Quaternionic Lorentzian space

Quaternionic Lorentzian (n + 1)-space is the inner product space consisting of the right H-module Hⁿ⁺¹ together with the (n + 1)-dimensional Lorentzian inner product.

The Lorentzian inner product is defined by

hx, yi = ¯x₁y_n+1+ ¯x_n+1y₁+ ¯x₂y₂+ · · · + ¯x_ny_n.

where x = (x1, . . . , xn+1) and y = (y1, . . . , yn+1). We denote by H^n,1 the quaternionic Lorentzian (n + 1)-space.

Definition 6.1. A vector x is said to be spacelike (resp. timelike and lightlike) if hx, xi > 0 (resp. hx, xi < 0 and hx, xi = 0).

Example 6.2 (Spacelike, timelike and lightlike vectors). We give examples of spacelike, timelike and lightlike vectors. First, (0, 1, 0) is a spacelike vector with positive norm.

An example of timelike vector is (i√

2/2, 0, −i√

2/2) whose norm is −1 < 0. Lastly, (1, 0, 0) is lightlike vector whose norm is 0.

Let 0 denote the origin vector (0, 0, . . . , 0) in Hⁿ⁺¹. obvious. Thus we complete the proof.

Lemma 6.4. For x, y ∈ H^n,1, hx, yi = hy, xi.

If hx, yi ∈ R, then hx, yi = hx, yi and thus hy, xi = hx, yi by Lemma

6.4. We have

the following corollary.

Corollary 6.5. If hx, yi ∈ R, then hy, xi = hx, yi.

Let v be a nonzero vector in H^n,1. Denote by v^⊥ the set of all vectors x ∈ H^n,1 with hv, xi = 0 i.e.,

v^⊥= {(x₁, . . . , x_n) ∈ Hⁿ⁺¹ | ¯v₁x_n+1+ ¯v_n+1x₁+ ¯v₂x₂+ · · · ¯v_nx_n= 0}.

By Lemma

4.6, v

^⊥ is a right H-submodule of Hⁿ⁺¹ of dimension n. From now on, we will focus on the case of n = 2.

Lemma 6.6. Let x be a nonzero timelike vector in H^2,1. Then x /∈ x^⊥ and every nonzero vector of x^⊥ is spacelike.

Proof. First of all, since x is timelike, it follows that hx, xi < 0. Thus obviously x /∈ x^⊥. Suppose that y = (y1, y2, y3) ∈ x^⊥. Then yq ∈ x^⊥ for all quaternions q ∈ H.

This follows from hx, yqi = hx, yiq = 0. Furthermore by observing hyq, yqi = ¯qhy, yiq = |q|²hy, yi

in order to prove that y is spacelike, it is sufficient to prove that yq is spacelike for some non-zero quaternion q.

Let x = (x₁, x₂, x₃). From the assumption that x is timelike, it follows that x₁6= 0.

By scaling x, we may assume that x1 = 1. The conditions of hx, xi < 0 and hx, yi = 0 give

2Re(x3) + |x2|²< 0, (6.1) y3+ ¯x2y2+ ¯x3y1= 0. (6.2) If y1 = 0, then hy, yi = |y2|². If y1 = 0 and y2 = 0, (6.2) forces y3 = 0, contrary to y 6= (0, 0, 0). Hence if y₁ = 0, then y₂ 6= 0, which implies that hy, yi = |y₂|² is positive.

In other words, y is spacelike.

From now on, we suppose that y₁ 6= 0. Then by scaling y, we can assume that y₁= 1. Then (6.2) is written by y₃+ ¯x₂y₂+ ¯x₃ = 0 and

hy, yi = 2Re(y₃) + |y₂|²

= −2Re¯x₂y₂− 2Re¯x₃+ |y₂|²

> −2Re¯x₂y₂+ |x₂|²+ |y₂|² = |x₂− y₂|² ≥ 0.

This leads to a conclusion that every non-zero vector in x^⊥ is spacelike.

