Start with a convex polygonal shape in the plane with an external angle of about $ \pi/16$ at one vertex. Remove two very thin spiral channels from the polygon that begin on adjacent edges of the polygon and spiral around one another, as in the left image of Fig. 4. If the spirals are sufficiently intertwined, then the spiral flaps on the developed naive cap will overlap. The right side of Fig. 4 shows the spirals reflected across the edges of the polygon.

Figure 4 Left : a piece of a convex polygon (lying above the red and green line segments) minus two narrow spiral channels (shown in orange and blue ) that begin from adjacent edges of the polygon. Each channel cut from the polygon is so narrow that we depict it as a curve. Right : a piece of its naive cap (again, above the red and green segments) with the two spiral flaps reflected outward, illustrating a non-planar Euclidean development (color figure online).

For the harmonic cap, it is possible to construct an example similar to that of Sect. 2.2. Indeed, very skinny channels removed from any planar shape will have negligible harmonic measure, and so we can arrange for overlapping spirals in the cap.

Figure 5 Left : two narrow spiral channels (shown in orange and blue ) cut from the interior of a square planar shape (a segment of which is shown in green ). Each channel cut from the polygon is so narrow that we depict it as a curve. Right : the two spirals on the exterior of the clover-shaped harmonic cap of the square, illustrating a non-planar Euclidean development. A complete and accurate picture of the harmonic cap of the square is shown in Fig. 10 (color figure online).

More precisely, begin with a square planar shape and choose a tiny $ \eps>0$. The harmonic cap for the square is shown in Fig. 10. Now remove two very skinny spiral channels from the square, emanating from a single edge, as in the left image of Fig. 5; the openings of each channel should have width smaller than $ \eps$. The openings of the two spiral channels can be placed at a specified distance apart from one another, so that the harmonic measure of the interval between them is approximately equal to $ 1/32$ of the total mass. (The number $ 1/32$ is chosen because it is $ 1/4\pi$ times the curvature of $ \pi/8$ for the polygon vertex shown in Fig. 4). We can choose $ \eps>0$ as small we wish so that the harmonic measure along the spiral boundaries is almost 0. Indeed, as the width of the spiral channels shrinks to 0, the domains $ \Chat{\setminus} P$ are converging in the Carath[U+00B4]eodory sense to the complement of the square; see, e.g., ([Duren1983], [U+00A7]3.1).

Recall that the boundary of the cap development is parameterized by the formula of Theorem 1.2. The parametrization of the spirals on the cap, which will lie outside the clover-like harmonic cap for the square, will be essentially equal to a reflection of their original parametrizations (because $ \kappa$ will be essentially constant along their boundaries, having chosen the harmonic measure of the spirals to be near 0). On the other hand, the non-trivial portion of harmonic measure on the boundary of the square between the spiral-channel openings will curve the boundary of the cap so the spirals overlap. The change in tangent direction of the clover cap between the two attaching points of the spirals will be $ \pi/8$, by construction. See Fig. 5.

Now assume that $ f: \Cak\to\Cak$ is a complex polynomial of degree $ d\geq 2$. Its filled Julia set is \begin{eqnarray*} K(f) = \left\{z \in \Cak: \sup_n |f^n(z)| < \infty\right\}. \end{eqnarray*} Assume that $ K(f)$ is connected, so it is a planar shape. A planar development of its cap is given by the formula of Theorem 1.1. We can parameterize the boundary of the cap's development for smooth or polygonal approximations to $ K(f)$ using Theorem 1.2.

The Green function for $ K(f)$ can be computed dynamically, as \begin{eqnarray*} G_f(z) = \lim_{n\to\infty} \frac{1}{d^n} \log^+|f^n(z)|. \end{eqnarray*} The harmonic measure $ \mu_f = \frac{1}{2\pi} \Delta G_f$ is the unique measure of maximal entropy for $ f$, and its support is equal to the Julia set $ J(f) = \del K(f)$ ([Brolin1965, Ljubich1983, Freire et al.1983]). The metric on $ \Chat$ is defined by \begin{eqnarray*} \rho_f = e^{-2G_f(z)}|dz| \end{eqnarray*} for $ z\in \Cak$, with curvature distribution $ \omega_f = - \Delta \log \rho_f(z) = 4\pi \mu_f$.

