The equation and homogeneous biadditive forms

The Equation F(x) + M( x)G(l/x) = 0 and Homogeneous Biadditive Forms C. T.zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA Ng* zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA Department of Pure Mathematics University of W aterloo Waterloo, Ontario, zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPO Canada N2L 3Gl Submitted by Robert Guralnick ABSTRACT We determine all additive F, G and multiplicative M satisfying the functional equation F(r) + M(x)G(l/ r) = 0 on a commutative field k of characteristic + 2. Let U, V be k-vector spaces. A functional (form) 7’: U + k is functionally homogeneous if, for some scalar function M: zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONM k + k, T(h) = M(A)T(u) for all X E k, u E U. A simple application of the equation in the title leads to a complete description of all biadditive forms T: V X V+ k which are functionally homogeneous. Some results of Baker, Gleason, and Kurepa concerning the Halperin problem on quadratic forms are generalized and unified. 1. INTRODUCTION Let k be a field. We assume, once and for all, that k is commutative characteristic all x, y E k, and is multiplicative purpose with f 2. A map zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIH f:k + k is additive if f(x + y) = f(x)+ f(y) for if f(xy) = f(x)f(y) for all x, y E k. The of this paper is to determine all additive F, G and multiplicative on the field k satisfying the functional F(r)+ for all x E k * = k \ (0). M equation M (x)G(l/x) =0 (FE) Since the value of M at x = 0 plays no role in (FE), *Supported by NSERC of Canada Grant A 8212. LINEAR ALGEBRA AND ITS APPLICATIONS 93:2.X%279 (1987) 255 0 Elsevier Science PublishingCo., Inc., 1987 52 V anderbilt Ave., New York, NY 10017 00243795/87/$3.50 256 zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA C. zyxwvutsrqponm T. zyxwvutsrqpon NG and since every map M: zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLK k * + k with the multiplicativity M( xy) = M(x)M(y) for all X, y E k * can be extended to a multiplicative map on k by defining M(0) = 0, there is no loss of generality in assuming M(0) = 0. (1.0.1) In his work on quadratic functionals, Kurepa [14] came across the functional equation (FE) with M(x) = x2 on the reals [w . Through this equation he obtained the general form of functionals Q on R-vector spaces satisfying the parallelogram law and the homogeneity Q(Ax) = X20(x), thus answering a question raised by Israel Halperin in 1963 in Paris. With this result over R he further solved [15] the problem on vector spaces over the complex C or the field of quatemions under the homogeneity Q(Ax) = 1X]2Q(~). VrbovH [ 171 managed to solve (FE) with F = G and M(X) = ]x12 on C and gave a new proof for the result of Kurepa on complex spaces. Baker [6] enriched Kurepa’s result on complex spaces and solved the problem under the homogeneity Q( Ax) = x” Q(x), or Q( hx) = x20(x). Gleason [9] determined all functionals Q on k-vector spaces satisfying the parallelogram law and the homogeneity Q(Xx) = x” Q(x). D avison [8] extended the result of Gleason to modules over a ring under appropriate homogeneity on Q, which incidently reduces to Q(Ax) = X2Q(x) when the ring happens to be k. Here we generalize the result of Baker, Gleason, and Kurepa not by broadening the scope of k, but by allowing a very general notion of homogeneity on Q. To be precise, we lay down some definitions and terminology. Let U, V be vector spaces over the field k (commutative and is of characteristic # 2). A biadditive zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPON form T on V is a map of V X V into k which is additive in each of its two variables. There is a one to one correspondence between sy mmetric biadditive forms S on V and functionals Q : V + k satisfying the parallelogram law Q(x + y ) + Q(x - y ) = 2Q(x) + 2Q(y), which is provided by Q(x) = S(x,x) and 4S(x,y) = Q(x+ y) - Q(x - y) [5, 111. We refer to Q as the diagonal of S, and S as the sy mmetric polarization of Q. The diagonal of a biadditive form T always satisfies the parallelogram law, and T is referred to as a polarization of its diagonal. A functional f: U + k is (functionally ) homogeneous if, for some scalar function M: k + k, f(Au) = M(X)f(u) for all A E k, u E U, and we shall say that f is M-homogeneous [lo]. A functional Q: V + k satisfying the parallelogram law is M-homogeneous if, and only if, its symmetric polarization S on U = V zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA x V is M-homogeneous. Thus the study of (functionally) homogeneous Q satisfying the parallelogram law is reduced to the study of (functionahy) homogeneous symmetric biadditive forms. A functional f: V X V + k is (functionully ) bihomogeneous if there exist a pair of functions M ,, M ,: k + k HOMOGENEOUS 257 BIADDITIVE FORMS such that f(Xx, zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFE py ) = M ,(h)M ,(p)f(x,y ) for all X, p E k, x,y E V. If so, we say that f is (M,, Ma>bihomogeneous. Obviously, zyxwvutsrqponmlkjihgfedcbaZY (M,, Ma)-bihomogeneity implies M,M,-homogeneity. A functional f: V X V + zyxwvutsrqponmlkjihgfedcbaZYX k is homogeneous if, and only if, both its symmetric part and its skew-symmetric part are. Hence the symmetric part and skew-symmetric part of a bihomogeneous f: V X V + k are both homogeneous. Generalizing the Halperin problem, we may ask the following question: For given M: k + k, and k-vector space V, is every symmetric M-homogeneous biadditive T: V X V + k the symmetric part of some bihomogeneous biadditive form? (1.0.2) We shall prove that M of interest must be one of the following three exclusive types: M = C#B~ where $J is a morphism on k, M = $xj where C#B and # are distinct morphisms on k, or M = $6 where + is a nontrivial embedding of k into a quadratic field extension k(G). The structure of M-homogeneous biadditive forms will be made clear so that many questions concerning them can be readily answered. Aczel observed the presence of real derivations in his work on the fundamental equation of information and its generalization [2,3]. It is not surprising that (FE) also finds its application in information theory [16]. There is much interesting reading material in [l]. 2. ON THE EQUATION (FE) WITH F = G LEMMA 2.1. Let additive A z 0 and multiplicative M on k satisfy the equation A(x)+ M(x)A(l/x) = 0 (2.1.1) for all x # 0 in k. Then the following hold for all x, y E k: M(x)A(x-‘y)+ M(1) = 1, A(xy) M(y)A(y-lx) M(-r)=M(r), =B(x)A(y)+A(x)B(y), =O, and x, y # 0, A(l)=O, (2.1.1.E) (2.1.2) (2.1.3) 258 zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA C. zyxwvutsrqponml T. NG zyxwvutsr where B(x) = 2-‘[1+ M (x) - M (l- X)]) (2.1.4) B is additive, B(xY) = B(+(Y)+ B(x) cAb)A(y) forsome constant - M (x)B(l/x) = 0, x # 0, M(x) = Proof. (2.1.1.E) W ith (2.1.5) a multiplicative B’(X) c E k, zyxwvutsrq (2.1.6) (2.1.7) -CA’(X). (2.1.8) M, the equivalence between (2.1.1) and is evident. Since A # 0 in (2.1.1), the multiplicative map M cannot be identically zero on k *. This implies M(1) = 1. Replacing x by - x in (2.1.1) and use the fact that A is odd, we obtain M( - x)A(l/ x) = M(x)A(l/ x). By fixing x = r0 # 0 with A(l/ x,) # 0, we get M( -x0) = M(x,). Since M is multiplicative, M( - x) = M( - x,)M(x/ x,) = M(r,)M(x/ x,) = M(x) for all r E k. Putting in (2.1.1) x = 1, we get 2A(l) = 0; and as char(k) f 2, A(1) = 0 follows. This proves (2.1.2). Consider the simple algebraic identity (l-n:)-l(l-y)~‘-(l-x)~‘=(l-x)-‘xy(l-y)~’+Y(l-Y) l> (2.1.9) which holds for all x f 1, y f 1 in k, and apply to it the additive map A term by term to get A[(l-r)-‘(l-y)-‘] -A[(l-x)-r] =A[(l-x)-‘xy(l-y)-‘]+A[&-Y)-r] (2.1.10) for all x # 1, y # 1. We multiply (2.1.10) by M ((lx)(1 - y)); use the additivity of A, the multiplicativity of M, and Equations (2.1.1.E) and HOMOGENEOUS (2.1.2); 259 zyxwvutsrqp BIADDITIVE FORMS and carry out the following sequence of computations: - A[(1 - x)(1 - y)] + M(l=- y)A(l -x) M(xy)A[(l-x)x-‘y-‘(l-y)] -M((l-~)Y)A[Y-‘(l-y)], A(x)+A(y)-A-M(l-y)A(x) = -&+y)A(x-‘y-‘-x-‘-y-‘+l)-M((l-x)y)A(yP’-1) = - M(xy)A(r-‘yP’)+M(y)i+)A(x-‘) + zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCB M (x)M (y )A(y - ‘) - M (l+W Y)A(Y- ‘), A(x)+ A(Y) - A = A(ry ) - M (l- - M (y )A(r) y )A(x) - M (r)A(y )+ M (l- ~)A(Y). Solving for the A(xy) term from the last equation, and using the assumption char(k) # 2, we get (2.1.3) for all x, y # O,l, where B is defined by (2.1.4). Equation (2.1.3) also holds when x = 0,l or y = 0,l because of (l.O.l), which implies B(0) = 0 and B(1) = 1. The additivity (2.1.5) of B follows from that of A f 0 in (2.1.3). We compute A(xyz) first as A((xy)z) and then as A(x(yz)) using (2.1.3), and compare the results. This comparison and A # 0 lead to (2.1.6) [12]. We verify (2.1.7) with the following computations: M(x)B(x-‘)=2-‘M(x)[l+M(x&-M(l-x-l)] =2P[M(x)+M(xx-‘)44(x(1-r-‘))] =2-r[M(x)+l-M(x-l)] =2-94(x)+144(1-x)] =B(x). The following computations use (2.1.1), (2.1.3), (2.1.6), and (2.1.7): M(x)A(y) = M(x)A(r+(xy)) = - A(x)[B(x)B(y )+ = M(x)A(x-l)B(xy)+ cA(x)A(y )l + B(~)[A(~)B(Y)+B(~)A(Y)~ = [B’(X)- cA2(x)]A(y ). M (r)B(x- ‘)A(xy ) C. zyxwvutsrqponml T. zyxwvutsrqpon NG 260 Then the final elimination of the factor A(y) yields (2.1.8) Equation (2.1.8) holds when 1c= 0 because of (1.0.1). for x # 0. n THEOREM 2.2. Let additive A # 0 and multiplicative M on k satisfy the functional equution (2.1.1) A(r)+ M(x)A(l/ x) = 0 for all x # 0. Let c E k be the unique constant in Lemma 2.1 such that (2.1.6) and (2.1.8) hold. Then, based on the nature of c, we can describe A and M as follows: Case (i). Suppose c = 0. Then for some (field) morphism +: k + k, + + 0, A is a nontrivial Here, a @derivation +derivation is an additive (2.2.1.a) and M = $I~. map A satisfying A(xy) = +(x)A( y) + for all x, Y E k. Case (ii). Suppose c # 0 and c = a2 for some a E k. Then for some morphisms +,$:k-,k, $,I+Q#O, and +ZI/J, zyxwvutsrqponmlkjihgfedcbaZYXWVUTSR +(y)A(x) A= (2a)-‘(+ -rC/) Case (iii). Suppose c # 0 and a = 6 ding +: k -+ k(a), (Pz & A=(2a)-‘(c)--s) Here the conjugate x+ya=r-yaforallx,yEk. operation and (2.2.1.b) M=+$. 