In this article the notation has been kept the same as in the previous article.

• ‖   ⋅   ‖ F is the Frobenius Norm.

• P [ θ ] is some projection operator in G P ( [ θ ] ) .

• I 2 is the 2 × 2 identity matrix.

• R n k ( P [ θ ] ) is the Rank of a projection operator.

• σ ( P [ θ ] ) is the spectrum of a projection operator.

• S t a b S O ( 2 ) ( P [ θ ] ) is the Stabilizer subgroup for some fixed element of G P ( [ θ ] ) .

• R ( α ) is an element of S O ( 2 ) for angle α .

• V e c is the Vectorisation Operator.

• ⊙ is the Hadamard Product.

• ⊗ is the Kronecker Product.

From [1] [2] we can get the following results.

Lemma 1. The Frobenius morn ‖   ⋅   ‖ F of a projection matrix is its trace. That is

‖ P [ θ ] ‖ F = T r ( P [ θ ] ) = 1

Proof. It is well known that

‖ P [ θ ] ‖ F 2 : = ∑ i , j | p i j ( θ ) | 2 = T r ( P [ θ ] T P [ θ ] )

Given that P [ θ ] is both symmetric and idempotent implies that

T r ( P [ θ ] T P [ θ ] ) = T r ( P [ θ ] P [ θ ] ) = T r ( P [ θ ] 2 ) = T r ( P [ θ ] )

Hence,

‖ P [ θ ] ‖ F = T r ( P [ θ ] )

therefore,

T r ( [ cos 2 [ θ ] cos [ θ ] si n ′ [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ) = cos 2 [ θ ] + sin 2 [ θ ] = 1, ∀ [ θ ] ∈ T

Theorem 1 The necessary and sufficient condition for P [ θ ] ∈ G P ( [ θ ] ) is given by

P [ θ ] 2 = P [ θ ] , ∀ [ θ ] ∈ T

Proof. We know, from the previous article that P [ θ ] is constructed as follows

Φ ( e i [ θ ] ) : = e i [ θ ] ⊗ ( e i [ θ ] ) T = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = P [ θ ]

P [ θ ] 2 = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] 2 = [ cos 4 [ θ ] + cos 2 [ θ ] sin 2 [ θ ] cos 3 [ θ ] sin [ θ ] + cos [ θ ] sin 3 [ θ ] cos 3 [ θ ] sin [ θ ] + cos [ θ ] sin 3 [ θ ] sin 4 [ θ ] + cos 2 [ θ ] sin 2 [ θ ] ] = [ cos 2 [ θ ] ( cos 2 [ θ ] + sin 2 [ θ ] ) cos [ θ ] sin [ θ ] ( cos 2 [ θ ] + sin 2 [ θ ] ) cos [ θ ] sin [ θ ] ( cos 2 [ θ ] + sin 2 [ θ ] ) sin 2 [ θ ] ( sin 2 [ θ ] + cos 2 [ θ ] ) ] = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = P [ θ ]

Theorem 2 The orthogonal complement of P [ θ ] is given by

P [ θ ± π / 2 ] = I 2 − P [ θ ]

where I 2 is the 2 × 2 identity matrix. This is known as a Vector Rejection.

Proof. We first show that

P [ θ ± π / 2 ] = I 2 − P [ θ ]

the RHS of the equation gives us

I 2 − P ( [ θ ] ) = [ 1 0 0 1 ] − [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = [ 1 − cos 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] 1 − sin 2 [ θ ] ] = [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ]

Now, the LHS of th equation gives us the following result

P [ θ ± π / 2 ] = [ cos 2 ( [ θ ] ± π / 2 ) cos ( [ θ ] ± π / 2 ) sin ( [ θ ] ± π / 2 ) sin ( [ θ ] ± π / 2 ) cos ( [ θ ] ± π / 2 ) sin 2 ( [ θ ] ± π / 2 ) ] = [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ]

since

cos 2 ( [ θ ] ± π / 2 ) = ( cos [ θ ] cos ( π / 2 ) ∓ sin [ θ ] sin ( π / 2 ) ) 2 = ( ∓ sin [ θ ] ) 2 = sin 2 [ θ ]

sin 2 ( [ θ ] ± π / 2 ) = ( sin [ θ ] cos ( π / 2 ) ± cos [ θ ] sin ( π / 2 ) ) 2 = ( ± cos [ θ ] ) 2 = cos 2 [ θ ]

Secondly, we show the effect these operators have on some vector x ∈ ℝ 2 . We easily show this in the following way

〈 P [ θ ] x , P [ θ ± π / 2 ] x 〉 = 0

We proceed as follows

〈 P [ θ ] x , P [ θ ± π / 2 ] x 〉 = 〈 P [ θ ] x , ( I 2 − P [ θ ] ) x 〉 = 〈 P [ θ ] x , I 2 x − P [ θ ] x 〉 = 〈 P [ θ ] x , x − P [ θ ] x 〉 = 〈 P [ θ ] x , x 〉 − 〈 P [ θ ] x , P [ θ ] x 〉 = ‖ P [ θ ] x ‖ ‖ x ‖ cos θ − ‖ P [ θ ] x ‖ 2 = ‖ P [ θ ] x ‖ ‖ P [ θ ] x ‖ − ‖ P [ θ ] x ‖ 2 = 0

Lemma 2 The product of the operators P [ θ ] and P [ θ ± π / 2 ] is zero. We shall call P [ θ ± π / 2 ] the Orthogonal Complement Operator of P [ θ ] which I shall denote as P [ θ ] ⊥ .

P [ θ ] P [ θ ] ⊥ = 0, ∀ [ θ ] ∈ T

We say that P [ θ ] ⊥ P [ θ ] ⊥

Proof.

