有限群表示习题解答

1、Solutions to Linear Representations of Finite GroupsSteven V SMay 18, 2010Contents2Character theory13Subgroups, products, induced representations55Examples76The group algebra97Induced representations; Mackeys criterion128Examples of induced representations169Artins theorem2110A theorem

2、of Brauer2611Applications of Brauers theorem2812Rationality questions2913Rationality questions: examples3114The groups RK (G), Rk(G), and Pk(G)3615The cde triangle3916Theorems4017Proofs4118Modular characters42References44by Jean-Pierre Serre12 CHARACTER THEORY22 Character theory1. Exercise. Letand0

3、be the characters of two representations. Prove the formulas:( + 0)2 =2 + 02 +0; Solution. By Proposition 3,2 (s) = 12 ( (s)2 + (s2) for all s 2 G, whence( + 0)2 (s) = 12( + 0)(s)2 + ( + 0)(s2)= 12( (s)2 + 2 0(s) + 0(s)2 + (s2) + 0(s2)= 12( (s)2 + (s2) + 12( 0(s)2 + 0(s2) + 0(s)= 2 (s) + 02(s) + 0(s

4、);for all s 2 G, which gives the rst formula. The second formula follows in exactly the same way. 2. Exercise. Let X be a nite set on which G acts, let be the corresponding permutation representation, and X be the character of . Let s 2 G; show that X (s) is the number of elements of X xed by s.Solu

5、tion. By de nition, the permutation representation is the vector space V with basis(ex)x2X , with s for s 2 G de ned by sex = esx. Put an ordering on X. If we write s as a matrix with respect to this basis, there is a 1 in the nth diagonal if the nth element in X is xed by s, and 0 otherwise. Thus t

6、he number of elements of X xed by s is the trace of s. 3. Exercise. Let : G ! GL(V ) be a linear representation with character and let V 0 be the dual of V , i.e., the space of linear forms on V . For x 2 V , x0 2 V 0 let hx; x0i denote the value of the linear form x0 at x. Show that there exists a

7、unique linear representation 0 : G ! GL(V 0), such thath sx; 0sx0i = hx; x0i for s 2 G; x 2 V; x0 2 V 0:This is called the contragredient (or dual) representation of ; its character is . Solution. De ne 0s := ( Ts ) 1 to be the inverse of the transpose of s. Thenh sx; 0sx0i = hx; Ts 0sx0i = hx; x0i:

8、ss2V and x02V 0, we haveIf 00 also satis es the conditions of 0 , then for any xhx; ( 000)x0i=hx; 0 x0i hx; 00x0i= 0:sssssssFixing x0 and letting x vary shows that ( s0s00)x0= 0 for all x0, so s0 = s00. The last remarkfollows from Proposition 1(ii) and the fact that the trace of a matrix equals the

9、trace of its transpose. 4. Exercise. Let 1 : G ! GL(V1) and 2 : G ! GL(V2) be two linear representations withcharacters 1 and 2. Let W = Hom(V1; V2), the vector space of linear mappings f : V1 ! V2. For s 2 G and f 2 W let sf = 2;s f 1;s1; so sf 2 W . Show that this de nes a2 CHARACTER THEORY3linear

10、 representation : G ! GL(W ), and that its character is 1 2. This representation is isomorphic to 01 2 where 01 is the contragredient of 1.Solution. Pick s; t 2 G and f 2 W . Thenstf = 2;stf1;st1 = ( 2;s2;t) f ( 1;t11;s1) = s tf;so this is a linear representation.Now de ne : V10 V2 ! W in the follow

11、ing way. For any element v0 v, let (v0 v) be the linear map V1 ! V2 that sends x 2 V1 to hv0; xi v. Extend this linearly to get . For any f 2 W , choosing bases for V1 and V2 and writing f in matrix notation makes it clear that is surjective. Since V10 V2 and W have the same dimension, is an isomorp

12、hism.We claim that s = ( 01;s2;s) for all s 2 G. To check this, pick x y 2 V10 V2. Then( s)(xy) = 2;s(xy)1;s1= 2;s (v 7!x;h vi y) 1;s1= 2;s (v 7!x;h 1;s1(v)i y)= v 7!x;h 1;s1(v)i 2;s(y)= v 7! h01;s(x); vi 2;s(y):On the other hand,( 01;s2;s)(x y) = ( 01;s(x)2;s(y) = v 7! h01;s(x); vi 2;s(y):Since thi

13、s equality holds for xy, it holds for a basis of V10V2, so it holds for a general element.This shows that the representationis isomorphic to 012 via 1.By Proposition 2(ii) and the previous exercise, we have the statement about the character.5. Exercise. Letbe a linear representation with character.

