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Symmetry Plane Causes Relativity and Rotation

Hiroaki Fujimori

e-mail: fuji@spatim.sakura.ne.jp; 2021.1.31

Abstract: Euclidean geometry, inherited from ancient Greece, was modeled on axiomatic methods

in modern science. Hilbert's “Foundations of Geometry” supplemented the lacking axioms and

seemed to have reached the stage of completion of plane geometry, but still a question remains

what the ultimate nature of a plane is. Looking back on the history of special relativity, Lorentz

and Poincaré were on their way to finding a theory to prove the results of Michelson-Morley

experiment. Meanwhile, Einstein published the theory of relativity. It seems that all things have

been done, but this is not enough. Digging into why the relativity principle holds, we arrive at a

deeper symmetry of spacetime. A paragraph of Hannyashin Sutra “空即是色 Kuu soku ze shiki”

is interpreted as “emptiness contains the form of cosmos". When a four-dimensional linear

spacetime is cut by a plane, a plane geometry is born. The basis of this plane geometry is linearity

and front-back symmetry from which rotation transformations of Euclidean geometry and Lorentz

transformations of the special theory of relativity are deduced.

Keywords: symmetry plane; Lorentz transformation; special relativity; arrow of time

1. Introduction

1.1. Three basic questions for mathematics and physics

For plane geometry

Why does the axiom system not depart from the existence condition of a plane itself [1], [2]?

For linear algebra

Why is the inner product not deduced from a Euclidean plane, but defined on a vector space?

A Euclidean plane is not an artifact but a gift of nature.

For the theory of special relativity

The condition of unidirectional of time is hidden behind the relativity principle, and as a result,

this condition is not used explicitly to derive the Lorentz transformation.

1.2. What does the symmetry of a plane deduce?

Put right-hand oblique coordinate systems on both face sides of a plane and make them

coincide with their origins. We define a 2×2 rear surface coordinate transformation Ｂ as an inside

out transformation, then detＢ< 0. Because it is not possible to distinguish which side of a plane is

the back or front, the symmetry plane equation is Ｂ=Ｂ ‒1. We obtain an oblique reflection

transformation Ｂ with two degrees of freedom:

Ｂ= (−𝑎 −𝑏𝑘𝑏 𝑎

), where detＢ=－a2＋kb2 =－1. (1)

Then we derive a special isodiagonal transformation Ｆ fromＢ such that Ｆ=ＭＢ= (−1 00 1

)Ｂ

= ( 𝑎 𝑏𝑘𝑏 𝑎

) which represents the coordinate transformation between right-hand systems from x1-y1

to xF-yF. It is found that k is a commutative coefficient, and the sign of k represents the type of a

symmetry plane with completely isotropic or semi-isotropic (see Equations (46)－(55)). If k=－1, then

Ｆ is a rotation transformation and Ｂ is a reflection transformation. If k< 0, k= 0, or k> 0, then Ｆ is

referred to as elliptic transformation, Galilean transformation, or Lorentz transformation

respectively. When the coefficient k is fixed on a symmetry plane, then matrices Ｆ and Ｂ have a

common invariant function φ (Ｂp)=φ (Ｆp)=φ (p)=－kx2＋y2 and they create a geometry of

isometric transformation group based on the invariant function φ(p) with the metric ‖Ｂp‖=

mailto:fuji@spatim.sakura.ne.jp

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‖Ｆp‖=‖p‖=φ(p)1/2 (see Eqations (37)－(39)). Thus, the symmetry of a plane gives rise not only

to Euclidean geometry when k=－1, but also to the principle of relativity when k ≥ 0.

2. Preliminaries

Terms:

An oblique reflection plane has two kind of intersecting lines such that one fold line and

innumerable parallel isotropic lines. A point p is transformed to a point Ｂp in a same isotropic

line by an oblique reflection transformation Ｂ, and their middle point is in the fold line (see

Figure 1.). This plane is semi-isotropic centered a fold line.

Definitions:

Invariant function: Let Ａ=(𝑎 𝑏𝑐 𝑑

) be a matrix, p=(𝑥𝑦) be a point, and f be a function of p. If f(Ａp)

= f(p), then this function f(p) is called an invariant function of Ａ.

Spacetime is a four-dimensional unified entity of space and time without considering all of the

matter from the universe.

Space is continuous, infinite, homogeneous, three-dimensional, and isotropic.

Time is continuous, infinite, homogeneous, one-dimensional, unidirectional, and irreversible.

A line in spacetime is one-dimensional, and it is inversion invariant for any point on itself.

A plane in spacetime is two-dimensional, and it is inversion invariant for any axis of a fold line

passing through any two points on itself.

An asymmetric plane is a plane with distinguishable back and front surfaces.

A symmetry plane or a symmetric plane is a plane with indistinguishable back and front surfaces.

