经典线性群

$$\begin{array}{rl}S{L}_n& =\{P\in G{L}_n(\mathbb{R}):detP=1\}\\ O_n& =\{P\in G{L}_n(\mathbb{R}):P^{t}P=I\}\stackrel{detP=1}{⟶}S{O}_n\\ U_n& =\{P\in G{L}_n(\mathbb{R}):P^{\ast }P=I\}\stackrel{detP=1}{⟶}S{U}_n\\ S{P}_{2n}& =\{P\in G{L}_{2n}(\mathbb{R}):P^{t}SP=S\},\text{where }S={\left(\begin{array}{cc}0& I_n\\ -I_n& 0\end{array}\right)}_{2n\times 2n}\end{array}$$

特殊酉群 $S{U}_2$

$$\begin{array}{rl}S{U}_2& ⟷{\mathbb{S}}^3=\{x_0^2+x_1^2+x_3^2+x_4^2=1\}\\ P=\left[\begin{array}{rr}{x}_0+x_{1}i& x_2+x_{3}i\\ -x_2+x_{3}i& x_0-x_{1}i\end{array}\right]& ⟷(x_0,x_1,x_2,x_3)\end{array}$$$$\begin{array}{rl}\underset{\text{4元组}}{\underbrace{I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\mathbf{i}=\left[\begin{array}{cc}i& 0\\ 0& -i\end{array}\right],\mathbf{j}=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right],\mathbf{k}=\left[\begin{array}{cc}0& i\\ -i& 0\end{array}\right]}}& ⟷\underset{{\mathbb{R}}^4\text{的基}}{\underbrace{e_0,e_1,e_2,e_3}}\end{array}$$

• The [[equator of SU_2|equator]] of ${\displaystyle S{U}_2}$.

The following conditons are equivalent:

• ${\displaystyle A}$ is the equator of ${\displaystyle S{U}_2}$.
• the eigenvalues of ${\displaystyle A}$ are $i$ , $-i$.
• ${\displaystyle A^2=-I}$.

The longitudes are conjugate subgroups of ${\displaystyle S{U}_2}$ .

Let $W$ be a two-demensional subspace of ${\displaystyle {\mathbb{R}}^4}$ that contains ${\displaystyle I}$, and let ${\displaystyle I}$ be the longitude of unit vectors in $W$.

• $L$ meets the equator ${\displaystyle \mathbb{E}}$ in two points. If ${\displaystyle A}$ is one of them, the other is ${\displaystyle -A}$. Moreover, ${\displaystyle (I,A)}$ is an orthonormal basis of ${\displaystyle W.}$
• The elements of $L$ can be written in the form ${\displaystyle P_{\theta }=(\cos \theta )I+(\sin \theta )A}$, with ${\displaystyle A}$ on ${\displaystyle \mathbb{E}}$ and ${\displaystyle 0\le \theta \le 2\pi }$.

[!EXAMPLE]

• the longitude ${\displaystyle (\cos \theta )I+(\sin \theta )\mathbf{i}}$ is the group of diagonal matrices in ${\displaystyle S{U}_2}$. We denote this longitude by $T$. Its elements have the form

$$\cos \theta \left[\begin{array}{cc}1& \\ & 1\end{array}\right]+\sin \theta \left[\begin{array}{cc}i& \\ & -i\end{array}\right]=\left[\begin{array}{cc}{e}^{i\theta }& \\ & e^{-i\theta }\end{array}\right]$$

[!EXAMPLE]

• the longitude ${\displaystyle (\cos \theta )I+(\sin \theta )\mathbf{j}}$ is the group of real matrices in ${\displaystyle S{U}_2}$, the rotation group ${\displaystyle S{O}_2}$. The matrix ${\displaystyle cI+s\mathbf{i}}$ represents rotation of the plane through the angle ${\displaystyle -\theta }$.

$$\cos \theta \left[\begin{array}{cc}1& \\ & 1\end{array}\right]+\sin \theta \left[\begin{array}{cc}& 1\\ -1& \end{array}\right]=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$$

![[Algebra, Second Edition (Michael Artin) (Z-Library).pdf#page=281&rect=73,313,380,540|Algebra, Second Edition (Michael Artin) (Z-Library), p.281]]

旋转群 $S{O}_3$

[[equator--conjugate class|equator is a conjuate class]]. Consider the rotation of ${\displaystyle S{U}_2}$, the spin ${\displaystyle (A,\alpha )}$ denotes the rotation with angle ${\displaystyle \alpha }$ about the pole ${\displaystyle A\in \mathbb{E}}$. Then ${\displaystyle (A,\alpha )}$ and ${\displaystyle (-A,-\alpha )}$ happen to be the same. So a rotation can be defined by considering ${\displaystyle (A,\alpha )}$ and ${\displaystyle (-A,-\alpha )}$ to be the same, which follows that the rotation group ${\displaystyle S{O}_3\cong S{U}_2/\{\pm I\}}$. So ${\displaystyle S{U}_2}$ is called the double covering of ${\displaystyle S{O}_3}$. From a geometric point, since ${\displaystyle S{U}_2\cong S_3}$, ${\displaystyle S{O}_3}$ can be obtained by identifying antipodal points of ${\displaystyle S{U}_2}$, i.e. ${\displaystyle S_3}$. So ${\displaystyle S{O}_3\cong {\mathbb{P}}^3}$, the (real) projective 3-space.

[!EXAMPLE] ${\displaystyle {\mathbb{P}}^1}$ is obtained by identifying the antipodal points of ${\displaystyle S_1}$, so this is obviously ${\displaystyle S_1}$ again. However, it's much harder to imagine higher dimensional projective spaces.