Figure 6 A Euclidean development of the cap for the real interval $ P = [-2,2]$ equipped with its harmonic measure. The figure shown is the image of the unit circle under $ g(z) = z - z^3/3$, so the cusps lie at $ z=\pm 2/3$. To form the metrized sphere, the cap is folded in half along the segment joining the cusp points, and the interval $ P$ forms the seam. The resulting convex body is degenerate. See Example 2.2 .

Figure 7 In blue , a polygonal approximation to the filled Julia set of $ f(z)=z^2+1/4$ with $ 2^{11}$ vertices, the preimages of $ z=0.5$ under $ f^{11}$. The approximation to harmonic measure puts equal weight on each of the $ 2^{11}$ vertices. In orange , the polygonal cap for this discrete curvature distribution. A single attaching point is shown in black . There is a unique realization of the glued shapes as the boundary of a convex body in $ \R^3$. Image generated with Mathematica (color figure online).

A convex polyhedron in $ \R^3$ is the intersection of finitely many closed halfspaces. It is said to be degenerate if it lies in a plane. When the polyhedron is non-degenerate and bounded, its boundary surface is topologically a sphere, and the Euclidean metric from $ \R^3$ induces an intrinsic path metric on the sphere. If the polyhedron is degenerate and bounded, but not contained in a line, we will still view its boundary as a topological sphere, doubling the planar polygon and gluing along the polygonal boundary.

Abstractly, a convex polyhedral metric on a 2-dimensional sphere is an intrinsic metric with non-negative curvature concentrated at finitely many points. In other words, in a small neighborhood of all but finitely many points, the surface is isometric to a region in $ \R^2$. In a neighborhood of each of the finitely many cone points , the surface is isometric to the point of a cone. The curvature of a cone point is equal to the angle deficit at the point; that is, if the circumference of any small circle of radius $ r$ centered at the cone point is equal to $ C(r)$, then the curvature is equal to $ (2\pi r - C(r))/r$. By the Gauss-Bonnet formula, the sum of the curvatures over all cone points on the sphere is equal to $ 4\pi$. [Alexandrov2005] examines the geometry of convex polyhedra in detail. He presents his proof from [Alexandroff1942] that any abstract polyhedral metric on a sphere is isometric to the boundary of a (possibly degenerate) convex polyhedron. Furthermore, the polyhedron in $ \R^3$ is unique, up to Euclidean isometries. Given a polyhedral metric on the sphere, and a simply-connected subset $ U$ of the sphere minus its cone points, a Euclidean development of $ U$ is a local isometry $ U \to \R^2$. Suppose we are given the image $ \mathcal{I}\subset \R^2$ of a Euclidean development of a full-area, simply-connected subset $ U$ of the sphere. Then, as a consequence of Alexandrov's theorem, the convex polyhedron in $ \R^3$ is uniquely determined by $ \mathcal{I}$ and the gluing along its boundary (that reconstructs the topological sphere). In particular, the planar development and the gluing information will uniquely determine the edges of the polyhedron and their dihedral angles in $ \R^3$---information that is not locally apparent.

Curvature is carefully treated by Alexandrov. It is defined by an additive set function $ \omega$ as follows. The curvature of a point is, as for a polyhedron, $ 2\pi$ minus the cone angle of the point. That is,

Reshetnyak, who was a student of Alexandrov, reformulated Alexandrov's theory of metrics and curvature on a surface in complex-analytic language, expressing curvature as a finite Borel measure ([Reshetnyak1993]). We exploit this useful point of view in Sect. 4.

