A complete classification of three-dimensional algebras over {\mathbb R} and {\mathbb C}
Yuji Kobayashi, Kiyoshi Shirayanagi, Sin-Ei Takahasi, Makoto Tsukada
aa r X i v : . [ m a t h . R A ] M a r A complete classification ofthree-dimensional algebras over R and C – OnKoChiShin (visiting old, learn new)
Yuji Kobayashi, Kiyoshi Shirayanagi, Sin-Ei Takahasi and Makoto Tsukada
Abstract
We provide a complete classification of three-dimensional associativealgebras over the real and complex number fields based on a completeelementary proof. We list up all the multiplication tables of the alge-bras up to isomorphism. We compare our results with those given bymathematicians in the 19th century.
Since the discovery of the quaternions by Hamilton [1], many mathematicianshad been trying to extend number systems to more general systems such as tes-sarines, coquaternions, biquaternions, and algebras introduced by Grassmannand by Clifford etc. (see Kolmogorov and Yushkevich [4] and van der Waerden[10]). These are finite-dimensional associative algebras over the real numberfield R , which were called ”hypercomplex number systems” at that time. It wasPeirce who first studied associative algebras systematically ([5]). He enumerated”pure” algebras of low dimension using so called Peirce’s decomposition. How-ever, not only was his terminology ambiguous and were his proofs not rigorous,but he also did not give proofs for non-isomorphism between the listed algebras.Hawkes [2] and Taber [9] tried to give rigid reasoning for the methods of Peirce,but their resulting proofs are not satisfactory from the viewpoint of modernmathematics. On the other hand, Scheffers [6] and Study [8] determined unitalalgebras of low dimension over R and the complex number field C . Keywords : Associative algebras, Unital algebras, Curled algebras, Waved algebras,Straight algebras, Peirce’s classification. : Primary 16B99; Secondary 16U99Research of the first, third and fourth authors was supported in part by JSPS KAKENHIGrant Number JP25400120.Research of the second author was supported in part by JSPS KAKENHI Grant NumberJP15K00025.
1n this paper, we classify (associative but not necessarily unital) algebras ofdimension 3 over R and C up to isomorphism. Our classification is completeand the methods are very different from Peirce’s. The proofs are rigorous andconcrete. We first treat unital algebras in a classical way. Then, we divide non-unital algebras into three types, curled , waved and straight algebras. An algebra A is curled if x and x are linearly dependent for any x ∈ A , it is waved if itis not curled but x, x and x are linearly dependent for any x ∈ A , and it isstraight otherwise. We look into subalgebras of A of dimension 2, and considerall combinations of them to list up all possible multiplication tables of algebrasof dimension 3. Finally, we make a complete list of algebras of dimension 3 thatare not isomorphic to each other. Our methods are elementary and we use idealtheory as little as possible.We appreciate the researches of the abovementioned past mathematicians,in particular, we respect Peirce for his pioneering work. By presenting our newcomplete classification, we attempt to make the paper as an homage to his work.We believe that the spirit of this paper can be best expressed in the subtitle, con-sisting of four Chinese characters which are pronounced as ”OnKoChiShin”. Itis a word from the Analects of Confucius [11], which is commonly used in Japanas an idiom meaning “visiting old, learn new”, or more precisely, discoveringnew things by studying the past through scrutiny of the old.Our result is as follows. Over C , up to isomorphism, we have 5 unital algebras U , U , U , U , U , and among the non-unital algebras, we have 5 curled algebras C , C , C , C , C , S , S , S , S and 9 waved algebras W , W , W , W , W , W , W , W , W and one infinite family (cid:8) W ( k ) (cid:9) k ∈H of waved algebras, where the symbols U i , C i , S i and W i denote specifiedmultiplication tables and H is the right half-plane { x + y i | x > x = 0 , y ≥ } of the complex plane. Over R , in addition to the above algebras, we have oneunital algebra U − , one non-unital straight algebra S − and one infinite family (cid:8) W − ( k ) (cid:9) k ≥ of non-unital waved algebras. The concrete description of theabove multiplication tables is given in Section 10.The rest of this paper is organized as follows. First, we give a classification oftwo-dimensional algebras in Section 3, because it will be utilized to achieve theaim of classifying three-dimensional algebras. Next, we determine unital alge-bras in Section 4. We classify curled algebras in Section 5 and straight algebras2n Section 6. In the subsequent three sections, we treat the most difficult casewhere algebras are waved. In this case we focus on two subalgebras having theirone-dimensional intersection. We explain the strategy for the classification inSection 7, enumerate all possible waved algebras in Section 8, and check if thealgebras enumerated are isomorphic or not in Section 9. In the final section, wesummarize the results and give a complete list of algebras modulo isomorphism,and compare it with the lists given by Peirce, Scheffers and Study.Throughout this paper, we assume that an algebra is associative. Let K = R or K = C . In this paper, we express an algebra over K by the mul-tiplication table. Let A be an algebra over K of dimension n where { e , . . . , e n } is a linear bases of A . Then, the structure of A is determined by the table whichis an n × n matrix a · · · a n ... . . . ... a n · · · a nn (1)with a ij ∈ A , where a ij = e i e j for all i ∈ [1 , n ] and j ∈ [1 , n ]. That is, for p = P ni =1 x i e i , q = P ni =1 y i e i ∈ A with x i , y i ∈ K , the product pq is defined by pq = X i,j ∈ [1 ,n ] x i y j a ij . We say that A is the algebra on { e , . . . , e n } defined by (1).We refer to A as a zero algebra if its multiplication table is the zero matrixwith a ij = 0 for all i ∈ [1 , n ] and j ∈ [1 , n ], or equivalently, if pq = 0 for all p, q ∈ A . Related to zero algebras, we call A a zeropotent algebra if p = 0 forall p ∈ A . Obviously a zero algebra is zeropotent, but in general the converseis not true. In fact, it is easy to show that A is zeropotent if and only if A isdefined by an alternative matrix with a ij = − a ji for all i ∈ [1 , n ] and j ∈ [1 , n ].For a classification of zeropotent nonassociative algebras, see [3, 7].Let us describe a criterion for two algebras to be isomorphic. Let A and B bealgebras over K of dimension n defined by tables ( n × n matrices), ( a ij ) and ( b ij ),where { e , . . . , e n } and { f , . . . , f n } are linear bases of A and B , respectively.Moreover, let e i e j = a ij = P ns =1 α sij e s and f i f j = b ij = P nt =1 β tij f t for all i ∈ [1 , n ] and j ∈ [1 , n ]. Suppose that ϕ : A → B is an algebra homomorphismand M ϕ = ( m st ) with m st ∈ K is the n × n matrix of ϕ as a linear mappingwith respect to the linear bases: ϕ ( e s ) = n X t =1 m st f t . Let us call M ϕ a transformation matrix from A to B . For each ( i, j ), the equality ϕ ( e i ) ϕ ( e j ) = ϕ ( e i e j ) = ϕ ( a ij ) holds. Calculating the both sides, we see that A
