In the previous note Constructing the conformal infinity we have started with one-point compactification of the Euclidean space and ended up with what is called the conformally compactified Minkowski space. In this note we will start with the end of the previous note and study the emerging structure in some detail. Therefore, let V denote R6 with Cartesian coordinates x,v,w,
x= (x1,x2,x3,x4) ∈ R4,
endowed with the quadratic form
q(x,v,w) = x2+v2-w2,
where
x2=(x1)2+ (x2)2+ (x3)2- (x4)2.
We define the “cone” C as the set of all (x,v,w) satisfying
q(x,v,w)=0, (x,v,w) ≠ 0.
So we have removed the origin x=v=w=0.
C inherits its topology from R6: open sets in C are intersections of open sets in R6 with C.
There are two equivalence relations in C that are of interest, we will denote then R and R+. They are defined as follows
(x,v,w) R (x',v',w') if and only if (x',v',w') = a(x,v,w), a∈R, a≠ 0
(x,v,w) R+ (x',v',w') if and only if (x',v',w') = a(x,v,w), a∈R, a>0
We take the quotients of C by these equivalence relations and denote them
Mc and M2c respectively:
Mc = C/R,
M2c = C/R+
We call Mc the compactified Minkowski space, and M2c the double compactified Minkowski space.
We endow both sets with the quotient topology. Thus open sets in Mc are defined as those of which the preimages by the natural map from C to C/R are open in C. Similarly with M2c .
We have a natural covering map from M2c to Mc . Let us denote by [(x,v,w)] the equivalence class of (xv,w) with respect to R, and by
[(x,v,w)]+ its equivalence class with respect to R+. Then [(x,v,w)] is the set
[(x,v,w)] ={a(x,v,w): a ≠ 0}
while
[(x,v,w)]+ ={a(x,v,w): a > 0}
We can say that [(x,v,w)]+ is a half-ray through (x,v,w), while [(x,v,w)] is the whole ray (excluding 0) through (x,v,w). Every half-ray is contained in a unique ray, and half-rays [(x,v,w)]+ and half-rays [(x,v,w)]+ and [-(x,v,w)]+ are contained in the same ray. Therefore the natural map M2c to Mc is 2:1.
The group O(4,2) is the group of linear transformations of R6 leaving invariant the quadratic form q. Its subgroup SO(4,2) consist of transformations given by 6x6 matrices of determinant 1. Since the transformations are linear, they transform equivalence classes of R and of R+ into equivalence classes, therefore they define transformations of Mc and M2c.
We now show that the action of SO(4,2) on M2c is transitive and that M2c is (pathways) connected.
Let [x,v,w]+ and [x',v',w']+ be two different points in M2c. We have q(x,v,w)=0 and q(x',v',w')=0. Therefore
(x1)2+ (x2)2+ (x3)2 + v2 = w2+(x4)2
and the expression above must be >0 since the origin is excluded. We can now replace (x,v,w) by the unique element in the same equivalence class for which the right hand side =1. We do the same with the second point. So we have
(x1)2+ (x2)2+ (x3)2 + v2 = w2+(x4)2=1
(x'1)2+ (x'2)2+ (x'3)2 + v'2 = w'2+(x'4)2=1
Assuming only one space dimension, therefore neglecting x2 and x3, here is what I got for a graphic representation of M2c , and the double conformal infinity (v=w):
Note: The numbers on the graphics are irrelevant.
Mathematica code used for the graphics:
torus2[a_, b_][fi_,psi_] := {(a + b Cos[fi]) Cos[psi], (a + b Cos[fi]) Sin[psi], b Sin[fi]}
mc = ParametricPlot3D[torus2[8, 3][fi, psi], {fi, 0, 2 Pi}, {psi, 0, 2 Pi}, Mesh -> None, PlotStyle -> Directive[Opacity[0.6], LightGray, Specularity[White, 10]]]
inf1 = ParametricPlot3D[torus2[8, 3][fi, fi], {fi, 0, 2 Pi}, PlotStyle -> {Black, Thick}]
inf2 = ParametricPlot3D[torus2[8, 3][fi, Pi-fi], {fi, 0, 2 Pi}, PlotStyle -> {Black, Thick}]
Show[{mc, inf1, inf2}]
Notice that M2c is, topologically, S1x S3 rather than S1x S1 as on the picture, where space is one- instead of 3-dimensional.
In the next post we will draw images of the Minkowski space on the torus, and we will see how time translations are represented there.
P.S.1. Something else to think about, that Laura brought my attention to:
By the way a very good example of how to give a talk!
Bjab -> Ark:
ReplyDelete"ended up with is called the conformally Minkowski space."
Mój kompilator naprawił sam dwa warningi. (Jest w sumie niezłej jakości.)
You can be really proud of your compiler. Fixed. Thanks.
DeleteBjab -> Ark
ReplyDeleteSkąd bierze się nazwa "równokątne uzwarcenie"? Co to znaczy równokątne? Czy mnożenie przez (1 + X^2) z poprzedniej notki zachowuje równokątność?
Dużo by mówić... I will remember your question about the origin of the term "conformal" in "conformal compactification". And I will also have to explain "compactification" in the case of Minkowski space.
DeleteAnd yes, multiplying all components of two tangent vectors by the same number(which may depend on a point) preserves "angles" between these vectors. For the Euclidean case and the sphere it is intuitive. In the case of Minkowski space - it goes deeper because we did not yet define any geometry on the compactification. Till now we have considered only its topology, and even so, not deeply enough, because we did not explain the not so evident "compactification" in this case.
Thank you for the question.
Bjab -> Ark
DeleteThank you for your answer.