Imagine a square centered at the origin of three-space, with corners on the x- and y- axes. Now picture those corners being pulled out along those axes to infinity, forming ribbons twisting as they stretch. Now pull up on the centers of the edges in quadrants one and three and down on the other two so that the surface has constant curvature −1. Is this possible?
If the surface exists, it should be represented near the origin by a function z = f(x, y) that is symmetric and odd, i.e., f(x, y) = f(y, x), f(–x, y) = –f(x, y) = f(x, –y). When the surface is the graph of z as a function of x and y, the curvature is (z11z22 − z122)/(1 + z12 + z22)2. Using this equation, we can calculate the power series of the surface starting with the assumption that its constant and first degree terms are 0. We get z(x, y) = xy + x3y/3 + xy3/3 + 2x5y/15 + 2x3y3/3 + 2xy5/15 + 17x7y/315 + 7x5y3/9 + 7x3y5/9 + 17xy7/315 + ... Unfortunately, the fact that we can calculate coefficients does not mean the series converges, so this is no proof that our surface exists. Note: in case some of the coefficients look familiar, be patient; we’ll return to this later.
Oddly, the approach that led to a proof of existence was the proof of non-existence of a complete hyperbolic plane in three-space (see [Willmore]. IV §7). This approach uses the so-called asymptotic lines, whose direction at each point has 0 normal curvature. It is shown in that proof that for a surface of constant negative curvature, these lines (curves, actually) can be chosen as a parameter system (x, y, z) = p(s1, s2) such that p1 and p2 are unit vectors, and that the angle θ between them satisfies the differential equation θ12 = sin(θ).
Step 1 is to show such a θ exists: For our surface, we should have θ = π/2 on both axes. We can solve the equation by iteration, starting with θ(0) = π/2, θ(1) = π/2 + xy and in general θ(n+1) = π/2 + ∫∫sin(θ(n))dx dy. Since |sin a – sin b| ≤ |a – b|, we find inductively that |θ(n+1) – θ(n)| ≤ (xy)n+1/(n + !)!2, so we get uniform convergence on any bounded region, and θ = ∫∫sin(θ)dx dy. Therefore θ12 = sin(θ).
Another way to construct θ is to note that we can write θ(x, y) = φ(xy) + π/2, where φ is a function of one variable satisfying φ' + tφ" = cos(φ), with φ(0) = 0, φ'(0) = 1, φ"(0) = 0. Although it is not relevant to this discussion, the behavior of φ as a complex function is quite interesting. (See a color-coded complex graph of this function, with π/2 subtracted to make it symmetric.)
A function (x1, x2, x3)
= p(s1, s2)
for which the matrix (gij) = (pi∙pj)
has non-zero determinant represents a coordinate system on a surface
(at least
locally). The matrix entries are often
denoted E = g11, F = g12 = g21, and G
= g22,
and the determinant EG − F2 is called the first
fundamental
form. Along with the second
fundamental form, LN − M2,
with coefficients L = p11·n,
M = p12·n, and N = p22·n, it determines the curvature K = (LN
− M2)/(EG – F2) .
Suppose p1 and p2 are unit vectors, so that E = G = 1. Then F = cos(θ), where θ is the angle between p1 and p2. If we can find a surface p(s, t) = (x(s, t), y(s, t), z(s, t)) with |p1| = |p2| = 1 and p1∙p2 = F, and with second fundamental form L = N = 0, M = √(1 – F2), then its curvature is (LN − M2)/(EG – F2) = −(1 – F2)/(1 – F2) = −1. A necessary and sufficient condition for the existence of the surface is that the equations of Gauss and Codazzi be satisfied.
Given the assumptions on E, F, G, L, M, N, and θ, the Gauss equation becomes
(1 – F2)F12 + (1 – F2)M2 + F1F2 = 0,
and the Codazzi equations become Mi /M = FFi /(1 – F2).
Recalling that F = cos(θ), and that θ12 = sin(θ), we find that
F12 = – sin2(θ) – F1F2 cos(θ)/sin2(θ).
Substituting F for cos(θ) and 1 – F2 for sin2(θ), this becomes
(1 – F2)F12 + (1 – F2)2 + F1F2 = 0.
Therefore, the Gauss equation is equivalent to M2 = 1 – F2. On the other hand, integrating the Codazzi equations tells us that ln M – ln(1 – F2)/2 is independent of each of the coordinates, which says that M is a constant multiple of √(1 – F2). Thus, the value we chose for M not only says the curvature is −1, but that Gauss and Codazzi are satisfied, so the surface exists.
REFERENCE
Willmore, T. J., An Introduction to Differential Geometry, Oxford, 1964