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Definition 2.1: Let A R and B R , t = Definition 2.5: Let = (V, ) be a k-dimensional vector ∈ m×n ∈ p×q A ∗ lcm(n,p) be the least common multiple of n and p. Then space with e = i , i , , i as a set of basis. Denote { 1 2 ··· k} the STP of A and B, denoted by A ⋉ B, is defined by k i i = cs i , i,j =1, 2, , k. (8) ⋉ i ∗ j i,j s ··· A B := A It/n B It/P , (1) s=1 ⊗ ⊗ X where is Kronecker product. Then the product matrix of is defined as ⊗ A It is easy to see that STP is a generalization of conventional c1 c1 c1 c1 1,1 1,2 ··· 1,k ··· k,k matrix product. That is, when n = p, the STP is degenerated to 2 2 2 2 c1,1 c1,2 c1,k ck,k P := ··· ··· . (9) the conventional matrix product, i.e.,, A ⋉ B = AB. Because A .. .. . . of this, in most cases the symbol ⋉ is omitted. ck ck ck ck 1,1 1,2 ··· 1,k ··· k,k One of the most important advantages of STP is that STP keeps most important properties of conventional matrix k Assume x = xiij , is expressed in a column vector form product available, including association, distribution, etc. In j=1 as x = (x , x ,P , x )T . Similarly, y = (y ,y , ,y )T . the following we introduce some additional properties of STP, 1 2 ··· k 1 2 ··· k Then which will be used in the sequel. Theorem 2.6: In vector form two product of two hypercom- Define a swap matrix W[m,n] mn×mn as follows: ∈M plex numbers x, y is computable via following formula. ∈ A W := I δ1 , I δ2 , , I δm, . [m,n] n ⊗ m n ⊗ m ··· n ⊗ m ∈ Lmn×mn (2) x y = PAxy. (10) Proposition 2.2: Let x Rm and y Rn be two column ∗ ∈ ∈ vectors. Then Using formula (10) and the properties of STP yields the following results, which are fundamental for our further in- ⋉ ⋉ W[m,n]x y = y x. (3) vestigation. Theorem 2.7: The following proposition “swaps” a vector with a matrix: Proposition 2.3: Let x Rt be a column vector, and A be (i) is commutative, if and only if, ∈ A an arbitrary matrix. Then P I W =0. (11) A k − [k,k] ⋉ ⋉ x A = (It A) x. (4) (ii) is associative, if and only if, ⊗ A Throughout this paper the default matrix product is assumed 2 PA = PA (Ik PA) . (12) to be STP, and the symbol ⋉ is omitted if there is no possible ⊗ confusion. III. HYPERCOMPLEX NUMBERS A. Perfect Hypercomplex Algebra on R B. Matrix Expression of an Algebra Definition 3.1: [11] A number p is called a hypercomplex R We are only interested in algebras over . number, if it can be expressed in the form Definition 2.4: [7] p = p + p i + + p i , (13) (i) An algebra over R is a pair, denoted by = (V, ), 0 1 1 ··· n n A ∗ where V be a real vector space, and : V V V , where p R, i = 0, 1, ,n, i , i = 1, 2, ,n are called ∗ × → i ∈ ··· i ··· satisfies hyperimaginary units. (ax + by) z = ax z + by z, Remark 3.2: A hypercomplex number may belong to dif- ∗ ∗ ∗ x (ay + bz)= ax y + bx z, x,y,z V, a,b R. ferent algebras, depending on their product structure matrices. ∗ ∗ ∗ ∈ ∈ A hypercomplex algebra, denoted by , is an algebra over R (5) A with basis e = i0 := 1, i1, , in . (ii) An algebra = (V, ) is said to be