Suppose that a planar shape $ P$ is the closure of a Jordan domain with a piecewise-differentiable boundary. Fix a nonnegative Borel measure $ \mu$ on the boundary of $ P$. Let $ L$ be the length of $ \del P$. Let $ s$ be a piecewise-differentiable parametrization by arclength of the boundary of $ P$, in the counterclockwise direction, and write \begin{eqnarray*} s'(t) = e^{i \alpha(t)} \end{eqnarray*} for a piecewise-continuous function $ \alpha: [0,L] \to \R$. For $ t \in [0, L]$, we define a curvature function $ \kappa: [0,L] \to [0, 4\pi]$ by $ \kappa(0) = 0$ and

Proof of Theorem 1.2.

If the boundary of the planar domain $ P$ and the measure $ \mu$ are smooth enough, then the curvature of Sect. 3.2 satisfies the following relation, as a consequence of Theorem 1.2.

It is interesting to compare the statement of Proposition 3.1 to the formula (3.1) for the Alexandrov curvature of a point, \begin{eqnarray*} \omega(\{x\}) = \lim_{r\to 0^+} \frac{2\pi r - C(x,r)}{r}, \end{eqnarray*} and to the Bertrand--Puiseux formula for the Gaussian curvature $ \kappa$ when the metric on a surface is smooth, \begin{eqnarray*} \kappa(x) = \; \lim_{r\to 0^+} \; 3 \, \frac{2\pi r - C(x,r)}{\pi r^3} \end{eqnarray*} ([Spivak1979], page 147). In our setting, the curvature of the surface is supported on a 1-dimensional curve, so the circumference discrepancy is proportional to $ r^2$.

We begin with a simple geometric lemma.

Proof.

Proof of Proposition 3.1.

A smooth conformal metric on a domain in $ \Cak$ can be expressed as \begin{eqnarray*} \rho(z) |dz| \end{eqnarray*} for a smooth and positive function $ \rho$. The metric has non-negative curvature if $ U(z) = -\log \rho(z)$ is a subharmonic function. Working with a more general class of metrics, we will only require that $ U$ be subharmonic, not necessarily differentiable or everywhere finite. We will also require that all pairs of points have finite distance from one another. These requirements can be formulated in terms of the curvature of the metric, as we explain below.

Formally, a conformal metric $ \rho$ on $ \Chat$ is a (singular) Hermitian metric on the tangent bundle $ T\Chat \simeq \mathcal{O}_{\P^1}(2)$, and the curvature form of the metric is the positive measure given in local coordinates by \begin{eqnarray*} \omega_\rho = - \Delta \log \rho \end{eqnarray*} (with $ \Delta = 2i \del \delbar$ taken in the sense of distributions), so that \begin{eqnarray*} \int_{\Chat} \omega_\rho = 4\pi. \end{eqnarray*} In more classical terms, for a smooth metric $ \rho$, the Gaussian curvature is computed locally as \begin{eqnarray*} \kappa_\rho = \frac{-\Delta \log \rho}{\rho^2}. \end{eqnarray*} See, for example, ([Ahlfors1973], [U+00A7]1.5) or ([Hubbard2006], [U+00A7]2.2).

That $ U = -\log\rho$ is subharmonic guarantees that the curvature form $ \omega_\rho \geq 0$ as a distribution. Finite diameter is guaranteed by the assumption that $ \omega_\rho(\{z_0\}) < 2\pi$ for all $ z_0 \in \Chat$ ([Reshetnyak1993], p. 100). Recall from Sect. 3.4 that concentrated curvature, at points $ z_0 \in\Chat$ where $ 0 < \omega_{\rho}(\{z_0\}) < 2\pi$, corresponds to cone points in the local geometry. Also in this setting, a computation shows that the circumference $ C(z_0, r)$ of a small circle around $ z_0$ of radius $ r>0$ will satisfy ([Reshetnyak1993], Lemma 8.1.1) \begin{eqnarray*} \lim_{r \to 0^+} \frac{2\pi r - C(z_0,r)}{r} = \omega_{\rho}(\{z_0\}). \end{eqnarray*}

