## 数学代写|表示论代写Representation theory代考|MAST90017

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|Some Algebras of Small Dimensions

One might like to know how many $K$-algebras there are of a given dimension, up to isomorphism. In general there might be far too many different algebras, but for small dimensions one can hope to get a complete overview. We fix a field $K$, and we consider $K$-algebras of dimension at most 2. For these, there are some restrictions.
Proposition 1.28. Let $K$ be a field.
(a) Every 1 -dimensional $K$-algebra is isomorphic to $K$.
(b) Every 2-dimensional $K$-algebra is commutative.
Proof. (a) Let $A$ be a 1-dimensional $K$-algebra. Then $A$ must contain the scalar multiples of the identity element, giving a subalgebra $U:=\left{\lambda 1_{A} \mid \lambda \in K\right} \subseteq A$. Then $U=A$, since $A$ is 1-dimensional. Moreover, according to axiom (Alg) from Definition $1.1$ the product in $U$ is given by $\left(\lambda 1_{A}\right)\left(\mu 1_{A}\right)=(\lambda \mu) 1_{A}$ and hence the map $A \rightarrow K, \lambda 1_{A} \mapsto \lambda$, is an isomorphism of $K$-algebras.

(b) Let $A$ be a 2-dimensional $K$-algebra. We can choose a basis which contains the identity element of $A$, say $\left{1_{A}, b\right}$ (use from linear algebra that every linearly independent subset can be exlended to a basis). The basis elements clearly commute; but then also any linear combinations of basis elements commute, and therefore $A$ is commutative.

We consider now algebras of dimension 2 over the real numbers $\mathbb{R}$. The aim is to classify these, up to isomorphism. The method will be to find suitable bases, leading to ‘canonical’ representatives of the isomorphism classes. It will turn out that there are precisely three $\mathbb{R}$-algebras of dimension 2, see Proposition $1.29$ below.

So we take a 2-dimensional $\mathbb{R}$-algebra $A$, and we choose a basis of $A$ containing the identity. say $\left{1_{A}, b\right}$, as in the above proof of Proposition $1.28$. Then $b^{2}$ must be a linear combination of the basis elements, so there are scalars $\gamma, \delta \in \mathbb{R}$ such that $b^{2}=\gamma 1_{A}+\delta b$. We consider the polynomial $X^{2}-\delta X-\gamma \in \mathbb{R}[X]$ and we complete squares,
$$X^{2}-\delta X-\gamma=(X-\delta / 2)^{2}-\left(\gamma+(\delta / 2)^{2}\right)$$

## 数学代写|表示论代写Representation theory代考|Definition and Examples

A vector space over a field $K$ is an abelian group $V$ together with a scalar multiplication $K \times V \rightarrow V$, satisfying the usual axioms. If one replaces the field $K$ by a ring $R$, then one gets the notion of an $R$-module. Although we mainly deal with algebras over fields in this book, we slightly broaden the perspective in this chapter by defining modules over rings. We always assume that rings contain an identity element.

Definition 2.1. Let $R$ be a ring with identity element $1_{R}$. A left $R$-module (or just $R$-module ) is an abelian group $(M,+)$ together with a map
$$R \times M \rightarrow M, \quad(r, m) \mapsto r \cdot m$$
such that for all $r, s \in R$ and all $m, n \in M$ we have
(i) $(r+s) \cdot m=r \cdot m+s \cdot m$;
(ii) $r \cdot(m+n)=r \cdot m+r \cdot n$;
(iii) $r \cdot(s \cdot m)=(r s) \cdot m$;
(iv) $1_{R} \cdot m=m$.

Exercise 2.1. Let $R$ be a ring (with zero element $0_{R}$ and identity element $1_{R}$ ) and $M$ an $R$-module with zero element $0_{M}$. Show that the following holds for all $r \in R$ and $m \in M$ :
(i) $0_{R} \cdot m=0_{M}$
(ii) $r \cdot 0_{M}=0_{M}$;
(ii) $-(r \cdot m)=(-r) \cdot m=r \cdot(-m)$, in particular $-m=\left(-1_{R}\right) \cdot m$.
Remark 2.2. Completely analogous to Definition $2.1$ one can define right $R$-modules, using a map $M \times R \rightarrow M,(m, r) \mapsto m \cdot r$. When the ring $R$ is not commutative the behaviour of left modules and of right modules can be different; for an illustration see Exercise $2.22$. We will consider only left modules, since we are mostly interested in the case when the ring is a $K$-algebra, and scalars are usually written to the left.

Before dealing with elementary properties of modules we consider a few examples.
Example 2.3.
(1) When $R=K$ is a field, then $R$-modules are exactly the same as $K$-vector spaces. Thus, modules are a true generalization of the concept of a vector space.
(2) Let $R=\mathbb{Z}$, the ring of integers. Then every abelian group can be viewed as a $\mathbb{Z}$-module: If $n \geq 1$ then $n \cdot a$ is set to be the sum of $n$ copies of $a$, and $(-n) \cdot a:=-(n \cdot a)$, and $0_{\mathbb{Z}} \cdot a=0$. With this, conditions (i) to (iv) in Definition $2.1$ hold in any abelian group.
(3) Let $R$ be a ring (with 1 ). Then every left ideal $I$ of $R$ is an $R$-module, with $R$-action given by ring multiplication. First, as a left ideal, $(I,+)$ is an abelian group. The properties (i)-(iv) hold even for arbitrary elements in $R$.
(4) A very important special case of $(3)$ is that every ring $R$ is an $R$-module, with action given by ring multiplication.

## 数学代写|表示论代写Representation theory代考|Some Algebras of Small Dimensions

(a) 每一维 $K$-代数同构于 $K$.
(b) 每个二维 $K$-代数是可交换的。

$\mathrm{~ U : = I l e f t { l l a m b d a ~ 1 _ { A } ~ \ m i d ~ \ l a m b d a ~ \ i n ~ K}$ 公理 (Alg) 1.1产品在 $U$ 是 (谁) 给的 $\left(\lambda 1_{A}\right)\left(\mu 1_{A}\right)=(\lambda \mu) 1_{A}$ 因此地图 $A \rightarrow K, \lambda 1_{A} \mapsto \lambda$ ，是一个同构 $K$-代 数。
(b) 让 $A$ 是二维的 $K$-代数。我们可以选择一个包含恒等元素的基 $A$ ，说】left{1_{A}, b\right } （使用线性代数，每个 线性独立的子集都可以扩展为一个基)。基本元素明显通勤；但随后基元素的任何线性组合也可以通勤，因此 $A$ 是 可交换的。

$$X^{2}-\delta X-\gamma=(X-\delta / 2)^{2}-\left(\gamma+(\delta / 2)^{2}\right)$$

## 数学代写|表示论代写Representation theory代考|Definition and Examples

$$R \times M \rightarrow M, \quad(r, m) \mapsto r \cdot m$$

(-) $(r+s) \cdot m=r \cdot m+s \cdot m$;
(二) $r \cdot(m+n)=r \cdot m+r \cdot n$;
$(\xi) r \cdot(s \cdot m)=(r s) \cdot m$
(四) $1_{R} \cdot m=m$.

(二) $r \cdot 0_{M}=0_{M}$;
(二) $-(r \cdot m)=(-r) \cdot m=r \cdot(-m)$ ，尤其是 $-m=\left(-1_{R}\right) \cdot m$.

(1) 当 $R=K$ 是一个场，那么 $R$-modules 与 $K$-向量空间。因此，模块是向量空间概念的真正概括。
(2) 让 $R=\mathbb{Z}$, 整数环。那么每个阿贝尔群都可以看成一个 $\mathbb{Z}$-模块: 如果 $n \geq 1$ 然后 $n \cdot a$ 被设置为总和 $n$ 的副本 $a$ ，和 $(-n) \cdot a:=-(n \cdot a)$ ，和 $0_{\mathbb{Z}} \cdot a=0$. 这样，定义中的条件 (i) 至 (iv) $2.1$ 在任何阿贝尔群中成立。
(3) 让 $R$ 是一个环（芇有 1) 。那么每一个左理想 $I$ 的 $R$ 是一个 $R$-模块，与 $R$-由环乘法给出的动作。首先，作为左派理 想， $(I,+)$ 是一个阿贝尔群。属性 (i)-(iv) 甚至适用于 $R$.
(4) 一个非常重要的特例 $(3)$ 是每一个环 $R$ 是一个 $R$-模块，通过环乘法给出动作。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MATH4314

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|Group Algebras

Let $G$ be a group and $K$ a field. We define a vector space over $K$ which has basis the set ${g \mid g \in G}$, and we call this vector space $K G$. This space becomes a $K$-algebra if one defines the product on the basis by taking the group multiplication, and extends it to linear combinations. We call this algebra $K G$ the group algebra.
Thus an arbitrary element of $K G$ is a finite linear combination of the form $\sum_{g \in G} \alpha_{g} g$ with $\alpha_{g} \in K$. We can write down a formula for the product of two elements, following the recipe in Remark 1.4. Let $\alpha=\sum_{g \in G} \alpha_{g} g$ and $\beta=\sum_{h \in G} \beta_{h} h$ be two elements in $K G$; then their product has the form
$$\alpha \beta=\sum_{x \in G}\left(\sum_{g h=x} \alpha_{g} \beta_{h}\right) x$$
Since the multiplication in the group is associative, it follows that the multiplication in $K G$ is associative. Furthermore, one checks that the multiplication in $K G$ is distributive. The identity element of the group algebra $K G$ is given by the identity element of $G$.

Note that the group algebra $K G$ is finite-dimensional if and only if the group $G$ is finite, in which case the dimension of $K G$ is equal to the order of the group $G$. The group algebra $K G$ is commutative if and only if the group $G$ is abelian.

Example 1.10. Let $G$ be the cyclic group of order 3 , generated by $y$, so that $G=\left{1_{G}, y, y^{2}\right}$ and $y^{3}=1_{G}$. Then we have
$$\left(a_{0} 1_{G}+a_{1} y+a_{2} y^{2}\right)\left(b_{0} 1_{G}+b_{1} y+b_{2} y^{2}\right)=c_{0} 1_{G}+c_{1} y+c_{2} y^{2},$$
with
$$c_{0}=a_{0} b_{0}+a_{1} b_{2}+a_{2} b_{1}, c_{1}=a_{0} b_{1}+a_{1} b_{0}+a_{2} b_{2}, c_{2}=a_{0} b_{2}+a_{1} b_{1}+a_{2} b_{0}$$

## 数学代写|表示论代写Representation theory代考|Path Algebras of Quivers

Path algebras of quivers are a class of algebras with an easy multiplication formula, and they are extremely useful for calculating examples. They also have connections to other parts of mathematics. The underlying basis of a path algebra is the set of paths in a finite directed graph. It is customary in representation theory to call such a graph a quiver. We assume throughout that a quiver has finitely many vertices and finitely many arrows.

Definition 1.11. A quiver $Q$ is a finite directed graph. We sometimes write $Q=\left(Q_{0}, Q_{1}\right)$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of arrows.

We assume that $Q_{0}$ and $Q_{1}$ are finite sets. For any arrow $\alpha \in Q_{1}$ we denote by $s(\alpha) \in Q_{0}$ its starting point and by $t(\alpha) \in Q_{0}$ its end point.

A non-trivial path in $Q$ is a sequence $p=\alpha_{r} \ldots \alpha_{2} \alpha_{1}$ of arrows $\alpha_{i} \in Q_{1}$ such that $t\left(\alpha_{i}\right)=s\left(\alpha_{i+1}\right)$ for all $i=1, \ldots, r-1$. Note that our convention is to read paths from right to left. The number $r$ of arrows is called the length of $p$, and we denote by $s(p)=s\left(\alpha_{1}\right)$ the starting point, and by $t(p)=t\left(\alpha_{r}\right)$ the end point of $p$.
For each vertex $i \in Q_{0}$ we also need to have a trivial path of length 0 , which we call $e_{i}$, and we set $s\left(e_{i}\right)=i=t\left(e_{i}\right)$.

We call a path $p$ in $Q$ an oriented cycle if $p$ has positive length and $s(p)=t(p)$.

## 数学代写|表示论代写Representation theory代考|Group Algebras

$$\alpha \beta=\sum_{x \in G}\left(\sum_{g h=x} \alpha_{g} \beta_{h}\right) x$$

$$\left(a_{0} 1_{G}+a_{1} y+a_{2} y^{2}\right)\left(b_{0} 1_{G}+b_{1} y+b_{2} y^{2}\right)=c_{0} 1_{G}+c_{1} y+c_{2} y^{2},$$

$$c_{0}=a_{0} b_{0}+a_{1} b_{2}+a_{2} b_{1}, c_{1}=a_{0} b_{1}+a_{1} b_{0}+a_{2} b_{2}, c_{2}=a_{0} b_{2}+a_{1} b_{1}+a_{2} b_{0}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MTH4107

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|Definition and Examples

We start by recalling the definition of a ring: A ring is a non-empty set $R$ together with an addition $+: R \times R \rightarrow R,(r, s) \mapsto r+s$ and a multiplication $:: R \times R \rightarrow R$, $(r, s) \mapsto r \cdot s$ such that the following axioms are satisfied for all $r, s, t \in R$ :
(R1) (Associativity of $+) r+(s+t)=(r+s)+t$.
(R2) (Zero element) There exists an element $0_{R} \in R$ such that $r+0_{R}=r=0_{R}+r$.
(R3) (Additive inverses) For every $r \in R$ there is an element $-r \in R$ such that $r+(-r)=0_{R}$
(R4) (Commutativity of $+) r+s=s+r$.
(R5) (Distributivity) $r \cdot(s+t)=r \cdot s+r \cdot t$ and $(r+s) \cdot t=r \cdot t+s \cdot t$.
(R6) (Associativity of $\cdot) r \cdot(s \cdot t)=(r \cdot s) \cdot t$.
(R7) (Identity element) There is an element $1_{R} \in R \backslash{0}$ such that $1_{R} \cdot r=r=r \cdot 1_{R}$
Moreover, a ring $R$ is called commutative if $r \cdot s=s \cdot r$ for all $r, s \in R$.
As usual, the multiplication in a ring is often just written as $r s$ instead of $r \cdot s$; we will follow this convention from now on.