Lemma 6.7. Let x be a nonzero spacelike vector in H^2,1. Then x /∈ x^⊥and x^⊥contains timelike, spacelike and lightlike vectors. Furthermore, there are linearly independent two lightlike vectors in x^⊥.

Proof. First of all, since hx, xi > 0, it immediately follows that x /∈ x^⊥. We claim that there is a timelike vector in x^⊥. Suppose, contrary to our claim, there is no timelike vector in x^⊥. Then every vector in x^⊥ is either spacelike or lightlike. If there is a spacelike vector y ∈ x^⊥, then we can find an orthogonal basis {x, y, z} where z ∈ x^⊥∩ y^⊥. By hypothesis, hz, zi ≥ 0. Then every vector v ∈ H^2,1 is written by v = xa + yb + zc for some a, b, c ∈ H and

hv, vi = |a|²hx, xi + |b|²hy, yi + |c|²hz, zi > 0 which contradicts the existence of a lightlike vector in H^2,1.

Now we can assume that every vector in x^⊥ is lightlike. Then there are linear independent vectors y and z in x^⊥. Since dim_Hx^⊥ = 2, it follows that span_H{y, z} = x^⊥. From the assumption that every vector in x^⊥ is lightlike, for all a, b ∈ H,

0 = hya + zb, ya + zbi = |a|²hy, yi + |b|²hz, zi + 2Re(¯ahy, zib) = 2Re(¯ahy, zib) which leads to a conclusion that hy, zi = 0. Then {x, y, z} is an orthogonal basis of H^2,1. Since x is spacelike and, y and z are lightlike, it is derived that every vector has non-negative norm. This also contradicts the existence of a timelike vector in H^2,1. Therefore, the claim holds. In other words, there is a timelike vector y ∈ x^⊥.

Choosing a vector z ∈ x^⊥∩y^⊥, we have an orthogonal basis {y, z} of x^⊥. By scaling y and z, we may assume that hy, yi = −1 and hz, zi = 1. Since x^⊥ = span_H{y, z}, it can be easily seen that y + z and y − z are linearly independent lightlike vectors in x^⊥. Summarizing, there are a timelike vector y, a spacelike vector z and a lightlike vector y + z in x^⊥. Furthermore, there are linearly independent two lightlike vectors y + z and y − z in x^⊥.

Lemma 6.8. Let x be a nonzero lightlike vector in H^2,1. Then x ∈ x^⊥ and every vector of x^⊥ is either spacelike or lightlike. Furthermore, y is a lightlike vector in x^⊥ if and only if y = xq for some nonzero q ∈ H.

Proof. It follows from the assumption of hx, xi = 0 that x ∈ x^⊥. First observe that there are no timelike vectors in x^⊥. If there is a timelike vector perpendicular to x, then x must be spacelike, contrary to the hypothesis that x is lightlike. Thus every vector of x^⊥ is either spacelike or lightlike. It remains to prove the second statement of the Lemma.

Suppose that there is a lightlike vector y such that {x, y} is linearly independent.

Then x^⊥= span_H{x, y}. For any a, b ∈ H,

hxa + yb, xa + ybi = |a|²hx, xi + |b|²hy, yi + 2Re(¯ahx, yib) = 2Re(¯ahx, yib).

If hx, yi = 0, then

2Re(¯x1x3) + |x2|² = 0, (6.3) 2Re(¯y1y3) + |y2|² = 0, (6.4)

¯

x1y3+ ¯x2y2+ ¯x3y1 = 0, (6.5) where x = (x₁, x₂, x₃) and y = (y₁, y₂, y₃). If x₁ = 0, then x₂ = 0 by (6.3) and hence x₃ 6= 0. Then y₁ = 0 by (6.5) and y₂= 0 by (6.4). This implies that x, y ∈ span_H{e₃} and hence, x and y are linearly dependent, contrary to the hypothesis. Thus x1 6= 0 and y₁6= 0. By scaling x and y, we may assume that x₁ = y₁ = 1. Then

