Introduction to Group Theory Note 2 Theory of Representation

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1 Introducton to Grou Teory Note Teory of Reresentton August 5, 009 Contents Grou Reresentton. De nton of Reresentton Reducble nd Irreducble Reresenttons Untry Reresentton Scur s Lemm 4 Gret Ortogonlty Teorem 6 4 Crcter of Reresentton 7 4. Decomoston of Reducble Reresentton Regulr Reresentton Crcter Tble Product Reresentton (Kronecker roduct) 6 Drect Product Grou Grou Reresentton In yscl lcton, te grou reresentton lys very mortnt role n deducng te consequence of te symmetres of te system. Rougly sekng, reresentton of grou s just some wy to relze te sme grou oerton oter tn te orgnl de nton of te grou. Of rtculr nterest to most ycl lcton s te relzton of grou oerton by te mtrces wose multlcton oerton cn be nturlly ssocted wt grou multlcton.. De nton of Reresentton Gven grou G = fa ; = ng : If for ec A G; tere s n n n mtrx D (A ) suc tt D (A ) D (A j ) = D (A A j ) ten D s forms n-dmensonl reresentton of te grou G. In oter words, te corresondence A D (A ) s omomorsm. Te condton n Eq smly mens tt te mtrces D (A ) stsfy te sme multlcton lw s te grou elments. If ts omomorsm turns out to be n somorsm ( ) ten te reresentton sclled ftful. Note tt mtrce M j cn be vewed s lner oertors M ctng on some vector sce V wt resect to some coce of bss e ; Me = e j M j j One wy to generte suc mtrces for te symmetry of certn geometrc objects s to use te grou nduced trnsformtons, dscussed before. Recll tt ec grou element A wll nduce trnsformton of te coordnte vector ~r; ~r A ~r Ten we cn tke ny functon of ~r; sy ' (~r) nd for ny grou element A de ne new trnsformton P A by

2 P A ' (~r) = ' A ~r Among te trnsformed functons obtned ts wy, P A ' (~r) ; P A ' (~r) ; P An ' (~r) ; we select te lnerly ndeendent set ' (~r) ; ' (~r) '` (~r) :Ten t s cler tt P A ' cn be exressed s lner combnton of ' ; P A ' = ` ' b D b (A ) nd D b (A ) forms reresentton of G: Ts cn be seen s follows. P AA j ' = P A P Aj ' = P A b D b (A j ) = c D cb (A ) D b (A j ) b:c: On te oter nd, b b= P AA j ' = c ' c D (A A j ) c Ts gves D (A A j ) c = D cb (A ) D b (A j ) wc mens tt D (A ) 0 s form reresentton of te grou. Exmle: Grou D ; symmetry of te trngle. As we ve dscussed n te revous cter, coosng coordnte system on te lne, we cn reresent te grou elements by te followng mtrces, A = K = 0 0 Coose f (~r) = f (x; y) = x y ; we get ; B = ; L = ; E = ; M = 0 0 P A f (~r) = f A ~r = x + y x y = 4 4 x y + xy: We now ve new functon g (x; y) = xy: We cn oerte on g (r) to get, P A g (~r) = g A ~r = x + y x y = x y xy = x y (xy) Tus we ve P A (f; g) = (f; g) Te mtrx generted ts wy s te sme s A s gven bove. Smlrly P B f (~r) = f B ~r = x y x + y = 4 4 x y + ( xy) P B g (~r) = g B ~r r + = x y x + y = + x y xy = x y ( x sme s B s gven before. Remrks P B (f; g) = (f; g)