4 k. Then for some (field) and on the field (2.2.l.c) M=+s. extension embed- k(a) is defined by Conversely, if A and M are given by (2.2.l.a), (2.2.l.b), or (2.2.1.~) for some constant c E k, then A # 0 is additive, M is multiplicative, and (2.1.1) is satisfied. Proof. Let additive A # 0 and multiplicative M satisfy (2.1.1), and let c E k be as given in Lemma 2.1. There are three cases to consider (cf. [7]). Case (i): Suppose c = 0. Then (2.1.6) reduces to the multiplicativity of B. Thus $I = I3 is a field morphism on k. From (2.1.3) where A f 0, we get + # 0 and that A is a @derivation. Equation (2.1.8) reduces to M = G2 as claimed. Case (ii): Suppose c # 0, and c = a2 with a E k. Then a f 0. Equations (2.1.3) and (2.1.6) correspond to the multiplicativity of $ = B + uA and $ = B - aA. Since B and A are additive, so are + and I/ J.Hence 9, and # HOMOGENEOUS 281 zyxwvutsrq BIADDITIVE FORMS are field morphisms on k. Since a, A # 0 and char(k) # 2, we have $J# 4. From the definition of + and I/J,we get immediately A = (2a)-l(+ - JI) and B = 2-‘($I + JI). Knowing both B and A in terms of + and J/, Equation (2.1.8) then delivers M = +#. Case (iii): This case is very much like (ii). W e also define + = B + UA and 4 = B - aA, and observe that 4 = 6, the conjugate of +. The converse is easy to verify. COROLLARY 2.3. x z 0 with additive A = D, of The general solution A(r)+ zyxwvutsrqponmlkjihgfedcbaZYXW M (x)A(l/x) = 0 for all A # 0 and multiplicative M on the reaZs R is given by a nontrivial real derivation, and M(x) = x2, (2.3.1.a) or A=a-‘Im$ and (2.3.1.b) M(x)=~+(x)~~, where u # 0 is a real constant, + : R -+ C is a nontrivial embedding of the reals into the complex numbers, Im $I is the imaginary part of $J, and 1.1 denotes the usual norm on Q=. Proof. Take Theorem 2.2 with k = R. The only (nonzero) morphism (p: R + R is given by G(X) = x, the identity map. Thus (2.2.1.a) reduces to (2.3.1.a). Case (ii) and (2.2.1.b) will not materialize, as there are no such distinct morphisms (P and 4. In case (iii), c < 0, say c = - u2 with (I E R, u # 0. Thus A’ = UA and M form a solution of (2.1.1) with c’ = - 1, and we can take a’ = J-1 = i. The field extension R(i) is C, and (2.2.1.~) reduces to A’ = (Zi)-l(+ - 4) = Im+ and M = j+12. Thus A = a-‘A’ = u-l Im$, and n (2.3.1.b) is established. 3. ON THE FUNCTIONAL LEMMA 3.1. EQUATION (FE) W ITH F = - G Let additive B # 0 and multiplicative M a k satisfy the equution B(x) - M(x)B(l/x) = 0 (3.14 262 zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA C. zyxwvutsrqponm T. zyxwvutsrqpon NG for all x z 0 in k. Then the following hold for all x, y E k: and M (l)=1 B(x) M(-r)=M(x), = B(1)2-‘[1+ lf we further assume, M (x) X)]. (3.1.3) that = 1, (3.1.4) also hold fm all x, y E k: M (x) [2B2(x) (3.12) - M (l- just for convenience, B(1) then the following =o, zyxwvutsrqponmlkjihgfedcbaZY X, y # 0, (3.1.1E) -M(y)B(y-‘x) M(x)B(x-‘y) - B(x2)] [2B2(y) [2B2(x) = 2P(x) - B(y2)] - B(X2), = 2B2(xy) - B(X2)] B(y) (3.1.5) (3.1.6) - B(x2y2), = 2B(x)B(xy) (3.1.7) zyxwvutsrqponm - B(r2y), b(X2)- B2(41 - zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJI wB(Y)12= DbY) (3.1.8) x [B(Y2) zyxwvutsrqponmlkjihgfedcbaZYXWVUTS - B2(Y)l) [BbY) - = Proof. (3.1.1.E) e4B(Y)l Mu4 - B(Y2) M is multiplicative, Since - JwB(Y)l %4Wl[ is obvious. The property same way (2.1.2) [WY) (3.1.9) B2(Y)]. the equivalence (3.1.2) follows from (2.1.1), - between follows from (3.1.1) with the exception (3.1.1) in exactly and the that we cannot say much about B(1). If we take the identity similar calculations obtain with (3.1.3). y=2-‘, (2.1.9), apply B to it term by term, to those done in Lemma In fact, for the current purpose, namely 2(1- r)-‘- we it is sufficient to take (2.1.9) 2 = 2x(1 - x)-l. can drop the factor 2 prior to applying and follow 2.1, which lead to (2.1.3), Since char(k) # 2, we B to it term by term. The following HOMOGENEOUS BIADDITIVE 263 zyxwvutsrq FORMS sequence of computations shows all the details: B[(l-x)-l] M(l-r)B[(l-Lx-‘] -B(l)=B[x(l-x)-l], -M(l-r)B(l)=M(l-x)R[(l-r)-lr], B(l- x) - M(l- x)B(l) = M(x)B[x_‘(1 B(1) - B(x) - M(l- +3(l) = M(x)B(x_1) B(1) - B(x) - M(l- x)B(l) = B(x) -x)], - M(x)B(l), - M(r)B(l). Solving for the B(x) term in the last equation gives (3.1.3) for x Z 0,l; it also holds for x = 0,l by (1.0.1) and (3.1.2). Since I3 # 0, from (3.1.3) we get B( 1) + 0. As the equation (3.1.1) is linear in B, we may assume the normalization (3.1.4) for convenience. We verify (3.1.5) with the following calculations using (3.1.3) and (3.1.4): 2B2(x)-B(x2)=2-‘[1+M(r)-M(1-x)]2 -2-‘[l+M(Xs)-M(l-x2)] = 2-‘[1+ P(z)+ - 2M(l- M2(1- x)+2M(x) x) - 2M(r)M(l- x)] -2-‘[l+Ms(X)-M(l-x)M(l+x)] = M(x)+2-‘M(l- x) x[ -2-2M(x)+M(l-x)+M(l+x)] = M(x)+2-‘M(l- Lx){ [l-r M(l- x) - M(x)] - [1+ M( -x) = M(x)+2-‘M(l- x)[2B(l = M(x)+2-‘M(l- x)[2B(l) -x) - M(lS - 2B( -x) x)] - 2) - 21 - 21 = M(x). Equation (3.1.6) is but a reminder that M in (3.1.5) is multiplicative. We replace y in (3.1.6) by 1 - y and subtract the resulting equation from (3.1.6) to obtain (3.1.7). C.T.zyxwvutsrqpo NG zyxwvutsrq 264 Replace y in (3.1.7) by y2 to get [2B2(x) - B(x2)]B(y2) - B(x2y2). We subtract it from (3.1.6) to get [2P(x)-B(x2)][B2(y)- = 2B(x)B(xy2) B(y2)]=B2(xy)-B(x)B(ry2). In Iigbt of (3.1.7), we can replace the factor B(xy2) in the last term to get bB2(x)- B(X2)1 [B2(Y) = B2bY) B(Y2)1 - W{2B(YMY4 - [2B2(Y) - ~(Y2)]W}. After simplication we obtain (3.1.8). Equation (3.1.9) is the symmetric polarization of (3.1.8) in the variable r. n THEOREM 3.2. Zf zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIH additive B # 0 and multiplicatiue M satisfy the equution (3.1.1) B(x) - M (x)B(l/x) = 0 zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQP forall x # 0 in k, then they zyxwvutsrqponmlkjihg have one of the following three exclusive forms: Case (i). Forsome stunt b#O ink. (field) morphism B=b+ Case (ii). Forsomemorphisms sameconstantb~Oink, and Case (iii). For some constant b # 0 in k, B=b2-‘(++G) Here 4: k 4 k(a) is an embedding (the conjugate of +). and cek +ZO, andsomecon- (3.2.1.a) M=+2. +,J/:k+k, B=b2-‘(++#) constant +:k+k, $,$#O and (3.2.1.b) M=+#. with a=&Gk, and and @#II/, and for some M=+$. (3.2.1.~) of k into its field extensiun k(a), Conuersely, if B and M are of the forms (3.2.la), (3.2.l.b) then B # 0 is additive, M is multiplicative, and (3.1.1) holds. up# 6 or (3.2.l.c), Proof. Let B # 0, and M be a solution of (3.1.1). Then by Lemma 3.1, B(1) f 0. We can normalize B and assume B(1) = 1. We need to show that HOMOGENEOUS BIADDITIVE FORMS 265 b = B(1) = 1. Consider the (3.2.l.a), (3.2.l.b), or (3.2.1.~) must hold with zyxwvutsrqponmlkjihgfedcbaZYXW equations (3.1.2) to (3.1.9) and begin with the last. There are two exclusive possibilities: Case 1. Suppose B(y’) - B2(y) = 0 for all y E zyxwvutsrqponmlkjihgfedcbaZYXW k. This is equivalent to the symmetric biadditive map B(xy) - B(x)B(y) = 0, i.e., B is multiplicative. Thus I#I= B is a (field) morphism on zyxwvutsrqponmlkjihgfedcbaZYXWVUT k and + # 0. Equation (3.1.5) then yields M(x) = +“ (x). This confirms (3.2.1.a) with b = 1. Case 2. Suppose for some ya, I?(yt) - B2(y0) # 0. Then by fixing such a y in (3.1.9), we get B(uv) - B(u)B(v) =cA(u)A(v), all u,vek, (3.2.2) is a nonzero constant in k, and A(u) = where c = [B(yt) - B2(y0)]-’ B( uy,) - B( u)B( y,,). Since B(W) - B( u)B( v) does not vanish identically, we obtain from (3.2.2) that A is additive and A # 0. (3.2.3) Putting (3.2.2) back in (3.1.7), we have - zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONM [B(x2)B(y)+ cA(x2)A(y)l. It can be simplified to 0 = 2cA(x)B(r)A(y) factor cA( y), we obtain A(r2) = 2A(x)B(x). - cA(x2)A(y). Canceling the Polarizing it symmetrically, we get Aby) From (3.1.5) and (3.2.2) = A(+(Y)+ BAG. (3.2.4) cA2(r). (3.2.5) we get M(x) = B2(x) At this point, we have constructed equations (2.1.2) to (2.1.8). Referring Theorem 2.2 which correspond to the and obtain either (3.2.1.b) or (3.2.l.c), The converse is easy to verify. an A and reinstated the system of to cases (ii) and (iii) in the proof of case c + 0, we observe the form of B with b = 1. W C. zyxwvutsrqponmlk T. NC 266 The deliberations that led to Corollary 2.3 from Theorem repeated for Theorem 3.2 with k = W, and give the following: 2.2 can be COROLMY 3.3. The general solution of B(x) - M(r)B(l/x) = 0 for all x # 0 with additive B # 0 and multiplicative M on the reals Iw is given by B=bx B=bRe$ and and M(;r)=r2 (3.3.1.a) (3.3.1.b) M(x)=l$(x)12, where b + 0 is an a&tray real constant, + : IR + C is a nontrivial embedding, Re $J is the real part of +, and 1.1 denotes the usual norm on Q=. 4. ON THE EQUATION (FE) AND ITS GENERAL SOLUTION We consider the functional equation F(x) + M(x)G(l/ x) = 0 forall x#Oin k, (FE) where F, G : k + k are additive and M: k + k is multiplicative. As pointed out in the introduction, there is no loss of generality in assuming (1.0.1) M(0) = 0, and to avoid trivial cases, we suppose that (4.0.1) F,G,M#O. In particular, M(r) # 0 and M(x-‘) x by x-l in (FE) and arrive at G(x) + M(x)F(l/x) = M(x)-’ = 0 for all x # 0 in k. We replace forall x+Oin k. (4.0.2) Forming the sum and difference of (FE) and (4.0.2), we transform them to the system of equations M(x)A(l/ x) = 0, (4.0.3) B(x) - M(x)B(l/ x) = 0, (4.0.4) A(x)+ HOMOGENEOUS BIADDITIVE 267 zyxwvutsrqp FORMS where A = 22’(F + G) and B = 2-‘(F - G) are again additive. Because of (4.0.1), either A # 0 or B # 0. We use Theorem 2.2 and Theorem 3.2 to obtain the solution of the system (4.0.3) and (4.0.4), which in turn gives the general solution of (FE) where zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPON F = A + B and G = A - B. Their combined use leads to the following result. