A symbolic product contains factors of seven types, viz. :- (i.) ,
(ii.) ,
(iii.) ,
(iv.) ,
(v.) ,
(vi.) , (vii) ,
and, if it is to have a real expression, each letter ... must occur m times, and each letter ... n times. An important proposition is the expression of a_{x}^{m} a_{y}^{n} in a series of polars of forms, of type (aa)^a^, each polar being multiplied by a power of (ay). Gordan's formula is ALGEBRAIC m an ay = Σ(m+n=k+1) { (aa) ²a_ - *a ~k), -k(ay)}" ; m-k_n-kn-kr Jy k(xy)*; and and the right-hand side is a simultaneous covariant of the forms k_m-k_n-k (aa) ka Cbx ax which involve but one series of variables. Other useful series are (*)(*) (^+μ+n-k {(aa) ² ax ax 1-K(wy)k; kλ-k_n-kn+μ-k Jy axa ay = Σ n n (ama² ) = = Σ( - )² (+) (aa) ² a (-)" (aa) kam-kn-k₁ As an application consider the second transvectant of the quartic a with its Hessian, (ab)a=A; Expand a member of the third polar of (ab)ab, viz. :- (ab)ažby=(ab) {a_b²}, += {(ab)²a²b²} (xy) + n-k+1) { (aa) ²a 9 + 10(ab)³ {a_bz}} (ay)² + (ab)*(xy)* ; the first and third terms vanish, and hence (ab)a2b² = {41} (xy)+(ab)*(xy)³. ² By a previous theorem if we transform from y to c and multiply by cz, {41} (ay) becomes c²(Ac)²4²=(4,4)². C Hence — (ab)(bc)³a²c₂=(A$,a£)ª+(ab)*c+, - (ab)(be) ³a²e₂={(ac) (be)³a²b₂ − (ab) (be)³a²c₂} = (be)³a₂ { (ac)b₂ − (ab)c₂} = (bc)³a;= (ab)*c₂. ... (4,²)² = (ab)*c+=i.f. n Summary of Results. We will now give a short account of the results to which the foregoing processes lead. Of any form a there exists a finite number of invariants and covariants, in terms of which all other covariants are rational and integral functions (cf. Gordan, Band ii. § 21). This finite number of forms is said to constitute the complete system. Of two or more binary forms there are also complete systems containing a finite number of forms. There are also algebraic systems, as above mentioned, involving fewer covariants which are such that all other covariants are rationally expressible in terms of them; but these smaller systems do not possess the same mathematical interest as those first mentioned. The Binary Quadratic.-The complete system consists of the form itself, a, and the discriminant, which is the second trans- vectant of the form upon itself, viz.: (f,f')² = (ab)²; or, in real coefficients, 2(aa-a). The first transvectant, (ff)¹=(ab)azbe, vanishes identically. Calling the discriminant D, the solution of the quadratic a=0 is given by the formula FORMS 301 If the covariant (f, p)¹ vanishes f and are clearly proportional and if the second transvectant of (f, p)¹ upon itself vanishes, f and possess a common linear factor; and the condition is both necessary and sufficient. In this case (f, p)¹ is a perfect square since its discriminant vanishes. If (f,)¹ be not a perfect square, and rz, Sz be its linear factors, it is possible to express f and in the canonical forms A₁ (ra) + A₂(Sz)², μ₁(ra)² + m₂(Sx) respectively. In fact, if f and have these forms, it is easy to verify that (f, p)¹=(u)(rs)rs. The fundamental system connected with quadratic forms consists of (i.) the n forms themselves fi,f2,...fm, (ii.) the (2) functional determinants (fi.f), (iii.) the (+¹) invariants (fifk), (iv.) the (3) forms (fi, fkfm))², each such form remaining unaltered for any permutations of i, k, m. Between these forms various relations exist (cf. Gordan, § 134). The Binary Cubic.-The complete system consists of n f=a,(f₁f)=(ab) ³a₂b₂=4, (,4)=(ab)²(ca)b₂c²=Q², If the form a be written as the product of its linear factors Palas the discriminant takes the form -(pq)². The vanishing of this invariant is the condition for equal roots. The simultaneous system of two quadratic forms a, say fand , consists of six forms, viz.: the two quadratic forms f, p; the two discriminants (ff), (p,p'), and the first and second transvectants of fupon , (,) and (f, o), which may be written (aa)azaz and (aa) These fundamental or ground forms are connected by the relation - 2{(f,)¹}=f(p,p')² - 2føƒ‚¢)² + ¢²(ƒ‚ƒ') ². and (A, A')=(ab)²(cd)(ad) (bc)=R. To prove that this system is complete we have to consider (S,A),(A, A')¹, (f, Q¹, Q², (S,Q)³, (A, Q)¹, (A, Q)², and each of these can be shown either to be zero or to be a rational integral function f, A, Q and R. These forms are connected by the relation 20² +4³+Rf2=0. The discriminant of f is equal to the discriminant of A, and is therefore (A, A')2=R; if it vanishes both fand A have two roots equal, A is a rational factor off and Q is a perfect cube; the cube root being equal, to a numerical factor près, to the square root of A. The Hessian A=A2 is such that (f, A)³=0, and if f is ex- pressible in the form (px)³+(q)³, that is as the sum of two perfect cubes, we find that A2 must be equal to Pxqx for then {(px)³ +μ(z)³, Pxqx}=0. Hence, if pz, qz be the linear factors of the Hessian A, the cubic can be put into the form X(pz)³ +μ(qz)³ and immediately solved. This method of solution fails when the discriminant R vanishes, for then the Hessian has equal roots, as also the cubic f. The Hessian in that case is a factor of f, and Q is the third power of the linear factor which occurs to the second power in f. If, moreover, A vanishes identically f is a perfect cube. The Binary Quartic.-The fundamental system consists of five forms af; (f.f)=(ab) ³ab²=4; (ff)=(ab)=i; (ƒ,4)¹ =(as)a³4³=(ab)²(cb)abc=t; (f,A)=(as)=(ab)²(bc)²(ca)² =j, viz., two invariants, two quartics, and a sextic. They are connected by the relation 21²=1ƒªa - 4³ - jƒ³. The discriminant, whose vanishing is the condition that f may possess two equal roots, has the expression -; it is nine times the discriminant of the cubic resolvent k³-ik-, and has also the expression 4(t, t')". The quartic has four equal roots, that is to say, is a perfect fourth power, when the Hessian vanishes identically; and conversely. This can be verified by equating to zero the five coefficients of the Hessian (ab)ab. Gordan has also shown that the vanishing of the Hessian of the binary ne is the necessary and sufficient condition to ensure the form being a perfect nth power. The vanishing of the invariants i and j is the necessary and sufficient condition to ensure the quartic having three equal roots. On the one hand, assuming the quartic to have the form 4,, we find i=j=0, and on the other hand, assuming i=j=0, we find that the quartic must have the form 01 a²= 1/(a +2²²2-2²₂ √P) ( 2²5 + 12²3 +2²₂ N a² + a2²2 +2²2 N-T). +40, which proves the proposition. The quartic will (² have two pairs of equal roots, that is, will be a perfect square, if it and its Hessian merely differ by a numerical factor. For it is easy to establish the formula (yx)=2f-f-2(f)² connecting the Hessian with the quartic and its first and second polars; now a, a root of f, is also a root of A, and consequently the first af af polar f must also vanish for the root a, and thence +y8x2 af a of and must also vanish for the same root; which proves dx₂ that a is a double root of f, and f therefore a perfect square. When f=6 it will be found that A= -f. The simplest form
