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# 数学代考|李群李代数代考Lie groups and Lie algebras代写|COMP5328 Exceptional Lie Algebra G2

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## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Exceptional Lie Algebra G2

Let $\mathcal{G}=G_2,\left(a_{i j}\right)=\left(\begin{array}{cc}2 & -1 \ -3 & 2\end{array}\right),\left(\alpha_1, \alpha_1\right)=3\left(\alpha_2, \alpha_2\right)=-2\left(\alpha_1, \alpha_2\right)$.
We choose $\left(\alpha_2, \alpha_2\right)=2$, then $\left(\alpha_1, \alpha_1\right)=6,\left(\alpha_1, \alpha_2\right)=-3$ and by Lemma $8 \alpha_1+\alpha_2, \alpha_1+$ $2 \alpha_2, \alpha_1+3 \alpha_2$ are positive roots (but not $\alpha_1+k \alpha_2, k \geq 4$ ). Checking for negative products among these roots we find only one more possibility (from Lemma 8): $2 \alpha_1+3 \alpha_2$.
Thus,
$$\Delta^{+}=\left{\alpha_1, \alpha_2, \alpha_1+\alpha_2, \alpha_1+2 \alpha_2, \alpha_1+3 \alpha_2, 2 \alpha_1+3 \alpha_2\right}$$

or in terms of the orthonormal basis $\varepsilon_1, \varepsilon_2, \varepsilon_3$
$$\Delta^{+}=\left{\varepsilon_1-\varepsilon_2, \varepsilon_2-\varepsilon_3, \varepsilon_1-\varepsilon_3, 2 \varepsilon_1-\varepsilon_2-\varepsilon_3, \varepsilon_1+\varepsilon_3-2 \varepsilon_2, \varepsilon_1+\varepsilon_2-2 \varepsilon_3\right} .$$
Thus, $G_2$ is $14-\operatorname{dim}$ ensional $\left(14=|\Delta|+\operatorname{rank} G_2\right)$.
For the simple roots we may choose
$$\alpha_1=\varepsilon_1+\varepsilon_3-2 \varepsilon_2, \quad \alpha_2=\varepsilon_2-\varepsilon_3 .$$
With the chosen normalization the roots $\alpha_1, \alpha_1+3 \alpha_2, 2 \alpha_1+3 \alpha_2$ have length 6, while $\alpha_2, \alpha_1+\alpha_2, \alpha_1+2 \alpha_2$ have length 2. The dual roots are: $\alpha_1^{\vee}=a_1 / 3$, $\alpha_2^{\vee}=a_2, \quad\left(\alpha_1+\alpha_2\right)^{\vee}=\alpha_1+\alpha_2=3 \alpha_1^{\vee}+\alpha_2^{\vee}, \quad\left(\alpha_1+2 \alpha_2\right)^{\vee}=\alpha_1+2 \alpha_2=3 \alpha_1^{\vee}+2 \alpha_2^{\vee}, \quad\left(\alpha_1+3 \alpha_2\right)^{\vee}=$ $\left(\alpha_1+3 \alpha_2\right) / 3=\alpha_1^{\vee}+\alpha_2^{\vee},\left(2 \alpha_1+3 \alpha_2\right)^{\vee}=\left(2 \alpha_1+3 \alpha_2\right) / 3=2 \alpha_1^{\vee}+\alpha_2^{\vee}$.

The Weyl group of $G_2$ is the dihedral group of order 12 . This follows from the fact that $\left(s_1 s_2\right)^6=1$, where $s_1, s_2$ are the two simple reflections (cf. next Section, (2.256)).

## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Exceptional Lie Algebra F4

