TITLE: What's wrong with my approach to drawing this model?
QUESTION [0 upvotes]: An object of mass 1.8kg is attached to the ends of two light elastic strings having the same modulus of elasticity. One of the strings has natural length 0.8m and the other has a natural length of 1.1m. The longer string is attached at A and the shorter string is attached at B on the same horizontal level. The object hangs 0.85m below O, a point on the same level as A and B, 1.4m from A and 0.8m from B. Find the modulus of elasticity of the strings.

I'm having a bit of trouble with modelling the question with a diagram. I know what the correct answer is, and the model needed to help solve it (1) (bear in mind I've omitted some info to draw them faster), but I thought (2) will also be valid, just using the extension of the string at B as 0.05 instead of using Pythagoras' theorem.
Why not 2?

REPLY [0 votes]: (2) is not valid. The reason is that (2) is not an equilibrium case. The string to $A$ has a horizontal component of the elastic force. In order to be in equilibrium, the elastic force of the string to $B$ has to have the horizontal component of the same magnitude, in opposite direction


text:
\begin{document}

\title{A geometric interpretation of Kirillov's conjecture}
\date{June 2018}
\thanks{The authors thank David Vogan for suggesting this problem.}
\author{Gang Liu and Jun Yu}

\address[Gang Liu]{Universit\'{e} de Lorraine,
Institut Elie Cartan de Lorraine,  3 rue Augustin Fresnel,
57073 Metz, France.}
\email{gang.liu@univ-lorraine.fr}

\address[Jun Yu]{BICMR, Peking University, No. 5 Yiheyuan Road, Haidian District, Beijing 100871, China.}
\email{junyu@bicmr.pku.edu.cn}

\abstract{In this article, we give a natural geometric interpretation of Kirillov's conjecture for tempered
representations in the framework of orbit method. This interpretation can also be considered
as a first generalization of Duflo's conjecture to non-discrete series representations.}
\endabstract

\maketitle

\noindent {\bf Mathematics Subject Classification (2010).} 22E46 (17B08, 53D20).

\noindent {\bf Keywords.} Kirillov's conjcture,  Duflo's conjecture, orbit method, moment map, tempered representations.

\tableofcontents

\section{Introduction}

Let $G=\GL(n, \mathbb{K})$, where $\mathbb{K}=$ $\mathbb{R}$ or $\mathbb{C}$, and let $P$ be
the subgroup of  $G$ with matrices whose last row is $(0,0,\dots,0,1)$. Let $\pi$ be an
irreducible unitary representation of $G$. Then we have the following theorem known as
"Kirillov's conjeture":

\begin{theorem}\label{Kirillov conjecture}
The restriction of $\pi$ to $P$, $\pi\vert_{P}$ is irreducible.
\end {theorem}

Kirillov's conjecture has a quite long story,  among the main contributors of its proof,
are Sahi, Sahi-Stein and  Baruch (who finally proved the conjecture in all generality in \cite{Baruch}).
It is noted that not only the statement of Kirillov's conjecture is of purely analytic
nature, the final proof of Baruch is also essentially analytic.

Kirillov's conjecture is of course a branching problem. Concerning branching problems for
reductive groups, significant progress has been made during last twenty years, notably by
Kobayashi (and his collaborators). However in these works, subgroups are almost always
reductive as well.

On the other hand, by works of Kostant, Kirillov, Souriau, Duflo and others, we know that
geometric methods have played an important role in the development of representation theory
and harmonic analysis. Then we may ask,  despite the analytic nature of branching problems,
can we interpret the branching laws in some geometric manner? Or can branching laws be
essentially determined by some geometric data (e.g. coadjoint orbits) related to
representations? The answer to this problem is positive in some context, more precisely,
in some cases, we can describe the branching laws in the framework of \emph{orbit method},
or more generally of \emph{geometric quantization} . In this direction, the first success
concerned the case where the group $G$ is compact as initiated in the thesis of Heckman
in the framework of orbit method, and generalized by  the Guillemin-Sternberg conjecture
(proved by Meinrenken and also Tian-Zhang)  in the framework of geometric quantization
under the celebrated slogan  \emph{quantization commutes with reduction}. Since then,
along this direction, the theory was quite generalized developed notably by Paradan,
Vergne, also Ma and Zhang (especially the proof of Vergne's conjecture). Nevertheless,
it should be pointed out that in almost all these works, the groups and subgroups are
reductive. Encouraged by all these developments, more recently, Duflo formulated a
conjecture which aims at giving a geometric interpretation for the branching problem
$(G,H,\pi)$, where $G$ is an (almost) algebraic group, $H$ is an (almost) algebraic
subgroup of $G$, and  $\pi$ is a  \emph{discrete series representation} of $G$.
For more details concerning Duflo's conjecture, see the next section.

The main purpose of this article is to give a geometric interpretation of Kirillov's
conjecture for \emph{tempered representations} in the spirit of Duflo's conjecture.
Since Duflo's conjecture in its initial formulation concerns only discrete series
representations, and in our case (namely $G=\GL(n, \mathbb{K})$), there are no any
discrete series representations, our work can also be considered as a (and to the
best of our knowledge, a first) generalization of Duflo's conjecture.

\section{Duflo's conjecture} \

In this section, we will state Duflo's initial conjecture. For the reader's sake, we will begin by outlining some essential ingredients in Duflo's orbit method. For more details and the general setting about the theory of Duflo's orbit method, we refer to Duflo's "CIME lectures" \cite{Duflo1}.

Let $G$ be an almost algebraic real group with Lie algebra $\mathfrak{g}$.  Denote by $\mathfrak{g}^*$ the algebraic dual of $\mathfrak{g}$.  In the framework of Duflo's theory,  a fundamental  notion is \emph{admissible} (in the sense of Duflo) and   \emph{well polarizable}  (in the sense of Pukanszky) $G$-coadjoint orbits
(in $\mathfrak{g}^*$).  Each such orbit $\mathcal{O}$ is attached to at least one irreducible unitary representation of $G$. Moreover, two different such orbits are associated to non-equivalent representations.  In general, the set of irreducible unitary $G$-representation  associated to admissible and well polarizable $G$-codajoint orbits is not the whole unitary dual $\hat{G} $. However, it is sufficient to describe the Plancherel formula of $G$ (namely, the decomposition of $L^2(G)$) by Duflo's theory. Especially all \emph{discrete series} representations of $G$ (i.e., those appearing in the discrete part of the Plancherel formula of $G$), and "almost" all  \emph{ tempered } representations of $G$  (i.e., those appearing in the spectral decomposition of $L^2(G)$) are attached to admissible \emph{stronlgy regular} $G$-adjoint orbits.  Recall that an element $f\in \mathfrak{g}^*$ is called strongly reagular, if $f$ is regular (i.e., the coadjoint orbit of $f$ is of maximal dimension) and its "reductive factor"
$\mathfrak{s}(f):= \{X\in \mathfrak{g}(f):\text{ad} X \  \text{is semisimple}  \}$ is  of maximal dimension among the reductive factors of all the
regular elements in  $\mathfrak{g}^*$.  A coadjoint orbit $\mathcal{O}$ is called strongly regular, if there exists
an element $f\in \mathcal{O}$ (then each $f\in \mathcal{O}$ ) which is strongly regular. Notice that each strongly regular coadjoint orbit is automatically well polarizable.

Now let $H$ be an almost algebraic subgroup of $G$ with Lie algebra $\mathfrak{h}$.  Let $\mathcal{O}$ be a $G$-cadjoint orbit (in $\mathfrak{g}^*$).  It is well known that equipped with the Kirillov-Kostant-Souriau symplectic form $\mathcal{ \omega}$, $\mathcal{O}$ becomes a $H$-Hamiltonian space. The corresponding moment map is just the natural projection $\text{p}: \mathcal{O} \rightarrow\mathfrak{h}^{\ast}$

Let $\pi$ be a discrete series of $G$, as stated above, it is attached to a strongly regular $G$-coadjoint orbit $\mathcal{O}_{\pi}$.  Consider the restriction of $\pi$ to $H$, $\pi\vert_{H}$. Then in the context, Duflo's conjecture states as follows:\\

\begin{itemize}

  \item[i)] $\pi\vert_{H}$ is $H$-admissible (in the sense of Kobayashi) if and only if the moment map $\text{p}: \mathcal{O}_{\pi}\rightarrow\mathfrak{h}^{\ast}$ is \textit{weakly proper}.

 \item[ii)]  If $\pi\vert_{H}$ is $H$-admissible, then each irreducible $H$-representation $\sigma$ which appears in $\pi\vert_{H}$ is attached to a \textit{strongly regular} $H$-coadjoint orbit $\Omega$ (in the sense of Duflo) which is contained in $\text{p}(\mathcal{O}_{\pi})$.

\item[iii)]  If $\pi\vert_{H}$ is $H$-admissible, the multiplicity of each such $\sigma$  can be expressed geometrically on the \textit{reduced space}  of $\Omega$ (with respect to the moment map $\text{p}$).

 \end{itemize}

Let us give some more explanations for Duflo's conjecture. Firstly, the notion "$H$-admissible" above is due to Kobayashi, which means that $\pi\vert_{H}$ decomposes discretely and with finite multiplicities.

 The "weak properness" in i) means that the preimage (for $\text{p}$) of each compact subset which is contained in $\text{p}(\mathcal{O}_{\pi})\cap\Upsilon_{sr}$ is compact in $\mathcal{O}_{\pi}$. Here $\Upsilon_{sr}$ is the set of all strongly regular elements in  $\mathfrak{h}^*$.

  For (ii), as we already mentioned above,  each discrete series of $G$ (resp. $H$) is attached to a strongly regular $G$ (resp. $H$)-coadjoint orbit. Moreover according to Duflo-Vargas's work (\cite{Du-Va1}, \cite{Du-Va2}), each irreducible $H$-representation $\widetilde{\sigma}$ which appears in the integral decomposition of  $\pi\vert_{H}$ (which is not necessarily $H$-admissible) is attached to a strongly regular $H$-coadjoint orbit. Note that $\widetilde{\sigma}$ is not necessarily a discrete series. However, if $\pi\vert_{H}$ is $H$-admissible, then each $H$-irreducible representation appearing in $\pi\vert_{H}$ must be a discrete series. Thus  (ii) has a nice geometric meaning.

  Despite some progress (see for example \cite{Liu}), Duflo's conjecture is still not fully established. Nevertheless, it seems reasonable  to ask if we can generalize Duflo's conjecture (possibly with "adapted" modifications) to larger family of unitary irreducible representations of $G$ ( which are not necessarily discrete series). A first such attempt could be naturally for tempered representations, since tempered representations are closely related to discrete series, and almost all of them are attached to strongly regular coadjoint orbits.  In fact, we will prove the generalization in this direction under the setting of Kirillov's conjecture, which, in return, gives a geometric interpretation of Kirillov's conjecture for tempered representations.

\section{Geometry of the moment map $p:\mathcal{O}_{f}:\rightarrow\mathfrak{p}^{\ast}$}\label{S:moment}

\subsection{Coadjoint action, the dual map and the moment map}\label{SS:Porbit}

Let $n\geq 1$, and $k=\mathbb{R}$ or $\mathbb{C}$. Write $G_{n}(k)=\GL(n,k)$, $$P_{n}(k)=
\{\left( \begin{array}{cc}A&\alpha\\0_{1\times(n-1)}&1\\\end{array}\right):A\in\GL(n-1,k),
\alpha\in k^{n-1}\}.$$ In the literature $P_{n}(k)$ is called a microbolic subgroup. Write
$\mathfrak{g}_{n}(k)=\mathfrak{gl}(n,k)$, which is the algebra of $G_{n}(k)$. Write
$$\mathfrak{p}_{n}(k)=\{\left(\begin{array}{cc}A&\alpha\\0_{1\times(n-1)}&0\\\end{array}
\right):A\in\mathfrak{gl}(n-1,k),\alpha\in k^{n-1}\},$$ which is the Lie algebra of $P_{n}(k)$.

Write $\mathfrak{g}_{n}(k)^{\ast}$ (or $\mathfrak{p}_{n}(k)^{\ast}$) for the dual space of
$\mathfrak{g}_{n}(k)$ (or $\mathfrak{p}_{n}(k)$). Then, $G=G_{n}(k)$ (or $P_{n}(k)$) acts on
$\frg^{\ast}=\mathfrak{g}_{n}(k)^{\ast}$ (or $\mathfrak{p}_{n}(k)^{\ast}$) through $$(g\cdot f)\xi
=f(g^{-1}\cdot\xi),\ \forall g\in G,\forall f\in\frg^{\ast},\forall\xi\in\frg.$$ This is called
the {\it coadjoint action}, and a $G$ orbit in $\frg^{\ast}$ is called a coadjoint orbit.

