This post is a continuation of The infinity ab initio., which ended with:
M∞ = {[X]: X0 = 0} = {[X]: X5 = X6}. (17)
M+ Blue, M- Black, M∞, Red.
In the next post we see that PN is a disjoint union of three sets
PN = M+ ∪ M- ∪ M∞, (18)
where
M+ = {[τ(x)]: x ∈ M}, (19)
M- = {[-τ(x)]: x ∈ M}. (20)
Thus PN, the doubly compactified Minkowski space, consists of two copies of M, and of the infinity set M∞.
So, here comes the promised 'next post'. We will discuss (18), and then, in the next post, we will move to the group of automorphisms of PM - a very important concept.
It should be clear that, with N = {X∈V: Q(X) = 0}, PN is a disjoint union of the following three sets
PN = {[X]: X∈N, X5 > X6} ∪ {[X]: X∈N, X5 < X6} ∪ {[X]: X∈N, X5 = X6}. (21)
We will now show that
M+ = {[X]: X∈N, X5 > X6} (22)
and
M- = {[X]: X∈N, X5 < X6}. (23)
First we recall the definition of the map τ: M → PN
τ(x) = [X(x)] = [ ( x, ½(1 - q(x, t)), -½(1 + q(x, t)) )]. (24)
Note. While working this out I realized that there were signs errors in the formula (9) of The infinity ab initio.
Invaluable Bjab seems to be, I hope, temporally, unavailable -
otherwise He would have noticed the the original X(x), in (9) therein,
was not in N. I corrected (9) and then (15).
Thus X5(x) - X6(x) = 1, and 1 is certainly >0. Therefore
M+ ⊂{[X]: X∈N, X5 > X6}. (25)
In order to prove (22) we thus need to show that {[X]: X∈N, X5 > X6}⊂M+. So, let X∈N,, and assume X5 > X6, thus
X5 - X6 > 0. We can choose a unique representative of the equivalence class [X] of X for which
X5 - X6 = 1. (26)
Define the real number q by
X5 + X6 = -q. (27)
From (26) and (27) we have
X5=(1-q)/2, X6 = -(1+q)/2. (28)
Thus X is of the form (x, (1-q)/2, -(1+q)/2), where x is in ℝ4.
But X is in N, i.e. X·X = x·x + (X5)2 - (X6)2 = 0. That means
x·x + (½(1-q))2 - (-½(1+q))2 = 0.
But (½(1-q))2 - (-½(1+q))2= -q, and so q = x·x =q(x). Therefore [X] = τ(x), and is in M+ . It is then evident that -[X] ≐ [-X] is in
M- . QED
"we will move to the group of automorphisms of PM" -->
ReplyDelete"we will move to the group of automorphisms of PN", ok?
"From (25) and (26) we have" -->
ReplyDelete"From (26) and (27) we have", right?
Fixed. Thanks.
Delete"Thus X is of the form (x, (1-q)/2, -(1+q)/2), where x is in ℝ4"
ReplyDeleteArk, why are you proving this fact, although we assumed that it is so by definition (24)?
I should have written it in a more clear way. The logic is this:
ReplyDeleteX(x) is of a very special form. While X1,...,X4 are arbitrary, X5 and X6 are very specific functions of the four first coordinates. The reasoning is to show than ANY X in N, such that X5-X6 is positive, is necessarily of this particular form.
Now I have got the idea, thank you!
ReplyDeleteWith this special form of X(x):
for X = (x, 1/2(1 - x^2), -1/2(1 + x^2)) we always have X5 > X6
and
for -X = (-x, -1/2(1 - x^2), 1/2(1 + x^2)), we always have X5 < X6
But we cannot make X5 = X6 whatever is x.
This is the weird M∞ region, which has no preimage in M.
M∞ is "at infinity", so it is beyond M+, because M+ consists only of finite points. But we will see that geodesics in M+ can be naturally extended, so that they can go to M∞ (and even beyond!).
ReplyDelete