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONM THEOREM 4.1. Let additive F, G : k + k and multiplicative M : k + k be rwnzero maps satisfy ing (FE). Then they have one of the following three exclzrsive representations: (i) For some (field) morphism $J: k - + k, cp# 0, some $derivation which is an additive map satisfy ing D(xy ) = +(x)D(y )+ +(y )D(r) D, fm all x, y in k, and a constant b E k with (D, b) f (O,O), we have F=D+b$, (ii) M =G~. G=D- b+, For some morphisms +,$:k+k, (p,$# O and @ # I/J, (4.1.1.a) and some constants cl, cz E k with (c,, c2) # (O,O), F = cl+ - czJl> (iii) k(a) G=c,+-c,$, Forsomeconstantc~kwitha=6Ek, M =+# . someembedding (4.1.1.b) +:k+ with $I f 5, and some nonzero constant X E k(a), F=h++ Here, conjugacy A+, G= - % # A& M =&j. (4.1.l.c) in k(a) is defined by x + y a= x - y a for all x, y E k. The converse is also true. Proof. Either A or B in (4.0.3) and (4.0.4) must be nonzero. If A = 0, or if B = 0, then (4.1.1.a) to (4.1.1.~) hold because of Theorem 3.2, or Theorem 2.2. Suppose A # 0 and B # 0 in (4.0.3) and (4.0.4). Then Theorem 2.2 and Theorem 2.3 are both applicable, along with their supporting lemmas. Compare (2.1.4) with (3.1.3), and observe that the B in the system (2.1.2) to (2.1.8) corresponding to Equation (4.0.3) coincides with the B in (4.0.4) when the latter is normalized. Hence (4.0.4) after normalization coincides with (2.1.7), and so (4.0.4) can be regarded as part of (4.0.3). In this case, there is no need to develop (3.1.5) to (3.1.9), and we can obtain the form of A and M from Theorem 2.2, and at the same time obtain B from its proof. Namely, B = (p with (2.2.l.a), I?= 2-‘(+ + 4) with (2.2.l.b), and B = C. zyxwvutsrqponml T. NG 268 2-‘($I + 6) with (2.2.1.~). We can put in a constant factor b # 0 next to the above forms of B to compensate the normalization. The final result for F = A + B and G = A - B is precisely (4.1.1.a) to (4.1.1.~) with D # 0, b#O, c,#fc,, A=2-‘(a-‘+b). Th us in ah situations, F, G, and M must be of the forms (4.1.1.a) to (4.1.1.~). No solutions wilI be lost if we restrict X to those of the form X = b or A = 2-‘(a-’ + b) where b E k (cf. Remark 4.4). The converse is easy to justify. n We combine Corollary 2.3 and Corollary 3.3 to give the general solution of (FE) over the real field: zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJI COROLLARY 4.2. The general solution of (FE) with F, G and multiplicative M on the reals 08 is given by F(x) = D(r) + bx, G(x) F=b,Im++b,Re+, = D(x) - br and G=b,Im$-b,Re$ nonzero additive M(x) =x2 and (4.2.1.a) M=l+l’. (4.2.1.b) Here D is a derivation on W, b E R is a constant, and (D, b) # (0,O); +: R -+ C is a nontrivial embedding, b,, b, E R are constants, and (b,, b,) + (0,O). uniqueness result for (i) For given M, the morphism (4.1.1.a) is unique. (ii) For given M, the distinct M = C#B#in (4.1.1.b) are unique up { C#B, I/J} is unique. (iii) Let us define the following kl&Zk} c k*: c-c’ ToeuchcEX, of Theorem 4.1, we have the representations (4.1.l.a) (4.1.l.b), and Under the hypothesis THEOREM 4.3. following (4.1.l.c): iff C#J ( # 0) on k generating M via M = +2 in morphisms 9, +!J( # 0) generating M via to a permutation, i.e., the unordered set equivalence c=d2c’ letS(c)={(F,G,M)jF=A++&G= relation on the set X = {c E forsome (4.3.1) d Ek. -&#B-& M=& 269 HOM OGENEOUS BIADDITIVE FORM S us zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA in (4.1.1.~)) be the associated family of solutions of (FE) over all X and +. It turns out that S(c) and S(c’) are either identical or disjoint, according to whether c or c’ are equivakmt or not. Furthermore, zyxwvutsrqponmlkjihgfedcbaZ forgiven M and c E X, the embedding +: k - - , k(G), generating M via M = & in (4.1.l.c), is unique up to conjugation. Proof. (i): Suppose 9, +’ are morphisms such that M = # = (+‘)‘. Then the symmetric biadditive maps +(x)+(y) and +‘(x)+‘(y) have equal diago- nal, and thus #(x)$(y) =+‘(x)+‘(y) for all x, y E k. Since +,# # 0, we have ~(1) = cp’(1) = 1, and so by setting y = 1 we get 4(r) p+‘(r) for all r E k, i.e., C$= $‘. This proves the uniqueness of $I in the representation M = (p”. (ii): Suppose M = +J, = +‘\I/‘.This implies the symmetric biadditive maps +(X)+(Y)+ J/(~)+(Y) and +‘(X)+‘(Y)+ #‘(x)+‘(~) are equal. Specifying y = 1, we get C#I + \c,= $I’ + JI’. This linear dependence between the four nonzero multiplicative functions (p, \I/,+‘, #’ implies they cannot be pairwke distinct; some two of them must be equal. Since + # 4, 9’ # #‘, and $I + # = +’ + 1+5’,we must have (c#J’,I/‘) = (+, #) or ( I+!J #I), , proving that the distinct morphisms (p, # generating M via M = +Ic, are unique up to a permutation. (iii): W ith reasonings analogous to (ii) above, we conclude that if nonzero +, +‘: k + k(A) +’ # J, are morphisms such that M = +? = +‘&‘, where + # 4 and then I#J = +’ or Cp= 3. This proves the uniqueness of (p up to