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 $\overline{\frak a^+}$.
Your root system on $\mathfrak{a}^*$ is induced by an inner product (Killing form, say) on $\mathfrak{a}$. So we might as well think of a root system as being in $\mathfrak{a}$. By Knapp $6.57$, the action of the Weyl group here (by reflections) is realised exactly by the $\operatorname{Ad}$ action of $$N_K(\mathfrak{a})/Z_K(\mathfrak{a}). $$ General root system theory says any $H\in \mathfrak{a}$ can be sent to a unique element in the closed Weyl chamber by the above action (see for example Bump Theorem $20.1$ (iii)).
What I don't know is why two elements $H_1, H_2$ in the closed Weyl Chamber can't be conjugate by some $k \in K \smallsetminus N_K(\mathfrak{a})$.