### 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。