§11.3 Topological Equivalence and Biholomorphic Equivalence

11.3 Topological Equivalence and Biholomorphic Equivalence

Lemma 11.3.1.

𝑷

is a permutation matrix, and

𝑫

is a diagonal matrix, then there exists a diagonal matrix

𝑫^{'}

such that

𝑷 𝑫 = 𝑫^{'} 𝑷

. Similarly, there exists a diagonal matrix

𝑫^{″}

such that

𝑫 𝑷 = 𝑷 𝑫^{″}

Proof.

Let $𝑫 = diag (𝑑_{1}, \dots, 𝑑_{𝑛})$ and let $𝜎$ be the permutation corresponding to $𝑷$ , $𝑷 𝒆_{𝑖} = 𝒆_{𝜎 (𝑖)}$ for each standard basis vector $𝒆_{𝑖}$ . Define

𝑫^{'} = diag (𝑑_{𝜎^{- 1} (1)}, \dots, 𝑑_{𝜎^{- 1} (𝑛)}) .

Then for every $𝑖$ ,

𝑷 𝑫 𝒆_{𝑖} = 𝑷 (𝑑_{𝑖} 𝒆_{𝑖}) = 𝑑_{𝑖} 𝒆_{𝜎 (𝑖)},

while

𝑫^{'} 𝑷 𝒆_{𝑖} = 𝑫^{'} 𝒆_{𝜎 (𝑖)} = 𝑑_{𝑖} 𝒆_{𝜎 (𝑖)} .

Hence

𝑷 𝑫 𝒆_{𝑖} = 𝑫^{'} 𝑷 𝒆_{𝑖}

for all $𝑖$ , so

𝑷 𝑫 = 𝑫^{'} 𝑷 .

Now apply this result to

𝑷^{⊤}

, and thus,

{(𝑷^{⊤} 𝑫)}^{⊤} = {(𝑫^{″} 𝑷^{⊤})}^{⊤} ⟺ 𝑫^{⊤} 𝑷 = 𝑷 𝑫^{″ ⊤} ⟺ 𝑫 𝑷 = {𝑷 𝑫}^{″}

since diagonal matrices are invariant under transposition. ⁠

□

Theorem 11.3.2 (Poincaré).

For any

𝑛 \geq 2

, the

𝑛

-dimensional unit ball

𝐵^{𝑛}

and the

𝑛

-dimensional polydisk

𝔻^{𝑛}

are not biholomorphically equivalent.

Proof.

Suppose, for the sake of contradiction, that there exists a biholomorphism $𝝋 : 𝔻^{𝑛} \to 𝐵^{𝑛}$ . Let $𝜶 = 𝝋 (𝟎) \in 𝐵^{𝑛}$ , and define $𝚽 = 𝝋_{‖ 𝜶 ‖} \circ 𝑼 \circ 𝝋$ , where $𝑼$ is a unitary matrix such that $𝑼 𝜶 = (‖ 𝜶 ‖, 0, \dots, 0)$ and $𝝋_{‖ 𝜶 ‖}$ is as in Proposition 11.2.16 $Proposition 11.2.16 Proposition 11.2.16$ .

The definition of $𝚽$ ensures that $𝚽 : 𝔻^{𝑛} \to 𝐵^{𝑛}$ and $𝚽 (𝟎) = 𝟎$ . Then $𝚽^{- 1} \circ Aut (𝐵^{𝑛}) \circ 𝚽$ consists of functions mapping $𝔻^{𝑛}$ to $𝔻^{𝑛}$ , or that

𝚽^{- 1} \circ Aut (𝐵^{𝑛}) \circ 𝚽 \subseteq Aut (𝔻^{𝑛}) ⟹ Aut (𝐵^{𝑛}) \subseteq 𝚽 \circ Aut (𝔻^{𝑛}) \circ 𝚽^{- 1} .

