Lie Sphere Geometry Part 14: Q and Q+ as homogeneous spaces

 

The Erlanger Programm, delivered by Felix Klein in 1872 during his inauguration at the University of Erlangen, revolutionized mathematics by proposing that geometries should be classified by their underlying transformation groups. Klein's unifying framework emphasized that each geometry is defined by the group of transformations preserving its fundamental properties.

The Erlanger Programm, delivered by Felix Klein in 1872

We will be moving now in this direction.

14.1 The space infinity point [e∞]

In Part. 13 we have introduced, instead of e₀ and e₅ , the null vectors e₊ and e₋ . The vector e₋ is sometimes denoted e_∞ , and we will now see the reason. Let us take a sphere Sρ(c). It is represented in Q (and in Q⁺ ) by the formula:

τ_s (c, ρ) = [e₊ + q(c, ρ)e₋ + c + ρ e₄ ].                 (14.1)

We now fix the signed radius ρ, and examine the limit |c| → ∞. As far as |c| is finite and large enough so that |c| > |ρ|, q(c, ρ) is positive, we can rewrite (14.1) as (Why?)

τ_s(c, ρ) = [e₊/q(c, ρ) + e₋ + c/q(c, ρ) + ρ/q(c, ρ) e₄ ].                 (14.2)

In the limit |c| → ∞ the first, third and fourth term become zero, and we get

lim_|c|→∞ τ_s(c, ρ) = [e₋ ], (14.3)

independently of the (fixed) value of ρ. We have obtained a single point of Q, which we may call the spatial infinity and denote by [e_∞ ]. This result seems to contradict the fact that in Q and in Q⁺ every point is as good as any other point, since Q and Q⁺
are homogeneous space for the action of the group SO₀(4, 2), as we will see in the next section. To understand this apparent contradiction will require some work, and we will postpone it for later.

In Exercise 1 of Part 13, I was suggesting to do the previous reasoning with a plane instead of a sphere. Anna attempted to guess the answer. I wasn't thinking carefully at that time. With the sphere we were dividing by q(c,ρ), which for |c| large enough is a positive number. But for planes we would have to divide by h, which can be positive or negative. This works perfectly for Q, but for Q⁺ it would create a strange result: e_- or -e_-, which are two different points in Q⁺. depending on the sign of h, which I do not fully understand yet.

14.2 The action of O(4, 2) on Q and Q⁺

Let G be the matrix G = diag(1, 1, 1, 1, −1, −1). Thinking of u ∈ R^4,2 as column vectors we can then write the scalar product as:

u · u'= u^TGu',                 (14.4)

where · ^T stands for the transposition.
We define the pseudo-orthogonal group O(4, 2) as the group of all real 6 × 6 matrices L satisfying

L^TGL = G.                 (14.5)

This is the isometry group of R^4,2 , and any two orthonormal basis in R^4,2 are related by a certain matrix from O(4, 2).
It will be convenient to write matrices L in a block form

L =[A, B;C, D]                , (14.6)

with A, B, C, D being, respectively, 4 × 4, 4 × 2, 2 × 4, and 2 × 2 matrices. Using this block matrix form Eq. (14.5) takes the form:

A^TA − C^TC = 1,                 (14.7)
A^TB − C^TD = 0,                 (14.8)
B^TB − D^TD = −1,                 (14.9)
where 1, 0, −1 on the right-hand side are of dimensions 4×4, 4×2, and 2×2 respectively.

We do not need the fourth condition, since it is the transpose of the second condition, and so it is not independent of the three conditions above. As an immediate consequence of these conditions we have the following Lemma:

Lemma 14.1. For every L = [ A, B; C, D] ∈ O(4, 2) we have

| det(A)| ≥ 1, | det(D)| ≥ 1.                 (14.10)

Proof. See the Notes or Ch. 14 alone below.

Definition 14.1. We denote by SO₀(4, 2) the connected component of the identity of the group O(4, 2), that is the set of all elements of O(4, 2) which can be connected to the identity element by a continuous path within the group.

It follows from the definition that SO₀(4, 2)  is a group, a proper subgroup of O(4, 2). (Why?).

As a consequence of Lemma 14.1 we obtain

Proposition 14.1. For every L = [ A, B; C, D] ∈ SO₀(4, 2) we have

det(A) ≥ 1, det(D) ≥ 1.                 (14.12)

Proof. See the Notes or Ch. 14 alone below.

Remark 14.1. The group SO(4) × SO(2) is a subgroup of SO₀(4, 2) corresponding to B = 0 and C = 0, so that

L =[A, 0;0, D],                 (14.13)

where A ∈ SO(4), D ∈ SO(2), with SO(4) and SO(2) being connected.

In line with the Erlangen program of Felix Klein (it was in 1872) geometry consists of the study of invariants under a group of transformations. We have our group–that is SO₀(4, 2). We will now look at it as a group of transformations of Q and Q⁺ , and then study constructions that are invariant under this action. Notice that the action of O(4, 2) on R^4,2 is linear, and preserves the null cone {u ∈ R^4,2 : u² = 0}, therefore it acts on Q and on Q⁺ by L[u] = [Lu].

Proposition 14.2. The actions of SO₀(4, 2) on Q and on Q⁺ are transitive.

Proof. See the Notes or Ch. 14 alone below.

Exercise 1. The action of SO(2) on S¹ is transitive. Why?

Exercise 2. The action of SO(4) on S³ is transitive. Why?

P.S. 13-05-25 14;27 My third substack post (repost) Talking about Science: 3 Tony Smith and “arXiv.org”, Carlos Castro Perelman and the tide