=0.
\end{IEEEeqnarray*}
\end{proof}
\section{Stochastic Optimal Control}\label{stochastic}
\subsection{Preliminaries}
In this subsection we provide some definitions and known results borrowed from Metivier \cite{metivier} and Aldous \cite{ald}.
\begin{definition}
Let $(\mathbb{S},\rho)$ be a separable and complete metric space. Let $u\in \D(0,T;\mathbb{S})$ and let $\delta>0$ be given. A modulus of $u$ is defined by
\begin{equation}
w_{[0,T],\mathbb{S}}(u,\delta):=\inf_{\Pi_\delta}\max_{t_i\in\overline{\omega}}\sup_{t_i\leq s<t<t_{i+1}\leq T}\rho(u(t)u(s)),
\end{equation}
where $\Pi_\delta$ is the set of all increasing sequences $\overline{\omega}=\{0-t_0<t_1<\ldots<t_n=T\}$ with the following property
\begin{equation*}
t_{i+1}-t_{i}\geq \delta,\qquad i=0,\ldots ,n-1.
\end{equation*}
\end{definition}
\begin{theorem}
\label{cmpct}
A set $A\subset \D(0,T;\mathbb{S})$ has compact closure iff it satisfies the following two conditions:\\
\begin{itemize}
\item[(a)] there exists a dense subset $I\subset [0,T]$ such that for every $t\in I$ the set $\{u(t),u\in A\}$ has compact closure in $\mathbb{S}$,
\item[(b)]$\lim_{\delta\rightarrow 0}\sup_{u\in A}w_{[0,T],\mathbb{S}}(u,\delta)=0$.
\end{itemize}
\end{theorem}
Taking the path space $\mathcal{Z} = \D(0,T;\mathbb{H}^{-1}(\mathcal{O}))_J\cap L^\infty(0,T;\Ll(\mathcal{O}))_{w^*}\cap L^2(0,T;\h(\mathcal{O}))_w\cap L^2(0,T;\Ll(\mathcal{O}))$ into account, we call $\tau_1 := \D(0,T;\mathbb{H}^{-1}(\mathcal{O}))_J$, where $J$ denotes the extended Skorokhod topology, $\tau_2 := L^\infty (0,T;\Ll(\mathcal{O}))_{w^*}$, where $w^*$ denotes the weak-star topology and $\tau_3 := L^2 (0,T;\h(\mathcal{O}))_w$, where $w$ denotes the weak topology and $\tau_4$ as the strong topology of $L^2(0,T;\Ll(\mathcal{O}))$. Note that the spaces $\D(0,T;\mathbb{H}^{-1}(\mathcal{O}))_J,\; L^\infty(0,T;\Ll(\mathcal{O}))_{w^*}.\; L^2(0,T;\h(\mathcal{O}))_w$ are completely regular and continuously embedded in $L^2(0,T;\mathbb{H}^{-1}(\mathcal{O}))_w$. Let $\tau$ be the supremum of four topologies, that is, $\tau = \tau_1\vee\tau_2\vee\tau_3\vee\tau_4$.
\begin{theorem}
\label{cpt}
A set $K\subset\mathcal{Z}$ is $\tau$-relatively compact if the following three conditions hold
\begin{itemize}
\item[(a)]$\forall u\in K$ and all $t\in [0,T],u(t)\in\Ll(\mathcal{O})$ and $\sup_{u\in K}\sup_{s\in [0,T]}\|u(s)\|_{\Ll}<\infty$,
\item[(b)]$\sup_{u\in K}\int_{0}^T\|u(s)\|_{\h}^2 ds<\infty$,
\item[(c)]$\lim_{\delta\rightarrow 0}\sup_{u\in K} w_{[0,T],\mathbb{H}^{-1}(\mathcal{O})}(u,\delta)=0$.
\end{itemize}
\end{theorem}
\begin{proof}
We can assume that $K$ is a closed subset of $\mathcal{Z}$. By condition (a) the weak-star topology in $L^\infty(0,T;\Ll(\mathcal{O}))_{w^*}$ induced on $\mathcal{Z}$ is metrizable. Similarly by condition (b) the weak topology in $L^2(0,T;\h(\mathcal{O}))_w$ induced on $\mathcal{Z}$ is metrizable. Thus the compactness of a subset of $\mathcal{Z}$ is equivalent to its sequential compactness. Let $(u_n)$ be a sequence in $K$. By the Banach-Alaoglu theorem $K$ is compact in $L^\infty(0,T;\Ll(\mathcal{O}))_{w^*}$ as well as $L^2(0,T;\h(\mathcal{O}))_w$. \\
We need to prove that $(u_n)$ is compact in $\D(0,T;\mathbb{H}^{-1}(\mathcal{O}))$. By condition (a) for every $t\in [0,T]$ the set $\{u_n(t),n\in\mathbb{N}\}$ is bounded in $\Ll(\mathcal{O})$. Since the embedding $\Ll(\mathcal{O})\subset \mathbb{H}^{-1}(\mathcal{O})$ is compact, $\{u_n(t),n\in\mathbb{N}\}$ is compact in $\mathbb{H}^{-1}(\mathcal{O})$. Using this and condition (c), Theorem \ref{cmpct} implies compactness of the sequence $(u_n)$ in the space $\D(0,T;\mathbb{H}^{-1}(\mathcal{O}))$. Therefore, there exists a subsequence denoted again by $(u_n)$ such that
\begin{equation*}
u_n\rightarrow u \text{ in } \D(0,T;\mathbb{H}^{-1}(\mathcal{O}))\text{ as } n\rightarrow\infty .
\end{equation*}
Since $u_n\rightarrow u$ in $\D(0,T;\mathbb{H}^{-1}(\mathcal{O})),\;u_n\rightarrow u$ in $\mathbb{H}^{-1}(\mathcal{O})$ for all continuity points of $u$. Hence, by condition (a) and Lebesgue dominated convergence theorem
\begin{equation*}
u_n\rightarrow u\text{ in } L^2(0,T;\mathbb{H}^{-1}(\mathcal{O}))\text{ as } n\rightarrow\infty .
\end{equation*}
To complete the proof, we need to show that
\begin{equation*}
u_n\rightarrow u\text{ in } L^2(0,T;\Ll(\mathcal{O}))\text{ as } n\rightarrow\infty .
\end{equation*}
Since embeddings in the Gelfand triple are compact, by Lions \cite{lions}, for every $\epsilon>0$, there exists a constant $C_\epsilon>0$ such that
\begin{equation*}
\|v\|_{\Ll(\mathcal{O})}^2\leq \epsilon\|v\|_{\h(\mathcal{O})}^2 + C_\epsilon\|v\|_{\mathbb{H}^{-1}(\mathcal{O})}^2\qquad\forall v\in\h(\mathcal{O}).
\end{equation*}
Thus
\begin{equation*}
\|u_n-u\|_{L^2(0,T;\Ll(\mathcal{O}))}^2\leq \epsilon\|u_n-u\|_{L^2(0,T;\h(\mathcal{O}))}^2 + C_\epsilon\|u_n-u\|_{L^2(0,T;\mathbb{H}^{-1}(\mathcal{O}))}^2.
\end{equation*}
By condition (b)
\begin{equation*}
\|u_n-u\|_{L^2(0,T;\h(\mathcal{O}))}^2 \leq 2(\|u_n\|_{L^2(0,T;\h(\mathcal{O}))}^2 + \|u\|_{L^2(0,T;\h(\mathcal{O}))}^2 )\leq 4c,
\end{equation*}
where $c=\sup_{u\in K}\|u\|_{L^2(0,T;\h(\mathcal{O}))}^2 $. Hence
\begin{equation*}
\limsup_{n\rightarrow\infty}\|u_n-u\|_{L^2(0,T;\Ll(\mathcal{O}))}^2\leq 4c\epsilon + \limsup_{n\rightarrow\infty}C_\epsilon\|u_n-u\|_{L^2(0,T;\mathbb{H}^{-1}(\mathcal{O}))}^2.
\end{equation*}
As $\epsilon>0$ is arbitrary,
\begin{equation*}
\lim_{n\rightarrow\infty}\|u_n-u\|_{L^2(0,T;\Ll(\mathcal{O}))}^2=0.
\end{equation*}
\end{proof}
\begin{definition}
Let $(\mathbb{S},\rho)$ be a separable and complete metric space. Let $(\Omega ,\mathcal{F},P)$ be a probability space with the filtration $F:=(\mathcal{F}_t)_{t\in[0,T]}$ satisfying the usual hypotheses, and let $(X_n)_{n\in\mathbb{N}}$ be a sequence of c\`adl\`ag, $F$-adapted and $\mathbb{S}$-valued processes. $(X_n)_{n\in\mathbb{N}}$ is said to satisfy the Aldous condition iff $\forall\epsilon>0,\;\forall\eta>0,\;\exists\delta>0$ such that for every sequence $(\tau_n)_{n\in\mathbb{N}}$ of stopping times with $\tau_n\leq T$
\begin{equation*}
\sup_{n\in\mathbb{N}}\sup_{0\leq\theta\leq\delta}P\{\rho(X_n(\tau_n+\theta),X_n(\tau_n))\geq\eta\}\leq\epsilon.
\end{equation*}
\end{definition}
\begin{lemma}
\label{aldt}
Let $(X_n)$ satisfy the Aldous condition. Let $P_n$ be the law of $X_n$ on $\D(0,T;\mathbb{S}),n\in\mathbb{N}$. Then for every $\epsilon>0$ there exists a subset $A_\epsilon\subset \D(0,T;\mathbb{S})$ such that
\begin{equation*}
\sup_{n\in\mathbb{N}}P_n(A_\epsilon)\geq 1-\epsilon,
\end{equation*}
and
\begin{equation*}
\lim_{\delta\rightarrow 0}\sup_{u\in A_{\epsilon}}w_{[0,T]}(u,\delta)=0.
\end{equation*}
\end{lemma}
\begin{lemma}
\label{ald}
Let $(E,\|\cdot\|_{E})$ be a separable Banach space and let $(X_n)_{n\mathbb{N}}$ be a sequence of $E$-valued random variables. Assume that for every sequence $(\tau_n)_{n\in\mathbb{N}}$ of $F$-stopping times with $\tau_n\leq T$ and for every $n\in\mathbb{N}$ and $\theta\geq 0$ the following condition holds
\begin{equation}
\label{sc5}
\mathbb{E}[(\|X_n(\tau_n+\theta)-X_n(\tau_n)\|_{E}^\alpha]\leq C\theta^{\beta},
\end{equation}
for some $\alpha,\beta>0$ and some constant $C>0$. Then the sequence $(X_n)_{n\in\mathbb{N}}$ satisfies the Aldous condition in the space $E$.
\end{lemma}
We use the tightness condition for the Prokhorov-Varadarajan theorem which states that a sequence of measures $(P^n)_{n\in\mathbb{N}}$ is tight on a topological space $E$ if for every $\epsilon>0$ there exists a compact set $K_\epsilon\subset E$ such that $\sup_n P^n(E\setminus K_{\epsilon})\leq\epsilon$. Hence the tightness of measure in $\mathcal{Z}$ is given by the following theorem.
\begin{theorem}
\label{sdcthm1}
Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of c\`adl\`ag $F$-adapted $\mathbb{H}^{-1}(\mathcal{O})$-valued processes such that
\begin{itemize}
\item[(a)] there exists a positive constant $C_1$ such that
\begin{equation*}
\sup_{n\in\mathbb{N}}\mathbb{E}[\sup_{s\in[0,T]}\|X_n(s)\|_{\Ll}]\leq C_1,
\end{equation*}
\item[(b)] there exists a positive constant $C_2$ such that
\begin{equation*}
\sup_{n\in\mathbb{N}}\mathbb{E}[\int_0^T \|X_n(s)\|_{\h}^2 ds]\leq C_2,
\end{equation*}
\item[(c)] $(X_n)_{n\in\mathbb{N}}$ satisfies the Aldous condition in $\mathbb{H}^{-1}(\mathcal{O})$.
\end{itemize}
Let $P_n$ be the law of $X_n$ on $\mathcal{Z}$. Then for every $\epsilon>0$ there exists a compact subset $K_\epsilon$ of $\mathcal{Z}$ such that
\begin{equation}
P_n(K_\epsilon)\geq 1-\epsilon,
\end{equation}
and the sequence of measures $\{P_n,n\in\mathbb{N}\}$ is said to be tight on $(\mathcal{Z},\tau)$.
\end{theorem}
\begin{proof}
Let $\epsilon>0$. By the Chebyshev inequality and by (a), we see that for any $r>0$
\begin{equation*}
P(\sup_{s\in[0,T]}\|X_n(s)\|_{\Ll}>r)\leq\dfrac{\mathbb{E}[\sup_{s\in[0,T]}\|X_n(s)\|_{\Ll}]}{r}\leq \dfrac{C_1}{r}.
\end{equation*}
Let $R_1$ be such that $\frac{C_1}{R_1}\leq\frac{\epsilon}{3}$. Then
\begin{equation*}
P(\sup_{s\in[0,T]}\|X_n(s)\|_{\Ll}>R_1)\leq\dfrac{\epsilon}{3}.
\end{equation*}
Let $B_1:=\{u\in\mathcal{Z}:\sup_{s\in[0,T]}\|u(s)\|_{\Ll}\leq R_1\}$.\\\\
By the Chebyshev inequality and by (b), we see that for any $r>0$
\begin{equation*}
P(\|X_n\|_{L^2(0,T;\h)}>r)\leq \dfrac{\mathbb{E}[\|X_n\|_{L^2(0,T;\h)}^2]}{r^2}\leq \dfrac{C_2}{r^2}.
\end{equation*}
Let $R_2$ be such that $\frac{C_2}{R_2^2}\leq \frac{\epsilon}{3}$. Then
\begin{equation*}
P(\|X_n\|_{L^2(0,T;\h)}>R_2)\leq\dfrac{\epsilon}{3}.
\end{equation*}
Let $B_2:=\{u\in\mathcal{Z}:\|u\|_{L^2(0,T;\h)}\leq R_2\}$.\\\\
By Lemma \ref{aldt}, there exists a subset $A_{\frac{\epsilon}{3}}\subset \D(0,T;\mathbb{H}^{-1})$ such that $P_n(A_{\frac{\epsilon}{3}}) \geq1-\frac{\epsilon}{3}$ and
\begin{equation*}
\lim_{\delta\rightarrow 0}\sup_{u\in A_{\frac{\epsilon}{3}}}w_{[0,T]}(u,\delta)=0.
\end{equation*}
It is sufficient to define $K_\epsilon$ as the closure of the set $B_1\cap B_2\cap A_{\frac{\epsilon}{3}}$ in $\mathcal{Z}$. By Theorem \ref{cpt}, $K_\epsilon$ is compact in $\mathcal{Z}$.
\end{proof}
\subsection{Martingale Problem}
We now consider the stochastic tidal dynamics equation with L\'evy forcing as defined in Section \ref{setting} with initial value control as
\begin{IEEEeqnarray}{lCl}
\label{sc1}
du + [Au+B(u)+g\nabla\z]dt
=f(t)dt+\s(t,u(t))dW(t)+\int_Z H(u,z) \n ,\nonumber\\
\\
\label{sc2}
\dot{\z}+Div(hu)=0,\\
\label{sc3}
u(0)=u_0+U,\qquad \z(0)=\z_0 ,
\end{IEEEeqnarray}
where $f\in L^2(0,T;\Ll(\mathcal{O}))$, $u_0,U\in\Ll(\mathcal{O})$ and $\z_0\in L^2(\mathcal{O})$. We also assume that $\E[\|U\|_{\Ll}^2]<C_c$.\\
We assume that $\s$ and $H$ satisfy the following hypotheses:
\begin{enumerate}
\item[S.1] $\s\in C([0,T]\times\h(\mathcal{O});L_Q(H_0,\Ll)), H\in\mathbb{H}^2_\lambda([0,T]\times Z;\Ll(\mathcal{O}))$,
\item[S.2] For all $t\in(0,T)$, there exists a positive constant $K$ such that for all $u\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)\|_{L_Q}^2+\int_Z\|H(u,z)\|_{\Ll}^2\la\leq K(1+\|u\|_{\Ll}^2),
\end{equation*}
\item[S.3] For all $t\in(0,T)$, there exists a positive constant $L$ such that for all $u,v\in\Ll(\mathcal{O})$
\begin{equation*}
\|\s(t,u)-\s(t,v)\|_{L_Q}^2+\int_Z \|H(u,z)-H(v,z)\|_{\Ll}^2\la
\leq L(\|u-v\|_{\Ll}^2).
\end{equation*}
\end{enumerate}
\begin{definition}
A martingale solution of \eqref{sc1}-\eqref{sc3} is a system \\
$(\overline{\Omega},\overline{\mathcal{F}},\overline{F},\overline{P},\overline{u},\overline{z},\overline{U},\overline{N},\overline{W})$, where
\begin{itemize}
\item $(\overline{\Omega},\overline{\mathcal{F}},\overline{F},\overline{P})$ is a filtered probability space with a filtration $\overline{F}=\{\overline{\mathcal{F}}_t\}_{t\geq 0}$,
\item $\overline{N}$ is a time homogeneous Poisson random measure over $(\overline{\Omega},\overline{\mathcal{F}},\overline{F},\overline{P})$ with the intensity measure $\lambda$,
\item $\overline{W}$ is a cylindrical Wiener process over $(\overline{\Omega},\overline{\mathcal{F}},\overline{F},\overline{P})$,
\item For all $t\in [0,T]$
\begin{equation*}
\overline{u}\in \D(0,T;\mathbb{H}^{-1}(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O}))\cap L^2(0,T;\h(\mathcal{O})),
\end{equation*}
\begin{equation*}
\overline{z}\in L^2(0,T;L^2(\mathcal{O})),
\end{equation*}
\begin{equation*}
\overline{U}\in L^2(0,T;L^2(\mathcal{O})),
\end{equation*}
such that for all $v\in\h(\mathcal{O})$ and for all $w\in L^2(\mathcal{O})$, the following identities hold $\overline{P}$-a.s.
\begin{IEEEeqnarray}{llr}
&(\overline{u}(t),v)_{\Ll}+\int_0^t (A(\ou(s)),v)_{\Ll}ds+\int_0^t(B(\ou(s)),v)_{\Ll}ds+\int_0^t(g\nabla\overline{z}(s) ,v)_{\Ll}ds&\nonumber\\
&=(u_0,v_{\Ll})+(\overline{U},v)_{\Ll}+\int_0^t(f(s),v)_{\Ll}ds\nonumber\\
&+\int_0^t(\s(s,\ou(s)),v)dW(s)_{\Ll}
+\int_0^t\int_Z (H(\ou(s-),z),v)_{\Ll} \tilde{\overline{N}}(ds,dz) .\\
&(\overline{z}(t),w)_{L^2}+\int_0^t (Div(h\ou(s)),w)_{L^2} ds=(z_0,w)_{L^2}.
\end{IEEEeqnarray}
\end{itemize}
\end{definition}
\subsection{Existence of Martingale Solution}
We prove the existence of a martingale solution using the Galerkin approximations as explained in Section \ref{estimate}. Hence
\begin{IEEEeqnarray*}{lrl}
\label{sc4}
u^n(t)&=&u^n_0+U^n-\int_0^t (A(u^n(s))+B(u^n(s))+g\nabla \z^n(s)-f(s))ds\\
&&+\int_0^t \s^n(s,u^n(s))dW^n(s)+\int_0^t \int_Z H^n(u^n(s-),z)\ns .\yesnumber\\
\z^n(t)&=&\z^n_0 -\int_0^tDiv(hu^n(s))ds .\yesnumber
\end{IEEEeqnarray*}
For each $n\in\mathbb{N}$, we use the measure $\mathcal{L}(u^n)$ defined on $(\mathcal{Z},\tau)$ by the solution $u^n$ of the Galerkin equation.
\begin{lemma}
\label{tight}
The set of measures $\{\mathcal{L}(u^n),n\in\mathbb{N}\}$ is tight on $(\mathcal{Z},\tau)$
\end{lemma}
\begin{proof}
We prove the tightness of the measures using the tightness criterion given in Theorem \ref{sdcthm1}. By Proposition \ref{prop} , conditions (a) and (b) are satisfied. We use Lemma \ref{ald} to prove the Aldous condition for the sequence $(u^n)_{n\in\mathbb{N}}$ in the space $\mathbb{H}^{-1}(\mathcal{O})$. Let $(\tau_n)_{n\in\mathbb{N}}$ be a sequence of stopping times where $0\leq\tau_n\leq T$. Then using \eqref{sc4}
\begin{IEEEeqnarray*}{lrl}
u^n(t)=u^n_0+U^n-\int_0^t A(u^n(s))ds-\int_0^t B(u^n(s)) ds-\int_0^t g\nabla\z^n(s) ds\\
+\int_0^t f(s)ds+\int_0^t \s^n(s,u^n(s))dW^n(s)+\int_0^t \int_Z H^n(u^n(s-),z)\ns\\
=J^n_1+J^n_2+J^n_3(t)+J^n_4(t)+J^n_5(t)+J^n_6(t)+J^n_7(t)+J^n_8(t),\qquad t\in [0,T].
\end{IEEEeqnarray*}
Let $\theta>0$. We need to show that \eqref{sc5} holds for each $J^n_i,\; i\in\{1,2,\ldots,6\}$. Since $J^n_1$ and $J^n_2$ are independent of time, they clearly satisfy \eqref{sc5}.\\\\
Since $A:\h(\mathcal{O})\rightarrow\mathbb{H}^{-1}(\mathcal{O})$, for all $v\in\h(\mathcal{O})$
\begin{equation*}
\langle A(u),v\rangle =(A(u),v)_{\Ll}=a(\nabla u,\nabla v)_{\Ll}\leq C_1\|u\|_{\h}\|v\|_{\h}.
\end{equation*}
Hence using the above inequality
\begin{equation*}
\| A(u)\|_{\mathbb{H}^{-1}}\leq C_1\|u\|_{\h}.
\end{equation*}
Therefore
\begin{IEEEeqnarray*}{lrl}
\mathbb{E}[\| J^n_3(\tau_n+\theta)-J^n_3(\tau_n)\|_{\mathbb{H}^{-1}}]&=&\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta}A(u^n(s))ds\|_{\mathbb{H}^{-1}}]\\
&\leq &C_1\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta}\| u^n(s)\|_{\h}ds]\\
&\leq &C_1\mathbb{E}[\theta^{1/2}\left(\int_0^T \|u^n(s)\|^2_{\h} ds\right)^{1/2}]\\
&\leq &c_2\theta^{1/2}.
\end{IEEEeqnarray*}
Thus $J^n_3$ satisfies $\eqref{sc5}$ with $\alpha=1$ and $\beta=\frac{1}{2}$.\\
Since $\Ll(\mathcal{O})\hookrightarrow\mathbb{H}^{-1}(\mathcal{O})$, we get for $J^n_4$
\begin{IEEEeqnarray*}{lrl}
\mathbb{E}&[&\| J^n_4(\tau_n+\theta)-J^n_4(\tau_n)\|_{\mathbb{H}^{-1}}]\\
&=&\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta}B(u^n(s))ds\|_{\mathbb{H}^{-1}}]\\
&\leq &c\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta}\|B(u^n(s))\|_{\Ll}ds]\\
&\leq &c\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta}\| B(u^n(s))\|_{\Ll}ds]\\
&\leq &cC_2\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta}\|u^n+w^0(s)\|_{\mathbb{L}^4}^2 ds]\\
&\leq &\sqrt{2}cC_2\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta} \|u^n+w^0(s)\|_{\Ll}\|u^n+w^0(s)\|_{\h} ds]\\
&\leq &c_3\mathbb{E}[\left(\int_{\tau_n}^{\tau_n+\theta} \|u^n+w^0(s)\|_{\Ll}^2 ds\right)^{1/2} \left(\int_{\tau_n}^{\tau_n+\theta} \|u^n + w^0(s)\|_{\h}^2 ds\right)^{1/2} ]\\
&\leq &c_3\mathbb{E}[\left(\sup_{0\leq s\leq T}\|u^n+w^0(s)\|_{\Ll}^2 \theta\right)^{1/2} \left(\int_{0}^{T} \|u^n + w^0(s)\|_{\h}^2 ds\right)^{1/2} ]\\
&\leq &c_4\theta^{1/2}.
\end{IEEEeqnarray*}
Thus $J^n_4$ satisfies $\eqref{sc5}$ with $\alpha=1$ and $\beta=\frac{1}{2}$.\\
Consider the operator $C:L^2(\mathcal{O})\rightarrow\mathbb{H}^{-1}(\mathcal{O})$ given by
\begin{equation*}
C(\z)=g\nabla \z.
\end{equation*}
For all $v\in\h(\mathcal{O})$
\begin{equation*}
\langle C(\z),v\rangle=(C(\z),v)_{\Ll}\leq -g(\z,Div(v))_{L^2}\leq g\|\z\|_{L^2}\|v\|_{\h}.
\end{equation*}
Hence
\begin{equation*}
\| C(\z)\|_{\mathbb{H}^{-1}}\leq g\|\z\|_{L^2}.
\end{equation*}
So we have
\begin{IEEEeqnarray*}{lrl}
\mathbb{E}[\| J^n_5(\tau_n+\theta)-J^n_5(\tau_n)\|_{\mathbb{H}^{-1}}]&=&\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta} g\nabla \z^n(s)ds\|_{\mathbb{H}^{-1}}]\\
&\leq &g\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta}\|\z^n(s)\|_{\Ll}ds]\\
&\leq &g\mathbb{E}[\theta^{1/2}\left(T\sup_{0\leq s\leq T}\|\z^n(s)\|_{\Ll}^2\right)^{1/2}]\\
&\leq & c_5\theta^{1/2}.
\end{IEEEeqnarray*}
Thus $J^n_5$ satisfies $\eqref{sc5}$ with $\alpha=1$ and $\beta=\frac{1}{2}$.\\
Since $\Ll(\mathcal{O})\hookrightarrow\mathbb{H}^{-1}(\mathcal{O})$, by H\"older inequality, we have
\begin{IEEEeqnarray*}{lrl}
\mathbb{E}[\| J^n_6(\tau_n+\theta)-J^n_6(\tau_n)\|_{\mathbb{H}^{-1}}]&=&\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta} f(s)ds\|_{\mathbb{H}^{-1}}]\\
&\leq &c\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta}\|f(s)\|_{\Ll}ds]\\
&\leq &c\mathbb{E}[\theta^{1/2}\left(\int_0^T\|f(s)\|_{\Ll}^2 ds\right)^{1/2}]\\
&=&c_6\theta^{1/2}.
\end{IEEEeqnarray*}
Thus $J^n_6$ satisfies $\eqref{sc5}$ with $\alpha=1$ and $\beta=\frac{1}{2}$.\\
Since $\Ll(\mathcal{O})\hookrightarrow\mathbb{H}^{-1}(\mathcal{O})$, using the hypothesis S.2 and the It\^o isometry,
\begin{IEEEeqnarray*}{lrl}
\mathbb{E}[\| J^n_7(\tau_n+\theta)-J^n_7(\tau_n)\|^2_{\mathbb{H}^{-1}}]&=&\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta} \s^n(s,u^n(s)) dW^n(s)\|^2_{\mathbb{H}^{-1}}]\\
&\leq &c\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta} \s^n(s,u^n(s)) dW^n(s)\|_{\Ll}^2]\\
&=&c \mathbb{E}[\int_{\tau_n}^{\tau_n+\theta} \|\s^n(s,u^n(s))\|_{L_Q}^2 ds]\\
&\leq & cK \mathbb{E}[\int_{\tau_n}^{\tau_n+\theta} (1+\| u^n(s))\|_{\Ll}^2) ds]\\
& \leq & cK\theta(1+\mathbb{E}[\sup_{0\leq s\leq T}\|u^n(s)\|_{\Ll}^2])\\
& \leq & c_6\theta.
\end{IEEEeqnarray*}
Thus $J^n_7$ satisfies $\eqref{sc5}$ with $\alpha=2$ and $\beta=1$.\\
Since $\Ll(\mathcal{O})\hookrightarrow\mathbb{H}^{-1}(\mathcal{O})$, using the hypothesis S.2, and the isometry \cite{rudiger}
\begin{equation}
\label{iso}
\mathbb{E}[\|\int_{0}^{t}\int_Z H^n(u^n(s-),z)\ns\|_{\Ll}^2]=\mathbb{E}[\int_{0}^{t}\int_Z \| H^n(u^n(s-),z)\|_{\Ll}^2\lambda(dz)ds],
\end{equation}
we get
\begin{IEEEeqnarray*}{lrl}
\mathbb{E}[\| J^n_8(\tau_n+\theta)-J^n_8(\tau_n)\|^2_{\mathbb{H}^{-1}}]&=&\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta} \int_Z H^n(u^n(s-),z)\ns\|_{\mathbb{H}^{-1}}^2]\\
&\leq & c\mathbb{E}[\|\int_{\tau_n}^{\tau_n+\theta}\int_Z H^n(u^n(s-),z)\ns\|_{\Ll}^2]\\
&=& c\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta}\int_Z \| H^n(u^n(s-),z)\|_{\Ll}^2\lambda(dz)ds]\\
& \leq & cK\mathbb{E}[\int_{\tau_n}^{\tau_n+\theta} (1+\| u^n(s))\|_{\Ll}^2) ds]\\
& \leq & cK\theta(1+\mathbb{E}[\sup_{0\leq s\leq T}\|u^n(s)\|_{\Ll}^2])\\
& \leq & c_6\theta.
\end{IEEEeqnarray*}
Thus $J^n_8$ satisfies $\eqref{sc5}$ with $\alpha=2$ and $\beta=1$.\\
Hence by Lemma \ref{ald}, the sequence $(u^n)$ satisfies the Aldous condition in the space $\mathbb{H}^{-1}(\mathcal{O})$, which completes the proof.
\end{proof}
\begin{lemma}
\label{tight2}
The set of measures $\{\mathcal{L}(\z^n),n\in\mathbb{N}\}$ is tight on $L^2(0,T;L^2(\mathcal{O}))_{w}$.
\end{lemma}
\begin{proof}
Using Proposition \ref{prop}, $\E[\|\z^n(t)\|_{L^2}^2]\leq C_1$, hence using Fubini's theorem
\begin{equation*}
\E[\|\z^n\|_{L^2(0,T;L^2)}^2]\leq C_1T.
\end{equation*}
Using the Chebychev inequality, we see that for any $r>0$
\begin{equation*}
P(\|\z^n\|_{L^2(0,T;L^2)}>r)\leq \dfrac{\mathbb{E}[\|\z^n\|_{L^2(0,T;L^2)}^2]}{r^2}\leq \dfrac{C_1T}{r^2}.
\end{equation*}
Let $R_1$ be such that $\frac{C_1T}{R_1^2}\leq \epsilon$. Then
\begin{equation*}
P(\|\z^n\|_{L^2(0,T;L^2)}>R_1)\leq \epsilon.
\end{equation*}
Define
\begin{equation*}
B_\epsilon = \{\z\in L^2(0,T;L^2(\mathcal{O})): P(\|\z\|_{L^2(0,T;L^2)}\leq R_1\}.
\end{equation*}
Hence $P(B_\epsilon)\geq 1-\epsilon$.
\end{proof}
\begin{lemma}
\label{tight3}
The set of measures $\{\mathcal{L}(\overline{U}^n),n\in\mathbb{N}\}$ is tight on $L^2(0,T;L^2(\mathcal{O}))_{w}$.
\end{lemma}
\begin{proof}
Using the assumption $\E[\|U^n(t)\|_{L^2}^2]\leq C_c$, hence using Fubini's theorem
\begin{equation*}
\E[\|U^n\|_{L^2(0,T;L^2)}^2]\leq C_cT.
\end{equation*}
Using the Chebychev inequality, we see that for any $r>0$
\begin{equation*}
P(\|U^n\|_{L^2(0,T;L^2)}>r)\leq \dfrac{\mathbb{E}[\|U^n\|_{L^2(0,T;L^2)}^2]}{r^2}\leq \dfrac{C_cT}{r^2}.
