Constructing the Conformal Infinity

 

This post is thought to be an answer to a comment by Bjab from the previous post.

I do not know who has discovered the stereographic projection and when? I don't know, But it was certainly a great discovery. If we are to believe Wikipedia:

“The stereographic projection was known to Hipparchus, Ptolemy and probably earlier to the Egyptians. It was originally known as the planisphere projection.[2] Planisphaerium by Ptolemy is the oldest surviving document that describes it. One of its most important uses was the representation of celestial charts.[2] The term planisphere is still used to refer to such charts.”

Here is the picture

Here we embed the real axis with the coordinate x into the unit circle. The point P' on the x-axis is projected, here from the north pole N,  onto the point P on the circle. If we denote by X the coordinate of P' and by x,y the two coordinates of P, we have the following transformation formulae (taken from Wikipedia):

We are going to make one change here: we do not like the north pole! It is on the ocean. We prefer to stay on the land, therefore we will choose the south pole as the projection origin. The formulae will be almost the same, only the sign of the z-axis with change into the opposite. Thus we will use

(X,Y) = (x/(1+z), y/(1+z))

(x,y,z) = ( 2X/(1+X²+Y²), 2Y/(1+X²+Y²), (1-X²-Y²)/(1+X²+Y²) )

Notice that th image of the whole plane is the sphere minus the south pole. The south pole, z=-1, escapes to “infinity” on the plane. Thus the sphere is the plane plus one-point – the infinity. It is called a one-point compactification of the Cartesian plane.

Now nothing prevents us from considering a one point compactification of the three-space with coordinates X,Y,Z. We simply need to add on dimension. Thus we consider for-dimensional space with coordinates x,y,z,v and use a straightforward extension of the above formulae:

(X,Y,Z) = (x/(1+v), y/(1+v), z/(1+v))

(x,y,z,v) = ( 2X/(1+X²+Y²+Z²), 2Y/(1+X²+Y²+Z²), 2Z/(1+X²+Y²+Z²), (1-X²-Y²-Z²)/(1+X²+Y²+Z²⁾)

We may be having problems with imagining the three-sphere in four dimensions, but the algebra is as simple as before. Algebra rules!

We are able now to construct a one-point compactification of Rⁿ for any n. It will be a unit sphere in Rⁿ⁺¹. We simply add one coordinate and use a straightforward generalization of the formulae above.

But: it is not yet the end of the story. For our story to have a happy end, as any good story should, we have to go through yet another adventure.

Let us go back to only one coordinate x. The line becomes a circle. But there, spontaenously, comes the observation that a circle comes from an intersection of the cone with the plane, as on the graphics below:

Do not pay attention to any detail on the picture (I have borrowed it from a random paper on Researchgate), except that the circle is created by the intersection of the light cone with a constant t=1 plane.

Guided by this new brave idea we add an extra variable playing the role of "time". But since it is just an extra variable, not a real "time", we call it w, and we will set w=-1. Why "-1" instead of 1? Well both are good, but to be in agreement with a certain number of conventions used in the literature, lest us agree for -1. Ok?

We also simplify our notation to easily cover a general case. So we will write X for the vector with coordinates X¹,..., Xⁿ, and we write X.X or X² for the sum of squares of the coordinates X¹,...,  Xⁿ . Similarly we write x for the vector with coordinates x¹,..., xⁿ . Our formula for the embedding reads now

(x,v,w) = ( 2X/(1+X²), (1-X²)/(1+X²), -1 )

The sum of squares of the first n+1 coordinates, x²+v² is now automatically equal 1 (it can be verified independently though), so the point (x,v,w) is on the surface of the cone

x²+v²-w² = 0.

The coordinate w=1 intersects this cone, the intersection is the sphere Sⁿ.

We are not quite happy yet. In the formula above we have denominators (1+X²), and we do not like them, even if here they are doing no harm. Our cone is made of generator lines. These are straight lines from the origin, along the cone. It is these lines that are important, not the particular intersection. Thus we replace the formula above by multiplying all coordinates on the right by the 1/2 of the common denominator for n+1 first coordinates. We end up with a new embedding formula:

(x,v,w) = ( X, ½ (1-X²), -½ (1+X²) )

We still have a point on the cone x²+v²-w² = 0. The mapping above is one to-one. We will look at it as a mapping from Rⁿ to generator lines of the cone in Rⁿ⁺².

Soon we will specify n that is of interest for us to be n=4, thus the total space Rⁿ⁺² will be (4+2=6) six-dimensional.

To feel more "at home" with our formulas let us do a little exercise. Namely, let us take a generator line on the cone x²+v²-w² = 0 and find the point X to which it corresponds. In the future instead of "generator line" we will simply use the term "line". Thus we should have

(x,v,w) = a ( X, ½ (1-X²), -½ (1+X²) ),

where a is a proportionality factor, telling us that (x,v,w) and a ( X, ½ (1-X²), -½ (1+X²) ) are on the same line. So, we should have

x = aX,

v = a(1-X²)/2,

w = -a(1+X²)/2.

