Recently I'm learning some basic theories about symmetric spaces from "Differential geometry, Lie groups, and symmetric space" written by Sigurdur Helgason, and I have some confusion. I Hope someone can help me, and thanks in advance.
Suppose $(M,g)$ is a Riemannian symmetric, and $G$ is the identity component of the isometry group of $(M,g)$. Then for any $p\in M$, if we use $K$ to denote the isotropic group $G_p$, then $(G,K)$ gives a Riemannian symmetric pair. Here are some of my questions:
- In the definition of Riemannian symmetric pair $(G,K)$, we request the Lie group $G$ to be connected, but why? Is there any advantage to doing this?
- Is there only one way to realize a Riemannian symmetric space as a Riemannian symmetric pair? In other words, Do there exist two Riemannian symmetric pairs corresponding to the same Riemannian symmetric space?
On the other hand, any Riemannian symmetric gives an effective orthogonal symmetric Lie algebra $(\mathfrak{g},s)$ with Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$. The effectivity is define by $\mathfrak{z}\cap\mathfrak{k}=0$, where $\mathfrak{z}$ is the center of $\mathfrak{g}$. Some other authors also define it by $\mathfrak{k}$ having no nonzero idea of $\mathfrak{g}$. Here are some of my questions:
- I cannot understand why "effective" is defined like these, and the relationship between these two definitions. I think it may be related to the action of $G$ on $G/K$ is effective, since some authors say "$\mathfrak{k}$ having no nonzero idea of $\mathfrak{g}$" if and only if $G$ acts on $G/K$ almost effectively (I'm not quite sure what is "almost effective", maybe only finite elements acts trivially).
- Since any orthogonal symmetric Lie algebra coming from Riemannian symmetric space is effective, why don't we only consider effective ones?
Given a symmetric pair $(G,K)$ with Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{m}$, we know that $K$ can act on $\mathfrak{m}$ via adjoint representation. On the other hand, $K$ can also act on $T_pM$ via isotropic representation, that is $k\mapsto (dL_k)_p$. My question is that
- This link said that if we identify $T_pM\cong\mathfrak{m}$, we will see that the adjoint representation and isotropic representation are the same, but why?