Each term on the right-hand side may be shown by permutation of a, b, c to be the symbolical representation of the same covariant; they are equivalent symbolic products, and we may accordingly write
2(ac) (bc)ai -1 bi -1 cx 2 =(ab)2a:-2b:-2c:,
a relation which shows that the form on the left is the product of the two covariants
n (ab) ay 2 by 2 and cZ.
The identities are, in particular, of service in reducing symbolic products to standard forms. A symbolical expression may be always so transformed that the power of any determinant factor (ab) is even. For we may in any product interchange a and b without altering its signification; therefore
(ab) 2m+1 4) 1 = - (ab) 2 " 4)2,
where 4,1 becomes by the interchange, and hence
(ab)2m+14)1= Z (ab) 2m+1 (4) 1 - 02);
and identity (I.) will always result in transforming 01-02 so as to make it divisible by (ab).
Ex. gr.
(ab)(ac)bxcx = - (ab)(bc)axcx = 2(ab)c x {(ac)bx-(bc)axi = 1(ab)2ci;
so that the covariant of the quadratic on the left is half the product of the quadratic itself and its only invariant. To obtain the corresponding theorem concerning the general form of even order we multiply throughout by (ab)2' 2c272 and obtain (ab)2m-1(ac)bxc2:^1=(ab)2mc2
Paying attention merely to the determinant factors there is no form with one factor since (ab) vanishes identically. For two factors the standard form is (ab) 2; for three factors (ab) 2 (ac); for four factors (ab) 4 and (ab) 2 (cd) 2; for five factors (ab) 4 (ac) and (ab) 2 (ac)(de) 2; for six factors (ab) 6, (ab) 2 (bc) 2 (ca) 2 , and (ab) 2 (cd) 2 (ef) 2 . It will be a useful exercise for the reader to interpret the corresponding covariants of the general quantic, to show that some of them are simple powers or products of other covariants of lower degrees and order.
The Polar Process.—The �th polar of ax with regard to y is
n-� a aye i.e. of the symbolic factors of the form are replaced by IA others in which new variables y1, y2 replace the old variables x1, x 2 . The operation of taking the polar results in a symbolic product, and the repetition of the process in regard to new cogredient sets of variables results in symbolic forms. It is therefore an invariant process. All the forms obtained are invariants in regard to linear transformations, in accordance with the same scheme of substitutions, of the several sets of variables.
An important associated operation is a ? 32 ax l ay 2 ax2ay1' which, operating upon any polar, causes it to vanish. Moreover, its operation upon any invariant form produces an invariant form. Every symbolic product, involving several sets of cogredient variables, can be exhibited as a sum of terms, each of which is a polar multiplied by a product of powers of the determinant factors ( xy), (xz), (yz),... Transvection. - We have seen that (ab) is a simultaneous invariant of the two different linear forms a x, bx, and we observe that (ab) is equivalent to where f =a x, 4)=b. If f =ay, 4 = b' be any two binary forms, we generalize by forming the function (m-k)! (n-k)! of a4) of a 4) k m! l ax 2 2 ax i l This is called the kth transvectant of f over 4); it may be conveniently denoted by (f, (15)k. (a m b n) k (ab) kamkbn-k x, x - x it is clear that the k th transvectant is a simultaneous covariant of the two forms.
It has been shown by Gordan that every symbolic product is expressible as a sum of transvectants.
If m > n there are n +1 transvectants corresponding to the values o, t, 2,... n of k; if k = o we have the product of the two forms, and for all values of k>n the transvectants vanish. In general we may have any two forms 01/1X1+ 'II � Yy + 02x2) p Y'x =, / / being the umbrae, as usual, and for the kth transvectant we have (4)1,,, 4)Q) k = (4)) k 4)2 -krk, a simultaneous covariant of the two forms. We may suppose of, 4 ,2 to be any two covariants appertaining to a system, and the process of transvection supplies a means of proceeding from them to other covariants.
The two forms ax, bx, or of, 0, may be identical; we then have the kth transvectant of a form over itself which may, or may not, vanish identically; and, in the latter case, is a covariant of the single form. It is obvious that, when k is uneven, the kth transvectant of a form over itself does vanish. We have seen that transvection is equivalent to the performance of partial differential operations upon the two forms, but, practically, we may regard the process as merely substituting (ab) k, (OW for azbx, 4x t ' respectively in the symbolic product subjected to transvection. It is essentially an operation performed upon the product of �two forms. If, then, we require the transvectants of the two forms f+Xf', 0+14', we take their product fc5+xf'95+,-ifct'+atif'cb', and the kth transvectant is simply obtained by operating upon each term separately, viz.
(f, 4)) k +(f, 4)) k +�(f, 4/) k +a�(1, 4)')k; and, moreover, if we require to find the kth transvectant of one linear system of forms over another we have merely to multiply the two systems, and take the k th transvectant of the separate products.
The process of transvection is connected with the operations 12; for ?k (a m b n) = (ab)kam-kbn-k, (x y x y or S 2 k (a x by) x = 4))k; so also is the polar process, for since f k m-k k k n - k k y = a x by, 4)y = bx by, if we take the k th transvectant of f i x; over 4 k, regarding y,, y 2 as the variables, (f k, 4)y) k (ab) ka x -kb k (f, 15)k; or the k th transvectant of the k th polars, in regard to y, is equal to the kth transvectant of the forms. Moreover, the kth transvectant (ab) k a m-k b: -k is derivable from the kth polar of ax, viz. ai by substituting for y 1, y 2 the cogredient quantities b2,-b1, and multiplying by by-k.
First and Second Transvectants.- A few words must be said about the first two transvectants as they are of exceptional Interest. Since, If F = An, 4) = By, 1 = I
(Df A4) Of A?) Ab A"'^1Bz 1=, (F, Mn Ax I Ax 2 Axe Ax1) J
The First Transvectant Differs But By A Numerical Factor From The Jacobian Or Functional Determinant, Of The Two Forms. We Can Find An Expression For The First Transvectant Of (F, �) 1 Over Another Form Cp. For (M N)(F,4)), =Nf.4Y Mfy.4), And F,4, F 5.4)= (Axby A Y B X) A X B X 1= (Xy)(F,4))1; (F,Ct)1=F5.D' 7,(Xy)(F4)1. Put M 1 For M, N I For N, And Multiply Through By (Ab); Then { (F ,C6) } = (Ab) A X 2A Y B X 1 M N I 2 (Xy) ,?) 2, = (A B)Ax 1B X 2B Y L I Multiply By Cp 1 And For Y L, Y2 Write C 2, C1;
Then The Right Hand Side Becomes
(Ab)(Bc)Am Lbn 2Cp 1 M I C P (F?) 2 M { N2 X, Of Which The First Term, Writing C P = ,,T, Is Mn 2 A B (Ab)(Bc)Axcx 1 M 2 N 2 P 2 2222 2 2 _2 A X B X C (Bc) A C Bx M N 2 2 2 M2°N 2 N 2 M 2 2 A X (Bc) B C P C P (Ab) A B B(Ac) Ax Cp 2 = 2 (04) 2 1 (F,0) 2.4 (F,Y') 2 �?;
And, If
(F,4)) 1 = Km " 2, (F??) 1 1 M N S X X X Af A _Af A Ax, Ax Ax Ax1