3s isomorphic to B over K if and only if there exists a non-singular matrix M ϕ such that X s α sij m st = X k,l β tkl m ik m jl for all t, i, j ∈ [1 , n ]. · · · · · · · · · ( ∗ )Assume that A is an algebra over K of dimension 3. An element e ∈ A is curled (resp. waved ) if { e, e } (resp. { e, e , e } ) is linearly dependent over K .An algebra A is called curled if every element of A is curled. A is called waved if A is not curled but every element of A is waved. Note that in a waved algebrathere is an element e such that { e, e } is linearly independent but every suchelement e satisfies that { e, e , e } is linearly dependent. We call A straight , if itis not curled nor waved, that is, if { e, e , e } forms a linear base of A for some e ∈ A .When B is an algebra of dimension 2, B is curled if every element of B iscurled, otherwise B is straight .For two algebras A and B over K , A ⊕ B denotes their direct sum; for p = ( x, y ) , q = ( x ′ , y ′ ) ∈ A ⊕ B their product is defined by pq = ( xx ′ , yy ′ ). Before entering the discussion on two-dimensional algebras, let us consider one-dimensional algebras. Let A be an algebra over K of dimension 1 with a linearbase { e } . Then, e = ke for some k ∈ K . Depending on whether k is 0 or not,it follows that A is the algebra defined by e = 0 or is isomorphic to the algebradefined by e = e . We denote the former algebra by A and the latter by A . A is a zero algebra and A is a unital algebra with the identity element e .Now let A be an algebra over K of dimension 2 with a linear base { e, f } .First, suppose that A is curled. Then, e = ke and f = ℓf (2)for some k, ℓ ∈ K . Here if k = 0 (resp. ℓ = 0), replacing e (resp. f ) by e/k (resp. f /ℓ ), we may assume that k and ℓ are 0 or 1 in (2).We can write ef = ae + bf and f e = ce + df with a, b, c, d ∈ K . We have kae + kbf = kef = e f = e ( ef ) = e ( ae + bf ) = ae + bef = a ( k + b ) e + b f. It follows that ab = 0 and b = kb. (3)Similarly we have cd = 0 , a = ℓa, c = ℓc and d = kd. (4)4e have kce + ade + bdf = e ( ce + df ) = ef e = ( ae + bf ) e = kae + bce + bdf. It follows that k ( a − c ) = ad − bc. Similarly we have ℓ ( d − b ) = ad − bc. Because A is curled, xe + yf and( xe + yf ) = x e + xyef + xyf e + y f = ( kx + axy + cxy ) e + ( ℓy + bxy + dxy ) f. are linearly dependent over K for any x, y ∈ K . Hence, (cid:12)(cid:12)(cid:12)(cid:12) x kx + ( a + c ) xyy ℓy + ( b + d ) xy (cid:12)(cid:12)(cid:12)(cid:12) = (cid:0) ( b + d − k ) x + ( ℓ − a − c ) y (cid:1) xy = 0 . It follows that k = b + d and ℓ = a + c. (5)It is easy to solve the equations (3), (4) and (5) for k, ℓ ∈ { , } and a, b, c, d ∈ K . We obtain 7 solutions:( k, ℓ, a, b, c, d ) = (0 , , , , , , (0 , , , , , , (0 , , , , , , (1 , , , , , , (1 , , , , , , (1 , , , , , , (1 , , , , , . In correspondence to them we have 7 curled algebras A , A , A , A ′ , A ′ , A ′ and A ′ defined by (cid:18) (cid:19) , (cid:18) e f (cid:19) , (cid:18) e f (cid:19) , (cid:18) e f (cid:19) , (cid:18) e f (cid:19) , (cid:18) e fe f (cid:19) and (cid:18) e ef f (cid:19) , respectively. Using the isomorphism criterion in the previous section, the alge-bras A and A ′ are isomorphic via the transformation matrix M ϕ = (cid:18) (cid:19) which is a non-singular matrix satisfying ( ∗ ). Similarly, A and A ′ are iso-morphic. The algebra A ′ is isomorphic to A via the transformation matrix (cid:18) (cid:19) . Similarly, A ′ is isomorphic to A . On the other hand, it is easyto see that there is no non-singular matrix satisfying ( ∗ ) between A and A .Consequently we have three non-isomorphic curled algebras A , A , and A .Next, we suppose that A is straight, that is, there is g ∈ A such that { g, g } is a linear base of A . Then, A is commutative, and we can write g = bg + cg (6)with b, c ∈ K . 5f b = c = 0, that is, g = 0, then letting e = g and f = g , A is defined bythe table (cid:18) e (cid:19) . (7)Next, if b = 0 and c = 0, then letting f = g/b and e = f , A is defined by thetable (cid:18) e ee e (cid:19) . (8)Lastly, suppose that c = 0. Let e = 1 c ( g − bg ) . (9)Then, using (6), we see eg = ge = g , and so e is the identity element of A .Moreover, by (9), { g, e } is a linear base of A and g = bg + ce (10)holds. Let D = b + 4 c , and let h = 2 g − be . Then, by (10) we have h = 4 g − bg + b e = De. (11)If K = C and D = 0, or K = R and D >
0, let f = √ D h . Then, by (11) wehave f = e . Hence, A is defined by (cid:18) e ff e (cid:19) . (12)If K = R and D <
0, let f = √− D h . Then, f = − e by (11). Hence, A isdefined by (cid:18) e ff − e (cid:19) . (13)If D = 0, let f = h , then, f = 0 by (11), and A is defined by (cid:18) e ff (cid:19) . (14)If A is defined by the table in (8), then by replacing f by f − e , A is definedalso by (cid:18) e
00 0 (cid:19) . (15)If A is defined by the table in (12), then by replacing e by e + f and f by e − f , A is defined by (cid:18) e f (cid:19) . (16)We denote the algebras defined by the tables in (7), (15), (12), (13) and(14) by A , A , A , A − and A , respectively. Using the isomorphism criterionin Section 2 it is easy to show that these algebras are not isomorphic to eachother. 6 Unital algebras
In this section, A is a unital algebra over K of dimension 3 with identity element1. First, suppose that there is an element h of A such that { , h, h } formsa linear base of A , in this case we say that A is unitally straight . Then, A iscommutative, and we can write h = ah + bh + c with a, b, c ∈ K . Thus, the algebra A is generated by h and is isomorphic to theresidue algebra of the polynomial algebra K [ X ] modulo the ideal generated by P ( X ) = X − aX − bX − c. Let α, β and γ be the roots of P in C ; P ( X ) = ( X − α )( X − β )( X − γ ) . (i) Suppose that α, β and γ are different from each other.(i1) Suppose that α, β, γ ∈ K . We have an isomorphism φ : A → K ⊕ K ⊕ K of algebras defined by φ ( h ) = ( α, β, γ ) . Let e = φ − (1 , , f = φ − (0 , ,
0) and g = φ − (0 , , { e, f, g } , A is defined by e f
00 0 g . (17)(i2) Next, suppose that K = R , and α is real but β and γ are not real. Wehave an isomorphism φ : A → K ⊕ B of algebras defined by φ ( h ) = ( α, ¯ h ) , where B = K [ X ] / (( X − β )( X − γ )) and ¯ h is the image of h by the naturalsurjection from A to B . By the results in the previous section, B is defined by (cid:18) f ′ g ′ g ′ − f ′ (cid:19) of B on a suitable base { f ′ , g ′ } . Let e = φ − (1 , f = φ − (0 , f ′ )and g = φ − (0 , g ′ ), then on the base { e, f, g } , A is defined by e f g g − f . (18)(ii) Next, suppose that α = β = γ , then α, β ∈ R and P ( X ) = ( X − α )( X − β ) .
7e have an isomorphism φ : A → K ⊕ B defined by φ ( h ) = ( α, ¯ h ) , here B = K [ X ] / (( X − β ) ) and ¯ h is the natural image of h in B . By the results inthe previous section B is defined by (cid:18) f ′ g ′ g ′ (cid:19) on the base { f ′ , g ′ } with f ′ = 1and g ′ = h − β . Thus, in the same way as above A is defined by e f g g . (19)on a suitable base { e, f, g } .(iii) Lastly, suppose that α = β = γ , that is, P ( X ) = ( X − α ) . Let e = 1, f = h − α and g = f . Then, f g = gf = g = 0, and so A is defined by e f gf g g . (20)Using the isomorphism criterion it is easy to show that the algebras definedby (17), (19) and (20) are not isomorphic to each other. When K = R , thealgebra defined by (18) is not isomorphic to the algebras above either.Next, suppose that A is not unitally straight, that is, there is no h ∈ A such that { , h, h } forms a linear base of A . Let { , f, g } be a linear base of A . Then { , f } is a linear base of the unital subalgebra B of A generated by f .As discussed in the previous section, B is defined by (12) or (14) when K = C .Hence, by changing f by a suitable element of B , we may suppose that f = 0or f = 1. When K = R , there is another possibility f = −
1, because B may be defined by (13). Similarly, we may suppose that g = 0 or g = 1 (or g = − K = R ).Let f g = a + bf + cg and gf = a ′ + b ′ f + c ′ g for a, b, c, a ′ , b ′ , c ′ ∈ K . If f = 0, then we have0 = f g = f ( a + bf + cg ) = af + c ( a + bf + cg ) = ac + ( a + bc ) f + c g. and0 = gf = ( a ′ + b ′ f + c ′ g ) f = a ′ f + c ′ ( a ′ + b ′ f + c ′ g ) = a ′ c ′ + ( a ′ + b ′ c ′ ) f + c ′ g. It follows that a = c = a ′ = c ′ = 0 . (21)Similarly, if g = 0, we have a = b = a ′ = b ′ = 0 . (22)If f = 1, then g = f g = f ( a + bf + cg ) = af + b + c ( a + bf + cg ) = ac + b + ( a + bc ) f + c g. ac + b = a + bc = c − . Similarly, we have a ′ c ′ + b ′ = a ′ + b ′ c ′ = c ′ − . Hence, we have c = ± , b = ∓ a (double-sign corresponds) and c ′ = ± , b ′ = ∓ a ′ (d-s.c.) . (23)Similarly, if g = 1, we have b = ± , c = ∓ a (d-s.c.) and b ′ = ± , c ′ = ∓ a ′ (d-s.c.) . (24)Finally, if K = R and f = −
1, then we have − g = f g = f ( a + bf + cg ) = af − b + c ( a + bf + cg ) = ac − b + ( a + bc ) f + c g. Hence, c = −
1, but this is impossible in R . Similarly, g = − f = g = 0, then by (21) and (22) we have a = b = c = a ′ = b ′ = c ′ = 0. Hence, A is defined by e f gf g (25)on the base { e, f, g } with e = 1.(ii) If f = 0 and g = 1, then by (21) and (24) a = c = a ′ = c ′ = 0 , b = ± b ′ = ± . Here, if b = b ′ = 1, that is, f g = gf = f , then 1 , f + g and( f + g ) = f + f g + gf + g = 2 f + 1are linearly independent. Hence, A is unitally straight. Similarly, the case b = b ′ = − b = 1 and b ′ = −
1, then A is defined by e f gf fg − f e (26)with e = 1. This is not unitally straight, because 1 , x + yf + zg and( x + yf + zg ) = x + z + 2 x ( yf + zg )are linearly dependent for any x, y, z ∈ K .The case b = − b ′ = 1 gives the algebra isomorphic to the previousalgebra defined by (26) via the transformation matrix − − .9iii) The case f = 1 , g = 0 is symmetric to (ii), and yields no new algebra.(iv) Suppose that f = g = 1. Because A is not unitally straight, 1 , h = x + yf + zg and h = x + y + z + 2 xyf + 2 xzg + yz ( f g + gf )= x + y + z + yz ( a + a ′ ) + y (2 x + z ( b + b ′ )) f + z (2 x + y ( c + c ′ )) g are linearly dependent for any x, y, z ∈ R . Hence, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x x + y + z + yz ( a + a ′ )0 y y (2 x + z ( b + b ′ ))0 z z (2 x + y ( c + c ′ )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = yz ( y ( c + c ′ ) − z ( b + b ′ )) = 0for any x, y, z ∈ R . It follows that b + b ′ = c + c ′ = 0 . (27)We have 4 solutions in a, b, c, a ′ , b ′ , c ′ satisfying (23), (24) and (27):( a, b, c, a ′ , b ′ , c ′ ) = ( − , , , − , − , − , (1 , − , , , , − , (1 , , − , , − , , ( − , − , − , − , , . Now, if ( a, b, c, a ′ , b ′ , c ′ ) = ( − , , , − , − , − f g = − f + g and gf = − − f − g, then let f ′ = f + g and g ′ = g . If ( a, b, c, a ′ , b ′ , c ′ ) = (1 , − , , , , − f ′ = f − g and g ′ = − g . If ( a, b, c, a ′ , b ′ , c ′ ) = (1 , , − , , − , f ′ = f − g and g ′ = g . If ( a, b, c, a ′ , b ′ , c ′ ) = ( − , − , − , − , , f ′ = f + g and g ′ = − g . In any case we have f ′ = 0 , g ′ = 1 , f ′ g ′ = f ′ and g ′ f ′ = − f ′ . Therefore, replacing e, f and g by 1 , f ′ and g ′ , respectively, A is defined by (26).The algebras defined by (25) and (26) are not isomorphic, because in thealgebra defined by (25), 1 and − x satisfying x = 1, butin the algebra defined by (26), g = 1.In summary, over C we have exactly 5 non-isomorphic unital algebras U , U , U , U and U defined by (25), (26), (17), (19) and (20), respectively. Over R we have exactly 6 non-isomorphic unital algebras, the algebras defined by thesame tables as above and another algebra U − defined by (18). In this section we classify curled algebras over K of dimension 3.Let C , C , C , C and C be the algebras on a base { e, f, g } defined by , e − e , e f g , e f g and e f g , (28)10espectively. It is easy to check that these are actually curled (associative)algebras.We claim that these algebras are not isomorphic to each other. For analgebra A over K , define the square and the left (and right) annihilator of A by A = { P ni =1 x i y j | x i , y j ∈ A, n > } , and la( A ) = { x ∈ A | xA = 0 } , ra( A ) = { x ∈ A | Ax = 0 } , respectively. Obviously these are subalgebras of A . Let α ( A ) = dim K A , β ( A ) = dim K la( A ) , γ ( A ) = dim K ra( A ) . Clearly, if A and A ′ are isomorphic, then α ( A ) = α ( A ′ ), β ( A ) = β ( A ′ ) and γ ( A ) = γ ( A ′ ). By an easy calculation, we see α ( C ) = 0 , α ( C ) = 1 and α ( C ) = α ( C ) = α ( C ) = 3 ,β ( C ) = 1 , β ( C ) = 2 and β ( C ) = 0 , and γ ( C ) = 1 , γ ( C ) = 0 and γ ( C ) = 2 . Thus, our algebras can not be isomorphic to each other, although in this casewe need not the information about the values of γ .Next, we shall prove that any curled algebra is isomorphic to one of thealgebras listed in (28). We divide cases depending on whether { e, f, ef } or { e, f, f e } forms a base of A or not.(a) Suppose that there are elements e and f of A such that { e, f, ef } islinearly independent. Because A is curled, e = ke and f = ℓf (29)for some k, ℓ ∈ K . If k = 0 (resp. ℓ = 0), by replacing e (resp. f ) by e/k (resp. f /ℓ ), we may suppose that k and ℓ are equal to 0 or 1. Set g = ef, (30)then { e, f, g } is a linear base of A . Because A is curled( e + f ) = ke + ℓf + g + f e = m ( e + f )for some m ∈ K . Hence, f e = ( m − k ) e + ( m − ℓ ) f − g. (31)Moreover, we have eg = eef = kef = kg, (32) gf = ef f = ℓef = ℓg, (33)11herefore, ge = ef e = e (( m − k ) e + ( m − ℓ ) f − g ) = k ( m − k ) e + ( m − k − ℓ ) g (34) f g = f ef = (( m − k ) e + ( m − ℓ ) f − g ) f = ℓ ( m − ℓ ) f + ( m − k − ℓ ) g (35)and g = gef = ( k ( m − k ) e + ( m − k − ℓ ) g ) f = (( k + ℓ )( m − k ) + ℓ ) g. (36)Because e + g and( e + g ) = k ( m − k + 1) e + (( k + ℓ )( m − k ) + m − ℓ − ℓ ) g are linearly dependent, we have k ( m − k + 1) = ( k + ℓ )( m − k ) + m − ℓ − ℓ, that is, ( ℓ + 1)( m − k − ℓ ) = 0 . Because ℓ = −
1, we have m = k + ℓ . Thus, (31), (34), (35) and (36) become f e = ℓe + kf − g, ge = kℓe, f g = kℓf, and g = kℓg. (37)(a1) If k = ℓ = 0, then by (29), (30), (32), (33) and (37) we have ef = g, f e = − g and e = f = eg = gf = ge = f g = g = 0 . Let e ′ = − g and g ′ = e . Then we have e ′ = e ′ f = e ′ g ′ = f e ′ = g ′ e = 0, f g ′ = e ′ and g ′ f = − e ′ . Hence, replacing e by e ′ and g by g ′ , A is defined bythe second table in (28), and is isomorphic to C .(a2) If k = 0 and ℓ = 1, then we have f = f, ef = gf = g, f e = e − g and e = eg = ge = f g = g = 0 . Let e ′ = e − g , then we have e ′ = ( e − g ) = e − eg − ge + g = 0 , e ′ f = ef − gf = 0 , e ′ g = eg − g = 0and f e ′ = f e − f g = e − g = e ′ , ge ′ = ge − g = 0 . Hence, replacing e by e ′ , A is defined by the third table in (28), and is isomorphicto C .(a3) If k = 1 and ℓ = 0, then we have e = e, ef = eg = g, f e = f − g and f = gf = ge = f g = g = 0 . Let e ′ = g, f ′ = e, g ′ = f − g . Then, we have e ′ = g = 0 , e ′ f ′ = ge = 0 , e ′ g ′ = gf − g = 0 , f ′ e ′ = eg = g = e ′ , ′ = e = e = f ′ , f ′ g ′ = ef − eg = 0 , g ′ e ′ = f g − g = 0 , g ′ f ′ = f e − ge = f − g = g ′ and g ′ = f − f g − gf + g = 0 . Thus, A is isomorphic to C again.(a4) Finally, if k = ℓ = 1, then we have e = ge = e, f = f g = f, ef = eg = gf = g = g and f e = e + f − g. Let e ′ = e − g and g ′ = g − f . Then, we have e ′ = e ′ f = e ′ g ′ = f g ′ = g ′ e ′ = g ′ = 0 , f e ′ = e ′ and g ′ f = g ′ . Therefore, A is isomorphic to C in this case too.(b) Suppose that { e, f, f e } is linearly independent for some e, f ∈ A . As weget (34) above, we can get ef = k ′ e + ℓ ′ f − f e with k ′ , ℓ ′ ∈ K . Thus, { e, f, ef } is also linearly independent. Therefore, this case is included in case (a).(c) Suppose that for any linearly independent elements e and f of A , both { e, f, ef } and { e, f, f e } are linearly dependent. Let { e, f, g } be a linear base of A , and let B be the subalgebra of A generated by { e, f } . Because ef and f e liein the space spanned by { e, f } , B is spanned by { e, f } and is two-dimensional.As discussed in Section 3, there are exactly three non-isomorphic curledalgebras of dimension 2; they are defined by (cid:18) (cid:19) , (cid:18) e f (cid:19) and (cid:18) e f (cid:19) . (38)Hence, we may assume that the subalgebra B is defined by one of the tables in(38). Because A is curled, g = kg (39)for some k ∈ K . Here we may suppose that k is equal to 0 or 1.(c1) Suppose that B is defined by the first table in (38) on the base { e, f } .Since { e, g, eg } is linearly dependent, eg = ae + bg for some a, b ∈ K . Because e = 0, we have 0 = e g = e ( ae + bg ) = b ( ae + bg ) = abe + b g. Hence, b = 0 and so eg = ae . By this last equality and (39) we have kae = keg = eg = aeg = a e. Hence, a = 0 or a = k . Similarly, we see that ge = a ′ e, f g = bf, gf = b ′ f, and a ′ , b and b ′ are equal to 0 or k . Thus, if k = 0, then a = a ′ = b = b ′ = 0, andhence, eg = ge = f g = gf = g = 0. Hence, A is isomorphic to the zero algebra C .Next, suppose that k = 1. Then, g = g and a, a ′ , b, b ′ ∈ { , } . Because A is curled, we have ℓ ( e + g ) = ( e + g ) = ( a + a ′ ) e + g ℓ ∈ K . Hence, a + a ′ = ℓ = 1. Similarly, we have b + b ′ = 1. So we wehave four possibilities( a, a ′ , b, b ′ ) = (0 , , , , (0 , , , , (1 , , ,
