Lý thuyết nhóm và ứng dụng trong vật lý lượng tử

8/2/2019 L thuyt nhm v ng dng trong vt l lng t

1/127

L thuyt nhm v ng dng trong vt l

lng t

Bi:GS. VS. Nguyn Vn Hiu


2/127


3/127

L thuyt nhm v ng dng trong vt llng t

Bi:GS. VS. Nguyn Vn Hiu

Phin bn trc tuyn:< http://voer.edu.vn/content/col10429/1.2/ >

Th vin Hc liu M Vit Nam


4/127

Ti liu ny v vic bin tp ni dung c bn quyn thuc v GS. VS. Nguyn Vn Hiu. Ti liu ny tun th theogiy php Creative Commons Attribution 3.0 (http://creativecommons.org/licenses/by/3.0/).Ti liu c hiu nh ngy: September 16, 2009Ngy to PDF: December 3, 2010 bit thng tin v ng gp cho cc module c trong ti liu ny, xem tr. 118.


5/127

Ni dung

1 C s l thuyt nhm

1.1 Khi nim v nhm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Cc v d v nhm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Nhm SO(3) cc php quay khng gian Euclide thc ba chiu . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 71.4 Nhm SU(2) cc bin i unita vi nh thc bng 1 trong khng gian Euclide

phc 2 chiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Nhm Lie v i s Lie . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.6 Ph lc c s l thuyt nhm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 C s l thuyt biu din nhm

2.1 Khi nim v biu din nhm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Cc php tnh i vi cc biu din . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.3 Hm c trng ca biu din . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4 Ph lc c s l thuyt biu din nhm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 Cc nhm im tinh th hc

3.1 Phn loi cc nhm im tinh th hc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2 H cc im Cn, Cnh, Cnv, Ci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3 H cc nhm im Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.4 H cc nhm im Dn, Dnh, Dnd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.5 H cc nhm im T, Th, Td . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.6 H cc nhm im O , Oh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.7 S i xng ca cc phn t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.8 S i xng ca cc tinh th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 903.9 Nhm i xng ca cc im c bit trong cc Wigner - Seitz ca cc mng

h lp phng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Ch mc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117Tham gia ng gp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 8


6/127

iv


7/127

Chng 1

C s l thuyt nhm

1.1 Khi nim v nhm1

1.1.1 nh ngha nhm

Tp hp G cc yu t a, b, c,. . . c gi l mt nhm nu c cc tnh cht sau y:1) Trn tp hp G tn ti mt php tnh gi l php nhn ca nhm. Php tnh ny t tng ng vi

mi cp hai yu t av b bt k ca tp hp G mt yu t c cng thuc tp hp ny, gi l tch ca avbav k hiu l ab :

a G, b G{a, b} ab = c G2) Php nhn ca nhm c tnh cht kt hp, ngha l vi mi yu t a, b, cca tp hp G ta lun c(ab) c = a (bc)3) Trn tp hp G tn ti mt yu t e, gi l yu t n v, m vi mi yu t a G ta lun lun ce a = a e = a4) Vi mi yu t a

G bao gi cng c mt yu t a1

G , gi l nghch o ca

a, sao cho

a-1a = a a-1 = eDo tnh cht kt hp ca php nhn ta c th vit

1.1.2 Cc nh ngha khc

Nu php nhn ca nhm G c tnh cht giao hon, ngha l vi mi cp yu t a G, b G ta lun lunc h thc a b = b a,

th nhm G c gi l nhm giao hon hay nhm Abel.Nu nhm G ch c mt s hu hn cc yu t khc nhau th nhm ny c gi l nhm hu hn, cn

s lng cc yu t khc nhau c gi l cp ca nhm. Trong trng hp ngc li, khi nhm G c v syu t khc nhau, n c gi l nhm v hn.Vi cc nhm hu hn ta c th trnh by quy tc phn nhm mt cch c th di dng mt bng nhn

nhm c thit lp nh sau. Ta k mt bng vung vi s hng v s ct bng s yu t ca nhm. phatri ca bng, u mi hang, v trn ca bng, u mi ct, ta ghi tt c cc yu t khc nhau ca nhmtheo mt th t no g1, g2, . . ., gn. Sau trn chung ca hang th j v ct th j ta ghi yu t l tchgjgi. Nhn bng phn nhm ta thy ngay quy tc nhn nhm i vi tng cp yu t.

Bng nhn nhm1Phin bn trc tuyn ca ni dung ny c .

1


8/127

2 CHNG 1. C S L THUYT NHM

Nu trong nhm G c mt tp hp cc yu t a1, a2, . . ., an vi tnh cht sau y: mi yu t ca nhmG u c th vit di dng mt tch m mi tha s l mt trong cc yu t ny (mt yu t c th cdng lm tha s nhiu ln), th ta ni rng nhm G c sinh ra bi cc yu t a1, a2, . . ., an, cn cc yut ny c gi l cc yu t sinh. Nhm hu hn c sinh ra bi mt yu t a, ngha l gm cc yu tc dng a, a2, . . ., an = e, c gi l nhm vng.

Nu mi yu t ca nhm G u l mt hm lin tc ca nhng thng s no v hon ton c xcnh bi gi tr ca nhng thng s ny, th nhm G gi l nhm lin tc. Ta quy c rng cc thng s nyl cc bin s c lp. Nu mi yu t ca nhm lin tc G u l hm kh vi ca cc thng s c lp, thnhm G c gi l nhm Lie.

T nh ngha nhm pht biu trn suy ra ngay mt s mnh sau y.1. Mi nhm ch c mt yu t n v.2. Nghch o ca tch ca hai yu t bng tch cc nghch o ca chng theo th t ngc li, ngha l(a b)-1 = b-1 a-13. Mi yu t ca nhm ch c mt yu t nghch o.

1.1.3 nh ngha yu t lin hp

Yu t a ca nhm G c gi l lin hp vi yu t b ca nhm ny nu c mt yu t no c G m

a c a-1 = bC th chng minh c rng quan h lin hp l mt quan h tng ng, ngha l1o) Nu a lin hp vi b th b lin hp vi a (tnh cht i xng).2o) Yu t a lin hp vi chnh n (tnh t lin hp).3o) Nu a lin hp vi b, b lin hp vi c th a lin hp vi a (tnh chuyn tip) .

1.1.4 Lp cc yu t lin hp

V rng mi quan h lin hp l mt quan h tng ng cho nn tt c cc yu t ca nhm G lin hpvi mt yu t xc nh no u lin hp vi nhau, v do ta c th chia nhm G thnh cc tp hp


9/127

3

con m tt c cc yu t trong mi tp hp con u lin hp vi nhau. Mi tp hp con cc yu t lin hpvi nhau ca nhm G c gi l mt lp cc yu t lin hp. Ch rng hai lp khc nhau khng c mtyu t chung no c, ngh l khng giao nhau.

1.1.5 nh ngha nhm con

Mt tp hp con G1 ca nhm G c gi l nhm con ca nhm G nu i vi php nhn ca nhm G tphp G1 ny cng to thnh mt nhm, ngha l nu G1 tha mn nhng iu kin sau y:

1) Nu a v b l hai yu t ca G1 th tch ab cng l mt yu t ca G1:a G1 ,b G1 a b G1Ta ni rng tp hp con G1 l kn i vi php nhn nhm:G 1 G 1 G 12) Tp hp con G1 cha yu t n v eca nhm G:e G13) Nu a l mt yu t ca G1 th a-1 cng l mt yu t ca G1:a G1 a-1 G 1Ta ni rng tp hp G1 l kn i vi php nghch o:G11 G1D thy rng t cc iu kin 1) v 3) suy ngay ra iu kin 2). Thc vy, ly mt yu t ty a ca

tp hp con G1. Theo iu kin 3) ta ca G1 a-1 G1Theo iu kin 1) tha G1 , a-1 G1 e = a a-1 G1 l iu kin 2). Ch rng php nhn ca nhm con G1 chc chn c tnh cht kt hp, v l

php nhn ca nhm G.

1.1.6 nh ngha tch trc tip ca hai nhm

Cho hai nhm G1 v G2 hon ton c lp vi nhau, vi cc yu t a1, b1, c1, . . . G1 v a2, b2, c2, . . .G2. Xt tp hp G1 G2 m mi yu t l mt cp {a1, a2}, {b1, b2}, {c1, c2}, . . . gm hai yu t ca hainhm. Ta nh ngha tch ca hai yu t ca G1 G2 nh nhau:

{a1, a2}{b1, b2} = {a1b1, a2b2}Gi e1 v e2 l hai yu t n v ca G1 v G2, a1-1 v a2-1 l hai yu t nghch o ca a1 v a2 trong

G1 v G2. Ta coi l yu t n v ca G1 G2 , {a1-1 v a21} l yu t nghch o ca {a1, a2}trong G1 G2. Tp hp G1 G2 vi php nhn nhm, vi yu t n v v yu t nghch o nh ngha nh vy tothnh mt nhm, gi l tch trc tip G1 G2 ca hai nhm cho. Tnh cht kt hp ca php nhn trnG1 G2 suy ra t tnh cht kt hp ca php nhn trn tng nhm G1 v G2.

C nhng nhm m cc yu t c bn cht khc nhau nhng cc php tnh ton di dc l cc yut ca nhm th li tng t nhau. S tng t c pht biu nh sau.

1.1.7 nh ngha s ng cu v s ng cu

Nhm G1 gi l ng cu vi nhm G2 nu c mt php tng ng gia cc yu t a1, b1, c1. . . ca G1 vicc yu t a2, b2, c2. . . ca G2,

G1 a1 a2 G2 ,

Sao cho ng vi mi yu t a1 G1 c mt yu t duy nht a2 G2 gi l nh hng ca a1 trong G2,mi yu t a2 G2 l nh hng ca t nht mt yu t a1 G1, v php tng ng ny bo ton phpnhn nhm, ngha l nu tng ng vi a1 G1 c a2 G2, tng ng b1 G1 c b2 G2 , th tng ngc a1b1 G1 , c a2b2 G2 :


10/127


Nu s tng ng ni trn l duy nht theo c hai chiu, ngha l nu mi yu t a2 G2 ch l nhhng ca mt yu t duy nht a1 G1, v do c php tng ng ngc li

G2 a2 G1 a1 (1.1)

th gi l c php tng ng 2 chiuG1 a1 a2 G2

th ta gi hai nhm G1 v G2 l ng cu.T iu kin bo ton php nhn nhm suy ra rng nh hng ca yu t n v e1 trong G1 phi l

yu t n v e2 trong G2, nh hng ca hai yu t nghch o vi nhau a1 v a2-1 ca G2.V phng din cu trc i s th hai nhm ng cu c cu trc i ging ht nhau v c th xem l

cng mt nhm, ngha l ta khng phn bit cc nhm ng cp vi nhau khi ta ch quan tm n cu trci s ca chng.

1.2 Cc v d v nhm2

1. Tp hp R cc s thc to thnh cc nhm vi php nhn nhm l php cng thng thng: tng x + yca hai s thc xv yc xem l tch ca hai yu t xv yca nhm. Ta bit rng php cng cc s thcc tnh cht kt hp. Yu t n v ca nhm l s 0. Nghch o ca yu t x l yu t -x. V php cngcc s thc c tnh cht giao hon nn R l mt nhm giao hon. Tng t nh vy, tp hp Rn cc vecttrong khng gian vect thc n chiu to thnh nhm vi php nhn nhm l php cng cc vect: tng x+yca hai vect c xem l tch ca hay yu t x v y, yu t n v l vect 0 (tt c cc thnh phn ubng khng), nghch o ca yu t x l yu t -x. y l mt nhm giao hon. Nhm cc s nguyn l mtnhm con ca nhm cc s thc i vi php cng.

2. Tp hp R {0} cc s thc khc khng cng nh tp hp C {0} cc s phc khc khng u tothnh nhm i vi php nhn nhm l php nhn thng thng c tnh kt hp. Yu t n v ca nhml s 1. Nghch o ca x l 1

x. Cc nhm ny cng l cc nhm giao hon. Nhm cc s dng khc khng

l nhm con ca nhm cc s thc khc khng i vi php nhn, nhm cc s thc khc khng l nhmcon ca nhm cc s phc khc khng i vi php nhn.

3. Tp hp cc ma trn n x n c nghch o to thnh nhm i vi php nhn ma trn. Ta nhc lirng nu A v B l hai ma trn vi cc yu t ma trn Aij v Bij, i, j = 1, 2, . . ., n, th AB l ma trn vicc yu t ma trn

(AB)ik =n

k=1

AikBkj AikBkj (1.2)

Php nhn ma trn c tnh cht kt hp, nhng ni chung khng giao hon. Yu t n v ca nhm l matrn n v I m cc yu t ma trn bng

Iij = ij (1.3)

Yu t nghch o ca ma trn A l ma trn nghch o A1

A1A = AA1 = I (1.4)

Ch rng ma trn tch AB c nghch o l(AB)1 = B1A1

Ty theo cc yu t ma trn l cc s thc hay cc s phc m nhm ny c k hiu l GL(n, R) hayGL (n, C). V cc ma trn trn c th thay i lin tc cho nn cc nhm ny nhng nhm lin tc. Khikhng cn ch r cc yu t ma trn l cc s thc hay s phc th ta vit GL(n). Nhm GL (n, R) l nhmcon ca nhm GL (n, C).

2Phin bn trc tuyn ca ni dung ny c .


11/127

5

Tp hp cc ma trn n x n vi nh thc bng 1 cng to thnh nhm i vi php nhn ma trn, vrng nu A c nh thc bng 1 th A-1 cng vy,

detA1 =1

detA= 1 (1.5)

nu A v B u c nh thc bng 1 th tch AB cng vy,det (AB) = (det A) (det B) = 1Ty theo cc yu t ma trn l cc s thc hay s phc m ta k hiu nhm ny l SL (n, R) hay SL (n,

C), cn khi khng cn ch r s thc hay s phc th ta k hiu l SL (n). Nhm SL (n) l nhm con canhm GL (n).

4. Tp hn cc ma trn trc giao n x n to thnh nhm i vi php nhn ma trn. Ta nhc li rng matrn chuyn v AT ca ma trn A c cc yu t ma trn sau y

(AT)ij= AijT nh ngha ny suy ra rng(A B)T = BT ATMa trn thc n x n, k hiu l O, c tnh cht

OT

O = O OT

= Igi ma trn trc giao. T y ta c ngay(O-1)T = O = (O-1)-1,Ngha l O-1 cng l ma trn trc giao. D th li rng nu O1 v O2 l hai ma trn trc giaoOT1 = O11 , OT2 = O

12

Th tch O1O2 cng l ma trn trc giao(O1O2)T = OT2 OT1 = O12 O

11 = (O1O2)-1.