Note that axioms ( $\mathrm{R} 1)-(\mathrm{R} 4)$ say that $(R,+)$ is an abelian group. We assume by Axiom (R7) that all rings have an identity element; usually we will just write 1 for $1_{R}$. Axiom (R7) also implies that $1_{R}$ is not the zero element. In particular, a ring has at least two elements.
We list some common examples of rings.
(1) The integers $\mathbb{Z}$ form a ring. Every field is also a ring, such as the rational numbers $\mathbb{Q}$, the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, or the residue classes $\mathbb{Z}{p}$ of integers modulo $p$ where $p$ is a prime number. (2) The $n \times n$-matrices $M{n}(K)$, with entries in a field $K$, form a ring with respect to matrix addition and matrix multiplication.
(3) The ring $K[X]$ of polynomials over a field $K$ where $X$ is a variable. Similarly, the ring of polynomials in two or more variables, such as $K[X, Y]$.

Examples (2) and (3) are not just rings but also vector spaces. There are many more rings which are vector spaces, and this has led to the definition of an algebra.

## 数学代写|表示论代写Representation theory代考|Division Algebras

A commutative ring is a field precisely when every non-zero element has an inverse with respect to multiplication. More generally, there are algebras in which every non-zero element has an inverse, and they need not be commutative.

Definition 1.7. An algebra $A$ (over a field $K$ ) is called a division algebra if every non-zero element $a \in A$ is invertible, that is, there exists an element $b \in A$ such that $a b=1_{A}=b a$. If so, we write $b=a^{-1}$. Note that if $A$ is finite-dimensional and $a b=1_{A}$ then it follows that $b a=1_{A}$; see Exercise $1.8$.

Division algebras occur naturally, we will see this later. Clearly, every field is a division algebra. There is a famous example of a division algebra which is not a field, this was discovered by Hamilton.

Example 1.8. The algebra $\mathbb{H}$ of quaternions is the 4-dimensional algebra over $\mathbb{R}$ with basis elements $1, i, j, k$ and with multiplication defined by
$$i^{2}=j^{2}=k^{2}=-1$$
and
$$i j=k, j i=-k, j k=i, k j=-i, k i=j, i k=-j$$
and extending to linear combinations. That is, an arbitrary element of $\mathbb{H}$ has the form $a+b i+c j+d k$ with $a, b, c, d \in \mathbb{R}$, and the product of two elements in $\mathbb{H}$ is given by
\begin{aligned} &\left(a_{1}+b_{1} i+c_{1} j+d_{1} k\right) \cdot\left(a_{2}+b_{2} i+c_{2} j+d_{2} k\right)= \ &\left(a_{1} a_{2}-b_{1} b_{2}-c_{1} c_{2}-d_{1} d_{2}\right)+\left(a_{1} b_{2}+b_{1} a_{2}+c_{1} d_{2}-d_{1} c_{2}\right) i \ &+\left(a_{1} c_{2}-b_{1} d_{2}+c_{1} a_{2}+d_{1} b_{2}\right) j+\left(a_{1} d_{2}+b_{1} c_{2}-c_{1} b_{2}+d_{1} a_{2}\right) k \end{aligned}
It might be useful to check this formula, see Exercise $1.11$.
One can check directly that the multiplication in $\mathrm{H}$ is associative, and that it satisfies the distributive law. But this will follow more easily later from a different construction of $\mathbb{H}$, see Example $1.27$.

## 数学代写|表示论代写Representation theory代考|Definition and Examples

(R1) (结合性 $+) r+(s+t)=(r+s)+t$.
(R2) (零元素) 存在一个元素 $0_{R} \in R$ 这样 $r+0_{R}=r=0_{R}+r$.
(R3) (加法逆) 对于每个 $r \in R$ 有一个元素 $-r \in R$ 这样 $r+(-r)=0_{R}$
(R4) (交换律 $+) r+s=s+r$.
(R5) (分配性) $r \cdot(s+t)=r \cdot s+r \cdot t$ 和 $(r+s) \cdot t=r \cdot t+s \cdot t$.
(R6) (结合性.) $r \cdot(s \cdot t)=(r \cdot s) \cdot t$.
(R7) (标识元素) 有一个元素 $1_{R} \in R \backslash 0$ 这样 $1_{R} \cdot r=r=r \cdot 1_{R}$

(1) 整数 $\mathbb{Z}$ 形成一个环。每个域也是一个环，比如有理数 $\mathbb{Q}$ ，实数 $\mathbb{R}$ ，复数 $\mathbb{C}$ ，或剩余类 $\mathbb{Z} p$ 整数模 $p$ 在哪里 $p$ 是一个素 数。(2) $n \times n$-矩阵 $M n(K)$ ，在字段中包含条目 $K$ ，关于矩阵加法和矩阵乘法形成一个环。
(3) 戒指 $K[X]$ 域上的多项式 $K$ 在哪里 $X$ 是一个变量。类似地，两个或多个变量中的多项式环，例如 $K[X, Y]$.

## 数学代写|表示论代写Representation theory代考|Division Algebras

$$i^{2}=j^{2}=k^{2}=-1$$

$$i j=k, j i=-k, j k=i, k j=-i, k i=j, i k=-j$$

$$\left(a_{1}+b_{1} i+c_{1} j+d_{1} k\right) \cdot\left(a_{2}+b_{2} i+c_{2} j+d_{2} k\right)=\quad\left(a_{1} a_{2}-b_{1} b_{2}-c_{1} c_{2}-d_{1} d_{2}\right)+\left(a_{1} b_{2}+b_{1} a_{2}\right.$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MATHS735

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|Intertwining Operators of the Restriction

We return to the setting of a simply connected nilpotent Lie group $G=\exp g$ and a closed connected subgroup $H=\exp \mathrm{h}$. Given an irreducible unitary representation $\pi$ of $G$, we present an explicit disintegration of the restriction $\pi_{\mid H}$ of $\pi$ to $H$, which is based on a precise description of the space of double cosets $H \backslash G / B$, where $B$ is any closed connected subgroup of $G$, and the well-known smooth disintegration of monomial representations of nilpotent Lie groups. The aim is still to write down a smooth intertwining operator for the decomposition of $\pi_{\mid H}$ into irreducibles. As an application we produce a concrete disintegration of tensor products of irreducible representations of $G$ and a criterium for the irreducibility of these representations.
One should first study the general problem of describing a concrete disintegration of the restriction of an irreducible unitary representation of $G$ to a closed connected subgroup $H=\exp$ b. Since by Kirillov’s theory every $\pi \in \hat{G}$ is of the form $\pi=\pi_{l, \mathfrak{b}}=$ ind $_{B}^{G} \chi_{l}$, where $l \in \mathrm{g}^{}$ and $\mathfrak{b} \subset \mathfrak{g}$ is a polarization at $l$, it is known from Mackey [124] that the restriction of $\pi$ to $H$ disintegrates over the set of double cosets $H \backslash G / B$ and that the integrands are of the form ind ${ }{B(x)}^{H} \chi{I(x)}$, where $B(x)=H \cap \psi(x) B \psi(x)^{-1}, x \in H \backslash G / B$ and $l(x)=\operatorname{Ad}^{}(\psi(x)) l_{\mid \mathfrak{h}}$, $x \in H \backslash G / B$ and $\psi: H \backslash G / B \rightarrow G$ is a section for the double cosets. The idea is to describe an open dense subset of $H \backslash G / B$ and a section $\psi$ which give us an explicit description of $\pi_{\mid H}$ in term of an integral over $H \backslash G / B$ of the representations ind ${ }{B(x)}^{H} \chi{l(x)}$ (see Proposition 3.5.27). The results concerning the explicit disintegration of monomial representations are used to obtain a concrete disintegration of the restriction. This is somehow needed to get an “abstract” disintegration of the restriction into irreducibles to connected closed subgroups of simply connected nilpotent Lie groups. ‘Abstract’ here means that the measure class in $\hat{G}$ for the disintegration of the restriction and the multiplicities of the irreducibles appearing in the disintegration are given. These constructions will be applied to the disintegration of the tensor product of two irreducible representations $\pi$ and $\pi^{\prime}$.

## 数学代写|表示论代写Representation theory代考|Double-Coset Space

Let $\mathfrak{g}$ be a nilpotent Lie algebra, $\mathfrak{b}$ any subalgebra, $B \subset G$ their simply connected Lie groups. Recall that the exponential mapping exp : $\mathfrak{g} \rightarrow G$ is a diffeomorphism. Given a sequence of ideals
$$\mathfrak{g}{n+1}:={0} \varsubsetneqq \ldots \varsubsetneqq \mathfrak{g}{i} \varsubsetneqq \ldots \varsubsetneqq \mathfrak{g}{1}=\mathfrak{g}, \operatorname{dim}\left(\mathfrak{g}{i} / \mathfrak{g}_{i+1}\right)=1,$$ denote for every $i=1, \ldots, n, G_{i}:=\exp \mathfrak{g}{i}$ and choose a vector $Z{i} \in \mathfrak{g}{i} \backslash \mathfrak{g}{i+1}$, so that $\mathfrak{g}{i}=\mathbb{R}$-span $\left(Z{i}, \ldots, Z_{n}\right)$. One obtains in this way a Jordan-Hölder basis $\mathscr{Z}:=\left(Z_{1}, \ldots, Z_{n}\right)$ of $\mathfrak{g}$. To simplify the notations, let
$$V_{1} \cdot V_{2} \cdots V_{k}:=\exp \left(V_{1}\right) \cdot \exp \left(V_{2}\right) \cdots \exp \left(V_{k}\right) \in G$$
for given vectors $V_{1}, \ldots, V_{k} \in \mathfrak{g}$. Denote as before $d g$ the Haar measure on $G$. Using the basis $\mathscr{Z}$, one can express $d g$ in the following way.
$$\int_{G} f(g) d g=\int_{\mathbb{R}^{n}} f\left(z_{1} Z_{1} \cdots z_{n} Z_{n}\right) d z,\left(f \in L^{1}(G)\right) .$$
Since $G$ is nilpotent, the quotient space $G / B$ has a $G$-invariant measure which is unique up to a positive scalar multiple. This measure (denoted by $d \dot{g}$ ) is described in Chap. 1, Sect. 1.2.2. Let us recall such construction. Let
$$\mathscr{I}^{8 / b}=\left{k \in{1, \ldots, n}, Z_{k} \notin \mathfrak{b}+\mathfrak{g}{k+1}\right}=:\left{k{1}<\ldots<k_{p}\right} .$$
One obtains the sequence of subalgebras
$$\mathfrak{b}^{p+1}:=\mathfrak{b} \varsubsetneqq \ldots \varsubsetneqq \mathfrak{b}^{j}:=\mathbb{R} Z_{k_{j}} \oplus \mathfrak{b}^{j-1} \varsubsetneqq \ldots \ldots \mathfrak{b}^{1}=\mathfrak{g}$$
and the Malcev basis $\mathscr{M}:=\left(Z_{k_{1}}, \ldots, Z_{k_{p}}\right)$ of $\mathfrak{g}$ relative to $\mathfrak{b}$. The invariant measure $d \dot{g}$ is then given for $\varphi \in \mathscr{C}{c}(G / B)$ by: $$\mu{\mathcal{M}}(\varphi)=\mu_{\mathrm{g} / \mathrm{b}}(\varphi)=\int_{G / B} \varphi(\dot{g}) d \dot{g}:=\int_{\mathbb{R}^{p}} \varphi\left(w_{1} Z_{k_{1}} \cdots w_{p} Z_{k_{p}} \cdot B\right) d w$$
where $\mathscr{C}{c}(G / B)$ denotes the space of complex-valued continuous functions with compact support on $G / B$. This is a consequence of the fact that the mapping \begin{aligned} E{\mathscr{M}}^{}: \mathbb{R}^{p} & \longrightarrow \ w=\left(w_{1}, \ldots, w_{p}\right) & \longmapsto w_{1} Z_{k_{1}} \cdots w_{p} Z_{k_{p}} \cdot B=: E_{\mathscr{A}}^{}(w) \end{aligned}
is a diffeomorphism. If $\mathfrak{h}$ is another subalgebra of $\mathfrak{g}$, then denote by $\mathbb{I}^{\mathfrak{h}} \subset{1, \ldots, n}$ the index set
$$\mathbb{I}^{\mathfrak{h}}:=\left{i \in{1, \ldots, n} ; \mathfrak{h}+\mathfrak{g}{i+1}=\mathfrak{h}+\mathfrak{g}{i}\right}={1, \ldots, n} \backslash \mathbb{I}^{\mathfrak{g} / \mathfrak{h}}$$
One can then assume that the vectors $Z_{i}, i \in \mathbb{I}^{\mathfrak{h}}$, lie in $\mathfrak{h}$ so that $\mathfrak{h}=\mathbb{R}$-span $\left{Z_{i}, i \in\right.$ $\left.\mathbb{I}^{\mathrm{h}}\right}$