\end{equation*}
Let $R_1$ be such that $\frac{C_cT}{R_1^2}\leq \epsilon$. Then
\begin{equation*}
P(\|U^n\|_{L^2(0,T;L^2)}>R_1)\leq \epsilon.
\end{equation*}
Define
\begin{equation*}
B_\epsilon = \{U\in L^2(0,T;L^2(\mathcal{O})): P(\|U\|_{L^2(0,T;L^2)}\leq R_1\}.
\end{equation*}
Hence $P(B_\epsilon)\geq 1-\epsilon$.
\end{proof}
\begin{theorem}
\label{martexist}
There exists a martingale solution of \eqref{sc1}-\eqref{sc3} provided the assumptions S.1-S.3 are satisfied.
\end{theorem}
\begin{proof}
By Lemma \ref{tight}, the set of measures $\{\mathcal{L}(u^n),n\in\mathbb{N}\}$ is tight on the space $(\mathcal{Z},\tau)$, by Lemma \ref{tight2}, the set of measures $\{\mathcal{L}(\z^n),n\in\mathbb{N}\}$ is tight on the space $L^2(0,T;L^2(\mathcal{O}))_w$, and by Lemma \ref{tight3}, the set of measures $\{\mathcal{L}(U^n),n\in\mathbb{N}\}$ is tight on the space $L^2(0,T;L^2(\mathcal{O}))_w$. Define $N^n=N,\;\forall n\in\mathbb{N}$. The set of measures $\{\mathcal{L}(N^n),n\in\mathbb{N}\}$ is tight. Define $W^n=W,\;\forall n\in\mathbb{N}$. The set of measures $\{\mathcal{L}(W^n),n\in\mathbb{N}\}$ is tight. Thus the set $\{\mathcal{L}(u^n,\z^n,U^n,N^n,\\ W^n),n\in\mathbb{N}\}$ is tight.
By the Skorokhod Embedding theorem \cite{motyl2011}, there exists a subsequence $(n_k)_{k\in\mathbb{N}}$, a probability space $(\overline{\Omega},\overline{\mathcal{F}},\overline{P})$, and on this space random variables $(u^*,z^*,U^*,N^*,W^*)$, $\{(\ou^k,\oz^k,,\overline{U}^k,\on^k,\ow^k),k\in\mathbb{N}\}$ such that
\begin{itemize}
\item[(i)] $\mathcal{L}((\ou^k,\oz^k,\overline{U}^k,\on^k,\ow^k))=\mathcal{L}((u^{n_k},\z^{n_k},U^{n_k},N^{n_{k}},W^{n_k}))$ for all $k\in\mathbb{N}$,
\item[(ii)]$(\ou^k,\oz^k,\overline{U}^k,\on^k,\ow^k)\rightarrow (u^*,z^*,U^*,N^*,W^*)$ with probability 1 on\\ $(\overline{\Omega},\overline{\mathcal{F}},\overline{P})$ as $k\rightarrow\infty$,
\item[(iii)]$(\on^k(\overline{\omega}),\ow^k(\overline{\omega}))=(N^*(\overline{\omega}),W^*(\overline{\omega}))$ for all $\overline{\omega}\in\overline{\Omega}$.
\end{itemize}
We denote these sequences again by $((u^n,\z^n,U^n,N^n,W^n))_{n\in\mathbb{N}}$ and \\
$((\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n))_{n\in\mathbb{N}}$. Using the definition of the space $\mathcal{Z}$, we have $\overline{P}-a.s.$
\begin{equation}
\ou^n\rightarrow u^*\;\text{in}\;L^2(0,T;\h(\mathcal{O}))_{w}\cap L^2(0,T;\Ll(\mathcal{O}))\cap \D(0,T;\mathbb{H}^{-1}(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O}))_{w^*},
\end{equation}
\begin{equation}
\oz^n\rightarrow z^*\quad\text{in}\quad L^2(0,T;L^2(\mathcal{O}))_w,
\end{equation}
and
\begin{equation}
\overline{U}^n\rightarrow U^*\quad\text{in}\quad L^2(0,T;L^2(\mathcal{O}))_w
\end{equation}
Since the random variables $\ou^n$ and $u^n$ are identically distributed, we have
\begin{equation}
\label{ineq1}
\sup_{n\geq 1}\overline{\E}[\sup_{0\leq s\leq T}\|\ou^n(s)\|_{\Ll}^2]\leq C_{1(2)},
\end{equation}
and
\begin{equation}
\label{ineq2}
\sup_{n\geq 1}\overline{\E}[\int_0^T\|\ou^n(s)\|_{\h}^2 ds]\leq C_{2(2)}.
\end{equation}
Since the random variables $\oz^n$ and $\z^n$ are also identically distributed, we have
\begin{equation}
\label{ineq3}
\sup_{n\geq 1}\overline{\E}[\sup_{0\leq s\leq T}\|\oz^n(s)\|_{L^2}^2]\leq C_{1(2)}.
\end{equation}
Using the assumption $\E[\|U\|_{L^2}^2]<C_c$, we have
\begin{equation}
\sup_{n\geq 1}\overline{\E}[\sup_{0\leq s\leq T}\|\overline{U}^n(s)\|_{L^2}^2]\leq C_c.
\end{equation}
Define for all $v\in\h(\mathcal{O})$
\begin{IEEEeqnarray*}{lrl}
K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)(t)&=&(\ou^n_0,v)_{\Ll}+(\overline{U}^n,v)_{\Ll}+\int_0^t (A(\ou^n(s)),v)_{\Ll}ds\\
&&+\int_0^t (B(\ou^n(s)),v)_{\Ll}ds-\int_0^t(\nabla \oz^n(s),v)_{\Ll} ds\\
&&+\int_0^t (\s^n(s,\ou^n(s)),v)_{\Ll}d\ow^n(s)\\
&&+\int_0^t\int_Z (H^n(\ou^n(s-),z),v)_{\Ll}\tilde{\on}^n(ds,dz),\yesnumber
\end{IEEEeqnarray*}
and for all $w\in L^2(\mathcal{O})$
\begin{IEEEeqnarray*}{lrl}
K^2(\ou^n,\oz^n,w)(t)=(\oz^n_0,w)_{L^2}-\int_0^t (Div(h\ou^n(s)),w)_{L^2} ds.
\end{IEEEeqnarray*}
Hence for all $v\in\h(\mathcal{O})$
\begin{IEEEeqnarray*}{lrl}
K^1(u^*,z^*,U^*,N^*,W^*,v)(t)&=&(u^*_0,v)_{\Ll}+(U^*,v)_{\Ll}+\int_0^t (A(u^*(s)),v)_{\Ll}ds\\
&&+\int_0^t (B(u^*(s)),v)_{\Ll}ds-\int_0^t(\nabla z^*(s),v)_{\Ll} ds\\
&&+\int_0^t (\s(s,u^*(s)),v)_{\Ll}dW^*(s)\\
&&+\int_0^t\int_Z (H(u^*(s-),z),v)_{\Ll}\tilde{N}^*(ds,dz),\yesnumber
\end{IEEEeqnarray*}
and for all $w\in L^2(\mathcal{O})$
\begin{IEEEeqnarray*}{lrl}
K^2(u^*,z^*,w)(t)&=&(z^*_0,w)_{L^2}-\int_0^t (Div(hu^*(s)),w)_{L^2} ds.
\end{IEEEeqnarray*}
We need to prove that for all $v\in\h(\mathcal{O})$
\begin{equation}
\label{pf1}
\lim_{n\rightarrow\infty}\|(\ou^n,v)_{\Ll}-(u^*,v)_{\Ll}\|_{L^2(0,T\times \overline{\Omega})}=0,
\end{equation}
and
\begin{equation}
\label{pf2}
\lim_{n\rightarrow\infty}\|K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)-K^1(u^*,z^*,U^*,N^*,W^*,v)\|_{L^2(0,T\times \overline{\Omega})}=0.
\end{equation}
We also need to show that for all $w\in\Ll(\mathcal{O})$
\begin{equation}
\label{pf3}
\lim_{n\rightarrow\infty}\|(\oz^n,w)_{L^2}-(z^*,v)_{L^2}\|_{L^2(0,T\times \overline{\Omega})}=0,
\end{equation}
and
\begin{equation}
\label{pf4}
\lim_{n\rightarrow\infty}\|K^2(\ou^n,\oz^n,w)-K^2(u^*,z^*,w)\|_{L^2(0,T\times \overline{\Omega})}=0.
\end{equation}
To prove \eqref{pf1} we see that
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\int_0^T |(\ou^n(t)-u^*(t),v)_{\Ll}|^2 dt&\leq &\|v\|_{\Ll}^2\lim_{n\rightarrow\infty}\int_0^T \|\ou^n(t)-u^*(t)\|_{\Ll}^2 dt.
\end{IEEEeqnarray*}
Since $\ou^n\rightarrow u^*$ in $L^2(0,T;\Ll(\mathcal{O}))$,
\begin{equation}
\label{vit1}
\lim_{n\rightarrow\infty}\int_0^T |(\ou^n(t)-u^*(t),v)_{\Ll}|^2 dt=0.
\end{equation}
Since both $\ou^n$ and $u^*$ satisfy the inequality \eqref{ineq1}, by \eqref{vit1} and the Vitali theorem
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\|(\ou^n,v)_{\Ll}-(u^*,v)_{\Ll}\|_{L^2(0,T\times \overline{\Omega})}&=&\lim_{n\rightarrow\infty}\overline{\E}[\int_0^T |(\ou^n(t)-u^*(t),v)_{\Ll}|^2 dt]\\
&=&0.
\end{IEEEeqnarray*}
To prove \eqref{pf2}, we see by Fubini's theorem
\begin{IEEEeqnarray*}{lrl}
\|K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)-K^1(u^*,z^*,U^*,N^*,W^*,v)\|_{L^2(0,T\times \overline{\Omega})}\\
=\overline{\E}[\int_0^T|K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)-K^1(u^*,z^*,U^*,N^*,W^*,v)|^2 dt]\\
=\int_0^T \overline{\E}[|K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)-K^1(u^*,z^*,U^*,N^*,W^*,v)|^2] dt.
\end{IEEEeqnarray*}
We show the term by term convergence of the above equation in ${L^2(0,T\times \overline{\Omega})}$.\\
Since $\ou^n\rightarrow u^*$ in $L^\infty(0,T;\Ll(\mathcal{O}))_{w^*}$, we have $\overline{P}$-a.s.
\begin{equation*}
(\ou^n(0),v)_{\Ll}\rightarrow (u^*(0),v)_{\Ll}.
\end{equation*}
Hence by \eqref{ineq1} and the Vitali theorem
\begin{equation*}
\lim_{n\rightarrow\infty}\overline{\E}[|(\ou^n(0)-u^*(0),v)_{\Ll}|^2]=0.
\end{equation*}
Hence
\begin{equation}
\lim_{n\rightarrow\infty}\|(\ou^n(0)-u^*(0),v)_{\Ll}\|_{L^2(0,T\times \overline{\Omega})}^2=0.
\end{equation}
Since $\overline{U}^n\rightarrow U^*$ in $\Ll(\mathcal{O})_w$, we have $\overline{P}$-a.s.
\begin{equation*}
(\overline{U}^n,v)_{\Ll}\rightarrow (U^*,v)_{\Ll}.
\end{equation*}
Since $\|\overline{U}^n\|_{\Ll}^2<\infty$, by the Vitali theorem
\begin{equation*}
\lim_{n\rightarrow\infty}\overline{\E}[|(\overline{U}^n-U^*,v)_{\Ll}|^2]=0.
\end{equation*}
Hence
\begin{equation}
\lim_{n\rightarrow\infty}\|(\overline{U}^n-U^*,v)_{\Ll}\|_{L^2(0,T\times \overline{\Omega})}^2=0.
\end{equation}
Since $\ou^n\rightarrow u^*$ in $L^2(0,T;\h(\mathcal{O}))_w$,
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\int_0^t (A(\ou^n(s)),v)_{\Ll}ds&=&\lim_{n\rightarrow\infty}\int_0^t (\ou^n(s),v)_{\h} ds\\
&=&\int_0^t (u^*(s),v)_{\h} ds\\
&=&\int_0^t (A(u^*(s)),v)_{\Ll}ds.
\end{IEEEeqnarray*}
By \eqref{ineq2}
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|\int_0^t (A(\ou^n(s)),v)_{\Ll} ds|^2]&\leq &\overline{\E}[\int_0^t |(\ou^n(s),v)_{\h}|^2 ds]\\
&\leq &\|v\|^2_{\h}\overline{\E}[\int_0^t \|\ou^n(s)\|^2_{\h}ds]\\
&<&\infty.
\end{IEEEeqnarray*}
Therefore by Vitali theorem, for all $t\in [0,T]$
\begin{equation*}
\lim_{n\rightarrow\infty}\overline{\E}[|\int_0^t (A(\ou^n(s))-A(u^*(s)),v)_{\Ll} ds|^2]=0.
\end{equation*}
Hence by the dominated convergence theorem
\begin{equation*}
\lim_{n\rightarrow\infty}\int_0^T \overline{\E}[|\int_0^t (A(\ou^n(s))-A(u^*(s)),v)_{\Ll} ds|^2]=0.
\end{equation*}
Since $\ou^n\rightarrow u^*$ in $L^2(0,T;\Ll(\mathcal{O}))$, we have by the continuity of the non-linear term $B(\cdot)$ given by \eqref{e1}
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\int_0^t(B(\ou^n(s))-B(u^*(s)),v)_{\Ll} ds &\leq & \lim_{n\rightarrow\infty}\int_0^t \|B(\ou^n(s))-B(u^*(s))\|_{\Ll}\|v\|_{\Ll} ds\\
&\leq &0.
\end{IEEEeqnarray*}
By \eqref{ineq1} and \eqref{ineq2}
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|\int_0^t(B(\ou^n(s)),v)_{\Ll} ds|]\\
\leq \|v\|_{\Ll} \overline{\E}[\int_0^t \|B(\ou^n(s))\|_{\Ll}ds]\\
\leq C_2\|v\|_{\Ll} \overline{\E}[\int_0^t \|\ou^n(s) +w^0(s)\|_{\mathbb{L}^4}^2 ds]\\
\leq \sqrt{2}C_2\|v\|_{\Ll} \overline{\E}[\int_0^t \|\ou^n(s) +w^0(s)\|_{\Ll}\|\ou^n(s) +w^0(s)\|_{\h}ds]\\
\leq \sqrt{2}C_2\|v\|_{\Ll} \overline{\E}[\left(\int_0^t \|\ou^n(s) +w^0(s)\|_{\Ll}^2 ds\right)^{1/2}\left(\int_0^t \|\ou^n(s) +w^0(s)\|_{\h}^2ds\right)^{1/2}]\\
\leq \sqrt{2}C_2\|v\|_{\Ll} \overline{\E}[\left(t\sup_{0\leq t\leq T}\|\ou^n(s) +w^0(s)\|_{\Ll}^2\right)^{1/2}\left(\int_0^t \|\ou^n(s) +w^0(s)\|_{\h}^2ds\right)^{1/2}]\\
<\infty.
\end{IEEEeqnarray*}
Hence by Vitali theorem, for all $t\in [0,T]$
\begin{equation*}
\lim_{n\rightarrow\infty}\overline{\E}[|\int_0^t (B(\ou^n(s))-B(u^*(s)),v)_{\Ll} ds|^2]=0.
\end{equation*}
Hence by the dominated convergence theorem
\begin{equation*}
\lim_{n\rightarrow\infty}\int_0^T \overline{\E}[|\int_0^t (B(\ou^n(s))-B(u^*(s)),v)_{\Ll} ds|^2]dt=0.
\end{equation*}
Since $\oz^n\rightarrow z^*$ in $L^2(0,T;L^2(\mathcal{O}))$, we have for all $t\in[0,T]$
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\int_0^t (g\nabla(\oz^n(s)-z^*(s)),v)_{\Ll} ds &\leq &\lim_{n\rightarrow\infty}\int_0^t |(g(\oz^n(s)-z^*(s)),\nabla v)_{L^2}| ds\\
&\leq &0.
\end{IEEEeqnarray*}
By \eqref{ineq3}
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|\int_0^t (g\nabla(\oz^n(s),v)_{\Ll} ds|^2]& \leq &g\overline{\E}[\int_0^t |(\oz^n(s),\nabla v)_{L^2}|^2 ds]\\
& \leq &g\|v\|_{\h}^2\overline{\E}[\int_0^t \|\oz^n(s)\|_{L^2}^2 ds]\\
& \leq &g\|v\|_{\h}^2\overline{\E}[t\sup_{0\leq s\leq T} \|\oz^n(s)\|_{L^2}^2]\\
&<& \infty.
\end{IEEEeqnarray*}
Hence by Vitali theorem, for all $t\in [0,T]$
\begin{equation*}
\lim_{n\rightarrow\infty}\overline{\E}[|\int_0^t (g\nabla(\oz^n(s)-z^*(s)),v)_{\Ll} ds|^2]=0.
\end{equation*}
Hence by the dominated convergence theorem
\begin{equation*}
\lim_{n\rightarrow\infty}\int_0^T\overline{\E}[|\int_0^t (g\nabla(\oz^n(s)-z^*(s)),v)_{\Ll} ds|^2]dt=0.
\end{equation*}
Let $v\in\h(\mathcal{O})$. Then using the hypothesis S.3 and that $\ou^n\rightarrow u^*$ in $L^2(0,T;\Ll(\mathcal{O}))$,
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\int_0^t |(\s^n(s,\ou^n(s))-\s(s,u^*(s)),v)_{\Ll}|^2 ds\\
\lim_{n\rightarrow\infty}\leq \|v\|_{\Ll}^2 \int_0^t \|\s^n(s,\ou^n(s))-\s(s,u^*(s))\|_{L_Q}^2 ds\\
\lim_{n\rightarrow\infty} \leq L\|v\|_{\Ll}^2 \int_0^t \|\ou^n(s)-u^*(s)\|_{\Ll}^2 ds\\
=0.
\end{IEEEeqnarray*}
Using hypothesis S.2 and \eqref{ineq1}
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|\int_0^t |(\s^n(s,\ou^n(s))-\s(s,u^*(s)),v)_{\Ll}|^2 ds|]\\
\leq \|v\|_{\Ll}^2 \overline{\E}[\int_0^t \|\s^n(s,\ou^n(s))\|_{L_Q}^2+\|\s(s,u^*(s))\|_{L_Q}^2 ds]\\
\leq K\|v\|_{\Ll}^2 \overline{\E}[\int_0^t (2+\|\ou^n(s)\|_{\Ll}^2+\| u^*(s)\|_{\Ll}^2) ds]\\
\leq KT\|v\|_{\Ll}^2 \overline{\E}[ (2+\sup_{0\leq s\leq T}\|\ou^n(s)\|_{\Ll}^2+\sup_{0\leq s\leq T}\| u^*(s)\|_{\Ll}^2)]\\
< \infty .
\end{IEEEeqnarray*}
Thus by Vitali's theorem
\begin{equation}
\lim_{n\rightarrow\infty}\overline{\E}[\int_0^t |(\s^n(s,\ou^n(s))-\s(s,u^*(s)),v)_{\Ll}|^2 ds]=0.
\end{equation}
Using the It\^o isometry and $\ow^n=W^*$,
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|(\int_0^t (\s^n(s,\ou^n(s))-\s(s,u^*(s)))dW^*(s),v)_{\Ll}|^2]\\
=\overline{\E}[\int_0^t |(\s^n(s,\ou^n(s))-\s(s,u^*(s)),v)_{\Ll}|^2 ds].
\end{IEEEeqnarray*}
Hence by the dominated convergence theorem
\begin{equation}
\lim_{n\rightarrow\infty}\int_0^T\overline{\E}[|(\int_0^t (\s^n(s,\ou^n(s))-\s(s,u^*(s)))dW^*(s),v)_{\Ll}|^2]dt=0.
\end{equation}
Let $v\in\h(\mathcal{O})$. Then using the hypothesis S.3 and that $\ou^n\rightarrow u^*$ in $L^2(0,T;\Ll(\mathcal{O}))$,
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\int_0^t \int_Z |(H^n(\ou^n(s-),z)-H(u^*(s-),z),v)_{\Ll}|^2 \lambda(dz)ds\\
\leq \lim_{n\rightarrow\infty}\|v\|_{\Ll}^2 \int_0^t\int_Z |(H^n(\ou^n(s-),z)-H(u^*(s-),z),v)_{\Ll}|^2 \lambda(dz)ds\\
\leq\lim_{n\rightarrow\infty} L\|v\|_{\Ll}^2 \int_0^t \|\ou^n(s)-u^*(s)\|_{\Ll}^2 ds\\
=0.
\end{IEEEeqnarray*}
Using hypothesis S.2 and \eqref{ineq1}
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|\int_0^t \int_Z |(H^n(\ou^n(s-),z)-H(u^*(s-),z),v)_{\Ll}|^2 \lambda(dz)ds|]\\
\leq \|v\|_{\Ll}^2 \overline{\E}[\int_0^t \int_Z \|H^n(\ou^n(s-),z)\|_{\Ll}^2+\| H(u^*(s-),z)\|_{\Ll}^2 \lambda(dz)ds]\\
\leq K\|v\|_{\Ll}^2 \overline{\E}[\int_0^t (2+\|\ou^n(s)\|_{\Ll}^2+\| u^*(s)\|_{\Ll}^2) ds]\\
\leq KT\|v\|_{\Ll}^2 \overline{\E}[ (2+\sup_{0\leq s\leq T}\|\ou^n(s)\|_{\Ll}^2+\sup_{0\leq s\leq T}\| u^*(s)\|_{\Ll}^2)]\\
<\infty .
\end{IEEEeqnarray*}
Thus by Vitali's theorem
\begin{equation}
\lim_{n\rightarrow\infty}\overline{\E}[\int_0^t \int_Z |(H^n(\ou^n(s-),z)-H(u^*(s-),z),v)_{\Ll}|^2 \lambda(dz)ds]=0.
\end{equation}
Using the isometry \eqref{iso} and $\on^n=N^*$,
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|(\int_0^t \int_Z(H^n(\ou^n(s-),z)-H(u^*(s-),z),v)_{\Ll}\tilde{N}^*(ds,dz)|^2]\\
=\overline{\E}[\int_0^t \int_Z |(H^n(\ou^n(s-),z)-H(u^*(s-),z),v)_{\Ll}|^2 \lambda(dz)ds].
\end{IEEEeqnarray*}
Hence by the dominated convergence theorem
\begin{equation}
\lim_{n\rightarrow\infty}\int_0^T\overline{\E}[|(\int_0^t \int_Z(H^n(\ou^n(s-),z)-H(u^*(s-),z),v)_{\Ll}\tilde{N}^*(ds,dz)|^2] dt=0.
\end{equation}
To prove \eqref{pf3} we see that since $\oz^n\rightarrow z^*$ in $L^2(0,T;L^2(\mathcal{O}))_w$,
\begin{equation}
\lim_{n\rightarrow\infty}\int_0^T |(\ou^n(t)-u^*(t),w)_{L^2}|^2 dt=0.
\end{equation}
Since both $\oz^n$ and $z^*$ satisfy the inequality \eqref{ineq3}, by the Vitali theorem
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\|(\oz^n,w)_{L^2}-(z^*,w)_{L^2}\|_{L^2(0,T\times \overline{\Omega})}&=&\lim_{n\rightarrow\infty}\overline{\E}[\int_0^T |(\oz^n(t)-z^*(t),w)_{L^2}|^2 dt]\\
&=&0.
\end{IEEEeqnarray*}
To prove \eqref{pf4}, we see by Fubini's theorem
\begin{IEEEeqnarray*}{lrl}
\|K^2(\ou^n,\oz^n,w)-K^2(u^*,z^*,w)\|_{L^2(0,T\times \overline{\Omega})}\\
=\overline{\E}[\int_0^T|K^2(\ou^n,\oz^n,w)-K^2(u^*,z^*,w)|^2 dt]\\
=\int_0^T \overline{\E}[|K^2(\ou^n,\oz^n,w)-K^2(u^*,z^*,w)|^2] dt.
\end{IEEEeqnarray*}
We show the term by term convergence of the above equation in ${L^2(0,T\times \overline{\Omega})}$.\\
Since $\oz^n\rightarrow z^*$ in $L^\infty(0,T;L^2(\mathcal{O}))_{w}$, we have $\overline{P}$-a.s.
\begin{equation*}
(\oz^n_0,w)_{\Ll}\rightarrow (z^*_0,w)_{L^2}.
\end{equation*}
Hence by \eqref{ineq1} and the Vitali theorem
\begin{equation*}
\lim_{n\rightarrow\infty}\overline{\E}[|(\oz^n_0-z^*_0,w)_{L^2}|^2]=0.
\end{equation*}
Hence
\begin{equation}
\lim_{n\rightarrow\infty}\|(\oz^n_0-z^*_0,w)_{\Ll}\|_{L^2(0,T\times \overline{\Omega})}^2=0.
\end{equation}
Since $\ou^n\rightarrow u^*$ in $L^\infty(0,T;\h(\mathcal{O}))_{w}$,
\begin{IEEEeqnarray*}{lrl}
\lim_{n\rightarrow\infty}\int_0^t (Div(h\ou^n(s)),w)_{L^2}ds
&=&\int_0^t (Div(hu^*(s)),w)_{L^2}ds.
\end{IEEEeqnarray*}
By \eqref{ineq2}
\begin{IEEEeqnarray*}{lrl}
\overline{\E}[|\int_0^t (Div(h\ou^n(s)),w)_{L^2} ds|^2]&\leq &\|w\|_{\Ll}^2\overline{\E}[\int_0^t \|\ou^n(s)\|_{\h}^2 ds]\\
&<&\infty.
\end{IEEEeqnarray*}
Therefore by Vitali theorem, for all $t\in [0,T]$
\begin{equation*}
\lim_{n\rightarrow\infty}\overline{\E}[|\int_0^t (Div(h\ou^n(s)),w)_{L^2}-(Div(hu^*(s)),w)_{L^2} ds|^2]=0.
\end{equation*}
Hence by the dominated convergence theorem
\begin{equation*}
\lim_{n\rightarrow\infty}\int_0^T \overline{\E}[|\int_0^t (Div(h\ou^n(s)),w)_{L^2}-(Div(hu^*(s)),w)_{L^2} ds|^2]=0.
\end{equation*}
Since $u^n$ is a solution of the Galerkin equation, we have for all $t\in [0,T]$
\begin{equation*}
(u^n(t),v)_{\Ll}=K^1(u^n,\z^n, U^n,N^n,W^n,v)(t)\qquad P\text{-a.s.}.
\end{equation*}
Hence
\begin{equation*}
\int_0^T \E[|(u^n(t),v)_{\Ll}-K^1(u^n,\z^n, U^n,N^n,W^n,v)(t)|^2]dt=0.
\end{equation*}
Since $\mathcal{L}((\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n))=\mathcal{L}((u^n,\z^n,U^n,N^n,W^n))$
\begin{equation*}
\int_0^T \overline{\E}[|(\ou^n(t),v)_{\Ll}-K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)(t)|^2]dt=0.
\end{equation*}
Using \eqref{pf1} and \eqref{pf2}
\begin{IEEEeqnarray*}{lrl}
\int_0^T \overline{\E}[|(u^*(t),v)_{\Ll}-K^1(u^*,z^*,U^*,N^*,W^*,v)(t)|^2]dt\\
\leq \int_0^T \overline{\E}[|(u^*(t),v)_{\Ll}-(\ou^n(t),v)_{\Ll}|^2]dt \\
+\int_0^T \overline{\E}[|(\ou^n(t),v)_{\Ll}-K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)(t)|^2]dt \\
+ \int_0^T \overline{\E}[|K^1(\ou^n,\oz^n,\overline{U}^n,\on^n,\ow^n,v)(t)-K^1(u^*,z^*,U^*,N^*,W^*,v)(t)|^2]dt\\
=0.
\end{IEEEeqnarray*}
Hence for all $t\in [0,T]$
\begin{equation*}
(u^*(t),v)_{\Ll}-K^1(u^*,z^*,U^*,N^*,W^*,v)(t)=0 \qquad P\text{-a.s.}.
\end{equation*}
Similarly we get for all $t\in [0,T]$
\begin{equation*}
(z^*(t),w)_{\Ll}-K^2(u^*,z^*,w)(t)=0 \qquad P\text{-a.s.}.
\end{equation*}
Taking $\ou=u^*,\oz=z^*,\overline{U}=U^*,\on=N^*$and $\ow=W^*$. we see that $(\overline{\Omega},\overline{\mathcal{F}},\overline{F},\overline{P},\overline{u},\overline{z},\overline{U},\overline{N},\overline{W})$ is a martingale solution of \eqref{sc1}-\eqref{sc3}.
\end{proof}
The pathwise uniqueness of the martingale solution is obvious from proof of uniqueness in Theorem \ref{thmuniq}.
\subsection{Existence of Optimal Control}
We define the cost functional as
\begin{equation}
F(u,U)=\overline{\E}\left[\int_0^T \int_{\mathcal{O}}L(t,u,U)dxdt\right],
\end{equation}
where $L(\cdot,\cdot,\cdot):[0,T]\times\h(\mathcal{O})\times\Ll(\mathcal{O})\rightarrow\mathbb{R}$ is such that
\begin{enumerate}
\item $L(\cdot,\cdot,\cdot):[0,T]\times\h(\mathcal{O})\times\Ll(\mathcal{O})$ is measurable,
\item $L(t,\cdot,\cdot):\h(\mathcal{O})_w\times\Ll(\mathcal{O})$ is lower semicontinuous $\forall t\in [0,T]$ where $\h(\mathcal{O})_w$ is the space $\h(\mathcal{O})$ endowed with the weak topology, and
\item $L(t,u,U)\geq k(U), \forall (t,u,U)\in[0,T]\times\h(\mathcal{O})\times\Ll(\mathcal{O})$ where $k(U)$ is non-negative and
\begin{equation}
\int_0^T\int_{\mathcal{O}}k(U)dxdt\rightarrow\infty\text{ as }\|U\|_{\Ll}\rightarrow\infty .
\end{equation}
\end{enumerate}
The task here is to find the optimal control $U$ which minimizes the cost functional $J(u,U)$.\\\\
\begin{lemma}
\label{lowersemcont}
For $u^n\rightarrow u$ in the $\tau$-topology and $U^n\rightarrow U$ in $L^2(0,T;\Ll(\mathcal{O}))$-weak
\begin{equation}
\liminf\limits_{n\rightarrow\infty}\int_0^T\int_{\mathcal{O}}L(t,u^n,U^n)dxdt\geq\int_0^T\int_{\mathcal{O}}L(t,u,U)dxdt.
\end{equation}
\end{lemma}
\begin{proof}
Let $L_N(t,u,U)=L(t,u,U)\wedge N$. Then
\begin{IEEEeqnarray*}{lrl}
\liminf\limits_{n\rightarrow\infty}&\int_0^T &\int_{\mathcal{O}}L(t,u^n,U^n)dxdt\\
&\geq &\liminf\limits_{n\rightarrow\infty}\int_0^T\int_{\mathcal{O}}L_N(t,u^n,U^n)dxdt\\
&\geq &-\limsup\limits_{n\rightarrow\infty}\int_0^T\int_{\mathcal{O}}\left(L_N(t,u^n,U^n)-L_N(t,u,U^n)\right)^- dxdt\\
&&+\liminf\limits_{n\rightarrow\infty}\int_0^T\int_{\mathcal{O}}L_N(t,u,U^n)dxdt,\IEEEyesnumber
\end{IEEEeqnarray*}
where $f(x)^-=(-f(x))\wedge 0$. The first integral on the right-hand side is zero due to the lemma given below. Due to the lower semicontinuity of $L_N$, we have
\begin{equation}
\liminf\limits_{n\rightarrow\infty}\int_0^T\int_{\mathcal{O}}L_N(t,u,U^n)dxdt\geq\int_0^T\int_{\mathcal{O}}L_N(t,u,U)dxdt.
