I'm trying to understand equation 6.65b from Knapp's 'Lie groups,' 2ed.
Setup: Let $\mathfrak{g}_0$ be a real semisimple Lie algebra with an involution $\theta$. Let $B$ be a bilinear, symmetric, non-degenerate, invariant, $\theta$-invariant form on $\mathfrak{g}_0$. Assume that $B_\theta(X,Y):= -B(X,\theta Y)$ is positive definite on $\mathfrak{g}_0$. We can decompose $\mathfrak{g}_0=\mathfrak{k}_0\oplus\mathfrak{p}_0$ as $\pm 1$ eigenspaces for $\theta$. Let $\mathfrak{g}$ be the complexification $\mathfrak{g}_0\otimes \mathbb{C}$, and by extension of scalars, extend $B$ and $\theta$ to $\mathfrak{g}$. We extend $B_\theta$ to a Hermitian inner product on $\mathfrak{g}$ by $B_\theta(X,Y):=-B(X,\theta\bar{Y})$.
Say $\mathfrak{h}_0 \subset \mathfrak{g}_0$ is a "Cartan subalgebra". That is, say that $\mathfrak{h}:=\mathfrak{h}_0 \otimes \mathbb{C} \subset \mathfrak{g}$ is a Cartan subalgebra. Assume that $\mathfrak{h}_0$ is $\theta$-stable so that we can write $\mathfrak{h}_0 = \mathfrak{a}_0 \oplus \mathfrak{t}_0$ where the components are in $\mathfrak{p}_0$ and $\mathfrak{k}_0$ respectively. Let $\Delta:= \Delta(\mathfrak{g}, \mathfrak{h}) \subset \text{Hom}_\mathbb{C}(\mathfrak{h}, \mathbb{C})$ be the set of roots. Recall, the roots are positive on $\mathfrak{a}_0 \oplus i\mathfrak{t}_0$ and that $B$ is positive definite here. We can thus transfer to an inner product on $V:=\text{span}_\mathbb{R} \Delta$ via the isomorphism $ \mathfrak{a}_0 \oplus i\mathfrak{t}_0 \to V;\ H \mapsto B(H, \cdot)$. This we denote by $\langle \cdot, \cdot \rangle$ and the norm we denote by $|\cdot|$.
Assume we have a root $\beta \in \Delta$ which is "imaginary". That is, it vanishes on $\mathfrak{a}_0$. This shows that the $\beta$-eigenspace satisfies $\theta \mathfrak{g}_\beta = \mathfrak{g}_\beta$ so that $\mathfrak{g}_\beta$ is contained in either $\mathfrak{p}:= \mathfrak{p}_0 \otimes \mathbb{C}$ or in $\mathfrak{k}:= \mathfrak{k}_0\otimes \mathbb{C}$. Assume that $\beta$ is "non-compact", that is, the eigenspace $\mathfrak{g}_\beta$ is contained in $\mathfrak{p}$.
Take any nonzero $E_\beta \in \mathfrak{g}_\beta$. The fact that $\beta$ is imaginary implies that $\overline{E}_\beta \in \mathfrak{g}_{-\beta}$. We can also assume that $B(E_\beta,\overline{E}_\beta) = 2 |\beta|^{-2}$ since we have $$ 0 < B_\theta(E_\beta, E_\beta) = - B(E_\beta, \theta \overline{E}_\beta) = B(E_\beta, \overline{E}_\beta). $$ If we define $H_\beta$ by the property $\beta(H) = B(H, H_\beta)\ \ \forall H \in \mathfrak{h}$ and $H'_\beta:= 2|\beta|^{-2} H_\beta$, we see that $[E_\beta, \overline{E}_\beta] = H'_\beta$ so that the triple $H'_\beta, E_\beta, \overline{E}_\beta$ generates a subalgebra isomorphic to $\mathfrak{sl}_2(\mathbb{C})$.
Now define the operator $$ c_\beta := \exp ad_{\frac{\pi}{4}\left(\overline{E}_\beta - E_\beta\right)} \in \text{Aut} \left(\mathfrak{g}\right) $$ which can be seen to satisfy \begin{equation}\label{use}\tag{1} c_\beta\left(H'_\beta\right) = E_\beta + \overline{E}_\beta. \end{equation}
Question: How do I see from \eqref{use} the identity \begin{equation}\label{ques}\tag{2} \mathfrak{g}_0 \cap c_\beta(\mathfrak{h}) = \text{ker}\left(\beta|_\mathfrak{h_0}\right) \oplus \mathbb{R} \left(E_\beta + \overline{E}_\beta\right). \end{equation}
Attempt: The definition of $c_\beta$ and the fact that $[\overline{E}_\beta - E_\beta, H] = 0$ for every $H\in \ker(\beta)$ shows that $\ker\left(\beta|_\mathfrak{h_0}\right) \subset \mathfrak{g}_0 \cap c_\beta(\mathfrak{h})$. And equation \eqref{use} shows that $\mathbb{R} \left(E_\beta + \overline{E}_\beta\right) \subset \mathfrak{g}_0 \cap c_\beta(\mathfrak{h})$.
It remains to show L.H.S $\subset$ R.H.S in \eqref{ques} which I'm unable to do. I was trying to show that the (real) dimension of the L.H.S is the same as the (real) dimension of $\mathfrak{h}_0$ which would suffice.
Please send help.
