Is $\mathfrak{o}(n)$ a subalgebra of $\mathfrak{u}(n)$?

Question

A quick simple question to start the weekend (I hope).

The Lie algebra $\mathfrak{u}(n)$ is the set of $n\times n$ skew-Hermitian matrices over $\mathbb{C}$ and the Lie algebra $\mathfrak{o}(n)$ is the set of $n\times n$ skew-symmetric matrices over $\mathbb{R}$. Is $\mathfrak{o}(n)$ a subalgebra of $\mathfrak{u}(n)$? If we restrict the field in $\mathfrak{u}(n)$ to $\mathbb{R}$, doesn't this just reproduce $\mathfrak{o}(n)$?

Cheers all!

Answer 1

The Lie algebra $\mathfrak{o}(n)$ is the subalgebra of $\mathfrak{su}(n)$ consisting of skew-symmetric matrices over $\mathbb{R}$, and $\mathfrak{su}(n)$ itself can be viewed as subalgebra of $\mathfrak{u}(n)$. So $\mathfrak{o}(n)$ can be viewed as a subalgebra of $\mathfrak{u}(n)$.