Notes on Complex Analysis

9.3 Topological Equivalence and Biholomorphic Equivalence

Lemma 9.3.1.

If is a permutation matrix, and is a diagonal matrix, then there exists a diagonal matrix such that . Similarly, there exists a diagonal matrix such that .

Proof.

Let and let be the permutation corresponding to , for each standard basis vector . Define

Then for every ,

while

Hence

for all , so

Now apply this result to , and thus, since diagonal matrices are invariant under transposition.

Theorem 9.3.1 (Poincaré).

For any , the -dimensional unit ball and the -dimensional polydisk are not biholomorphically equivalent.

Proof.

Suppose, for the sake of contradiction, that there exists a biholomorphism . Let , and define , where is a unitary matrix such that and is as in Proposition 9.2.3.

The definition of ensures that and . Then consists of functions mapping to , or that

Similarly, . Therefore, , and

defines a group isomorphism between and . Let and be subgroups fixing . Therefore, (9.3.1) induces a group isomorphism between and as well.

By Theorem 9.2.1, every element of may be uniquely identified with a matrix in the form of

where is a permutation matrix and . Hence is isomorphic to the group of unitary monomial matrices. The structure of is given by Proposition 9.2.4, and each element corresponds uniquely to a unitary matrix. Thus there is a natural isomorphism , the unitary group.

For , the spectral theorem allows it to be expressed in the form of . Hence, for any positive integer , we have

and

The adjacent products of simplify to the identity and the entire expression then simplifies to . Hence the unitary group is divisible.

Consider the unitary monomial matrix

inducing the permutation , swapping the first and second entries. Assume that there exists some unitary monomial matrix (where is diagonal and is a permutation matrix corresponding to the permutation ) such that . This is equivalent to

where and are diagonal matrices, where the former existence are given by Lemma 9.3.1. Thus, (and ) since their permutation parts must match. This is an impossibility since corresponds to an even permutation, while corresponds to an odd permutation. Thus, the unitary monomial group is not divisible.

By Proposition 9.2.5, the two groups cannot be isomorphic to each other.

This contradicts the existence of (9.3.1), and therefore, no such biholomorphism exists.

Remark.

A more succinct proof of the nonexistence of an isomorphism in the proof of Theorem 9.3.1 can be briefly described by means of topology:

Let denote the subgroup of all monomial matrices in , or the subgroup of unitary matrices with exactly one nonzero entry in each row and each column, and those nonzero entries lying in .

For each permutation (the symmetric group of permutations) let be the corresponding permutation matrix and define

Each is homeomorphic to the torus , and every element of lies in exactly one . Hence

a disjoint union of copies of .

Each is clopen in by their pairwise disjointness, the topology of the torus, and the fact that their union is . Therefore each is a connected component of . Because each is connected, has connected components.

The elements of may be unitarily diagonalized (by the spectral theorem) into , where

is a diagonal unitary matrix and . Then there exists a connected path connecting to , which corresponds to the matrix . Because every matrix is path-connected to the identity, is connected.

Consequently for , the subgroup (which has more than one connected component) cannot be isomorphic to 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 . 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.

Esc