Lie algebra of a matrix Lie group

The automorphisms of the Lie algebra $L(G)$ of a given linear Lie group $G$ in turn form a linear Lie group $\operatorname {Aut}(L(G))$. If $\Gamma_{t}$ is some curve in $\operatorname {Aut}(L(G))$, then $\Gamma_{t}[\mathbf{a}, \mathbf{b}]=\left[\Gamma_{t} \mathbf{a}, \Gamma_{t} \mathbf{b}\right]$. Differentiation at $t=0$ leads, with the notation $\mathcal{D}=\left(\frac{d}{d t} \Gamma_{t}\right)_{t=0}$ (we assume $\Gamma_{0}=E$), to the relation

which allows one to call $\mathcal{D}$ a derivation of the Lie algebra $L(G)$. We have already met this notion in [ВА II].

Prove that if $\mathcal{D}$ is a derivation of the Lie algebra $L(G)$, then $\exp \mathcal{D}$ is an automorphism of it.

The unitary matrices are elements of the Lie group $U(n)$. The corresponding Lie algebra $u(n)$ is the set of matrices with the condition $X^{*} = -X$. An important subgroup of $U(n)$ is the Lie group $SU(n)$ with the condition that $\operatorname {det}\left(U\right) = 1$. The unitary matrices

are not elements of the Lie group $SU(2)$ since the determinants of these unitary matrices are $-1$. The corresponding Lie algebra $su(n)$ of the Lie group $SU(n)$ are the $n \times n$ matrices given by

Let $\sigma_{1}, \sigma_{2}, \sigma_{3}$ be the Pauli spin matrices. Then any unitary matrix in $U(2)$ can be represented by

where $0 \leq \alpha < 2\pi$, $0 \leq \beta < 2\pi$, $0 \leq \gamma \leq \pi$ and $0 \leq \delta < 2\pi$. Calculate the right-hand side.

Consider the Lie group $T_{3}$ consisting of all $4 \times 4$ matrices

$SU(2)$ is the Lie group of all $2 \times 2$ unitary unimodular matrices, i.e., the group of all matrices of the form

where $a, b \in \mathbb {C}$ and $\bar{a}$ is the complex conjugate of $a$.

Consider the Lie group $O(2,\mathbb {R})$ and its Lie algebra $\mathfrak {o}(2,\mathbb {R})$. Show that $O(2,\mathbb {R})$ is not commutative. Discuss the properties of $\mathfrak {o}(2,\mathbb {R})$.