Perplexed by AI

 The last post, "Flow of time and flow of space", ended with "In the next post we take a look at what kind of life can reside on the two  infinity circles?".  That was my plan. Trying to bring it to life I started calculating and researching. One thing was leading unavoidably to another, while I was intensively using AI (mainly Perplexity) for digging into "what has been already published on the subject?" My search was mostly unsuccessful. Perplexity could not help itself but to start hallucinating. It was referring me to sources, papers, textbooks, online resources, that did not contain the stuff it was claiming to have. So I was changing my prompts until finally I was led to a paper that made me very very happy. It was like discovering a wonderful meadow with lots of daisies and grazing unicorns. The title of the paper is:   "A primer on the (2 + 1) Einstein universe".

A primer on the (2 + 1) Einstein universe

It was exactly what I needed. The paper is some fifty pages long, and I will be studying it slowly while at the same time describing what I have already learned here. After that we will return to the point where we have left, but with a much broader perspective. The paper is written by five mathematicians, T. Barbot, V. Charette, T. Drumm, M. Goldmann, K. Melnick, while they were visiting the Schrodinger Institute in Vienna. You can download it from arxiv here. It was published in print in"Recent Developments in Pseudo-Riemannian Geometry", (ESI Lectures in Mathematics and Physics), Dmitri V. Alekseevsky and Helga Baum eds.,  EMS - Publishing House, Zürich 2008, pp. 179-229.

Keywords: Minkowski space, spacetime, Lorentzian manifold, conformal structure, Lie algebra, symplectic vector space
"(...) The purpose of this article is to develop the synthetic geometry of the Einstein universe in terms of its homogeneous submanifolds and causal structure..."

I will start with the math in the next post. This post is just an announcement. But I also want to tell you that my view on Artificial Intelligence has evolved. The AI is not "intelligent" at all. It is a useful companion, it has access to huge databases, but occasionally it is simply dumb! It can be also dangerous when you try to rely on it.  Just yesterday I asked Perplexity to sketch a proof of a certain result quoted in the paper I am advertising, but without providing a proof. Perplexity happily "proved" the statement and suggested a reference to look at. I looked and there was nothing even roughly similar there. Then I thought for a while, and decided to ask if another statement, contradicting the first one is true. The answer was: yes, the second statement is true, which means that the first one, the "proven" one, is false. Sure, both statements look "similar", but we are dealing with mathematics. The statements x=y and x≠y look similar! But, while using two-valued logic, only one of them can be true at a time.

Here is another example. I asked Perplexity to analyze the picture below and describe what kind of optical illusion it provides:

Here is the reply:

The attached picture shows a classic impossible object illusion, specifically related to the genre of the Reutersvärd Triangle or Penrose Triangle.

In the image, you see a group of cubes arranged in a triangular configuration on an isometric grid. At first glance, each cube's shape appears normal, but as you mentally connect the arrangement as a 3D object, you realize the structure can't exist in real 3D space: the cubes fit together in a way where the geometry of connections between them is self-contradictory.

This is an example of an impossible figure illusion: the brain tries to interpret the arrangement as a three-dimensional object, but the spatial relationships between the cubes force contradictory perspectives. The drawing leverages local plausibility with global impossibility—meaning each small part looks realistic, but the whole configuration violates Euclidean geometry when viewed as a solid object.

Such illusions are used to illustrate how perception can be tricked by carefully designed two-dimensional images that depict impossible three-dimensional shapes—one of the pioneering styles introduced by Oscar Reutersvärd and later made famous by Penrose and Escher.

Well, I do not see anything "impossible" in the picture!
So I asked the same question Yandex AI, Alice. The answer was much much more satisfactory:

The optical illusion in the attached picture is an example of a 3D isometric illusion (also known as a multistable perception illusion or illusion of depth).

Explanation:

The image shows a group of cubes drawn on a grid using isometric perspective — a technique where parallel lines remain parallel, creating a 3D effect on a 2D surface.

The shading and positioning of the cubes trick the brain into perceiving depth and volume, even though the drawing is flat.

The illusion works by:

Using consistent angles (typically 30° or 45° from horizontal) to represent the three dimensions (length, width, height).

Applying shading to simulate light and shadow, which enhances the 3D effect.

Arranging the cubes in a way that mimics spatial relationships (some appear in front, others behind).