2Re(x3) + |x2|² = 0, (6.6)

2Re(y3) + |y2|² = 0, (6.7)

y3+ ¯x2y2+ ¯x3 = 0. (6.8) If we take the real part on both sides of (6.8), we have

0 = 2Re(y₃) + 2Re(¯x₂y₂) + 2Re(¯x₃) = −|y₂|²+ 2Re(¯x₂y₂) − |x₂|²= −|x₂− y₂|². Therefore x2 = y2. By (6.6) and (6.7), we have Re(x3) = Re(y3). Taking the imaginary part on both sides of (6.8),

0 = Im(y3) + Im(¯x2y2) + Im(¯x3) = Im(y3) + Im(|x2|²) − Im(x3) = Im(y3) − Im(x3).

Since Re(x₃) = Re(y₃) and Im(x₃) = Im(y₃), we conclude that x₃ = y₃. Therefore x = y, contrary to the assumption that x and y are linearly independent. Finally, there are no lightlike vectors in x^⊥ which are linearly independent with x.

Lemma 6.9. Let V be a 2-dimensional right H-submodule of H³. Then there is a nonzero vector x ∈ H³ such that V = x^⊥.

Proof. Since V is a 2 dimensional right H submodule of H³, there exists a basis {u, v}.

V = {uq₁+ vq₂ | q₁, q₂∈ H} = span_H{u, v}.

Find an element x ∈ H^2,1 such that x^⊥∈ V.

Since hu, xi = hv, xi = 0, we have huq₁ + vq₂, xi = q₁hu, xi + q₂hv, xi = 0 for any q1, q2∈ H. Let u = (u1, u2, u3), v = (v1, v2, v3), then

u₃x₁+ u₂x₂+ u₁x₃ = 0.

v3x1+ v2x2+ v1x3 = 0.

We can find a soluton of these equations. Because the number of x_i is larger than the number of equations.

Corollary 6.10. Let V be a 2-dimensional right H-submodule of H³. Then

(1) if V has linearly independent two lightlike vectors, then V = x^⊥ where x is space-like,

(2) if V has only one lightlike vectors up to scaling by a nonzero quaternion, then V = x^⊥ where x is lightlike.

(3) if V has no lightlike vectors, then V = x^⊥ where x is timelike.

Definition 6.11. A basis {v₁, v₂, v₃} of H^2,1 is said to be orthonormal if |hv_i, v_ii| = 1 for i = 1, 2, 3 and hvi, vji = 0 for all distinct i, j.

Example 6.12. An example of an orthonormal basis of H^2,1 is

ne1√+e3 con-sists of two spacelike vectors and one timelike vector.

Proof. If everty vector vi is spacelike, for any v = v1a1+ v2a2+ v3a3, hv, vi = |a₁|²hv₁, v1i + |a₂|²hv₂, v2i + |a₃|²hv₃, v3i ≥ 0

which means that every vector in H^2,1 is spacelike, contrary to the fact that there is a timelike vector in H^2,1. If every vector v_i is timelike, every vector in H^2,1is timelike in a similar way, contrary to the fact that there is a spacelike vector. Thus any orthonormal basis of H^2,1 has at least one spacelike and one timelike vector.

An example of orthonormal basis of H^2,1 is {^e1√+e3

2 , e₂,^e1√−e3

2 }. Here ^e1√+e3

2 and e2 are spacelike vectors and ^e1√−e3

2 is a timelike vector. Let v1 be a spacelike vector and v₂, v₃ timelike vectors. Then span_H{^e1√+e3 of H^2,1 has two spacelike vectors and one timelike vector.

Let f : H^2,1→ H^2,1 be a right H-module homomorphism preserving the Lorentzian inner product i.e.

hf (x), f (y)i = hx, yi.

