§11.2 The Group of Holomorphic Automorphisms on $𝔻^{𝑛}$ and $𝐵^{𝑛}$

11.2 The Group of Holomorphic Automorphisms on $𝔻^{𝑛}$ and $𝐵^{𝑛}$

A function $𝒇 : Ω \subseteq ℂ^{𝑚} \to ℂ^{𝑛}$ is called holomorphic iff each of its component functions is holomorphic. It is important to allow for vector-valued outputs, since we are interested in automorphisms on complex domains in higher dimensions.

For the aforesaid purpose, we require a generalization of the Schwarz Lemma (Lemma 3.5.1 $Lemma 3.5.1 Lemma 3.5.1$ ), which is equivalent to several results of Cartan.

In preparation, we will introduce several relevant concepts.

Definition 11.2.1 (Multi-Index Notation).

A multi-index is an $𝑛$ -tuple of nonnegative integers $𝒌 = (𝑘_{1}, \dots, 𝑘_{𝑛}) \in ℤ_{\geq 0}^{𝑛}$ . We define

| 𝒌 | = \sum_{𝑗 = 1}^{𝑛} 𝑘_{𝑗}, 𝒛^{𝒌} = \prod_{𝑗 = 1}^{𝑛} 𝑧_{𝑗}^{𝑘_{𝑗}}, 𝜕^{𝒌} = \frac{𝜕^{| 𝒌 |}}{𝜕 𝑧_{1}^{𝑘_{1}} \dots 𝜕 𝑧_{𝑛}^{𝑘_{𝑛}}} = \prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}},

where

𝒛 = (𝑧_{1}, \dots, 𝑧_{𝑛}) \in ℂ^{𝑛} .

Definition 11.2.2.

A polynomial $𝝍 : ℂ^{𝑛} \to ℂ^{𝑚}$ of several variables is said to be homogeneous of degree $𝑑$ iff

𝝍 (𝜆 𝒛) = 𝜆^{𝑑} 𝝍 (𝒛) \forall 𝜆 \in ℂ, 𝒛 \in ℂ^{𝑛},

or equivalently, iff $𝝍$ can be written as

𝝍 (𝒛) = \sum_{| 𝒌 | = 𝑑} 𝒂_{𝒌} 𝒛^{𝒌}

where

𝒌 \in ℤ_{\geq 0}^{𝑛}

is a multi-index.

Proposition 11.2.3.

Let $𝝍 : ℂ^{𝑛} \to ℂ^{𝑚}$ be a homogeneous polynomial of degree $𝑑$ (not identically zero).

For any multi-index $𝜶 = (𝛼_{1}, \dots, 𝛼_{𝑛})$ with $| 𝜶 | \leq 𝑑$ ,
$𝜕^{𝜶} 𝝍 (𝒛) = \frac{𝜕^{| 𝜶 |} 𝝍}{𝜕 𝑧_{1}^{𝛼_{1}} \dots 𝜕 𝑧_{𝑛}^{𝛼_{𝑛}}} (𝒛)$
is a homogeneous polynomial of degree $𝑑 - | 𝜶 |$ .
If $| 𝜶 | = 𝑑 \neq 0$ , then $𝜕^{𝜶} 𝝍$ is constant (and there exists a multi-index $𝜶^{'}$ with $| 𝜶^{'} | = 𝑑$ such that $𝜕^{𝜶^{'}} 𝝍$ is nonzero).
If $| 𝜶 | > 𝑑$ , then $𝜕^{𝜶} 𝝍 \equiv 0$ .

Proof.

Writing $𝝍 (𝒛) = \sum_{| 𝒌 | = 𝑑} 𝒂_{𝒌} 𝒛^{𝒌}$ with coefficients $𝒂_{𝒌} \in ℂ^{𝑚}$ , we compute

𝜕^{𝜶} 𝝍 (𝒛) = \sum_{| 𝒌 | = 𝑑} 𝒂_{𝒌} \prod_{𝑗 = 1}^{𝑛} \frac{𝑘_{𝑗}!}{(𝑘_{𝑗} - 𝛼_{𝑗})!} 𝑧_{𝑗}^{𝑘_{𝑗} - 𝛼_{𝑗}}, 𝒌 = (𝑘_{1}, \dots, 𝑘_{𝑛}),

where the terms with $𝑘_{𝑗} < 𝛼_{𝑗}$ vanish. For each remaining term, the total degree is

(𝑘_{1} - 𝛼_{1}) + \dots + (𝑘_{𝑛} - 𝛼_{𝑛}) = 𝑑 - | 𝜶 | .

Hence, $𝜕^{𝜶} 𝝍$ is a homogeneous polynomial of degree $𝑑 - | 𝜶 |$ , establishing Part 1 $Part 1 Part 1$ .

If $| 𝜶 | = 𝑑$ , every surviving monomial has degree $0$ , so $𝜕^{𝜶} 𝝍$ is constant. Moreover, since $𝝍$ has degree exactly $𝑑$ , there exists some multi-index $𝒌$ with $| 𝒌 | = 𝑑$ and $𝒂_{𝒌} \neq 𝟎$ ; choosing $𝜶^{'} = 𝒌$ yields a nonzero constant derivative. This proves Part 2 $Part 2 Part 2$ .

Finally, if $| 𝜶 | > 𝑑$ , then for every term in the expansion, at least one $𝑘_{𝑗} < 𝛼_{𝑗}$ , so all summands vanish identically. Thus $𝜕^{𝜶} 𝝍 \equiv 0$ , verifying Part 3 $Part 3 Part 3$ . ⁠ $□$

Lemma 11.2.4 (Cartan).

Let $Ω \subset ℂ^{𝑛}$ be a bounded region, and suppose that $𝒇 = (𝑓_{1}, \dots, 𝑓_{𝑛}) : Ω \to Ω$ is holomorphic. If $\exists 𝒂 \in Ω$ such that $𝒇 (𝒂) = 𝒂$ and the complex Jacobian at $𝒂$ is the identity matrix, or equivalently, if

𝑱_{𝒇} (𝒂) = (\begin{matrix} \frac{𝜕 𝑓_{1}}{𝜕 𝑧_{1}} (𝒂) & \dots & \frac{𝜕 𝑓_{1}}{𝜕 𝑧_{𝑛}} (𝒂) \\ ⋮ & ⋱ & ⋮ \\ \frac{𝜕 𝑓_{𝑛}}{𝜕 𝑧_{1}} (𝒂) & \dots & \frac{𝜕 𝑓_{𝑛}}{𝜕 𝑧_{𝑛}} (𝒂) \end{matrix}) = 𝑰 = (\begin{matrix} 1 & \dots & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & 1 \end{matrix}), (11.2 . 1)

then

𝒇 (𝒛) \equiv 𝒛

is the identity map.

Proof.

