TITLE: Does solving Schroedinger Equation means finding Eigenstates?
QUESTION [0 upvotes]: While one day I was in the student room. One of my colleagues made this statement:

There are many ways to solve Schroediner's Equation. It is just because finding eigenvalues is easy so we solve the Schroedinger's Equation by finding the eigenvalues of the observable.

Is that true?

REPLY [2 votes]: In this fundamental example, we will study a particle in a potential well using SE but without introducing the methods of eigenstates or eigenvalues.
We will only work with a unidimensional model, consider a potential well of width $a$ ($V(x)=0$ for $x$ in $[0,a]$ and $\infty$ elsewhere) and the particle has energy $E>0$:
Then using SE in stationary state we have:
$$
-\frac{{\hbar}^2}{2m}\frac{d^2\phi}{dx^2}+V(x)\phi(x) = E\phi(x)\\
\frac{{\hbar}^2}{2m}\frac{d^2\phi}{dx^2}+(E-V(x))\phi(x) =0
$$
For $x<0$ and $x>a$, the potential $V$ is infinite, so, if we want to make sense of $\phi$ and its second derivative, then they must be finite since they are continuous, hence, we find that $\phi(x)=0$, and by continuity we conclude that:
$\phi(a)=\phi(0)=0$
Now let's move to the domain $[0,a]$, in this latter the SE simplifies to:
$$
\frac{{\hbar}^2}{2m}\frac{d^2\phi}{dx^2}+E\phi(x)=0\\
\frac{d^2\phi}{dx^2}+\frac{2mE}{{\hbar}^2}\phi(x)=0
$$
We put $k^2=\frac{2mE}{{\hbar}^2}>0$, we observe that SE reduces to a simple seconder order DE, and its general solution is:
$$
\phi(x) = A\cos(kx)+B\sin(kx)
$$
Using the conditions we found earlier, we'll determine the constants:
$$
\phi(0)=A=0
$$
So $\phi$ becomes:
$\phi(x) = B\sin(kx)$
And using the second one we obtain:
$\phi(a) = Bsin(ka)=0$
Clearly the case $B=0$ doesn't satisfy the normalzation condition, so:
$\sin(ka)=0$
Which yields:
$k_n=\frac{n\pi}{a}$
So, our solution is:
$\phi(x) = B\sin(k_{n}x)$
And you can get $B$ from the normalization constant.
I hope this helps!


TITLE: Value of Definite Integral
QUESTION [2 upvotes]: Since $\sin\left(\frac{1}{x}\right)$ is bounded and continuous on $(0, 1]$ the proper (Riemann-)integral $\int_0^1\sin\left(\frac{1}{x}\right)dx$ exists and is independent of the value prescribed at $x=0$. What is the value of this integral?

REPLY [0 votes]: There is no solution in terms of elementary functions. Mathematica gives :
$$\int_0^1\sin({1\over x})\,dx=\sin(1)−\hbox{CosIntegral}(1),$$ where CosIntegral($x$) is defined as $−\int_x^{\infty}{\cos t\over t}\,dt$


TITLE: Limit of sum of squares of sine
QUESTION [1 upvotes]: I am trying to get $\lim \limits_{n \to \infty} {\sin^{2}1 + \sin^{2}2 + \dots \sin^{2}n \over n}$. I get stuck on rewriting the numerator using trig identities. Any suggestions appreciated.

REPLY [3 votes]: I'll explain in more detail what Ian Miller said. The limit is $\frac 12$ as the sine is bounded. You get the first equality by using $cos(x+y)=\text{cos} (x)\text{cos} (y)-\text{sin}(x)\text{sin}(y)$. The second equality comes from Lagrange's trigonometric identity for sums of cosines.


TITLE: Number of ways can the 8 people occupy the seats if no two sit in the same pair of seats
QUESTION [1 upvotes]: A bus has 30 seats arranged in 15 pairs of seats.In how many ways can the 8 people occupy the seats if no two sit in the same pair?

My Solution:
$\binom{15}{8}$*${2}^{8}$
Choose 8 pairs of seats from 15. In each pair, the person can be seated in 2 possible ways.
Is this correct?

REPLY [1 votes]: Assuming that the people are interchangable (i.e. if you take two people and just change one for the other), you are correct.
Otherwise, you would have to multiply the solution by $8!$, and you get $66,421,555,200$ total ways (10 times more than the population of earth).


TITLE: Friedmann equations with $w <-1$
QUESTION [5 upvotes]: Let's consider a flat universe with an FRW metric with scale factor $a(t)$, with some matter content. The continuity equation $\nabla_\mu T^{\mu\nu}=0$ combined with assumptions of isotropy and homogeneity implies the equation $\frac{d\rho}{dt}=-3\frac{\dot a}{a}(\rho+p)$. Now, if we consider equations of state of the form $p=w\rho$, we can solve the above equation for $\rho$ to obtain $\rho\propto a^{-3(1+w)}$. One usually considers the cases $w=-1,0,\frac{1}{3}$, and there aren't many problems there (except $\rho=const.$ for $w=-1$). But if one puts $w<-1$, it appears that the energy density increases with the scale factor! What's worse, a somewhat straightforward calculation would imply that the scale factor decreases with time. This kind of substance violates all the energy conditions, so I'm not sure how "physical" it is. I've tried integrating the Friedmann equations for $K\neq 0$, but they seem a bit complex for an analytical solution. Does this imply that a substance with $w<-1$ can't exist in a FRW universe, or is one of my assumptions wrong ($K=0$ comes to mind)?

REPLY [3 votes]: Under the assumptions that $a > 0$ and that the universe is expanding, we can derive some interesting results about the fate of such a universe.
From the Friedmann equations alone, we may derive
$$
\frac{d}{d \tau} (\rho a^3) = - P \frac{d}{d \tau} (a^3).
$$
For $P = w \rho$, as long as $w \neq -1$, this yields
$$
\rho \propto \frac{1}{a^{3(1 + w)}},
$$
exactly as you stated in your question. So, yes, if the universe is expanding and $w < -1$, then the energy density does increase with time!
In the flat case with no cosmological constant, $\Lambda = K = 0$, we may integrate the first Friedmann equation $3 H^2 = 8 \pi \rho$, with this expression for $\rho$, to yield
$$a(\tau) \propto \tau^{\frac{2}{3(1+w)}},$$
exactly as in the case $w > -1.$ The Hubble parameter is thus given by
$$
H = \frac{\dot{a}}{a} = \frac{3}{3 (1 + w) \tau}.
$$
In an expanding universe, we have $H > 0.$ Since $1 + w < 0$, we must have $\tau < 0$ in order for the universe to be expanding. But since $a(\tau) \propto \tau^{\frac{2}{3(1+w)}}$ and $w < -1,$ the scale factor diverges at $\tau = 0.$ So the universe will suffer a "big rip" singularity at some finite time in the future.
In a contracting universe, we have $H < 0,$ and so $\tau > 0.$ Since this is past the $\tau=0$ singularity, such a universe must have originated at a "big rip" at some finite time in the past.
So if we assume that the universe has existed for finite time, then it must be contracting (as you stated in your question), and it must have originated with a divergent scale factor. On the other hand, if we assume that the universe is expanding, then it will meet a singularity in finite time as the scale factor diverges.


TITLE: What is the probability that the mouse ever gets to eat the cheese?
QUESTION [0 upvotes]: I'm doing an exercise about Markov chain.

A merry mouse moves in a maze. If it is at time $n$ in a room with $k$ horizontal or vertical adjacent rooms, it will be at time $n+1$ in one of the $k$ adjacent rooms, choosing one at random, each with probability $1 / k$. A fat lazy cat remains all the time in room $3,$ and a piece of cheese waits for the mouse in room $5$ The mouse starts in room $1$. See the following figure:

The cat is not completely lazy: If the mouse enters the room inhabited by the cat, the cat will eat it. Also, if the mouse eats the cheese, it rests forever. Let $X_{n}$ be the position of the mouse at time $n$.
What is the probability that the mouse ever gets to eat the cheese?

From the graph, the transition matrix is as follows:
$$P=\begin{pmatrix}0 & 1/2 & 0 & 1/2 & 0 \\
1/2 & 0 & 1/2 & 0 & 0 \\
0 & 1/2 & 0 & 1/2 & 0 \\
1/3 & 0 & 1/3 & 0 & 1/3 \\
0 & 0 & 0 & 1 & 0 \\
\end{pmatrix}$$
Then the probability that the mouse ever gets to eat the cheese is $$\mathbb P \left (\forall n \in \mathbb N:X_n \neq 5 \right )$$
Could you please leave me some hints to compute this probability? Thank you so much!

REPLY [2 votes]: The transition matrix should actually be
$$
P = \begin{pmatrix}
0 & 1/2 & 0 & 1/2 & 0\\
1/2 & 0 & 1/2 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
1/3 & 0 & 1/3 & 0 &1/3\\
0 & 0 & 0 & 0 & 1
\end{pmatrix}.
$$
Let $X_n$ be the state the mouse is in at time $n$, then $\{X_n:n=0,1,2,\ldots\}$ is a Markov chain on $\{1,2,3,4,5\}$ with the above transition probabilities. We want to compute $\mathbb P(\tau_1)$, where
$$
\tau_i = \inf\{n>0: X_n = 5 \mid X_0 = i\},\quad i\in\{1,2,3,4,5\}.
$$
It is clear that $\mathbb P(\tau_3)=0$ and $\mathbb P(\tau_5)=1$. To compute the remaining $\tau_i$ we have the system of linear equations
\begin{align}
\mathbb P(\tau_1) &= P_{12}\mathbb P(\tau_2) + P_{14}\mathbb P(\tau_4)\\
\mathbb P(\tau_2) &= P_{21}\mathbb P(\tau_1)\\
\mathbb P(\tau_4) &= P_{41}\mathbb P(\tau_1) + P_{45}.
\end{align}
Substituting in the transition probabilities, we have
\begin{align}
\mathbb P(\tau_1) &= 1/2\mathbb P(\tau_2) + 1/2\mathbb P(\tau_4)\\
\mathbb P(\tau_2) &= 1/2\mathbb P(\tau_1)\\
\mathbb P(\tau_4) &= 1/3\mathbb P(\tau_1) + 1/3.
\end{align}
This yields
$$
\mathbb P(\tau_1) = \frac27,\quad \mathbb P(\tau_2) = \frac17,\quad \mathbb P(\tau_4) = \frac37.
$$
Indeed, if we compute $\lim_{n\to\infty}P^n$ (which exists since $P$ is aperiodic), we get
$$
\left(
\begin{array}{ccccc}
 0 & 0 & \frac{5}{7} & 0 & \frac{2}{7} \\
 0 & 0 & \frac{6}{7} & 0 & \frac{1}{7} \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & \frac{4}{7} & 0 & \frac{3}{7} \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right),
$$
and the desired $\mathbb P(\tau_i)$ values can be read off the last column of this matrix.


TITLE: What is wrong with my algorithm for finding how many positive integers are divisible by a number d in range [x,y]?
QUESTION [1 upvotes]: I have been solving basic counting problems from Kenneth Rosen's Discrete Mathematics textbook (6th edition).  These come from section 5-1 (the basics of counting), pages 344 - 347.
This question is not specifically about finding an answer to a problem or being given the correct equation, but whether my reasoning is sound.  Therefore I would find it hard to argue this is a duplicate of seemingly similar questions like this one or this one.
The problems I have been dealing with come of the form How many positive integers in range [x,y] are divisible by d?  All additional questions are based on the composition of the information learned in these, e.g. how many positive integers in range [x,y] are divisible by d or e?
To answer the simple question I wrote this "equation/algorithm," which takes as input an inclusive range of positive integers $[x,y]$ and a positive integer $d$, and returns $n$, the total number of positive integers in range $[x,y]$ which are divisible by $d$.
(1) $n = \left \lfloor{\frac{y}{d}}\right \rfloor - \left \lfloor{\frac{x}{d}}\right \rfloor$
The idea is that in order to count how many positive integers are divisible by $d$ from $[1,m]$, we simply calculate $\left \lfloor{\frac{m}{d}}\right \rfloor$, because every $dth$ positive integer must be divisible by $d$.  However, this does not work when given a range $[x,y]$ where $x \not= 1$ or when $x > 1$.  So we need to subtract the extra integers we counted, which is $\left \lfloor{\frac{x}{d}}\right \rfloor$, i.e. the number of positive integers divisible by $d$ from $[1,x]$.
For a sanity check, I also wrote a brute force algorithm that does a linear search over every positive integer in the range $[x,y]$ and counts it if $x \text{ mod } d == 0$.  It also can list out the integers it picked, in case I am feeling really paranoid.
With (1) I've been getting the correct answers except on this problem/input: How many positive integers between 100 and 999 inclusive are odd? My solution was to calculate how many are even, and subtract this from the total number of positive integers in range $[100,999]$.  To find the evens I simply use the algorithm in (1):
$\left \lfloor{\frac{999}{2}}\right \rfloor - \left \lfloor{\frac{100}{2}}\right \rfloor = 499 - 50 = 449$
But this answer is wrong, since there actually $450$ even numbers in range $[100,999]$ by the brute force algorithm.  (1) is somehow counting off by 1.  My question is, why is (1) failing for this input of $(2, [100,999])$ but so far it's worked on every other input?  What do I need to do to fix (1) so it produces the correct answer for this case?  Perhaps I'm actually over counting because $x$ should actually be $x - 1$?
(1') $n = \left \lfloor{\frac{y}{d}}\right \rfloor - \left \lfloor{\frac{x - 1}{d}}\right \rfloor$
(1') returns the correct answer for this specific input now, but I am not sure if it will break my other solutions.

REPLY [1 votes]: After computing the number of positive multiples of $d$ less than or equal to $y,$ you correctly want to subtract the multiples that are not actually in the range $[x,y].$ Those are the multiples that are less than $x.$ When $x$ is divisible by $d,$ then $\left\lfloor \frac xd \right\rfloor$ counts all multiples of $d$ up to and including $x$ itself. So as you surmised, you want to subtract $\left\lfloor \frac {x-1}d \right\rfloor$ instead. This is true for any divisor, not just $d=2.$


text:
{\bf Problem.} When the vectors $\begin{pmatrix} -5 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ are both projected onto the same vector $\mathbf{v},$ the result is $\mathbf{p}$ in both cases.  Find $\mathbf{p}.$
{\bf Level.} Level 5
{\bf Type.} Precalculus
{\bf Solution.} Note that the vector $\mathbf{p}$ must lie on the line passing through $\begin{pmatrix} -5 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 3 \end{pmatrix}.$  This line can be parameterized by
\[\begin{pmatrix} -5 \\ 1 \end{pmatrix} + t \left( \begin{pmatrix} 2 \\ 3 \end{pmatrix} - \begin{pmatrix} -5 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} -5 \\ 1 \end{pmatrix} + t \begin{pmatrix} 7 \\ 2 \end{pmatrix} = \begin{pmatrix} 7t - 5 \\ 2t + 1 \end{pmatrix}.\][asy]
usepackage("amsmath");

unitsize(1 cm);

pair A, B, O, P;

A = (-5,1);
B = (2,3);
O = (0,0);
P = (O + reflect(A,B)*(O))/2;

draw((-6,0)--(3,0));
draw((0,-1)--(0,4));
draw(O--A,Arrow(6));
draw(O--B,Arrow(6));
draw(O--P,Arrow(6));
draw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);

label("$\begin{pmatrix} -5 \\ 1 \end{pmatrix}$", A, N);
label("$\begin{pmatrix} 2 \\ 3 \end{pmatrix}$", B, N);
label("$\mathbf{p}$", P, N);
[/asy]

The vector $\mathbf{p}$ itself will be orthogonal to the direction vector $\begin{pmatrix} 7 \\ 2 \end{pmatrix},$ so
\[\begin{pmatrix} 7t - 5 \\ 2t + 1 \end{pmatrix} \cdot \begin{pmatrix} 7 \\ 2 \end{pmatrix} = 0.\]Hence, $(7t - 5)(7) + (2t + 1)(2) = 0.$  Solving, we find $t = \frac{33}{53}.$  Hence, $\mathbf{p} = \boxed{\begin{pmatrix} -34/53 \\ 119/53 \end{pmatrix}}.$

set_name: MATH


text:
\begin{document}

\title{Symmetric pairs and branching laws}

\author{Paul-Emile PARADAN
\footnote{Univ. Montpellier, CNRS, Montpellier, France, {\em paul-emile.paradan@umontpellier.fr}}}

\maketitle

\begin{abstract}Let $G$ be a compact connected Lie group and let $H$ be a subgroup fixed by an involution. A classical result assures that the $H_\Cbb$-action on
the flag variety $\Fcal$ of $G$ admits a finite number of orbits. In this article we propose a formula for the branching coefficients of the symmetric pair
$ (G,H)$ that is parametrized by $H_\Cbb\backslash \Fcal$.