Then the quaternion 3 × 3 matrix A associated to f satisfies A^∗J A = J where

J =

Let Sp(2, 1) denote the set of all quaternion 3 × 3 matrices A with A^∗J A = J . Lemma 6.14. Every element of Sp(2, 1) is invertible.

Proof. For any A ∈ Sp(2, 1), A^∗J A = J and hence J · A^∗J A = J · J = I. Then it follows that A⁻¹ = J A^∗J . In other words every element of Sp(2, 1) is invertible.

Proposition 6.15. The set Sp(2, 1) is a group.

Proof. Let A, B and C be matrices in Sp(2, 1). The associativity of quaternions gives A · (B · C) = (A · B) · C. This implies that Sp(2, 1) satisfies the associative law.

Obviously, the identity matrix I is in Sp(2, 1) and thus it is the identity of Sp(2, 1).

By Lemma

6.14, each matrix A in Sp(2, 1) has an inverse A

⁻¹ = J A^∗J . Moreover, (J A^∗J )^∗· J · (J A^∗J ) = J A(J A^∗J ) = J AA⁻¹= J.

Thus A⁻¹∈ Sp(2, 1). Therefore Sp(2, 1) is a group.

Let V0 be the set of lightlike vectors of H^2,1. Let V+ (resp. V−) be the space of spacelike (resp. timelike) vectors whose norms are 1 (resp. −1).

There is a natural Sp(2, 1)-action the space of bases of H^2,1 as follows: Let A ∈ Sp(2, 1) and {v₁, v₂, v₃} be an orthonormal basis of H^2,1. Since A preserves the Lorentzian inner product, it immediately follows that {Av1, Av2, Av3} is also an orthonormal basis of H^2,1.

Definition 6.16 (transitivity). A group action G × X → X is transitive if it possesses only a single group orbit, i.e., for every pair of elements x and y, there is a group element g such that gx = y. In this case, X is isomorphic to the left cosets of the isotropy group, X ∼ G/G_x. The space X, which has a transitive group action, is called a homogeneous space when the group is a Lie group.

If, for every two pairs of points x₁, x₂ and y₁, y₂, there is a group element g such that gxi = yi, then the group action is called doubly transitive. Similarly, a group action can be triply transitive and, in general, a group action is k-transitive if every set {x₁, ..., yk} of 2k distinct elements has a group element g such that gx_i = yi.

Theorem 6.17. The Sp(2, 1)-action on the set of orthonormal bases of H^2,1 is tran-sitive.

Proof. Let u1 = (e1 + e3)/√

2, u2 = e2 and u3 = (e1− e₃)/√

2. Then we have shown that {u₁, u₂, u₃} is an orthonormal basis of H^2,1. To prove the theorem, it suffices to show that for a given orthonormal basis {v1, v2, v3} ∈ F₁, there exists an element

A ∈ Sp(2, 1) such that A · {u1, u2, u3} = {v₁, v2, v3}. Let |v₁| = |v₂| = 1 and |v₃| = −1.

. Then by a direct computation,

Au1 = A e₁+ e₃

0. By scaling y, we may assume that hx, yi = 1. According to corollary

6.10, there is a

spacelike vector z such that hz, zi = 1 and z^⊥= span_H{x, y}. Define A = (x, z, y). Theorem

6.17, there exists an element A ∈ Sp(2, 1) such that Ae

₂ = y. Hence the Sp(2, 1)-action on V+ is transitive. Similarly, one can prove that the Sp(2, 1)-action on V− is transitive.

Proposition 6.19. The Sp(2, 1)-action on V₀×V₀\∆ is transitive where ∆ = {(x, x) | x ∈ V0}.

Proof. In the proof of Lemma

6.18, we have shown that for any linearly independent

vectors x and y in V0, there exists an element A ∈ Sp(2, 1) such that Ae1 = x and Ae₃= y, which implies the transitivity of the Sp(2, 1)-action on V₀× V₀\ ∆.

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문서에서 사원수 선형 대수에 관한 연구 (페이지 33-41)