Group Theory in Physics

Transcription

1 Group Theory n Physcs Lng-Fong L (Insttute) Representaton Theory / 4

2 Theory of Group Representaton In physcal applcaton, group representaton s mportant n deducng the consequence of the symmetres of the system. Most mportant one s realzaton of group operaton by matrces. De nton of Representaton Gven a group G = fa, = ng. If for each A G, there are an n n matrces D (A ) such that D (A ) D (A j ) = D (A A j ) () then D s forms a n-dmensonal representaton of the group G.e. correspondence A! D (A ) s a homomorphsm. If ths s an somorphsm, representaton s called fathful. A matrx M j can be vewed as lnear operator M actng on some vector space V wth repect to some bass e, Me = e j M j j Note that the choce of the bass s not unque. If we make a change to a new bass, e = f n S n, S non-sngular j then n Mf n S n = f k S kj M j k,j (Insttute) Representaton Theory / 4

3 Multply by S m, Mf m = f k S kj M j Sm k,j, = f k k SMS km = f k k M 0 km where M 0 = SMS To generate such matrces for the symmetry of certan geometrc objects s to use the group nduced transformatons, dscussed before. Recall that each element A a nduces a transformaton of the coordnate vector ~r, ~r! A a ~r Then for any functon of ~r, say ϕ (~r ) and for any group element A a de ne a new transformaton P Aa by P Aa ϕ (~r ) = ϕ A a ~r Among the transformed functons obtaned ths way, P A ϕ (~r ), P A ϕ (~r ), P An ϕ (~r ), we select the lnearly ndependent set ϕ (~r ), ϕ (~r ) ϕ` (~r ).Then t s clear that P A ϕ a can be expressed as lnear combnaton of ϕ.ths s because Thus we can wrte P Ab ϕ a = P Ab P Aa ϕ (~r ) = P Ab ϕ Aa ~r = ϕ Aa Ab ~r = ϕ (A b A a ) ~r (Insttute) Representaton Theory 3 / 4

4 ` P A ϕ a = ϕ b D ba (A ) b= and D ba (A ) forms a representaton of G. Ths can be seen as follows. P A A j ϕ a = P A P Aj ϕ a = P A b φ b D ba (A j ) = φ c D cb (A ) D ba (A j ) b.c. On the other hand, Ths gves P A A j ϕ a = ϕ c D (A A j ) ca c D (A A j ) ca = D cb (A ) D ba (A j ) whch means that D (A ) 0 s form representaton of the group. (Insttute) Representaton Theory 4 / 4

5 Example: Group D 3, symmetry of the trangle. As seen before, choosng a coordnate system on the plane, we can represent the group elements by the followng matrces, A = K = p! 3 p 3, B = p 3 0, L = 0 Choose f (~r ) = f (x, y ) = x y, we get P A f (~r ) = f A ~r = 4 p! 3 0, E = 0 p! 3 p 3, M = p 3 p! 3 x + p p 3y 3x y = 4 x y + p 3xy. We now have a new functon g (x, y ) = xy showng up. We can operate on g (r ) to get, P A g (~r ) = g A ~r = = hp 3 x + p p 3y 3x p 3 x y xy = y x y (xy ) Thus we have P A (f, g ) = (f, g ) p 3 p! 3 (Insttute) Representaton Theory 5 / 4

6 The matrx generated ths way s the same as A f we change g to -g. Smlarly same as B Remarks P B f (~r ) = f B ~r = p p x 3y 3x + y = 4 4 x y p 3xy P B g (~r ) = g B ~r p p = x 3y 3x + y r h p = x y p 3 xy = x y (xy ) P B (f, g ) = (f, g ) p 3 p! 3 If D () (A) and D () (A) are both representaton of the group, then D (3) D (A) = () (A) 0 0 D () (A) (block dagonal form) also forms a representaton. We wll denote t as a drect sum, D (3) (A) = D () (A) D () (A) drect sum (Insttute) Representaton Theory 6 / 4

7 If D () (A) and D () (A) are representatons of G wth same dmenson and 9 a non-sngular matrx U such that D () (A ) = UD () (A ) U for all A G. then D () and D () are equvalent representatons. (Insttute) Representaton Theory 7 / 4

