9.2 The Group of Holomorphic Automorphisms on and
A function 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), which is equivalent to several results of Cartan.
In preparation, we will introduce several relevant concepts.
Definition 9.2.1 (Multi-Index Notation).
A multi-index is an -tuple of nonnegative integers . We define
where
Definition 9.2.2.
A polynomial of several variables is said to be homogeneous of degree iff
or equivalently, iff can be written as
where is a multi-index.
Proposition 9.2.1.
Let be a homogeneous polynomial of degree .
-
For any multi-index with ,
is a homogeneous polynomial of degree .
- If , then is constant (and there exists a multi-index with such that is nonzero).
- If , then .
Proof.
Writing with coefficients , we compute
where terms with vanish. For each remaining term, the total degree is
Hence, is a homogeneous polynomial of degree , establishing Part 1.
If , every surviving monomial has degree , so is constant. Moreover, since has degree exactly , there exists some multi-index with and ; choosing yields a nonzero constant derivative. This proves Part 2.
Finally, if , then for every term in the expansion, at least one , so all summands vanish identically. Thus , verifying Part 3.
Lemma 9.2.1 (Cartan).
Let be a bounded region, and suppose that is holomorphic. If such that and the complex Jacobian at is the identity matrix, or equivalently, if
then is the identity map.
Proof.
By Theorem 9.1.2, we have the expansion
which is absolutely convergent on some polydisk centered at , where and . The terms have been rearranged (from absolute convergence) so that the inner summation is a homogeneous polynomial with a zero at and degree .
Trivially, by (9.2.1). Similarly, . Hence, the linear homogeneous polynomial of (9.2.2) equals
and the entire expansion is thus equal to
Define a sequence of holomorphic functions by
Assume the existence of some , the smallest such that is not identically zero. Because
it then follows that
Assume, for induction, that
Then we have
Since for any , the sequence is uniformly bounded on . By Montel’s Theorem (Theorem 9.1.6), there exists a subsequence that converges locally uniformly to some holomorphic function by virtue of Weierstrass (Theorem 9.1.5).
Since , there exists satisfying such that
is a nonzero constant by Proposition 9.2.1. Consequently,
is a homogeneous polynomial with degree and thus vanishes as . Similarly, is homogeneous with degree and thus vanishes. Therefore,
which diverges as . Weierstrass’ Convergence Theorem (Theorem 9.1.5) gives that which must be finite by holomorphy, contradicting the divergence. Hence, the assumed value for cannot exist and hence for all . Thus, on some polydisk centered at . By the Identity Theorem (Theorem 9.1.3), on .
Definition 9.2.3 (Reinhardt Domain).
An open domain is a Reinhardt domain centered at iff ,
is fully contained in . In other words, is invariant under all rotations about the center in each coordinate.
Definition 9.2.4.
A Reinhardt domain centered at is said to be complete iff , the polydisk
is contained in .
Definition 9.2.5 (Circular Domain).
An open domain is a circular domain centered at iff ,
is fully contained in .
Definition 9.2.6.
A circular domain centered at is said to be complete iff ,
is contained in .
Proposition 9.2.2.
Let be open domains with for each , and let and be holomorphic maps. Define the composition by . Then for every , the complex Jacobian matrix of at is
Proof.
Fix and let . Write
where each is holomorphic for . Similarly, write
where each and each is holomorphic for and . Then for each . By the chain multivariable rule, the complex Jacobian of at is the matrix
Lemma 9.2.2 (Cartan).
Let be a bounded complete circular domain centered at , and suppose that is a biholomorphism. If , then is linear.
Proof.
Let for all and suppose that . By Proposition 9.2.2, we must have that
By Lemma 9.2.1, on . Hence, for all . Together with Theorem 9.1.2, write
on a polydisk centered at . Thus,
On the other hand, composing with with (9.2.3) gives
Hence, by the uniqueness of power series expansions, we must either have that , , or equivalently, that ,
This is only possible when by irrationality, and thus for all . Therefore, must be linear.
Remark.
If , then for some and any automorphism with a fixed point is a rotation in the form of , hence linear, the effective statement of the Schwarz Lemma (Lemma 3.5.1).
Theorem 9.2.1 (The Holomorphic Automorphism Group on ).
The holomorphic automorphism group of the polydisk consists solely of biholomorphisms in the form of
where is a permutation matrix (for coordinate permutations), , and . Moreover, every such map is indeed an automorphism.