Conversely, every probability measure $ \mu$ on $ \Chat$ with $ \mu(\{z\}) < 1/2$ for all $ z$ gives rise to a conformal metric of finite diameter with curvature distribution $ 4\pi \mu$, unique up to scale. Indeed, there is a one-to-one correspondence between probability measures $ \mu$ on $ \Chat$ and their potentials, up to an additive constant, which can be viewed as logarithmically-homogeneous, plurisubharmonic functions $ G_\mu$ on the tautological line bundle $ \Cak^2{\setminus} \{(0,0)\} \to \P^1$; see, e.g., ([Forn[U+00E6]ss and Sibony1993], Theorem 5.9) and ([DeMarco2003], Section 12). The function $ G_\mu$ will satisfy $ (2\pi)^{-1} \Delta G_\mu(z,1) = \mu$ in local coordinates $ z$ on $ \Chat$, and the conformal metric is expressed as \begin{eqnarray*} \rho_\mu = e^{-2 G_\mu(z,1)} |dz|. \end{eqnarray*} The identification between measures and their potentials is continuous, taking the $ L^1_{loc}$ topology on potentials and the weak topology on measures. Moreover, convergence of curvatures implies convergence of the metrics ([Reshetnyak1993], Theorem 7.3.1).

Let $ P$ be a compact, connected set in $ \Cak$ containing at least 2 points, so that $ P$ is a planar shape as defined in Sect. 1. Let $ G_P: \Cak\to \R$ be the Green function for $ P$; it is the unique continuous function on $ \Cak$ satisfying (1) $ G_P \equiv 0$ on $ P$, (2) $ G_P(z) = \log|z| + O(1)$ for $ z$ near $ \infty$, and (3) $ G_P$ is harmonic on $ \Cak{\setminus} P$. Then define a metric on $ \Cak$ by \begin{eqnarray*} \rho_P = e^{-2G_P(z)} |dz|. \end{eqnarray*} By elementary potential theory, the function $ G_P$ satisfies $ G_P(z) = \log(z) + \gamma + o(1)$ for $ z$ near $ \infty$ for some real number $ \gamma$, so the metric extends uniquely by continuity across $ z=\infty$. Note that this metric is flat (with 0 curvature) away from the boundary $ \del P$. Its curvature form $ \omega_P = 2 \Delta G_P$ is equal to ($ 4\pi$ times) the harmonic measure on $ \del P$ (more precisely, the harmonic measure for the domain $ \Chat{\setminus} P$, relative to the point $ \infty$).

Let $ P$ be any planar shape. Let $ \Phi$ be the Riemann map from the complement of the unit disk to the complement of $ P$, sending infinity to infinity. Consider the holomorphic 1-form \begin{eqnarray*} \eta = \frac{1}{(\Phi^{-1}(z))^2} \; dz \end{eqnarray*} on the complement of $ P$. Since the Green function satisfies \begin{eqnarray*} G_P(z) = \log|\Phi^{-1}(z)| \end{eqnarray*} on $ \Chat{\setminus} P$, we see that $ |\eta|$ is precisely the conformal metric $ \rho_P$ defined above, when restricted to the complement of $ P$. Recall that Theorem 1.1 asserts that a Euclidean development of the harmonic cap of $ P$ is given by the locally univalent function $ g: \D\to\Cak$ defined by \begin{eqnarray*} g(z) = \int_0^z \Phi'(1/x) \, dx. \end{eqnarray*} It also asserts that there exist examples where the locally univalent $ g$ fails to be univalent.

Proof of Theorem 1.1.

Here we present an alternative proof of the cap parametrization in Theorem 1.2, in the special setting of harmonic measure.

