Every biholomorphic self-map of \(\DD\) has the form \(e^{i\theta}\psi_a\) for some \(a\in\DD\) and \(\theta\in\RR\), where \(\psi_a(z) \coloneqq (a-z)/(1-\bar a z)\) is the Blaschke factor.
Blaschke factors
Blaschke factors and finite Blaschke products (Tag D-MFPYG)
For \(a\in\DD\), the Blaschke factor, or hyperbolic translation, at \(a\) is \[ \psi_a(z)\coloneqq\frac{a-z}{1-\bar a z},\qquad z\in\CC,\ \bar a z\neq1. \] For \(n\ge1\), points \(a_1,\ldots,a_n\in\DD\), and \(\lambda\in\CC\) with \(\abs{\lambda}=1\), the finite Blaschke product with zeros \(a_1,\ldots,a_n\) is \[ B(z)\coloneqq\lambda\prod_{k=1}^{n}\psi_{a_k}(z)=\lambda\prod_{k=1}^{n}\frac{a_k-z}{1-\bar a_k z}. \]
Some authors take the Blaschke factor to be \(\phi_a(z)\coloneqq\frac{z-a}{1-\bar a z}=-\psi_a(z)\); the two conventions differ by the unimodular constant \(-1\), which is absorbed into \(\lambda\) in a Blaschke product.
Let \(a\in\DD\).
For \(\abs{z}\le1\), \(1-\abs{\psi_a(z)}^2=\frac{(1-\abs{a}^2)(1-\abs{z}^2)}{\abs{1-\bar a z}^2}\); hence \(\psi_a(\DD)\subseteq\DD\) and \(\psi_a(S^1)\subseteq S^1\).
\(\psi_a\circ\psi_a=\operatorname{id}\) on \(\overline{\DD}\), so \(\psi_a\) is a biholomorphism \(\DD\to\DD\) exchanging \(0\) and \(a\).
\(\psi_a\) preserves the hyperbolic metric \(\frac{\abs{dz}}{1-\abs{z}^2}\) on \(\DD\): \(\frac{\abs{\psi_a'(z)}}{1-\abs{\psi_a(z)}^2}=\frac{1}{1-\abs{z}^2}\) for \(z\in\DD\).
Every finite Blaschke product \(B\) is holomorphic on a neighborhood of \(\overline{\DD}\), maps \(\DD\) into \(\DD\) and \(S^1\) into \(S^1\), and its zeros in \(\DD\) are exactly \(a_1,\ldots,a_n\).
Since \(\abs{a}<1\) and \(\abs{z}\le1\), \(\abs{\bar a z}<1\), so \(\psi_a\) is holomorphic on the disc \(\abs{z}<1/\abs{a}\), which contains \(\overline{\DD}\).
Expanding, \(\abs{1-\bar a z}^2-\abs{a-z}^2=1+\abs{a}^2\abs{z}^2-\abs{a}^2-\abs{z}^2=(1-\abs{a}^2)(1-\abs{z}^2)\); divide by \(\abs{1-\bar a z}^2\). The right side is positive for \(\abs{z}<1\) and zero for \(\abs{z}=1\).
For \(\abs{z}\le1\), \[ \psi_a(\psi_a(z))=\frac{a(1-\bar a z)-(a-z)}{(1-\bar a z)-\bar a(a-z)}=\frac{z(1-\abs{a}^2)}{1-\abs{a}^2}=z. \] With (i), \(\psi_a\colon\DD\to\DD\) is holomorphic with holomorphic inverse \(\psi_a\), and \(\psi_a(0)=a\), \(\psi_a(a)=0\).
By the quotient rule, \(\psi_a'(z)=-\frac{1-\abs{a}^2}{(1-\bar a z)^2}\); divide by the expression for \(1-\abs{\psi_a(z)}^2\) in (i).
Each factor is holomorphic near \(\overline{\DD}\), has modulus less than \(1\) on \(\DD\) and equal to \(1\) on \(S^1\) by (i), and vanishes on \(\overline{\DD}\) only at its own \(a_k\); multiply, using \(\abs{\lambda}=1\).