8 Reducble and Irreducble Representatons A representaton D of a group G s called rreducble representaton (rrep) f t s de ned on a vector space V (D ) whch has no non-trval nvarant subspace. Otherwse, t s reducble. In other words, all group actons can be realzed n ths subspace. Suppose representaon D s reducble on the vector space V. Then 9 subspace S nvarant under D. For any vector v V, we can wrte, v = s + s? where s S and s? belongs to the complement S? of S. If we wrte, v = s s? then the representaton matrx s D (A) Av = D (A) v = D 3 (A) Snce S s nvarant under group operators, we get D (A) D 4 (A) D 3 (A ) = 0, 8 A G (Insttute) Representaton Theory 8 / 4

9 .e. D (A ) are all of the upper trangular form, D (A D (A ) = ) D (A ) 0 D 4 (A ), 8 A G () Note that D D D = D 0 D 0 0 D 4 0 D4 0 = D D 0 D D 0 + D D D 4 D 0 4 Thus nvarant subspace s stll nvarant f operate on t many tmes. A representaton s completely reducble f all matrces D (A ) can be smultaneously brought nto block dagonal form by the same smlarty transformaton U, UD (A ) U D (A = ) 0, for all A 0 D (A ) G Theorem: Any untary reducble representaton s completely reducble. Proof: For smplcty we assume that the vector space V s equpped wth a scalar product (u, v ). We can choose the complement space S? to be perpendcular to S,.e. (u, v ) = 0, f u S, v S? Snce scalar product s nvarant under the untary transformaton, 0 = (u, v ) = (D (A ) u, D (A ) v ) (Insttute) Representaton Theory 9 / 4

10 Thus f D (A ) u S, then D (A ) v S? whch mples that S? s also nvarant under the group operaton. In physcal applcatons, we deal mostly wth untary representatons and they are completely reducble. Untary Representaton Snce untary operators preserve the scalar product, representaton by untary matrces wll smplfy the analyss of group theory. For nte groups, we can show that rep can always be transformed nto untay one. Fundamental Theorem Every rrep of a nte group s equvalent to a untary rrep (rep by untary matrces) Proof: Let D (A r ) be a representaton of the group G = fe, A A n g Consder the sum n H = D (A r ) D (A r ) then H = H r = Snce H s postve semde nte, de ne squre root h by h = H, h + = h Ths can be acheved by dagonzng ths hermtan matrx H by untary transformaton and then transformng t back after takng the square root of the egvalues. De ne new set of matrces by smlarty transformaton D (A r ) = h D (A r ) h r =,,, n D (A r ) s equvalent to D (A r ). We wll now show that D (A r ) s untary, (Insttute) Representaton Theory 0 / 4

11 D (A r ) D (A r ) = h D (A r ) h h hd (A r ) h = h D (A r ) = h " n where we have used rearrangement theorem. n h D (A S ) D (A S ) D (A r ) h s= # D (A r A s ) D (A r A s ) h s= = h n D (A S 0 ) D (A s 0 ) h = h h h = s 0 = (Insttute) Representaton Theory / 4

12 Schur s Lemma an mportant theorems n rreducble reprentaton () Any matrx whch commutes wth all matrces of rrep s a multple of dentty matrx. Proof: Assume 9 M MD (A r ) = D (A r ) M 8 A r G Hermtan conjugate D (A r ) M = M D (A r ) Snce D (A r ) s untary, we get M = D (A r ) M D (A r ) or M D (A r ) = D (A r ) M =) M also commutes wth all D s and so are M + M and M M, whch are hermtan. Thus, we take M to be hermtan. Start by dagonalzng M by untary matrx U, M = UdU d : dagonal De ne D (A r ) = U D (A r ) U, then we have d D (A r ) = D (A r ) d (Insttute) Representaton Theory / 4

13 or n terms of matrx elements, Snce d s dagonal, we get d αβ D βr (A s ) = β β D αβ (A s ) d βγ (d αα d γγ ) D αγ (A s ) = 0 =) f d αα 6= d γγ, then D αγ (A s ) = 0 _ =) f dagonal elements d are all d erent, then o -dagonal elements of D _ are all zero. The only possble non-zero o -dagonal elements of D can arse when some of dααs 0 _ are equal. For example, f d = d, then D can be non-zero. Thus D s n block dagonal form,.e. (Insttute) Representaton Theory 3 / 4