Similarly, $𝚽 \circ Aut (𝔻^{𝑛}) \circ 𝚽^{- 1} \subseteq Aut (𝐵^{𝑛})$ . Therefore, $Aut (𝐵^{𝑛}) = 𝚽 \circ Aut (𝔻^{𝑛}) \circ 𝚽^{- 1}$ , and

𝝍 \mapsto 𝚽 \circ 𝝍 \circ 𝚽^{- 1} (11.3 . 1)

defines a group isomorphism between $Aut (𝔻^{𝑛})$ and $Aut (𝐵^{𝑛})$ . Let ${Aut}^{'} (𝔻^{𝑛}) < Aut (𝔻^{𝑛})$ and ${Aut}^{'} (𝐵^{𝑛}) < Aut (𝐵^{𝑛})$ be subgroups fixing $𝟎$ . Therefore, (11.3.1) $(11.3.1) (11.3.1)$ induces a group isomorphism between ${Aut}^{'} (𝔻^{𝑛})$ and ${Aut}^{'} (𝐵^{𝑛})$ as well.

By Theorem 11.2.11 $Theorem 11.2.11 Theorem 11.2.11$ , every element of ${Aut}^{'} (𝔻^{𝑛})$ may be uniquely identified with a matrix in the form of

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

where $𝑷$ is a permutation matrix and $(𝜃_{1}, \dots, 𝜃_{𝑛}) \in {[0, 2 π)}^{𝑛}$ . Hence ${Aut}^{'} (𝔻^{𝑛})$ is isomorphic to the group of unitary monomial matrices. The structure of ${Aut}^{'} (𝐵^{𝑛})$ is given by Proposition 11.2.17 $Proposition 11.2.17 Proposition 11.2.17$ , and each element corresponds uniquely to a unitary matrix. Thus there is a natural isomorphism ${Aut}^{'} (𝐵^{𝑛}) ≅ U (𝑛)$ , the $𝑛 \times 𝑛$ unitary group.

For $𝑼 \in U (𝑛)$ , the spectral theorem allows it to be expressed in the form of $𝑽 diag (e^{i 𝜃_{1}}, \dots, e^{i 𝜃_{𝑛}}) 𝑽^{†}$ . Hence, for any positive integer $𝑚$ , we have

𝑽 diag (e^{i \frac{𝜃_{1}}{𝑚}}, \dots, e^{i \frac{𝜃_{𝑛}}{𝑚}}) 𝑽^{†} \in U (𝑛)

and

\begin{matrix} {(𝑽 diag (e^{i \frac{𝜃_{1}}{𝑚}}, \dots, e^{i \frac{𝜃_{𝑛}}{𝑚}}) 𝑽^{†})}^{𝑚} \\ \begin{matrix}  \end{matrix} = 𝑽 diag (e^{i \frac{𝜃_{1}}{𝑚}}, \dots, e^{i \frac{𝜃_{𝑛}}{𝑚}}) 𝑽^{†} \dots 𝑽 diag (e^{i \frac{𝜃_{1}}{𝑚}}, \dots, e^{i \frac{𝜃_{𝑛}}{𝑚}}) 𝑽^{†} . \end{matrix}

The adjacent products of $𝑽^{†} 𝑽$ simplify to the identity and the entire expression then simplifies to $𝑼$ . Hence the unitary group is divisible.

Consider the unitary monomial matrix

𝑷_{𝜏} = (\begin{matrix} 0 & 1 & 0 & 0 & \dots & 0 \\ 1 & 0 & 0 & 0 & \dots & 0 \\ 0 & 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & 0 & 0 & \dots & 1 \end{matrix}) .

inducing the permutation $𝜏$ , swapping the first and second entries. Assume that there exists some unitary monomial matrix $𝑸 = 𝑷_{𝜎} 𝑫$ (where $𝑫$ is diagonal and $𝑷_{𝜎}$ is a permutation matrix corresponding to the permutation $𝜎$ ) such that $𝑸^{2} = 𝑷_{𝜏}$ . This is equivalent to