By Theorem 11.1.3 $Theorem 11.1.3 Theorem 11.1.3$ , we have the expansion

\begin{aligned} 𝒇 (𝒛) & = \sum_{| 𝒌 | = 0}^{\infty} 𝒂_{𝒌} {(𝒛 - 𝒂)}^{𝒌} = \sum_{𝑗 = 0}^{\infty} 𝝍_{𝑗} (𝒛 - 𝒂) \\ = 𝒂 + \sum_{𝑗 = 1}^{\infty} \sum_{| 𝒌 | = 𝑗} 𝒂_{𝒌} {(𝒛 - 𝒂)}^{𝒌}, (11.2 . 2) \end{aligned}

which is absolutely convergent on some polydisk centered at $𝒂$ , where $𝒂_{𝒌} = \frac{𝜕^{𝒌} 𝒇 (𝒂)}{\prod_{𝑗 = 1}^{𝑛} 𝑘_{𝑗}!}$ and $𝒌 = (𝑘_{1}, \dots, 𝑘_{𝑛})$ . The terms have been rearranged (from absolute convergence) so that the inner summation is a homogeneous polynomial $𝝍_{𝑗}$ with a zero at $𝒛 = 𝒂$ and degree $𝑗$ .

Trivially, $𝒂_{1, 0, \dots, 0} = \frac{𝜕 𝒇}{𝜕 𝑧_{1}} (𝒂) = (1, 0, \dots, 0)$ by (11.2.1) $(11.2.1) (11.2.1)$ . Similarly, $𝒂_{0, 1, 0, \dots, 0} = (0, 1, 0, \dots, 0), \dots, 𝒂_{0, \dots, 0, 1} = (0, \dots, 0, 1)$ . Hence, the linear homogeneous polynomial of (11.2.2) $(11.2.2) (11.2.2)$ equals

(𝑧_{1} - 𝑎_{1}, \dots, 𝑧_{𝑛} - 𝑎_{𝑛}) = 𝒛 - 𝒂,

and the entire expansion is thus equal to

𝒇 (𝒛) = 𝒛 + \sum_{𝑗 = 2}^{\infty} \sum_{| 𝒌 | = 𝑗} 𝒂_{𝒌} {(𝒛 - 𝒂)}^{𝒌} .

Define a sequence of holomorphic functions ${𝒇_{𝑘} (𝒛)}_{𝑘 \in ℕ}$ by

𝒇_{1} = 𝒇, 𝒇_{𝑘 + 1} = 𝒇_{𝑘} \circ 𝒇 \forall 𝑘 \in ℕ .

Assume the existence of some $𝑚 \in ℕ$ , the smallest $𝑗 \geq 2$ such that $𝝍$ is not identically zero. Because

𝒇_{1} (𝑧) = 𝒛 + 𝝍_{𝑚} (𝒛 - 𝒂) + \sum_{𝑗 > 𝑚} 𝝍_{𝑗} (𝒛 - 𝒂),

it then follows that

\begin{aligned} 𝒇_{2} (𝒛) & = 𝒛 + 𝝍_{𝑚} (𝒛 - 𝒂) + \sum_{𝑗 > 𝑚} 𝝍_{𝑗} (𝒛 - 𝒂) \\ \begin{matrix}  \end{matrix} + 𝝍_{𝑚} (𝒛 - 𝒂 + \sum_{𝑗 \geq 𝑚} 𝝍_{𝑗} (𝒛 - 𝒂)) + \sum_{𝑗 > 𝑚} 𝝍_{𝑗} (𝒇 (𝒛) - 𝒂) \\ = 𝒛 + 2 𝝍_{𝑚} (𝒛 - 𝒂) \\ \begin{matrix}  \end{matrix} + (homogeneous polynomials of degree > 𝑚) (𝒛 - 𝒂) . \end{aligned}

Assume, for induction, that

\begin{aligned} 𝒇_{𝑘} (𝒛) & = 𝒛 + 𝑘 𝝍_{𝑚} (𝒛 - 𝒂) \\ \begin{matrix}  \end{matrix} + (homogeneous polynomials of degree > 𝑚) (𝒛 - 𝒂) . \end{aligned}

Then we have

\begin{aligned} 𝒇_{𝑘 + 1} (𝒛) & = 𝒛 + \sum_{𝑗 \geq 𝑚} 𝝍_{𝑗} (𝒛 - 𝒂) \\ \begin{matrix}  \end{matrix} + 𝑘 𝝍_{𝑚} (𝒛 - 𝒂 + \sum_{𝑗 \geq 𝑚} (degree 𝑗 hom. polynomial) (𝒛 - 𝒂)) \\ \begin{matrix}  \end{matrix} + \sum_{𝑗 > 𝑚} (homogeneous polynomial of degree 𝑗) (𝒇 (𝒛) - 𝒂) \\ = 𝒛 + (𝑘 + 1) 𝝍_{𝑚} (𝒛 - 𝒂) \\ \begin{matrix}  \end{matrix} + (degree > 𝑚 homogeneous polynomials) (𝒛 - 𝒂) . \end{aligned}

Since $𝒇_{𝑘} (Ω) \subseteq Ω$ for any $𝑘$ , the sequence ${𝒇_{𝑘}}_{𝑘 \in ℕ}$ is uniformly bounded on $Ω$ . By Montel’s Theorem (Theorem 11.1.7 $Theorem 11.1.7 Theorem 11.1.7$ ), there exists a subsequence ${𝒇_{𝑘_{𝑙}}}_{𝑙 \in ℕ}$ that converges locally uniformly to some holomorphic function $\tilde{𝒇}$ by virtue of Weierstrass (Theorem 11.1.6 $Theorem 11.1.6 Theorem 11.1.6$ ).

Since $𝝍_{𝑚} ≢ 0$ , there exists $𝜶$ satisfying $‖ 𝜶 ‖ = 𝑚$ such that

𝜕^{𝜶} 𝝍_{𝑚} \equiv 𝒄 \neq 𝟎

is a nonzero constant by Proposition 11.2.3 $Proposition 11.2.3 Proposition 11.2.3$ . Consequently,

𝜕^{𝜶} (homogeneous polynomials of degree > 𝑚) (𝒛 - 𝒂)

is a homogeneous polynomial with degree $\geq 1$ and thus vanishes as $𝒛 \to 𝒂$ . Similarly, $𝒛 \mapsto 𝒛$ is homogeneous with degree $1 < 𝑚$ and thus $𝜕^{𝜶} 𝑧$ vanishes. Therefore,

𝜕^{𝜶} 𝒇_{𝑘} (𝒂) = 𝑘 𝒄,

which diverges as $𝑘 \to \infty$ . Weierstrass’ Convergence Theorem (Theorem 11.1.6 $Theorem 11.1.6 Theorem 11.1.6$ ) gives that $𝜕^{𝜶} 𝒇_{𝑘_{𝑙}} (𝒂) \to 𝜕^{𝜶} \tilde{𝒇} (𝒂)$ which must be finite by holomorphy, contradicting the divergence. Hence, the assumed value for $𝑚$ cannot exist and hence $𝝍_{𝑗} \equiv 0$ for all $𝑗 \geq 2$ . Thus, $𝒇 (𝒛) \equiv 𝒛$ on some polydisk centered at $𝒂$ . By the Identity Theorem (Theorem 11.1.4 $Theorem 11.1.4 Theorem 11.1.4$ ), $𝒇 (𝒛) \equiv 𝒛$ on $Ω$ . ⁠ $□$

Definition 11.2.5 (Reinhardt Domain).