commutative, if { ··· } A ∗ Proposition 3.3: Assume x y = y x, x, y V. (6) ∗ ∗ ∈ = p0 + p1i1 + + pnin p0,p1, ,pn R . A { ··· | ··· ∈ } (iii) An algebra = (V, ) is said to be associative, if A ∗ Then its product matrix (x y) z = x (y z), x,y,z V. (7) P := [M ,M , ,M ], ∗ ∗ ∗ ∗ ∈ A 0 1 ··· n 3 where M R , i = 0, 1, ,n, satisfies the The ξ(x , x , , x ) is called the characteristic function i ∈ (n+1)×(n+1) ··· 0 1 ··· n following condition. of . A (i) Example 3.8: Consider C = R(i). Calculating right hand side of (17) for PC, we have M0 = In+1 (14) 2 2 xi(x1, x2)= x + x . is an identity matrix. 1 2
(ii) Hence, ξ(x1, x2)=0, if and only if, x1 = x2 =0. It follows that PC is jointly non-singular. Col (M )= δj+1 , j =1, 2, , n. (15) 1 j n+1 ··· Summarizing above arguments, we have the following re- Definition 3.4: A hypercomplex algebra is called a PHA, sult. A denoted by H, if it is commutative and associative. Proposition 3.9: Let be a finite dimensional algebra over A ∈ A Example 3.5: Consider C. It is easy to calculate that its R. Then is a field, if and only if, A product structure matrix is (i) is commutative, that is, (11) holds; A 1 0 0 1 (ii) is associative, that is, (12) holds; PC = . (16) A − (iii) Each 0 = x is invertible, that is, P is jointly "011 0 # 6 ∈ A A invertible. A straightforward computation verifies (11) and (12), hence it is a PHA. Unfortunately, it is well known that the only finite dimen- sional algebra over R, which is a field, is C. In this paper we are particularly interested in commutative and associative B. Invertibility of elements in PHA algebra H. Then unless calA = C, there must be some Now for a PHA, say, = (V, ), if every 0 = x V has A ∈ A ∗ 6 ∈ elements x =0, which are not invertible. its inverse x−1 such that x x−1 = x−1x = 1, then is 6 ∗ A Definition 3.10: Let H. Its zero set is defined by a field. Unfortunately, according to Weierstrass, if = C, it A ∈ A 6 is not a field. Hence, when an element x V (we also say A := z det (PAz)=0 (18) ∈ Z { ∈ A | } x , which means x V .) is invertible. ∈ A ∈ It is clear that To answer this question, we need some new concepts, which (i) if = C, then = 0 ; are firstly discussed in [5]. A ZA { } (ii) if = C, then 0 , and 0 = . Definition 3.6: A 6 { } ∈ ZA ZA\{ } 6 ∅
(i) Let A1, A2, , Ar be a set of square real matrices. ··· C. Coordinate Transformation of Hypercomplex Algebra A , A , , A are said to be jointly nonsingular, if their 1 2 ··· r non-trivial linear combination is non-singular. That is, Since the product matrix PA of a hypercomplex algebra assume depends on the basis, it is necessary to consider different r forms of PA under a change of basis, which is commonly det ciAi =0, i=1 ! called a coordinate transformation. Let i1, i2, , in be the X ··· then c = c = = c =0. hyperimaginary units of and 1 2 ··· r A (ii) Let A Rk×k2 . A is said to be jointly nonsingular, if ∈ (1, j1, j2, , jn)=(1, i1, i2, , in)T, (19) A = [A , A , , A , where A R 2 and A i = ··· ··· 1 2 ··· k s ∈ k×k { ik 1, 2, , k are jointly nonsingular. ··· } 1 E R Then we have the following result: T = (n+1)×(n+1), (20) "0 T0# ∈ Proposition 3.7: [5] A Rk×k2 is jointly non-singular, if ∈ where T is non-singular. and only if, one of the following two equivalent conditions is 0 Then x can be expressed as satisfied: ∈ A