单参量群

$$\begin{array}{rl}A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]& ⟷e^{tA}=\left[\begin{array}{cc}1& t\\ & 1\end{array}\right]\\ A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]& ⟷e^{tA}=\left[\begin{array}{cc}\cos t& -\sin t\\ \sin t& \cos t\end{array}\right]\end{array}$$

若 ${\displaystyle A}$ 反对称，则 ${\displaystyle e^A}$ 正交，若 ${\displaystyle A}$ 反 Hermite，则 ${\displaystyle e^A}$ 酉.

For any matrix $A$, $e^{\mathrm{Tr}(A)}=det{e}^A$.

Proof: 假设 $A$ 的特征值为 ${\lambda }_1,\dots ,{\lambda }_n$，则 $e^A$ 的特征值为 $e^{{\lambda }_1},\dots ,e^{{\lambda }_n}$ (容易验证).

$$e^{\mathrm{Tr}(A)}=e^{{\lambda }_1+\cdots +{\lambda }_n}=e^{{\lambda }_1}\cdot \cdots \cdot e^{{\lambda }_n}=det{e}^A$$

李代数

Explanation

矩阵群 $G$ 在单位元处的切向量空间被称为群 $G$ 的李代数，记为 $\text{Lie}(G)$.

[!EXAMPLE] When we represent the circle group as the unit circle in ${\displaystyle \mathbb{C}}$, the Lie algebra is the space of real multiples of $i$.

Similar to the tangent space of a path in ${\displaystyle {\mathbb{R}}^n}$, the tangent space of a path in linear groups ${\displaystyle G}$ is a matrix, and the path is a matrix-value function. If ${\displaystyle \phi (t)}$ is a path in ${\displaystyle G}$, and ${\displaystyle \phi (0)=I}$, then ${\displaystyle {\phi }^{\prime }(0)}$ lies in Lie algebra.

${\displaystyle O_n}$ 的 Lie 代数包含反对称矩阵.

Let ${\displaystyle \phi }$ be a [[path in GL_n|path]] in ${\displaystyle G{L}_n}$ with ${\displaystyle \phi (0)=I}$ and ${\displaystyle {\phi }^{\prime }(0)=A}$. Then ${\displaystyle {(\frac{d}{dt}(det\phi ))}_{t=0}=\mathrm{Tr}(A)}$.

Proof: we write the matrix entries of ${\displaystyle \phi }$ as ${\displaystyle {\phi }_{ij}}$. Then

$$\begin{array}{rl}det\phi & =\sum _{p\in S_n}\text{sgn}(p)\cdot {\phi }_{1,p(1)}{\phi }_{2,p(2)}\dots {\phi }_{n,p(n)}\\ \frac{d}{dt}det\phi & =\sum _{p\in S_n}\text{sgn}(p)\sum _{i=1}^n{\phi }_{1,p(1)}\dots {\phi }_{i,p(i)}^{\prime }\dots {\phi }_{n,p(n)}\end{array}$$

Since ${\displaystyle \phi (0)=0}$, then ${\displaystyle {\phi }_{i,j}(0)={\delta }_{ij}}$. Since ${\displaystyle {\phi }^{\prime }(0)=A}$, then ${\displaystyle {\phi }_{i,j}^{\prime }(0)=a_{ij}}$. Hence

$$\begin{array}{rl}{(\frac{d}{dt}det\phi (t))}_{t=0}& =\sum _{p\in S_n}\text{sgn}(p)\sum _{i=1}^n{\phi }_{1,p(1)}(0)\dots {\phi }_{i,p(i)}^{\prime }(0)\dots {\phi }_{n,p(n)}(0)\\ & =\sum _{p\in S_n}\text{sgn}(p)\sum _{i=1}^n{\delta }_{1,p(1)}\dots a_{i,p(i)}\dots {\delta }_{n,p(n)}\\ & \stackrel{p=1}{=}\sum _{i=1}^n{a}_{ii}=\mathrm{Tr}(A)\end{array}$$

The Lie algebra of ${\displaystyle S{L}_n}$ consists of the trace-zero matrices.

That's because

$S{L}_2$ 的正规子群

The center of group ${\displaystyle S{L}_2(\mathbb{F})}$ is ${\displaystyle \{\pm I\}}$. The quotient group ${\displaystyle S{L}_2(\mathbb{F})/\{\pm I\}}$ is called the projective group, and is denoted by ${\displaystyle PS{L}_2(\mathbb{F})}$. Its elements are the cosets ${\displaystyle \{\pm P\}}$. If ${\displaystyle \mathbb{F}}$ is a field of order ${\displaystyle \ge 4}$, then the only proper normal subgroup of ${\displaystyle S{L}_2(\mathbb{F})}$ is its center ${\displaystyle Z=\{\pm I\}}$. And the projective group is a simple group. Let $q$ be a power of a prime. The order of ${\displaystyle S{L}_2({\mathbb{F}}_q)}$ is ${\displaystyle q^3-q}$. If $q$ is not a power of 2 , the order of $PS{L}_2\left({\mathbb{F}}_q\right)$ is $\frac{1}{2}(q^3-q)$. If $q$ is a power of 2 , then $PS{L}_2\left({\mathbb{F}}_q\right)\approx S{L}_2\left({\mathbb{F}}_q\right)$, and the order of $PS{L}_2\left({\mathbb{F}}_q\right)$ is $q^3-q$.

[!EXAMPLE] ${\displaystyle PS{L}_2(\mathbb{F}2)\cong S_3}$, the symmetric group. ${\displaystyle PS{L}_2(\mathbb{F}3)\cong A_4}$.