\end{abstract}

{\small
\tableofcontents}

\section{Introduction}

Let $G$ be a compact connected Lie group equipped with an involution $\theta$. Let  $G^{\theta}:=\{g\in G, \theta(g)=g\}$ be the subgroup
fixed by the involution. We consider a subgroup $H\subset G$ such that $(G^{\theta})_0\subset H\subset G^{\theta}$.
The purpose of this paper is the study of the branching laws between $G$ and $H$.

Let $T$ be a maximal torus of $G$ that we choose $\theta$-invariant. Let $\tgot$ be the Lie algebra of $T$.
Let $\Lambda\subset\tgot^*$ be the lattice of weights, and let $\tgot_{+}^*$
be a Weyl chamber. The irreducible representations of $G$ are parametrized by the semi-group $\Lambda_G^+:=\Lambda\cap\tgot^*_{+}$
of dominant weights.

Let $\lambda\in \Lambda_G^+$. In order to study the restriction $V_{\lambda}^{G}\vert_{H}$ of the irreducible $G$-representation $V_{\lambda}^{G}$, we consider the $H$-action on the flag variety $\Fcal=G/T$ of $G$. An important object is the $H$-invariant subset
$$
Z_\theta\subset \Fcal
$$
formed of the elements $x\in \Fcal$ for which the stabilizer subgroup $G_x:=\{g\in G, gx=x\}$  is stable under $\theta$. In orther words,
$gT\in Z_\theta$ if and only if $g^{-1}\theta(g)$ belongs to the normalizer subgroup $N(T)$.  A well-known result tells us that the group $H$ has finitely many
orbits in $Z_\theta$, and that the finite set $H\backslash Z_\theta$
parametrizes the $H_\Cbb$-orbits in $\Fcal$ \cite{Matsuki79,Rossmann79,Richardson-Springer90,Mi-Vi-Uz}.

Let $x\in Z_\theta$.  The stabilizer subgroup $G_{x}$ is a maximal torus in $G$ with Lie algebra $\ggot_{x}$. We will also consider the abelian subgroup
$H_{x}:=G_x\cap H$ (that is not necessarily connected). Any weight  $\mu\in\Lambda$ determines a character $\Cbb_{\mu_x}$ of the torus
$G_{x}$ by taking $\mu_x=g\cdot\mu$ if $x=gT\in \Fcal$.

We denote by $\Rgot_x\subset \ggot^*_x$ the set of roots relative to the action of the Cartan subalgebra $\ggot_x$ on $\ggot\otimes\Cbb$. The map
$\mu\in\Rgot\mapsto \mu_x\in\Rgot_x$ is an isomorphism, and we take $\Rgot_x^+\subset\Rgot_x$ as the image of $\Rgot^+\subset\Rgot$ through this isomorphism.

The involution $\theta$ leaves the set $\Rgot_x$ invariant, and $\alpha \in\Rgot_x$ is an {\em imaginary root} if $\theta(\alpha)=\alpha$.
If $\alpha$ is imaginary, the subspace $(\ggot\otimes\Cbb)_\alpha$ is $\theta$-stable. There are two cases. If the action of $\theta$ on
$(\ggot\otimes\Cbb)_\alpha$ is trivial then $\alpha$ is {\em compact imaginary}.  If the action of $-\theta$ on
$(\ggot\otimes\Cbb)_\alpha$ is trivial, then $\alpha$ is {\em non-compact imaginary}. We denote respectively by $\Rgot_x^{\hbox{\scriptsize ci}}$ and
by $\Rgot_x^{\hbox{\scriptsize nci}}$ the subsets of compact imaginary and non-compact imaginary roots, and we introduce the following $G_x$-modules
$$
\mathbb{E}^{\hbox{\scriptsize ci}}_{x}:=\sum_{\alpha\in\Rgot_x^{\hbox{\scriptsize ci}}\cap\Rgot^{+}_{x}}(\ggot\otimes\Cbb)_\alpha,
\quad
\quad
\mathbb{E}^{\hbox{\scriptsize nci}}_{x}:=\sum_{\alpha\in\Rgot_x^{\hbox{\scriptsize nci}}\cap\Rgot^{+}_{x}}(\ggot\otimes\Cbb)_\alpha.
$$

The weight
$$
\delta(x):=\frac{1}{2}\sum_{\stackrel{\alpha\in\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})}{\theta(\alpha)\neq\alpha}}\alpha
$$
defines a character $\Cbb_{\delta(x)}$ of the abelian group $H_x$. Let
$$
m_x= \frac{1}{2}| \Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap\{\theta(\alpha)\neq\alpha\}| + \dim \mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}.
$$

We denote by $R(H)$ and by $R(H_{x})$ the representations rings of the compact Lie groups $H$ and $H_{x}$. An element $E\in R(H)$ can be represented
as a {\em finite} sum $E=\sum_{V\in\widehat{H}}m_V V$, with $m_V\in\Zbb$. We denote by $\widehat{R}(H)$ (resp. $\widehat{R}(H_{x})$) the space
of $\Zbb$-valued functions on $\widehat{H}$ (resp. $\widehat{H_{x}}$). An element $E\in \widehat{R}(H)$ can be represented as an {\em infinite} sum
$\sum_{V\in\widehat{H}}m_V V$, with $m_{V}\in\Zbb$. The induction map $\mathrm{Ind}_H^K: \widehat{R}(H_{x})\to \widehat{R}(H)$ is the dual of the restriction morphism $R(H)\to R(H_{x})$.

The main result of this paper is the following theorem.

\begin{theo}\label{theo-principal}
Let $\lambda\in\Lambda_{G}^+$. We have the decomposition
\begin{equation}\label{eq:theoreme-principal}
V_{\lambda}^{G}\vert_{H}=\sum_{Hx\,\in\, H\backslash Z_{\theta}} Q_{Hx}(\lambda)
\end{equation}
 where the terms $Q_{Hx}(\lambda)\in \widehat{R}(H)$ are defined by the following relation :
$$
Q_{Hx}(\lambda)=(-1)^{m_{x}} {\rm Ind}^{H}_{H_{x}}\left(\Cbb_{\lambda_x+\delta(x)}\otimes
\det(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x})\otimes {\rm Sym}(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x})\otimes\bigwedge
\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}\right).
$$
Here ${\rm Sym}(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x})$, which is the symmetric algebra of
$\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}$, is an admissible representation of $H_x$ and $\bigwedge \mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}=
\bigwedge^{+}\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}\ominus\bigwedge^{-} \mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}$ is a virtual representation of $H_{x}$.
\end{theo}

We give now another formulation for decomposition (\ref{eq:theoreme-principal}) using the (right) action of the Weyl group $W=N(T)/T$ on the flag variety $\Fcal$. If
$x=gT\in \Fcal$ and $w\in W$ we take $xw:=gwT$. We notice immediately that $Z_\theta$ is stable under the action of $W$.

We associate to an element $x=gT\in Z_\theta$ the subgroup $W^H_x\subset W$ defined by the relation $w\in W^H_x\Longleftrightarrow Hxw=Hx$.
We denote by $H\backslash Z_{\theta}\slash W$ the quotient of $Z_\theta$ by the action of $H\times W$, and by $\bar{x}\in H\backslash Z_{\theta}\slash W$ the
image of $x\in Z_\theta$ through the quotient map. We associate to $\bar{x}\in H\backslash Z_{\theta}\slash W$ the element $Q_{\bar{x}}(\lambda)\in \widehat{R}(H)$ defined as follows
$$
Q_{\bar{x}}(\lambda)=\sum_{\bar{w}\in W^H_x\backslash W} Q_{Hxw}(\lambda).
$$
Theorem \ref{theo-principal} says then that $V_{\lambda}^{G}\vert_{H}=\sum_{\bar{x}\,\in\, H\backslash Z_{\theta}\slash W}Q_{\bar{x}}(\lambda)$. Here is a new formulation of Theorem \ref{theo-principal}.

\begin{theo}\label{theo-principal-bis}
We have  $V_{\lambda}^{G}\vert_{H}=\sum_{\bar{x}\,\in\, H\backslash Z_{\theta}\slash W}Q_{\bar{x}}(\lambda)$
where $Q_{\bar{x}}(\lambda)\in \wR(H)$ has the following description
$$
Q_{\bar{x}}(\lambda)=
{\rm Ind}^{H}_{H_{x}}\left(\mathbb{M}_x(\lambda)\otimes \Cbb_{\delta(x)}\otimes\bigwedge \mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}\right),
$$
for some\footnote{The precise expression of  $\mathbb{M}_x(\lambda)$ is given in Proposition \ref{prop:formule-M-lambda}.} $\mathbb{M}_x(\lambda)\in \wR(H_x)$.
\end{theo}

\bigskip

We finish this section by giving two basic examples associated to the group $SU(2)$. Here the flag variety of $SU(2)$ is the $2$-dimensional sphere $\mathbb{S}^2$.
For $n\geq 0$, we denote by $V_n$ the irreducible representation of $SU(2)$ of dimension $n+1$.

\medskip

{\bf Example 1}. $G=SU(2)$ and the involution $\theta$ is the conjugaison by the matrix $\left(\begin{array}{cc} 1& 0 \\ 0 & -1\end{array}\right)$. The subgroup fixed by $\theta$ is the torus $T\simeq U(1)$ and the
critical set $Z_\theta\subset\mathbb{S}^2$ is composed by the poles $S,N$ and the equator $E$, so that $T\backslash Z_\theta$ has three terms. We take $\lambda=n$ in $\widehat{SU(2)}\simeq \mathbb{N}$.

For $Hx=E$, we have  $\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}=\mathbb{E}^{\hbox{\scriptsize\rm ci}}_{x}=\{0\}$, $H_x\simeq \mathbb{Z}_2$, and
$\Cbb_{\lambda_x+\delta(x)}=\Cbb_n\vert_{\mathbb{Z}_2}$.  The contribution of $E$ is then
${\rm Ind}^{U(1)}_{\mathbb{Z}_2}(\Cbb_n\vert_{\mathbb{Z}_2})=\Cbb_{n}\otimes\sum_{k\in\mathbb{Z}} \Cbb_{2k}$.

For $Hx=N$, we have $H_x=T$, $\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}=\Cbb_2$, $\mathbb{E}^{\hbox{\scriptsize\rm ci}}_{x}=\{0\}$, and
$\Cbb_{\lambda_x+\delta(x)}=\Cbb_{n}$. The contribution of $N$ is then $-\Cbb_{n+2}\otimes{\rm Sym}(\Cbb_{2})$.

For $Hx=S$, we have $H_x=T$, $\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}=\Cbb_{-2}$, $\mathbb{E}^{\hbox{\scriptsize\rm ci}}_{x}=\{0\}$, and
$\Cbb_{\lambda_x+\delta(x)}=\Cbb_{-n}$. The contribution of $S$ is then $-\Cbb_{-n-2}\otimes{\rm Sym}(\Cbb_{-2})$.

Finally, Relations (\ref{eq:theoreme-principal}) become
\begin{eqnarray*}
V_n\vert_T&=& \Cbb_{n}\otimes\sum_{k\in\mathbb{Z}} \Cbb_{2k}  - \Cbb_{-n-2}\otimes{\rm Sym}(\Cbb_{-2}) - \Cbb_{n+2}\otimes{\rm Sym}(\Cbb_{2})\\
&=&\sum_{k=-n}^{0} \Cbb_{2k+n}.
\end{eqnarray*}

\medskip

{\bf Example 2}. $G=SU(2)\times SU(2)$ and the involution $\theta$ is the map $(a,b)\mapsto (b,a)$. The subgroup fixed by $\theta$ is $SU(2)$ embedded diagonally and the critical set $Z_\theta\subset\mathbb{S}^2\times \mathbb{S}^2$ is equal to
the union of the orbits $SU(2)\cdot(N,N)$ and $SU(2)\cdot(S,N)$. Let $\lambda=(n,m)\in \widehat{G}$.

For $x=(N,N)$ or $x=(S,N)$ we have $\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}=\mathbb{E}^{\hbox{\scriptsize\rm ci}}_{x}=\{0\}$ and $H_x\simeq T$. For
$x=(N,N)$ we have $\lambda_x+\delta(x)=m+n+2$, and for $x=(S,N)$ we have $\lambda_x+\delta(x)=m-n$. Relations \ref{eq:theoreme-principal} give then
$$
V_n\otimes V_m=   \ind_{T}^{SU(2)}(\Cbb_{m-n}) - \ind_{T}^{SU(2)}(\Cbb_{m+n+2}).
$$
It is not difficult to see that the previous identities correspond to the classical Clebsch-Gordan relations (see Example \ref{SU(2)-Clebsch-Gordan}).

\begin{center}
\bf Notations
\end{center}

Throughout the paper :
\begin{itemize}
\item $G$ denotes a compact connected Lie group with Lie algebra $\ggot$.
\item $T$ is a maximal torus in $G$ with Lie algebra $\tgot$.
\item $\Lambda\subset \tgot^*$ is the weight lattice  of $T$ : every $\mu\in \Lambda$ defines a $1$-dimensional
$T$-representation, denoted by $\Cbb_\mu$, where $t=\exp(X)$ acts by $t^\mu:= e^{i\langle\mu, X\rangle}$.

\item The coadjoint action of $g\in G$ on $\xi\in \ggot^*$ is denoted by $g\cdot\xi$.
\item When a Lie group $K$ acts on set $X$, the stabilizer subgroup of $x\in X$ is denoted by $K_x:=\{k\in K\ \vert\ k\cdot x=x\}$ and the Lie algebra of $K_x$ is denoted by $\kgot_x$.
\item  When a Lie group $K$ acts on a manifold $M$, we denote by $X\cdot m:=\frac{d}{dt} e^{tX}\cdot m\vert_{t=0}$, $m\in M$, the vector field generated by $X\in \kgot$.
\end{itemize}

 \section{Non abelian localization}

Our main result is obtained by means of a non-abelian localization of the Riemann-Roch character on the flag variety $\Fcal$ of $G$.
For that we will use the family $(\Omega_r)_r$ of symplectic structure parametrized by the interior of the Weyl chamber $\tgot^*_+$.
The symplectic structure $\Omega_r$ comes from the identication $gT\to g\cdot r$ of $\Fcal$ with the coadjoint orbit $Gr$.
The moment map $\Phi_r:\Fcal\to\ggot^*$ associated to the action of $G$ on $(\Fcal,\Omega_r)$ is the map
$gT \mapsto g\cdot r$.

At the level of Lie algebras we have $\ggot=\hgot\oplus\qgot$ where
$\hgot=\ggot^\theta$ and $\qgot=\ggot^{-\theta}$. For any $\xi\in \ggot=\hgot\oplus\qgot$,
we denote by $\xi^+$ his $\hgot$-part and by $\xi^-$ his $\qgot$-part. We use a $G$-invariant scalar product $(-,-)$ on $\ggot$ such that
the involution $\theta$ is an orthogonal map. It induces identifications $\ggot^*\simeq\ggot$,  $\hgot^*\simeq\hgot$ and $\qgot^*\simeq\qgot$.

The moment map $\Phi^H_r:\Fcal\to\hgot^*$ associated to the action of $H$ on $(\Fcal,\Omega_r)$ is the map
$gT \mapsto (g\cdot r)^+$.