1) and (1 , , , . In correspondence to them we have the algebras defined by e f g , fe g , e f g and e f g . (40)The algebra defined by the first table in (40) is nothing but C and thealgebra defined by the last table is C . The algebra defined by the second tableis isomorphic to the algebra C via the transformation matrix . Thealgebra defined by the third table is also isomorphic to C via .(c2) Suppose that B is defined by the second table in (38). As above wehave eg = ae, ge = a ′ e, a + a ′ = k with a, a ′ ∈ { , k } . (41)Because { f, g, f g } is linearly dependent, f g = bf + cg for some b, c ∈ K . Wehave bf + cg = f g = f g = f ( bf + cg ) = bf + c ( bf + cg ) = b ( c + 1) f + c g and kbf + kcg = kf g = f g = ( bf + cg ) g = b ( bf + cg ) + kcg = b f + ( b + k ) cg. Hence, bc = c ( c −
1) = b ( b − k ) = 0 . (42)Since 0 = ef g = e ( bf + cg ) = ace and a ′ e = a ′ f e = f ge = ( bf + cg ) e = be + a ′ ce by (41), we have ac = 0 and b = a ′ (1 − c ) . (43)Let gf = b ′ f + c ′ g with b ′ , c ′ ∈ K . As above we have b ′ c ′ = c ′ ( c ′ −
1) = b ′ ( b ′ − k ) = 0 . (44)and ac ′ = 0 and b ′ = a ′ (1 − c ′ ) . (45)14ecause A is curled, for any y, z ∈ K , yf + zg and( yf + zg ) = y f + kz g + yz ( bf + cg ) + yz ( b ′ f + c ′ g )= ( y + ( b + b ′ ) z ) yf + ( kz + ( c + c ′ ) y ) zg are linearly dependent. Hence( y + ( b + b ′ ) z ) yz = ( kz + ( c + c ′ ) y ) yz holds. Because y and z are arbitrary, it follows that b + b ′ = k and c + c ′ = 1 . (46)Now, if k = 0, then by (41) – (46), we see that a = a ′ = b = b ′ = 0 and either c = 0 , c ′ = 1 or c = 1 , c ′ = 0 . In the first case, eg = ge = f g = gg = 0 and gf = g , In the second case, eg = ge = gf = gg = 0 and f g = g . Hence, A is defined by e f g or e f g . (47)In the first case A is nothing but C . In the second case, A is isomorphic to C via .Next, if k = 1, then again by (41) – (46), we see that a = 0 , a ′ = 1 and either ( b = c ′ = 0 , b ′ = c = 1) or ( b = c ′ = 1 , b ′ = c = 0) . Hence, A is defined by e f ge f g or e f fe g g . (48)The algebra defined by the first table in (48) is isomorphic to C via − .The algebra defined by the second table in (48) is isomorphic to C via − .(c3) Suppose that B is defined by the last table. This case is symmetricwith (c2), and from (47) and (48) we have four curled algebras defined by e f g , e f g , e e f f g g and e e f g f g . (49)15he algebra defined by the first table in (49) is isomorphic to C via .The algebra defined by the last table is also isomorphic to C via the isomor-phism via − . The algebras defined by the second and third tablesare isomorphic to C in the same way as we see above in (c2).(d) By the above case classification, we have proved that there are exactlyfive non-isomorphic curled algebras C , C , C , C and C defined by the tablesin (28). Let A be a straight algebra over K of dimension 3, and let h be an element of A such that { h, h , h } forms a linear base of A . Then, A is commutative, andwe can write h = ah + bh + ch (50)with a, b, c ∈ K .If c = 0, let e = 1 c ( h − ah − bh ) , then he = eh = h . Hence, e is the identity element and A is a unital algebra.Because we already have classified unital algebras in Section 4, we suppose that c = 0. Then (50) becomes h = ah + bh . (51)Here, if a = b = 0, then h = 0. Let e = h, f = h and g = h , then on thebase { e, f, g } , A is defined by f g g . (52)Next, suppose that a = 0 and b = 0. Then h = ah . Let e = h a , then wesee that h n a n = e for all n ≥
3. Let f ′ = ha and g ′ = f ′ = h a . Then, we have e = ef ′ = f ′ e = g ′ = eg ′ = g ′ e = f ′ g ′ = g ′ f ′ = e. Set f = f ′ − e and g = g ′ − e . We thus have ef = ef ′ − e = 0 , eg = eg ′ − e = 0 , f e = f ′ e − e = 0 ,f = f ′ − f ′ e − ef ′ + e = g ′ − e = g, f g = f ′ g ′ − f ′ e − eg ′ + e = 016nd ge = g ′ e − e = 0 , gf = g ′ f ′ − g ′ e − ef ′ + e = 0 , g = g ′ − g ′ e − eg ′ + e = 0 . Hence, A is defined by e g
00 0 0 . (53)Finally, suppose that b = 0. Let g = h − ah − bh, then hg = gh = 0 by (51), and so Ag = gA = 0.Let e = ( a + b ) h − ah b , (54)then, by (51) and (54) we have eh = h e = h , e = e and( h ) = ( a + 2 b ) h − b e. (55)Hence, the subspace B = K { h , h } spanned by h and h is a unital straightalgebra of dimension 2 over K generated by the element h satisfying (55). Let D = a + 4 b .If K = C and D = 0 or K = R and D >
0, then by the discussion inSection 3, with a suitable element f of B , B is defined by (12) on the base { e, f } . Therefore, A is defined by e f f e
00 0 0 . (56)If K = R and D <
0, then B is defined by (13), and so A is defined by e f f − e
00 0 0 . (57)If D = 0, B is defined by (14) and A is defined by e f f . (58)We denote the algebras defined by (52), (53), (56), (57) and (58) by S , S , S , S − and S , respectively. The algebra S has no nonzero idempotent, butthe other algebras have one. Their left annihilators are the same I = Kg which17s a two-sided ideal. The residue algebras of S , S , S and S − by I is thetwo-dimensional algebras defined by (cid:18) e
00 0 (cid:19) , (cid:18) e ff e (cid:19) , (cid:18) e ff (cid:19) and (cid:18) e ff − e (cid:19) , (59)respectively. The algebras defined by the first three tables in (59) are notisomorphic over C to each other. Over R , the algebra defined by the last tableis not isomorphic to any other algebras. Therefore, S , S , S and S arenon-isomorphic over C , and S − is another non-isomorphic algebra over R . In this section and the following two sections we study waved algebras. Let A be a waved algebra over K of dimension 3 which is not unital. Then, there isa non-curled element f ∈ A , such that { f, f , f } is linearly dependent, that is, { f, f } is a linear base of the subalgebra A ′ of A generated by f .Take g ∈ A so that { f, f , g } forms a linear base of A . Here, if g is curled,that is, g = kg for some k ∈ K , then let g ′ = ℓf + g with ℓ ∈ K . Write f g + gf = af + bf + cg with a, b, c ∈ K . Choose ℓ so that ℓ = 0 and ℓ = − b ,then g ′ and g ′ = ℓ f + ℓ ( f g + gf ) + g = ℓ ( ℓ + b ) f + ℓaf + ( ℓc + k ) g are linearly independent, that is, g ′ is not curled. Hence, from the beginning wemay assume that g is not curled and { g, g } is a linear base of the subalgebra A ′′ of A generated by g . The subalgebras A ′ and A ′′ are straight algebras ofdimension 2, and the intersection A ′ ∩ A ′′ is a subalgebra of A of dimension 1.From the results in Section 3, we have exactly 4 straight non-isomorphicalgebras A , A , A and A of dimension 2 over C defined by(a) (cid:18) e (cid:19) , (b) (cid:18) e
00 0 (cid:19) , (c) (cid:18) e ff e (cid:19) , and (d) (cid:18) e ff (cid:19) , respectively. Over R , in addition to the above algebras, we have the algebra A − defined by ( r ) (cid:18) e ff − e (cid:19) . Note that these algebras are all commutative. As mentioned in Section 3, A is also defined by (16).The algebra A has the unique subalgebra of dimension 1, which is thesubalgebra Ke generated by e and is isomorphic to A . A has the uniquesubalgebra Kf of dimension 1 isomorphic to A and the unique subalgebra Ke of dimension 1 isomorphic to A . From easy calculations A has the exactly 3subalgebras Ke , K ( e + f ) and K ( e − f ) of dimension 1, which are all isomorphicto A . If we choose the table in (16) for A , K ( e + f ), Ke and K f are the three18ubalgebras isomorphic to A . A has the exactly 2 subalgebras Ke ∼ = A and Kf ∼ = A of dimension 1. A − has the unique subalgebra Ke ∼ = A of dimension1. Our strategy for classifying waved algebras is that we check through allpossible combinations of the subalgebras A ′ and A ′′ . Lemma 7.1.