A line of a coordinate axis on a plane in spacetime is one of two types, namely, a space line is

isotropic, or a time line is unidirectional.

A plane in spacetime is one of two types, namely, space × space type or space × time type.

A space × space type plane is completely isotropic, since the space line is isotropic. This type of

plane exists as a subspace of a three-dimensional space in an inertial system.

A space × time type plane is semi-isotropic, since the time line is of unidirectional and the

space line is isotropic. This type of plane exists when we think of one-dimensional space in

which all inertial systems move on one line and inertial coordinate systems with space-time

axes coexist in one common space × time plane. Note that “spacetime” means four- dimensional

space of space-time, and “space × time” means two-dimensional plane.

Axioms:

Inertial system axiom: There are an infinite number of empty inertial coordinate systems (or

inertial systems) in empty spacetime. Each has its own four-dimensional spacetime and keeps

its uniform motion on a straight line.

Symmetry plane axiom: It is not possible to distinguish which side of a plane in spacetime is the

back or which is the front.

Theorem:

A symmetry plane in spacetime is a linear space.

Mathematical preparations:

An invariant line f(p) of a 2×2 matrix Ｂ= (𝑎 𝑏𝑐 𝑑

) is the solution to a first order invariant

function equation in the form of f(Ｂp) ≡ f(p) = ux＋vy

⇔ u(ax＋by)＋v(cx＋dy) = ux＋vy ⇔ [(a－1)u＋cv] x＋[bu＋(d－1)v] y = 0

⇔ x,y are arbitrary, and in order to have a non-self-explanatory solution of u and v,

the determinant = (a－1)(d－1)－bc = 0. (2)

This means when the matrix Ｂ has an eigen value λ=1, we obtain an invariant line f(p) by

substituting u=c, v=－(a－1) on f(p) = ux＋vy. Thus, an invariant line is

f(Ｂp) ≡ f(p) = cx－(a－1)y. (3)

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Quadratic invariant function φ(p) of a 2×2 matrix Ａ= (𝑎 𝑏𝑐 𝑑

) is the solution of a second order

invariant function equation in the form of

φ(Ａp) ≡ φ(p) = ux2＋vy2＋wxy ⇔

φ(Ａp) =φ(ax＋by, cx＋dy) = u(ax＋by)2＋v(cx＋dy)2＋w(ax＋by)(cx+dy) = ux2＋vy2＋wxy

⇔ [(a2－1)u＋c2v＋acw] x2＋[b2u＋(d2－1)v＋bdw] y2＋[2abu＋2cdv＋(ad＋bc－1)w] xy = 0.

(4)

Since x2, y2, and xy are arbitrary, each coefficient must be equal to 0, and we obtain a

simultaneous equation of u, v, and w,

(𝑎2－1 𝑐2 𝑎𝑐

𝑏2 𝑑2 − 1 𝑏𝑑2𝑎𝑏 2𝑐𝑑 𝑎𝑑 + 𝑏𝑐 − 1

) (𝑢𝑣𝑤

) = (000

). (5)

In order to have a non-self-explanatory solution, the determinant of the matrix in Equation (5)

must be equal to 0. Thus,

the determinant = (ad－bc－1)[(ad－bc＋1)2－(a＋d)2] = 0

⇔ ad－bc = 1 or ad－bc =－1 and a＋d = 0. (6)

Here, by using the solution of u=－c, v=b, w=a－d, we obtain an identity:

φ(Ａp) ≡ detＡ･φ(p), where Ａ is a matrix Ａ= (𝑎 𝑏𝑐 𝑑

), p is a point p=(𝑥𝑦),

and φ(p) is a quadratic function given by φ(p) =－cx2＋by2＋(a－d)xy.

(7)

(1) From Equations (6) and (7), if detＡ= ad－bc =1, then

φ(Ａp) =φ(p) =－cx2＋by2＋(a－d)xy. (8)

In this case, φ(p) is the quadratic invariant function of the matrix Ａ. φ(p) is allowed to be

multiplied by a scale factor, such that

if Φ(p) = rφ(p), then Φ(Ａp) = rφ(Ａp) = r detＡφ(p) = rφ(p) = Φ(p). (9)

(2) Moreover, if detＡ≠1, then φ(p) is called a relative invariant function of matrix Ａ.

(3) We also obtain another invariant function from Equation (6). If detＡ= ad－bc =－1 and traceＡ=a

＋d=0 ⇔ if eigen values of Ａ are λ=±1, then change the notation of matrix Ａ into Ｂ for

convenience, and the invariant function has the same part of Equation (8) with the cross term xy

cancelled, such that

when Ｂ= (−𝑎 −𝑏𝑐 𝑎

), and detＢ=－a2＋bc =－1, then φ(Ｂp) =φ(p) =－cx2＋by2. (10)

The following assumes that commutative coefficients k and h are fixed on a coordinate plane.