\subsection{Matsuki duality}

Consider the complex reductive groups $G_\Cbb$ and $H_\Cbb$ associated to the compact Lie groups $G$ and $H$.
Let $L\subset G_\Cbb$ be the real form such that $H\subset L$ is a maximal compact subgroup of $L$.

Matsuki duality is the statement that a one-to-one correspondence exists between the $H_\Cbb$-orbits and the $L$-orbits in $\Fcal$; two orbits are in duality
when their intersection is a single orbit of $H$.

Uzawa, and Mirkovic-Uzawa-Vilonen  \cite{Uzawa90,Mi-Vi-Uz} proved the Matsuki correspondence by showing that both $H_\Cbb$-orbits and $L$-orbits in
$\Fcal$ are parametrized by the $H$-orbits in the set of critical points of the function $\|\Phi^H_r\|^2: \Fcal\to \Rbb$.

First we recall the elementary but fundamental fact that the subset $Z_\theta$ is equal to the set of critical points of the function $\|\Phi^H_r\|^2$
\cite{Mi-Vi-Uz,Bremigan-Lorch}.

\begin{lem}\label{lem-Uzawa}
Let $x=gT\in \Fcal$ and $r\in {\rm Interior}(\tgot^*_+)$. The following statements are equivalent:
\begin{itemize}
\item[i)] the subalgebra $\ggot_x$ is invariant under $\theta$ (i.e. $x\in Z_\theta$),
\item[ii)] $g^{-1}\theta(g)\in N(T)$,
\item[iii)] $x$ is a critical point of the function $\|\Phi^H_r\|^2$,
\item[iv)] $(g\cdot r)^+$ and $(g\cdot r)^-$ commutes.
\end{itemize}
\end{lem}

\begin{proof} Let $n_g= g^{-1}\theta(g)$ and let $r$ be a regular element of $\tgot^*\simeq \tgot$. Since $\ggot_x=Ad(g)\tgot$ we see that
\begin{eqnarray*}
 \theta(\ggot_x)=\ggot_x  &\Longleftrightarrow &n_g\in N_G(T)\\
 &\Longleftrightarrow & [n_g\cdot\theta(r),r]=0\\
 &\Longleftrightarrow & [\theta(g\cdot r),g\cdot r]=0\\
 &\Longleftrightarrow & [(g\cdot r)^+,(g\cdot r)^-]=0.
\end{eqnarray*}

A small computation shows that for any $X\in\ggot$ the derivative of the function $t\mapsto \|\Phi^H_r(e^{tX}x)\|^2$ at $t=0$
is equal to $(X,[g\cdot r,\theta(g\cdot r)])$. Hence $x=gT$ is a critical point of the function $\|\Phi^H_r\|^2$ if and only if $[g\cdot r,\theta(g\cdot r)]=0$.
Finally we have proved that the statements $i),ii),iii)$ and $iv)$ are equivalent. $\Box$
\end{proof}

\medskip

Let us check the other easy fact.

\begin{lem}
The set $H\backslash Z_\theta$ is finite.
\end{lem}

\begin{proof}
Let $x=gT\in Z_\theta$. A neighborhood of $x$ is defined by elements of the form $e^{X}e^{Y}x$ where $X\in\hgot$ and $Y\in\qgot$. Now we see that
$e^{X}e^{Y}gT\in Z_\theta$ if and only if $e^{-2g^{-1}Y}\in N(T)$. If $Y$ is sufficiently small the former relation is equivalent to $g^{-1}Y\in \tgot$,
and in this case $e^{X}e^{Y}x=e^{X}x$. We have proved that any element in $H\backslash Z_\theta$ is isolated. As $H\backslash Z_\theta$ is compact, we can conclude that
$H\backslash Z_\theta$ is finite. $\Box$
\end{proof}

\subsection{Borel-Weil-Bott theorem}

We first recall the Borel-Weil-Bott theorem. The flag manifold $\Fcal$ is equipped with the $G$-invariant complex structure such that
$$
\T_{eT}\Fcal\simeq \sum_{\alpha\in \Rgot^+}(\ggot\otimes\Cbb)_\alpha
$$
is an identity of $T$-modules. Let us consider the tangent bundle $\T\Fcal$ as a complex vector bundle on $\Fcal$ with the invariant
Hermitian structure $h_\Fcal$ induced by the invariant scalar product on $\ggot$.

Any weight $\lambda\in \Lambda$ defines a line bundle $\Lcal_\lambda\simeq G\times_T\Cbb_\lambda$ on $\Fcal$.

\begin{defi}
 We associated to a weight $\lambda\in \Lambda$

 $\bullet$ the spin-c bundle on $\Fcal$
$$
\Scal_\lambda:= \bigwedge_{\Cbb}\T \Fcal\otimes \Lcal_\lambda,
$$

$\bullet$ the Riemann-Roch character $\RR_G(\Fcal,\Lcal_\lambda)\in R(G)$ which is the equivariant index of the
Dirac operator $D_\lambda$ associated to the spin-c structure $\Scal_\lambda$.
\end{defi}

The Borel-Weil-Bott theorem asserts that $V_\lambda^G=\RR_G(\Fcal,\Lcal_\lambda)$ when $\lambda$ is dominant.
Now we consider the restriction $V_\lambda^G\vert_H=\RR_H(\Fcal,\Lcal_\lambda)$. In the next section we will explain
how we can localize the $H$-equivariant Riemann-Roch character $\RR_H(\Fcal,\Lcal_\lambda)$ on the critical set of the function $\|\Phi^H_r\|^2$ \cite{pep-RR}.

\subsection{Localization of the Riemann-Roch character}\label{sec:localisation}

In this section we explain how we perform the ``Witten  non-abelian localization'' of the Riemann-Roch character with the help of
 the moment map $\Phi^H_r:\Fcal\to\hgot^*$ attached to a  regular element $r$ of the Weyl chamber  \cite{pep-RR,Ma-Zhang14,pep-vergne:witten}.

Let us denote by $X\mapsto [X]_{\ggot/\tgot}$ the projection $\ggot\to\ggot/\tgot$. The Kirwan vector field $\kappa_r$ on $\Fcal$ is defined as follows:
$$
\kappa_r(x)=-\Phi^H_r(x)\cdot x \in \T_x\Fcal.
$$
Through the identification $\ggot/\tgot\simeq\T_x\Fcal, X\mapsto \frac{d}{dt}\vert_{t=0} ge^{tX}T$, the vector $\kappa_r(x)\in\T_x\Fcal$ is equal to
$[g^{-1}\theta(g)\cdot r]_{\ggot/\tgot}$. Hence the set $Z_\theta\subset \Fcal$ is exactly the set where $\kappa_r$ vanishes.

Lat $D_0$ be the Dirac operator associated to the spin-c structure $\Scal_0= \bigwedge_{\Cbb}\T \Fcal$. The principal symbol of the elliptic
operator $D_0$ is the bundle map $\sigma(\Fcal)\in \Gamma(\T^* \Fcal, \hom(\bigwedge^+_\Cbb\T\Fcal,\bigwedge^-_\Cbb \T\Fcal))$
defined by the Clifford action
$$
\sigma(\Fcal)(x,\nu)=\clif_{x}(\tilde{\nu}): {\bigwedge}^+_\Cbb\T_x\Fcal\to {\bigwedge}^-_\Cbb\T_x\Fcal.
$$
where $\nu\in \T^*_x \Ocal\simeq \tilde{\nu}\in \T_x \Ocal$ is the one to one map associated to the identification
$\ggot^*\simeq \ggot$ (see \cite{B-G-V}).

Now we will deform the elliptic symbol $\sigma(\Fcal)$ by means of the vector field $\kappa_r$ \cite{pep-RR,pep-vergne:witten}.

\begin{defi}\label{def:pushed-sigma}
The symbol  $\sigma(\Fcal)$ shifted by the vector field $\kappa_r$ is the
symbol on $\Fcal$ defined by
$$
\sigma_r(\Fcal)(x,\nu)=\clif_{x}(\tilde{\nu}-\kappa_r(x))
$$
for any $(x,\nu)\in\T^* \Fcal$.
\end{defi}

Consider an $H$-invariant open subset $\Ucal\subset \Fcal$ such that $\Ucal\cap Z_\theta$
is compact in $\Fcal$. Then the restriction $\sigma_r(\Fcal)\vert_\Ucal$ is a $H$-transversally elliptic symbol on $\Ucal$,
and so its equivariant index is a well defined element in $\wR(H)$ (see \cite{Atiyah74,pep-RR,pep-vergne:witten}).

Thus we can define the following localized equivariant indices.

\begin{defi}\label{def:indice-localise}
Let $Hx\subset Z_\theta$. We denote by
$$
\RR_H(\Fcal,\Lcal_\lambda,\Phi_r^H, Hx)\ \in\ \widehat{R}(H)
$$
the equivariant index of $\sigma_r(\Fcal)\otimes\Lcal_\lambda\vert_\Ucal$ where
$\Ucal$ is an invariant neighbourhood of $Hx$ so that $\Ucal\cap Z_\theta=Hx$.
\end{defi}

We proved in \cite{pep-RR} that we have the decomposition
$$
\RR_H(\Fcal,\Lcal_\lambda)=\sum_{Hx\in H\backslash Z_\theta}
\RR_H(\Fcal,\Lcal_\lambda, \Phi_r^H, Hx) \quad \in\, \wR(H).
$$

The computation of the characters $\RR_H(\Fcal,\Lcal_\lambda, \Phi_r^H, Hx)$ will be handle in Section \ref{sec:computation-Q-Hx-lambda}. To undertake these calculations we
need to describe geometrically  a neighborhood of $Hx$ in $\Fcal$. This is the goal of the next section.

\subsection{Local model near $Hx\subset Z_\theta$}

Let $x=gT\in Z_\theta$. We need to compute a symplectic model of a neighborhood of $Hx$ in $(\Fcal,\Omega_r)$. Here we use the identification
$\ggot\simeq \ggot^*$ given by the choice of an invariant scalar product. Let $\mu=g\cdot r$ that we write $\mu=\mu^+ +\mu^-$ where
$\mu^+\in \hgot$ and $\mu^-\in \qgot$.

The tangent space $\T_x \Fcal$ is equipped with the symplectic two form
$\Omega_r\vert_x$:
$$
\Omega_r\vert_x(X\cdot x,Y\cdot x)=(\mu,[X,Y]),\quad X,Y\in \ggot.
$$

We need to understand the structure of the symplectic vector space $(\T_x\Fcal,\Omega_r\vert_x)$.
If $\agot\subset \ggot$ is a vector subspace we denote by $\agot\cdot x:=\{X\cdot x, X\in\agot\}$ the corresponding subspace of $\T_x\Fcal$. The symplectic orthogonal of $\agot\cdot x$ is denoted by $(\agot\cdot x)^{\perp,\Omega}$.

If $\agot,\bgot$ are two subspaces, a small computation gives that
\begin{equation}\label{eq:orthogonal-a-x}
(\agot\cdot x)^{\perp,\Omega}\cap \bgot\cdot x \simeq \agot^\perp\cap [\bgot,\mu],
\end{equation}
where $\agot^\perp\subset \ggot$ is the orthogonal of $\agot$ relatively to the scalar product.

We denote by $\ggot_{\mu^+}=\hgot_{\mu^+}\oplus\qgot_{\mu^+}$ the subspaces fixed by $ad(\mu^+)$. Notice that
$\ggot_{\mu}=\ggot_x$ is an abelian subalgebra containing $\mu^+$ since $[\mu^+,\mu^-]=0$. It follows that $\ggot_x\subset \ggot_{\mu^+}$.

\begin{lem}$\ggot_{\mu^+}\cdot x$ and $[\hgot,\mu^+]\cdot x$ are symplectic subspaces of $\T_x\Fcal$.
\end{lem}
\begin{proof}
It is a direct consequence of (\ref{eq:orthogonal-a-x}). $\Box$
\end{proof}

\medskip

We consider now the symplectic subspace $V_x \subset \T_x\Fcal$ defined by the relation
\begin{equation}\label{eq:V-x}
V_x= ([\hgot,\mu^+]\cdot x)^{\perp,\Omega}\cap [\ggot,\mu^+]\cdot x.
\end{equation}
A small computation shows that $X\cdot x\in V_x$ if and only if $[X,\mu]\subset [\qgot,\mu^+]$.

We have the following important Lemma.

\begin{lem}\label{lem:decomposition-T-x-O}
\begin{itemize}
\item We have the following decomposition
\begin{equation}\label{eq:T-x-F}
\T_x\Fcal    = \ggot_{\mu^+}\cdot x \stackrel{\perp}{\oplus} \left[\hgot,\mu^+\right]\cdot x\stackrel{\perp}{\oplus} V_x
\end{equation}
where $\perp$ stands for the orthogonal relative to $\Omega_r\vert_x$.
\item $\ggot_{\mu^+}\cdot x$  is  symplectomorphic to $\hgot_{\mu^+}/\hgot_x\oplus (\hgot_{\mu^+}/\hgot_x)^*$.
\item $\left[\hgot,\mu^+\right]\cdot x$  is symplectomorphic to $\hgot/\hgot_{\mu^+}$ equipped
with the symplectic structure $\Omega_{\mu^+}(\bar{u},\bar{v})=(\mu^+,[u,v])$.
\item $V_x$  is symplectomorphic to $(\hgot\cdot x)^{\perp,\Omega}\slash (\hgot\cdot x)^{\perp,\Omega}\cap \hgot\cdot x$.
\end{itemize}
\end{lem}

\begin{proof}
If we use the decomposition $\ggot= \ggot_{\mu^+}\oplus[\ggot,\mu^+]$ and the fact that the abelian subalgebra $\ggot_x$ is contained in $\ggot_{\mu^+}$ we obtain
$$
\T_x\Fcal=  \ggot_{\mu^+}\cdot x \oplus [\ggot,\mu^+]\cdot x.
$$
It is obvious to check that the subspaces $[\ggot,\mu^+]\cdot x$ and $\ggot_{\mu^+}\cdot x$ are orthogonal relatively to the symplectic form
$\Omega_r\vert_x$. Since $[\hgot,\mu^+]\cdot x$ is a symplectic subspace we have $[\ggot,\mu^+]\cdot x=[\hgot,\mu^+]\cdot x \stackrel{\perp}{\oplus} V_x$ where
$V_x$ is defined by (\ref{eq:V-x}). The first point is proved.

The identities $\ggot_x=\theta(\ggot_x)=\ggot_{\theta(x)}$ imply the decompositions $\ggot_x=\hgot_x\oplus\qgot_x$ and $[\ggot_{\mu^+},x]=[\qgot_{\mu^+},x]\oplus[\hgot_{\mu^+},x]$. The vector subspace $[\hgot_{\mu^+},x]$ is isomorphic to $\hgot_{\mu^+}/\hgot_x$, and the map $v\mapsto \Omega_r\vert_x(v,-)$ defines an isomorphism between $[\qgot_{\mu^+},x]$ and the dual of $[\hgot_{\mu^+},x]$.
The second point is proved.

For the third point we use the isomophism $j:[\hgot,\mu^+]\to \hgot/\hgot_{\mu^+}$ induces by the projection $\hgot \to \hgot/\hgot_{\mu^+}$. Then the map
$\bar{u}\mapsto j(\bar{u})\cdot x$ defines a symplectomorphism between $(\hgot/\hgot_{\mu^+},\Omega_{\mu^+})$ and $[\hgot,\mu^+]\cdot x$.

