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Let $I_0(M)$ be the connected component of the identity of the isometry group $I(M)$. If $M$ is a compact globally symmetric space, then $M \cong I_0(M)/K$ where $K$ is the isotropy group of a point $p$.

Then, we have that $\mathfrak{g}=\mathfrak{k} \oplus \mathfrak{p}$, where $\mathfrak{g}$ is the lie algebra of $I_0(M)$, $\mathfrak{k}$ is the lie algebra of $K$ and $\mathfrak{p} \cong T_pM$.

There exists a way to visualize elements of $\mathfrak{g}$ as killing vector fields, but I cannot understand how this is done. I don't even understand how to transform elements of $\mathfrak{g}$ into vector fields. How is this done?

Aloizio Macedo
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    The answer to this is essentially my answer here. –  Sep 22 '15 at 01:20
  • Yes, thank you! Just one thing... What about taking the connected component of the identity? How does this affect things? – Aloizio Macedo Sep 22 '15 at 01:58
  • Well, we can't even reach the other parts of $I(M)$. A killing field induces a one-parameter subgroup $\Bbb R \to I(M)$, and because $\Bbb R$ is connected, this lives in $I_0(M)$. (The Lie algebra of $I(M)$ is the Lie algebra of $I_0(M)$ anyway - all that matters is stuff near the identity.) –  Sep 22 '15 at 02:16
  • Right. I realized I didn't get one thing, though... you say that taking the derivative of a $1$-parameter subgroup $\mathbb{R} \rightarrow I(M)$ gives a killing field. But what is "taking the derivative at $e$"? We have a map from $\mathbb{R}$ to $I(M)$, where are you taking the derivative to arrive at a map $X: M \rightarrow TM$? – Aloizio Macedo Sep 22 '15 at 02:21
  • This is actually a general thing. Whenever you have a group $G$ acting smoothly on $M$, you get a homomorphism of Lie algebras $f: \mathfrak g \to \mathfrak{X}(M)$. Exponentiate $g \in \mathfrak g$ to get a one-parameter subgroup $\varphi_t$; for all $p \in M$ this defines a curve $\Bbb R \to M$ by $t \mapsto \varphi_t(p)$. Taking the derivative of this at $t=0$ gives you $f(g)_p$. If the action is effective, this is an isomorphism onto its image. You can see the basic properties of this written down somewhere in the first chapter of Kobayashi-Nomizu. –  Sep 22 '15 at 02:27
  • So what I'm really doing here is identifying $\mathfrak g$ with its image in $\mathfrak X(M)$. –  Sep 22 '15 at 02:28

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"The answer to this is essentially my answer here" - Mike Miller, in comments.

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Aloizio Macedo
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