Explanation of the relationship between geometric and algebraic qualities of Globally Riemannian Symmetric Spaces.

Question

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!

Answer 1

I think there are a few assumptions here that don't hold water.

Firstly, there isn't in general an $\mathfrak{m}$. Certainly not a specific choice. That is unique to reductive homogeneous spaces. It's fine in this case since symmetric spaces are reductive. However, you should be careful in saying how it "generates" $G/K$

Secondly, the adjoint action is not transitive and the one parameter subgroups do not cover the group (even under the assumption that $G$ is connected). Note that $\exp$ is not a surjective map in general so not every element is in a 1-parameter subgroup. What you get by taking the unions of the 1-parameter subgroups is just the image of $\exp$. I'm not totally sure that the adjoint action on a single 1-parameter subgroup generates even this much, since there are many different adjoint orbits in the Lie algebra. For example a nilpotent element and a semisimple element live in different orbits

Thirdly, you can't split the $G$-orbits into $K$ and $G/K$ orbits. The second isn't a group and most likely will not have a well-defined action on $G$. Even if we assume $K$ is a normal subgroup so $G/K$ is a group and $G= K \times G/K$, I still don't believe that every orbit would be either a $K$-orbit or a $G/K$ orbit

Answer 2

The geometric definition of a symmetric space $X$ is that each point $p$ admits a geodesic symmetry $S_p \colon X \to X$. A good definition of geodesic symmetry is that $S_p$ fixes $p$ and acts by $-\mathrm{Id}$ on the tangent space at $p$.

Now there is an induced involution on $G = \mathrm{Isom}_0(X)$, the connected component of the isometry group containing the identity, given by $\sigma_p(g) = S_p \circ g \circ S_p$. The derivative of $\sigma_p$ at the identity gives a Lie algebra involution $\theta_p \colon \mathfrak{g} \to \mathfrak{g}$. It has a $+1$-eigenspace, denoted $\mathfrak{k}$, and a $-1$-eigenspace, denoted $\mathfrak{p}$ (or called $\mathfrak{m}$ in the OP). The relations $$ [\mathfrak{k},\mathfrak{k}]\subset \mathfrak{k}, \quad [\mathfrak{k},\mathfrak{p}]\subset \mathfrak{p}, \quad [\mathfrak{p},\mathfrak{p}]\subset \mathfrak{k} $$ follow immediately by applying this definition. For example, to see the third relation, consider any $X,Y \in \mathfrak{p}$. Then $\theta[X,Y] = [\theta X,\theta Y]=[-X,-Y]=[X,Y]$.

There is an alternative way to prove the second relation, and this proof works in a wider setting than symmetric spaces. It turns out that the action of $G$ on $X$ is transitive and that $\mathfrak{k}$ is the Lie algebra of the subgroup $K$ stabilizing the point $p$. Since $G$ preserves a Riemannian metric, $K$ is necessarily compact. The adjoint action of $K$ on $\mathfrak{g}$ preserves some inner product, and also preserves $\mathfrak{k}$, so it admits a complement $\mathfrak{m}$ (it turns out that this complement must equal $\mathfrak{p}$ when $X$ is a symmetric space). By definition, $K$ preserves $\mathfrak{m}$, and this implies that $[\mathfrak{k},\mathfrak{m}] \subset \mathfrak{m}$.

The point I want to make is that the three Lie algebra relations encode the structure in a sort of increasingly special way. The relation $[\mathfrak{k},\mathfrak{k}]\subset \mathfrak{k}$ just says that $\mathfrak{k}$ is a subalgebra; but this is just the Lie algebra of our point stabilizer, so this makes sense. The second relation $[\mathfrak{k},\mathfrak{p}]\subset \mathfrak{p}$ is the infinitesimal version of finding a $K$-invariant complement to $\mathfrak{k}$ in $\mathfrak{g}$: This is the setting of reductive homogeneous spaces, as Callum mentions. Note that the differential of the orbit map identifies $\mathfrak{p}$ with $T_pX$ as representations of $K$.

The final relation $[\mathfrak{p},\mathfrak{p}]\subset \mathfrak{k}$ is actually very special, and already places you in the world of symmetric spaces. If you have a Lie algebra $\mathfrak{g}$ and a decomposition satisfying all three of these relations, then you can define an involution $\theta$ by declaring it to be the identity on $\mathfrak{k}$ and $-\mathrm{Id}$ on $\mathfrak{p}$, and these relations can be used to check that such a $\theta$ respects the bracket. This gives $\mathfrak{g}$ the structure of a symmetric Lie algebra. To get a Riemannian symmetric space, we also require $\mathfrak{k}$ to be a compact Lie algebra. For more details, see the books by Helgason or Eberlein.