Hua Luogeng's definition of automorphism group for Hermitian symmetric space I'm trying to make sense of a definition appearing in Hua Luogeng's book "Harmonic Analysis of Functions of Several Complex Variables in the Classical Domains".
Consider the Hermitian symmetric space of noncompact type
$$
D=\{z\in M_n(\mathbb C) \, | \, 1-z^*z>0 \mbox{ and } z^\text{t}=-z\}.
$$
Hua says the automorphism group of this domain is composed of matrices of the form
$$
\begin{pmatrix}
A & B \\ -\bar B & \bar A
\end{pmatrix}\in M_{2n}(\mathbb C)
$$
with
$$
A^\text{t} B=-B^\text{t}A,\quad A^*A-B^*B=\mathbb I_n,
$$
acting on $z\in D$ as
$$
(Az+B)(-\bar B z+\bar A)^{-1}.
$$
However, we know from Cartan's classification that
$$\mathrm{Aut}(D)=\mathrm{SO}^*(2n) = \{ M \in \mathrm{SU}(n,n) \,|\, M^\text{t}\gamma M = \gamma\},
\quad
\gamma=
\begin{pmatrix}
0  & \mathbb I_n \\
\mathbb I_n & 0
\end{pmatrix},
$$
i.e. consists of the matrices of the form
$$
g=
\begin{pmatrix}
A & B\\C & D
\end{pmatrix}\in M_{2n}(\mathbb C)
$$
acting on $z\in D$ as
$$
gz=(Az+B)(Cz+D)^{-1}
$$
where $\det(g)=1$ and
$$
\begin{align}
A^*A - C^*C &=\mathbb I_n \\
D^*D - B^*B &=\mathbb I_n \\
A^\text{t}D + C^\text{t} B &=\mathbb I_n
\end{align}
$$
$$
\begin{align}
A^*B &= C^*D \\
A^\text{t} C &= - C^\text{t}A \\
B^\text{t}D &= - D^\text{t}B.
\end{align}
$$
Why does Hua's definition of the automorphism group define $C$ and $D$ repsectively as $-\bar B$ and $\bar A$?
The book is very concise and doesn't explain what is going on, nor does it have any reference for this definition. Am I missing something? Note that although I am using $\mathrm{SO}^*(2n)$ and antisymmetric $z$, the same thing happens with respectively $Sp(n,\mathbb{R})$ and symmetric $z$, with some sign changes.
Update #1
So far what I was able to find is that the constraints on the matrices $A, B,C,D$ guarantee that both $A$ and $D$ are invertible, and that
$$
\det(g) = 1
\quad\Rightarrow
\quad \det(D)=\det(\bar A).
$$
This of course doesn't prove that $D=\bar A$, but at least is consistent with it. Similarly I was able to find that
$$
\det(C)=(-1)^n \det(\bar B).
$$
 A: I was able to prove that, indeed, it must necessarily be $C=-\bar B$ and $D=\bar A$.
First consider Woodbury identity
$$
(A-BD^{-1}C)^{-1}=A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1};
$$
using the constraints on $A,B,C,D$, it reduces to
$$
D^\text{t}=A^{-1}(\mathbb I_n - BC^\text{t})\quad \Rightarrow \quad AD^\text{t}+BC^\text{t}=\mathbb I_n.
$$
Now, we have
$$
CD^\text{t}A=C(\mathbb I_n-B^\text{t}C)=(\mathbb I_n-CB^\text{t})C=DA^\text{t}C
$$
so that
$$
D^{-1}C D^\text{t}=A^\text{t} C A^{-1}=-C^\text{t}AA^{-1}=-C^\text{t}.
$$
Finally, we have that
$$
\begin{split}
C^*&=A^* B D^{-1}\\
&= (\bar D^{-1}-C^*\bar B \bar D^{-1})BD^{-1}\\
&= (\bar D^{-1}+C^*(D^*)^{-1}B^*)BD^{-1}\\
&=\bar D^{-1}BD^{-1}+C^*(D^*)^{-1}B^*B D^{-1}\\
&=\bar D^{-1}BD^{-1}+C^*(D^*)^{-1}(D^*D -\mathbb I_n)D^{-1}\\
&=\bar D^{-1}BD^{-1}+C^*-C^*(D^*)^{-1} D^{-1}
\end{split}
$$
so that
$$
\bar D^{-1}BD^{-1}=C^*(D^*)^{-1} D^{-1} 
\quad \Rightarrow\quad
C=D B^* (D^\text{t})^{-1};
$$
using the previous result we get
$$
C^\text{t}=-D^{-1}C D^\text{t}=-B^*.
$$
The analogous result for $D$ follows easily now, as
$$
A^\text{t}D-B^*B=\mathbb I_n=D^*D -B^*B
\quad \Rightarrow\quad
A^\text{t}=D^*.
$$