3 . If (A) nd (A) re bot reresentton of te grou, ten t s cler tt (A) = (A) 0 0 (A) lso forms reresentton. We wll denote t s drect sum ; (A) = (A) (A) (block dgonl form) drect sum. If (A) nd (A) re reresenttons of G wt sme dmenson nd tere exsts squre mtrx U suc tt (A ) = U (A ) U for ll A G: ten nd re sd to be equvlent reresenttons. Recll tt f we cnge te bss used to reresent te lner oertors, te corresondng mtrces undergo smlr trnsformton. Snce tey reresent te sme oertors, we consder tem te "sme" reresentton but wt resect to d erent coce of bss.. Reducble nd Irreducble Reresenttons A reresentton D of grou G s clled rreducble f t s de ned on vector sce V (D) wc s no nontrvl nvrnt subsce. Oterwse, t s reducble. In essence ts de nton smly mens tt for reducble reresentton, te lner oertors correondng to te grou elements wll leve some smller vector sce nvrnt. In oter words, ll te grou ctons cn be relzed n some subsce. We need to convert tese sttment nto more rctcl crteron. Suose te reresenton D s reducble on te vector sce V. Ten tere exsts subsce S wc s nvrnt under D. For ny vector v V; we cn decomos t s, v = s + s? were s S nd s? belongs to te comlement S? of S: If we wrte te vector v n te block form, v = ten te reresentton mtrx cn be wrtten s D (A) Av = D (A) v = D (A) s s? For te sce S to be nvrnt under grou oertors mens tt D (A ) = 0;.e. te mtrces D (A ) re ll of te uer trngulr form, D (A D (A ) = ) D (A ) 0 D 4 (A ) D (A) D 4 (A) 8A G s s? ; 8A G A reresentton s comletely reducble f ll te mtrces n te reresenttons D (A ) cn be smultneously brougt nto block dgonl form by te sme smlrty trnsformton U; UD (A ) U D (A = ) 0 ; for ll A 0 D (A ) G.e. D (A ) = 0 n te uer trngulr mtrces gven n Eq. In oter words, te sce comlement to S s lso nvrnt under te grou oerton. Ts wll be te cse f te reresnetton mstrces re untry s stted n te teorem; Teorem: Any untry reducble reresentton s comletely reducble. Proof: For smlcty we ssume tt te vector sce V s equed wt sclr roduct (u; v) : It s esy to see n ts cse we cn coose te comlement sce S? to be erendculr to S;.e. (u; v) = 0; f u S; v S? Recll tt te sclr roduct s nvrnt under te untry trnsformton, 0 = (u; v) = (D (A ) u; D (A ) v) Tus f D (A ) u S; ten D (A ) v S? wc mles tt S? s lso nvrnt under te grou oerton. In yscl lctons, we del mostly wt untry reresenttons nd tey re comletely reducble.

4 . Untry Reresentton Snce untry oertors reserve te sclr roduct of vector sce, reresentton by untry mtrces wll smlfy te nlyss of grou teory. In te relm of nte grous, t turns out tt we cn lwys trnsform te reresentton nto unty one. Ts s te content of te followng teorem. Fundmentl Teorem Every rre of nte grou s equvlent to untry rre (re by untry mtrces) Proof: Let D (A r ) be reresentton of te grou G = fe; A A n g Consder te sum n H = D (A r ) D y (A r ) ten H y = H r= Snce H s ostve semde nte, we cn de ne squre root by De ne new set of mtrces by = H; y = D (A r ) = D (A r ) r = ; ; ; n Snce ts s smlrty trnsformton, D (Ar ) lso forms re wc s equvlent to D (A r ) : demonstrte tt D (A r ) s untry, We wll now D (A r ) D y (A r ) = D (A r ) D y (A r ) = D (A r ) " n # = D (A r A s ) D y (A r A s ) = s= were we ve used te rerrngement teorem. Scur s Lemm n D (As ) D y (A s ) D y (A r ) s= n s 0 = D (A s 0) D y (A s 0) = = One of te most mortnt teorems n te study of te rreducble rerentton s te followng lemm. Scur s Lemm Any mtrx wc commutes wt ll mtrces of rre s multle of dentty mtrx. Proof: Assume 9 M suc tt ten by tkng te ermtn conjugte, we get MD (A r ) = D (A r ) M D y (A r ) M y = M y D y (A r ) As sown bove, we cn tke D (A r ) to be untry, so we cn wrte 8 A r G M y = D (A r ) M y D y (A r ) or M y D (A r ) = D (A r ) M y Ts mens tt M y lso commutes wt ll D s nd so re te combnton M + M y nd M M y ; wc re ermtn. Tus, we only ve to consder te cse were M s ermtn. Strt by dgonlzng M by untry mtrx U; M = UdU y De ne D (Ar ) = U y D (A r ) U; ten we ve d : dgonl d D (A r ) = D (A r ) d or n terms of mtrx elements, d _ D r (A s ) = _ D (A s ) d 4

