Travaux sur les symétries de Lie des équations aux ... - DMA - Ens

Our first task is to compute the determinant in the denominator of (5.8).Recalling that we make the notational identification y 0 ≡ x, it will be convenientto reexpress the B k l 1in a slightly compacter form, using the totaldifferentiation operator D:(5.9) ⎧⎪⎨⎪⎩B k l 1= −X y l 1 Y kx + Y ky l 1X x +m∑l 2 =1= Y ky l 1· DX − X y l 1 · DY k .[]y l 2x · −X y l 1 Y ky l 2+ Y ky l 1X y l 2 =81Lemma 5.10. We have the following expression for the determinant of thematrix (B k l 1) 1km1l 1 m :(5.11){ ∣ ∣ ∣ ∣Y ky 1 · DX − X y 1 · DY k | · · · |Y ky m · DX − X y m · DY k∣ ∣ ∣ ∣ == [DX] m−1 · ∆ ( x|y 1 | · · · |y m) .Proof. By multilinearity, we may develope the determinant written in thefirst line of (5.11). Since it contains two terms in each columns, we shouldobtain a sum of 2 m determinants. However, since the obtained determinantsvanish as soon as the column DY k (multiplied by various factors X y l) appearsat least two different places, it remains only (m + 1) nonvanishingdeterminants, those for which the column DY k appears at most once:(5.12)⎧∣ ∣ Yk⎪⎨ y · DX − X 1 y 1 · DY k | · · · |Yy k · DX − X m y m · DY k∣ ∣ == [DX] m · ∣∣ ∣ ∣ ∣Y k ky1| · · · |Yy ∣ − [DX] m−1 X⎪⎩m y 1 · ∣∣ ∣ ∣ ∣DY k |Y k ky2| · · · |Yy − · · · −m− [DX] m−1 X y m · ∣∣ ∣ Yk ky1| · · · |Yy m−1|DY k∣ ∣ .To establish the desired expression appearing in the second line of (5.11),we factor out by [DX] m−1 and we develope all the remaining total differentiationoperators D. Since y 0 ≡ x, we have yx 0 = 1, and this enables us tocontract X x + ∑ ml 1y l 1 x X y l 1 as ∑ ml 1 =0 yl 1 x X y l 1 . So, we achieve the following