\end{equation}
So
\begin{equation}
\liminf\limits_{n\rightarrow\infty}\int_0^T\int_{\mathcal{O}}L(t,u^n,U^n)dxdt\geq \int_0^T\int_{\mathcal{O}}L_N(t,u,U)dxdt.
\end{equation}
Now using the Beppo-Levi theorem on the bounded measurable functions $L_N$ we get the required semicontinuity.
\end{proof}
\begin{lemma}
Let $u^n\rightarrow u$ in the $\tau$-topology and $U^n\rightarrow U$ in $L^2(0,T;\Ll(\mathcal{O}))$-weak. Let $\varphi(\cdot,\cdot,\cdot):[0,T]\times\h(\mathcal{O})\times\Ll(\mathcal{O})\rightarrow \mathbb{R}_+$ be a bounded measurable function such that, $\forall t\in (0,T), \varphi(t,\cdot,\cdot):\h(\mathcal{O})\times\Ll(\mathcal{O})\rightarrow \mathbb{R}_+$ is lower semicontinuous. Then
\begin{equation}
\label{cost}
\lim\limits_{n\rightarrow\infty}\int_0^T\int_{\mathcal{O}}\left(\varphi(t,u^n,U^n)-\varphi(t,u,U^n)\right)^- dxdt=0.
\end{equation}
\end{lemma}
\begin{proof}
Define $\Theta(t,z,U):=\varphi(t,z,U)-\varphi(t,u,U)$. For $\gamma>0$ and $y\in\mathbb{H}^{-1}(\mathcal{O})$, define
\begin{equation}
Y^m:=\left\{(t,U)\in[0,T]\times\Ll(\mathcal{O});\inf\limits_{|\langle y,u\rangle-\langle y,z\rangle|\leq 1/m}\Theta(t,z,U)\leq -\gamma\right\},
\end{equation}
and
\begin{equation}
Y^m_n:=\left\{(t,U^n)\in[0,T]\times\Ll(\mathcal{O});\inf\limits_{|\langle y,u\rangle-\langle y,z\rangle|\leq 1/m}\Theta(t,z,U)\leq -\gamma\right\}.
\end{equation}
Clearly, $Y^{m+1}\subseteq Y^m$. \\\\
As $u^n\rightarrow u$, we have
\begin{equation}
\liminf\limits_{n\rightarrow\infty}\Theta(t,u^n,U)\geq 0.
\end{equation}
Hence
\begin{equation}
\cap_m Y^m=\emptyset.
\end{equation}
The lower semicontinuity of $\varphi$ implies that each $t$-section of $Y^m$ is closed. Hence we have
\begin{IEEEeqnarray*}{lrl}
\lim\limits_{n\rightarrow\infty}\limsup\limits_{n\rightarrow\infty}\int_0^T\int_{Y^m_n}1\;dxdt\;&\leq &\limsup\limits_{n\rightarrow\infty}\int_0^T\int_{Y^m}1\;dxdt\\
&\leq &\int_0^T\int_{\cap_{r>0}\cup_{m>r}Y^m}1\;dxdt\\
&=&\int_0^T\int_{\cap_m Y^m}1\;dxdt\\
&=&\int_0^T\int_\emptyset 1\;dxdt=0.\IEEEyesnumber
\end{IEEEeqnarray*}
Define
\begin{equation}
\hat{Y}^n:=\left\{(t,U);\Theta(t,u^n(t),U)^->\gamma\right\}=\left\{(t,U);\Theta(t,u^n(t),U)<-\gamma\right\}.
\end{equation}
Since $u^n\rightarrow u$ in $L^2(0,T,\h(\mathcal{O}))$, we have for large enough $n$
\begin{equation}
\int_0^T\int_{\hat{Y}^n}1\;dxdt\leq \int_0^T\int_{Y^m}1\;dxdt.
\end{equation}
Thus,
\begin{equation}
\limsup\limits_{n\rightarrow\infty}\int_0^T\int_{\hat{Y}^n}1\;dxdt.
\end{equation}
Therefore by \cite{jacod}
\begin{equation}
\lim\limits_{n\rightarrow\infty}\int_0^T\int_\mathcal{O}\Theta(t,u^n,U^n)1\;dxdt=0,
\end{equation}
which proves \eqref{cost}.
\end{proof}
\begin{theorem}
Given $u_0\in\Ll(\mathcal{O})$, there exists a pair
\begin{equation*}
(\hat{u},\hat{U})\in (L^2(0,T;\h(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O})\cap \D(0,T;\mathbb{H}^{-1}(\mathcal{O})))\times L^2(0,T;\Ll(\mathcal{O})),
\end{equation*}
which gives a martingale solution to equations \eqref{sc1}-\eqref{sc3}, and
\begin{IEEEeqnarray*}{rcl}
J(\hat{u},\hat{U})=\min\{&&J(u,U);(u,U)\in (L^2(0,T;\h(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O}))\\
&&\cap \D(0,T;\mathbb{H}^{-1}(\mathcal{O})))\times L^2(0,T;\Ll(\mathcal{O})),\\
&&\text{where the pair }(u,U)\text{ gives a martingale solution}\\
&&\text{to equations }\eqref{sc1}-\eqref{sc3}\}.
\end{IEEEeqnarray*}
\end{theorem}
\begin{proof}
We restrict ourselves to admissible pairs $(u,U)\in (L^2(0,T;\h(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O})\cap \D(0,T;\mathbb{H}^{-1}(\mathcal{O})))\times L^2(0,T;\Ll(\mathcal{O}))$ which satisfy \eqref{sc1}-\eqref{sc3} in the martingale sense such that $F(u,U)<+\infty$\\\\
We have $F(u,U)\geq 0$ and
\begin{equation}
F(u,U)\rightarrow +\infty\;\;\text{as}\;\;\|U\|_{\Ll}\rightarrow\infty .
\end{equation}
Now,
\begin{equation*}
F(u^n,U^n)\leq R,
\end{equation*}
implies that
\begin{equation*}
\overline{E}\left[\|U^n\|_{\Ll(\mathcal{O})}\right]\leq c(R),
\end{equation*}
for some constant $c(R)$. Hence by Lemma \ref{tight3} there exists a sequence of tight measures $\mathcal{L}(\overline{U}^n)$ in $L^2(0,T;L^2(\mathcal{O}))_{w}$. By Theorem \ref{martexist}, there exists a corresponding sequence $u^n\in L^2(0,T;\h(\mathcal{O}))\cap L^\infty(0,T;\Ll(\mathcal{O})\cap \D(0,T;\mathbb{H}^{-1}(\mathcal{O}))$ such that the pair $(u^n,U^n)$ gives a martingale solution to \eqref{sc1}-\eqref{sc3}, where $u^n\rightarrow \hat{u}$ in the $\tau$-topology and $U^n\rightarrow \hat{U}$ weakly in $\Ll(\mathcal{O})$. By the same theorem we also have that $(\hat{u},\hat{U})$ solves \eqref{sc1}-\eqref{sc3}.\\
By Lemma \ref{lowersemcont}, $\int_0^T\int_{\mathcal{O}}L(t,u^n,U^n)dxdt$ is lower semi-continuous and hence by Theorem 55 in Chapter III of \cite{dell}, so is $F$. So,
\begin{equation*}
F(\hat{u},\hat{U})\leq \liminf_n F(u^n,U^n)\leq \lim_n F(u^n,U^n)=\inf F(u,U).
\end{equation*}
\end{proof}
\medskip\noindent
{\bf Acknowledgements:} Pooja Agarwal would like to thank Department of Science and Technology, Govt. of India, for the INSPIRE fellowship. Utpal Manna's work has been supported by the National Board of Higher Mathematics of Department of Atomic Energy, Govt. of India, under Grant No. NBHM/RP46/2013/Fresh/421. The authors would like to thank Indian Institute of Science Education and Research(IISER) - Thiruvananthapuram for providing stimulating scientific environment and resources. | 0.02116 |
30 January 2013
In this video Burkhard Schaefer, Co-founder and President of BSSN Software, tells SelectScience about the AnIML Data Standard, which has been designed for the storing and sharing of experimental data. Burkhard discusses some of the key benefits of this technology for managing analytical data. The AnIML Data Standard enables the integration of data from a large number of sources, which increases the productivity of any workflow. Interview filmed by SelectScience at SLAS2013. | 0.85985 |
PESHAWAR – American Consulate Peshawar’s Regional Security Office hosted a seminar today at Pearl Continental Hotel to raise awareness of trafficking in persons (TIP). Around 60 participants from the legislative, law enforcement, and NGO communities attended the event, which focused on how Americans and Pakistanis can work together to stop this crime.
The goal of the seminar, titled “A Whole Village Approach,” was to create a forum for pursuing legislative improvements, prosecution, and victim protection related to TIP. Speakers from the National Assembly and civil society talked about TIP in Peshawar and how law enforcement, legislators, and NGOs can collaborate to confront. Find out more at:. | 0.701271 |
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Pregnant women may want to stay away from caffeine
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By Nicole Boone
WBTW News13 Anchor
Published: November 4, 2008 two or more cups of coffee daily are at twice the risk of having a miscarriage.
The commenting period has ended or commenting has been deactivated for this article. | 0.039341 |
TITLE: What role does Cauchy's determinant identity play in combinatorics?
QUESTION [19 upvotes]: When studying representation theory, special functions or various other topics one is very likely to encounter the following identity at some point:
$$\det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)}$$
This goes under the name of Cauchy's determinant identity and has various generalizations and analogous statements. There is also a lot of different proofs using either analysis or algebra. In my case I have always seen it introduced (or motivated) as an identity that plays an important role in combinatorics, but I realized that I haven't really seen this identity in a combinatorial context before. In this question I'm asking for a combinatorial interpretation of the above identity. A bonus to someone who can give such an interpretation to Borchardt's variation:
$$\det \left(\frac{1}{(x _i+y _j)^2}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)} \cdot \text{per}\left(\frac{1}{x _i +y _j}\right) _{1 \le i,j \le n}$$
(This seems a little too ambitious though, and I would be happy to accept an answer of just the first question)
REPLY [16 votes]: See also pp. 397--398 of Enumerative Combinatorics, vol. 2. Cauchy's determinant is given in a slightly different but equivalent form. It is explained there that the evaluation of the determinant is equivalent to the fundamental identity $\prod(1-x_iy_j)^{-1} =\sum_\lambda s_\lambda(x)s_\lambda(y)$ in the theory of symmetric functions. | 0.090234 |
2007: Third-team All-Conference USA honoree ... Started all 19 games in which he played ... Had eight points on two goals and four assists (third-most assists on SMU team) ... Scored in SMU wins over UCF (10/6/07) and Central Arkansas (11/9/07) ... Has 24 points on five goals and 14 assists in his career.
2006: Played in 21 games, including seven starts ... recorded two goals and eight assists (12 points) on 25 shots.
2005: Saw action in 16 games, including 11 starts ... Scored one goal and added two assists on three shots on goal ... Selected to the Conference USA All-Freshman team.
High School: Three-year letterwinner while at Jesuit College Prep ... Named Varsity MVP in 2004 and 2005 and served as team captain ... First-team All-State selection as a freshman ... All-District selection as a junior ... 2005 Parade All-America selection ... Gatorade High School Player of the Year nominee ... Current member of the Dallas Texans club team and the United States U-20 Men's National Team Player Pool ... USYSA ODP National All-Star Team selection (2005) and named to the Region III team ... Selected to the Dallas Morning News All-State team as a junior and earned District Offensive MVP honors as well.
Personal: Majoring in psychology ... Enjoys golf, basketball and music ... Would most like to visit Italy ... Favorite non-soccer athlete is Steve Nash ... Listens to Indie rock, alternative country and bluegrass music ... Grandfather and mother are SMU graduates ... Has one younger brother ... Parents are Jeff Harwell and Lauren Embrey ... Full name is Jeffery Daniel Dean Harwell Jr. | 0.004397 |
Multiple reports indicate that the Angels have agreed to a two-year, $15 million contract with free agent starting pitcher Joe Blanton. The 31-year-old is not going to win a Cy Young, but he will stomach innings and he’s a nice addition to a club that lacked rotation depth in 2012.
Blanton has tossed 175+ innings in seven of the past eight seasons. He isn’t the most effective starter in baseball, but he stays healthy and he doesn’t walk a lot of batters. Angel Stadium and the Angels’ defense are also pretty good fits for Blanton. The Big A is known for suppressing home run rates, and the Angels have above-average potential at every defensive position.
Throughout his career Blanton has been more of a flyball pitcher with some strikeout ability. He’s developed excellent control of all of his pitches, but he struggled to keep the ball in the park this past season. 15.3 percent of his fly balls landed in the seats, well above league-average (which is typically around 10 percent). In 2012, he also had a higher-than-average line drive rate (23.4 percent!); Mike Trout and Peter Bourjos are pretty good at making those into outs.
This might be my favorite deal of Jerry Dipoto’s tenure as the Angels’ GM. It shows that the Angels don’t need to spend like maniacs to be successful. $7.5 million per year for a pitcher of Blanton’s caliber is chump change, and it still leaves plenty of cash available to fill different needs.
Every deal the Angels have made could be scrutinized heavily if Zack Greinke walks. People will complain about how spending money on Hanson, Madson, Blanton, and Burnett put Greinke’s asking price above the Halos’ budget, but the front office isn’t stupid. They know how high they’re willing to jump for Greinke, and if he’s already balked at that high it just makes sense for them to search for cheaper depth. | 0.000949 |
\begin{document}
\title[Isoclasses Determined by Automorphisms]{On Isoclasses of Maximal Subalgebras Determined by Automorphisms}
\author{Alex Sistko}
\thanks{{2010 \textit{Mathematics Subject Classification}. Primary 16S99; Secondary 16W20, 16Z99, 05E15, 05C60}\\
{\bf Keywords} finite-dimensional algebra, maximal subalgebra, subalgebra variety, automorphism group, presentation, isoclass, type A, Dynkin quiver}
\address{Department of Mathematics, University of Iowa,
Iowa City, IA 52242}
\email{alexander-sistko@uiowa.edu}
\maketitle{}
\date{}
\begin{abstract}
Let $k$ be an algebraically-closed field, and let $B = kQ/I$ be a basic, finite-dimensional associative $k$-algebra with $n := \dim_kB < \infty$. Previous work shows that the collection of maximal subalgebras of $B$ carries the structure of a projective variety, denoted by $\msa (Q)$, which only depends on the underlying quiver $Q$ of $B$. The automorphism group $\Aut_k(B)$ acts regularly on $\msa (Q)$. Since $\msa (Q)$ does not depend on the admissible ideal $I$, it is not necessarily easy to tell when two points of $\msa (Q)$ actually correspond to isomorphic subalgebras of $B$. One way to gain insight into this problem is to study $\Aut_k(B)$-orbits of $\msa (Q)$, and attempt to understand how isoclasses of maximal subalgebras decompose as unions of $\Aut_k(B)$-orbits. This paper investigates the problem for $B = kQ$, where $Q$ is a type $\mathbb{A}$ Dynkin quiver. We show that for such $B$, two maximal subalgebras with connected Ext quivers are isomorphic if and only if they lie in the same $\Aut_k(B)$-orbit of $\msa (Q)$.
\end{abstract}
\section{Introduction}\label{s.intro}
\noindent Let $B$ be a finite-dimensional, unital, associative algebra over an algebraically-closed field $k$. Then the celebrated \emph{Wedderburn-Malcev Theorem} states that there exists a $k$-subalgebra $B_0 \subset B$ such that $B_0 \cong B/J(B)$ and $B = B_0 \oplus J(B)$, where $J(B)$ denotes the Jacobson radical of $B$. Furthermore, for any subalgebra $B_0' \subset B$ isomorphic to $B_0$, there exists a $x \in J(B)$ such that $(1+x)B_0(1+x)^{-1} = B_0'$. For more details, see for instance \cite{Farn} or Theorem 11.6 of \cite{Pi}. Of course, the collection of all $k$-algebra automorphisms of $B$, which we denote by $\Aut_k(B)$, acts on the set of subalgebras of $B$. For any $x \in J(B)$, the map $y \mapsto (1+x)y(1+x)^{-1}$ is an automorphism of $B$. So another way to state the second half of the Wedderburn-Malcev Theorem is to say that the \emph{isoclass} of $B_0$ in $B$, i.e. the set of all subalgebras of $B$ isomorphic to $B_0$, is a single $\Aut_k(B)$-orbit.
Unsurprisingly, this statement is false for general subalgebras $A$ of $B$. Nevertheless, recent investigations into maximal subalgebras of finite-dimensional algebras suggest that examples of such $A$ are not necessarily rare \cite{IS}. It is therefore natural to ask what conditions we can impose on $A$ to ensure that its isoclass in $B$ is an orbit of $\Aut_k(B)$. More generally, one can ask whether there is any way to classify the $\Aut_k(B)$-orbits of subalgebras of $B$, and relate them to isoclasses of subalgebras. This is one source of inspiration for the current paper.
Another source of inspiration comes from the study of varieties of subalgebras, as the author has done recently in \cite{Sis}. For any $1 \le m \le \dim_kB$, the collection of all $m$-dimensional subalgebras of $B$ carries the structure of a projective $k$-variety, which we call $\AlgGr_m(B)$. The linear algebraic group $\Aut_k(B)$ acts regularly on this variety. Neither $m$ or $B$ are enough to specify $\AlgGr_m(B)$ up to equivalence of varieties: in fact, if $B = kQ/I$ is a basic algebra and $m = \dim_kB - 1$, then $\AlgGr_m(B)$ \emph{only} depends on $Q$. So it will be difficult in general to choose an admissible ideal $I$ of $kQ$, and determine whether two points of $\AlgGr_{\dim_kkQ/I-1}(kQ/I)$ actually represent isomorphic subalgebras of $kQ/I$. Thankfully, the automorphism group $\Aut_k(kQ/I)$ \emph{is} sensitive to the data contained in $I$. So, provided that one can impose reasonable conditions on the relationship between orbits and isoclasses, one can expect that orbits under this group action will yield significant information on isoclasses of subalgebras. In \cite{Sis} we discuss one possible version of ``reasonable conditions,'' where the variety is a finite union of orbits.
The purpose of this paper is to carry out this program as far as possible for a suitable ``test class'' of algebras. For us, these will be path algebras of type $\mathbb{A}$ Dynkin quivers and their maximal subalgebras. As it turns out, many maximal subalgebras of such algebras will have isoclasses that are single $\Aut_k(B)$-orbits. However, we will show that even for such a nicely-behaved class, isoclasses differ from orbits in at least some circumstances.
This paper is organized as follows. In Section \ref{s.bg} we review the basic notions associated to path algebras and their automorphisms. We also discuss the major results from \cite{IS}, \cite{Sis} which will be used to prove our main result. In Section \ref{s.pres}, we discuss the problem of presenting maximal subalgebras of basic algebras. In particular, Propositions \ref{p.seppres} and \ref{c.pathalg} provide explicit presentations for maximal subalgebras of hereditary algebras. The results of this section will be used in Section \ref{s.main}, where we prove the main result of this article:
\begin{theorem}\label{t.main}
Let $k$ be an algebraically closed field, $Q$ a type $\mathbb{A}$ Dynkin quiver, and $B = kQ$. Suppose that $A, A' \in \msa (Q)$ have connected Ext quivers. Then $A$ and $A'$ lie in the same $\Aut_k(B)$-orbit if and only if $A \cong A'$ as $k$-algebras.
\end{theorem}
\noindent This is essentially done by showing that the Ext quivers of $A$ and $A'$ are nonisomorphic whenever they lie in different $\Aut_k(B)$-orbits. We note that this theorem can be rephrased as follows: if the underlying graph of $Q$ is an oriented tree with maximum degree $2$ and $B = kQ$, then the isoclass of a connected maximal subalgebra of $B$ coincides with its $\Aut_k(B)$-orbit. The author does not currently know whether similar statements hold for all trees with maximum degree $3$ or higher. We end on an example which shows that if the Ext quiver of $A$ is not connected, then its isoclass can differ from its $\Aut_k(B)$-orbit.
\section{Background}\label{s.bg}
Unless otherwise stated, $k$ will denote an algebraically-closed field. All algebras are unital, associative, finite-dimensional $k$-algebras, and our terminology essentially comes from \cite{ASS}. Let $Q$ be a finite quiver with vertex set $Q_0$, arrow set $Q_1$, and source (resp. target) function $s$ (resp. $t$) $: Q_1\rightarrow Q_0$. The \emph{underlying graph} of $Q$ is obtained by forgetting the orientations on the arrows. Let $kQ$ denote the path algebra of $Q$, and let $J(Q)$ denote the two-sided ideal in $kQ$ generated by $Q_1$. For $n \geq 2$, we let $T_nQ := kQ/J(Q)^n$ denote the \emph{$n^{th}$ truncated path algebra} associated to $Q$. By a slight abuse of notation, for any $u, v \in Q_0$ we let $uQ_1v$ denote the set of arrows in $Q$ with source $u$ and target $v$, and we let $ukQ_1v$ denote their $k$-span inside $kQ$. Note that if $uQ_1v = \emptyset$, then $ukQ_1v = \{ 0 \}$ and $\GL(ukQ_1v)$ is the trivial group. Similar to \cite{GS2}, we define $V^2(Q) = \{ (u,v) \in Q_0\times Q_0\mid uQ_1v \neq \emptyset\}$. A \emph{basic algebra} is an algebra of the form $B = kQ/I$, where $I$ is an \emph{admissible ideal} of $kQ$, i.e. an ideal satisfying $J(Q)^2 \supset I \supset J(Q)^{\ell}$ for some $\ell \geq 2$. Note that $B = kQ_0 \oplus J(B) = kQ_0 \oplus J(Q)/I$, and that $kQ_0 \cong B/J(B) \cong k^{|Q_0|}$.
We let $\Aut_k(B)$ denote the group of all $k$-algebra automorphisms of $B$. It is a Zariski-closed subgroup of $\GL(B)$, and hence a linear affine algebraic group. Our notation for subgroups of $\Aut_k(B)$ is borrowed from the notation in \cite{Poll}, \cite{GS1}, \cite{GS2}. If $G$ is a subgroup of $\Aut_k(B)$, we say that two subalgebras $A$ and $A'$ are \emph{$G$-conjugate} (in $B$) if there exists a $\phi \in G$ such that $A' = \phi (A)$. For a unit $u \in B^{\times}$, we let $\iota_u$ denote the corresponding \emph{inner automorphism}, i.e. the map $\iota_u(x) = uxu^{-1}$ for all $x \in B$. We let $\Inn (B)$ denote the group of all inner automorphisms, and $\Inn^*(B) = \{ \iota_{1+x} \mid x \in J(B) \}$ denote the group of \emph{unipotent inner automorphisms}. If $B = kQ/I$ is basic, we let $\hat{H}_B = \{ \phi \in \Aut_k(B) \mid \phi (Q_0) = Q_0 \}$ and $H_B = \{ \phi \in \Aut_k(B) \mid \phi\mid_{Q_0} = \id_{Q_0} \}$. By Theorem 10.3.6 of \cite{HGK}, $\Inn(B)$ acts transitively on complete collections of primitive orthogonal idempotents. Since inner automorphisms induced by units of the form $\sum_{v \in Q_0}{\lambda_vv}$ (where $\lambda_v \in k^{\times}$ for each $v$) fix vertices, $\Inn^*(B)$ is also transitive on this set and we have a decomposition $\Aut_k(B) = \Inn^*(B) \cdot \hat{H}_B = \hat{H}_B\cdot \Inn^*(B)$. If the underlying graph of $Q$ is a tree, then $\Aut (Q)$ can be considered a subgroup of $\Aut_k(T_nQ)$ for any $n$, and it is easy to see that we have a decomposition $\hat{H}_{T_nQ} = \Aut (Q) \cdot H_{T_nQ} = H_{T_nQ} \cdot \Aut (Q)$. Since $J(Q)^n = 0$ for large powers $n$, this statement includes the fact that $\Aut(Q)$ is a subgroup of $\Aut_k(kQ)$.
In Theorem 4.1 of \cite{IS}, the author and M. C. Iovanov proved the following classification Theorem for maximal subalgebras of basic algebras:
\begin{theorem}\label{t.basic}
Let $B = kQ/I$ be a basic algebra over an algebraically-closed field $k$. Let $A \subset B$ be a maximal subalgebra. Consider the following two classes of maximal subalgebras of $B$:
\item For a two-element subset $\{ u,v \} \subset Q_0$, we define
\begin{equation*}
A(u+v) := k(u+v)\oplus \left( \bigoplus_{w \in Q_0\setminus\{ u,v\}}{kw} \right) \oplus J(B).
\end{equation*}
\item For an element $(u,v) \in V^2(Q)$ and a codimension-$1$ subspace $U \le ukQ_1v$, we define
\begin{equation*}
A(u,v,U) := kQ_0 \oplus U \oplus \left( \bigoplus_{(w,y) \in Q_0^2\setminus\{(u,v)\}}{wkQ_1y} \right) \oplus J(B)^2.
\end{equation*}
\noindent Then there exists a unipotent inner automorphism $\iota_{1+x} \in \Inn^*(B)$ such that either $\iota_{1+x}(A) = A(u+v)$ or $\iota_{1+x}(A) = A(u,v,U)$, for some appropriate choice of $u$, $v$, and possibly $U$.
\end{theorem}
\noindent As in \cite{IS}, if $A$ is $\Inn^*(B)$-conjugate to a subalgebra of the form $A(u+v)$, then we say that $A$ is of \emph{separable type}. If $A$ is $\Inn^*(B)$-conjugate to an algebra of the form $A(u,v,U)$, then we say that $A$ is of \emph{split type}. As an immediate consequence of Theorem \ref{t.basic}, all maximal subalgebras of a basic algebra are basic, have codimension $1$, and contain the radical square. In fact, we have the following easy corollary, which first appeared in \cite{Sis}:
\begin{corollary}\label{c.inherit}
Let $B$ be a basic $k$-algebra of dimension $n$, and let $A \subset B$ be a subalgebra. Then $A$ satisfies the following:
\begin{enumerate}
\item $A$ is also a basic algebra.
\item If $A$ is a maximal subalgebra, then $\dim_kA = n-1$.
\item If $A$ is a maximal subalgebra, then $J(A)$ is a $B$-subbimodule of $J(B)$, $J(A) = A \cap J(B)$, and $J(B)^2 \subset J(A)$.
\item More generally, if $m = \dim_kA$, then $J(B)^{2(n-m)} \subset A$.
\end{enumerate}
\end{corollary}
If $B$ is any $k$-algebra and $m$ is a positive integer $1 \le m \le \dim_kB$, then the collection $\AlgGr_{m}(B)$ of all $m$-dimensional subalgebras of $B$ is a Zariski-closed subset of the usual Grassmannian $\Gr_m(B)$. In particular, it is a projective variety over $k$. For any $A \in \AlgGr_m(B)$, we let $\Iso (A,B)$ denote the set of all $A' \in \AlgGr_m(B)$ such that $A\cong A'$ as $k$-algebras. Clearly $\Iso (A,B)$ is $\Aut_k(B)$-invariant, and hence a union of $\Aut_k(B)$-orbits. Suppose that $B = kQ/I$ is a basic algebra of dimension $n$. By the remarks above, it follows that $\AlgGr_{n-1}(B)$ is the variety of maximal subalgebras of $B$, and that there is a (biregular) bijection between maximal subalgebras of $B$ and maximal subalgebras of $B/J(B)^2 \cong T_2Q$. In other words, $\AlgGr_{n-1}(B)$ only depends on the underlying quiver $Q$, and so we define $\msa (Q) := \AlgGr_{n-1}(B)$. We can think of $\msa (Q)$ as the variety of maximal subalgebras of \emph{any} basic algebra with Ext quiver $Q$. See \cite{Sis} for more details.
Suppose that the underlying graph of $Q$ is a tree and $B = kQ$. Theorem \ref{t.basic} classifies $\Inn^*(B)$-orbits of $\msa (Q)$, and it is easy to see that for any such $B$, $\phi (A) = A$ for all $\phi \in H_B$. So classification of $\Aut_k(B)$-orbits boils down to determining which $\Inn^*(B)$-orbits of $\msa (Q)$ are related by elements of $\Aut (Q)$. More specifically, it is equivalent to classifying $\Aut (Q)$-orbits on the finite sets $V^2(Q)$ (for split type) and $\{ \{ u,v\} \subset Q_0 \mid u \neq v\}$ (for separable type). Although this may represent an intractable problem for general $Q$, it at least implies that every $B$-isoclass of $\msa (Q)$ is a finite union of $\Aut_k(B)$-orbits. In Section \ref{s.main} we will show that if $Q$ is a type $\mathbb{A}$ Dynkin quiver, then each $B$-isoclass consisting of connected algebras is a \emph{single} $\Aut_k(B)$-orbit. The first step will be to find presentations for each maximal subalgebra as a bound quiver algebra, which we do below.
\section{Presentations of Maximal Subalgebras}\label{s.pres}
Corollary \ref{c.inherit} (1) states that if $B$ is basic, then all of its maximal subalgebras are also basic. In particular, they can be presented as bound quiver algebras. Ideally, one hopes for explicit presentations of maximal subalgebras in terms of a given presentation for $B$. More specifically, if $B$ is given as $B = kQ/I$ and $A$ is a maximal subalgebra of $B$, one would like a combinatorial procedure to obtain the Ext quiver of $A$, call it $\Gamma$, from $Q$, and another procedure to find generators for the kernel of the projection map $k\Gamma \rightarrow A$. As it currently stands, if $I \subset kQ$ is an arbitrary admissible ideal, and $A \subset kQ/I$ is a maximal subalgebra of split type, then it is not clear to the author how one can explicitly reconstruct $\Gamma$ from $Q$. Nevertheless, Theorem \ref{t.basic} provides us with some insight into the presentation problem. In fact, it is good enough to give us a full description for separable type subalgebras, as well as explicit presentations for \emph{all} maximal subalgebras in the hereditary case, i.e. $I = \{ 0 \}$.
We start by describing presentations for maximal subalgebras of separable type. Take a $2$-element subset $\{ u,v\} \subset Q_0$, and let $\Gamma$ be the quiver obtained from $Q$ by gluing $u$ and $v$ together. More explicitly, $\Gamma$ has vertex set $Q_0\setminus\{ u,v\} \cup \{u+v\}$, and for all $w,y \in Q_0\setminus\{ u,v\}$ we have
\begin{equation*}
(u+v)\Gamma_1y = uQ_1y \cup vQ_1y,
\end{equation*}
\begin{equation*}
w\Gamma_1(u+v) = wQ_1u\cup wQ_1v,
\end{equation*}
\begin{equation*}
(u+v)\Gamma_1(u+v) = uQ_1u \cup uQ_1v \cup vQ_1u \cup vQ_1v.
\end{equation*}
\noindent In other words, $\Gamma_1$ is just a re-partitioning of $Q_1$ into arrows with possibly new endpoints. This induces a bijective map $\phi : Q_1 \rightarrow \Gamma_1$. Hence, if $p = \alpha_1\cdots \alpha_d$ is a path in $Q$, then $\phi (p) := \phi (\alpha_1)\cdots \phi(\alpha_d)$ is a well-defined path in $\Gamma$. We can extend this to an algebra map $\phi : kQ \rightarrow k\Gamma$ by defining
\begin{equation*}
\phi(w) = w \text{ for all $w \in Q_0\setminus\{ u,v\}$,}
\end{equation*}
\begin{equation*}
\phi(u) = \phi(v) = u+v,
\end{equation*}
\noindent and extending to $k$-linear combinations of arbitrary paths.
\begin{proposition}\label{p.seppres}
Let $B = kQ/I$, and $A$ a maximal subalgebra of separable type. Suppose that $A$ is $\Inn^*(B)$-conjugate to $A(u+v)$, for some two-element subset $\{u,v\} \subset Q_0$. Then $A \cong k\Gamma/I'$, where
\begin{enumerate}
\item $\Gamma$ is obtained from $Q$ by gluing vertices $u$ and $v$.
