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''.
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| 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⟘ .
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| 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.
Ark, I just started reading the entire document, but I have a tricky question about the Afternote:
ReplyDelete"Make the radius of this sphere approach zero when you approach p or -p".
And what about the maximum radius of this imaginary sphere? Minkowski space has a metric, so the size question seems reasonable. Moreover, I expected the speed of light to appear at the scene somehow. We always set "c" = 1 and loose it completely in this proportional world. But perhaps it is present in the proportion between the torus radii and the maximum size of the spheres?
It gives you only the topology. In fact it gives you more, it gives you a differentiable structure, as a manifold with corners. Forget metric - it is degenerate on the infinity manifold.
DeleteOkay, I see:
ReplyDelete"...this graphic representation is only topologically correct. Metrically it is totally misleading."
This is to be expected. Point "o" could be anywhere inside the blue (or yellow) pentagon in Figure 6, right?
Yes. "o" can be anywhere.
DeleteNow i am chewing the Proof of Lemma 4.1. Could you help me with a technical question:
ReplyDelete"Then B(e2,e2) = 2B(x,y) = 1, B(e4,e4) = −2B(x,y) = −1, <...> Vectors e2 and e4 span a subspace of V, and B restricted to this subspace is of signature (1,1)"
Why B restricted to is of signature (1, 1) if B(e2, e2) = 1 and
B(e4, e4) = -- 1? Shouldn't the signature be (1, -1)?
By signature of a nondegenerate quadratic form (or "metric" we usually mean a pair of nonnegative numbers (p,q), where p is the number of pluses and q number of minuses in the diagonal form of the metric. Alternatively we write (+++...,---), listing pluses p time and minuses q times.
DeleteIf the metric is degenerate we can say "signature (r,p,q), where is the number of zeros. For nondegenerate metric we use only (p,q). I am using this convention: p the number of (+1)'s, q - the number of (-1)'s.
Ok, thanks! Indeed, p=1 and q=1. This denotation is convenient, I don't know why it slipped my mind at that moment.
DeleteArk, the essay "Projective Geometry of the Null Cone in R2,2" has indeed become even more comprehensive than previously. Many points are perfectly clarified and even more pedagogically formulated. Thank you very much.
ReplyDeleteI am reading the whole text from the very beginning. Will report typos while meeting them:
p.3:
submanifotld --> submanifold
continuoslt --> continuosly
rotate by 180 degrees the first two coordinates --> rotate by 180 degrees the plane of the first two coordinates (?)
p.4: Wikipedia article on Grassmannians has only four
lines, very incomplete, about the oriented case --> ... the UNoriented case (?)
p.4
ReplyDeletewith graphics --> with graphics.
Consider the oriented case first case. --> Consider the oriented case first.
p.5
What is then the unoriented one. --> What is then the unoriented one?
Foe instance --> For instance
p.6
Note 3.1. . Well, --> Note 3.1. Well,
p.7
ReplyDeleteIf we set z = eiα, w = eiβ, , --> If we set z = eiα, w = eiβ,
Definition 4.1. . --> Definition 4.1.
Note 4.1. . --> Note 4.1.
p.8
For this w --> For this we
p.9
The case of x3 = x1 --> The case of x3 = x1.
The case of x3 = −x1 --> The case of x3 = −x1.
p.10
The case of x3 = x1 --> The case of x3 = x1.
p.11
The case of x3 = −x1 --> The case of x3 = −x1.
p.12
Figure 8: p⊥. The unoriented case. Square diagram/ --> Figure 8: p⊥. The unoriented case. Square diagram.
We will find now the stability group Gp < SO0(V ) of p --> We will find now the stability group Gp < SO0(V ) of p.
as we expact --> as we expect
p.13
The subgroup Ts(a) --> The subgroup Ts(a).
The subgroup L(ζ) --> The subgroup L(ζ).
The subgroup D(α) --> The subgroup D(α).
This is the expected dimesion. --> This is the expected dimension.
p.14
et o an arbitrary point in M+ --> let o be an arbitrary point in M+.
The coordinates of y transforms --> The coordinates of y transform
as well a --> as well as
very point q --> every point q
p.15
(for the group SO0(V) --> (for the group SO0(V))
Fixed. Thanks!
Delete"It (p⊥) contains at least these two points: p and −p. The construction
ReplyDeleteis O(V )-invariant: L(p⊥) = (Lp)⊥. Thus through each point of PN+ we have two special curves"
Sorry, I cannot grasp the conclusion above. We have p⊥ with two opposite points p and -p on it. Due to O-invariance, we can rotate the construction and obtain a new curve (p⊥)' = L(p⊥), which is perpendicular to the image of point p: p' = Lp. But where do these 'two special curves' come from?
p⊥ is the union of these two curves, with p and -p being the two common points. p⊥ consist of nothing more than these two curves.
DeleteOne more misprint has been found:
ReplyDeletep. 12
For x1 and x2 constant --> For x2 and x3 constant
Thank you! Fixed.
Delete"Since they all belong to O(V), they leave invariant the set p⊥"
ReplyDeleteDoes it mean that p⊥ is an orbit of point p under O(V)-transformation?
Ark, thanks, I have seen myself, p⊥ is of course an O(V) orbit, namely, it is the orbit on which the form Q = inv = 0.
DeleteO(V) acts on PN+ transitively. So the O(V) orbit of any point is the whole PN+.t
DeleteOk, the whole PN+ is the factor-group of SO(V) by the p-point stability group Gp: SO(V)/Gp, right? But p⊥ is a very special trajectory. Isn't it an orbit of some group? Is there a group transform that moves a point along its ⊥-complement?
Deletep⊥ consists of several orbits of Gp. Two of them are one-point orbits {p} and {-p}, the rest of p⊥ are half-circle orbits. Probably four such.
DeleteThank you, Ark! Probably, you've been planning to discuss this in the following post, and I am running ahead of the train.
DeleteThank you. Fixed.
ReplyDelete