Qu tht cc ma trn trc giao n x n to thnh nhm, k hiu l O(n).Tng t, cc ma trn trc giao n x n vi nh thc bng 1 cng to thnh nhm k hiu l SO(i).5. Tp hp cc ma trn unita n x n to thnh nhm i vi php nhn ma trn. Ta nhc li rng ma

trn lin hp hermitic A+ ca ma trn A c cc yu t ma trn sau y(A+)ij = (Aji)*,ngha l

A + = (A T ) *T nh ngha ny suy ra rng(A B) + = B + A + .Ma trn phc n x n, k hiu l U, c tnh chtU+ U = UU+ = Ingha lU+ =U-1gi l ma trn ca unita. T y ta c ngay(U-1)+ = U = (U-1)-1,Ngha l U-1 cng l m trn unita. D th li nu U1 v U2 l hai ma trn unita,U+1 = U

11 , U

+2 = U

12 ,

th tch U1U2 cng l ma trn unita,(U1U2)+ = U+2 U

+1 = U

12 U

11 = (U1U2)-1

Qu tht l cc ma trn unita n x n to thnh nhm, gi l nhm U(n).Tng t, cc ma trn unita n x n vi nh thc bng 1 cng to thnh nhm, gi l nhm SU(n). NhmSU(n) l nhm con ca nhm U(n).

6. Tp hp cc php tnh tin ca mt khng gian thc n chiu to thnh nhm i vi php nhn nhmnh ngha nh sau: thc hin lin tip hai php tnh tin, ta c php tnh tin gi l tch ca chng. Khiu l Ta v Tb hai php tnh tnh khng gian trong im r bt k chuyn thnh r+ av r+ b,

Ta: r r + a,Tb: r r + b.Thc hin lin tip hai php tnh tin ny, ta c


12/127


Tb Ta : r r+ ar + b + aHai php tnh tin ny cho kt qu tng ng vi php tnh tin Tb+aTb+a: r r+ b + a.Vy ta c

T b+a = T b T aYu t n v ca nhm lT 0 = ID th li rngT -a = (T a ) -1Cc nhm tnh tin khng gian thc n chiu to thnh nhm tnh tin T(n). l mt nhm giao hon.

Nhm tnh tin ng cu vi nhm cc vect trong khng gian m php nhn nhm l php cng cc vect.7. Tp hp cc php bin i tuyn tnh khng k d ca mt khng gian vect n chiu to thnh nhm

i vi php nhn nhm nh ngha nh sau: thc hin lin tip hai php bin i tuyn tnh khng k d Ari n B, ta c kt qu tng ng vi thc hin mt php bin i tuyn tnh khng k d k hiu lBA v c coi l tch ca A v B. Xt h cc vect c s c lp tuyn tnh e1, e2, . . ., en ca khng giann chiu cho. Bin i tuyn tnh A chuyn cc vect ny thnh cc vect e1, e2, .., en

Aei=

e

igi l khng k d nu c bin i tuyn tnh k hiu l A-1 chuyn ngc li cc vect ei thnh ei,A-1ei = einh ngha tch ca hai php bin i m ta pht biu vn tt trn c c th ha nh sau. Xt mt

vect bt k rtrong khng gian vect n chiu cho. Php bin i tuyn tnh khng k d A chuyn vectny thnh vect r:

r

Ar = Ar.Tip theo sau php bin i A ta thc hin php bin i tuyn tnh khng k d B. Php bin i ny

chuyn thnh vect r thnh vect r

r

Br = Br = B (Ar).Kt qu l ta thu c mt php bin i tuyn tnh chuyn vect r thnh vect r k hiu l (BA):r

ABr = Br = B (Ar) = (BAr)Ta coi bin i (BA) ny l tch ca hai bin i A v B v cn k hiu n l BA. Yu t n v ca

nhm l php ng nht I:Iei = eiV cc vect c s e1, e2, . . ., en l c lp tuyn tnh cho nn tt c cc vect ei u c th c biu

din di dng cc t hp tuyn tnh ca cc vect c s nyei = ej AjiMa trn vi cc yu t ma trn Aij hon ton xc nh php bin i AAe i = e j A jiTa cng k hiu ma trn ny l A. Tng t nh vy, php bin i B c din t bi ma trn B vi

cc yu t ma trn Bkj,Bej= ek BkjTc dng lin tip hai php bin i A v B, ta cB Ae i = B (e j A ji ) = (Be j ) A ji = e k B kj A ji = e k (B A) ki

Vy bin i tch B A c ma trn l tch ca hai ma trn ca hai php bin i B v A. Bin i ngnht c ma trn l ma trn n v. Bin i nghch o c ma trn l ma trn nghch o. Vy nhm ccbin i tuyn tnh khng k d ca khng gian vect n chiu ng cu vi nhm GL(n) cc ma trn n x nc nghch o m ta xt trn. Ta cng gi l nhm GL (n).

8. Tp hp cc php quay ca khng gian Euclide thc n chiu quanh gc ta to thnh nhm ivi php nhn nhm nh ngha nh sau: thc hin hai php quay lin tip ta c mt php quay th bal tch ca chng. Php bin i ng nht (khng quay t no c) l yu t n v. Php quay ngc li lyu t nghch o. Ta nhc li rng trong khng gian Euclide thc n chiu ta c th chn h vect n vc s e1, e2, . . ., en trc giao chun ha, ngha l tha mn iu kin

(ei, ej) = ij, i, j = 1, 2, . . ., n


13/127

7

Trong php quay O cc vect ny chuyn thnh e1, e2, .., en cng trc giao chun ha(e1, e2) = ijThay vo y cc biu thc vit trn biu din ei qua ej v dng tnh cht trc giao chun ha ca

cc vect ei , ta thu c h thcOki Okj = ijVy ma trn O vi cc yu t ma trn Oij tha mn iu kinOT O = INhn t bn phi c hai v vi O-1, ta cOT = O-1ngha l ma trn ca php quay phi l ma trn trc giao. T iu kin ma trn trc giao cn suy ra

rngdet OT det O = (det O)2 = 1ngha ldet O = 1V m trn ca php bin i ng nht c nh thc bng +1, m cc php quay li l cc php bin

i lin tc, cho nn nh thc khng th nhy t +1 sang -1. Vy ta phi cdet

O= 1

Tm li, nhm cc php quay trong khng gian Eucide thc n chiu ng cu vi nhm SO(n). Ta cnggi nhm quay ny l nhm SO(n).

9. Trong khng gian Euclide phc n chiu vi tch v hng xc nh dng c cc tnh cht sau y(b, a1 + a2) = (b, a1) + (b, a2)(b1 + b2, a) = (b1, a) + (b2, a)(b, a) = (a, b)*,(b, a) = (b, a)vi mi s phc v do ( b, a) = * (b, a)tp hp cc php bin i tuyn tnh t u bo ton tch v hng ca hai vect bt k(ua, ub) = (a, b)To thnh nhm i vi php nhn ca hai php bin i c nh ngha l s thc hin lin tip hai

php bin i .Trong khng gian vect ang xt ta hy chn h vect n v c s trc giao chun ha e1, e2, . . ., en,(ei, ej) = ij, i, j = 1, 2, . . ., nPhp bin i t U chuyn cc vect ny thnh cc vect n v mi e1, e2, .., enei= U ei= ejujiV bin i U bo ton cc tnh v hng cho nn(e i e j ) = ( U ei , U ej )= (e k , e i ) U ki U ij = U ki U kj = U

+ik u k j = ij

Trong U+ l ma trn lin hp hermitic ca U. Do ta c h thcU+U = Ihay lU+ = U -1Ma trn ca cc php bin i Ubo ton tch v hng trong khng gian Euclide phc n chiu l cc ma

trn unita n x n. Vy nhm cc php bin i tuyn tnh bo ton tch v hng trong khng gian Euclide

phc n chiu ng cu vi nhm U (n). Ta cng gi l nhosmm U (n). Nu ta t thm iu kin nhthc ca cc php bin i phi bng 1 th ta c nhm SU(n).

1.3 Nhm SO(3) cc php quay khng gian Euclide thc ba chiu3

Trong mc ny ta kho st chi tit nhm SO(3) cc php quay khng gian Euclide thc ba chiu, v y lnhm i xng rt thng gp trong nhiu lnh vc vt l lng t: vt l nguyn t, vt l ht nhn, vt l



14/127


ht s cp. Ta bt u t vic nghin cu cc php quay ca mt phng xOy quanh gc ta , to thnhnhm SO(2). chnh l nhm quay khng gian ba chiu quanh trc c nh Oz, mt nhm con ca nhmSO(3). Mi php quay ca mt phng xOy c c trng bi gc quay v k hiu l O(). Thc hinlin tip hai php quay cc gc 1 v 2, ta c php quay gc 1 + 2 l tch ca hai php quay ni trn

O(1) O(2) = O (1 + 2)Tt c cc php quay ny giao hon vi nhau cho nn SO(2) l nhm giao hon. Mi yu t O() ca

nhm ny u hon ton c xc nh bi gi tr ca thng s thay i lin tc t 0 n 2 . Do SO(2) l nhm lin tc mt thng s. Trong php quay O() cc vect n v c s i v j chuyn thnh vectn v mi i v j lin h vi i v j bi cc h thc (xem hnh 1.1)

i = i cos + j sin j = -i sin + j cos Hai cng thc ny c th vit li di dng ma trn nh sau:

(i j )=(i j )

cos() (sin())

sin() cos ()

Vy ma trn ca php bin i O () l

O ()=

cos() (sin())

sin() cos ()

D dng th li rng O () l ma trn trc giaoO ()TO () = O ()O ()T = Ic nh mc bng 1,

det O () = 1,v tha mn iu kinO (1) O (2) = O (1 + 2)Ma trn O () hon ton xc nh php quay tng ng. V cc yu t ma trn ca n l cc hm kh

vi ca cho nn O () l nhm Lie.Trong php quay O ()vect r vi cc thnh phn xv y,r= xi + yj,chuyn thnh vect r vi cc thnh phn x v y,r = x i + y j .


15/127

9

Mt khc, v r , i , j thu c t r, i, j sau cng mt php quay cho nn h thc gia r v i , j cdng ging ht nh h thc gia rv i, j, c th l

r = x i + y j Thay vo y cc biu thc din t i v j theo iv j, ta suy rax = cos x - sin y

y = sin x + cos yCc cng thc ny cn c vit di dng ma trn nh sau x

y

=

cos() (sin())

sin() cos ()

x

y

Cc php quay mt phng xOy xung quanh gc ta O ng thi cng l cc php quay ca khnggian ba chiu quanh trc Oz. K hiu cc vect n v c s ca khng gian Euclide ba chiu l i, j, k, phpquay gc quanh trc Oz l Cz(). Php quay ny chuyn cc vect n v c s ni trn thnh cc vectn v c s mi sau y.

i = i cos + j sin ,j = -i cos + j cos ,

k = kDo ma trn ca php quay Cz() c dng

Cz()=

cos() (sin()) 0

sin() cos () 0

0 0 1

Tng t nh vy, ma trn ca cc php quay gc quanh cc trc Ox v Oy, k hiu l Cz() vCy(), c dng

Cx()=

1 0 0

0 cos () (sin())

0 sin () cos ()

Cy()= cos(

) 0 sin ()0 1 0

(sin()) 0 cos ()

Xt nhm quay trong khng gian ba chiu SO(3). Mi php quay khng gian ba chiu quanh gc ta

O u c th c thc hin di dng t hp ca ba php quay lin tip sau y: php quay gc quanhtrc Oz chuyn cc trc ta Ox v Oy thnh Ox v Oy, php quay gc quanh trc mi Ox chuyncc trc mi Oy v Ox thnh Oy v Oz, php quay gc quanh trc mi Oz (xem hnh 1.2). Ba thngs , , gi l ba gc Euler. K hiu php quay vi ba gc Euler.

, , l O( , ,). Ma trn ca php quay ny l tch ca ba ma trn tng ng vi cc php quayquanh cc trc Oz, Ox v Oz, c th l

O( , ,) = Cz( ) Cx( ) Cz().Thay vo y cc biu thc ca Cx(), Cz( ) v Cx( ), ta thu cO (, , ) =

Cc gc v thay i t 0 n 2 , cn gc thay i t 0 n . Nhm SO(3) l nhm Lie ba thngs.


16/127


Trong on trc ta nh ngha cc yu t lin hp. By gi ta hy chng minh tnh cht lin hpca hai php quay cng mt gc quanh hai trc khc nhau.

Mnh . Trong nhm quay SO(3) hai php quay cng mt gc quanh hai trc quay khc nhau lunlun lin hp vi nhau.