## 数学代写|表示论代写Representation theory代考|The Set of Double Cosets

For $g \in G$, denote by $\bar{g}$ its double coset $H \cdot g \cdot B={h g b,(h, b) \in H \times B}$. The aim is to find an open dense subset of $H \backslash G / B$ which will support the measure $d \gamma(\bar{g})$ and which is diffeomorphic to a Zariski open subset of $\mathbb{R}^{d}$ for some $d \in \mathbb{N}^{*}$. The following example illustrates this fact:

Example 3.5.1 Let $\mathfrak{g}$ be the 7-dimensional Lie algebra spanned by the JordanHölder basis $\mathscr{Z}=\left(Z_{1}, \ldots, Z_{7}\right)$, equipped with the brackets
$$\left[Z_{1}, Z_{4}\right]=Z_{6}, \quad\left[Z_{1}, Z_{5}\right]=Z_{7}, \quad\left[Z_{2}, Z_{3}\right]=Z_{7}$$
Consider its Abelian subalgebras $h=\mathbb{R}-\operatorname{span}\left(Z_{4}, Z_{5}, Z_{7}\right)$ and $\mathfrak{b}=\mathbb{R}-\operatorname{span}\left(Z_{3}, Z_{4}\right.$, $Z_{7}$. Since many products commute, the element $g=: z_{1} Z_{1} \cdots z_{7} Z_{7} \in G$, $\left(z_{1}, \ldots, z_{7}\right) \in \mathbb{R}^{7}$, can be described in the following way:
\begin{aligned} g &=\left(z_{1} Z_{1} \cdot z_{5} Z_{5}\right) \cdot z_{2} Z_{2} \cdot z_{6} Z_{6} \cdot\left(z_{3} Z_{3} \cdot z_{4} Z_{4} \cdot z_{7} Z_{7}\right) \ &=\left(z_{1} z_{5} Z_{7} \cdot z_{5} Z_{5} \cdot z_{1} Z_{1}\right) \cdot z_{2} Z_{2} \cdot z_{6} Z_{6} \cdot\left(z_{3} Z_{3} \cdot z_{4} Z_{4} \cdot z_{7} Z_{7}\right) \ & \in H \cdot z_{1} Z_{1} \cdot z_{2} Z_{2} \cdot z_{6} Z_{6} \cdot B \end{aligned}
This implies that $\bar{g}=H \cdot g \cdot B=H \cdot z_{1} Z_{1} \cdot z_{2} Z_{2} \cdot z_{6} Z_{6} \cdot B$. On the other hand if $z_{1} \neq 0$, the element $z_{1} Z_{1} \cdot z_{2} Z_{2} \cdot z_{6} Z_{6}$ can also be written as $z_{1} Z_{1} \cdot z_{2} Z_{2}$ conjugated by $\exp \left(-\frac{26}{z_{1}}\right) Z_{4}$, which is contained in $H \cap B$. Hence if $z 1 \neq 0$, then $\bar{g}=H \cdot z_{1} Z_{1} \cdot z_{2} Z_{2} \cdot B$. As a conclusion,
$H \backslash G / B=\left{H \cdot z_{1} Z_{1} \cdot z_{2} Z_{2} \cdot B,\left(z_{1}, z_{2}\right) \in \mathscr{V}\right} \dot{\cup}\left{H \cdot z_{2} Z_{2} \cdot z_{6} Z_{6} \cdot B,\left(z_{2}, z_{6}\right) \in \mathbb{R}^{2}\right}$
$=\quad \mathbf{p}\left(G \backslash G_{2}\right) \quad \dot{u} \quad \mathbf{p}\left(G_{2}\right)$
where $\mathbf{p}: g \longmapsto \tilde{g}$, is the canonical projection of $G$ on $H \backslash G / B$, and $\mathscr{V}:=$ $\mathbb{R}^{\star} \times \mathbb{R}$. Hence the space $H \backslash G / B$ is the disjoint union of two subsets, the first is the projection of a Zariski open subset of $G$ and the second of a Zariski closed subset. The measure $d \gamma(\bar{g})$ is shown to be supported on the first set.

## 数学代写|表示论代写Representation theory代考|Double-Coset Space

Gn+1:=0⫋…⫋G一世⫋…⫋G1=G,暗淡⁡(G一世/G一世+1)=1,表示每个一世=1,…,n,G一世:=经验⁡G一世并选择一个向量从一世∈G一世∖G一世+1， 以便G一世=R-跨度(从一世,…,从n). 以这种方式获得 Jordan-Hölder 基从:=(从1,…,从n)的G. 为了简化符号，让

∫GF(G)dG=∫RnF(和1从1⋯和n从n)d和,(F∈大号1(G)).

\mathscr{I}^{8 / b}=\left{k \in{1, \ldots, n}, Z_{k} \notin \mathfrak{b}+\mathfrak{g}{k+1}\右}=:\left{k{1}<\ldots<k_{p}\right} 。\mathscr{I}^{8 / b}=\left{k \in{1, \ldots, n}, Z_{k} \notin \mathfrak{b}+\mathfrak{g}{k+1}\右}=:\left{k{1}<\ldots<k_{p}\right} 。

bp+1:=b⫋…⫋bj:=R从ķj⊕bj−1⫋……b1=G

μ米(披)=μG/b(披)=∫G/乙披(G˙)dG˙:=∫Rp披(在1从ķ1⋯在p从ķp⋅乙)d在

\mathbb{I}^{\mathfrak{h}}:=\left{i \in{1, \ldots, n} ; \mathfrak{h}+\mathfrak{g}{i+1}=\mathfrak{h}+\mathfrak{g}{i}\right}={1, \ldots, n} \反斜杠 \mathbb{I} ^{\mathfrak{g} / \mathfrak{h}}\mathbb{I}^{\mathfrak{h}}:=\left{i \in{1, \ldots, n} ; \mathfrak{h}+\mathfrak{g}{i+1}=\mathfrak{h}+\mathfrak{g}{i}\right}={1, \ldots, n} \反斜杠 \mathbb{I} ^{\mathfrak{g} / \mathfrak{h}}

## 数学代写|表示论代写Representation theory代考|The Set of Double Cosets

[从1,从4]=从6,[从1,从5]=从7,[从2,从3]=从7

G=(和1从1⋅和5从5)⋅和2从2⋅和6从6⋅(和3从3⋅和4从4⋅和7从7) =(和1和5从7⋅和5从5⋅和1从1)⋅和2从2⋅和6从6⋅(和3从3⋅和4从4⋅和7从7) ∈H⋅和1从1⋅和2从2⋅和6从6⋅乙

H \反斜杠 G / B=\left{H \cdot z_{1} Z_{1} \cdot z_{2} Z_{2} \cdot B,\left(z_{1}, z_{2}\right) \in \mathscr{V}\right} \dot{\cup}\left{H \cdot z_{2} Z_{2} \cdot z_{6} Z_{6} \cdot B,\left(z_{2 }, z_{6}\right) \in \mathbb{R}^{2}\right}H \反斜杠 G / B=\left{H \cdot z_{1} Z_{1} \cdot z_{2} Z_{2} \cdot B,\left(z_{1}, z_{2}\right) \in \mathscr{V}\right} \dot{\cup}\left{H \cdot z_{2} Z_{2} \cdot z_{6} Z_{6} \cdot B,\left(z_{2 }, z_{6}\right) \in \mathbb{R}^{2}\right}
=p(G∖G2)在˙p(G2)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MATH4031

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|A Rational Disintegration of L2 for an Exponential

Let $G$ be an exponential solvable Lie group with Lie algebra $\mathfrak{g}$ and left-regular representation $\lambda_{G}=$ ind $_{{e}}^{G} 1$. Here $f=0$ and $\Gamma_{f}=\mathrm{g}^{\star}$. Take a good sequence of subalgebras of $\mathrm{g}$
$$a_{0}={0} \subset a_{1} \subset a_{2} \subset \cdots \subset a_{n}=\mathfrak{g}$$
from which we extract a Malcev basis $\left{X_{1}, \ldots, X_{n}\right}$ of $\mathrm{g}, X_{i} \in \mathrm{a}{i} \backslash \mathfrak{a}{i-1}$. In this case and as in Sect. $3.3 .1, K^{{e}}$ is the set of all $j \in{1, \ldots, n}$ such that all $A_{j}$-orbits are saturated with respect to $a_{j-1}$, which implies $V=\left{\phi \in \mathfrak{g}^{*}:\left\langle\phi, X_{j}\right\rangle=0, j \in\right.$ $\left.K^{{e}}\right}$. Let $\phi \in V$ and set $\phi_{i}=\phi_{\mid a_{j}}$. Let
$$\mathrm{b}(\phi)=\sum_{i=1}^{n} a_{i}\left(\phi_{i}\right)$$
be the Vergne polarization at $\phi$ with respect to the Jordan-Hölder sequence (3.3.20) and $B(\phi)$ its associated Lie group. In addition, we have from the Pukanszky condition that
$$\operatorname{Ad}^{\star}(B(\phi)) \phi=\phi+\mathfrak{b}(\phi)^{\perp}$$

Let $\mu_{G}$ be the Haar measure on $G$. We have the following rational disintegration of $L^{2}(G)$
$$\left(L^{2}(G), \mu_{G}\right) \simeq \int_{V}^{\oplus}\left(L^{2}(G / B(\phi)), \phi\right) d \lambda(\phi)$$
The isometry is given by:
$$U(\xi)(\phi)(g)=\int_{B(\phi)} \xi(g u) \chi_{\phi}(u) \Delta_{B(\phi), G}^{-\frac{1}{2}}(u) d_{B(\phi)}(u), g \in G$$
where $\xi \in C_{c}^{\infty}(G)$ is the set of $C^{\infty}$ functions with compact support in $G$ and $\phi \in V$, $d_{B(\phi)}$ is the Haar measure on $B(\phi)$.

## 数学代写|表示论代写Representation theory代考|Intertwining of Representations Induced from Maximal

Definition 3.4.1 Let $G$ be a Lie group. A subgroup $H$ of $G$ is said to be a maximal subgroup if $H \neq G$ and for every subgroup $K$ such that $H \subset K \subset G$, then either $K=H$ or $K=G$.

Remark 3.4.2 If $G$ is a simply connected solvable Lie group and $H$ is a maximal subgroup of $G$, then $H$ has codimension one or two. In the latter case $H$ cannot be a normal subgroup of $G$.

The following result describes the structure of maximal subalgebras of exponential solvable algebras, a proof of which can be found in [106].

Theorem 3.4.3 Let $G=\exp g$ be an exponential solvable Lie group and $H=$ exph a non-normal maximal subgroup of $G$. Then

1. if $\mathrm{h}$ is a hyperplane, there exist a codimension-one subalgebra $\mathrm{g}{0}$ of $\mathrm{h}$ which is a codimension- $t$ wo ideal in $\mathfrak{g}$, plus two elements $A \in \mathfrak{h} \backslash \mathfrak{g}{0}, X \in \mathfrak{g} \backslash \mathfrak{h}$ such that
$$[A, X]=X \bmod \mathfrak{g}_{0}$$
2. If $\mathrm{h}$ has codimension two, there exists a codimension-one subalgebra $\mathrm{g}{0}$ of $\mathrm{h}$ which is a codimension-three ideal in $\mathrm{g}$, plus three nonzero vectors $A, X, Y$ and a nonzero real number $\alpha$ such that $$\begin{gathered} \mathfrak{g}=\mathfrak{h} \oplus \mathbb{R} X \oplus \mathbb{R} Y, \quad \mathfrak{h}=\mathfrak{g}{0} \oplus \mathbb{R} A \ {[A, X]=X+\alpha Y \bmod \mathfrak{g}{0},[A, Y]=Y-\alpha X \bmod \mathfrak{g}{0},} \end{gathered}$$ and$$[X, Y]=0 \bmod \mathfrak{g}_{0} .$$We now prove the following disintegration formula, which basically stems from Theorem 3.4.3.

## 数学代写|表示论代写Representation theory代考|Construction of an Intertwining Operator

In this section we construct an intertwining operator between the induced representation $\tau=\operatorname{ind}{H}^{G} \chi{f}$ and its decomposition into irreducibles explicitly. Let $s=\left(a_{j}\right){j=0}^{n}$ be a good sequence of subalgebras of $g$ passing through $g{0}$, where $\mathfrak{g}_{0}$ is defined as in Theorem 3.4.3. With the notations above, we can choose s as follows:

1. If $h$ is an ideal of $g$ we have codim $h=1$, then $a_{n-1}=h=g_{0}$.
2. If $h$ is not an ideal and $\operatorname{codim} h=1$, then $\mathfrak{a}{n-2}=\mathfrak{g}{0}$ and $\mathfrak{a}{n-1}=\mathfrak{g}{0} \oplus \mathbb{R} X$.
3. If $\operatorname{codim} h=2$, then $\mathfrak{a}{n-3}=\mathfrak{g}{0}, \mathfrak{a}{n-2}=\mathfrak{g}{0} \oplus \mathbb{R} X$ and $\mathfrak{a}{n-1}=\mathfrak{g}{0} \oplus \mathbb{R} X \oplus \mathbb{R} Y$.
In the sequel, we shall identify $\mathscr{O}(\tau)$ with the set $\left{\phi_{t} \in \Gamma_{f}: t \in \mathscr{O}(\tau)\right}$. For $l$ in $\mathscr{O}(\tau)$, let $\mathrm{b}[l]$ be the Vergne polarization of $l$ associated to $s$ and $B[l]=\exp \mathrm{b}[l]$. We prove first the following

Lemma 3.4.5 For any $l$ in $\mathscr{O}(\tau)$, there exist a coexponential basis $\mathscr{y}$ of $\mathfrak{b}[l] \cap \mathfrak{h}$ in $\mathfrak{b}[l]$, a coexponential basis $\mathcal{Z}$ of $\mathfrak{b}[l] \cap \mathrm{h}$ in $\mathrm{h}$ and a coexponential basis $\mathscr{X}$ of $\mathrm{b}[l]$ in $\mathrm{g}$ which do not depend on $l$.