A 2×2 special linear transformation matrixＳ has commutative coefficients k and h, and is

disassembled as

Ｓ=(𝑎 𝑏𝑐 𝑑

) = (𝑚 + ℎ𝑏 𝑏

𝑘𝑏 𝑚 − ℎ𝑏),

where detＳ= m2－Δb2 = 1, Δ= h2＋k, m = (a＋d)/2, k = c/b, 2h = (a－d)/b, and b≠ 0. (11)

When Ｓ1=(𝑚1 + ℎ𝑏1 𝑏1

𝑘𝑏1 𝑚1 − ℎ𝑏1), andＳ2= (

𝑚2 + ℎ𝑏2 𝑏2𝑘𝑏2 𝑚2 − ℎ𝑏2

), then matrices Ｓ1, Ｓ2, and their

products Ｓ1Ｓ2 have common coefficients k, h, such that

Ｓ1Ｓ2 =Ｓ2Ｓ1 =

(𝑚1𝑚2 + (ℎ

2 + 𝑘)𝑏1𝑏2 + ℎ(𝑚1𝑏2 + 𝑚2𝑏1) 𝑚1𝑏2 + 𝑚2𝑏1𝑘(𝑚1𝑏2 + 𝑚2𝑏1) 𝑚1𝑚2 + (ℎ

2 + 𝑘)𝑏1𝑏2 − ℎ(𝑚1𝑏2 + 𝑚2𝑏1)).

(12)

From Equations (8), (9) and (11), the matrix Ｓ has a normalized invariant function,

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φ(Ｓp) =φ(p) =－kx2＋y2＋2hxy. (13)

The matrix Ｓ and the invariant function φ(p) is classified into three types based on the sign

of the discriminant Δ=h2＋k. If Δ< 0, then they are of an elliptic type. If Δ> 0, then they are of a

hyperbolic type. If Δ= 0, then they are of a linear type.

Thus, we represent a form of 2×2 special matrixＳ such thatＳ= (𝑎 𝑏𝑐 𝑑

) = (𝑚 + ℎ𝑏 𝑏𝑘𝑏 𝑚 − ℎ𝑏

),

where detＳ=m2－Δb2=1, using argument θ, and commutative coefficient k, h as follows:

For Δ< 0, Ｓ=Ｓ(θ,k,h) = (cosθ +

ℎ

√−𝛥sinθ

1

√−𝛥sinθ

𝑘

√−𝛥sinθ cosθ −

ℎ

√−𝛥sinθ

), elliptic type. (14)

For Δ> 0, Ｓ=Ｓ(θ,k,h) = (coshθ +

ℎ

√𝛥sinhθ

1

√𝛥sinhθ

𝑘

√𝛥sinhθ coshθ −

ℎ

√𝛥sinhθ

), hyperbolic type. (15)

For Δ= 0, Ｓ=Ｓ(b,h) = (𝑚 + ℎ𝑏 𝑏

−ℎ²𝑏 𝑚 − ℎ𝑏), where m =±1, linear type. (16)

Any 2×2 non-diagonal regular matrix Ａ is represented by the polar form of

Ａ= (detＡ)1/2Ｓ(θ,k,h) or Ａ= (detＡ)1/2Ｓ(b,h)． (17)

However, when detＡ< 0, the matrix Ａ represents an inside out transformation, and then the orbit

of the invariant function as shown Equation (13) branches off to a conjugate curve of φ(p), and

complex number of argument θ appears.

We obtain the addition theorems of argument θ or b from Equations (14)－(16) as follows:

Ｓ(θ1,k,h)Ｓ(θ2,k,h) =Ｓ(θ1＋θ2,k,h), Ｓ(θ,k,h)n =Ｓ(nθ,k,h), Ｓ(θ,k,h)－1 =Ｓ(－θ,k,h),

Ｓ(b1,h)Ｓ(b2,h) =Ｓ(b1＋b2,h), Ｓ(b,h)n =Ｓ(nb,h), Ｓ(b,h)－1 =Ｓ(－b,h).

(18)

(19)

The norm ‖p‖ of a vector p is defined by the invariant functionφ(p) such that

‖p‖2=φ(p) =－kx2＋y2＋2hxy, and the norm ‖p‖=φ(p)1/2. (20)

The inner product (p,q) of vector p and q is defined by the invariant function φ(p) as follows:

p = (x1,y1), q = (x2,y2) =Ａp = (detＡ)1/2Ｓ(θ,k,h) p, (21)

‖p‖2 =φ(p) =－kx12＋y12＋2hx1y1, ‖q‖2 =φ(q) =－kx22＋y22＋2hx2y2,

‖p＋q‖2 =φ(p＋q) =－k(x1＋x2)2＋(y1＋y2)2＋2h(x1＋x2)(y1＋y2),

=‖p‖2＋‖q‖2＋2(－kx1x2＋y1y2＋h(x1y2＋x2y1))．

Thus, we induce the inner product and the cosine theorems from Equations (14), (15), and (21),

