Báo cáo toán học: "Permutations with Kazhdan-Lusztig polynomial Pid,w (q) = 1 + q h" pptx

... follows that H u,π 2 (q) = P u,w (q) + q h 2 E u (q) . Since H u,π 2 (q) − P u,w (q) has nonnegative coefficients and deg P u,w (q) ≤ (h − 1) /2 < h − 1, P u,w (q) = 1 + · · · + q s 1 4 For tho se readers ... w]: I:  (y + 1) z · · · 1( y + z + 2) · · · (y + 2), (y + z + 2)(y + 1) y · · · 2(y + z + 1)...

Ngày tải lên: 08/08/2014, 01:20

... 2, 1, −2, 1, 2, 1, −2 F 1, −2, −2, 1 , 2, 2, 1 D 1, −2, 1, 2, 2, 1, −2 P 2, 1, − 2, 1, 2, 1, −2 I 1, −2, 1, 2 , 1, −2, 1 E 2, 1, −2, −2, 1, 2, 1 R 2, 1, −2, 1, 2, 1, −2 K 1, −2, 1, 2 ... 2 , 1, −2, 1 H 2, 1, −2, −2, 1 , 2, 2 C 2, 1, −2, 1, 2, 2, 1 M 1, −2, 1, 2, 1, −2, 1 J 2, 1 , −2, 1, 2, 1, −2 G 2, 2, 1, −2, −2, 1, 2 O 1, −2, 1, 2 , 1, −2, 1 L 2, 1, −2,...

Ngày tải lên: 08/08/2014, 11:20

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Ngày tải lên: 05/08/2014, 15:20

... class="bi x0 y0 w1 h1" alt=""

Ngày tải lên: 05/08/2014, 15:21

... properties of known polynomial projectors, and in the 252 Dinh Dung that µ j (1) = 1, j =1 , 2 and µ j (u α )=0 , 1 ≤|α|≤d, j =1 , 2. Fix two multi-indices α 1 ,α 2 with |α 1 | = |α 2 | = k − 1. W e have D α j µ j (u β )= α j β ,j =1 , ... equations (HPDE) of degree k if for every f ∈ H(C n ) and every homogeneous polynomial of degree k, q( z )=  |α|=k...

Ngày tải lên: 06/08/2014, 05:20

... we have (3 .11 ) = 4 cos 2 θ  α,k,l (B 2n+l α,n+k B 3n+l αk − B 3n+l α,n+k B 2n+l αk ) − 2cos 2 θ  α,β B(e α ,e β ) 2 ≤ 2cos 2 θ  α,k,l {(B 2n+l α,n+k ) 2 +( B 3n+l αk ) 2 +( B 3n+l α,n+k ) 2 +( B 2n+l αk ) 2 } − ... ∇ e β B(e α ,e n+k ),e 2n+k  = ∇ ⊥ e β (B(e α ,e n+k ),e 2n+k  + B(e α ,e n+k ), ∇ ⊥ e β e 2n+k  = (∇ e β B)(e α ,e n+k )+B(∇ e β e α ,e n+k...

Ngày tải lên: 06/08/2014, 05:20

... t 0 (n)+t 1 (n )+ + t n (n) = a n + a n 1 + b n−2 + + b 0 Rewriting Lemmas 8 and 9 we obtain a n +2 a n 1 +3 b n−2 +4 b n−3 + +( n +1 ) b 0 =  2n n  a n 1 + b n−2 + b n−3 + + b 0 =  2n − 2 n − 1  the ... are then PQR ∼ 312 , PQR ∼ 213 , PQR ∼ 13 2, PQR ∼ 2 31 respectively. We also adopt this language for sets of 4 or mor...

Ngày tải lên: 07/08/2014, 06:22

... from w C (q) by (iii), then w C (q )= w 1 R j = w 1 RL j = u, so q = w 1 C (w 1 )(k − j +1 ) (k − j +2 ) k, w 1 C (u)=w 1 C (w 1 )(k − j +2 )(k − j +3 ) (k +1 ) (k − j +1 ) , and to get q from w 1 C (u)wejustremovek ... ( |q m | 1) that agrees with s n ( {13 2, 213 ,q 1 ,q 2 , ,q m }) for all n ≥ ( |q 1 | 1) ( |q 2 | 1) ( |q m | 1...

Ngày tải lên: 07/08/2014, 07:21

... 6 Proof. In (16 ) set x 1 = xq λ 1 and x i = q λ i for all i ≥ 2toobtain  π∈S (12 43, 214 3) q f(π) x |π| =1 + xq λ 1 1 − xq λ 1 − xq λ 1 + 2 1 − xq λ 1 + 2 − xq λ 1 +2 λ 2 + 3 1 − xq λ 1 +2 λ 2 + 3 − ... prove ( 21) , we apply (16 ) with x i =  j 1 q A ij j for all i ≥ 1. We have C BA (q )= 1+ x 1 1 − x 1 −...

Ngày tải lên: 07/08/2014, 07:21

... derivative with respect to a of this equation, that is of x(−(t +1 ) a t +2 a +1 ) +( (t +2 )a t +1 − 4a +1 )=0 . Eliminating x between these two equations gives: a t +1 (a t +1 − ta 2 +( t −3)a +2 )− 3a 2 +4 a − 1= 0 . the ... still further through multiplication by g 1 yielding 0=g t+2 t − xg t +1 t +( x − 2)g 2 t +( x +1 ) g t − x = x(−g t...

Ngày tải lên: 07/08/2014, 07:21