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

82computation (further explanations and comments just afterwards):(5.13) ⎧ {∑ m= [DX] m−1 · y l 1x X y l 1 · ∣∣ ∣ ∣ ∣Y k k1| · · · |Yy ∣ − m⎪⎨⎪⎩= [DX] m−1 ·l 1 =0−{ m∑m∑l 1 =0− · · · −l 1 =0y l 1y l 1x X y 1 ·m∑l 1 =0y∣ ∣∣∣Y ky l 1|Y k k ∣∣ ∣∣y2| · · · |Yy − my l 1x X y m ·∣∣∣Y k k1| · · · |Yyy m−1|Y ky l 1x X y l 1 · ∣∣ ∣ ∣ ∣Y k k1| · · · |Yy ∣ − X m y 1 · ∣∣ ∣ ∣Y ky}∣x |Y ky∣ k2| · · · |Yy ∣ − m−yx 1 X y 1 · ∣∣ ∣ Yky 1|Y ∣ k ky2| · · · |Yy ∣ − · · · −m−X y m · ∣∣ ∣ Yk ky1| · · · |Yy m−1|Y xk ∣ ∣ − ymx X y m · ∣∣ ∣ Yk ky1| · · · |Y= [DX] m−1 · {Xx · ∣∣ ∣ ∣ Yk ky1| · · · |Y ∣y − m Xy 1 · ∣∣ ∣ ∣ Ykx |Y k ky2| · · · |Y ∣y − · · · −m−X y m · ∣∣ ∣ Yky 1| · · · |Yy k m−1|Y xk ∣ ∣ }= [DX] m−1 · {∆(x|y 1 | · · · |y m ) } .For the passage to the equality of line 4, using the fact that a determinanthaving two identical columns vanishes, we observe that in each of the msums ∑ ml 1 =0appearing in lines 2 and 3 (including the cdots), there remainsonly two non-vanishing determinants. For the passage to the equality of line7, we just sum up all the linear combinations of determinants appearing inlines 4, 5 and 6. Finally, for the passage to the equality of line 9, we recognizethe development of the fundamental Jacobian determinant (3.2) alongits first line (X x , X y 1, . . .,X y m), modulo some permutations of columns inthe m × m minors. The proof is complete.Our second task, similar but computationnally more heavy, is to computethe determinant in the numerator of (5.8). First of all, we have to re-expressthe A k defined implicitely between (5.3) and (5.5) using the total differentiationoperator to contract them as follows(5.14)⎧⎪⎨⎪⎩A k = DX · Y kxx − DY k · X xx + 2+m∑m∑l 1 =1 l 2 =1m∑l 1 =1y m−1|Y k[]y l 1x · DX · Y kxy l 1− DY k · X xy l 1 +[]y l 1x y l 2x · DX · Y ky l 1y l 2− DY k · X y l 1y l 2 .y m ∣ ∣ ∣ ∣}