14 f d = d d d d... 3 d then D = 6 4 D 0 0 D Ths s true for every matrx n the rep. Thus all the matrces n rep are block dagonal. But D s rreducble =) not all matrces can be block dagonal. Thus all d s have to be equal d = ci. or M = UdU = duu = d = ci (Insttute) Representaton Theory 4 / 4

15 () If the only matrx that commutes wth all the matrces of a representaton s a multple of dentty, then the representaton s rrep. Proof: Suppose D s reducble, then we can transform them nto construct M = D D (A ) = () (A ) I 0 0 I then D () (A ) for all A G D (A ) M = MD (A ) for all But M s not a multple of dentty (contradcton). rreducble. Therefore D must be Remarks. Any rrep of Abelan group s dmensonal. Because for any element A, D (A) commutes wth all D (A ). Then Schur s lemma =) D (A) = ci 8A G. But D s rrep, so D has to be matrx.. In any rrep, the dentty element E s always represented by dentty matrx. Ths follows Schur s lemma. 3. From D (A) D A = D (E ) = I., =) D A = [D (A)] and for untary representaton D A = D (A). (Insttute) Representaton Theory 5 / 4

16 () If D () and D () are rreps of dmenson l, and dmenson l and MD () (A ) = D () (A ) M. (3) then (a) f l 6= l M = 0 (b) f l = l, then ether M = 0 or det M 6= 0 and reps are equvalent. Proof: : take l l. Hermtan conjugate of Eq(3) gves D () M = M D (), MM D () (A ) = MD () (A ) M = D () (A ) MM or MM D () (A ) = D () (A ) MM 8 (A ) G Then from Schur s lemma () we get MM = ci,where I s a l dmensonal dentty matrx. Frst consder the case l = l, where we get jdet M j = c `. Then ether det M 6= 0, ) M s non-sngular and from Eq(3) D () (A ) = M D () (A ) M 8 (A ) G Ths means D () (A ) and D () (A ) are equvalent. Otherwse f the determnant s zero, det M = 0 =) c = 0 or MM = 0 =) M αγ M βγ = 0 γ 8α.β. (Insttute) Representaton Theory 6 / 4

17 In partcular, for α = β γ jm αγ j = 0 M αγ = 0 for all α.γ =) M = 0. Next, f l < l, then M s a rectangular l l, matrx.. M = l.. {z } l De ne a square matrx by addng colums of zeros l z } { N = [M, 0]gl l l square matrx then M N = 0 M and NN = (M, 0) = MM = ci 0 where I s the l l dentty matrx. But from constructon det N = 0. Hence c = 0, =) NN = 0 or M = 0 dentcally. (Insttute) Representaton Theory 7 / 4

18 Great Orthogonalty Theorem most mportant theorem for the representaton of the nte group. Theorem(Great orthogonalty theorem): Suppose G a group wth n elements, fa, =,, ng, and D (α) (A ), α =, are all the nequvalent rreps of G wth dmenson l α. Then n D (α) j (A α ) D (β) k` (A α ) = n δ α= l αβ δ k δ j` α Proof: De ne M = a D (α) (A a ) XD (β) Aa where X s an arbtrary l α l β matrx. Multplyng M by representaton matrx, D (α) (A b ) M = D (α) (A b ) a D (α) (A a ) XD (β) Aa h D (β) Ab D (β) (A b ) = D (α) (A b A a ) XD (β) (A b A a ) D (β) (A b ) = MD (β) (A b ) a (Insttute) Representaton Theory 8 / 4

19 () If α 6= β, then M = 0 from Schur s lemma, we get M = a D (α) r (A a ) X rs D (β) sk Aa = a D (α) r (A a ) X rs D (β) ks (A a ) = 0 Choose X rs = δ rj δ sl (.e. X s zero except the jl element). Then a D (α) j (A α ) D (β) k` (A α ) = 0 =) for d erent rreps, the matrx elements, after summng over group elements, are orthogonal to each other. () α = β then we can wrte M = D (α) (A a ) XD (α) A a.ths mples a Then Take h T r a D (α) (A a ) M = MD (α) (A b ) =) M = ci D (α) (A a ) XD (α) Aa X rs = δ rj δ s` then T r X = δ j` and = cl or nt r X = cl, or c = (T r X ) n l α D (α) (A a ) j D (α) (A a ) k` = n δ k δ a l j` α Ths gves orthogonalty for d erent matrx elements wthn a gven rreducble representaton. (Insttute) Representaton Theory 9 / 4