Now we see that (\ref{eq:T-x-F}) together with the decomposition $\hgot\cdot x=$\break  $[\hgot,\mu^+]\cdot x +\hgot\cdot x$   leads to
\begin{eqnarray*}
(\hgot\cdot x)^{\perp,\Omega}&=& ([\hgot,\mu^+]\cdot x)^{\perp,\Omega}\cap(\hgot_{\mu^+}\cdot x)^{\perp,\Omega}\\
&=&([\hgot,\mu^+]\cdot x)^{\perp,\Omega}\cap(\ggot_{\mu^+}\cdot x) \oplus V_x\\
&=&(\hgot\cdot x)^{\perp,\Omega}\cap \hgot\cdot x\oplus V_x.
\end{eqnarray*}
The last point follows. $\Box$
\end{proof}

\medskip

We denote by $\Omega_{V_x}$ the restriction of $\Omega_r\vert_x$ on the symplectic vector subspace $V_x$. The action of $H_x$ on $(V_x,\Omega_{V_x})$
is Hamiltonian, with moment map $\Phi_{V_x}: V_x\to \hgot_x^*$ defined by the relation
$$
\langle\Phi_{V_x}(v), A\rangle= \frac{1}{2}\Omega_{V_x}(v,Av),\quad v\in V_x,\ A\in \hgot_x.
$$

Thanks to Lemma \ref{lem:decomposition-T-x-O}, we know that the $H_x$-symplectic vector space $(\T_x\Fcal,\Omega_r\vert_x)$ admits the following decomposition
$$
\T_x\Fcal \simeq \hgot_{\mu^+}/\hgot_x\oplus (\hgot_{\mu^+}/\hgot_x)^*
\stackrel{\perp}{\oplus}\hgot/\hgot_{\mu^+}
\stackrel{\perp}{\oplus} V_x
$$

Thanks to the normal form Theorem of Marle \cite{Marle85} and Guillemin-Sternberg \cite{Guillemin-Sternberg84}, we get the following result.

\begin{coro}
An $H$-equivariant symplectic model of a neighborhoood of $Hx$ in $\Fcal$ is
$\Fcal_x:= H\times_{H_{\mu^+}} Y_x$
where
$$
Y_x= H_{\mu^+}\times_{H_x}\left((\hgot_{\mu^+}/\hgot_x)^*\times V_x\right).
$$
The corresponding moment map on $\Fcal_x$ is
$$
\Phi_{\Fcal_x}([h; \eta,v])= h(\eta + \mu^+ + \Phi_{V_x}(v))
$$
for $[h; \eta,v]\in H\times_{H_x}\left((\hgot_{\mu^+}/\hgot_x)^*\times V_x\right)$.
\end{coro}

\bigskip

We finish this section by computing a compatible complex structure on $V_x$.

By definition, the map that sends $X\cdot x$ to $[X,\mu]$ defines an isomorphism $i:V_x\to [\qgot,\mu^+]$.
The adjoint map $ad(\mu)$ defines also an automorphism of $[\ggot,\mu^+]$: for any $X\in[\ggot,\mu^+]$ we denote by
$\tilde{X}\in [\ggot,\mu^+]$ the unique element such that $ad(\mu)\tilde{X}=X$.

The symplectic structure $\Omega_\mu:=(i^{-1})^*\Omega_{V_x}$ satisfies the relations
$$
\Omega_\mu(X,Y)=(\mu,[\tilde{X},\tilde{Y}])=(X,\tilde{Y})=-(\tilde{X},Y),\quad \forall X,Y\in [\qgot,\mu^+].
$$

We consider the one to one map $-ad(\mu)ad(\theta(\mu)):[\ggot,{\mu^+}]\to [\ggot,{\mu^+}]$ and the $H_x$-invariant
complex structure $J_{\mu^+}= ad(\mu^+)(-ad(\mu^+)^2)^{-1/2}$ on $[\ggot,{\mu^+}]$. It
restricts to a one to one map $T_x:[\qgot,{\mu^+}]\to [\qgot,{\mu^+}]$ and a complex structure
on $[\qgot,{\mu^+}]$ (still denoted by $J_{\mu^+}$).

Let $S_x:=(T_x^2)^{-1/2} T_x$. The map $J_{V_x}:= J_{\mu^+}\circ S_x$ defines a $H_x$-invariant complex structure on $[\qgot,{\mu^+}]$.

\medskip

\begin{lem}\label{lem:V-x-complexe}The $H_x$-symplectic space $(V_x,\Omega_{V_x})$ is isomorphic to $[\qgot,{\mu^+}]$ equipped with the symplectic form
$\Omega_{\mu}^1(v,w)= (J_{V_x}v,w)$.
\end{lem}

\begin{proof} We know already that $(V_x,\Omega_{V_x})\simeq([\qgot,{\mu^+}],\Omega_\mu)$. If one takes $L=T_x\circ (-ad(\mu^+)^2)^{-1/4}\circ (T_x^2)^{-1/4}$,
we check easily that $\Omega_{\mu}(L(v),L(w))=(J_{V_x}v,w)$. $\Box$
\end{proof}

 \section{Proof of the main theorem}

 We start with the following lemma.

\begin{lem}
The quantity $\RR_H(\Fcal,\Lcal_\lambda,\Phi_r^H, Hx)$ does not depend on the choice of the regular element $r$ in the Weyl chamber.
In the following we will denote it by $Q_{Hx}(\lambda)\in\wR(H)$.
\end{lem}

\begin{proof}
Let $r_0,r_1$ be two regular elements of the Weyl chamber. For $t\in [0,1]$, we consider the regular element $r(t)=tr_1+(1-t)r_0$: the Kirwan vector field
$\kappa_{r(t)}$ vanishes exactly on $Z_\theta$ for any $t\in [0,1]$.  If $\Ucal$ is an invariant neighbourhood of $Hx$ so that $\Ucal\cap Z_\theta=Hx$, then
$t\in [0,1]\mapsto \sigma_{r(t)}(\Fcal)\otimes\Lcal_\lambda\vert_\Ucal$ defines an homotopy of transversally elliptic symbols. Accordingly, the equivariant index of
$\sigma_{r_0}(\Fcal)\otimes\Lcal_\lambda\vert_\Ucal$ and $\sigma_{r_1}(\Fcal)\otimes\Lcal_\lambda\vert_\Ucal$ are equal. $\Box$

\end{proof}

 \subsection{Computation of $Q_{Hx}(\lambda)$}\label{sec:computation-Q-Hx-lambda}

The computation of $Q_{Hx}(\lambda)$ is done in three steps.

 \subsubsection{Step 1: holomorphic induction}

Let $H_{\mu^+}\subset H$ be the stabilizer subgroup of $\mu^+:=\Phi^H_r(x)$. We know that an $H$-equivariant symplectic model of a neighborhoood of $Hx$ in
$\Fcal$ is the manifold $H\times_{H_{\mu^+}} Y_x$ where
$$
Y_x= H_{\mu^+}\times_{H_x}\left((\hgot_{\mu^+}/\hgot_x)^*\times V_x\right).
$$
The symplectic two form on $Y_x$ is built from the canonical symplectic structure on $H_{\mu^+}\times_{H_x}(\hgot_{\mu^+}/\hgot_x)^*\simeq \T^*
(H_{\mu^+}\slash {H_x})$ and the symplectic structure on $V_x$. The moment map relative to the action of $H_{\mu^+}$ on
$Y_x$ is
$$
\Phi_{Y_x}([h; \eta,v])= h(\eta + \mu^+ + \Phi_{V_x}(v))\in \hgot^*_{\mu^+},
$$
for $[h; \eta,v]\in H_{\mu^+}\times_{H_x}\left((\hgot_{\mu^+}/\hgot_x)^*\times V_x\right)$.

Let $\kappa_{Y_x}$ the Kirwan vector field on $Y_x$. It is immediate to check that $[h; \eta,v]\in \{\kappa_{Y_x}=0\}$ if and only if $\eta=0$ and
$(\mu^+ + \Phi_{V_x}(v))\cdot v=0$. The map $v\mapsto \mu^+\cdot v$ is bijective and $v\mapsto \Phi_{V_x}(v)\cdot v$ is homogeneous of degree equal to $3$. Then
there exists $\epsilon>0$ such that
$$
(\mu^+ + \Phi_{V_x}(v))\cdot v=0 \quad {\rm and} \quad \|v\|\leq \epsilon \  \Longrightarrow \ v=0.
$$
In $Y_x$, we still denote by $x$ the point $[e,0,0]$. We equip $Y_x$ with an invariant almost complex structure that is compatible with the symplectic structure, and
we denote by $\RR_{H_{\mu^+}}(Y_{x},\Lcal_\lambda\vert_{Y_x},\Phi_{Y_x}, H_{\mu^+} x)$ the Riemann-Roch character on $Y_x$ localized on the component
$H_{\mu^+} x\subset \{\kappa_{Y_x}=0\}$.

The quotient $\hgot/\hgot_{\mu^+}$, which is equipped with the invariant complex structure $J_{\mu^+}:=ad(\mu^+)(-ad(\mu^+)^2)^{-1/2}$, is  a complex $H_{\mu^+}$-module.

In \cite{pep-RR}[Theorem 7.5], we proved that $Q_{Hx}(\lambda)=\RR_H(\Fcal,\Lcal_\lambda,\Phi_r^H, Hx)$ is equal to
\begin{equation}\label{eq:holomorphic induction}
{\rm Ind}_{H_{\mu^+}}^H\left( \RR_{H_{\mu^+}}(Y_{x},\Lcal_\lambda\vert_{Y_x},\Phi_{Y_x}, H_{\mu^+} x)\otimes \bigwedge \hgot/\hgot_{\mu^+}\right).
\end{equation}

 \subsubsection{Step 2: cotangent induction}

The map $\Phi_x(v):= \mu^+ + \Phi_{V_x}(v)$ is a moment map for the Hamiltonian action of $H_x$ on $V_x$. The moment map on the $H_{\mu^+}$-manifold
$$
Y_x= H_{\mu^+}\times_{H_x}\left((\hgot_{\mu^+}/\hgot_x)^*\times V_x\right)
$$
is $\Phi_{Y_x}([h; \eta,v])= h(\eta +\Phi_x(v))\in \hgot^*_\mu$.

Let $\kappa_{V_x}(v)=-\Phi_x(v)\cdot v$ be the Kirwan vector field on $V_x$. We are interested in the connected component $\{0\}$ of
$\{\kappa_{V_x}=0\}$. We choose a compatible almost complex structure on the symplectic vector space and we denote by
$\RR_{H_x}(V_x,\Phi_{x}, \{0\})\in \wR(H_x)$ the
Riemann-Roch character localized on $\{0\}\subset \{\kappa_{V_x}=0\}$.

In Section 3.3 of \cite{pep-vergne:witten} we have proved that
\begin{equation}\label{eq:cotangent induction}
\RR_{H_{\mu^+}}(Y_{x},\Lcal_\lambda\vert_{Y_x},\Phi_{Y_x}, H_{\mu^+} x)
={\rm Ind}_{H_x}^{H_{\mu^+}}\left(\RR_{H_x}(V_x,\Phi_{x}, \{0\}) \otimes \Lcal_\lambda\vert_{x}\right).
\end{equation}

 \subsubsection{Step 3: linear case}

We write $\qgot/\qgot_{\mu^+}$ for the vector space $[\qgot,\mu^+]$ equipped with the complex structure $J_{\mu^+}$.
So $\qgot/\qgot_{\mu^+}$ is a $H_{\mu^+}$-module and we denote by ${\rm Sym}(\qgot/\qgot_{\mu^+})$ the corresponding symmetric algebra.

We need to compare the virtual $H_x$-modules $\bigwedge_{J_{V_x}} V_x$ and $\bigwedge_{-J_{\mu^+}} V_x$.
The weight
$$
\delta(x):=\frac{1}{2}\sum_{\stackrel{\alpha\in\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})}{\theta(\alpha)\neq\alpha}}\alpha \
$$
defines a character $\Cbb_{\delta(x)}$ of the abelian group $H_x$. Recall that $m_x\in\mathbb{N}$ corresponds to the quantity
$ \frac{1}{2}| \Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap\{\theta(\alpha)\neq\alpha\}| + \dim \mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}$.

The following lemma will be proved in Section \ref{sec:Q-Hx-Lambda}.

\begin{lem}\label{lem:wedge-J-compare}
The following identity holds :
$$
\bigwedge_{J_{V_x}} V_x\simeq (-1)^{m_x}\, \Cbb_{\delta(x)}\otimes \det(\mathbb{E}_x^{\hbox{\scriptsize\rm nci}})\otimes\bigwedge_{-J_{\mu^+}} V_x.
$$
\end{lem}

On the vector space $V_x$, we can work with two localized Riemann-Roch characters:
\begin{itemize}
\item $\RR_{H_{x}}(V_{x},\Phi_x,\{0\})$ is defined with the complex structure $J_{V_x}$,
\item $\widetilde{\RR}_{H_{x}}(V_{x},\Phi_x,\{0\})$ is defined with the complex structure $-J_{\mu^+}$.
\end{itemize}
The previous Lemma gives that $\RR_{H_{x}}(V_{x},\Phi_x,\{0\})$ is equal to
$(-1)^{m_x}\, \Cbb_{\delta(x)}\otimes \widetilde{\RR}_{H_{x}}(V_{x},\Phi_x,\{0\})$.

\begin{prop}
We have
\begin{equation}\label{eq:RR-V-x}
\RR_{H_{x}}(V_{x},\Phi_x,\{0\})=(-1)^{m_x} \Cbb_{\delta(x)} \otimes \det(\mathbb{E}_x^{\hbox{\scriptsize\rm nci}})\otimes {\rm Sym}(\qgot/\qgot_{\mu^+}).
\end{equation}
\end{prop}

\begin{proof}
For $s\in [0,1]$, we consider the $H_x$-equivariant map $\Phi^s:V_x\to\hgot_x^*$ defined by the relations
$\Phi^s(v)= \mu^+ + s\Phi_{V_x}(v)$. The corresponding Kirwan vector field on $V_x$ is
$\kappa^s(v)= -\Phi^s(v)\cdot v$.
It is not difficult to see that there exists $\epsilon>0$ such that $\{\kappa^s=0\}\cap\{\|v\|\leq \epsilon\}=\{0\}$ for any
$s\in [0,1]$. Then a simple deformation argument gives that
$\widetilde{\RR}_{H_{x}}(V_{x},\Phi^s,\{0\})$ does not depend on $s\in [0,1]$. We have proved that
$$
\widetilde{\RR}_{H_{x}}(V_{x},\Phi^s,\{0\})=\widetilde{\RR}_{H_{x}}(V_{x},\mu^+,\{0\})
$$
where $\mu^+$ denotes the constant map $\Phi^0$. Standard computations  gives
$\widetilde{\RR}_{H_{x}}(V_{x},\mu^+,\{0\})={\rm Sym}(\qgot/\qgot_{\mu^+})$ (see \cite{pep-RR}[Proposition 5.4]). Our proof is completed. $\Box$
\end{proof}

 \subsubsection{Conclusion}

If we use the formulas  (\ref{eq:holomorphic induction}), (\ref{eq:cotangent induction}) and (\ref{eq:RR-V-x}) we obtain
the following expression
$$
Q_{Hx}(\lambda)=(-1)^{m_x}
{\rm Ind}_{H_{x}}^H\left( \Cbb_{\lambda_x+\delta(x)} \otimes \det(\mathbb{E}_x^{\hbox{\scriptsize\rm  nci}})\otimes {\rm Sym}(\qgot/\qgot_{\mu^+})\otimes\bigwedge_\Cbb \hgot/\hgot_{\mu^+}\right)
$$
in $\wR(H)$. Here $\Cbb_{\lambda_x}$ is the character of $G_x$ associated to the weight $\lambda_x=g\lambda$.

The previous formula  depends on a choice of a regular element $r$ in the Weyl chamber. In the next section we will propose another expression for $Q_{Hx}(\lambda)$ that does not depend on this choice.

 \subsection{Another expression for $Q_{Hx}(\lambda)$}\label{sec:Q-Hx-Lambda}

Let $\Rgot_x\subset \ggot_x^*$ be the roots for the action of the torus $G_x$ on $\ggot\otimes \Cbb$. The involution $\theta:\tgot^*\to\tgot^*$ leaves
the set $\Rgot_x$ invariant and a root $\alpha\in\Rgot_x$ is called
{\em imaginary} if $\theta(\alpha)=\alpha$. We denote respectively by $\Rgot_x^{\hbox{\scriptsize ci}}$ and by $\Rgot_x^{\hbox{\scriptsize nci}}$
the subsets of {\em compact imaginary} and {\em non-compact imaginary} roots.

