Garnir Polynomial and their Properties

Transcription

1 Univrsity of Cliforni, Dvis Dprtmnt of Mthmtis Grnir Polynomil n thir Proprtis Author: Yu Wng Suprvisor: Prof. Gorsky Eugny My 8, 07

2 Grnir Polynomil n thir Proprtis Yu Wng mil: uywng@uvis.u. In this ppr, w r stuying th polynomils in n vrils whih hv nily unr prmuttion of vrils. Th stuy in n vrils polynomils whih oul wily ppli in mny rs inluing rprsnttion thory, lgr n physis. W strt with som si notions lik symmtri polynomils whih o not hng th vlu unr prmuttions (for xmpl: x + x ), n nti-symmtri polynomils whih only hng th sign of th vlu unr prmuttions (for xmpl: x x ) to illustrt how prmuttions work s wll s som othr proprtis of ths polynomils. Thn w will fous on mor intrsting xmpls lik Grnir polynomils whih n xprss in trms of Young igrms. Grnir polynomils r polynomils of th form (x i x j )... (x s x t ), thy hv mny intrsting proprtis. In this ppr, w minly r out Grnir polynomils whih only involv four vrils t tim, n stuy th imnsion of thir spn. Symmtri polynomils Symmtri polynomils hv th sm vlu t ny prmuttions. Dfinition.0.. A polynomil f(x,..., x n ) is symmtri polynomil if for ll prmuttions φ, f(x,...x n ) = f(x φ(),...x φ(n) ).

3 W fin th polynomils of k s follows: k (x, x,..., x n ) = j j...j k n x j...x jk For k from to n, w n fin th following lmntry symmtri polynomils: (x, x,..., x n ) = x j j n (x, x,..., x n ) = x j x k... jk n n (x, x,..., x n ) = x x...x n Clrly, k is symmtri for ll k n n. Thorm.0.. Evry symmtri funtion in x... x n is polynomil in th lmntry symmtri polynomils k. Proof. W first fin n orr on monomils tht x x...x n n x x...x n n if n n ; or n = n n n n ; or n = n, n = n n n n ; n so on... For ny s = x x...x n n n t = x x...x n n, s n t r lwys omprl, so ithr s = t, or s > t or s t. For givn symmtri polynomil, w n lwys fin mximl monomil x j x j...x jn n, n w n lwys fin prout of lmntry symmtri polynomils k suh tht x j x j...xn jn is lso th mximl in it. For xmpl, givn x x x is th mximl lmnt, w n onstrut polynomil y multiplying (x, x, x ) (x, x, x ) (x, x, x ) (x, x, x )=(x x x ) (x x + x x + x x ) (x + x + x ). Multiplying th ling trm in h lmntry symmtri polynomil, w gt (x x x ) (x x ) (x ) = x x x, whih givs th mximl lmnt s w sir.

4 For symmtri polynomil f(x,..., x n ), if x x...x n n is th ling trm, w know tht... n. Sin f is symmtri, w n lwys mk this inqulity hol. If w sutrt this prout from th originl symmtri polynomil, it is ovious tht th mximl lmnt in th nw polynomil is stritly smllr thn th mximl lmnt in th ol on. Sin th numr of lmnts smllr thn th mximl lmnt is finit, y sutrting ths polynomils gin n gin, th originl polynomil will vntully om 0 in finit numr of stps. Hn, ll symmtri polynomils r polynomils in k. Antisymmtri polynomils Antisymmtri polynomils hng th sign n kp th vlu t ny prmuttions. In othr wors:f(x,..., x i,..., x j,..., x n )= f(x,..., x j,..., x i,..., x n ). Thorm.0.. W = ij(x i x j ) is ntisymmtri. Proof. Lt f(x,..., x s,..., x t,..., x n )= ij(x i x j ) n w prmut x s n x t. Dnot th nw polynomil y f(x,..., x t,..., x s,..., x n ). W n fin out tht th two polynomils hv som ommon ftors whih o not involv x s, x t, n thos whih hv on of thm ut th othr vril is for oth of x s, x t or ftr oth of thm. Dnot thm s C = (x i x j ) n i,j s,t C = (x m x n ), C = (x m x n ) Rpl th ommon ftors n w gt m=s or t,n>s,t n=s or t,ms,t f(x,..., x s,..., x t,..., x n )=C C C [(x s x s+ )...(x s x t )][(x s+ x t )...(x t x t )](x s x t ) n f(x,..., x t,..., x s,..., x n )=C C C [(x t x s+ )...(x t x t )][(x s+ x s )...(x t x s )](x t x s ). Thr r xtly (t s + ) ftors in th first rkt n son rkt, n thy r ngtiv vlu of h othr, this lvs us th lst ftor (x s x t ) n (x s x t ). So,f(x,..., x s,..., x t,..., x n )=( ) (t s+)+ f(x,..., x t,..., x s,..., x n ). It givs f(x,..., x s,..., x t,..., x n )= f(x,..., x t,..., x s,..., x n ) s w sign to hv.