\item $I'$ is generated by relations in $\phi (I)$, along with elements of the form $\phi(\alpha) \phi(\beta)$, where either $\alpha \in Q_1u$ and $\beta \in vQ_1$, or $\alpha \in Q_1v$ and $\beta \in uQ_1$.
\end{enumerate}
\end{proposition}
\begin{proof}
There is a map $ k\Gamma \rightarrow A(u+v)$ which acts as the identity on $Q_0\setminus \{u,v\}$, sends $u+v \in \Gamma_0$ to $u+v \in A(u+v)$, and which acts on $\Gamma_1$ via the bijection $\Gamma_1 \leftrightarrow Q_1$. The kernel of this map is precisely the admissible ideal $I'$.
\end{proof}
\begin{proposition}\label{c.splitpres}
Let $B = kQ/I$, and $A$ a maximal subalgebra of split type. Suppose that $A$ is $\Inn^*(B)$-conjugate to $A(u,v,U)$, for some $(u,v) \in V^2(Q)$ and codimension-$1$ subspace $U\le ukQ_1v$. Write $A \cong k\Gamma/I'$ for a certain quiver $\Gamma$ and admissible ideal $I'$. Then $\Gamma_0 = Q_0$, and for all $w,x \in \Gamma_0$,
\begin{equation*}
\dim_kw\left(J(A)/J(A)^2\right)x = \dim_kw\left(J(A)/J(B)^2\right)x + \dim_kw\left(J(B)^2/J(A)^2\right)x.
\end{equation*}
\noindent In particular:
\begin{enumerate}
\item For all $w \neq u$ and $x \neq v$, there are $w\left( J(B)/J(B)^2\right)x$ arrows from $w$ to $x$ in $\Gamma$,
\item There are $\dim_ku\left(J(B)/J(B)^2\right)v -1$ arrows from $u$ to $v$, and
\item There are at least $\dim_ku\left(J(B)/J(B)^2\right)x$ (resp. $\dim_kw\left(J(B)/J(B)^2\right)v$) arrows from $u$ to $x$ (resp. from $w$ to $v$).
\end{enumerate}
\noindent Furthermore, $J(B)^4 \subset J(A)^2$, so that any arrows in $\Gamma$ that do not appear as arrows in $Q$ arise from elements of $J(B)^2$ or $J(B)^3$.
\end{proposition}
\begin{proof}
$A(u,v,U)/J(A(u,v,U)) = kQ_0$ implies that $\Gamma_0 = Q_0$. The dimension formula follows from the $kQ_0$-bimodule isomorphism $J(A)/J(A)^2/\left( J(B)^2/J(A)^2 \right) \cong J(A)/J(B)^2$, and claims (1)-(3) follow from the formula
\begin{equation*}
J(A) = J(B)\cap A = U \oplus \left( \bigoplus_{(w,x)\neq (u,v)}{wkQ_1x} \right) \oplus J(B)^2.
\end{equation*}
\noindent The final claim follows from Corollary \ref{c.inherit} (3).
\end{proof}
\noindent Although this corollary does not give us an explicit form for $I'$, we can use it to present maximal subalgebras of split type in the hereditary case, i.e. when $I = \{ 0\}$.
\begin{proposition}\label{c.pathalg}
Let $B = kQ$ for an acyclic quiver $Q$, and $A \subset B$ a maximal subalgebra conjugate to $A(u,v,U)$, for some $(u,v) \in V^2(Q)$. Write $A \cong k\Gamma / I'$, for a finite quiver $\Gamma$ and admissible ideal $I' \subset k\Gamma$. Then $\Gamma_0 = Q_0$, and $\Gamma_1$ can be obtained from $Q$ as follows:
\begin{enumerate}
\item Replace the $|uQ_1v|$ arrows from $u$ to $v$ in $Q$ with $|uQ_1v|-1$ arrows, indexed by a fixed basis $\{ \alpha_1,\ldots , \alpha_d\}$ of $U$;
\item For each arrow $\gamma$ with target $u$, add an arrow $\overline{\gamma} : s(\gamma) \rightarrow v$;
\item For each arrow $\gamma$ with source $v$, add an arrow $\underline{\gamma}: u \rightarrow t(\gamma)$.
\end{enumerate}
\noindent Furthermore, $I'$ can be taken to be the ideal generated by the relations $\overline{\beta}\gamma - \beta\underline{\gamma}$, for all arrows $\beta$ and $\gamma$ in $Q$ with $t(\beta) = u$ and $s(\gamma) = v$.
\end{proposition}
\begin{proof}
We may assume without loss of generality that $A = A(u,v,U)$. Find $\alpha_{d+1} \in ukQ_1v$ such that $U \oplus k\alpha_{d+1} = ukQ_1v$. Then for each $\gamma \in Q_1$ with $t(\gamma ) = u$, $\gamma\alpha_{d+1} \in J(A)\setminus J(A)^2$. If $w = s(\gamma)$, then clearly the paths of the form $\gamma \alpha_{d+1}$, along with the arrows in $Q_1$ from $w$ to $v$, form a basis for $w(J(A)/J(A)^2)v$. Define $\overline{\gamma} = \gamma\alpha_{d+1}$. A similar argument exhibits a basis for $u(J(A)/J(A)^2)x$, and allows us to define $\underline{\gamma} = \alpha_{d+1}\gamma$ if $\gamma$ is an arrow in $Q$ from $v$ to $x$. The form for $\Gamma$ then follows from Proposition \ref{c.splitpres}. For all arrows $\beta, \gamma \in Q_1$ with $t(\beta) = u$ and $s(\gamma) = v$, $\overline{\beta}\gamma = \beta\alpha_{d+1}\gamma = \beta\underline{\gamma}$. Hence, $\overline{\beta}\gamma-\beta\underline{\gamma}$ is in the kernel of the projection map $k\Gamma \rightarrow A$. If $I'$ is the ideal generated by these commutation relations, then it is straightforward to check that the induced $k$-algebra projection $k\Gamma/I' \rightarrow A$ has an inverse, so that the desired isomorphism holds.
\end{proof}
\begin{example}\label{e.A4}
Let $B = kQ$, where
\[
Q=
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, scale = 2]
\node[point,label={below:$v_1$}] (1) at (0,0) {} ;
\node[point,label={below:$v_2$}] (2) at (1,0) {};
\node[point,label={below:$v_3$}] (3) at (2,0) {};
\node[point,label={below:$v_4$}] (4) at (3,0) {};
\path[->] (1) edge node[above] {$\alpha$} (2) ;
\path[->] (2) edge node[above] {$\beta$} (3) ;
\path[->] (3) edge node[above] {$\gamma$} (4) ;
\end{tikzpicture} }}
\]
\noindent is an equioriented Dykin quiver of type $\mathbb{A}_4$. Then any maximal subalgebra of separable type must be $\Inn^*(B)$-conjugate to one of the six bound quiver algebras displayed below: \newline
\begin{center}
$
\begin{array}{| c | c | c |}
\hline k\Gamma / I & \Gamma & I \\
\hline A(v_1+v_2) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3, scale =2]
\node[point] (1) at (0,0) {} edge[in = -290, out = -250, loop] node[above] {$\alpha$} () ;
\node[point] (2) at (1,0) {};
\node[point] (3) at (2,0) {};
\path[->] (1) edge node[above] {$\beta$} (2) ;
\path[->] (2) edge node[above] {$\gamma$} (3) ;
\end{tikzpicture} & (\alpha^2) \\
\hline A(v_1+v_3) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3, scale=2]
\node[point] (1) at (0,0) {} ;
\node[point] (2) at (1,0) {} ;
\node[point] (3) at (2,0) {} ;
\path[->] (2) edge node[above] {$\gamma$} (1) ;
\path[->] (2.25) edge[bend left=25] node[above] {$\alpha$} (3.25) ;
\path[->] (3.-25) edge[bend right=-25] node[below] {$\beta$} (2.-25) ;
\end{tikzpicture} & (\beta \alpha) \\
\hline A(v_1+v_4) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3, scale=2]
\node[point] (1) at (0,0) {} ;
\node[point] (2) at (1,.5) {} ;
\node[point] (3) at (2,0) {} ;
\path[->] (1) edge node[above] {$\alpha$} (2)
(2) edge node[above] {$\beta$} (3)
(3) edge node[below] {$\gamma$} (1);
\end{tikzpicture}
& (\gamma \alpha) \\
\hline A(v_2+v_3) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3, scale=2]
\node[point] (2) at (1,0) {} edge[in = -290, out = -250, loop] node[above] {$\beta$} () ;
\node[point] (1) at (0,0) {};
\node[point] (3) at (2,0) {};
\path[->] (1) edge node[above] {$\alpha$} (2) ;
\path[->] (2) edge node[above] {$\gamma$} (3) ;
\end{tikzpicture}
& (\beta^2) \\
\hline A(v_2+v_4) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3, scale=2]
\node[point] (1) at (0,0) {} ;
\node[point] (2) at (1,0) {} ;
\node[point] (3) at (2,0) {} ;
\path[->] (1) edge node[above] {$\alpha$} (2) ;
\path[->] (2.25) edge[bend left=25] node[above] {$\beta$} (3.25) ;
\path[->] (3.-25) edge[bend right=-25] node[below] {$\gamma$} (2.-25) ;
\end{tikzpicture}
& (\gamma \beta) \\
\hline A(v_3+v_4) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point] (3) at (2,0) {} edge[in = -290, out = -250, loop] node[above] {$\gamma$} () ;
\node[point] (2) at (1,0) {};
\node[point] (1) at (0,0) {};
\path[->] (1) edge node[above] {$\alpha$} (2) ;
\path[->] (2) edge node[above] {$\beta$} (3) ;
\end{tikzpicture} & (\gamma^2) \\
\hline
\end{array}
$
\end{center}
\noindent Any maximal subalgebra of split type must be $\Inn^*(B)$-conjugate to one of the three bound quiver algebras displayed below:
\begin{center}
$
\begin{array}{| c | c | c |}
\hline k\Gamma / I & \Gamma & I \\
\hline A(v_1,v_2,\{ 0 \} ) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point,label={below:$v_1$}] (1) at (0,-.5) {} ;
\node[point,label={above:$v_2$}] (2) at (0,.5) {} ;
\node[point,label={above:$v_3$}] (3) at (1,0) {} ;
\node[point,label={above:$v_4$}] (4) at (2,0) {} ;
\path[->] (1) edge node[below] {$\underline{\beta}$} (3)
(2) edge node[above] {$\beta$} (3)
(3) edge node[above] {$\gamma$} (4);
\end{tikzpicture}
& \{ 0 \} \\
\hline A(v_2,v_3,\{ 0 \} ) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point,label={left:$v_1$}] (1) at (0,0) {} ;
\node[point,label={above:$v_3$}] (3) at (1,.5) {} ;
\node[point,label={below:$v_2$}] (2) at (1,-.5) {} ;
\node[point,label={right:$v_4$}] (4) at (2,0) {} ;
\path[->] (1) edge node[above] {$\overline{\alpha}$} (3)
(1) edge node[below] {$\alpha$} (2)
(3) edge node[above] {$\gamma$} (4)
(2) edge node[below] {$\underline{\gamma}$} (4);
\end{tikzpicture}
& (\overline{\alpha}\gamma - \alpha\underline{\gamma}) \\
\hline A(v_3,v_4.\{ 0 \} ) & \begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point,label={left:$v_1$}] (1) at (0,0) {} ;
\node[point,label={above:$v_2$}] (2) at (1,0) {} ;
\node[point,label={right:$v_3$}] (3) at (2,.5) {} ;
\node[point,label={right:$v_4$}] (4) at (2,-.5) {} ;
\path[->] (1) edge node[above] {$\alpha$} (2)
(2) edge node[above] {$\beta$} (3)
(2) edge node[below] {$\overline{\beta}$} (4);
\end{tikzpicture}
& \{ 0 \} \\
\hline
\end{array}
$
\end{center}
\end{example}
\noindent {\bf{Note:}} If $I \neq \{ 0 \}$, then $I \subset J(Q)^2 \subset A$ allows us to consider relations in $I$ as ``generalized relations'' in $\Gamma$. We say ``generalized'' because these elements may not actually lie in $J(A)^2$. In other words, we can always realize $A$ as a \emph{generally non-admissible} quotient of $k\Gamma/I$. This level of detail is sufficient for the purposes of this paper. \newline
\begin{example}
Consider $B = kQ/I$, where
\[
Q=
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, scale = 2]
\node[point,label={below:$v_1$}] (1) at (0,0) {} ;
\node[point,label={below:$v_2$}] (2) at (1,0) {};
\node[point,label={below:$v_3$}] (3) at (2,0) {};
\node[point,label={below:$v_4$}] (4) at (3,0) {};
\path[->] (1) edge node[above] {$\alpha$} (2) ;
\path[->] (2.75) edge node[above] {$\beta_1$} (3.75)
(2.-75) edge node[below] {$\beta_2$} (3.-75);
\path[->] (3) edge node[above] {$\gamma$} (4) ;
\end{tikzpicture} }}
\]
\noindent and $I = (\alpha\beta_1-\alpha\beta_2 )$. Then the maximal subalgebra of $kQ$ corresponding to the triple $(v_2,v_3, k\beta_1)$ can be presented as $A= k\Gamma / I'$, where
\[
\Gamma =
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point,label={left:$v_1$}] (1) at (0,0) {} ;
\node[point,label={above:$v_3$}] (3) at (1,.5) {} ;
\node[point,label={below:$v_2$}] (2) at (1,-.5) {} ;
\node[point,label={right:$v_4$}] (4) at (2,0) {} ;
\path[->] (1) edge node[above] {$\overline{\alpha}$} (3)
(1) edge node[below] {$\alpha$} (2)
(3) edge node[above] {$\gamma$} (4)
(2) edge node[below] {$\underline{\gamma}$} (4)
(2) edge node[left] {$\beta_2$} (3);
\end{tikzpicture}
}}
\]
\noindent and $I' = (\overline{\alpha}\gamma - \alpha\underline{\gamma})$. Therefore, $A(v_2,v_3,k\beta_1) = A/I = A/(\overline{\alpha} - \alpha\beta_2)$. It follows that $\overline{\alpha}$ is in the radical square of $A(v_2,v_3,k\beta_1)$. Hence, to present $A(v_2,v_3, k\beta_1)$, we must actually bound the quiver
\[
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point,label={left:$v_1$}] (1) at (0,0) {} ;
\node[point,label={above:$v_3$}] (3) at (1,.5) {} ;
\node[point,label={below:$v_2$}] (2) at (1,-.5) {} ;
\node[point,label={right:$v_4$}] (4) at (2,0) {} ;
\path[->] (1) edge node[below] {$\alpha$} (2)
(3) edge node[above] {$\gamma$} (4)
(2) edge node[below] {$\underline{\gamma}$} (4)
(2) edge node[left] {$\beta_2$} (3);
\end{tikzpicture}
}}
\]
\noindent by the relation $\alpha\beta_2\gamma - \alpha \underline{\gamma}$.
\end{example}
\section{Type $\mathbb{A}$ Path Algebras}\label{s.main}
In this section we prove theorem \ref{t.main}. Unless otherwise stated, let $Q$ be a type-$\mathbb{A}$ Dynkin quiver on $n$ vertices, in other words a quiver whose underlying graph is of the form:
\begin{figure}[th]\label{f.dynkin}
\[
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point] (1) at (0,0) {} ;
\node[point] (2) at (1,0) {} ;
\node[point] (3) at (2,0) {} ;
\node (4) at (3,0) {$\cdots$} ;
\node[point] (5) at (4,0) {} ;
\node[point] (6) at (5,0) {} ;
\path (1) edge (2)
(2) edge (3)
(3) edge (4)
(4) edge (5)
(5) edge (6) ;
\end{tikzpicture}
}}
\]
\caption{A type $\mathbb{A}$ Dynkin diagram.}
\end{figure}
Let $B = kQ$. If $n = 2m$ is even, label the vertices of Figure 1 from left to right as $v_{-m}$, $v_{-m+1}$,$\ldots$, $v_{-1}$, $v_1$,$\ldots$, $v_{m-1}$, $v_m$. If $n = 2m+1$ is odd, label the vertices $v_{-m}$,$\ldots$, $v_{-1}$, $v_0$, $v_1$,$\ldots$, $v_m$ in an analogous manner. Whenever it is understood that we are talking about vertices, we may abbreviate ``$v_i$'' as ``$i$.'' We call $v_{-i}$ the \emph{predecessor} of $v_{-i+1}$, and $v_{-i+1}$ the \emph{successor} of $v_{-i}$. Note that if $n$ is even, $v_{-1}$ is the predecessor of $v_1$. If we need to talk about the predecessor (resp. successor) of a vertex $v$, we simply denote it by $\operatorname{pred}(v)$ (resp. $\operatorname{succ}(v)$). We recursively define $\operatorname{pred}^i(v)$ and $\operatorname{succ}^i(v)$ as follow: $\operatorname{pred}^1(v) = \operatorname{pred}(v)$ and $\operatorname{succ}^1(v) = \operatorname{succ}^1(v)$. If $\operatorname{pred}^i(v)$ and $\operatorname{succ}(v)$ have already been defined, then $\operatorname{pred}^{i+1}(v)$ is the predecessor of $\operatorname{pred}^i(v)$, and $\operatorname{succ}^{i+1}(v)$ is the successor to $\operatorname{succ}^i(v)$, whenever these vertices are defined. Let $\underline{w}$ denote the binary word of length $n-1$ such that $\underline{w}(i) = +1$ if the edge from $v_i$ to its successor starts at $v_i$, and $\underline{w}(i) = -1$ if the edge starts at the successor to $v_i$. We treat $\underline{w}$ as a function $\{ -m,\ldots , m-1\} \rightarrow \{ -1, +1\}$, or as an ordered $(m-1)$-tuple $\underline{w} = w_{-m}w_{-m+1}\cdots w_{m-1}$, where each $w_i \in \{ \pm 1\}$. Let $\alpha_i$ denote the edge between $v_i$ and its successor. To ease notation slightly, we use the following shorthand for maximal subalgebras of $kQ$: \newline
\begin{equation*}
A_i := A(v_i, \operatorname{succ}(v_i), \{ 0\}) \text{ for all $i< m$,}
\end{equation*}
\begin{equation*}
A_{i,j} := A(v_i+v_j), \text{ for all $v_i, v_j \in Q_0$ with $i \neq j$.}
\end{equation*} \newline
Since any automorphism of $Q$ induces an automorphism of its underlying Dynkin diagram, $\Aut (Q) \le C_2$, the cyclic group of order 2. Define $\underline{w}^*$ to be the binary word of the quiver obtained from $Q$ by applying the unique non-identity automorphism of the underlying Dynkin graph to $Q$. In other words, $\underline{w}^*(i) = -\underline{w}(\operatorname{pred}(-i))$ for all $-m \le i \le m-1$. Clearly, $\underline{w}^{**} = \underline{w}$ and $\Aut (Q) = C_2$ if and only if $\underline{w}^* = \underline{w}$. The following lemma will be used extensively throughout our proofs:
\begin{lemma}\label{l.autos}
Let $i$ and $j$ be integers with $-m \le i,j \le m-1$. Suppose that $A_i$ has a connected Ext quiver. If $A_i \cong A_j$, then there exists an isomorphism $\psi : A_i \rightarrow A_j$ such that $\psi(Q_0) = Q_0$. If $\psi (\ell ) = -\ell$ for all $\ell \in Q_0$, then $\Aut (Q) = C_2$.
\end{lemma}
\begin{proof}
Let $\psi' : A_i \rightarrow A_j$ be any $k$-algebra isomorphism. Since $Q_0$ is a complete set of primitive orthogonal idempotents for both $A_i$ and $A_j$, $\psi' (Q_0)$ is a complete set of primitive orthogonal idempotents for $A_j$. Since $\Inn(A_j)$ acts transitively on such sets, there exists a $\iota_u \in \Inn(A_j)$ such that $\iota_u\psi' (Q_0) = \psi (Q_0)$. Setting $\psi = \iota_u\circ \psi'$ demonstrates the first claim. For the second, suppose that $\psi (\ell) = -\ell$ for all $\ell \in Q_0$. Then for all $\ell \neq i$, Proposition \ref{c.pathalg} implies $\dim_k\ell J(B)/J(B)^2\operatorname{succ}(\ell) = \dim_k \ell J(A_i)/J(A_i)^2 \operatorname{succ}(\ell) = \dim_k (-\ell) J(A_j)/J(A_j)^2 \operatorname{pred}(-\ell)$. But $\underline{w}(\ell) = +1$ if and only if $\dim_k\ell J(B)/J(B)^2 \operatorname{succ}(\ell) = 1$. But $Q$ is a tree: since $-\ell$ is adjacent to $\operatorname{pred}(-\ell)$ in $Q$ and $J(A_j) \subset J(B)$, the equality $\dim_k(-\ell)J(A_j)/J(A_j)^2\operatorname{pred}(-\ell )=1$ implies $\operatorname{pred}(-\ell)J(B)/J(B)^2(-\ell) = 0$ and $\underline{w}(\operatorname{pred}(-\ell)) = -1$. In other words, for all $\ell \neq i$ we have $\underline{w}(\ell) = - \underline{w}(\operatorname{pred}(-\ell) )$. Suppose by way of contradiction that $\underline{w}(i) = \underline{w}(\operatorname{pred}(-i))$. Without loss of generality we may assume $\underline{w}(i) = +1$. Then Proposition \ref{c.pathalg} and the assumption that $A_i$ has a connected Ext quiver together imply that at least one of the two conditions must hold:
\begin{enumerate}
\item $\operatorname{pred}(i)$ exists and $\underline{w}(\operatorname{pred}(i)) = +1$, or
\item $\operatorname{succ}^2(i)$ exists and $\underline{w}(\operatorname{succ}^2(i)) = +1$.
\end{enumerate}
Suppose that the first condition holds. Then by Corollary \ref{c.inherit} (3), $\alpha_{\operatorname{pred}(i)}\alpha_i \in J(A_i)$ and the vector space $\operatorname{pred}(i)J(A_i)\operatorname{succ}(i)$ is non-zero. Therefore, the vector space $\operatorname{succ}(-i)J(A_j)\operatorname{pred}(-i)$ is non-zero as well. But note that since $Q$ is a type $\mathbb{A}$ Dynkin quiver, $\operatorname{succ}(-i)J(A_j)\operatorname{pred}(-i) = \operatorname{succ}(i)J(B)(-i)J(B)\operatorname{pred}(-i)$. But then in particular $(-i)J(B)\operatorname{pred}(-i) \neq 0$, which holds if and only if we have $\operatorname{pred}(-i)J(B)(-i) = 0$. In turn, this holds if and only if $\underline{w}(\operatorname{pred}(-i)) = -1$, a contradiction. So we must have $\underline{w}(i) = -\underline{w}(\operatorname{pred}(-i))$. If the second condition holds, a similar argument allows us to conclude again that $\underline{w}(i) = -\underline{w}(\operatorname{pred}(-i))$. But then we must have $\underline{w}^* = \underline{w}$ and hence $\Aut (Q) = C_2$.
\end{proof}
To prove the main theorem, we prove it separately for maximal subalgebras of split type and separable type. For split type, we use Proposition \ref{c.pathalg} to distinguish three essential cases, depending on the form of the Ext quiver of $A_i$. If $w_{\operatorname{pred}(i)}w_iw_{\operatorname{succ}(i)} = (+1)(+1)(+1)$ or $(-1)(-1)(-1)$, then the Ext quiver will contain a commutative square:
\begin{center}
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=1.5]
\node[point] (i-1) at (0,0) {};
\node[point] (i) at (1,0.5) {};
\node (0) at (1,0) {$\circlearrowright$} ;
\node[point,] (i+1) at (1,-0.5) {};
\node[point] (i+2) at (2,0) {};
\path[->] (i-1) edge (i)
(i-1) edge (i+1)
(i) edge (i+2)
(i+1) edge (i+2);
\end{tikzpicture},
\end{center}
\noindent where the two middle vertices are $i$ and $\operatorname{succ}(i)$. In all other cases, $A_i$ will be hereditary. Again, Proposition 3.3 implies that the underlying graph of the Ext quiver of $A_i$ has the form:
\begin{figure}[th]
\[
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=1.5]
\node[point,label={below:$-m$}] (-m) at (0,0) {};
\node[point,label={below:$\operatorname{pred}(i)$}] (i-1) at (2,0) {};
\node (b1) at (1,0) {$\cdots$};
\node[point,label={below:$i$}] (i) at (3,0) {};
\node[point,label={above:$\operatorname{succ}(i)$}] (i+1) at (4,0) {};
\node[point,label={above:$\operatorname{succ}^2(i)$}] (i+2) at (5,0) {};
\node (b2) at (6,0) {$\cdots$};
\node[point,label={below:$m$}] (m) at (7,0) {};
\path (-m) edge (0.5,0)
(1.5,0) edge (i-1)
(i-1) edge (i)
(i+1) edge (i+2)
(i+2) edge (5.5,0)
(6.5,0) edge (m);
\path[dotted] (i-1) edge[bend left=25] (i+1)
(i) edge[bend right=25] (i+2);
\end{tikzpicture},
}}
\]
\caption{}\label{f.quiverform}
\end{figure}
\noindent where for $i = -m$ we take this graph to only include the edges to the right of $i$, for $i = m-1$ we delete the section containing $\operatorname{succ}^2(i)$, and where dotted lines indicate that an edge may or may not be present. By the connectivity hypothesis, both dotted edges cannot be absent. Note that $A_{-m+1}$ and $A_{m-2}$ may themselves be path algebras over type $\mathbb{A}$ Dynkin quivers, say if $w_{-m}w_{-m+1}w_{-m+2} = (+1)(+1)(-1)$. This will be our second case. Otherwise, this graph will necessarily contain a trivalent vertex, and a unique leaf adjacent to this trivalent vertex. This will be our third and final case.
We now prove our first case, where $A_i$ is a non-hereditary algebra:
\begin{lemma}\label{l.splitnonher}
Let $Q$ and $B$ be as before. Let $A$ be a non-hereditary maximal subalgebra of split type whose Ext quiver is connected. Then $\Iso (A,B) = \Aut_k(B)\cdot A$.
\end{lemma}
\begin{proof}
Suppose first $-m \le i \le -1$ and that $A_i$ is not hereditary. Then $i>-m$ and $w_{\operatorname{pred}(i)}w_iw_{\operatorname{succ}(i)} = (+1)(+1)(+1)$ or $(-1)(-1)(-1)$. By inspecting the full sub-quiver from $-m$ to $\operatorname{pred}(i)$ and the full sub-quiver from $\operatorname{succ}^2(i)$ to $m$, we conclude that $A_i \cong A_j$ implies either $j = i$ or $j = -i-1$. If $A_i \cong A_{-i-1}$, use Lemma \ref{l.autos} to find an isomorphism $\psi : A_i \rightarrow A_{-i-1}$ such that $\psi(Q_0) = Q_0$ and $\dim_k(uJ(A_i)/J(A_i)^2v) = \dim_k(\psi(u)J(A_{-i-1})/J(A_{-i-1})^2\psi(v))$ for all $u,v \in Q_0$. In other words, we may assume without loss of generality that $\psi$ induces an automorphism of the underlying quivers of $A_i$ and $A_{-i-1}$. But then we must have $\psi (m) = -m$, $\psi (\operatorname{pred}(i)) = \operatorname{succ}(-i)$, and $\psi(\operatorname{succ}^2(i)) = \operatorname{pred}(-i-1)$. This forces $\psi (\ell) = -\ell$ for all $\ell \in Q_0$, so that again by Lemma \ref{l.autos} we have $\Aut (Q) = C_2$. In this case, the non-identity automorphism of $Q$ sends $A_i$ to $A_{-i-1}$. Therefore, $\Aut_k(B)\cdot A_i = \Aut_k(B)\cdot A_{-i-1}$ and so $\Iso (A_i,B) = \Aut_k(B)\cdot A_i$. Otherwise $A_i\not\cong A_{-i-1}$, and again $\Iso (A_i,B) = \Aut_k(B)\cdot A_i$. The $i>-1$ case follows from replacing $\underline{w}$ with $\underline{w}^*$ and repeating the argument above.
\end{proof}
We now prove our second case, where $A_i$ is hereditary but does not contain a trivalent vertex. We note again that this forces $i \in \{-m+1, m-2\}$.
\begin{lemma}\label{l.splitnotri}
Let $Q$ and $B$ be as before, and let $A = A_{-m+1}$ or $A_{m-2}$. Suppose that $A$ is hereditary, that the Ext quiver of $A$ is connected, and that it does not contain a trivalent vertex. Then $\Iso (A,B) = \Aut_k(B)\cdot A$.
\end{lemma}
\begin{proof}
It suffices to prove the claim for $A_{-m+1}$. Without loss of generality, we may assume that $w_{-m+1} = +1$. For $1 \le m\le 3$ the claim can be verified through straightforward, but tedious computations. So we will assume $m>3$ throughout. By hypothesis, we must have $w_{-m}w_{-m+1}w_{-m+2} = (+1)(+1)(-1)$. From Proposition \ref{c.pathalg}, the Ext quiver of $A_{-m+1}$ is a type $\mathbb{A}$ Dynkin quiver, whose valence-$1$ vertices are $-m+1$ and $m$. The binary word associated to this quiver, choosing $-m+1$ to be the left-most vertex, is then simply $\underline{u} = (-1)(+1)(-1)w_{-m+3}\cdots w_{m-1}$. The only other $i$ for which $A_i$ can have a type $\mathbb{A}$ Dynkin quiver as its Ext quiver is $i = m-2$. Therefore, we have an inclusion $\Iso (A_{-m+1},B) \subset \Inn (B)\cdot A_{-m+1} \cup \Inn (B)\cdot A_{m-2}$. We claim that this inclusion is an equality if and only if $\Aut (Q) = C_2$. If it is not an equality then $A_{-m+1}\not\cong A_{m-2}$, and so in particular $\Aut (Q) = 1$. But then $\Iso (A,B) = \Inn (B)\cdot A = \Aut_k(B)\cdot A$, and the claim of the lemma is true in this case. So, suppose instead that $A_{m-2} \cong A$. Then $-m$ and $m-1$ are the leaves in the quiver of $A_{m-2}$. Comparing the full subquivers of $A_{m-2}$ and $A$ on $\{-m, -m+1, -m+2, -m+3\}$, we find that no isomorphism $A \rightarrow A_{m-2}$ can carry $-m+1$ to $-m$. But then this implies that we can find an isomorphism $\psi : A \rightarrow A_{m-2}$ which permutes $Q_0$ and carries $-m+1$ to $m-1$. This forces $\psi(j) = -j$ for all $j \in Q_0$. By Lemma \ref{l.autos} we have $\Aut (Q) = C_2$. In this case the non-identity element of $\Aut (Q)$, carries $A_{-m+1}$ to $A_{m-2}$, and so $\Iso (A,B) = \Aut_k(B)\cdot A$ in this case as well.
\end{proof}
Now it only remains to prove the hereditary trivalent case. We break the proof into the cases where $|Q_0|$ is even or odd. For the next lemma, let $i$ be chosen such that $-m \le i \le m-1$. Then according to Figure \ref{f.quiverform}, exactly one of the dotted arrows must represent an arrow in the quiver of $A_i$. We refer to $A_i$ as $L_i$ if the arrow from $\operatorname{pred}(i)$ to $\operatorname{succ}(i)$ is present, and $R_i$ if the arrow from $i$ to $\operatorname{succ}^2(i)$ is present. For brevity, we write $A_i = L_i$ if the former holds, and $A_i = R_i$ if the latter holds. In either case, there is a unique trivalent vertex in the quiver of $A_i$, and a unique univalent vertex adjacent to it. We refer to this univalent vertex as the \emph{root} of the quiver. We call the smallest connected full subquiver containing this univalent vertex and $-m$ as the \emph{left path}. Similarly, the smallest connected full subquiver containing the univalent vertex and $m$ is called the \emph{right path}. Note that the left path and right path are just type $\mathbb{A}$ Dynkin quivers. The \emph{length} of the left/right path is just the number of arrows in it.
\begin{lemma}\label{l.splittri}
Let $G$ and $B$ be as before. Let $A$ be a hereditary maximal subalgebra of split type, whose Ext quiver is connected and contains a trivalent vertex. Then $\Iso (A,B) = \Aut_k(B)\cdot A$.