(p,q) = p・q =－kx1x2＋y1y2＋h(x1y2＋x2y1) = p・(detＡ)1/2Ｓ(θ,k,h)p

= 1

2(‖p＋q‖2－‖p‖2－‖q‖2)=

1

2(φ(p＋q)－φ(p)－φ(q))

= (detＡ)1/2φ(p)cos θ = ‖p‖‖q‖cos θ, when Ｓ is an elliptic type,

= (detＡ)1/2φ(p)cosh θ = ‖p‖‖q‖cosh θ, when Ｓ is a hyperbolic type.

(22)

(23)

Furthermore, when d=a ⇔ h=0 on a special linear transformation Ｓ, we define a special

isodiagonal linear transformation Ｆ and invariant function φ(p) from Equations (11) and (13)，

Ｆ = (𝑎 𝑏𝑐 𝑎

) = (𝑎 𝑏

𝑘𝑏 𝑎), detＦ= a2－kb2 = 1, and

φ(Ｆp) =φ(p) =－kx2＋y2, where k = c/b is a commutative coefficient. (24)

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In this case, we define the norm ‖p‖ and the inner product (p,q) as follows:

‖p‖2 =φ(p) =－kx2＋y2, and the norm ‖p‖=φ(p)1/2, (25)

(p,q) = p・q =－kx1x2＋y1y2.

If (p,q) = 0 ⇔ (y1/x1)(y2/x2) = k, then the vectors of p and q are defined as orthogonal.

(26)

We obtain the polar form of matrix Ｆ from Equations (14)－(16)．

When k< 0, then Ｆ= (𝑎 𝑏

𝑘𝑏 𝑎) = (

cos 𝜃 sin 𝜃 /√−𝑘

−√−𝑘 sin 𝜃 cos 𝜃), θ is an elliptic angle. (27)

When k=－1, then Ｆ is called a rotation transformation.

When k> 0, then Ｆ= (𝑎 𝑏

𝑘𝑏 𝑎) = (

cosh 𝜃 sinh 𝜃 /√𝑘

√𝑘 sinh 𝜃 cosh 𝜃), θ is a hyperbolic angle. (28)

This Ｆ is called a Lorentz transformation.

When k = 0, then Ｆ= ( 𝑎 𝑏𝑘𝑏 𝑎

) = (𝑎 𝑏0 𝑎

), a = ±1. (29)

This Ｆ is called a Galilean transformation.

3. Geometric structure of a line

Put two number lines 1 and 2 on one line and make them coincide with their origins. The

relation between their x-coordinates of x1 and x2 is x2=rx1 ⇔ x1=r－1x2, where r is a proportional

constant. For two equivalent number lines, we must have

r = r－1 ⇔ r2 = 1 ⇔ r = ±1.

When r=－1, then the two number lines are inversion of each other, and this type of line is

isotropic. A space line fits into this category.

When r=1, then the two number lines coincide with each other, and this type of line is one way.

A time line fits into this category.

When r≠±1, then the two number lines are similar.

4. Geometric structure of a plane

Put right-hand oblique coordinate systems on both face sides of a plane and make them

coincide with their origins. We define a 2×2 rear surface coordinate transformation Ｂ as an inside

out transformation, then detＢ< 0. Because it is not possible to distinguish which side of a plane is

the back or front, the symmetry plane equation is

Ｂ=Ｂ－1 ⇔ Ｂ2=Ｅ, where detＢ< 0. (30)

We obtain an oblique reflection transformation matrix Ｂ with two degrees of freedom:

Ｂ= ±(−𝑎 −𝑏𝑐 𝑎

)= ±(−𝑎 −𝑏𝑘𝑏 𝑎

), detＢ=－a2＋kb2 =－1, k=c/b, eigen values λ= ±1. (31)

The matrix Ｂ has the following properties (we shall treat the negative solution－Ｂ later).

As Ｂ has an eigen value of 1, and substituting a→－a in Equation (3), then Ｂ has an

invariant line f(p) given by,

f(Ｂp) ≡ f(p) = cx＋(a＋1)y. (32)

For λ=1 ⇔ a fixed-point equation Ｂp=p, this eigen line is called a fold line f:

cx＋(a－1)y = 0. (33)

For λ=－1 ⇔ an inversion equation Ｂp=－p, this eigen line is called an isotropic line g:

cx＋(a＋1)y = 0. (34)

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p0

p

Ｂp

invariant lines

f(Ｂp)=f(p)=y－x/2 ∥ line g

x

y

r

Ｂp0

f(p)

Figure 1. Oblique reflection plane

f(p) f: fold line

y=2x

Ｂp

p

p0

Ｂp0

r

yF =y2

x2

xF r0

r0

This line g is isotropic regarding its origin and is parallel to an invariant line f(p) given as per

Equation (32).