We choose a generic element $r\in\tgot^*_+$ such that $\mu^+=(g\cdot r)^+$ satisfies the following relation : for any $\alpha\in\Rgot_x$, we have
$$
(\alpha,\mu^+)=0\Longleftrightarrow \theta(\alpha)=-\alpha.
$$

Notice that an imaginary roots $\alpha$ is positive if and only if $(\alpha,\mu^+)>0$.

\begin{defi}
We consider the subset  $\Agot_x\subset \Rgot_x$
$$
\alpha\in \Agot_x \ \Longleftrightarrow\  \alpha(\mu^+)>0,\ \theta(\alpha)\neq \alpha
$$
\end{defi}

The involution $\theta$ defines a free action of $\Zbb_2$ on the set $\Agot_x$. We denote by $\Agot_x/\Zbb_2$ its quotient.
For any $\alpha\in \Rgot_x$, we denote by $\Cbb_{\alpha}$ the corresponding $1$-dimensional representation of $G_x$,
and $\Cbb_{\alpha}\vert_{H_x}$ its restriction to the subgroup $H_x$. We have a natural map
$[\alpha]\in \Agot_x/\Zbb_2 \longmapsto \Cbb_{\alpha}\vert_{H_x}\in R(H_x)$.

For any $\alpha\in\Rgot_x$ we define
$$
\tilde{\alpha}=\pm \alpha
$$
where $\pm$ is the sign of $\alpha(\mu)\alpha(\theta(\mu))$.

We consider the $H_x$-modules $\hgot/\hgot_{\mu^+}:=([\hgot,\mu^+], J_{\mu^+})$, $\qgot/\qgot_{\mu^+}:=(V_{x}, J_{\mu^+})$  and $(V_x, J_{V_x})$.
\begin{lem}
We have the following isomorphisms of $H_x$-modules
\begin{eqnarray*}
\hgot/\hgot_{\mu^+}&\simeq& \bigoplus_{[\alpha]\in \Agot_x/\Zbb_2} \Cbb_{\alpha}\vert_{H_x}\oplus \bigoplus_{\alpha\in\Rgot_x^{\hbox{\scriptsize \rm ci}}\cap\Rgot^{+}_{x}}\Cbb_{\alpha}\vert_{H_x}\qquad \qquad[A],\\
\qgot/\qgot_{\mu^+}&\simeq& \bigoplus_{[\alpha]\in \Agot_x/\Zbb_2} \Cbb_{\alpha}\vert_{H_x}\oplus \bigoplus_{\alpha\in\Rgot_x^{\hbox{\scriptsize \rm nci}}\cap\Rgot^{+}_{x}} \Cbb_{\alpha}\vert_{H_x}\qquad \qquad[B],\\
(V_{x}, J_{V_x})&\simeq &\bigoplus_{[\alpha]\in \Agot_x/\Zbb_2} \Cbb_{\tilde{\alpha}}\vert_{H_x}\oplus \bigoplus_{\alpha\in\Rgot_x^{\hbox{\scriptsize \rm nci}}\cap\Rgot^{+}_{x}}
\Cbb_{\tilde{\alpha}}\vert_{H_x}\qquad \qquad [C].
\end{eqnarray*}
\end{lem}

\begin{proof}
Thanks to Lemma \ref{lem:V-x-complexe}, we know that the $H_x$-module $(V_{x}, J_{V_x})$ is isomorphic to the vector space $[\qgot,{\mu^+}]$ equipped with the complex structure $J_{V_x}:= J_{\mu^+}\circ S_x$.
We consider the vector spaces $[\qgot, \mu^+]$ and $[\ggot, \mu^+]$ equipped with the complex structure $J_{\mu^+}$.
The projection (taking the real part)
${\bf r}:\ggot\otimes \Cbb\to \ggot$ induces an isomorphism of $G_x$-modules
$$
{\bf r} : \bigoplus_{\alpha(\mu^+)>0} (\ggot\otimes \Cbb)_\alpha\quad \longrightarrow\quad [\ggot, \mu^+].
$$

The orthogonal projections ${\bf p}_1: [\ggot, \mu^+]\to  [\qgot, \mu^+]$ and ${\bf p}_2: [\ggot, \mu^+]\to  [\hgot, \mu^+]$ commutes with the $H_x$-action, so
the maps
\begin{eqnarray*}
{\bf p}_1\circ {\bf r}:\bigoplus_{\alpha(\mu^+)>0} (\ggot\otimes \Cbb)_\alpha &\longrightarrow&   [\qgot, \mu^+],\\
{\bf p}_2\circ {\bf r}:\bigoplus_{\alpha(\mu^+)>0} (\ggot\otimes \Cbb)_\alpha &\longrightarrow&  [\hgot, \mu^+]
\end{eqnarray*}
are surjective morphisms of $H_x$-modules.

Let $V^1_x(\alpha)={\bf p}_1\circ {\bf r}((\ggot\otimes \Cbb)_\alpha)$. We notice that $\dim_\Cbb V^1_x(\alpha)\in \{0,1\}$:
$V^1_x(\alpha)=\{0\}$ only if $\alpha$ is a non-compact imaginary root and $V^1_x(\alpha)\simeq \Cbb_{\alpha}\vert_{H_x}$ when $V^1_x(\alpha)\neq \{0\}$.
We notice also that $V^1_x(\alpha)=V^1_x(\theta(\alpha))$, hence
$$
\qgot/\qgot_{\mu^+}=([\qgot,{\mu^+}],J_{\mu^+})\simeq\bigoplus_{[\alpha]\in \Agot_x/\Zbb_2} V^1_x(\alpha)\oplus \bigoplus_{\alpha\in\Rgot_x^{\hbox{\scriptsize nci}}\cap\Rgot^{+}_{x}} V^1_x(\alpha).
$$
The identity $[B]$ is proved.

Similarly we consider $V^2_x(\alpha)={\bf p}_2\circ {\bf r}((\ggot\otimes \Cbb)_\alpha)$. We notice that $\dim_\Cbb V^2_x(\alpha)\in \{0,1\}$:
$V^2_x(\alpha)=\{0\}$ only if $\alpha$ is a compact imaginary root and $V^2_x(\alpha)\simeq \Cbb_{\alpha}\vert_{H_x}$ when $V^2_x(\alpha)\neq \{0\}$.
We notice also that $V^2_x(\alpha)=V^2_x(\theta(\alpha))$, hence
$$
\hgot/\hgot_{\mu^+}=([\hgot,{\mu^+}],J_{\mu^+})\simeq\bigoplus_{[\alpha]\in \Agot_x/\Zbb_2} V^2_x(\alpha)\oplus \bigoplus_{\alpha\in\Rgot_x^{\hbox{\scriptsize ci}}\cap\Rgot^{+}_{x}} V^2_x(\alpha).
$$
The identity $[A]$ is proved.

Finally we check that the complex structures $J_{\mu^+}$ and $J_{V_x}$ preserve each $V^1_x(\alpha)$ and that $(V_x(\alpha),J_{V_x})\simeq \Cbb_{\tilde\alpha}\vert_{H_x}$ when
$(\alpha,\mu^+)>0$. The identity $[C]$ follows. $\Box$
\end{proof}

\medskip

We consider the $H_x$-module $\mathbb{V}_x:= \sum_{[\alpha]\in \Agot_x/\Zbb_2}\Cbb_{\alpha}\vert_{H_x}$,
and the $G_x$-modules
$\mathbb{E}^{\hbox{\scriptsize nci}}_{x}:=\sum_{\alpha\in\Rgot_x^{\hbox{\scriptsize nci}}\cap\Rgot^{+}_{x}}\Cbb_{\alpha}$ and
$\mathbb{E}^{\hbox{\scriptsize ci}}_{x}:=\sum_{\alpha\in\Rgot_x^{\hbox{\scriptsize ci}}\cap\Rgot^{+}_{x}}\Cbb_{\alpha}$.
In the previous lemma we have proved that $H_x$-modules $\hgot/\hgot_{\mu^+}$ and $\qgot/\qgot_{\mu^+}$ are respectively isomorphic to
$\mathbb{V}_x\oplus \mathbb{E}^{\hbox{\scriptsize ci}}_{x}$ and  $\mathbb{V}_x\oplus \mathbb{E}^{\hbox{\scriptsize nci}}_{x}$. If we use the fact that
${\rm Sym}(\mathbb{V}_x)\otimes \bigwedge \mathbb{V}_x=1$, we get the
following corollary.

\begin{coro}\label{coro:simplification}
We have the following identity of virtual $H_x$-modules:
$$
{\rm Sym}(\qgot/\qgot_{\mu^+})\otimes \bigwedge \hgot/\hgot_{\mu^+}\simeq {\rm Sym}(\mathbb{E}^{\hbox{\scriptsize \rm nci}}_{x})\otimes \bigwedge \mathbb{E}^{\hbox{\scriptsize \rm ci}}_{x}.
$$
\end{coro}

\medskip

Let us now prove Lemma \ref{lem:wedge-J-compare}.

Let $\Bcal:=\Agot_x/\Zbb_2\bigcup (\Rgot_x^{\hbox{\scriptsize nci}}\cap\Rgot^{+}_{x})$. We see that
$\bigwedge_{J_{V_x}} V_x = \prod_{\alpha\in \Bcal}(1-t^{\tilde{\alpha}})$ whereas $\bigwedge_{-J_{\mu^+}} V_x = \prod_{\alpha\in \Bcal}(1-t^{-\alpha})$. Accordingly
we get $\bigwedge_{J_{V_x}} V_x\simeq (-1)^{|\Bcal' |}\, \Cbb_{\eta}\otimes \bigwedge_{-J_{\mu^+}} V_x$ where $\Bcal'=\{\alpha\in \Bcal,\tilde{\alpha}=\alpha\}$ and
$\eta=\sum_{\alpha\in\Bcal'}\alpha$. Now it is easy to check that an element $\alpha\in\Bcal$ belongs to $\Bcal'$ if and only if $\alpha$ and $\theta(\alpha)$ both belong to
$\Rgot_x^+$. In other words
$$
\Bcal'= \left\{\alpha\in\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+}),\ \theta(\alpha)\neq\alpha\right\}\slash \Zbb_2 \bigcup\Rgot_x^{\hbox{\scriptsize nci}}\cap\Rgot^{+}_{x}.
$$
We have proved that
$$
\bigwedge_{J_{V_x}} V_x\simeq (-1)^{m_x}\, \Cbb_{\delta(x)}\otimes\det(\mathbb{E}^{\hbox{\scriptsize \rm nci}}_{x})\otimes \bigwedge_{-J_{\mu^+}} V_x.
$$
$\Box$

\medskip

Finally, thanks to Lemma \ref{lem:wedge-J-compare} and Corollary \ref{coro:simplification}, we obtain the final formula for $Q_{Hx}(\lambda)$ (that does not depend on the choice of $r$):
$$
Q_{Hx}(\lambda)=(-1)^{m_x}
{\rm Ind}_{H_{x}}^H\left(\Cbb_{\lambda_x+\delta(x)} \otimes\det(\mathbb{E}^{\hbox{\scriptsize \rm nci}}_{x})\otimes  {\rm Sym}(\mathbb{E}^{\hbox{\scriptsize \rm nci}}_{x})\otimes\bigwedge \mathbb{E}^{\hbox{\scriptsize ci}}_{x}\right).
$$

\subsection{Computation of the virtual module  $\mathbb{M}_x(\lambda)$} \label{sec:formule-M-x}

According to Theorem \ref{theo-principal}, we have the decomposition $V_{\lambda}^{G}\vert_{H}=\sum_{\bar{x}} Q_{\bar{x}}(\lambda)$
where $Q_{\bar{x}}(\lambda)={\rm Ind}^{H}_{H_{x}}\left( \mathbb{A}_{x}(\lambda)\right)$, and $\mathbb{A}_{x}(\lambda)\in \wR(H_x)$ has the following description
$$
\mathbb{A}_{x}(\lambda)=
\frac{1}{|W^H_x|}\sum_{w\in W}(-1)^{m_{xw}} \Cbb_{\lambda_{xw}+\delta(xw)}
\otimes \det(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{xw})\otimes {\rm Sym}(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{xw}) \otimes \bigwedge
\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{xw}.
$$

The aim of this section is to simplify the expression of the virtual $H_x$-module $\mathbb{A}_{x}(\lambda)$. We start by comparing the $G_x$-modules $\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{xw}$ and
$\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}$. We use the decomposition
$\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}=\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w\oplus \left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^-_w$
where
$$
\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w:=\sum_{\alpha\in \Rgot_x^{\hbox{\tiny \rm nci}}\cap\Rgot^{+}_x\cap \Rgot^{+}_{xw}}\Cbb_{\alpha},\quad{\rm and}\quad
\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^-_w=\sum_{\alpha\in \Rgot_x^{\hbox{\tiny \rm nci}}\cap\Rgot^{+}_x\cap -\Rgot^{+}_{xw}}\Cbb_{\alpha}.
$$

We have the following basic lemma (see Lemma \ref{lem:formulation-M-lambda}).

\begin{lem} The $G_x$-module $\vert\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\vert_w:=\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w\oplus \overline{\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^-_w}$ is isomorphic to
$\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{xw}$.

\end{lem}

Let $\rho=\tfrac{1}{2}\sum_{\alpha\in\Rgot^+}\alpha$. We denote by $w\bullet\lambda=w(\lambda+\rho)-\rho$ the affine action of the Weyl group on the lattice
$\Lambda$.

The main result of this section is the following proposition.

\begin{prop}\label{prop:formule-M-lambda} Let $x\in Z_\theta$. We have
$$
\mathbb{A}_{x}(\lambda)=\mathbb{M}_x(\lambda)\otimes\Cbb_{\delta(x)} \otimes \bigwedge
\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}
$$
 where $\mathbb{M}_x(\lambda)\in \wR(H_x)$ is defined by the following expression

$$
\mathbb{M}_{x}(\lambda)=
\frac{(-1)^{n_x}}{|W^H_x|}\sum_{w\in W}(-1)^{k_{x,w}}\ \Cbb_{(w\bullet\lambda)_{x}}\otimes \det(\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w)
\otimes {\rm Sym}(|\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}|_w),
$$
and
\begin{itemize}
\item $k_{x,w}= |\Rgot^{+}_x\cap \Rgot^{+}_{xw}\cap\{\theta(\alpha)\neq \pm \alpha\}| +|\Rgot^{+}_x\cap \Rgot^{+}_{xw}\cap \Rgot_x^{\hbox{\scriptsize \rm ci}} |$,
\item $n_x:=| \theta(\Rgot^{+}_x)\cap \Rgot^{+}_x| - \frac{1}{2}| \theta(\Rgot^{+}_x)\cap \Rgot^{+}_x\cap\{\theta(\alpha)\neq\alpha\}|$.
\end{itemize}
\end{prop}

\begin{rem}
We can describe $Q_{\bar{x}}(\lambda)$ differently by taking $\{w_1,\cdots,w_p\}\subset W$ such that $W/W^H_x\simeq \{\bar{w}_1,\cdots,\bar{w}_p\}$. We have
$Q_{\bar{x}}(\lambda)={\rm Ind}^{H}_{H_{x}}\left( \tilde{\mathbb{A}}_{x}(\lambda)\right)$ with
$$
\tilde{\mathbb{A}}_{x}(\lambda)=\tilde{\mathbb{M}}_x(\lambda)\otimes\Cbb_{\delta(x)}\otimes\bigwedge
\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}
$$
and where $\tilde{\mathbb{M}}_x(\lambda)\in \wR(H_x)$ is defined by the following expression
$$
\tilde{\mathbb{M}}_{x}(\lambda)=(-1)^{n_{x}}\,\sum_{k=1}^p
(-1)^{k_{x,w_k}}\ \Cbb_{(w_k\bullet\lambda)_{x}}\otimes \det(\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_{w_k})
\otimes {\rm Sym}(|\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}|_{w_k}).
$$
\end{rem}