conjugation in the representation M = ~$6 when c is fixed. W e now proceed to examine the families S(c) over all c E X. Suppose S(c) and S(c’) meet. Then there exist embeddings +: k + k(h) and +‘:k+k(@ ) such that M =&=+‘$. W e write #=B+Afi and 9’ = B’ + A’@ , where (B, A) and (B’, A’) : k + k are additive pairs satisfy- ing the relations (2.1.3) to (2.1.8). In view of (2.1.4) and (2.1.8), we have B = B’ and B2 - cA2 = BT2 - c’At2; i.e, B = B’ and cA2 - c’Ar2. Since A, A’ # 0, the relation cA2 = c’A’~ implies c - c’, where - is defined by (4.3.1). This proves that S(c) and S(c’) meet only if c - c’. Suppose c - c’; we proceed to show that S(c) = S( c’). By symmetry it is sufficient to show S(c) c S(c’). Let (F, G, M ) E S(c) be arbitrarily given. Then there exist constant A and an embedding @ : k + k(6), I# # I 4, such that (4.1.1.~) holds. W e write $I = B + Afi, where B, A are additive maps on k. Then the relations (2.1.3) to (2.1.8) hold. Since c - c’, we can write c = d2c’ for some d E k. W e define B’ = B, A’ = ~54. Then (2.1.3) to (2.1.8) hold for B’, A’, c’, and M . Thus 9’ = B’ + A’@ defines an embedding of k into k(c), and M =+‘i. Let h=X,+h,fi with X,,X,Ek; then FA+ +q 2(A,B is equivalent to F= + c’. dh,. dA) = 2(A,B’ 2(h,B+ &,A). + c’. dX,.A’), But c =d2c’, and so F = which is equivalent to C. zyxwvutsrqponml T. NC 270 F = 2~’ + A’+‘, where A’ is defined by x’ = X 1 + dh,@ . Similarly, G = - A+ - h6 translates into G = -E+’ - 22. Obviously +’ ~2 and x’ # 0. Hence (F, G, M ) E S( c’). This proves S(c) c S( c’) as claimed. n REMARK4.4. We pointed out in the proof of Theorem 4.1 that there is no loss of solutions to (FE) if we restrict A E k(a) to be of the form X = 2-‘b or X = 22’(a-’ + b) where b E k. By allowing X E k(u) to be free of this restriction, we have created some redundancy in the representation (4.1.1.~). The reason for creating the redundancy in the first place is so that we can now remove it by a suitable choice of c. To be precise, in light of Theorem 4.3, we can restrict the choice of c E k with fi P k to those that are not mutually equivalent under the definition (4.3.1). For example, Corollary 4.2 is formulated with this idea in mind. When k=[W,wehaveX={cEkl~4IW}=]-00,0[andthereisonlyoneequivalence class on X, as c - - 1 for alI c E X. It is sufficient to consider just one field extension W( - 1) = 6. In general, the set Y = { d2 1d E k * } is a subgroup of k * under multiplication, and it is the image of k * under the multiplicative map d + d ‘. The set k * is the disjoint union of X and Y. The equivalence classes on X under the equivalence relation (4.3.1) are members of the quotient k */Y with Y removed. A choice of representatives from each of the equivalence classes on X corresponds to a lifting I (restricted to (k */Y )\{ Y }) from the quotient group k */Y back to k *. When k is a finite field, the Lagrange theorem implies ]Y 1= 2- ‘1k * 1and k */Y is cyclic of order 2, and so there is only one equivalence class on X. Hence for finite fields, it is sufficient to use one field extension in generating solutions in (4.4.l.c). In general, we can restrict the choice of c in (4.1.1.~) to the image of an arbitrarily fixed lifting 1. 5. ON THE EQUATIONS THEIR INVERSIONS (2.1.1) AND (3.1.1) WITH GIVEN M AND In retrospect, we solved (FE) by a reduction to the equations (2.1.1) and (3.1.1), which are tied by a natural duality. We have shown in Corollary 2.3 the presence of nontrivial additive A and multiplicative M other than derivations and the square. In the course of solving (2.1.1) and (3.1.1), we have established some important correlations between the maps A, B, and M , which are contained in Lemma 2.1 and Lemma 3.1. Some of these served only a passing role; e.g. (2.1.2), (3.1.2), (3.1.4), (3.1.6) to (3.1.9). The rest deserve more attention, as they offer a basis for the construction of the functions, and provide transformation rules between them. 271 HOMOGENEOUS BIADDITIVE FORMS In many applications the map M is given and we want to solve for A and B in (2.1.1) and (3.1.1). Conversely, given A # 0 or B # 0, we can reconstruct M. The functional relations reported in Lemma 2.1 and Lemma 3.1 can be used effectively in such computations, as summarized in Figures 1 and 2. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA EXAMPLE 5.1. To find the general solution of F(z) + l.z12G(Z) = 0 on C with additive F, G, we simply follow the flow chart in Figure 1 beginning with M(z) = 1~1~.The logical steps in the reasoning are listed below: M = 1~1~; B,(z) = 2-‘(2 + 5); B, is additive; the general solution of B is B = C(Z + Z), with arbitrary c E C; B,,(zi) - @(~a) # 0 is true, say with ~a_= i; A,(z) = B,(zi) - B,(z)&,(i) = 22’(zi +zi) = 2-‘i(z - 5); the general solution of A is A = c( z - Z), with arbitrary c E C. Therefore the general solution of (FE) with M(z) = lz12 on C is given by F = A + B = az - bZ, G(z) = A - B = bz - aZ, where a, b E C are arbitrary constants. 