5 Thorm.0.. Evry ntisymmtri polynomil f n writtn s f(x, x,..., x n ) = W g(x, x,..., x n ) whr W is fin in Thorm.0. n g is symmtri. Proof. First, w wnt to prov tht f is ivisil y (x i x j ) for ll i j. Assum tht x i = x j, thn f(x,..., x i,..., x j,..., x n ) = f(x,..., x j,..., x i,..., x n ). In ition f is ntisymmtri, w hv: f(x,..., x i,..., x j,..., x n ) = f(x,..., x j,..., x i,..., x n ). Comining ov two qutions givs f(x,..., x i,..., x i,..., x n ) = 0. So f is ivisil y (x i x j ) n thrfor ivisil y W. W rwrit th qution s f(x,x,...,x n) W = g(x, x,..., x n ), sin oth f(x, x,..., x n ) n W r ntisymmtri, w know tht g(x, x,..., x n ) os not hng th vlu unr prmuttions. Hn g is symmtri. Young Digrms A Young igrm is omintoril ojt usful in rprsnttion thory. A Young igrm is sust of (Z + ), suh tht if (i, j) is ontin in it, thn ll (, ) r ontin for i, j. (,) As w n s, sin (,) is ontin in this Young igrm, thn (, ) is lso ontin for,. Dfinition.0.. A Young tlu is wy to fill th igrm with numrs from to n (n is th numr of oxs). Dfinition.0.. A stnr Young tlu is Young tlu suh tht th numrs in th sm row n olumn r in inrsing orr from ottom to top n lft to right.

6 Th following is n xmpl of Young igrm (,,) n on of its Young tlu n on of its stnr Young tlu Young igrm Young tlu stnr Young tlu Grnir Polynomils Suppos tht T is stnr Young tlu. Dfinition.0.. W fin Grnir polynomil y th qution G T = In th following s, G(,,, ) = (x x )(x x ) i ov j (x i x j ). In th following thorm, w only onsir Young igrms with rows of lngth (n, ). Thorm.0.. If w prmut th vrils G(,,, ) G(,,, ) linr spn V n of ll G(,,, ) is prsrv y prmuttions. (so V n is rprsnttion of S n ). Thorm.0.. Th sis in V n is givn y G(,,, ) suh tht th orrsponing Young tlux r stnr. (so,,, n w n hv svrl s) Thorm.0.. Th numr of stnr Young tlux is qul to n(n ). Thorm.0. is lr. Lt s prov Thorm.0.: 5

7 Proof. In orr to prov Thorm.0., w wnt to prov tht th Grnir polynomils w gnrt from ny four numrs in th first y squr n writtn s th spn of Grnir polynomils whih orrspon to stnr Young tlux. W hv th following quivln rltions:.exhng numrs in th sm olumn is ntisymmtri. > G T = G T T T.Exhng th ntir olumn is symmtri. > G T = G T T T By using ths two rltions, w oul t lst nsur, n in th following Young tlu.? If, thn this Young tlu is stnr; w n to isuss th s >. W onstrut rltion s follow: + = 0 (x x )(x x ) + (x x )(x x ) (x x )(x x ) = 0 Sin w know tht > > >, on n hk tht th son n th thir Young tlux r stnr. From th prvious stps, w oul lwys mk > > n > >. 6

8 ? Lst, w n to hk whthr. If, it is stnr, so w n onsir th s tht. W onstrut rltion s follow: + = 0 (x x )(x x ) (x x )(x x ) + (x x )(x x ) = 0 Sin w know tht > >, > > >, on n hk tht th son n th thir Young tlux r stnr. Hn, Thorm.0. is prov. Nxt, w wnt to prov Thorm.0.: Proof. W will prov it y inution. First, w hk th s s tht only hs vrils. As shown low, thr r stnr Young tlux of vrils. Assum thr r n(n ) stnr Young tlux for n vrils. W now hk th numr of stnr Young tlux of (n + ) vrils. It is tru tht th iggst numr (n+) n only pl on th rightmost lnk, so w ivi it into two ss. 7

