*Definition*. A group $(G,\cdot,{}^{-1},e)$ is a *Lie group* if $G$ is also a differentiable manifold and the binary operation $\cdot: G\times G\longrightarrow G$ and the unary operation (inverse) ${}^{-1}: G\longrightarrow G$ are smooth maps.

A subgroup of a Lie group is not necessarily a Lie subgroup.

*Theorem*. [C. Chevalley] Every closed subgroup of a Lie group is a Lie subgroup.

*Examples of Lie Groups*.

- Let $M(m,n)=\{m\times n-\mbox{matrices over}\ \mathbb{R}\}\cong\mathbb{R}^{mn}$. Let $A=(a_{ij})\in M(m,n)$. Define an identification map\begin{align*}M(m,n)&\longrightarrow\mathbb{R}^{mn}\\(a_{ij})&\longmapsto(a_{11},\cdots,a_{1n};\cdots;a_{m1},\cdots,a_{mn}).\end{align*} We can naturally define topology on $M(m,n)$ by the identification map. $M(m,n)$ is covered by a single chart and the identification map is the coordinate map.
- The
*General Linear Group*${\rm GL}(n)$: Let $\mathrm{GL}(n)=\{\mbox{non-singular}\ n\times n-\mbox{matrices}\}$. Define a map\begin{align*}\mathrm{GL}(n)&\longrightarrow\mathbb{R}\\A&\longmapsto\det A.\end{align*} This map is onto and continuous since $\det A$ is a polynomial function of entries $a_{ij}$ of $A$. $\mathrm{GL}(n)=\det^{-1}(\mathbb{R}-\{0\})$is an open subset of $\mathbb{R}^{n^2}$, so that it is a submanifold of $\mathbb{R}^{n^2}$. This group is called the general linear group. The set of all $n\times n$ non-singular real (complex) matrices is denoted by $\mathrm{GL}(n;\mathbb{R})$ ($\mathrm{GL}(n;\mathbb{C})$, resp.). More generally, the set $n\times n$ non-singular matrices whose entries are the elements of a field $F$ is denoted by $\mathrm{GL}(n;F)$ or $\mathrm{GL}(V)$ where $V$ is the vector space isomorphic to $F^n$. Note that $\mathrm{GL}(V)$ is also the set of all linear isomorphisms of $V$. - The
*Orthogonal Group*$\mathrm{O}(n)$: The orthogonal group $\mathrm{O}(n)$ is defined to be the set $$\mathrm{O}(n)=\{n\times n-\mbox{orthogonal matrices}\},$$ i.e., $$A\in\mathrm{O}(n)\Longleftrightarrow A\cdot{}^tA=I,$$ where ${}^tA$ is the transpose of $A$ and $I$ is the $n\times n$ identity matrix. - The
*Special Orthogonal Group*$\mathrm{SO}(n)$: The special orthogonal group is defined to be the following subgroup of $\mathrm{O}(n)$: $$\mathrm{SO}(n)=\{A\in\mathrm{O}(n): \det A=1\}.$$ - The
*Special Linear Group*$\mathrm{SL}(n)$: The special linear group is defined to be the following subgroup of $\mathrm{GL}(n)$ $$\mathrm{SL}(n)=\{A\in\mathrm{GL}(n): \det A=1\}.$$ - The
*Unitary Group*$\mathrm{U}(n)$: The unitary group $\mathrm{U}(n)$ is the set of all $n\times n$-unitary matrices, i.e. $$\mathrm{U}(n)=\{U\in\mathrm{GL}(n;\mathbb{C}): UU^\ast=I\},$$ where $U^\ast={}^t\bar U$. Physicists often write $U^\ast$ as $U^\dagger$. $\mathrm{U}(n)$ is a Lie subgroup of $\mathrm{GL}(n;\mathbb{C})$. - The
*Special Unitary Group*$\mathrm{SU}(n)$: The special unitary group $\mathrm{SU}(n)$ is a Lie subgroup of $\mathrm{U}(n)$ and $\mathrm{SL}(2;\mathbb{C})$ $$\mathrm{SU}(n)=\{U\in\mathrm{SL}(2;\mathbb{C}):UU^\ast=I\}.$$

*Proposition*. For any $n\times n$ real or complex matrix $X$,

$$e^X:=\sum_{m=0}^\infty\frac{X^m}{m!}$$ converges and is a continuous function.

*Proof*. For the proof of the proposition click here.

*Definition*. Let $G$ be a matrix Lie group. The Lie algebra of $G$, denoted by $\mathfrak{g}$, is the set of all matrices $X$ such that $e^{tX}\in G$ for all $t\in\mathbb{R}$.

*Definition*. A function $A:\mathbb{R}\longrightarrow\mathrm{GL}(n;\mathbb{C})$ is called a *one-parameter subgroup* of $\mathrm{GL}(n;\mathbb{C})$ if

- $A$ is continuous;
- $A(0)=I$;
- $A(t+s)=A(t)A(s)$ for all $t,s\in\mathbb{R}$.

*Theorem*. If $A$ is a one-parameter subgroup of $\mathrm{GL}(n;\mathbb{C})$, then there exists uniquely an $n\times n$-complex matrix $X$ such that $A(t)=e^{tX}$ for all $t\in\mathbb{R}$.

In differential geometry, the Lie algebra $\mathfrak{g}$ is defined to be the tangent space $T_eG$ to $G$ at the identity $e$. The two definitions coincide if $G$ is $\mathrm{GL}(n;\mathbb{C})$ or its Lie subgroup. If $X\in\mathfrak{g}$ then by definiton $e^{tX}\in G$ for all $t\in\mathbb{R}$. The one-parameter subgroupÂ $\{e^{tX}:t\in\mathbb{R}\}$ of $G$ can be regarded as a differentiable curve $\gamma:\mathbb{R}\longrightarrow G$ such that $\gamma(0)=e$ where $e$ is the $n\times n$ identity matrix $I$. Thus $\dot\gamma(0)=X$ is the tangent vector to $G$ at the identity $e$, i.e. $X\in T_eG$. Conversely, $X\in T_eG$. Let $\{\phi_t:G\longrightarrow G\}_{t\in\mathbb{R}}$ be the flow generated by $X$, i.e.

$$\frac{d}{dt}\phi_t(p)=X_{\phi_t(p)}.$$ Then $\phi_t$ is smooth, $\phi_0=e$, and $\phi_t\circ \phi_s=\phi_{t+s}$. That is, $\{\phi_t:G\longrightarrow G\}_{t\in\mathbb{R}}$ is a one-parameter subgroup of $\mathrm{GL}(n;\mathbb{C})$. Hence by the above Theorem, there exists uniquely an $n\times n$-complex matrix $Y$ such that $A(t)=e^{tY}$. Since $\dot A(0)=Y$, $Y=X$ i.e. $A(t)=e^{tX}\in G\leq\mathrm{GL}(n;\mathbb C)$. Therefore $X\in\mathfrak{g}$.

*Physicists’ convention*: In the physics literature, the exponential map $\exp:\mathfrak{g}\longrightarrow G$ is usually given by $X\longmapsto e^{iX}$ instead of $X\longmapsto e^X$. The reason for that comes from quantum mechanics and it will be discussed later.

**References:**

[1] Andrew Baker, Matrix Groups, An Introduction to Lie Group Theory, Springer 2001

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer-Verlag 2004

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