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. 



automorphism groups of unit disk $\mathbf{D}^n $ and unit ball $ B^n $
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 $1,\ldots,n$.