(i) The matrix P x R is non-singular for all x =0. x = x0 + x1i1 + + xnin = x0 +¯x1j1 + +¯xnjn, A ∈ k×k 6 ··· ··· (ii) The following homogeneous polynomial which is called a coordinate change on . A ξ x , , x Ax T T ( 1 k) = det( ) Denote x = (x0, x1, , xn) , x¯ = (¯x0, x¯1, , x¯n) . ···k k ··· ··· Then = µi1,··· ,ik xi1 xik =0, x =0. i1=1 ··· ik =1 ··· 6 ∀ 6 P P (17) x¯ = T −1x. (21) 4
Definition 3.11: Let and be two n +1 dimensional IV. LOWER DIMENSIONAL PHAS A A hypercomplex algebras. and are called isomorphic, if A A This section considers some examples of various dimen- there exists a bijective mapping Ψ: , satisfying A → A sional algebras. (i)
Ψ(1) = 1 (22) A. Structure of H A ∈ 2 H (ii) Consider 2. According to Proposition3.3, its PSM is A ∈ Ψ(ax + by)= aΨ(x)+ bΨ(y), x,y ,a,b R; 1 0 0 α ∈ A ∈ PA = (27) (23) "0 1 1 β# (iii) Consider a coordinate change
Ψ(x y)=Ψ(x) Ψ(y), x,y . (24) 1 s ∗ ∗ ∈ A T = , t =0. Ψ is called an isomorphism. "0 t# 6 A straightforward verification shows the following result Using formula (25), we have immediately. −1 P = T PA (T T ) Proposition 3.12: Assume A ⊗ 1 0 0 αt2 s(s + tβ) (28) n = − = x0 + xiii x0, x1 , xn R "0 1 1 s + tβ # A | ··· ∈ i=1 Pn Now we may choose = x¯0 + x¯iji x¯0, x¯1 , x¯n R . A i=1 | ··· ∈ s = tβ, − and are isomorphic,P if and only if, there is a non-singular A A Then we have matrix T R(n+1)×(n+1) of the form (20), such that (21) ∈ 2 holds true. 1 0 0 αt PA = (29) Proposition 3.13: Assume , H , with their PSMs "011 0 # A A ∈ n+1 as P and P respectively. and are isomorphic, if and We consider them case by case. A A A A only if, there exists a non-singular matrix T as in (20) such • If α =0, we have that 1 0 00 −1 PA = (30) P = T PA (T T ) . (25) 0 1 10 A ⊗ " # 1 Proof: (Necessary) Let T be constructed as is (20), such • If α> 0, choosing t = , then we have that that √|α| x¯ = T −1x. 1 00 1 PA = . (31) Then we have "0 11 0# 1 P xy = T P x¯y,¯ x,y . (26) • If α< 0, choosing t = yields A A ∈ A √|α| The right hand side (RHS) of (26) becomes 1 0 0 1 P . (32) −1 −1 A = − RHS(26) = = T PAT xT y "011 0 # −1 −1 = T PAT In+1 T xy. We conclude that up to isomorphism there are three ⊗ A ∈ Since x, y are arbitrary, we have H2, they are −1 −1 P = T P T I T . • set of dual numbers ( D), which corresponds to (30); A A n+1 ⊗ A • set of hyperbolic numbers ( ), which corresponds to Hence, AH −1 (31); P = T PA (In+1 T ) T A ⊗ C = T −1P (T T ) • set of complex numbers ( ), which corresponds to (32). A ⊗ (Sufficiency) If (26) holds true, it is easy to verify that Next, using (17), we can calculate their characteristic func- tions. x¯ = T −1x • is an isomorphism. 2 ξAD = x0. (33) 5
Then Then we have the following result:
