📂Representation Theory

Lie Algebra of the Lie Group of Matrices

Definition¹

$G$ shall denote a matrix Lie group. The set of all elements satisfying the following is called the Lie algebra of $G$, and is denoted by $\mathfrak{g}$.

$$ \mathfrak{g} = \left\{X : e^{tX} \in G \text{ for all } t \in \mathbb{R} \right\} $$

Here $e^{tX}$ is the matrix exponential.

Remarks

A matrix Lie algebra, as in the definition, is conventionally denoted by the corresponding matrix Lie group in lowercase Fraktur.

$$ \operatorname{GL}(n, \mathbb{R}) \leftrightarrow \mathfrak{gl}(n, \mathbb{R}) $$

$$ \operatorname{SL}(n, \mathbb{R}) \leftrightarrow \mathfrak{sl}(n, \mathbb{R}) $$

On a Lie algebra there exists an antisymmetric, linear binary operation $[\cdot, \cdot]$; for the Lie algebra of a matrix Lie group this is defined in terms of matrix multiplication as $[X, Y] = XY - YX$. This is called the bracket or commutator.

Properties

Let $G$ be a matrix Lie group and $\mathfrak{g}$ its Lie algebra. For $X, Y \in \mathfrak{g}$ the following hold.

(a) $\forall A \in G$,$\quad AXA^{-1} \in \mathfrak{g}$

(b) $\forall s \in \mathbb{R}$, $\quad sX \in \mathfrak{g}$

(c) $X + Y \in \mathfrak{g}$

(d) $[X, Y] = XY - YX \in \mathfrak{g}$

Proof

(a)

By the definition of $\mathfrak{g}$, to show that $AXA^{-1} \in \mathfrak{g}$ it suffices to verify that $e^{t(AXA^{-1})} \in G$. By properties of the matrix exponential (../3758), if $A \in G$ then $e^{t(AXA^{-1})} = e^{A(tX)A^{-1}} = Ae^{tX}A^{-1}$. Since $A, e^{tX}, A^{-1} \in G$ and $G$ is closed under matrix multiplication, we have $Ae^{tX}A^{-1} \in G$. Therefore $AXA^{-1} \in \mathfrak{g}$.

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(b)

$e^{t(sX)} = e^{(ts)X}$, and for every $t \in \mathbb{R}$ we have $e^{tX} \in G$, hence for every $s$ it follows that $sX \in \mathfrak{g}$.

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(c)

By the Baker–Campbell–Hausdorff / Lie bracket formula the following holds.

$$ e^{t(X+Y)} = \lim\limits_{m \to \infty} (e^{tX/m} e^{tY/m})^{m} $$

Here $X, Y \in G$ and since $G$ is closed under matrix multiplication, for every $m$ we have $(e^{tX/m}e^{tY/m})^{m}\in G$. Also, by the definition, sequences in $G$ converge to elements of $G$. Hence $e^{t(X+Y)} \in G$, and therefore $X + Y \in \mathfrak{g}$.

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(d)

First, by differentiation of matrix-valued functions and properties of the matrix exponential we have

$$ \begin{align*} \left.\dfrac{d}{dt} (e^{tX}Ye^{-tX}) \right|_{t=0} &= Xe^{0X}Ye^{0X} + e^{0X}Y(-X)e^{0X} \\ &= XY - YX \end{align*} $$

On the other hand, by (a), for every $t$ we have $e^{tX}Ye^{-tX} \in \mathfrak{g}$. Further, by (b) and (c) and since it is closed under addition and scalar multiplication, $\mathfrak{g}$ is a real subspace of $M_{n}(\mathbb{C})$. Equipped with the standard norm on $M_{n}(\mathbb{C})$, $\mathfrak{g}$ is a finite-dimensional normed space and therefore complete. Thus since it is a complete subspace, $\mathfrak{g} \subset M_{n}(\mathbb{C})$ is closed.

As seen above, $e^{hX}Ye^{-hX} \in \mathfrak{g}$ and since $\mathfrak{g}$ is a subspace, for each $h \ne 0$ the difference is $\dfrac{e^{hX}Ye^{-hX} - Y}{h} \in \mathfrak{g}$. Therefore, because $\mathfrak{g}$ is closed, the limit of these differences — the value below — also belongs to $\mathfrak{g}$.

$$ XY - YX = \lim_{h \to 0} \dfrac{e^{hX}Ye^{-hX} - Y}{h} $$

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