An open connected domain $Ω \subseteq ℂ^{𝑛}$ is a Reinhardt domain centered at $𝒂 = (𝑎_{1}, \dots, 𝑎_{𝑛}) \in ℂ^{𝑛}$ iff $\forall 𝜻 = (𝜁_{1}, \dots, 𝜁_{𝑛}) \in Ω$ ,

{(𝑧_{1}, \dots, 𝑧_{𝑛}) \in ℂ^{𝑛} : | 𝑧_{𝑘} - 𝑎_{𝑘} | = | 𝜁_{𝑘} - 𝑎_{𝑘} |, 1 \leq 𝑘 \leq 𝑛}

is fully contained in

Ω

. In other words,

Ω

is invariant under all rotations about the center

𝒂

in each coordinate.

Definition 11.2.6.

A Reinhardt domain $Ω \subseteq ℂ^{𝑛}$ centered at $𝒂 = (𝑎_{1}, \dots, 𝑎_{𝑛})$ is said to be complete iff $\forall 𝜻 = (𝜁_{1}, \dots, 𝜁_{𝑛}) \in Ω$ , the polydisk

{(𝑧_{1}, \dots, 𝑧_{𝑛}) \in ℂ^{𝑛} : | 𝑧_{𝑘} - 𝑎_{𝑘} | \leq | 𝜁_{𝑘} - 𝑎_{𝑘} |, 1 \leq 𝑘 \leq 𝑛}

is contained in

Ω

Definition 11.2.7 (Circular Domain).

An open connected domain $Ω \subseteq ℂ^{𝑛}$ is a circular domain centered at $𝒂 \in ℂ^{𝑛}$ iff $\forall 𝜻 \in Ω$ ,

{𝒂 + e^{i 𝜃} (𝜻 - 𝒂) : 0 \leq 𝜃 < 2 π}

is fully contained in

Ω

Definition 11.2.8.

A circular domain $Ω \subseteq ℂ^{𝑛}$ centered at $𝒂 = (𝑎_{1}, \dots, 𝑎_{𝑛})$ is said to be complete iff $\forall 𝜻 \in Ω$ ,

{𝒂 + 𝜇 (𝜻 - 𝒂) : \forall 𝜇 \in \overline{𝔻}}

is contained in

Ω

Proposition 11.2.9.

Let $𝑈_{0} \subseteq ℂ^{𝑛_{0}}, 𝑈_{1} \subseteq ℂ^{𝑛_{1}}, 𝑈_{2} \subseteq ℂ^{𝑛_{2}}$ be open domains with $𝑛_{𝑖} \geq 1$ for each $𝑖$ , and let $𝒇 : 𝑈_{1} \to 𝑈_{2}$ and $𝒈 : 𝑈_{0} \to 𝑈_{1}$ be holomorphic maps. Define the composition $𝒉 : 𝑈_{0} \to 𝑈_{2}$ by $𝒉 (𝒛) = 𝒇 (𝒈 (𝒛))$ . Then for every $𝒛 \in 𝑈_{0}$ , the complex Jacobian matrix of $𝒉$ at $𝒛$ is

𝑱_{𝒉} (𝒛) = 𝑱_{𝒇} (𝒈 (𝒛)) \cdot 𝑱_{𝒈} (𝒛) .

Proof.

Fix $𝒛 \in 𝑈_{0}$ and let $𝒘 = 𝒈 (𝒛) \in 𝑈_{1}$ . Write

𝒉 (𝒛) = (ℎ_{1} (𝒛), \dots, ℎ_{𝑛_{2}} (𝒛)),

where each $ℎ_{𝑙} : 𝑈_{0} \to ℂ$ is holomorphic for $𝑙 = 1, \dots, 𝑛_{2}$ . Similarly, write

𝒈 (𝒛) = (𝑔_{1} (𝒛), \dots, 𝑔_{𝑛_{1}} (𝒛)), 𝒇 (𝒛) = (𝑓_{1} (𝒛), \dots, 𝑓_{𝑛_{2}} (𝒛)),

where each $𝑔_{𝑝} : 𝑈_{0} \to ℂ$ and each $𝑓_{𝑙} : 𝑈_{1} \to ℂ$ is holomorphic for $𝑝 = 1, \dots, 𝑛_{1}$ and $𝑙 = 1, \dots, 𝑛_{2}$ . Then $ℎ_{𝑙} (𝒛) = 𝑓_{𝑙} (𝒈 (𝒛))$ for each $𝑙$ . By the chain multivariable rule, the complex Jacobian of $𝒉$ at $𝒛$ is the $𝑛_{2} \times 𝑛_{0}$ matrix

\begin{aligned} 𝑱_{𝒉} & = (\begin{matrix} \frac{𝜕 ℎ_{1}}{𝜕 𝑧_{1}} & \dots & \frac{𝜕 ℎ_{1}}{𝜕 𝑧_{𝑛_{0}}} \\ ⋮ & ⋱ & ⋮ \\ \frac{𝜕 ℎ_{𝑛_{2}}}{𝜕 𝑧_{1}} & \dots & \frac{𝜕 ℎ_{𝑛_{2}}}{𝜕 𝑧_{𝑛_{0}}} \end{matrix}) \\ = (\begin{matrix} \sum_{𝑝 = 1}^{𝑛_{1}} \frac{𝜕 𝑓_{1}}{𝜕 𝑔_{𝑝}} (𝒈) \frac{𝜕 𝑔_{𝑝}}{𝜕 𝑧_{1}} & \dots & \sum_{𝑝 = 1}^{𝑛_{1}} \frac{𝜕 𝑓_{1}}{𝜕 𝑔_{𝑝}} (𝒈) \frac{𝜕 𝑔_{𝑝}}{𝜕 𝑧_{𝑛_{0}}} \\ ⋮ & ⋱ & ⋮ \\ \sum_{𝑝 = 1}^{𝑛_{1}} \frac{𝜕 𝑓_{𝑛_{2}}}{𝜕 𝑔_{𝑝}} (𝒈) \frac{𝜕 𝑔_{𝑝}}{𝜕 𝑧_{1}} & \dots & \sum_{𝑝 = 1}^{𝑛_{1}} \frac{𝜕 𝑓_{𝑛_{2}}}{𝜕 𝑔_{𝑝}} (𝒈) \frac{𝜕 𝑔_{𝑝}}{𝜕 𝑧_{𝑛_{0}}} \end{matrix}) \\ = (\begin{matrix} \frac{𝜕 𝑓_{1}}{𝜕 𝑔_{1}} (𝒈) & \dots & \frac{𝜕 𝑓_{1}}{𝜕 𝑔_{𝑛_{1}}} (𝒈) \\ ⋮ & ⋱ & ⋮ \\ \frac{𝜕 𝑓_{𝑛_{2}}}{𝜕 𝑔_{1}} (𝒈) & \dots & \frac{𝜕 𝑓_{𝑛_{2}}}{𝜕 𝑔_{𝑛_{1}}} (𝒈) \end{matrix}) (\begin{matrix} \frac{𝜕 𝑔_{1}}{𝜕 𝑧_{1}} & \dots & \frac{𝜕 𝑔_{1}}{𝜕 𝑧_{𝑛_{0}}} \\ ⋮ & ⋱ & ⋮ \\ \frac{𝜕 𝑔_{𝑛_{1}}}{𝜕 𝑧_{1}} & \dots & \frac{𝜕 𝑔_{𝑛_{1}}}{𝜕 𝑧_{𝑛_{0}}} \end{matrix}) = 𝑱_{𝒇} (𝒈) \cdot 𝑱_{𝒈} . □ \end{aligned}

Lemma 11.2.10 (Cartan).