When a point p is in an invariant line f(p), and the point r is the intersection point of a fold line

f and an invariant line f(p), then in the fold line f, Ｂr=r, and in the invariant line, f(Ｂp)

=f(p)=f(r). Obtained by translating the vector (p－r) onto the isotropic line g,

Ｂ(p－r) =－(p－r) ⇔ Ｂp－r =－p＋r ⇔Ｂp＋p = 2r. (35)

Since the fixed-point r is the middle point of the

point p and Ｂp, and each invariant line f(p) is

parallel to the isotropic line g, then the invariant

lines f(p) are isotropic. The inner product of f and g

is −𝑐

𝑎−1

−𝑐

𝑎+1 =

𝑐2

𝑎2−1 =

𝑐2

𝑏𝑐 =

𝑐

𝑏 = k. From Equation (26),

these two lines of f and g are orthogonal, but

commonly seem not perpendicular. This eigen

plane with eigen lines of f and g is called an oblique

reflection plane, and it is semi-isotropic. Figure 1.

shows the case of Ｂ = 1

3(

−5 4−4 5

), where k=1. The

point Ｂp on the back side is hidden behind the

point p on the front side.

When Ｂ≠Ｂ－１, where h≠0, then we have an

asymmetric plane geometry based on the

invariant function φ(Ｓp) =φ(p) =－kx2＋y2＋

2hxy on which special linear matrices Ｓ

create an isometric and commutative group.

When Ｂ = Ｂ － １ on a symmetry plane, we

derive a special isodiagonal linear transformation Ｆ from an oblique reflection transformation

Ｂ and the reflection transformation Ｍ such that

Ｆ=ＭＢ = Ｍ (−𝑎 −𝑏𝑐 𝑎

) = (𝑎 𝑏𝑐 𝑎

) = (𝑎 𝑏

𝑘𝑏 𝑎)，where Ｍ =(−1 0

0 1), detＦ= 1, k = c/b. (36)

Ｆ is a coordinate transformation between the right-hand systems in which the left-hand x2-y2

system on the back side is reflected in the right-hand xF-yF system on the front side by the reflection

transformation Ｍ . Note that Ｆ=±ＢＭ or Ｆ=±ＭＢ are equivalent as to left to right-hand

coordinate system inversion.

Since Ｂ=ＭＦ, φ(Ｍp) =φ(p), and φ(Ｆp) =φ(p) =－kx2＋y2 as per Equation (24), then Ｂ has

the same invariant function φ(p) of Ｆ which is already implied in Equation (10),

φ(Ｂp) =φ(Ｍ(Ｆp)) =φ(Ｆp) =φ(p) =－kx2＋y2. (37)

When the commutative coefficient k is fixed, then we observe that any combination of Ｂ and

Ｆ has the common invariant function φ(p), and their joint operation is closed in the orbit of φ(p)

such that

φ(ＢＦ2…Ｂ－1Ｆ－1p) =φ(Ｆ2…Ｂ－1Ｆ－1p) =φ(Ｆ…Ｂ－1Ｆ－1p) =φ(Ｂ－1Ｆ－1p)

=φ(ＢＦ－1p) =φ(Ｆ－1p) =φ(Ｆp) =φ(p) =－kx2＋y2, where Ｂ－1 =Ｂ. (38)

Thus, we conclude that any combination of Ｂ and Ｆ creates an isometric transformation group

based on the orbit of invariant functionφ(p) with the metric

‖Ｂp‖2=‖Ｆp‖2 =‖p‖2 =φ(Ｂp) =φ(Ｆp) =φ(p) =－kx2＋y2. (39)

The oblique reflection transformation Ｂ transforms a point p on the front side to the

corresponding rear point q on the back side as

q1=Ｂp1, q2 =Ｂp2. (40)

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However, on the front side, a figure transformation Ｘ transforms a point from p1 to p2, and also on

the back side, a figure transformationＹ transforms a point from q1 to q2 as

p2 =Ｘp1, detＸ> 0, q2=Ｙq1, detＹ> 0. (41)

Consequently, from these four equations, we obtain the relation,

q2 =Ｙq1=ＹＢp1 =Ｂp2 =ＢＸp1. (42)

Since the point p1 is arbitrary and Ｂ=Ｂ－1, then we obtain

ＹＢ =ＢＸ ⇔ Ｙ =ＢＸＢ ⇔ＢＹ =ＸＢ, where detＹ = detＸ> 0, traceＹ = traceＸ. (43)

Therefore, Ｙ is similar to Ｘ, and they are of the same type of transformation. Then, substituting

Ｂ=Ｍ (Ｍ is one of the solutions of Ｂ=Ｂ－1) and Ｂ=ＭＦ into Equation (43),

ＹＭ = ＭＸ and ＹＭＦ = ＭＦＸ = ＭＸＦ. (44)

Comparing the second and third sides, ＦＸ=ＸＦ. In the same way ＦＹ=ＹＦ.