5 Snce te mtrx d s dgonl, we get (d d ) D (A s ) = 0 =) f d 6= d ; ten D (A s ) = 0 Ts mens f dgonl elements d re ll d erent, ten te o -dgonl elements of D re ll zero. In ts _ cse, D 0 s re ll dgonl nd ence ll reducble. Te only ossble non-zero o -dgonl elements of D cn rse wen some of d 0 _ s re equl. For exmle, f d = d ; ten D cn be non-zero. Tus D wll be n te block dgonl form,.e. d d f d = d d... ten 6 D = 4 d _ D 0 0 D _ Ts s true for every mtrx n te reresentton. Tus ll te mtrces n te reresentton re n te block dgonl form. But D s rreducble wc mens tt not ll mtrces cn be brougt nto block dgonl form. Tus ll d s ve to be equl d = ci: or M = UdU y = duu y = d = ci If te only mtrx tt commutes wt ll te mtrces of reresentton s multle of dentty, ten te reresentton s rre. Proof: construct M = Suose D s reducble, ten we cn trnsform tem nto I 0 0 I D D (A ) = (A ) ten clerly (A ) for ll A G D (A ) M = MD (A ) for ll But M s not multle of dentty (contrdcton). Terefore D must be rreducble. Remrks. Any rre of Abeln grou s dmensonl. Ts s becuse for ny element A; D (A) commutes wt ll D (A ) : Ten Scur s lemm =) D (A) = ci 8A G: But D s rre, so D s to be mtrx.. In ny rre, te dentty element E s lwys reresented by dentty mtrx. Ts follows Scur s lemm.. From D (A) D A = D (E) = I:; we see tt D A = [D (A)] nd for untry reresentton D A = D + (A) : If nd re rres of dmenson l ; nd dmenson l nd ten () f l 6= l M = 0 M (A ) = (A ) M :for ll A G (b) f l = l ; ten eter M = 0 or det M 6= 0 nd res re equvlent. Proof: : Wtout loss of generlty we cn tke l l : Note tt snce Eq s true for ll elements we cn relce A by A ; wc cn be wrtten s Hermtn conjugte of Eq gves M A = D A M M (A ) y = (A ) y M y M y = M y y ; MM y (A ) y = M (A ) y M y = (A ) y MM y 5

6 or MM y (A ) = (A ) MM y 8 (A ) G Ten from Scur s lemm we get MM y = ci;were I s l dmensonl dentty mtrx. Frst consder te cse l = l ; were we get jdet Mj = c` : Ten eter det M 6= 0;wc mles M s non-sngulr nd from Eq (A ) = M (A ) M 8 (A ) G Ts mens (A ) nd (A ) re equvlent. Oterwse f te determnnt s zero, In rtculr, for = det M = 0 =) c = 0 or MM y = 0 =) P jm j = 0 M = 0 for ll : =) M = 0: Next, f l < l ; ten M s retngulr l l ; mtrx : : M = : : {z } l we cn de ne de ne squre mtrx by ddng colums of zeros l M M = 0 8:: N = l z } { [M; 0]gl l l squre mtrx ten M N y y = 0 M nd NN y y = (M; 0) = MM y = ci 0 were I s te l l dentty mtrx. But from constructon we see tt det N = 0; Hence c = 0; =) NN y = 0 or M = 0 dentclly. Gret Ortogonlty Teorem Te most useful teorem for te reresentton of te nte grou s te followng one. Teorem(Gret ortogonlty teorem): Suose G s grou wt n elements,fa ; = ; ; ng ; nd (A ) ; = ; re ll te nequvlent rres of G wt dmenson l : Ten n j (A ) k` (A ) = n k j` l = Proof: De ne M = (A ) A were s n rbtrry l l mtrx. Ten multlyng M by reresentton mtrces, we get (A b ) M = (A b ) (A ) A A b (A b ) = (A b A ) (A b A ) (A b ) = M (A b ) If 6= ; ten M = 0 from Scur s lemm, we get M = r (A ) rs sk A = r Coose rs = rj sl (.e. s zero excet te jl element). Ten we ve j (A ) k` (A ) = 0 (A ) rs ks (A ) = 0 Ts sows tt for d erent rreducble reresenttons, te mtrx elements, fter summng over grou elements, re ortogonl to ec oter. 6