𝑷_{𝜎} 𝑫 𝑷_{𝜎} 𝑫 = 𝑷_{𝜏} ⟹ 𝑷_{𝜎} 𝑷_{𝜎} 𝑫^{'} 𝑫 = 𝑷_{𝜏} ⟹ 𝑷_{𝜎}^{2} 𝑫^{″} = 𝑷_{𝜏},

where $𝑫^{'}$ and $𝑫^{″}$ are diagonal matrices, where the former existence is given by Lemma 11.3.1 $Lemma 11.3.1 Lemma 11.3.1$ . Thus, $𝑷_{𝜎}^{2} = 𝑷_{𝜏}$ (and $𝑫^{″} = 𝑰$ ) since their permutation parts must match. This is an impossibility since $𝑷_{𝜎}^{2}$ corresponds to an even permutation, while $𝑷_{𝜏}$ corresponds to an odd permutation. Thus, the unitary monomial group is not divisible.

By Proposition 11.2.19 $Proposition 11.2.19 Proposition 11.2.19$ , the two groups cannot be isomorphic to each other.

This contradicts the existence of (11.3.1) $(11.3.1) (11.3.1)$ , and therefore, no such biholomorphism $𝝋$ exists. ⁠ $□$

Remark.

A more succinct proof of the nonexistence of an isomorphism in the proof of Theorem 11.3.2 $Theorem 11.3.2 Theorem 11.3.2$ can be briefly described by means of topology:

Let $𝑀_{𝑛}$ denote the subgroup of all monomial matrices in $U (𝑛)$ , or the subgroup of unitary matrices with exactly one nonzero entry in each row and each column, and those nonzero entries lying in $U (1)$ .
For each permutation $𝜎 \in 𝑆_{𝑛}$ (the symmetric group of permutations) let $𝑷_{𝜎}$ be the corresponding permutation matrix and define
$𝑇_{𝜎} = {𝑫 𝑷_{𝜎} : 𝑫 = diag (e^{i 𝜃_{1}}, \dots, e^{i 𝜃_{𝑛}}) \in U {(1)}^{𝑛}} .$
Each $𝑇_{𝜎}$ is homeomorphic to the torus $U {(1)}^{𝑛}$ , and every element of $𝑀_{𝑛}$ lies in exactly one $𝑇_{𝜎}$ . Hence
$𝑀_{𝑛} = ⋃_{𝜎 \in 𝑆_{𝑛}} 𝑇_{𝜎},$
a disjoint union of $| 𝑆_{𝑛} | = 𝑛!$ copies of $U {(1)}^{𝑛}$ .
Each $𝑇_{𝜎}$ is clopen in $𝑀_{𝑛}$ by their pairwise disjointness, the topology of the torus, and the fact that their union is $𝑀_{𝑛}$ . Therefore each $𝑇_{𝜎}$ is a connected component of $𝑀_{𝑛}$ . Because each $𝑇_{𝜎}$ is connected, $𝑀_{𝑛}$ has $𝑛!$ connected components.
The elements of $U (𝑛)$ may be unitarily diagonalized (by the spectral theorem) into $𝑽 𝑫 𝑽^{†}$ , where
$𝑫 = diag (e^{i 𝜃_{1}}, \dots, e^{i 𝜃_{𝑛}})$
is a diagonal unitary matrix and $𝑽 \in U (𝑛)$ . Then there exists a connected path connecting $(𝜃_{1}, \dots, 𝜃_{𝑛})$ to $(0, \dots, 0)$ , which corresponds to the matrix $𝑽 𝑰 𝑽^{†} = 𝑰$ . Because every matrix is path-connected to the identity, $U (𝑛)$ is connected.
Consequently for $𝑛 \geq 2$ , the subgroup $𝑀_{𝑛}$ (which has more than one connected component) cannot be isomorphic to $U (𝑛)$ as a topological group.

Although we do not justify these topological claims in detail here, it is worth noting, heuristically, why such topological considerations naturally arise.

A biholomorphism between two domains induces a homeomorphism with respect to their natural topologies (the compact-open topology) by $𝝍 \mapsto 𝚽 \circ 𝝍 \circ 𝚽^{- 1}$ . Hence any induced topological invariant of an automorphism group, such as connectivity, is to be preserved under this equivalence. Of course, we are yet to verify the rigor and intuition used within the topology, but the intuitive picture already hints to the validity of the connectivity argument.