Write $$(\xi,\eta)=\tr(\xi\eta),\ \forall\xi,\eta\in\mathfrak{g}_{n}(k).$$ This gives a $G_{n}(k)$
conjugation invariant nondegenerate bilinear form on $\mathfrak{g}_{n}(k)$. It gives a $G_{n}(k)$
equivariant isomorphism $$\pr:\mathfrak{g}_{n}(k)\rightarrow\mathfrak{g}_{n}(k)^{\ast},\quad\xi
\mapsto f$$ defined by $$f(\eta)=\tr(\xi\eta),\ \forall\eta\in\mathfrak{g}_{n}(k).$$ Through $\pr$,
the above bilinear form on $\mathfrak{g}_{n}(k)$ induces a nondegenerate bilinear form on
$\mathfrak{g}_{n}(k)^{\ast}$ defined by $$(f,g)=(\pr^{-1}(f),\pr^{-1}(g)),\forall f,g\in
\mathfrak{g}_{n}(k)^{\ast}.$$

Write $$\overline{\mathfrak{p}}_{n}(k)=\{\left(\begin{array}{cc}A&0_{(n-1)\times 1}\\\alpha^{t}&0
\\\end{array}\right):A\in\mathfrak{gl}(n-1,k),\alpha\in k^{n-1}\}.$$ Define $\pr':\mathfrak{g}_{n}(k)
\rightarrow\mathfrak{p}_{n}(k)^{\ast}$ by $$(\pr'(\xi))(\eta)=\tr(\xi\eta),\ \forall\eta\in
\mathfrak{p}_{n}(k).$$ It is easy to show that $$\ker(\pr')=\{\left(\begin{array}{cc}
0_{(n-1)\times(n-1)}&\alpha\\0_{1\times(n-1)}&t\\\end{array}\right):\alpha\in k^{n-1},t\in k.\}$$
It is clear that $\mathfrak{g}_{n}(k)=\ker(\pr')\oplus\overline{\mathfrak{p}}_{n}(k)$ as a linear
space. Thus, $\pr'|_{\overline{\mathfrak{p}}_{n}(k)}:\overline{\mathfrak{p}}_{n}(k)\rightarrow
\mathfrak{p}_{n}(k)^{\ast}$ is a linear isomorphism. In this way, any element in $f\in
\mathfrak{p}_{n}(k)^{\ast}$ could be represented by $$f=\pr'(\xi)=\pr(\xi)|_{\mathfrak{p}_{n}(k)}$$
for a unique $\xi\in\overline{\mathfrak{p}}_{n}(k)$.

The {\it moment map} $p:\mathfrak{g}_{n}(k)^{\ast}\rightarrow\mathfrak{p}_{n}(k)^{\ast}$ is defined
by $$f\mapsto f'=f|_{\mathfrak{p}_{n}(k)}.$$ For any $\xi\in\mathfrak{g}_{n}(k)$, we have
$$p(\pr(\xi))=\pr'(\xi).$$

Note also that for any Lie group $G$, the coadjoint action of $\mathfrak{g}$ on $\mathfrak{g}^{\ast}$
(which is the differential of the coadjoint action of $G$ on $\mathfrak{g}^{\ast}$) is determined by
$$(\ad(X)f)(Y)=-f([X,Y]),\ \forall X,Y\in\mathfrak{g},\forall f\in\mathfrak{g}^{\ast}.$$

\subsection{The classification of $P$ coadjoint orbits}

Write $$N_{n}(k)=\{\left(\begin{array}{cc}I_{n-1}&\alpha\\0_{1\times(n-1)}&1\\\end{array}\right):
\alpha\in k^{n-1}\},$$ and $$\mathfrak{n}_{n}(k)=\{\left(\begin{array}{cc}0_{n-1}&\alpha
\\0_{1\times(n-1)}&0\\\end{array}\right):\alpha\in k^{n-1}\}.$$ Then, $N_{n}(k)$ is the unipotent
radical of $P_{n}(k)$, and $\mathfrak{n}_{n}(k)$ is its Lie algebra (=nilpotent radical of
$\mathfrak{p}_{n}(k)$). Write $$L_{n}(k)=\{\left(\begin{array}{cc}A&0_{(n-1)\times 1}\\0_{1\times(n-1)}
&1\\\end{array}\right):A\in G_{n-1}(k)\},$$ and $$\mathfrak{l}_{n}(k)=\{\left(\begin{array}{cc}A&
0_{(n-1)\times 1}\\0_{1\times(n-1)}&0\\\end{array}\right):A\in\mathfrak{g}_{n-1}(k).\}.$$ Then,
$L_{n}(k)$ (or $\mathfrak{l}_{n}(k)$) is a Levi subgroup (or Levi subalgebra) of $P_{n}(k)$ (or of
$\mathfrak{l}_{n}(k)$).

We could identify $\mathfrak{l}_{n}(k)$ with $\mathfrak{p}_{n}(k)/\mathfrak{n}_{n}(k)$. By this, there is
an exact sequence of $P_{n}(k)$ modules, $$0\rightarrow\mathfrak{n}_{n}(k)\rightarrow\mathfrak{p}_{n}(k)
\rightarrow\mathfrak{l}_{n}(k)\rightarrow 0.$$ Dually, there is an exact sequence of $P_{n}(k)$ modules,
$$0\rightarrow\mathfrak{l}_{n}(k)^{\ast}\rightarrow\mathfrak{p}_{n}(k)^{\ast}\rightarrow
\mathfrak{n}_{n}(k)^{\ast}\rightarrow 0.$$

\begin{lemma}\label{L:Porbit-inductive}
For any $n\geq 1$, there is an identification $$\mathfrak{p}_{n}(k)^{\ast}/P_{n}(k)=
\mathfrak{l}_{n}(k)^{\ast}/L_{n}(k)\bigsqcup\mathfrak{p}_{n-1}(k)^{\ast}/P_{n-1}(k).$$
\end{lemma}
\begin{proof}
From the exact sequence $0\rightarrow\mathfrak{l}_n(k)^{\ast}\rightarrow\mathfrak{p}_n(k)^{\ast}
\rightarrow\mathfrak{n}_n(k)^{\ast}\rightarrow 0$, we get $$\mathfrak{p}_{n}(k)^{\ast}/P_{n}(k)=
\mathfrak{l}_{n}(k)^{\ast}/P_n(k)\bigsqcup(\mathfrak{p}_n(k)^{\ast}-
\mathfrak{l}_n(k)^{\ast})/P_{n}(k).$$ As $N_{n}$ acts trivially on $\mathfrak{l}_{n}(k)^{\ast}$,
we get $$\mathfrak{l}_{n}(k)^{\ast}/P_n(k)=\mathfrak{l}_{n}(k)^{\ast}/L_n(k).$$

Choose an element $\overline{h}\in\mathfrak{p}_{n}(k)^{\ast}$ such that $0\neq h=
\overline{h}|_{\mathfrak{n}_{n}(k)^{\ast}}\in\mathfrak{n}_{n}(k)^{\ast}$ and
$\overline{h}|_{\mathfrak{l}_{n}(k)}=0$. Due to the fact that $L_{n}(k)$ acts transitively on
$\mathfrak{n}_{n}(k)^{\ast}-\{0\}$, any $P_{n}(k)$ orbit in $\mathfrak{p}_n(k)^{\ast}-
\mathfrak{l}_n(k)^{\ast}$ intersects with $\mathfrak{l}_{n}(k)^{\ast}+\overline{h}$. From this we
get $$(\mathfrak{p}_{n}(k)^{\ast}-\mathfrak{l}_{n}(k)^{\ast})/P_n(k)\cong(\mathfrak{l}_{n}(k)^{\ast}
+\overline{h})/P_n(k)^{h},$$ where $P_{n}(k)^{h}=\Stab_{P_n(k)}(h)$. Moreover we write $L_{n}(k)^{h}
=\Stab_{L_n(k)}(h)$, and write $\mathfrak{p}_{n}(k)^{h}$ (or $\mathfrak{l}_{n}(k)^{h}$) for the Lie
algebra of $P_{n}(k)^{h}$ (or of $L_{n}(k)^{h}$). As $N_{n}(k)$ is abelian, we have $N_{n}(k)
\subset P_n(k)^{h}$. Thus, $$P_n(k)^{h}=N_{n}(k)\rtimes L_n(k)^{h}.$$

Since $N_{n}(k)$ acts trivially on $\mathfrak{l}_{n}(k)^{\ast}$, its action on
$\mathfrak{l}_{n}(k)^{\ast}+\overline{h}$ is through translations. Actually, $$\exp(X)\cdot(g+
\overline{h})=g+\overline{h}+\ad(X)\overline{h}$$ for any $X\in\mathfrak{n}_{n}(k)$ and any $g\in
\mathfrak{l}_{n}(k)^{\ast}.$ In the below we show that \begin{equation}\label{Eq1}
\mathfrak{l}_{n}(k)^{\ast}/\ad(\mathfrak{n}_{n}(k))\overline{h}=(\mathfrak{l}_{n}(k)^{h})^{\ast}.
\end{equation} From this it follows that $$(\mathfrak{l}_n(k)^{\ast}+ h)/P_n(k)^{h}\cong
\mathfrak{l}_{n}(k)^{h}/L_{n}(k)^{h}.$$ It is easy to check directly that $L_n(k)^{h}\cong P_{n-1}
(k)$. Then it follows that $$(\mathfrak{p}_{n}(k)^{\ast}-\mathfrak{l}_{n}(k)^{\ast})/P_{n}(k)\cong
\mathfrak{p}_{n-1}(k)^{\ast}/P_{n-1}(k).$$ Therefore, $$\mathfrak{p}_{n}(k)^{\ast}/P_{n}(k)=
\mathfrak{l}_{n}(k)^{\ast}/L_{n}(k)\bigsqcup\mathfrak{p}_{n-1}(k)^{\ast}/P_{n-1}(k).$$

Now we show $\mathfrak{l}_{n}(k)^{\ast}/\ad(\mathfrak{n}_{n}(k))\overline{h}=(\mathfrak{l}_{n}
(k)^{h})^{\ast}.$ Since the pairing between $\mathfrak{l}_{n}(k)$ and $\mathfrak{l}_{n}(k)^{\ast}$ is
nondegenerate. It is equivalent to show \begin{equation}\label{Eq2}\{\xi\in\mathfrak{l}_{n}(k):
(\ad(\eta)\overline{h})(\xi)=0,\forall\eta\in\mathfrak{n}_{n}(k)\}=\mathfrak{l}_{n}(k)^{h}.
\end{equation} This follows from $$(\ad(\eta)\overline{h})(\xi)=-\overline{h}([\eta,\xi])=
\overline{h}([\xi,\eta])=-(\ad(\xi)\overline{h})(\eta).$$
\end{proof}

\begin{lemma}\label{L:Porbit-invariant}
Assume $\overline{h}|_{\mathfrak{l}_{n}(k)^{\ast}}\!=\!0$ and $0\!\neq\!h\!=\!
\overline{h}|_{\mathfrak{n}_{n}(k)^{\ast}}\!\in\!\mathfrak{n}_{n}(k)^{\ast}$. Then, \begin{itemize}
\item[(1),] $\ad(\mathfrak{l}_{n}(k)^{h})\overline{h}=0$.
\item[(2),] the map $$\mathfrak{n}_{n}(k)\rightarrow\mathfrak{l}_{n}(k)^{\ast},\xi\mapsto
(\ad\xi)\overline{h}$$ is injective.
\end{itemize}
\end{lemma}

\begin{proof}
For assertion (1), let $\xi\in\mathfrak{l}_{n}(k)^{h}$. For any $\eta\in\mathfrak{n}_{n}(k)$,
$(\ad(\xi)\overline{h})(\eta)=0$ since $\ad(\xi)\overline{h}\in\mathfrak{l}_{n}(k)^{\ast}$. For any
$\eta\in\mathfrak{l}_{n}(k)$, $$(\ad(\xi)\overline{h})(\eta)=-\overline{h}([\xi,\eta])=0$$ since
$\overline{h}|_{\mathfrak{l}_{n}(k)^{\ast}}=0$. Thus, $\ad(\xi)\overline{h}=0.$ Therefore,
$\ad(\mathfrak{l}_{n}(k)^{h})\overline{h}=0$.

For assertion (2), we may assume that $\overline{h}=\pr'(\xi)$, where $\xi'=\left(\begin{array}{cc}
0_{n-1}&0_{(n-1)\times 1}\\\beta'^{t}&0\end{array}\right)$, $\beta'^{t}=(0,\dots,0,1)$. Then, a
direct calculation of $\ad(\eta)\overline{h}=\pr'([\eta,\xi'])$ ($\eta\in\mathfrak{n}_{n}(k)$)
shows the assertion.
\end{proof}

Now assume $\overline{h}|_{\mathfrak{l}_{n}(k)^{\ast}}=0$ and $0\neq h=
\overline{h}|_{\mathfrak{n}_{n}(k)}\in\mathfrak{n}_{n}(k)^{\ast}$. Choose a complement of
$(\ad(\mathfrak{n}_{n}(k))\overline{h}$ in $\mathfrak{l}_{n}(k)^{\ast}$, denoted by $V_{h}$.

\begin{lemma}\label{L:Porbit-stabilizer}
Assume $\overline{h}|_{\mathfrak{l}_{n}(k)^{\ast}}\!=\!0$ and $0\!\neq\!h\!=\!
\overline{h}|_{\mathfrak{n}_{n}(k)^{\ast}}\!\in\!\mathfrak{n}_{n}(k)^{\ast}$. Then, we have the following
assertions, \begin{itemize}
\item[(1),]each $P_{n}(k)$ orbit intersecting with $\mathfrak{l}_{n}(k)^{\ast}+\overline{h}$ has a
representative of the form $f=g+\overline{h}$ where $g\in V_{h}.$
\item[(2),] Two elements $g_1+\overline{h}$ and $g_2+\overline{h}$ ($g_1,g_2\in V_{h}$) are in one $P_{n}(k)$
orbit if and only if $[g_1]$ and $[g_2]$ are in one $L_{n}(k)^{h}$ orbit, where$$[g_{i}]\!=
\!g_{i}\!+\!\ad(\mathfrak{n}_{n}(k))\overline{h}\in\mathfrak{l}_{n}(k)^{\ast}/\ad(\mathfrak{n}_{n}(k))
\overline{h}\!=\!(\mathfrak{l}_{n}(k)^{\overline{h}})^{\ast}$$ ($i=1,2$) are considered as elements in
$(\mathfrak{l}_{n}(k)^{h})^{\ast}$.
\item[(3),] Assume $g\in V_{h}$. Then, the map $$\Stab_{P_{n}(k)}(g+\overline{h})\rightarrow L_{n}(k),\
nl\mapsto l$$ ($n\in N_{n}(k)$, $l\in L_{n}(k)$) gives an isomorphism $$\Stab_{P_{n}(k)}(g+\overline{h})
\cong\Stab_{L_{n}(k)^{h}}([g]).$$
\end{itemize}
\end{lemma}

\begin{proof}
The assertion $(1)$ follows from Equation $(1)$.