20 Geometrc Interpretaton Imagne a complex n dm vector space, axes (or componenets) are labeled by group elements E, A.A 3 A n (Group element space). Components of vector are made out of matrx element of rreducble representaton matrx D (α) (A a ) j.each vector n ths n dm space s labeled by 3 ndces,, µ.ν ~D µν () = D µν () (E ), D µν () (A ), D µν () (A n ) Great orthogonalty theorem =) these vectors are? to each other. As a result l n because no more than n mutually? vectors n n-dm vector space. As an example, we take the -dmensonal representaton we have work out before, E = K = 0, A = 0 0, L = 0 p! 3 p 3, B = p 3 p! 3 p 3, M = p 3 p! 3 p! 3 (4) (5) (Insttute) Representaton Theory 0 / 4

21 Label the axses by the groupl elements n the order (E, A, B, K, L, M ). Then we can construct four 6-dmensonal ectors from these matrces, D () = (,,,,, ) D () p p p p = (0, 3 3, 0, 3, 3 ) D () p p p p = (0, 3, 3, 0 3, 3 ) D () = (,,,,, ) It s straghtforward to check that these 4 vectors are perpendcular to each other. Note that the other two vectors whch are orthogonal to these vectors are of the form, D E = (,,,,, ) D A = (,,,,, ) (6) comng from the dentty representaton and other -dmensonal representaton. (Insttute) Representaton Theory / 4

22 Character of Representaton Matrces n rrep are not unque because of smlarty transformaton. But trace of matrx s nvarant under smlarty transformaton, Tr SAS = TrA We can use trace, or character, to characterze the rrep. χ (α) (A ) T r hd (α) (A ) = D aa (α) (A ) a Useful Propertes If D (α) and D (β) are equvalent, then χ (α) (A ) = χ (β) (A ) 8 A G If A and B are n the same class χ (α) (A) = χ (α) (B ) Proof: If A and B are n same class =) 9x G such that xax = B =) D (α) (x ) D (α) (A) D (α) x = D (α) (B ) Usng D (α) x = D (α) (x ) (Insttute) Representaton Theory / 4

23 we get h T r D (α) (x ) D (α) (A) D (α) (x ) = T r hd (α) (B ) or χ (α) (A) = χ (α) (B ) Hence χ (α) s a functon of class, not of each element 3 Denote χ = χ (C ), character of th class. Let n c : number of classes n G, and n : number of group elements n C. From great orthogonalty theorem whch mples or r D (α) j (A r ) D (β) k` (A r ) = n δ l αβ δ k δ jl α χ (α) (A r ) χ (β) (A r ) = n δ r l αβ l α = nδ αβ α n χ (α) (C ) χ (β) (C ) = nδ αβ Ths s the great orthogonalty theorem for the characters. (Insttute) Representaton Theory 3 / 4

24 De ne U α = q n n χ(α) (C ), then great orthogonalty theorem mples, n c = U α U β = δ αβ Thus, f we consder U α as components n n c dm vector space, ~ Uα = (U α, U α, U αnc ), then ~ Uα α =,, 3 n r (n r : # of ndep rreps)form an othornormal set of vectors,.e. n c U β U α = U α U β = δ αβ = Ths mples that n r n c As an llustraton, characters for rep gven n Eqs(4,5) are χ (3) (E ) =, χ (3) (A) = χ (3) (B ) =, χ (3) (K ) = χ (3) (L) = χ (3) (M ) = 0 From these form a 3-dm vector, χ (3) = (,, 0) (7) Smlarly for rep n Eq(6) we get χ () (E ) =, χ () (A) = χ () (B ) =, χ () (K ) = χ () (L) = χ () (M ) = (Insttute) Representaton Theory 4 / 4