Proof As above, we distinguish two cases. We keep the same notations as in Proposition 3.4.4. If $\mathfrak{g}{\theta}=\mathfrak{h}$, then $\tau$ is irreducible and $\mathscr{O}(\tau)={\phi}$ where $\phi \in p^{-1}({f})$. Hence $\operatorname{dim} \mathfrak{b}[\phi]=\operatorname{dim} \mathfrak{b}$. We are going to prove that in this situation $\mathfrak{b}[\phi]=\mathfrak{b}$, which implies $$\mathscr{y}=\mathscr{Z}=\emptyset$$ Suppose for starters that $H$ is a codimension-one subgroup of $G$. Then $\mathrm{g}=$ $\mathfrak{h} \oplus \mathbb{R} X$. If $H$ is a normal subgroup of $G$, then as $\mathfrak{g}(\phi) \subset \mathfrak{g}{\theta}=\mathfrak{h}$, we already get that $\mathfrak{b}[\phi]=$ h. Now we suppose that $H$ is a non-normal subgroup of $G$. It follows from the definition of the Vergne polarization $\mathfrak{b}[\phi]$ and for all $i=$ $0, \ldots, \operatorname{codim} h, \quad \mathfrak{a}{n-i}\left(\phi{\mid a_{n-i}}\right) \subset a_{n-i} \cap \mathfrak{g}{\theta} \subset h$, that $\mathfrak{b}[\phi] \subset h$, which implies $\mathfrak{b}[\phi]=\mathfrak{h}$. We conclude that if codim $\mathfrak{h}=1$, we have $$\mathscr{C}={X}$$ and if $\operatorname{codim} h=2$, we have $$\mathscr{Q}={X, Y}$$ We now look at the case where $\mathfrak{g}{\theta}=\mathfrak{g}$. We have $X \in \mathfrak{b}[l]$ and $\mathfrak{g}_{0} \subset \mathfrak{g}(l)$, for all $l$ in $\mathscr{O}(\tau)$. Suppose first that $H$ is a codimension-one subgroup of $G$. If $H$ is a normal subgroup of $G$, then $\mathfrak{g}(l)=\mathfrak{g}$ and then $\mathfrak{b}[l]=\mathfrak{g}$. Therefore,
$$\mathscr{Y}={X}, \mathscr{Z}=\mathscr{X}=\emptyset .$$
Assume then that $H$ is a non-normal subgroup of $G$, so $\mathfrak{g}\left(\phi_{s_{1}}\right)=\mathfrak{g}\left(\phi_{s_{2}}\right)=\mathfrak{g}{0}$ from Eq. (3.4.2) and hence $\mathfrak{b}\left[\phi{s_{1}}\right]=\mathfrak{b}\left[\phi_{s_{2}}\right]=\mathfrak{g}_{0} \oplus \mathbb{R} X$. This implies that
$$\mathscr{Y}={X}, \mathscr{Z}={A} \text { and } \mathscr{X}={A}$$

## 数学代写|表示论代写Representation theory代考|A Rational Disintegration of L2 for an Exponential

b(φ)=∑一世=1n一个一世(φ一世)

(大号2(G),μG)≃∫在⊕(大号2(G/乙(φ)),φ)dλ(φ)

## 数学代写|表示论代写Representation theory代考|Intertwining of Representations Induced from Maximal

1. 如果H是一个超平面，存在一个余维子代数G0的H这是一个codimension-吨我的理想在G, 加上两个元素一个∈H∖G0,X∈G∖H这样
[一个,X]=X反对G0
2. 如果H有余维二，存在余维一子代数G0的H这是一个余维三理想G，加上三个非零向量一个,X,是和一个非零实数一个这样G=H⊕RX⊕R是,H=G0⊕R一个 [一个,X]=X+一个是反对G0,[一个,是]=是−一个X反对G0,和[X,是]=0反对G0.我们现在证明下面的分解公式，它基本上源于定理 3.4.3。

## 数学代写|表示论代写Representation theory代考|Construction of an Intertwining Operator

1. 如果H是一个理想的G我们有codimH=1， 然后一个n−1=H=G0.
2. 如果H不是一个理想和科迪姆⁡H=1， 然后一个n−2=G0和一个n−1=G0⊕RX.
3. 如果科迪姆⁡H=2， 然后一个n−3=G0,一个n−2=G0⊕RX和一个n−1=G0⊕RX⊕R是.
接下来，我们将确定○(τ)与套装\left{\phi_{t} \in \Gamma_{f}: t \in \mathscr{O}(\tau)\right}\left{\phi_{t} \in \Gamma_{f}: t \in \mathscr{O}(\tau)\right}. 为了l在○(τ)， 让b[l]是 Vergne 极化l关联到s和乙[l]=经验⁡b[l]. 我们首先证明以下

C=X而如果科迪姆⁡H=2， 我们有

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MATH7333

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|A Base Space of the Disintegration of Induced Representations

We fix in the whole section an exponential solvable Lie group $G=\exp g$ with Lie algebra $\mathrm{g}$. Let $f$ be an element of $\mathrm{g}^{*}$ and $H=\exp \mathrm{h}$ a normal subgroup of $G$. Recall the monomial representation $\tau=\operatorname{ind}{H}^{G} \chi{f}$, which is realized by left translations in the Hilbert space $\mathscr{H}_{2}$ of continuous functions $\xi$ on $G$ that satisfy the covariance relation (3.3.1) for all $g$ in $G$ and $h$ in $H$ and are square-integrable on $G / H$ for the $G$-invariant measure. A result on the disintegration of $\tau$ was obtained earlier (cf. Theorem 1.4.2). We first recall the precise disintegration formula:

Theorem 3.3.1 Let $G=\exp g$ be an exponential solvable group and $H=\exp$ h $a$ normal subgroup of $G$. Then
$$\tau \simeq \int_{f+\mathbf{b}^{\perp} / H}^{\oplus} \pi_{l} d \mu(I)$$
where $\mu$ is the image under the Kirillov-Bernat map of a finite positive measure on $\Gamma_{f} \subset \mathrm{g}^{\star}$ equivalent to the Lebesgue measure. On the other hand, the multiplicities involved in this decomposition are identically 1 or $+\infty$, depending on whether
$$\operatorname{dim}(H \cdot l)=\operatorname{dim}\left(G \cdot l \cap \Gamma_{f}\right)$$
or not, for l generic in $\Gamma_{f}$. Equivalently, we might have
$$2 \operatorname{dim}(H \cdot l)=\operatorname{dim}(G \cdot l)$$
or not. In either case the multiplicity of $\pi_{l}$ in $\tau$ is the number of $H$-orbits in $G \cdot l \cap \Gamma_{f} \cdot$.

## 数学代写|表示论代写Representation theory代考|Construction of the Intertwining Operator

Let $\mathfrak{s}=\left(a_{j}\right){j=0}^{n}$ be a good sequence of subspaces of $\mathfrak{g}$ adapted to $h$, and extract a coexponential basis $B=\left{X{1}, \ldots, X_{r}\right}$ to $\mathfrak{h}$ in $\mathfrak{g}$. Consider also the disintegration space $V$ endowed with the Lebesgue measure $d \lambda$ as in Sect.3.3.1 and formula (3.3.5). The basis $B$ defines an invariant measure on $G / H$, which allows to fix the norm
$$|\xi|_{L^{2}(G / H, f)}=\left(\int_{G / H}|\xi(g)|^{2} d_{G, H}(g)\right)^{1 / 2}$$
on $\mathscr{H}{r}=L^{2}(G / H, f)$ of $\tau$. We now build a Zariski open set $V{0}$ of $V$ and for $\phi \in V_{0}$ the Vergne polarization $\mathfrak{b}(\phi)$ at $\phi$ relatively to the good sequence $\mathfrak{s}$ as in Theorem 1.2.4. Since this good sequence is adapted to $\mathfrak{h}$, we must have $\mathfrak{h} \subset \mathfrak{b}(\phi)$ for all $\phi \in V_{0}$. We next construct, for $\phi \in V_{0}$, a coexponential basis $X(\phi)$ to $b(\phi)$ in $\mathfrak{g}$ and a coexponential basis $Y(\phi)$ to $\mathfrak{h}$. All these bases vary continuously on $V_{0}$. For $\phi \in V^{\prime}$ and $j=1, \ldots, n$, we set
$$J_{j}(\phi)=\left{k \in{1, \ldots, j}: \mathfrak{a}{j}\left(\phi{j}\right)+\mathfrak{a}{k-1} \varsubsetneqq \mathfrak{a}{j}\left(\phi_{j}\right)+\mathfrak{a}{k}\right} .$$ The set of indices $J{j}(\phi)$ is typically not constant for $\phi \in V^{\prime}$, but its cardinality is constant and equal to $d_{j}$ for all $j=1, \ldots, n$. Note then
$$J_{j}(\phi)=\left{i_{1}(\phi)<\cdots<i_{d_{j}}(\phi)\right},$$
We endow the set $\left{J_{j}(\phi), \phi \in V^{\prime}\right}$ with the lexicographic order defined by
$$\left{i_{1}(\phi)<\cdots<i_{d_{j}}(\phi)<i_{1}\left(\phi^{\prime}\right)<\cdots<i_{d_{j}}\left(\phi^{\prime}\right)\right},$$
if there exists $\sigma \in\left{1, \ldots, d_{j}\right}$ such that $i_{1}(\phi)=i_{1}\left(\phi^{\prime}\right), \ldots, i_{\sigma-1}\left(\phi^{\prime}\right), i_{\sigma}(\phi)<$ $i_{\sigma}\left(\phi^{\prime}\right)$. Using this order, let
$$J_{j}=\min {\phi \in V^{\prime}} J{j}(\phi)=\left{i_{1}<\cdots<i_{d_{j}}\right}$$

## 数学代写|表示论代写Representation theory代考|The Inverse Operator

We now prove that formulas (3.3.13) and (3.3.14), established in the proof of the theorem on $C_{c}^{\infty}(G / H, f)$, actually hold on $L^{2}(G / H, f)$. We will resume the cases studied in the previous theorem.
In the first case, it is clear that $L^{2}(G / H, f)=L^{2}\left(G_{0} / H, f_{0}\right)$ and that
$$\int_{V} L^{2}(G / B(\phi), \phi) d \lambda(\phi)=\int_{V_{0}} L^{2}\left(\mathbb{R}, L^{2}\left(G_{0} / B\left(\phi_{0}\right), \phi_{0}\right)\right) d \lambda^{0}\left(\phi_{0}\right)$$
Hence $U=\tilde{U}{0} \circ W$, where $W: L^{2}(G / H, f) \rightarrow L^{2}\left(\mathbb{R}, L^{2}\left(G{0} / H, f\right)\right)$ is the operator field defined by
\begin{aligned} &W(\xi)(t)\left(g_{0}\right)=\xi\left(\exp (t X) \cdot g_{0}\right)=\xi_{t}\left(g_{0}\right), g_{0} \in G_{0} \ &\text { and } \tilde{U}{0}(\xi)(t)\left(g{0}\right)=U_{0}\left(\xi_{t}\right)\left(g_{0}\right) \text {. } \end{aligned}
We move to the second case, so let $\phi \in V_{0}$ and $\phi_{0}=\phi_{\mid g_{0}}$. Then $\phi=\phi_{s}=$ $\phi_{0}+s X^{\star}$ for some $s \in \mathbb{R}$. For $\eta \in C_{c}^{\infty}\left(G / B\left(\phi_{0}\right)\right.$, $\left.\phi_{0}\right)$, let $\eta^{s}$ be the function defined on $G$ by
$$\eta^{s}(g)=\int_{\mathbb{R}} \eta\left(g \exp \left(t B_{n}(\phi)\right)\right) e^{-i t s} \Delta_{B(\phi), G}^{-1 / 2}\left(\exp \left(t B_{n}(\phi)\right)\right) e^{-i t \phi_{0}}\left(Z_{0}(\phi)\right) d t, g \in G$$

## 数学代写|表示论代写Representation theory代考|A Base Space of the Disintegration of Induced Representations

τ≃∫F+b⊥/H⊕圆周率ldμ(我)

2暗淡⁡(H⋅l)=暗淡⁡(G⋅l)

## 数学代写|表示论代写Representation theory代考|Construction of the Intertwining Operator

|X|大号2(G/H,F)=(∫G/H|X(G)|2dG,H(G))1/2

J_{j}(\phi)=\left{k \in{1, \ldots, j}: \mathfrak{a}{j}\left(\phi{j}\right)+\mathfrak{a}{ k-1} \varsubsetneqq \mathfrak{a}{j}\left(\phi_{j}\right)+\mathfrak{a}{k}\right} 。J_{j}(\phi)=\left{k \in{1, \ldots, j}: \mathfrak{a}{j}\left(\phi{j}\right)+\mathfrak{a}{ k-1} \varsubsetneqq \mathfrak{a}{j}\left(\phi_{j}\right)+\mathfrak{a}{k}\right} 。索引集Ĵj(φ)通常不是恒定的φ∈在′, 但它的基数是恒定的并且等于dj对所有人j=1,…,n. 然后注意

J_{j}(\phi)=\left{i_{1}(\phi)<\cdots<i_{d_{j}}(\phi)\right}，J_{j}(\phi)=\left{i_{1}(\phi)<\cdots<i_{d_{j}}(\phi)\right}，

