# constant-curvature Riemannian metric for Bring's surface

There is a well-known and very symmetric space that is called either "Bring's curve" or "Bring's surface", depending upon the context. (Bring was a Swedish mathematician in the 18th century.) Let's here call it simply $B$. In brief, I am looking for an explicit formula for a constant-curvature Riemannian metric on $B$.

If we treat quintuples of complex numbers $[z_0,...,z_4]$ as the homogeneous coordinates of points in complex projective 4-space, then $B$ consists of all solutions of the three simultaneous equations $\sum_i z_i=0$, $\sum_i z_i^2=0$, and $\sum_i z_i^3=0$. The first of those equations is linear, cutting out a projective 3-space. The second equation gives a quadric surface in that 3-space, while the third gives a cubic surface. Bring's curve $B$ is the complete intersection of those two surfaces, which is a nonsingular sextic curve in complex projective 3-space. So $B$ is a complex $1$-manifold. Since we can permute the five coordinates freely, $B$ has at least an $S_5$ of automorphisms -- in fact, it has exactly those 120 automorphisms. We can also take the complex conjugates of all five coordinates; so there are also 120 antiautomorphisms.

If we work over the real numbers, rather than over the complexes, we view $B$ as a smooth, orientable manifold of real dimension 2 and of genus 4. It has 240 automorphisms, 120 of which preserve orientation and 120 of which reverse it.

​As a complex 1-manifold, Bring's curve $B$ comes to us with a conformal structure. Unless I am confused, it follows from the Uniformization Theorem that $B$ can be equipped with a Riemannian metric, compatible with that conformal structure, under which its Gaussian curvature is the constant $-1$.

Question 1: Is that metric unique? To remain compatible with the conformal structure, our only freedom is a positive-real magnification factor at each point. And it might be the case that the smooth function giving this magnification factor is uniquely determined by the requirement that the resulting Gaussian curvature turn out to be everywhere $-1$. But maybe not.

Question 2: Whatever the answer to Question 1, I am trying to find an explicit description of some constant-curvature metric. If there are multiple such metrics, then choose some simple one.

If we equip $B$ with such a metric, we can then lay $B$ out in the hyperbolic plane, getting a repeating pattern. Given the structure of the automorphism group, we can tile a fundamental domain of that pattern with 240 hyperbolic triangles, each of which has vertex angles of $\pi/5$, $\pi/4$, and $\pi/2$ --- that is, 36, 45, and 90 degrees. Let's call those triangles "basic triangles". The page http://en.wikipedia.org/wiki/Orbifold_notation#mediaviewer/File:H2checkers_245.png shows the hyperbolic plane tiled by basic triangles; but note that there are various subsets of 240 of them that constitute a fundamental domain. The area of each basic triangle is $\pi/20$, leading to a total area for the fundamental domain of $12\pi$, as the Gauss-Bonnet Theorem requires for a surface of genus 4. And any one of the basic triangles can be taken to any other, either by an automorphism, if the two triangles are the same color in the picture above, or by an antiautomorphism, if they are opposite colors.

I know explicit formulas, in terms of the coordinates $[z_i]$, for the vertices and edges of all of the basic triangles, as I will describe in a moment. But that's all that I have managed to do so far. Given the coordinates $[z_0,...,z_4]$ of some point along one of the edges of a basic triangle, I would like to be able to compute the distances from that point to the two vertices at the endpoints of the edge. Finally, I would like to compute distances also for points in the interiors of the triangles.

Here are further details, in case anyone is still reading.

The 36-degree vertex of any basic triangle is, in fact, the 36-degree vertex of ten basic triangles. So there are 240/10 = 24 such vertices, and we'll call them Type A. The 45-degree vertex of any triangle is the 45-degree vertex of eight triangles. So there are 240/8 = 30 such, which we'll call Type B. And the 90-degree vertex of any triangle is the 90-degree vertex of four triangles. So there are 240/4 = 60 such, which we'll call Type C.

