A mapping between geometric spaces is one that maps smooth trajectories to smooth trajectories and preserves angles between them where they meet one another. Although such a mapping may distort the geometry – expanding some parts and contracting others, turning some straight lines into curved ones and vice versa – it preserves just enough, locally, to ensure some nice properties.
Consider, by way of example, a standard mapping of a two-dimensional plane onto a sphere. Pick a point on the plane as origin, construct a unit sphere centred on that point and a radius of this perpendicular to the plane. Connect the point, P, where this radius meets the sphere to any point of the plane and see where the straight line between them cuts the sphere: map the point on the plane to that point on the sphere.
The equator
of the sphere, in which it meets the plane, is mapped to
itself. The point opposite P is mapped to the centre of the sphere. P itself
has no image, although its neighbourhoods are mapped to neighbourhoods of
infinity
, i.e. the complements of bounded regions of the plane.
Consider a radius of the sphere that makes angle turn/2 −u with the radius to P. A plane containing these two radii cuts the sphere in a circle. Connecting P and the point opposite it to the sphere's end, E, of this radius, we form a right-angle triangle whose edge opposite P subtends angle u at the centre of the sphere hence (as the edge is a chord of the circle) angle u/2 at P. The ray from P to E thus cuts the equatorial plane at distance Tan(u/2) from the origin.
Using u as a lattitude angle, with an angle w for longitude around the equator of our sphere, we can define Cartesian co-ordinates for which a [u, w] point of the sphere is described by [Sin(u).Cos(w), Sin(u).Sin(w), Cos(u)] and corresponds to the point on the plane Tan(u/2).[Cos(w), Sin(w), 0]. Writing r = Tan(u/2), to make the point on the plane r.[Cos(u), Cos(w), 0], we can rewrite the point on the sphere as [2.r.Cos(w), 2.r.Sin(w), 1 −r.r]/(1 +r.r). With r = Tan(u/2), we get 2.radian.dr/du = 1 +r.r = power(−2, Cos(u/2)) = 2/(Cos(u) +1), which I'll use later, along with r.r = 2/(Cos(u) +1) −1 = (1 −Cos(u))/(Cos(u) +1) = power(2, Sin(u)/(Cos(u) +1)).
In the plane, let [x, y] = r.[Cos(w), Sin(w)], giving metric
In the three-dimensional cartesian co-ordinates for which the sphere is [Sin(u).Cos(w), Sin(u).Sin(w), Cos(u)], we can introduce a spherical radial co-ordinate R for which [X, Y, Z] = R.[Sin(u).Cos(w), Sin(u).Sin(w), Cos(u)], with dR = 0, to give us a 3-space metric's restriction to the sphere R = 1 (with a scaling to save repetition later):
Thus the metric induced on the sphere by the projection of the plane is just the metric of the sphere itself scaled down by power(2, Cos(u) +1). Since the angles between trajectories at a point are obtained by dividing the inner product of their tangents at that point by the product of those tangents' lengths, of form g(t, s)/√(g(t, t).g(s, s)) for the cosine of the angle between s and t, an over-all scaling to the metric g cancels itself out and gives the same value for the cosine of the angle.
Thus this mapping from the sphere to the plane is conformal. It is, indeed, the Riemann sphere transformation between the complex plane and its extension to include an infinite value that corresponds to the point P on our sphere. Speaking of the complex plane, this leads into …
For a function from the complex plane to itself to be differentiable, its derivative's action on a small displacement of the input must correspond to multiplying that displacement by a complex number. It turns out this is quite a strong condition.
If we think of the plane as the Cartesian real plane, with [x, y] representing the complex value x +i.y, our typical function will be expressed in terms of two real-valued functions, f(x +i.y) = s(x, y) +i.t(x, y). As s and t are functions from the real 2-plane to reals, we can differentiate them using real derivatives, which shall deliver ∂s/∂x, ∂s/∂y, ∂t/∂y and ∂t/∂x describing the derivative. If we make a small perturbation to the input, adding h to x and k to y, we get
ignoring second-order terms, as usual. For this to be a complex derivative, we require that there be some complex a +i.b, with a, b real (and depending only on x, y, not h, k), for which:
for all real h, k. Comparing terms, e.g. for [k, h] in {[0, 1], [1, 0]}, we get ∂s/∂x = a = ∂t/∂y and ∂t/∂x = b = −∂s/∂y.
So a function from the complex plane to itself is only complex-differentiable if: its real part varies with the real part of its input the same way its imaginary part varies with the imaginary part of its input; and its imaginary part varies with the real part of its input as the negation of its real part varies with the imaginary part of its input.
Let's now look at how this transforms the metric. The metric of the output is
which is, again, just a scaling of the metric of the input, so preserves angles and thus is conformal. So every complex-differentiable function from the complex plane to itself is conformal.
Written by Eddy.