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How does one compute the group of biholomorphisms of $\mathbf{D}^n = \{(z_1, \ldots, z_n) \in \mathbb{C}^n: \forall_i \; |z_i| \leq 1\}$, i.e., the unit polydisk, and of the unit ball $B^n = \{(z_1, \ldots, z_n) \in \mathbb{C}^n: |z_1|^2 + \ldots |z_n|^2 \leq 1\}$, and show these automorphism groups are not isomorphic? Apparently Poincaré proved that the unit disk $\mathbf{D}^n $ and the unit ball $ B^n $ are not biholomorphic using this fact. There are of course other proofs available.

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    $\begingroup$ Did you in fact mean the unit polydisk which is an $n$-fold cartesian product of unit disks $D \subset \mathbb{C}$? If so, then referring to it as the unit disk and using the notation $\Delta^n$ could be a little confusing. $\endgroup$
    – Todd Trimble
    Jan 15, 2014 at 20:20
  • $\begingroup$ yes i meant that $\endgroup$
    – Koushik
    Jan 16, 2014 at 8:08

1 Answer 1

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All this is explained in the book: MR1192135 Shabat, B. V. Introduction to complex analysis. Part II. Functions of several variables. AMS, Providence, RI, 1992. And I am sure it is in many other textbooks.

The group of automorphisms of the ball $B$ consists of fractional-linear transformations $$w_j=\frac{a_{j,0}+\sum_{i=1}^n a_{j,i}z_i}{a_{0,0}+\sum_{i=1}^na_{0,i}z_i},$$ which satisfy $$\sum_{j=1}^na_{j,i}\overline{a_{j,k}}=a_{0,i}\overline{a_{0,k}},\; i\neq k,$$ and $$\sum_{j=1}^n|a_{j,i}|^2-|a_{0,i}|^2=-\sum_{j=1}^n|a_{j,0}|^2+|a_{0,0}|^2\neq 0.$$ While the group of automorphisms of the polydisc $U^n$ consists of maps which act as one-dimensional automorphisms of the unit disc $U$ coordinatewise, namely $$z_j\mapsto e^{i\alpha_{\sigma(j)}}\frac{z_{\sigma(j)}-a_{\sigma(j)}}{1-\overline{a_{\sigma(j)}}z_{\sigma(j)}},$$ where $\sigma$ is a permutation of $1,\ldots,n$.

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    $\begingroup$ In other words, the group PU$(n,1)$ is the group of all automorphisms of the holomorphic $n$-ball. To show this, it suffices to see the ball as all negative points in the complex projectivization of a $\mathbb C$-linear space with hermitian form of signature $(n,1)$. As to the polydisc, its group of automorphisms is PU$(1,1)^n$ extended by the $n$th symmetric group. $\endgroup$ Jan 15, 2014 at 21:01

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