The paper "Bring's curve", by W. L. Edge, published in volume 2-18 of the Journal of the London Math. Society in 1978, pages 539-545, gives explicit formulas for the coordinates of these various vertices. For vertices of Type A, let $\epsilon:=e^{2 \pi i/5}$ be a primitive fifth root of unity, and then freely permute these coordinates: $$[1, \epsilon, \epsilon^2, \epsilon^3, \epsilon^4].$$ There are 120 ways to permute those five coordinates, of course, but the homogeneity of the coordinates means that we can multiply by any nonzero complex number, and multiplying by powers of $\epsilon$ puts those 120 permutations into groups of size 5, of which there must be 24.

The vertices of Type B result from freely permuting the coordinates: $$[0, 1, i, -1, -i].$$ Again, there are 120 ways to permute those coordinates, but multiplying by any power of $i$ puts them into groups of size 4, of which there must be 30.

The vertices of Type C are a bit more complicated. Let $\alpha$, $\beta$, and $\gamma$ denote the three roots of the cubic polynomial $z^3+2z^2+3z+4$. Then, the Type C vertices result from permuting the coordinates: $$[1, 1, \alpha, \beta, \gamma].$$ The 120 permutations come in pairs that differ only in swapping the two copies of 1; so we get precisely the 60 vertices of Type C.

Now, these vertices lie along various hyperbolic lines. Let's refer to the edges of a basic triangle as being of Type a, Type b, or Type c, where the edge of Type a is opposite the vertex of Type A, and so forth. So the edges of Type c are the hypotenuses of their triangles, those of Type b are the longer legs, and those of Type a are the shorter legs.

In the resulting tiling, there are lines of two types, say Type 1 and Type 2. A line of Type 1 passes through vertices in the pattern $(A B A C)^4$. So it passes through eight vertices of Type A, four of Type B, and four of Type C. In the process, it travels along eight edges of Type c and eight of Type b, in the repeating pattern $(c c b b)^4$. So there must be 15 lines of Type 1, which travel along a total of 120 edges of Type c and 120 of Type b --- each such edge acting as an edge for two of the 240 basic triangles.

A line of Type 2 passes through vertices in the pattern $(B C)^6$. In the process, it travels along twelve edges, all of Type a. So there must be 10 lines of Type 2, which travel along the 120 edges of Type a.

The formulas for these lines must involve something beyond complex arithmetic, and that extra something is complex conjugation. Let consider lines of Type 2 first. One such line consists of all points for which $[z_0,...,z_4]$ coincides, homogeneously, with $[\bar z_1,\bar z_0,\bar z_2,\bar z_3,\bar z_4]$. That is, we swap the first two coordinates and we conjugate all five coordinates. If that combined process leaves us where we started, homogeneously, then the point where we started lies on our current example of a line of Type 2, which we might call "the (0,1) line of Type 2". The ten lines of Type 2 are determined by the ten ways of choosing which two coordinates to swap.

Lines of Type 1 are similar, but we swap two pairs of coordinates, rather than just one. There are fifteen ways to choose two separate pairs of coordinates to swap, and those choices give rise to the fifteen lines of Type 1.

As we move along any line of either type, consider the ratio $r:=z_i/z_j$, for some fixed indices $i$ and $j$. The ratio $r$ moves along one of five algebraic curves in the complex plane, which one depending upon how the indices $i$ and $j$ relate to the pairs of coordinates that get swapped -- a single pair for a line of Type 2 or two pairs for a line of Type 1. The five resulting curves turn out to be of degrees 1, 2, 3, 6, and 12, and it might be helpful to study how they cut up the Riemann sphere.

But those curves, along with all of the other structure discussed so far, are forced by how the automorphisms behave on $B$. Is the entire constant-curvature metric also forced? And, if so, how?

