📂Representation Theory

Special Unitary Lie Algebra

Definition¹

The set of all $n \times n$ complex matrices whose trace is $0$ and which satisfy $X^{\ast} = - X$ is the Lie algebra of the special unitary group $\operatorname{SU}(n)$.

$$ \mathfrak{su}(n) = \left\{ X \in M_{n}(\mathbb{C}) : \trace(X) = 0 \text{ and } X^{\ast} = - X \right\} $$

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

Explanation

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

$$ \begin{align*} \mathfrak{su} &= \left\{ X : e^{tX} \in \operatorname{SU} \text{ for all } t \in \mathbb{R} \right\} \\ &= \left\{ X : (e^{tX})^{\ast}(e^{tX}) = I \text{ and } \det(e^{tX})=1 \text{ for all } t \in \mathbb{R} \right\} \\ \end{align*} $$

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

Meanwhile, note that although $\mathfrak{su}(n)$ is a set of complex matrices, it is an $(n^{2}-1)$-dimensional $\mathbb{R}$-vector space. Multiplying a skew-Hermitian matrix by the complex number $i$ gives a Hermitian matrix as follows, so scalar multiplication is allowed only over the real field $\mathbb{R}$. Let $X$ be a skew-Hermitian matrix.

$$ (iX)^{\ast} = \overline{i} X^{\ast} = (-i)(-X) = iX $$

Therefore $iX \notin \mathfrak{su}(n)$.

Proof

From the cases of the unitary Lie algebra and the special linear Lie algebra, one can see that the conditions for the Lie group and for the Lie algebra correspond as below.

	Condition for the Lie group	Condition for the Lie algebra
$\operatorname{SL}$ and $\mathfrak{sl}$	$\det(X) = 1$	$\trace(X) = 0$
$\operatorname{U}$ and $\mathfrak{u}$	$X^{\ast} = X^{-1}$	$X^{\ast} = - X$

Now, since the conditions for $\operatorname{SU}$ are $X^{\ast} = X^{-1}$ and $\det(X) = 1$, we see that its Lie algebra is the set of matrices satisfying $X^{\ast} = -X$ and $\trace(X) = 0$.

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