“... there is something still lacking in this picture—the 'conformal
infinity,' the 'absolutes' of projective geometry. Points where
parallel lines meet...” That’s what I claimed in the previous post.
It seems, however, that the idea of parallels meeting "somewhere"
doesn’t sit well with the otherwise grounded tone of our series on the
Clifford geometric algebra of space. Naturally, this stirred the
curiosity of one particular reader (Bjab), who came armed with questions
sharper than Occam's razor:
- "What does 'meet' mean? (Different) Parallel lines don't have common points."
- "What does 'intersect' mean?
- "Meridians are not circles. They are semicircles."
- "Are they really (in the common sense of the word) intersecting, or are those circles just tangent to each other?"
Ah, questions! The perpetual call to adventure for anyone foolish—or
ambitious—enough to write about math. Since the cosmos obliges, so must
I. Today, we’re taking a small detour (I really intend a small one) to
tackle these questions head-on. And as luck would have it, this
diversion ties neatly into our main subject, because it showcases the
adaptability of the mathematical tools we’ve been exploring. By
repurposing the same algebraic framework, we’ll reveal not just the
versatility of mathematics but its elegance as well. Prepare for a
deeper dive into the sublime weirdness of projective and conformal
geometry!
Circles, Spheres, and Stereographic Projections
Let’s ease into this with a concept we all know and love: the humble circle. Actually, we should start with the slightly more ambitious sphere—it’s a better stand-in for the idea of "space." But hey, 3D geometry is cumbersome to scribble on a napkin or screen, so let’s stick to 2D for now. Circles are easier to draw, and, more importantly, we can rely on algebraic tools we’ve already mastered.
For this introductory phase, we’ll forego Clifford algebra. Don’t worry,
it’ll return when we shift from static descriptions to dynamic
transformations—that’s where Clifford algebra shines. After all, it’s
not just about understanding objects but also how they morph, twist, and
turn under various operations. These transformations form groups, and
the Clifford group, as you’ll see later, is the MVP of this show. But
for now, let's lay the groundwork with simpler tools.
Enter the Klein absolute. This concept will guide us into the
realms of projective and conformal geometry, which are surprisingly
practical—think computer vision and robotics. The journey begins with a
stereographic projection on the 2D plane, using Cartesian coordinates x=(x1,x2).
Ready? Here’s the formula for (inverse) stereographic projection through the South Pole (because why not switch hemispheres for variety?):
σ(x) = (2x, (1-x·x))/(1+x·x).
Looks innocent enough, right? But let’s imagine we’re afflicted with a
peculiar condition—denominatophobia. Yes, it’s a (fictional) fear of
denominators. Even when they’re calm and friendly, like the one above,
they still make us squirm. Now, as long as x is a real vector, there’s
no reason to panic. But what if x develops a taste for the imaginary?
After all, we’ve seen this happen before—constructing the Clifford
algebra of space brought us to complex Minkowski spacetime! And who’s to
say we won’t one day construct a Clifford algebra of a Clifford
algebra? (Math is an infinite playground.)
To calm our nerves, we’ll take preventive measures. Mathematics offers a clever remedy for denominatophobia: projective space.
Projective Space: Fearless Math for Fearful Minds
We introduce a fourth coordinate, x0, alongside x1,x2,x3.
Initially, we set x0 = 1. This embeds our 2D point into 4D space:
x ⟼ ( 1, 2x/(1+x·x), (1-x·x)/(1+x·x) ).
In projective geometry, equivalence reigns supreme. Two points ζ and ζ′
are equivalent (ζ ∼ζ′) if there exists a scalar λ>0 such that ζ′
= λζ. These equivalence classes—essentially half-lines in R4—are denoted by [ζ]. With this setup, we can rewrite the embedding as:
[( 1, 2x/(1+x·x), (1-x·x)/(1+x·x) )] = [((1+x·x)/2, x, (1-x·x)/2 )].
Victory! The denominators are gone, leaving us free to bask in mathematical serenity. But wait—there’s more.
The Bonus: A Light Cone Emerges
The point ζ = ((1+x·x)/2, x, (1-x·x)/2 ) lies on the light cone of Minkowski space. Let’s verify this. Define:
ζ0 = (1+x·x)/2,
ζ1 = x1,
ζ2 = x2,
ζ3 = (1-x·x)/2.