Chng minh. K hiu cc vect n v c s i, j, k ca h ta Descartes l ei, i = 1, 2, 3 v gi sn v n l hai vect n v c chung im u l gc ta O. C mt php quay R no chuyn vectn thnh vect n v gi s rng trong php quay ny cc vect n v c s ei chuyn thnh ei. Cc phpquay gc quanh cc trc n v n k hiu l Cn() v Cn(). Trong h ta vi cc vect n v c sei php quay Cn() c cc yu t ma trn ging ht nh cc yu t m trn ca php quay Cn() trongh ta vi cc vect n v c s ei. Ni khc i, nu

Cn() ei=ej A jithCn () ei =ej A jiThaye i = R eivo h thc (4)

Cn()

R ei=

Rej Ajiri nhn c hai v vi R-1 t bn tri, ta thu cR-1Cn() Rei = ej AjiSo snh vi h thc (3), ta suy ra rngR-1Cn () R = Cn()hay lCn() = RCn()R-1Ta cn vit li h thc ny nh sauCRn ()=RCn ()R -1Vy CRn () v Cn () l hai yu t lin hp vi nhau ca nhm SO(3).By gi bng nhng lp lun tng qut chng ta hy thit lp biu thc ca php quay Cn() mt gc

v cng b quanh trc quay hng theo vect n v n trong php gn ng cp 1 theo . Ta hy ctrng php quay gc quanh trc quay hng theo vect n bng vect c gi tr bng v hng theo

trc quay, = n.Ma trn Cn () phi quy v ma trn n v I khi t = 0, cho nn n c dngCn () = I + X ()Trong ma trn X () l i lng b cp 1 theo . B qua s hng cp 2, ta c

[Cn ()]1

= I X() (1.6)

Mt khc

[Cn ()]T

= I + [X()]T (1.7)

T iu kin ma trn Cn( ) l ma trn trc giao

[Cn ()]T

= [Cn ()]1

(1.8)suy ra rng ma trn X () phi l ma trn phn i xng

[X()]T

= X() T

Ta thy rng trong s chn yu t ma trn ca X () th ba yu t cho phi bng khng

[X()]ii = 0 (1.9)


17/127

11

su yu t khng nm trn ng cho chia thnh ba cp, mi cp gm hai yu t bng nhau v ln vngc du nhau,

[X()]ij = [X()]ji, i = j (1.10)

Vy ma trn X( ) ch cha ba thng s c lp. Ta c th chn ba thn phn k, k= 1, 2, 3, ca vect lm ba thng s c lp ny v vit

X( ) = - i S = - i k S ktrong Sk, k = 1, 2, 3, l ba ma trn phn i xng 3 x 3 c lp tuyn tnh vi nhau. Ta a them

n v o i vo cng thc va vit cho thun tin sau ny. V cc yu t ma tr ca X( ) phi l ccs thc cho nn cc yu t ma trn ca cc ma trn Sk phi l cc s o.

T cc biu thc va vit trn ca Cn () v X () suy ra rng cc php quay gc v cng b quanh cc trc Ox, Oy, v Oz c cc ma trn sau y

Cx ()= I - i S1Cy ()= I - i S2Cz ()= I - i S3Cc ma trn Sk, k = 1, 2, 3, gi l cc vi t ca cc php quay quanh ba trc ta . Ta li cng

bit cc biu thc (1a) - (1c) ca cc php quay Cx( ), Cy( ), Cz( ) vi cc gc quay bt k. Dngcc biu thc ny ri thay bng v cng b v ch gi li cc s hng cp 1 theo , ta suy ra

Cx()=

1 0 0

0 1

0 1

,

Cy()=

1 0

0 1 0

0 1

,

Cz()=

1 0

1 00 0 1

,

So snh cc biu thc ny vi cc cng thc biu din cc ma trn Cx( ), Cy( ) v Cz( ) qua ccvi t S1, S2, S3 m ta vit trn, ta thu c

S1 =

0 0 0

0 0 i

0 i 0

S2 =

0 0 i

0 0 0

i 0 0

S3 =

0 i 0

i 0 0

0 0 0

D th li rng ba ma trn S1, S2, S3 tha mn cc h thc giao hon sau y[S1, S2]= i S3, [S2, S3]= i S1, [S3, S1]= i S2hay l di dng thu gn[Si, Sj ]= i ijk Sk,


18/127


Trong ijk l tenx hon ton phn i xng hng 3 trong khng gian ba chiu, vi

123 = 1 (1.11)

1.4 Nhm SU(2) cc bin i unita vi nh thc bng 1 trongkhng gian Euclide phc 2 chiu4

Trong on ny chng ta kho st chi tit v nhm SU(2) cc bin i tuyn tnh bo ton tch v hngv c nh thc bng 1 ca khng gian Euclide phc hai chiu. Nhiu cng thc v mt s lp lun trnhby di y thng hay c p dng khi nghin cu nhng vn trong nhiu lnh vc ca vt l lngt. Trong h vect n v c s giao chun hoa smooix php bin i thuc nhm SU(2) c din t bimt ma trn 2 x 2 unita U.

U+ = U-1,V c nh thc bng 1,

det U = 1Yu t n v ca nhm l ma trn n v I. Yu t c ma trn bng U-1 l nghch o ca yu t c matrn bng U.

tm cc tham s c lp cng nh cc vi t tng ng ta hy xt cc bin i v cng gn yu tn v, ngha l cc php bin i m cc ma trn c dng.

U( i) = I i X ( i),Trong ma trn X( i) l i lng cp 1 i vi cc tham s thc v cng b i. B qua cc s

hng cp 2, ta c[U(i)]

-1 = I + i X ( i),Mt khc,[U(i)]

+ = I + [X(i)]+T iu kin ma trn U( i) phi l ma trn unitaU( i)-1 = [U(i)]+

suy ra rng ma trn X ( i) phi t lin hp, ngha lX ( i)+ = X ( i).Do hai yu t cho ca ma trn X ( i) phi l hai s thc[X(i)]jj = [X(i)]jj, j = 1, 2Cn hai yu t khng cho ca ma trn ny th phi lin hp phc vi nhau[X(i)]12 = [X(i)]21Nu khng t thm iu kin g khc th ma trn t lin hp X ( i) cha bn tham s thc c lp

vi nhau. Nhng ta cn c iu kin nh thc ca U( i) phi bng 1. B qua cc s hng cp cao ta cdet [U(i)] = 1 i Tr [X(i)].Vy ma trn X ( i) phi c vt bng khngTr [X(i)] = 0hai yu t cho phi c ln bng nhau v ngc du. Tm li, ma trn 2 x 2 t lin hp v c vt

bng khng X ( i) biu din qua bat ham s thc c lp vo cng b i, i = 1, 2, 3 v ta cU ( i) = I i isi = I i sTrong cc vi t si, i = 1, 2, 3 l ma trn 2 x 2 t lin hp c lp tuyn tnh v c vt bng khng,

l vect ba chiu vi cc thnh phn i, s l ma trn vect ba chiu vi cc thnh phn si. cho sauny c thun tin ta chn cc ma trn si bng cc ma trn Pauli i nhn vi 12 :

si = 12i4Phin bn trc tuyn ca ni dung ny c .


19/127

13

1=

0 1

1 0

, 2=

0 i

i 0

, 3=

1 0

0 1

D th li rng cc ma trn Pauli u c bnh phng bng ma trn n v21 = 22 = 23 = I ,hai ma trn Pauli khc nhau phn giao hon vi nhau v c tch bng12 = - 21 = i3, 23= - 32 = i 1, 31 = - 13 = i 2.Cc h thc ca bnh phng ma trn Pauli v tch hai ma trn Pauli khc nhau c th vit gp li nh

sauij = ijI + iijkkT y suy ra cc h thc giao hon[i, j ] = 2iijkkChia cc ma trn i cho 2 ta c cc ma trn si tha mn cc h thc giao hon ging nh cc h thc

giao hon gia cc vi t Si ca nhm SO(3), c th l[si, sj] = iijkskCoi cc ma trn si, i = 1, 2, 3 l cc yu t v giao hon t [si, sj ] l tch ca hai yu t si v sj, ta thit

lp c mt i s Lie ca nhm SU(2). Ta thy cng chnh l i s Lie ca nhm SO(3) thnhlp trn.

Sau khi thu c biu thc ca cc bin i U( i) rt gn yu t n v, vi cc tham s v cngb i, by gi cc hy m rng cc lp lun trn thit laapjbieeru thc ca php bin i bt k U(i) ca nhm SU(2) ph thuc vo cc tham s thc i, c cc gi tr hu hn. Ta cng s thy rng c batham s c lp. Trc ht ta ch rng mi ma trn unita U( i) u c th vit di dng

U(i) = eiX(i),

Trong X( i) l ma trn t lin hp[X(i)]

+= X(i)

Lm mt php bin i thch hp ca h ta a ma trn t lin hp X( i) v dng cho, ta cth chng minh rng nh thc ca ma trn U( i) biu din qua vt ca ma trn X(i)nh sau

det [U(i)] = e1Tr[X(i)]

T iu kin nh thc ca U( i) phi bng 1 suy ra rng vt ca ma trn X( i) phi bng khng

[X(i)] = 0V rng c ba ma trn 2 x 2 c lp tuyn tnh, t lin hp v c vt bng khng, cho nn ma trn 2 x 2t lin hp c vt bng khng X( i) ph thuc vo ba tham s thc i, i = 1, 2, 3 v c th vit nh sau

[X(i)] =12

ii =12

Vy ma trn ca php bin i bt k thuc nhm SU(2) c dng tng qutU( i) = e

12ii = e

12

Xt cc trng hp khi m ch c mt tham s k trong ba tham s 1, 2, 3 l khc khng, cn haitham s kia bng khng. Ta c

U( k = , i=k = 0) = U(k)( ) = e12 [U+03D5]k

Khai trin hm m thnh chui ly tha v dng tnh cht 2k

= 1,ta thu c

U(k)( ) = n=1

(1)n

(2n)!22n

- i k

n=1

(1)n

(2n+1)!22n+1

= cos 2

- i k

sin 2

.

Vi k= 1, 2 ta c


20/127


Cn vi k= 3

Mi lot cc php bin i U(k)( ) vi k c nh to thnh mt nhm con mt tham s ca nhm SU(2).Cc ma trn U(k)( ) l cc hm kh vi ca cho nn nhm SU(2) l mt nhm Lie.

By gi ta quay li yu t c dng tng qut U (i) v k hiu n l vect n v hng theo vect n =

.

Dng cc h thc vit trn i vi tch ca cc ma trn Pauli, d th li rng(n)

2= 1.

Khai trin hm m e12(n) thnh chui ly tha, ta li thu c

U( i) =

n=1(1)n

(2n)!

2

2n- i (n)

n=1

(1)n

(2n+1)!

2

2n+1= cos 2 - i k sin

2

Vy biu thc sau y ca yu t bt k ca nhm SU(2)U( i) = cos 2 - i

sin 2 .Cc ma trn thuc nhm SU(2) c nh thc bng 1. Nu ta khng i hi nh thc ca ma trn 2 x

2 ca php bin i unita phi bng 1 th ta c nhm U(2). By gi ma trn X( i) khng nht thit phic vt bng khng v do ph thuc bn tham s, ba tham s l thnh phn ca vect ba chiu xt

trn v mt tham s mi 0. Ma trn 2 x 2 t lin hp ( i) ph thuc bn tham s c biu din quabn ma trn 2 x 2 t lin hp c lp tuyn tnh l ba ma trn Pauli i, i = 1, 2, 3 v ma trn n v I khiu l 0,

X( i) = 1200 +12.

Ngoi ba nhm con mt tham s gm cc bin i U (k)( ) xt trn by gi ta c thm mt nhmcon mt tham s na l nhm U(1) vi cc php bin i

U(0) ( ) = e 12 .Cc bin i ny giao hon vi cc bin i ca nhm SU(2) v do nhm U(2) l tch trc tip ca

nhm U(1) v nhm SU(2).U(2) = U(1) SU(2).By gi ta dn ra y mt s cng thc i vi cc ma trn Pauli m ta thng dng khi nghin cu

cc vn thuc nhiu lnh vc ca vt l lng t. Trc ht ta ch rng vt ca cc ma trn Pauli bngkhng

Tr ( i) = (i) = 0cn tch ca hai ma trn Pauli c vt bngTr ( ij) = (i)(i)= 2 ijBa ma trn Pauli iv ma trn n v I to thnh bn ma trn n x n c lp tuyn tnh. Mi ma trn

2 x 2 u c th trin khai theo bn ma trn ny nh sauA = A0 + Ai(i) = A0 + A () , , = 1, 2hay lA = A 0 I + A i i = A 0 I + A


21/127

15

Ly vt c hai v h thc trn, ta cA0 =

12

A =12Tr (A).

Cn nu nhn c hai v h thc vi k xong ri mi ly vt th ta thu c

Ak = 12

A(k) =12

Tr (Ak) (1.12)

hay lA = 12A () =

12Tr (A) .

Cc ma trn 1 v 3l i xng( 1)T= 1, ( 3)T = 3ngha l(1) = (1) , (3) = (3) ,cn ma trn 2l phn i xng( 2)T= - 2,ngha l(2) = - (2) .

T cc tnh cht i xng hoc phn i xng ny ca c ma trn Pauli v tnh cht phn giao honca cc ma trn Pauli khc nhau suy ra h thc

2i2 = Ti .Nhn c hai v ca h thc ny vi 2 t bn phi hoc t bn tri v thc hin cc php tnh ton thch

hp tip theo, ta s c2i = Ti 2 =

Ti

T2 = (21)

T,i2 = 2Ti =

T2

Ti = (i2)

T.Vy cc ma trn 2i v i2l cc ma trn i xng,(2i)T = ( 2i), ( i2)T = ( i2),ngha l(2i)= (2i) , (i2)= (i2) ,So snh cc kt qu va thu c i vi nhm SU(2) v cc kt qu trnh by trn v nhm quay

SO(3), ta thy c mt s tng t: c hai nhm u l cc nhm Lie bat ham s, cc vi t ca chng thamn nhng h thc giao hon ging ht nhau. Ta hy chng minh rng nhm SO(3) ng cu vi nhmSU(2).

Xt mt vect r trong khng gian ba chiu. T ba thnh phn r1 = x, r2 = y, r3 = z ca vect ny tahy lp ra ma trn R sau y

Dng cc tnh cht ca cc ma trn Pauli i m ta trnh by trn, d thy rng cc thnh phnca vect r c biu din ngc li qua ma trn R nh sau

ri =1

2Tr (Ri) (1.13)

hay lr = 12Tr (R)Tnh nh thc ca ma trn R, ta thu cdet R = - r2 .Cho U l mt yu t ca nhm SU(2) v xt php bin i tuyn tnh sau y ca ma trn RRR = U RU +.K hiu vect trong khng gian ba chiu tng ng vi ma trn R l r:R = r.Trong php bin i R thnh R, vect r chuyn thnh r


22/127


R R r r.Ta k hiu php bin i ny ca khng gian ba chiu l O,r = O r,v thit lp c s tng ng gia mi yu t U ca nhm SU(2) vi mt php bin i O ca khng

gian ba chiuU O.Trc ht, ta hy chng minh rng php bin i O bo ton chiu di ca cc vect trong khng gian

ba chiu. Thc vy, ta cr 2 = - det R = - det(U RU +) = - (detU) (det R) (det U+) = - det R = r2Vy O l php quay hoc l t hp ca php quay v php nghch o hoc / v php phn x gng.