However, since the coordinate transformation Ｆ and the figure transformations Ｘ and Ｙ

have a same commutative coefficient k, then Ｘ andＹ have a common relative invariant function

φ(p) from Equations (7) and (24), such that

φ(Ｘp) = detＸφ(p) =φ(Ｙp) = detＹφ(p), where φ(Ｆp) =φ(p) =－kx2＋y2. (45)

Thus, when we fix the commutative coefficient k on a symmetry plane, we conclude that any

combination of the matrices Ｆ , Ｘ , and Ｙ creates a commutative transformation group.

Furthermore, any combination of the matrices Ｂ, Ｆ, Ｘ, and Ｙ creates a transformation group

based on the orbit of invariant function φ(p) on both sides of a plane. This super group geometry

of a symmetry plane involves both Euclidean geometry and Minkowski plane geometry. An

example is shown in Figure A.1. of Appendix A.

On the other hand, based on the sign of k, we obtain the existing direction of fold and isotropic

lines that vary on the coordinate system.

(1) If k< 0, then Ｆ and the invariant function φ(Ｆp)=φ(p)=－kx2＋y2 are elliptic type, and

from Equation (27), we express Ｂ=ＭＦ given by

Ｂ =ＭＦ= (−𝑎 −𝑏𝑐 𝑎

) = (− cos 𝜃 − sin 𝜃 /√−𝑘

−√−𝑘 sin 𝜃 cos 𝜃), where Ｍ = (

−1 00 1

), detＢ =－1. (46)

From Equations (33) and (34), we have the fold line f and the isotropic line g：

f: y =−𝑐

𝑎−1𝑥 = √−𝑘

sin𝜃

cos𝜃−1 𝑥 = － √−𝑘cot

𝜃

2･ 𝑥 =ux, g: 𝑦 =

−𝑐

𝑎+1𝑥 = √−𝑘tan

𝜃

2･𝑥＝vx. (47)

The existing directions of lines f and g are

－∞＜u＜∞, －∞＜v＜∞, for θ=－∞→∞. (48)

We observe that this oblique reflection plane made of a fold line f and an isotropic line g exists

in all directions centered around the origin. Similar is the case of a negative solution of the matrix－

Ｂ. Therefore, we conclude that this symmetry plane made of oblique reflection planes is completely

isotropic. This type of symmetry plane which has the oblique reflection transformation Ｂ with k<

0 fits in the space × space type plane and forms an elliptic type of plane geometry. When k=－1, then

Euclidean geometry on a Euclidean plane is given.

(2) If k> 0, then matrix Ｆ and the invariant function φ(Ｆp)=φ(p)=－kx2＋y2 are hyperbolic

type, and from Equation (28), we express the matrix Ｂ=ＭＦ given by

Ｂ= (−𝑎 −𝑏𝑐 𝑎

) = (− cosh 𝜃 − sinh 𝜃 /√𝑘

√𝑘 sinh 𝜃 cosh 𝜃), detＢ=－1, a=cosh θ, c= √𝑘sinh θ. (49)

From Equations (33) and (34), we have the fold line f and the isotropic line g：

f: y =−𝑐

𝑎−1𝑥 = －√𝑘

sinh𝜃

cosh𝜃−1 𝑥 = －√𝑘coth

𝜃

2･ 𝑥＝ux, g: 𝑦 =

−𝑐

𝑎+1𝑥 = － √𝑘tanh

𝜃

2･𝑥＝vx. (50)

8 of 10

The coordinate plane is divided by asymptotes into four quadrant regions. The asymptotes are

y =±√𝑘x. ( for θ→±∞) (51)

The existing directions of the lines f are

－∞＜u＜－√𝑘 and √𝑘＜u＜∞, upper and lower quadrant. (52)

The existing directions of the lines g are

－√𝑘＜v＜√𝑘, left and right quadrant.

From (𝑥2𝑡2

)=Ｂ(𝑥𝑡

), the x2-y2 axis on the back side of a symmetry plane is (53)

y2 axis: x2=－ax－by=0 ⇔ y=－√𝑘coth θ･x, x2 axis: y2=cx＋ay=0 ⇔ y=－√𝑘tanh θ･x (54)

The direction of the fold line f exists in the upper and lower quadrant regions, and the

direction of the isotropic line g exists in the left and right quadrant regions on the coordinate system.

