# The effect of a Cayley transform on a Cartan subalgebra

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 $$$$\label{use}\tag{1} c_\beta\left(H'_\beta\right) = E_\beta + \overline{E}_\beta.$$$$

Question: How do I see from \eqref{use} the identity $$$$\label{ques}\tag{2} \mathfrak{g}_0 \cap c_\beta(\mathfrak{h}) = \text{ker}\left(\beta|_\mathfrak{h_0}\right) \oplus \mathbb{R} \left(E_\beta + \overline{E}_\beta\right).$$$$

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.

Er, I guess one can write $$\mathfrak{h}_0 = \mathbb{R}(iH_\beta') \oplus \text{ker}(\beta|_{\mathfrak{h}_0})\ \text{ so that }\ \mathfrak{h}= \mathbb{C}(iH'_\beta) \oplus\mathbb{C}\cdot\text{ker}(\beta|_{\mathfrak{h}_0}).$$ Then $$$$\begin{split} \mathfrak{g}_0\cap c_\beta(\mathfrak{h}) &= \mathfrak{g}_0 \cap \left(\mathbb{C}\left(\overline{E}_\beta + E_\beta\right) \oplus \mathbb{C}\cdot \text{ker}(\beta|_{\mathfrak{h}_0})\right) \\ &= \mathfrak{g}_0\cap \mathbb{C}\cdot\left(\left(\overline{E}_\beta + E_\beta\right) \oplus \text{ker}(\beta|_{\mathfrak{h}_0})\right) \\ &= \mathbb{R}\left(\overline{E}_\beta + E_\beta\right) \oplus \text{ker}(\beta|_{\mathfrak{h}_0}). \end{split}$$$$ wefnawiebv;gwbeiwebg!