Then, check the relation:
Exercise 1: Prove this formula.
This light cone is the so-called Klein absolute for our 2D plane. But
don’t pack up just yet—there’s more excitement in store. In the next
post, circles and parallel lines will make their dramatic entrance. For
now, the stage is set, the props are in place, and the mathematical
actors are ready for action.
References:
[1[ Wilson, Mitchell A, "Meeting at a far meridian", "Встреча на далеком меридиане" - Movie (1977) here.
[2] Cecil, Thomas E. "Lie Sphere Geometry", Springer 2000
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteYes. I have fixed it. I think....
DeleteThis comment has been removed by the author.
ReplyDeleteIndeed it was because of copy-paste.
DeleteIf you agree, we can delete these comments, as far as I'm concerned, because they don't bring anything new to the table.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThanks. Fixed.
DeleteNeeds to be said, great post! You are a natural for presenting heavy math in a funny and entertaining manner. Thank you!
ReplyDeleteHere’s the formula for stereographic ->
ReplyDeleteHere’s the formula for inverse stereographic
projection from the South ->
projection through the South
Corrected. Thank you.
DeleteUnfortunately, i did not understand the main idea...:( Why do we "embed our 2D point" at once "into 4D space", why not into a 3D space for beginning? Projections usually link spaces with adjacent dimensions (diminish the dimension of space by one), but may be my view is old-fashioned.
ReplyDeleteAnyhow, Ark, many thanks for the impetus, I should read about the Klein absolute and think it over myself. It seems to me that absolutes are important since they remain invariant under the movements of spaces which they encompass. But it is only a conjecture.
But I did it in two steps! First step to 3D to get to stereographic projection, second step adding one more dimension to get rid of denominators.
DeleteAnd a propos your comments about Lounesto book: you may like to read my post:
Deletehttps://ark-jadczyk.blogspot.com/2024/07/dangerous-math-tale-of-perilous-pursuits.html
Thank you, Ark, for this addition. Now, it appears to be clearer. Yes, I see your post where Lounesto book is mentioned, i will read it attentively, thanks a lot.
Delete(1) the post https://ark-jadczyk.blogspot.com/2024/07/dangerous-math-tale-of-perilous-pursuits.html
Deleteis scaring.
(2) The words appearing at the picture are all appealing ('dimensions' are my special favourites)
(3) 'He showed me how quantum fractals connect to conformal geometry', this phrase definitely catalyzes my reading of your QF book.
(4) Those tragedies evidence that some research paths can swirl their followers taking them too close to the absolute of human knowledge.
(4) Well, maybe not the "absolute", but perhaps into certain areas that have been 'forbidden'.
DeleteAgain, like with Pertti Lounesto, I was discussing with Berezin (coherent states) at the Alushta conference
Deletehttps://inspirehep.net/conferences/1415174
in April, a year before his tragic accident.
Alushta in April - an ideal time and place for a meeting. Gardens in blossom, spring air is fresh and warm.
DeleteOh, such famous names: Faddeev, Blokhintsev, Slavnov...
And such sounding titles: mostly about gauge theories.
I hope, our science will resurge some day in the former splendor
Indeed, I remember all these flowers quite vividly. And I remember a trip to Simferopol on a bus, when the door trapped the arm of my friend, J. Lukierski, and he cursed aloud in Polish! The driver understood!
Delete"I hope, our science will resurge some day in the former splendor"
DeleteIt certainly will!
The relation:
ReplyDelete(ζ0)2- (ζ1)2 - (ζ2)2 - (ζ3)2 = 0.
would be also true if ζ0 were opposite.
Indeed. It would be better to start with Minkowski space and the projective light cone and see what can get this way. Then the ambiguity of signs would show its geometrical meaning. I have chosen the "pedagogical" way.
DeleteWhat is important is that with my choice ζ3-ζ0 = 1>0. Sign of ζ3-ζ0 is important. With your choice it would be sign of ζ3+ζ0. We will come to this point in the next post.
Delete'But I did it in two steps! First step to 3D to get to stereographic projection, second step adding one more dimension to get rid of denominators'.
DeleteArk, a schematic drawing would be of much help here. At least for 2d and 3d, i am not asking for more.
You are right. I added a drawing taken from Ref. 2 in the P.S. to the note.
Delete