\medskip

We need to introduce some notations. To $x\in Z_\theta$, we associate~:

\begin{itemize}

\item The polarized roots : to $\alpha\in\Rgot_x$ and $w\in W$, we associate $|\alpha|_w\in\Rgot_x$ defined as follows
$$
|\alpha|_w=
\begin{cases}
\alpha \qquad \ \ {\rm if} \ \ \alpha \in \Rgot_{xw}^+,\\
-\alpha \qquad {\rm if} \ \ \alpha \notin \Rgot_{xw}^+.
\end{cases}
$$

\item The following $G_x$-weights :
$$
\gamma^{\hbox{\scriptsize \rm ci}}_{x,w}:=\sum_{\stackrel{\alpha\in \Rgot_x^{\hbox{\tiny \rm ci}}\cap \Rgot_x^+}{|\alpha|_w\neq\alpha}}\alpha, \quad
\gamma^{\hbox{\scriptsize \rm nci}}_{x,w}:=\sum_{\stackrel{\alpha\in \Rgot_x^{\hbox{\tiny \rm nci}}\cap \Rgot_x^+}{|\alpha|_w\neq\alpha}}\alpha, \quad
\gamma_{x,w}:=\sum_{\stackrel{\alpha\in \Rgot_x^+}{|\alpha|_w\neq\alpha}}\alpha.
$$

\end{itemize}
\medskip

The proof of Proposition \ref{prop:formule-M-lambda} is based on the following Lemma.

\begin{lem}\label{lem:formulation-M-lambda} Let $x\in Z_\theta$ and $w\in W$. Let $d^{\hbox{\scriptsize \rm ci}}_{x,w}$ be the cardinal of the set \break
$\{\alpha\in \Rgot_x^{\hbox{\scriptsize \rm ci}}\cap \Rgot_x^+ ,\ |\alpha|_w\neq\alpha\}$. We have the following relations
\begin{enumerate}
\item
$
\mathbb{E}^{\hbox{\scriptsize \rm nci}}_{xw}\simeq \sum_{\alpha\in \Rgot_x^{\hbox{\tiny \rm nci}}\cap \Rgot_x^+}\Cbb_{|\alpha|_w}
\quad {\rm and} \quad
\mathbb{E}^{\hbox{\scriptsize \rm ci}}_{xw}\simeq \sum_{\alpha\in \Rgot_x^{\hbox{\tiny \rm ci}}\cap \Rgot_x^+}\Cbb_{|\alpha|_w},
$

\item $\det(\mathbb{E}^{\hbox{\scriptsize \rm nci}}_{xw})=\Cbb_{-\gamma^{\hbox{\tiny \rm nci}}_{x,w}}\otimes \det(\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w)$,

\item $\bigwedge \mathbb{E}^{\hbox{\scriptsize \rm ci}}_{xw}=
(-1)^{d^{\hbox{\tiny \rm ci}}_{x,w} }\,\Cbb_{-\gamma^{\hbox{\tiny \rm ci}}_{x,w}}\otimes
\bigwedge \mathbb{E}^{\hbox{\scriptsize \rm ci}}_x$.
\item The $H_x$-weight $\delta(xw)-\delta(x)$ is equal to the restriction of  $G_x$-weight
$\gamma^{\hbox{\scriptsize \rm nci}}_{x,w}+\gamma^{\hbox{\scriptsize \rm ci}}_{x,w}- \gamma_{x,w}$ to $H_x$.

\end{enumerate}

\end{lem}

\begin{proof}
We remark that $\Rgot_x=\Rgot_{xw}$, $\Rgot_x^+=g(\Rgot^+)$ and $\Rgot_{xw}^+=g(w\Rgot^+)$. The first point follows and points 2. and 3. derive from the first.

Let us check the last point. The term $\rho_x:=\frac{1}{2}\sum_{\alpha\in\Rgot_x^+}\alpha$ is the image of
$\rho:=\frac{1}{2}\sum_{\alpha'\in\Rgot^+}\alpha'$ through the map $\mu\mapsto\mu_x$. We see that
$$
\rho_x+\theta(\rho_x)=\sum_{\alpha\in\Rgot_x^+\cap\theta(\Rgot_x^+)}\alpha=2\delta(x)+2\rho^{\hbox{\scriptsize \rm nci}}_x+2\rho^{\hbox{\scriptsize \rm ci}}_x
$$
where $\rho^{\hbox{\scriptsize \rm nci}}_x=\frac{1}{2}\sum_{\alpha\in\Rgot^{\hbox{\tiny \rm nci}}_x\cap\Rgot_x^+}\alpha$ and $\rho^{\hbox{\scriptsize \rm ci}}_x=
\frac{1}{2}\sum_{\alpha\in\Rgot^{\hbox{\tiny \rm ci}}_x\cap\Rgot_x^+}\alpha$. Similarly we have
$$
\rho_{xw}+\theta(\rho_{xw})=2\delta(xw)+2\rho^{\hbox{\scriptsize \rm nci}}_{xw}+2\rho^{\hbox{\scriptsize \rm ci}}_{xw}.
$$

Thus the $H_x$-weight $\delta(xw)-\delta(x)$ is equal to the restriction to $H_x$ of the $G_x$-weight
$$
\beta(x,w):= \rho_{xw}-\rho_{x}+ (\rho^{\hbox{\scriptsize \rm nci}}_{x}-\rho^{\hbox{\scriptsize \rm nci}}_{xw})+
(\rho^{\hbox{\scriptsize \rm ci}}_{x}-\rho^{\hbox{\scriptsize \rm ci}}_{xw}).
$$
We notice that $ \rho_{xw}-\rho_{x}= (w\rho-\rho)_x= -\gamma_{x,w}$. Furthermore, small computations gives that $\rho^{\hbox{\scriptsize \rm nci}}_{x}-\rho^{\hbox{\scriptsize \rm nci}}_{xw}= \gamma^{\hbox{\scriptsize \rm nci}}_{x,w}$ and
$\rho^{\hbox{\scriptsize \rm ci}}_{x}-\rho^{\hbox{\scriptsize \rm ci}}_{xw}= \gamma^{\hbox{\scriptsize \rm ci}}_{x,w}$. We have proved that
$\beta(x,w)= \gamma^{\hbox{\scriptsize \rm nci}}_{x,w}+\gamma^{\hbox{\scriptsize \rm ci}}_{x,w}- \gamma_{x,w}$. The last point follows. $\Box$
\end{proof}

\bigskip

Now, we can finish the proof of the Proposition \ref{prop:formule-M-lambda}.  We must check that the virtual $H_x$-module
$$
{\bf A}:=(-1)^{m_{xw}} \Cbb_{\lambda_{xw}+\delta(xw)}\otimes\det(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{xw})
\otimes \bigwedge \mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{xw}
$$
is equal to the virtual $H_x$-module
$$
{\bf B}:=(-1)^{n_x+k_{x,w}}\ \Cbb_{(w\bullet\lambda)_{x}+\delta(x)}\otimes \det(\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w)\otimes \bigwedge \mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}.
$$

If we use the previous Lemma, we get
$$
{\bf A}=
(-1)^{m_{xw}+d^{\hbox{\tiny \rm ci}}_{x,w}} \Cbb_{(w(\lambda +\rho)-\rho)_x +\delta(x)}\otimes \det(\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w)\otimes \bigwedge \mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}.
$$
Thus the equality ${\bf A}={\bf B}$ follows from the following lemma.

\begin{lem}
For any $x\in Z_\theta$ and $w\in W$, we have $n_x+k_{x,w}=m_{xw}+d^{\hbox{\tiny \rm ci}}_{x,w}\ {\rm mod}\ 2$.
\end{lem}

\begin{proof} In order to simplify our notations, we write $a\equiv b$ for $a=b \ {\rm mod}\ 2$.

We have $\dim \mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}= \dim \mathbb{E}^{\hbox{\scriptsize\rm nci}}_{xw}$ and
$\dim \mathbb{E}^{\hbox{\scriptsize\rm ci}}_{x}= \dim \mathbb{E}^{\hbox{\scriptsize\rm ci}}_{xw}$, then
\begin{eqnarray*}
\lefteqn{ m_{xw}-m_x}\\
&=& \frac{1}{2}\left(| \Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})\cap\{\theta(\alpha)\neq\alpha\}| -
| \Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap\{\theta(\alpha)\neq\alpha\}|\right)\\
&=& \frac{1}{2}\left(| \Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})| -
| \Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})|\right).
\end{eqnarray*}
We remark now that
$$
\Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})= A_{++}\cup A_{--}\cup A_{+-}\cup A_{-+}
$$
with $A_{++}=\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap \Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})$,
$A_{--}=-\Rgot_{x}^{+}\cap-\theta(\Rgot_{x}^{+})\cap \Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})$,
$A_{+-}=\Rgot_{x}^{+}\cap\theta(-\Rgot_{x}^{+})\cap \Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})$ and
$A_{-+}=-\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap \Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})$.

Similarly we have
$$
\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})= B_{++}\cup B_{--}\cup B_{+-}\cup B_{-+}
$$
with $B_{++}=\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap \Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})$,
$B_{--}=\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap -\Rgot_{xw}^{+}\cap\theta(-\Rgot_{xw}^{+})$,
$B_{+-}=\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap \Rgot_{xw}^{+}\cap\theta(-\Rgot_{xw}^{+})$ and
$B_{-+}=\Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap -\Rgot_{xw}^{+}\cap\theta(\Rgot_{xw}^{+})$.

We have the obvious relations : $A_{++}=B_{++}$, $A_{--}=-B_{--}$, $\theta(A_{+-})=A_{-+}$, $\theta(B_{+-})=B_{-+}$ and
$A_{++}=B_{++}$. So we get
$$
m_{xw}-m_x\equiv |A_{+-}| +|B_{+-}|.
$$
Let consider $\Acal:=\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}$ and $\Bcal:= \Rgot_{x}^{+}\cap -\Rgot_{xw}^{+}$. We have
\begin{eqnarray*}
m_{xw}-m_x &\equiv& |\Acal\cap \theta(\Bcal)|+ |\Acal\cap -\theta(\Bcal)| \\
&\equiv& |\Acal | +  |\Acal\cap \theta(\Acal)|+ |\Acal\cap -\theta(\Acal)|.
\end{eqnarray*}
Now we remark that
\begin{eqnarray*}
|\Acal\cap \theta(\Acal)| &\equiv& |\Acal\cap \theta(\Acal)\cap\{\theta(\alpha)= \alpha\}| \\
&\equiv& |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)= \alpha\}|.
\end{eqnarray*}
Similarly
\begin{eqnarray*}
|\Acal\cap -\theta(\Acal)| &\equiv& |\Acal\cap -\theta(\Acal)\cap\{\theta(\alpha)= -\alpha\}|\\
&\equiv& |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)= -\alpha\}| .
\end{eqnarray*}

At this stage we have proved that
\begin{eqnarray*}
\lefteqn{m_{xw}-m_x}\\
& \equiv & | \Rgot_{x}^{+}\cap \Rgot_{xw}^{+}| +
|\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)= \alpha\}|  +
|\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)= -\alpha\}|\\
&\equiv& |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)\neq -\alpha\}| + |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)= \alpha\}|.
\end{eqnarray*}

As $d^{\hbox{\scriptsize \rm ci}}_{x,w}= |\Rgot_{x}^{+}\cap -\Rgot_{xw}^{+}\cap \Rgot_x^{\hbox{\scriptsize ci}}|$, we have
$|\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)= \alpha\}|+ d^{\hbox{\scriptsize \rm ci}}_{x,w}$ is equal to
$\dim \mathbb{E}^{\hbox{\scriptsize ci}}_x+ |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap \Rgot_x^{\hbox{\scriptsize nci}}|$. This implies that
$m_{xw}+ d^{\hbox{\scriptsize \rm ci}}_{x,w}$ is equal, modulo $2$, to
$$
m_x + \dim \mathbb{E}^{\hbox{\scriptsize ci}}_x+ |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap \Rgot_x^{\hbox{\scriptsize nci}}|
+ |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)\neq -\alpha\}|
$$
$$
\equiv m_x + \dim \mathbb{E}^{\hbox{\scriptsize ci}}_x+ |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap \Rgot_x^{\hbox{\scriptsize ci}}|
+ |\Rgot_{x}^{+}\cap \Rgot_{xw}^{+}\cap\{\theta(\alpha)\neq \pm \alpha\}|.
$$
By definition $m_x=\frac{1}{2}| \Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap\{\theta(\alpha)\neq\alpha\}| + \dim \mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}$ and then
\begin{eqnarray*}
m_x + \dim \mathbb{E}^{\hbox{\scriptsize ci}}_x
&\equiv& \frac{1}{2}| \Rgot_{x}^{+}\cap\theta(\Rgot_{x}^{+})\cap\{\theta(\alpha)\neq\alpha\}|+
|\Rgot_{x}^{+}\cap \{\theta(\alpha)=\alpha\}|\\
&\equiv& n_x.
\end{eqnarray*}

Finally we have proved that $m_{xw}+ d^{\hbox{\scriptsize \rm ci}}_{x,w}$ is equal, modulo $2$, to $n_x+k_{x,w}$. $\Box$

\end{proof}

\section{Examples}

In this section we will study in details some examples of our formula
$$
V_{\lambda}^{G}\vert_{H}=\sum_{\bar{x}\,\in\, H\backslash Z_{\theta}\slash W} Q_{\bar{x}}(\lambda)
$$
where $Q_{\bar{x}}(\lambda)={\rm Ind}^{H}_{H_{x}}\left(\mathbb{M}_x(\lambda)\otimes\Cbb_{\delta(x)} \otimes \bigwedge
\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x}\right)$ and
$$
\mathbb{M}_{x}(\lambda)=
\frac{(-1)^{n_x}}{|W^H_x|}\sum_{w\in W}(-1)^{k_{x,w}}\ \Cbb_{(w\bullet\lambda)_{x}}\otimes \det(\left(\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}\right)^+_w)
\otimes {\rm Sym}(|\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}|_w).
$$

Here the integers $k_{x,w}$ and $n_x$ are defined as follows:
\begin{itemize}
\item $k_{x,w}= |\Rgot^{+}_x\cap \Rgot^{+}_{xw}\cap\{\theta(\alpha)\neq \pm \alpha\}| +|\Rgot^{+}_x\cap \Rgot^{+}_{xw}\cap \Rgot_x^{\hbox{\scriptsize \rm ci}} |$,
\item $n_x=| \theta(\Rgot^{+}_x)\cap \Rgot^{+}_x| - \frac{1}{2}| \theta(\Rgot^{+}_x)\cap \Rgot^{+}_x\cap\{\theta(\alpha)\neq\alpha\}|$.
\end{itemize}

\subsection{$K\subset K\times K$}

Let $K$ be a connected compact Lie group. Here we work with the Lie group $G=K\times K$ and the involution $\theta(k_1,k_2)=(k_2,k_1)$.
The subgroup $H=G^\theta$ is the group $K$ embedded diagonally in $G$.

Let $T$ be a maximal torus of $K$ and let $W_K=N_K(T)/T$ be the Weyl group of $K$. We denote by $\Rgot_K$ the set of roots for $(K,T)$, and we make the choice of a set $\Rgot_K^+$ of positive roots.

In the next lemma we describe the critical set $Z_\theta$ in the flag manifold $\Fcal=K/T\times K/T$ of $G$.

\begin{lem}
We have $Z_\theta= \bigcup_{w\in W_K} Z_w$ with $Z_w=K\cdot(wT,T)$. In other words, the set $H\backslash Z_{\theta}\slash W$ is a singleton.
\end{lem}

\begin{proof}
The element $x=(aT,bT)\in \Fcal$ belongs to $Z_\theta$ if and only if $=(a^{-1}b,b^{-1}a)\in W\times W$. If $b^{-1}a=w\in W$ then
$(aT,bT)\in Z_w$. $\Box$
\end{proof}

\bigskip

We take $x=(T,T)\in Z_\theta$. For each $w\in W_K$, we write $xw=(wT,T)$. We take
$\lambda=(a,b)\in \Lambda_K^+\times\Lambda^+_K=\widehat{G}$.