Let A ′ and A ′′ be straight subalgebras of dimension 2 of an algebra A of dimension 3. Let { e, f } and { e ′ , g } be linear bases of A ′ and A ′′ , respec-tively. Suppose that { e, f, g } forms a linear base of A and A ′ ∩ A ′′ = Ke = Ke ′ holds, that is, e = ke ′ for some k ∈ K \ { } . Then we have(1) eg = ge = 0 if e ′ g = 0 .(2) eg = ge = e if e ′ g = e ′ .(3) eg = ge = kg if e ′ g = g .Proof. (1) We have eg = ke ′ g = 0 if e ′ g = 0. Moreover, ge = kge ′ = ke ′ g = 0because A ′′ is commutative from the tables (a), (b), (c), (d) and (r).(2) We have eg = ge = ke ′ g = ke ′ = e if e ′ g = e ′ .(3) We have eg = ge = ke ′ g = kg if e ′ g = g . Lemma 7.2.
Suppose the same situation as in Lemma 7.1.(1) If e = 0 , then e ′ = 0 , and ef, f e, eg, ge ∈ Ke .(2) If e = e and e ′ = e ′ , then k = 1 and e = e ′ .Proof. (1) Because e = ke ′ with k = 0, we see that e = 0 if and only if e ′ = 0.Because ef ∈ A ′ we can write ef = ae + bf with a, b ∈ K . If e = 0, we have0 = e f = e ( ae + bf ) = b ( ae + bf ) = abe + b f. It follows that b = 0 and we see ef ∈ Ke . Similarly, f e ∈ Ke and eg, ge ∈ Ke ′ = Ke .(2) If e = e and e ′ = e ′ , we have ke ′ = e = e = kee ′ = k e ′ = k e ′ . Because k = 0, we get k = 1 and e = e ′ . Corollary 7.3. If e = e , e ′ = e ′ , ef = f and e ′ g = g , then A is unital.Proof. Since A ′ is commutative, f e = f . By Lemma 7.2, (2) and Lemma 7.1,(3) we have eg = ge = g . These imply that e is the identity element of A . Lemma 7.4.
In the same situation as in Lemma 7.1, let f g = ae + bf + cg and gf = a ′ e + b ′ f + c ′ g with a, b, c, a ′ , b ′ , c ′ ∈ K .(1) If f = 0 , then c = c ′ = 0 , and moreover, if ef = 0 or e = 0 , then a = a ′ = 0 . If g = 0 , then b = b ′ = 0 , and moreover, if eg = 0 or e ′ = 0 , then a = a ′ = 0 .(2) If f = e and eg ∈ Ke (in particular e = 0 ), then c = c ′ = 0 and eg = aef + be = a ′ ef + b ′ e . If g = e ′ and ef ∈ Ke (in particular e = 0 ), then b = b ′ = 0 and ef = kaeg + ce = ka ′ ef + c ′ e .(3) If f = f , then c = 0 or c = 1 , and c ′ = 0 or c ′ = 1 . If g = g , then b = 0 or b = 1 , and b ′ = 0 or b ′ = 1 . roof. We have f g = f ( ae + bf + cg ) = af e + bf + cae + cbf + c g. (60)(1) If f = 0, then (60) becomes0 = af e + cae + cbf + c g. Since af e + cae + cbf ∈ A ′ and { e, f, g } is linearly independent, we have c = 0and af e = 0. Hence, a = 0 if ef (= f e ) = 0. If ef = 0 and e = 0, then0 = ef g = e ( ae + bf ) = ae and we see a = 0. Similarly, we have c ′ = 0, and a ′ = 0 if ef = 0 or e = 0.Similarly, if g = 0, then b = b ′ = 0, and moreover, if eg = 0 or e ′ = 0, then a = a ′ = 0.(2) If f = e , then (60) becomes eg = af e + ( b + ca ) e + cbf + c g. Here, if eg ∈ Ke (this holds if e = 0 by Lemma 7.2, (1)), we see c g ∈ A ′ .Hence, c = 0 and eg = aef + be . Similarly, we have c ′ = 0 and eg = ge = a ′ ef + b ′ e . The case g = e ′ is similar.(3) If f = f , then by (60) we have ae + bf + cg = f g = f g = af e + ace + b ( c + 1) f + c g. (61)Hence, c ( c − g ∈ A ′ and we have c = 0 or c = 1. Similarly, we have c ′ = 0 or c ′ = 1. If g = g , then we have ae + bf + cg = f g = aeg + abe + c ( b + 1) g + b f, (62)and hence b = 0 or b = 1. Lemma 7.5.
In the same situation as in Lemma 7.1, if ef = 0 and e ′ g = g orif e ′ g = 0 and ef = f , then f g = gf = 0 .Proof. If ef = 0, and e ′ g = g , then by Lemma 7.1, (3) we have 0 = ef g = f eg = kf g and 0 = gef = kgf. Because k = 0, we see f g = gf = 0. The casewhere e ′ g = 0 and ef = f is similar. Lemma 7.6.
Let B be a straight subalgebra of A of dimension 2 and let g bean element of A \ B such that Bg = gB = { } . Then, A is a straight algebra, if(1) g = g , or(2) g = 0 and B is isomorphic to A , A or A − .Proof. Because B is straight, it is defined by (a), (b), (c), (d) or (r) above.(1) Suppose that g = g . Note that there is an element x in B such that x − x and x − x are linearly independent. In fact, f , 2 e + f , 2 f , 2 e + f and f aresuch elements in the algebras defined by (a), (b), (c), (d) and (r), respectively.For such an element x of B , we claim that x + g is not waved. In fact, if a ( x + g ) + b ( x + g ) + c ( x + g ) = ax + bx + cx + ( a + b + c ) g = 020or a, b, c ∈ K , then, a + b + c = 0 because g / ∈ B . Hence, b ( x − x )+ c ( x − x ) = 0.Consequently, a = b = c = 0. Hence, A is straight.(2) Suppose that g = 0 and B isomorphic to A , A or A − . Note that B has an element x such that x and x are linearly independent. In fact, f , e + f and f are such elements in the algebras defined by (c), (d) and (r), respectively.For such an element x of B , x + g , ( x + g ) = x and ( x + g ) = x are linearlyindependent because g / ∈ B . Hence, A is straight. Let A ′ and A ′′ be straight subalgebras of dimension 2 of a waved algebra A suchthat A ′ ∩ A ′′ is one-dimensional, which is isomorphic to either A or A . Weshall advance the discussion by case division as follows: (a) A ′ ∼ = A and itssubdivisions (aa) A ′′ ∼ = A , (ab) A ′′ ∼ = A , (ac) A ′′ ∼ = A and (ad) A ′′ ∼ = A ;(b) A ′ ∼ = A and its subdivisions (bb) A ′′ ∼ = A , (bc) A ′′ ∼ = A and (bd) A ′′ ∼ = A ; (c) A ′ ∼ = A and its subdivisions (cc) A ′′ ∼ = A and (cd) A ′′ ∼ = A ;(dd) A ′ ∼ = A ′′ ∼ = A ; and finally in the case where K = R (r) A ′ ∼ = A − and itssubdivisions (ra) A ′′ ∼ = A , (rb) A ′′ ∼ = A , (rc) A ′′ ∼ = A , (rd) A ′′ ∼ = A and (rr) A ′′ ∼ = A − .(a) Suppose that A ′ is isomorphic to A . Let { e, f } be a linear base of A ′ such that e = ef = f e = 0 and f = e .(aa) Suppose that A ′′ are also isomorphic to A , and let { e ′ , g } be linearbase of A ′′ such that e ′ = e ′ g = ge ′ = 0 and g = e ′ . Because Ke and Ke ′ are the unique subalgebras of dimension 1 of A ′ and A ′′ , respectively, we see A ′ ∩ A ′′ = Ke = Ke ′ . Hence, e ′ = ℓe for some ℓ ∈ K \ { } .By Lemma 7.1, (1) we have eg = ge = 0 , and by Lemma 7.4, (2) we see f g = ae and gf = a ′ e for some a, a ′ ∈ K .Set g ′ = af − g , then we have eg ′ = aef − eg = 0 and g ′ e = af e − ge = 0 , (63) g ′ = a f + g − a ( f g + gf ) = a e + ℓe − a ( a + a ′ ) e = ( ℓ − aa ′ ) e, (64) f g ′ = af − f g = ae − ae = 0 (65)and g ′ f = af − gf = ( a − a ′ ) e. (66)(aa1) If aa ′ = ℓ and a = a ′ , then by (63) – (66) we have eg ′ = g ′ e = g ′ = f g ′ = g ′ f = 0 . Thus, replacing g by g ′ , A is defined by e
00 0 0 . (67)21aa2) If aa ′ = ℓ and a = a ′ , then replacing g by g ′ / ( a − a ′ ), A is defined by e e . (68)Now suppose that aa ′ = ℓ .(aa3) If K = C , or K = R and ℓ − aa ′ >
0, then letting g ′′ = g ′ / √ ℓ − aa ′ ,we have g ′′ = e, eg ′′ = g ′′ e = f g ′′ = 0 and g ′′ f = a − a ′ √ ℓ − aa ′ e. Hence, replacing g by g ′′ and letting k = ( a − a ′ ) / √ ℓ − aa ′ , A is defined by e ke e , (69)where k can be an arbitrary element of K .(aa4) If K = R and ℓ − aa ′ <