Properties of the Blaschke factors $\psi_a$ (Tag PR-ULJAJ)
For \(a\in\DD\) let \(\psi_a(z)=\frac{a-z}{1-\bar az}\) be the Blaschke factor. Then:
\(\psi_a\in\Aut(\DD)\), and \(\psi_a\) maps the unit circle \(S^1\) onto itself;
\(\psi_a(0)=a\) and \(\psi_a(a)=0\);
\(\psi_a\circ\psi_a=\id_\DD\), so \(\psi_a^{-1}=\psi_a\);
\(\psi_a'(z)=\frac{\abs{a}^2-1}{(1-\bar az)^2}\);
\(\psi_a(\lambda z)=\lambda\,\psi_{\bar\lambda a}(z)\) for every \(\lambda\in\CC\) with \(\abs\lambda=1\).
The denominator \(1-\bar az\) is nonzero for \(\abs{z}\le1\), since \(\abs{\bar az}<1\). For \(\abs{z}=1\), \(\abs{1-\bar az}=\abs{\bar z-\bar a}=\abs{a-z}\), so \(\abs{\psi_a(z)}=1\). By the maximum modulus principle \(\abs{\psi_a}<1\) on \(\DD\) (\(\psi_a\) is nonconstant because \(\abs a<1\)). The values at \(0\) and \(a\) are immediate. A direct computation gives \[ \psi_a(\psi_a(z))=\frac{a(1-\bar az)-(a-z)}{(1-\bar az)-\bar a(a-z)}=\frac{z(1-\abs a^2)}{1-\abs a^2}=z, \] so \(\psi_a\) is a bijection of \(\DD\) equal to its own inverse, hence in \(\Aut(\DD)\), and likewise a bijection of \(S^1\). By the quotient rule, \[ \psi_a'(z)=\frac{-(1-\bar az)+\bar a(a-z)}{(1-\bar az)^2}=\frac{\abs a^2-1}{(1-\bar az)^2}. \] Finally, for \(\abs\lambda=1\), \(\lambda\bar\lambda=1\) gives \[ \lambda\,\psi_{\bar\lambda a}(z)=\lambda\,\frac{\bar\lambda a-z}{1-\lambda\bar az}=\frac{a-\lambda z}{1-\bar a\lambda z}=\psi_a(\lambda z). \]
Automorphisms of the unit disc (Tag FF-4GUYL)
The biholomorphisms \(\DD\to\DD\) are exactly the maps \[ f(z)=\lambda\,\frac{\alpha-z}{1-\bar\alpha z},\qquad \alpha\in\DD,\ \lambda\in\CC,\ \abs{\lambda}=1. \]
Write \(\psi_\alpha(z)\coloneqq\frac{\alpha-z}{1-\bar\alpha z}\) for the Blaschke factor at \(\alpha\in\DD\); it is a biholomorphism \(\DD\to\DD\) with \(\psi_\alpha\circ\psi_\alpha=\operatorname{id}\), so \(\lambda\psi_\alpha\) is a biholomorphism of \(\DD\) for \(\abs{\lambda}=1\).
Conversely, let \(f\colon\DD\to\DD\) be a biholomorphism, let \(\alpha\coloneqq f^{-1}(0)\in\DD\), and put \(g\coloneqq f\circ\psi_\alpha\). Then \(g\) is a biholomorphism of \(\DD\) with \(g(0)=f(\alpha)=0\). The Schwarz lemma applied to \(g\) and to \(g^{-1}\) gives \(\abs{g'(0)}\le1\) and \(\abs{(g^{-1})'(0)}=1/\abs{g'(0)}\le1\), so \(\abs{g'(0)}=1\) and \(g(z)=\lambda z\) with \(\abs{\lambda}=1\). Hence \(f=g\circ\psi_\alpha=\lambda\psi_\alpha\).
For \(a\neq 0\), \(\psi_a(z) = (a-z)/(1-\bar a z)\) has a simple zero at \(z=a\) and a simple pole at \(z = 1/\bar a\), the reflection of \(a\) in the unit circle.
(Properties)
For \(\abs z=1\), \(\abs{1-\bar a z} = \abs{\bar z}\abs{1-\bar a z} = \abs{\bar z - \bar a} = \abs{a-z}\), so \(\abs{\psi_a(z)} = 1\) and \(\psi_a\) maps \(S^1\) to \(S^1\). Since \(\psi_a\) is holomorphic on a neighborhood of \(\overline\DD\) (its pole \(1/\bar a\) lies outside \(\overline\DD\)) and \(\abs{\psi_a}=1\) on \(S^1\), the maximum modulus principle gives \(\psi_a(\DD)\subseteq\DD\).