Let $\mathcal{G}=F_4$ :
$$\left(a_{i j}\right)=\left(\begin{array}{cccc} 2 & -1 & 0 & 0 \ -1 & 2 & -1 & 0 \ 0 & -2 & 2 & -1 \ 0 & 0 & -1 & 2 \end{array}\right)$$
and $\left(\alpha_1, \alpha_1\right)=\left(\alpha_2, \alpha_2\right)=2\left(\alpha_3, \alpha_3\right)=2\left(\alpha_4, \alpha_4\right)$.
Next we construct the root system.
We choose $\left(\alpha_3, \alpha_3\right)=1$, then $\alpha_1^{\vee}=2 \alpha_1 /\left(\alpha_1, \alpha_1\right)=\alpha_1, \alpha_2^{\vee}=\alpha_2, \alpha_3^{\vee}=2 \alpha_3, \alpha_4^{\vee}=2 \alpha_4$; $\left(\alpha_1, \alpha_2\right)=\left(\alpha_2, \alpha_3\right)=-1,\left(\alpha_3, \alpha_4\right)=-1 / 2$. Thus
$$\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_2+2 \alpha_3, \alpha_3+\alpha_4 \in \Delta^{+} .$$
(We note that for $\beta, y \in \Delta^{+},(\beta+y)^{\vee}=\beta^{\vee}+y^{\vee}$ iff $(\beta, \beta)=(y, y)=(\beta+y, \beta+y)$.) Further one has: $\left(\alpha_1,\left(\alpha_2+\alpha_3\right)^{\vee}\right)=-2,\left(\alpha_1, \alpha_2+2 \alpha_3\right)=-1 \Rightarrow$
$$\begin{gathered} \alpha_1+\alpha_2+\alpha_3, \alpha_1+2 \alpha_2+2 \alpha_3, \alpha_1+\alpha_2+2 \alpha_3 \in \Delta^{+} ; \ \left(\alpha_2,\left(\alpha_3+\alpha_4\right)^{\vee}\right)=-2 \Rightarrow \ \alpha_2+\alpha_3+\alpha_4, \alpha_2+2 \alpha_3+2 \alpha_4 \in \Delta^{+} ; \ \left.\left(\alpha_1+\alpha_2,\left(\alpha_3+\alpha_4\right)^{\vee}\right)=-2, \alpha_2+\alpha_3,\left(\alpha_3+\alpha_4\right)^{\vee}\right)=-1 \Rightarrow \ \alpha_1+\alpha_2+\alpha_3+\alpha_4, \alpha_1+\alpha_2+2 \alpha_3+2 \alpha_4, \alpha_2+2 \alpha_3+\alpha_4 \in \Delta^{+} ; \end{gathered}$$

$\left(\alpha_1+\alpha_2+2 \alpha_3, \alpha_4^{\vee}\right)=-2,\left(\alpha_4^{\vee}, \alpha_1+2 \alpha_2+2 \alpha_3\right)=-2 \Rightarrow$
$\alpha_1+\alpha_2+2 \alpha_3+\alpha_4, \alpha_1+\alpha_2+2 \alpha_3+2 \alpha_4$,
$\alpha_1+2 \alpha_2+2 \alpha_3+\alpha_4, \alpha_1+2 \alpha_2+2 \alpha_3+2 \alpha_4 \in \Delta^{+}$;
(F4e)
$\left(\alpha_3^{\vee}, \alpha_1+2 \alpha_2+2 \alpha_3+\alpha_4\right)=-1,\left(\alpha_3^{\vee}, \alpha_1+2 \alpha_2+2 \alpha_3+2 \alpha_4\right)=-2 \Rightarrow$
$\alpha_1+2 \alpha_2+3 \alpha_3+\alpha_4, \alpha_1+2 \alpha_2+3 \alpha_3+2 \alpha_4, \alpha_1+2 \alpha_2+4 \alpha_3+2 \alpha_4 \in \Delta^{+} ;$
$\left(\alpha_2, \alpha_1+2 \alpha_2+4 \alpha_3+2 \alpha_4\right)=-1 \Rightarrow$
$\alpha_1+3 \alpha_2+4 \alpha_3+2 \alpha_4 \in \Delta^{+}$;
$\left(\alpha_1, \alpha_1+3 \alpha_2+4 \alpha_3+2 \alpha_4\right)=-1 \Rightarrow$
$\tilde{\alpha}=2 \alpha_1+3 \alpha_2+4 \alpha_3+2 \alpha_4 \in \Delta^{+}$.