Our data are as follows:

$\bullet$ the group $G_{x}$ is the maximal torus $T\times T\subset K$,

$\bullet$ the group $H_{x}$ is the maximal torus $T\subset K$,

$\bullet$ $\Cbb_{(w\bullet\lambda)_x+\delta(x)}=\Cbb_{w(a+\rho)+b+\rho}$ as a character of $T$,

$\bullet$ $n_{x}=|\Rgot_K^+|$,

$\bullet$ $k_{x,w}$ is equal to $|w\Rgot_K^+\cap\Rgot_K^+|+ |\Rgot_K^+|$, so $(-1)^{k_{x,w}}=(-1)^w$,

$\bullet$ the vector spaces $\mathbb{E}^{\hbox{\scriptsize\rm ci}}_{x},\mathbb{E}^{\hbox{\scriptsize\rm nci}}_{x}$ are reduced to $\{0\}$.

\medskip

In this context we recover the classical relation
\begin{equation}\label{eq:general-Clebsch-Gordan}
V^K_a\otimes V_{b}^K=(-1)^{\dim(K/T)/2}\sum_{w\in W_K}(-1)^w\, \ind_T^K\left(\Cbb_{w(a+\rho)+b+\rho} \right).
\end{equation}

\medskip

\begin{example}\label{SU(2)-Clebsch-Gordan}
The irreducible representation $SU(2)$ are parametrized by $\Nbb$. If $n\geq 0$, the irreducible representation
$V_n$ of $SU(2)$ satisfies
$$
V_n=\ind_{U(1)}^{SU(2)}((\Cbb_0-\Cbb_2)\otimes\Cbb_n).
$$
If we take $m\geq n\geq 0$, then (\ref{eq:general-Clebsch-Gordan}) gives
\begin{eqnarray*}
V_n\otimes V_m&=& \ind_{U(1)}^{SU(2)}(\Cbb_{m-n}) - \ind_{U(1)}^{SU(2)}(\Cbb_{m+n+2})\\
&=& \sum_{k=0}^n\ind_{U(1)}^{SU(2)}((\Cbb_0-\Cbb_2)\otimes\Cbb_{m+n-2k})\\
&=& \sum_{k=0}^n \, V_{m+n-2k}.
\end{eqnarray*}
We recognize here the classical Clebsch-Gordan relations.
\end{example}

\subsection{$U(p)\times U(q)\subset U(p+q)$}

Let $p\geq q\geq 1$ and $n=p+q$. We take $G=U(n)$ with maximal torus $T\simeq U(1)^{n}$ the subgroup formed by the diagonal matrices.
We use the canonical map $\tau$ from the symmetric group $\mathfrak{S}_n$ into $G$. It induces an isomorphism between $\mathfrak{S}_n$ and the
Weyl group $W$ of $G$.

We work with the involution $\theta(g)=\Delta g \Delta^{-1}$ where $\Delta:={\rm diag}(I_p,-I_q)$: the subgroup fixed by $\theta$ is $H=U(p)\times U(q)$.

In the next section we describe the critical set $Z_\theta \subset \Fcal$. For another type of parametrization of
$H_\Cbb\backslash \Fcal$, see Section 5 of \cite{Richardson-Springer93}.

\subsubsection{The  critical set}

We consider the following elements of $O(2)$:
$$
R= \left(\begin{array}{cc}
\tfrac{1}{\sqrt{2}}& \tfrac{1}{\sqrt{2}}\\
\tfrac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{array}
\right),
\quad
S=\left(\begin{array}{cc} 0& 1\\ 1& 0\end{array}\right),
\quad
J= \left(\begin{array}{cc}
0& -1\\
1 &0
\end{array}
\right).
$$

The element $R$ is of order $8$, $R^2=-J$ and
$R^{-1}\left(\begin{array}{cc} 1& 0\\ 0& -1\end{array}\right)R=S$.

To any $j\in\{0,\ldots,q\}$ we associate :
\begin{itemize}
\item $g_j:={\rm diag}(\underbrace{1,\ldots,1}_{p-j\ {\rm times}},\underbrace{R,\ldots,R}_{j\ {\rm times}},\underbrace{1,\ldots,1}_{q-j\ {\rm times}})\in G$,

\item the permutation $w_j\in \mathfrak{S}_n$ that fixes the elements of $[1,\cdots, p-j]$\break $\cup[p+j+1,\cdots,n]$ and such that
$$
w_j(p-j+2k-1)=p-j+k,\quad w_j(p-j+2k)=p+k,\quad {\rm for}\quad 1\leq k\leq j,
$$

\item $k_j= \tau_j g_j\in G$, where $\tau_j\in N(T)$ is the image of $w_j$ by $\tau$,

\item $x_j=k_j T\in \Fcal$.
\end{itemize}

We notice that the adjoint map $Ad(\tau_j):G\to G$ sends the matrix
${\rm diag}(a_1,\ldots,a_{p-j},b_1,\ldots,b_{2j},c_1,\ldots,c_{q-j})$ to the matrix
$$
{\rm diag}(a_1,\ldots,a_{p-j},b_1,b_3,\ldots,b_{2j-1},b_2,b_4,\ldots,b_{2j},c_1,\ldots,c_{q-j}).
$$
We see then that
$$
\sigma_j:=k_j^{-1}\Delta k_j=
{\rm diag}(\underbrace{1,\ldots,1}_{p-j\ {\rm times}},\underbrace{S,\ldots,S}_{j\ {\rm times}},\underbrace{-1,\ldots,-1}_{q-j\ {\rm times}})
$$
and $k_j^{-1}\theta(k_j)=\sigma_j \Delta$ belong to $N(T)$. Thus the elements $x_0,\ldots,x_q$ belongs to $Z_\theta$.

\begin{lem}
In the flag manifold $\Fcal$ the set $Z_\theta$ has the following description:
$$
Z_\theta=\bigcup_{0\leq j\leq q} \ \bigcup_{\bar{w}\in W_{x_j}\backslash W} H x_j w
$$
So we have  $H\backslash Z_{\theta}\slash W=\{\bar{x}_0,\ldots,\bar{x}_q\}$.
\end{lem}
\begin{proof} If $1\leq a<b\leq n$, we denote by $\tau_{a,b}\in N(T)$ the permutation matrix associated to the transposition $(a,b)$.

Let $gT\in Z_{\theta}$. Then $k:=g^{-1}\theta(g)\Delta=g^{-1}\Delta g$ is an element of order two in $N(T)$.
The Weyl group element $\bar{k}\in W$ is of order two,
then there exists $0\leq l\leq n/2$, and a family $(a_1<b_1),\ldots,(a_l<b_l)$ of disjoint couples in $\{1,\ldots,n\}$  such that
$k T= \tau_{a_1,b_1}\ldots \tau_{a_l,b_l}T$.

Now, if we use the fact that the characteristic polynomial of $k\in G$ is equal to $(X-1)^p(X+1)^q$ with  $p\geq q\geq 1$, we see that

$\bullet $ $l\leq q$,

$\bullet$ there exists $n\in N(T)$ such that $n^{-1}kn= \sigma_l=k_l^{-1} \Delta k_l$.

If we take $w=\bar{n}\in W$, the previous identity says that $g\in Hk_l w T$. $\Box$
\end{proof}

\medskip

\subsubsection{Localized indices}

We work with the groups $T\subset H=U(p)\times U(q)\subset G=U(n)$ and the corresponding Lie algebras
$\tgot\subset \hgot\subset \ggot$. Let $\Rgot=\{\varepsilon_r-\varepsilon_s\}$ be the set of non-zero roots for
the action of $T$ on $\ggot\otimes\Cbb$.
We choose the Weyl chamber so that $\Rgot^+:=\{\varepsilon_r-\varepsilon_s,\ 1\leq r<s\leq n\}$.

Let $j\in\{0,\ldots,q\}$. The aim of this section is to compute the localized index $Q_{\bar{x}_j}(\lambda)\in \wR(H)$.
In order to have a fairly simple expression we will rewrite the terms of the form
${\rm Ind}^{H}_{H_{x_j}}(\Cbb_{\beta}\otimes \bigwedge
\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x_j})$.

We write $\{1,\ldots, n\}=I^1_j\cup I^2_j\cup I^3_j\cup I^4_j$ where
$I^1_j=\{1\leq k\leq p-j\}$, $I^2_j=\{p-j+1\leq k\leq p\}$,
$I^3_j=\{p+1\leq k\leq p+j\}$, and $I^4_j=\{p+j+1\leq k\leq n\}$.

For the maximal torus $T\subset G$ we have a decomposition
$$
T\simeq T^1_{j}\times T^2_{j}\times T^3_{j}\times T^4_{j}
$$
where $T^p_{j}=\{(t_k), k\in I^p_j\}$. Let $T_{j} \subset T^2_{j}\times T^3_{j}$ be the subtorus defined by the relations :
$((t_k)_{k\in I^2_j},(s_{k'})_{k'\in I^3_j}) \in T_j$
if and only if $t_{p-j+k}=s_{p+k}$ for all $1\leq k\leq j$.

\medskip

The elements of order two $\sigma_j\in G$ induce involutions on $G$ (by conjugation) that we still denote by $\sigma_j$. We start with a basic lemma whose proof is left to the reader.

\begin{lem} \label{lem:calcul_x_j}
Let $x_j=k_j T\in \Fcal$.

$\bullet$ The adjoint map $Ad(k_j):\ggot\to\ggot$ realizes an isomorphism between the vector space $\tgot$ equipped with the involution induced by $\sigma_j$ and
the vector space $\ggot_{x_j}$ equipped with the involution $\theta$.

$\bullet$ The group $N(T)^{\sigma_j}/T^{\sigma_j}$ is isomorphic with
$\mathfrak{S}_{p-j}\times\mathfrak{S}_{q-j}\times\mathfrak{S}_{j}\times \{\pm\}^j$.

$\bullet$ The adjoint map $Ad(k_j):G\to G$ induces an isomorphism $N(T)^{\sigma_j}/T^{\sigma_j}$ $\simeq W_{x_j}$.

$\bullet$ The stabilizer subgroup $H_{x_j}$ is equal to $T^1_{j}\times T_{j}\times T_{j}^4 \subset T$.

$\bullet$ If $\Cbb_\alpha$ is a character of $T$, then $\Cbb_{k_j\alpha}$ is a character of $G_{x_j}$ and $\Cbb_{\tau_j\alpha}$ is  a character of
$T$. We have the relation
$$
\Cbb_{k_j\alpha}\vert_{H_{x_j}}=\Cbb_{\tau_j\alpha}\vert_{H_{x_j}}.
$$

$\bullet$ The set of roots $\Rgot_{x_j}^{\hbox{\scriptsize \rm ci}}$ is equal to
$$
 k_j\cdot\left\{\varepsilon_r-\varepsilon_s, 1\leq r<s\leq p-j\right\}
\bigcup k_j\cdot\left\{\varepsilon_r-\varepsilon_s, p+j+1\leq r<s\leq n\right\}
$$
and $\Rgot_{x_j}^{\hbox{\scriptsize \rm nci}}= k_j\cdot\left\{\varepsilon_r-\varepsilon_s, 1\leq r\leq p-j\ \&\ p+j+1\leq s\leq n \right\}$.

\end{lem}

We denote by $\mathbb{M}_j$ the $T$-module $\Cbb^{p-j}\otimes (\Cbb^{q-j})^*$ where the subgroup $T^2_{j}\times T^3_{j}$ acts trivially and
the $T^1_{j}\times T^4_{j}$-action is the canonical one. Thank to Lemma \ref{lem:calcul_x_j}, we have the following isomorphisms of $H_{x_j}$-modules:
$$
\mathbb{E}^{\hbox{\scriptsize \rm nci}}_{x_j}\simeq \mathbb{M}_j.
$$

We consider the Lie group

$$
K_j:=U(p-j)\times U(q-j)
$$
that we view as a subgroup of $H$ in such a way that $T^1_{j}\times T^4_{j}$ is a maximal torus of $K_j$.
A set of positive roots for $(K_j,T^1_{j}\times T^4_{j})$ is
$\varepsilon_r-\varepsilon_s$ for $1\leq r<s\leq p-j$ and $p+j+1\leq r<s\leq n $. We equip $\kgot_j\slash\tgot^1_j\times \tgot^4_j$
with a complex structure
such that
$$
\mathbb{E}^{\hbox{\scriptsize \rm ci}}_{x_j}\simeq \kgot_j\slash \tgot^1_j\times \tgot^4_j
$$
is an isomorphism of $T_j^1\times T_j^4$-modules.

The holomorphic induction map ${\rm Hol}^{K_j}_{T^1_{j}\times T^4_{j}}: R(T^1_{j}\times T^4_{j})\to R(K_j)$ is defined as follows:
$$
{\rm Hol}^{K_j}_{T^1_{j}\times T^4_{j}}(V):={\rm Ind}^{K_j}_{T^1_{j}\times T^4_{j}}(V\otimes \bigwedge\kgot_j\slash \tgot^1_j\times \tgot^4_j).
$$
If $a=(a_1\geq \cdots\geq a_{p-j})\in \Zbb^{p-j}$ and
$b= (b_1\geq \cdots\geq b_{q-j})\in \Zbb^{q-j}$, then $\Cbb_{(a,b)}$ defines a character of $T^1_{j}\times T^4_{j}$ and
$$
{\rm Hol}^{K_j}_{T^1_{j}\times T^4_{j}}\left(\Cbb_{(a,b)}\right)= V_a^{U(p-j)}\otimes V_b^{U(q-j)}
$$
is the irreducible representation of $K_j$ with highest weight $(a,b)$.

A character $\Cbb_{\beta}$ of the torus $T$ can we written
$\Cbb_{\beta}= \Cbb_{\beta^{14}}\otimes \Cbb_{\beta^{23}}$ where $\Cbb_{\beta^{14}}$ is a character of
$T^1_j\times T_j^4$ and $\Cbb_{\beta^{23}}$ is a character of $T^2_j\times T_j^3$. Note that
$\Cbb_{\tau_j\beta}\vert_{H_{x_j}}= \Cbb_{\beta^{14}}\otimes \Cbb_{\beta'}$ where
$\beta'=\tau_j\beta^{23}$ defines a character of $T_j\subset T^2_j\times T_j^3$.

\begin{lem}
Let $\Cbb_{\beta}$ be a character of
$T$. Then ${\rm Ind}^{H}_{H_{x_j}}(\Cbb_{\beta}\vert_{H_{x_j}}\otimes \bigwedge
\mathbb{E}^{\hbox{\scriptsize\rm  ci}}_{x_j})$ is equal to
$$
{\rm Ind}^{H}_{K_j\times T^2_{j}\times T^3_{j}}\left( {\rm Hol}^{K_j}_{T^1_{j}\times T^4_{j}}\left(\Cbb_{\beta^{14}}\right)\otimes \Cbb_{\beta^{23}}\otimes
L^2(T^2_{j}\times T^3_{j}/T_j)\right),
$$
where $L^2(T^2_{j}\times T^3_{j}/T_j)={\rm Ind}^{T^2_{j}\times T^3_{j}}_{T_j}(1)\in \widehat{R}(T^2_{j}\times T^3_{j})$.
\end{lem}

\begin{rem}To gain some space in our formulas, we will write ${\rm Hol}^{K_j}_{T^1_{j}\times T^4_{j}}\left(\Cbb_{\beta}\right)$ instead of
${\rm Hol}^{K_j}_{T^1_{j}\times T^4_{j}}\left(\Cbb_{\beta^{14}}\right)\otimes \Cbb_{\beta^{23}}$
\end{rem}

We need to fix some notations.

\begin{defi}
$\bullet$ Let $\chi: H\to\Cbb$ be the character $(A,B)\mapsto \det(A)\det(B)^{-1}$.