For assertion $(2)$, we show the necessarity first. Assume $g_2+\overline{h}=x\cdot(g_1+\overline{h})$
for some $x\in P_{n}(k)$. Then, $x\in P_{n}(k)^{h}=N_{n}(k)\rtimes L_{n}(k)^{h}$. Write $x=nl=
\exp(\xi)l$ for some $\xi\in\mathfrak{n}_{n}(k)$ and $l\in L_{n}(k)^{h}$. Then, $$\exp(-\xi)\cdot
(g_2+\overline{h})=l\cdot(g_1+\overline{h}).$$ By Lemma \ref{L:Porbit-invariant}, we have $l\cdot
\overline{h}=0$. As $\mathfrak{n}_{n}(k)$ is an abelian ideal of $\mathfrak{n}_{n}(k)$, we have
$\exp(-\xi)\cdot(g_2+\overline{h})=(g_2-\ad(\xi)\overline{h})+\overline{h}$. Thus, $g_2-\ad(\xi)
\overline{h}=l\cdot g_1$. This just means, $[g_1]$ and $[g_2]$ are in one $L_{n}(k)^{h}$ orbit. The
sufficiency could be shown with similar facts used in showing the necessarity.

For assertion $(3)$, write $n=\exp(\xi)$ ($\xi\in\mathfrak{n}_{n}(k)$). By the above proof for (2),
we see that $nl\in\Stab_{P_{n}(k)}(g+\overline{h})$ if and only if $l\in L_{n}(k)^{h}$ and $$g-
\ad(\xi)\overline{h}=l\cdot g.$$ The last is just the condition for $l\in\Stab_{L_{n}(k)^{h}}([g])$.
Thus, the map $$\Stab_{P_{n}(k)}(g+\overline{h})\rightarrow L_{n}(k),\ nl\mapsto l$$ gives a
surjection $\Stab_{P_{n}(k)}(g+\overline{h})\rightarrow\Stab_{L_{n}(k)^{h}}([g]).$ On the other
hand, suppose $l=1$. Then, $\ad(\xi)\overline{h}=0$. By Lemma \ref{L:Porbit-invariant}(2), this
implies that $\xi=0$. Thus, the above surjection $\Stab_{P_{n}(k)}(g+\overline{h})\rightarrow
\Stab_{L_{n}(k)^{h}}([g])$ is an isomorphism.
\end{proof}

We have some remarks regarding Lemma \ref{L:Porbit-stabilizer}(3). \begin{itemize}
\item[(1),] the statement is valid for any $g\in\mathfrak{l}_{n}(k)^{\ast}$.
\item[(2),] $(\ad(\mathfrak{n}_{n}(k))\overline{h}$ is stable under the action of $L_{n}(k)^{h}$, but
$L_{n}(k)^{h}$ is not a reductive subgroup. Actually one can verify that we couldn't take $V_{h}$ stable
under $L_{n}(k)^{h}$.
\item[(3),] we have $$\Stab_{L_{n}(k)^{h}}(g)\subset\Stab_{P_{n}(k)}(g+\overline{h})$$ and
$$\Stab_{L_{n}(k)^{h}}(g)\subset\Stab_{L_{n}(k)^{h}}([g]),$$ but neither is an equality in general.
\end{itemize}

The following lemma is easy to show.
\begin{lemma}\label{L:Porbit-stabilizer2}
When $h=0$, we have $f\in\mathfrak{l}_{n}(k)^{\ast}\subset\mathfrak{n}_{n}(k)^{\ast}$. Then,
$$\Stab_{P_{n}(k)}(f)=N_{n}(k)\rtimes\Stab_{L_{n}(k)}(f).$$
\end{lemma}

\begin{theorem}\label{T:Porbit}
Any $P_{n}(k)$ coadjoint orbit $P_{n}(k)\cdot f$ is characterized by an integer $j$ ($0\leq j\leq n-1$)
and an $L_{n-j}(k)$ coadjoint orbit $L_{n-j}(k)\cdot g$. Moreover, we have $\Stab_{P_{n}(k)}f\cong
N_{n-j}(k)\rtimes\Stab_{L_{n-j}(k)}(g)$.
\end{theorem}

\begin{proof}
Applying Lemma \ref{L:Porbit-inductive} inductively, we get $$\mathfrak{p}_{n}(k)^{\ast}/P_{n}(k)=
(\bigsqcup_{0\leq j\leq n-2}\mathfrak{l}_{n-j}(k)^{\ast}/L_{n-j}(k))\bigsqcup\mathfrak{p}_{1}
(k)^{\ast}/P_{1}(k).$$ As both $P_{1}(k)$ and $L_{1}(k)$ are the trivial group, we get $\mathfrak{p}_{1}
(k)^{\ast}/P_{1}(k)=\mathfrak{l}_{1}(k)^{\ast}/L_{1}(k)=\textrm{singleton}.$ Thus, \begin{equation}
\label{Eq:depth}\mathfrak{p}_{n}(k)^{\ast}/P_{n}(k)=\bigsqcup_{0\leq j\leq n-1}\mathfrak{l}_{n-j}
(k)^{\ast}/L_{n-j}(k).\end{equation} For any $j$, $L_{n-j}(k)\cong\G_{n-1-j}(k)$. Thus, we get the
first statement of the theorem.

If a $P_{n}(k)$ coadjoint orbit $P_{n}(k)\cdot f$ is characterized by an integer $j$ ($0\leq j\leq n-1$)
and an $L_{n-j}(k)$ coadjoint orbit $L_{n-j}\cdot g$, then $\Stab_{P_{n}(k)}f\cong\Stab_{L_{n-j}}(g)$ by
Lemma \ref{L:Porbit-stabilizer}(3) and Lemma \ref{L:Porbit-stabilizer2}.
\end{proof}

Note that $N_{n-j}(k)\cong k^{n-1-j}$. With Theorem \ref{T:Porbit}, we not only classified all $P$ coadjoint
orbits, but also calculated their {\it stabilizers}.

\begin{definition}\label{D:depth}
According to Equation (\ref{Eq:depth}), any $P_{n}(k)$-coadjoint orbit $P_{n}(k)\cdot f\subset
\mathfrak{p}_{n}(k)^{\ast}$ is parameterized by an integer $j$ ($0\leq j\leq n-1$) and a $L_{n-j}(k)$
coadjoint orbit $L_{n-j}(k)\cdot g\subset \mathfrak{l}_{n-j}(k)^{\ast}$. We call $j+1$ the {\it depth}
of this orbit (and points in it).
\end{definition}

By Theorem \ref{T:Porbit}, $\Stab_{P_{n}(k)}(f)\cong N_{n-j}(k)\rtimes\Stab_{L_{n-j}(k)}(g)$. We call $f$
a {\it semisimple element} (or $P_{n}(k)\cdot f$ a {\it semisimple orbit}) if $g$ is a semisimple element
in $\mathfrak{l}_{n-j}(k)^{\ast}$ (or $L_{n-j}(k)\cdot g$ is a semisimple orbit). The latter means
$\Stab_{L_{n-j}(k)}(g)$ is an a reductive (algebraic) subgroup of $L_{n-j}(k)\cong\GL(n\!-\!1\!-\!j,k)$.
In this case we call the $(n\!-\!1\!-\!j)$ eigenvalues of $\pr^{-1}(g)\in\mathfrak{gl}(n\!-\!1\!-\!j,k)$
the eigenvalues of $f$.

In the case of $j=n-1$, we have $\Stab_{P_{n}(k)}(f)=\Stab_{L_{n-j}(k)}(g)=\{1\}$, this gives an open orbit
in $\mathfrak{p}_{n}(k)^{\ast}$. In the case of $j<n-1$, we have $\dim\Stab_{P_{n}(k)}(f)=
\dim\Stab_{L_{n-j}(k)}(g)\geq n-1-j\geq 1$, none of these orbits is open. Again, by the proof of Lemma
\ref{L:Porbit-inductive} and Theorem \ref{T:Porbit}, we see that the complement of the above open orbit in
$\mathfrak{p}_{n}(k)^{\ast}$ is a closed subset of codimension one. This shows the following.

\begin{proposition}\label{P:openOrbit}
There is a unique open $P_{n}(k)$-coadjoint orbit in $\mathfrak{p}_{n}(k)^{\ast}$, which we denote by
$\mathcal{O}_{0}$. It is dense and its complement is a codimension one closed subset. For any $f\in
\mathcal{O}_{0}$, we have $\Stab_{P_{n}(k)}(f)=1$.

In particular , $\mathcal{O}_{0}$ is the only strongly regular coadjoint orbit of $P_{n}(k)$ in
$\mathfrak{p}_{n}(k)^{\ast}$. Consequently, $P_{n}(k)$ has one and only one discrete series representation,
which we denote by $\tau$. Moreover $\tau$ is attached to $\mathcal{O}_{0}$, and $L^{2}(P_{n}(k))\cong\tau
\otimes\tau$.
\end{proposition}

We give some more precise information about the unique open $P_{n}(k)$-orbit in $\mathfrak{p}_{n}(k)^{\ast}$
below.

\begin{example}\label{E:openOrbit1}
Let $$\xi_{n}=\left(\begin{array}{ccccc}0&0&\ldots&0&0\\1&0&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots
\\0&0&\ldots&0&0\\0&0&\ldots&1&0\\\end{array}\right),$$ and $f_{n}=\pr'(\xi_{n})\in\mathfrak{p}_{n}(k)^{\ast}$.
Write $h$ for the projection of $f_{n}$ to $\mathfrak{n}_{n}(k)^{\ast}$. By calculation one sees that
$L_{n}(k)^{h}=P_{n-1}(k)$, and the element $f_{n}+\ad(\mathfrak{n}_{n}(k))\overline{h}\in
(\mathfrak{l}_{n}(k)^{h})^{\ast}$ is equal to $\pr'(\xi_{n-1})\in\mathfrak{p}_{n-1}(k)^{\ast}=
(\mathfrak{l}_{n}(k)^{h})^{\ast}$. Taking induction, one sees that $f_{n}$ represents the unique open
$P_{n}(k)$ orbit in $\mathfrak{p}_{n}(k)^{\ast}$.
\end{example}

\begin{example}\label{E:openOrbit2}
Let $k=\mathbb{C}$. Let $a_1,\dots,a_{n-1},b_1,\dots,b_{n-1}\in\mathbb{C}$ such that $a_{i}\neq a_{j}$
($\forall i,j$, $1\leq i<j\leq n-1$) and $b_{j}\neq 0$ ($\forall j$, $1\leq j\leq n-1$). Set $$\xi'_{n}
=\left(\begin{array}{ccccc}a_1&0&\ldots&0&0\\0&a_2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&
\ldots&a_{n-1}&0\\b_1&b_2&\ldots&b_{n-1}&0\\\end{array}\right)$$ and $f'_{n}=\pr'(\xi_{n})\in
\mathfrak{p}_{n}(k)^{\ast}$. We show that $f'_{n}$ represents the unique open $P_{n}(k)$ orbit in
$\mathfrak{p}_{n}(k)^{\ast}$.

Suppose $$g=\left(\begin{array}{cc}B&\beta\\0_{1\times(n-1)}&1\\\end{array}\right)\in\Stab_{P_{n}}
(f'_{n}),$$ where $B\in\GL(n-1,\mathbb{C})$ and $\beta\in\mathbb{C}^{n-1}$. Write $A=\diag\{a_1,\dots,
a_{n-1}\}$ and $\alpha=(b_1,\dots,b_{n-1})^{t}$. Then, $$\xi'_{n}=\left(\begin{array}{cc}A&0_{(n-1)
\times 1}\\\alpha^{t}&0\\\end{array}\right).$$ Due to $g\cdot f'_{n}=g\pr'(\xi'_{n})=\pr'(g\xi'_{n}g^{-1})$,
we see that $$g\in\Stab_{P_{n}}(f_{n})\Leftrightarrow g\xi'_{n}g^{-1}-\xi'_{n}\in\mathfrak{n}'_{n},$$
where \begin{eqnarray*}&&\mathfrak{n}'_{n}\\=&&\{\eta\in\mathfrak{gl}(n,\mathbb{C}):\tr(\eta\theta)=0,
\forall\theta\in\mathfrak{p}_{n}\}\\=&&\{\left(\begin{array}{cc}0_{n-1}&\alpha\\
0_{1\times(n-1)}&t\\\end{array}\right):\alpha\in\mathbb{C}^{n-1},t\in\mathbb{C}\}.\end{eqnarray*}
By calculation $$g\xi'_{n}g^{-1}= \left(\begin{array}{cc}BAB^{-1}+\beta\alpha^{t}B^{-1}&-BAB^{-1}\beta
-\beta\alpha^{t}B^{-1}\beta\\\alpha^{t}B^{-1}&-\alpha^{t}B^{-1}\beta\\\end{array}\right).$$ Thus,
$$g\in\Stab_{P_{n}}(f_{n})\Leftrightarrow BAB^{-1}+\beta\alpha^{t}B^{-1}=A\textrm{ and }\alpha^{t}B^{-1}
=\alpha^{t}.$$ This is equivalent to: $$AB-BA=\beta\alpha^{t}\textrm{ and }\alpha^{t}B=\alpha^{t}.$$
Write $$B=(x_{i,j})_{(n-1)\times(n-1)}\textrm{ and }\beta^{t}=(y_{j})_{1\leq j\leq(n-1)}.$$ From
$AB-BA=\beta\alpha^{t}$, we get $$(a_{i}-a_{j})x_{i,j}=b_{j}y_{i},\forall i,j,1\leq i,j\leq n-1.$$
Let $i=j$, from $b_{j}\neq 0$ we get $y_{j}=0$. Thus, $\beta=0$. Let $i\neq j$, from $y_{i}=0$ we get
$x_{i,j}=0$. Thus, $B$ is a diagonal matrix. From $\alpha^{t}B=\alpha^{t}$ and all entries of $\alpha$
are not equal to $0$, we get $B=I_{n-1}$. Therefore, $\Stab_{P_{n}}(f'_{n})=1$.
\end{example}

Remark:  After we finished this article, Prof. Ra\"{i}s informed us that some results in this sub-section were also obtained in his article \cite{Rais}. Especially in his article, the existence of open $P$-coadjoint orbit was established, and a representative element of the open $P$-orbit was given. However, our method is different from his, and our results are more explicit. Moreover, our method is useful for us to interpret Sahi's result in the framework of Duflo's orbit method in next section (see Section \ref{SS: Duflo-Sahi}).

