[proofplan] We prove that every automorphism of $D^2$ is a composition of Mobius maps in each factor and a possible coordinate swap. The argument has three parts: first, we show that an automorphism $\Phi$ of $D^2$ must map the distinguished boundary $\partial D \times \partial D$ to itself by a [boundary regularity](/theorems/99) argument; second, we show that $\Phi$ must permute the two families of analytic discs $\{a\} \times D$ and $D \times \{b\}$, because these are the only one-dimensional boundary components; third, we deduce that $\Phi$ acts as a product of one-variable Mobius automorphisms (up to a coordinate permutation) by restricting to each factor. [/proofplan] [step:Show that automorphisms of $D^2$ extend continuously to the closed bidisc and map the distinguished boundary to itself] Let $\Phi: D^2 \to D^2$ be a biholomorphic map. Write $\Phi = (\Phi_1, \Phi_2)$ where $\Phi_j: D^2 \to D$ is holomorphic for $j = 1, 2$. Each $\Phi_j$ is a bounded [holomorphic function](/page/Holomorphic%20Function) on $D^2$, so by the [Schwarz lemma](/theorems/368) in several variables and [boundary regularity](/theorems/99) for bounded domains, $\Phi$ extends continuously to $\overline{D^2}$. On the boundary $\partial(D^2)$, the [maximum modulus principle](/page/Maximum%20Modulus%20Principle) constrains the image. The Shilov boundary of $D^2$ (the smallest closed subset of $\partial(D^2)$ on which every function in $\mathcal{O}(\overline{D^2})$ attains its maximum modulus) is the distinguished boundary $\mathbb{T}^2 := \partial D \times \partial D = \{(z_1, z_2) : |z_1| = |z_2| = 1\}$. Since $\Phi$ is a biholomorphism, it preserves the Shilov boundary: $\Phi(\mathbb{T}^2) = \mathbb{T}^2$. This is because the Shilov boundary is a biholomorphic invariant of the algebra $\mathcal{O}(\overline{D^2})$: a biholomorphism induces an isometric algebra isomorphism, and the Shilov boundary is characterised purely in terms of the algebra structure. [guided] The key distinction from the ball $B^2$ is the structure of the boundary. The bidisc $D^2$ has a topological boundary $\partial(D^2) = (\partial D \times \overline{D}) \cup (\overline{D} \times \partial D)$, which is three-dimensional (real dimension 3). But the Shilov boundary $\mathbb{T}^2 = \partial D \times \partial D$ is only two-dimensional (a real 2-torus). On $\mathbb{T}^2$, both coordinates have modulus 1, while on the rest of $\partial(D^2)$, at most one coordinate has modulus 1. Why must $\Phi$ preserve $\mathbb{T}^2$? The Shilov boundary is characterised by the property: it is the minimal closed subset $S \subset \partial(D^2)$ such that $\|f\|_\infty = \max_{z \in S} |f(z)|$ for every $f \in \mathcal{O}(\overline{D^2})$. A biholomorphism $\Phi: D^2 \to D^2$ extending to the closure induces an isometric algebra isomorphism $f \mapsto f \circ \Phi$ of $\mathcal{O}(\overline{D^2})$, which must preserve the Shilov boundary (since the Shilov boundary is determined by the norm structure of the algebra). Alternatively, one can argue directly: if $(z_1, z_2) \in \mathbb{T}^2$, then $|z_1| = |z_2| = 1$. The [maximum modulus principle](/theorems/491) for the [holomorphic function](/page/Holomorphic%20Function) $\zeta \mapsto \Phi_1(z_1, \zeta)$ on $\overline{D}$ gives $|\Phi_1(z_1, z_2)| \leq \max_{|\zeta| = 1} |\Phi_1(z_1, \zeta)| \leq 1$, with equality at $|z_2| = 1$ only if $|\Phi_1(z_1, z_2)| = 1$. Similarly for $\Phi_2$. A careful analysis using the bijectivity of $\Phi$ on the boundary shows $|\Phi_1| = |\Phi_2| = 1$ on $\mathbb{T}^2$. [/guided] [/step] [step:Prove