P [ θ ] P [ θ ] ⊥ = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ] = [ cos 2 [ θ ] sin 2 [ θ ] − cos 2 [ θ ] sin 2 [ θ ] − cos 3 [ θ ] sin [ θ ] + cos 3 [ θ ] sin [ θ ] sin 3 [ θ ] cos [ θ ] − sin 3 [ θ ] cos [ θ ] − cos 2 [ θ ] sin 2 [ θ ] + sin 2 [ θ ] cos 2 [ θ ] ] = [ 0 0 0 0 ]

Lemma 3 P [ θ ] + P [ θ ] ⊥ = I 2

Proof.

P [ θ ] + P [ θ ] ⊥ = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] + [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ] = [ 1 0 0 1 ] = I 2

Theorem 3 A projection operator of dimension onto V ⊂ ℝ 2 where dim ( V ) = r then

P [ θ ] = T Δ r T − 1 (1)

where

Δ r : = [ 1 0 0 0 ] (2)

The matrix T is formed by the columns of the basis vectors which span the subspaces V and W respectively.

Proof. We know that V , W ⊂ ℝ 2 , furthermore we also know that dim ( V ) = dim ( W ) = 1 ⇒ r = 1 .

Suppose the vector a = ( cos θ , sin θ ) T is a basis for subspaces V i.e. S p n { a } = V then the matrix T can be computed in the following way.

Now, taking into account the fact that W = V ⊥ implies that b is given by

b = ( cos ( θ ± π / 2 ) , sin ( θ ± π / 2 ) ) = ( ∓ sin θ , ± cos θ )

Hence, T is computed in the following way

T = [ cos [ θ ] ∓ sin [ θ ] sin [ θ ] ± cos [ θ ] ] ∴ det ( T ) = ± cos 2 [ θ ] ± sin 2 [ θ ] = ± 1

Hence, T − 1 is given by

T − 1 = ± 1 [ ± cos [ θ ] ± sin [ θ ] − sin [ θ ] cos [ θ ] ] = [ cos [ θ ] sin [ θ ] ∓ sin [ θ ] ± cos [ θ ] ] = T T ⇒ T , T T ∈ O (2,ℝ)

Hence, P [ θ ] is

T Δ r T − 1 = [ cos [ θ ] ∓ sin [ θ ] sin [ θ ] ± cos [ θ ] ] [ 1 0 0 0 ] ± 1 [ ± cos [ θ ] ± sin [ θ ] − sin [ θ ] cos [ θ ] ] = ± 1 [ cos [ θ ] ∓ sin [ θ ] sin [ θ ] ± cos [ θ ′ ] ] [ ± cos [ θ ] ± sin [ θ ] 0 0 ] = ± 1 [ ± cos 2 [ θ ] ± cos [ θ ] sin [ θ ] ± sin [ θ ] cos [ θ ] ± sin 2 [ θ ] ] = P [ θ ]

We can see that T , T − 1 ∈ O ( 2, ℝ ) and are orthogonal reflection matrices.

Lemma 4 Let P [ θ ] be some projection matrix for some [ θ ] ∈ T then we have

R n k ( P [ θ ] ) = T r ( P θ ) (3)

Proof. We know that

P [ θ ] = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ⇒ T r ( P [ θ ] ) = cos 2 [ θ ] + sin 2 [ θ ] = 1

We can now calculate its rank. The column space of P [ θ ] is given by

C l n ( P [ θ ] ) = { [ cos 2 [ θ ] sin [ θ ] cos [ θ ] ] , [ cos [ θ ] sin [ θ ] sin 2 [ θ ] ] }

where C l n ( P [ θ ] ) is the column space of P [ θ ] . R n k ( P [ θ ] ) = 1 iff ∃ α 1 , α 2 ≠ 0 ∈ ℝ such that

α 1 [ cos 2 [ θ ] sin [ θ ] cos [ θ ] ] + α 2 [ cos [ θ ] sin [ θ ] sin 2 [ θ ] ] = [ 0 0 ]

This gives us the following linear system

[ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ α 1 α 2 ] = [ 0 0 ]

Given that det ( P [ θ ] ) = 0 implies an infinite set of solutions exists.

Writing the homogeneous linear system we find the general set of vector solutions as follows

α 1 cos 2 [ θ ] + α 2 cos [ θ ] sin [ θ ] = 0

α 1 sin [ θ ] cos [ θ ] + α 2 sin 2 [ θ ] = 0

therefore

α 1 sin [ θ ] cos [ θ ] + α 2 sin 2 [ θ ] = 0

⇒ α 1 cos [ θ ] + α 2 sin [ θ ] = 0

∴ α 1 = − α 2 sin [ θ ] cos [ θ ] = − α 2 tan [ θ ]

Hence, the solution vector

U = { α 2 ∈ ℝ | α = α 2 ( − tan [ θ ] ,1 ) }

This implies that R n k ( P [ θ ] ) = T r ( P [ θ ] ) = 1 .

Theorem 4 Let A = [ a 1 ] such that V = S p n ( A ) . Then we have

P [ θ ] = A ( A T A ) − 1 A T (4)

Proof. a 1 = ( cos [ θ ] , sin [ θ ] ) T = A , then we have

A ( A T A ) − 1 A T = A ( a 1 T a 1 ) − 1 A = A A T = a 1 a 1 T = a 1 ⊗ a 1 = P [ θ ]

Lemma 5 The spectrum σ ( P [ θ ] ) of the P [ θ ] for some [ θ ] ∈ T is given by

σ ( P [ θ ] ) { λ 1 = 0, λ 2 = 1, ∀ P [ θ ] ∈ G P ( [ θ ] ) }

Proof. We simply perform the following calculation

det ( P [ θ ] − λ I 2 ) = | cos 2 θ − λ cos θ sin θ sin θ cos θ sin 2 θ − λ | = ( cos 2 θ − λ ) ( sin 2 θ − λ ) − cos 2 θ sin 2 θ = cos 2 θ sin 2 θ − λ cos 2 θ − λ sin 2 θ + λ 2 − cos 2 θ sin 2 θ = λ 2 − λ

therefore we see that

det ( P [ θ ] − λ I 2 ) = 0 ⇒ λ 2 − λ = 0 ⇒ λ ( λ − 1 ) = 0 ⇒ λ = 0 ∨ λ = 1

Hence, σ ( P [ θ ] ) = { 0 , 1 } as claimed.

