Intrinsic geometry is determined by the infinitessimal distance between neighbouring points. Such rule is called the metric.

Given a metric, we can compute a finite length by concatenating small segments.

We map a surface into a plane. The complex plane ℂ.

Consider two points Q and Z on the surface S. They will be mapped to q and z on the complex plane.

The distance between Q and Z in S is dS. The distance between q and z in ℂ is ds. The metric is the rule that gives dS in terms of ds:

dS = (metric)ds

So the metric is a stretching factor. A similar concept to an eigenvalue in this regard.

Projective map, central projection: from the centre of a sphere, trace rays outwards towards a plane tangent to the sphere at the south pole.
	- a circle in S becomes an ellipse in ℂ
	- a geodesic becomes a straight line
	- angles are sacrificed, but circles of longitude and latitude remain perpendicular
	- we'll see later how it's usually better to sacrifice straight lines and preserve angles

The same surface, with the same intrinsic geometry, can be described by different metrics.

For example, for a sphere like Earth we can use the central projection, as commented before, with the plane being tangent at the south pole (polar coordinates r,θ), or use latitude and longitude φ and θ. The first case gives dS² = A*dr² + B*dθ². The second, dS² = C*dφ² + D*dθ². Both describe the same intrinsic geometry.

Notice how dS² is the square of dS, the real distance between two points in a surface.

Even in the flat plane, there are several choices of coordinates to get the Euclidean distance. Not all is dS² = dx² + dy². For example, dS² = dr² + r²dθ².

General form of dS²: assume two families of curves, u and v, on the surface, such that a given δS can be decomposed into δS_1 + δS_2. In our complex plane this maps to u + iv.

So, for δu on the plane there is a corresponding δS_1 on the surface. Similarly for δv on the plane and δS_2 on the surface. Thus, our (local) scale factors are

A = δS_1/δu
B = δS_2/δv

Notice how A=A(u,v) and B=B(u,v).

If ω is the angle between the two increments on the surface, and applying Pythagoras,

dS² = A²du² + B²dv² + 2Fdudv with F=ABcosω

Warning: this is usually written as dS² = Edu² + Gdv² + 2Fdudv, leading to lots of √.

Locally, we can always choose the v curves ortogonal to the u's, so

dS² = A²du² + B²dv²

Not possible to cover all the surface with a single (u,v) coordinate system. Two u (or v) curves will end up intersecting. For example, let u be latitude curves on a sphere => v curves will be longitude circles. But then, these meridians will cross at the poles, and these poles are assigned many v-values. This is unavoidable.

If we are given a particular dS² = A²du² + B²dv² without knowledge of the surface, this formula tells us everything about its intrinsic geometry. But how to extract this info? For example, can we know the curvature from the metric? Yes, with the metric curvature formula, given here as an act of faith:

κ = -1/(AB)(∂_v[∂_vA/B] + ∂_u[∂_uB/A])

If applied to the Euclidean metric, we get κ = 0. To the sphere, κ = 1/R².

We know dS² gives us a conversion from distance on the flat map to distance on the surface, but what about converting areas? A region in the map has area dudv, and this will be converted to

d(area) = sqrt((AB)² - F²)dudv

but with an orthogonal system, we get

d(area) = ABdudv

We saw before that a given projection may preserve straight lines and sacrifice angles or the other way around. Well, it's better to keep angles: a map that preserves the magnitude and the sense (clock or anticlockwise) is called conformal. If it preserves the magnitude but reverses the sense, anticonformal.

The angle between two curves at a point is the angle between their tangents.

A map is conformal iff the scale factor Λ does not depend on the direction of the vector on the plane:

conformal map <=> dS = Λ(z)ds  (Λ only depending on position on the plane)

This has a great advantage: if the map is conformal, an infinitesimal shape on the surface has the same shape on the plane, and both shapes can only differ in size. Any length of the shape on the surface is Λ times bigger than on the plane.

18th century version of the word conformal = similar in the small.

Imagine a big triangle on the plane. It's conformal image on the surface is, in general, a curved triangle with the same angles than the plane version. If we reduce the size of the original triangle, it's image will be ultimately similar to the original.

Since the scale factor does not depend on the direction, A = B = Λ, and infinitesimal circles are mapped to infinitesimal circles. Then,

dS² = Λ²(du² + dv²)

If that's the case, (u,v) are conformal (or isothermal) coordinates.

