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I'm trying to understand a statement on page 49 in the book "Analysis on Symmetric Cones" by Faraut and Koranyi.

The situation is as follows. Let $\Omega$ be a symmetric cone in $V$ and denote by $G$ the connected component of the identity in $G(Ω)$ (the automorphisms of the cone $\Omega$). Now consider $K := G \cap O(V)$ where $O(V) := \{T \in \mathrm{GL(V)} \colon T^* = T^{-1}\}$. There is an element $e \in \Omega$ such that $K$ equals the stabilizer subgroup $G_e$ of $G$. For $G$ and $K$ we have the corresponding Lie algebras $\mathfrak{g}$ of $G$ and $\mathfrak{k}$ of $K$, respectively. It turns out that $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ where $\mathfrak{p} := \{T \in \mathfrak{g} \colon T^* = T\}$. Since $\Omega$ is symmetric, $G$ acts transitively on $\Omega$, that is $G \cdot e = \Omega$. Furthermore, as $\mathfrak{k} \cdot e = 0$, we have $\mathfrak{g}\cdot e = \mathfrak{p}\cdot e$.

Could someone please help me understand why this implies $\exp \mathfrak{g} \cdot e = \exp \mathfrak{p} \cdot e$? If possible using the least amount of Lie theory. I'm not an expert in the theory of Lie groups or Lie algebras.

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    $\begingroup$ Isn't it just that, if $X \in \mathfrak g$ and $Y \in \mathfrak p$ are such that $X\cdot e = Y\cdot e$, then $\exp(X)\cdot e = \exp(Y)\cdot e$, just by thinking in terms of power series in $\operatorname{End}(V)$? $\endgroup$
    – LSpice
    Commented Sep 15 at 17:22
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    $\begingroup$ @LSpice Thank you for your comment. I don't understand why $X \cdot e = Y \cdot e$ would imply that $X^n \cdot e = Y^n \cdot e$? $\endgroup$ Commented Sep 15 at 17:44
  • $\begingroup$ Re, I'm sorry, you are right. If $X \in \mathfrak g$ equals $X_+ + X_-$, with $X_+ \in \mathfrak k$ and $X_- \in \mathfrak p$, then one can move all $X_+$ terms in $X^n$ to the right by using the Lie bracket and the fact that $[\mathfrak k, \mathfrak p]$ is contained in $\mathfrak p$, but I don't see an obvious reason why the "surviving" terms should be exponentials (or, as I assume is meant, products of exponentials) of anything. $\endgroup$
    – LSpice
    Commented Sep 15 at 18:37

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