### 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。