Dng cc biu thc cho trn ca cc yu t U(k)(), k= 1, 2, 3, ca cc nhm con mt tham s trongnhm SU(2) ri thc hin php nhn ma trn tm cc ma trn

U (k)( ) RU (k) ( ) +ta thu c ngay ma trn ca cc php bin i bin i O tng ng ca khng gian ba chiu. Kt qu

l ta c s tng ng sau y gia cc yu t U(k)( ), k = 1, 2, 3 v cc php quay Cx (), Cy (), Cz(),:

U

(1) ()Cx

(),U(2) () Cy (),U(3) () Cz ().D th li rng s tng ng ni trn gia cc yu t ca hai nhm bo ton php nhn nhm. Vy ta

thit lp c s ng cu ca nhm SU(2) ln nhm SO(3). Ch rng nu tat hay U bng U th tavn c cng mt php quay O. Vy trong php ng cu ca nhm SU(2) ln nhm SO(3) hai yu t tridu nhau ca nhm SU(2) tng ng vi cng mt yu t ca nhm SO(3). Nhm SO(3) ng cu nhngkhng ng cu vi nhm SU(2).

1.5 Nhm Lie v i s Lie5

Khi nghin cu v cc nhm SO(3) v SU(2) chng ta thit lp cc h thc giao hon gia cc vi t cami nhm ny v thy rng cc vi t to thnh i s Lie. By gi chng ta m rng cc l lun trnhby khi nghin cu v cc nhm SO(3) v SU(2) ra cho trng hp mt nhm Lie G gm cc php bin ituyn tnh tha mn nhng iu kin nhy nh ca mt khng gian vect no v chng minh rng ccv t ca nhm ny to thnh mt i s Lie. Trc ht ta hy gii thiu nhng khi nim c bn v i sLie.

1.5.1 i s Lie

Cho mt khng gian vect V trn ng R cc s thc hoc trng C cc s phc. K hiu cc yu t caV l X, Y, Z. . . cc yu t ca trng R hoc C l , , . . . Gi s rng trn tp hp V c mt quy tcgi l php nhn cho php ta t hai yu t X, Y bt k ca V xc nh c mt v ch mt yu t th baca V k hiu l X Y v gi l tch ca X v Y, m

X ( Y) = ( X) Y = (XY),(X + Y) Z = X Z + Y Z,X (Y + Z) = X Y + X + Z. (38)Khng gian vect V vi php nhn hai yu t c nh ngha nh th c gi l mt i s A. Nu

php nhn cc yu t ca mt i s c tnh cht kt hpX (Y Z) = (X Y) Zth i s A c gi l i s kt hp.Mt i s A vi php nhn hai yu t{X, Y} (X Y)



23/127

17

tha m cc iu kin(X Y) = (Y X) (phn giao hon) (39)(X (Y Z)) + (Y (Z X)) + (Z (X Y)) = 0 (40)(ng nht thc Jacobi)c gi l mt i s Lie. Cho mt i s kt hp A vi tch ca hai yu t X v Y c k hiu l X

Y. Trn tp hp A ta hy a ra mt nh ngha khc ca php nhn hai yu t{X, Y} (X Y) [X, Y] = X Y Y X (41)

Vi nh ngha mi ny ca tch hai yu t i s A tr thnh mt i s Lie L. Thc vy, d dng thli rng nh ngha (41) ca tch hai yu t tha mn cc iu kin (39) v (40).

Xem nh mt khng gian vect mi i s Lie c mt h cc vect c s Xi, i = 1, 2,. . ., s, m mi yut X ca L u c th vit mt cch n gi di dng

X =s

i=1 iXi (42)vi cc h s i trong trng s cho. Xt hai yu t Xi v Xjty ca h c s ca mt i s Lie L

v tch (Xi Xj) ca chng. V (Xi Xj) cng l mt yu t ca i s L cho nn n cng li phi l mtt hp tuyn tnh ca cc yu t ca h c s, ngha l phi c dng

(XiXj) =

si=1 ijkXk . (43)

Cc h s ijk

c gi l cc hng s cu trc ca i s LieL

. T cc iu kin (39) v (40) suy rarng cc hng s cu trc ijk tha mn cc h thc sau y:

jik = ijk (44)ilmikl + jlmkil + klmijl= 0 (45)Cho hai i s Lie L v L vi cc yu t k hiu l X, Y, Z v.v. v X, Y, Z v.v.. Ta ni rng i s

Lie L ng cu vi i s Lie L nu c php nh x tuyn tnh khng gian vect L ln khng gian vect L,LL,c tnh cht bo ton php nhn ca i s Lie, ngha l tXX , YY suy ra(X Y) (X Y)Nu php nh x tuyn tnh ca i s Lie L ln i s Lie L l n gi theo c hai chiuLL

v bo ton php nhn ca i s Lie, th ta ni rng hai i s lie L v L ng cp vi nhau. Sau nychng ta s khng phn bit cc i s Lie ng cu.

1.5.2 Lin h gia nhm Lie cc php bin i v i s Lie

Sau khi bit mt s khi nim c bn v i s Lie by gi chng ta thit lp mi lin h gia mi nhmLie cc php bin i tuyn tnh ca mt khng gian vect v i s Lie tng ng. Trong khng gian vect ta hy chn mt h c s v biu din mi php bin i Tbng mt ma trn cng k hiu l T v t

T = e - iX . (46)T nh ngha nhm G suy ra nhng iu kin m ma trn T phi tha mn, ri t nhng iu kin

ny suy ra nhng iu kin m ma trn X phi tha mn. Th d nh nu G l nhm cc bin i trc giaotrong khng gian Euclide th cc yu t ca n phi l nhng ma trn trc giao O tha mn iu kin

O T = O -1

v do cc ma trn X trong h thcO = e -iXphi l cc ma trn phn giao honXT = -X.Tng t nh vy, nu G l nhm cc bin i unita trong mt khng gian phc th cc yu t ca n

phi l nhng m trn unita U tha mn iu kinU + = U -v do cc ma trn X trong biu thcU = e -iXphi l cc ma trn t lin hp


24/127


X + = XNgoi ra, nu cc ma trn O hoc U c nh thc bng 1, ngha l nudet O = 1hocdet U = 1th cc ma trn X phi c vt bng khng,Tr X = 0Trong khng gian vect cc ma trn X tha mn cc iu kin suy ra t nh ngha ca nhm G cho

ta hy chn mt h c s gm cc ma trn c lp tuyn tnh Xi, i = 1, 2, . . ., s, m mi ma trn X angxt u c th c biu din di dng mt t hp tuyn tnh (42) ca cc ma trn Xi ca h c s nyvi cc h s i. Ta xt trng hp cc h s i l cc tham s thc. Cc ma trn X v T tng ng vicc tham s thc i, i = 1, 2, . . ., sc k hiu l X( 1, 2,. . ., s) v T ( 1, 2,. . ., s) . Ta c

X( 1, 2,. . ., s) =s

i=1 iXi (42)v theo cng thc (46)T ( 1, 2,. . ., s) = ei

PiiXi (47)

D th li rng

i

T(1,2,...,s)

i |1=2=...=s=0= Xi (48)cho nn Xi, i = 1, 2, . . ., s, l cc vi t ca nhm bin i G ang xt. Vi nhng gi tr v cng b cacc tham s 1, 2,. . ., sma trn T ( 1, 2,. . ., s) rt gn ma trn n v v c dng gn ng

T ( 1, 2,. . ., s) I - i

j jXi. (49)Cho hai ma trn T ( 1, 2,. . ., s) v T ( 1, 2,. . ., s) l hai yu t ca nhm G v hy thit lp ma

trnT ( 1, 2,. . ., s) T ( 1, 2,. . ., s) T ( 1, 2,. . ., s)-1 v T ( 1, 2,. . ., s)-1cng l mt yu t trong nhm G. Bng cch tnh trc tip c th th li rng vi nhng tham s 1 ,...,s

v 1,...,s tt c u l v cng b ta c biu thc gn ngT ( 1, 2,. . ., s) T ( 1, 2,. . ., s) T ( 1, 2,. . ., s)-1 v T ( 1, 2,. . ., s)-1I + (-

i)2s

i,k=1 ji[Xj , Xk] . (50)V ma trn ny l mt yu t ca nhm G rt gn ma trn n v cho nn theo cng thc (49) n phi

c dng gn ng

T ( 1, 2,. . ., s) T ( 1, 2,. . ., s) T ( 1, 2,. . ., s)-1 v T ( 1, 2,. . ., s)-1 I i

sl=1 fl (1, 1, ..., s; 1, 2, ..., s) Xl

trong fl (1, 1, ..., s; 1, 2, ..., s) l hm ca cc tham s 1, . . ., s v 1, . . ., s trit tiu khi cctham s 1, 2,. . ., s hoc 1, 2,. . ., sng thi bng khng. Trong php gn ng cp thp nht theocc tham s v cng b 1, . . ., s v 1, . . ., s ta c th vit biu thc ca fl (1, 1, ..., s; 1, 2, ..., s )di dng tng qut

fl (1, 1, ..., s; 1, 2, ..., s) is

j,k=1 jkjkl

vi cc h s khng i jkl, thnh thT ( 1, 2,. . ., s) T ( 1, 2,. . ., s) T ( 1, 2,. . ., s)-1 T ( 1, 2,. . ., s)-1I

sj,k,l=1 jkjklXl. (51)

So snh hai biu thc trong v phi cc h thc (50) v (51), ta thu c[Xj , Kk] =

sl=1 iklXk. (52)

Cng thc ny chng t rng cc vi t Xi, i = 1, 2 , . . ., s, ca nhm bin i G to thnh mt i sLie vi nh ngha tch ca hai yu t ca i s l giao hon t ca hai ma trn tng ng.

1.6 Ph lc c s l thuyt nhm6

Trong ph lc ny chng ta trnh by thm mt s khi nim v chng minh mt s mnh chung trongl thuyt nhm.



25/127

19

T nh ngha nhm suy ra ngay mt s mnh sau y.

1.6.1 Mnh 1

Mi nhm ch c mt yu t n v.Chng minh. Gi s trong nhm G c hai yu t n v l e1 v e2. Theo nh ngha yu t n v thvi mi yu t aG ta lun c

e 1 a = a,a e 2 = aTrong h thc th nht hy ly a = e2 v ce1 e2 = e2 ,cn li trong h thc th hai hy ly a = e1 v ce 1 e 2 = e 1Vy e1 phi trng vi e2 , e1 = e2

1.6.2 Mnh 2

Nghch o ca tch ca hai yu t bng tch cc nghch o ca chng theo th t ngc li, ngha l(a b) -1 = b -1a-1Chng minh. Thc vy, ta cb-1a-1a b = b -1b = ea b b -1a-1 = a a-1 = eVy b -1a-1 l nghch o ca a b.

1.6.3 Mnh 3

Nu a v b l hai yu t khc nhau ca nhm G,a = bth vi mi yu t c G ta lun lun cc a=c b, a c=b cChng minh. Ta thy gi s ngc li rng c mt yu t cno mc a = c b.Nhn c hai v h thc ny vi c -1 t bn tri v ch rng c-1c= e, ta cc-1ca= ea= c-1c b = e bngha l a = b, tri vi gi thit. Tng t nh vy, nu c mt yu t cno ma c = b cth sau khi nhn c hai v ca h thc ny t bn phi vi c-1, ch rng c c -1 = e, ta s ca c c-1 = a e = b c c-1 = b engha l li c a = b, tri vi gi thit.

1.6.4 Mnh 4

Mi yu t ca nhm ch c mt yu t nghch o.Chng minh. Ta gi s rng mt yu t ano ca nhm G c hai yu t nghch o k hiu la11 a = e,aa12 = e.Nhn c hai v ca h thc th nht vi a12 t bn phi v nhn c hai v ca h thc th hai vi a

11

t bn tri, ta ca11 aa

12 = ea

12 = a

12 ,

a11 aa12 = a

11 e= a

11 .

Vy ta phi c a11 = a12


26/127


1.6.4.1 nh ngha yu t lin hp

Yu t aca nhm G c gi l lin hp vi yu t b ca nhm ny nu c mt yu t no cG mc a c-1 = b

1.6.5 Mnh

Quan h lin hp l mt quan h tng ng, ngha l1) Nu a lin hp vi b th b lin hp vi a (tnh i xng).2) Yu t a lin hp vi chnh n (tnh t lin hp),3) Nu a lin hp vi b, b lin hp vi c th a lin hp vi c (tnh chuyn tip).Chng minh.1) Yu t alin hp vi yu t b c ngha l c mt yu t cno mc a c-1 = bKhi c-1b (c-1) -1 = a,ngha l a-1 lin hp vi b -1

2) Vi mi yu t aG ta lun ce a e-1 = a,ngha l at lin hp vi chnh n.3) Nu alin hp vi b, b lin hp vi c th c hai yu t d v f no md a d -1 = b, f b f -1 = c,Khi f d a d -1f -1 = c, (f d) a(f d) -1 = c,ngha l alin hp vi c.

1.6.6 nh ngha lp ln cn ca nhm con

Gi s nhm G c mt nhm con G1 gm nhng yu t g0 = e, g1, g2, . . ., v cho al mt yu t bt kca nhm G. Tp hp cc yu t a, ag1, ag2,. . ., thu c bng cch nhn tt c cc yu t ca G1 vi a t

bn tri c gi l lp ln cn tri ca G1 v k hiu laG1 : {agi | gi G1}.Tng t nh vy, tp hp cc yu t a, g1 a, g2 a. . ., thu c bng cch nhn tt c cc yu t ca G1

vi a t bn phi c gi l lp ln cn phi ca G1 v k hiu lG1 a: {gia | gi G1}

1.6.7 Mnh

Hai lp ln cn tri (phi) hoc khng c mt lp yu t chung no, hoc hon ton trng nhau.Chng minh. Ta chng minh mnh i vi lp ln cn tri. Vi lp ln cn phi c th lp li l lun

tng t. Gi s hai lp ln cn tri aG1 v b G1 ca nhm con G1 c mt yu t chung, ngha l c haiyu t g1 v g2 ca G1 m

a g1 = b g2 , g1g2G1Nhn c hai v ca h thc ny vi g11 t bn phi, ta ca= bg2g11Mi yu t ca lp ln cn tri a G1 c dnga gk = b g2 g11 gk, gkG1.Vg 2 g11 g k G 1cho nn b g2g11 gk l mt yu t ca lp ln cn b G1. Vy mi yu t ca a G1 u thuc vo b G1,

ngha la G 1 b G 1


27/127

21

Tng t nh vyb G 1 a G 1Hai h thc ny chng t rng hai lp ln cn a G1 v b G1 phi trng nhaua G1 = b G1Ngc li, nu chng khng trng nhau th chng khng th c yu t chung no.Xt mt nhm hu hn G v cp n v gi s n c mt nhm con G1 cp n1. T Mnh va chng

minh suy ra rng nhm G c tch ra thnh cc lp ln cn khng giao nhau ca nhm con G1, mi lpu c cng mt s yu t bng s yu t n1 ca nhm con G1. Nu c r lp ln cn th s yu t ca nhmG l

n = r n 1Vy ta c mnh sau,

1.6.8 Mnh (nh l Lagrange)

Cp ca nhm con G 1 ca nhm hu hn G l c s ca cp ca nhm G

1.6.9 nh ngha nhm con bt binNhm con H ca nhm G c gi l nhm con bt bin nu vi mi yu t a ca nhm G lp ln cn triaH trng vi lp ln cn phi H a:

a H = H a.Ta cn vita H a-1 = HH thc ny chng t rng vi mi yu t b ca nhm con H,bH,ta lun lun ca b a-1 Hvi bt k mt yu t ano ca nhm G,aG.Vy theo nh ngha, nu nhm con bt bin H cha mt yu t b no th n cng cha tt c cc

yu t lin hp vi b. Ni khc i, mt nhm con bt bin bao gi cng cha gn ton b tng lp cc yut lin hp.