\subsection{The moment map in the $\GL(n,\mathbb{C})$ case}\label{SS:GL(C)}

Let $G=\GL(n,\mathbb{C})$, and $$P=P_{n}(\mathbb{C})=\{\left(\begin{array}{cc}A&\alpha\\0_{1\times(n-1)}
&1\\\end{array}\right):A\in\GL(n-1,\mathbb{C}),\alpha\in\mathbb{C}^{n-1}\}.$$ Let $T$ be the maximal
torus of $G$ consisting of all diagonal matrices in $G$. Let $\vec{a}=(a_1,..., a_n)\in\mathbb{C}^{n}$
with $a_{i}\neq a_{j}$ ($1\leq i<j\leq n$). Write $$\xi=\xi_{\vec{a}}=\diag\{a_1,..., a_n\}\in\mathfrak{t}
\subset\mathfrak{g},$$ and $$f=f_{\vec{a}}=\pr(\xi)\in\mathfrak{g}^{\ast}.$$ Then $f$ is a regular
semisimple element in $\mathfrak{g}^{\ast}$, and any regular semisimple orbit $\mathcal{O}\subset
\mathfrak{g}^{\ast}$ is of the form $\mathcal{O}=\mathcal{O}_{f}=G\cdot f$ with $f=f_{\vec{a}}$ as
above.

From now on, we fix $\vec{a}$, $\xi$ and $f$ in this subsection.

\begin{lemma}\label{L:double1}
There are exactly $2^{n}\!-\!1$ $P$ orbits in $\mathcal{O}_{f}$. One of them is Zariski open and dense,
and the union of the rest is a codimension one Zariski closed subset.
\end {lemma}

\begin{proof}
With the assumption, $\Stab_{G}(f)=T$. Thus, $P\backslash\mathcal{O}_{f}=P\backslash G/T$. It is clear
that the map $$PgT\mapsto Tg^{-1}P$$ gives a bijection $P\backslash G/T\cong T\backslash G/P.$ Consider
the transitive $G$-action on $\mathbb{C}^{n}\!-\!\{0\}$, $$g\cdot(x_1,..., x_n)^{t}=(g^{-1})^{t}(x_1,
\dots,x_n)^{t},\ \forall g\in G,\forall(x_1,\dots,x_{n})\in\mathbb{C}^n\!-\!\{0\}.$$ Let $v_0=(0,...,
0,1)^{t}$. Then, $\Stab_{G}(v_0)=P$. Thus, $G/P=\mathbb{C}^n-\{0\}$ and $T\backslash G/P=
T\backslash(\mathbb{C}^n-\{0\})$.

For any $I=\{i_1,...,i_k\}$ where $1\leq i_1<i_2<...<i_k\leq n$ and $1\leq k\leq n$, let $v_{I}=(x_1,
\dots,x_{n})^{t}\in\mathbb{C}^n-\{0\}$ be defined by $x_{i}=1$ if $i\in\{i_1,\dots,i_{k}\}$, and $x_{i}=0$
if $i\not\in\{i_1,\dots,i_{k}\}$. Note that $T$ acts on $\mathbb{C}^{n}-\{0\}$ through $$\diag\{\lambda_1,
\dots,\lambda_n\}\cdot(x_1,..., x_n)^{t}=(\lambda_1^{-1}x_1,..., \lambda_n^{-1}x_n)^{t}.$$ It is easy
to see that $\{v_{I}:\emptyset\neq I\subset\{1,...,n\}\}$ represent all different $T$ orbits in
$\mathbb{C}^{n}-\{0\}$. This shows the lemma.
\end{proof}

Write $I_0=\{1,2,...,n\}$. For any $\emptyset\neq I\subset I_0$, introduce a matrix $g_{I}\in G$. In the
case of $n\in I$, define $g_{I}$ by $g_{I}=\left(\begin{array}{cc}I_{n-1}&0_{(n-1)\times 1}\\\beta^{t}&
0\\\end{array}\right)$, where $\beta^{t}=(x_1,\dots,x_{n-1})$ with $x_{i}=1$ if $i\in I$, and $x_{i}=1$
if $i\not\in I$. In the case of $n\not\in I$, let $k\!=\!\max\{1\leq\!i\leq\!n\!-\!1\!:\!i\in I\}.$ Define
$g_{I}$ by $$g_{I}=\left(\begin{array}{cccc}I_{k-1}&0_{(k-1)\times 1}&0_{(k-1)\times(n-1-k)}&0_{(k-1)
\times 1}\\0_{1\times(k-1)} &0&0_{1\times(n-1-k)}&1\\0_{(n\!-\!1\!-\!k)\times(k-1)}&0_{(n\!-\!1\!-\!k)
\times 1}&I_{n\!-\!1\!-\!k}&0_{(n\!-\!1\!-\!k)\times 1}\\\beta'^{t}&1&0_{1\times(n-1-k)}&0\\\end{array}
\right),$$ where $\beta'^{t}=(x_1,\dots,x_{k-1})$ with $x_{i}=1$ if $i\in I$, and $x_{i}=0$ if $i\not
\in I$.

\begin{proposition}\label{P:double2}
$\{g_{I}\cdot f:\emptyset\neq I\subset I_0\}$ represent all different $P$ orbits in $\mathcal{O}_{f}$.
Among these orbits, $Pg_{I_{0}}\cdot f$ is a Zarisk open and dense $P$ orbit and its complement is a
Zariski closed subset of codimension one.
\end{proposition}
\begin{proof}
One can show that $g_{I}^{-1}\cdot v_0=v_{I}$ for any $I$. By the proof of Lemma \ref{L:double1}, $\{g_{I}
\cdot f:\emptyset\neq I\subset I_0\}$ represent all different $P$ orbits in $\mathcal{O}_{f}$.

It is clear that $Tg_{I_{0}}^{-1}\cdot v_0$ is a Zarisk open and dense subset in $\mathbb{C}^{n}-\{0\}$,
and its complement is a Zariski closed subset of codimension one. Thus, $Pg_{I_{0}}\cdot f$ is a Zarisk
open and dense $P$ orbit and its complement is a Zariski closed subset of codimension one.
\end{proof}

\begin{lemma}\label{L:moment1}
For any $\emptyset\neq I\subset I_0$, the element $p(g_{I}\cdot f)$ is semisimple, its depth is equal to
$\#I$ and its eigenvalues are $\{a_{i}: i\in I_{0}-I\}$.
\end{lemma}
\begin{proof}
We have $$p(g_{I}\cdot f)=p(g_{I}\cdot\pr(\xi))=p(\pr(g_{I}\cdot\xi))=\pr'(g_{I}\cdot\xi).$$

In the case of $n\in I$, by calculation we have $\pr'(g_{I}\cdot\xi)=\pr'(\xi_{I})$, where $$\xi_{I}=  \left(\begin{array}{cc}\diag\{a_1,\dots,a_{n-1}\}&0_{(n-1)\times 1}\\\beta^{t}&0\end{array}\right))$$
with $\beta^{t}=(x_1,\dots,x_{n-1})$, $x_{i}=a_{i}-a_{n}$ if $i\in I$, and $x_{i}=0$ if $i\not\in I$.
Separating $I_0=\{1,2,\dots,n\}$ into the disjoint union of two subsets, $I$ and $I_0-I$, we see that
$\xi_{I}$ is a block diagonal matrix. The one with rows and columns indexed by $I$ is of the form in
Example \ref{E:openOrbit2}; the one with rows and columns indexed by $I_0-I$ is a diagonal matrix with
eigenvalues $\{a_{i}: i\in I_{0}-I\}$.

We know that there is a unique open $P$ orbit in $\mathfrak{p}^{\ast}$, and any matrix of the form in
Example \ref{E:openOrbit1} or \ref{E:openOrbit2} is in this orbit. From this, substituting $\xi_{I}$
by a $P$ conjugate matrix $\xi'_{I}$, we could make $\xi'_{I}$ still a block diagonal matrix with two
blocks indexed by $I$ and $I_0-I$ respectively, with the part indexed by the set $I$ of the form in
Example \ref{E:openOrbit1} (with degree of $\#I$, instead of $n$), and the part indexed by $I_0-I$
a digonal matrix with eigenvalues $\{a_{i}: i\in I_{0}-I\}$. Applying the reduction \ref{T:Porbit}
$\#I-1$ times, we arrive at a diagonal matrix in $\mathfrak{l}_{n\!+\!1\!-\!\#I}(\mathbb{C})$ with
eigenvalues $\{a_{i}: i\in I_{0}-I\}$. That just means, $g_{I}\cdot f$ is semisimple, with depth
equal to $\#I$, and its eigenvalues are $\{a_{i}: i\in I_{0}-I\}$.

In the case of $n\in I$, let $k=\max\{i: i\in I\}$. By calculation we have $g_{I}\cdot f=
\pr'(g_{I}\cdot\xi)=\pr'(\xi_{I})$, where $$\xi_{I}=\left(\begin{array}{cc}\diag\{a_1,\dots,a_{k-1},
a_{n},a_{k+1},\dots,a_{n-1}\}&0_{(n-1)\times 1}\\\beta^{t}&0\end{array}\right))$$ with $\beta^{t}=
(x_1,\dots,x_{k-1},0,\dots,0)$, where $x_{i}=a_{i}-a_{n}$ if $i\in I$, and $x_{i}=0$ if $i\not\in I$.
Write $$I'=(I-\{k\})\cup\{n\}.$$ Separating $I_0=\{1,2,\dots,n\}$ into the disjoint union of two
subsets, $I'$ and $I_0-I'$, we see that $\xi_{I}$ is a block diagonal matrix. The one with rows and
columns indexed by $I'$ is of the form in Example \ref{E:openOrbit2}; the one with rows and columns
indexed by $I_0-I'$ is a diagonal matrix with eigenvalues $\{a_{i}: i\in I_{0}-I\}$. The proof of
the rest is the same in the case of $n\in I$.
\end{proof}

\begin{theorem}\label{T:moment1}
The set $p(\mathcal{O}_{f})$ consists of exactly $2^{n}-1$ semisimple $P$ orbits, with the unique open $P$
coadjoint orbit in $\mathfrak{p}^{\ast}$ among them. Moreover,  we have:
\begin{itemize}
\item[(1),] the moment map $p:\mathcal{O}_{f}\rightarrow\mathfrak{p}^{\ast}$ is \emph{weakly proper} .
\item[(2),] the \emph{reduced space} of the unique open $P$ orbit in $\mathfrak{p}^{\ast}$ with respect to
the moment map $p$ is a single point.
\end{itemize}
\end{theorem}

\begin{proof}
By Lemma \ref{L:double1}, $\mathcal{O}_{f}$ is the union of exactly $2^{n}-1$ $P$ orbits. In Lemma
\ref{L:moment1}, we described the image of the moment map $p:\mathfrak{g}^{\ast}\rightarrow
\mathfrak{p}^{\ast}$ for each of these $P$ orbits. Particularly, we see that: each of them is semisimple,
and different $P$ orbits in $\mathcal{O}_{f}$ are mapped to different $P$ orbits in $\mathfrak{p}^{\ast}$
(due to $a_{i}\neq a_{j}$), and the unique open $P$ orbit in $\mathfrak{p}^{\ast}$ is among them (for
$I=I_0$). This shows the first statement of the theorem.