Theorem 4.3: H3, if and only if, PA has the form of AD = x0 + x1i D x0 =0 . (34) A ∈ Z { ∈ A | } (39) with parameters satisfying • a = ce + f 2 bf cr, 2 2 − − ξAH = x0 x1. (35) d = cq ef, (42) − − p = e2 + fq bq er. Then − − Proof: (Necessity) Comparing both sides of (??),(??) AD = x0 + x1i H x0 = x1 . (36) Z { ∈ A | ± } shows that a necessary condition for (40) holds true is (refer • to the 6th and 8th blocks of both sides) 2 2 ξC = x0 + x1; . (37) AB = BA. (43) Then Then it is easy to verify that (42) provides necessary and C = 0 . (38) sufficient condition for (43) to be true. Z { } (Sufficiency) A careful computation shows as long as (42) Remark 4.1: holds, the RHS of (40) and the LHS of (40), shown in (41), (i) It is obvious that all , , and C are associative and AD AH are equal. commutative, that is they are all PHAs. (ii) They have minimum polynomials x2, x2 x2, and 0 0 − 1 Remark 4.4: Theorem 4.3) provides an easy way to con- x2 +x2 respectively. According to Galois theory, only the 0 1 struct H . In fact, the parameters b,c,e,f,q,r can be minimum polynomial of i is irreducible, the extension is A ∈ 3 assigned freely, then a,d,p can be obtained by (42). It is easy a field. Hence, only C is a fields. to see that there are uncountably many algebras of dimension (iii) It is easily seen that their zero sets are all zero measure 3, which are commutative and associative. set. This is always true for all PHA. Because their Next, we give a numerical example. zero set are zero set of algebraic equations, which are Example 4.5: Construct H by setting b = c = f = called algebraic numbers. The set of algebraic numbers A ∈ 3 q = r = 0 and e = 1. Then we have d = a = 0 and p = 1. is always zero measure set. The PSM of is A B. Structure of Triternions 100000001 P = . (44) Definition 4.2: An algebra of dimension 3 is called a A 010101010 H A 001000100 triternion if 3. A ∈ It is easy to see that if H is symmetric, then its PSM In fact, when x is expressed into standard form as A ∈ 3 ∈ A is x = x0 + x1i1 + x2i2, x0, x1, x2 R, 1 0 0 0 a d 0 d p ∈ then we have PA = 0 1 0 1 b e 0 e q (39) i2 = 0; i2 =1, 0 0 1 0 c f 1 f r 1 2 i1 i2 = i2 i1 = i1. Next, we consider when is associative. According to ∗ ∗ A Theorem 2.7, the necessary and sufficient condition is Then it is easy to calculate that
2 2 P = P (I P ) . (40) ξA = (x0 x2)(x0 + x2) . (45) A A 3 ⊗ A −
Denote I = I3, Hence,
0 a d 0 d p = (x , x , x ) R3 x = x . (46) ZA { 0 1 2 ∈ | 0 ± 2} A = 1 b e , B = 0 e q . 0 c f 1 f r C. Structure of Perfect Quaternions A direct computation shows that This section considers some some algebras in H4. It seems LHS of (40) = (I,A,B,A,aI + bA + cB, not easy to provide a general description for algebras in H4. dI + eA + fB,B,dI + eA + fB,pI + qA + rB), (41) The principle argument is similar to triternions. We give some 2 2 RHS of (40) = (I, A, B, A, A ,AB,B,BA,B ). simple examples. 6
