# How to tell if a simply-connected curve is the complex plane or disk

Suppose I have an analytic function $f: \mathbb{C}^2 \to \mathbb{C}$ whose zero locus $V=\{(z,w) \in \mathbb{C}^2 : f(z,w)=0\}$ is smooth and simply connected. By uniformization, $V$ is conformally equivalent to either $\mathbb{C}$ or the open unit disk $\Delta \subset \mathbb{C}$. But which one? Is this "easily" determined by some properties of the function $f$?

• Do you know any examples where it's the upper half-plane? I have a suspicion that it might always be $\mathbb{C}$. – Qiaochu Yuan Jan 16 '18 at 7:19
• @QiaochuYuan There do exist proper holomorphic embeddings of the open unit disk into $\mathbb{C}^2$ (e.g. arxiv.org/abs/0808.0791). And I'm under the impression that every smooth, codimension one holomorphic submanifold of $\mathbb{C}^2$ is the zero locus of a holomorphic function. Is this off? – Kyle Jan 17 '18 at 13:14
• Did you check Ahlfors and Sario "Riemann Surfaces"? They spend considerable effort on discussing the problem of type for Riemann surfaces, check if they consider the case of complex curves given as subvarieties in $C^2$. – Moishe Kohan Jan 17 '18 at 16:50
• As for your question if a complex curve in $C^2$ can be defined via a single equation, the answer is positive, since $H^2(C^2, {\mathbb Z})=0$, see math.stackexchange.com/questions/1237192/…. – Moishe Kohan Jan 17 '18 at 20:51
• Here is an easy example of a holomorphic embedding of the unit disc into $\mathbb{C}^2$: take a complex torus $T$ with $\dim_{\mathbb{C}}(T) = 2$ and any holomorphic curve curve $C \subset T$ of genus $g >2$ (for example, one might take $g = 2$ and $T$ to be the Jacobian of $T$). The holomorphic map $C \to T$ induces a holomorphic map of universal coverings: $\tilde{C} = \Delta \to \mathbb{C}^2$ –  V. Rogov Feb 7 at 17:13