As in Theorem 1.2, assume that $ P$ has a piecewise-differentiable boundary which is a Jordan curve parameterized by arclength by $ s: [0,L] \to \Cak$. Recall that $ s^{\prime}(t) = e^{i \alpha(t)}$ for some piecewise continuous function $ \alpha:[0,L] \rightarrow \R$. Let $ \Phi$ be a Riemann map from the complement of the unit disk to the complement of $ P$, sending infinity to infinity. Then $ \Phi$ extends to a homeomorphism from the unit circle to the boundary of $ P$. Define the conformal angle $ \theta: [0,L] \rightarrow \R$ by \begin{eqnarray*} \theta(t) := \mathrm{arg}(\Phi^{-1}(s(t))). \end{eqnarray*} Without loss of generality, we may assume $ \theta(0)=0$ so that $ \theta$ defines a homeomorphism from $ [0,L]$ to $ [0,2\pi]$. It follows that the curvature function of (3.2) for the harmonic measure $ \mu$ on $ \del P$ is equal to \begin{eqnarray*} \kappa(t) = 4\pi \mu(s(0,t]) = 2\theta(t). \end{eqnarray*} Therefore, from Theorem 1.2, we know that the parametrization of the boundary of the harmonic cap is given by

Theorem 1.1 grants an alternate proof of (4.1). Indeed, with the $ g: \D \to \Cak$ of Theorem 1.1, a parametrization of the boundary of the harmonic cap is given by \begin{eqnarray*} \hat{s}(t) = -g(1/\Phi^{-1}(s(t))) = -g(e^{-i\theta(t)}). \end{eqnarray*} Moreover, the derivative of $ g$ is $ g'(z) = \Phi'(1/z)$, and therefore, \begin{eqnarray*} \hat{s}'(t) &=& -g'(1/\Phi^{-1}(s(t)))\frac{-(\Phi^{-1})'(s(t)) \; s'(t)}{\Phi^{-1}(s(t))^2} \\ &=& \frac{-\Phi'(\Phi^{-1}(s(t)))}{-\Phi'(\Phi^{-1}(s(t)))} \; \frac{s'(t)}{\Phi^{-1}(s(t))^2} \\ &=& e^{i(\alpha(t) - 2\theta(t))}. \end{eqnarray*}

We conclude by returning to our original problem about the existence of a metric $ \rho(P,\mu)$, for the case where $ P$ is a planar shape with Jordan curve boundary and the probability measure $ \mu$ is arbitrary.

Suppose that $ J$ is a Jordan curve in $ \Chat$, cutting the sphere into Jordan domains $ A$ and $ B$. We may assume that $ 0 \in A$ and $ \infty \in B$. Suppose that $ \nu$ is a probability measure supported on $ J$, and let \begin{eqnarray*} U(z) = \int_\Cak \log|z-w| \, d\nu(w) \end{eqnarray*} be a potential function for $ \nu$ with logarithmic singularity at $ \infty$. The conformal metric \begin{eqnarray*} e^{-2U(z)}|dz| \end{eqnarray*} on $ \Cak$ extends to $ \Chat$ and has curvature distribution equal to $ 4\pi\nu$. Since $ A$ is simply connected, there exists a non-vanishing analytic function $ \phi: A \to \Cak$ so that \begin{eqnarray*} U(z) = \log|\phi(z)|. \end{eqnarray*} The function $ \phi$ is determined uniquely, up to postcomposition by a rotation. Set \begin{eqnarray*} f_\nu(z) = \int_0^z \frac{dz}{\phi(z)^2} \end{eqnarray*} for $ z \in A$. Then $ f_\nu: A \to \Cak$ is a locally-univalent Euclidean development of $ A$ into the plane. It extends continuously to the boundary curve $ J$. This proves the following proposition.

When $ \mu$ is the harmonic measure on $ \del P$, observe that we may take $ J = \del P$ and $ \nu = \mu$ in the statement of Proposition 4.2. Indeed, the potential function for harmonic measure satisfies $ U \equiv 0$ on $ P$ so that $ f_\nu = \mathrm{Id}$.