Inverting: set \(\psi_a(z) = w\) and solve. \[ \begin{aligned} {a-z \over 1 - \bar{a}z} &= w \\ \implies a-z - w(1-\bar{a} z) &= 0 \\ \implies z&= {w-a \over \bar a w - 1} = {a-w\over 1-\bar a w} \end{aligned},\] so \(\psi_a\circ\psi_a=\id\), and \(\psi_a\) is an automorphism of \(\DD\).
Differentiating, by the quotient rule: \[ \psi'_a(z) = {-(1-\bar a z) + \bar a(a-z) \over \qty{1-\bar a z}^2} = {-1 + \abs{a}^2 \over \qty{1-\bar a z}^2} .\]
Scaling: for \(\abs\lambda = 1\), insert \(1 = \bar\lambda\lambda\), \[ \begin{aligned} \psi_a(\lambda z) &= {a - \lambda z \over 1 - \bar a \lambda z}\\ &= {\lambda\bar\lambda a - \lambda z \over 1 - \bar a \lambda z} \\ &= \lambda {\bar\lambda a - z \over 1 - \bar{\bar\lambda a} z} \\ &= \lambda \psi_{\bar \lambda a}(z) \end{aligned}.\]
The automorphism group of the disc
Automorphisms of the unit disc as rotated Blaschke factors (Tag T-W26VL)
For \(a\in\DD\) let \(\psi_a(z)\coloneqq\dfrac{a-z}{1-\overline{a}z}\) be the Blaschke factor at \(a\). The group \(\Aut_\CC(\DD)\) of biholomorphisms \(\DD\to\DD\) is \[ \Aut_\CC(\DD)=\{\lambda\psi_a : a\in\DD,\ \lambda\in S^1\}, \] so every automorphism of \(\DD\) is a Blaschke factor followed by a rotation [@SS03].
(Sketch)
The maps \(\lambda\psi_a\) are automorphisms, being compositions of the rotation \(z\mapsto \lambda z\) with \(\psi_a\).
Let \(f \in \Aut(\DD)\) and let \(a\coloneqq \inverseof{f}(0)\).
Then \(g\coloneqq f\circ \psi_a \in \Aut(\DD)\) fixes \(0\). The Schwarz lemma applied to \(g\) and to \(\inverseof{g}\) gives \(\abs{g(z)}\leq\abs z\) and \(\abs z = \abs{\inverseof{g}(g(z))}\leq\abs{g(z)}\), so \(g(z) = \lambda z\) with \(\abs\lambda = 1\).
Hence \(f = g\circ\psi_a = \lambda\psi_a\).
Automorphisms of the upper half-plane (Tag T-VGDFW)
Let \(\HH\coloneqq\{z\in\CC : \operatorname{Im}z>0\}\) be the upper half-plane. The group \(\Aut_\CC(\HH)\) of biholomorphisms \(\HH\to\HH\) is \[ \Aut_\CC(\HH)=\Bigl\{z\mapsto\frac{az+b}{cz+d} : a,b,c,d\in\RR,\ ad-bc=1\Bigr\}, \] and sending \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\SL_2(\RR)\) to this map induces a group isomorphism \(\PSL_2(\RR)\cong\Aut_\CC(\HH)\).
See [@SS03].
Exercises
A self-map of $\DD$ is bounded by the Blaschke product of its zeros (Tag E-IVHVW)
Let \(f: \DD\to \DD\) with \(\ts{a_k}_{k\leq n}\) the zeros of \(f\) in \(\DD\). Show that \[ \abs{f(z)}\leq \prod_{k\leq n} \abs{ \psi_{a_k}(z) } .\]
Solution
Here \(\psi_a(z)\da{a-z\over1-\bar az}\), and the zeros \(a_k\) are listed with multiplicity. Define \(\Psi(z) \da \prod_{k\leq n} \psi_{a_k}(z)\) and \(g(z) \da f(z)/\Psi(z)\). Each zero of \(\Psi\) in \(\DD\) is a zero of \(f\) of at least the same order, so \(g\) has removable singularities and is holomorphic on \(\DD\). The claim is that \(\abs{g(z)} \leq 1\), which implies the result directly. Since \(\Psi\) is continuous on \(\overline\DD\) with \(\abs{\Psi} = 1\) on \(\abs{z} = 1\), \(m(r)\da\min_{\abs z=r}\abs{\Psi(z)}\to1\) as \(r\to1^-\). Fix \(z\in\DD\). For \(\abs z<r<1\) close enough to \(1\) that \(m(r)>0\), the maximum modulus principle on \(\abs\zeta\le r\) gives \[ \abs{g(z)}\le\max_{\abs\zeta=r}\abs{g(\zeta)} = \max_{\abs\zeta=r}{ \abs{f(\zeta)} \over \abs{\Psi(\zeta)} } \leq {1\over m(r)} \convergesto{r\to 1^-} 1 .\]
Automorphisms of $\CC$ (Tag E-XQ4BA)
Show that \(\Aut(\CC) = \ts{ z \mapsto az+b\st a\in \unitsof{\CC}, b\in \CC }\).