Example 4.6: Consider an H . Assume (ii) Consider : A ∈ 4 A8 It is easy to calculate that = p + p i + p i + p i p ,p ,p ,p R , A { 0 1 1 2 2 3 3 | 0 1 2 3 ∈ } ξA8 = det(PA8 x) satisfying = x4 + x4 + x4 + x4 0 1 2 3 (49) 2 2 2 2 2 2 2 2 3 2 2 i , i , i 1, 0, 1 , i1 i2 = i2 i1 = i3, 2(x x + x x + x x + x x 1 2 3 ∈ {− } ∗ ∗ ± 0 1 0 2 0 3 1 − 2 2 2 2 i2 i3 = i3 i2 = i1, i3 i1 = i1 i3 = i2. +x x + x x )+8x x x x . ∗ ∗ ± ∗ ∗ ± 1 3 2 3 0 1 2 3 To save space, we denote It follows that T R4 A8 = (x0, x1, x2, x3) PAi = [I4,Qi]. Z ∈ | (50) ξ (x , x , x , x )=0 . A8 0 1 2 3 Using MATLAB for an exhausting searching, we got eight } algebras as follows: D. Some Other Examples • 0 −1 0 00 0 −100001 H To see there are also n, for n > 4, such examples 10 0000 010010 Q1 = A ∈ 00 0110 000100 are presented as follows. First example is a set of simplest 0 0 −1 0 0 −1 0 01000 • PHAs, which are called trivial PHAs. 0 0 −10 0 00 −1000 01 Example 4.8: Define an n+1 dimensional algebra as n+1 1000000 −1 0 0 −1 0 A Q2 = 0 0 0 −110 0 0 0 −1 0 0 follows: Let i , k = 1, 2, ,n be its hyperimaginary units. k 0010010 010 00 ··· Set •
0 −1000010000 −1 is it =0, s,t =1, 2, , n. 10 00000 −1 0 0 −1 0 ∗ ··· Q3 = 00 011000010 0 0 Then it is easy to verify that H . Moreover, its 0 0 −1 0 0 −10 0 10 0 0 An+1 ∈ n+1 • PSM, P 0 can be determined by the following: 0 −10 0 00100 0 0 −1 |calAn+1 100000010010 Q4 = 0 0 0 −110000 −1 0 0 i δ , i =1, 2, ,n + 1; 001001001000 n+1 ··· r+1 Coli P 0 = δ , i = r(n +1)+1, r =1, 2, , n, • An+1 n+1 ··· 010 000 −100 0 0 −1 0, Otherwise. 100 000 010010 Q5 = 0 0 0 −11 0 0 00 −1 0 0 (51) 0 0 −1 0 0 −1 0010 0 0 Its characteristic function is • 010000 −1 0 00 0 −1 100000 0 −1 0 0 −1 0 n+1 n Q6 = ξ = x . (52) 0001100 0010 0 AD 0 0010010 0100 0 Hence, •
010 0001000 01 n A = x0 + x1i1 + + xnin x0 =0 . (53) 100 0000 −1 0 0 −1 0 Z D { ··· | } Q7 = 0 0 0 −11 0 0 0 0 −1 0 0 − − c 0 0 1 0 0 10010 00 If x An , say ∈ Z D • 010000100001 x = x0 + x1i1 + + xnin, x0 =0, 100000010010 ··· 6 Q8 = 000110000100 then 001001001000 n −1 1 xi Next, choose some H for further study. x = ii. 4 x − x2 A ∈ 0 i=1 0 Example 4.7: Recall Example 4.6). X Next, we give an example for n =5. (i) Consider : A3 Example 4.9: Consider a hypercomplex algebra , which It is easy to calculate that A has its PSM as ξA = det(PA x) 3 3 PA := δ5[1, 2, 3, 4, 5, 2, 0, 0, 3, 0, 3, 0, 2 2 2 2 2 2 (54) = (x0 x2) + (x1 x3) 0, 0, 0, 4, 3, 0, 0, 0, 5, 0, 0, 0, 0], − −2 2 +2(x0x1 + x2x3) + 2(x0x3 + x1x2) . where δ0 = 0 . (47) 5 5 A straightforward computation shows that is commutative It follows that A and associative, hence H5. Then it is easy to calculate = (x , x , x , x )T R4 A ∈ ZA3 0 1 2 3 ∈ | that (x0 = x2) (x1 = x3) or (x0 = x2) (x1 = x3) . ∩ − − ∩ } ξ (x)= x3(x2 x x ). (55) (48) A 0 0 − 1 3 7
So its zero set is
= (x , x , x , x , x )T R5 x =0, or x2 = x x . ZA { 0 1 2 3 4 ∈ | 0 0 1 3} (56)
V. CONCLUSION In this paper the perfect hypercomplex algebra is consid- ered. Using STP, necessary and sufficient conditions on the structure matrix of an algebra to be a PHA is proposed. Based on the matrix expression of homomorphisms between algebras, certain lower dimensional PHAs are classified up to isomorphism. Their characteristic functions and zero sets are discussed.
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