If you google search Globally Riemannian Symmetric Spaces, you will receive several links which provide a geometric understanding of these spaces. I mean specifically you will get stuff about geodesics, geodesic symmetry, etc. I feel as though I have a sufficient understanding of this. However, an alternate algebraic picture comes into play as well.

If the manifold is a Lie group, $\mathbb{G}$, with a subgroup, $\mathbb{K}$, the algebra can be decomposed into a direct sum of that which generates $\mathbb{K}$ and that which generates $\frac{\mathbb{G}}{\mathbb{K}}$ like this, $\mathfrak{g}=\mathfrak{k}$ $\oplus$ $\mathfrak{m}$. If $\frac{\mathbb{G}}{\mathbb{K}}$ is a Globally Riemannian Symmetric Space, elements of $\mathfrak{k}$ and $\mathfrak{m}$ satisfy these relationships:

$\begin{equation}
\hspace{100pt}[\mathfrak{k},\mathfrak{k}]\subset \mathfrak{k}\hspace{20pt}[\mathfrak{m},\mathfrak{m}]\subset \mathfrak{k}\hspace{20pt}[\mathfrak{m},\mathfrak{k}]= \mathfrak{m}
\end{equation}$

**How can I derive the latter two relationships from the geometric definition?**

Here is what I have tried:

Any geodesic through the identity element of a Lie group $\mathbb{G}$ is a 1-parameter subgroup, $g(\alpha)=e^{\alpha X}$ for $X\in \mathfrak{g}$. We know that the adjoint action of the group on itself leaves the identity element in place, i.e. $Ad\_{g}(id\_\mathbb{G})=g\hspace{2pt}id\_\mathbb{G}\hspace{2pt}g^{-1}=g\hspace{2pt}g^{-1}=id\_\mathbb{G}$. So, the adjoint action of any $g\in\mathbb{G}$ on any geodesic through the identity will just swivel the geodesic around. In other words, all 1-param subgroups are linked via the adjoint action.

From this I claim that I can recreate the entire group with the adjoint action orbit of a 1-parameter subgroup, i.e. $\mathbb{G}\subset\cup\_{h\in\mathbb{G}}Ad\_h(g(\alpha))$. Now, if $\mathbb{K}\subset\mathbb{G}$ is a Lie subgroup, then the adjoint orbit can be split into a union of $Ad\_\mathbb{K}$-orbits and the $Ad\_\mathbb{\frac{\mathbb{G}}{\mathbb{K}}}$-orbits,

$\hspace{100pt}\mathbb{G}\subset\cup\_{k\in\mathbb{K}}Ad\_k(g(\alpha))\bigcup\cup\_{m\in\mathbb{\frac{\mathbb{G}}{\mathbb{K}}}}Ad\_m(g(\alpha))$

If $g\in\mathbb{K}$, then $\mathbb{G}=\mathbb{K}\otimes\mathbb{\frac{\mathbb{G}}{\mathbb{K}}}\subset\mathbb{K}\bigcup\cup\_{m\in\mathbb{\frac{\mathbb{G}}{\mathbb{K}}}}Ad\_m(g)$. By identifying each part of the group with its algebra, we write $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{m}\subset Ad\_g(\mathfrak{k})\oplus Ad\_g(\mathfrak{m})$. Therefore, $ad\_\mathfrak{k}(\mathfrak{k})=[\mathfrak{k},\mathfrak{k}]\subset\mathfrak{k}$ and $ad\_\mathfrak{m}(\mathfrak{l})=[\mathfrak{m},\mathfrak{l}]\subset\mathfrak{m}$

Similarly, if $g\in\mathbb{\frac{\mathbb{G}}{\mathbb{K}}}$, then $\mathbb{G}\subset\cup\_{k\in\mathbb{K}}Ad\_k(g)\bigcup\cup\_{m\in\mathbb{\frac{\mathbb{G}}{\mathbb{K}}}}Ad\_m(g)$, but this is not so simple. Since $\mathbb{K}$ is a group on its own, the $\frac{\mathbb{G}}{\mathbb{K}}$-adjoint orbit of elements inside of $\mathbb{K}$ would necessarily take them out of $\mathbb{K}$, so, via the same argument above, I could identify $\cup\_{k\in\mathbb{K}}Ad\_k(g)$ with $\frac{\mathbb{G}}{\mathbb{K}}$, and then say $ad\_\mathfrak{l}(\mathfrak{m})=[\mathfrak{l},\mathfrak{m}]\subset\mathfrak{m}$. However, the not so simple part is that I can't think of a single argument that would identify $\cup\_{m\in\mathbb{\frac{\mathbb{G}}{\mathbb{K}}}}Ad\_m(g)$ with $\mathbb{K}$ to give $[\mathfrak{m},\mathfrak{m}]\subset\mathfrak{k}$.

Any help would be appreciated!