📂Representation Theory

Special Orthogonal Lie Algebra

Definition¹

The set of all $n \times n$ real matrices satisfying $X^{\mathsf{T}} = - X$ is the Lie algebra of the orthogonal group $\operatorname{O}(n)$ and of the special orthogonal group $\operatorname{SO}(n)$.

$$ \mathfrak{so}(n) = \mathfrak{o}(n) = \left\{ X \in M_{n}(\mathbb{R}) : X^{\mathsf{T}} = - X \right\} $$

Here, $X^{\mathsf{T}}$ denotes the transpose of $X$. The property $X^{\mathsf{T}} = - X$ is called being anti-symmetric.

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 algebras of $\operatorname{O}$ and $\operatorname{SO}$ are defined respectively as follows.

$$ \begin{align*} \mathfrak{o} &= \left\{ X : e^{tX} \in \operatorname{O} \text{ for all } t \in \mathbb{R} \right\} \\ \mathfrak{so} &= \left\{ X : e^{tX} \in \operatorname{SO} \text{ for all } t \in \mathbb{R} \right\} \end{align*} $$

What the definition says is that the set of all anti-symmetric matrices is both $\mathfrak{o}$ and $\mathfrak{so}$, so the two coincide. As a notation, $\mathfrak{so}(n)$ is more commonly used.

$$ \mathfrak{so}(n) = \mathfrak{o}(n) $$

Proof

• If $e^{tX}$ is an orthogonal matrix, then $X$ is anti-symmetric.

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

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

  Likewise, by the properties of the matrix exponential, the following also holds.

  $$ (e^{tX})^{\mathsf{T}} = e^{tX^{\mathsf{T}}} $$

  Therefore, $X^{\mathsf{T}} = -X$ holds.

• If $X$ is anti-symmetric, then $e^{tX}$ is an orthogonal matrix.

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

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

So far we have shown that the set of anti-symmetric matrices is $\frak{o}(n)$. Now, by definition, all diagonal entries of an anti-symmetric matrix are $0$. That is, its trace is $0$, which means that the condition on the corresponding Lie group includes "matrices with determinant $1$", so the set of anti-symmetric matrices is also the Lie algebra of $\operatorname{SO}(n)$.

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