\end{lemma}
\begin{proof}
\emph{Case 1:} Suppose that $n$ is even. Then in particular, $\operatorname{succ}(-1) = 1$. We start by showing that for any $i$ and $j$ with $-m \le i , j \le -1$ and $i \neq j$, $A_i \not\cong A_j$. To see this, first note that for such $i$, the length of the left path in $L_i$ is $i+m$, and the length of the right path is $m-i-1$. Similarly, length of the left path in $R_i$ is $m+i+2$, and the length of the right path is $m+i-2$. In particular, the difference between the lengths of the left and right paths in $L_i$ is odd, whereas the difference between the left and right paths in $R_i$ is even. It follows that if either $A_i = L_i$ and $A_j = R_j$, or $A_i = R_i$ and $A_j = R_j$, then $A_i \not\cong A_j$. Furthermore, $L_i \not\cong L_j$, since for $i \neq j$, the corresponding sets of left and right path lengths are disjoint. It follows that if $A_i \cong A_j$ for $i,j \le -1$, then necessarily $A_i = R_i$ and $A_j = R_j$. In fact, by inspection of path lengths, the only possibility is $m \geq 3$ and $\{ i,j\} = \{ -1,-3\}$. Without loss of generality, we may assume that $w_{-1} = (+1)$. Then necessarily $w_{-2}w_{-1}w_1 = (-1)(+1)(+1)$, so that $1$ is the root of $A_{-1}$ and it is a source in the quiver of $A_{-1}$. Since $A_{-3} = R_{-3}$ and $w_{-2} = -1$, we must also have $w_{-3} = -1$. But then $-2$ is the root of $A_{-3}$, and it is a sink in its quiver. This is a contradiction, and so $A_{-1}\not\cong A_{-3}$.
The above argument implies that if $-m \le i \le -1$ and $j$ is chosen such that $A_i \cong A_j$, then $j \geq 1$. Note that if $j \geq 1$, then the lengths of the left and right paths of $L_j$ have an even difference, whereas they have an odd difference in $R_j$. Comparing path lengths, we find the following:
\begin{enumerate}
\item If $i \in \{ -1,-3\}$ and $A_i = R_i$, then $A_i \cong A_j$ implies $A_j = L_j$ and $j \in \{ -1,2\}$.
\item If $i \not\in \{ -1,-3\}$ and $A_i = R_i$, then $A_i \cong A_j$ implies $A_j = L_{-i-1}$.
\item If $A_i = L_i$, then $A_i \cong A_j$ implies $A_j = R_{\operatorname{pred}(-i)}$.
\end{enumerate}
We have already shown that $A_{-1}\not\cong A_{-3}$. Similar computations show that if the root of $R_{-1}$ (resp. $R_{-2}$) is a source, then the root of $L_2$ (resp. $L_1$) is a sink, and vice-versa. We conclude $R_{-1} \not\cong L_2$ and $R_{-2}\not\cong L_1$. The remaining cases from statements (1)-(3) can be rephrased as follows: for all $i$, either $\Iso (A_i,B) = \Inn (B)\cdot A_i$ or $\Iso (A_i,B) = \Inn (B)\cdot A_i \cup \Inn (B)\cdot A_{\operatorname{pred}(-i)}$. If $\Iso (A_i,B) = \Inn (B)\cdot A_i$, there is nothing to show. So suppose $i \neq \operatorname{pred}(-i)$ and $A_i \cong A_{\operatorname{pred}(-i)}$, for some negative integer $i$. In particular $i \neq -1$, and if $i = -2$ we may assume $A_{-2} = L_{-2}$. Note that this implies that the left path of $A_i$ must have a shorter length than the right path, whereas the left path of $A_{-i-1}$ must have a longer length that its right path. Since any isomorphism $A_i \rightarrow A_{-i-1}$ permuting $Q_0$ must send the trivalent vertex (resp. root) of $A_i$ to the trivalent vertex (resp. root) of $A_{-i-1}$, it follows that such an isomorphism satisfies $j \mapsto -j$ for all $j \in Q_0$. By Lemma \ref{l.autos} we have $\Aut (Q) = C_2$, a contradiction (since $n$ is even). Hence $\Iso (A_i,B) = \Inn(B)\cdot A_i$, as we wished to show.
\emph{Case 2:} Suppose that $n$ is odd. Then $Q_0$ is just the interval $[-m,m]$ of $\ZZ$. Let $i$ be chosen such that $-m \le i \le -1$. The the left path of $L_i$ has length $i+m$, and the right path has length $m-i+1$. The left path of $R_i$ has length $m+i+2$, and the right path has length $m-i-1$. As before, we want to start by showing that if $j$ is any number between $-m$ and $-1$ with $i \neq j$, then $A_i \not\cong A_j$. Suppose that $A_i = L_i$. Then by comparing path sizes, we see that if $A_i \cong A_j$, then $A_j = R_{i-2}$. Without loss of generality, we may assume $w_i = +1$. Then $w_{i-1}w_iw_{i+1} = (+1)(+1)(-1)$. But since $A_{i-2} = R_{i-2}$, we must have $w_{i-2} = +1$ as well. But then the root of $A_i$ is $i$, which is a sink, whereas the root of $A_{i-2}$ is $i-1$, which is a source. This is a contradiction, and so $A_i \not\cong A_j$ if $A_i = L_i$. Otherwise $A_i = R_i$. If $A_i \cong A_j$ and $A_j = L_j$ we are in the previous case, so assume $A_j = R_j$ as well. By comparing path lengths, we find that the only possibility is $\{ i,j\} = \{ -1,-2\}$. Suppose without loss of generality that $w_{-1} = +1$. Then since $A_{-1} = R_{-1}$ and is connected hereditary, we must have $w_{-2}w_{-1}w_{0} = (-1)(+1)(+1)$. But then $A_{-2} \neq R_{-2}$, since $A_{-2} = R_{-2}$ requires $w_{-2}$ and $w_{-1}$ to have the same sign. Hence $R_{-1} \not\cong R_{-2}$, and it follows that for all $-m \le i,j \le -1$, $A_i \not\cong A_j$.
Now we show that if $A_i \cong A_j$ for $j\geq 0$, then $j = {-i-1}$ and $\Aut (Q) = C_2$. Suppose first that $A_i = L_i$. Then the other algebras $A_j$ for $j\geq 0$ which have the same set of path lengths as $A_i$ are $R_{-i-1}$ and $L_{-i+1}$. We first show that $L_i \not\cong L_{-i+1}$. Without loss of generality, we let $w_i = +1$. The root of $L_i$ is $i$, and the right path has a larger length than the left path. Treating the right path as a type $\mathbb{A}$ Dynkin quiver starting at $i$, its associated binary word is $(-1)(+1)w_{i+1}\cdots w_{m-1}$. Now, the larger path in $L_{-i+1}$ is the left path, and its root is vertex $-i+1$. Since the root of $L_i$ is a sink, $-i+1$ must be a sink too. This forces $w_{-i}w_{-i+1}w_{-i+2} = (+1)(+1)(-1)$. Treating the left path of $L_{-i+1}$ as a type $\mathbb{A}$ Dynkin quiver starting at $-i+1$, its associated binary word is $(-1)(-w_{-i-1})\cdots (-w_{-m})$. If $L_i \cong L_{-i+1}$, then we must have $(-1)(+1)w_{i+1}\cdots w_{m-1} = (-1)(-w_{-i-1})\cdots (-w_{-m})$. In particular, $w_{-i-1} = -1$ and for all $i+1 \le j \le m-1$, we have $w_j = -w_{-(j+1)}$. But then setting $j = -i$ we find $+1= w_{-i} = -w_{-(-i+1)} = -w_{i-1} = -(+1) = -1$, a contradiction. So $L_i\not\cong L_{-i+1}$, as we wished to show. Hence, we assume $L_i \cong R_{-i-1}$. Since the root of $L_i$ is a sink, it must be a sink in $R_{-i-1}$ as well. This implies that if we start at the root, the longer path in $R_{-i-1}$ has binary word $(-1)(+1)(-1)(-w_{-i-2})\cdots w_{-m}$. Therefore, we have $(-1)(+1)(-w_{-i-2})\cdots (-w_{-m}) = (-1)(+1)(w_{i+1})\cdots w_{m-1}$. This implies that $w_j = -w_{-(j+1)}$ for $i+1 \le j \le m-1$. Similarly, the binary word for the short path in $R_{-i-1}$ is $w_{-i}w_{-i+1}\cdots w_{m-1}$, whereas it is $(-w_{i-1})(-w_{i-2})\cdots (-w_{-m})$ for $L_i$. Hence, $w_j = -w_{-(j+1)}$ for $-m \le j \le i-1$. But $w_i= +1$ by hypothesis, and $w_{-i-1} = -1$ since $R_{-i-1}$ is a subalgebra of $B$ isomorphic to $L_i$. In other words, $\underline{w}^* = \underline{w}$, and $\Aut (Q) = C_2$.
This shows that if $A_i \cong A_j$ for $j \geq 0$, then $j = -i-1$ and $\Aut (Q) = C_2$ for the $A_i = L_i$ case. Note that by replacing $\underline{w}$ by $\underline{w}^*$ if necessary, it only remains to consider the case when $A_i = R_i$ and $A_j = R_j$. By comparison of path lengths, $A_j = R_{-i-3}$. Note that we may assume $i \le -3$, for otherwise this reduces to a case that has been previously discussed. Comparing the binary words for the longer paths in $R_i$ and $R_{-i-3}$, we see $w_{i+1}\cdots w_{m-1} = (+1)(-1)(-w_{-i-4})\cdots (-w_{-m})$. Therefore, $w_j = -w_{-(j+1)}$ for $i+3 \le j \le m-1$. Since $i \le -3$, $i+3 \le 0$ and so this suffices to show that $\underline{w}^* = \underline{w}$. But since $w_i = +1$ and $A_i = R_i$, $w_{i+1} = +1$. Since we also assume $A_{-i-3} = R_{-i-3} \cong R_i$, comparison of roots yields $w_{-i-2} = +1$. But then $w_{i+1} \neq -w_{-(i+1+1)}$, and so $\underline{w}^* \neq \underline{w}$. This contradiction shows that $R_i \not\cong R_{-i-3}$.
Putting this all together, we see that if $A_i \cong A_j$, then $j = -i-1$ and $\Aut (Q) = C_2$. But in this case $A_i$ and $A_{-i-1}$ lie in the same $\Aut_k(B)$-orbit, and so the lemma is proved.
\end{proof}
Lemmas \ref{l.splitnotri}, \ref{l.splitnonher}, and \ref{l.splittri} combine to prove the following proposition:
\begin{proposition}\label{p.splitorbit}
Let $Q$ and $B$ be as before. If $A \subset B$ is a maximal subalgebra of split type, and the Ext quiver of $A$ is connected, then $\Iso (A,B) = \Aut_k(B)\cdot A$.
\end{proposition}
To finish the proof of Theorem \ref{t.main}, we need to show that $\Iso (A_{i,j}, B) = \Aut_k(B)\cdot A_{i,j}$ for all $i$ and $j$. If $i<j$, then Proposition \ref{p.seppres} asserts that the underlying graph of the Ext quiver of $A_{i,j}$ looks as follows: \newline
\begin{figure}[th]
\[
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=1.5]
\node[point,label={below:$-m$}] (-m) at (0,0) {};
\node (b1) at (1,0) {$\cdots$};
\node[point,label={below:$\operatorname{pred}(i)$}] (i-1) at (2,0) {};
\node[point,label={below:$i+j$}] (i+j) at (3,0) {};
\node (b2) at (2.5,0.5) {};
\node (b3) at (3,1) {};
\node (b4) at (3.5,0.5) {};
\node[point,label={below:$\operatorname{succ}(j)$}] (j+1) at (4,0) {};
\node (b5) at (5,0) {$\cdots$} ;
\node[point,label={below:$m$}] (m) at (6,0) {};
\path (-m) edge (0.5,0)
(1.5,0) edge (i-1)
(i-1) edge (i+j)
(i+j) edge[bend left=25] (b2)
(i+j) edge[bend right=25] (b4)
(i+j) edge (j+1)
(j+1) edge (4.5,0)
(5.5,0) edge (m);
\path[dotted] (b2) edge[bend left=25] (b3)
(b3) edge[bend left=25] (b4);
\end{tikzpicture},
}}
\]
\caption{}\label{f.sepform}
\end{figure}
\noindent where if $i = -m$ (resp. $j = m$) we delete the subgraph including $\operatorname{pred}(i)$ and all vertices to its left (resp. $\operatorname{succ}(j)$ and all vertices to its right). Examining the paths from $-m$ to $i+j$ and $i+j$ to $m$, we see $\Iso (A_{i,j},B) \subset \Aut_k(B)\cdot A_{i,j} \cup \Aut_k(B)\cdot A_{-i,-j}$. In particular, $\Iso (A_{i,j},B) = \Aut_k(B)\cdot A_{i,j}$ if $i=-j$, and so without loss of generality we may replace the $i<j$ assumption by the assumption that $|i|<|j|$ throughout (allowing now for the possibility that $i\geq j$). If $\Aut (Q) = C_2$, then the unique non-identity automorphism of $Q$ extends to an automorphism of $B$ which sends $v_i\mapsto v_{-i}$, $v_j \mapsto v_{-j}$. Therefore $\Iso (A_{i,j},B) = \Aut_k(B)\cdot A_{i,j}$ in this case as well, and so we may also assume $\Aut (Q) = 1$.
Under these assumptions, we will be done if we can show $A_{i,j} \not\cong A_{-i,-j}$ for the remaining cases. In other words, we must prove the following lemma:
\begin{lemma}\label{l.evenodd}
Under the assumptions above, $A_{i,j} \not\cong A_{-i,-j}$.
\end{lemma}
\noindent Again we break this up into two arguments, depending on whether $n$ is even or odd. First suppose that $Q$ has $n=2m$ vertices. Note that if $i,j \le -1$ or $i,j \geq 1$, then $A_{i,j} \not\cong A_{-i,-j}$ by examination of the edge between $v_{-1}$ and $v_1$ in the quivers of $A_{i,j}$ and $A_{-i,-j}$. Therefore, $\Iso (A_{i,j},B) = \Inn(B)\cdot A_{i,j}$ in this case. Hence, we may assume without loss of generality that $-m \le i \le -1 < 1 \le j \le m$, so that combined with our reductions above, we have $1 \le -i < j \le m$. Write $\underline{w} = \underline{w}_1\cdot\underline{w}_2\cdot\underline{w}_3\cdot\underline{w}_4\cdot\underline{w}_5$, where the $\underline{w}_i$ are defined as follows (see Figure \ref{f.wdef}):
\begin{enumerate}
\item $\underline{w}_1$ is the binary word from $v_{-m}$ to $v_{-j}$,
\item $\underline{w}_2$ is the binary word from $v_{-j}$ to $v_i$,
\item $\underline{w}_3$ is the binary word from $v_i$ to $v_{-i}$,
\item $\underline{w}_4$ is the binary word from $v_{-i}$ to $v_j$, and
\item $\underline{w}_5$ is the binary word from $v_j$ to $v_m$.
\end{enumerate}
\begin{figure}[th]
\[
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=1.5]
\node[point,label={below:$-m$}] (-m) at (0,0) {};
\node (b1) at (1,0) {$\cdots$};
\node (w1) at (1,-0.25) {$\underline{w}_1$};
\node[point,label={below:$-j$}] (i-1) at (2,0) {};
\node (w2) at (2.5,-0.25) {$\underline{w}_2$};
\node[point,label={below:$i+j$}] (i+j) at (3,0) {};
\node (b2) at (2.5,0.5) {};
\node (w3) at (2.25,0.5) {$\underline{w}_3$};
\node[point,label={above:$-i$}] (b3) at (3,1) {};
\node (b4) at (3.5,0.5) {};
\node (w4) at (3.75,0.5) {$\underline{w}_4$};
\node (b5) at (4,0) {$\cdots$} ;
\node (w5) at (4,-0.25) {$\underline{w}_5$};
\node[point,label={below:$m$}] (m) at (5,0) {};
\path (-m) edge (b1)
(b1) edge (i-1)
(i-1) edge (i+j)
(i+j) edge[bend left=25] (b2)
(i+j) edge[bend right=25] (b4)
(i+j) edge (b5)
(b5) edge (m);
\path[dotted] (b2) edge[bend left=25] (b3)
(b3) edge[bend left=25] (b4);
\end{tikzpicture},
}}
\]
\caption{}\label{f.wdef}
\end{figure}
\newpage
\begin{proofEven}
Suppose there was an isomorphism $A_{i,j} \cong A_{-i,-j}$ by way of contradiction. Then the quivers of $A_{i,j}$ and $A_{-i,-j}$ must be isomorphic. Since $|i|<|j|$, this can only be true if the following three conditions hold:
\begin{enumerate}
\item $\underline{w}_1 = (\underline{w}_5)^*$,
\item $\underline{w}_2 = (\underline{w}_4)^*$, and
\item either $\underline{w}_3\cdot\underline{w}_4 = \underline{w}_3^*\cdot\underline{w}_2^*$ or $\underline{w}_3\cdot \underline{w}_4 = (\underline{w}_3^*\cdot \underline{w}_2^*)^* = \underline{w}_2\cdot \underline{w}_3$.
\end{enumerate}
\noindent If $\underline{w}_3\cdot \underline{w}_4 = \underline{w}_3^*\cdot \underline{w}_2^*$, then by applying condition (2) and cancelling $\underline{w}_4$ from both sides, $\underline{w}_3 = \underline{w}_3^*$. Then $\underline{w}^* = ( \underline{w}_1\cdot \underline{w}_2 \cdot \underline{w}_3 \cdot \underline{w}_4 \cdot \underline{w}_5)^* = \underline{w}_5^* \cdot \underline{w}_4^* \cdot \underline{w}_3^*\cdot \underline{w}_2^* \cdot \underline{w}_1^* = \underline{w}_1\cdot \underline{w}_2 \cdot \underline{w}_3 \cdot \underline{w}_4 \cdot \underline{w}_5 = \underline{w}$, a contradiction. Hence, $\underline{w}_3\cdot \underline{w}_2^* = \underline{w}_2\cdot \underline{w}_3$.
\emph{Case 1:} Suppose that the length of $\underline{w}_2$ is less than or equal to the length of $\underline{w}_3$. Then since $\underline{w}_3\cdot \underline{w}_2^* = \underline{w}_2\cdot \underline{w}_3$, we may write $\underline{w}_2 = \e_1\cdots \e_d$, $\underline{w}_3= \e_1\cdots \e_d\e_{d+1}\cdots \e_{d+f}$, where each $\e_a \in \{ \pm 1\}$. Then
\begin{equation}
(\e_1\cdots \e_d\e_{d+1}\cdots \e_{d+f})(-\e_d)\cdots (-\e_1)= (\e_1\cdots \e_d)(\e_1\cdots\e_d\e_{d+1}\cdots \e_{d+f}).
\end{equation}
Suppose that $d > f$. Then by comparing terms, we have $\e_{d+a} = \e_a$ for all $1 \le a \le f$ and $\e_{f+b} = -\e_{d-b+1}$ for all $1 \le b \le d$. Since $d+f$ is odd, $(d+f+1)/2$ is a natural number, and $f<d$ implies that $(d+f+1)/2 \le d$. Now, $d-b+1 = (d+f+1)/2$ precisely when $b = (d-f+1)/2$. Then $\e_{f+b} = -\e_{d-b+1}$ yields $\e_{(d+f+1)/2} = -\e_{(d+f+1)/2}$ for this value of $b$, a contradiction. Note that $d \neq f$ since $d+f$ is odd, and so the only remaining possibility is that $d< f$. After cancelling the $\e_1\cdots \e_d$ term from both sides, we find that
\begin{equation}
\e_{d+1}\cdots \e_{d+f}(-\e_d)\cdots (-\e_1) = (\e_1\cdots \e_f )(\e_{f+1} \cdots \e_{d+f}).
\end{equation}
\noindent For $1\le a \le f$, define $q = q(a)$ to be the largest non-negative integer such that $dq+a \le f$. Now, comparing terms on the left-hand-side and the right-hand-side of the above equation, we see $\e_a = \ldots = \e_{dq+a} = \e_{d(q+1)+a}$. Since $d(q+1)+a > f$, it follows that $\e_{d(q+1)+a} = -\e_{d+f+1-(d(q+1)+a)} = -\e_{f-qd-a+1} = -\e_{f-(dq+a-1)}$. Notice that $q = \lfloor \frac{f-a}{d} \rfloor$. If we can choose $a$ such that $a = f-(dq+a-1)$, we will have the desired contradiction.
\emph{Sub-Case 1(a):} Say that $f$ is odd. Then $d$ is even, and for all $s$, $(f-ds+1)/2$ is an integer. Define $r$ to be the largest positive integer such that $dr+1 \le f$. Then $f < d(r+1)+1 < d(r+2)+1$. Let $a = (f-dr+1)/2$. It is clear that $a \le f$. The inequality $dr+1 \le f < d(r+2)+1$ can be re-arranged to say that $r \le \frac{f-\frac{f-rd+1}{2}}{d} < r+1$. In other words, $\lfloor \frac{f-a}{d} \rfloor = r$. But $a = f-dr-a+1 = f-d\lfloor \frac{f-a}{d} \rfloor -a + 1$ by definition, and we have found a choice of $a$ which works.
\emph{Sub-Case 1(b):} Say that $f$ is even. Then $d$ is odd, and for all odd $s$, $(f-sd+1)/2$ is an integer. Let $r$ be the same as in Sub-Case 1(a). If $r$ is odd, then again $a = (f-rd+1)/2$ works. Otherwise $r$ is even, so that $r-1$ is odd and $a = (f-d(r-1)+1)/2$ is an integer. Notice that $d(r-1)+1 \le f < d((r-1)+2)+1$, so that $\lfloor \frac{f-a}{d} \rfloor = r$. Now, for this choice of $a$, we have $a = f-d(r-1) - a + 1 = (f - d\lfloor\frac{f-a}{d}\rfloor -a + 1) + d$. But since $r \geq 2$, $d(r-1)+a - 1 \geq 0$ and $f-d(r-1)-a+1 \le f$. Hence, $-\e_{f-(dr+a-1)} = -\e_{f-(dr+a-1)+d} = -\e_{f-(d(r-1)+a-1)}$, and we obtain our desired contradiction. The proof of Case 1 is now complete.
\emph{Case 2:} Suppose that the length of $\underline{w}_2$ is greater than or equal to the length of $\underline{w}_3$. Note that they cannot be equal, for otherwise $\underline{w}_3\cdot \underline{w}_2^* = \underline{w}_2\cdot \underline{w}_3$ would imply $\underline{w}_3 = \underline{w}_2$, and hence $\underline{w}_3^* = \underline{w}_3$, a contradiction. So then, we may write $\underline{w}_3 = \e_1\cdots \e_d$ and $\underline{w}_2 = \e_1\cdots \e_d\e_{d+1}\cdots \e_{d+f}$, for some $d,f>0$. As before, $d \neq f$. Hence, our equation reads
\begin{equation}
\e_1\cdots \e_d(-\e_{d+f})\cdots (-\e_{d+1})(-\e_d)\cdots (-\e_1) = \e_1\cdots \e_d\e_{d+1}\cdots \e_{d+f}\e_1\cdots \e_d.
\end{equation}
\noindent Comparing the last $d$-terms on both sides of this equation, we conclude that $\e_a = -\e_{d-a+1}$, for $1 \le a \le d$. Now, $d$ is odd, so $(d+1)/2$ is an integer $\le d$. Therefore, for $a = (d+1)/2$ we have $\e_{(d+1)/2} = -\e_{d-(d+1)/2 +1} = -\e_{(2d - d - 1 + 2)/2} = -\e_{(d+1)/2}$, a contradiction.
\end{proofEven}
The remaining case left to consider is when $Q$ has $n = 2m+1$ vertices, $m \geq 1$. Again, we assume $\Aut (Q) = 1$ and $|i| < |j|$. If $i,j \le 0$ or $i,j \geq 0$ and $A_{i,j} \cong A_{-i,-j}$, then inspection of the long path in Figure \ref{f.sepform} implies that $\underline{w}(\ell) = -\underline{w}(\operatorname{pred}(-\ell ))$ for either $-m \le \ell \le 0$ or $0 \le \ell \le m-1$. In either case we conclude $\underline{w}^* = \underline{w}$, contradicting the triviality of $\Aut (Q)$. So we can assume $i<0 $ and $0\le j$ in addition to $|i| < |j|$. Then we can decompose $\underline{w}$ into $\underline{w} = \underline{w}_1\cdot \underline{w}_2 \cdot \underline{w}_3 \cdot \underline{w}_4 \cdot \underline{w}_5$ as in Figure \ref{f.wdef}. Under the assumptions that $A_{i,j} \cong A_{-i,-j}$ and $n$ is odd, the length of $\underline{w}_3$ must be even, and $\underline{w}_3\cdot \underline{w}_2^* = \underline{w}_2 \cdot \underline{w}_3$.
With this modified setup, we can now finish the proof of Lemma \ref{l.evenodd}:
\begin{proofOdd}
\emph{Case 1:} Say that the length of $\underline{w}_3$ is greater than the length of $\underline{w}_2$. Write $\underline{w}_2 = \e_1\cdots \e_d$ and $\underline{w}_3 = \e_1\cdots \e_d\e_{d+1}\cdots \e_{d+f}$. Again, we find that equation (2) holds, and so $\e_{d+a} = \e_a$ for $1 \le a \le f$ and $\e_{f+b} = -\e_{d-b+1}$ for $1\le b \le d$.
\emph{Sub-Case 1(a):} Suppose that $f\le d$. If $d$ and $f$ are both odd, then $a= (f+1)/2$ is an integer $\le f$. Since $d+a\geq f+a > f$, we have $\e_a = \e_{d+a} = \e_{f+(d+a-f)} = -\e_{d-(d+a-f)+1} = -\e_{f-a+1}$. But for this choice of $a$, $a = f-a+1$, and we arrive at a contradiction. So we may assume that $d$ and $f$ are both even. Applying the same logic, we find that for all $a \le f$, we have a sequence of equalities: $\e_a = \e_{d+a} = -\e_{f-a+1} = -\e_{d+f-a+1} = \e_a$. Since $f$ and $d$ are both even, $\{ a,d+a\} \cap \{f-a+1, d+f-a+1\} = \emptyset$. The equalities $\e_a = -\e_{d+f - a + 1}$ and $\e_{d+a} = -\e_{f-a+1}$ are equivalent to the statement that $\underline{w}_3^* = \underline{w}_3$, which contradicts our hypotheses. Hence, the $f\le d$ case is complete.
\emph{Sub-Case 1(b):} Suppose that $f>d$. For any $a \le f$, define $q = q(a)$ to be the largest non-negative integer such that $dq + a \le f$. Then $\e_a = \ldots = \e_{dq+a} = \e_{d(q+1)+a} = -\e_{f-(dq+a-1)}$. If $f$ and $d$ are odd, then define $r$ to be the largest positive integer such that $dr +1 \le f$. If $r$ is even, then $a = (f-dr+1)/2$ is an integer $\le f$ which satisfies $\lfloor \frac{f-a}{d} \rfloor = r$ and $i = f-dr+a-1$. This implies $\e_a = -\e_a$ as before. Otherwise $r$ is odd. Then $a = (f-d(r-1)+1)/2$ satisfies $\lfloor\frac{f-a}{d}\rfloor = r$, $f-d(r-1)-a+1 \le f$, and $a=(f - dr -a + 1) + d$, which implies $\e_a = -\e_{f-d(r-1) -a +1} = -\e_a$ as before. It remains to consider the case that $f$ and $d$ are even. Then we have $\e_a = \ldots = \e_{dq+a} = \e_{d(q+1)+a} = -\e_{f-(dq+a-1)} = \ldots = -\e_{f-a+1} = -\e_{d+f-a+1}$ for all $1 \le a \le f$, so $\e_a = -\e_{d+f - (a-1)}$. Finally, for each $1 \le a \le d$, there is a unique $f< s \le f+d$ with $s \equiv a$ $(\operatorname{mod}$ $d)$. But then, $s = d(q+1) + a$, and the equality $\e_{d(q+1)+a} = -\e_{f-(dq+a-1)}$ tells us that $\e_b = -\e_{d+f-(b-1)}$ for all $1 \le b \le f+d$. This implies $\underline{w}_3 = \underline{w}_3^*$, contrary to our assumption that $\Aut (Q) = 1$. Hence, the $f>d$ case is complete.
\emph{Case 2:} Say that the length of $\underline{w}_3$ is less than or equal to the length of $\underline{w}_2$. As before, $\ell (\underline{w}_3) \neq \ell (\underline{w}_2)$, so we may assume $\ell (\underline{w}_3) < \ell (\underline{w}_2)$. Write $\underline{w}_3 = \e_1\cdots \e_d$ and $\underline{w}_2 = \e_1\cdots \e_d\e_{d+1}\cdots \e_f$. As before, equation (3) holds. If $d$ is odd, repeat the same argument given for the even-vertices case. Otherwise $d$ is even, and so the equation $\e_a = -\e_{d-a+1}$ for $1 \le a \le d$ tells us $\underline{w}_3^* = \underline{w}_3$, contrary to our hypotheses.
\end{proofOdd}
\noindent Lemma \ref{l.evenodd} immediately implies the following:
\begin{proposition}\label{p.seporbit}
Let $A \subset B$ be a connected maximal subalgebra of separable type, where $B = kQ$ and $Q$ is a type $\mathbb{A}$ Dynkin quiver. Then $\Iso (A,B) = \Aut_k(B)\cdot A$.
\end{proposition}
\noindent Finally, Propositions \ref{p.splitorbit} and \ref{p.seporbit} combine to yield Theorem \ref{t.main}.
We end with an example to show that the connectedness hypothesis is necessary:
\begin{example}
We note that the conclusion of Theorem \ref{t.main} is false if $A$ has a disconnected Ext quiver. For instance, suppose that $Q$ is the type $\mathbb{A}_4$ Dynkin quiver
\[
\vcenter{\hbox{
\begin{tikzpicture}[point/.style={shape=circle,fill=black,scale=.3pt,outer sep=3pt},>=latex, baseline=-3,scale=2]
\node[point,label={below:$v_{-2}$}] (1) at (0,0) {} ;
\node[point,label={below:$v_{-1}$}] (2) at (1,0) {} ;
\node[point,label={below:$v_1$}] (3) at (2,0) {} ;
\node[point,label={below:$v_2$}] (4) at (3,0) {} ;
\path[->] (1) edge (2)
(3) edge (2)
(3) edge (4) ;
\end{tikzpicture}
}}.
\]
Then $\Aut_k(B) = \Inn (B)$ and $\Iso (A_{-2}, B) = \Inn(B)\cdot A_{-2} \cup \Inn(B)\cdot A_1$. Of course, $A_{-2}$ and $A_1$ lie in different $\Inn(B)$-orbits, since their Jacobson radicals are distinct.
\end{example} | 0.002231 |
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Syntocinon dosage in massive postpartum haemorrhage
Report By:
Neelima Deshpande -
SPR O&G
Search checked by
tbc -
Institution:
Birmingham Women's Hospital
Date Submitted:
30th October 2002
Last Modified:
1st November 2002
Status:
Blue (submitted but not checked)
Three Part Question
In patients with massive postpartum haemorrhage what is the appropriate dose of syntocinon and what is the maximum dose that can be used safely, to arrest haemorrhage or prevent further uterine bleeding?
Clinical Scenario
In patients suffering with massive postpartum haemorrhage syntocinon is often used intravenously to control atonic uterine bleeding. Currently, clinical units have protocols for administering up to 10 units intravenously stat, repeated up to once, followed by an infusion of 40 units in 40 mls of diluent administered over 4 hours and repeated if required. There are variations of this protocol in use in different hospitals. There is a need to support the use of such doses of syntocinon / oxytocin with evidence from proper studies on the effectiveness of different uterotonics and reports of any adverse effects.