Cho mt nhm G v mt nhm con bt bin H ca n. Mi yu t akhng thuc vo H hon ton xcnh lp ln cn a H v c th xem l yu t i din ca lp ny. Cho hai lp a H v b H i din bi haiyu t av b v xt tp hp cc yu t l tch ca mt yu t ca lp a H v mt yu t ca b H. Tp hpny c k hiu l a H b H. V H l nhm con bt bin cho nn

H b = b H, a H b H = a b H Hv do a H b H a b H.Vy tt c cc tch ang xt u l cc yu t c lp a b H m i din l yu t a b.

1.6.10 nh ngha nhm thng

Cho nhm G v nhm con bt bin H ca n. Trn tp hp cc lp ln cn ca nhm H ta nh ngha phpnhn nh sau: tch ca hai lp ln cn a H v b H l lp ln cn a b H. Yu t n v ca php nhn ny lchnh nhm con H. Yu t nghch o ca lp ln cn a H l lp ln cn a-1H. Vi php nhn, yu t nv v yu t nghch o nh ngha nh vy, tp hp cc lp ln cn ca nhm con H to thnh mt nhmgi l nhm thng G/H ca nhm G i vi nhm con bt bin H. Tnh cht kt hp ca php nhn trnG/H suy ra t tnh cht kt hp ca php nhn trn G.


28/127



29/127

Chng 2

C s l thuyt biu din nhm

2.1 Khi nim v biu din nhm1

2.1.1 nh ngha biu din nhm

Cho mt nhm G gm cc yu t e, a, b, c,. . . m bn cht l ty v mt nhm Tcc php bin i tuyntnh trong mt khng gian vect L. Ta gi nhm T cc php bin i trong khng gian L l mt biu dinca nhm G nu c mt php ng cu ca nhm G ln nhm T, ngha l nu ng vi mi yu t a, b, c,. . .ca nhm G c php bin i T(a), T(b), T(c),. . . trong nhm Tm s tng ng ny bo ton php nhnnhm:

Ta ni c mt biu din ca nhm G trong khng gian L v khng gian L thc hin biu din T canhm G. Th nguyn ca khng gian L gi l th nguyn ca biu din T. Ma trn ca cc php bin i

T(a) i vi mt h c s no trong khng gian L cng c k hiu l T(a). T nh ngha suy ra cctnh cht sau y.1) ng vi yu t n v e ca nhm G l php bin i ng nht I trong khong gian L:T(e) = I.Thc vy, ta c

Vi a G. Nhng ae= ea= a, do T(ae) = T(ea) = T(a).VyT(a), T(e) = T(e), T(a) = T(a),

bin i T(e) phi l yu t n v trong nhm T.2) Bin i T(a-1) ng vi yu t nghch o a-1 ca al nghch o ca bin i T(a) ng vi yu ta:

T(a-1) = [T (a)] -1.Thc vy,


23


30/127

24 CHNG 2. C S L THUYT BIU DIN NHM

Vi aG. Nhng aa-1 = a-1a= e, do T(a) T(a-1) = T(a-1) T(a) = T(e) = IVy bin i T(a-1) phi l nghch o [T (a)]-1 ca T(a).Ta bit rng nhm SU(2) ng cu vi nhm SO(3). Ta thy cc bin i ca nhm SO(3) to thnh

biu din ca nhm SU(2). Trong vt l ngi ta m rng khi nim biu din v cn coi nhm SU(2) lbiu din ca nhm quay SO(3). Trong trng hp ny ng vi mt yu t ca nhm SO(3) c hai bin ikhc nhau thuc nhm SU(2). Ta ni rng nhm SU(2) l biu din lng tr ca nhm SO(3).

Mi nhm u c nhiu biu din, trong c nhng biu din tng ng vi nhau theo nh nghasau y.

2.1.2 nh ngha biu din tng ng

Cho hai biu din T(1) v T(2) ca cng mt nhm G trn hai khng gian vect L1 v L2. Hai biu dinny c gi l tng ng vi nhau nu gia hai khng gian vect L1 v L2 c mt php nh x tuyntnh n gi theo c hai chiu

X : L1L2,X-1 : L2L1(ng vi mt vect ca L1 c mt vect ca L2 v ng vi mt vect ca L2 c mt vect ca L1) m vi

mi yu t aca nhm G hai php bin i tuyn tnh ca T(1)(a) vT(2)(a) lin h vi nhau bi cng thc

Cc biu din tng ng c mt s ging nhau su sc c din t trong mnh di y.

2.1.3 Mnh

Nu T (1) v T (2) l hai biu din tng ng th ta c th chn hai vect c s trong hai khng gianvect L 1 v L 2 thc hin hai biu din ny th no cc yu t ma trn ca cc php bin i T (1)(a) v T (2) (a) hon ton trng nhau vi mi a G.

Do khi nghin cu biu din nhm ta khng phn bit cc biu din tng ng vi nhau v coi ttc cc biu din tng ng vi nhau ch l mt biu din.

Khng gian L thc hin biu din T ca nhm G c th qu ln v bao gm mt khng gian con btbin L1 theo ngha sau y. Tt c cc php bin i T(a) vi mi aG khi tc dng ln mt vect bt kca L1 u cho kt qu l cc vect hon ton nm trong L1:

aG, r1L1 : T(a)r1L1Khi cc php bin i T(a) u c th c xem l cc php bin i T1(a) ca khng gian L1T1(a) r1n T(a)r1, r1L1.Cc php bin i T1(a) ng vi tt c cc yu t aca nhm G to thnh biu din T1 ca nhm ny

trong khng gian L1. Ta ni rng trn khng gian con bt bin L1 biu din T quy v biu din T1, v cnh ngha sau y,

2.1.4 nh ngha biu din kh quy v biu din ti gin

Cho mt biu din T ca nhm G trong khng gian vect L. Nu trong L c mt khng gian con L1 btbin i vi tt c cc php bin i T(a) ca biu din T, vi mi yu t aca nhm G, th ta ni rng Tl mt biu din kh quy. Trong trng hp ngc li nu trong khng gian L khng c mt khng gian conno bt bin i vi tt c cc php bin i vi tt c cc php bin i T(a), tr hai khng gian con tmthng l chnh khng gian L v khng gian con bng khng, th ta ni rng T l biu din ti gin.

Xt biu din kh quy T trong khng gian n chiu L, trong c khng gian con bt bin m chiu L1,m < n. Khng gian L l tng ca khng gian con L1 v mt khng gian con (n m) chiu L2 no .

L = L1 + L2Khng gian con L2 c th khng phi l khng gian con bt bin, m cng c th l khng gian con bt

bin. Gi e1, e2, . . ., en l h vect c s trong L1, v em+1, em+2, . . ., en l h vect c s trong L2. Ta hy


31/127

25

chn cc vect ny lm h vect c s trong L v xt tc dng ca cc php bin i T(a) ln cc vect .K hiu cc yu t ma trn l Tij (a), ta c

Vi i m tt c cc vect T(a)ei u nm trong L1, do trong v phi cng thc (2) va vit trnch c th c cc vect ej vi j m. VyTji (a) = 0 nu jm v j > m,ngha l cc ma trn T(a) ca biu din kh quy ang xt phi c dng sau y:

Tt c cc yu t ma trn nm trong bn tri pha di phi bng khng.Gi s rng khng gian con L2 li cng l khng gian con bt bin. Khi , vi mi ei m i > m, tt c

cc vect T(a)ei u nm trong L2 v do biu thc khai trin ca vect ny theo ej khng th cha ccvect ej vi j m . Vy

Tji (a) = 0 Nu i >m v jm,ngha l cc ma trn T(a) by gi c dng

Trn hai khng gian con bt bin L1 v L2 biu din T quy v hai biu din T(1) v T(2) gm cc phpbin i hon ton c lp vi nhau. Nu cc biu din ny hoc mt trong hai biu din ny l kh quy,ngh l c L1 ln L2 hoc mt khng gian con trong s hai khng gian ny li cha khng gian con bt bin,th ta lp li cc lp lun trn. Nu ln no khng gian thc hin biu din kh quy cng tch thnh haikhng gian con bt bin nh va trnh by trn, th c tip tc tch cc khng gian con cho n khi khngth tch c na ta s i n kt qu cui cng sau y: Khng gian vect L tch thnh cc khng giancon bt bin L1, L2, . . ., Lf m trn mi khng gian con Li ny biu din T quy v mt biu din ti ginT(i) . Ta i n nh ngha sau y.

2.1.5 nh ngha biu din hon ton kh quyBiu din T ca nhm G trong khng gian vect L c gi l hon ton quy nu L l tng ca cc khnggian con bt bin L1, L2, . . ., Lf m trn mi khng gian con L i ny biu din T quy v mt biu din tigin T(i).

Ma trn ca cc bin i T(a) ca mt biu din hon ton kh quy c dng tng qut sau y, gi ldng cho ,


32/127


trong cc cho l cc ma trn ca cc biu din ti gin, cn tt c cc khng cho u c cc yut ma trn bng khng.

Gi s khng gian vect L thc hin biu din T ca nhm G l mt khng gian Euclide phc m trn ta c nh ngha tch v hng ca hai vect l dng song tuyn tnh xc nh dng ca cc thnh phnca cac vect ny (tuyn tnh i vi mt vect v phn tuyn tnh i vi vect kia). Bit tch v hngca hai vect bt k, ta c th nh ngha hai vect trc giao vi nhau. Cho mt khng gian con L1 ca L.

Tt c cc vect trong L trc giao vi L1 to thnh khong gian con L2 gi l phn ph trc giao ca L1.Khng gian L l tng ca L1 v L2. Ta vit L = L1L2. Bit tch v hng ca hai vect bt k, ta cn cth nh ngha ton t unita l ton t thc hin php bin i tuyn tnh bo ton tch v hng ca ttc cc vect. C c cc ton t unita, by gi ta c th nh ngha biu din unita.

2.1.6 nh ngha biu din unita

Biu din T ca nhm G trong khng gian Euclide phc L gi l biu din unita nu vi tt c cc yu taca nhm G tt c cc php bin i T(a) u l cc ton t unita:

[T (a)]+ = [T (a)]-1, aG.Cc biu din unita c cc tnh cht c bit sau y.1. Trong khng gian L thc hin biu din unita T ca nhm G phn ph trc giao L 2 ca mi khng

gian con bt bin L 1

T(a) L1L1, aG.L = L1L2,cng l mt khng gian bt bin,T(a) L2L2, aG,2. Mi biu din unita kh quy u hon ton kh quy.C th chng minh c rng mi biu din ca mt nhm hu hn u tng ng vi mt biu din

unita. Trong gio trnh ny chng ta ch xt cc biu din unita cho nn khi khng tht cn thit th takhng nhc n t unita na.

2.1.7 Biu din ca nhm Lie

Xt trng hp G l mt nhm Lie m mi yu t aca n c xc nh bi n tham s c lp jc thnhn nhng gi tr thc thay i lin tc, yu t n v l yu t m tt c cc tham s jc gi tr bng

khng. Xt mt biu din T ca nhm Lie ny trong khng gian vect L. ng vi yu t c nhm m cctham s c lp c cc gi tr j ta c mt ton t T( 1, 2, ..., s) m cc yu t ma trn i vi mt hvect c s bt k trong khng gian L l cc hm kh vi c cc tham s 1, 2, ..., s. Khi tt c cc thams bng khng th ton t T( 1, 2, ..., s) l hon t n v:

T(0, 0, . . .,0) = I.Xt cc ton t T( 1, 2, ..., s) ng vi c gi tr v cng b ca cc tham s c lp j . Ch gi li

cc s hng cp mt v b qua cc s hng cp cao hn, ta c th vitT (1, 2, ..., s) I+

nj=1

T(1,2,...,s)j

|1=...=s=0j .

t


33/127

27

Cc ton t

c gi l cc vi t ca biu din T ca nhm Lie G trong khng gian L.

2.1.8 Biu din nhm Lie v biu din i s Lie

Cho mt nhm Lie G cc php bin i tuyn tnh T( 1, 2, ..., s) ph thuc s tham s c lp1, 2, ..., sca mt khng gian vect V v gi s c mt biu din ca nhm ny trong khng gianvect V, ngha l c mt nhm Lie Gcc php bin i T( 1, 2, ..., s) ca khng gian Vv mt phpng cu ca G ln G.

T( 1, 2, ..., s) T( 1, 2, ..., s)

K hiu Xiv Xi, i = 1, 2, . . ., s l cc vi t ca cc nhm bin i G v G. Php ng cu ca G lnGko theo php nh x i s Lie cc vi t Xi ln i s Lie cc vi t Xi,Xi Xi, i = 1, 2, . . ., sm ng vi mt vi t Xi ch c mt vi t duy nht Xic nh l Xi. Xt yu t

ca nhm G. Trong php ng cu ca G ln Gyu t ny c nh l yu t

V hai yu t (8) v (9) biu din ging nh nhau qua cc giao hon t [Xi, Xj ] v

Xi, Xj

, theo th

t, cho nn gio ha t [Xi, Xj ] c nh l giao hon t

Xi, Xj

:

Vy php ng cu ca nhm G ln nhm Gc h qu l php ng cu gia i s Lie cc vi t Xi canhm G v i s Lie cc vi t

Xi, X

j

ca nhm G. p dng s ng cu ny cho trng hp mi nhm

Lie cc php bin i v mi biu din ca n ta c th ni rng i s Lie cc vi t ca mt biu din canhm Lie G cc php bin i ca mt khng gian vect ng cu vi i s Lie ca chnh nhm G.

