In chapter 14, the author shows that the octonionic projective plane is the quotient of exceptional Lie group $F_{4}$ by the group $Spin(9)$ (Theorem 14.99). When he is leading with an element that fixes diagonal matrix $E_{1} = diag(1,0,0)$, the action of such automorphism $g$ is by blocks and he claims $h \in SO(9)$, but I can't see why. The print:
EDITED: Some calculations show that $h \in O(9)$ the orthogonal group since preserves the bilinear form in a subset of Jordan Algebra of hermitian matrices with order 2, which coincides with the dot product in $\mathbb{R}^{9}$. If we assume that the isotropy group in $E_{1}$ is connected (see below *) and we use the fact that there exists a copy of $Spin(9)$ in this isotropy group, we must have $h \in SO(9)$, I suppose. But I think there is a short path.
- Exceptional Lie Groups - Yokota: https://arxiv.org/abs/0902.0431 (p. 58)
Thanks in advance!
A possible solution:
One can show that $h \in O(9)$. If $h \notin SO(9)$, $-h^{-1} \in SO(9)$. Take $\tilde{g} \in$ (a copy of) $Spin(9)$ such that $\chi_{\tilde{g}} = -h^{-1}$, that is, the adjoint representation. Thus $\tilde{g} \circ g$ fixes $I_{2}$, indeed, if a element fixes $E_{1}$ then fixes the block $E_{2} + E_{3}$ where $E_{2} = diag(0,1,0)$ and $E_{3} = diag(0,0,1)$. On the other hand $\tilde{g} \circ g$ acts on $I_{2}$ by $-h^{-1} \circ h (I_{2}) = -I_{2}$, which is a contradiction.