The second statement also follows from the description of the moment map.
\end{proof}

In Theorem \ref{T:moment1}, for any $\emptyset\neq I\subset I_0$, write $j=\#I-1$. Then, $$\Stab_{P}(g_{I}
\cdot f)=P\cap g_{I}Tg_{I}^{-1}=g_{I}(T\cap g_{I}^{-1}Pg_{I})g_{I}^{-1},$$ and $$T\cap g_{I}Pg_{I}^{-1}=
\Stab_{T}(g_{I}^{-1}\cdot v_0),$$ which is a torus isomorphic to $(\mathbb{C}^{\times})^{n-1-j}$. From
the description of $p(g_{I}\cdot f)$ in Lemma \ref{L:moment1}, by Lemma \ref{L:Porbit-stabilizer} and
Lemma \ref{L:Porbit-stabilizer2} we have $$\Stab_{P}(p(g_{I}\cdot f))\cong\mathbb{C}^{n-1-j}\rtimes
(\mathbb{C}^{\times})^{n-1-j}.$$

\subsection{The $\GL(n,\mathbb{R})$ case}\label{SS:GL(R)}

Now let $G=\GL(n,\mathbb{R})$, and $$P=P_{n}(\mathbb{R})=\{\left(\begin{array}{cc}A&\alpha\\
0_{1\times(n-1)}&1\\\end{array}\right):A\in\GL(n-1,\mathbb{R}),\alpha\in\mathbb{R}^{n-1}\}.$$
Let $0\leq k\leq\frac{n}{2}$. Let $z_{1},\dots,z_{n}$ be $n$ distinct complex numbers with
$z_{2j-1}=a_{j}+\mathbf{i}b_{j}$ ($1\leq j\leq k$), $z_{2j}=a_{j}-\mathbf{i}b_{j}$ ($1\leq j
\leq k$), and $z_{2k+j}=a_{2k+j}$ ($1\leq j\leq n-2k$), where $a_1,\dots,a_{k},a_{2k+1},\dots,
a_{n},b_1,\dots,b_{k}\in\mathbb{R}$ and $b_1\cdots b_{k}\neq 0$. Write $$\vec{z}=(z_1,...,z_{n})
\in\mathbb{C}^{n}.$$ Set $$J=\left(\begin{array}{cc}0&1\\-1&0\\\end{array}\right).$$ Write $$\xi
=\xi_{\vec{z}}=\diag\{a_1I_{2}+b_1J,\dots,a_{k}I_{2}+b_{k}J,a_{2k+1},\dots,a_{n}\}\in\mathfrak{g},$$
and $$f=f_{\vec{z}}=\pr(\xi)\in\mathfrak{g}^{\ast}.$$ Write \begin{eqnarray*}T_{k}&=&
\{\diag\{\lambda_1I_{2}\!+\!\mu_1J,\dots,\lambda_{k}I_{2}\!+\!\mu_{k}J,\lambda_{2k\!+\!1},\dots,
\lambda_{n}\}:\\&&\lambda_{1},\dots,\lambda_{k},\lambda_{2k\!+\!1},\dots,\lambda_{n},\mu_1,\dots,
\mu_{k}\in\mathbb{R},\\&&(\lambda_{1}^2\!+\!\mu_{1}^{2})\cdots(\lambda_{k}^{2}\!+\!\mu_{k}^{2})
\lambda_{2k\!+\!1}\cdots\lambda_{n}\neq 0\},\end{eqnarray*} which is a maximal torus in $G$.
It is clear that $\Stab_{G}(f)=T_{k}$. Thus, $f$ is a regular semisimple element in
$\mathfrak{g}^{\ast}$. Any regular semisimple orbit $\mathcal{O}\subset\mathfrak{g}^{\ast}$ is
of the form $\mathcal{O}=\mathcal{O}_{f}=G\cdot f$ for some $0\leq k\leq\frac{n}{2}$, $a_1,\dots,
a_{k},a_{2k+1},\dots,a_{n},b_1,\dots,b_{k}\in\mathbb{R}$ with $b_1\cdots b_{k}\neq 0$, and
$f=f_{\vec{z}}$ as above. From now on, we fix $\vec{z}$, $\xi$ and $f$ in this subsection.

\begin{lemma}\label{L:double3}
There are exactly $2^{n-k}\!-\!1$ $P$ orbits in $\mathcal{O}_{f}$. One of them is open and dense,
and the union of the rest is a codimension one (if $k<\frac{n}{2}$) or two (if $k=\frac{n}{2}$)
closed subset.
\end {lemma}

\begin{proof}
Let $G$ act on $\mathbb{R}^{n}-\{0\}$ though $$g\cdot(x_1,\dots,x_{n})^{t}=(g^{-1})^{t}(x_1,\dots,
x_{n})^{t},\ \forall g\in G,\forall (x_1,\dots,x_{n})^{t}\in\mathbb{R}^{n}-\{0\}.$$ Write $v_0=
(0,\dots,0,1)^{t}\in\mathbb{R}^{n}-\{0\}$. Then, $\Stab_{G}(v_0)=P$ and $\mathbb{R}^{n}-\{0\}=G/P$.
Similar as in the proof of Lemma \ref{L:moment1}, we have identifications $$P\backslash\mathcal{O}_{f}
=P\backslash G/T_{k}\cong T_{k}\backslash G/P=T_{k}\backslash(\mathbb{R}^{n}-\{0\}).$$

Write $I_{0}^{(1)}=\{1,\dots,k\}$, $I_{0}^{(2)}=\{2k+1,\dots,n\}$. For any $I_1\subset I_{0}^{(1)}$
and $I_2\subset I_{0}^{(2)}$ with $\#I_{1}+\#I_{2}>0$, let $v_{I_1,I_2}=(x_1,\dots,x_{n})\in
\mathbb{R}^n-\{0\}$ be defined by $(x_{2i-1},x_{2i})=(0,1)$ if $i\in I_{1}$, $(x_{2i-1},x_{2i})=(0,0)$
if $i\in I_{0}^{(1)}-I_{1}$, $x_{i}=1$ if $i\in I_{2}$, and $x_{i}=0$ if $i\in I_{0}^{(2)}-I_{2}$.
Note that $T_{k}$ acts on $\mathbb{R}^{n}-\{0\}$ through \begin{eqnarray*}&&\diag\{\lambda_1I_{2}+
\mu_{1}J,\dots,\lambda_{k}I_{2}+\mu_{k}J,\lambda_{2k+1},\dots,\lambda_n\}\cdot(x_1,..., x_n)^{t}
\\&=&((\lambda_{1}^{2}\!+\!\mu_{1}^{2})^{-1}(x_1\lambda_1\!+\!y_{1}\mu_1),(\lambda_{1}^{2}\!+\!
\mu_{1}^{2})^{-1}(-x_1\mu_1\!+\!y_{1}\lambda_1),\dots,\\&&(\lambda_{k}^{2}\!+\!\mu_{k}^{2})^{-1}
(x_{k}\lambda_{k}\!+\!y_{k}\mu_{k}),(\lambda_{k}^{2}\!+\!\mu_{k}^{2})^{-1}(-x_{k}\mu_{k}\!+\!
y_{k}\lambda_{k}),\\&&\lambda_{2k\!+\!1}^{-1}x_{2k\!+\!1},\dots,\lambda_n^{-1}x_n)^{t}.
\end{eqnarray*} It is easy to see that $\{v_{I_1,I_2}:\emptyset\neq I\subset\{1,...,n\}\}$
represent all different $T_{k}$ orbits in $\mathbb{R}^{n}-\{0\}$. This shows the lemma.
\end{proof}

For $(I_1,I_2)$ as in the above proof, define a matrix $g_{I_1,I_2}\in G$. In the case of $n\in I_{2}$,
let $$g_{I_1,I_2}=\left(\begin{array}{cc}I_{n-1}&0_{(n-1)\times 1}\\\beta^{t}&1\\\end{array}\right),$$
where $\beta^{t}=(x_1,\dots,x_{n-1})$ with $(x_{2i-1},x_{2i})=(0,1)$ if $i\in I_1$, $(x_{2i-1},x_{2i})
=(0,0)$ if $i\in I_{0}^{(1)}-I_1$, $x_{i}=1$ if $i\in I_{2}$, and $x_{i}=0$ if $i\in I_{0}^{(2)}-I_2$.

In the case of $I_{2}\not=\emptyset$ and $n\in I_{0}^{(2)}-I_2$, set $k'\!=\!\max\{1\leq\!i\leq\!n\!-
\!1\!:\!i\in I_2\}.$ Let $$g_{I_1,I_2}=\left(\begin{array}{cccc}I_{k'-1}&0_{(k'-1)\times 1}&0_{(k'-1)
\times(n-1-k')}&0_{(k'-1)\times 1}\\0_{1\times(k'-1)} &0&0_{1\times(n-1-k')}&1\\0_{(n\!-\!1\!-\!k')
\times(k'-1)}&0_{(n\!-\!1\!-\!k')\times 1}&I_{n\!-\!1\!-\!k'}&0_{(n\!-\!1\!-\!k')\times 1}\\\beta'^{t}
&1&0_{1\times(n-1-k')}&0\\\end{array}\right),$$ where $\beta'^{t}=(x_1,\dots,x_{k'-1})$ with $(x_{2i-1},
x_{2i})=(0,1)$ if $i\in I_1$, $(x_{2i-1},x_{2i})=(0,0)$ if $i\in I_{0}^{(1)}-I_1$, $x_{i}=1$ if
$i\in I_{2}$, and $x_{i}=0$ if $i\not\in I_{0}^{(2)}-I_2$.

In the case of $I_{2}=\emptyset$ and $\frac{n}{2}\in I_1$, let $$g_{I_1,I_2}=\left(\begin{array}{cc}
I_{n-1}&0_{(n-1)\times 1}\\\beta^{t}&1\\\end{array}\right),$$ where $\beta^{t}=(x_1,\dots,x_{n-1})$
with $(x_{2i-1},x_{2i})=(0,1)$ if $i\in I_1$ ($1\leq i<\frac{n}{2}$), $(x_{2i-1},x_{2i})=(0,0)$ if
$i\in I_{0}^{(1)}-I_1$ ($1\leq i<\frac{n}{2}$), and $x_{n-1}=0$.

In the case of $I_{2}=\emptyset$ and $\frac{n}{2}\not\in I_1$, set $k'\!=\!\max\{1\leq\!i<\!
\frac{n}{2}\!:\!i\in I_1\}.$ Let $$g_{I_1,I_2}=\left(\begin{array}{cccc}I_{2k'-1}&0_{(2k'-1)\times 1}
&0_{(2k'-1)\times(n-1-2k')}&0_{(2k'-1)\times 1}\\0_{1\times(2k'-1)} &0&0_{1\times(n-1-2k')}&1\\
0_{(n\!-\!1\!-\!2k')\times(2k'-1)}&0_{(n\!-\!1\!-\!2k')\times 1}&I_{n\!-\!1\!-\!2k'}&0_{(n\!-
\!1\!-\!2k')\times 1}\\\beta'^{t}&1&0_{1\times(n-1-2k')}&0\\\end{array}\right),$$ where $\beta'^{t}
=(x_1,\dots,x_{2k'-1})$ with $(x_{2i-1},x_{2i})=(0,1)$ if $i\in I_1$, $(x_{2i-1},x_{2i})=(0,0)$ if
$i\in I_{0}^{(1)}-I_1$, and $x_{2k'-1}=0$.

\begin{proposition}\label{P:double4}
$\{g_{I_1,I_2}\cdot f:(\emptyset,\emptyset)\neq (I_1,I_2)\subset(I_{0}^{(1)},I_{0}^{(2)})\}$ represent
all different $P$ orbits in $\mathcal{O}_{f}$. Among these orbits, $Pg_{I_{0}^{(1)},I_{0}^{(2)}}
\cdot f$ is an open and dense $P$ orbit, and its complement is a closed subset of codimension one
(in the case of $k\neq\frac{n}{2}$) or two (in the case of $k=\frac{n}{2}$).
\end{proposition}

\begin{proof}
One can show that $g_{I_1,I_2}^{-1}\cdot v_0=v_{I_1,I_2}$ for any $(I_1,I_2)$. By the proof of Lemma
\ref{L:double1}, this indicates that $$\{g_{I_1,I_2}\cdot f:(\emptyset,\emptyset)\neq (I_1,I_2)
\subset(I_{0}^{(1)},I_{0}^{(2)})\}$$ represent all different $P$ orbits in $\mathcal{O}_{f}$.

