mpham26uchicago / scaling-train

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k-subalgebra of Type AI involution #6

Open mpham26uchicago opened 1 year ago

mpham26uchicago commented 1 year ago

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 commented 1 year ago

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

mpham26uchicago commented 1 year ago

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 commented 1 year ago

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