The inverse relation of g and f is the negative solution of the matrix－Ｂ. Therefore, we conclude that

this symmetry plane made of oblique reflection planes is semi-isotropic, such that the time axes are

f, y, y2 and the space axes are g, x, x2. This type of symmetry plane which has the oblique reflection

matrix Ｂ with k> 0 fits in the space × time type plane and forms a hyperbolic type of plane

geometry. However, when k=1/c2 and y=t, we call this hyperbolic type of plane geometry a

Minkowski plane geometry. On our space × time plane, we conclude that k> 0.

(3) If k= 0, then matrix Ｆ is linear type, and from Equation (29), Ｆ=(1 𝑏0 1

) and Ｂ=(−1 −𝑏0 1

)

are obtained. This type of plane fits in the space × time type plane which is called a Newtonian

plane, since the invariant function φ(Ｆp)=φ(p)=y2 represents Newton’s absolute time (y=time).

5. Expected conclusions

5.1. Conceptual answer to the relativity principle

According to Poincaré’s conjecture [3], [4], the

relativity principle should be proved strictly and

all at once from the relative space [and time].

←v S2

g

S1 o v→

We think the two inertial systems S1 and S2 move

on one line, going away from each other at the

speed of v m/sec. In the Figure 2., the space-time

axes of inertial coordinate systems S1 and S2 are

x1-t1 on the front side, and x2-t2 on the back side of a

space × time plane. Make them coincide with their

origins. The transformation Ｆ with k> 0 is a

Lorentz transformation which rules a space × time

type symmetry plane. From Equation (28) and

substituting y=t, we have Lorentz transformation Ｌ: xL=ax1+bt1, tL=kbx1+at1. xL-tL represents the

Lorentz coordinate axis. From the first equation, a is a unitless constant, and b is a velocity constant.

As the motion of S2 is represented by x1=vt1 and tL axis is represented by xL= ax1+bt1=0 on the front

side, then v=－b/a is derived. From the second equation, k represents reciprocal of the velocity

squared in which we put k=1/c2>0 by convention, and c is a velocity constant.

If inertial systems of 1,2,...,n move by the speed of v12, v23,..., vn1, respectively on a straight line,

then the following equation holds, (c－v12)(c－v23)・・・(c－vn 1) = (c+v12)(c+v23)・・・(c+vn 1).

This proves that there must be the maximum invariant speed c (=1/√𝒌). The more about the velocity

c are discussed in the web site [5].

t2 = tL

x2

t1

x1

Ｂp

isotropic line g

fold

line f

p p Ｂp xL=－x2

1/γ

－γv

1

γ

v

γ

γv v

v/γ v/γ

φ(p) =1

time dilation

ationlength contraction

1

L*p L*p

1/γ

－γv γv

v

θ

Figure 2. Symmetry Plane Causes Relativity

γv 1

x=ct

v/γ

f(L*p)= f(ＢL*p)

9 of 10

Since detＬ=1, then we obtain a=γ=1／√1－𝑣2/𝑐2 ≥1. Thus, the Lorentz transformation Ｌ and its

oblique reflection transformation Ｂ are defined as follows:

Ｌ= 𝛾 (1 −𝑣

−𝑣/𝑐2 1) =ＭＢ, detＬ= 1, Ｍ= (

−1 00 1

)=ＬＢ, γ= 1／√1－𝑣2/𝑐2 ≥1

Ｂ= 𝛾 (−1 𝑣

−𝑣/𝑐2 1) =ＭＬ, detＢ=－1, (

𝑥𝐿𝑡𝐿

) =Ｌ(𝑥1𝑡1

), Ｌ(01

) = (−𝛾𝑣

𝛾 ),

t2-axis: x2=γ(－x1＋vt1)=0, x2-axis: t2=γ(－vx1/𝑐2＋t1)=0,

φ(ＢＬp) =φ(Ｌp) =φ(L*p) =φ(Ｂp) =φ(p) =φ(𝑥1𝑡1

) =φ(𝑥2𝑡2

) =φ(𝑥𝐿𝑡𝐿

) =－kx2＋t2, k=1／c2>0,

S1→S2: x1=vt1, S2→S1: x2=vt2, x1-axis∥Ｂp―L*p, g∥p―Ｂp∥f(Ｂp)=f(p).

*Note that Ｂp or L*p means figure transformation and Ｂp orＬp means coordinate transformation.

The fold line f and the isotropic line g of matrix Ｂ are drawn perpendicular. It is proposed

that the matrix Ｂ is the same case of figure 1. as Ｂ= 1

3(

−5 4−4 5

) for example, and, c=1, v=4/5, k=1, γ=

5/3, and θ =－1.0986 which is hyperbolic angle between axes y1=t1 and y2=t2 as per Equation (54). We

see the spacetime reduction from S1 to S2 and also from S2 to S1 reciprocally. Each reduction is

caused by the projection between both coordinate systems.

The time dilation of a stationary clock on the tL axis is given by

Ｌ－1(0

𝟏/𝜸) = (

𝑣𝟏

) and the length contraction of a moving rod is given by Ｌ－1(𝜸

−𝛾𝑣𝑐2

⁄)= (𝟏

0).

