Euler's triangle and the decomposition of tensor powers of adjoint representation of A 1 Lie algebra
aa r X i v : . [ m a t h . G M ] F e b Euler’s triangle and the decomposition of tensor powers ofadjoint representation of A Lie algebra
A.M. Perelomov
Institute for Theoretical and Experimental Physics , , Moscow , Russia . Abstract
We consider the relation between Euler’s trinomial problem and the problem of decom-position of tensor powers of adjoint representation of A Lie algebra. By using thisapproach, some new results for both problems are obtained.
1. Introduction.
In 1765 Euler [Eu 1765/1767] investigated the coefficients of trinomial(1 + x + x ) n = n X k = − n a ( k ) n x n + k . (1)For central trinomial coefficients a (0) n he found the generating function and two term recur-rence relation. For discussion of properties a ( k ) n see [Ri 1974].Let us change x to exp( iθ ) and rewrite the left hand side of (1) as(1 + x + x ) n = x n X n , X = 1 + 2 cos θ . (2)Note that X is character of adjoint representation of Lie algebra A . So, Euler’s problemis equivalent to the problem of multiplicities of weights of representation X n . We consideralso related to this problem of decomposition of X n into irreducible representations of Liealgebra A .
2. Euler’s triangle.
It is evident that a ( − k ) n = a ( k ) n . So, it is sufficient to consider only quantities a ( k ) n , k ≥ n = 10: n/k
16 10 4 15
45 30 15 5 16
126 90 50 21 6 17
357 266 161 77 28 7 18 a ( k ) n +1 = a ( k − n + a ( k ) n + a ( k +1) n . (3)Let us introduce the generating function F ( t ) for central trinomial coefficient a n = a (0) n F ( t ) = ∞ X n =0 a n t n . (4) Theorem (Euler 1765) . The generating function F ( t ) has the form F ( t ) = 1 p (1 − t − t ) (5) and we have two term recurrence relation for coefficients a n n a n = (2 n − a n − + 3( n − a n − . (6)We give here a very short proof of the first statement different from Euler’s one. Proof.
Note that a n = 1 π Z π X n dθ, X = (1 + 2 cos θ ) . (7)So, F ( t ) = 1 π Z π dθ − t − t cos θ . (8)Evaluating this integral we obtain formula (5).The second statement is a special subcase of more general statement. Theorem 1.
We have two term recurrence relation for coefficients a ( k ) n ( n − k ) a ( k ) n = n (2 n − a ( k ) n − + 3 n ( n − a ( k ) n − . (9) Proof.
We have a ( k ) n = 1 π Z π X n cos kθ dθ, (10)and Z π X n (cid:20)(cid:18) d dθ + k (cid:19) cos kθ (cid:21) dθ = 0 = Z π cos kθ (cid:20)(cid:18) d dθ + k (cid:19) X n (cid:21) dθ. (11)But, d X n dθ = − n X n + n (2 n − X n − + 3 n ( n − X n − . (12)From this, it follows equation (9). Theorem 2.
We have another two term recurrence relations for coefficients a ( k ) n ( n + 1) (cid:16) a ( k − n − a ( k +1) n (cid:17) = k a ( k ) n +1 . (13)( n − k + 1) a ( k − n = ka ( k ) n + ( n + k + 1) a ( k +1) n , (14)( n − k + 1) a ( k ) n +1 = ( n + 1)( a ( k ) n + 2 a ( k +1) n ) , (15)( n + k + 1) a ( k ) n +1 = ( n + 1) ( a ( k ) n + 2 a ( k − n ) . (16)2 roof. From the identity Z π (cid:20) ddθ ( X n sin kθ ) (cid:21) dθ = 0 , (17)we obtain equation (13). Combining this relation with (3), we obtain equations (14)–(16).Note that from (3) it follows a (1) n = 12 ( a n +1 − a n ) , a (2) n = 12 ( a n +2 − a n +1 − a n ) ,a (3) n = 12 ( a n +3 − a n +2 + 2 a n ) , a (4) n = 12 ( a n +4 − a n +3 + 2 a n +2 + 4 a n +1 − a n ) . (18) Corollary 1.