Search Strategy
OVID interface through BMA site 1966 to date included Embase. PUBmed through world wide web. CINAHL, Biological absracts, SIGLE (grey literature in europe), hand search and cross references.
exp (third stage OR 3rd stage) OR (post partum h$emorrhage OR postpartum h$emorrhage OR post-partum h$emorrhage) OR PPH OR blood loss OR blood-loss AND (exp oxytocin OR syntocinon)AND NOT (induc* OR rip* OR prelab* or cerv* or dystocia OR augment*)
Search Outcome
There were three case reports from 1970-75 about water intoxication following the use of syntocinon. The pharmaceutical company Alliance, which manufactures syntocinon sent us their protocol for the use of a syntocinon infusion and the rate of administration but no evidence on which it was based. There is one systematic review on the use of uterotonics in the control of postpartum haemorrhage but does not have any information about evidence for the most appropriate dosage of syntocinon to be used or any known complications from currently used dosages.
Syntocinon / oxytocin is a widely used parenteral uterotonic for the control of post partum haemorrhage. In massive haemorrhage, protocols with infusions of up to 40 units over 4 hours have been advocated. There have been no large studies (observational or otherwise) to quantify the actual amount or rate that is effective in controlling atonic uterine haemorrhage or preventing uterine atony once it has been controlled. Case reports about water intoxication following large doses of syntocinon suggested that Dextrose infusion in large quantities was to blame but this was never proven. There is a need for further studies looking at the effectiveness of current protocols of syntocinon used in the post partum period and especially in cases with massive postpartum haemorrhage where there is great complexity in managing infused fluid volumes and monitoring output.
Clinical Bottom Line
The recommended rate of infusion in post partum haemorrhage of 5 IU syntocinon slowly IV followed by intravenous infusion at a rate of 20 - 40 mU/min or higher is stated by the Clinical guideline for induction of labour (NICE 2001) and by the RCOG evidence based clinical guideline 2001. There is no evidence that is quoted to support the current protocols in use for massive post partum haemorrhage. | 0.690771 |
Support Our Regional Trail Stewards
Trails exist all over, but that doesn’t necessarily mean they’re accessible to all forms of recreation and travel. It takes a lot of love and a lot of work to obtain trail access for mt. bikers, to create new trails and most of all to maintain trails over time. Since 1998, Northwest Trail Alliance has been striving for Portland-area trail access. They were instrumental in the foundation of two new trails systems: Gateway Green and Sandy Ridge.
Over the years the organization has lived varied lives, with different leaders at the helm, various goals and a plethora of projects. But their mission has remained true: help get more people on the trails, on bikes.
The Northwest Trail Alliance Fall Membership Drive is underway and this is a great time to join or renew your membership. Their goal is to reach 2,000 members this year through individual and family memberships.
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Membership is what powers this grassroots organization’s efforts like trail building, social rides, Take A Kid Mountain Biking Day and trail advocacy efforts. NWTA is a strong organization founded on a strong base of partnerships with local land managers, elected officials, the bike industry and boots-on-the-dirt volunteers.
Your membership support will allow them to realize the many respectable trail goals they have for the upcoming year, and will help keep mt. biking strong in the region. | 0.857289 |
TITLE: If $M=M^{\perp\perp}$ for every closed subspace $M$ of a pre-Hilbert space then $H$ is complete
QUESTION [7 upvotes]: It is well-known that if $H$ is a Hilbert space then for every closed subspace $M$ we have $M=M^{\perp\perp}$ see here
I would like to prove the converse by showing that: if $M=M^{\perp\perp}$ for every closed subspace $M$ of a pre-Hilbert space then $H$ is complete
REPLY [10 votes]: We will show that the embedding $i:H\hookrightarrow \hat H$ of $H$ into its completion is onto. Given $z\in \hat H\setminus \{0\}$, the subspace $M=\{x\in H:\langle i(x),z\rangle =0\}$ is closed in $H$ and its orthogonal complement $M^\perp$ is different from $\{0\}$ since otherwise $M=M^{\perp\perp}=\{0\}^\perp=H$ which implies $z=0$. For fixed $y\in M^\perp\setminus \{0\}$ and the linear functionals $\varphi(x)= \langle x,y\rangle$ and $\psi(x)=\langle i(x),z\rangle$ we have kern$(\psi)=M\subseteq$ kern$(\varphi)$ which implies $\varphi=a\psi$ for some scalar $a$ (indeed, if $e\in H$ satisfies $\psi(e)=1$ we get $x-\psi(x)e\in$ kern$(\psi)$ so that $0=\varphi(x-\psi(x)e)=\varphi(x)-a\psi(x)$ for all $x\in H$ with $a=\varphi(e)$). Now, the functionals on $\hat H$ given by $\langle\cdot,i(y)\rangle$ and $\langle\cdot, az\rangle$ coincide on the dense subspace $i(H)$ and hence on $\hat H$ which implies $i(y)=az$. Since $a\neq 0$ this gives $z\in i(H)$. | 0.006984 |
\begin{document}
\begin{center} {\Large \bf Coupling for some partial differential equations driven by white noise}
\vspace{5mm}
Giuseppe Da Prato, Scuola Normale Superiore di Pisa,
Piazza dei Cavalieri 7, 56126, Pisa, Italy\vspace{2mm}
Arnaud Debussche, Ecole Normale Sup\'erieure de Cachan,
antenne de Bretagne,
Campus de Ker Lann, 35170 Bruz, France.\vspace{2mm}
Luciano Tubaro, Department of Mathematics, University of Trento,
Via Sommarive 14, 38050 Povo, Italy.
\end{center}\vspace{5mm}
\begin{abstract} We prove, using coupling arguments, exponential convergence to equilibrium for reaction--diffusion and
Burgers equations driven by space-time white
noise. We use a coupling by reflection.
\end{abstract}
\noindent{\it 2000 Mathematics Subject Classification}: 60H15, 35K57, 35Q53
\noindent{\it Key words}: Coupling, Reaction--Diffusion equations, Burgers equation, Exponential convergence to equilibrium.
\section{Introduction}
We are concerned with a stochastic differential equation in a separable Hilbert space $H$,
with inner product
$(\cdot,\cdot )$ and norm $|\cdot|$,
\begin{equation}
\label{e1.1}
dX=(AX+b(X))dt+dW(t),\quad X(0)=x\in H,
\end{equation}
where $A\colon D(A)\subset H\to H$ is linear, $b\colon D(b)\subset H\to H$ is nonlinear
and $W$ is a cylindrical Wiener process defined in some probability space $(\Omega, \mathcal F,\P)$ in $H$.
Concerning $A$ we shall assume that
\begin{Hypothesis}
\label{h1.1}
\begin{enumerate}
\item[]
\item[(i)] $A\colon D(A)\subset H\to H$ is the infinitesimal
generator of a strongly continuous semigroup $e^{tA}$.
\item[(ii)] For any $t> 0$ the linear operator $Q_{t},$ defined as
\begin{equation}
\label{e1.2}
Q_{t}x=\int_{0}^{t}e^{sA}e^{sA^{*}}xds,\;\;x\in H,\;t\ge 0,
\end{equation}
is of trace class.
\end{enumerate}
\end{Hypothesis}
We shall consider situations where \eqref{e1.1} has a unique mild solution
$X(\cdot,x)$, that is a mean square adapted process
such that
\begin{equation}
\label{e1.3}
X(t,x)=e^{tA}x+\int_0^te^{(t-s)A}b(X(s,x))ds+z(t),\quad\P\mbox{\rm--a.s.},
\end{equation}
where $z(t)$ is the stochastic convolution
\begin{equation}
\label{e1.4}
z(t)=\int_0^te^{(t-s)A}dW(s).
\end{equation}
It is well known that, thanks to Hypothesis \ref{h1.1}, for each $t> 0$, $z(t)$ is a
Gaussian random variable with mean $0$ and
covariance operator
$Q_t.$
We will also assume that the solution has continuous trajectories. More precisely, we assume
\begin{equation}
\label{e1.4bis}
X(\cdot,x)\in L^2(\Omega;C([0,T];H)),\mbox{ for any } x\in H.
\end{equation}
In this paper we want to study the exponential convergence to equilibrium of the transition semigroup
$$
P_t\varphi(x)=\E[\varphi(X(t,x))],\quad x\in H,\;t\ge 0,
$$
where $ \varphi\colon H\to \R$. We wish to use coupling arguments. It is well known that exponential convergence to equilibrium
implies the uniqueness of invariant measure.
It seems that the first paper using a coupling method to prove uniqueness of the invariant
measure and mixing
property for a stochastic partial differential equation is \cite{mueller}. There, an equation with globally Lipschitz coefficients is considered,
some of them are also assumed to be monotone.
Coupling argument have also been used recently to prove ergodicity and exponential convergence to
equilibrium in the context of the Navier-Stokes equation driven by very degenerate noises
(see \cite{KS1}, \cite{ks}, \cite{mattingly1}, \cite{mattingly2}). The method has also
been studied in \cite{Hairer1} for reaction--diffusion equations and in \cite{odasso}
for Ginzburg--Landau equations.
Our interest here is different since we are interested in space-time white noises as in
\cite{mueller} but without the strong restrictions on the coefficients. Ergodicity
is well known in the cases considered here. It can be proved
by the Doob theorem (see \cite{DPZ2}). Indeed,
since the noise is non degenerate, it is not diffiucult to prove that the transition semigroup
is strong Feller and irreducible. However, this argument does not imply exponential
convergence to equilibrium and we think that it is important to study this question.
In this paper we shall follow the construction of couplings introduced in \cite{LR}
(see also \cite{CL}) to treat both reaction diffusion
and Burgers equations driven by white noise and obtain exponential
convergence to equilibrium.
Note that exponential convergence to equilibrium for reaction--diffusion equations is well
known. Anyway, we have chosen to treat this example because we think that the method
presented here provides a very simple proof. Moreover, we recover the spectral gap property
obtained in \cite{DPDG} by a totally different
- and simpler - argument.
In the case of the Burgers equation driven by space-time white noise, it seems that
our result is new.
The coupling method based on Girsanov transform introduced in
\cite{KS1}, \cite{ks}, \cite{mattingly1}, \cite{mattingly2} can easily be used
if the noise is nuclear. It is also possible
it could be extended to our case. However, the extension is not straightforward and
the method used here seems to be simpler.
Moreover, it is not clear that, in the case of the reaction-diffusion
equation, it is possible to prove the spectral gap property with this method.
Next section is devoted to describing the construction of the coupling used here, we follow \cite{CL}.
Note that the coupling is constructed as the solution of a stochastic differential
equation with discontinuous coefficients. In \cite{CL}, the existence of the coupling is
straightforward. It is easy to see that the corresponding martingale problem has a solution.
This argument is difficult in infinite dimension and we have preferred to prove directly
the existence of a strong solution.
Section 3 is devoted to application to reaction--diffusion equations and
section 4 to the Burgers equation driven by white noise.
We finally remark that our method extends to other equations such as reaction-diffusion equations
or the stochastic Navier-Stokes equation in space dimension two with non degenerate noise. We
have chosen
to restrict our presentation to these two examples for clarity of the presentation.
\section{Construction of the coupling}
We shall consider the following system of stochastic differential equations:
\begin{equation}
\label{e1.5}
\left\{\begin{array}{lll}
dX_1=(AX_1+b(X_1))dt+\frac{1}{\sqrt{2}}\;dW_1+\frac{1}{\sqrt{2}}\;\left(1-2\frac{(X_1-X_2)\otimes(X_1-X_2)}{|X_1-X_2|^2}\right)dW_2\\
\\
dX_2=(AX_2+b(X_2))dt+\frac{1}{\sqrt{2}}\;\left(1-2\frac{(X_1-X_2)\otimes(X_1-X_2)}{|X_1-X_2|^2}\right)dW_1+\frac{1}{\sqrt{2}}\;dW_2\\
\\
X_1(0)=x_1,\quad X_2(0)=x_2,
\end{array}\right.
\end{equation}
where $W_1,W_2$ are independent cylindrical Wiener processes.
This corresponds to a coupling with reflection, see \cite{CL}. Equation
\eqref{e1.5} is associated to the Kolmogorov operator
in $H\times H$ defined by
$$
\begin{array}{lll}
\mathcal K \;\Phi(x_1,x_2)\\
\\
=\ds{\frac12\;\mbox{\rm Tr}\;\left[ \left(\begin{array}{cc}
1&{\ds 1-2\frac{(x_1-x_2)\otimes(x_1-x_2)}{|x_1-x_2|^2}}\\
\ds{1-2\frac{(x_1-x_2)\otimes(x_1-x_2)}{|x_1-x_2|^2}}&1
\end{array} \right)
D^2 \Phi(x_1,x_2)\right]}\\
\\
+\left(\left(\begin{array}{c}
Ax_1+b(x_1)\\
Ax_2+b(x_2)
\end{array} \right),D \Phi(x_1,x_2) \right),
\end{array}
$$
or, equivalently,
\begin{equation}
\label{e1.6}
\begin{array}{lll}
\mathcal K \;\Phi
&=&\ds{\frac12\;\mbox{\rm Tr}\;\left[\Phi_{x_1x_1}+\left(2-4\frac{(x_1-x_2)\otimes(x_1-x_2)}{|x_1-x_2|^2}\right)
\Phi_{x_1x_2}+\Phi_{x_2x_2}
\right]}\\
\\
&&+( Ax_1+b(x_1),\Phi_{x_1} )+( Ax_2+b(x_2),\Phi_{x_2} ).
\end{array}
\end{equation}
The following formula will be useful in the sequel.
\begin{Lemma}
\label{l1.1}
Let $f\colon\R\mapsto\R$ be a ${\mathcal C}^2$ function and let $\Phi$
be defined by $\Phi(x_1,x_2)=f(|x_1-x_2|),\; x_1,x_2\in H,\;f\in C^2(\R)$, then
\begin{equation}
\label{e1.13}
\begin{array}{lll}
\mathcal K \;\Phi(x_1,x_2)
&=&\ds{2\;f''(|x_1-x_2|)+ \frac{f'(|x_1-x_2|)}{|x_1-x_2|}}\\
\\
&&\times ( A(x_1-x_2)+ b(x_1)-b(x_2),x_1-x_2 ).
\end{array}
\end{equation}
\end{Lemma}
{\bf Proof:}
We have
$$
\Phi_{x_1}=-\Phi_{x_2},\quad \Phi_{x_1x_1}=\Phi_{x_2x_2}=-\Phi_{x_1x_2},
$$
$$
\Phi_{x_1}(x_1,x_2)=f'(|x-y|)\;\frac{x_1-x_2}{|x_1-x_2|},
$$
and
$$
\begin{array}{lll}
\Phi_{x_1x_1}(x_1,x_2)&=&\ds{f'(|x_1-x_2|)\;\frac{ |x_1-x_2|^2-(x_1-x_2)\otimes(x_1-x_2)}{|x_1-x_2|^3}}\\
\\
&&\ds{+f''(|x_1-x_2|)\; \frac{(x_1-x_2)\otimes(x_1-x_2)}{|x_1-x_2|^2},}
\end{array}
$$
The result follows.
\hfill$\Box$\bigskip
We will use functions $f$
such that, for a suitable positive constant $\kappa$, we have
\begin{equation}
\label{e1.12}
\mathcal K\Phi(x_1,x_2)\le -\kappa,\quad \mbox{\rm for all}\;x_1,x_2\in H.
\end{equation}
Thanks to Lemma \ref{l1.1}, we have to solve the following basic inequality,
in the unknown $f$ (notice that $f$ has to be nonnegative),
\begin{equation}
\label{e1.14}
\ds{2\;f''(|x_1-x_2|)}
\ds{+ \frac{f'(|x_1-x_2|)}{|x_1-x_2|}\; ( A(x_1-x_2)+ b(x_1)-b(x_2),x_1-x_2 )}
\le -\kappa.
\end{equation}
We now study problem \eqref{e1.5}.
In our applications, it will be easy to verify that,
for any $\varepsilon >0$, it has a unique mild solution
$X(t,x_1,x_2)=(X_1(t,x_1,x_2),X_2(t,x_1,x_2))$ on the random interval
$[0,\tau_{x_1,x_2}^\varepsilon]$ where
$\tau_{x_1,x_2}^\varepsilon=\inf\{t\in [0,T]\,|\, |X_1(t)-X_2(t)|\le \varepsilon\}$.
Clearly $\tau_{x_1,x_2}^\varepsilon$ is increasing as $\varepsilon\to 0$ so that we can
define $\tau_{x_1,x_2}=\lim_{\varepsilon\to 0}\tau_{x_1,x_2}^\varepsilon$ and
get a unique mild solution
$X(t,x_1,x_2)=(X_1(t,x_1,x_2),X_2(t,x_1,x_2))$ on the random interval
$[0,\tau_{x_1,x_2})$.
\begin{Lemma}
\label{l1.2}
$X(t,x_1,x_2)$ has a limit in $L^2(\Omega)$ when $t\to \tau_{x_1,x_2}$. Moreover,
if $\tau_{x_1,x_2}<T$, we have $X_1(\tau_{x_1,x_2},x_1,x_2)=X_2(\tau_{x_1,x_2},x_1,x_2)$.
\end{Lemma}
{\bf Proof:}
Let us define
$$
X_1^\varepsilon(t)= \left\{
\begin{array}{l}
X_1(t,x_1,x_2),\; t\le \tau_{x_1,x_2}^\varepsilon,\\
\\
X(t,\tau_{x_1,x_2}^\varepsilon,X_1(\tau_{x_1,x_2}^\varepsilon,x_1,x_2),\;
t\ge \tau_{x_1,x_2}^\varepsilon.
\end{array}
\right.
$$
We have denoted by $X(\cdot,s,x)$ the solution of \eqref{e1.1} with the condition $X(s,s,x)=x$
at time $s$ instead of $0$. It is not difficult to check that $X_1$ and $X(\cdot,x_1)$ have
the same law. Let us write for $\eta_1,\eta_2>0$:
$$
\begin{array}{l}
\E(|X_1(\tau_{x_1,x_2}-\eta_1,x_1,x_2)-X_1(\tau_{x_1,x_2}-\eta_2,x_1,x_2)|^2)\\
\\
=\lim_{\varepsilon\to 0} \E(|X_1(\tau_{x_1,x_2}^\varepsilon-\eta_1)
-X_1(\tau_{x_1,x_2}^\varepsilon-\eta_2)|^2)\\
\\
=\lim_{\varepsilon\to 0} \E(|X_1^\varepsilon(\tau_{x_1,x_2}^\varepsilon-\eta_1)
-X_1^\varepsilon(\tau_{x_1,x_2}^\varepsilon-\eta_2)|^2)\\
\\
\le \lim_{\varepsilon\to 0}\E(\sup_{t\in[0,T]} |X_1^\varepsilon(t-\eta_1,x_1,x_2)
-X_1^\varepsilon(t-\eta_2,x_1,x_2)|^2).
\end{array}
$$
Since, $X_1$ and $X(\cdot,x_1)$ have
the same law, we can write
$$
\E(\sup_{t\in[0,T]} |X_1^\varepsilon(t-\eta_1)
-X_1^\varepsilon(t-\eta_2)|^2)
= \E(\sup_{t\in[0,T]} |X(t-\eta_1,x_1)
-X(t-\eta_2,x_1)|^2).
$$
By \eqref{e1.4bis}, we know that this latter term goes to zero so that we prove that
$X_1(t)$ has a limit. We treat $X_2(t)$ exactly in the same way.
Finally, if $\tau_{x_1,x_2}<T$, then $|X_1(\tau_{x_1,x_2}^\varepsilon,x_1,x_2)
-X_2(\tau_{x_1,x_2}^\varepsilon,x_1,x_2)|=\varepsilon$ for any $\varepsilon >0$.
Letting $\varepsilon\to 0$ we deduce the last statement.
\hfill$\Box$\bigskip
We also consider the following equation
\begin{equation}
\label{e1.15}
\left\{\begin{array}{lll}
dX_1=(AX_1+b(X_1))dt+\frac{1}{\sqrt{2}}\;\mbox{1{\kern-2.7pt}l}_{[0,\tau_{x_1,x_2}]}(t)dW_1\\
\\
+\frac{1}{\sqrt{2}}\;\mbox{1{\kern-2.7pt}l}_{[0,\tau_{x_1,x_2}]}(t)\left(1-2\frac{(X_1-X_2)\otimes(X_1-X_2)}{|X_1-X_2|^2}\right)dW_2+\frac{1}{\sqrt{2}}
(dW_1+dW_2)(1-\mbox{1{\kern-2.7pt}l}_{[0,\tau_{x_1,x_2}]}(t))\\
\\
dX_2=(AX_2+b(X_2))dt+\frac{1}{\sqrt{2}}\;\mbox{1{\kern-2.7pt}l}_{[0,\tau_{x_1,x_2}]}(t)\left(1-2\frac{(X_1-X_2)\otimes(X_1-X_2)}{|X_1-X_2|^2}\right)dW_1\\
\\
+\frac{1}{\sqrt{2}}\;\mbox{1{\kern-2.7pt}l}_{[0,\tau_{x_1,x_2}]}(t)dW_2+
\frac{1}{\sqrt{2}}(dW_1+dW_2)(1-\mbox{1{\kern-2.7pt}l}_{[0,\tau_{x_1,x_2}]}(t)),\\
\\
X_1(0)=x_1,\quad X_2(0)=x_2.
\end{array}\right.
\end{equation}
It is clear that for $t\le \tau_{x_1,x_2}$ the solutions of \eqref{e1.5} and \eqref{e1.15} do coincide, whereas for
$t\ge \tau_{x_1,x_2}$ \eqref{e1.15} reduce to
\begin{equation}
\label{e1.16}
\left\{\begin{array}{lll}
dX_1=(AX_1+b(X_1))dt+ \frac{1}{\sqrt{2}}(dW_1(t)+ dW_2(t)),\quad t\ge \tau_{x_1,x_2},\\
\\
dX_2=(AX_2+b(X_2))dt+ \frac{1}{\sqrt{2}}(dW_1(t)+ dW_2(t)),\quad t\ge \tau_{x_1,x_2}.
\end{array}\right.
\end{equation}
Using Lemma \ref{l1.2}, we easily prove that \eqref{e1.15} has a unique solution.
Moreover, since $\frac{1}{\sqrt{2}}(W_1(t)+W_2(t))$ is
a cylindrical Wiener process, it follows that $X_1$ and $X_2$ have the same law as
$X(\cdot,x_1)$ and $X(\cdot,x_2)$. In other words, $(X_1,X_2)$ is a coupling
of the laws of $X(\cdot,x_1)$ and $X(\cdot,x_2)$.
We are interested in the first time $\tau_{x_1,x_2}$ when $X_1(t,x_1,x_2))$ and $X_2(t,x_1,x_2))$ meet.
That is $\tau_{x_1,x_2}$ is the
stopping time
\begin{equation}
\label{e1.9}
\tau_{x_1,x_2}=\inf\{t>0:\;X_1(t,x_1,x_2)=X_2(t,x_1,x_2)\}.
\end{equation}
Our goal is first to show that
\begin{equation}
\label{e1.10}
\E(\tau_{x_1,x_2})<+\infty.
\end{equation}
To prove \eqref{e1.10} we look, following \cite{CL}, for a Lyapunov function $f$
such that \eqref{e1.14} holds. This is motivated by next Proposition.
\begin{Proposition}
\label{p2.4}
Assume that there exists a ${\mathcal C}^2$ function such that \eqref{e1.14} holds.
Let $x_1,x_2\in H$ with $x_1\neq x_2$. Then we have
\begin{equation}
\label{e2.8}
\P(\tau_{x_1,x_2}=+\infty)=0.
\end{equation}
Moreover,
\begin{equation}
\label{e2.9}
\E(\tau_{x_1,x_2})\le f(|x_1-x_2|).
\end{equation}
\end{Proposition}
{\bf Proof}. We shall write for simplicity,
$$
X_1(t)=X_1(t,x_1,x_2),\quad X_1(t)=X_1(t,x_1,x_2).
$$
Also, we can assume without loss of generality that $\kappa=1$.
Then we introduce the following stopping times:
$$
S_N=\inf\left\{t\ge 0: |X_1(t)-X_2(t)|>N\right\},\quad N\in \N,
$$
and
$$
\tau_{n,N}=\tau_{x_1,x_2}^{1/n}\wedge S_N.
$$
By the It\^o formula{\footnote{The application of the It\^o formula can be justified rigorously
thanks to a regularization argument. This can be done easily in the applications considered
hereafter.}, Lemma \ref{l1.1} and \eqref{e1.14}, we have
\begin{equation}
\label{e2.11}
\begin{array}{l}
f(|X_1(t\wedge \tau_{n,N})-X_2(t\wedge \tau_{n,N})|)
\le f(|x_1-x_2|)\\\\
\ds+\int_0^{t\wedge \tau_{n,N}} \frac{f'(|X_1(s)-X_2(s)|)}{|X_1-X_2|}
\tfrac1{\sqrt{2}} (X_1-X_2,d(W_1-W_2))\\\\
- \;(t\wedge \tau_{n,N}).
\end{array}
\end{equation}
It follows that
$$
\E\big(f(|X_1(t\wedge \tau_{n,N})-X_2(t\wedge \tau_{n,N})|)\big)
\le f(|x-y|)- \E(t\wedge \tau_{n,N}),
$$
and
$$
\E(t\wedge \tau_{n,N})\le f(|x-y|) .
$$
Consequently as $t\to \infty$ we find
$$
\E(\tau_{n,N})\le f(|x-y|)
$$
which yields as $n\to \infty$
$$
\E(\tau_{x_1,x_2}\wedge S_N)\le f(|x-y|).
$$
By \eqref{e1.4bis}, we easily prove that $S_N\to \infty$ as $N\to \infty$ so that we get
$$
\E(\tau_{x_1,x_2})\le f(|x-y|).
$$
$\Box$\bigskip
\section{Dissipative systems with white noise}
We consider the case when there exist $\lambda\ge 0, a>0$ such that
\begin{equation}
\label{e2.1}
(A(x_1-x_2)+ b(x_1)-b(x_2),x_1-x_2 )\le \lambda|x_1-x_2|^2-a|x_1-x_2|^4,
\end{equation}
for all $x_1,x_2\in H.$ We also assume that Hypothesis \ref{h1.1} and \eqref{e1.4bis} hold.
A typical equation satisfying such assumptions is the following
stochastic reaction-diffusion equation on $[0,1]$
$$
\left\{
\begin{array}{l}
dX=(\partial_{\xi\xi} X-\alpha X^3 +\beta X^2+\gamma X+\delta)dt +dW, \; t>0,\; \xi\in (0,1),\\
\\
X(t,0)=X(t,1)=0,\; t>0,\\
\\
X(0,\xi)=x(\xi), \xi\in (0,1),
\end{array}
\right.
$$
where $\alpha>0$.
In this case, we take $A=D^2_{\xi}$ on the domain $D(A)=H^2(0,1)\cap H^1_0(0,1)$,
$b(x)=-\alpha x^3 +\beta x^2+\gamma x+\delta$. We could also consider the more
general example where $b$ is a polynomial of degree $2p+1$ with negative leading
coefficient. Note that this equation is gradient, the invariant measure is known explicitly.
However, we shall not use this fact. We could treat as well perturbation of this equation
which are not gradient but satisfy \eqref{e2.1}
Following the above discussion, we look for a positive function $f$
such that
\begin{equation}
\label{e2.2}
2\;f''(r)+ f'(r)\left(\lambda r-a\;r^3\right) =-1.
\end{equation}
Setting $f'=g$, \eqref{e2.2} becomes
\begin{equation}
\label{e2.3}
g'(r)- \frac12\;g(r) (ar^3-\lambda r) =-\frac{1}{2}.
\end{equation}
whose general solution is given by
$$
g(r)=e^{\frac{1}{8}\;(ar^4-2\lambda r^2)}g(0)-\frac{1}{2}\int_0^re^{\frac{1}{8}\;(ar^4-as^4-2\lambda r^2+
2\lambda s^2)}ds.
$$
Finally, we have
\begin{equation}
\label{e2.4}
\begin{array}{lll}
f(r)&=&\ds{f(0)+\int_0^re^{\frac{1}{8}\;(as^4-2\lambda s^2)}ds\;f'(0)}\\
\\
&&\ds{-\frac{1}{2}\;\int_0^re^{\frac{1}{8}\;(as^4-2\lambda s^2)}
\left[\int_0^se^{-\frac{1}{8}\;(a\sigma^4-
2\lambda \sigma^2)}d\sigma \right]\;ds.}
\end{array}
\end{equation}
Setting $f(0)=0$ and $f'(0)=\frac{1}{2}\;\int_0^\infty e^{-\frac{1}{8}\;(a\sigma^4-
2\lambda \sigma^2)}d\sigma$ we obtain
\begin{equation}
\label{e2.5}
f(r)=\frac{1}{2}\;\int_0^re^{\frac{1}{8}\;(as^4-2\lambda s^2)}
\left[\int_s^{+\infty}e^{-\frac{1}{8}\;(a\sigma^4-
2\lambda \sigma^2)}d\sigma \right]\;ds
\end{equation}
and
\begin{equation}
\label{e2.6}
f'(r)=\tfrac{1}{2}\; e^{\frac{1}{8}\;(ar^4-2\lambda r^2)}
\int_r^{+\infty}e^{-\frac{1}{8}\;(a\sigma^4-
2\lambda \sigma^2)}d\sigma .
\end{equation}
We need some properties on $f$.
\begin{Lemma}
\label{l2.1}
We have $(a r^3-\lambda r) f'(r)< 1$ for any $r\ge 0$.
\end{Lemma}
\medskip
{\sf Proof:} The function $r\mapsto a r^3-\lambda r$
is increasing and positive if $r>\delta:=\sqrt{\frac{\lambda}{a}}\;$ so that in this case
$$
(a r^3-\lambda r) f'(r)< \tfrac12\;e^{\frac{1}{8}\;(ar^4-2\lambda r^2)}
\int_r^{+\infty} (a \sigma^3-\lambda \sigma) e^{-\frac{1}{8}\;(a\sigma^4-
2\lambda \sigma^2)}d\sigma=1.
$$
Since for $0\le r\le \delta$ we have $(a r^3-\lambda r) f'(r)\le 0$, the conclusion follows. $\Box$\bigskip
\begin{Corollary}
\label{c2.2}
$f' $ is a decreasing positive function.
\end{Corollary}
\medskip
{\sf Proof:} Since $f$ satisfies \eqref{e2.2} , we deduce from Lemma \ref{l2.1} that
$f''<0$ so that $f'$ decreases. Moreover, $f'$ is positive by \eqref{e2.6}. $\Box$\bigskip
\begin{Lemma}
\label{l2.3}
There exists $\Lambda$ depending only on $a, \lambda$
such that for any $r>0$
\begin{itemize}
\item[i)] $f'(r)\le \Lambda$
\item[ii)] $f(r)\le \Lambda$
\end{itemize}
\end{Lemma}
\medskip
{\sf Proof:} (i) follows obviously from Corollary \ref{c2.2}. Let us show (ii). Fix
$r>r_0:=\sqrt{\frac{\lambda}{a}}.$
Since $f$ is increasing, we have
$$
f(r)\le f(r_0),\quad r\le r_0.
$$
If $r>r_0$ we have by Lemma \ref{l2.1},
$$
\begin{array}{l}
\ds f(r)= f(r_0)+
\int_{r_0}^r
f'(s)\;ds \\\\
\ds\hphantom{f(r)}\le f(r_0)+
\int_{r_0}^\infty
\tfrac{ds}{a s^3-\lambda s}= f(r_0)+ \tfrac1{2\lambda} \ln(1+\tfrac{\lambda}{ar_0^2-\lambda}) .
\end{array}
$$
Therefore $f(\infty)<\infty$ and ii) follows provided
$
\Lambda\ge \max\{f(\infty),f(r_0)\}.
