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## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Cartan Subalgebra

Let $\mathcal{G}0$ be a finite-dimensional reductive Lie algebra over $\mathbb{C}$ or $\mathbb{R}, \mathcal{G}_0=\mathcal{G} \oplus \mathcal{Z}$, where $\mathcal{G}=\left[\mathcal{G}_0, \mathcal{G}_0\right]$ is a semisimple ideal of $\mathcal{G}_0, \mathcal{Z}$ is the center of $\mathcal{G}_0$. The description of the structure of $\mathcal{G}_0$ is based on a special class of commutative subalgebras $\mathcal{H}_0 \subset \mathcal{G}_0$. A subalgebra $\mathcal{H}_0 \subset \mathcal{G}_0$ is called a Cartan subalgebra of $\mathcal{G}_0$ if: (1) $\mathcal{H}_0$ is a maximal Abelian subalgebra of $\mathcal{G}_0$; (2) $a d X$ are diagonal in $\mathcal{G}_0$ for all $X \in \mathcal{H}_0$. It is clear that $\mathcal{H}_0 \supset \mathcal{Z}, \mathcal{H}_0=\mathcal{H} \oplus \mathcal{Z}$, where $\mathcal{H}$ is a Cartan subalgebra of $\mathcal{G}$. A theorem of Cartan shows that each semisimple Lie algebra has a nontrivial Cartan subalgebra. We give an idea of the proof in order to introduce the so-called regular elements. For $X \in \mathcal{G}$, let $$\mathcal{G}_X^0 \equiv\left{Y \in \mathcal{G}:(\operatorname{ad} X)^s(Y)=0 \text { for some } s \in \mathbb{N}\right} .$$ $X$ is called a regular element of $\mathcal{G}$ if $$\operatorname{dim} \mathcal{G}_X^0=\min {Z \in \mathcal{G}} \operatorname{dim} \mathcal{G}_Z^0$$
then one shows that regular elements exist and that $\mathcal{G}_X^0$ is a Cartan subalgebra for everv regular element.

## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Lemmas on Root Systems

Further some results will be formulated as simple lemmas.
Lemma 1: (a) $B\left(\mathcal{G}\alpha, \mathcal{G}\beta\right)=0$ if $\alpha \neq-\beta$; (b) $B\left(\mathcal{H}, \mathcal{G}\alpha\right)=0 \forall \alpha \in \Delta$; (c) $B$ restricted to $\mathcal{H}$ is nondegenerate; (d) if $\alpha \in \Delta$, then $-\alpha \in \Delta$ and $B$ is a nondegenerate pairing of $\mathcal{G}\alpha$ with $\mathcal{G}{-\alpha}$. For further use let us denote $E\alpha \in \mathcal{G}\alpha, E{-\alpha} \in \mathcal{G}{-\alpha}$ as elements for which $B\left(E\alpha, E_{-\alpha}\right)=1$.

Lemma 2: $\Delta$ spans $\mathcal{H}^$. We can define for $\lambda \in \mathcal{H}^$, unique $H_\lambda: B\left(H_\lambda, H\right)=\lambda(H) \forall H \in \mathcal{H}$.
Then $\lambda \rightarrow H_\lambda$ is an isomorphism between $\mathcal{H}^$ and $\mathcal{H}$. We define inner products in $\mathcal{H}$ and $\mathcal{H}^$ by
\begin{aligned} \langle X, Y\rangle & \equiv B(X, Y), \quad X, Y \in \mathcal{H} \ \langle\lambda, \mu\rangle & \equiv\left\langle H_\lambda, H_\mu\right\rangle\left(=\lambda\left(H_\mu\right)=\mu\left(H_\lambda\right)\right), \lambda, \mu \in \mathcal{H}^* \end{aligned}
Note that $H_\alpha \neq 0$ for $\forall \alpha \in \Delta$.
Lemma 3: Let $X \in \mathcal{G}\alpha, Y \in \mathcal{G}{-\alpha}$, then
$$[X, Y]=B(X, Y) H_\alpha .$$
It is clear that $\left[\mathcal{G}\alpha, \mathcal{G}{-\alpha}\right]=\mathbb{C} H_\alpha$ because $\left[E_\alpha, E_{-\alpha}\right]=H_\alpha$. Then $a d H_\alpha=\left[\operatorname{ad} E_\alpha\right.$, ad $\left.E_{-\alpha}\right]$.
Lemma 4: tr $\left.a d H_\alpha\right|V=0$ for any subspace $V \subset \mathcal{G}$ invariant under both $a d E\alpha$, ad $E_{-\alpha}$ (and consequently under $a d H_\alpha$ ).

Lemma 5: Let $\alpha, \beta \in \Delta$. Then $\exists$ a rational number $q$ such that $\langle\beta, \alpha\rangle=q\langle\alpha, \alpha\rangle$; moreover $\langle\alpha, \alpha\rangle$ is a rational number greater than zero.
Lemma 6: $\operatorname{dim} \mathcal{G}\alpha=1, \alpha \in \Delta$. If $\alpha \in \Delta, k \alpha \in \Delta$, with $k \in \mathbb{Z}$, then $k=\pm 1$. Lemma 7: Let $$\mathcal{H}_R=\sum\alpha \mathbb{R} H_\alpha .$$
(1) Then $\operatorname{dim}_R \mathcal{H}_R=\operatorname{dim} \mathcal{H}\left(\equiv \operatorname{dim}_C \mathcal{H}\right)$. Moreover, $\langle$,$\rangle is a positive definite scalar$ product on $\mathcal{H}_R \times \mathcal{H}_R$. Each root is real-valued on $\mathcal{H}_R$.
(2) $\mathcal{H}=\mathcal{H}_R+i \mathcal{H}_R$
The positive number $\langle\alpha, \alpha\rangle$ will be called the length of the root $\alpha$, or root length.

## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Lemmas on Root Systems

$$\langle X, Y\rangle \equiv B(X, Y), \quad X, Y \in \mathcal{H}\langle\lambda, \mu\rangle \quad \equiv\left\langle H_\lambda, H_\mu\right\rangle\left(=\lambda\left(H_\mu\right)=\mu\left(H_\lambda\right)\right), \lambda, \mu \in \mathcal{H}^*$$

$$[X, Y]=B(X, Y) H_\alpha .$$

$$\mathcal{H}R=\sum \alpha \mathbb{R} H\alpha .$$
(1) 然后 $\operatorname{dim}_R \mathcal{H}_R=\operatorname{dim} \mathcal{H}\left(\equiv \operatorname{dim}_C \mathcal{H}\right)$. 而且， 〈〉isapositivedefinitescalar产品在 $\mathcal{H}_R \times \mathcal{H}_R$. 每个根在 $\mathcal{H}_R$.
(2) $\mathcal{H}=\mathcal{H}_R+i \mathcal{H}_R$

## 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。