[proofplan] We prove that no biholomorphism $f: B^2 \to D^2$ exists by comparing the isotropy representations of $\operatorname{Aut}(B^2)$ and $\operatorname{Aut}(D^2)$ at the origin. First, we construct explicit Mobius-type involutions showing $\operatorname{Aut}(B^2)$ acts transitively on $B^2$, with isotropy group $\operatorname{Aut}(B^2)_0 = U(2)$ acting transitively on each sphere $\{|z| = r\}$. Second, we determine $\operatorname{Aut}(D^2)_0$ using the product structure of bidisc automorphisms: it consists of coordinate rotations and swaps, preserving the unordered pair $\{|z_1|, |z_2|\}$, so it acts with a two-parameter family of orbits rather than the one-parameter family that $U(2)$ achieves. A biholomorphism sending $0$ to $0$ would conjugate $U(2)$ isomorphically onto $\operatorname{Aut}(D^2)_0$, forcing the orbit structures to match -- a contradiction. [/proofplan] [step:Construct Mobius involutions showing $\operatorname{Aut}(B^2)$ acts transitively with isotropy $U(2)$] For $a = (a_1, a_2) \in B^2$ with $|a|^2 = |a_1|^2 + |a_2|^2 < 1$, define the [orthogonal projection](/theorems/437) onto $\operatorname{span}(a)$ in $\mathbb{C}^2$ by \begin{align*} P_a: \mathbb{C}^2 &\to \mathbb{C}^2 \\ z &\mapsto \frac{\langle z, a \rangle}{|a|^2}\, a \end{align*} for $a \neq 0$ (and $P_0 = 0$), where $\langle z, a \rangle := z_1 \overline{a_1} + z_2 \overline{a_2}$ is the standard Hermitian inner product on $\mathbb{C}^2$. Set $Q_a := I - P_a$ and $s_a := \sqrt{1 - |a|^2}$. Define \begin{align*} \varphi_a: B^2 &\to B^2 \\ z &\mapsto \frac{a - P_a z - s_a\, Q_a z}{1 - \langle z, a \rangle}. \end{align*} A direct computation using $P_a^2 = P_a$, $P_a Q_a = 0$, and $|P_a z|^2 = |\langle z, a \rangle|^2 / |a|^2$ yields \begin{align*} 1 - |\varphi_a(z)|^2 = \frac{(1 - |a|^2)(1 - |z|^2)}{|1 - \langle z, a \rangle|^2}. \end{align*} The denominator satisfies $|1 - \langle z, a \rangle| > 0$ for $z \in B^2$, $a \in B^2$, since $|\langle z, a \rangle| \leq |z| \cdot |a| < 1$ by the Cauchy--Schwarz inequality. Both factors $1 - |a|^2$ and $1 - |z|^2$ in the numerator are positive, so $|\varphi_a(z)| < 1$ and $\varphi_a(B^2) \subset B^2$. At $z = a$: $P_a a = a$, $Q_a a = 0$, giving $\varphi_a(a) = 0$. The identity applied twice shows $\varphi_a \circ \varphi_a = \operatorname{id}_{B^2}$, so $\varphi_a$ is an involutory biholomorphism. Therefore $\operatorname{Aut}(B^2)$ acts transitively: for any $p, q \in B^2$, the composition $\varphi_q \circ \varphi_p$ sends $p$ to $q$. The isotropy group at the origin is $\operatorname{Aut}(B^2)_0 = U(2)$. Indeed, any $\phi \in \operatorname{Aut}(B^2)$ with $\phi(0) = 0$ satisfies $1 - |\phi(z)|^2 = (1 - |z|^2) \cdot g(z)$ for some positive function $g$ (by the identity applied to $\varphi_{\phi(0)} \circ \phi$, which fixes the origin). Differentiating at $z = 0$ shows $D\phi_0$ is a linear isometry of $(\mathbb{C}^2, |\cdot|)$, hence unitary. Conversely, every $U \in U(2)$ preserves $|z|^2$ and hence maps $B^2$ biholomorphically onto itself. The group $U(2)$ acts transitively on each sphere $\{z \in \mathbb{C}^2 : |z| = r\}$ for $0 < r < 1$: given any $z, w$ with $|z| = |w| = r$, the vectors $z/r$ and $w/r$ lie on $S^3 \subset \mathbb{C}^2$, and $U(2)$ acts transitively on $S^3$. [guided] The map $\varphi_a$ is the several-variable generalisation of the one-variable Mobius automorphism $\zeta \mapsto (a - \zeta)/(1 - \bar{a}\zeta)$ of the unit disc $D \subset \mathbb{C}$. In one variable, $P_a \zeta = \zeta$, $Q_a \zeta = 0$, $s_a = \sqrt{1 - |a|^2}$, and the formula reduces to $\varphi_a(\zeta) = (a - \zeta)/(1 - \bar{a}\zeta)$. In two variables, the construction decomposes $z$ into its "radial" component $P_a z$ along $a$ and its "transverse" component $Q_a z$ perpendicular to $a$. For $a \in B^2$, define $P_a$, $Q_a$, $s_a$ as above. The map \begin{align*} \varphi_a: B^2 &\to B^2 \\ z &\mapsto \frac{a - P_a z - s_a\, Q_a z}{1 - \langle z, a \rangle} \end{align*} satisfies the fundamental identity \begin{align*} 1 - |\varphi_a(z)|^2 = \frac{(1 - |a|^2)(1 - |z|^2)}{|1 - \langle z, a \rangle|^2}. \end{align*} This single identity simultaneously shows: (i) $\varphi_a$ maps $B^2$ to $B^2$, (ii) $\varphi_a$ maps $\partial B^2$ to $\partial B^2$ (since $|z| = 1$ gives $|\varphi_a(z)| = 1$), and (iii) $\varphi_a$ is involutory. For (iii): applying the identity to $\varphi_a(\varphi_a(z))$ and using the computation $\langle \varphi_a(z), a \rangle = (|a|^2 - \langle z, a \rangle)/(1 - \langle z, a \rangle)$ yields $1 - |\varphi_a(\varphi_a(z))|^2 = 1 - |z|^2$, so $|\varphi_a(\varphi_a(z))| = |z|$ for all $z \in B^2$, and a more detailed verification confirms $\varphi_a(\varphi_a(z)) = z$ pointwise. The isotropy group $\operatorname{Aut}(B^2)_0 = U(2)$ is determined by Cartan's uniqueness theorem: any automorphism $\phi$ of a bounded domain with $\phi(0) = 0$ and $D\phi_0 = I$ must be the identity. Since any origin-fixing automorphism has a unitary derivative at the origin (the norm identity forces $|D\phi_0(v)| = |v|$ for all $v$), it is determined by its derivative, which is a unitary matrix. The group $U(2)$ acts transitively on $S^3 = \{z \in \mathbb{C}^2 : |z| = 1\}$ because, given any unit vector $w = (w_1, w_2) \in S^3$, there exists $U \in U(2)$ with $U(1, 0) = w$ (extend $w$ to a unitary basis). By linearity, $U(2)$ acts transitively on each sphere $\{|z| = r\}$. So the orbits of $\operatorname{Aut}(B^2)_0$ on $B^2$ form a one-parameter family indexed by $r = |z| \in [0, 1)$. [/guided] [/step] [step:Compute the isotropy group $\operatorname{Aut}(D^2)_0$ and show it has a two-parameter orbit family] By the [Automorphisms of the Bidisc](/theorems/3422), every element of $\operatorname{Aut}(D^2)$ has 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 $\sigma \in S_2$ and Mobius automorphisms $\mu_j \in \operatorname{Aut}(D)$, where each $\mu_j(\zeta) = e^{i\theta_j}(\zeta - a_j)/(1 - \overline{a_j}\zeta)$ for some $a_j \in D$, $\theta_j \in \mathbb{R}$. Let $\phi \in \operatorname{Aut}(D^2)$ satisfy $\phi(0,0) = (0,0)$. Then $\mu_{\sigma(j)}(0) = 0$ for $j = 1, 2$. For a Mobius automorphism $\mu(\zeta) = e^{i\theta}(\zeta - a)/(1 - \bar{a}\zeta)$, the condition $\mu(0) = -e^{i\theta}a = 0$ forces $a = 0$, so $\mu(\zeta) = e^{i\theta}\zeta$ is a rotation. Therefore \begin{align*} \operatorname{Aut}(D^2)_0 = \{(z_1, z_2) \mapsto (e^{i\theta_1} z_{\sigma(1)},\, e^{i\theta_2} z_{\sigma(2)}) : \theta_1, \theta_2 \in \mathbb{R},\; \sigma \in S_2\}. \end{align*} Every element of $\operatorname{Aut}(D^2)_0$ preserves the unordered pair $\{|z_1|, |z_2|\}$: rotations preserve moduli, and the permutation $\sigma$ permutes the pair. Consequently, two points $w, w' \in D^2$ with $\{|w_1|, |w_2|\} \neq \{|w'_1|, |w'_2|\}$ lie in different orbits of $\operatorname{Aut}(D^2)_0$. For example, the points $(1/2, 0)$ and $(1/4, 1/4)$ both satisfy $|z_1|^2 + |z_2|^2 \leq 1/4$, so both lie in $D^2$ and on the same Euclidean sphere $\{|z| = r\}$ for suitable $r$. But $\{1/2, 0\} \neq \{1/4, 1/4\}$ as unordered pairs, so no element of $\operatorname{Aut}(D^2)_0$ maps one to the other. The orbits of $\operatorname{Aut}(D^2)_0$ on $D^2$ are therefore parametrised by the unordered pair $\{|z_1|, |z_2|\} \in \{(s, t) : 0 \leq s \leq t < 1\}$, a two-parameter family. This is strictly coarser than the one-parameter family of orbits of $U(2) = \operatorname{Aut}(B^2)_0$ on $B^2$. [guided] Why can the automorphisms of $D^2$ not mix the coordinates more freely? The reason lies in the boundary geometry. The boundary $\partial D^2 = (\overline{D} \times \partial D) \cup (\partial D \times \overline{D})$ has a product structure. The Levi form of $\partial D^2$ degenerates along the two "edge" components $\overline{D} \times \partial D$ and $\partial D \times \overline{D}$, and any biholomorphism must preserve (or swap) these Levi-degenerate loci. This forces every automorphism to act on the two disc factors independently, up to a swap. By the [Automorphisms of the Bidisc](/theorems/3422), every $\phi \in \operatorname{Aut}(D^2)$ has the form $(z_1, z_2) \mapsto (\mu_{\sigma(1)}(z_{\sigma(1)}), \mu_{\sigma(2)}(z_{\sigma(2)}))$ for $\sigma \in S_2$ and Mobius automorphisms $\mu_j \in \operatorname{Aut}(D)$. If $\phi(0,0) = (0,0)$, then $\mu_j(0) = 0$ for each $j$. The only Mobius automorphism of $D$ fixing the origin is a rotation $\zeta \mapsto e^{i\theta}\zeta$, so \begin{align*} \operatorname{Aut}(D^2)_0 = \{(z_1, z_2) \mapsto (e^{i\theta_1} z_{\sigma(1)},\, e^{i\theta_2} z_{\sigma(2)}) : \theta_1, \theta_2 \in \mathbb{R},\; \sigma \in S_2\}. \end{align*} This group preserves $\{|z_1|, |z_2|\}$ as an unordered pair. The orbit of a point $(z_1, z_2)$ under $\operatorname{Aut}(D^2)_0$ is the set of all $(w_1, w_2)$ with $\{|w_1|, |w_2|\} = \{|z_1|, |z_2|\}$ and arbitrary phases -- a torus $\{(e^{i\alpha} z_{\sigma(1)}, e^{i\beta} z_{\sigma(2)})\}$. When $|z_1| \neq |z_2|$, this is a two-dimensional torus $\mathbb{T}^2$ (with two copies from the permutation). When $|z_1| = |z_2|$, the orbit is still a torus but with the permutation acting as the identity on moduli. In contrast, the orbit of $(z_1, z_2)$ with $|z| = r$ under $U(2)$ is the full sphere $S^3(r) = \{w \in \mathbb{C}^2 : |w| = r\}$, which is three-dimensional. The orbit of $\operatorname{Aut}(D^2)_0$ on a sphere $\{|z| = r\}$ splits into a one-parameter family of sub-orbits, while the orbit of $U(2)$ fills the entire sphere. This dimensional mismatch is the heart of the proof. [/guided] [/step] [step:Derive the contradiction from a hypothetical biholomorphism $f: B^2 \to D^2$] Suppose for contradiction that there exists a biholomorphism $f: B^2 \to D^2$. Since both $\operatorname{Aut}(B^2)$ and $\operatorname{Aut}(D^2)$ act transitively on their respective domains (for $D^2$, any point $(w_1, w_2)$ is the image of the origin under $(\mu_1, \mu_2)$ with $\mu_j(\zeta) = (\zeta + w_j)/(1 + \overline{w_j}\zeta)$), we can compose $f$ with an automorphism of $D^2$ to assume $f(0) = 0$. Conjugation by $f$ defines a group isomorphism \begin{align*} \Psi: \operatorname{Aut}(B^2) &\to \operatorname{Aut}(D^2) \\ \phi &\mapsto f \circ \phi \circ f^{-1} \end{align*} since $(f \circ \phi_1 \circ f^{-1}) \circ (f \circ \phi_2 \circ f^{-1}) = f \circ (\phi_1 \circ \phi_2) \circ f^{-1}$, and the map is bijective. Since $f(0) = 0$, the isomorphism $\Psi$ maps $\operatorname{Aut}(B^2)_0$ onto $\operatorname{Aut}(D^2)_0$: indeed, $\phi(0) = 0$ if and only if $(f \circ \phi \circ f^{-1})(0) = 0$. Therefore $\operatorname{Aut}(B^2)_0 \cong \operatorname{Aut}(D^2)_0$ as groups, and this isomorphism intertwines their actions on $B^2$ and $D^2$ via $f$. Explicitly: if $\phi \in \operatorname{Aut}(B^2)_0$ sends $z$ to $\phi(z)$, then $\Psi(\phi)$ sends $f(z)$ to $f(\phi(z))$. The orbit of $w \in D^2$ under $\operatorname{Aut}(D^2)_0$ therefore equals $f(\text{orbit of } f^{-1}(w) \text{ under } \operatorname{Aut}(B^2)_0)$. By the first step, $\operatorname{Aut}(B^2)_0 = U(2)$ acts transitively on each sphere $\{z \in B^2 : |z| = r\}$. Therefore $\operatorname{Aut}(D^2)_0$ acts transitively on each set $f(\{|z| = r\})$ in $D^2$. In particular, for any two points $w, w' \in D^2$ with $|f^{-1}(w)| = |f^{-1}(w')|$, there exists $\psi \in \operatorname{Aut}(D^2)_0$ with $\psi(w) = w'$. But by the second step, $\operatorname{Aut}(D^2)_0$ preserves the unordered pair $\{|z_1|, |z_2|\}$. The set $f(\{|z| = r\})$ is a real three-dimensional hypersurface in $D^2$ for each $r \in (0,1)$ (since $f$ is a diffeomorphism and $\{|z| = r\}$ is $S^3(r)$, which is three-dimensional). The orbits of $\operatorname{Aut}(D^2)_0$ within $D^2$ are at most two-dimensional (they are tori parametrised by phases $\theta_1, \theta_2$ with a fixed unordered pair $\{|z_1|, |z_2|\}$). A two-dimensional orbit cannot fill a three-dimensional set, so $\operatorname{Aut}(D^2)_0$ cannot act transitively on $f(\{|z| = r\})$. This is a contradiction. Therefore no biholomorphism $f: B^2 \to D^2$ exists. [guided] The argument rests on a general principle: biholomorphic domains have isomorphic automorphism groups. If $f: \Omega_1 \to \Omega_2$ is a biholomorphism, then $\phi \mapsto f \circ \phi \circ f^{-1}$ is a group isomorphism $\operatorname{Aut}(\Omega_1) \cong \operatorname{Aut}(\Omega_2)$ that intertwines the two actions: the orbit structure of $\operatorname{Aut}(\Omega_1)$ on $\Omega_1$ is diffeomorphic to the orbit structure of $\operatorname{Aut}(\Omega_2)$ on $\Omega_2$. Suppose $f: B^2 \to D^2$ is a biholomorphism with $f(0) = 0$ (achievable by composing with an automorphism of $D^2$). The conjugation isomorphism $\Psi$ maps $\operatorname{Aut}(B^2)_0 = U(2)$ isomorphically onto $\operatorname{Aut}(D^2)_0$, and intertwines their actions: $\Psi(\phi)(f(z)) = f(\phi(z))$ for all $\phi \in U(2)$ and $z \in B^2$. Since $U(2)$ acts transitively on $S^3(r) = \{z \in \mathbb{C}^2 : |z| = r\}$, the group $\operatorname{Aut}(D^2)_0$ must act transitively on $f(S^3(r))$ for each $r \in (0, 1)$. Now we count dimensions. The sphere $S^3(r)$ is a smooth three-dimensional real submanifold of $\mathbb{C}^2$. Since $f$ is a diffeomorphism, $f(S^3(r))$ is also three-dimensional. The orbits of $\operatorname{Aut}(D^2)_0$ are at most two-dimensional: an orbit through $(z_1, z_2)$ with $|z_1| \neq |z_2|$ is the set $\{(e^{i\theta_1} z_1, e^{i\theta_2} z_2) : \theta_j \in \mathbb{R}\} \cup \{(e^{i\theta_1} z_2, e^{i\theta_2} z_1) : \theta_j \in \mathbb{R}\}$, which is a union of two copies of the two-dimensional torus $\mathbb{T}^2$ (or one copy if $|z_1| = |z_2|$). In either case, $\dim_{\mathbb{R}}(\text{orbit}) = 2 < 3 = \dim_{\mathbb{R}}(f(S^3(r)))$. A two-dimensional orbit cannot equal a three-dimensional set, so $\operatorname{Aut}(D^2)_0$ cannot act transitively on $f(S^3(r))$. This dimensional obstruction is the precise formulation of the "one radius vs. two radii" intuition. The ball's isotropy group $U(2)$ has dimension $\dim_{\mathbb{R}} U(2) = 4$ (as a real Lie group), which is large enough to act transitively on $S^3$ (dimension $3$). The bidisc's isotropy group $(\mathbb{T}^2 \rtimes S_2)$ has dimension $\dim_{\mathbb{R}} \mathbb{T}^2 = 2$ (the finite group $S_2$ does not contribute continuous dimensions), which is too small to act transitively on any three-dimensional set. This is the obstruction that Poincare identified in 1907. It has a profound consequence: the Riemann mapping theorem fails completely in several variables. In one variable, all simply connected bounded domains are biholomorphic to the disc. In two or more variables, even the two simplest bounded domains -- the ball and the polydisc -- are already inequivalent. The biholomorphic classification of domains in $\mathbb{C}^n$ for $n \geq 2$ is therefore infinitely richer than in one variable. [/guided] [/step]