We can calculate the associated Eigenspaces. For λ = 0 we get

P [ θ ] E 0 = λ E 0 = 0

Hence, it is clear that given some P [ θ ] its Eigenspace for λ = 0 is K e r ( [ P [ θ ] ] ) , that is, in this case, the Eigenspace is spanned by the following vectors

E 0 ( [ θ ] ) : = ( cos ( [ θ ] + π / 2 ) , sin ( [ θ ] + π / 2 ) ) T = [ − sin ( [ θ ] ) cos ( [ θ ] ) ]

Therefore, we can say that is I m ( P [ θ ] ) is V ⊂ ℝ 2 then S p n { E 0 ( [ θ ] ) } = V ⊥ .

For λ = 1 we get the following Eigenvectors

P [ θ ] E 1 = λ E 1 = E 1

P [ θ ] E 1 − E 1 = ( P [ θ ] − I 2 ) E 1 = 0

In matrix form, this gives us

[ cos 2 θ − 1 cos θ sin θ sin θ cos θ sin 2 θ − 1 ] [ E 1 E 2 ] = [ 0 0 ]

[ − sin 2 θ cos θ sin θ sin θ cos θ − cos 2 θ ] [ E 1 E 2 ] = [ 0 0 ]

This gives us the following system

− sin 2 θ E 1 + cos θ sin θ E 2 = 0

sin θ cos θ E 1 − cos 2 θ E 2 = 0

therefore, from (24) we get

sin θ cos θ E 1 = cos 2 θ ⇒ E 2 tan θ E 1 = E 2

Therefore,

E 2 E 1 = tan θ ⇒ E 1 ( θ ) = E 2 cot θ

Therefore, the Eigenvector is given

E 1 = E 2 [ cot θ 1 ]

Lemma 6 The Eigenvectors E 0 and E 1 are linearly independent. Moreover, S p n { E 0 , E 1 } = ℝ 2 .

Proof. For γ 0 , γ 1 ∈ ℝ we have

γ 0 E 0 + γ 1 E 1 = 0

Clearly, E 0 , E 1 are linearly independent iff γ 0 = γ 1 = 0

− γ 0 sin θ + γ 1 cot θ = 0

γ 0 cos θ + γ 1 = 0

This can be written matrix form in the following way

[ − sin θ cot θ cos θ 1 ] [ γ 0 γ 1 ] = [ 0 0 ]

Now,

| − sin θ cot θ cos θ 1 | = − sin θ − cot θ cos θ ≠ 0 , ∀ θ

We can clearly see that, the determinant is non-zero hence, E 0 and E 1 are linearly independent.

We can now talk about the Diagonalizability of Projection Matrices P [ θ ] . Given that P [ θ ] has distinct eigenvalues implies that it is diagonalizable i.e. P [ θ ] = P D P − 1 where D is a diagonal matrix

Lemma 7

P [ θ ] = P [ 0 0 0 1 ] P − 1

where

P = [ E 0 , E 1 ] = [ − sin θ cot θ cos θ 1 ] , P − 1 = − 1 sin θ + cot θ cos θ [ 1 − cot θ − cos θ − sin θ ]

Proof.

P [ θ ] = − 1 sin θ + cot θ cos θ [ − sin θ cot θ cos θ 1 ] [ 0 0 0 1 ] [ 1 − cot θ − cos θ − sin θ ] = − 1 sin θ + cot θ cos θ [ − sin θ cot θ cos θ 1 ] [ 0 0 − cos θ − sin θ ] = − 1 sin θ + cot θ cos θ [ − cot θ cos θ − cot θ sin θ − cos θ − sin θ ] = [ cos 2 θ cos θ sin θ sin θ cos θ sin 2 θ ] = P [ θ ]

Theorem 5 The projection matrix P [ θ ] with spectrum σ = { λ 1 , λ 2 } is diagonalizable, hence there exists matrices { G 1 , G 2 } such that

P [ θ ] = λ 1 G 1 + λ 2 G 2 (5)

where G 1 , G 2 are projection matrices onto N ( P [ θ ] − λ i I 2 ) along R ( P [ θ ] − λ i I 2 ) . Furthermore,

1) G i , G j = 0 whenever i ≠ j .

2) ∑ i = 1 m k   G i = I 2 .

Proof. We know that the Eigenvalues are λ 1 = 0 ∧ λ 2 = 1 this obviously implies that

P [ θ ] = λ 2 G 2 = G 2

P [ θ ] = P D P − 1 = ( X 1 | X 2 ) ( λ 1 I 0 0 λ 2 I ) ( Y 1 T Y 2 T ) = λ 1 X 1 Y 1 T + λ 2 X 2 Y 2 T = λ 2 X 2 Y 2 T = X 2 Y 2 T = − 1 sin θ + cot θ cos θ [ cot θ 1 ] [ − cos θ − sin θ ] = − 1 sin θ + cot θ cos θ [ − cot θ cos θ − cot θ sin θ − cos θ − sin θ ] = P [ θ ]

This implies that G 2 = X 2 Y 2 T .

Lemma 8 Let G 0 and G 1 be the spectral projectors for the eigenvalues λ 1 and λ 2 respectively. Then we show that

1)

G 0 G 1 = 0

2)

G 0 + G 1 = I 2

Proof.

G 0 G 1 = X 1 Y 1 T = [ − sin θ cos θ ] [ 1 − cot θ ] [ cot θ 1 ] [ − cos θ − sin θ ] = [ − sin θ cot θ sin θ cos θ − cos θ cot θ ] [ − cot θ cos θ − sin θ cot θ − cos θ − sin θ ] = [ 0 0 0 0 ]

As claimed.

