k-subalgebra of Type AI involution

[mpham26uchicago]

We have the following table

Remark ($\mathfrak{k}$-subalgebra of $\mathfrak{su}(n) $):

Cartan Type	$\mathfrak{k}$-Subalgebra	Properties	$K$-Subgroup	Properties	Dimension
AI	$\mathfrak{so}(n)$	Real Skew-Symmetric	$SO(n)$	Special Orthogonal	$\frac{n(n-1)}{2}$

Observe that $\mathfrak{k}$-subalgebra of Type AI Cartan involution satisfies

$$u \in \mathfrak{k}{\mathbf{AI}} \iff \theta{\mathbf{AI}}(u) = -u^T = u \iff u^T = -u \iff u\text{ is skew-symmetric} \iff u \in \mathfrak{so}(n)$$

How do we show that $u$ is real as well?

[mpham26uchicago]

Only need to prove this for su(n) anyway. Maybe show that this works for operator basis and then we can generalize?

[mpham26uchicago]

Proof. We show that $\mathfrak{k} = \mathfrak{so}(n)$.

$(\subseteq)$ For $u \in \mathfrak{k}$, we have that $\theta(u) = -u^T = u$. Then, $u^T = -u$, so we have that $\mathfrak{k}$ is skew-symmetric (and traceless). If $u$ is not real, then $\exp(u) \notin SO(n)$. So $u$ is real. Thus $u \in \mathfrak{so}(n)$.

$(\supseteq)$ For $u \in \mathfrak{so}(n)$, we have that $u$ is real and $u^T = -u$. Then, $\theta(u) = u$. Thus $u \in \mathfrak{k}$.

[mpham26uchicago]

Show that if $u$ is not real, then $\exp(u) \notin SO(n)$.