why can we assume that f is linear when proving open balls and open polydiscs are not biholomorphically equivalent as n>1?

Question

I'm reading Kaup's Holomorphic functions of several variables.

I have some trouble in understanding Proposition 3.11 which proves that open balls and open polydiscs are not biholomorphically equivalent as $n>1$. It says "Moveover,by 3.9 and the Chain rule,we may assume that $f$ is linear." Why?Can you help me explain the reason why we can assume that $f$ is linear? Thanks in advance!

Answer 1

Before lemma 3.9, Kaup and Kaup write

For the nonequivalence of balls and polydisks, we show first that each holomorphic mapping between them induces a linear mapping between them.

And that is indeed what lemma 3.9 shows, for every holomorphic $f\colon \mathbf{P}^n\to\mathbf{B}^m$ with $f(0) = 0$, the differential $Df\lvert_0$ maps $\mathbf{P}^n$ into $\mathbf{B}^m$, and conversely, for every holomorphic $g\colon \mathbf{B}^m\to \mathbf{P}^n$ with $g(0) = 0$, the differential $Dg\lvert_0$ maps $\mathbf{B}^m$ into $\mathbf{P}^n$.

So if we have a biholomorphism $F \colon \mathbf{P}\to \mathbf{B}$ with $F(0) = 0$, the derivative $DF\lvert_0$ is a linear isomorphism of $\mathbb{C}^n$ with $DF\lvert_0(\mathbf{P}) \subset \mathbf{B}$, and since the inverse $F^{-1}$ maps $\mathbf{B}$ into $\mathbf{P}$, by 3.9 so does its derivative, hence $\bigl(D(F^{-1})\lvert_0\bigr)(\mathbf{B}) = \bigl(DF\lvert_0\bigr)^{-1}(\mathbf{B}) \subset \mathbf{P}$. So $DF\lvert_0$ actually maps $\mathbf{P}$ bijectively to $\mathbf{B}$.

Thus every biholomorphism between the ball and polydisk that fixes $0$ induces a linear biholomorphism between the two. Since the automorphism group of $\mathbf{P}$ acts transitively, every biholomorphism between the ball and the polydisk induces one that fixes $0$.

Hence, if there exists any biholomorphism between the ball and the polydisk, then there exists a linear biholomorphism between them. The proof of 3.11 now says, given any biholomorphism $F$ between ball and polydisk, "uh, that may be complicated, let me look at the induced linear biholomorphism instead, that's easier to analyse".

Since a linear biholomorphism exists only in dimension one, where ball and polydisk are the same, the ball and the polydisk are not biholomorphically equivalent in dimensions $\geqslant 2$.

Thus in hindsight, every biholomorphism between a ball and a polydisk that fixes $0$ is indeed linear (an automorphism of the unit disk $\mathbb{D}$ that fixes $0$ is of the form $z\mapsto e^{i\varphi}z$ for some $\varphi\in [0,2\pi)$), but a priori, that was not known, only that if there exists any biholomorphism, then there exists a linear biholomorphism.