2.1.9 Cc nh l v biu din ti gin

Cc biu din ti gin c tnh cht sau y.

2.1.9.1 B Shur

Nu trong khng gian L thc hin biu din ti gin T ca nhm G c mt ton t A khc khng v giaohon vi tt c cc ton t T(a)ca biu din T, a G , th ton t A phi l bi ca ton t n v

A = I (2.1)

Trong trng hp G l mt nhm Lie, T l mt biu din vi cc vi t X i ,i=1,2, . . .,s,


34/127


2.1.9.2 B Shur i vi nhm Lie

Nu trong khng gian L thc hin biu din ti gin T ca nhm G c mt ton t A khc khng v giaohon vi tt c c ton t A khc khng v giao hon vi t c cc vi t X i , i = 1, 2, . . ., s ca biu din

T, th ton t A phi l bi ca ton t n vA = I .Cui cng ta dn ra y hai nh l v biu din nhm hu hn thng c s dng trong vt l.Gi s nhm hu hn G cp N chia lm Nk lp cc yu t lin hp. Ta c nh l sau y.

2.1.10 nh l v s cc biu din ti gin khng tng ng

S f cc biu din ti gin khng tng ng vi nhau ca mt nhm hu hn G bng s N k lp cc yut lin hp ca nhm ny.

f = N kGi s nhm hu hn G cp Nc f biu din ti gin khng tng ng vi nhau T(), biu din T()

c th nguyn d. Cc gi tr d phi tun theo nh l sau y.

2.1.11 nh l v th nguyn ca cc biu din ti gin khng tng ngCc th nguyn d ca tt c cc biu din ti gin khng tng ng ca mt nhm hu hn G cp N

phi tha mn h thc.f=1 d

2 = N

2.2 Cc php tnh i vi cc biu din2

T hai biu din T(1) v T(2) ca mt nhm G, ta c th thit lp c mt biu din gi l tch ca chngv k hiu l T(1)T(2). T mt biu din Tno ca nhm G ta c th thit lp c mt biu din

~

Tgi l biu din lin hp vi biu din T. Cc biu din ny c cc nh ngha nh sau.

2.2.1 nh ngha tch ca hai biu dinCho hai biu din T(1) v T(2) ca mt nhm hu hn G trong cc khng gian vect L1 v L2 vi cc hvect c s e(1)1 , e

(1)2 , . . ., e

(1)d1 , v e

(2)1 , e

(2)2 , . . ., e

(2)d2 , d1 v d2 l th nguyn ca L1 v L2. Tch ca hai biu

din T(1) v T(2) l biu din T trong khng gian L1L2 th nguyn d1d2 vi h vect c s.

m ton t T( ) tng ng vi yu t ca nhm G c xc nh nh sau

trong T(1) (a) v T(2) (a) l hai ton t trong hai khng gian L1 v L2 tng ng vi yu t acanhm G. Ta vit

T = T(1)T(2). chng minh rng cc ton t T(a) to thnh mt biu din ca nhm G, ngha l tha mn iu kin

bo ton php nhn nhmT(a) T(b) = T(ab),ta ch cn dng nh ngha (12) v tnh cht bo ton php nhn nhm ca cc biu din T(1) v T(2),

c th lT()(a) T()(b) = T()(ab), = 1, 2



35/127

29

K hiu cc yu t ma trn ca ton t T(1)(a) v T(2)(a) trong cc h vect c s cho e(1)1 , e(1)2 , . . .,

e(1)d1 v e

(2)1 , e

(2)2 , . . ., e

(2)d2 l T

1ij(a) v T2kl(a):

T(1) (a) e(1)i = e(1)j T

(1)(ji)( )

T(1)(a) e(2)k = e(2)l T(2)(lk)( )Ta cT(a)f(ik) = T(a)( e

(1)i e

(2)k ) = (T(1)(a) e

(1)i ) (T(2)(a) e

(2)k ) = ( e

(1)j e

(2)l ) T

(1)(ji) ( ) T

(2)(lk)( ) =

f(jl)T(1)(ji) ( ) T(2)(lk)( )

So snh hai biu thc ca T(a) f(ji), ta thu c h thc din t cc yu t ma trn ton t T(a) quacc yu t ma trn cc nhm ton t T(1)(a) v T(2)(a)

Cho hai biu din (unita) ti gin T() v T() ca mt nhm G no trn cc khng gian L() vL() vi th nhim d() v d(). Tch

T = T()T()

ca hai biu din ny l mt biu din unita trn khng gianL = L()L()Nu Tkhng phi l ti gin th n hon ton kh quy v c th phn tch thnh tng trc giao ca cc

biu din ti gin T() trn cc khng gian L( ) th nguyn d( ). Trong s cc biu din ti gin Tny cth c cc biu din tng ng vi nhau. Khng gian L thc hin biu din T l tng trc giao ca cckhng gian con L( ) thc hin cc biu din ti gin T()

L =

L().

Th nguyn ca L ld =

d

().Mt khcd = d()d()Vy ta c h thc

K hiu cc h vect n v c s trong cc khng gian L(), L(), L( ) , v.v. . . le()i , ia= 1, 2, . . ., d()

e()i , ia= 1, 2, . . ., d()

e()i , ia= 1, 2, . . ., d()

v.v. . . Trong khng gian L cc vect c dnge()i e

()i

to thnh mt vect c s trc giao chun ha Tp hp tt c cc vect e()i , iy = 1, 2, . . ., d( ), vi mich s c mt trong v phi cng thc (14) cng l mt h cc vect c s trc giao chun ha khc trongkhng gian L. Gia cc vect n v ca hai h ny ta c cc php bin i unita sau y

Cc h s Ciii

trong cc php bin i (15) v (16) gi l cc h s Clebsh-Gordan.

By gi ta a vo khi nim biu din~

T lin hp vi mt biu din T cho. Gi s T(a) l cc tont tuyn tnh ca biu din T ca nhm G trong khng gian vect L. Vi mi yu t a ca nhm G ta hy


36/127


thit lp ton t sau y

Vi cc yu t ma trn

Ta hy th li bng s tng ng gia cc yu t a ca nhm G v cc ton t~

T(a) bo ton php nhnnhm. Thc vy, ta c

~

T(ab) =

T (ab)1T

=

T

b1

T

a1T

=

T

a1T

T

b1T

=~

T(a)~

T(b).

Vy ton t~

T(a) cng to thnh mt biu thc biu din ca nhm G. Ta c nh ngha sau y.

2.2.2 nh ngha biu din lin hp

Cho hai biu din T v~

T ca cng mt nhm G trong hai khng gian vect L v~L. Nu trong hai khng

gian L v~

L ta c th chn hai h vect c s mt cch thch hp cc yu t ma trn Tij(a) v~

Tij(a) cacc ton t T(a) v

~

T(a) ca hai bin i ny lin h vi nhau bi cng thc~

Tij(a) = Tji(a-1),

th ta gi T v~

T l hai biu din lin hp vi nhau.

Vic xt ng thi hai biu din lin hp vi nhau Tv~

T cho php ta thit lp c mt i lng bt

bin i vi php bin i ca nhm G. Thc vy, trong hai khng gian L v~

Lthc hin hai biu din lin

hp vi nhau T v~

T ta hy chn cc h vect c s e1, e2, . . ., edv f1, f2, . . .., f d co cc yu t matrn ca cc ton t T(a) v

~T(a) tha mn h thc (18). Trong khng gian vect d2 chiu L

~L ta hy xt

vect sau y.

K hiu tch ca hai biu din T v~

T v T~

T. Cc hon t ca biu din ny tc dng ln cc vect

c s ca khng gian tch L~L nh sau

(T~

T) (a) (eifj) = (T(a)ei) (~

T(a)fj) = (ekfl) Tki(a)~

Tlj(a)Tc dng ca cc hon t ln vect i xc nh bi cng thc (19) v dng h thc (18) gia cc yu

t ma trn ca cc hon t T(a) v~

T(a), ta c

(T ~

T ) (a) i = (T(a)e m ) (~

T (a)f m ) = e k f l T km (a)~

T lm (a) = e k f l Tkm (a)T ml (a-1) = ek f l T kl (e) = ek f k = i

Vy ta c nh l saunh l. Vecti =

dm=1 em fm

trong khng gian L ~L thc hin biu din T

~T ca nhm G bt bin i vi mi php bin

i (T ~

T )(a) ca biu din tch T ~

T . Do khng gian con mt chiu vi vect n v i thc

hin mt biu din ti gian mt chiu cha trong biu din T ~

T .H qu. Biu din T

~T, l tch ca mt biu din T v biu din

~T lin hp vi n, bao gi cng cha

biu din ti gin mt chiu.


37/127

31

Biu din ti gin mt chiu c thit lp trong khi chng minh nh l va trnh by trn thngdin t cc i lng vt l bin i vi cc php bin i nhm i xng. Do trong cc bi ton vt lta thng s dng khi nim biu din lin hp.

Tch ca hai biu din ca nhm Lie. Cho hai biu din T(1) v T(2) ca nhm Lie G trong haikhng gian vect L1 v L2, T l tch ca hai biu din ny. Cc ton t T( 1, 2, ..., s) ca biu din Tc dng

K hiu cc vi t ca cc biu din T(1) v T(2) l X1j v X2j , j = 1, 2, . . ., s ca biu din T l Xj, j =1, 2, . . ., s. Vi cc thng s jv cng b ta c

trong I(1)

v I(2)

l cc ton t n v trong cc khng gian L1 v L2. Thay cc biu thc (21) votrong v phi cng thc (20) v ch gi li cc s hng cp mt theo cc thng s j , ta cT( 1, 2, ..., s) = I - i

sj=1 j

X

(1)j I

(2) + I(1) X(2)j

,

trong I = I (1) I (2)l ton t n v trong khng gian L = L(1)L(2). So snh vi nh ngha ca cc vi t Xj ,

ta suy ra

vit h thc ny di dng cha tng minh cc yu t ma trn trong hai khng gian L1 v L2 tahy chn hai h vect c s e1m, m = 1, 2, . . ., d1 v e2p, p = 1, 2, . . ., d2, sau ta ly cc vect sau ye(mp) = e

(1)m e

(2)p

trong khng gian L = L1L2, m = 1, 2,. . ., d1, p = 1, 2, . . ., d2, lm h c s ca khng gian ny. Khiu cc yu t ma trn ca cc ton t X(1)j , X

(2)j v Xj i vi cc h c s tng ng ni trn vect l (

X(1)j )mm, ( X

(2)j )pp, v ( Xj)(mp)(mp). Cng thc (23) cho ta

Cui cng, ta xt hai biu din lin hp vi nhau T v~

T ca mt nhm Lie G v k hiu cc ton tca hai biu din ny l T( 1, 2, ..., s) v

~

T( 1, 2, ..., s), k hiu cc vi t tng ng vi cc tham s

thc c lp j l Xj v~

Xj . Ch rng nu al mt yu t ca G vi cc tham s v cng b j th trong

php gn ng cp mt yu t vi cc tham s - j s l nghch o a-1 ca a. Do ta c cc cng thc

T(a-1) T( 1, 2, ..., s) I + in

j=1 jXj

v do


38/127


Mt khc

Theo nh ngha cc biu din lin hp vi nhau ta phi c~

T(a) =

T

a-1T

Thay vo y cc biu thc (26) v (27), ta thu c h thc lin h cc vi t Xj v~

Xj ca hai biudin lin hp vi nhau:

Nu bit cc vi t ca mt biu din T no , dng h thc (28) ta thit lp c ngay cc vi t ca

biu din~

T lin hp vi T.

2.3 Hm c trng ca biu din3

Cho mt biu din T ca nhm G trong khng gian vect L th nguyn d. Trong khng gian L hy chnmt vect n v c s e1, e2, . . ., ed v k hiu cc yu t ma trn ca cc ton t tuyn tnh T(a), aG,i vi h n v c s ny l Tij(a), i, j = 1, 2, . . ., d,

Thc hin mt php bin i tuyn tnh X, ta chuyn h vect n v c s cho e1, e2, . . ., edthnhmt h vect mi e1, e2, . . ., ed. i vi h mi ny cc ton t tuyn tnh T(a) c cc yu t ma trnTij (a),

Hy tm mi lin h gia cc yu t ma trn Tij(a) v Tij(a). K hiu cc yu t ma trn ca ton t Xi vi h vect n v c s e1, e2, . . ., ed l Xij. Ta c

Thay biu thc (31) ca eivo c hai v ca h thc (30), ta thu c

Dng biu thc (29) ca T(a) ej, ta vit li cng thc (32) nh saue k Tkj (a) X ji = e k Tkj (a) Tji (a) .Vy

Nhn c hai v ca h thc (33) vi (X-1)lk ri cng theo k t 1 n d, ta thit lp c h thc giaTij(a) v Tij (a):

Vy trong hai h vect n v c s lin h vi nhau bi h thc (31), ton t T(a) c cc yu t ma

trn khc nhau Tij(a) v T

ij (a) lin h vi nhau bi cng thc (34).T cc yu t ma trn khc nhau Tij (a) v Tij (a) ca cng mt ton t T(a) ta c th thit lp cmt i lng c trng cho biu din T m khng ph thuc v s la chn h c s. Thc vy, t l= itrong c hai v ca h thc (34) ri cng theo i t 1 n d, ta c

TiI(a) = (X -1 ) ik T kj (a) X ji = T kj (a) X ji (X -1 ) tk = T kj (a) jk = T kk (a)Vy vt ca ma trn ca php bin i T(a) khng ph thuc s la chn h vect n v c s v c

th c dng lm i lng c trng cho biu din T m ta ang xt. Ta c nh ngha sau y.3Phin bn trc tuyn ca ni dung ny c .


39/127

33

2.3.1 nh ngha hm c trng ca biu din

Cho mt biu din T ca nhm G trong khng gian vect L. Vt ca cc ma trn php bin i T(a) cabiu din ny, aG, khng ph thuc s la chn h vect n v c s trong khng gian L v c gi l

hm c trng (a)ca biu din T:

2.3.2 Cc mnh v hm c trng

T nh ngha ca hm c trng ca biu din suy ra mt s mnh c bn.