Let

Ω \subset ℂ^{𝑛}

be a bounded complete circular domain centered at

𝟎

, and suppose that

𝒇 = (𝑓_{1}, \dots, 𝑓_{𝑛}) : Ω \to Ω

is a biholomorphism. If

𝒇 (𝟎) = 𝟎

, then

𝒇

is linear.

Proof.

Let $𝝆_{𝜃} (𝒛) = e^{i 𝜃} 𝒛$ for all $𝜃 \in ℝ$ and suppose that $𝝋 = 𝝆_{- 𝜃} \circ 𝒇^{- 1} \circ 𝝆_{𝜃} \circ 𝒇$ . By Proposition 11.2.9 $Proposition 11.2.9 Proposition 11.2.9$ , we must have that

\begin{aligned} 𝑱_{𝝋} (𝒛) & = 𝑱_{𝝆_{- 𝜃}} (𝒇^{- 1} \circ 𝝆_{𝜃} \circ 𝒇 (𝒛)) \cdot 𝑱_{𝝆_{- 𝜃} \circ 𝒇^{- 1}} (𝝆_{𝜃} \circ 𝒇 (𝒛)) \\ \begin{matrix}  \end{matrix} \cdot 𝑱_{𝝆_{- 𝜃} \circ 𝒇^{- 1} \circ 𝝆_{𝜃}} (𝒇 (𝒛)) \cdot 𝑱_{𝝆_{- 𝜃} \circ 𝒇^{- 1} \circ 𝝆_{𝜃} \circ 𝒇} (𝒛) \\ 𝑱_{𝝋} (𝟎) & = (\begin{matrix} e^{- i 𝜃} & \dots & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & e^{- i 𝜃} \end{matrix}) \cdot 𝑱_{𝒇^{- 1}} (𝟎) \cdot (\begin{matrix} e^{i 𝜃} & \dots & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & e^{i 𝜃} \end{matrix}) \cdot 𝑱_{𝒇} (𝟎) \\ = e^{- i 𝜃} e^{i 𝜃} (𝑱_{𝒇^{- 1}} \cdot 𝑱_{𝒇}) (𝟎) = 𝑰 . \end{aligned}

By Lemma 11.2.4 $Lemma 11.2.4 Lemma 11.2.4$ , $𝝋 (𝒛) \equiv 𝒛$ on $Ω$ . Hence, $𝒇 \circ 𝝆_{𝜃} = 𝝆_{𝜃} \circ 𝒇$ for all $𝜃 \in ℝ$ . Together with Theorem 11.1.3 $Theorem 11.1.3 Theorem 11.1.3$ , write

𝒇 (𝒛) = \sum_{𝒌 : | 𝒌 | = 0}^{\infty} 𝒂_{𝒌} 𝒛^{𝒌} (11.2 . 3)

on a polydisk centered at $𝟎$ . Thus,

𝒇 \circ 𝝆 (𝒛) = \sum_{𝒌 : | 𝒌 | = 0}^{\infty} 𝒂_{𝒌} {(e^{i 𝜃} 𝒛)}^{𝒌} = \sum_{𝒌 : | 𝒌 | = 0}^{\infty} 𝒂_{𝒌} e^{i 𝜃 | 𝒌 |} 𝒛^{𝒌} .

On the other hand, composing with $𝝆_{𝜃}$ with (11.2.3) $(11.2.3) (11.2.3)$ gives

𝝆_{𝜃} \circ 𝒇 (𝒛) = e^{i 𝜃} \sum_{𝒌 : | 𝒌 | = 0}^{\infty} 𝒂_{𝒌} 𝒛^{𝒌} = \sum_{𝒌 : | 𝒌 | = 0}^{\infty} 𝒂_{𝒌} e^{i 𝜃} 𝒛^{𝒌} .

Hence, by the uniqueness of power series expansions, we must either have that $𝒂_{𝒌} = 𝟎$ , $e^{i 𝜃} \equiv e^{i 𝜃 | 𝒌 |}$ , or equivalently, that $\forall 𝜃 \in ℝ$ ,

𝜃 | 𝒌 | \equiv 𝜃 mod 2 π ⟹ | 𝒌 | \equiv 1 mod 2 π (by letting 𝜃 = 1) .

This is only possible when

| 𝒌 | = 1

by irrationality, and thus

𝒂_{𝒌} = 𝟎

for all

| 𝒌 | \neq 1

. Therefore,

𝒇

must be linear. ⁠

□

Remark.

𝑛 = 1

, then

Ω = 𝐷 (0, 𝑅)

for some

𝑅 > 0

and any automorphism

𝑓

with a fixed point

0

is a rotation in the form of

𝑧 \mapsto e^{i 𝜃} 𝑧

, hence linear, the effective statement of the Schwarz Lemma (Lemma 3.5.1

Lemma 3.5.1 Lemma 3.5.1

Theorem 11.2.11 (The Holomorphic Automorphism Group on $𝔻^{𝑛}$ ).

The holomorphic automorphism group of the polydisk $𝔻^{𝑛}$ consists solely of biholomorphisms in the form of

𝒛 = (𝑧_{1}, \dots, 𝑧_{𝑛}) \mapsto 𝑷 (e^{i 𝜃_{1}} \frac{𝑧_{1} - 𝑎_{1}}{1 - \overline{𝑎_{1}} 𝑧_{1}}, \dots, e^{i 𝜃_{𝑛}} \frac{𝑧_{𝑛} - 𝑎_{𝑛}}{1 - \overline{𝑎_{𝑛}} 𝑧_{𝑛}}), (11.2 . 4)

where

𝑷

is a

𝑛 \times 𝑛

permutation matrix (for coordinate permutations),

(𝜃_{1}, \dots, 𝜃_{𝑛}) \in ℝ^{𝑛}

, and

(𝑎_{1}, \dots, 𝑎_{𝑛}) \in 𝔻^{𝑛}

. Moreover, every such map is indeed an automorphism.

Proof.

Let $𝒇 \in Aut (𝔻^{𝑛})$ be arbitrary, and set $𝜶 = (𝛼_{1}, \dots, 𝛼_{𝑛}) = 𝒇 (𝟎)$ . Define the Möbius transformation $𝝋 (𝑧_{1}, \dots, 𝑧_{𝑛}) = (\frac{𝑧_{1} - 𝛼_{1}}{1 - \overline{𝛼_{1}} 𝑧_{1}}, \dots, \frac{𝑧_{𝑛} - 𝛼_{𝑛}}{1 - \overline{𝛼_{𝑛}} 𝑧_{𝑛}}) \in Aut (𝔻^{𝑛})$ . It follows that $𝝋 \circ 𝒇 (𝟎) = 𝟎$ and $𝝋 \circ 𝒇 \in Aut (𝔻^{𝑛})$ .