0, then letting g ′′ = g ′ / √ aa ′ − ℓ , we have g ′′ = − e, eg ′′ = g ′′ e = f g ′′ = 0 and g ′′ f = a − a ′ √ aa ′ − ℓ e. Hence, replacing g by g ′′ and letting k = ( a − a ′ ) / √ aa ′ − ℓ , A is defined by e ke − e , (70)where k ∈ R .(ab) Suppose that A ′′ is isomorphic to A and let { e ′ , g } a linear base of A ′′ such that e ′ = e ′ g = ge ′ = 0 and g = g . Then, A ′ ∩ A ′′ = Ke = Ke ′ .By Lemma 7.1, (1), eg = ge = 0 , and by Lemma 7.4, (2) we have f g = ae + bf with a, b ∈ K and 0 = eg = aef + be = be. Hence, b = 0. Moreover, we have ae = f g = f g = aeg = 0 . Hence, a = 0 and so f g = 0. Similarly, we have gf = 0.Thus, A ′ g = gA ′ = { } and g = g . Hence, A is not waved by Lemma 7.6,(1), and we exclude this algebra (actually, A is isomorphic to S ).(ac) Suppose that A ′′ is isomorphic to A . However, this case is impossiblebecause A ′ has the unique subalgebra of dimension 1 isomorphic to A but A ′′ has only subalgebras of dimension 1 isomorphic to A .22ad) Suppose that A ′′ is isomorphic to A , and let { e ′ , g } be a linear baseof A ′′ such that e ′ = 0, e ′ g = ge ′ = e ′ and g = g . Because Ke ′ is the uniquesubalgebra of A ′′ of dimension 1 isomorphic to A , we have Ke = Ke ′ .By Lemma 7.1, (2) we have eg = ge = e, and by Lemma 7.4, (2) we have f g = ae + bf with a, b ∈ K and e = eg = aef + be = be, Hence, b = 1, that is, f g = ae + f . We have ae + f = f g = f g = ( ae + f ) g = 2 ae + f. Hence, a = 0 and we see f g = f . Similarly, we have gf = f. These imply that g is the identity element and A is unital, and we canexclude this algebra.(b) Suppose that A ′ is isomorphic to A .(bb) Suppose that A ′′ is also isomorphic to A .(bb1) Suppose that A ′ ∩ A ′′ ∼ = A . Let { e, f } and { e ′ , g } be linear bases of A ′ and A ′′ such that e = e, ef = f e = f = 0 and e ′ = e ′ , e ′ g = ge ′ = g = 0.Then, A ′ ∩ A ′′ = Ke = Ke ′ . By Lemma 7.1, (1) we have eg = ge = 0. Moreover,by Lemma 7.4, (1), we see f g = gf = 0. Therefore, A is defined by e (71). (bb2) Suppose that A ′ ∩ A ′′ ∼ = A . Let { e, f } and { e ′ , g } be linear bases of A ′ and A ′′ such that e = ef = f e = 0 , f = f and e ′ = e ′ g = ge ′ = 0 , g = g .Then, A ′ ∩ A ′′ = Ke = Ke ′ . By Lemma 7.1, (1) we have eg = ge = 0 . Let f g = ae + bf + cg with a, b, c ∈ K , then by Lemma 7.4, (3) we have c = 0 or c = 1, and b = 0 or b = 1. If c = 0, then by (61), we have ae = af e = 0. Hence a = 0. If c = 1, then by (61) we have bf = − af e = 0. Hence b = 0. Because ae = aeg = 0 by (62), again we have a = 0. Therefore, f g = bf + cg . Similarly,we have gf = b ′ f + c ′ g with b ′ , c ′ ∈ K .Thus B = k { f, g } is a subalgebra of A of dimension 2 such that eB = Be = { } and e = 0. Hence, B is curled or isomorphic to A or A , otherwise A is straight by Lemma 7.6, (2). Because f = f and g = g , B cannot beisomorphic to A nor A , and the only possibility is that B ∼ = A or B ∼ = A .Therefore, choosing a suitable base { f, g } of B , A can be defined by f g or f g . (72)(bc) Suppose that A ′′ is isomorphic to A . Let { e, f } be a linear base of A ′ such that e = e and ef = f e = f = 0. Then, A ′ ∩ A ′′ = Ke ∼ = A . Asobserved in the previous section, A ′′ has exactly three subalgebras (1) Ke , (2)23 ( e + f ) and (3) K ( e − f ) isomorphic to A for the table (12). Accordingly wehave the following three possiblities.(bc1) Let { e ′ , g } be a linear base of A ′′ such that e ′ = g = e ′ and e ′ g = ge ′ = g . Suppose that A ′ ∩ A ′′ = Ke = Ke ′ . By Lemma 7.2, (2) and Lemma 7.1,(3), we have e ′ = e and eg = ge = g . By Lemma 7.5 we have f g = f g = 0.Because the subalgebra A ′′ is isomorphic to A , and f A ′′ = A ′′ f = { } and f = 0, A is not waved by Lemma 7.6, (2).(bc2) Replacing e ′ by e + f and g by e − f , let { e ′ , g } be a linear base of A ′′ such that e ′ = e ′ , g = g and e ′ g = ge ′ = 0 as in (16), and suppose that Ke = Ke ′ . By Lemma 7.1, (1) we have eg = ge = 0 , and by Lemma 7.4,(1) we see f g = bf and gf = b ′ f with b, b ′ ∈ K . Therefore, B = K { f, g } isa subalgebra of dimension 2. Because eB = Be = { } and e = e , B mustbe curled by Lemma 7.6, (1), because otherwise A would be straight. Because g = g , B is isomorphic to either A or A . Hence, choosing a suitable base { f, g } of B , A is defined by e f g or e f g (73)(bc3) Similarly to (bc2), let { e ′ , g } be a linear base of A ′′ such that e ′ = e ′ , g = g and e ′ g = ge ′ = 0. Suppose that Ke = Kg . Because the automor-phism of A ′′ interchanging e ′ and g maps the subalgebra Kg onto the subalgebra Ke ′ , the situation is essentially the same as (bc2) and we can omit this case.(bd) Suppose that A ′′ is isomorphic to A .(bd1) Suppose that A ′ ∩ A ′′ ∼ = A . Let { e, f } be a linear base of A ′ suchthat e = ef = f e = 0 and f = f , and { e ′ , g } be a linear base of A ′′ such that e ′ = 0 , e ′ g = ge ′ = e ′ and g = g . Then, A ′ ∩ A ′′ = Ke = Ke ′ . By Lemma 7.1,(2) we have eg = ge = e. Let f g = ae + bf + cg with a, b, c ∈ K . By Lemma 7.4,(3) we see b = 0 or b = 1. We have0 = ef g = e ( ae + bf + cg ) = ce. Hence, c = 0. We have ae + bf = f g = f g = f ( ae + bf ) = bf. Hence, a = 0, and we see f g = bf , where b = 0 or b = 1. Similarly, gf = b ′ f and b ′ = 0 or b ′ = 1. Because bf = bf = f gf = b ′ f = b ′ f, we see b = b ′ . Therefore, we see f g = gf = 0 or f g = gf = f . Consequently, A is defined by e f e g or e f fe f g . (74)24bd2) Suppose that A ′ ∩ A ′′ ∼ = A . Let { e, f } be a linear base of A ′ suchthat e = e and ef = f e = f = 0. Let { e ′ , g } be a linear base of A ′′ such that e ′ = e ′ , e ′ g = ge ′ = g and g = 0. Then, Ke = Ke ′ . Then by Lemma 7.1,(3) and Lemma 7.2, (2) we have eg = ge = g. Moreover, by Lemma 7.5, wesee f g = gf = 0. Because the subalgebra A ′′ is isomorphic to A , and f A ′′ = A ′′ f = { } and f = 0, A is not waved by Lemma 7.6, (2).(c) Suppose that A ′ is isomorphic to A .(cc) Suppose that A ′′ is also isomorphic to A . As mentioned in (bc), A ′′ hasexactly three subalgebras (1) Ke , (2) K ( e + f ) and (3) K ( e − f ) isomorphic to A for (12). Therefore, A ′ ∩ A ′′ ∼ = A and we have the three possible combinations:(1) and (1), (1) and (2), and (2) and (2), where (3) is similar to (2) and so canbe neglected. Accordingly, we have the following three cases.(cc1) Let { e, f } and { e ′ , g } be linear bases of A ′ and A ′′ , respectively suchthat e = f = e , ef = f e = f , e ′ = g = e ′ and e ′ g = ge ′ = g . If A ′ ∩ A ′′ = Ke = Ke ′ , then A is unital by Corollary 7.3, and we can exclude thiscase.(cc2) Let { e, f } and { e ′ , g } be linear bases of A ′ and A ′′ , respectively suchthat e = f = e , ef = f e = f , e ′ = e ′ , g = g and e ′ g = ge ′ = 0. Supposethat Ke = Ke ′ . By Lemma 7.1, (1) we have eg = ge = 0 , and by Lemma 7.5we see f g = gf = 0. Thus, A is not waved by Lemma 7.6, (1), because thesubalgebra A ′ is straight, and gA ′ = A ′ g = { } and g = g .