\left{i_{1}(\phi)<\cdots<i_{d_{j}}(\phi)<i_{1}\left(\phi^{\prime}\right)<\cdots<i_{ d_{j}}\left(\phi^{\prime}\right)\right},\left{i_{1}(\phi)<\cdots<i_{d_{j}}(\phi)<i_{1}\left(\phi^{\prime}\right)<\cdots<i_{ d_{j}}\left(\phi^{\prime}\right)\right},

J_{j}=\min {\phi \in V^{\prime}} J{j}(\phi)=\left{i_{1}<\cdots<i_{d_{j}}\right}J_{j}=\min {\phi \in V^{\prime}} J{j}(\phi)=\left{i_{1}<\cdots<i_{d_{j}}\right}

## 数学代写|表示论代写Representation theory代考|The Inverse Operator

∫在大号2(G/乙(φ),φ)dλ(φ)=∫在0大号2(R,大号2(G0/乙(φ0),φ0))dλ0(φ0)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MAST90017

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|Disintegration of Monomial Representations

Let $B=\exp b$ be a connected closed subgroup of $G$ and let $\left{Y_{1}, Y_{2}, \ldots, Y_{k}\right}$ be a Jordan-Hölder basis of $\mathbf{b}$. Then the map
$$\psi: \mathbb{R}^{k} \longrightarrow B,\left(t_{1}, \ldots, t_{k}\right) \longrightarrow \exp t_{1} Y_{1} \cdots \exp t_{k} Y_{k}$$

is a diffeomorphism. We recall from Sect. 1.2.2 how to choose normalized measures $d b$ on $B$ and $G / B$ : for any $C^{\infty}$-function with compact support $\eta$ in $B$,
$$\int_{B} \eta(b) d(b)=\int_{\mathbb{R}^{d}} \eta\left(\exp \left(t_{1} Y_{1}\right) \ldots \exp \left(t_{k} Y_{k}\right)\right) d t_{1} \ldots d t_{k} .$$
This measure is left-invariant and therefore a Haar measure. On the other hand, if we choose a Malcev basis $\left{X_{1}, \ldots, X_{d}\right}$ of $\mathfrak{g}$ relative to $b$, ie
$$\mathfrak{g}=\mathbb{R} X_{1} \oplus \ldots \oplus \mathbb{R} X_{d} \oplus \mathfrak{b} \text { and } \sum_{k=j}^{d} \mathbb{R} X_{k}+\mathfrak{b}$$
is a subalgebra of $\mathfrak{g}$ for all $j$, then the measure $d_{G, B}$ on $G / B$ is defined so that
$$\int_{G / B} \tilde{a}(g) d_{G, B}(g)=\int_{\mathbb{R}^{d}} \tilde{a}\left(\exp \left(t_{1} X_{1}\right) \cdots \exp \left(t_{d} X_{d}\right)\right) d t_{1} \cdots d t_{d}$$
is left-invariant for any continuous function with compact support $\tilde{a}$ on $G / B$. By normalizing one of the vectors $X_{j}$, we have that
$$\int_{G} q(g) d g=\int_{G / B}\left(\int_{B} q(x b) d b\right) d_{G, B}(x)$$
for any continuous function with compact support $q$ on $G$. We always choose the invariant measures $d_{G, B}$ on the quotient spaces $G / B$ in such a way that this identity holds.

Let $\mathfrak{b}{1}, \mathfrak{b}{2}$ be two polarizations at the point $\phi \in \mathfrak{g}^{\star}$, and $B_{1}, B_{2}$ the two associated subgroups. Notice
$$S\left(G / B_{i}, \phi\right)=\mathscr{H}{\mathrm{ind}{B_{i}}^{G} \chi_{\phi}}^{\infty}, i \in{1,2},$$
the space of $C^{\infty}$-vectors of the representation spaces ind ${ }{B{i}}^{G} \chi_{\phi}, i \in{1,2}$, which are $\pi_{\phi, B}$ denote the representation ind $G_{B}^{G} \chi_{\phi}$. If $d_{B_{2}, B_{2} \cap B_{1}}$ denotes the $B_{2}$-left-invariant measure on $B_{2} / B_{2} \cap B_{1}$, for any function $\bar{k}$ of $S\left(G / B_{1}, \phi\right)$ the integral
$$T_{B_{2}, B_{1}} \tilde{k}(g)=\int_{B_{2} / B_{2} \cap B_{1}} \tilde{k}(g b) \chi \phi(b) d_{B_{2}, B_{2} \cap B_{1}}(b)$$
defined for every $g \in G$ is absolutely convergent and defines an isomorphism of $S\left(G / B_{1}, \phi\right)$ on $S\left(G / B_{2}, \phi\right)$ which extends by continuity into an intertwining operator between $\pi_{\phi, B_{1}}$ and $\pi_{\phi, B_{2}}$. Furthermore, if the measures on the homogeneous spaces $G / B_{1}$ and $G / B_{2}$ are suitably normalized, $T_{B_{2}, B_{1}}$ is an isometry.

## 数学代写|表示论代写Representation theory代考|Construction of the Intertwining Operator

We now specify a flag $\mathscr{A}$ of ideals of $\mathfrak{g}$. The peculiarity comes from the fact that if $\mathscr{C}=\left{Z_{1}, \ldots, Z_{n}\right}$ is the Jordan-Hölder basis of $\mathfrak{g}$ extracted from $\mathscr{A}$, then $\mathscr{C}$ contains a Jordan-Hölder basis
$$\mathscr{D}=\left{V_{1}=Z_{l_{1}}, \ldots, V_{n-r}=Z_{l_{n-r}}\right}$$
of $\mathfrak{h}$. We also extract from the latter the Malcev basis $\mathscr{B}=\left{B_{1}, \ldots, B_{r}\right}$ of $\mathfrak{g}$ relative to $h$ as above. The basis $\mathscr{C}$ gives us the index sets $I^{H}$ and $L^{H}$ and allows to choose a family $R=\left{R_{1}, \ldots, R_{r}\right}$ of real affine functions on $\mathbb{R}^{k}$ having the properties defined above for $L^{H}$. Moreover, by what we saw earlier the basis $\mathscr{B}$ gives us a $G$-invariant measure $d_{G, H}$ on $G / H$, which allows to fix the norm
$$|\xi|_{L^{2}(G / H, f)}=\left(\int_{G / H}|\xi(g)|^{2} d_{G, H}(g)\right)^{\frac{1}{2}},$$
for $\xi \in \mathscr{H}{\tau}=L^{2}(G / H, f)$. Let $\mathscr{V}=\mathscr{V} R, \mathscr{B}$. We will now construct a Zariski open set $\mathscr{V}{0}$ of $\mathscr{V}$ on which all of the following objects will be well defined. For $\phi \in \mathscr{V} 0$, we will construct a polarization $\mathfrak{b}(\phi)$ at $\phi$, a Malcev basis $\mathscr{X}^{\prime}(\phi)=\left{X_{1}(\phi), \ldots, X_{l}(\phi)\right}$ of $\mathfrak{g}$ relative to $\mathfrak{b}(\phi)$, a Jordan-Hölder basis $\mathscr{D}(\phi)=$ $\left{V_{1}(\phi), \ldots, V_{q}(\phi)\right}$ of $\mathfrak{h} \cap \mathfrak{b}(\phi)$, a Malcev basis $\mathscr{Y}(\phi)=\left{Y_{1}(\phi), \ldots, Y_{m}(\phi)\right}$ of $\mathfrak{b}(\phi)$ relative to $\mathfrak{h} \cap \mathfrak{b}(\phi)$ and finally a basis $\mathscr{U}(\phi)=\left{U_{1}(\phi), \ldots, U_{p}(\phi)\right}$ of $\mathbf{h}$ relative to $\mathfrak{h} \cap \mathfrak{b}(\phi)$. Here the numbers $l, m, p$ do not depend upon $\phi \in \mathscr{V}{0}$. In addition, all vectors $X{j}(\phi), V_{j}(\phi), Y_{j}(\phi)$ and $U_{j}(\phi)$ vary rationally and smoothly on $\phi \in \mathscr{V}_{0}$.

The vectors $\left(X_{j}(\phi)\right)$ ) will define a $G$-invariant measure on $G / B(\phi)$, and hence the norm of the space $\pi_{\phi}$. Likewise, the vectors $\left(Y_{j}(\phi)\right){j}$ determine the $B(\phi)$-invariant measure on $B(\phi) / H \cap B(\phi)$, hence the infinitesimal intertwining operator $T{B(\phi), H}$ and the vectors $\left(U_{j}(\phi)\right)_{j}$ will determine the measure on $H / H \cap B(\phi)$. All objects mentioned above are going to be constructed inductively, step by step. Step $s=0$ consists in defining the bases using certain subalgebras, set of indices and Zariski open sets of $\mathscr{V}$. We will also introduce the tools at this stage, while the objects will be given at the intermediate step $s \in{0, \ldots, \operatorname{dim}(\mathfrak{g})}$. In (3.2.3) we will explain in detail the techniques for passing from $s$ to $s+1$ and exhibit the newly constructed objects. At the very end we will show that the procedure actually stops, at some $s_{0} \in{0, \ldots, \operatorname{dim}(\mathrm{g})}$, and that the outcome bases are convenient. Note that our constructions depend only upon $8, f$ and $h$.

## 数学代写|表示论代写Representation theory代考|The Case of Exponential Solvable Groups

We study in this section exponential solvable Lie groups $G=\exp g$ for which the inducing subgroup $H=\exp \mathrm{h}$ is normal. We still consider a monomial representation $\tau=\operatorname{ind}{H}^{G} \chi{f}$, where $\chi_{f}$ denotes a character of $H$. Starting from a good sequence of subalgebras $\mathfrak{s}=\left(\mathfrak{a}{i}\right){i=0}^{n}$ passing through $\mathfrak{h}$, we determine an affine subspace $V$ of $\Gamma_{f}$ and a measure $d \lambda$ on $V$ such that
$$\tau \simeq \int_{V}^{\oplus} \pi_{\phi} d \lambda(\phi),$$
where $\pi_{\phi}$ are the irreducible representations associated to $\phi$. We next construct an explicit unitary intertwining operator and we find its inverse. The construction of such an operator $U$ goes through the following steps. We start from the good sequence 5 , we construct a coexponential basis $B$ of $\mathrm{h}$ in $\mathrm{g}$, we obtain our disintegration space $V$ with Lebesgue measure $d \lambda$. The basis $B$ defines an invariant measure on $G / H$, hence the norm on the space $\mathscr{H}{\tau}$ of $\tau$. We build next a Zariski open set $V{0}$ of $V$, and for each $\phi \in V_{0}$ the Vergne polarization $\mathfrak{b}(\phi)$ at $\phi$ relative to the good sequence $\mathfrak{5}$. These polarizations obviously contain $\mathfrak{h}$. We determine then for all $\phi \in V_{0}$ a coexponential basis $X(\phi)$ of $\mathfrak{b}(\phi)$ in $\mathfrak{g}$, which fixes a $G$ invariant positive form $v_{G, B(\phi)}$ on the space $K(G, B(\phi)):=\mathscr{E}(G, B(\phi))$ as in Sect. 1.2.2. The latter is the space of continuous numerical functions on $G$, with compact support modulo $B(\phi)=\exp (\mathfrak{b}(\phi))$ and satisfying:
$$F(g b)=\Delta_{B(\phi), G}(b) F(g)(g \in G, b \in B(\phi)),$$
Then we have a norm on the space $\mathscr{H}{\phi}$ of the irreducible representation $\pi{\phi}=$ ind $_{B(\phi)}^{G} \chi_{\phi}$. We also construct a coexponential basis $Y(\phi)$ to $\mathfrak{h}$ in $\mathfrak{b}(\phi)$ then an invariant measure $d_{B(\phi), H}$ on $B(\phi) / H$. All these bases vary continuously with $\phi \in V_{0}$ and allow to define the set
$$\mathscr{H}=\int_{V}^{\oplus} \mathscr{H}_{\phi} d \lambda(\phi)$$ of the disintegration of $\tau$. Now we associate, to each smooth function $\xi$ on $G$ with compact support modulo $H$ satisfying the generalized covariance relation (1.2.2):
$$\xi(g h)=\chi_{f}\left(h^{-1}\right) \Delta_{H, G}^{1 / 2}(h) \xi(g),(g \in G, h \in H)$$
and to each $\phi \in V_{0}$, the $C^{\infty}$-vector
$$T_{b(\phi), h} \xi(g)=\int_{B(\phi) / H} \xi(g b) \chi_{\phi}(b) \Delta_{B(\phi), G}^{-1 / 2}(b) d_{B(\phi), H}(b), \quad g \in G$$
of $\mathscr{H}_{\phi}$. We prove next that this operator is invertible. We will also examine some examples and describe a smooth disintegration of $L^{2}(G)$ for an exponential solvable Lie group $G$.

## 数学代写|表示论代写Representation theory代考|Disintegration of Monomial Representations

ψ:Rķ⟶乙,(吨1,…,吨ķ)⟶经验⁡吨1是1⋯经验⁡吨ķ是ķ

∫乙这(b)d(b)=∫Rd这(经验⁡(吨1是1)…经验⁡(吨ķ是ķ))d吨1…d吨ķ.