By Lemma 11.2.10 $Lemma 11.2.10 Lemma 11.2.10$ , the map $𝝋 \circ 𝒇$ is linear, so $𝝋 \circ 𝒇 (𝒛) = 𝑨 𝒛$ for some invertible constant matrix $𝑨 = (\begin{matrix} 𝜁_{1, 1} & \dots & 𝜁_{1, 𝑛} \\ ⋮ & ⋱ & ⋮ \\ 𝜁_{𝑛, 1} & \dots & 𝜁_{𝑛, 𝑛} \end{matrix})$ , hence $𝑨 \in Aut (𝔻^{𝑛})$ . Thus,

| \sum_{𝑗 = 1}^{𝑛} 𝜁_{𝑘, 𝑗} 𝑧_{𝑗} | < 1 \forall 𝒛 \in 𝔻^{𝑛}, \forall 𝑘 \in {1, \dots, 𝑛},

which implies $| 𝜁_{𝑘, 𝑗} | \leq 1$ for all $𝑗, 𝑘 \in {1, \dots, 𝑛}$ (for if $| 𝜁_{𝑘, 𝑗} | > 1$ , then choosing $𝑧_{𝑗} = \frac{1}{| 𝜁_{𝑘, 𝑗} |} + 𝜀$ with $0 < 𝜀 < 1 - \frac{1}{| 𝜁_{𝑘, 𝑗} |}$ and $𝑧_{𝑙} = 0$ for $𝑙 \neq 𝑗$ yields a contradiction).

For each $𝑗 \in {1, \dots, 𝑛}$ , define the sequence ${𝒛_{𝑗, 𝑘}}_{𝑘 \in ℕ}$ for each $𝑘 \in ℕ$ by

𝒛_{𝑗, 𝑘} = (𝑧_{𝑗, 𝑘, 1}, \dots, 𝑧_{𝑗, 𝑘, 𝑛}) = ((1 - \frac{1}{𝑘}) \frac{| 𝜁_{𝑗, 1} |}{𝜁_{𝑗, 1}}, \dots, (1 - \frac{1}{𝑘}) \frac{| 𝜁_{𝑗, 𝑛} |}{𝜁_{𝑗, 𝑛}}) \in 𝔻^{𝑛},

where we informally let $\frac{| 𝜁_{𝑗, 𝑖} |}{𝜁_{𝑗, 𝑖}} = 0$ if $𝜁_{𝑗, 𝑖} = 0$ . Then, for all $𝑗 \in {1, \dots, 𝑛}$ and $𝑘 \in ℕ$ ,

𝝋 \circ 𝒇 (𝒛_{𝑗, 𝑘}) = (1 - \frac{1}{𝑘}) (\begin{matrix} \sum_{𝑖 = 1}^{𝑛} (\frac{| 𝜁_{𝑗, 𝑖} |}{𝜁_{𝑗, 𝑖}}) 𝜁_{1, 𝑖} \\ ⋮ \\ \sum_{𝑖 = 1}^{𝑛} (\frac{| 𝜁_{𝑗, 𝑖} |}{𝜁_{𝑗, 𝑖}}) 𝜁_{𝑗, 𝑖} \\ ⋮ \\ \sum_{𝑖 = 1}^{𝑛} (\frac{| 𝜁_{𝑗, 𝑖} |}{𝜁_{𝑗, 𝑖}}) 𝜁_{𝑛, 𝑖} \end{matrix}) \in 𝔻^{𝑛} .

In particular, the $𝑗$ -th component is

(1 - \frac{1}{𝑘}) \sum_{𝑖 = 1}^{𝑛} | 𝜁_{𝑗, 𝑖} | \in 𝔻 .

As $𝑘 \to \infty$ ,

\sum_{𝑖 = 1}^{𝑛} | 𝜁_{𝑗, 𝑖} | \leq 1 \forall 𝑗 \in {1, \dots, 𝑛} . (11.2 . 5)

Now consider, for each $𝑗 \in {1, \dots, 𝑛}$ , the sequence $𝒛_{𝑗, 𝑘}^{'} = (0, \dots, 0, 1 - \frac{1}{𝑘}, 0, \dots, 0)$ , where $1 - \frac{1}{𝑘}$ is in the $𝑗$ -th position. Then

𝝋 \circ 𝒇 (𝒛_{𝑗, 𝑘}^{'}) = (1 - \frac{1}{𝑘}) (𝜁_{1, 𝑗}, \dots, 𝜁_{𝑛, 𝑗}) .

As $𝑘 \to \infty$ , $𝒛_{𝑗, 𝑘}^{'} \to 𝒆_{𝑗} \in 𝜕 (𝔻^{𝑛})$ (the $𝑗$ -th unit basis vector), so the limit is

𝜻_{𝑗} = (𝜁_{1, 𝑗}, \dots, 𝜁_{𝑛, 𝑗}) \in \overline{𝔻^{𝑛}} .

Because the function $𝑨$ is injective on all of $ℂ^{𝑛}$ , if $𝜻_{𝑗} \in 𝔻^{𝑛}$ (within the interior), then $𝑨^{- 1} \in Aut (𝔻^{𝑛})$ would map $𝜻_{𝑗}$ to $𝒆_{𝑗} \in 𝜕 (𝔻^{𝑛})$ , which is an impossibility. Hence, $𝜻_{𝑗} \in 𝜕 (𝔻^{𝑛})$ , and consequently, ${max}_{𝑖 \in {1, \dots, 𝑛}} | 𝜁_{𝑖, 𝑗} | = 1$ . Combined with (11.2.5) $(11.2.5) (11.2.5)$ , this forces exactly one entry in the $𝑗$ -th column of $𝑨$ to have absolute value $1$ (of the form $e^{i 𝜃_{𝑗}}$ ), with all others zero.

Invertibility of $𝑨$ ensures each column has at least one nonzero entry, so $𝑨$ is a monomial matrix, which factors to

𝑨 = 𝑷 diag (e^{i 𝜃_{1}}, \dots, e^{i 𝜃_{𝑛}})

for some permutation matrix $𝑷$ . Therefore,

𝒇 (𝒛) = 𝝋^{- 1} \circ (𝑷 diag (e^{i 𝜃_{1}}, \dots, e^{i 𝜃_{𝑛}}) 𝒛) .

Let $𝜎 : ℕ_{\leq 𝑛} \to ℕ_{\leq 𝑛}$ be the permutation induced by $𝑷$ . The map $𝑨$ multiplies the $𝑚$ -th input coordinate $𝑧_{𝑚}$ by $e^{i 𝜃_{𝑚}}$ and permutes to place it in the $𝜎 (𝑚)$ -th output position, so the $𝜎 (𝑚)$ -th coordinate of $𝑨 𝒛$ is $e^{i 𝜃_{𝑚}} 𝑧_{𝑚}$ . Applying $𝝋^{- 1}$ component-wise then gives, for the $𝑘$ -th output coordinate,

{(𝒇 (𝒛))}_{𝑘} = 𝜑_{𝛼_{𝑘}}^{- 1} (e^{i 𝜃_{𝜎^{- 1} (𝑘)}} 𝑧_{𝜎^{- 1} (𝑘)}) = \frac{e^{i 𝜃_{𝜎^{- 1} (𝑘)}} 𝑧_{𝜎^{- 1} (𝑘)} + 𝛼_{𝑘}}{1 + \overline{𝛼_{𝑘}} e^{i 𝜃_{𝜎^{- 1} (𝑘)}} 𝑧_{𝜎^{- 1} (𝑘)}} .