Key characteristics:

Ambiguity: The viewer’s brain struggles to reconcile the 2D lines with the 3D interpretation.

Forced perspective: The grid lines help maintain the illusion by suggesting a consistent spatial grid.

Depth perception: The shading and cube arrangement create a convincing sense of depth.

Similar illusions: This type of illusion is related to:

Necker cube: A classic example of a reversible 3D cube drawing.

Impossible objects: Like the Penrose triangle, which uses similar isometric techniques but creates geometrically impossible shapes.

Conclusion: This is a classic example of how 2D drawings can deceive the visual system into perceiving three-dimensional space, relying on perspective, shading, and spatial arrangement.

For a perplexed Reader here is the link to  Oscar Reutersvärd's "impossible art"

Shape of infinity

 Our model toy universe is a torus It is a homogeneous space for the group SO₀(2,2). It does not carry any natural metric. But it carries, as we will see later, a natural conformal structure. Thus there are no ``geodesics'' -- ``shortest'' lines connecting points, but there are natural isotropic lines, or ``null'' lines - they represent ``light rays''. If we select a point p, there are two light rays emanating from p, one ``to the right'', and one ``to the left''. 

Fig. 1 Two Minkowski spacetimes with their boundaries at infinity.

They form two circles. So we have, automatically two circular light rays intersecting at p, and also in the opposite point (-p). These two light rays form p^⟘ .

Shape of infinity

These are red and blue circles on Fig. 1. These two circles of light form the ``infinity'' boundary for the complement, which splits into two disjoint open regions -- red and blue. Each region carries a natural structure of an affine space. It can be identify with Minkowski space-time, endowed with Minkowski space-time quadratic form (or ``metric''). But this Minkowski metric depends on the choice of p. Thus the torus as a whole does not have a natural metric. The stability group of $p$ consists of time and space translations, Lorentz boosts, and dilations. It We can arbitrarily select an ``origin'' in one of these two Minkowski spaces. This further breaks the symmetry to just Lorentz boosts and dilations. Every light ray through a point in one Minkowski space intersects with two light rays at infinity, somewhere in the middle between (p) and (-p). But no such light ray passes through the points (p) or (-p). This opens the following question: which paths in Minkowski space goes to (p) and (-p)?

We will address this question in the forthcoming post.

The whole document containing the details is shown below. You can also download the pdf here. The old notes have been essentially expanded and updated.

The Spin Chronicles (Part 24): Conformal dig

 Continuing from Part 23,  in the grand tradition of curious explorers, let’s keep digging..

We dig up diamonds by the score

A thousand rubies, sometimes more

Though we don't know what we dig 'em for

We dig dig dig a-dig dig

A thousand rubies, sometimes more

Though we don't know what we dig 'em for

We are left with exp(tX) for yet unexplored candidates for X, namely e₁, e₂, ie₁, ie₂. Here the calculations can become cumbersome, and, in fact, this what they are. But they do not have to be such if we are clever enough. I was not that clever at the beginning, so I lost quite some time, until I remembered something vaguely, and an enlightenment suddenly occurred to me. After that I checked, and rechecked, and here comes you rabbit from the hat - the Eureka great discovery!  The trick is extremely simple ..... once you know it!

We have these e₁, e₂, ie₁, ie₂ that we want to exponentiate. But why exactly THEM? They are part of the "Lie algebra" (you don't have to know what the Lie algebra is), they satisfy the fundamental requirement X+ ν(X) = 0, but so is any linear combination of them! So, let's take these linear combination, but in a clever way. And what is this "clever way" I am telling you right away, now.

Here are the linear combinations that work for us:

1. e¹ + ie²

2. e² - ie¹

3. e¹ - ie²

4. e² + ie¹.

We will take 1/2 of them. Today, we’ll focus on 1 and 2. Combinations 3 and 4? I leave those as a rewarding exercise for you, dear Reader.

Now, what is so special about these linear combinations? What is this "great discovery"? Take, for instance

X = (e¹ + ie²)/2

We have that e¹ and e² anti-commute and their squares are 1, while i² = -1. As the result X² = 0. Check it! As the result the whole series for the exponential exp(tX) terminates after the linear term:

exp(tX) = 1 + tX. = 1 + t(e¹ + ie²)/2

Then τ(X) = (e¹ - ie²)/2,

and

τ(exp(tX)) = 1 + t(e¹ - ie²)/2.