G=RX1⊕…⊕RXd⊕b 和 ∑ķ=jdRXķ+b

∫G/乙一个~(G)dG,乙(G)=∫Rd一个~(经验⁡(吨1X1)⋯经验⁡(吨dXd))d吨1⋯d吨d

∫Gq(G)dG=∫G/乙(∫乙q(Xb)db)dG,乙(X)

## 数学代写|表示论代写Representation theory代考|Construction of the Intertwining Operator

\mathscr{D}=\left{V_{1}=Z_{l_{1}}, \ldots, V_{nr}=Z_{l_{nr}}\right}\mathscr{D}=\left{V_{1}=Z_{l_{1}}, \ldots, V_{nr}=Z_{l_{nr}}\right}

|X|大号2(G/H,F)=(∫G/H|X(G)|2dG,H(G))12,

## 数学代写|表示论代写Representation theory代考|The Case of Exponential Solvable Groups

τ≃∫在⊕圆周率φdλ(φ),

F(Gb)=Δ乙(φ),G(b)F(G)(G∈G,b∈乙(φ)),

H=∫在⊕Hφdλ(φ)的解体τ. 现在我们关联到每个平滑函数X上G带紧凑支撑模H满足广义协方差关系（1.2.2）：

X(GH)=χF(H−1)ΔH,G1/2(H)X(G),(G∈G,H∈H)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MTH4107

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|The Key Point Is the Convergence

We keep all previous notation. In this last section we shall see that the key point is the convergence of the integral
$$\left(I_{h_{2} h_{1}}^{G} \varphi\right)(g):=\left(I_{h_{2} h_{1}} \varphi\right)(g)=\oint_{H_{2} /\left(H_{1} \cap H_{2}\right)} \varphi(g h) \chi_{f}(h) \Delta_{H_{2}, G}^{-1 / 2}(h) d v(h)(g \in G)$$
At least formally, it is clear that the function $I_{\mathfrak{h}{2} \mathfrak{h}{1}} \varphi$ satisfies the necessary covariance condition for belonging in the space $\mathscr{H}{\pi{2}}$, and also that $I_{\mathfrak{h}{2} \mathfrak{h}{1}}$ commutes with left translations. We can therefore assert that the convergence of the integral is one major issue to deal with. We call elements of $I(f, \mathfrak{g})$ Pukanszky polarizations.
In this section we look at all pairs $\left(\mathfrak{h}{1}, \mathfrak{h}{2}\right)$ of Pukanszky polarizations at $f$ which meet the following convergence property:
$(C P):$ The integral
$$\oint_{H_{2} /\left(H_{1} \cap H_{2}\right)}|\varphi(g h)| \Delta_{H_{2}, G}^{-\frac{1}{2}}(h) d v(h), g \in G$$
converges on a $G$-invariant and $\mathscr{U}(\mathrm{g})$-invariant dense subspace $\mathscr{H}$ of $\mathscr{H}{\pi{1}}^{\infty}$.
Proposition 2.8.1 Suppose that the pair of Pukanszky polarizations $\left(\mathfrak{h}{1}, \mathbf{h}{2}\right)$ satisfies property $(C P)$. Then for any function $\varphi \in \mathscr{K}\left(f, \mathfrak{h}{1}, G\right)$ we have that $$\oint{H_{2} /\left(H_{1} \cap H_{2}\right)}|\varphi(g h)| \Delta_{H_{2}, G}^{-\frac{1}{2}}(h) d v(h)<\infty, g \in G$$
and the function $g \mapsto\left(I_{b_{2} h_{1}} \varphi\right)(g)$ is continuous.

## 数学代写|表示论代写Representation theory代考|Intertwining Operators of Induced Representations

Let $G$ be a connected and simply connected nilpotent Lie group with Lie algebra g. Consider, as in (1.6.1), the monomial representation $\tau=$ ind $_{H}^{G} \chi$ induced by the unitary character $\chi$ of an analytic subgroup $H$ of $G$. Recall that $h$ is the Lie subalgebra corresponding to $H$, and $\chi$ is written in the form $\chi(\exp X)=$ $e^{i\langle f, X\rangle}\left(X \in\right.$ h) with $f \in \mathfrak{g}^{\star}$. Let $\Gamma_{f}=f+\mathfrak{h}^{\perp}$ be as in Eq. (1.4.8). In parallel to decomposition formula (1.6.2), the disintegration of $\tau$ into irreducibles reads:
$$\tau \simeq \int_{\Gamma_{f} / H}^{\oplus} \pi_{l} d l$$

where $d l$ is some natural measure on the space $\Gamma_{f} / H$ of $H$-orbits (cf. Theorem 1.4.2).

We construct in this section an intertwining operator of decomposition (3.2.1) for an arbitrary subgroup $H$. The idea is to make formula (3.2.1) explicit through the construction of certain affine subspaces
$$\mathscr{V}^{R, \mathfrak{B}}=\left{\sum_{i=1}^{r} R_{i}(T) f_{i} ; T \in \mathbb{R}^{k}\right}$$
of $\Gamma_{f}$, and a measure $d \lambda=d \lambda^{R, \mathfrak{B}}$ on $\mathscr{V} R, \mathfrak{B}$, so that
$$\tau \simeq \int_{\mathscr{V} R, \mathfrak{g}}^{\oplus} \pi_{\phi} d \lambda^{R, \mathfrak{B}}(\phi) .$$
Here $R$ denotes a family $\left{R_{1}, \ldots, R_{r}\right}$ of affine functions defined on $\mathbb{R}^{k}$, for some non-negative integer $k \in \mathbb{N}$, and $\mathfrak{B}^{\star}=\left{f_{r}, \ldots, f_{1}\right}$ denotes the dual of a Malcev basis $\mathfrak{B}$ relative to $\mathfrak{h}$. Such a basis defines an invariant measure on $G / H$ and so the norm on the space $\mathscr{H}{\mathrm{r}}$ of $\tau$. We then determine a Zariski open set $\mathscr{V}{0}$ of $\mathscr{V}^{R, \mathfrak{B}}$, a polarization $\mathfrak{b}(\phi)$ in $\phi$ for each $\phi \in \mathscr{V} 0$, a Malcev basis $\mathscr{X}(\phi)$ of $\mathfrak{b}(\phi)$, a Malcev basis $\mathscr{Y}(\phi)$ of $\mathfrak{b}(\phi)$ relative to $\mathfrak{h} \cap \mathfrak{b}(\phi)$ and a Malcev basis $\mathscr{U}$ of $\mathfrak{h}$ relative to $\mathfrak{h} \cap \mathfrak{b}(\phi)$. All these bases vary continuously in $\phi \in \mathcal{V}{0}$, which allows to fix the invariant measure $d{G, B(\phi)}$ on $G / B(\phi)$, where $B(\phi)=\exp (\mathfrak{b}(\phi)$ ) (and hence the norm on the space $\mathscr{H}{\phi}$ of the irreducible representation $\pi{\phi}=$ ind $\left.{B(\phi)}^{G} \chi{\phi}\right), d_{B(\phi), B(\phi) \cap H}$ on $B(\phi) / B(\phi) \cap H$, and $d_{H, H \cap B(\phi)}$ on $H / B(\phi) \cap H$. This permits to define the disintegration space
$$\mathscr{H}=\int_{\mathscr{y},, \mathfrak{B}}^{\oplus} \mathscr{H}{\phi} d \lambda(\phi)$$ of $\tau$. We now associate, to any sufficiently regular function $\xi$ of $\mathscr{H}{\tau}$ and any $\phi \in \mathscr{V}{0}$, the $C^{\infty}$-vector of $\mathscr{H}{\phi}$ :
$$T_{B(\phi), H} \xi(g)=\int_{B(\phi) / B(\phi) \cap H} \xi(g b) \chi_{\phi}(b) d_{B(\phi), B(\phi) \cap H(b), g \in G}$$
We will show that
$$\int_{\mathscr{Y}{0}}\left|T{B(\phi), H} \xi\right|_{\mathscr{H}{\phi}}^{2} d \phi=|\xi|{\mathscr{H}_{\mathrm{T}}}^{2}$$
therefore obtaining an isometric intertwining operator of (3.2.2). A (smooth) disintegration of $L^{2}(G)$ is also generated.

## 数学代写|表示论代写Representation theory代考|Notation and Backgrounds

Let $G$ be a connected and simply connected nilpotent Lie group with Lie algebra g. Let exp denote, as earlier, the exponential map, so that $G=\exp g$. Let $V, W$ be real vector spaces of finite dimension with $W \subset V$. We denote by $V^{}$ the dual vector space of $V$ and by $W^{\perp}$ the orthogonal to $W$ in $V^{}$. If $u_{1}, \ldots, u_{p},(p \in \mathbb{N})$ indicate linearly independent vectors in $V$, we denote by $\mathbb{R}-\operatorname{span}\left(u_{1}, \ldots, u_{p}\right)$ the vector subspace of $V$ they span, and we say the basis $\left{u_{1}, \ldots, u_{p}\right}$ generates this space. Given $l \in \mathrm{g}^{*}$ and $X \in \mathfrak{g}$, we denote by $\langle l, X\rangle$ the image of $X$ under $l$. Recall that the kernel of the bilinear form $B_{l}$, defined in Sect. 1.2.4 by $B_{l}(X, Y)=$ $\langle l,[X, Y]\rangle$, is denoted by $\mathrm{g}(l)=\left{X \in \mathfrak{g} ; B_{l}(X, Y)=0\right.$ for all $\left.Y \in \mathfrak{g}\right}$. The largest ideal contained in $\mathrm{g}(I)$ will be denoted by a $(l)$. Clearly
$$\mathrm{a}(l)=\bigcap_{\phi \in G-l} \mathrm{~g}(\phi)$$
Let $\mathfrak{h} \in S(l, \mathfrak{g})$ and let $\chi_{l}$ be the unitary character of the analytic subgroup $H=$ exp $h$ associated to $l$ by
$$\chi_{l}(\exp X)=e^{-i\langle l, X\rangle}$$
for $X \in 6$
Let $V$ be a real vector space of finite dimension and $\rho: G \longrightarrow \operatorname{End}(V)$ a unipotent action of $G$ on $V$. We designate by
$$(0)=V_{n} \subset V_{n-1} \subset \ldots \subset V_{1} \subset V_{0}=V$$
a Jordan-Hölder sequence for $G$ and call $\mathscr{L}=\left{v_{1}, \ldots, v_{n}\right}$ an associated JordanHölder basis $\left(v_{j} \in V_{j-1} \backslash V_{j}\right.$ ). If $v \in V$, we write $\rho(x) v=x \cdot v$ for all $x \in G$ and
$$X \cdot v=\left.\frac{d}{d t}{\rho(\exp (t X)) \cdot v}\right|{t=0},(X \in \mathfrak{g})$$ The set of indices $e^{\mathscr{L}}(v)=\left{i{1}<\ldots<i_{d}\right}$ of an element $v$ in $V$ relative to $\mathscr{L}$ is the following subset of ${1, \ldots, n}$ :
$$i \in e^{\mathscr{L}}(v) \Leftrightarrow \exists X \in \mathfrak{g}: X \cdot v \in V_{i-1} \backslash V_{i}$$
The cardinality $d$ of $e^{\mathscr{L}}(v)$ is the dimension of the $G$-orbit $\Omega$ of $v$ and does not vary with $v$ in $\Omega$.

## 数学代写|表示论代写Representation theory代考|The Key Point Is the Convergence

(我H2H1G披)(G):=(我H2H1披)(G)=∮H2/(H1∩H2)披(GH)χF(H)ΔH2,G−1/2(H)d在(H)(G∈G)

(C磷):积分

∮H2/(H1∩H2)|披(GH)|ΔH2,G−12(H)d在(H),G∈G

∮H2/(H1∩H2)|披(GH)|ΔH2,G−12(H)d在(H)<∞,G∈G

## 数学代写|表示论代写Representation theory代考|Intertwining Operators of Induced Representations

τ≃∫ΓF/H⊕圆周率ldl

\mathscr{V}^{R, \mathfrak{B}}=\left{\sum_{i=1}^{r} R_{i}(T) f_{i} ; T \in \mathbb{R}^{k}\right}\mathscr{V}^{R, \mathfrak{B}}=\left{\sum_{i=1}^{r} R_{i}(T) f_{i} ; T \in \mathbb{R}^{k}\right}

τ≃∫在R,G⊕圆周率φdλR,乙(φ).

H=∫是,,乙⊕Hφdλ(φ)的τ. 我们现在将任何足够规则的函数联系起来X的Hτ和任何φ∈在0， 这C∞-向量Hφ :

∫是0|吨乙(φ),HX|Hφ2dφ=|X|H吨2

## 数学代写|表示论代写Representation theory代考|Notation and Backgrounds

χl(经验⁡X)=和−一世⟨l,X⟩

(0)=在n⊂在n−1⊂…⊂在1⊂在0=在
Jordan-Hölder 序列为G并打电话\mathscr{L}=\left{v_{1}, \ldots, v_{n}\right}\mathscr{L}=\left{v_{1}, \ldots, v_{n}\right}相关的 JordanHölder 基(在j∈在j−1∖在j）。如果在∈在， 我们写ρ(X)在=X⋅在对所有人X∈G和

X⋅在=dd吨ρ(经验⁡(吨X))⋅在|吨=0,(X∈G)索引集e^{\mathscr{L}}(v)=\left{i{1}<\ldots<i_{d}\right}e^{\mathscr{L}}(v)=\left{i{1}<\ldots<i_{d}\right}一个元素的在在在关系到大号是以下子集1,…,n :

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|MATH4314

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|The General Case

Take $\mathfrak{h}{j} \in I(f, \mathfrak{g})$ and $H{j}=\exp \left(h_{j}\right)(j=1,2)$. We shall construct an intertwining operator $T_{\mathfrak{b}{2} \mathfrak{h}{1}}$ so that, given $\mathfrak{h}_{j} \in I(f, \mathfrak{g})(1 \leq j \leq 3)$, the composition formula
holds. In order to define this operator we use a third polarization, a Vergne polarization to be precise.

