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

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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。

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