Thus we need to calculate

e'⁰ = (1 + t(e¹ + ie²)/2) e⁰ (1 + t(e¹ - ie²)/2),

e'¹ = (1 + t(e¹ + ie²)/2) e¹ (1 + t(e¹ - ie²)/2),

e'² = (1 + t(e¹ + ie²)/2) e² (1 + t(e¹ - ie²)/2),

e'³ = (1 + t(e¹ + ie²)/2) e³ (1 + t(e¹ - ie²)/2).

It’s a straightforward algebraic exercise—boring, yes, but manageable. I used the powerful and free computer algebra system "Reduce" to automate these tedious steps. If you’re not using tools like this, you’re missing out! Anyway, you tell the program the rules, and then it does the job. My rules were

Operator e$
Noncom e$
for all i let e(i)*e(i)=1$
for all i,j such that i>j let e(i)*e(j)=-e(j)*e(i)$
let e(1)*e(2)=i*e(3)$
let e(2)*e(3)=i*e(1)$
let e(1)*e(3)=-i*e(2)$

Note: Alternatively, one can play with the Pauli matrices instead.

In my mind I set e⁰=1, as it is an exception. Then you give it expressions to calculate, and then, without even a blink of the eye, it spits out the results. You tell the program to write them to a file. Here they are, rewritten using our notation, where I manually replaced the scalar part by e⁰:

e'⁰ = (1+t²/2) e⁰ + te¹ + (t²/2)e³,
e'¹ = te⁰ + e¹ + te³,
e'² = e²,
e'³ = -(t²/2)e⁰ - te¹ + (1-t²/2)e³.

The transformation matrix, in Mathematica notation, is

{{1 + t^2/2,  t,  0,  t^2/2},
{t,  1,  0,  t},
{0,  0,  1, 0},
{-t^2/2,  -t,  0,  1 - t^2/2}}

The coordinates transformation matrix is the transposed matrix,

m={{1 + t^2/2, t, 0, -(t^2/2)},
{t, 1, 0, -t},
{0, 0, 1, 0},
{t^2/2, t, 0, 1-t^2/2}}

We are almost done. We recall from Part 18 the formula for embedding the x¹=x, x²=y plane into the null cone of Minkowski space:

x ⟼ ζ={ζ₀,ζ₁,ζ₂,ζ₃}={1, 2 x/(1 + q), 2 y/(1 + q), (1 - q)/(1 + q)},

where I set x·x = x² + y² = q.

We act with the matrix  m on vector ζ to get:

ζ'={(1 + q + q t^2 + 2 t x)/(1 + q), (2 (q t + x))/(1 + q), (2 y)/(1 + q), (1 + q (-1 + t^2) + 2 t x)/(1 + q)}

From this we can read x' = ζ'₁/(ζ'₀+ζ^'₃) and y' = ζ'₂/(ζ'₀+ζ^'₃). The result is:

x' = (x + q t)/(1 + q t^2 + 2 t x), 

y' = y/(1 + q t^2 + 2 t x).

Now we can repeat the same procedure with

Y = (e² - ie¹)/2.

This time we get

x' = x/(1 + q t^2 + 2 t y), 

y' = (y +qt)/(1 + q t^2 + 2 t y).

We can then check the very important property: X and Y commute! We replace t by "a" for tX, and by "b" in tY. Then

g = exp(aX)exp(bY)=exp(bY)exp(aX)=exp(aX+bY).

The transformation implemented by such a g is then

x' =(x + qa)/(1+ q (a²+b²) + 2(xa+yb)),
y' =(y + qb)/(1+ q (a²+b²) + 2(xa+yb)).

This is known as a "special conformal transformation". Conformal transformations are known to preserve angles, and to transform circles into circles (straight lines are considered as infinite radius circles).

Exercise 1. Do the same with 3) and 4). This should be even easier. Why? You will see when you try.

Spacetime and anti-spacetime

 Here is the graphics representing the Minkowski space (yellow) and Minkowski anti-space (blue) embedded in the double cover of the conformal compactification. They are separated by (double cover of) the conformal infinity. Can light travel between spacetime and anti-spacetime? I am going to find an answer to this fascinating question.

And from the yellow side:

To be continued.....

