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

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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|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

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

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

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

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