📂Representation Theory

Unitary Lie Algebra

Definition¹

The set of all $n \times n$ complex matrices satisfying $X^{\ast} = - X$ is the Lie algebra of the unitary group $\operatorname{U}(n)$.

$$ \mathfrak{u}(n) = \left\{ X \in M_{n}(\mathbb{C}) : X^{\ast} = - X \right\} $$

Here $X^{\ast}$ denotes the conjugate transpose of $X$. The property $X^{\ast} = - X$ is called skew-Hermitian.

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 $\operatorname{U}$ is defined as follows.

$$ \begin{align*} \mathfrak{u} &= \left\{ X : e^{tX} \in \operatorname{U} \text{ for all } t \in \mathbb{R} \right\} \\ &= \left\{ X : (e^{tX})^{\ast}(e^{tX}) = I \text{ for all } t \in \mathbb{R} \right\} \\ \end{align*} $$

What the definition says is that if one collects all skew-Hermitian matrices whose trace is $0$, this becomes $\mathfrak{su}$.

Proof

• If $e^{tX}$ is unitary, then $X$ is skew-Hermitian.

  First, suppose that $e^{tX}$ is a unitary matrix. Then, by the properties of the matrix exponential and the definition of a unitary matrix, the following holds.

  $$ (e^{tX})^{\ast} = (e^{tX})^{-1} = e^{-tX} $$

  Also, likewise by the properties of the matrix exponential, the following holds.

  $$ (e^{tX})^{\ast} = e^{tX^{\ast}} $$

  Therefore $-X = X^{\ast}$ holds.

• If $X$ is skew-Hermitian, then $e^{tX}$ is unitary.

  Conversely, assume that $-X = X^{\ast}$. Then the following holds.

  $$ \begin{align*} && e^{tX^{\ast}} &= e^{-tX} \\ \implies && (e^{tX})^{\ast} &= (e^{tX})^{-1} \\ \implies && (e^{tX})^{\ast} (e^{tX}) &= I \\ \implies && e^{tX} &\text{ is unitary} \end{align*} $$

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