For the second part of the proof, we simply add the matrices

G 0 + G 1 = − 1 sin θ + cot θ cos θ { [ − sin θ cot θ sin θ cos θ − cos θ cot θ ] + [ − cot θ cos θ − sin θ cot θ − cos θ − sin θ ] } = − 1 sin θ + cot θ cos θ [ − ( sin θ + cot θ sin θ ) 0 0 − ( cos θ cot θ + sin θ ) ] = I 2

From [3] [4] we have

Definition 1 Two squares matrices A and B are said to be Congruent if there exists an invertible matrix R such that

B = R T A R (6)

The form of the equation would tend to suggest that R T , R ∈ S O ( 2, ℝ ) such that R T R = R R T = I 2 .

Let α = θ ′ − θ be the rotation by some angle α such that α ∈ [ 0, π ) . α > 0 implies the rotation is counter-clockwise, α < 0 implies the rotation is clockwise.

Theorem 6 The action of S O ( 2 ) on G P ( [ θ ] ) defines a congruence between two elements P [ θ ] and P [ θ ′ ] in G P ( [ θ ] ) i.e.

ξ : S O ( 2 ) × G P ( [ θ ] ) → G P ( [ θ ] )

such that

ξ ( R ( α ) , P [ θ ] ) = R T ( α ) P [ θ ] R ( α ) = P [ θ ′ ] (7)

Proof. A point on S 1 can be represented as e i θ = ( cos θ , sin θ ) T .

Hence, we can choose two points on S 1 , e i [ θ ] = ( cos [ θ ] , sin [ θ ] ) T and e i [ θ ′ ] = ( cos [ θ ′ ] , sin [ θ ′ ] ) T .

ξ ( R ( α ) , P [ θ ] ) = R T ( α ) P [ θ ] R ( α ) = ( R T e i [ θ ] ) ⊗ ( ( e i [ θ ] ) T R ) = ( ( e i [ θ ] ) T R ) T ⊗ ( ( e i [ θ ] ) T R ) = e [ i θ ′ ] ⊗ ( e [ i θ ′ ] ) T = P [ θ ′ ]

It should be clear this action corresponds to the projector in the direction of [ θ ] + α , α < | π | which, as described in the first article, consistent with the topological structure. This is a clockwise transformation of the projection operator.

This, of course, implies that all projection operator are congruent matrices since it is always possible to find some R ( α ) ∈ S O ( 2 ) . Moreover, we know that I is the identity of S O ( 2 ) hence we get the following result

ξ ( I , P [ θ ] ) = I T P [ θ ] I = I P [ θ ] I = P [ θ ] I = P [ θ ]

Finally, given some P [ θ ] ∈ G P ( [ θ ] ) the mapping ξ is bijective on [ θ ] + α < | π | which implies that the mapping ξ is invertible and can be defined as

ξ − 1 ( R ( α ) , P [ θ ] ) : = ξ ( R ( − α ) , P [ θ ] ) (8)

We can see that this is equal to

ξ − 1 ( R ( − α ) , P [ θ ] ) = R T ( − α ) P [ θ ] R ( − α ) = R ( α ) P [ θ ] R T (α)

such that

ξ − 1 ∘ ξ = R ( α ) ( R T ( α ) P [ θ ] R ( α ) ) R T = I P [ θ ] I = P [ θ ]

Note that ξ − 1 is an counter-clockwise transformation of the projection operator.

Lemma 9 Let R ( α ) , R ( α ′ ) ∈ S O ( 2 ) and let P [ θ ] ∈ G P ( [ θ ] ) , then

ξ ( R ( α ′ ) , ξ ( R ( α ) , P [ θ ] ) ) = R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = ξ ( R ( α ) R ( α ′ ) , P [ θ ] ) (9)

Proof.

ξ ( R ( α ′ ) , ξ ( R ( α ) , P [ θ ] ) ) = ξ ( R ( α ′ ) , R T ( α ) P [ θ ] R ( α ) ) = R T ( α ′ ) ( R T ( α ) P [ θ ] R ( α ) ) R ( α ′ ) = R T ( α ′ ) R T ( α ) P [ θ ] R ( α ) R ( α ′ ) = R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = ξ ( R ( α ) R ( α ′ ) , P [ θ ] )

Lemma 10 Let R ( α ) and P [ θ ] be elements of S O ( 2 ) and G P ( [ θ ] ) respectively. Then we have

ξ ( R ( 0 ) , P [ θ ] ) = ξ ( R ( k π ) , P [ θ ] ) = P [ θ ] , ∀ k ∈ ℤ (10)

Proof. Beginning with the case where φ = 0 we get the following result

ξ ( R ( 0 ) , P [ θ ] ) = R T ( 0 ) P [ θ ] R ( 0 ) = [ 1 0 0 1 ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ 1 0 0 1 ] = P [ θ ]

Now choosing φ = k π for some k ∈ ℤ leads to

ξ ( R ( k π ) , P [ θ ] ) = R T ( k π ) P [ θ ] R ( k π ) = [ cos ( k π ) sin ( k π ) − sin ( k π ) cos ( k π ) ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ cos ( k π ) − sin ( k π ) sin ( k π ) cos ( k π ) ] = [ ( − 1 ) k 0 0 ( − 1 ) k ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ ( − 1 ) k 0 0 ( − 1 ) k ] = [ ( − 1 ) k cos 2 [ θ ] ( − 1 ) k cos [ θ ] sin [ θ ] ( − 1 ) k sin [ θ ] cos [ θ ] ( − 1 ) k sin 2 [ θ ] ] [ ( − 1 ) k 0 0 ( − 1 ) k ] = [ ( − 1 ) 2 k cos 2 [ θ ] ( − 1 ) 2 k cos [ θ ] sin [ θ ] ( − 1 ) 2 k sin [ θ ] cos [ θ ] ( − 1 ) 2 k sin 2 [ θ ] ] = P [ θ ]

This shows us that projection matrices have a period of π .

We can now calculate the Stabilizer of G P ( [ θ ] ) as follows.