P.S.1 Currently I am recalculating everything and finding really annoying misprints in my old papers. They drive me crazy!  Yesterday I have started an exchange with a well known mathematician in Novosibirsk, one of the authors of a Springer monograph about Riemannian geometry and geodesics,  also interested in conformal infinity. He tried to read my old papers and wasn't very happy, so I am preparing a CLEARLY WRITTEN DOCUMENT FOR HIM. Being very careful this time.

P.S.2.My new friend, mentioned in P.S.1, send me today his new paper - and introduction to Segal's chronometric theory (in Russian). There are some points when we agree (topology) and some when he disagrees with me (differentiable structure). We have to agree on everything before the publication of the paper. So I am now busy so that I do not have time to eat!

P.S.3. In his paper I have found new to me realization of the universal covering of the group U(2) in terms of quaternions. How could I not see it all by myself?! So finally we have some good use of quaternions in physics!!! Progress!

P.S.4. We now have snow in the sothern France. Yesterday is also snowing. Looks like we have a climate change? A while ago one of our intertnet cables got disconnected from the pole. Last time it happened on February 24 last year. It was the time of the start of the Russian "denazification" of Ukraine. Is it going to be a sign of a new decisive Russian offensive?

P.S.5. Valera (first name of my Russian friend, mathematician) sends me a fourth version of his 21 pages long paper - after his midnight!. It becomes more and more clear (the paper). His approach is through the unitary group  U(2), while I prefer to play with isotropic subspaces in R^6. Now even one result from my Quantum Fractals book is being quoted as useful, though it is concentrated on somewhat different issues.

P.S.6. I have to learn about "manifolds with corners". Perhaps the conformal infinity has what is called a "corner"? A corner of the universe? (Ark Fleet Ship B is being used  there). I don't know yet for sure.

P.S.7. Probably the best book on smooth manifolds: John M. Lee, Introduction to Smooth Manifolds, 2nd edition, Springer 2012. Table of content can be perused here. Includes concise appendices on linear algebra, on topology, sections on categories and functors, manifolds with corners, lot of examples.

.P.S.8. While working on the final version of one of the papers (in order to make the reviewers and the editor happy), and it will be sent, updated,  to the publisher today or tomorrow, so I will be able to return to the problem of light travelling along the conformal infinity on the boundary between space and anti-space) I have found (by chance of course) a very interesting paper that opens for me some new doors I was looking for a long time:

"Explicit construction of a time superoperator for quantum unstable systems"

November 2001 Chaos Solitons & Fractals 12:2591-2601

DOI: 10.1016/S0960-0779(01)00074-1

Gonzalo E Ordonez, Tomio Petrosky, Evgueni Karpov, I. Prigogine

Abstract: 

A time superoperator T conjugate to the Liouville superoperator LH=[H,] is constructed for a quantum system with one excited state or unstable particle. While there is no time operator conjugate to the Hamiltonian in the wave function space due to the positivity of energy, T may exist in the density matrix space as the spectrum of LH covers all the real axis. This is the first example of an observable that can only be formulated in the Liouville–von Neumann space of density matrices. In our example the expectation value of T gives the lifetime of the unstable particle. Once the time superoperator is obtained it is easy to define an entropy superoperator.

P.S.9. A mathematically more precise paper on the same fascinating subject:

Adolfo R. Ordonez, "Rigged Hilbert Spaces associated with Misra-Prigogine-Courbage Theory of Irreversibility", published in ; Physsica A 252 (1998) 362-376

P.S.10. In fact Prigogine's idea of considering "superoperators" fits perfectly the subject of this note. In finite-dimensional case the space in which superoperators act is H⊗H*, where H* is the dual Hilbert space (anti-space), and the spectrum of super-Hamiltonians made of ordinary Hamiltonians consists of the differences of the eigenvalues, thus is always symmetric: if a1-a2 is positive then a2-a1 is negative.

P.S.11. In fact I was playing with "superoperators" probably before the term was made famous by Ilya Prigogine. Here is a Remark I have made in my second published paper (that became a part of  my PhD Thesis) :

:

P.S.12 And today (Thursday January 26) we will have a close encounter: "In fact, this is one of the closest approaches by a known near-Earth object ever recorded"

 "The asteroid will be closest to Earth at 4:17 p.m. EST (2117 GMT) that day. At that point, it will be about 2,200 miles (3,600 kilometers) above the planet's surface, according to NASA."

P.S.13. I am still trying to find an aesthetically satisfactory way to justify mathematically the two images from this post. I am getting closer and closer.

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.