Set $𝑎_{𝜎^{- 1} (𝑘)} = - 𝛼_{𝑘} e^{- i 𝜃_{𝜎^{- 1} (𝑘)}} \in 𝔻$ . Then

{(𝒇 (𝒛))}_{𝑘} = \frac{e^{i 𝜃_{𝜎^{- 1} (𝑘)}} 𝑧_{𝜎^{- 1} (𝑘)} + 𝛼_{𝑘}}{1 + \overline{𝛼_{𝑘}} e^{i 𝜃_{𝜎^{- 1} (𝑘)}} 𝑧_{𝜎^{- 1} (𝑘)}} = e^{i 𝜃_{𝜎^{- 1} (𝑘)}} \frac{𝑧_{𝜎^{- 1} (𝑘)} - 𝑎_{𝜎^{- 1} (𝑘)}}{1 - \overline{𝑎_{𝜎^{- 1} (𝑘)}} 𝑧_{𝜎^{- 1} (𝑘)}} .

Hence,

{(𝒇 (𝒛))}_{𝜎 (𝑘)} = e^{i 𝜃_{𝑘}} \frac{𝑧_{𝑘} - 𝑎_{𝑘}}{1 - \overline{𝑎_{𝑘}} 𝑧_{𝑘}} ⟺ 𝒇 (𝒛) = 𝑷 (e^{i 𝜃_{1}} \frac{𝑧_{1} - 𝑎_{1}}{1 - \overline{𝑎_{1}} 𝑧_{1}}, \dots, e^{i 𝜃_{𝑛}} \frac{𝑧_{𝑛} - 𝑎_{𝑛}}{1 - \overline{𝑎_{𝑛}} 𝑧_{𝑛}}),

as in (11.2.4) $(11.2.4) (11.2.4)$ . Finally, each automorphism of this form lies in $Aut (𝔻^{𝑛})$ trivially. ⁠ $□$

Definition 11.2.12.

The conjugate transpose or Hermitian transpose of a complex matrix

𝑼

is defined as

𝑼^{†} = {\overline{𝑼}}^{⊤}

, or the transpose of the matrix with each element replaced with its complex conjugate.

Definition 11.2.13.

A matrix

𝑼

is said to be unitary iff its inverse is its conjugate transpose, or iff

𝑼^{†} 𝑼 = 𝑼 𝑼^{†} = 𝑰

Definition 11.2.14.

A matrix

𝑴

is said to be monomial iff it has exactly one nonzero entry in each row and each column.

Theorem 11.2.15 (Spectral Theorem).

For any unitary matrix

𝑼

, there exists a unitary matrix

𝑽

such that

𝑼 = 𝑽 𝑫 𝑽^{†}

, where

𝑫

is a diagonal matrix whose diagonal entries are all of unit modulus.

Proof.

Because ${‖ 𝑼 𝒛 ‖}^{2} = 𝒛^{†} 𝑼^{†} 𝑼 𝒛 = {‖ 𝒛 ‖}^{2}$ for any $𝒛 \in ℂ^{𝑛}$ , any eigenvalue $𝜆_{1}$ (existence given by the Fundamental Theorem of Algebra in Theorem 3.3.1 $Theorem 3.3.1 Theorem 3.3.1$ on the characteristic equation) of $𝑼$ must satisfy

𝑼 𝒗_{1} = 𝜆_{1} 𝒗_{1} ⟹ ‖ 𝑼 𝒗_{1} ‖ = ‖ 𝒗_{1} ‖ = | 𝜆_{1} | ‖ 𝒗_{1} ‖ ⟹ | 𝜆_{1} | = 1,

where $‖ 𝒗_{1} ‖ = 1$ is the corresponding eigenvector in $ℂ^{𝑛}$ . Then

𝑼^{- 1} 𝑼 𝒗_{1} = 𝑼^{- 1} 𝜆_{1} 𝒗_{1} ⟹ \frac{1}{𝜆_{1}} 𝒗_{1} = 𝑼^{- 1} 𝒗_{1} ⟹ \overline{𝜆_{1}} 𝒗_{1} = 𝑼^{†} 𝒗_{1} .

Let $𝒗_{1}^{⟂} = {𝒘 : 𝒗_{1}^{†} 𝒘 = 𝟎} \subset ℂ^{𝑛}$ be an $(𝑛 - 1)$ -dimensional subspace. For any $𝒘 \in 𝒗_{1}^{⟂}$ ,

𝒗_{1}^{†} 𝑼 𝒘 = {(𝑼^{†} 𝒗_{1})}^{†} 𝒘 = {(\overline{𝜆_{1}} 𝒗_{1})}^{†} 𝒘 = 𝜆_{1} 𝒗_{1}^{†} 𝒘 = 0,

so $𝑼 𝒘 \in 𝒗_{1}^{⟂}$ . Hence $𝒗_{1}^{⟂}$ is invariant under $𝑼$ . The restriction of $𝑼$ to $𝒗_{1}^{⟂}$ , $𝑼 |_{𝒗_{1}^{⟂}}$ , yields another eigenvalue $𝜆_{2} \in 𝜕 𝔻$ with eigenvector $𝒗_{2} \in 𝒗_{1}^{⟂}$ satisfying $| 𝜆_{2} | = 1$ and $‖ 𝒗_{2} ‖ = 1$ . Similarly, we may define $𝒗_{2}^{⟂} \subset 𝒗_{1}^{⟂}$ , which is an $(𝑛 - 2)$ -dimensional subspace invariant under $𝑼$ . Repeating this process inductively, we obtain an orthonormal basis ${𝒗_{1}, \dots, 𝒗_{𝑛}}$ of eigenvectors of $𝑼$ with corresponding eigenvalues $𝜆_{1}, \dots, 𝜆_{𝑛} \in 𝜕 𝔻$ . Setting

𝑽 = (\begin{matrix} 𝒗_{1} & \dots & 𝒗_{𝑛} \end{matrix}), 𝑫 = diag (𝜆_{1}, \dots, 𝜆_{𝑛})

gives that

𝑽^{†} 𝑼 𝑽 = 𝑽^{†} (\begin{matrix} 𝑼 𝒗_{1} & \dots & 𝑼 𝒗_{𝑛} \end{matrix}) = 𝑽^{†} (\begin{matrix} 𝜆_{1} 𝒗_{1} & \dots & 𝜆_{𝑛} 𝒗_{𝑛} \end{matrix}) = 𝑽^{†} 𝑽 𝑫 .