25 χ () (E ) =, χ () (A) = χ () (B ) =, χ () (K ) = χ () (L) = χ () (M ) = Form another two 3-dmensonal vectors, χ () = (,, ) χ () = (,, ) The orthogonalty relatons n Eq (??) these 3-dmensonal vectors are orthogonal to each other when weghted by # of elements n the class. For example, χ (), χ (3) = + ( ) + 0 ( ) 3 = 0 χ (), χ (3) = + ( ) = 0 (Insttute) Representaton Theory 5 / 4

26 Decomposton of Reducble Representaton For a reducble rep, we can wrte D = D () D () D.e. D (A ) = () (A ) D () (A ) 8A G The trace s χ (A ) = χ () (A ) + χ () (A ) Denote by D (α), α =, n r, all nequvalent untary rrep. decomposed as In terms of traces, Then any rep D can be D = c α D (α) c α : some nteger, # of tme D (α) appears α χ (C ) = c α χ (α) (C ) α The coe cent can be calculated as follows (by usng orthogonty theorem). Multply by n χ (β) and sum over or χ χ (β) n = c α χ (α) α c β = n χ (β) χ χ (β) n n = c α nδ αβ = nc β α (Insttute) Representaton Theory 6 / 4

27 From ths, n χ χ = n c α χ (α) c β χ (β) α,β = n jc α j α Ths leads to the followng theorem: Theorem: If rep D wth character χ sat es, then the representaton D s rreducble. n χ χ = n (Insttute) Representaton Theory 7 / 4

28 Regular Representaton Gven a group G = fa = E, A A n g.we can construct the regular rep as follows: Take any A G. If AA = A 3 = 0A + 0A + A 3 + 0A 4 +.e. we wrte the product "formally" as lnear combnaton of group elements, n n AA s = C rs A r = A r D rs (A),.e. C rs = D rs (A) s ether 0 or. (8) r = r =.e. D rs (A) = f AA s = A r or A = A r As = 0 otherwse Strctly speakng, the sum over group elments s unde ned. But here only one group element shows up n the rght-hand sde n Eq(8). No need to de ne the sum of group elements. Then D (A) s form a rep of G : regular representaton wth dmensonal n. Ths can be seen as follows: or r A r D rs (AB ) = ABA s = A t A t D ts (B ) = D ts (B ) A r D rt (A) tr D rs (AB ) = D rt (A) D ts (B ) (Insttute) Representaton Theory 8 / 4

29 From de nton of regular representaton dagonal elements are, D rs (A) = AA s = A r D rr (A) = AA r = A r or A = E Then every character s zero except for dentty class, Reduce D (reg ) to rreps. then Wrte c α = n χ (reg ) (C ) = 0 6= χ (reg ) (C ) = n = (9) D (reg ) = c α D (α) α χ (reg ) χ (α) n = n χ(reg ) χ (α) = n nl α = l α =) D reg contans rreps as many tmes as ts dmenson, χ (reg ) n r = l α χ (α) or χ (reg ) n r = χ (α) χ (α) = nδ α α= (Insttute) Representaton Theory 9 / 4

30 For dentty class χ reg = n, χ (α) = l α, then α l α = n Ths constrants the possble dmensonaltes of rreps becuase both n and l α have to be ntegers. For D 3, wth n = 6, the only possble soluton for α l α = 6 s l =. l =. l 3 =, and ther permutatons. (Insttute) Representaton Theory 30 / 4

31 We now want to show that n c = n r.e. # of classes = # of rreps. We wll derve another orthogonal relatons χ (α) wth summaton over rreps rather than classes, χ (α) χ (α) j α The dervaton s separated nto several steps. De ne D (α) by D (α) = AC D (α) (A) Then we can show that D (α) D (α) (A j ) D (α) D (α) Aj s a multple of dentty. Frst, = D (α) (A j ) D (α) (A) D (α) Aj AC = D (α) A j AAj = D (α) AC Usng D (α) Aj = D (α) (A j ) (Insttute) Representaton Theory 3 / 4