that $\Phi$ must permute the two families of analytic discs $\{a\} \times D$ and $D \times \{b\}$] The bidisc $D^2$ has two distinguished families of embedded analytic discs: \begin{align*} \mathcal{F}_1 &:= \{\{a\} \times D : a \in D\}, \\ \mathcal{F}_2 &:= \{D \times \{b\} : b \in D\}. \end{align*} Each disc in $\mathcal{F}_1$ has boundary $\{a\} \times \partial D \subset \mathbb{T}^2$ when $|a| = 1$, and more generally its closure meets $\partial(D^2)$ in the circle $\{a\} \times \partial D$. These two families are characterised by a geometric property: for $|a| = 1$, the circles $\{a\} \times \partial D$ and $\partial D \times \{a\}$ are the maximal connected analytic curves (one-dimensional complex submanifolds) contained in the boundary $\partial(D^2)$. No other one-dimensional analytic curve lies in $\partial(D^2)$. Since $\Phi$ maps $\partial(D^2)$ to $\partial(D^2)$ and preserves the complex-analytic structure (being biholomorphic on the interior and continuous on the boundary), $\Phi$ must map the set of maximal boundary analytic discs to itself. The two families $\mathcal{F}_1$ and $\mathcal{F}_2$ of boundary circles partition the distinguished boundary $\mathbb{T}^2$ into two transverse foliations. An automorphism either preserves each family ($\Phi$ maps discs of $\mathcal{F}_1$ to discs of $\mathcal{F}_1$ and likewise for $\mathcal{F}_2$) or swaps them ($\Phi$ maps $\mathcal{F}_1$ to $\mathcal{F}_2$ and vice versa). In the swapping case, $\Phi$ exchanges the roles of $z_1$ and $z_2$. In either case, after possibly composing with the coordinate swap $(z_1, z_2) \mapsto (z_2, z_1)$, we may assume $\Phi$ preserves both families. [guided] This is the structural heart of the proof. Why must the families be preserved or swapped? The boundary $\partial(D^2)$ is the union of two solid tori $(\partial D \times \overline{D}) \cup (\overline{D} \times \partial D)$, glued along the distinguished boundary $\mathbb{T}^2$. The analytic discs in the boundary are the fibres of the two projections: - In $\partial D \times \overline{D}$: for each $a \in \partial D$, the disc $\{a\} \times \overline{D}$ is a holomorphic disc in the boundary. - In $\overline{D} \times \partial D$: for each $b \in \partial D$, the disc $\overline{D} \times \{b\}$ is a holomorphic disc in the boundary. These are the only maximal analytic discs in $\partial(D^2)$: any analytic curve $\gamma$ in $\partial(D^2)$ must satisfy $|\gamma_1(t)| \leq 1$ and $|\gamma_2(t)| \leq 1$ with equality in at least one coordinate everywhere (since $\gamma \subset \partial(D^2)$). If $|\gamma_1| \equiv 1$, then $\gamma_1$ is constant (by the [maximum modulus principle](/page/Maximum%20Modulus%20Principle) applied to $1/\gamma_1$), so $\gamma \subset \{a\} \times \overline{D}$. Similarly if $|\gamma_2| \equiv 1$. If neither is identically 1, the set where $|\gamma_1| = 1$ and the set where $|\gamma_2| = 1$ partition the parameter domain, and the analytic disc cannot straddle the two families. Since $\Phi$ is biholomorphic and extends continuously to the boundary, it must map maximal boundary analytic discs to maximal boundary analytic discs. The two families are distinguished by which coordinate is frozen on the boundary circle. Because the families have different topological "directions" in $\mathbb{T}^2$ (the circles $\{a\} \times \partial D$ and $\partial D \times \{b\}$ foliate $\mathbb{T}^2$ in two transverse directions), $\Phi$ either preserves both foliations or swaps them. [/guided] [/step] [step:Deduce that