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"](http://www.ams.org/bookstore-getitem/item=MMONO-6 "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
$$
\det(C)=(-1)^n \det(\bar B).
$$



Hua Luogeng's definition of automorphism group for Hermitian symmetric space
I was able to prove that, indeed, it must necessarily be $C=-\bar B$ and $D=\bar A$.


First consider [Woodbury identity](https://en.wikipedia.org/wiki/Woodbury_matrix_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
$$
A^\text{t}D-B^\*B=\mathbb I\_n=D^\*D -B^\*B
\quad \Rightarrow\quad
A^\text{t}=D^\*.
$$