6. REPRESENTATION OF FUNCTIONALLY BIADDITIVE FORMS HOMOGENEOUS THEOREM 6.1. Let V be a vector space over k. Let T:V XV+ nonzero M-homogeneous biadditive form for some M: k + k, i.e. k be a forall XEk, x,y~V. (6.1.1) T(h zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA Xy) = M(A)T(x,y) Then M is multiplicative, and is of one of the following three exclusive types: M = c$~ where C$: k + k is a nonzero ( field) M=NJ where C#B, II/: k + k are distinct rumzero morphisms, M=& where +: k ---*k(G) is a nontrivial embedding of k into an extension of k obtained by adjoining a square root which is not in k. morphism, (6.1.2.1) (6.1 .Z.II) (6.1.2.111) Furthermore, 4.3). such representations of M are essentially unique (cf. Theorem C. T. NG w A=0 and B=O are the general solutions of (A) and (B) is the general solution of (B) w BO is a field morphism and A=a B,-derivation is the general solution of (A) Yes FIG. 1. Flow chart for finding the general solutions of (A) A(x)+ M(x)A(l/x) = 0 or (B) B(r) - M(r)B(l/r) = 0 with additive A, B when multiplicative M is given. HOMOGENEOUS 273 BIADDITIVE FORMS Given E! Given A There exists no solution M M,(x) = 2B’o (x) - B,lx2) No ) There exists no solution M 1Yes M = M, is the unique solution for = 0 FIG. 2. Flow chart for finding the general solution of A(x) + M( r)A(l/r) or B(x) - M( r)B(l/r) = 0 with multiplicative M when additive A + 0 or B f 0 is given. C. zyxwvutsrqponm T. NG 274 W e fix arbitrarily x0, y,, E V such that T(x,, yO)# 0. Since T( Xp. Proof. x0, XP.Y,) = M ($)T(x,,yJ and T(Ae~lx~, X.pyO) = M (X)T(P~~, PY~) = th e multiplicativity M ( AW L)= M ( A)M (p) follows. Since zyxwvutsrqpon M (~)M (P)T(x,,YlA VAx,,y,) T(Xx,,y,) A-‘y,), = WX)?‘(x,, and the maps G(X) = - T(x,, Xy,,) F, G: k + k induced satisfy (FE). by F(X) = Clearly F and G are additive and nonzero, and so by Theorem 3.2, M must be of one of the cited w forms. The next three theorems disclose the structure of all M -homogeneous biadditive forms based on the three M types. THEOREM 6.2. Let nonzero 9: k + k be a morphism. The following statements concerning biadditive functions T on a vector space V over k are equivalent: (1) T is G2-homogeneous. (2) T satisfies the following T(hy) laws: = D(Lx,y)+ d+W(x>y), T(x, hy) = - D(b,y)+ where D: k x V x V + k satisfies the following D(.,x,y) isa @derivationon (6.2.1) $@)T(x,y), description: kforeachfixedr,yEV, D( A,. , . ) is a (+, +)-bihomogeneous biadditive form on V for each fixed X E k. (3) T can be constructed through the following (6.2.2) (6.2.3) steps: (i) Choose an arbitrary basis Q = (e,), E I for V over k. (ii) Znitiate T on Q X Q (into k) arbitrarily. (iii) Initiate for each (e,e’) E G x Q a @derivation on k, labeled by D( . , e, e’), arbitrarily. (iv) Extend T and D( A,. , .) above to V X V by the definitions T(x,y) = CG(Pj)D(Xi,ei,ej) zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJI ij - D(Xtx,y) = +(Xi)D(pj,ei,ej)+ G(Ai)G(Pj)T(ei,ej), (6.2.5) C~(hi)~(l-lj)D(h,ei,ej), ij where x = Xihiei and y = Cjpjej, (6-2.4) Xi, pj E k. HOMOGENEOUS BIADDITIVE 275 FORMS zyxwvutsrqp Proof. To show that (1) implies (Z), let zyxwvutsrqponmlkjihgfedcbaZYXWVU T be +2-homogeneous. For each fixed x, y E V, the functions (6.2.6) G (X) = - T(x, Xy ) F(h) = T(hy), zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPO satisfy (FE) with M = $I~ because F(h) = T(Ax,y ) = ~~(h)T(x, A- ‘y ) = Thus by Theorem 4.1 and Theorem 4.3, there exist a @derivation D and a constant b such that F(h) = D(h) + b+(h) and G(X)=D(X)- b+(A). Since F and G depend on x and y, we explicitly write F(X) = D(A,x,y )+ b(x,y )+(X) and G(X) = D(X,x,y ) - b(x,y )+(A). Hence, from (6.2.6) we get the following laws governing T: - c$~(A)G(X- ‘). Thy) = D(b>y>+ b(x,y)$(X), T(x,Ay) = - D(X,x,y)+ b(x,y )+(X). Since D( 1, x, y) = 0 and (p(1) = 1, we get b(x, y ) = T(x, y ), and so T(hy) = D(Lx,y)+ Tb, xy> = - D(Lx,y) +(X)T(x,y)> + cp(h)T(x,y), (6.2.7) (6.2.8) proving (6.2.1) and (6.2.2). The additivity of D in its second and third variables follow from (6.2.7) and the additivity of T in both x and y. What is left in (6.2.3) is the +-homogeneity of D in x and y. To obtain this we perform the following calculations using (6.2.7): Th-y) = D(kx, Y)+ ~h$‘(x,y) = @)D(PAY)+ T(hx,y) G(dD(Lx>y) + +(%(~)T(x,y), (6.2.9) = D(k /-my) + +(A)T(px,y) =D(X,~x,y )+~(X)[D(~,x,y )+~(~1)T(x,y )]. (6.2.10) Comparing the right sides of the above, we get D(L my) = @(p)D(b,y), which is the +-homogeneity of D in its second variable. +-homogeneity of D in its third variable follows from (6.2.8). (6.2.11) Similarly, the 276 C. T. NG TO show zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA that (2) implies (l), we display the following calculation: T(Xx, hy)=D(A 9x9hy)++(A)W, Xy)=~(h)D(h,x,y)+~(X)[-D(X,x,y) + +(~)W,y)l = &vw,y). The equivalence between (2) and (3) is obvious. n THEOREM 6.3. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHG Let nonmo $, 4: k + k be distinct mmphisms. The following two statements concerning biadditive fm T on a vector space V over k are equivalent: (1) T is q+homogeneous. (2) T has the decomposition zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIH Thy) = c,(x,y) - C,(x,y) (6.3.1) for some unique biadditive C,, C, : V X V + k, with (+, #>bihomogeneous C, and (4, +>bihomogeneous C,. Proof. Let T be &!+homogeneous. Then for each fixed x,Y E V, the functions F(h) =T(Xx,y), G(X) = - T(x, Xy) satisfy (FE) with M = ++ because F(X) = T(Xx,y) = +(A)$(h)T(x, A-‘y) = - +(h)#(X)G(h-‘). By Theorem 4.1 and Theorem 4.3, there exist constants C,(x, y) and zyxwvutsrqponmlkjihgf C,(x, y) such that WXPY) =c,(wM~) T(x, AY) = - Cs(xJ)+O)+ - C,(x,y)J/(X), C,(x,y>$(h). (6.3.2) (6.3.3) Since + and 4 are linearly independent, the above equations and the biadditivity of T imply that C, and Cs are biadditive. We compute T(Apx, y) under (6.3.2) first as T(Xp . x, y) = C,(x, y)$(Xp) - Cz(x, y)J/(Ap) = C,(x, Y)+(~)+(P) - &(x9 y)$(h)$(p), and then as T(h. CLX,y) = C,(/wY)9(A) - CZ(cLX>Y)W). c om Parison leads to Cr(x, y)+( X)+(p) C&,y M AM pL) = C1(px,y )$(h) - C&x,y )# (~). The independence of G(X) and +(A) yields W x,y ) = C1(x,y M (~) ad C&x,y ) = C,(x,y )W ). Similarly, using (6.3.3) we get C,(x, py) = C,(x,Y)$+), and Ca(x, py) = n o er words, C, is (I$, #)-bihomogeneous, and C, is (4, +)C,(%Y)@(EL). 1 th bihomogeneous. The representation (6.3.1) comes from (6.3.2) with X = 1. This proves that (1) implies (2). The converse is obvious. The uniqueness of n C, and C, is due to (6.3.2) and (6.3.3) where + and J, are independent. THEOREM6.4. Let c E k be such that 6 P k, and let +: k + k(6) be a nontrivial embedding, The following two statements concerning biadditive HOMOGENEOUS 277 zyxwvutsrqp BIADDITIVE FORMS forms T zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA on a vector space V over k are equivalent: (1) T is &&homogeneous. (2) T hu.s the unique decomposition Thy) where A: V x V + k(h) (6.4.1) = zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFE Nx,y) + &,y) is ($, &)&homogeneous biudditive. Proof. By similar arguments to those employed for the previous two theorems, F(X) = T( Ax,y) zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIH and G(A) = - T(x, Xy)zyxwvutsrqponmlkjihgfedcbaZYXWVU satisfy (FE) with M = $16, and so by Theorem 4.1 and Theorem 4.3, there exist constants A (x, y) zyxwvutsrqponmlkjih E k (6) such that Thy) = Nd#O)+ ~(x,y&i(A), (6.4.2) (6.4.3) T(x, AY) = x(x,y)@(A)+ zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIH Aby)?( The discussion on the uniqueness of (4.1.1.~) in Theorem 4.3 and Remark 4.4 is used here to support the fact that c and the embedding $I: k + k(G) can be chosen independent of x and y. The rest is a rerun of the proof of n Theorem 6.3. The meaning of bihomogeneity is naturally broadened. REMARK 6.5. A functional T: V X V --) k is M-homogeneous and biadditive if, and only if, its symmetric and skew-symmetric parts are. It is easy to derive from the above three theorems their symmetric and skew-symmetric counterparts. In Theorem 6.2, T is symmetric (skew-symmetric) if, and only if, D( X, . , - ) is skew-symmetric (symmetric). Hence, to construct symmetric (skew-symmetric) T, it is necessary and sufficient that the initializations in (3)(ii), (iii) be symmetric (skew-symmetric) and skew-symmetric (symmetric) respectively. In Theorem 6.3, T is symmetric [skew-symmetric] if, and only if, C,(x,y) + C,(y,x) = 0 [CXx, y) - C2(y,x) = 01. In Theorem 6.4, T is symmetric (skew-symmetric) if, and only if, zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPO A(x, y) = h(y,x) [ A(x, y) = - @y,x)]. Such decompositions have the same spirit as those results reported in [lo]. One can easily answer questions such as (1.0.2), using the above results. REMARK6.6. The above result can be routinely applied to determine all biadditive M-homogeneous transforms T: V X V + W where W is a k-vector space. All it requires is a coordinatization of W with respect to some basis, and we pass from the properties of T to those of its coordinate functions. 278 C. T. NG T: zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA V X V --, k zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHG is cohomogeneous if there exists a function N such T(x, Ay) = N(X)T(x, zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPON y). We can immediately deduce from Theorems 6.2-6.4 that homogeneous biadditive T are all cohomogeneous. On A map that T(Xz, y) + the other hand, it is easy to show that cohomogeneity implies homogeneity, because T(Xx, Ay )+ T(x, A.hy ) = N(A)T(x, Ay ), T(Xx, Ay )+ T(h.Xx,y )= N(X)T(Xx, zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA y)> Vh)[T(h y) + T(x, Ay)] = N2(A)T(x, y), T(x, A2y ) + T( A2x, y ) = N( A2)T(x, y ). Subtracting the last from the sum of the first three, we get the M-homogeneity of T with M (X) = 2- ‘[ N 2(A) - N( X2)]. 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