9 x n + n + s s For s, th numr of stnr Young tlux r th sm s th s for n vrils, whih is n(n ). For s, th smllst numr is finitly fill in th lft ottom ornr, n w only n to hoos th numr for x. As long s numr x is hosn, th rmining numrs will utomtilly fill in th rmining lnks in lin in n inrsing orr. Hn, it is (n ) hois. Sum up ths two ss, n(n ) + (n ) = (n+)(n ). W gt (n+)(n ) stnr Young tlux s w sir to hv. 5 Hook Lngth Bfor w introu Hook Lngth formul, w fin λ (i, j) n l λ (i, j) s follow: Suppos tht (i, j) is th ox in Young igrm, λ (i, j)=numr of oxs to th right of (i, j); l λ (i, j)=numr of oxs ov (i, j). Dfinition Th Hook Lngth formul for th Young tlu is xprss y: H λ = n! hλ (i,j), whr h λ(i, j) = λ (i, j) + l λ (i, j) +. Exmpl For th following Young igrm (6,,,), w hv th h λ (i, j) in th oxs s follow: 8

10 Thn th Hook Lngth formul H λ = n! hλ (i,j) 6! = =50050 Thorm Th numr of stnr Young tlux of shp λ is qul to H λ. This thorm is known in gnrl, hr w prov it for ny two-row Young igrm. Proof. Consir Young igrm with row lngths k n l (k l).in th Pi 5., n = k +l, th prout of hook lngth for th first row is h λ (, j) = k! n th prout of th hook lngth for th son row is h λ (, j) = (l+)!. By lultion, l k+ H λ = (k + l)! (l k + ) k!(l + )!. Lt im(k, l) th numr of stnr Young tlux with row lngths k n l. Knowing tht th lrgst numr n only fill in th rightmost oorint for stnr Young tlux, w n gt th rltion tht: im(k, l) = im(k, l) + im(k, l ). Chk th rltion y rpling y Hook Lngth formul, Right Hn Si= (k+l)! (l k+) k!(l+)! Lft Hn Si= (k+l )! (l k+) + (k+l )! (l k) (k )!(l+)! k!(l)! Now hking RHS-LHS, first w multiply h ftor y (k )!l! (k+l )!, thn (k )!l! (k+l)(l k+) (RHS-LHS)= l k+ l k = 0 (k+l )! k(l+) l+ k Sin (k )!l! (k+l )! 0, w hv RHS=LHS. Lst, w hk th s s k = : RHS= (l+)! (l +) (l+)! = l 9

11 LHS= (+l )! (l +) + (+l )! (l ) 0!(l+)!!(l)! = (l+)! + l! (l ) (l+)! (l)! = + (l ) = l It lso givs RHS=LHS, whih shows this rltion is orrt. k l + l-k+ Pi 5. Corollry For th Young igrm with row lngth (l, l,..., l n ),(l i > l j for i j), th numr of stnr Young tlux is y pplying th Hook Lngth Formul. ( n l i )! i l j + j i) i= ij(l n ((l i + n i)!) i= On n pply Hook Lngth Formul to prov th ov orollry. 6 Applition n Rlt Qustions Hr is qustion tht mny of us r fmilir. Suppos thr is y squr n w not th top-lft ornr A, ottom-right ornr B, () how mny wys to wlk from A to B? () How mny wys to wlk from A to B suh tht th whol pth is unr th igonl AB? A B For qustion(), w hv th following 6 wys: 0

12 () () () () (5) (6) For qustion (), w hv wys, thy r () n (5) in qustion (). How is this prolm rlt to Young tlux? Consir th y Young igrm, fill th numrs,,,. Thn w fin h numr from to, if th numr is in th top row, w mov right, n if th numr is in th ottom row, w mov own. Sin w r moving stps right n stps own, w will vntully gos to B. Following r th orrsponing Young tlux to qustion (). (Thr r mor thn on orrsponing Young tlu to h pth in (), ut w rrng th orr, so tht h row is in inrsing orr from lft to right. () () () () (5) (6) W n fin tht (),(5) is stnr.

13 Thorm For n n y n squr, th pth from top-lft ornr to ottom-right ornr unr th igonl is orrsponing to th y n stnr Young tlu. Proof. W not h vrtx (, ) for is th numr of oxs to th lft n is th numr of oxs ov it. Hn, w hv top-lft vrtx s (0,0), n ottom-right vrtx s (n, n), whil moving from (0,0) to (n, n) tks n stps right n n stps own, totlly n stps. As w wnt th whol pth unr th igonl, th vrtis (, ) w pss shoul hv th proprty tht. In th stnr Young tlux, in th sm olumn, th top numr is lwys lrgr thn th ottom numr. So th numr of stps w mov right nnot lrgr thn th numr of stps w mov own, this is quivlnt to.