1.5: Cartan's Uniqueness Theorem
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The following theorem is another analogue of Schwarz’s lemma to several variables. It says that for a bounded domain, it is enough to know that a self mapping is the identity at a single point to show that it is the identity everywhere. As there are quite a few theorems named for Cartan, this one is often referred to as the Cartan's uniqueness theorem. It is useful in computing the automorphism groups of certain domains. An automorphism of \(U\) is a biholomorphic map from \(U\) onto \(U\). Automorphisms form a group under composition, called the automorphism group. As exercises, you will use the theorem to compute the automorphism groups of \(\mathbb{B}_{n}\) and \(\mathbb{D}^n\).
Suppose \(U \subset \mathbb{C}^n\) is a bounded domain, \(a \in U\), \(f \colon U \to U\) is a holomorphic mapping, \(f(a) = a\), and \(Df(a)\) is the identity. Then \(f(z) = z\,\) for all \(z \in U\).
Find a counterexample to the theorem if \(U\) is unbounded. Hint: For simplicity take \(a=0\) and \(U=\mathbb{C}^n\).
Before we get into the proof, let us write down the Taylor series of a function in a nicer way, splitting it up into parts of different degree.
A polynomial \(P \colon \mathbb{C}^n \to \mathbb{C}\) is homogeneous of degree \(d\) if \[P(s z) = s^d P(z)\] for all \(s \in \mathbb{C}\) and \(z \in \mathbb{C}^n\). A homogeneous polynomial of degree \(d\) is a polynomial whose every monomial is of total degree \(d\). For instance, \(z^2w-iz^3+9zw^2\) is homogeneous of degree 3 in the variables \((z,w) \in \mathbb{C}^2\). A polynomial vector-valued mapping is homogeneous of degree \(d\), if each component is. If \(f\) is holomorphic near \(a \in \mathbb{C}^n\), then write the power series of \(f\) at \(a\) as \[\sum_{j=0}^{\infty} f_j(z-a) ,\] where \(f_j\) is a homogeneous polynomial of degree \(j\). The \(f_j\) is called the degree j homogeneous part of \(f\) at \(a\). The \(f_j\) would be vector-valued if \(f\) is vector-valued, such as in the statement of the theorem. In the proof, we will require the vector-valued Cauchy estimates (exercise below)\(^{1}\).
Prove a vector-valued version of the Cauchy estimates. Suppose \(f \colon \overline{\Delta_r(a)} \to \mathbb{C}^m\) is continuous function holomorphic on a polydisc \(\Delta_r(a) \subset \mathbb{C}^n\). Let \(\Gamma\) denote the distinguished boundary of \(\Delta\). Show that for any multi-index \(\alpha\) we get
\[\left| \left|\frac{\partial^{|\alpha|}f}{\partial z^{\alpha}}(a)\right| \right| \leq\frac{\alpha !}{r^{\alpha}}\sup_{z\in\Gamma} ||f(z)||.\]
Proof of Cartan’s uniqueness theorem.
Without loss of generality, assume \(a=0\). Write \(f\) as a power series at the origin, written in homogeneous parts: \[f(z) = z + f_k(z) + \sum_{j=k+1}^\infty f_j(z) ,\] where \(k \geq 2\) is an integer such that \(f_2(z),f_3(z),\ldots,f_{k-1}(z)\) is zero. The degree-one homogeneous part is simply the vector \(z\), because the derivative of \(f\) at the origin is the identity. Compose \(f\) with itself \(\ell\) times:
\[f^\ell(z) = \underbrace{f \circ f \circ \cdots \circ f}_{\ell\text{ times}} (z) .\] As \(f(U) \subset U\), then \(f^\ell\) is a holomorphic map of \(U\) to \(U\). As \(U\) is bounded, there is an \(M\) such that \(||z|| \leq M\) for all \(z \in U\). Therefore, \(||f(z)|| \leq M\) for all \(z \in U\), and \(||f^\ell(z)|| \leq M\) for all \(z \in U\).
If we plug in \(z + f_k(z) + {}\)higher order terms into \(f_k\), we get \(f_k(z) + {}\)some other higher order terms. Therefore, \(f^2(z) = z + 2 f_k(z) + {}\)higher order terms. Continuing this procedure, \[f^\ell(z) = z + \ell f_k(z) + \sum_{j=k+1}^\infty \tilde{f}_j(z) ,\] for some other degree \(j\) homogeneous polynomials \(\tilde{f}_j\). Suppose \(\Delta_r(0)\) is a polydisc whose closure is in \(U\). Via Cauchy estimates, for any multinomial \(\alpha\) with \(|\alpha|=k\),
\[\frac{\alpha !}{r^{\alpha}}M\geq \left| \left|\frac{\partial^{|\alpha |}f^{\ell}}{\partial z^{\alpha}}(0)\right| \right| =\ell\left| \left| \frac{\partial^{|\alpha |}f}{\partial z^{\alpha}}(0)\right| \right|.\]
The inequality holds for all \(\ell \in \mathbb{N}\), and so \(\frac{\partial^{|\alpha|} f}{\partial z^\alpha}(0) = 0\). Therefore, \(f_k \equiv 0\). On the domain of convergence of the expansion, we get \(f(z) = z\), as there is no other nonzero homogeneous part in the expansion of \(f\). As \(U\) is connected, then the identity theorem says \(f(z) = z\) for all \(z \in U\).
As an application, let us classify all biholomorphisms of all bounded circular domains that fix a point. A circular domain is a domain \(U \subset \mathbb{C}^n\) such that if \(z \in U\), then \(e^{i\theta} z \in U\) for all \(\theta \in \mathbb{R}\).