Definition 2 The rotation group S O ( 2 ) acts on G P ( [ θ ] ) then the Stabilizer of some element P [ θ ] ∈ G P ( [ θ ] ) denoted S t a b S O ( 2 ) is defined as

S t a b S O ( 2 ) ( P [ θ ] ) : = { R ( α ) ∈ S O ( 2 ) : ξ ( R ( α ) , P [ θ ] ) = P [ θ ] }

From the above definition it is clear that

S t a b S O ( 2 ) ( P [ θ ] ) : = { R [ k π ] : α = k π , k ∈ ℤ }

Lemma 11

S t a b S O ( 2 ) ( P [ θ ] ) ≤ S O (2)

Proof. Choosing k = 0 implies R ( 0 ) = I 2 ⇒ R ( 0 ) P [ θ ] = I 2 P [ θ ] = P [ θ ] . We know that for every R ( k π ) ∃ R z − 1 ( k π ) = R T ( k π ) s.t. R ( k π ) R T ( k π ) = R ( k π ) T R ( k π ) = I 2 .

hence, for some k , k ′ ∈ ℤ we have

R ( k π ) ( R T ( k π ) P [ θ ] R ( k π ) ) R T ( k π ) = R ( k π ) ( P [ θ ] ) R T ( k π ) = P [ θ ]

This implies that

R ( k π ) ( R T ( k π ) R ( k π ) ) R T ( k π ) = R ( k π ) I 2 R T ( k π ) = I 2 ∈ S t a b S O ( 2 ) ( P [ θ ] )

Theorem 7. Let P [ θ ] , P [ θ ′ ] ∈ G P ( [ θ ] ) . Let θ ′ = α + α ′ then

P [ θ + θ ′ ] = R T ( α + α ′ ) P [ θ ] R ( α + α ′ )

Proof.

R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = R T ( α ) R T ( α ′ ) P [ θ ] R ( α ′ ) R ( α ) = R T ( α ) P [ θ + α ′ ] R ( α ) = P [ θ + ( α ′ + α ) ] = P [ θ + θ ′ ]

Hence, we can conclude that

R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = P [ θ + θ ′ ]

Lemma 12 The group of action of S O ( 2, ℝ ) on G P ( [ θ ] ) not faithful but it is ∞-transitive.

Proof. for a group action to be faithful implies that for every pair of distinct elements of S O ( 2, ℝ ) there is some P [ θ ] such that R ( α ) P [ θ ] R T ( α ) ≠ R ( α ′ ) P [ θ ] R T ( α ′ ) , ∀ P [ θ ] ∈ G P ( [ θ ] ) .

Consider the set S O ( 2, ℝ ) × S O ( 2, ℝ ) and let us choose some arbitrary projector in G P ( [ θ ] ) . Specifically, let us consider the pair ( R ( α ) , R ( α ′ ) ) ∈ S O ( 2, ℝ ) × S O ( 2, ℝ ) such that α ′ = α ± π . Then we demonstrate that it is impossible to find an element of G P ( [ θ ] ) such that the above definition is satisfied.

R T ( α ′ ) P [ θ ] R ( α ′ ) = R T ( α ± π ) P [ θ ] R ( α ± π ) = − R T ( α ) P [ θ ] − R ( α ) = R T ( α ) P [ θ ] R ( α ) (11)

Since α ′ ≠ α and P [ θ ] is arbitrary, we conclude that this action is not faithful.

However, we are going to demonstrate that it is n-transitive. To show this we can consider two pairwise distinct projector sequences of form

( { P [ θ i ] } , { P [ θ r ] } ) , i , r = 1 , ⋯ , n

each sequence is pairwise distinct, that is, P [ θ i ] ≠ P [ θ j ] , ∀ i , j = 1 , ⋯ , n and P [ θ r ] ≠ P [ θ s ] , ∀ r , s = 1 , ⋯ , n . Suppose that each sequence is chosen so that ϕ ( P [ θ i ] ) , ϕ ( P [ θ r ] ) ∈ T and form two arithmetic sequences such that the quotient metric satisfies d T ( [ θ i ] , [ θ r ] ) = inf k ∈ ℤ { | θ i − θ r + 2 k π | } = β < π , ∀ i , r = 1, ⋯ , n . Then we can have the following result

R T ( β ) P [ θ i ] R ( β ) = P [ θ r ] , ∀ i , r = 1, ⋯ , n

we can define a refinement of the sequence in the following way

σ ( β ) = β ′ , β ′ < β ⇒ σ ( { P [ θ i ] } i = 1 n ) = { P [ θ i ] } i = 1 n ′ , n < n ′

As β → 0 ⇒ n → ∞ we have lim n → ∞ ( { P [ θ i ] } i = 1 n ∪ { P [ θ r ] } r = 1 n ) = P α ⊆ G P ( [ θ ] ) ∈ τ G P ( [ θ ] ) .

where, it was shown that τ G P ( [ θ ] ) is the topology G P [ θ ] . Hence, this group action is ∞-transitive.

Lemma 13 The Kernel of ξ is given by

K e r ξ : = { R ( α ) ∈ S O ( 2, ℝ ) : R T ( α ) P [ θ ] R ( α ) = P [ 0 ] } = { R ( α ) ∈ S O ( 2, ℝ ) : α : = − θ }

for some P [ θ ] ∈ G P ( [ θ ] ) .

Proof. Let P [ θ ] be some element of G P ( [ θ ] ) and choose α = − θ the we have

ξ ( R ( − θ ) P [ θ ] ) = R T ( − θ ) P [ θ ] R ( − θ ) = R ( θ ) P [ θ ] R T ( θ ) = ξ − 1 ( R ( θ ) , P [ θ ] ) = P [ 0 ] = I d G P ( [ θ ] )

Lemma 14 The Lie Group G P ( [ θ ] ) has exactly one orbit.