The

𝑘

-th diagonal entry of

𝑽^{†} 𝑽

is equal to

𝒗_{𝑘}^{†} 𝒗_{𝑘} = {‖ 𝒗_{𝑘} ‖}^{2} = 1

, while the non-diagonal entries correspond to

𝒗_{𝑘}^{†} 𝒗_{𝑙}

for some

𝑘 \neq 𝑙

, which vanish by orthogonality in construction. Thus,

𝑽^{†} 𝑽 = 𝑰

(unitary) and

𝑽 𝑫 𝑽^{†} = 𝑼

. ⁠

□

A unitary transformation is a map in the form of $𝒛 \mapsto 𝑼 𝒛$ , where $𝑼$ is a unitary matrix.

Proposition 11.2.16.

For any $𝑎 \in 𝔻$ ,

\begin{aligned} 𝒘 & = (𝑤_{1}, \dots, 𝑤_{𝑛}) = 𝝋_{𝑎} (𝒛) \\ = (\frac{𝑧_{1} - 𝑎}{1 - \overline{𝑎} 𝑧_{1}}, 𝑧_{2} \frac{\sqrt{1 - {| 𝑎 |}^{2}}}{1 - \overline{𝑎} 𝑧_{1}}, 𝑧_{3} \frac{\sqrt{1 - {| 𝑎 |}^{2}}}{1 - \overline{𝑎} 𝑧_{1}}, \dots, 𝑧_{𝑛} \frac{\sqrt{1 - {| 𝑎 |}^{2}}}{1 - \overline{𝑎} 𝑧_{1}}) (11.2 . 6) \end{aligned}

lies in

Aut (𝐵^{𝑛})

, where

𝒛 = (𝑧_{1}, \dots, 𝑧_{𝑛})

. Moreover,

𝝋_{𝑎}^{- 1} = 𝝋_{- 𝑎}

Proof.

For $𝒛 = (𝑧_{1}, \dots, 𝑧_{𝑛}) \in 𝐵^{𝑛}$ , because $\sum_{𝑘 = 2}^{𝑛} {| 𝑧_{𝑘} |}^{2} < 1 - {| 𝑧_{1} |}^{2}$ ,

\begin{aligned} {‖ 𝝋_{𝑎} (𝒛) ‖}^{2} & = \frac{1}{{| 1 - \overline{𝑎} 𝑧_{1} |}^{2}} [{| 𝑧_{1} - 𝑎 |}^{2} + \sum_{𝑘 = 2}^{𝑛} (1 - {| 𝑎 |}^{2}) {| 𝑧_{𝑘} |}^{2}] \\ < \frac{1}{(1 - \overline{𝑎} 𝑧_{1}) (1 - 𝑎 \overline{𝑧_{1}})} [(𝑧_{1} - 𝑎) (\overline{𝑧_{1}} - \overline{𝑎}) + (1 - {| 𝑧_{1} |}^{2}) (1 - {| 𝑎 |}^{2})] \\ = \frac{{| 𝑧_{1} |}^{2} + {| 𝑎 |}^{2} - 2 ℜ𝔢 (\overline{𝑎} 𝑧_{1}) + 1 + {| 𝑎 𝑧_{1} |}^{2} - {| 𝑎 |}^{2} - {| 𝑧_{1} |}^{2}}{1 + {| 𝑎 𝑧_{1} |}^{2} - 2 ℜ𝔢 (\overline{𝑎} 𝑧_{1})} = 1 . \end{aligned}

Hence, $𝝋_{𝑎}$ maps $𝐵^{𝑛}$ to $𝐵^{𝑛}$ . A simple calculation shows that

𝑤_{1} = \frac{𝑧_{1} - 𝑎}{1 - \overline{𝑎} 𝑧_{1}} ⟹ 𝑧_{1} = \frac{𝑤_{1} + 𝑎}{1 + \overline{𝑎} 𝑤_{1}}, 𝑧_{𝑘} = 𝑤_{𝑘} \frac{1 - \overline{𝑎} 𝑧_{1}}{\sqrt{1 - {| 𝑎 |}^{2}}} = 𝑤_{𝑘} \frac{\sqrt{1 - {| 𝑎 |}^{2}}}{1 + \overline{𝑎} 𝑤_{1}},

and hence

𝝋_{𝑎}

is bijective, admitting the inverse

𝝋_{- 𝑎}

. Therefore,

𝝋_{𝑎} \in Aut (𝐵^{𝑛})

. ⁠

□

Proposition 11.2.17.

A function

𝒇

is a unitary transformation iff

𝒇 \in Aut (𝐵^{𝑛})

and

𝒇 (𝟎) = 𝟎

Proof.

Because $𝐵^{𝑛}$ is a bounded complete circular domain centered at $𝟎$ , from Lemma 11.2.10 $Lemma 11.2.10 Lemma 11.2.10$ we have that $𝒇 \equiv 𝑼$ for some constant invertible matrix

𝑼 = (\begin{matrix} 𝜁_{1, 1} & \dots & 𝜁_{1, 𝑛} \\ ⋮ & ⋱ & ⋮ \\ 𝜁_{𝑛, 1} & \dots & 𝜁_{𝑛, 𝑛} \end{matrix}) .

Similarly, we have $𝒇^{- 1} = 𝑼^{- 1}$ , so $‖ 𝒛 ‖ = ‖ 𝑼^{- 1} 𝑼 𝒛 ‖$ . Observe that

‖ \frac{1}{‖ 𝒛 ‖} 𝒇 (𝒛) ‖ = ‖ 𝒇 (\frac{𝒛}{‖ 𝒛 ‖}) ‖ = 1 ⟹ {‖ 𝑼 𝒛 ‖}^{2} = {‖ 𝒛 ‖}^{2} .

More explicitly, we have

𝑼 𝒛 = (\sum_{𝑘 = 1}^{𝑛} 𝜁_{1, 𝑘} 𝑧_{𝑘}, \dots, \sum_{𝑘 = 1}^{𝑛} 𝜁_{𝑛, 𝑘} 𝑧_{𝑘}) ⟹ {‖ 𝑼 𝒛 ‖}^{2} = \sum_{𝑗 = 1}^{𝑛} {| \sum_{𝑘 = 1}^{𝑛} 𝜁_{𝑗, 𝑘} 𝑧_{𝑘} |}^{2} .

Letting $𝒛 = 𝒆_{𝑖}$ ( $1 \leq 𝑖 \leq 𝑛$ ) be the $𝑖$ -th unit basis vector, we obtain

‖ 𝑼 𝒛 ‖ = 1 = {‖ (𝜁_{1, 𝑖}, \dots, 𝜁_{𝑛, 𝑖}) ‖}^{2} = \sum_{𝑘 = 1}^{𝑛} {| 𝜁_{𝑘, 𝑖} |}^{2} = \sum_{𝑘 = 1}^{𝑛} 𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑖}} . (11.2 . 7)

