Step in the Bruhat decomposition for reductive Lie groups

Question

Err, not research but if anyone has read this part of Knapp's book recently, I'd be obliged if they could help me out. Also posted on MSE.

I'm stuck on a line in the proof of Theorem 7.40 in Knapp's 'Lie Groups: Beyond an Introduction 2ed'.

Notation: $G$ is a real 'reductive' Lie group with Lie algebra $\mathfrak{g}$. Here 'reductive' not only means that the Lie algebra $\mathfrak{g}$ is reductive in the sense that it is semisimple + center but also that $G$ has certain topological properties compatible with the algebra $\mathfrak{g}$. One of the properties especially important for my present question is that $\operatorname{Ad}(G)\otimes_{\mathbb{R}} 1 \subset \operatorname{Int}(\mathfrak{g}\otimes_{\mathbb{R}} \mathbb{C})$ (note, $G$ is not assumed to be connected). The rest are related to the Cartan decomposition and I only mention page 446 where they are listed.

We have a Cartan involution $\theta$ along with a decomposition into $\pm 1$ eigenspaces $$ \mathfrak{g} = \mathfrak{k}\oplus\mathfrak{p} $$ and we let $\mathfrak{a} \subset \mathfrak{p}$ be a maximal abelian subspace, we let $\mathfrak{m}$ be the centralizer of $\mathfrak{a}$ in $\mathfrak{k}$ and we denote the center of $\mathfrak{g}$ by $Z_\mathfrak{g}$. We have an inner product on $\mathfrak{g}$ compatible with $\theta$ and so we diagonalize $\operatorname{ad}_\mathfrak{a}$ simultaneously to get roots $\Sigma$ in the dual of $\mathfrak{a}$. We fix a notion of positivity and let $\mathfrak{n}$ denote the set of positive roots. We say an element $H \in \mathfrak{a}$ is regular if $\lambda(H) \neq 0$ for all $\lambda \in \Sigma$.

Question: On page 463 of the book, 6 lines above (7.47), we run into a situation where we have an element $g \in G$ and a regular element $H \in \mathfrak{a}$ such that \begin{equation}\label{eq1}\tag{1} Z:= \operatorname{Ad}(g)H \in (\mathfrak{m} \cap Z_{\mathfrak{g}}) \oplus \mathfrak{a} \oplus \mathfrak{n}. \end{equation} Why does it then follow what $Z \in \mathfrak{m} \oplus \mathfrak{a}$?

Attempt: The author says this follows because $\operatorname{Ad}(g)^{-1}$ fixes vectors in $Z_{\mathfrak{g}}$ (since the automorphism $\operatorname{Ad}(g)$ is inner by our definitions). I don't see what's going on here. Sure, the above fact means we can write $$ H = M + \operatorname{Ad}(g)^{-1}(A + N) $$ according to the decomposition in \eqref{eq1}. Then perhaps we can look at the adjoint action to see that $$ \operatorname{ad}_{H} = \operatorname{Ad}(g)^{-1}\circ\operatorname{ad}_{A+N}\circ\operatorname{Ad}(g), $$ but why does it imply that $N=0$? Perhaps we can choose a basis of $\mathfrak{g}$ to see that $\operatorname{ad}_{A+N}$ is upper triangular?

I even tried looking at Harish-Chandra's paper 'On a Lemma of F. Bruhat' on which this argument is based. But there doesn't seem to be any analogous step there to help my confusion. (Note, this paper only deals with semisimple groups.)

Answer 1

Er, the author kindly pointed out that this is a typo and that the conclusion (in the 8th line of the Proof of Existence in Theorem 7.40) should read $$ Z \text{ is in } \mathfrak{a}_0 \oplus \mathfrak{n}_0. $$ I guess one can see this by writing our element (in the language of the question above) $$ Z:= \operatorname{Ad}(g)H = M + A + N \text{ with } (M, A, N) \in (\mathfrak{m}\cap Z_{\mathfrak{g}}) \times \mathfrak{a} \times \mathfrak{n} $$ and then computing, in the inner product $B_\theta$ associated to the bilinear form and Cartan involution, \begin{equation} \begin{split} B_\theta(M,M) &= B_\theta(Z, M)\\ &= -B(Z,\theta M)\\ &=-B(\operatorname{Ad}(g)(H),M)\\ &=-B(H, \operatorname{Ad}(g)^{-1}M) \\ &=-B(H,M)\\ &=0. \end{split} \end{equation} The last equality holds since $H \in \mathfrak{a}$.