Proof. We know that the action is transitive since for all pairs on G P ( [ θ ] ) × G P ( [ θ ] ) of the form ( P [ θ ′ ] , P [ θ ] ) there exists R ( α + k π ) , k ∈ ℤ such that P [ θ ′ ] = R T ( α + k π ) P [ θ ] R ( α + k π ) , ∀ k ∈ ℤ .

We just need to show that for some fixed P [ θ ] ∈ G P ( [ θ ] ) we get R T ( α ) P [ θ ] R ( α ) = G P ( [ θ ] ) . It is clear that choosing − π 2 < α ≤ π 2 we get

{ R T ( π 2 ) P [ θ ] R ( π 2 ) = P [ θ + π 2 ] = P [ θ ] ⊥ R T ( − π 2 ) P [ θ ] R ( − π 2 ) = P [ θ − π 2 ] = R ( π 2 ) P [ θ ] R T ( π 2 ) = P [ θ ] ⊥

Due to the fact that projector repeat every π we get see that R T ( α ) P [ θ ] R ( α ) = G P ( [ θ ] ) .

Definition 3 The Vec operator is a linear transformation which converts a matrix into a column vector. That is

V e c : ℝ n × m → ℝ n m

defined as

V e c ( A ) = [ a 1,1 , ⋯ , a m ,1 , ⋯ , a 1,2 , ⋯ , a m ,2 , ⋯ , a m , n ] T , A ∈ ℝ n × m

Theorem 8 The group S O ( 2, ℝ ) defines a homeomorphism group on G P ( [ θ ] ) . That is

S O ( 2, ℝ ) : G P ( [ θ ] ) → G P ( [ θ ] )

For some P [ θ ] ∈ G P ( [ θ ] ) and for some translation angle α ∈ [ 0, π ) this mapping can be defined as follows

V e c − 1 { ( R z ( α ) ⊗ R z ( α ) ) V e c ( P [ θ ] ) }

Proof.

( R ( α ) ⊗ R ( α ) ) V e c ( P [ θ ] ) = [ cos 2 α − cos α sin α − sin α cos α − sin 2 α cos α sin α cos 2 α − sin 2 α − sin α cos α sin α cos α − sin 2 α cos 2 α − cos α sin α sin 2 α sin α cos α cos α sin α cos 2 α ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = [ cos 2 α cos 2 [ θ ] − 2 cos α sin α cos [ θ ] sin [ θ ] + sin 2 α sin 2 [ θ ] cos α sin α cos 2 [ θ ] + cos 2 α cos [ θ ] sin [ θ ] − sin 2 α sin [ θ ] cos [ θ ] − sin α cos α sin 2 [ θ ] cos α sin α cos 2 [ θ ] + cos 2 α cos [ θ ] sin [ θ ] − sin 2 α sin [ θ ] cos [ θ ] − sin α cos α sin 2 [ θ ] sin 2 α cos 2 [ θ ] + 2 sin α cos α cos θ sin [ θ ] + cos 2 α sin 2 [ θ ] ]

= [ ( cos α cos [ θ ] − sin α sin [ θ ] ) 2 ( cos [ θ ] sin α ) ( cos [ θ ] cos α − sin α sin [ θ ] ) + cos α sin θ ( cos α cos [ θ ] − sin α sin [ θ ] ) cos α sin θ ( cos α cos [ θ ] − sin α sin [ θ ] ) ( sin α cos [ θ ] + cos α sin [ θ ] ) 2 ] = [ cos 2 ( α + [ θ ] ) sin ( α + [ θ ] ) cos ( α + [ θ ] ) cos ( α + [ θ ] ) sin ( α + [ θ ] ) sin 2 ( α + [ θ ] ) ]

Let θ ′ = α + [ θ ] then we have

( R ( α ) ⊗ R ( α ) ) V e c ( P [ θ ] ) = [ cos 2 [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ′ ] ]

Knowing the size of the original matrix i.e. a 2 × 2 matrix and using the properties of the V e c Operator i.e. the isomorphism ℝ 2 × 2 ≅ ℝ 4 we find that

V e c − 1 ( R ( α ) ⊗ R ( α ) ) V e c ( P [ θ ] ) = V e c − 1 ( [ cos 2 [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ′ ] ] ) = [ cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] ] = P [ θ ′ ]

We have already seen that the group operation on G ( [ θ ] ) is a follows

φ ( P [ θ ] , P [ θ ′ ] ) = P [ θ + θ ′ ] , ∀ P [ θ ] , P [ θ ′ ] ∈ G P ( [ θ ] ) (12)

P [ θ ] = e i [ θ ] ⊗ ( e i [ θ ] ) T , P [ θ ′ ] = e i [ θ ′ ] ⊗ ( e i [ θ ′ ] ) T

Hence, P [ θ + θ ′ ] = e [ θ + θ ′ ] ⊗ ( e i [ θ + θ ′ ] ) T .

Definition 4 ⊙ is the Hadamard Product defined as follows.

Let A and B be two matrices the Hadamard product is

( A ⊙ B ) i , j : = ( A ) i , j ( B ) i , j

Theorem 9

P [ θ + θ ′ ] = φ ( P [ θ ] , P [ θ ′ ] ) = [ φ 11 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 22 ( P [ θ ] , P [ θ ′ ] ) ] (13)

where

φ ( P [ θ ] , P [ θ ′ ] ) : = { φ 11 ( P [ θ ] , P [ θ ′ ] ) : = 1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ) φ 12 ( P [ θ ] , P [ θ ′ ] ) = φ 21 ( P [ θ ] , P [ θ ′ ] ) : = 1 + T V e c ( P [ θ ] ⊙ Q P [ θ ′ ] Q * ) φ 22 ( P [ θ ] , P [ θ ′ ] ) : = 1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ⊥ )

where 1 − T = [ 1, − 1, − 1,1 ] T and 1 + T = [ 1,1,1,1 ] T . The matrices Q and Q * are defined as

Q : = [ 0 1 1 0 ] , Q * = [ 1 0 0 − 1 ]

Proof.