Suppose \(U, V \subset \mathbb{C}^n\) are bounded circular domains with \(0 \in U\), \(0 \in V\), and \(f \colon U \to V\) is a biholomorphic map such that \(f(0) = 0\). Then \(f\) is linear.
For example, \(\mathbb{B}_{n}\) is circular and bounded. So a biholomorphism of \(\mathbb{B}_{n}\) (an automorphism) that fixes the origin is linear. Similarly, a polydisc centered at zero is also circular and bounded.
- Proof
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The map \(g(z) = f^{-1}\bigl(e^{-i\theta}f(e^{i\theta} z)\bigr)\) is an automorphism of \(U\) and via the chain rule, \(g'(0) = I\). Therefore, \(f^{-1}\bigl(e^{-i\theta}f(e^{i\theta} z)\bigr) = z\), or in other words \[f(e^{i\theta} z) = e^{i\theta}f(z) .\] Write \(f\) near zero as \(f(z) = \sum_{j=1}^\infty f_j(z)\) where \(f_j\) are homogeneous polynomials of degree \(j\) (notice \(f_0 = 0\)). Then \[\sum_{j=1}^\infty e^{i\theta} f_j(z) = e^{i\theta} \sum_{j=1}^\infty f_j(z) = \sum_{j=1}^\infty f_j(e^{i\theta} z) = \sum_{j=1}^\infty e^{ij\theta}f_j(z) .\] By the uniqueness of the Taylor expansion, \(e^{i\theta} f_j(z) = e^{ij\theta} f_j(z)\), or \(f_j(z) = e^{i(j-1)\theta} f_j(z)\), for all \(j\), all \(z\), and all \(\theta\). If \(j\not=1\), we obtain that \(f_j \equiv 0\), which proves the claim.
Show that every automorphism \(f\) of \(\mathbb{D}^n\) (that is, a biholomorphism \(f \colon \mathbb{D}^n \to \mathbb{D}^n\)) is given as \[f(z) = P \left( e^{i\theta_1} \frac{z_1-a_1}{1-\bar{a}_1z_1} , e^{i\theta_2} \frac{z_2-a_2}{1-\bar{a}_2z_2} , \ldots, e^{i\theta_n} \frac{z_n-a_n}{1-\bar{a}_nz_n} \right)\] for \(\theta \in \mathbb{R}^n\), \(a \in \mathbb{D}^n\), and a permutation matrix \(P\).
Given \(a \in \mathbb{B}_{n}\), define the linear map \(P_a z = \frac{\langle z,a\rangle }{\langle a,a\rangle}a\) if \(a \not= 0\) and \(P_0z = 0\). Let \(s_a = \sqrt{1-||a||^2}\). Show that every automorphism \(f\) of \(\mathbb{B}_{n}\) (that is, a biholomorphism \(f \colon \mathbb{B}_{n} \to \mathbb{B}_{n}\)) can be written as \[f(z) = U \frac{a-P_az-s_a(I-P_a)z}{1-\langle z,a\rangle }\] for a unitary matrix \(U\) and some \(a \in \mathbb{B}_{n}\).
Using the previous two exercises, show that \(\mathbb{D}^n\) and \(\mathbb{B}_{n}\), \(n \geq 2\), are not biholomorphic via a method more in the spirit of what Poincaré used: Show that the groups of automorphisms of the two domains are different groups when \(n \geq 2\).
Suppose \(U \subset \mathbb{C}^n\) is a bounded open set, \(a \in U\), and \(f \colon U \to U\) is a holomorphic mapping such that \(f(a) = a\). Show that every eigenvalue \(\lambda\) of the matrix \(Df(a)\) satisfies \(|\lambda| \leq 1\).
Find a domain \(U \subset \mathbb{C}^n\) such that the only biholomorphism \(f \colon U \to U\) is the identity \(f(z) = z\). Hint: Take the polydisc (or the ball) and remove some number of points (be careful in how you choose them). Then show that \(f\) extends to a biholomorphism of the polydisc. Then see what happens to those points you took out.
- Show that Cartan’s uniqueness theorem is not true in the real case, even for rational functions. That is, find a rational function \(R(t)\) of a real variable \(t\), such that \(R\) that takes \((-1,1)\) to \((-1,1)\), \(R'(0) = 1\), and \(R(t)\) is not the identity. You can even make \(R\) bijective.
- Show that also Exercise \(\PageIndex{6}\) is not true in the real case. For any \(\alpha \in \mathbb{R}\) find a rational function \(R(t)\) of a real variable \(t\), such that \(R\) takes \((-1,1)\) to \((-1,1)\) and \(R'(0) = \alpha\).
Suppose \(U \subset \mathbb{C}^n\) is an open set, \(a \in U\), \(f \colon U \to U\) is a holomorphic mapping, \(f(a) = a\), and suppose that \(|\lambda| < 1\) for every eigenvalue \(\lambda\) of \(D f(a)\). Prove that there exists a neighborhood \(W\) of \(a\), such that \(\lim_{\ell \to \infty} f^{\ell}(z) = a\) for all \(z \in W\).
Let \(U \subset \mathbb{C}^n\) be a bounded open set and \(a \in U\). Show that the mapping \(\varphi \mapsto \bigl(\varphi(a),D\varphi(a)\bigr)\) from the set \(\text{Aut}(U)\) of automorphisms of \(U\) to \(\mathbb{C}^n \times \mathbb{C}^{n^2}\) is injective.
Footnotes
[1] The normal Cauchy estimates could also be used in the proof of Cartan by applying them componentwise.