Gauss discovered it is always locally possible to draw such a map on a general surface. The proof of this theorem requires complex numbers. There is a strong link between complex numbers and conformal maps.

If conformal, the curvature formula becomes

κ = -(1/Λ²)∇²lnΛ

where ∇² is the Laplacian, ∇² = ∂²_u + ∂²_v.

Conformal maps are very interesting, even if they map the plane to itself, and they are strongly related to complex numbers. For more detail, see the VCA book.

If S is the surface and ℂ the complex plane, we call F a conformal mapping F:ℂ→S. And we also have dS = Λ(u,v)ds.

We can also have f:ℂ→ℂ, so that z=u+iv is mapped to f(z)=f(u+iv) or Z=U+iV. Then, we can compose these maps with F, so that the result is still a conformal map.

All the familiar functions (in the complex plane) like exp(z), sin(z), etc are conformal. This is the magic of complex analysis. For example, f(z) = z² (or any other power), is conformal. This means every small shape is ultimately similar to its image.

If a function f(z) is analytic (f'(z) exists), then

δZ ≍ f'(z)δz = aexp(iτ)δz

This transformation implies an expansion factor a(z) and a rotation angle τ(z). This implies that every arrow δz emanating from z, when mapped by f, undergoes the same expansion a and the same rotation τ to get the image arrow δZ emanating from Z.

Conclusion: all differentiable complex mappings are conformal.

The derivative f'(z) consists in an amplification a plus a twist τ => amplitwist.

Derivative = amplitwist

f'(z) = amplitwist of f at z = (amplification)e^(i*(twist)) = aexp(iτ)

For example, and if z=rexp(iθ), f(z) = z² implies an amplification = 2r and a twist θ, so

(z²)' = 2rexp(iθ) = 2z

This is equivalent to (x²)' = 2x, but the complex version has a more powerful geometrical meaning.

Stereographic projection: historically to plot the position of bodies on the celestial sphere. It consists in a spherical surface Σ and a complex plane that passes through its equator. To project a point of the sphere on the complex plane we connect the north pole N with the point on the sphere and extend the straight line until it intersects the plain. This will give us the stereographic image of the point. We can also do the process in reverse, and still call it "stereographic image". We always need to be aware of the context of the projection, whether it is Σ to ℂ or ℂ to Σ.

- The south pole S is mapped to the origin of ℂ.
- The southern hemisphere is mapped to the interior of the circle |z|=R.
- The equator points are mapped to themselves.
- The exterior of the circle |z|=R is mapped to the northern hemisphere (except N).
- As a point is farther on the plane, the projection gets closer to N.
- Σ is usually taken as the unit sphere, so R=1. And called the Riemann sphere.

The stereographic image of a line in the plane is a circle on Σ that passes through N in a direction parallel to the original line.

Stereographic projections preserve angles.

If we define the sense of the angle on Σ by its appearance to an observer insider Σ, and anticlockwise being positive, then stereographic projection is conformal.

Then, a small circle of radius δs on the plane is mapped to a small circle of radius δS on the sphere, and δS = Λ δs. The problem is now to find Λ, but since it does not depend on direction, we choose the easiest one. After some algebra,

Λ = 2/(1 + (r/R)²)

with r = |z|.

We could apply the curvature formula to get κ = 1/R². We would need the polar form of the Laplacian:

∇² = ∂²_r + (1/r)∂_r + (1/r²)∂²_θ

Stereographic formulas (from here, R=1):

Let z=x+iy be the coordinates of a point of the plane and (X,Y,Z) the coordinates on the sphere, with x aligned with X and y aligned with Y.

X + iY = 2z/(1+|z|²)

Z = (|z|² - 1)/(|z|² + 1)

In spherical coordinates, with θ being the angle around the Z axis and θ=0 aligned with the positive X, and with ϕ the angle wrt the Z axis (this is the American convention -- the British is the opposite),

z = cot(ϕ/2)exp(iθ)

If two points P and Q in Σ are antipodal, then their stereographic projections p and q are such that

q = -(1/q*)
p = -(1/p*)

Very important: stereographic projection preserves circles. Not just infinitesimal ones, but circles of any size! But the centre of the circle is not mapped to the centre of the mapped circle!