Proof.
Let be arbitrary, and set . Define the Möbius transformation . It follows that and .
By Lemma 9.2.2, the map is linear, so for some invertible constant matrix , hence . Thus,
which implies for all (for if , then choosing with and for yields a contradiction).
For each , define the sequence for each by
where we informally let if . Then, for all and ,
In particular, the -th component is
As ,
Now consider, for each , the sequence , where is in the -th position. Then
As , (the -th unit basis vector), so the limit is
Because the function is injective on all of , if (within the interior), then would map to , which is an impossibility. Hence, , and consequently, . Combined with (9.2.5), this forces exactly one entry in the -th column of to have absolute value (of the form ), with all others zero.
Invertibility of ensures each column has at least one nonzero entry, so is a monomial matrix, which factors to
for some permutation matrix . Therefore,
Let be the permutation induced by . The map multiplies the -th input coordinate by and permutes to place it in the -th output position, so the -th coordinate of is . Applying componentwise then gives, for the -th output coordinate,
Set . Then
Hence,
as in (9.2.4). Finally, each automorphism of this form lies in trivially.
Definition 9.2.7.
The conjugate transpose or Hermitian transpose of a complex matrix is defined as , or the transpose of the matrix with each element replaced with its complex conjugate.
Definition 9.2.8.
A matrix is said to be unitary iff its inverse is its conjugate transpose, or iff .
Definition 9.2.9.
A matrix is said to be monomial iff it has exactly one nonzero entry in each row and each column.
Theorem 9.2.2 (Spectral Theorem).
For any unitary matrix , there exists a unitary matrix such that , where is a diagonal matrix whose diagonal entries are all of unit modulus.
Proof.
Because for any , any eigenvalue (existence given by the Fundamental Theorem of Algebra in Theorem 3.3.1 on the characteristic equation) of must satisfy
where is the corresponding eigenvector in . Then
Let be an -dimensional subspace. For any ,
so . Hence is invariant under . The restriction of to , , yields another eigenvalue with eigenvector satisfying and . Similarly, we may define , which is an -dimensional subspace invariant under . Repeating this process inductively, we obtain an orthonormal basis of eigenvectors of with corresponding eigenvalues . Setting
gives that
The -th diagonal entry of is equal to , while the non-diagonal entries correspond to for some , which vanish by orthogonality in construction. Thus, (unitary) and .
A unitary transformation is a map in the form of , where is a unitary matrix.
Proposition 9.2.3.
For any ,
lies in , where . Moreover, .
Proof.
For , because ,
Hence, maps to . A simple calculation shows that
and hence is bijective, admitting the inverse . Therefore, .
Proposition 9.2.4.
A function is a unitary transformation iff and .
Proof.
Because is a bounded complete circular domain centered at , from Lemma 9.2.2 we have that for some constant invertible matrix
Similarly, we have , so . Observe that
More explicitly, we have
Letting () be the -th unit basis vector, we obtain
Letting (), we have
which implies that . Similarly, letting gives
which implies that . Therefore, by (9.2.7), for all , observe that
where is the Kronecker delta. Hence, we have , and thus is unitary.
Conversely, if for some unitary matrix , then for any ,
so maps to . Since is invertible with unitary inverse , the map is bijective with inverse , which also maps to . Therefore, and .
Definition 9.2.10.
A group (under juxtaposition) is said to be divisible iff for every and every positive integer , there exists some such that .
Proposition 9.2.5.
The divisibility of a group is preserved under group isomorphisms.
Proof.
Let be a group isomorphism between groups and with juxtaposition.
Assume is divisible. Fix and a positive integer . Since is bijective there is with . By divisibility of there exists with . Applying and using the homomorphism property gives
Thus every element of has an -th root, so is divisible.
Conversely, if is divisible then the same argument applied to shows is divisible. Therefore divisibility is preserved under group isomorphisms.
Theorem 9.2.3 (The Holomorphic Automorphism Group on ).
The holomorphic automorphism group consists solely of biholomorphisms in the form of
where are unitary matrices, , and is defined as in (9.2.6) (and every such function lies in ).
Proof.
Let be arbitrary, and set . Then there exists a unitary matrix such that .
Now let be as in Proposition 9.2.3, mapping to . Then, the map fixes , so by Proposition 9.2.4 it is a unitary transformation, say . Therefore,
The converse is trivial.