It is clear that $Tg_{I_{0}^{(1)},I_{0}^{(2)}}^{-1}\cdot v_0$ is an open and dense subset in
$\mathbb{R}^{n}-\{0\}$, and its complement is a Zariski closed subset of codimension one (in the
case of $k\neq\frac{n}{2}$) or two (in the case of $k=\frac{n}{2}$). Thus, $Pg_{I_{0}^{(1)},
I_{0}^{(2)}}\cdot f$ is an open and dense $P$ orbit, and its complement is a closed subset of
codimension one (in the case of $k\neq\frac{n}{2}$) or two (in the case of $k=\frac{n}{2}$).
\end{proof}

\begin{lemma}\label{L:moment2}
For any $(\emptyset,\emptyset)\neq (I_1,I_2)\subset(I_{0}^{(1)},I_{0}^{(2)})$, the element $p(g_{I_1,
I_2}\cdot f)$ is semisimple, its depth is equal to $2\#I_1+\#I_2$, and its eigenvalues are
$\{z_{2i-1},z_{2i},z_{j}:i\in I_{0}^{(1)}-I_{1},j\in I_{0}^{(2)}-I_2\}$.
\end{lemma}

\begin{proof}
Regard $\mathfrak{p}_{n}(\mathbb{R})$ as a real form of $\mathfrak{p}_{n}(\mathbb{C})$. Then, $\mathfrak{p}_{n}
(\mathbb{R})^{\ast}$ is naturally contained in $\mathfrak{p}_{n}(\mathbb{C})^{\ast}$ as a real form
of it. By the proof of Theorem \ref{T:Porbit}, we see that this imbedding does not change the depth
and eigenvalues. It is convenient to do conjugation regarding $P_{n}(\mathbb{C})$ to see the depth
and eigenvalues of $p(g_{I_1,I_2}\cdot f)$, and hence shows the lemma.
\end{proof}

The following theorem could be shown similarly as Theorem \ref{T:moment1}.
\begin{theorem}\label{T:moment2}
The set $p(\mathcal{O}_{f})$ consists of exactly $2^{n-k}-1$ semisimple $P$ orbits, with the unique
open $P$ orbit in $\mathfrak{p}^{\ast}$ among them. Moreover,  we have:
\begin{itemize}
\item[(1),] the moment map $p:\mathcal{O}_{f}\rightarrow\mathfrak{p}^{\ast}$ is \emph{weakly proper}.
\item[(2),] the \emph{reduced space} of the unique open $P$ orbit in $\mathfrak{p}^{\ast}$ with respect
to the moment map $p$ is a single point.
\end{itemize}
\end{theorem}

Analogous to the $\GL(n,\mathbb{C})$ case, in Theorem \ref{T:moment2} we write $$j=2\#I_1+\#I_{2}-1\in
[0,n-1]$$ for any $(\emptyset,\emptyset)\neq (I_1,I_2)\subset(I_{0}^{(1)},I_{0}^{(2)})$. Then, with
Lemma \ref{L:moment2} one can show that $$\Stab_{P}(g_{I_1,I_2}\cdot f)\cong\U(1)^{k-\!\#I_{1}}\!\times
\!(\mathbb{R}^{\times})^{n\!-\!k\!-\!\#I_1\!-\!\#I_2}.$$ On the other hand, Lemma \ref{L:Porbit-stabilizer}
and Lemma \ref{L:Porbit-stabilizer2} indicate that $$\Stab_{P}(p(g_{I_1,I_2}\cdot f))\cong\mathbb{R}^{n\!-
\!1\!-\!j}\!\rtimes(\U(1)^{k\!-\!\#I_1}\!\times\!(\mathbb{R}^{\times})^{n\!-\!k-\#I_1\!-\!\#I_2}).$$

\section{Kirillov's conjecture and orbit method}\label{Kirillov}

\subsection{General setting}

Let $G=\GL(n, \mathbb{K})$ with Lie algebra $\mathfrak{g}$. Let $$P=\{\left(\begin{array}
{cc}A&\alpha \\0&1\end{array}\right): A\in \GL(n_1,\mathbb{K}),\alpha\in\mathbb{K}^{n-1}\}$$
with Lie algebra $\mathfrak{p}$. Identify $\mathfrak{g}$ with its algebraic dual $\mathfrak{g}^{\ast}$
via the trace function $\text{tr} $. Let $T$ be the diagonal torus of $G$ with Lie algebra $\mathfrak{t}$.

Recall that by  Proposition \ref{P:openOrbit}, there is one and only one open $P$-coadjoint
orbit in $\mathfrak{p}^*$. For now on, we denote by $\Omega$, the unique open $P$-coadjoint orbit in
$\mathfrak{p}^{\ast}$; and by $\tau$, the (discrete series) representation of $P$ attached to $\Omega$.

Let $\pi$ be a tempered representation (with regular infinitesimal character) of $G$.
Then $\pi$ is associated to a strongly regular coadjoint orbit $\mathcal{O}_{\pi}$ (in the sense of Duflo).
Notice that since $G$ is reductive, all regular $G$-coadjoint orbits are strongly regular.  Let $\text{p}:
\mathcal{O}_{\pi}\rightarrow\mathfrak{p}^{\ast}$ be the moment map of the $P$-Hamiltonian space
$\mathcal{O}_{\pi}$. Then the following theorem serves not only as a geometric interpretation of Kirillov's
conjecture for tempered representations, but as a generalization of Duflo's conjecture (to tempered
representations):

\begin{theorem}\label{main theorem}

There are only finitely many $P$-orbits in $\text{p}(\mathcal{O}_{\pi})$, and the
unique open $P$- coadjoint orbit $\Omega$ is contained in $\text{p}(\mathcal{O}_{\pi})$. Moreover, we have

\begin{itemize}
\item[i)] The moment map $\text{p}: \mathcal{O}_{\pi}\rightarrow\mathfrak{p}^{\ast}$ is
\emph{weakly proper} .
\item[ii)] The restriction of $\pi$ to $P$, $\pi\vert_{P}$ (which is irreducible) is attached
to $\Omega$ (in the sense of Duflo). In other words, $\pi\vert_{P}=\tau$.
\item[iii)] The \emph{reduced space} of $\Omega$  (with respect to the moment map $\text{p}$)
is a single point.
\end{itemize}
\end {theorem}

Remarks: \begin{enumerate}
\item The weak properness in the above theorem "predicts" the $P$-admissibility (in the sense
of Kobayashi) of $\pi$.
\item Since there is  one and only one open $P$-coadjoint orbit (namely $\Omega$) in
$\text{p}(\mathcal{O}_{\pi})$,  it is also the unique strongly regular $P$-coadjoint orbit
contained in in $\text{p}(\mathcal{O}_{\pi})$.  This geometric fact "implies" that only the
irreducible representation of $P$ attached to  $\Omega$, namely  $\tau$, can appear in the
restriction $\pi\vert_{P}$.
\item The fact that reduced space of  $\Omega$ is a single point "signifies" that the multiplicity
of $\tau$ in $\pi\vert_{P}$ is no more than one (so is exactly one in our context).
\end{enumerate}

These three points together mean exactly $\pi\vert_{P}=\tau$. \\

We already proved, in previous section (Theorem \ref{T:moment1} and Theorem \ref{T:moment2}),
that except  for $\text{ii)}$, the Theorem  \ref{main theorem} is true for all regular
$G$-coadjoint orbits $\mathcal{O}\subset  \mathfrak{g}^*$ (which are not necessarily attached
to $G$-representations).  Then we only need to treat $\text{ii)}$ of the Theorem  \ref{main theorem}.

\subsection{Sahi's results on Kirillov's conjecture}

A first success for Kirillov's conjecture was accomplished by Sahi. He established (in \cite{Sahi})
Kirillov's conjecture completely for $\GL(n,\mathbb{C})$ and partially for $\GL(n,\mathbb{R})$.
In particular he proved the conjecture for all tempered representations $\pi$ of any $\GL(n,\mathbb{K})$
(for $\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$). In fact, for $\pi$ tempered, he determined more or less
explicitly  $\pi\vert_{P}$ as an irreducible representation of $P$ based on two functors "$I$" and
"$E$". Especially, it turns out that for any tempered representation $\pi$ of $\GL(n,\mathbb{K})$,
$\pi\vert_{P}$ is a same irreducible representation of $P$,  which is denoted  by "$I^{n-1}E 1$"
in Sahi's article. Let us briefly explain the construction of $I^{n-1}E 1$. For more details,
the reader is referred to \cite{Sahi}.

So write $G_n:=\GL(n,\mathbb{K})$ and $P_n$ the subgroup in question.  We have the two facts:

\begin{itemize}
\item[1)] $P_n \cong \mathbb{K}^{n-1}\rtimes G_{n-1}$.
\item[2)] $ G_{n-1}$ has exactly two orbits in $ (\mathbb{K}^{n-1})^*$: $\{0\}$ and $(\mathbb{K}^{n-1})^{\ast}
\setminus\{0\}$. Moreover if we fix a character $\xi\in (\mathbb{K}^{n-1})^*\setminus \{0\}$ by $\xi((x_1,\dots,
x_{n-1}))=x_{n-1}$, then $\text{Stab}_{ G_{n-1}}(\xi)\cong P_{n-1} $.
\end{itemize}

Then we deduce by Mackey's classic theory that each irreducible unitary representation of
$P_n$ is obtained in one of the two ways as follows:

\begin{itemize}
\item[a)] by trivially extending an irreducible unitary representation of $G_{n-1}$.
\item[b)] by extending an irreducible unitary representation of $P_{n-1}$ to $\mathbb{K}^{n-1}
\rtimes P_{n-1}$ by the a character $\xi$ and then inducing to $P_{n}$.
\end{itemize}

We use $E$ and $I$ for the above constructions a) and b) respectively. We can actually check that they are functors. Then we have $\widehat{P_n}=E(\widehat{G_{n-1}}) \bigsqcup I(\widehat{P_{n-1}})$. Moreover, using the convention that $P_1=G_0=$ the trivial group, we have the following fact: \\

Each irreducible unitary representation $\rho$ of $P_n$ is of the form: $ \rho=I^{k-1}E \sigma$
for some integer $k\geq 1$ and $ \sigma \in \hat{G_{n-k}}$. Moreover, $k$ and $\sigma$ are uniquely
determined. The integer $k$ is called by Sahi the \emph{depth} of $\rho$.

We have the following theorem due to Sahi:

\begin{theorem}\label{T: Sahi}

 Let $\pi$ be a tempered representation of $G_n$. Then we have $\pi\vert_{P_n}=I^{n-1}E 1$.

\end{theorem}

So in order to prove ii) of the Theorem \ref{main theorem}, we only need to prove that $I^{n-1}E 1$
is attached to the unique open $P_n$-coadjoint orbit $\Omega$, namely $I^{n-1}E 1=\tau$. We will
show it in the next subsection.

\subsection{Construction of $\tau$ in the framework of Duflo's orbit method}\label{SS: Duflo-Sahi}

As we mentioned, for any almost algebraic real groups, Duflo associated admissible and well
polarizable coadjoint orbits to irreducible unitary representations. However, the general
construction of irreducible representations attached to given orbits is carried out in an
indirect and inductive manner, and involves some non-trivial ingredients (e.g., some "metaplectic"
two-fold coverings, see  \cite{Duflo1}). Nevertheless, in our context, the construction of $\tau$,
namely the representation attached to the unique open $P_n$-orbit $\Omega$, is quite transparent
(though still by induction) in the framework of Duflo's theory, and it coincides with the classic
Mackey theory.

Now Let us explain how to construct $\tau$. We retain the notation in the preceding subsection.
Recall that $ P_n \cong \mathbb{K}^{n-1}\rtimes G_{n-1}$ and $\mathfrak{p}_{n}\cong
\mathbb{K}^{n-1} +  \mathfrak{g}_{n-1}$, with $\mathbb{K}^{n-1}$ the (abelian) nilradical of
$ \mathfrak{p}_n $.

Firstly, we will choose an element $f\in \Omega$, it is known that  the construction doesn't depend
on the choice of $f$. However, in order to be adapted to the construction of $I^{n-1}E 1$ by Sahi,
we choose a $f\in \Omega$ such that $f\vert_{\mathbb{K}^{n-1}}=\xi$.  According to the proof of Lemma
\ref{L:Porbit-inductive},  we know such $f$ exists, and  $\text{Stab}_{ P_{n}}(\xi)\cong
\mathbb{K}^{n-1}\rtimes P_{n-1} $. Note that in our case, the nilradical $\mathbb{K}^{n-1}$
(of $\mathfrak{p}_n$) is contained in $\text{Stab}_{ P_{n}}(\xi)$ (which implies that
$\text{Stab}_{ P_{n}}(\xi).\mathbb{K}^{n-1}=\text{Stab}_{ P_{n}}(\xi)$). Then in the framework of
Duflo's theory, we have

$$\tau=\text{Ind}^{P_{n}}_{ \mathbb{K}^{n-1}\rtimes P_{n-1}}(\xi \otimes \tilde{\tau}).$$

Here $\tilde{\tau} \in \widehat{P_{n-1}}$ is attached to the $P_{n-1}$-coadjoint orbit $P_{n-1}.\tilde{f}$,
with $\tilde{f}=f\vert_{\mathfrak{p}_{n-1}}$. However, again by the proof of Lemma \ref{L:Porbit-inductive},
we see that $P_{n-1}.\tilde{f}$ is the unique open $P_{n-1}$-coadjoint orbit in $(\mathfrak{p}_{n-1})^*$.
Then according to the construction of $I^{n-1}E 1$  and by a direct inductive argument, we obtain that
$\tau=I^{n-1}E 1$. \\

Remarks: \begin{enumerate}
\item As  $I^{n-1}E 1$ is attached to  the open orbit $\Omega$, it is a discrete series.
\item Sahi defined the \emph{depth} for $ \rho=I^{k-1}E \sigma$.  In previous section, for any
$P_n$-coadjoint orbit $\Xi$ , we also defined an integer called the \emph{depth} of $\Xi$. Actually
we can check that the two notions are compatible, in the sense that if $\Xi$ is attached to an irreducible
unitary representation $\rho$ (in the sense of Duflo), then the depth of  $\rho$  is that of $\Xi$.
\end{enumerate}

config: arxiv


TITLE: some manual calculations
QUESTION [0 upvotes]: I have a ratio between two probabilities
$$ \frac{(1-p)^{1000}+1000*p*(1-p)^{999}}{ \binom{1000}{2}*p^2*(1-p)^{998}}$$
and it is asked to show that this ratio is >1 (without a calculator)
$p=2/1000$
I arrived to this passage.
$$ \frac{(1-p)^{1000}+1000*p*(1-p)^{999}}{ \binom{1000}{2}*p^2*(1-p)^{998}}=
\frac{(1-p)^{2}+1000*p*(1-p)}{ 500*999*p}$$
Then the suggested solution says
$$ \frac{(1-p)^{2}+2*p*(1-p)}{ 999*p}$$
but i don't understant where does it come from?
it there something that I'm ignoring?

