Poincaré's theorem that the ball and polydisc are not biholomorphic

Remarks

One proof compares automorphism groups. The holomorphic automorphisms of $\mathbb{B}^n$ act transitively, so a biholomorphism $F \colon \Delta^n \to \mathbb{B}^n$ may be assumed to satisfy $F(0) = 0$. Then conjugation by $F$ maps the automorphisms of $\Delta^n$ fixing $0$ isomorphically onto the automorphisms of $\mathbb{B}^n$ fixing $0$. By Cartan's uniqueness theorem both groups consist of linear maps. For $\mathbb{B}^n$ it is the unitary group $U(n)$, which is connected and not abelian for $n \ge 2$. For $\Delta^n$ it consists of the maps $z \mapsto (e^{i\theta_1} z_{\sigma(1)}, \dots, e^{i\theta_n} z_{\sigma(n)})$ with $\sigma$ a permutation, and its identity component is abelian. So the two groups are not isomorphic as topological groups, a contradiction. For $n = 1$ both sets are the unit disc, so the hypothesis $n \ge 2$ is needed.

Going deeper

For a student

The theorem shows that there is no Riemann mapping theorem in several complex variables. It opened the study of biholomorphic invariants of domains, such as automorphism groups, invariant metrics and the geometry of the boundary. Another proof uses the boundary. The boundary of $\Delta^n$ contains nonconstant holomorphic discs, while the sphere bounding $\mathbb{B}^n$ contains none. A limiting argument applied to a would-be biholomorphism turns this difference into a contradiction. Poincaré proved the result for $n = 2$ by computing automorphism groups.