• I guess you don't have an explicit description of the cover $p:\mathbf H\to B$, perhaps as a concrete map $\mathbf H\to\Bbb{CP}^4$? – Olivier Bégassat Jan 14 '15 at 18:24
• I doubt this will be of any help, but if you think of the $[z_0:\cdots:z_4]$ as roots of polynomials, then the equations defining Bing's surface (modulo the $S_5$-symmetry) and Newton's identities for symmetric polynomials tell you that $$\prod_{i=0}^4(X-z_i)=X^5+\lambda X+\mu$$ with at least one of $\lambda,\mu$ nonzero, and thus you get a ramified covering $B/S^5\to\mathbb{CP}^1$ by the formula $$[z_0:\cdots:z_4]\mapsto\left[\left(\sum_{i=0}^4\prod_{j\neq i}z_j\right)^5:\left(\prod_{i=0}^4z_i\right)^4\right]$$ The ramification points are related to the vertices of type A and B you describe. – Olivier Bégassat Jan 14 '15 at 18:57

Question 1: The constant-curvature metric on $B$ is unique. To see this, note that equipping $B$ with any metric of constant curvature $-1$ gives us a biholomorphim (aka a conformal isomorphism) between the universal covering surface of $B$ and the unit disk in the complex plane. Combining two such metrics would give us a biholomorphism from the disk to itself. But the only such biholomorphims, by the uniqueness part of the Riemann Mapping Theorem, are those Möbius transformations that preserve the disk, and all of those are hyperbolic isometries. So the metric that we get on $B$ is uniquely determined.

Question 2 -- a start at an answer: It seems that getting an explicit formula for the unique constant-curvature metric on $B$ is actually a problem in conformal mapping; so Schwarz-Christoffel transformations and the like will probably be involved.

We begin by finding some conformal map from $B$ to $\Bbb{CP}^1$. The image, under that map, of any of the "basic triangles" on $B$ will be a curvilinear triangle in the complex plane with 36, 45, and 90-degree angles. Our challenge, then, will be to map that triangle conformally onto a hyperbolic triangle with those angles, viewed as a curvilinear triangle in Poincare model of the hyperbolic plane. Given such a conformal map, we can then pull the hyperbolic metric back onto whichever basic triangle of $B$ we started from, and then replicate it by symmetry to the rest of $B$.

One way to get a map from $B$ to $\Bbb{CP}^1$ was suggested at the end of the original posting. Given a point $[z_0,...,z_4]$ in $B$, we can simply map it to the ratio of some two coordinates, say the ratio $z_1 / z_0$. The result realizes $B$ as a six-sheeted covering of the Riemann sphere, with a bunch of order-2 branch points. Here is a picture:

The red dots are the images, under the coordinate-ratio map, of vertices of Type A; the green dots are Type B; and the blue dots are Type C. (The point at infinity is a green dot.) And the curves shown are the images of the lines of Type 1 and Type 2. The Riemann sphere is thus divided into $240/6=40$ curvilinear triangles, each with vertex angles of 36, 45, and 90 degrees -- except that six of the blue dots are the images of branch points of order 2, so the 90-degree angles at those vertices have been widened to 180-degree angles. As commented at the end of the original posting, the curves that make up this picture are of degrees 1, 2, 3, 6, and 12. The curves of degrees 1 and 12 are carried into themselves by inversion in the unit circle, which is the curve of degree 2. The curves of degrees 3 and 6 are interchanged by that inversion.

The problem that remains would then be to take one of those curvilinear triangles and map it conformally onto one of the basic triangles in the hyperbolic-plane picture. I'm not sure how hard that will be. Note that at least one of the edges of the starting triangle will be a curve of degree at least 3, while all three edges of the destination triangle will be either straight lines or circles. So it will take some work to get a conformal map, as shouldn't be surprising.

Olivier Bégassat has suggested another map from $B$ to $\Bbb{CP}^1$, a map that may well work better than the coordinate-ratio map discussed above. Bégassat's map realizes $B$ as a 120-sheeted covering of the Riemann sphere, with branch points of orders 2, 4 and 5. So the image of $B$ under his map consists of just two basic triangles, one filling up the entire upper half-plane, the other the lower half-plane. All of the vertices of Type A map to the origin, all of those of Type B map to infinity, and all of those of Type C map to the real number $-3125/256$; and all of the lines of Types 1 and 2 map to portions of the real axis.

Using Bégassat's map, the only problem that remains is to get a conformal map from a basic triangle in the Poincare hyperbolic plane to the upper half-plane, or, equivalently, to the unit disk. Since the Poincare triangle has at least one curved edge, this can't be done with the most standard flavor of Schwarz-Christoffel mapping. But the problem seems straightforward enough that it has probably been solved in the literature on conformal mapping.