2.3.2.1 Mnh 1

Cc biu din tng ng c cng mt hm c trng.Chng minh. Gi s c hai biu din tng ng T(1) v T(2) ca cng mt nhm G trong hai khng

gian vect L1 v L2. Khi c mt ton t tuyn tnh X chuyn cc vect ca khng gian L1 thnh cacvect khng gian L2 sao cho

K hiu (i)(a) l cc hm c trng ca hai biu din cho(1)(a) = Tr

T(1) (a)

,

(2)(a) = Tr

T(2) (a).

Tnh vt ca cc ma trn ca cc ton t trong hai v ca h thc (36) i vi cc vect n v c sbt k v dng tnh cht sau y ca vt ca tch hai ton t A v B.

Tr [AB] = Tr [BA],ta thu c(2)(a) = Tr

T(2) (a)

= Tr

XT(1) (a) X1

= Tr

X1XT(1) (a)

= Tr

T(1) (a)] = (1)(a)

Vy hm c trng (1)(a) v (2)(a) ca hai biu din tng ng T(1) v T(2) bng nhau.Cho mt biu din hon ton kh quy T trong khng L th nguyn d, l tng trc giao ca hai biu

din T(1) v T(2) trong hai khng gian con bt bin L1 v L2 th nguyn d1 v d2, d = d1 + d2. K hiu

cc hm c trng ca cc biu din T, T(1) v T(2) l (1)(a) v (2)(a). Cc hm c trng ny khngph thuc s la chn cc h vect n v c s trong cc khng gian vect L, L1 v L2. thun tin khithit lp gia cc hm c trng ny hy chn cc h vect n v c s e1, e2, . . ., ed trong khng gian L1v f1, f2, . . ., fd2trong khng gian L2 ri chn cc vect e1, e2, . . ., ed, f1, f2, . . ., fd2 lm h n v c strong khng gian L. i vi h ny ma trn ca cc php bin i T(a) c dng cho theo nh sau

T y suy ra rng(a) = Tr [T (a)] = Tr

T(1) (a)

+

T(2) (a)

= (1)(a) + (2)(a)M rng lp lun trn cho trng hp biu din T l tng trc tip ca cc biu din ti gin khng

tng ng T() vi = 1, 2,. . ., m biu din ti gin T() c cha nln trong biu din T, ta cmnh sau y.

2.3.2.2 Mnh 2

Nu biu din hon ton kh quy T l tng trc giao ca cc biu din ti gin khng tng ng T()vi = 1, 2, . . ., m biu din ti gin T() c cha nln trong biu din T, th hm c trng ()(a)


40/127


ca cc biu din T() nh sau:() =

n

() (a).Hm c trng () ca mt biu din T l mt hm trn nhm. Xt gi tr ca hm ny trn hai yu

t lin hp vi nhau av b a b -1, ta c mnh sau.

2.3.2.3 Mnh 3

Trn hai yu t lin hp vi nhau av b a b -1, trong av b l hai yu t ty ca nhm G, hm ctrng () ca mt biu din T c cng mt gi tr, ngha l

() = (b a b -1), a G, b G.Chng minh. Theo nh ngha ca hm c trng ta c() = Tr [T (a)],(b a b-1) = Tr

T

bab1

= Tr

T (b) T (a) T

b1

= Tr {T (b) T (a)

T

b1

} = Tr{

T

b1

T (b) T (a)} = Tr [T (a)] = ()

Vy mnh c chng minh.Theo mnh ny trn tt c cc yu t ca mt lp cc yu t lin hp hm c trng c cng mt

gi tr. Vy hm c trng cng c th xem l trn tp hp cc lp K cc yu t lin hp.K= {bab1 | b G}Ta vit() = (K ).C mt nh l thng dng v hm c trng ca cc biu din ti gian khng tng ng ca nhm

hu hn. Gi s c nhm hu hn G v k hiu ()(a) l cc hm c trng ca cc biu din ti ginkhng tng ng T(). Ta hy coi N gi tr ()(a) l N thnh phn ca mt vect trong khng gianEuclide phc N chiu v nh ngha tch v hng ca hai hm c trng () v () nh l tch v hngca hai vect chia cho N

2.3.3 nh l v tnh trc giao chun ha ca cc hm c trng

Cc hm c trng ca cc biu din ti gin khng tng ng vi nhau T() ca nhm hu hn Gtha mn iu kin trc giao chun ha

nh l ny c mt s h qu thng c s dng. Gi s c mt nhm hu hn G v ta bit ttc cc hm c trng ()(a) ca tt c cc biu din ti gin khng tng ng vi nhau T() ca nhmny. Cho mt biu din T bt k ca nhm G v gi s rng ta bit c hm c trng ()(a) cabin T. Khi ta c th xc nh c ngay rng biu din T c cha biu din ti gin T() hay khng,v nu c cha t cha bao nhiu ln. Thc vy, theo Mnh 2, nu T cha T()nln, th

(a) =

a n() (a)

Ly tch v hng c hai v ca h thc ny vi hm c trng ()(a) no v dng cng thc (37),ta thu c

n =

()

, .

2.3.3.1 H qu 1

Cho () (a) l cc hm c trng ca cc biu din ti gin T() ca mt nhm hu hn G, T l mtbiu din no vi hm c trng (a) . Biu din T cha bin din T ( ) mt s ln bng

n =

(), .

Hy xt tch v hng ca hm c trng ca mt biu din ty T vi chnh n v gi l i lngthu c l bnh phng v hng ca hm c trng. T Mnh 2 v nh l v tnh trc giao chunha ca cc hm hm c trng suy ra rng.


41/127

35

(, ) =

n(),

n

()

=

n2.

Nu T l mt biu din ti gin th trong s cc s nguyn nch c mt s khc khng v bng 1. Khi

(, ) = 1.Cn nu T l mt biu din kh quy th t nht c hai s n ln hn hoc bng 1.

2.3.3.2 H qu 2

Nu mt biu din ca nhm hu hn G l ti gin th bnh phng v hng ca hm c trng ca nbng 1, cn nu biu din l kh quy th bnh phng v hng ca n ln hn 1.

2.4 Ph lc c s l thuyt biu din nhm4

Trong ph lc ny chng ta pht biu v chng minh mt s nh l c bn trong l thuyt biu din nhm.

2.4.1 Biu din tng ng u chng ta c biu thc (1) lin h hai php bin i tng ng. Cc biu din tng ng cmt s ging nhau su sc c din t trong mnh di y.

2.4.1.1 Mnh

Nu T(1) v T(2) l hai biu din tng ng th ta c th chn hai h vect c s trong hai khng gianvect L1 v L2 thc hin hai biu din ny th no cc yu t ma trn ca cc php bin i T(1)(a) vT(2)(a) hon ton trng nhau vi mi aG.

Chng minh. Gi s e1, . . ., en l h vect c s trong khng gian L1 v trong h ny php bin iT(1)(a) c cc yu t ma trn D(1)ij (a):

Trong khng gian vect L2 ta hy chn h vect c s: f i, i = 1, 2, . . ., n nh sau

v k hiu cc yu t ma trn ca php bin i T(2)(a) i vi h vect c s ny l D(2)ij (a):

Thay T(2)(a) bng biu thc (1) lin h n vi T(1)(a)X T(1)(a) X-1f i = f jD(2)ji (a)ri nhn c hai v cng thc ny vi X-1 , ta c

Nhng theo nh ngha (40) ca fi , ta li c

Vy cng thc (42) tr thnh

So snh hai biu thc (39) v (40) ca T(1)(a) ei , ta suy ra ngayD

(2)ji (a)= D

(1)ji (a),



42/127


ngha l cc yu t ma trn D(1)ij (a)v D(2)ij (a)ca hai php bin i T(1)(a) v T(2)(a) i vi cc h

vect c s e1, e2, . . ., en v f1, f2, . . ., fn hon ton trng nhau.Chnh v c th chn cc vect c s mt cch thch hp cho cc biu din tng ng c chung nhau

cc yu t ma trn, cho nn ta khng cn phn bit cc biu din tng ng v xem chng nh l mtbiu din. Ch c cc biu din khng tng ng mi thc s l nhng biu din khc nhau.

2.4.2 Biu din unita

Cc biu din unita c cc tnh cht c bit sau y.

2.4.2.1 nh l 1

Trong khng gian L thc hin biu din unita T ca nhm G phn ph trc giao L 2 ca mi khng gian conbt bin L 1

T(a) L1L1, a GL = L1L2cng l mt khng gian con bt bin,

T(a) L2L2, a GChng minh. Ta s dng tnh cht unita ca cc ton t T(a) vi mi yu t A G. K hiu tch v

hng ca hai vect xv y l (x, y), ta lun lun c ng thc

Gi s L1 l mt khng gian con bt bin i vi tt c cc ton t T(a):T(a) L1L1, A Gv k hiu L2 l phn ph trc giao ca L1 trong L:L = L1L2(x1, x2) = 0, x1 L1, x2 L2Ta hy chnx= T(a)-1x1, y = x2vi x1, x2 l hai vect bt k trong cc khng gian con L1 v L2. ng thc (45) vit trn tr thnh

V L1 l khng gian con bt bin cho nn T(a)-1x1 cng thuc vo L1 v do trc giao vi vect x2bt k ca L2,

(T(a)-1x1, x2) = 0Dng h thc (46) ta suy ra rng(x1, T(a)x2) = 0, x1 L1, x2 L2, A G.Vy tt c cc vect T(a)x2 vi mi yu t aca G v mi vect x2 ca L2 u trc giao vi tt c cc

vect x1 ca L1, ngha l u thuc vo L2,T(a) L2L2,L2 cng l khng gian con bt bin ca biu din T.Ta hy dng nh l 1 nh mt b chng minh nh l v tnh cht hon ton kh quy ca mi

biu din unita kh quy.

2.4.2.2 nh l 2

Mi biu din unita kh quy u hon ton kh quy.Chng minh. Cho biu din unita kh quy Ttrong khng gian L v gi s L1 l mt khng gian con bt

bin. Khi phn ph trc giao L2 ca L1 trong L cng l mt khng gian con bt bin, L l tng trc giaoca hai khng gian con bt bin L1 v L2. Trn hai khng gian con ny biu din T quy v hai biu dinT(1) v T(2) hon ton c lp vi nhau. Nu mt trong hai biu din ny hoc c hai biu din cn khquy th khng gian con tng ng li cha khng gian con bt bin nh hn v do li l tng trc giao


43/127

37

ca hai khng gian con bt bin nh hn. C tip tc thc hin vic tch mt khng gian thnh tng trcgiao ca hai khng gian con bt bin nh vy cho n khi khng cn c th tch c na, cui cng ta i

n vic tch khng gian L thnh tng trc giao ca cc khng gian con bt bin~

L1,~

L2, . . .,~

Lfthc hin

cc biu din ti gin~

T(1),~

T(2), . . .,~

T(f).

Ta cn ni rng biu din kh quy T l tng trc giao ca cc biu din ti gian~

T(1),~

T(2), . . .,~

T(f) vvit

T =~

T(1)~

T(2) . . . ~

T(f).V cc biu din unita c tnh cht din t bi nh l 2 cho nn khi nghin cu cc biu din unita ta

ch cn xt cc biu din ti gin. u chng ny ta nh ngha cc biu din tng ng v coi cc biu din tng ng vi nhau

ch l mt biu din. Do cc tnh cht c bit ca cc biu din unita, khi c mt biu din no camt nhm th ta hy tm xem n c tng ng vi mt biu din unita no hay khng. u tin ta hyxt trng hp G l mt nhm hu hn v chng minh nh l sau y.

2.4.2.3 nh l 3

Mi biu din ca mt nhm hu hn u tng ng vi mt biu din unita.Chng minh. Gi s c mt nhm hu hn G no cp N v mt biu din T ca nhm ny trong

mt khng gian Euclide phc L vi mt tch v hng c dng cho trc (x, y). u tin ta hy chng minhrng c th tm c mt nh ngha mi ca tch v hng, k hiu l {x, y}, m i vi tch v hng nyth tt c cc ton t T(a) vi mi aG u l cc ton t unita:

{T (a) x, T (a) y} = {x, y}, A G, x L, y LThc vy, ta t

vi tng trong v phi l tng theo tt c cc yu t b ca nhm G. Thay th x v y bng T(a)x v

T(a)y, ta c{T (a) x, T (a) y} = 1N

b T (b) T (a) x, T (b) T (a) y =1N

b T (ba) x, T (ba) y

Ta dng nh ngha ca biu dinT(b)T(a) = T(ba)Ch rng khi b chy mt vng theo tt c cc yu t ca nhm G th vi mi yu t a c nh tch b

a cng chy mt vng theo tt c cc yu t ca nhm ny, ch c iu l theo th t khc m thi. Do b T (ba) x, T (ba) y =

ba T (ba) x, T (ba) y =

c T (c) x, T (c) y

Vy ta cT (ba) x, T (ba) y = 1

N

c T (c) x, T (c) y = {x, y},

ngha l i vi tch v hng mi th tt c cc ton t T(a) u l hon t unita.Trong khng gian L ta hy chn hai h vect n v c s: h cc vect n v c s e1, . . ., en trc giao

chun ha i vi tch v hng cho t trc (x, y), c th l

v h cc vect n v c s f1, . . ., fn trc giao chun ha i vi tch, v hng mi, ngha l

K hiu php bin i tuyn tnh chuyn cc vect e1, e2, . . ., en thnh f1, f2, . . ., fn l X :

Vi hai vect bt kX = x i e i , y = y i e ita c


44/127


X x = xi fi , X y = yi fiDo

Vi mi yu t aca nhm G ta t

Ta bit rng T(a) l cc hon t unita i vi tch v hng mi. By gi ta chng minh rng tt c cc

ton t~

T (a) u l cc ton t unita i vi tch v hng {x, y} cho t trc. Thc vy, p dng ngthc (51) va thu c trn gia cc tch v hng ( x, y) v {Xx,Xy}, dng cng thc

X~

T (a) = T(a) Xsuy ra t nh ngha (52) v tnh cht unita ca cc ton t T(a) i vi tch v hng mi, ta c

(~

T (a) x,~

T (a) y) = {X~T (a) x, X

~T (a) y} = {

~T (a) Xx,

~T (a) Xy }

= {Xx,Xy} = (x, y)

Vy cc ton t~

T (a) l cc ton t unita i vi tch v hng cho t trc v to thnh mt biu

din unita~

T; biu din cho T tng ng vi biu din unita ny.Khi chng minh nh l 3 ta s dng mt i lng l tch v hng mi m i vi tch v hngny th biu din cho T l biu din unita. Rt d tm c tch v hng mi ny trong trng hpnhm G l nhm hu hn cp N. l gi tr trung bnh

1N

a fx,y (a)

ca hm trn nhm G

ngha l ca mt hm m bin s l yu t a chy trn tt c nhm G. Ta ch rng i vi nhm huhn ta c cng thc sau y i vi mi hm trn nhm f(a):

vi mi yu t b c nh ca nhm GMun chng minh nh l tng t nh nh l 3 i vi cc nhm v hn, k c cc nhm lin tc, taphi mi rng khi nim gi tr trung bnh ca hm trn nhm ra cho trng hp ny v s dng mt ilng gi l phim hm trung bnh.