$
\hfill$\Box$\bigskip
The following results strengthen Proposition \ref{p2.4}.
\begin{Proposition}
\label{p2.5}
Let $x_1,x_2\in H$ with $x_1\neq x_2$, then there exists $\Lambda>0$ such that
$$
\E\left(e^{{}^{\frac1{2\Lambda^2}\;\scriptstyle\tau_{x_1,x_2}}}\right)\le
e^{\frac1\Lambda}
$$
\end{Proposition}
{\sf Proof:} We use the same notations as in the proof of
Proposition \ref{p2.4}. By \eqref{e2.11} we have
$$
t\wedge \tau_{n,N}\le f(|x_1-x_2|)+\int_0^{t\wedge \tau_{n,N}} \frac{f'(|X_1-X_2|)}{|X_1-X_2|}
\tfrac1{\sqrt{2}} (X_1-X_2,d(W_1-W_2)),
$$
so that
\begin{equation}
\label{e2.17}
\E\left( e^{\alpha (t\wedge \tau_{n,N})}\right)\le
e^{ \alpha f(|x_1-x_2|)}\E\left( e^{{}^{ \alpha
\scriptstyle\int_0^{t\wedge \tau_{n,N}}
\frac{f'(|X_1-X_2|)}{|X_1-X_2|}
\frac1{\sqrt{2}} (X_1-X_2,d(W_1-W_2))}}
\right).
\end{equation}
On the other hand, since
$$
\E\left( e^{{}^{ \alpha
\scriptstyle\int_0^{t\wedge \tau_{n,N}}
\frac{f'(|X_1-X_2|)}{|X_1-X_2|}
\frac1{\sqrt{2}} (X_1-X_2,d(W_1-W_2))}}
\right)\le\left(\E\left( e^{{}^{ 2\alpha^2
\scriptstyle\int_0^{t\wedge \tau_{n,N}}
|f'(|X_1-X_2|)|^2 ds}}
\right)\right)^{1/2},
$$
we deduce from Lemma \ref{l2.3} that
$$
\E\left( e^{{}^{ \alpha
\scriptstyle\int_0^{t\wedge \tau_{n,N}}
\frac{f'(|X_1-X_2|)}{|X_1-X_2|}
\frac1{\sqrt{2}} (X_1-X_2,d(W_1-W_2))}}
\right)\le\left(\E\left( e^{2\alpha^2 \Lambda^2
(t\wedge \tau_{n,N})}
\right)\right)^{1/2}
$$
Substituting in \eqref{e2.17} yields
$$
\E\left( e^{\alpha (t\wedge \tau_{n,N})}\right)\le
e^{ \alpha f(|x_1-x_2|)}
\left(\E\left(e^{2\alpha^2 \Lambda^2
(t\wedge \tau_{n,N})}\right)\right)^{1/2}
$$
Choosing $\alpha=\frac1{2\Lambda^2}$, we deduce
$$
\E\left( e^{\frac1{2\Lambda^2} (t\wedge \tau_{n,N})}\right)\le
e^{\frac{1}{2 \Lambda^2}f(|x_1-x_2|)}\le
e^{1/\Lambda}
$$
Letting $n\to \infty$, $N\to\infty$ and arguing as in the proof of Proposition \ref{p2.4} we find
the conclusion.\\
\null\hfill$\Box$
\begin{Corollary}
\label{c2.6}
We have
$$
|P_t\varphi(x_1)-P_t\varphi(x_2)|\le
2\;\|\varphi\|_0\;e^{{}^\frac1\Lambda}\;
e^{{}^{-\frac1{2\Lambda^2}t}}
$$
\end{Corollary}
\medskip
{\sf Proof:} Let $x_1,x_2\in H.$ Since $(X_1,X_2)$ is a coupling of $(X(\cdot,x_1),X(\cdot,x_2))$,
we have
$$
|P_t\varphi(x_1)-P_t\varphi(x_2)|=|\E(\varphi(X_1(t))-\varphi(X_2(t)))|
\le \|\varphi\|_0\;\P(\tau_{x_1,x_2}\ge t).
$$
Now the conclusion follows Proposition \ref{p2.5}. $\Box$\bigskip
We end this section by proving that the spectral gap property holds. We thus recover a known
result (see for instance \cite{DPDG}) with a totally different method.
\begin{Proposition}
\label{p2.7}
Let $\nu$ be an invariant measure then, for any $p>1$, there exist
$c_p$, $\alpha_p$ such that
$$
|P_t\varphi-\bar\varphi|_{L^p(H,\nu)}\le
c_p e^{-\alpha_p t}\;|\varphi|_{L^p(H,\nu)}
$$
\end{Proposition}
\medskip
{\sf Proof:} By Corollary \ref{c2.6}, we have the result for $p=\infty$.
Using that $P_t$ is a contraction semigroup on $L^1(H,\nu)$
and an interpolation argument, we obtain the result. $\Box$\bigskip
\section{Burgers equation}
We take here $H=L^2(0,1)$ and denote by $\|\cdot\|$ the norm of the Sobolev space $H^1_0(0,1)$.
We consider the equation
\begin{equation}
\label{e3.1}
\left\{
\begin{array}{l}
dX=(AX+b(X))dt+dW,\\\\
X(0)=x,
\end{array}\right.
\end{equation}
where
$$
Ax=D^2_\xi x,\quad x\in D(A)=H^2(0,1)\cap H^1_0(0,1)
$$
and
$$
b(x)=D_\xi(x^2),\quad x\in H^1_0(0,1).
$$
It well known that problem \eqref{e3.1} has a unique solution for any $x\in L^2(0,1)$ which we denote by
$X(t,x)$, see \cite{DPDT}. It defines a transition semigroup $(P_t)_{t\ge 0}$. It is also known
that it has a unique invariant measure and is ergodic (see \cite{DPZ2}). The following
result states the exponential convergence to equilibrium.
\begin{Theorem}
\label{main-burgers}
There exist constants $C, \gamma >0$ such that for any $x_1,x_2\in L^2(0,1)$, $\varphi\in C_b(L^2(0,1))$,
$$
|P_t\varphi(x_1)-P_t\varphi(x_2)|\le c e^{-\gamma t}\|\varphi\|_0(1+|x_1|^4+|x_2|^4)
$$
\end{Theorem}
To prove this result,
we want to construct a coupling for equation \eqref{e3.1}. It does not seem possible
to apply directly the method of section 2. We shall first consider a cut off
equation
\begin{equation}
\label{e3.2}
\left\{
\begin{array}{l}
dX=(AX+D_\xi F_R(X))dt+dW,\\\\
X(0)=x,
\end{array}\right.
\end{equation}
where $F_R:L^4(0,1)\to L^2(0,1)$ is defined by
$$
F_R(x)=\left\{\begin{array}{l}
x^2 ,\quad \mbox{\rm if}\;|x|_{L^4}\le R,\\
\\
\frac{R^2x^2}{|x|^2_{L^4}},\quad \mbox{\rm if}\;|x|_{L^4}\ge R.
\end{array}\right.
$$
We have
\begin{equation}
\label{e3.3}
|F_R(x)-F_R(y)| \le 2R\; |x-y|_{L^4},\quad x,y\in L^4(0,1)
\end{equation}
and
\begin{equation}
\label{e3.4}
|F_R(x)|\le R^2,\quad x\in L^4(0,1).
\end{equation}
We have denoted by $|\cdot|_{L^4}$ the norm in $L^4(0,1)$. The norm in $L^2(0,1)$ is
still denoted by $|\cdot|$.
It is not difficult to check that Hypothesis \ref{h1.1} and \eqref{e1.4bis} hold so that the results
of section 2 can be applied.
We then need a priori estimates on the solutions of \eqref{e3.1} so that we can control
when the coupling for the cut-off equation can be used. These are given in section 4.2.
Then, we construct a coupling for the Burgers equation which enables us to prove
the result in section 4.4.
\subsection{Coupling for the cut--off equation}
Here $R>0$ is fixed. We denote by the same symbol $c_R$ various constants
depending only on $R$.
\begin{Lemma}
\label{l3.1}
There exists $\alpha>0,\beta>0, c_R>0$ such that
\begin{equation}
\label{e3.5}
\frac{2\alpha}{1-\beta}\in [1,2),
\end{equation}
and
\begin{equation}
\label{e3.6}
|F_R(x)-F_R(y)|\le c_R |x-y|^\alpha \|x-y\|^\beta,\quad x,y\in H^1_0(0,1).
\end{equation}
\end{Lemma}
{\bf Proof}. First notice that by \eqref{e3.3} and \eqref{e3.4} it follows that
$$
|F_R(x)-F_R(y)|\le (2R)^{2-\gamma}\; |x-y|_{L^4}^\gamma,\quad x,y\in L^4(0,1),
$$
for any $\gamma\in[0,1]$. Moreover, by the Sobolev embedding theorem
we have $H^{1/4}(0,1)\subset L^4(0,1)$ and using
a well known interpolatory inequality we find that
\begin{equation}
\label{e3.6bis}
|x-y|_{L^4}\le c|x-y|_{H^{1/4}}\le c|x-y|^{3/4}\;\|x-y\|^{1/4},\quad x,y\in H^1_0(0,1).
\end{equation}
Consequently
$$
|F_R(x)-F_R(y)|\le c(2R)^{2-\gamma}\;|x-y|^{3\gamma/4} \|x-y\|^{\gamma/4}
$$
Now setting $\alpha=3\gamma/4,\;\beta=\gamma/4$, the conclusion follows choosing
$\gamma\in [\frac47,1]$.
$\Box$\bigskip
We now construct the coupling for equation \eqref{e3.2}.
For any $x,y\in D(A)$ we have, taking into account Lemma \ref{l3.1},
$$
\begin{array}{lll}
(A(x-y)+D_\xi F_R(x)-D_\xi F_R(y),x-y)&\le& -\|x-y\|^2+|F_R(x)-F_R(y)|\;\|x-y\|\\\\
&\le&
-\|x-y\|^2+c_R\;|x-y|^\alpha\;\|x-y\|^{1+\beta}.
\end{array}
$$
Using the elementary inequality
$$
uv\le \tfrac{1+\beta}{2}\;(\epsilon u)^{\frac{2}{1+\beta}}
+\tfrac{1-\beta}{2}\;(\epsilon^{-1} v)^{\frac{2}{1-\beta}},\quad u,v,\epsilon>0,
$$
and choosing suitably $\epsilon$ we find
$$
\begin{array}{lll}
(A(x-y)+D_\xi F_R(x)-D_\xi F_R(y),x-y)&\le&
-\tfrac12 \|x-y\|^2+c_R\;|x-y|^{\frac{2\alpha}{1-\beta}}\\
\\
&\le& -\tfrac12\|x-y\|^2+\tfrac{\pi^2}{4}|x-y|^2+c_R\;|x-y|,
\end{array}
$$
since $\frac{2\alpha}{1-\beta}\in [1,2)$. By the Poincar\'e inequality we conclude that
$$
(A(x-y)+D_\xi F_R(x)-D_\xi F_R(y),x-y)\le
-\tfrac{\pi^2}{4}|x-y|^2+c_R\;|x-y|.
$$
Consequently \eqref{e1.14} (with $\kappa=1$) reduces to
$$
2f_R''(r)-f_R'(r)(\tfrac{\pi^2}4 r-c_R)=-1,
$$
$$
f_R''(r)-f_R'(r)(2\,a\, r-c_R)=-\tfrac12,\qquad a=\tfrac{\pi^2}{16}.
$$
Then
$$
f_R'(r)=e^{a r^2-c_R r}f_R'(0)-\tfrac12 \int_0^r
e^{a (r^2-s^2)-c_R (r-s)}ds
$$
and
$$
f_R(r)=f_R(0)+f_R'(0)\int_0^re^{a s^2-c_R s}ds-\tfrac12 \int_0^r ds\int_0^s
e^{a (s^2-u^2)-c_R (s-u)}du.
$$
Setting
$$
f_R(0)=0,\quad f_R'(0)=\tfrac12\;e^{a r^2-c_R r}\int_0^{+\infty}
e^{-a u^2+c_R u}du,
$$
we obtain the solution
\begin{equation}
\label{enuova}
f_R(r):=\tfrac12\int_0^r e^{as^2-c_Rs}\left(
\int_s^\infty e^{-a\xi^2+c_R\xi}d\xi\right)ds.
\end{equation}
We denote by $X_R(t,x)$ the solution of the cut-off equation \eqref{e3.2}.
The corresponding coupling constructed above is denoted by
$\big(X_{1,R}(t;x_1,x_2),X_{2,R}(t;x_1,x_2)\big)$. Then, setting
$$
\tau_R^{x_1,x_2}=\inf\{t>0 : X_{1,R}(t;x_1,x_2)=X_{2,R}(t;x_1,x_2)\}.
$$
By Proposition \ref{p2.4}, we have
\begin{equation}
\label{e3.8}
\E(\tau_R^{x_1,x_2})\le f_R(|x_1-x_2|)
\end{equation}
\begin{Remark}
Using similar arguments as in section 3, we can derive bounds on $f_R$ and $f_R'$ and
prove following result for the transition semigroup associated the the cut-off equation.
For all $\varphi\in C_b(H)$ we have
$$
|P^R_t\varphi(x_1)-P^R_t\varphi(x_2)| \le 2c\;\|\varphi\|_0\;
(1+|x_1-x_2|)^{1/\Lambda}\;
e^{{}^{-\frac1{2\Lambda^2}t}},\quad x_1,x_2\in H.
$$
\end{Remark}
\subsection{A priori estimates}
Next result is similar to Proposition \ref{p2.4} in \cite{dpd04}.
\begin{Lemma}
\label{l3.7}
Let $\alpha\ge 0$ and
$$
z^\alpha(t)=\int_{-\infty}^t e^{(A-\alpha)(t-s)} dW(s).
$$
Then, for any $p\in\N$, $\varepsilon>0$, $\delta>0$, there exists a random variable $K(\varepsilon,\delta,p)$ such that
$$
|z^\alpha(t)|_{L^p} \le K(\varepsilon,\delta,p) \;\alpha^{-\frac14+\varepsilon}(1+|t|^\delta)
$$
Moreover, all the moments of $K(\varepsilon,\delta,p)$ are finite.
\end{Lemma}
\medskip
{\sf Proof:} Proceeding as in the proof of Proposition 2.1 in \cite{dpd04}, we have
$$
z^\alpha(t)=\int_{-\infty}^t \Big[(t-\sigma)^{\beta-1}-\alpha
\int_\sigma^t (\tau-\sigma)^{\beta-1} e^{-\alpha(t-\tau)}d\tau\Big]
e^{A(t-\sigma)}Y(\sigma)d\sigma
$$
where
$$
Y(\sigma)=\frac{\sin \pi\beta}{\pi}\int_{-\infty}^\sigma (\sigma-s)^{-\beta}
e^{A(\sigma-s)}dW(s)
$$
and $\beta\in (0,1/4)$. It is proved in \cite{dpd04} that for $\gamma\in [0,1]$
$$
\begin{array}{l}
\ds\Big|(t-\sigma)^{\beta-1}-\alpha
\int_\sigma^t (\tau-\sigma)^{\beta-1} e^{-\alpha(t-\tau)}d\tau\Big| \\\\
\hspace{3cm}\le
c(\beta,\gamma)\;\alpha^{-\gamma}
(t-\sigma)^{\beta-1-\gamma}+
(t-\sigma)^{\beta-1} e^{-\alpha(t-\sigma)}.
\end{array}
$$
We deduce, by Poincar\'e inequality,
$$
|z^\alpha(t)|_{L^p} \le c \int_{-\infty}^t \!\!
\Big(\alpha^{-\gamma}
(t-\sigma)^{\beta-1-\gamma}+
(t-\sigma)^{\beta-1} e^{-\alpha(t-\sigma)}\Big)
e^{-\lambda_p (t-\sigma)} |Y(\sigma)|_{L^p}d\sigma,
$$
and, for $r>1$, $m\in\N$, by H\"older inequality we obtain if $\beta>\frac1{2m}$ and $\beta>\gamma+\frac1{2m}$
$$
\begin{array}{l}
|z^\alpha(t)|_{L^p} \le c \Big[ \alpha^{-\gamma}\Big(\ds\int_{-\infty}^t
(t-\sigma)^{\frac{2m}{2m-1}(\beta-1-\gamma)}(1+|\sigma|^r)^{\frac1{2m-1}}
e^{-\lambda_p(t-\sigma)}d\sigma\Big)^{\frac{2m-1}{2m}}\\\\
\hspace{2.5cm}+\Big(\ds\int_{-\infty}^t
(t-\sigma)^{(\beta-1)\frac{2m}{2m-1}}e^{-\frac{2m\alpha}{2m-1}(t-\sigma)}
(1+|\sigma|^r)^{\frac1{2m-1}}
d\sigma\Big)^{\frac{2m-1}{2m}} \Big]\\\\
\hspace{6.5cm}\ds\times\Big(\int_{-\infty}^t (1+|\sigma|^r)^{-1} |Y(\sigma)|_{L^p}^{2m}\,d\sigma\Big)^{\frac1{2m}}
\\\\
\hph{|Z^\alpha(t)|_{L^p} }\le c\Big(\alpha^{-\gamma}+\alpha^{-\beta+\frac1{2m}}\Big)(1+|t|^{\frac{r}{2m}})
\Big(\ds \int_{-\infty}^t (1+|\sigma|^r)^{-1} |Y(\sigma)|_{L^p}^{2m}\,d\sigma\Big)^{\frac1{2m}}
\end{array}
$$
and the first statement follows if $\gamma$, $\beta$, $m$ are chosen so that
$$
\tfrac1{2m}< \min(\tfrac{\delta}{r},\tfrac{\varepsilon}2),\qquad
\tfrac14-\tfrac{\varepsilon}2<\gamma+\tfrac{\varepsilon}2<\tfrac14.
$$
Indeed, proceeding as in \cite{dpd04}, we easily prove that
$$
K(\varepsilon,\delta,p)=\Big(\int_{-\infty}^\infty
(1+|\sigma|^r)^{-1} |Y(\sigma)|_{L^p}^{2m}\,d\sigma\Big)^{\frac1{2m}}
$$
has all moments finite. $\Box$
\bigskip
\begin{Proposition}
\label{p3.8}
Let $x\in L^2(0,1)$ and let $X(t,x)$ be the solution of \eqref{e3.2}.
\begin{itemize}
\item[i)] For any $\delta >0$, there exists a constant $K_1(\delta)$
such that for any $x\in L^4(0,1)$, $t\ge 0$
$$
\E\left(\sup_{s\in [0,t]} |X(s,x)|_{L^4}^4\right)\le 4 |x|_{L^4}^4+K_1(\delta)(1+t^\delta)
$$
\item[ii)] There exists a constant $K_2\ge 0$ such that for any $x\in L^4(0,1)$
and $t\ge 0$
$$
\E( |X(t,x)|_{L^4}^4)\le (e^{-\pi^2 t/16}|x|_{L^4}+K_2)^4.
$$
\item[iii)] There exists a constant $K_3$ such that for any $x\in L^2(0,1)$
and $t\in[1,2]$
$$
\E( |X(t,x)|_{L^4}^4)\le K_3 (1+|x|_{L^2}^4).
$$
\end{itemize}
\end{Proposition}
\medskip
{\sf Proof:} In the proof we shall denote by $c$ several different constant. Let us prove (i).
Fix $t>0, x\in L^4$ and $\delta>0$ and set
$Y(s)=X(s,x)-z^\alpha(s)$,
$Y(s)$ satisfies
$$
\left\{\begin{array}{l}
\ds\frac{dY(s)}{ds}=AY(s)+D_\xi(Y(s)+z^\alpha(s))^2+\alpha z^\alpha(s),\\\\
Y(0)=x-z^\alpha(0)
\end{array}\right.
$$
By similar computations as in \cite[Proposition 2.2]{dpd04}, we have that
$$
\tfrac14\; \frac{d\ }{ds}|Y(s)|_{L^4}^4+\tfrac32 \int_0^1|Y(s)|^2\,|D_\xi Y(s)|^2\,d\xi
\le c |z^\alpha(s)|_{L^4}^{\frac83}\,|Y(s)|_{L^4}^4+
c |z^\alpha(s)|_{L^4}^4+\alpha^4|z^\alpha(s)|_{L^4}^4,
$$
and, by the Poincar\'e inequality,
\begin{equation}
\label{e1}
\frac{d\ }{ds} |Y(s)|_{L^4}^4+\tfrac{3\pi^2}8 |Y(s)|_{L^4}^4\le
c |z^\alpha(s)|_{L^4}^{\frac83} |Y(s)|_{L^4}^4+(c+\alpha^4(s))|z^\alpha(s)|_{L^4}^4.
\end{equation}
We now choose $\alpha$ so large that
\begin{equation}
\label{e2}
c |z^\alpha(s)|_{L^4}^{\frac83}\le \tfrac{\pi^2}8,\quad s\in [0,t].
\end{equation}
For this we use Lemma \ref{l3.7} with
$$
\varepsilon=\tfrac18,\quad p=4 \quad\mbox{\rm and}\;\delta\;\mbox{\rm replaced by}\;\tfrac\delta 8.
$$
We see that there exists $ c>0$ such that \eqref{e2} holds provided
$\alpha= c \;\big( K(\tfrac18,\tfrac{\delta}8,4)(1+t^{\delta/8}\big)^8$.
So, by \eqref{e1} it follows that
$$
\frac{d\ }{ds}|Y(s)|_{L^4}^4+\tfrac{\pi^2}4 |Y(s)|_{L^4}^4\le
c\,\Big(1+K(\tfrac18,\tfrac{\delta}8,4)^8(1+t)^\delta\Big),\quad s\in [0,t].
$$
Consequently, by the Gronwall lemma, we see that
$$
\begin{array}{l}
|Y(s)|_{L^4}^4\le e^{-\frac{\pi^2}4s}|Y(0)|_{L^4}^4+c\ds\int_0^s
e^{-\frac{\pi^2}4(s-\sigma)}(1+K(\tfrac18,\tfrac{\delta}8,4)^8(1+\sigma)^\delta d\sigma\\\\
\hph{|Y(s)|_{L^4}^4}\le e^{-\frac{\pi^2}4s} |Y(0)|_{L^4}^4+c(1+K(\tfrac18,\tfrac{\delta}8,4)^8(1+t^\delta)),
\quad s\in [0,t],
\end{array}
$$
which yields by \eqref{e2}
\begin{equation}
\label{e3}
\begin{array}{l}
|X(s,x)|_{L^4}^4\le 4e^{-\frac{\pi^2}4s}|x|_{L^4}^4+4|z^\alpha(0)|^4_{L^4}+4|z^\alpha(s)|^4_{L^4}
+c\,(1+K(\tfrac18,\tfrac{\delta}8,4)^8(1+t^\delta) )\\\\
\hph{|X(s,x)|_{L^4}^4}\le
4e^{-\frac{\pi^2}4s}|x|_{L^4}^4+c(1+K(\tfrac18,\tfrac{\delta}8,4)^8(1+t^\delta)),
\quad s\in [0,t].
\end{array}
\end{equation}
Now, i) follows since $K(\tfrac18,\tfrac{\delta}8,4)$ has finite moments.\bigskip
To prove ii) we denote by $X(t,-t_0;x)$ the solution at time $t$ of the Burgers equation
with initial data $x$ at the time $-t_0$.
Since $X(t_0,x)$ and $X(0,-t_0;x)$ have the same
law, it suffices to prove
$$
\E(|X(0,-t_0;x)|_{L^4}^4)\le (e^{-\pi^2t_0/16} |x|_{L^4}+K_2)^4,\quad t_0\ge 0.
$$
We set $Y(t)=X(t,-t_0:x)-z^\alpha(t)$.
Proceeding as above we find
$$
|Y(t)|_{L^4}^4\le e^{-\frac{\pi^2}4(t-s)}|Y(s)|_{L^4}^4+c(1+K(\tfrac18,\tfrac{1}8,4)^8(1+|s|)),
\mbox{ for } 0\ge t\ge s.
$$
Since $(a+b)^{1/4}\le a^{1/4}+b^{1/4}$, we obtain
$$
|X(t,-t_0;x)|_{L^4}\le e^{-\frac{\pi^2}{16}(t-s)}|X(s,-t_0;x)|_{L^4}
+c(1+K(\tfrac18,\tfrac{1}8,4)^8(1+|s|))^{1/4},
\mbox{ for } 0\ge t\ge s.
$$
We choose $-s=-t+1=1,2,\dots,n_0$ with $n_0=[t_0]$ and then $-s=-t+1=t_0$, we obtain
$$
\begin{array}{ll}
|X(0,-t_0,x)|_{L^4} &\le e^{-n_0\pi^2(t-s)/16} |X(-n_0,-t_0,x)|_{L^4}\\
\\
&+c
\sum_{l=0}^{n-1}e^{-l\pi^2/16} (1+K(\tfrac18,\tfrac{1}8,4)^8(1+l))^{1/4}\\
\\
&\le e^{-t_0\pi^2(t-s)/16} |x|_{L^4}+c
\sum_{l=0}^{n-1}e^{-l\pi^2/16} (1+K(\tfrac18,\tfrac{1}8,4)^8(1+l))^{1/4}\\
\\
&+ e^{-n_0\pi^2/16} (1+K(\tfrac18,\tfrac{1}8,4)^8(1+t_0))^{1/4}\\
\\
&\le e^{-t_0\pi^2(t-s)/16} |x|_{L^4}+c (1+K(\tfrac18,\tfrac{1}8,4)^8)^{1/4},
\end{array}
$$
so, ii) follows.\bigskip
To prove iii), we use, as in i), $Y(s)=X(s,x)-z^\alpha(s)$ and have
$$
\tfrac12\;\frac{d\ }{ds}|Y(s)|_{L^2}^2+|D_\xi Y(s)|_{L^2}^2\le
c|z^\alpha(s)|_{L^4}^{\frac83} |Y(s)|_{L^2}^2+c|z^\alpha(s)|_{L^4}^4 +\alpha^2|z^\alpha(s)|_{L^2}^2
$$
so that, choosing $\alpha$ conveniently,
$$
\sup_{[0,2]}|Y(s)|_{L^2}^2+\int_0^2 |D_\xi Y(s)|_{L^2}^2 ds\le
|x|_{L^2}^2 +c(1+\alpha^2).
$$
For instance, we can take $\alpha=c K(\tfrac18,1,4)^8$.\\
Moreover
$$
\frac{d\ }{ds}\big( s|Y(s)|_{L^4}^4\big)+\tfrac{\pi^2}4 s |Y(s)|_{L^4}^4\le |Y(s)|_{L^4}^4
+c(1+\alpha^4)
$$
so that for $t\in[1,2]$
$$
|Y(t)|_{L^4}^4 \le \int_0^2 |Y(s)|_{L^4}^4 ds+c(1+\alpha^4).
$$
Using the inequality \eqref{e3.6bis}, we have
$$
|Y(t)|_{L^4} \le c\; |Y(t)|_{L^2}^{\frac34} \;|D_\xi Y(t)|_{L^2}^{\frac14}
$$
and we deduce
$$
\begin{array}{l}
|Y(t)|_{L^4}^4\le c\ds\int_0^2|Y(s)|_{L^2}^3|D_\xi Y(s)|_{L^2}ds +c(1+\alpha^4)\\\\
\hph{|Y(t)|_{L^4}^4}\le c(|x|_{L^2}^4 +(1+\alpha^2)^2)+c(1+\alpha^4)
\end{array}
$$
for $t\in[1,2]$ and iii) follows.
$\Box$
\bigskip
Next lemma is similar to Lemma 2.6 in Kuksin-Shirikyan \cite{ks}.
\begin{Lemma}\label{l3.9}
Let $\rho_0>0$ and $\rho_1>0$, there exist $\alpha(\rho_0,\rho_1)>0$ and $T(\rho_0,\rho_1)>0$
such that for $t\in[T(\rho_0,\rho_1),2T(\rho_0,\rho_1)]$
and $|x_1|_{L^4}\le \rho_0$, $|x_2|_{L^4}\le \rho_0$,
$$
\P\big(|X(t,x_1)|_{L^4} \le \rho_1
\mbox{ and }
|X(t,x_2)|_{L^4} \le \rho_1\big)\ge \alpha(\rho_0,\rho_1).
$$
\end{Lemma}
\medskip
{\sf Proof:} Let $X^0$ be the solution of the deterministic Burgers equation
$$
\left\{
\begin{array}{l}
\ds\frac{dX^0}{dt}(t,x)=AX^0(t,x) + b(X^0(t,x))\\\\
X^0(0,x)=x.
\end{array}
\right.
$$
Since, as easily checked, $(b(X^0),(X^0)^3)=0$, we obtain by standard computations
$$
|X^0(t,x)|_{L^4}^4\le e^{-\pi^2t/4 } |x|_{L^4}^4\le e^{-\pi^2t/4 } \rho_0^4,\quad\mbox{\rm for all}\;|x|_{L^4}\le \rho_0.
$$
We choose
$$
T(\rho_0,\rho_1)=\frac{16}{\pi^2}\; \ln\Big(\frac{2\rho_0}{\rho_1}\Big).
$$
Then, for $t\ge T(\rho_0,\rho_1)$
$$
|X^0(t,x)|_{L^4} \le \frac{\rho_1}2\quad\mbox{\rm for all}\;|x|_{L^4}\le \rho_0.
$$
let
$$
z(t)=\int_0^t e^{(t-s)A}dW(s)
$$
then for any $\eta>0$
$$
\P\Big(\sup_{[0,2T_0(\rho_0,\rho_1)]} |z(t)|_{L^4} \le \eta\Big) >0
$$
and since the solution of the stochastic equation is a continuous function
of $z$ we can find $\eta$ such that
$$
|X(t,x_i)|_{L^4} \le |X^0(t,x_i)| +\frac{\rho_1}{2},\; i=1,2,
$$
for $t\in [0,2 T(\rho_0,\rho_1)]$ provided $|z(t)|_{L^4} \le \eta$ on
$[0,2 T(\rho_0,\rho_1)]$.
The conclusion follows easily. $\Box$
\bigskip
\subsection{Construction of the coupling}
Let $x_1,x_2\in L^4(0,1)$, we construct
$\big(X_1(\cdot;x_1,x_2),X_2(\cdot;x_1,x_2)\big)$ a coupling of $X(\cdot;x_1)$ and $X(\cdot;x_2)$ as follows.
Fix $\rho_0>0$, $\rho_1>0$,
$R>\max\{\rho_0,\rho_1\}$, $T>T_0:=T(\rho_0,\rho_1)$ (defined in Lemma \ref{l3.9}), all to be chosen later.
We recall that $\big(X_{1,R}(t;x_1,x_2),X_{2,R}(t;x_1,x_2)\big)$ represents the coupling of $X_R(\cdot;x_1)$ and
$X_R(\cdot;x_2)$ constructed in section 4.1,
where $X_R(\cdot;x_1)$ and $X_R(\cdot;x_2)$ are the solutions of the cut-off equation
\eqref{e3.2}.
We shall need also the coupling of $X_R(\cdot,T_0;x_1)$ and $X_R(\cdot,T_0;x_2)$ when the initial time is any $T_0>0$ instead of 0.
We denote it by
$$
\big(X_{1,R}(t,T_0;x_1,x_2),X_{2,R}(t,T_0;x_1,x_2)\big)
$$
and in this case we shall write
$$
\tau_{R}^{x_1,x_2}=\inf\{t>0 : X_{1,R}(t+T_0,T_0;x_1,x_2)=X_{2,R}(t+T_0,T_0;x_1,x_2)\}.
$$
Notice that $\tau_{R}^{x_1,x_2}$ does not depend on $T_0$ thank to the Markov property because the
Burgers equation does not depend explicitely on time.\bigskip
\noindent First we shall construct the coupling on $[0,T]$, defining $(X_1(t;x_1,x_2),X_2(t;x_1,x_2))$ as follows.