Letting $𝒛 = \frac{\sqrt{2}}{2} (𝒆_{𝑖} + 𝒆_{𝑗})$ ( $𝑖 \neq 𝑗$ ), we have

\begin{aligned} ‖ 𝑼 𝒛 ‖ = 1 & = \frac{1}{2} {‖ (𝜁_{1, 𝑖} + 𝜁_{1, 𝑗}, \dots, 𝜁_{𝑛, 𝑖} + 𝜁_{𝑛, 𝑗}) ‖}^{2} = \frac{1}{2} \sum_{𝑘 = 1}^{𝑛} {| 𝜁_{𝑘, 𝑖} + 𝜁_{𝑘, 𝑗} |}^{2} \\ = \frac{1}{2} \sum_{𝑘 = 1}^{𝑛} (| 𝜁_{𝑘, 𝑖}^{2} | + | 𝜁_{𝑘, 𝑗}^{2} | + 2 ℜ𝔢 (𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑗}})) = 1 + \sum_{𝑘 = 1}^{𝑛} ℜ𝔢 (𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑗}}), \end{aligned}

which implies that $\sum_{𝑘 = 1}^{𝑛} ℜ𝔢 (𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑗}}) = 0$ . Similarly, letting $𝒛 = \frac{\sqrt{2}}{2} (𝒆_{𝑖} + i 𝒆_{𝑗})$ gives

\begin{aligned} ‖ 𝑼 𝒛 ‖ = 1 & = \frac{1}{2} {‖ (𝜁_{1, 𝑖} + i 𝜁_{1, 𝑗}, \dots, 𝜁_{𝑛, 𝑖} + i 𝜁_{𝑛, 𝑗}) ‖}^{2} = \frac{1}{2} \sum_{𝑘 = 1}^{𝑛} {| 𝜁_{𝑘, 𝑖} + i 𝜁_{𝑘, 𝑗} |}^{2} \\ = \frac{1}{2} \sum_{𝑘 = 1}^{𝑛} (| 𝜁_{𝑘, 𝑖}^{2} | + | 𝜁_{𝑘, 𝑗}^{2} | + 2 ℑ𝔪 (𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑗}})) = 1 + \sum_{𝑘 = 1}^{𝑛} ℑ𝔪 (𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑗}}), \end{aligned}

which implies that $\sum_{𝑘 = 1}^{𝑛} ℑ𝔪 (𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑗}}) = 0$ . Therefore, by (11.2.7) $(11.2.7) (11.2.7)$ , for all $𝑖, 𝑗 \in {1, \dots, 𝑛}$ , observe that

{(𝑼^{†} 𝑼)}_{𝑗, 𝑖} = \sum_{𝑘 = 1}^{𝑛} 𝜁_{𝑘, 𝑖} \overline{𝜁_{𝑘, 𝑗}} = 𝛿_{𝑗, 𝑖},

where $𝛿_{𝑗, 𝑖}$ is the Kronecker delta. Hence, we have $𝑼^{†} 𝑼 = 𝑰$ , and thus $𝑼$ is unitary.

Conversely, if $𝒇 (𝒛) = 𝑼 𝒛$ for some unitary matrix $𝑼$ , then for any $𝒛 \in 𝐵^{𝑛}$ ,

{‖ 𝒇 (𝒛) ‖}^{2} = {‖ 𝑼 𝒛 ‖}^{2} = 𝒛^{†} 𝑼^{†} 𝑼 𝒛 = 𝒛^{†} 𝒛 = {‖ 𝒛 ‖}^{2},

so $𝒇$ maps $𝐵^{𝑛}$ to $𝐵^{𝑛}$ . Since $𝑼$ is invertible with unitary inverse $𝑼^{†}$ , the map $𝒇$ is bijective with inverse $𝒇^{- 1} (𝒘) = 𝑼^{†} 𝒘$ , which also maps $𝐵^{𝑛}$ to $𝐵^{𝑛}$ . Therefore, $𝒇 \in Aut (𝐵^{𝑛})$ and $𝒇 (𝟎) = 𝟎$ . ⁠ $□$

Definition 11.2.18.

A group

𝐺

(under juxtaposition) is said to be divisible iff for every

𝑔 \in 𝐺

and every positive integer

𝑛

, there exists some

ℎ \in 𝐺

such that

ℎ^{𝑛} = 𝑔

Proposition 11.2.19.

The divisibility of a group is preserved under group isomorphisms.

Proof.

Let $𝜑 : 𝐺 \to 𝐻$ be a group isomorphism between groups $𝐺$ and $𝐻$ with juxtaposition.

Assume $𝐺$ is divisible. Fix $𝑦 \in 𝐻$ and a positive integer $𝑛$ . Since $𝜑$ is bijective there is $𝑥 \in 𝐺$ with $𝜑 (𝑥) = 𝑦$ . By divisibility of $𝐺$ there exists $ℎ \in 𝐺$ with $ℎ^{𝑛} = 𝑥$ . Applying $𝜑$ and using the homomorphism property gives

{𝜑 (ℎ)}^{𝑛} = 𝜑 (ℎ^{𝑛}) = 𝜑 (𝑥) = 𝑦 .

Thus every element of $𝐻$ has an $𝑛$ -th root, so $𝐻$ is divisible.

Conversely, if $𝐻$ is divisible then the same argument applied to $𝜑^{- 1} : 𝐻 \to 𝐺$ shows $𝐺$ is divisible. Therefore divisibility is preserved under group isomorphisms. ⁠ $□$

Theorem 11.2.20 (The Holomorphic Automorphism Group on $𝐵^{𝑛}$ ).

The holomorphic automorphism group $Aut (𝐵^{𝑛})$ consists solely of biholomorphisms in the form of

𝒛 \mapsto 𝑼^{- 1} 𝝋_{𝑎} \circ 𝑽 𝒛, (11.2 . 8)

where

𝑼, 𝑽

are unitary matrices,

𝑎 \in 𝔻

, and

𝝋_{𝑎}

is defined as in (11.2.6)

(11.2.6) (11.2.6)

(and every such function lies in

Aut (𝐵^{𝑛})

Proof.

Let $𝒇 \in Aut (𝐵^{𝑛})$ be arbitrary, and set $𝜶 = 𝒇 (𝟎)$ . Then there exists a unitary matrix $𝑼$ such that $𝑼 𝜶 = (‖ 𝜶 ‖, 0, \dots, 0)$ .

Now let $𝝋_{‖ 𝜶 ‖}$ be as in Proposition 11.2.16 $Proposition 11.2.16 Proposition 11.2.16$ , mapping $(‖ 𝜶 ‖, 0, \dots, 0)$ to $𝟎$ . Then, the map $𝝋_{‖ 𝜶 ‖} \circ 𝑼 𝒇 \in Aut (𝐵^{𝑛})$ fixes $𝟎$ , so by Proposition 11.2.17 $Proposition 11.2.17 Proposition 11.2.17$ it is a unitary transformation, say $𝑽$ . Therefore,

𝝋_{‖ 𝜶 ‖} \circ 𝑼 𝒇 \equiv 𝑽 ⟹ 𝒇 (𝒛) \equiv 𝑼^{- 1} 𝝋_{‖ 𝜶 ‖}^{- 1} \circ 𝑽 𝒛 .

The converse is trivial. ⁠ $□$

§11.2 The Group of Holomorphic Automorphisms on 𝔻𝑛 and 𝐵𝑛

11.2 The Group of Holomorphic Automorphisms on 𝔻𝑛 and 𝐵𝑛

§11.2 The Group of Holomorphic Automorphisms on $𝔻^{𝑛}$ and $𝐵^{𝑛}$

11.2 The Group of Holomorphic Automorphisms on $𝔻^{𝑛}$ and $𝐵^{𝑛}$