$$通过选择的归一化根 \alpha_1, \alpha_1+3 \alpha_2, 2 \alpha_1+3 \alpha_2 长度为 6 ，而 \alpha_2, \alpha_1+\alpha_2, \alpha_1+2 \alpha_2 长度为 2^2 。对偶根是: \alpha_1^{\vee}=a_1 / 3, \alpha_2^{\vee}=a_2, \quad\left(\alpha_1+\alpha_2\right)^{\vee}=\alpha_1+\alpha_2=3 \alpha_1^{\vee}+\alpha_2^{\vee}, \quad\left(\alpha_1+2 \alpha_2\right)^{\vee}=\alpha_1+2 \alpha_2=3 \alpha_1^{\vee}+2 \alpha_2^{\vee}, \quad\left(\alpha_1+3 \alpha_2\right)^{\vee}= \left(\alpha_1+3 \alpha_2\right) / 3=\alpha_1^{\vee}+\alpha_2^{\vee},\left(2 \alpha_1+3 \alpha_2\right)^{\vee}=\left(2 \alpha_1+3 \alpha_2\right) / 3=2 \alpha_1^{\vee}+\alpha_2^{\vee}. 外尔群 G_2 是 12 阶二面角群。这是因为 \left(s_1 s_2\right)^6=1 ，在哪里 s_1, s_2 是两个简单的反射cf.nextSection, (2.256). ## 数学代考|李群李代数代考LIE GROUPS AND LIE ALGEBRAS代奇|EXCEPTIONAL LIE ALGEBRA F4 让 \mathcal{G}=F_4 : 和 \left(\alpha_1, \alpha_1\right)=\left(\alpha_2, \alpha_2\right)=2\left(\alpha_3, \alpha_3\right)=2\left(\alpha_4, \alpha_4\right). 接下来我们构建根系统。 我们选择 \left(\alpha_3, \alpha_3\right)=1 ，然后 \alpha_1^{\vee}=2 \alpha_1 /\left(\alpha_1, \alpha_1\right)=\alpha_1, \alpha_2^{\vee}=\alpha_2, \alpha_3^{\vee}=2 \alpha_3, \alpha_4^{\vee}=2 \alpha_4 ;\left(\alpha_1, \alpha_2\right)=\left(\alpha_2, \alpha_3\right)=-1,\left(\alpha_3, \alpha_4\right)=-1 / 2. 因此$$
\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_2+2 \alpha_3, \alpha_3+\alpha_4 \in \Delta^{+} .

\begin{aligned}
&\alpha_1+\alpha_2+\alpha_3, \alpha_1+2 \alpha_2+2 \alpha_3, \alpha_1+\alpha_2+2 \alpha_3 \in \Delta^{+} ;\left(\alpha_2,\left(\alpha_3+\alpha_4\right)^{\vee}\right)=-2 \Rightarrow \alpha_2+\alpha_3+\alpha_4, \alpha_2+2 \alpha_3+2 \alpha_4 \in \Delta^{+} ;\left(\alpha_1+\alpha_2,\left(\alpha_3-\right.\right. \
&\left(\alpha_1+\alpha_2+2 \alpha_3, \alpha_4^{\vee}\right)=-2,\left(\alpha_4^{\vee}, \alpha_1+2 \alpha_2+2 \alpha_3\right)=-2 \Rightarrow \
&\alpha_1+\alpha_2+2 \alpha_3+\alpha_4, \alpha_1+\alpha_2+2 \alpha_3+2 \alpha_4 \
&\alpha_1+2 \alpha_2+2 \alpha_3+\alpha_4, \alpha_1+2 \alpha_2+2 \alpha_3+2 \alpha_4 \in \Delta^{+} \
&F 4 e \
&\left(\alpha_3^{\vee}, \alpha_1+2 \alpha_2+2 \alpha_3+\alpha_4\right)=-1,\left(\alpha_3^{\vee}, \alpha_1+2 \alpha_2+2 \alpha_3+2 \alpha_4\right)=-2 \Rightarrow \
&\alpha_1+2 \alpha_2+3 \alpha_3+\alpha_4, \alpha_1+2 \alpha_2+3 \alpha_3+2 \alpha_4, \alpha_1+2 \alpha_2+4 \alpha_3+2 \alpha_4 \in \Delta^{+} \
&\left(\alpha_2, \alpha_1+2 \alpha_2+4 \alpha_3+2 \alpha_4\right)=-1 \Rightarrow \
&\alpha_1+3 \alpha_2+4 \alpha_3+2 \alpha_4 \in \Delta^{+} \
&\left(\alpha_1, \alpha_1+3 \alpha_2+4 \alpha_3+2 \alpha_4\right)=-1 \Rightarrow \
&\tilde{\alpha}=2 \alpha_1+3 \alpha_2+4 \alpha_3+2 \alpha_4 \in \Delta^{+}
\end{aligned}


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