1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ) = [ 1, − 1, − 1,1 ] v e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ]         ⊙ [ cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] v e c ( [ cos 2 [ θ ] cos 2 [ θ ′ ] cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ] cos [ θ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ] sin 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] [ cos 2 [ θ ] cos 2 [ θ ′ ] sin [ θ ] cos [ θ ] sin [ θ ′ ] cos [ θ ′ ] cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ] sin 2 [ θ ′ ] ] = cos 2 [ θ ] cos 2 [ θ ′ ] − 2 sin [ θ ] cos [ θ ] sin [ θ ′ ] cos [ θ ′ ] + sin 2 [ θ ] sin 2 [ θ ′ ] = cos 2 ( [ θ + θ ′ ] )

For the anti-diagonal elements, it is clear that they are equal due to symmetry.

φ 12 = φ 21 = [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ]                               ⊙ [ 0 1 1 0 ] [ cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] ] [ 1 0 0 − 1 ] ) = [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ]       ⊙ [ sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] ] [ 1 0 0 − 1 ] )

= [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ]       ⊙ [ sin [ θ ′ ] cos [ θ ′ ] − sin 2 [ θ ′ ] cos 2 [ θ ′ ] − cos [ θ ′ ] sin [ θ ′ ] ] ) = [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] sin [ θ ′ ] cos [ θ ′ ] − cos [ θ ] sin [ θ ] sin 2 [ θ ′ ] sin [ θ ] cos [ θ ] cos 2 [ θ ′ ] − sin 2 [ θ ] cos [ θ ′ ] sin [ θ ′ ] ] ) = [ 1,1,1,1 ] [ cos 2 [ θ ] sin [ θ ′ ] cos [ θ ′ ] sin [ θ ] cos [ θ ] cos 2 [ θ ′ ] − cos [ θ ] sin [ θ ] sin 2 [ θ ′ ] − sin 2 [ θ ] cos [ θ ′ ] sin [ θ ′ ] ]

= cos 2 [ θ ] sin [ θ ′ ] cos [ θ ′ ] + sin [ θ ] cos [ θ ] cos 2 [ θ ′ ]         − cos [ θ ] sin [ θ ] sin 2 [ θ ′ ] − sin 2 [ θ ] cos [ θ ′ ] sin [ θ ′ ] = cos ( [ θ + θ ′ ] ) sin ( [ θ + θ ′ ] )

for the element p 22 we get the following result

1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ⊥ ) = [ 1, − 1, − 1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ]                                                         ⊙ [ sin 2 [ θ ′ ] − cos [ θ ′ ] sin [ θ ′ ] − cos [ θ ′ ] sin [ θ ′ ] cos 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] V e c ( [ cos 2 [ θ ] sin 2 [ θ ′ ] − cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] − sin [ θ ] cos [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ] cos 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] [ cos 2 [ θ ] sin 2 [ θ ′ ] − sin [ θ ] cos [ θ ] cos [ θ ′ ] sin [ θ ′ ] − cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ] cos 2 [ θ ′ ] ] = cos 2 [ θ ] sin 2 [ θ ′ ] + 2 sin [ θ ] cos [ θ ] cos [ θ ′ ] sin [ θ ′ ] + sin 2 [ θ ] cos 2 [ θ ′ ] = sin 2 ( [ θ + θ ′ ] )

Now choosing θ ′ = 0 implies that we have φ ( P [ θ ] , P [ 0 ] ) = P [ θ + 0 ] = P [ θ ] , that is we have

φ ( P [ θ ] , P [ 0 ] ) = [ φ 11 ( P [ θ ] , P [ 0 ] ) φ 21 ( P [ θ ] , P [ 0 ] ) φ 21 ( P [ θ ] , P [ 0 ] ) φ 22 ( P [ θ ] , P [ 0 ] ) ] = [ 1 − T V e c ( P [ θ ] ⊙ P [ 0 ] ) 1 + T V e c ( P [ θ ] ⊙ Q P [ 0 ] Q * ) 1 + T V e c ( P [ θ ] ⊙ Q P [ 0 ] Q * ) 1 − T V e c ( P [ θ ] ⊙ P [ 0 ] ⊥ ) ] = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = P [ θ ]

To be thorough let us choose θ ′ = − θ , we should expect to calculate φ ( P [ θ ] , P [ − θ ] ) = P [ θ ] + [ − θ ] = P [ 0 ] , that is

φ ( P [ θ ] , P [ − θ ] ) = [ φ 11 ( P [ θ ] , P [ − θ ] ) φ 21 ( P [ θ ] , P [ − θ ] ) φ 21 ( P [ θ ] , P [ − θ ] ) φ 22 ( P [ θ ] , P [ − θ ] ) ] = [ cos 2 ( θ − θ ) cos ( θ − θ ) sin ( θ − θ ) sin ( θ − θ ) cos ( θ − θ ) sin 2 ( θ − θ ) ] = [ 1 0 0 0 ] = P [ 0 ] = P e

Last but least we will check that operation is associative that we want to make sure that

φ ( P [ θ ″ ] , φ ( P [ θ ] , P [ θ ′ ] ) ) = φ ( φ ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )

Proof. Clearly the operation is associative if all component functions are also associative. Hence we shave to show that

φ 11 ( P [ θ ″ ] , φ 11 ( P [ θ ] , P [ θ ′ ] ) ) = φ 11 ( φ 11 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )

φ 11 ( P [ θ ″ ] , φ 11 ( P [ θ ] , P [ θ ′ ] ) ) = φ 11 ( P [ θ ″ ] , P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ P [ θ + θ ′ ] ) = [ 1, − 1, − 1,1 ] [ cos 2 θ ″ cos 2 ( θ + θ ′ ) cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) sin θ ″ cos θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) sin 2 θ ″ sin 2 ( θ + θ ′ ) ] = cos 2 θ ″ cos 2 ( θ + θ ′ ) − 2 cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ )       + sin 2 θ ″ sin 2 ( θ + θ ′ )