(cc3) Let { e, f } a linear base of A ′ with e = e, f = f and ef = f e = 0,and let { e ′ , g } be a linear base of A ′′ with e ′ = e ′ , g = g and e ′ g = ge ′ = 0.Suppose that A ′ ∩ A ′′ = Ke = Ke ′ , By Lemma 7.1, (1) we see eg = ge = 0 . Let f g = ae + bf + cg with a, b, c ∈ K . We have0 = ef g = e ( ae + bf + cg ) = ae, and hence a = 0, that is, f g = bf + cg . Similarly, gf = b ′ f + c ′ g with b ′ , c ′ ∈ K .Hence, B = { f, g } is a subalgebra of A of dimension 2 such that eB = Be = { } .Because e = e , B must be curled by Lemma 7.6, (1). Because f = f , B isisomorphic to A or A . Thus, choosing a suitable base { f, g } of B , A is definedby one of the tables in (72), and we get no new algebra here.(cd) Suppose that A ′′ is isomorphic to A . Then, A ′ ∩ A ′′ ∼ = A . Let { e ′ , g } be a linear base of A ′′ such that e ′ = e ′ , e ′ g = ge ′ = g and g = 0. Because Ke ′ is the unique subalgebra of dimension 1 isomorphic to A , we see A ′ ∩ A ′′ = Ke ′ .(cd1) Let { e, f } be a linear base of A ′ such that e = f = e and ef = f e = f ,and suppose that Ke = Ke ′ . Then by Corollary 7.3, A is unital.(cd2) Let { e, f } be a linear base of A ′ such that e = e , f = f and ef = f e = 0. Suppose that A ′ ∩ A ′′ = Ke = Ke ′ . By Lemma 7.1, (3) and Lemma 7.2,(2) we have eg = eg = g, and by Lemma 7.5 we see f g = gf = 0. But A isnot waved by Lemma 7.6, (1), because the subalgebra B = K { e, g } is straight, f B = Bf = { } and f = f .The case A ′ ∩ A ′′ = Kf is symmetric, which we can omit.(dd) Suppose that both A ′ and A ′′ are isomorphic to A . As observed inthe previous section, A has the exactly 2 subalgebras Ke ∼ = A and Kf ∼ = A .Therefore we have the following two cases.25dd1) Suppose that A ′ ∩ A ′′ ∼ = A . Let { e, f } and { e ′ , g } be linear basesof A ′ and A ′′ , respectively such that e = e, ef = f e = f and f = 0, and e ′ = e ′ , e ′ g = ge ′ = g and g = 0. Then, A ′ ∩ A ′′ = Ke = Ke ′ . By Corollary 7.3, A , is unital and we exclude this case.(dd2) Suppose that A ′ ∩ A ′′ ∼ = A . Interchanging e and f , and e ′ and g , let { e, f } and { e ′ , g } be linear bases of A ′ and A ′′ such that e = e ′ = 0 , f = f, ef = f e = e , g = g and e ′ g = ge ′ = e ′ . Then, A ′ ∩ A ′′ = Ke = Ke ′ . Then,By Lemma 7.1, (2) we have eg = ge = e. Let f g = ae + bf + cg with a, b, c ∈ K .We have e = eg = ef g = e ( ae + bf + cg ) = ( b + c ) e. It follows that b + c = 1 . (75)We have ae + bf + cg = f g = f f g = f ( ae + bf + cg ) = ae + bf + c ( ae + bf + cg ) . Hence, a = a ( c + 1) , b = b ( c + 1) and c = c. (76)Similarly, we have a = a ( b + 1) , c = c ( b + 1) and b = b, (77)By (75) – (77) we see that a = 0, and b = 1 , c = 0 or b = 0 , c = 1, that is, f g = f or f g = g . Similarly, we see gf = f or gf = g . Thus, A is defined by e ee f fe f g , e ee f fe g g , e ee f ge f g or e ee f ge g g . (78)(r) Finally, suppose that K = R and A ′ is isomorphic to A − . Let { e, f } bea linear base of A ′ such that e = − f = e and ef = f e = f . Then Ke is theunique subalgebra of A ′ of dimension 1, which is isomorphic to A .(ra) Suppose that A ′′ ∼ = A . However, this case is impossible because A hasthe unique subalgebra of dimension 1 isomorphic to A .(rb) Suppose that A ′′ ∼ = A . Let { e ′ , g } be a linear base of A ′′ such that e ′ = e ′ and e ′ g = ge ′ = g = 0. Then, A ′ ∩ A ′′ = Ke = Ke ′ . We have eg = ge = 0 , by Lemma 7.1, (1) and f g = gf = 0 by Lemma 7.5. Because gA ′ = A ′ g = { } and g = 0, A is not waved by Lemma 7.6, (2).(rc) Suppose that A ′′ ∼ = A .(rc1) Let { e ′ , g } be a linear base of A ′′ such that e ′ = g = e ′ and e ′ g = ge ′ = g . Suppose that A ′ ∩ A ′′ = Ke ′ . By Corollary 7.3 A is unital, and weexclude this case.(rc2) Let { e ′ , g } be a linear base of A ′′ such that e ′ = e ′ , g = g and e ′ g = ge ′ = 0. Suppose that A ′ ∩ A ′′ = Ke = Ke ′ . We have eg = eg = 0 byLemma 7.1, (1) and f g = gf = 0 by Lemma 7.5. Since gA ′ = A ′ g ′ = { } and g = g , A is not waved by Lemma 7.6, (1)26rd) Suppose that A ′′ = A . Let { e ′ , g } be a linear base of A ′′ such that e ′ = e ′ , e ′ g = ge ′ = g and g = 0. Then A ′ ∩ A ′′ = Ke = Ke ′ . By Corollary 7.3 A is unital. We exclude this case.(rr) Suppose that A ′′ is also isomorphic to A − . Let { e ′ , g } be a linear baseof A ′′ such that e ′ = − g = e ′ and e ′ g = ge ′ = g . Then, Ke = Ke ′ . ByLemmas 7.1 and 7.2 we have eg = ge = g. Let f g = ae + bf + cg . We have − g = − eg = f g = f ( ae + bf + cg ) = af − be + c ( ae + bf + cg )= ( ac − b ) e + ( bc + a ) f + c g. Hence, c = −
1, but there is no such c in R . In the previous section, we have enumerated all possible waved algebras ofdimension 3. In this section, we shall investigate whether each two of themare isomorphic or not.(aa1) Let W denote the algebra defined by (67). Interchanging f and g inthe base { e, f, g } , W is defined by e . (aa2) Next, let W be the algebra defined by (68). Let f ′ = f − g , then wehave ef ′ = ef − eg = 0 , f ′ e = f e − ge = 0 , f ′ = f − f g − gf + g = 0 , and f ′ g = f g − g = 0 , gf ′ = gf − g = e. Hence, replacing f by f ′ , W is defined by e . (aa3) We denote the algebra defined by (69) by W ( k ) for k ∈ K . First, wesee that W ( − k ) is isomorphic to W ( k ) via the isomorphism sending e to e , f to f and g to − g .To prove the converse we may assume that K = C , because if W ( k ) and W ( ℓ ) are not isomorphic over C , they are not isomorphic over R . Assume that φ : W ( k ) → W ( ℓ ) is an isomorphism. Let e ′ = φ ( e ) , f ′ = φ ( f ) and g ′ = φ ( g ).Because Ke is the left annihilator of W ( k ) (and of W ( ℓ )), we see Ke = Ke ′ ,that is, e ′ = me m ∈ K \ { } . Write f ′ = ze + xf + yg and g ′ = we + uf + vg with x, y, z, u, v, w ∈ K . Then, we have me = e ′ = φ ( e ) = φ ( f ) = f ′ = ( x + y + kxy ) e,me = e ′ = g ′ = ( u + v + kuv ) e f ′ g ′ = ( xu + yv + kyu ) e and mℓe = ℓe ′ = g ′ f ′ = ( xu + yv + kxv ) e. Thus replacing x, y, u and v by x/ √ m , y/ √ m , u/ √ m and u/ √ m , respectively,we have x + y + kxy = 1 , u + v + kuv = 1 , xu + yv + kyu = 0 (79)and xu + yv + kxv = ℓ. (80)Because we have the identity( y − u ) v = u v ( x + y + kxy −