7 = ten we cn wrte M = P (A ) A :Ts mles wc gves, Tke (A ) M = M (A b ) =) M = ci T r (A ) A = cl or nt r = cl ; or c = (T r) n l rs = rj s` ten T r = j` nd (A ) j (A ) k` = n k j` l Ts gves te ortogonlty for d erent mtrx elements wtn gven rreducble reresentton. Geometrc Interretton Imgne comlex n dmensonl vector sce n wc xes (or comonenets) re lbeled by grou elements E; A :A A n (Grou element sce). Consder te vector n ts sce wt comonets mde out of te mtrx element of rreducble reresentton mtrx (A ) j :Ec vector n ts n dmensonl sce s lbeled by ndces, ; : ~D = D (E) ; D (A ) ; D (A n ) (4) Gret ortogonlty teorem sys tt ll tese vectors re? to ec oter. As result l n becuse tere cn be no more tn n mutully? vectors n n-dmenson vector sce. As n exmle, we tke te -dmensonl reresentton we ve work out before, 0 E = ; A = ; B = 0 K = 0 ; L = 0 ; M = Lbel te xses by te groul elements n te order (E; A; B; K; L; M) : Ten we cn construct four 6-dmensonl ectors from tese mtrces, = ( ; = (0 ; ; ; ; ; ; ) ; 0 ; ; ) ; 0 ; ) = (0 ; = ( ; ; ; ; ; ) It s strgtforwrd to ceck tt tese 4 vectors re erendculr to ec oter. Note tt te oter two vectors wc re ortogonl to tese vectors re of te form, D E = ( ; ; ; ; ; ) D A = ( ; ; ; ; ; ) comng from te dentty reresentton nd oter -dmensonl reresentton. 4 Crcter of Reresentton Te mtrces n rre re not unque, becuse we cn generte noter equvlent rre by smlrty trnsformton. However, te trce of mtrx s nvrnt under suc trnsformton, T r SAS = T ra We cn use te trce, or crcter, to crcterze te rre. (A ) T r (A ) = (A ) Useful Proertes 7

8 . If nd re equvlent, ten (A ) = (A ) 8A G. If A nd B re n te sme clss, (A) = (B) Proof: If A nd B re n sme clss =) 9x G suc tt xax = B =) (x) (A) x = (B) Usng we get x = (x) T r (x) (A) (x) = T r (B) Hence s functon of clss, not of ec element. or (A) = (B). Denote = (C ) ; te crcter of t clss. Let n c be te number of clsses n G; nd n te number of grou elements n C : From gret ortogonlty teorem r j (A r ) k` (A r ) = n k jl l we get (A r ) (A r ) = n l = n l r or P n = n Ts s te gret ortogonlty teorem for te crcters. De ne U = n n (C ) ; ten gret ortogonlty teorem mles, n c = U U = Tus, f we consder U s comonents n n c dmensonl vector sce, ~ U = (U U U nc ) ; ten ~ U = ; ; n r (n r : # of nde rres)form n otornorml set of vectors,.e. Ts mles tt n c U U = U U = = n r n c. e. number of rres s smller tn te number of clsses. Ts gretly restrcts te number of ossble rres. 4. Decomoston of Reducble Reresentton For reducble reresentton, we cn wrte D = D.e. D (A ) = (A ) Ten we ve for te trce (A ) (A ) = (A ) + (A ) 8A G Denote by ; = ; n r ; ll te nequvlent untry rre. Ten ny re D cn be decomosed s D = c c : some nteger, # of tme ers In terms of trces, we get (C ) = c (C ) 8