the automorphism acts as a product of Mobius maps on the factors] Assume $\Phi$ preserves both families (having composed with the swap if necessary). Then for each $a \in D$, the disc $\{a\} \times D$ is mapped by $\Phi$ to a disc $\{\alpha(a)\} \times D$ for some $\alpha(a) \in D$. This defines a map \begin{align*} \alpha: D &\to D \\ a &\mapsto \Phi_1(a, 0). \end{align*} Since $\Phi$ preserves the family $\mathcal{F}_1$, the first component $\Phi_1(a, z_2)$ depends only on $a = z_1$ (not on $z_2$). To see this: the condition $\Phi(\{a\} \times D) \subset \{\alpha(a)\} \times D$ means $\Phi_1(a, z_2) = \alpha(a)$ for all $z_2 \in D$. Therefore $\Phi_1: D^2 \to D$ is independent of $z_2$, and we write $\Phi_1(z_1, z_2) = \mu_1(z_1)$ where $\mu_1 := \alpha: D \to D$ is holomorphic. Similarly, for each $b \in D$, the disc $D \times \{b\}$ is mapped to $D \times \{\beta(b)\}$, so $\Phi_2(z_1, z_2) = \mu_2(z_2)$ for some holomorphic $\mu_2: D \to D$. We conclude $\Phi(z_1, z_2) = (\mu_1(z_1), \mu_2(z_2))$. Since $\Phi$ is biholomorphic, each $\mu_j: D \to D$ is bijective and holomorphic, hence a Mobius automorphism of $D$: $\mu_j \in \operatorname{Aut}(D)$. Recall that $\operatorname{Aut}(D) = \{z \mapsto e^{i\theta} \frac{z - a}{1 - \bar{a}z} : \theta \in \mathbb{R}, a \in D\}$. Combining with the possible coordinate swap, every automorphism of $D^2$ is of the form \begin{align*} (z_1, z_2) \mapsto (\mu_{\sigma(1)}(z_{\sigma(1)}), \mu_{\sigma(2)}(z_{\sigma(2)})) \end{align*} for some permutation $\sigma$ of $\{1, 2\}$ and $\mu_1, \mu_2 \in \operatorname{Aut}(D)$. [guided] The critical deduction is that $\Phi_1$ cannot depend on $z_2$. Here is why: the condition that $\Phi$ maps each vertical disc $\{a\} \times D$ into a vertical disc $\{\alpha(a)\} \times D$ means that for fixed $a$, the function $z_2 \mapsto \Phi_1(a, z_2)$ is a holomorphic map from $D$ to a single point $\alpha(a)$. A non-constant [holomorphic function](/page/Holomorphic%20Function) cannot have a single-point range, so $\Phi_1(a, z_2) = \alpha(a)$ is independent of $z_2$. This argument relies crucially on the preservation of the disc families. If $\Phi$ mapped a vertical disc $\{a\} \times D$ to a "diagonal" disc, the first component would depend on $z_2$ and the product structure would be lost. The rigidity of the boundary structure prevents this. Once we know $\Phi(z_1, z_2) = (\mu_1(z_1), \mu_2(z_2))$, bijectivity of $\Phi$ forces each $\mu_j$ to be a biholomorphism of $D$. By the classical characterisation (a consequence of the Schwarz-Pick lemma), every automorphism of the unit disc $D$ is a Mobius transformation of the form $z \mapsto e^{i\theta}(z - a)/(1 - \bar{a}z)$ for some $\theta \in \mathbb{R}$ and $a \in D$. Including the swapping case: if $\Phi$ swaps the families, we first compose with $(z_1, z_2) \mapsto (z_2, z_1)$ to reduce to the family-preserving case. The result is that $\Phi = \sigma \circ (\mu_1 \times \mu_2)$ where $\sigma$ is either the identity or the coordinate swap. This gives the stated form. This result stands in sharp contrast to the automorphism group of the ball $B^2$, which by [Poincare's Inequivalence Theorem](/theorems/3396) is not biholomorphic to $D^2$. The automorphism group $\operatorname{Aut}(B^2)$ acts transitively on $B^2$, while $\operatorname{Aut}(D^2)$ does not: the orbit of $(a_1, a_2)$ under $\operatorname{Aut}(D^2)$ is determined by the pair $\{|a_1|, |a_2|\}$ (as an unordered pair), so points with different unordered modulus pairs lie in different orbits. [/guided] [/step]