2.4.2.4 nh ngha phim hm trung bnh trn nhm

Cho khng gian vect cc hm f (a) trn nhm G v hn (c th l nhm lin tc). Mt phim hm tuyntnh F(f) trn khng gian vect ny c gi l phim hm trung bnh nu n tn ti i vi mi hm giini trn nhm v tha mn cc iu kin sau y.

1) nu f(a) > 0, a G, th F(f) > 0.2) nu f(a) = 1, a G, th F(f) = 1.

3) nu f b(a) = f(ba), v~

fb (a) = f(ba), th F(f b) = F~

fb = f(b).iu kin 3) c ngha l khi ta x dch i s a ca hm trn nhm f (a) nh sauaab v aba,trong b l yu t ty ca G, th gi tr F (f) ca phim hm khng thay i. Do ta cn gi

phim hm ny l tch phn bt bin v dng k hiu tch phn

vi mt o d (a) no . Tnh cht bt bin ca phim hm c th hin tnh cht bt bin ca o:


45/127

39

vi mi yu t c nh bG.Dng tch phn bt bin ca hm trn nhm fx,y(a), tc l phim hm trung bnh F(fx,y), lm gi tr

trung bnh, by gi ta c th nh ngha tch v hng mi nh sau trong trng hp nhm G l nhm vhn

V o d (a) l bt bin cho nn i vi tch v hng mi n tt c cc ton t T(b) u l ton tunita:

{(T (b) x, T (b) y} =G

(T (a) T (b) x, (T (a) T (b) yd (a)

=G

(T (ab) x, (T (ab) yd (a) =G

(T (ab) x, (T (ab) yd (ab)

=G

(T (c) x, (T (c) yd (c) = {x, y}

Vy ta c nh l sau y.

2.4.2.5 nh l 4

Cho mt nhm v hn G (c th l nhm lin tc). Nu vi mi hm gii ni f(a) trn nhm G tn ti phimhm trung bnh F(f), tc l tn ti tch phn bt bin

G

(f(a) d (a) ,

th mi biu din ca nhm G u tng ng vi mt biu din unita.Chng minh. Ta dng phim hm trung bnh F(fx,y) lm tch v hng mi {x, y} ri lp li tt c cc

lp lun ging nh khi chng minh nh l 3.Trc khi kt thc on ny ta hy dn ra y mt vi th d v phim hm trung bnh.

Example 2.1G l nhm hu hn cp N. Ta nh ngha

F(f) = 1N

a f(a)R rng l F(f) tha mn cc iu kin 1) v 2). th li iu kin 3) ta ch cn dng nh

ngha ca f b,~

fb v h thc (54)a f(a) =

a f(ab)=

a f(ba) , b G.

Example 2.2G l nhm cc php quay khng gian ba chiu quanh mt trc no . Mi php quay c c

trng bi gc quay , hm trn nhm l hm tun hon ca vi chu k 2f() = f( + 2) .Ta nh nghaF(f) = 12

20

f() d

R rng l F(f) tha mn hai iu kin 1) v 2). Ch rng G l nhm giao hon cho nn

f () b=

~

f ()= f( + )T tnh cht tun hon ca hm f suy ra rng F(f) tha mn iu kin 3):F( f) = 12

20

f( + ) d = 1220

f() d = F(f)

Example 2.3G l nhm quay trong khng gian Euclide thc ba chiu. Mi php quay c c trng bi ba

gc Euler , , vi v thay i t 0 n 2 , cn thay i t 0 n . Hm trn nhm lhm f( , , ) ca ba gc Euler. Phim hm trung bnh c dng

F(f) = 182

f(, , ) sindddC th chng minh rng l mt tch phn bt bin trn nhm quay.


46/127


2.4.3 Biu din ti gin

2.4.3.1 nh l 5 (B Shur 1)

Nu trong khng gian L thc hin biu din ti gin T ca nhm G c mt ton t A khc khng v giaohon vi tt c cc ton t T(a) ca biu din T, a G , th ton t A phi l bi ca ton t n v

Chng minh. Ton t A khc khng c t nht mt vect ring r trong khng gian L:A r = rTp hp tt c cc vect ring tng ng vi cng mt gi tr ring to thnh mt khng gian con L

ca L. Ta hy chng minh rng L bt bin i vi mi php bin i T(a) ca biu din T:T(a) LL , a GThc vy, cho r l mt vect con ty trong L v hy xt tt c cc vect T(a)r, a G. Tc dng

ton t A ln cc vect ny, dng gi thit v s giao hon ca A vi tt c cc ton t T(a) v ch rngr l vect ring ca A vi gi tr ring , ta c

A(T(a)r) = AT(a)r= T(a)Ar= T(a) r= (T(a))r.

Kt qu ny chng t rng T(a)r cng l cc vect ring ca A cng mt gi tr ring . Vy L quthc l mt khng gian con bt bin. Nhng theo gi thit th biu din T trong khng gian L li l biudin ti gin, L khng th cha khng gian con bt bin no khc khng v khc L. Vy L khc khng thphi trng vi L, ngha l mi vect trong khng gian cho L l vect ring ca ton t A vi cngmt gi tr ring . A phi l bi ca ton t n v.

2.4.4 Hm trn nhm sinh ra bi biu din

Trong khng gian L thc hin biu din T ta hy chn mt h vect c s no . Khi mi php bini T(a) c din t bi mt ma trn vi cc yu t ma trn Tij (a) l cc hm trn nhm G, gi cc hmtrn nhm c sinh ra bi biu din T. Ta xt trng hp tch phn bt bin ca mi hm gii ni trnnhm u tn ti, v c nh ngha sau y

2.4.4.1 nh ngha tch v hng ca hai hm trn nhm

Cc hm trn nhm c th c coi l cc vect trong mt khng gian tuyn tnh v ta nh ngha tch vhng ca hai hm trn nhm v , tc l ca hai vect trong khng gian cc hm trn nhm, nh sau

p dng nh ngha ny cho cc hm trn nhm c sinh ra bi mt biu din ti gin, ta c nh lsau y.

2.4.4.2 nh l 6

Mt biu din T unita ti gin th nguyn d ca nhm G sinh ra d 2 hm trn nhm Tij (a), a G, thamn h thc

Chng minh. Hy ly mt ton t tuyn tnh B bt k trong khng gian L thc hin biu din ti ginT ri thit lp cc ton t

T(a) BT(a-1), aG,vi cc yu t ma trn l cc hm trn nhm, v ly tch phn (bt bin) ton t ny theo atrn nhm

G. Ta thu c ton t sau y


47/127

41

Ta hy chng minh rng ton t A giao hon vi mi php bin i T(a). Thc vy, vi mi yu t a

ca nhm G ta cT(a) AT(a)-1 = T(a) AT(a-1) =G

T (a) T (b) BT

b1

T

a1

d (b)

=G

T (ab) BT

b1a1

d (b) =

G

T (ab) BT

(ab)1

d (b)

V rng o d (b) l bt bin,d (b) = d (ab),cho nn ta c th thay n bng d (ab) ri i bin s tch phn, t ab = c, v cT(a) AT(a)-1 =

G

T (c) BT

c1

d (c)

ngha lT(a) AT(a)-1 = A, a GNhn c hai v ca h thc ny vi T(a) t bn phi, ta thu cT(a) A = AT(a), a GVy s giao hon ca A vi mi ton t T(a) c chng minh. By gi ta p dng B Shur 1.

Theo gi thit T l mt biu din ti gin. Ton t A giao hon vi tt c cc ton t T(a) ca biu dinny phi l bi ca ton t n v

A = IPhi hp h thc ny vi cng thc (60), ta thu c h thc

hay l di dng tng minh cc yu t ma trn

Biu din T c th nguyn bng d cho nn cc ch s i, j,k, l trong cng thc (62) chy theo cc snguyn dng t 1 n d. Ly vt ca c hai v cng thc (61), ngha l t i = ltrong c hai v ca cngthc (62) ri cng theo i t 1 n d ta c

BjkG

Tki

b1

Tij

b1b

d (b) = d

Ch rngT ki (b -1 )T ij (b) = T kj (b -1 b)=T kj (e) = kjvG

d (b) = 1

ta thu c= 1

dBii = 1d Tr B

Vy cng thc (62) c th vit li nh sau

Theo gi thit T l mt biu din unita, do T(b-1) = T(b)-1 = T(b)+ngha lTkl(b-1) = Tlk(b)*Thay vo v tri cng thc (63), ta thu c


48/127


Cng thc ny ng i vi mi ma trn B. Ta hy chn B l mt ma trn c bit c yu t ma trnBij bng 1 khi i = p v j = q vi hai s nguyn dng p v 1 no nh hn hoc bng d, cn tt c ccyu t ma trn khc bng khng,

Ta cTr B = pqKhi cng thc (64) tr thnhG

Tlq (b) Tip (b) d (b)= 1dilpq

l cng thc (59). Vy nh l c chng minh.nh l 6 c th c chng minh mt cch tng t i vi cc nhm hu hn. Trong trng hp ny

tch phn bt bin ca hm trn nhm c thay bng mt i lng tng t l gi tr trung bnh ca hmtrn nhm

GTij (a) Kkl (a) d (a)

1Na

Tij (a) Tkl (a),

trong N l s yu t ca nhm hu hn. Theo nh l 6 ta c d 2 hm trn nhm Tij(a) trc giao vinhau v do c lp tuyn tnh vi nhau. V s hm c lp tuyn tnh trn nhm hu hn cp N nhiunht cng ch bng N, cho nn d 2 N. Vy dnh l 6 i vi nhm hu hn c h qu sau y.

2.4.4.3 H qu

Th nguyn d ca mi biu din ti gin ca mt nhm hu hn cp N bao gi cng tho mn h thcd 2N.By gi ta hy m rng nh l 6 ra cho trng hp cc hm trn nhm c sinh ra bi hai biu din

ti gin khng tng ng. Ging nh khi chng minh nh l 6, ta cn b sau y.

2.4.4.4 nh l 7 (B Shur 2)

Cho T(1) v T(2) l hai biu din ti gin khng tng ng ca nhm G trong cc khng gian vect L1v L2, T(1)(a) v T(2)(a) l cc php bin i trong L1 v L2, tng ng vi yu t a G, A l mt tont tuyn tnh chuyn cc vect trongL2 thnh cc vect trong L1. Nu vi mi yu t a ca nhm G ton tA tho mn h thc:

th A phi bng khng.Chng minh. Gi th nguyn ca khng gian L1 l d1, th nguyn ca khng gian L2 l d2. C th xy

ra ba trng hp.1) d1 > d2. Gi M l min gi tr ca ton t A, ngha l tp hp tt c cc vect r1L1 c dng

trong r2 l mt vect bt k ca L2. Ta vit

Th nguyn ca M phi b hn hoc bng th nguyn ca khng gian L2 v do phi b hn thnguyn d1 ca khng gian L1. Ta hy chng minh rng M l mt khng gian con bt bin i vi tt ccc ton t T(1)(a), A G. Thc vy, mi vect r1 Mu c dng xc nh bi cng thc (66) v do theo h thc (65) ta c:

T(1)(a) r1 = T(1)(a) Ar2 = AT(2)(a) r2.Nhng T(2)(a) r2 cng l mt vect trong L2 cho nn theo nh ngha (67) ta cAT(2)(a) r2MVy vi mi vect r1M vect T(1)(a) r1 cng l mt vect thuc khng gian con M,


49/127

43

T(1)(a) r1MM,M l mt khng gian con bt bin i vi tt c cc php bin i T(1)(a). Ta li bit rng th nguyn

ca M nh hn th nguyn d1 ca khng gian L1. Nhng theo gi thit th biu din T(1) l mt biu dinti gin, khng gian L1 khng th c khng gian con bt bin no khc khng m c th nguyn nh hnd1. Vy ta phi c M = 0, ngha l A = 0.

2) d1 < d2. V ton t tuyn tnh A chuyn cc vect trong khng gian d2 chiu thnh cc vect trongmt khng gian c s chiu d1 nh hn, cho nn trong L2 phi c t nht mt vect r2 no m

Gi N l tp hp tt c cc vect trong khng gian L2 tho mn iu kin (68). V c t nht mt vecttho mn iu kin ny nn N=0. Ta hy chng minh rng N l khng gian con bt bin i vi biu dinT(2). Thc vy, cho r2 l vect bt k trong N. Theo gi thit (65) v nh ngha (68) ta c

AT(2)(a) r2 = T(1)(a) A r2 = 0vi mi yu t aca nhm G, tc l tt c cc vect T(2)(a) r2 cng thuc vo N. VyT(2)(a) NN, a G,N l khng gian con bt bin khc khng ca L2. Nhng theo gi thit biu din T(2) l biu din ti

gin. Vy N phi trng vi L2, ngha l mi vect r2 ca L2 u tha mn iu kin (68). Ta suy ra rngA=0.

3) d1=d2. u tin ta ch rng A khng th c nghch o, v nu A-1 tn ti th h thc (65) cho tangay

T(2) (a) = A1T(1) (a) A, a G,ngha l hai biu din T(1) v T(2) tng ng vi nhau, tri vi gi thit. V rng A khng c nghch

o cho nn min gi tr M ca A phi c th nguyn nh hn d2 = d1. L lun ging

Documents

Lý thuyết nhóm và ứng dụng trong vật lý lượng tử