Solution
\(\supseteq\): Clear, every affine function is bijective with inverse \(g(z) \da \inverseof{a}(z-b)\), and holomorphic since \[ g(f(z)) = z \implies g'(f(z)) f'(z) = 1 \implies g'(w) = {1\over f'(\inverseof{f}(w))} .\]
\(\subseteq\): Suppose \(f\) is entire and bijective. If \(f\) is bounded, then \(f(z) = b\) is constant by Liouville and we’re done, so suppose not. Then \(\abs{f(z)}\to \infty\) as \(\abs{z}\to \infty\), making \(z=\infty\) either an essential singularity or a pole, since it is isolated and not removable since \(f\) is unbounded in every small enough neighborhood of \(\infty\). If \(f(b) = 0\) without loss of generality we can translate to assume \(f(0) = 0\) by replacing \(f(z)\) with \(f(z) - b\). By bijectivity, \(z=0\) must be a zero of order 1, so \({1\over f(z)}\) has a pole of order 1 at \(z=0\). So define \(G(z) \da {z\over f(z)}\), which is now an entire function.
On a large enough disc, \(\abs{f(z)} > 1\) for all \(\abs{z} > R\), so \(\abs{G(z)} < 1\inverseof{R}\) on \(\abs{z} > R\). Since \(G\) is holomorphic and thus continuous on \(\abs{z}\leq R\), which is compact, \(G\) is bounded here too. Thus \(G\) is constant by Liouville, and \(G(z) = c \implies f(z) = \inverseof{c} z \da az\). Unwinding the initial translation that ensured \(f(0)=0\), we get \(f(z) = az + b\).
Automorphisms of $\CP^1$ (Tag E-CFTRQ)
Show that \[ \Aut(\CP^1) =\left\{z\longmapsto {az+b\over cz+d}:a,b,c,d\in\CC,\ ad-bc\neq0\right\}. \]
Solution
Let \[ A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\GL_2(\CC). \] The linear automorphism \(A:\CC^2\to\CC^2\) descends to a holomorphic automorphism of projective space \[ [z_0:z_1]\longmapsto[az_0+bz_1:cz_0+dz_1]. \] On the affine chart \(z=[z:1]\), this is \[ z\longmapsto\frac{az+b}{cz+d}. \] Its inverse is induced by \(A^{-1}\), so every fractional linear transformation with \(ad-bc\neq0\) lies in \(\Aut(\CP^1)\).
Conversely, let \(f\in\Aut(\CP^1)\). By the classification of meromorphic functions on \(\CP^1\), write \[ f(z)=\frac{p(z)}{q(z)} \] with coprime polynomials \(p,q\), and put \[ n=\max\{\deg p,\deg q\}. \] For every finite value \(w\) except possibly the ratio of the leading coefficients when \(\deg p=\deg q=n\), the equation \[ f(z)=w \] is equivalent to \[ p(z)-wq(z)=0, \] a polynomial equation of degree \(n\). Hence it has \(n\) solutions counted with multiplicity.
Because \(f\) is an automorphism, its inverse is holomorphic. Therefore \(f\) has no critical points: in local coordinates, \[ (f^{-1})'(f(z))f'(z)=1. \] Thus every point in every fiber has multiplicity one. Since \(f\) is also injective, a generic fiber contains exactly one point. Consequently \(n=1\).
Therefore \(p(z)=az+b\) and \(q(z)=cz+d\) for some \(a,b,c,d\in\CC\). The determinant \(ad-bc\) is nonzero, since otherwise the two linear polynomials are proportional and \(f\) is constant. Hence \(f\) is fractional linear.