REPLY [1 votes]: Write ${1000 \choose 2}=\frac 12 \cdot 1000 \cdot 999$.  Divide numerator and denominator by $(1-p)^{998}$ and use the extra factor of $p$ in the denominator to cancel the $\frac {1000}2$


TITLE: Fourier transform derivation from Laurent series
QUESTION [3 upvotes]: Using Laurent Series of a function $f(z)$ around a point $a \in \mathbb{C}$
$$f(z) = \sum^{\infty}_{n=-\infty} c_n(z-a)^n \ \ \ \ (1)$$
where
$$c_n = \frac{1} {2\pi i}\int\limits_{\gamma}\frac {f(z)} {(z-a)^{n+1}} dz \ \ \ \ (2)$$
where $\gamma$ is a closed curve around $a$.
And choosing $\gamma$ such that $z$ can be parameterized as $z=a+e^{it}\  t \in [-\pi,\pi] $ in order to obtain Fourier Series
$$ f(t) = \sum^{\infty}_{n=-\infty} c_ne^{int} \ \ \ \ (3) $$
where
$$ c_n = \frac{1} {2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}dt \ \ \ \ (4) $$
we can "intuitively" prove Fourier Series by just using introductory complex calculus without any advanced mathematical background in Hilbert-Banach Spaces, Functional Analysis etc.
My question is, how can we derive other forms of Fourier Series (also called Fourier Transforms) such as Discrete Fourier Transform (DFT) and Continuous Time Fourier Transform (CTFT) by a similar manner? The problem here is that DFT equations are not closed contour integrals anymore, just discrete finite sums. I could not wrap my head around how Laurent Series may still work in this discrete case. For the CTFT case we have to let integral limits in $(4)$ to go infinity, which corresponds to infinitely many contour integrals in $(2)$. This seems to be divergent, if I am not terribly mistaken?

REPLY [2 votes]: Continuous time Fourier transform and Laurent series:
I recall equations (1) and (2), for convenience set $a=0$, and substitute $z=e^{it/T}$. The function $g(t)=f(e^{it/T})$, with $t\in(-\pi T,\pi T)$, is periodic with period $2\pi T$, given by the Laurent series
$$g(t)=\sum_{n=-\infty}^\infty c_n e^{int/T},$$
with coefficients
$$c_n=\frac{1}{2\pi T}\int_{-\pi T}^{\pi T}g(t)e^{-int/T}\,dt.$$
For $T\gg t$ the sum over $n$ may be approximated by an integral over $\omega=n/T$ with coefficients $C(\omega)=Tc_{n=\omega T}$, giving
$$g(t)=\int_{-\infty}^\infty C(\omega) e^{i\omega t}d\omega,$$
$$C(\omega) =\frac{1}{2\pi}\int_{-\infty}^\infty g(t)e^{-i\omega t}\,dt.$$
In this way the Fourier integral can be obtained as the limit of the Laurent series when the periodicity of the function tends to infinity. Notice that the exponent $nt/T$ cannot be set to zero because, even though $t/T\ll 1$, the product $nt/T=\omega t$ need not be small.

Discrete Fourier transform and Laurent series:
We now start from a discrete time signal $x_n$ and construct the Z-transform
$$X(z)=\sum_{n=-\infty}^\infty x_{n}z^{-n}=\sum_{n=-\infty}^\infty x_{-n}z^n.$$
This is a Laurent series centered at $a=0$, with inversion formula
$$x_{n}=\frac{1}{2\pi i}\oint X(z)z^{n-1}\,dz.$$
For the discrete Fourier transform one has only $N$ distinct values of $x_n$, the set $\{x_0,x_1,x_2,\ldots x_{N-1}\}$. We extend this set periodically by $x_{n+N}=x_n$. The Z-transform is a series of $N$ Dirac delta functions with coefficients $X_k$,
$$X(z=e^{2\pi iq/N})=\sum_{k=0}^{N-1}X_k\delta(q-k),\;\;X_k=\sum_{n=0}^{N-1}x_n e^{-2\pi i kn/N}.$$
The delta function converts the integral in the inversion formula into a sum,
$$x_n=\frac{1}{N}\sum_{k=0}^{N-1}X_k e^{2\pi ikn/N}.$$


TITLE: Spectrum of the ring $k[T]/(T^2)$
QUESTION [4 upvotes]: Consider the the ring $B = k[T]/(T^2)$ where $k$ is a field.
If $I$ is a prime ideal in $B$ then $I = (a + bT)$ for some $a,b \in
k$ (with $b \ne 0$). Then $T = a^{-1}T \cdot (a + bT) \in I$. Hence $(T) \subset I$ and
$(T^2) \subset (T) \subset I$. Since $b$ is invertible we can write
$$I = \left( b^{-1}(a+bT) \right) = \left( b^{-1}a + bb^{-1}T \right) = \left( c + T \right) \ . $$
Since $(T) \subset I$ we have that $(c + T , T) = I$ but that implies that $1 \in I$:
$$ I \ni c^{-1} (c + T) - c^{-1}T = 1 + c^{-1}T -c^{-1}T = 1 $$
if $c \ne 0$ and clearly $I = (T)$ also if $c = 0$.
Hence, $k[T]/(T^2)$ is a local ring with single prime and maximal ideal $I = (T)$. Note that $(0)$ is not a prime ideal as $T^2 = 0$ and $T \notin (0)$. Thus,
$$ \operatorname{Spec} k[T]/(T^2) = \{ (T) \}  $$
is a single point (as a set).
Is my analysis correct?

REPLY [4 votes]: Why not making your life easier? There is no need to calculate with polynomials.
We have homeomorphisms
$\mathrm{Spec}(k[T]/(T^2)) \cong V(T^2) = V(T) \cong \mathrm{Spec}(k[T]/(T)) \cong \mathrm{Spec}(k) = \{\eta\}$
It maps $\eta$ to the kernel of $k[T]/(T^2) \to k[T]/(T) \cong k$, which is $(T)$.
In general, if $A$ is a commutative ring with a maximal ideal $\mathfrak{m}$ and $n \geq 1$, then $A/\mathfrak{m}^n$ has a single prime ideal, namely $\mathfrak{m}/\mathfrak{m}^n$. Direct proof: The prime ideals of $A/\mathfrak{m}^n$ correspond to the prime ideals of $A$ containing $\mathfrak{m}^n$. But these have to contain $\mathfrak{m}$, i.e. have to be $=\mathfrak{m}$.


TITLE: Predicate logic problem using natural deduction
QUESTION [2 upvotes]: I have the following clauses:
$1. \forall x ({Ax \rightarrow Bx})$
$2. \forall x (({{Cx \wedge Bx }) \rightarrow Dx})$
$3. \forall x \exists y ({Cy \wedge Ryx})$
$4. \forall x \forall y ({({Ryx \wedge Dy }) \rightarrow Dx})$
And I need to prove:
$\forall x (({{ \forall y ({Ryx \rightarrow Ay })}) \rightarrow Dx})$
Using natural deduction.
I've tried to manipulate #4 but could not get it to look like the expected clause. Any clue is appericiated.

REPLY [2 votes]: $$\newcommand{\DeductionBox} [1]{\begin{array} {l|} #1 \\ \hline \end{array}}$$
The basic idea of the proof is, loosely:

for a free x, there is a set of ys satisfying Ryx
for all of those ys, Ay holds by assumption, and thus By holds
for at least one of those ys
  Cy holds
  so for that y, Dy holds
If Dy holds for some y, then Dx holds for the y

$$\begin{array} {l}
\begin{array} {rlr}
(1) & \forall \langle x,~ Ax \implies Bx \rangle                                          & \text{Given} \\
(2) & \forall \langle x,~ Bx \land Cx \implies Dx \rangle                                 & \text{Given} \\
(3) & \forall \langle x,~ \exists \langle y,~ Cy \land Ryx \rangle ~ \rangle              & \text{Given} \\
(4) & \forall \langle x,~ \forall \langle y,~ Ryx \land Dy \implies Dx \rangle ~ \rangle  & \text{Given} \\
\end{array}\\
\\
\DeductionBox{
  \begin{array} {rlr}
  \quad (5) & \forall \langle y,~ Ryx \implies Ay \rangle      &  \text{New Assumption}              \\
  \end{array} \\
  \\
  \DeductionBox{
    \begin{array} {rlr}
    \quad \quad (6)  & Cy \land Ryx                                         & \text{New Assumption}                              \\
    \quad \quad (7)  & Ryx                                                  & \land-\text{Elimination of }(6)                    \\
    \quad \quad (8)  & Ryx \implies Ay                                      &  \forall-\text{Elimination of }(5)                 \\
    \quad \quad (9) & Ay                                                    & \text{Modus Ponens of }(7)  \text{ and } (8)       \\
    \quad \quad (10) & Ay \implies By                                       & \forall-\text{Elimination of }(1)                  \\
    \quad \quad (11) & By                                                   & \text{Modus Ponens of }(9) \text{ and } (10)       \\
    \quad \quad (12) & Cy                                                   & \land-\text{Elimination of }(6)                    \\
    \quad \quad (13) & By \land Cy                                          & \land-\text{Introduction of }(11) \text{ and }(12) \\
    \quad \quad (14) & By \land Cy \implies Dy                              &  \forall-\text{Elimination of }(2)                 \\
    \quad \quad (15) & Dy                                                   & \text{Modus Ponens of }(13) \text{ and } (14)      \\
    \quad \quad (16) & Ryx \land Dy                                         & \land-\text{Introduction of }(7) \text{ and }(15)  \\
    \quad \quad (17) & \forall \langle y,~ Ryx \land Dy \implies Dx \rangle & \forall-\text{Elimination of }(4)                  \\
    \quad \quad (18) & Ryx \land Dy \implies Dx                             & \forall-\text{Elimination of }(17)                 \\
    \quad \quad (19) & Dx                                                   & \text{Modus Ponens of }(16) \text{ and } (18)      \\
    \end{array}
  } \\
  \\
  \begin{array} {rlr}
  \quad (20) & \exists \langle y,~ Cy \land Ryx \rangle   &  \forall-\text{Elimination of }(3)                               \\
  \quad (21) & Dx                                         &  \exists-\text{Elimination of }(20) \text{ and } (6) \vdash (19) \\
  \end{array}
} \\
\\
\begin{array} {rlr}
(22) & \forall \langle y,~ Ryx \implies Ay \rangle \implies Dx                              &  \text{Deduction of }(5) \vdash (21) \\
(23) & \forall \langle x,~ \forall \langle y,~ Ryx \implies Ay \rangle \implies Dx \rangle  &  \forall-\text{Introduction of }(22) \\
\end{array}
\end{array}$$
The possible parsing of the original question, $\forall \langle x,~ \forall \langle y,~ (Ryx \implies Ay) \implies Dx \rangle ~\rangle$, is not provable, which can be established with the model:
$$\Omega = \{1,~ 2\}$$
$$R = \{(1,~ 1),~ (1,~ 2),~ (2,~ 1),~ (2,~ 2)\}$$
$$A = \{1\}$$
$$B = \{1\}$$
$$C = \{2\}$$
$$D = \{~\}$$


TITLE: Is this a vector?
QUESTION [0 upvotes]: Can anyone help me with this probability question?
Given two independent random variables $Y \sim N(\mu,\sigma)$ and $Z \sim U[0,3]$, we define
$X=Y+Z$
Prove or disprove: The vector $(Z,X)^T$ is a vector.
I would appreciate any help :)
Thanks

REPLY [0 votes]: The vector $V=(V_1,V_2)^T=(Z,-(-1)^I |Z|)^T=(Z,\pm |Z|)$ is not a 2-dimensional random Gaussian vector.
If it were, the linear combination $V_1-V_2$ would have a Gaussian distribution.  But with probability $1/4$ we have $I=1$ and $Z>0$, in which case $V_1=V_2$.  Hence $P(V_1=V_2)\ge 1/4$.  Similarly with probability  $1/4$ we have $I=0$ and $Z>0$, in which case $V_1=-V_2$, so $P(V_1\ne V_2)\ge 1/4$, so $P(V_1=V_2)\le 3/4$. These imply that $V_1-V_2$ is a non-degenerate random variable whose distribution has a discrete component.  Hence it is not Gaussian.  Hence the vector $V$ is not a bivariate Gaussian vector.
That is to say, since $I$ and $Z$ are independent, $P(I=1\text{ and }Z>0)=P(I=1)\times P(Z>0)$, which, by your assumptions, equals $1/2\times 1/2=1/4$. Thus, the chance that $V_1-V_2=0$ is at least $1/4$.  Similarly, the chance that $V_1\ne V_2$ is at least equal to the chance that $Z>0$ and $I=0$, which is also $1/4$. Gaussian distributions do not assign positive probability to singleton points like this.
Your distribution forces $|V_1|=|V_2|$; its support is the union of two lines in the plane.  Gaussian distributions in the plane, however, are supported on a singleton point, on a line, or on the  whole plane.


TITLE: Counting by prime factors and "base prime" notation
QUESTION [0 upvotes]: Question is whether clear patterns exist in the $p_n$ "digit" of a factorization notation where each digit stands for the power of the prime in that digit's position.
For example, one would count from 2 to 17 as follows:
1, 10, 2, 100, 11, 1000, 3, 20, 101, 10000, 12, 100000, 1001, 110, 4, 1000000
What pattern, if any, would we expect to see from the right-most digit in this notation (the power of 2 in the factorization). Or in the 2nd right-most digit (the power of 3 in the factorization).
If this notion pre-exists, would appreciate a pointer to it, as I am not sure how to describe it in an unknown unknown phraseology.