Theorem 2.5.1 Let $\mathfrak{h}{j} \in I(f, \mathfrak{g})(j=1,2)$ and take a Vergne polarization ho at $f \in \mathrm{g}^{*}$. Then the intertwining isometry $$T{\mathfrak{h}{2} \mathfrak{h}{1}}=a\left(\mathfrak{h}{1}, \mathfrak{h}{0}, \mathfrak{h}{2}\right) T{\mathfrak{h}{2} \mathfrak{h}{0}} \circ T_{\mathfrak{h}{0} \mathfrak{h}{1}},$$
where $T_{\mathfrak{h}{2} \mathfrak{h}{0}}, T_{\mathfrak{h}{0} \mathfrak{h}{1}}$ are the intertwining isometries defined in Theorem 2.4.2, does not depend on the choice of the Vergne polarization $\mathrm{h}{0}$. Moreover, $T{\mathrm{h}{2} \mathfrak{h}{1}}$ coincides with $I_{\mathfrak{h}{2} \mathfrak{h}{1}}$ if at least one of $\mathfrak{h}{1}, \mathfrak{h}{2}$ is a Vergne polarization.

Proof We proceed again by induction on $\operatorname{dim} G$. As in the previous proofs, we can at once suppose that there is no minimal ideal which is not central and $f$ does not vanish on any ideal of $\mathrm{g}$.
Let now $\mathbf{h}{3}, \mathbf{h}{4}$ be two Vergne polarizations at $f$. We claim
$$a\left(b_{1}, h_{3}, h_{2}\right) T_{h_{2} b_{3}} \circ T_{h_{3} h_{1}}=a\left(b_{1}, h_{4}, b_{2}\right) T_{b_{2} b_{4}} \circ T_{\mathfrak{h}{4} b{1}},$$
where $a\left(\mathfrak{h}{i}, \mathbf{h}{j}, \mathfrak{h}{k}\right)=e^{\frac{\mathbf{I}^{2}}{} \tau\left(\mathfrak{h}, \mathfrak{h}{j}, \mathbf{h}{k}\right)}$. If there is a minimal non-central ideal a contained in $\mathfrak{h}{3} \cap \mathbf{h}{4}$, then $\mathbf{h}{3} \cup \mathbf{h}{4} \subset a^{f}$ and we can apply the induction hypothesis to $\mathfrak{h}{i}^{\prime}=\mathfrak{h}{i} \cap \mathfrak{a}^{f}+\mathfrak{a}, i=1,2$. Thus $$a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{3}, \mathfrak{h}{2}^{\prime}\right) T_{\mathfrak{h}{2}^{\prime} \mathfrak{h}{3}} \circ T_{\mathfrak{h}{3} \mathfrak{h}{1}^{\prime}}=a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{4}, \mathfrak{h}{2}^{\prime}\right) T{\mathfrak{h}{2}^{\prime} \mathfrak{h}{4}} \circ T_{\mathfrak{h}{4} \mathfrak{h}{1}^{\prime}}$$
We deduce from this
$$a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{3}, \mathfrak{h}{2}^{\prime}\right) T{\mathfrak{h}{2} \mathfrak{h}{3}} \circ T_{\mathfrak{h}{3} \mathfrak{h}{1}}=a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{4}, \mathfrak{h}{2}^{\prime}\right) T{\mathfrak{h}{2} \mathfrak{h}{4}} \circ T_{\mathfrak{h}{4} \mathfrak{h}{1}}$$
Hence it suffices to show
$$a\left(h_{1}, h_{3}, h_{2}\right) a\left(h_{1}^{\prime}, h_{4}, h_{2}^{\prime}\right)=a\left(h_{1}, h_{4}, h_{2}\right) a\left(h_{1}^{\prime}, h_{3}, h_{2}^{\prime}\right)$$
But from Lemma 2.3.11 we have
$a\left(h_{1}, h_{3}, h_{2}\right)=a\left(h_{4}, h_{3}, h_{2}\right) a\left(h_{1}, h_{4}, h_{2}\right) a\left(h_{1}, h_{3}, b_{4}\right)$
$=a\left(h_{4}, \mathbf{h}{3}, \mathbf{h}{2}^{\prime}\right) a\left(h_{1}, \mathbf{h}{4}, \mathbf{h}{2}\right) a\left(\mathbf{h}{1}^{\prime}, \mathbf{h}{3}, \mathbf{h}_{4}\right)$

## 数学代写|表示论代写Representation theory代考|A Local Result

We denote by $\mathscr{H}{\rho}$ the Hilbert space of a unitary representation $\rho$ of $G$, by $\mathscr{H}{\rho}^{\infty}$ the space of $C^{\infty}$-vectors of $\rho$ equipped with the usual topology and by $\mathscr{H}{\rho}^{-\infty}$ the anti-dual space of $\mathscr{H}{\rho}^{\infty}$. Given a closed subgroup $K$ of $G$ and a character $\lambda$ of $K$, we set
$$\left(\mathscr{H}{\rho}^{-\infty}\right)^{K, \lambda}=\left{a \in \mathscr{H}{\rho}^{-\infty} ; \rho(k) a=\lambda(k) a, \forall k \in K\right}$$
Let $T_{\mathfrak{h}{2} \mathfrak{h}{1}}$ be the isometry defined above which intertwines $\pi_{1}$ and $\pi_{2}$. This section is devoted to the proof of the next theorem.
Theorem 2.6.1 There exists a positive form $v=v_{H_{2}, H_{1} \cap H_{2}}$ such that
$$T_{b_{2} h_{1}} \varphi(e)=\oint_{H_{2} /\left(H_{1} \cap H_{2}\right)} \varphi(h) \chi_{f}(h) \Delta_{H_{2}, G}^{-1 / 2}(h) d v(h)$$
for any function $\varphi \in \mathscr{H}{\pi{1}}^{\infty}$ whose support is sufficiently small modulo $H_{1}$.

## 数学代写|表示论代写Representation theory代考|The Case Where h1 + h2 Is a Subalgebra

Lemma 2.7.1 Let $\mathfrak{h}{j} \in I(f, \mathfrak{g})(j=1,2)$ be such that $\mathfrak{t}=\mathfrak{h}{1}+\mathfrak{h}{2}$ is a subalgebra of $\mathrm{g}$. Then $K=\exp \mathrm{E}=\mathrm{H}{2} \mathrm{H}{1}$ and there exists a coexponential basis of $\mathrm{h} 1 \mathrm{in} \mathrm{g}, a$ part of which is coexponential to $\mathfrak{h}{1} \cap \mathfrak{h}{2}$ in $\mathrm{h}{2}$. In particular, $H_{1} H_{2}$ is closed in $G$.
Proof Supposing $\mathfrak{g}=\mathfrak{h}{1}+\mathbf{h}{2}$, we shall verify $G=H_{1} H_{2}$. We already know that $H_{1} H_{2}$ is open in $G$. Consider a sequence of subalgebras
$$h_{1}=m_{0} \subset m_{1} \subset \cdots \subset m_{k}=\mathfrak{h}{1}+[\mathfrak{g}, \mathfrak{g}]$$ such that the adjoint action of $\mathrm{m}{j-1}$ on $\mathrm{m}{j} / \mathrm{m}{j-1}$ is irreducible for all $1 \leq j \leq k$. We put $M_{j}=\exp \left(m_{j}\right)$. Supposing $M_{j-1} \subset H_{1} H_{2}$, we shall prove the same inclusion for $M_{j}$. Suppose first $\operatorname{dim}\left(m_{j} / m_{j-1}\right)=1$. We take a coexponential basis ${X}$ of $\mathrm{m}{j-1}$ in $\mathrm{m}{j}$. By writing $g \in M_{j}$ in the form $g=\exp (x X) \cdot g_{0}$, where $x \in \mathbb{R}$ and $g_{0} \in M_{j-1}$, we immediately see that $g \in H_{2} H_{1}$ if and only if $\exp (x X) \in H_{2} H_{1}$.

On the other hand, the condition
$$\exp (x X) \in H_{2} H_{1}=\left{a \in G ; a \cdot\left(f+\left(\mathfrak{h}{1}\right)^{\perp}\right) \cap\left(f+\left(\mathbf{h}{2}\right)^{\perp}\right) \neq \emptyset\right}$$
is expressed by the vanishing of a real analytic function of $x$. Since $H_{2} H_{1}$ is open in $G$, we have rank $B(g) \leq \operatorname{rank} A(e)$ for all $g \in G$. Here, $B(g)$ and $A(e)$ denote the matrices introduced just after Lemma 2.3.11. On the other hand we have $\left(H_{2} H_{1}\right) \cap$ $M_{j}=\exp \left(\mathrm{h}{2} \cap \mathrm{m}{j}\right) H_{1}$, which is connected. These observations give us the desired inclusion $\mathrm{M}{j} \subset \mathrm{H}{2} \mathrm{H}_{1}$.

Suppose now $\operatorname{dim}\left(\mathrm{m}{j} / \mathrm{m}{j-1}\right)=2$. We take $\left{X, X^{\prime}\right}$ in $[\mathfrak{g}, \mathfrak{g}]$ so that $\left{X, X^{\prime}\right}$ is a coexponential basis of $\mathrm{m}{j-1}$ in $\mathrm{m}{j}$. When we write $X=Y_{1}+Y_{2}, X^{\prime}=Y_{1}^{\prime}+Y_{2}^{\prime}$ with $Y_{i}, Y_{i}^{\prime} \in \mathfrak{h}{i}(i=1,2)$, we may for instance assume $\left[Y{i}^{\prime}, \mathrm{m}{j}\right] \subset \mathrm{m}{j-1}(i=1,2)$. If, further, $\left[Y_{i}, \mathrm{~m}{j}\right] \subset \mathrm{m}{j-1}(i=1,2)$, then $\left{Y_{2}, Y_{2}^{\prime}\right}$ becomes a coexponential basis of $\mathrm{m}{j-1}$ in $\mathrm{m}{j}$. For otherwise $\left[Y_{i}, \mathrm{~m}{j}\right] \not \subset \mathrm{m}{j-1}$ and it suffices to replace $X$ by $\left[Y_{1}, X^{\prime}\right]-\left[X, Y_{2}^{\prime}\right]=\left[Y_{1}, Y_{1}^{\prime}\right]-\left[Y_{2}, Y_{2}^{\prime}\right]$ to go fall back into the previous case.

Hence $H_{1} \subset M_{k} \subset H_{2} H_{1}$. Since $\mathrm{m}{k} \supset[\mathrm{g}, \mathrm{g}]$, we can finally choose elements of $\mathfrak{h}{2}$ which constitute a coexponential basis of $m_{k}$ in $\mathfrak{g}$.

## 数学代写|表示论代写Representation theory代考|The General Case

=一个(H4,H3,H2′)一个(H1,H4,H2)一个(H1′,H3,H4)

## 数学代写|表示论代写Representation theory代考|A Local Result

\left(\mathscr{H}{\rho}^{-\infty}\right)^{K, \lambda}=\left{a \in \mathscr{H}{\rho}^{-\infty} ; \rho(k) a=\lambda(k) a, \forall k \in K\right}\left(\mathscr{H}{\rho}^{-\infty}\right)^{K, \lambda}=\left{a \in \mathscr{H}{\rho}^{-\infty} ; \rho(k) a=\lambda(k) a, \forall k \in K\right}

## 数学代写|表示论代写Representation theory代考|The Case Where h1 + h2 Is a Subalgebra

H1=米0⊂米1⊂⋯⊂米ķ=H1+[G,G]这样的伴随动作米j−1上米j/米j−1对所有人都是不可约的1≤j≤ķ. 我们把米j=经验⁡(米j). 假如米j−1⊂H1H2，我们将证明相同的包含米j. 假设首先暗淡⁡(米j/米j−1)=1. 我们以共指数为基础X的米j−1在米j. 通过写作G∈米j在表格中G=经验⁡(XX)⋅G0， 在哪里X∈R和G0∈米j−1，我们立即看到G∈H2H1当且仅当经验⁡(XX)∈H2H1.

