Hua Luogeng's definition of automorphism group for Hermitian symmetric space

Question

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 $$ \det(C)=(-1)^n \det(\bar B). $$

Answer 1

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 $$ A^\text{t}D-B^*B=\mathbb I_n=D^*D -B^*B \quad \Rightarrow\quad A^\text{t}=D^*. $$