Explicit expressions for quantities a ( n − k ) n for small k may be obtained from(9) and (14) and we have a ( n − k ) n = 1 k ! Q k ( n ) , (19)where Q k ( n ) is polynomial of degree k in n .The recurrence relation for these polynomials follows from (14) and we give the explicitexpression for first ten polynomials. Q k +1 ( n ) = ( n − k ) Q k ( n ) + k (2 n − k + 1) Q k − ( n ) . (20) Q = 1; Q = n ; Q = n ( n + 1); Q = ( n − n ( n + 4); (21) Q = ( n − n ( n + 7 n − Q = ( n − n − n ( n + 1)( n + 12); Q = ( n − n − n ( n + 18 n + 17 n − Q = ( n − n − n − n ( n + 27 n + 116 n − Q = ( n − n − n − n ( n + 1)( n + 10)( n + 23 n − Q = n ( n − n − n − n −
4) ( n + 46 n + 467 n + 86 n − Q = n ( n − n − n − n −
4) ( n + 55 n + 665 n − n − n + 15120) .
3. Decomposition of X n into irreducible representations. This problem is equivalent to expansion of X n in terms of characters of Lie algebra A . X n = n X k =0 b ( k ) n χ k ( θ ) . (22)These characters are well known (see for example [We 1939].) χ k = 1 + 2 cos( θ ) + 2 cos(2 θ ) + ... + 2 cos( kθ ) . (23)They are orthogonal 1 π Z π χ k ( θ ) χ l ( θ ) (1 − cos( θ )) dθ = δ k,l , (24)3nd we have b ( k ) n = 1 π Z π X n f k ( θ ) dθ , f k ( θ ) = cos( kθ ) − cos(( k + 1) θ ) . (25)From this it follows the basic relation b ( k ) n = a ( k ) n − a ( k +1) n , (26)the three term recurrence relation similar to relation (3) b ( k ) n +1 = b ( k − n + b ( k ) n + b ( k +1) n , for n ≥ , k ≥ , (27)and other relations b n = b (0) n = 12 (3 a n − a n +1 ) , b (1) n = b n +1 , b (2) n = b n +2 − b n +1 − b n ,b (3) n = b n +3 − b n +2 − b n +1 + b n , b (4) n = b n +4 − b n +3 + 3 b n − . (28)Here we have the triangle n/k
15 15 10 4 16
36 40 29 15 5 17
91 105 84 49 21 6 18
232 280 238 154 76 28 7 19
603 750 672 468 258 111 36 8 110
Theorem 3.
The generating function G ( t ) = P ∞ n =0 b n t n here has the form. G ( t ) = 12 t − √ − t p (1 + t ) ! . (29) Proof.
Take into account identity1 − cos( θ )1 − t − t cos( θ ) = 12 t (1 − − t − t cos( θ ) ) . (30)Then the proof is reduced to the proof for F ( t ) . We have also the recurrence relation thatfollows from (6) and b n = a n − a (1) n ( n + 1) b n = ( n − b n − + 3 b n − ) . (31) Theorem 4.
We have here four term recurrence relation A nk b ( k ) n + B nk b ( k ) n − + C nk b ( k ) n − + D nk b ( k ) n − + E nk b ( k ) n − = 0 , (32)4here A n,k = ( n − ( k + 1) )( n − k ); B n,k = − n (2 n − n + k )( n − k − C n,k = − n ( n − n − n + 3 − k ( k + 1)); (34) D n,k = 6 n ( n − n − n − E n,k = 9 n ( n − n − n − . (35) Proof.
We have b ( k ) n = 1 π Z π X n f k ( θ ) dθ (36)where X = (1 + 2 cos( θ )) , f k ( θ ) = cos( kθ ) − cos(( k + 1) θ ) (37)and A k f k ( θ ) = 0 , A k = ( d dθ + k )( d dθ + ( k + 1) ) (38)Integrating by part in (36) we get1 π Z π f k ( θ )( A k X n ) dθ = 0 and (32) − − (35) . (39) Theorem 5.
We have here three term recurrence relation ( k + 1)( n + 1 − k ) b ( k − n = ( k ( k + 1) − n − b ( k ) n + k ( n + k + 2) b ( k +1) n . (40) Proof.
This folows from (14) and the relation b ( k ) n = a ( k ) n − a ( k +1) n References [Eu 1765/67] Euler L.:
Observationes analyticae , Novi Comm. Acad. Sci. Petropolitanae , 124-143 (1765/1767)[Ri 1974] Riordan J.: An Introduction to Combinatorial Analysis , Reidel (1974)[We 1939] Weyl H.: