TITLE: Why do Ad(K) orbits in the $-1$ eigenspace of a Cartan decomposition intersect the Weyl chamber once?
QUESTION [5 upvotes]: Let $G$ be a semisimple Lie group and let $\frak p\oplus k$ be a Cartan decomposition of $\frak g$ and $K$ the connected subgroup with Lie algebra $\frak k$.  Choose a maximal abelian subalgebra $\frak a \subset p$, an ordering of $\frak a^\ast$, and let $\Sigma^+$ be the positive roots of $\frak a$ with respect to $\frak g$.  Denote by $\frak a^+$ the positive Weyl chamber, $$\mathfrak{a}^+=\{ H \in \mathfrak{a} : \lambda (H)>0 \hspace{3mm} \lambda \in \Sigma^+ \}$$
I read that every $Ad(K)$ orbit in $\frak p$ intersects $\frak \overline{a^+}$ exactly once.  Why is this?  
Edit: I'm aware that $$\bigcup_{k\in K} Ad(k)\frak a= p$$ so all I need to know is why any $H\in \frak a$ has a unique conjugate in $\overline{\frak a^+}$.

REPLY [3 votes]: For uniqueness, the main technical result is given in Eberlein as 2.20.18. There is actually a typo- the word "maximal" is missing from the phrase "maximal abelian subspaces" in the last sentence of the statement. 
This proof replaces $\mathfrak{g}=\mathfrak{k}+\mathfrak{p}$ with $\mathfrak{g}^\ast=\mathfrak{k}+i\mathfrak{p}$, a compact real form of the complexification of $\mathfrak{g}$. The proof takes advantage of the fact that a compact connected Lie group has surjective exponential map. 
It's not a complete surprise that uniqueness should be delicate. The full isometry group may contain isometries acting nontrivially on a given Weyl chamber.