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

350(5.4)+++++m∑l 1 ,l 2 =1m∑l 1 ,l 2 ,l 3 =1m∑l 1 ,l 2 =1m∑l 1 =1m∑l 1 ,l 2 =1n∑k 1 ,k 2 ,k 3 =1n∑k 1 ,k 2 ,k 3 ,k 4 =1n∑k 1 ,k 2 ,k 3 ,k 4 =1n∑k 1 ,k 2 ,k 3 =1n∑k 1 ,k 2 ,k 3 ,k 4 =1[δ k 1,k 2 ,k 3i 1 , i 2 , i 3Y jy l 1y l 2 + δk 3,k 1 ,k 2i 1 , i 2 , i 3Y jy l 1y l 2 + δk 2,k 3 ,k 1i 1 , i 2 , i 3Y jy l 1y l 2 −−δ j l 1δ k 2,k 3i 1 , i 2X k 1x i 3y l 2 − δj l 1δ k 2,k 3i 1 , i 3X k 1x i 2y l 2 − δj l 1δ k 2,k 3i 2 , i 3X k 1x i 1y l 2 −−δ j l 2δ k 3,k 1i 1 , i 2X k 2x i 3y l 1 − δj l 2δ k 3,k 1i 1 , i 3X k 2x i 2y l 1 − δj l 2δ k 3,k 1i 2 , i 3X k 2x i 1y l 1 −−δ j l 2δ k 1,k 2i 1 , i 2X k 3x i 3y l 1 − δj l 2δ k 1,k 2i 1 , i 3X k 3x i 2y l 1 − δj l 2δ k 1,k 2i 2 , i 3X k 3x i 1y l 1]y l 1k1y l 2k2 ,k 3+[−δ j l 3δ k 1,k 2 ,k 3i 1 , i 2 , i 3X k 4y l 1y l 2 − δj l 3δ k 2,k 3 ,k 1i 1 , i 2 , i 3X k 4y l 1y l 2 − δj l 3δ k 3,k 2 ,k 1i 1 , i 2 , i 3X k 4y l 1y l 2 −−δ j l 2δ k 3,k 4 ,k 1i 1 , i 2 , i 3X k 2y l 1y l 3 − δj l 2δ k 3,k 1 ,k 4i 1 , i 2 , i 3X k 2y l 1y l 3 −−δ j l 2δ k 1,k 3 ,k 4i 1 , i 2 , i 3X k 2y l 1y l 3[−δ j l 2δ k 1,k 2 ,k 3i 1 , i 2 , i 3X k 3y − l 1 δj l 2δ k 2,k 4 ,k 1i 1 , i 2 , i 3X k 3y − l 1 δj l 2δ k 4,k 1 ,k 2i 1 , i 2 , i 3X k 3y l 1]y l 1k1y l 2k2y l 3k3 ,k 4+]y l 1k1 ,k 2y l 2k3 ,k 4+[]δ k 1,k 2 ,k 3i 1 , i 2 ,ß 3Y jy − l 1 δj l 1δ k 1,k 2i 1 , i 2X k 3x − i 3 δj l 1δ k 1,k 2i 1 , i 3X k 3x − i 2 δj l 1δ k 1,k 2i 2 , i 3X k 3y l x i 11 k1 ,k 2 ,k 3+[−δ j l 2δ k 1,k 2 ,k 3i 1 , i 2 , i 3X k 4y l 1 − δj l 2δ k 4,k 1 ,k 2i 1 , i 2 , i 3X k 3y l 1 − δj l 2δ k 3,k 4 ,k 1i 1 , i 2 , i 3X k 2y l 1 −]−δ j l 1δ k 2,k 3 ,k 4i 1 , i 2 , i 3X k 1y l 1y l 2 k1y l 2k2 ,k 3 ,k 4.5.5. Final synthesis. To obtain the general formula for Y j i 1 ,...,i κ, we have toachieve the synthesis between the two formulas (3.74) and (4.19). We startwith (3.74) and we modify it until we reach the final formula for Y j i 1 ,...,i κ.We have to add the µ 1 +· · ·+µ d sums ∑ ml 1:1 =1 · · ·∑ml 1:µ1 =1 · · · · · ·∑ml d:1 =1 · · ·∑ml d:µd =1 ,together with various indices l α . About these indices, the only point whichis not obvious may be analyzed as follows.A permutation σ ∈ F (µ 1,λ 1 ),...,(µ d ,λ d )µ 1 λ 1 +···+µ d λ dyields the list:(5.6)σ(1:1:1), . . ., σ(1:1:λ 1 ), . . .σ(1:µ 1 :1), . . .,σ(1:µ 1 :λ 1 ), . . .. . .,σ(d:1:1), . . ., σ(1:1:λ d ), . . .σ(d:µ d :1), . . .,σ(d:µ d :λ d ),In the sixth line of (3.74), the last term σ(d : µ d : λ d ) of the above listappears as the subscript of the upper index k σ(d:µd :λ d ) of the term X k σ(d:µ d :λ d) .According to the formal rules of Theorem 4.19, we have to multiply thepartial derivative of X k σ(d:µ d :λ d) by a certain Kronecker symbol δ j l α. Thequestion is: what is the subscript α and how to denote it?