9 were we ndcte tt te trce s functon of clss C :Te coe cent cn be clculted s follows (by usng ortogonty teorem). Multly by n nd sum over n = c n = c n = nc or c = P n n From ts we lso get, n = n c c Ts leds to te followng teorem: Teorem: If te re D wt crcter st es te relton, n = n ten te reresentton D s rreducble. ; = n jc j 4. Regulr Reresentton Gven grou G = fa = E; A A n g :We cn construct te regulr re s follows: Tke ny A G: If AA = A = 0A + 0A + A + 0A 4 +.e. we wrte te roduct "formlly" s lner combnton of grou elements, AA s = n C rs A r = r= n A r D rs (A) ;.e. C rs = D rs (A) s eter 0 or: (5) r=.e. D rs (A) = f AA s = A r or A = A r As = 0 oterwse Note strctly sekng, te sum over grou elments s unde ned. But ere only one grou element sows u n te rgt-nd sde n Eq(5), we do not need to de ne te sum of grou elements. Ten D (A) s form re of G: regulr reresentton wt dmensonl n: Ts cn be seen s follows: A r D rs (AB) = ABA s = A A t D ts (B) = D ts (B) A r D rt (A) r t tr or D rs (AB) = D rt (A) D ts (B) From te de nton of te regulr reresentton D rs (A) = AA s = A r we see tt te dgonl elements re of te form, D rr (A) = ff AA r = A r or A = E Terefore every crcter s zero excet for dentty clss, (reg) (C ) = 0 6= (reg) (C ) = n = (6) From ts we cn work out ow D (reg) reduces to rres. Wrte D (reg) = c 9

10 ten c = n (reg) n = n (reg) = n nl = l Ts mens tt D reg contns te rres s mny tmes s ts dmenson; (reg) = n r l or (reg) For te dentty clss reg = n; = l ; ten we get P l = n n r = = = n Ts severely constrnts te ossble dmensonltes of rres becuse bot n nd l ve to be ntegers. For D, wt n = 6; te only ossble soluton for P l = 6 s l = : l = : l = ; nd ter ermuttons. Te relton n Eq(7) mles tt te vector sce formed by vectors de ned n Eq(4) s dmenson n; te number of elements n te grou. Snce tose vectors n Eq(4) re ortogonl to ec oter, ence lnerly ndeendent, nd tere re n suc vectors, tey must stsfy te comleteness relton, ;; l n D (A k ) (A l ) = kl comleteness relton (8) Te fctor l comes from te normlzton of te vectors n Eq(4). n We now wnt to sow tt n c = n r.e. # of clsses = # of rres. De ne by ddng u ll mtrces corresondng to elements n te sme clss C ; = AC (A) (7) Ten, (A j ) A j = A = A (A j ) (A) A j A j AA j = D Usng we get.e. A j = (A j ) (A j ) = (A j ) commutes wt ll mtrce n te rre. From Scur s lemm, we get Tkng te trce, we get = were s some number n = l or = n l = n were s te crcter of dentty clss. In te comleteness relton n Eq(8), we cn sum A k over grou elements n clss C r nd A l over clss C s to get ;; l n r s = n r rs (9) 0