32 we get.e. D (α) D (α) D (α) (A j ) D (α) = D (α) D (α) (A j ) commutes wth all matrce n the rrep. From Schur s lemma, we get = λ (α) where λ (α) s some number. Takng the trace, n χ (α) = λ (α) l α, or λ (α) = n χ (α) l α = n χ (α) χ (α) where χ (α) s the character of dentty class. From the property of the class multplcaton, we have C C j = C jk C k k LHS C C j s a collectons of group element of the type A A j where A C and A j C j. Map these products nto rrep matrces and sum over to get D (α) D (α) j = C jk D (α) k k or λ (α) λ (α) j = C jk λ (α) k k (Insttute) Representaton Theory 3 / 4

33 For example, n the group D 3, we have classes, C = fe g, C = fa, B g, C 3 = fk, L, M g. Then C C 3 = fa, B g fk, L, M g = fak, AL, AM, BK, BL, BM g = fk, L, M g = C 3 Map these elements to ther matrx representaton and sum over, LHS = D (AK ) + D (AL) + D (AM ) + D (BK ) + D (BL) + D (BM ) = D (A)D (K ) + D (A)D (L) + D (A)D (M ) + D (B )D (K ) + D (B )D (L) + D (B )D (M = [D (A) + D (B )] [D (L) + D (M ) + D (K ] and RHS = [D (L) + D (M ) + D (K ] Ths llustrates the relaton.usng the values of λ (α) n Eq(??), n χ (α) χ (α) n j χ (α) j χ (α) = k n k χ (α) k C jk χ (α) or n n j χ (α) χ (α) j = χ (α) C jk n k χ (α) k k Sum over the rrep α, α n n j χ (α) χ (α) j = α C jk n k χ (α) χ (α) k k (Insttute) Representaton Theory 33 / 4

34 3 We now compute the coe cents C jk. If any element A belongs to th class, let the class whch contans A be denoted by 0 th class. We then get 0 C j = n for j 6= 0 for j = 0 We make use of the property of regular rep, α χ α χα k = nδ k, n Eq(??) to get, α n n j χ (α) χ (α) j = α C jk n k χ (α) χk α 0 = nc j = nn k for j 6= 0 for j = 0 Then Snce rep s untary n r χ (α) χ (α) j = n δj 0 α= n j we get and D (α) (A ) = D (α) (A ) = D (α) A = D (α) (A 0 ) Ths s the orthogonal relaton for χ (α). χ (α) 0 = χ (α) n r χ (α) χ (α) j α= = n n j δj (Insttute) Representaton Theory 34 / 4

35 If we now consder χ (α) as a vector n n r dm space! χ = χ (), χ (),... χ (nr ) we get n c n r Combne ths wth the result n r n c, we have derved before, we get n r = n c (Insttute) Representaton Theory 35 / 4

36 Character Table For a nte group, the essental nformaton about the rrep can be summarzed n a table wtth characters of each rrep n terms of the classes. To construct such table, use the followng useful nformaton: # of columns = # of rows = # of classes α l α = n 3 n χ (α) χ (β) = nδ αβ and χ (α) α χ (α) j = n n δ j 4 If l α =, χ s tself a rep. 5 χ (α) A = T r D (α) A = T r D (α)+ A = χ (α) (A) If A and A are n the same class then χ (A) s real. 6 D (α) s a rep =) D (α) s also a rep so f χ (α) s are complex numbers, another row wll be ther complex conjugate 7 If l α >, χ (α) = 0 for at least one class. Ths follows from the relaton n jχ j = n and n = n 8 For physcal symmetry group, x.y and z form a bass of a rep. Example : D 3 character table (Insttute) Representaton Theory 36 / 4

37 E C 3 3C 0 x + y, z A R z, z. A (xz, yz) (x, y ) E 0 x y, xy (R x.r y ) Here, typcal bass functons up to quadratc n coordnate system are lsted. Remark: the bass functons are not necessarly normalzed. Usng the transformaton propertes of the coordnate, we can also nfer the transformaton propertes of any vectors. For example, the usual coordnates have the transformaton property, ~r = (x, y, z) A E n D 3 Ths means that electrc eld of ~ E or magnetc eld ~ B wll have same transformaton property, ~ B s ~ E s A E because they all transform the same way under the rotaton. (Insttute) Representaton Theory 37 / 4

