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The euclidean plane can conformally branch cover the sphere. This is witnessed by the tiled Peirce quincuncial projection.

Is there likewise a conformal branched covering map from the hyperbolic plane to the euclidean plane?

When I say conformal branch covering map, I mean a branch covering map that is conformal everywhere except at the branching points. (Ideally, each branch point should have degree two.)

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Yes, there is. Here is a geometric construction (which could easily be visualized if one wants to do so): Let $S$ be a square (i.e., a quadrilateral with 4 equal sides and 4 equal angles) in the hyperbolic plane with angles of $\pi/4$, and let $f:S \to [0,1]^2$ be the conformal map of $S$ to the unit square, mapping vertices to vertices. Now the map $f$ can be extended by reflection to a map of the hyperbolic plane onto the Euclidean plane. The resulting map will have branch points of degree two over all points in the integer lattice, since it is angle-doubling at those points.

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  • $\begingroup$ Is this the Order-8 square tiling? $\endgroup$ – PyRulez Apr 17 '18 at 16:54
  • $\begingroup$ Also, how do we know there is conformal map from the hyperbolic square to the euclidean one? $\endgroup$ – PyRulez Apr 17 '18 at 18:32
  • $\begingroup$ @PyRulez: Yes, that is the tiling. (Didn't know it had a name and a Wikipedia page...) We know that there is a conformal map from the Riemann mapping theorem (+ some boundary regularity and symmetry considerations.) $\endgroup$ – Lukas Geyer Apr 17 '18 at 19:20
  • $\begingroup$ doesn't the Riemann mapping theorem only apply to the interior. $\endgroup$ – PyRulez Apr 17 '18 at 19:21
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    $\begingroup$ @PyRulez: Yes, except for the corners. This follows from the reflection principle for analytic functions. $\endgroup$ – Lukas Geyer Apr 17 '18 at 20:16
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Sure. Here's a quick way to see this using some high-powered machinery. First, note that if $X$ and $Y$ are two smooth projective algebraic curves (let's say over $\mathbb{C}$ since that's what we care about but it doesn't really matter for this part), then there is a smooth projective curve $Z$ which is a branched cover of both $X$ and $Y$. Indeed, if $\mathbb{C}(X)$ is the function field of $X$ and $\mathbb{C}(Y)$ is the function field of $Y$, then both of these fields can be viewed as finite extensions of $\mathbb{C}(t)$. Letting $K$ be a compositum of these finite extensions, then $K$ is the function field of some smooth projective curve $Z$, and the embeddings of $\mathbb{C}(X)$ and $\mathbb{C}(Y)$ into $K$ make $Z$ a branched cover of $X$ and $Y$.

Now consider the case where $X$ is an elliptic curve and $Y$ has genus greater than $1$. Then we obtain a branched covering $p:Z\to X$ where $Z$ is also a branched cover of $Y$. By the Riemann-Hurwitz formula, $Z$ must also have genus greater than $1$.

But now considering these curves as compact Riemann surfaces, our elliptic curve $X$ has $\mathbb{C}$ (i.e., the Euclidean plane) as its universal cover and the curve $Z$ of genus greater than $1$ has $\mathbb{H}$ (i.e., the hyperbolic plane) as its universal cover. Lifting the map $p:Z\to X$ to the universal covers, we get a holomorphic (and therefore conformal) branched covering $\mathbb{H}\to\mathbb{C}$.

(Note that the Peirce quinuncial projection can be obtained in exactly the same way, by taking a branched cover of the Riemann sphere by an elliptic curve.)

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  • $\begingroup$ Good answer. Do you have an image of what it looks like, by any chance? $\endgroup$ – PyRulez Apr 17 '18 at 18:06

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