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

  • 1
    $\begingroup$ This formula is also given in Helgason's book "Differential Geometry, Lie Groups and Symmetric Spaces", AMS 2001, Exercise D.1 on page 526. $\endgroup$ Dec 7, 2015 at 20:30
  • $\begingroup$ @MikhailBorovoi: Thank you for the reference, I will check it out. $\endgroup$ Dec 8, 2015 at 14:32

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.