38 Product Representaton (Kronecker product) Let x be the bass for D (α),.e. x 0 = `α x j D (α) j (A) j= `β y` be the bass for D (β),.e. yk 0 = y`d (β) `k (A) `= then the products x j y l transform as x 0 y 0 k = D (α) j j` (A) D (β) k` (A) x j y` j` D (αβ) j`;k (A) x j y` where D (αβ) j`;k (A) = D (α) j (A) D (β) `k (A) Note that matrces, row and column are labelled by ndces, nstead of one. We can show that D (αβ) forms a rep of the group. h D (αβ) (A) D (αβ) (B ) = D (αβ) (A) j;k` j,st D (αβ) (B ) st;k` s.t = s.t D (α) s (A) D (β) jt (A) D (α) sk (B ) D (β) t` (B ) = D (α) k (AB ) D (β) j` (AB ) = D (αβ) (AB ) k;k` or D (αβ) (A) D (αβ) (B ) = D (αβ) (AB ) (Insttute) Representaton Theory 38 / 4

39 The bass functons for D (αβ) are x y j The character of ths rep can be calculated by makng the row and colum ndces the same and sum over, χ (αβ) (A) = j` D (αβ) j`;j` (A) = j` D (α) jj χ (αβ) (A) = χ (α) (A) χ (β) (A) (A) D (β) `` (A) = χ (α) (A) χ (β) (A) If α = β, we can further decompose the product rep by symmetrzaton or antsymmetrzaton; D fααg k,j` (A) = h D (α) j (A) D (α) (α) k` (A) + D ` (A) D (α) kj (A) bass p (x y k + x k y ) D [αα] k,j` (A) = h D (α) j (A) D (α) k` (A) D (α) ` (A) D (α) kj (A) These matrces also form rep of G and the characters are gven by bass p (x y k x k y ) χ fααg (A) = χ (α) (A) + χ (α) A, χ [αα] (A) = χ (α) (A) χ (α) A Example D 3 E. C 3 3C 0 Γ R z. z Γ (xz, yz) (x, y ) Γ 3 0 x y, xy Γ 3 Γ = Γ Γ Γ 3 (Γ 3 Γ 3 ) s 3 0 = Γ Γ 3 (Γ 3 Γ 3 ) a = Γ (Insttute) Representaton Theory 39 / 4

40 Drect Product Group Gven groups G = fe, A A n g, G = fe, B B m g, de ne product group as G G = fa B j ; = n, j = mg wth multplcaton law (A k B`) (A k 0 B`0 ) = (A k B k 0 ) (B`B`0 ) It turns out that rrep of G G are just drect product of rreps of G, and G. Let D (α) (A ) be an rrep of G and D (β) (B j ) an rrep of G then the matrces de ned by wll have the property D (αβ) (A B j ) ab;cd D (α) (A ) ac D (β) (B j ) bd h D (αβ) (A B j ) D (αβ) (A k B`) ab;cd h h = D (αβ) (A B j ) D (αβ) (A k B`) ab;ef ef ;cd ef = hd (α) (A ) ac D (α) (A k ) ec hd (β) (B j ) bf D (β) (B e ) fd ef = D (α) (A A k ) ac D (β) (B j B`) bd = D (αβ) (A A k. B j B`) ab;cd Thus matrce D (αβ) (A B j ) form a representaton of the product group G G. The characters are, χ (αβ) (A B j ) = D (αβ) (A B j ) ab;ab = ab D (α) (A ) aa D (β) (B j ) bb = χ (α) (A ) χ (β) (B j ) ab (Insttute) Representaton Theory 40 / 4

41 Then j χ (αβ) (A B j ) =! χ (α) (A ) j! χ (β) (B j ) = nm =) D (αβ) s rrep. Example, G = D 3 = fe, C 3, 3C 0 g, G = fe, σ h g = ϕ where σ h : re ecton on the plane of trangle. Drect product group s then D 3h D 3 ϕ = E, A, B = fe, C 3, 3C 0, σ h, C 3 σ h, 3C 0 σ hg Character Table ϕ E σ h Γ + Γ Character Table E C 3 C 0 σ h C 3 σ h C 0 σ h Γ + Γ + Γ Γ + Γ + Γ C 3 C 0 D 3 E AB KLM Γ Γ Γ 3 0 (Insttute) Representaton Theory 4 / 4