= ( cos θ ″ cos ( θ + θ ′ ) − sin θ ″ sin ( θ + θ ′ ) ) 2 = ( cos ( θ ″ + θ ′ + θ ) ) 2 = cos 2 ( θ ″ + θ ′ + θ )

It is known that the Hadamard Product is known to be associative we can conclude that

φ 11 ( P [ θ ″ ] , φ 11 ( P [ θ ] , P [ θ ′ ] ) ) = φ 11 ( P [ θ ″ ] , P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ ( P [ θ ′ ] ⊙ P [ θ ] ) ) = 1 − T V e c ( ( P [ θ ″ ] ⊙ P [ θ ′ ] ) ⊙ P [ θ ] ) = φ 11 ( φ 11 ( P [ θ ″ ] , P [ θ ] ) , P [ θ ] )

Next, we deal with the anti-diagonal elements. We need to show that

φ 12 ( P [ θ ″ ] , φ 12 ( P [ θ ] , P [ θ ′ ] ) ) = ( φ 12 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )

φ 12 ( P [ θ ″ ] , φ 12 ( P [ θ ] , P [ θ ′ ] ) ) = 1 + T V e c ( P [ θ ] ⊙ Q P [ θ + θ ′ ] Q * ) = 1 + T V e c ( [ cos 2 θ ″ cos θ ″ sin θ ″ sin θ ″ cos θ ″ sin 2 θ ″ ]       ⊙ [ sin ( θ + θ ′ ) cos ( θ + θ ′ ) − sin 2 ( θ + θ ′ ) cos 2 ( θ + θ ′ ) − cos ( θ + θ ′ ) sin ( θ + θ ′ ) ] ) = 1 + T V e c ( [ cos 2 θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) − cos θ ″ sin θ ″ sin 2 ( θ + θ ′ ) sin θ ″ cos θ ″ cos 2 ( θ + θ ′ ) − sin 2 θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) ] )

= [ 1,1,1,1 ] [ cos 2 θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) sin θ ″ cos θ ″ cos 2 ( θ + θ ′ ) − cos θ ″ sin θ ″ sin 2 ( θ + θ ′ ) − sin 2 θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) ] = cos 2 θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) + sin θ ″ cos θ ″ cos 2 ( θ + θ ′ )       − cos θ ″ sin θ ″ sin 2 ( θ + θ ′ ) − sin 2 θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ )

We now use the following the following results(without proof)

cos ( ∑ i = 1 3 ) = cos θ ′ cos θ ″ cos θ ‴ − cos θ ′ sin θ ″ sin θ ‴ − cos θ ″ sin θ ′ sin θ ‴                               − cos θ ‴ sin θ ′ sin θ ″

sin ( ∑ i = 1 3 ) = sin θ ′ cos θ ″ cos θ ‴ + sin θ ″ cos θ ′ cos θ ‴ + sin θ ‴ cos θ ″ cos θ ′                             − sin θ ′ sin θ ″ sin θ ‴

with some algebra, we can show that

cos ( ∑ i = 1 3 ) sin ( ∑ i = 1 3 ) = 1 + T V e c ( P [ θ ] ⊙ Q P [ θ + θ ′ ] Q * )

multiplication being commutative implies associativity for the anti-diagonal elements.

Finally, for element φ 22 we have to show that

φ 22 ( P [ θ ″ ] , φ 22 ( P [ θ ] , P [ θ ′ ] ) ) = φ 22 ( φ 22 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )

φ 22 ( P [ θ ″ ] , φ 22 ( P [ θ ] , P [ θ ′ ] ) ) = φ 22 ( P [ θ ″ ] , P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ P [ θ + θ ′ ] ⊥ ) = [ 1, − 1, − 1,1 ] [ cos 2 θ ″ sin 2 ( θ + θ ′ ) − cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) − sin θ ″ cos θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) sin 2 θ ″ cos 2 ( θ + θ ′ ) ] = cos 2 θ ″ sin 2 ( θ + θ ′ ) + 2 cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ )       + sin 2 θ ″ cos 2 ( θ + θ ′ )

= ( cos θ ″ sin ( θ + θ ′ ) + sin θ ″ cos ( θ + θ ′ ) ) 2 = ( sin ( θ ″ + θ ′ + θ ) ) 2 = sin 2 ( θ ″ + θ ′ + θ ) = 1 − T V e c ( P [ θ ″ + θ ′ ] ⊙ P [ θ ] ⊥ ) = φ 22 ( φ 22 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )

We can see, from the previous section, that

R z T ( α + α ′ ) P [ θ ] R z ( α + α ′ ) = P [ θ + θ ′ ] = φ ( P [ θ ] , P [ θ ′ ] ) = [ φ 11 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 22 ( P [ θ ] , P [ θ ′ ] ) ]

it is also worthwhile noting since the Hadamard Product is commutative confirms that φ ( P [ θ ] , P [ θ ′ ] ) is a commutative group operation which implies that the centre of G P ( [ θ ] ) is itself.

We have shown that the Lie Group S O ( 2, ℝ ) acts on the on the manifold G P ( [ θ ] ) to generate new elements in G P ( [ θ ] ) where many interesting properties of this group action have been demonstrated. The group operation φ ( P [ θ ] , P [ θ ′ ] ) in terms of matrix operations requires the use of the vectorisation operator and the Hadamard Product because it is not a traditional vector sum when the angles are added together. Adding the vectors in a traditional way would require the tensor product of the sum and a normalisation constant. I believe that projection matrices have more interesting structures that can be further studied. I hope that this article will raise some interest in what, I think, does deserve more investigation.

I wish to personally thank the Editor(s) and Mrs Eunice Du for all her help. I also wish to extend my gratitude to the referee for their comments.

The author declares no conflicts of interest regarding the publication of this paper.

Niglio, J.-F. (2019) A Follow-Up on Projection Theory: Theorems and Group Action. Advances in Linear Algebra & Matrix Theory, 9, 1-19. https://doi.org/10.4236/alamt.2019.91001