If
\begin{equation}
\label{condition}
x_1\not = x_2,\; |x_1|_{L^4}\le \rho_0,\; |x_2|_{L^4}\le \rho_0,\; |X(T_0;x_1)|_{L^4}\le \rho_1\mbox{ and }|X(T_0;x_2)|_{L^4}\le \rho_1,
\end{equation}
we set
$$
X_1(t;x_1,x_2)=X(t,x_1),\;\; X_2(t;x_1,x_2)=X(t,x_2),\quad\mbox{\rm for}\;t\in [0,T_0]
$$
and for $i=1,2$
$$
X_i(t;x_1,x_2)=\left\{\begin{array}{l}
X_{i,R}(t,T_0;X(T_0,x_1),X(T_0,x_2))\;
\mbox{\rm if}\;T_0\le t\le \min\{\widetilde{\tau}_R,T\},\\\\
X(t,\tilde\tau_R;X_{i,R}(\tilde\tau_R,T_0;X(T_0,x_1),X(T_0,x_2))\;
\mbox{\rm if}\; \min\{\widetilde{\tau}_R,T\}<t\le T,
\end{array}\right.
$$
where
$$
\widetilde{ \tau}_R=\inf\{t\ge T_0: \max\{|X_{i,R}(t,T_0;X(T_0,x_1),X(T_0,x_2))|_{L^4},\;i=1,2\}>R\}.
$$
If \eqref{condition} does not hold, we simply set
$$
X_1(t;x_1,x_2)=X(t,x_1),\;\; X_2(t;x_1,x_2)=X(t,x_2),\quad\mbox{\rm for}\;t\in [0,T]
$$
So, we have constructed the coupling on $[0,T]$.
The preceding construction can be obviously generalized considering a time interval $[t_0,t_0+T]$
and random initial data $(\eta_1,\eta_2)$, $\mathcal F_{t_0}$-measurable. In this case we
denote the coupling by
$$
(X_1(t,t_0,\eta_1,\eta_2),X_2(t,t_0,\eta_1,\eta_2)).
$$
Now we define the coupling
$(X_1(t;x_1,x_2),X_2(t;x_1,x_2))$ for all time, setting by recurrence
$$
X_i(t,x_1,x_2):=X_i(t,kT,X_1(kT,x_1,x_2),X_2(kT,x_1,x_2)),\quad i=1,2,\;t\in [kT,(k+1)T].
$$
Let us summarize the construction of the coupling on $[0,T]$. We first let the original
processes $X(\cdot,x_1)$, $X(\cdot,x_2)$ evolve until they are both
in the ball of radius $\rho_0$. Then, we let them evolve and if at time $T_0$ they both enter the
ball of radius $\rho_1$ we use the coupling of the truncated equation as long
as the norm do not exceed $R$ (so that if $\widetilde{\tau}_R\ge T-T_0$ we have
a coupling of the Burgers equation having good properties).
Then, if the coupling is successful, i.e. if $X_1(T;x_1,x_2)=X_2(T;x_1,x_2)$,
we use the original Burgers equation and the solutions remain equal. Otherwise,
we try again in $[T,2T]$ and so on.
\subsection{Exponential convergence to equilibrium for the Burgers equation}
We shall choose now $\rho_0$, $\rho_1$, $R$ and $T$ (recall that $T_0=T(\rho_0,\rho_1)$ is determined
by Lemma \ref{l3.9}).
We first assume that
$$
|x_1|_{L^4}\le \rho_0\qquad \mbox{and}\qquad |x_2|_{L^4}\le \rho_0
$$
and set
$$
\begin{array}{l}
\ds A=\left\{\tau_R^{X(T_0,x_1),X(T_0,x_2)}\le T-T_0\right\}\quad(\mbox{\rm the coupling is successfull in }\;[T_0,T]),\\
\\
\ds{B=\left\{\sup_{t\in [T_0,T],i=1,2}|X_{i,R}(t,T_0;X(T_0,x_1),X(T_0,x_2))|_{L^4}
\le R\right\},}\\
\\
\ds{C=\left\{\max_{ i=1,2} |X (T_0,x_i)|_{L^4}\le \rho_1 \right\}.}
\end{array}
$$
Then, we have
$$
\P\big( X_1(T;x_1,x_2)=X_2(T,x_1,x_2)\big)\ge\P(A\cap B\cap C).
$$
We are now going to estimate $\P(A\cap B\cap C)$.
Concerning $C$ we note that by Lemma \ref{l3.9} it follows that
$$
\P\left(\max_{ i=1,2}|X(T_0,x_i)|_{L^4}\le \rho_1\right)\ge \alpha(\rho_0,\rho_1).
$$
Moreover,
\begin{equation}
\label{e3.9}
\begin{array}{l}
\P(A\cap B\cap C)=\\Ê
\\
\ds{\hspace{-.8cm}\int\limits_{\hspace{1cm}|y_i|_{L^4}\le \rho_1,i=1,2}\hspace{-1.2cm}\P(A\cap B|
X(T_0,x_1)=y_1,X(T_0,x_2)=y_2)\P(X(T_0,x_1)\in dy_1,X(T_0,x_2)\in dy_2).}
\end{array}
\end{equation}
But
\begin{equation}
\label{e3.10}
\begin{array}{l}
\P(A\cap B|X(T_0,x_1)=y_1,X(T_0,x_2)=y_2)\\
\\
=\P(\tau_R^{y_1,y_2} \le T-T_0 \mbox{ and }
\ds\sup_{[T_0,T]} |X_{i,R}(t,T_0;y_1,y_2)|_{L^4}\le R\quad \mbox{for } i=1,2)\\\\
\ge 1-\P(\tau_R^{y_1,y_2}> T-T_0)-\ds\sum_{i=1,2}
\P\left(\sup_{[T_0,T]} |X_{i,R}(t,T_0;y_1,y_2)|>R\right)
\end{array}
\end{equation}
By the Chebyshev inequality and \eqref{e3.8} it follows that
$$
\P(\tau_R^{y_1,y_2}\ge T-T_0)\le \frac1{T-T_0}\;\E(\tau_R^{y_1,y_2})\le \frac1{T-T_0}\;f ^R(|y_1-y_2|)
$$
and, since $\mathcal L\big(X_{i,R}(\cdot,T_0;y_1,y_2)\big)=
\mathcal L\big(X_R(\cdot,T_0;y_i)\big)=\mathcal L\big(X_R(\cdot;y_i)\big)$, we have, taking into account Proposition
\ref{p3.8}-(i), that
$$
\begin{array}{l}
\ds\P\left(\sup_{[T_0,T]}|X_{i,R}(t,T_0;y_1,y_2)|_{L^4}> R\right)=
\P\left(\sup_{[0,T-T_0]}|X_R(t,y_i)|_{L^4}> R\right)\\
\\
\ds =\P\left(\sup_{[0,T-T_0]}|X(t,y_i)|_{L^4}> R\right)
\ds\le \frac{4\rho_1^4+K_1(\delta)(1+(T-T_0)^\delta)}{R^4}.
\end{array}
$$
Consequently, if $|y_i|_{L^4}\le \rho_1$ for $i=1,2$,
$$
\begin{array}{l}
\ds 1-\P(\tau_R^{y_1,y_2}> T-T_0)-\sum_{i=1,2}
\P\left(\sup_{[T_0,T]} |X_{i,R}(t,T_0;y_1,y_2)|>R\right)\\\\
\ds \hspace{3cm}\ge 1-\frac1{T-T_0}\; f_R(|y_1-y_2|)-
2\frac{4\rho_1^4+K_1(\delta)(1+(T-T_0)^\delta)}{R^4}\\\\
\ds \hspace{3cm}\ge 1-\frac1{T-T_0}\; f_R(2\rho_1)-
\frac{8\rho_1^4+2K_1(\delta)(1+(T-T_0)^\delta)}{R^4}
\end{array}
$$
We deduce by \eqref{e3.9} and \eqref{e3.10} that for $|x_i|_{L^4}\le \rho_0$, $i=1,2$,
\begin{equation}
\label{e3.11}
\begin{array}{l}
\P\big(X_1(T,x_1,x_2)=X_2(T,x_1,x_2)\big)\ge \big(1-\frac1{T-T_0}\;f_R(2\rho_1)\\
\\
-\frac{8\rho_1^4+2K_1(\delta)(1+(T-T_0)^\delta)}{R^4}\big) \alpha(\rho_0,\rho_1).
\end{array}
\end{equation}
We choose now $T-T_0=1$, $\rho_1\le 1$ and $R$ such that
\begin{equation}
\label{e3.12}
\frac{8+4K_1(\delta)}{R^4}\le \frac14.
\end{equation}
Then, we take $\rho_1$ such that
\begin{equation}
\label{e3.13}
f_R(2\rho_1)\le \frac14
\end{equation}
(this is possible since $f_R(0)=0$ and $f_R$ is continuous).\\
It follows
\begin{equation}
\label{e3.14}
\P\big(X_1(T;x_1,x_2)=X_2(T;x_1,x_2)\big)\ge \frac12\;\alpha(\rho_0,\rho_1),\quad\mbox{\rm for all }\;
|x_1|_{L^4}\le \rho_0,\;|x_2|_{L^4}\le \rho_0.
\end{equation}
\bigskip
To treat the case of arbitrary $x_1$, $x_2$, have to choose $\rho_0$. We proceed as in
Kuksin-Shirikyan \cite{ks} and introduce the following
Kantorovich functional
$$
F_k=\E\big( (1+\nu(|X_1(kT;x_1,x_2)|_{L^4}^4+|X_2(kT;x_1,x_2)|_{L^4}^4))
\mbox{1{\kern-2.7pt}l}_{X_1(kT;x_1,x_2)\neq X_2(kT;x_1,x_2)} \big),
$$
where $k\in \N\cup \{0\}$ and $\nu$ is to be chosen later.
\begin{Proposition}
\label{t3.11}
There exist positive numbers $\rho_0,\nu,\gamma$ such that
\begin{equation}
\label{t3.15}
\P(X_1(kT;x_1,x_2)\neq X_2(kT;x_1,x_2))\le e^{-\gamma k}
\;(1+\nu (|x_1|_{L^4}^4+|x_2|_{L^4}^4)),\quad x_1,x_2\in L^4(0,1).
\end{equation}
\end{Proposition}
{\bf Proof}. We shall denote by the same symbol $c$ several different constants.
Let us estimate $F_1$ in terms of $F_0=(1+\nu(|x_1|_{L^4}^4+|x_2|_{L^4}^4))\mbox{1{\kern-2.7pt}l}_{x_1\neq x_2}$.
If
$x_1=x_2$ then $X_1(T;x_1,x_2)=X_2(T;x_1,x_2)$ a.s. and so, $F_1=0$.
Let now $x_1\neq x_2$. If $|x_1|_{L^4}>\rho_0$ then $X_1(T;x_1,x_2)=X(T,x_1)$ and $X_2(T;x_1,x_2)=X(T,x_2)$.
Consequently, taking into account Proposition \ref{p3.8}--(ii),
$$
\begin{array}{l}
F_1=\E\left[(1+\nu(|X(T,x_1)|_{L^4}^4+|X(T,x_2)|_{L^4}^4))
\mbox{1{\kern-2.7pt}l}_{X(T,x_1)\neq X(T,x_2)}\right]\\\\
\hspace{1cm}\le \E\big[1+\nu(|X(T,x_1)|_{L^4}^4+|X(T,x_2)|_{L^4}^4)\big]\\\\
\hspace{1cm}\le 1+\nu\big((e^{-\pi^2T/16} |x_1|_{L^4}+K_2)^4+
(e^{-\pi^2T/16} |x_2|_{L^4}+K_2)^4 \big).
\end{array}
$$
Since $T\ge T-T_0=1$, there exists $c$ such that
\begin{equation}
\label{t3.16}
(e^{-\pi^2T/16} a+b)^4\le e^{-\pi^2T/8} a^4+c \;b^4,
\qquad \mbox{for any } a,b\ge 0
\end{equation}
and we deduce that, if $x_1\neq x_2$ and $|x_1|_{L^4}>\rho_0$ (or if $x_1\neq x_2$ and $|x_2|_{L^4}>\rho_0$),
\begin{equation}
\label{t3.17}
F_1\le 1+\nu\big(e^{-\pi^2T/8}
(|x_1|_{L^4}^4+|x_2|_{L^4}^4)+cK_2^4 \big)
\end{equation}
Let us now consider the case when $x_1\neq x_2$, $|x_1|_{L^4}\le \rho_0$ and $|x_2|_{L^4}\le \rho_0$.
Taking into account \eqref{e3.14} and Proposition \ref{p3.8}, we have
\begin{equation}
\label{t3.18}
\begin{array}{l}
F_1=\E\big[(1+\nu(|X_1(T;x_1,x_2)|_{L^4}^4+|X_2(T;x_1,x_2)|_{L^4}^4))
\mbox{1{\kern-2.7pt}l}_{X_1(T;x_1,x_2)\neq X_2(T;x_1,x_2)}\big]\\\\
\le \P(X_1(T;x_1,x_2)\neq X_2(T;x_1,x_2))+\nu
\E[|X_1(T;x_1,x_2)|_{L^4}^4+|X_2(T;x_1,x_2)|_{L^4}^4]\\\\
\ds \le 1-\frac12\; \alpha(\rho_0,\rho_1)+\nu \big(e^{-\pi^2T/8}
(|x_1|_{L^4}^4+|x_2|_{L^4}^4)+cK_2^4\big)
\end{array}
\end{equation}
since $\mathcal L(X_i(T;x_1,x_2))=\mathcal L(X(T;x_i))$ and thanks to
Proposition \ref{p3.8}--ii) and \eqref{t3.16}.
To conclude, we shall choose $\rho_0$ and $\nu$ such that
\begin{equation}
\label{t3.18bis}
\begin{array}{l}
q_1(\lambda):=
\ds\frac{1+\nu(e^{-\pi^2 T/8}\lambda+
cK_2^4)}{1+\nu\lambda}\le e^{-\gamma},\quad\mbox{\rm for }\lambda> \rho_0^4\\\\
q_2(\lambda):=
\ds\frac{1-\frac12\; \alpha(\rho_0,\rho_1)+\nu(e^{-\pi^2 T/8}\lambda+
cK_2^4)}{1+\nu\lambda}\le e^{-\gamma},\quad\mbox{\rm for }\lambda\le 2\rho_0^4.
\end{array}
\end{equation}
Note first that $q_1$ is decreasing in $\lambda$ and
tends to $e^{-\pi^2 T/8}$ as $\lambda \to \infty$.
We choose $\rho_0$ such that
$$
\rho_0^4= \frac{2(1+ cK_2^4)}{1- e^{-\pi^2T/8}},\quad T=1+T_0(\rho_0,\rho_1).
$$
(It is easy to see that this equation has
a solution). With this choice of
$\rho_0$, it is also easy to check that $q_1(\rho^4_0)<1$.\\
Choosing now
$$
\nu = \frac{\alpha(\rho_0,\rho_1)}{4cK_2^4}
$$
it is easy to check that
$$
q_2(\lambda)\le \max\{e^{-\pi^2T/8},1-\tfrac14\; \alpha(\rho_0,\rho_1)\}.
$$
Hence, choosing $\gamma$ such that
$$
e^{-\gamma} \ge \max\{1-\tfrac14\; \alpha(\rho_0,\rho_1),q_1(\rho_0)\},
$$
\eqref{t3.18bis} is fulfilled.
Now we continue the estimate of $F_1$. For any $\lambda\ge \rho_0^4$ we have from the first inequality in \eqref{t3.18bis}
that
$$
1+\nu( e^{-\pi^2T/8} \lambda+c K_2^4)\le e^{-\gamma}(1+\nu \lambda).
$$
Then from \eqref{t3.17} we deduce that if $x_1\neq x_2$ and $|x_1|_{L^4}>\rho_0$ or
$|x_2|_{L^4}>\rho_0$, we have
$$
F_1\le e^{-\gamma}(1+\nu (|x_1|_{L^4}^4+|x_2|_{L^4}^4 ))=e^{-\gamma} F_0.
$$
Moreover
$$
1-\frac12\; \alpha(\rho_0,\rho_1)+\nu \big(e^{-\pi^2T/8} \lambda+cK_2^4\big)
\le e^{-\gamma}(1+\nu \lambda)
$$
for any $\lambda \le 2 \rho_0^4$. Then, by \eqref{t3.18}, we obtain
for $x_1\neq x_2$, $|x_1|_{L^4}\le \rho_0$ and $|x_2|_{L^4}\le \rho_0$
$$
F_1\le e^{-\gamma}(1+\nu (|x_1|_{L^4}^4+|x_2|_{L^4}^4 ))=e^{-\gamma} F_0.
$$
Therefore, we have in any case
$$
F_1\le e^{-\gamma}\;F_0,\quad x_1,x_2\in L^4(0,1).
$$
It is not difficult to check that $(X_1(kT;x_1,x_2),X_2(kT;x_1,x_2))_{k\in\N}$
is a Markov chain so that we obtain for any $k\in\N$
$$
F_{k+1}\le e^{-\gamma}\;F_k,\quad x_1,x_2\in L^4(0,1).
$$
and, so
$$
F_{k}\le e^{-k\gamma}\;F_0,\quad x_1,x_2\in L^4(0,1).
$$
In particular
$$
\P(X_1(kT;x_1,x_2)\neq X_2(kT;x_1,x_2))\le
e^{-k\gamma}\;(1+\nu (|x_1|_{L^4}^4+|x_2|_{L^4}^4)),\quad x_1,x_2\in L^4(0,1).
$$
$\Box$\bigskip
By Proposition \ref{t3.11} the exponential convergence to equilibrium follows for
$x_1,x_2\in L^4(0,1)$.
If $x_1,x_2\in L^2(0,1)$, we write
$$
\begin{array}{l}
\P(X_1(kT;X(1,x_1),X(1,x_2))\neq X_2(kT;X(1,x_1),X(1,x_2))\\\\
\hspace{2cm}\le
e^{-(k-1)\gamma}\;\big(1+\nu \E(|X(1,x_1)|_{L^4}^4+|X(1,x_2)|_{L^4}^4)\big)
\end{array}
$$
and use Proposition \ref{p3.8}--(iii) to conclude the proof of Theorem \ref{main-burgers}. $\Box$ | 0.008907 |
Yesterday we heard fantastic news, Nao's brother in-law is safe and well. He turned up at their house in Kesennuma! Luckily he managed to escape the Tsunami in his car, whilst at his work in the next town. We are so relieved for Nao's sister who is expecting their first baby next month. It truly is a miracle that none of Nao's family is missing, or worse. There are so many 'what if...s' that don't bare thinking about. A massive sigh of relief in our house today, but now we are worried about the power station.
My heart breaks for those people who have lost their families...it is so horrific and I hope there will be people around to support them. It is so cruel.
So, we feel very selfish to be in a happier mood now, and i have been thinking a lot about how lucky we are not to have these kind of problems here in the UK.
It was both of our birthdays during this nightmare, mine just the day before. I got some lovely pressies, like these Narcissus sent all the way from the Scilly Isles! they smell gorgeous.
they arrived in this box, amazing! Thank you Marie & Blaise X
This book is fascinating, its all about the colours of Japan, from textiles and ceramics and paintings. I love it! Thank you Nao's Mum XXX
Nao bought me a mug, not sure why he is laughing at me?
maybe its my terrible bed hair?
some delicious home made goodness from our Rust colleague Eiko. Its a mystery how everything she makes taste so good? wish i had that ability.
and my Mum sent us some beautiful ceramics, from the Isle of Wight!
Nao is having the mug with a cod on it. I am having the one with sprats on it.
I really love the design, don't know who made these, but if they are by you, let me know and I'll post a link!
I got a pressie for Nao too, his birthday was on monday, I can't say it was a happy one, maybe we will celebrate it another day. {It was an old fishing satchel, bottle of wine and some Neal's Yard bath salts}
...and a big thanks to Dan and Tara for the hugemungous box of biscuits and cakes! Jammy Dodgers, Tunnocks Tea Cakes, Mr.Kipling, you name it! I could literally dive into the box and swim around. X
another pressie and something to cheer us up a bit...
spring is coming!
p.s. again, thank you all for such kind comments, i do believe in the power of positive thinking, and to have you all on our side, I'm sure it has helped!
Great news indeed about Nao's family. Happy Birthdays too!
XX
Happy Birthday! I don't think it's selfish to feel happy, I think it shows an attitude of gratitude. Glad to hear you've had good news.
so is that all of nao's family safe and well? i'm so pleased for you :)
i don't think you should feel selfish for appreciating lovely gifts from people who love you. enjoy them! happy birthday to the both of you!
So pleased to hear you have had fantastic news from Japan! Happy Birthday xxx
What a great news and what a relief for the family, with a little baby on her way too! It's a bit of light amongst all this desperation and tragedy.
fantastic news glad everyone is safe ~ rx
Its great to know his family are all safe. Love that tiger mug.
Wonderful birthday gifts. Those bowls and cups are beautiful.
On the Japan front, my housemate and I are wanting to help, but after giving so much money to Haiti and seeing that it hasn't been received we'd like to help in a different way - do you know any organisations we can donate clothes/toiletries/tents/towels to? we have all of these things that we can spare {and more - the housemate runs a clothing shop with a lot of 'faulty' stock} and want to send.
We have googled but seen nothing. If you know of any organisations can you email me ohgoshem at googlemail.com.
thanks
em
I am so pleased that you can feel relief now that you have better news.
I was thinking over the weekend how poignant that the present from Nao's mum you'd pictured in your post was sitting there, with you guys in limbo...
Happy (belated) Birthdays to you both xx
Happy Birthday to both of you! And so pleased it's all good news X X
Huge relief, I'm so pleased for you both that all are safe.
Hope you can now enjoy belated birthdays! x
Happy Birthday to you and Nao.
Ceramic with fish is great ♥
em...
The Red Cross are there in Sendai saving lives NOW, they have been the main source of relief so far together with the Japanese army and rescue services and they cannot run for free and their supplies will not last long. All ways are good ways to donate, but we cannot underestimate the need for the Red Cross to receive donations.
Here in the UK you can also support:
A much smaller but just as valuable charity.
thanks! X
Happy Birthday and I am so happy for your good news. I am fairly new to your blog and love it. Best wishes to you . x
Happy birthday to you both. I hope the news of Nao's family continues to be positive.
I'm so pleased to read your good news, and Happy Birthday to you both, but like you, and everyone else, there is still the sense of overwhelming dread for those people that we have never met, would never have really thought much about, but now, cannot stop thinking about. I have just seen on the news that snow and freezing conditions are now making everything even worse, as if that was possible. The memories of this will hang over the world for a long time to come.
oh what good news - I have been thinking of you both and the worry you must be going through. we are so lucky here in our UK bubble and hearing your story has made the pictures all the more real.
I can only donate which seems like so little.
Happy Birthday and I'm so pleased for your news even though I know thats not the end of all the worry for you both.
x x x
What wonderful news for you both, that's such a relief, I have been thinking about you.
Hi Artemis,
I am a long time fan of this blog - but a very quiet one.
I live in Sendai and just wanted to try and reassure you as much as I could about the nuclear situation.
I have been freaked out about it for several days - hearing all sorts of reports. But today we had a meeting with the Australian, New Zealand and Canadian embassy and we are feeling so reassured. The international media has been fear-mongering and creating panic. But many many experts (from all over the world) are here in Japan and are confident that it is under control. Even if the worst was to happen - it would only effect a small area - already evacuated.
I have been completely overwhelmed by the incredible spirit, strength and courage of the Japanese people. Never have I admired Japan so much.
Lots of love to you and Nao.
oh and happy birthday!!!
xx
I am so pleased for you both. I just cannot imagine the stress you have both been under whilst 'wondering' what had happened. Of course my thoughts are still with the families that have been devastated by loss. Homes can be re-built....
Oh dear! I'm sorry, forgot to wish you both belated happy birthday. Now is the time to celebrate :o)
You've certainly had some wonderful gifts - I love the mug!
Really happy for you and Nao. In celebration of your birthdays I've just made a further red-cross donation.
Alexis (not brother... the female one, who loves reading your blog)!
I'm so happy for you and Nao!! I've been reposting your news on my blog so that my readers can go and help if need be. My birthday was Saturday as well and it was hard to focus while all of this devastation was happening.
--julie
oh wow, so glad to hear everyone is alright!! happy birthday to you both.
my thoughts with you & Nao. happy birthdays!
Phew!
Happy Birthday, and such fab news.
Love the Isle of Wight pottery - where can I buy some?
I am delighted for you! What a relief that Naos family are well. We have been thinking of you in our household.
I love your blog, I look at it every day but have never left a comment until last week. Funny how awful events make people pluck up the courage to communicate....is that British?
Happy Birthday to you both!
Katherine.
That fantastic! realy! these website is way better then everything I ever saw.
I'm so happy that you have had good news. Happy birthday to both of you
I am very happy for his family and you guys, now lets wait that they managed to sort out the nuclear station. all the best from here to Japan.
That's such good news about Nao's family - really happy for you both.
Happy Birthday to both of you too, I love the mug! x
Oh great news! What a relief that must have been and happiest of birthdays. Will keep on hoping and wishing for Japan x
My internet connection was down all day! So frustrating I couldn't join everyone in rejoicing at the great news of Nao's family survival! This is the best Bday present you two could receive I guess! so, HAPPY BIRTHDAY to YOU TWO! Let us pray that such good news reach other survivors of this terrible catastrophy and that the issue with the Fukushima plant get solved rapidly.
Again, so happy for you and for Nao's family!
All the best!
Armide
Wow...it's all happening in your household at the moment. Happy Birthday to you both and enjoy each moment of relief and happiness after all the worry.
What a relief for you both! Can't imagine how harrowing it's been. Belated happy birthday too....:)
Don't know if you've seen this - someone else doing some good for Japan.
PS- FANTASTIC ceramics!! x
Artemis, I am so happy for you and Nao that everyone is safe and sound. I too feel slightly selfish at my happiness that M and his family are now back on UK soil after being in Japan, but then I also think, we should not feel guilty for being grateful for the safety of loved ones, as long as we spare a thought for those less lucky.
Beautiful narcissus, I think spring is here.
Such wonderful news! Happy Birthdays, i love your gifts your ceramic pieces are lovely xx
Happy Birthday to you both!!
So good to hear the family is safe.
just love the wonderful cookies and truffles and the packaging, wow!
Blessings and happy birthdays to you both. Wonderful news about your family! We truly are blessed and I'm sure our collective thoughts and wishes for Japan will help millions. Xx
It's wonderful to hear that Nao's family is safe. Sending you good wishes from Canada.
Wonderful news for your family, very glad to hear it. Happy birthday to you both!
such such good news. his wife must have just wept with joy. so happy for all of you!
Good news, happy times and such a cute tiger mug. Breathe...and enjoy the beginning of spring knowing your loved ones are safe.
Ingrid x
Hello Artemis,
I'm French and i'm a real fan of junkakaholique and Rust. There is no day for me without having a look at your blog. I love your pictures and little stories and today, i just want to thank you for your notes about Japan. I was there 2 weeks ago and i'm so sad for all Japanese people who are going through all of this. I'm really happy that Nao's sister and family are safe.
Greetings from the French Alps.
Anne-Lyse
NB:would you be able to send me the references of the Japanese book about colors?
So pleased to hear that Nao's family are all safe...probably the best birthday gift for both of you! lisa xox
I have to say that mug and you made me chuckle too! What wonderful news about Nao's family, it's been a really hard week for a lot of people.
Also, incase you haven't heard tomorrow will be a bloggers day of silence for Japan. Here's a link for more information.
xo
Oh I'm so happy for you both and your family. So happy! Happy Birthday to you both. Pisces childs are the best : )
P.S The flowers are beautiful x
Hope the best for Japan and Nao's family...
Love,
Débora
from France
Not selfish at all, so happy for you both. Out of all the sadness and despair I see on the news regarding Japan every night I come on here and read good news and it's just lovely.
What awesome presents!
What a relief! I'm sure you couldn't ask for better birthday presents. I am very happy for you.
Such relief you must feel. Phew. So happy for you, and so nice to read some good news. I hope you have managed to relax and enjoy your birthdays. Belated wishes to you both...gorgeous gifts! Love the cup! x
Hi Artemis, I so happy that Nao's family are safe. The events in Japan are so tragic and I don't know how the people are coping with events. I'm glad you could celebrate your birthday and your gifts are fab! I love your blog and I wish you, Nao,and all the people of Japan continued Love, Peace and Happiness. Ruth in Dublin
Happy belated birthday to you both... and so glad Nao's brother-in-law turned up safe x | 0.002294 |
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Aug 9, 2013; Minneapolis, MN, USA; Houston Texans quarterback Matt Schaub (8) talks with tight end Owen Daniels (81) on the bench during the second quarter against the Minnesota Vikings at the Metrodome. Mandatory Credit: Jesse Johnson-USA TODAY SportsAdded August 9th | 0.000863 |
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I stumbled upon Natural Order through a track that Stereofox posted this week. I then listened to his entire album, which was released at the end of June and was blown away.
I assume that he skillfully mixes samples with self-played stuff, but it’s hard to tell what he digged out of the record crate and what not. He combines it with mostly boombap-heavy but always varied drums, groovy basslines and synth melodies that you want to sing along. Often his productions somehow remind me of Freddie Joachim and as you probably noticed already, the producer from Boston has exactly met my taste.
Apparently he produced the album during the lockdown, mainly so his friends could listen to it. But I think that every Beat Lover should check out “Shibuya”. | 0.058985 |
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79.99
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49.99
PART#KU110001A
- Use with Neon Glow wire for a complete night Heli
- For all 400 size helis
- Complete package comes with everything you need
RJX Main Rotor Blades for Full Size Helicopters
3D performance at an affordable price
- Available in CF and FG
- Great lift and no drag
- Quality product for 3D performance
RJX Main Blades for Micro and Mini Helis
Available in multiple types and colors
- Choose from GF or CF
- Optimum lift with minimum drag
- Deliver value and great 3D performance
SAB 430mm Symmetrical blades
SAB 430mm carbon fiber main blades suitable for the T-Rex 500 and similar sized helicopters (ie. Lepton)
65.95
PART#MA3696-7
Specifications:
- Shape: Symmetrical wing
- Length: 430 x 42mm
- Root: 9mm (will need washers)
- Shaft Diameter: 3mm
- Weight: 55g each
CFC Night blades 325mm 4 5 3
The Brightest blades - MSC
64.99
PART#MS20325LI
- Recommended for T-Rex 450, TT mini Titan, and Zoom 400
- Spacers included
- By MSC
RJX Tail Blades and Paddles
Made with Quality Carbon composites
- Available in multiple sizes
- Quality composites made of high grade Carbon
- Work flawlessly with the RJX Main rotor blades
Lightning Power 325mm GF High-Performance Blades
325mm Glass Fiber Blades designed by Lightning Power to maximize your bird's lift. Made of High-Grade materials and tough as nails, you'll find these'll be the last pair of blades you buy for a while.
MS Composit GFC Blades
Glass Fiber - Ultra smooth with incredible cyclic performance
- Size Recommendations:
- 420mm for JR Voyager, Hirobo Lepton
- 470mm for ECO 8, Sniper
- 515mm for Concept 30, Logo 10, Shuttle 30
- 550mm for Logo 14, Logo 20, Raptor 30, Caliber 30, Sceadu 30
- 600mm for Raptor 50, Moskito Conquest, Freya 50, GPH 347
- 710mm for .60 and .90 size helis, Futura, Millenium, Raptor
- Spacers included
GFC main blades 325 mm - PAINTED
Stand out with this unique painted finish
37.00
PART#MS10325-3-P
- Racing design printed directly on the blades surface.
MAH Multi-Color Night Blades
Ultra Bright LEDs
- External battery mounts- Ultra bright LED technology- For 30 size to full size Helis 60+
V Blades 255mm Main Blades ( For Hornet )
Endorsed by the top fliers, V Blades from Vic Campbell have a reputation as the best blades available. Custom built and precision balanced. Now available for the micros! These are the BEST blades to use for the Hornet X and CPX.
43.75
PART#VB-255
V-Blades 325mm Main Rotor Blades (T-Rex)
For 400 Size Helis such as the 3DX and Trex
59.99
PART#VB-325
- Symmetrical
- Chord: 34mm
- Weight: 20gr
- White flash near tips highlights the rotor disc | 0.000896 |