Because the appearance from inertial system S1 to S2 is the same as the appearance from S2 to

S1, then the view from the front side is the same as the view from the back side. Any point p on the

front side is equivalent to point p on the back side such relations that

Front p Ｂ q p (front)→Ｂp= q (back)→Ｂq=p (back),

Symmetry plane Ｂ Ｂ and

Back q Ｂ p p (back)→Ｂp= q (front)→Ｂq=p (front).

We also observe that the fold line f, the isotropic line g, the invariant lines f(Ｂp)=f(p), and the

invariant function φ(p)=－x2＋y2 have the same shape from each side of the coordinate plane.

Therefore, the back and the front worlds are completelythe the same in terms of the coordinate

plane. This immediately means the relativity principle in which any basic law of nature with space

and time vectors (𝑥𝑡

) must be oblique reflection transformation Ｂ invariant, and therefore Lorentz

transformation Ｌ invariant, because Ｌ=ＭＢ and basic laws have x inversion invariance. This is

the same as the theorem of Euclidean geometry must be rotation transformation invariant.

5.2. One answer to the arrow of time problem

Why does the physical phenomenon only proceed in one fixed direction of time in spite of the

fact that fundamental law of physics has inversion symmetry with respect to time? This is the

"arrow of time" problem which has long been unresolved in physics. Physical phenomena occur in

linear spacetime in which unit scale intervals are regularly arranged, but its ±direction is not

determined. Although time progresses from the past to the future, the direction of time may be

positive or negative. Similarly, the direction of the eastward straight line can be positive or

negative.

To solve this problem, we have to think Minkowski plane from both the back and the front

sides, which represents two inertial systems of the same speed. When observing a pendulum

motion, if time progresses in the positive direction on the front side and in the negative direction on

the back side, then it can be distinguished which side of a plane you and I are on, but this is not the

case. Both times go in the negative or in the positive direction according to the future direction.

Nature is elaborate so that it cannot be distinguished which side of a plane is the front and which is

the back. The symmetry of a space × time plane in spacetime is the heart of the arrow of time

problem.

10 of 10

5.3. Pythagorean theorem on a Euclidean plane

When k=－1 on Equations (24) and (27), then a symmetry plane is a Euclidean plane and the

invariant function is φ(Ｆp) =φ(Ｂp) =φ(p) =x2＋y2. This is equivalent to the Pythagorean theorem.

Thus, the symmetry of rotation or relativity arises fom the symmetry of a plane in spacetime.

Appendix A. Nine-point circle theorem on symmetry planes

Nine‒point circle on both Minkowski plane and Euclidean plane hold the same logic.

Point O is the center of circumscribed circle on △ABC.

Point G is the center of gravity of △ABC.

Point H is the crossing point of perpendicular lines of AL,BM,CN.

Point K is the middle point of O and H, and the center of 9-point circle.

Line OGKH is the Euler line.

Figure A1. Symmetry plane geometry is constructed both on a Euclidean and a Minkowski plane.

References

[1] Tsuruichi Hayashi,“初等幾何学の体裁 Form of Elementary Geometry”, 哲学 Philosophy 296 号,

1911 http://fomalhautpsa.sakura.ne.jp/Science/Other/kikagaku-teisai.pdf [in Japanese]

[2] Jun Tosaka,“幾何学と空間 Geometry and Space”, 思想 Thought 57号, 1926

https://www.aozora.gr.jp/cards/000281/files/43263_35546.html [in Japanese, accessed 17 July 2020].

[3] Henri Poincaré,“Electricité et optique. Lalumiére et les théories élecrtodynamiques. Leçon

professée a la Sorbonne en 1889”,Paris,1901.

A well-developed theory should prove this [relativity] principle strictly and all at once, ･･･

[4] Henri Poincaré,“La Valuer de la Science”,1905 Chap3.1 Concept of relative space;

It is not possible to distinguish when one world changes the coordinate axis into another world, …

Poincaré was expecting a geometrical theory to prove the relativity principle.

[5] Hiroaki Fujimori “表裏対称平面の幾何(Geometry of a Symmetry Plane)”,

http://spatim.sakura.ne.jp/5syo.pdf [in Japanese]

A

D L

H

O

G K

P

Q R

Euclidean plane

circumscribed circle

9-point circle DEFLMNPQR

S N

F E

M

C B

A

Q

B

C

S

O

D

E F

G

K

P

N

L

(M)

R H

9-point circle FPE

circumscribed circle

Minkowski plane

x

y

9-point circle NQLDRM

http://fomalhautpsa.sakura.ne.jp/Science/Other/kikagaku-teisai.pdfhttps://www.aozora.gr.jp/cards/000281/files/43263_35546.htmlhttp://spatim.sakura.ne.jp/