REPLY [2 votes]: I believe Ross Millikan is correct.  But you did ask about notation (I think -- your last sentence says "notion" but the subject line said "notation"), as well.  So I wanted to answer that part -- the "unknown phraseology" you mention :)
There's a handful of systems people have used for this number system; https://www.reddit.com/r/baseprime/comments/id341b/other_posts_about_prime_notation_systems_base/ has some links to some (https://www.reddit.com/r/puremathematics/comments/f7ayn0/base_prime_written_in_base_prime/ and https://www.reddit.com/r/askscience/comments/58i1ww/would_a_baseprime_number_system_be_useful_in_any/ are maybe the two most interesting links there) but I've always favored using list notation for this, to relieve theambiguity pointed out by agarttha:
1 -> {}
2 -> {1}
3 -> {0,1}
4 -> {2}
5 -> {0,0,1}
etc.
I really want to submit a comment, too, suggesting alternate ways of looking for patterns, in case that would be helpful for whatever topic you are investigating, but you have to have a reputation of 50 to submit a comment... so I'm just going to leave this here as an addendum on my answer submission:
How about putting the natural numbers into a (nearly) binary tree?  Instead of constructing the "base prime" numbers from the naturals, go the other way around.  If you are looking for a relationship between a sequence of numbers and patterns in complex representations, sometimes it is helpful to approach from the opposite direction: create patterns in those complex representations which are bijective with your original sequence, and then work backwards to see how that maps to your original sequence.
For example, in "base prime," start with 1 and define two operations: (1) adding a "0" to the end (so shifting every prime exponent up to the next-higher prime factor), (2) increasing the first digit by 1 (raising the exponent in the 2's place).  If you build a tree this way, where each node has a right branch (the 0 concatenation) and a left branch (the increment of the 2's place), you'll (eventually) get all whole numbers greater than or equal to 2.  I have a diagram where I did a few rows of this, to help explain it visually; note that I used the aforementioned list notation (and threw in a 0 at the top, which is the "base prime" representation of 1), but you can view it at http://copland.udel.edu/~os/baseprime-bintree-ljpost-binarytree.png to see it.  Once you have this binary tree constructed, just walk horizontally across each row from top to bottom and you'll have an ordering which does some odd, pattern-like things, but not completely.  It starts out (if you walk from right to left) 2; 3, 4; 5, 6, 9, 8: 7, 10, 15, 12, 25, 18, 27, 16; 11, 14, 21, 20, 35, 30, 45, 24, 49, 50, 75, 36, 125, 54, 81, 32; ... (and you can calculate the mean of each row, for example: 2; 3.5; 7; 16.25; 43.875; ... which seem to follow a pattern of semi-steadily diverging ratios -- something which could warrant further study)


TITLE: Is $\sum_{k=0}^{\infty}\frac1{2^{k^2}}$ rational? Transcendental?
QUESTION [8 upvotes]: Is $\sum_{k=0}^{\infty}\frac1{2^{k^2}}$ rational?
Clearly this series is convergent (compare to geometric series with ratio 1/2). I'm sure it's irrational since a rational number written in base 2 will have either a terminating or repeating decimal representation. But the hard part is to show this representation in question doesn't repeat. (cf
https://www.google.com/search?q=periodic+rational+base&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a )
Can you show this number is transcendental?

REPLY [1 votes]: The question was asked, and a reference to a proof of transcendentality was given, at the accepted answer here at MO.
Actually, it seems it has also been answered here at m.se.


TITLE: The largest clique in $X\times Y$ is the smallest of the largest cliques in $X$ and $Y$
QUESTION [0 upvotes]: From the book Algebraic Graph Theory by Godsil and Royle

Show that $\omega(X\times Y)$ is the minimum of $\omega(X)$ and $\omega(Y)$.

Here $\omega(X)$ denotes the size of the largest clique in $X$.
This reminded me of the Hadetniemi conjecture
$\chi(X\times Y) = min\{\chi(X), \chi(Y)\}$
Since it is a conjuecture we will instead use the weaker (proven) statement:
$\chi(X\times Y) \le min\{\chi(X), \chi(Y)\}$
Clearly if $X$ and $Y$ are graphs such that $n = \chi(X) \le \chi(Y) = m$, we have that $\chi(X\times Y) \le n$ and therefore $\omega(X \times Y) \le n$.

This is as far as I got, the only bound I could get for $\omega(X\times Y)$, was that $\omega(X\times Y) \le \omega(X)$ and $\omega(X\times Y) \le \omega(Y)$. What I'm trying to do now is find that $\min\{\omega(X), \omega(Y)\}$ is a lower bound for $\omega(X\times Y)$, but I'm stumped.
How can I provide a lower bound for $\omega(X\times Y)$?

REPLY [1 votes]: This is just to show $\min\{\omega(X), \omega(Y)\} \leq \omega(X\times Y)$.
Hint: Consider a clique $J\leq X$ and a clique $K\leq Y$. Can you find a clique of size $\min\{|J|, |K|\}$ in the subgraph $J\times K$ of $X\times Y$?
You might find it helpful to label $J = \{u_1, u_2, \dots, u_j\}$ and $K = \{v_1, v_2, \dots, v_k\}$, and draw a grid $J\times K = \{(u_i, v_n): 1\leq i\leq k, 1\leq n\leq j\}$, noting that two vertices are adjacent if and only if they are not in the same row or column.
Answer in the box below:

 Let $|J| = j$ and $|K| = k$. Assume without loss of generality that $j\leq k$. Then the diagonal entries $\{(u_1, v_1), (u_2, v_2), \dots, (u_j, v_j)\}$ form a clique of size $j$.


TITLE: pairwise disjoint subsets of divisors of $ n $ (maximum number)
QUESTION [1 upvotes]: Let $ n \in \mathbb{N} $, $ n>1 $ and $ a_1,\ldots,a_k \in \mathbb{N} $ (not necessarily distinct!) with $ a_i \mid n $ for all $ i=1,\ldots,k $ be given. Assume that $ \sum_{i=1}^k a_i = K\cdot n $ for some $ K \in \mathbb{N} $. A subset $ M \subseteq \lbrace 1,\ldots,k \rbrace $ is called strong, if $ \sum_{i \in M} a_i \ge n$. Let $ m:=m(n,a) \in \mathbb{N} $ denote the maximum number of pairwise disjoint strong subsets.

Clearly $ m\le K $. In which cases we have $ m=K $? Is there a characterization?
Is it always possible to find at least $ K-1 $ pairwise disjoint strong subsets?
Is there a closed expression for $ m(n,a) $, depending on the input data $ n $ and the vector of divisors $ a=(a_1,\ldots,a_k)^{\top} $?

If anyone has an idea, please write it. Thanks for your help.
(This question arises e.g. in covering problems. With the help of some small pieces, the $ a_i $'s, as many items of length $ n $ as possible have to be covered.)

REPLY [0 votes]: As an initial step one could do the following:
Let us consider minimal strong subsets $ M \subseteq \lbrace 1,\ldots,k \rbrace $, i.e., no real subset $ M' \subset M $ is strong. For such a set obviously $ \sum_{i\in M} a_i  \le n-1+a^\star $ holds, where $ a^\star:= \max \left\lbrace a_i \, : \, a_i < \max_{j \in \lbrace 1\ldots,k \rbrace} \lbrace a_j \rbrace \right\rbrace $, i.e., the second largest (with resprect to "$ < $") number $ a_i $.
Thus we have:
$ m \ge \left\lfloor \frac{n\cdot K}{n-1+a^\star} \right\rfloor $
But this is only a beginning. I hope, there are other people who know stronger bounds for $ m $.


TITLE: Radon transform inversion formula
QUESTION [1 upvotes]: I have trouble following Deans's derivation of the inverse Radon transform formula for $n=2$ on this page of his book "The Radon Transform and Some of its Applications" (see snapshot)
Formulas (3.9) and (3.10) make sense, but I can't quite see the change of variables used to get (3.11) and (3.12)?
Any insight appreciated, thanks!
p.

REPLY [1 votes]: Let us represent points on the plane using polar coordinates $(p,\phi)$, where the radial coordinate $p$ is allowed to be negative. Then, for any function $F(p,\phi)$, it follows that $F(p,\phi+\pi) = F(-p,\phi)$. Using this property,
\begin{align}
\int_0^{2\pi} \int_{-\infty}^{\infty} F(p,\phi) \, dp \, d\phi &=
\int_0^{\pi} \int_{-\infty}^{\infty} F(p,\phi) \, dp \, d\phi
+ \int_0^{\pi} \int_{-\infty}^{\infty} F(p,\phi + \pi) \, dp \, d\phi \\
&= \int_0^{\pi} \int_{-\infty}^{\infty} F(p,\phi) \, dp \, d\phi
+ \int_0^{\pi} \int_{-\infty}^{\infty} F(-p,\phi) \, dp \, d\phi \\
&= \int_0^{\pi} \left[\int_{-\infty}^{\infty} F(p,\phi) \, dp +  \int_{-\infty}^{\infty} F(-p,\phi) \, dp \right] d\phi \\
&= \int_0^{\pi} \left[2 \int_{-\infty}^{\infty} F(p,\phi) \, dp \right] d\phi \\
&= 2 \int_0^{\pi} \int_{-\infty}^{\infty} F(p,\phi) \, dp \, d\phi.
\end{align}
Then (3.11) follows immediately if you set the integrand of (3.10) as $F(p,\phi)$ in the equation above. Equation (3.12) can be derived similarly using the relation above, except you will need to define the $\phi$-integral over $[-\pi/2,3\pi/2]$ and then split it at $\phi = \pi/2$.


TITLE: a.s. convergence and conditional expectation
QUESTION [2 upvotes]: I have a stochastic process $X(t,\omega)$ which is a martingale. It is showed that there exists a r.v. $X(\infty)$  such that in $L^1(\Omega)$, $\lim_{t \rightarrow \infty}X(t) = X(\infty)$. In my book it is then stated that:
$X(t) =_{a.s} E[\lim _{s\rightarrow \infty}X(s+t))|\mathcal{F}_t]=E[X(\infty)|\mathcal{F}_t]$. It is said that this is the case because the conditional expectation is continuous in $L^1(\Omega)$.
But I don't really see how. I get that In $L_1$ convergence we have that $E[X(s+t)|\mathcal{F}_t] \rightarrow_{s \rightarrow \infty} E[X(\infty)|\mathcal{F}_t]$. This follows from $E[|E[X(s+t)|\mathcal{F}_t]-E[X(\infty)|\mathcal{F}_t|]\le E[|X(s+t)-X(\infty)|]$, and the last converges to 0 by the $L^1$ convergence.
But how do we get the a.s. convergence?
Here is the argument in my book, I scanned it for you(For clarity $\mathcal{H}^2$ in the argument is the space of stochastic processes where $E[(\sup_t|X_t|)^2]<\infty$):.
http://i.stack.imgur.com/jI0WD.jpg
http://i.stack.imgur.com/DqAPD.jpg

REPLY [1 votes]: By the $L^1$-convergence, we have
$$L^1-\lim_{s \to \infty} \mathbb{E}(X(t+s) \mid \mathcal{F}_t) = \mathbb{E} \left( X(\infty) \mid \mathcal{F}_t \right). \tag{1}$$
On the other hand, as $(X(t))_{t \geq 0}$ is a martingale, we have
$$\mathbb{E}(X(t+s) \mid \mathcal{F}_t) = X(t)$$
for all $s \geq 0$. Hence, trivially,
$$L^1-\lim_{s \to \infty} \mathbb{E}(X(t+s) \mid \mathcal{F}_t) = X(t). \tag{2}$$
Since $L^1$-limits are unique (up to a null set), we find by combining $(1)$ and $(2)$ that
$$X(t) = \mathbb{E}(X(\infty) \mid \mathcal{F}_t) \quad \text{almost surely}$$
... and that's exactly what is claimed in your textbook.


TITLE: $p\ \text dq$ is the "tautological" one-form?
QUESTION [17 upvotes]: The one-form
$$\theta=\sum_i p_i\, \text dq^i$$
is a central object in hamiltonian mechanics. It has a bunch of applications: $\omega=\text d\theta$ is the symplectic structure on phase space, $S=\int\theta$ is the classical action, and so on and so forth. It is associated with the names Liouville one-form, Poincaré one-form, canonical one-form, and symplectic potential, none of which surprises me, but its Wikipedia entry informs me that the preferred[by whom?] name for it is actually "tautological" one-form, on the grounds that 'canonical' (which would be my natural choice) is 'already heavily loaded', and because of the risk of confusion with some algebraic thingammy.
This name completely mystifies me. Why was the name "tautological" chosen for this object? When, where, and by whom? Or was this name chosen because that's its name?

REPLY [1 votes]: Here's how it finally clicked for me. Let $\phi: M \rightarrow N$ be a smooth map, and $\alpha$ be a 1-form on $N$. Recall that the pullback of $\alpha$ by $\phi$ is a 1-form $\phi^*\alpha$ on $M$ defined by
$$(\phi^*\alpha)_x(v) = \alpha_{\phi(x)}(d\phi_x(v))$$
where $d\phi$ is the pushforward of $\phi$.
When $M$ is $T^*N$, something nice happens. We find that a point $(x, \alpha$) on $T^*N$ itself gives us a 1-form ($\alpha$) to pull back, and the canonical projection $\pi$ gives us something obvious to pull back by. ($\alpha$ isn't actually a 1-form, as another answer says, but a single covector. Nonetheless, it can be pulled back in the same way.) So we get a 1-form
$$\theta_{(x, \alpha)}(v) = \alpha(d\pi_{(x,a)}(v))$$
or in other words
$$\theta_{(x, \alpha)} = \alpha \circ d\pi_{(x,a)}$$
without having to make any additional choices. It is in this sense canonical (though why they call it "tautological" I don't know).