# Conformal branched cover from the hyperbolic plane to the euclidean plane

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.)

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.

• Is this the Order-8 square tiling? – PyRulez Apr 17 '18 at 16:54
• Also, how do we know there is conformal map from the hyperbolic square to the euclidean one? – PyRulez Apr 17 '18 at 18:32
• @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.) – Lukas Geyer Apr 17 '18 at 19:20
• doesn't the Riemann mapping theorem only apply to the interior. – PyRulez Apr 17 '18 at 19:21
• @PyRulez: Yes, except for the corners. This follows from the reflection principle for analytic functions. – Lukas Geyer Apr 17 '18 at 20:16

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.)

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