1) + uy ( x + ky )( u + v + kuv − y − uvx − u y − kuvy )( xu + yv + kyu ) , we find v ( y − u )( y + u ) = 0by (79)If v = 0, then by (79) we have u = 1 , x + ky = 0 , y = 1 and x = k . Hence, by (80) ℓ = xu = ± k. If y = u , then by (79) ( x + v + ku ) u = 0 . Here, if u = 0, then y = 0 and x = v = 1 by (79). Hence, by (80) ℓ = kxv = ± k. If x + v + ku = 0, then ℓ = xu + uv + kxv = − ( v + ku )( u + kv ) + uv = − k ( u + v + kuv ) = − k by (79) and (80). If y = − u , we can show ℓ = ± k in a similar manner.Thus, we have proved that W ( k ) and W ( ℓ ) are isomorphic if and only if ℓ = ± k . 28aa4) We denote the algebra defined by (70) by W − ( k ) for k ∈ R . For any k ∈ R , W − ( k ) and W − ( − k ) are isomorphic via the isomorphism sending e to e , f to f and g to − g .We shall show that W − ( k ) is not isomorphic to W ( k ′ ) for any reals k and k ′ over R . If they were isomorphic, we would have elements f ′ , g ′ ∈ W − ( k ) suchthat f ′ = g ′ = 0 and f ′ g ′ = 0. Let f ′ = ze + xf + yg and g ′ = we + uf + vg with x, y, z, u, v, w ∈ R . Then we have x − y + kxy = u − v + kuv = 0 and xu − yv + kyu = 0 (81)If y = 0, then by (81), x = u − v + kuv = 0 and xu = 0. Hence, u = 0 and x = − v ( = 0) (82)The left-hand side of (82) is positive but the right-hand side is negative, acontradiction. If y = 0, then by (81) v = ( x + ky ) uy and x − y + kxy = u − ( x + ky ) u y + k ( x + ky ) u y = − u y ( x − y + kxy ) = 0 . This is again impossible, and W − ( k ) and W ( k ′ ) can not be isomorphic.On the other hand, W − ( k ) is isomorphic to W ( i k ) over C via the trans-formation matrix − i . Hence, for distinct positive numbers k and k ′ , W − ( k ) and W − ( k ′ ) are not isomorphic over C , because W ( i k ) and W ( i k ′ )are not isomorphic, and hence they are not isomorphic over R , a fortiori.(bb1) We denote the algebra defined by the table in (71) by W .(bb2) The algebras defined by the first table and the second table in (72)are denoted by W and W , respectively.(bc2) We denote the algebras defined by the first and the second tables in(73) by W and W , respectively.(bd1) The algebra defined by the first table in (74) is not waved. In fact, e + f − g , ( e + f − g ) = − e + f + g and ( e + f − g ) = 3 e + f − g are linearlyindependent. The algebra defined by the second table in (74) is defined by thefirst one replacing g by g − f .(dd2) The algebra defined by the first (resp. last) table in (78), g (resp. f )is the identity element. Hence, they are unital. Let W (resp. W ) be thealgebra defined by the second (resp, third) table in (78). In the second table,let g ′ = g − f , the we have eg ′ = g ′ e = f g ′ = g ′ = 0 and g ′ f = g ′ . Thus,replacing g by g ′ , W is defined by e e f g .
29n the third table, let g ′ = g − f , then we have eg ′ = g ′ e = g ′ f = g ′ = 0 and f g ′ = g ′ . Replacing g by g ′ , W is defined by e e f g . In summary, any non-unital waved algebra A of dimension 3 over C isisomorphic to W , W , W , W , W , W , W , W , W or W ( k ) with k ∈ H = { x + y i ∈ C | x > x = 0 , y ≥ } . Over R , A can be isomor-phic to W − ( k ) with k ≥ W ( k ) and W ( k ′ ) arenot isomorphic for distinct k, k ′ ∈ H , W − ( k ) and W − ( k ′ ) are not isomorphicover R for any distinct nonnegative k and k ′ . Moreover, W ( k ) and W − ( k ′ )are not isomorphic over R for any nonnegative k and k ′ .We use the same notations as in Section 5 for an algebra A over K ;la( A ) = { x ∈ A | xA = 0 } , ra( A ) = { x ∈ A | Ax = 0 } , and α ( A ) = dim K A , β ( A ) = dim K la( A ) , γ ( A ) = dim K ra( A ) . We see α ( W ) = α ( W ) = α ( W ( k )) = α ( W − ( k )) = α ( W ) = 1 ,α ( W ) = α ( W ) = 2 and α ( W ) = α ( W ) = α ( W ) = α ( W ) = 3 . Moreover, β ( W ) = β ( W ) = β ( W ) = β ( W ) = 2 ,β ( W ( k )) = β ( W − ( k ) ) = β ( W ) = β ( W ) = β ( W ) = 1 , β ( W ) = β ( W ) = 0 ,γ ( W ) = γ ( W ) = γ ( W ) = γ ( W ) = 2 , and γ ( W ( k )) = γ ( W − ( k )) = γ ( W ) = γ ( W ) = γ ( W ) = 1 , γ ( W ) = γ ( W ) = 0 . These values are summarized in the table shown below. W W W ( k ) W W W W W W W W − ( k ) α β γ W ( k ), W − ( k ), W and W are not isomorphic toany of the others. W , W and W are not isomorphic to any algebra in theother group. W is not isomorphic to W nor to W because W and W arecommutative but W is not. W and W are not isomorphic because W has a30onzero idempotent e ; e = e but W has no nonzero idempotent. Finally, W , W , W and W are not isomorphic to any algebra in the other group. W and W have two zeropotent elements e and g ; e = g = 0 but W and W have no two zeropotent elements that are linearly independent. Hence, thesealgebras are not isomorphic to each other.We have proved that all the algebras are not isomorphic to each other.
10 Summary
By summarizing the results in Sections 4, 5, 6 and 9, we list up all the multipli-cation tables of three-dimensional algebras. It is easy to see that all the tablessatisfy associativity, that is, ( e i e j ) e k = e i ( e j e k ) for all i, j, k ∈ { , , } where e = e , e = f and e = g .Over C , we have, up to isomorphism, exactly 5 unital algebras U , U , U , U , U defined by e f gf g , e f gf fg − f e , e f
00 0 g , e f g g , e f gf g g , respectively, exactly 5 non-unital curled algebras C , C , C , C , C defined by , e − e , e f g , e f g , e f g , respectively, exactly 4 non-unital straight algebras S , S , S , S defined by f g g , e g
00 0 0 , e f f e
00 0 0 , e f f , respectively, and exactly 9 non-unital waved algebras W , W , W , W , W , W , W , W , W defined by e , e , e , f g , f g , e f g , e f g , e e f g , e e f g , respectively and one infinite family (cid:8) W ( k ) (cid:9) k ∈{ x + y i | x> or x =0 ,y ≥ } of non-unital waved algebras defined by e ke e . R , in addition to the above algebras, we have one unital algebra U − ,one non-unital straight algebra S − and one infinite family (cid:8) W − ( k ) (cid:9) k ≥ ofnon-unital waved algebras defined by e f g g − f , e f f − e
00 0 0 , e ke − e , respectively.We remark that zeropotent algebras are only C and C which are alternativematrices.An algebra is indecomposable , if it is not isomorphic to a direct sum ofnontrivial subalgebras. We will not go into the details, but it is possible to showthat the algebras U , U , U , C , C , C , C , S , W , W ( k ), W − ( k ), W , W are indecomposable, and the others are not.Peirce [5] listed five families of ”pure” algebras of dimension 3. They corre-spond to our U , S , W ( k ) , W and C . The list of unilal algebras of dimension3 given by Scheffer [6] and Study [8] is in accordance with our list. References [1] W. R. Hamilton, On quaternions; or a new system of imaginaries in algebra,The London, Edinburgh and Dublin Philosophical Magazine and Journalof Science (1844), 489–495.[2] H. E. Hawkes, On hypercomplex number systems, Trans. Amer. Math.Soc., (1902), 312–330.[3] Y. Kobayashi, K. Shirayanagi, S.-E. Takahasi and M. Tsukada, Classifica-tion of three-dimensional zeropotent algebras over an algebraically closedfield, Comm. Algebra, Vol. 45, Iss. 12 (2017), 5037–5052.[4] A. N. Kolmogorov and A.P. Yushkevich, Mathematics of the 19th cen-tury, Mathematical Logic, Algebra, Number Theory, Probability Theory,Birkh¨auser, 1992.[5] B. Peirce, Linear associative algebra, Amer. J. Math., (1881), 97–229.[6] G. Scheffers, Zur¨uckf¨uhrung complexer zahlensysteme auf typische formen,Math. Ann. (1891), 293–390.[7] K. Shirayanagi, S.-E. Takahasi, M. Tsukada and Y. Kobayashi: Classifi-cation of three-dimensional zeropotent algebras over the real number field ,Comm. Algebra, to appear.[8] E. Study, ¨Uber systeme complexer zahlen und ihre anwendung in der the-orie der transformationsgruppen, Monatsh. Math. u. Phisik (1890), 283–354. 329] H. Taber, On hypercomplex number systems, Trans. Amer. Math. Soc.,5