Here x,v,w are given, and we want to calculate a,X. Subtracting the two last equations we find

v-w = a.

Thus, as long as v≠ w we have a≠ 0 and from the first equation

X=x/(v-w).

When v=w, and we are on the cone x²+v²-w² = 0, x must be 0. This is one generator line (0,v,v), v∈R

.So far being brave did not lead us to any trouble. So now we dare to be even more brave. First of all we specify n=4. By X we mean a vector with coordinates X₁,X₂,X₃,X₄. But now we will think of X₄ as "time coordinate". Therefore by X² we will now mean

X² = (X₁)² + (X₂)² + (X₃)² - (X₄)².

Notice that we have changed the coordinate indices from upper to lower to avoid the confusion about what X² means.

So, we will keep the formula:

(x,v,w) = ( X, ½ (1-X²), -½ (1+X²) )

but with the above understanding what X and X² mean.

Altogether we thus have the signature (+++-) for spacetime, + for v, and – for w. The total signature of the six-dimensional space is (4,2), more specifically (+++-+-).

We will check now carefully if we got into some trouble this way?

Let us analyze the map (*) X ↦ (x,v,w) from M to V. We notice that it is one-to-one, we have an injection. Indeed if X≠X' then the images are also different, simply because, as it is evident form (*), x=X, and it is sufficient for just one coordinate of two points to be different for these two points to be different. Moreover, (x,v,w) and (x',v',w') are certainly not on the same line.

But it is not surjective. Let us find the set on which the inverse map is not defined. In the purely Euclidean case it was just one point, and now what are we going to get? To answer this question we follow the previous method and try to express X in terms of (x,v,w). Thus we write again

x = aX,

v = a(1-X²)/2,

w = -a(1+X²)/2.

And try to solve for a and X in terms of x,v,w, assuming x²+v²-w² =0. As before v-w=a, therefore as long as v≠ w, we have a≠ 0 and

X=x/(v-w).

The "infinity" now is defined as before by v=w. Before it was just one point (one generator line). But now? If v=w, from x²+v²-w² =0 we deduce x² = 0. We now need to find algebraic equations describing our (projective) manifold.

But now x² = (x₁)² + (x₂)² + (x₃)² - (x₄)², and we cannot deduce that x=0 in the Euclidean case. We have a 3-dimensional surface. Let us analyze this surface remembering that we are dealing with lines. We are on the quadric in described in coordinates x,v,w by the formula

x²+v²-w² = 0

In projective geometry these are called homogeneous coordinates. Writing explicitly:

(x₁)² + (x₂)² + (x₃)² - (x₄)² + v² - w² = 0

or

(x₁)² + (x₂)² + (x₃)² + v² = (x₄)² + w² = a

The common value a above cannot vanish, since if it vanishes, all x,v,w would be zero, and the origin is excluded, as it does not define any line. Thus a>0. We can therefore replace x,v,w by (x,v,w)/√ a , and get a pair of equations

(x₁)² + (x₂)² + (x₃)² + v² = 1

(x₄)² + w² = 1

Now we set w=v (our "infinity"):

(x₁)² + (x₂)² + (x₃)² + v² = 1

(x₄)² + v² = 1

We have two equations for five variables. Thus our "conformal infinity" now is not a point, it is a three-dimensional surface in a five-dimensional space of coordinates (x,v). We will explain why this term "conformal" in the coming posts. The point is that the conformal transformations which are singular in space time, are nicely realized by linear transformations from the group SO(4,2) acting in our 6D space of variables x,u,v.

 Skipping one space dimension, say x₃, we get a two-dimensional surface in a four-dimensional space

(x₁)² + (x₂)² + v² = 1

(x₄)² + v² = 1

Removing also the variable x₂ we have the intersection of two cylindrical surfaces in 3D:

(x₁)² + v² = 1

(x₄)² + v² = 1

And these were our equations form the previous two posts, though the names of the coordinates were different.

P.S.1. If we were to take care about the physical dimensions, we would have to replace "1" in all the above equations by some constant R (or R squared, as the case may be) - where R is the "radius of the universe". I. Segal et co  used this finite number R to avoid the need for renormalization of quantum field theory at low energies.

P.S.2. Why couldn't we take v=w and x^2=0 as the solution? The problem is that if v=w is different from 0, then x=0 belongs to the solution. While if v=w=0, we have to exclude x=0. This is not a simple algebraic equation. We have to proceed differently.

P.S.3. There is still one problem with our solution. If (x,v,w) satisfies our equations, then (-x,-v,-w) also satisfies them. But these two points are on the same line, so they should be identified. In other words: we have to take the quotient of our solution set by an equivalence relation. We will discuss it later when we will be talking about topology and differential structure on the compactified Minkowski space.