$\bullet$ Let $\psi_j$ be the character\footnote{Remark that $\psi_j$ is trivial $T^1_{j}\times T^3_{j}\times T^4_{j}$.}  of $T$ associated to the weight
$$
\sum_{1\leq k\leq j} (q-p+2+2j-4k)\varepsilon_{p-j+k}.
$$

$\bullet$ For any $(j,w)\in [0,q]\times W$, we define the integer $d_{j,w}$ by the relation
$$
d_{j,w}=\dim (\mathbb{M}_j)_w^+ + | \{1\leq k\leq j,\ w^{-1}(p-j+2k-1)<w^{-1}(p-j+2k)\} |.
$$
\end{defi}

A small computation gives the following lemma.

\begin{lem}
\begin{itemize}
\item The $H_{x_j}$-character $\Cbb_{\delta(x_j)}$ is equal to
$\chi^{\otimes j}\otimes \psi_j   \vert_{H_{x_j}}$.

\item For any $(j,w)\in [0,q]\times W$, we have
$$
(-1)^{n_{x_j}+k_{x_j,w}}= (-1)^{j(n+1)}(-1)^w (-1)^{d_{j,w}}.
$$
\end{itemize}
\end{lem}

The main result of this section is the following proposition.

\begin{prop}
$$
V_{\lambda}^{U(n)}\vert_{U(p)\times U(q)}=\sum_{j=0}^q Q_{\bar{x}_j}(\lambda)
$$
where $Q_{\bar{x}_j}(\lambda)\in \widehat{R}(U(p)\times U(q))$ is determined by the relation
$$
Q_{\bar{x}_j}(\lambda)=\frac{(-1)^{j(n+1)}}{|W_{x_j}|}\sum_{w\in W}(-1)^w(-1)^{d_{j,w}} \
{\rm Ind}^{U(p)\times U(q)}_{K_j\times T^2_{j}\times T^3_{j}}\left(\mathbb{A}_{j}^w(\lambda)\otimes \psi_j\right)\otimes \chi^{\otimes j}.
$$
Here the elements $\mathbb{A}_{j}^w(\lambda)\in \widehat{R}(K_j\times T^2_{j}\times T^3_{j})$ are defined as follows:
$$
\mathbb{A}_{j}^w(\lambda)={\rm Hol}^{K_j}_{T^1_{j}\times T^4_{j}}\left(\Cbb_{\tau_j(w\bullet\lambda)}
\otimes \det((\mathbb{M}_{j})^+_w)\otimes {\rm Sym}(|\mathbb{M}_{j}|_w)\right)\otimes L^2(T^2_{j}\times T^3_{j}/T_j).
$$
\end{prop}

We finish this section by considering particular situations.

\subsubsection{The extreme cases : $j=0$ or $j=q$}

When $j=0$, the torus $T_0^2$ and $T_0^3$ are trivial and $K_0=U(p)\times U(q)=H$. Moreover
$\mathbb{M}_0=\Cbb^{p}\otimes(\Cbb^q)^*$ and $d_{0,w}=\dim (\mathbb{M}_0)_w^+$. Thanks to Lemma \ref{lem:calcul_x_j}, we know also that
$W_{x_0}\simeq \mathfrak{S}_p\times \mathfrak{S}_q$.

So we get the formula
$$
Q_{\bar{x}_0}(\lambda)=\frac{1}{p!q!}
\sum_{w\in W}(\pm)_w\
{\rm Hol}^{H}_{T}\left(\Cbb_{w\bullet\lambda}
\otimes \det((\mathbb{M}_{0})^+_w)\otimes {\rm Sym}(|\mathbb{M}_{0}|_w)\right)
$$
where $(\pm)_w= (-1)^w(-1)^{\dim (\mathbb{M}_0)_w^+}$.

\begin{rem}An useful exercise is to consider the term
$$\mathbb{A}_w:=(\pm)_w\
{\rm Hol}^{H}_{T}\left(\Cbb_{w\bullet\lambda}
\otimes \det((\mathbb{M}_{0})^+_w)\otimes {\rm Sym}(|\mathbb{M}_{0}|_w)\right)$$ and  verify that
$\mathbb{A}_{w'w}=\mathbb{A}_w$ when $w'\in W_{x_0}$.
\end{rem}

\medskip

When $j=q$, the torus $T_q^4$ is trivial, $K_q=U(p-q)$ and $\mathbb{M}_q=\{0\}$. Moreover  $W_{x_q}\simeq \mathfrak{S}_{p-q}
\times \mathfrak{S}_q\times \{\pm\}^q$. In this case we obtain
$$
Q_{\bar{x}_1}(\lambda)= \frac{(-1)^{q(n+1)}}{(p-q)!q! 2^q}\sum_{w\in W}(-1)^w(-1)^{d_{q,w}} \ \chi^{\otimes q} \otimes Q_q^w(\lambda)
$$
with
$$
Q_q^w(\lambda)={\rm Ind}^{U(p)\times U(q)}_{U(p-q)\times T^2_{q}\times T^3_{q}}
\left({\rm Hol}^{U(p-q)}_{T^1_{q}}\left(\Cbb_{\tau_q(w\bullet\lambda)}\right)\otimes\psi_q \otimes L^2(T^2_{q}\times T^3_{q}/T_q)\right).
$$

\subsubsection{$U(n-1)\times U(1)\subset U(n)$}

Here we are in the case where $q=1$, and so
$$
V_{\lambda}^{U(n)}\vert_{U(n-1)\times U(1)}=Q_{\bar{x}_0}(\lambda)+ Q_{\bar{x}_1}(\lambda).
$$

To simplify the expression of $Q_{\bar{x}_0}(\lambda)$ we use the fact that the quotient $W_{x_0}\backslash W$ is represented by the class of the
elements $\tau_{k,n}\in G$ associated to the transposition $(k,n)$ for $1\leq k\leq n$. We write $T=T'\times U(1)$ where $T'$ is a maximal torus of $U(n-1)$. The $T'$-module
$\Cbb^{n-1}$ can be decomposed as $\mathbb{V}_k\oplus\mathbb{V}'_k$ where $\mathbb{V}_k=\sum_{j=1}^{k-1}\Cbb_{\varepsilon_j}$ and
$\mathbb{V}'_k=\sum_{j=k}^{n-1}\Cbb_{\varepsilon_j}$.

The $T$-module $\mathbb{M}_0$ is equal to $\Cbb^{n-1}\otimes \Cbb^*=\mathbb{V}_k\otimes \Cbb_{-\varepsilon_n}\oplus
\mathbb{V}'_k\otimes \Cbb_{-\varepsilon_n}$ and the polarized $T$-module $|\mathbb{M}_{0}|_{\tau_{k,n}}$ is equal to
$\mathbb{V}_k\otimes \Cbb_{-\varepsilon_n}\oplus
\overline{\mathbb{V}'_k}\otimes \Cbb_{\varepsilon_n}$.
We have $\dim (\mathbb{M}_{0})^+_{\tau_{k,n}}= k-1$ and $\det(\mathbb{M}_{0})^+_{\tau_{k,n}}= \Cbb_{\mu_k}\otimes \Cbb_{\varepsilon_n}^{\otimes 1-k}$ with
$\mu_k=\sum_{j=1}^{k-1}\Cbb_{\varepsilon_j}$.

So we obtain
\begin{eqnarray*}
   \lefteqn{Q_{\bar{x}_0}(\lambda)=} \\
   &&\sum_{\stackrel{i,j\geq 0}{1\leq k\leq n}}(\pm)_k
{\rm Hol}^{U(n-1)}_{T'}\left(\Cbb_{\tau_{k,n}\bullet\lambda+\mu_k}\otimes {\rm Sym}^j(\mathbb{V}_k)\otimes {\rm Sym}^i(\overline{\mathbb{V}'_k})
\right)\otimes \Cbb_{\varepsilon_n}^{\otimes 1+i-j-k},\nonumber
\end{eqnarray*}
where $(\pm)_k=(-1)^k$ if $k<n$ and $(\pm)_n=(-1)^{n-1}$.

We consider now the term $Q_{\bar{x}_1}(\lambda)$. When $j=q=1$, the torus $T_1^4$ is trivial, $K_1=U(n-2)$ and $\mathbb{M}_0=\{0\}$. Moreover
$W_{x_1}\simeq \mathfrak{S}_{n-2}\times \{\pm\}$, $\tau_1=Id$ and $\psi_1=(2-n)\varepsilon_{n-1}$. Here the quotient $W_{x_1}\backslash W$ is represented by the class of the
elements $\tau_{i,n-1}\tau_{j,n}\in G$ for $1\leq i<j\leq n$. We denote by $\lambda_{ij}$ the term $\tau_{i,n-1}\tau_{j,n}\bullet\lambda$.

In this case we obtain
$$
Q_{\bar{x}_1}(\lambda)=(-1)^{n}\sum_{1\leq i<j\leq n}\  (-1)^{|\{i,j\}\cap\{n-1,n\}|}\ \chi \otimes Q_1^{ij}(\lambda)
$$
with
\begin{eqnarray*}
Q_1^{ij}(\lambda)&=&{\rm Ind}^{U(n-1)\times T^2_{1}}_{U(n-2)\times T^2_{1}\times T^3_{1}}
\left({\rm Hol}^{U(n-2)}_{T^1_{1}}\left(\Cbb_{\lambda_{ij}}\right)\otimes\psi_1 \otimes L^2(T^2_{1}\times T^3_{1}/T_1)\right)\\
&=& \sum_{k\in\Zbb} {\rm Ind}^{U(n-1)}_{U(n-2)\times T^2_{1}}
\left({\rm Hol}^{U(n-2)}_{T^1_{1}}\left(\Cbb_{\lambda_{ij}}\right) \otimes \Cbb_{\varepsilon_{n-1}}^{\otimes k}\right)
\otimes \Cbb_{\varepsilon_{n}}^{\otimes 2-n-k}.
\end{eqnarray*}

\medskip

Let us finish this section by considering the simplest example: $U(1)\times U(1)\subset U(2)$. Take $\lambda=(\lambda_1\geq \lambda_2)\in\widehat{U(2)}$. We have $V_{\lambda}^{U(2)}\vert_{U(1)\times U(1)}=Q_{\bar{x}_0}(\lambda)+ Q_{\bar{x}_1}(\lambda)$ where
$$
Q_{\bar{x}_0}(\lambda)=-\ \Cbb_\lambda\otimes \sum_{-\infty}^{\lambda_2-\lambda_1-1}\Cbb_{\varepsilon_1-\varepsilon_2}^{\otimes k}
\ -\ \Cbb_\lambda\otimes  \sum_{k\geq 1}\Cbb_{\varepsilon_1-\varepsilon_2}^{\otimes k}
$$
and $Q_{\bar{x}_1}(\lambda)=\Cbb_\lambda\otimes \sum_{k\in\mathbb{Z}}\Cbb_{\varepsilon_1-\varepsilon_2}^{\otimes k}$. We recover the basic relation
$$
V_{\lambda}^{U(2)}\vert_{U(1)\times U(1)}=\Cbb_\lambda\otimes \sum_{k=\lambda_2-\lambda_1}^0\Cbb_{\varepsilon_1-\varepsilon_2}^{\otimes k} .
$$

{\small

config: arxiv


TITLE: Exact Differential Equation Geometry
QUESTION [2 upvotes]: In a variety of contexts, I have noticed hints of a strong connection between exact differential equations and machinery from multivariable calculus.
From another question, I have gathered that the geometry of finding the general solution to an exact differential equation $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$ includes finding the potential function of $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$, that is, the surface in $\mathbb{R^3}$ whose gradient is the vector field $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$.  What is missing from this description?  For example, we can write the LHS of a not-necessarily-exact $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$ as $\begin{bmatrix}M(x,\ y) \\ N(x,\ y) \end{bmatrix} \cdot \begin{bmatrix}dx \\ dy\end{bmatrix}$, which expresses the differential of work done by $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$, and thus shows promise for extending the above geometric picture.  In fact, since work is a path function, I suspect the Wikipedia quote

In mathematics, an integrating factor is a function that is chosen to
  facilitate the solving of a given equation involving differentials. It
  is commonly used to solve ordinary differential equations, but is also
  used within multivariable calculus when multiplying through by an
  integrating factor allows an inexact differential to be made into an
  exact differential (which can then be integrated to give a scalar
  field). This is especially useful in thermodynamics where temperature
  becomes the integrating factor that makes entropy an exact
  differential.

is misleading, in that the ordinary differential equations and multivariable calculus uses are actually identical, which if correct, might allow geometry from other path functions (such as the referenced thermodynamics example) to be leveraged.
A big part of what I'm trying to remedy is that it's not clear to me why finding the potential function of $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$ should solve the ODE.  I understand the algebraic rationale that the solution technique undoes the multivariable chain rule (divide through by $dx$ then integrate both sides), but is the geometric picture just an afterthought, or does it offer its own independent rationale for the solution technique?

REPLY [2 votes]: Interesting question, althought I am not exactly sure what kind of answer you want. If this answer is not fully satisfactory, please leave a comment.
Why finding the potential function solves the differential equation. Suppose we have a potential function $\phi$ for $(M(x,y),N(x,y))$. Then all the curves $\phi(x,y)=c$ with $c$ a constant form a solution of the ODE. Indeed, if we take the total derivative of $\phi(x,y)=c$ we get
$$
  \phi_x(x,y) dx + \phi_y(x,y) dy = 0, \\
  M(x,y) dx + N(x,y) dy = 0.
$$
This is a rather formal explanation.
The physics picture. The differential $Mdx + N dy$ can indeed be regarded as the infinitesimal amount of work done by a field $\vec F=(M(x,y),N(x,y))$. This picture can help you understand intuitively why $F(x,y)=c$ solves the ODE $Mdx + Ndy =0$.
Note that a potential in physics is a scalar function $\phi$ such that $-\nabla \phi =\vec{F}=(M,N)$; one adds a minus sign. If such a potential exists, we say that the vector field is exact or, more commonly in physics, conservative. Well-known examples of conservative vector fields are the gravitational force field and electric field in electrostatics.
Conservative vector fields are characterized by the fact that the line integral of them only depend on the end points. Physically this means that the work done by the fields does not depend on the path, only on the beginning and end position. So if $(M(x,y),N(x,y))=\vec{F} = -\nabla \phi$, then
$$
  \int_C M(x,y)dx + N(x,y)dy = \int_C \vec{F} = -(\phi(p_2) - \phi(p_1))
$$
where $C$ is a curve going from $p_1$ to $p_2$. This line integral can be interpreted as the work done by the field $\vec F$ along the path $C$. This line integral (i.e. the total amount of work) is zero iff the potential of the endpoints are the same.
Back to the ODE.
Now let us look back at the ODE $M(x,y)dx + N(x,y)dy = 0$. Physically this equation means that the infinitesimal work should be zero. The solution curves $\phi(x,y)=c$ are the equipotential curves. These curves are exactly the paths along which the work is zero for every infinitesimal step.


TITLE: Is every norm increasing?
QUESTION [11 upvotes]: Let $N$ be any norm on $\Bbb R^n$. Is it true that if $0 \leq a_i \leq b_i$ for $1 \leq i \leq n$, then $N(a_1, \ldots, a_n) \leq N(b_1, \ldots, b_n)$ ?
Clearly this is true for the norms $\| \cdot \|_p$ where $1 \leq p \leq \infty$, being defined as a composition of increasing functions. Any two norms are equivalent, but I don't see how my property is preserved under equivalence.

REPLY [17 votes]: The answer is no.
Consider the norm $\|(x,y)\| = |x| + |x-y|$ on $\mathbb{R}^2$.
We have $(1,0) \le (1,1)$ in the sense you defined but
$$\|(1,0)\| = 2, \quad \|(1,1)\| = 1$$

There is a nontrivial theorem characterizing this property:

Let $\|\cdot\|$ be a norm on $\mathbb{C}^n$. The following is equivalent:

$|x_i| \le |y_i|, \forall i=1, \ldots, n$ implies that $\|(x_1, \ldots, x_n)\| \le \|(y_1, \ldots, y_n)\|$.
$\|(x_1, \ldots, x_n)\| = \|(\left|x_1\right|, \ldots, \left|x_n\right|)\|, \forall (x_1, \ldots, x_n) \in \mathbb{C}^n$.