11 Usng vlue of n Eq(9) we ve n r r = Ts te comleteness relton for te crcters. ; ; : : : (nr) we get s = n n r rs comleteness If we now consder s vector n n r dm sce = n c n r Combne ts wt te result n r n c ; we ve derved before, we get 4. Crcter Tble n r = n c For nte grou, te essentl nformton bout te rreducble reresenttons cn be summrzed n tble wc lsts te crcters of ec rreducble reresentton n terms of te clsses. Ts tble s mny useful lctons. To construct suc tble we cn use te followng useful nformton:. # of columns = # of rows = # of clsses P. l = n. P n = n nd 4. If l = ; s tself re. P j = n n j 5. A = T r A = T r + A = (A) If A nd A re n te sme clss ten (A) s rel. 6. s re =) s lso re so f s re comlex numbers, noter row wll be ter comlex conjugte 7. If l > ; = 0 for t lest one clss. Ts follows from te relton n j j = n nd n = n 8. For yscl symmetry grou, x:y nd z form bss of re. Exmle : D crcter tble E C C 0 x + y ; z A R z; z: A (xz; yz) (x; y) E 0 x y ; xy (R x :R y ) In ts tble, te tycl bss functons u to qudrtc n coordnte system re lsted. Remrk: te bss functons lsted n te usul crcter tble re not necessrly normlzed. In rtculr, te qudrtc functons ve to be ndled crefully. Te dnger s tt f we use te bss functons gven n te crcter tble, we mgt not generte untry mtrces. Usng te trnsformton roertes of te coordnte, we cn lso nfer te trnsformton roertes of ny vectors. For exmle, te usul coordntes ve te trnsformton roerty, ~r = (x; y; z) A E n D Ts mens tt electrc eld of ~ E or mgnetc eld ~ B wll ve sme trnsformton roerty, ~B s ~ E s A E becuse tey ll trnsform te sme wy under te rotton.

12 5 Product Reresentton (Kronecker roduct) Let x be te bss for ;.e. y` be te bss for ;.e. ten te roducts x j y l trnsform s x 0 = P ` x j (A) x 0 y 0 k = j` j= y 0 k = `P j j y` `k `= (A) (A) k` (A) x jy` j` j`;k (A) x j y` were j`;k (A) = j (A) `k (A) Note tt n tese mtrces, row nd column re lbelled by ndces, nsted of one. It s esy to sow tt forms re of te grou. (A) (B) = (A) j;st (B) st;k` j;k` s:t = s:t s (A) D jt (A) sk (B) D t` (B) = k (AB) D j` (AB) = D (AB) k;k` or (A) (B) = (AB) Te bss functons for re x y j Te crcter of ts new re cn be clculted by mkng te row nd colum ndces te sme nd sum over, (A) = j` j`;j` (A) = j` jj (A) `` (A) = (A) (A) (A) = (A) (A) If = ; we cn furter decomose te roduct re by symmetrzton or ntsymmetrzton; D fg k;j` (A) = D [] k;j` (A) = j j (A) k` (A) k` (A) + D ` (A) D ` (A) kj (A) (A) kj (A) Tese mtrces lso form re of G nd te crcters re gven by bss bss (x y k + x k y ) (x y k x k y ) fg (A) = (A) + A ; [] (A) = (A) A Exmle D E: C C 0 R z: z (xz; yz) (x; y) 0 x y ; xy 4 0 = ( ) s 0 = ( ) =

13 6 Drect Product Grou Gven grous G = fe; A A n g ; G = fe; B B m g ; we cn de ne te roduct grou s G G = fa B j ; = n; j = mg wt multlcton lw (A k B`) (A k 0B`0) = (A k A k 0) (B`B`0) It turns out tt rre of G G re just drect roduct of rres of G; nd G : Let (A ) be n rre of G nd (B j ) n rre of G ten te mtrces de ned by wll ve te roerty (A B j ) b;cd (A ) c (B j ) bd (A B j ) (A k B`) b;cd = ef = ef (A B j ) (A k B`) b;ef ef;cd (A ) c (A k ) ec (B j ) bf (B e ) fd = (A A k ) c (B j B`) bd = (A A k: B j B`) b;cd Ts mens tt te mtrce (A B j ) form reresentton of te roduct grou G G : Te crcters cn be clculted, Ten (A B j ) = b (A B j ) = j (A B j ) b;b = b (A ) (B j ) bb = (A ) (B j ) 0 (A j (B j ) A = nm =) s rre. Exmle, G = D = fe; C ; C 0 g ; G = fe; g = ' were : re ecton on te lne of trngle. Drect roduct grou s tend D ' = E; A; B = fe; C ; C 0 ; ; C ; C 0 g Crcter Tble ' E + C C 0 D E AB KLM 0 Crcter Tble E C C 0 C C