📂Representation Theory

General Linear Lie Algebra

Definition¹

The set $M_{n}(\mathbb{C})$ of all $n \times n$ complex matrices is the Lie algebra of the general linear group $\operatorname{GL}(n, \mathbb{C})$.

$$ \mathfrak{gl}(n, \mathbb{C}) = M_{n}(\mathbb{C}) $$

$$ \mathfrak{gl}(n, \mathbb{R}) = M_{n}(\mathbb{R}) $$

Generalization

Let $V$ be a vector space. If the set of all linear transformations from $V$ to $V$ is equipped with the bracket given by the commutator, it becomes a Lie algebra, which is denoted by $\mathfrak{gl}(V)$.

$$ \mathfrak{gl}(V) = \operatorname{End} = \left\{ X : V \to V \mid X \text{ is linear} \right\} $$ $$ [X, Y] = XY - YX, \quad X, Y \in \mathfrak{gl}(V) $$

Explanation

The above is both a definition and a theorem. By the definition of the Lie algebra of a matrix Lie group, the Lie algebra of $\operatorname{GL}$ is defined as follows.

$$ \begin{align*} \mathfrak{gl} &= \left\{ X : e^{tX} \in \operatorname{GL} \text{ for all } t \in \mathbb{R} \right\} \\ &= \left\{ X : e^{tX} \text{ is invertible for all } t \in \mathbb{R} \right\} \\ \end{align*} $$

The point is that the set of all matrices is in fact equal to the set above.

Proof

• $M_{n}(\mathbb{R}) \subset \mathfrak{gl}(n, \mathbb{R})$

  Let $X \in M_{n}(\mathbb{R})$. By the properties of the matrix exponential, $e^{tX}$ is an invertible matrix. Therefore $X \in \mathfrak{gl}(n, \mathbb{R})$.

• $\mathfrak{gl}(n, \mathbb{R}) \subset M_{n}(\mathbb{R})$

  Let $X \in \frak{gl}(n, \mathbb{R})$. Then $e^{tX} \in \operatorname{GL}(n, \mathbb{R})$, that is, $e^{tX}$ is a real matrix. Since $X = \left. \dfrac{\d e^{tX}}{\d t} \right|_{t=0}$, $X$ is also a real matrix.

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