\exp (x X) \in H_{2} H_{1}=\left{a \in G ; 一个 \cdot\left(f+\left(\mathfrak{h}{1}\right)^{\perp}\right) \cap\left(f+\left(\mathbf{h}{2}\right)^ {\perp}\right) \neq \emptyset\right}\exp (x X) \in H_{2} H_{1}=\left{a \in G ; 一个 \cdot\left(f+\left(\mathfrak{h}{1}\right)^{\perp}\right) \cap\left(f+\left(\mathbf{h}{2}\right)^ {\perp}\right) \neq \emptyset\right}

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|表示论代写Representation theory代考|Intertwining Operators for Irreducible Representations

statistics-lab™ 为您的留学生涯保驾护航 在代写表示论Representation theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写表示论Representation theory代写方面经验极为丰富，各种代写表示论Representation theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|A Trace Relation

Let $G=\exp \mathfrak{g}$ be an exponential solvable Lie group with Lie algebra $\mathfrak{g}$. If h,s and $\mathfrak{h}{2}$ are two polarizations of $\mathfrak{g}$ at $f \in \mathfrak{g}^{*}$ which satisfy the Pukanszky condition, the orbit method asserts that the monomial representations $\pi{i}=$ ind $_{H_{i}}^{G} \chi_{f}\left(H_{i}=\right.$ $\exp \left(h_{i}\right), 1 \leq i \leq 2$ ) of $G$ are irreducible and mutually equivalent. We are interested in constructing an explicit intertwining operator between these representations. Let us sketch the idea, which goes back to Vergne [151]. If we have
$$\text { Tr } \mathrm{ad}{\mathfrak{h}{1} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} X+\operatorname{Tr} \mathrm{ad}{\mathbf{h}{2} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} X=0$$ for any $X \in \mathfrak{h}{1} \cap \mathfrak{h}{2}$, then
$$\Delta_{H_{1}, G}(h)=\Delta_{H_{2}, G}(h) \Delta_{H_{1} \cap H_{2}, H_{2}}(h)^{2}$$
for all $h \in H_{1} \cap H_{2}$, and, for $\phi$ in the Hilbert space $\mathscr{H}{\pi{1}}$ of $\pi_{1}$ and $g \in G$, the function $\Phi_{g}$ on $H_{2}$ given by
$$\Phi_{g}(h)=\phi(g h) \chi_{f}(h) \Delta_{H_{2}, G}^{-1 / 2}(h)$$
verifies the relation
$$\Phi_{g}(h x)=\Delta_{H_{1} \cap H_{2}, H_{2}}(x) \Phi_{g}(h)\left(h \in H_{2}, x \in H_{1} \cap H_{2}\right)$$
Thus we can, at least formally, consider the integral
$$\left(I_{b_{2} \mathfrak{b}} \phi\right)(g)=\oint_{H_{2} /\left(H_{1} \cap H_{2}\right)} \phi(g h) \chi_{f}(h) \Delta_{H_{2}, G}^{-1 / 2}(h) d v_{H_{2}, H_{1} \cap H_{2}}(h)$$
for $\phi \in \mathscr{H}{\Pi}$ and $g \in G$. If this integral converges for any $g \in G$, it is clear that $I{\mathfrak{h}{2} \mathfrak{h}{1}} \phi$ verifies the covariance relation required on the elements of the space $\mathscr{H}{\pi{2}}$, and $I_{\left.\mathfrak{h}_{2} \mathfrak{h}\right) 1}$ commutes with the action of $G$ by left translations. In fact, Vergne proved the following proposition. An ideal of $g$ is said to be minimal non-central if it is minimal among all non-central ideals of $\mathfrak{g}$. For an ideal $a$ of $\mathfrak{g}$, we put
$$\mathfrak{a}^{f}={X \in \mathfrak{g} ; f([X, \mathfrak{a}])={0}} .$$

## 数学代写|表示论代写Representation theory代考|Relations Between Two Polarizations

We keep the notations.
Remark 2.3.1 Without assuming the Pukanzsy condition Theorem $2.2 .2$ might fail, and we cannot write the integral of $(2.2 .1)$ even if $\pi_{1} \simeq \pi_{2}$. Indeed, let $\mathfrak{g}$ be the completely solvable Lie algebra of dimension 4 with basis $(T, P, Q, Z)$ satisfying
$$[T, P]=\frac{1}{2} P,[T, Q]=\frac{1}{2} Q,[T, Z]=[P, Q]=Z .$$

Let $f=Z^{} \in \mathfrak{g}^{}$. Then $\mathfrak{g}(f)={0}$ and
$$\mathbf{h}{1}=\mathbb{R} T+\mathbb{R} P+\mathbb{R} Z, \mathbf{h}{2}=\mathbb{R} T+\mathbb{R} Q+\mathbb{R} Z$$
belong to $M(f, \mathfrak{g})$, but neither $\mathfrak{h}{1}$ nor $\mathfrak{h}{2}$ verifies the Pukanszky condition.
There exist two open coadjoint orbits $O_{\pm}$of $G=\exp g$ :
$$O_{+}=\left{l \in \mathfrak{g}^{} ; l(Z)>0\right}, O_{-}=\left{l \in \mathfrak{g}^{} ; l(Z)<0\right}$$
Denoting by $\rho(\Omega)$ the irreducible unitary representation of $G$ corresponding to the orbit $\Omega \in \mathfrak{g}^{*} / G$, we know that $\pi_{1} \simeq \pi_{2} \simeq \rho\left(O_{+}\right) \oplus \rho\left(O_{-}\right)$(cf. [151]). However, it is evident that $\operatorname{Tr} \mathrm{ad}{\mathfrak{h}{1} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} T=\operatorname{Tr} \mathrm{ad}{\mathfrak{h}{2} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} T=1 / 2$.

When $G=\exp g$ is nilpotent and $K_{1}, K_{2}$ are analytic subgroups, the product $K_{1} K_{2}=\left{k_{1} k_{2} ; k_{1} \in K_{1}, k_{2} \in K_{2}\right}$ is always a closed subset of $G$ (cf. [97]). We denote by $\mathscr{H}\left(f, \mathfrak{h}{1}, G\right)$ the dense subspace of $\mathscr{H}{\pi_{1}}$ consisting of continuous functions with compact support modulo $H_{1}$. It follows that the integral (2.2.1) converges for $\phi \in \mathscr{H}\left(f, \mathbf{h}{1}, G\right)$. In fact, Lion proved that (2.2.1) gives an intertwining operator between $\pi{1}$ and $\pi_{2}$. On the other hand, when we pass to the exponential case $K_{1} K_{2}$ might not be closed, thus making the convergence of integral (2.2.1) a serious issue.
Example 2.3.2 Take
$$G_{2}=\exp g_{2}=\left{\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in M_{2}(\mathbb{R}) ; a>0\right}(a x+b \text { group })$$
The elements
$$e_{1}=\left(\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right), e_{2}=\left(\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right)$$
form a basis of $\mathfrak{g}{2}$ such that $\left[e{1}, e_{2}\right]=e_{2}$. Let $f=e_{2}^{} \in \mathfrak{g}{2}^{}, \mathfrak{t}{1}=\mathbb{R} e_{1}, \mathfrak{e}{2}=\mathbb{R}\left(e{1}+\right.$ $\left.e_{2}\right)$ and $K_{i}=\exp \mathfrak{t}{i}(i=1,2)$. Then $\mathfrak{t}{i} \in M\left(f, \mathfrak{g}{2}\right)$ but $\mathfrak{t}{i} \notin I\left(f, \mathfrak{g}{2}\right)(i=1,2)$. We immediately see $$K{2} K_{1}=\left{g=\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in G_{2} ; b>-1\right}$$
and $\mathfrak{t}{2}=g{0} \cdot \mathfrak{t}{1}$ with $g{0}=\left(\begin{array}{cc}1 & -1 \ 0 & 1\end{array}\right) \in G_{2}$.

## 数学代写|表示论代写Representation theory代考|Vergne Polarizations

We retain the notations $G=\exp g, f \in \mathfrak{g}^{*}, \mathfrak{h}{j} \in I(f, \mathfrak{g})$ and $H{j}=\exp \left(\mathfrak{h}_{j}\right)$ for $j=1,2$.

Lemma 2.4.1 If $h_{1}$ is a Vergne polarization, then there exists a coexponential basis of $\mathrm{h}{1} \cap \mathrm{h}{2}$ in $\mathrm{h}{2}$ which is a part of a coexponential basis of $\mathrm{h}{1}$ in $\mathrm{g}$. Likewise, there exists a coexponential basis of $\mathrm{h}{1} \cap \mathrm{h}{2}$ in $\mathrm{h}{1}$ which is a part of a coexponential basis of $h{2}$ in $\mathrm{g}$. In particular, $\mathrm{H}{2} \mathrm{H}{1}$ (hence $\mathrm{H}{1} \mathrm{H}{2}$, too) is closed in $G$.

Proof We proceed by induction on $\operatorname{dim} G$. If there exists a non-trivial ideal a of $\mathfrak{g}$ on which $f$ vanishes, everything passes to the quotient $G / A$ with $A=\exp a$. This case will be excluded in what follows. If there exists a minimal ideal $a$ which is not central, we know (cf. [24, Chap. VI]) that a is contained in any element of $I(f, \mathfrak{g})$. Hence for $j=1,2, \mathfrak{h}{j} \subset \mathfrak{a}^{f} \neq \mathfrak{g}$. Suppose that the Vergne polarization $\mathfrak{h}{1}$ is constructed starting from a good sequence $\mathfrak{s}=\left(a_{j}\right){0 \leq j \leq n}$ of subalgebras, namely: $$a{j-1} \subset a_{j}, \operatorname{dim}\left(a_{j}\right)=j, \mathbf{h}{1}=\sum{j=1}^{n} a_{j}\left(f_{j}\right), f_{j}=\left.f\right|{a{j}}(1 \leq j \leq n) .$$
Put $\mathfrak{b}{j}=\mathfrak{a}{j} \cap \mathfrak{a}^{f}$ and $f_{j}^{\prime}=\left.f\right|{\mathfrak{b}{j}}(0 \leq j \leq n)$. There exists an index $j_{0}\left(1 \leq j_{0} \leq n\right)$ such that $\mathbf{b}{j 0}=\mathfrak{b}{j_{0}-1}$ and it is clear that $\mathfrak{s}^{\prime}=\left(\mathfrak{b}{j}\right){0 \leq j \leq n, j \neq j_{0}}$ is a good sequence of subalgebras of $\mathfrak{a}^{f}$. Since $\mathbf{h}{1}=\sum{j=1}^{n} a_{j}\left(f_{j}\right) \subset \mathbf{a}^{f}$, we see $a_{j}\left(f_{j}\right) \subset b_{j}\left(f_{j}^{\prime}\right)$ for every $j$ and hence
$$\mathfrak{h}{1} \subset \sum{j=1}^{n} \mathfrak{b}{j}\left(f{j}^{\prime}\right)$$
Well, $\mathfrak{h}_{1}$ being a Lagrangian subspace, we necessarily have equality in (2.4.1). Now it suffices to apply the induction hypothesis to $A^{f}=\exp \left(a^{f}\right)$.

Suppose now that there exists no minimal ideal of $\mathfrak{g}$ which is not central. Take the good sequence $s$ which defines $\mathfrak{h}{1}$. With our hypothesis, we necessarily have $\mathfrak{a}{1}=\mathfrak{z}$ and $f_{1} \neq 0$. If $\mathfrak{a}{2}$ is an ideal of $\mathfrak{g}$, the above reasoning shows that $\mathfrak{h}{1}$ is a Vergne polarization of the subalgebra $\left(a_{2}\right)^{f}$. If $h_{2} \subset\left(a_{2}\right)^{f}$, it suffices to apply the induction hypothesis. If not, we modify $\mathfrak{h}{2}$ to $\mathfrak{h}{2}^{\prime}=\left(\mathfrak{h}{2} \cap\left(\mathfrak{a}{2}\right)^{f}\right)+\mathfrak{a}{2}$ and $H{2}$ to $H_{2}^{\prime}=\exp \left(\mathfrak{h}{2}^{\prime}\right)$. From the Pukanszky condition we can take (cf. [24, Chap. VI]) a coexponential basis ${X}$ of $\left(a{2}\right)^{f}$ in $g$ in such a way that $X$ belongs to $h_{2}$. Take also $Y$ in $a_{2} \backslash a_{1}$

On the other hand, the induction hypothesis says that there exists a coexponential basis $\left{X_{1}^{\prime}, \ldots, X_{k}^{\prime}\right}$ to $\mathfrak{h}{1}$ in $\left(a{2}\right)^{f}$ which contains a coexponential basis $\left{X_{i_{1}}^{\prime}, \ldots, X_{i_{m}}^{\prime}\right}$ of $\mathfrak{h}{1} \cap \mathfrak{h}{2}^{\prime}$ in $\mathfrak{h}{2}^{\prime}$ and even in $\mathfrak{h}{2} \cap\left(a_{2}\right)^{f}$. Then $\left{X, X_{1}^{\prime}, \ldots, X_{k}^{\prime}\right}$ is a coexponential basis of $h_{1}$ in $\mathfrak{g}$, whose part $\left{X_{i_{1}}^{\prime}, \ldots, X_{i_{m}}^{\prime}\right}$ is coexponential for $\mathfrak{h}{1} \cap \mathfrak{h}{2}$ in $\boldsymbol{h}_{2}$.

## 数学代写|表示论代写Representation theory代考|A Trace Relation

Tr 一个dH1/(H1∩H2)X+Tr⁡一个dH2/(H1∩H2)X=0对于任何X∈H1∩H2， 然后

ΔH1,G(H)=ΔH2,G(H)ΔH1∩H2,H2(H)2

(我b2bφ)(G)=∮H2/(H1∩H2)φ(GH)χF(H)ΔH2,G−1/2(H)d在H2,H1∩H2(H)

## 数学代写|表示论代写Representation theory代考|Relations Between Two Polarizations

[吨,磷]=12磷,[吨,问]=12问,[吨,从]=[磷,问]=从.

H1=R吨+R磷+R从,H2=R吨+R问+R从

O_{+}=\left{l \in \mathfrak{g}^{} ; l(Z)>0\right}, O_{-}=\left{l \in \mathfrak{g}^{} ; l(Z)<0\右}O_{+}=\left{l \in \mathfrak{g}^{} ; l(Z)>0\right}, O_{-}=\left{l \in \mathfrak{g}^{} ; l(Z)<0\右}

G_{2}=\exp g_{2}=\left{\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in M_{2}(\mathbb {R}); a>0\right}(a x+b \text { group })G_{2}=\exp g_{2}=\left{\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in M_{2}(\mathbb {R}); a>0\right}(a x+b \text { group })

K{2} K_{1}=\left{g=\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in G_{2} ; b>-1\右}K{2} K_{1}=\left{g=\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in G_{2} ; b>-1\右}

H1⊂∑j=1nbj(Fj′)

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