14、Show that the number of timesPthatcontains the unit representation is equal to ( j1) = (1=g)s2G(s).Solution. This is an immediate consequence of Theorem 4 and Corollary 2 to Theorem 4. In particular, ( j1) is the number of irreducible components of with character 1, and the only such representation

15、is the unit representation. 6. Exercise. Let X be a nite set on which G acts, let be the corresponding permutation representation and let be its character.(a) The set Gx of images under G of an element x 2 X is called an orbit. Let c be the numberof distinct orbits. Show that c is equal to the numbe

16、r of times that contains the unit representation 1; deduce from this that ( j1) = c. In particular, if G is transitive (i.e., if c = 1), can be decomposed into 1 and does not contain the unit representation. Ifis the character of , we have= 1 +and ( j1) = 0.(b) Let G act on the product X X of X by i

17、tself by means of the formula s(x; y) = (sx; sy). Show that the character of the corresponding permutation representation is equal to 2.(c) Suppose that G is transitive on X and that X has at least two elements. We say that G is doubly transitive if, for all x; y; x0; y0 2 X with x 6= y and x0 6= y0

18、, there exists s 2 G such that x0 = sx and y0 = sy. Prove the equivalence of the following properties:2 CHARACTER THEORY4(i) G is doubly transitive.(ii) The action of G on X X has two orbits, the diagonal and its complement.(iii) ( 2j1) = 2.(iv) The representation de ned in (a) is irreducible.Soluti

19、on. First suppose that the action of G on X is transitive. For any element x 2 X, the subgroup of elements in G that x x has index equal jXj, the cardinality of X. Thus, the numberPof pairs (s; x) where s 2 G and x 2 X such that sx = x is equal to jXj x2X g=jXj = g. For a general group action, there

20、 is an induced transitive group action on any orbit, so the number of pairs (s; x) as above is cg. Then1X1( j1) =(s) =cg = cg s2Ggby (Ex. 2.2), so by (Ex. 2.5), we get the rst claim in (a). The other statements are immediate.For s 2 G, (x; y) 2 X X is a xed point if and only if both x and y are xed.

21、 If n is the number of xed points of s on X, then the number of xed points of s on X X is n2. By (Ex. 2.2), we get (b).Now we prove (c). That (i) is equivalent to (ii) is immediate from the de nition of doubly transitive. From (b), we know that 2 is the character of the group action of G on X X, and

22、 from (a), ( 2j1) is the number of orbits of this action. This gives the equivalence of (ii)and (iii). Finally, by (a), 2 = 1 + 2 + 2, and ( j1) = 0. Assuming (iii), we deduce that (1j1) + ( 2j1) = 2, which is equivalent to ( 2j1) = 1. Since (s) is one less than the number ofxed points in X of s, it

23、 is a real number. In particular, this means that ( 2 j1) = ( j ), so by Theorem 5, we know that is irreducible if and only if ( 2j1) = 1, so we see that (iii) and (iv) are equivalent. 7. Exercise. Show that each character of G which is zero for all s 6= 1 is an integral multiple of the character rG

24、 of the regular representation.Solution. Let be the character of a representation of G which is zero for all s 6= 1. If 1 is the character of the unit representation, which has degree 1, then1X1h ; 1i =(s 1)1(s) =(1)gs2Ggis the number of times that the unit representation appears in . Since this mus

25、t be an integer, g divides (1), so is a multiple of rG. 8. Exercise. Let Hi be the vector space of linear mappings h: Wi ! V such that sh = h s for all s 2 G. Each h 2 Hi maps Wi into Vi.(a) Show that the dimension of Hi is equal to the number of times that Wi appears in V , i.e., to dim Vi= dim Wi.

26、(b) Let G act on Hi Wi through the tensor product of the trivial representation of G on Hi and the given representation on Wi. Show that the mapF : HiWi ! Vide ned by the formulai WiXiXF (h w ) =h (w )is an isomorphism of Honto V .2 CHARACTER THEORY5(c) Let (h1; : : : ; hk) be a basis of Hi and form

27、 the direct sum WiWi of k copies of Wi.The system (h1; : : : ; hk) de nes in an obvious way a linear mapping h of Wi Wi into Vi; show that it is an isomorphism of representations and that each isomorphism is thus obtainable. In particular, to decompose Vi into a direct sum of representations isomorp

28、hic to Wi amounts to choosing a basis for Hi.Solution. To show (a), rst choose a linear mapping h: Wi ! V such that sh = h s for all s 2 G. Then h maps Wi into Vi = Wi Wi, say the number of times Wi appears in V is ki. Composing this with the ki projection functions Vi ! Wi shows that h is a linear

29、combination of maps Wi ! Wi with the above commutativity condition. Thus, it is enough to calculate the dimension of the space of all such functions and show that it is 1. But this follows immediately from Schurs lemma (Proposition 4(2).The map F de ned in (b) is certainly surjective by the above co

30、mments, and the dimensions of Hi Wi and Vi are the same by (a), so F is an isomorphism of vector spaces. Let 0 denote the representation G ! GL(Hi Wi) described in (b). On the one hand,F ( 0s(hw ) = F (hs(w ) = h ( s(w ) = s(h (w )for all s 2 G, and on the other hand, s(F (h w ) = s(h (w ) for all s

31、 2 G. Since F 0s = s F for all generators, we can extend linearly to see that they agree on Hi Wi, soF is an isomorphism of representations.Now given a basis (h1; : : : ; hk), we can de ne h: WiWi ! Vi by (w1; : : : ; wk) 7!h1(w1) + hk(wk). To see that h is an isomorphism of vector spaces, it is eno

32、ugh to show that itPis surjective. But this follows from (b) because any element of Vi is of the form h (w ), and (h1; : : : ; hk) is a basis, so this sum can be expressed as a linear combination of these basis vectors. The proof that h is an isomorphism of representations is similar to the one give

33、n for F . Conversely, given an isomorphism h: Wi Wi ! Vi of representations, we can obtain linear maps hj : Wi ! Vi by letting hj = h j where j : Wi ,! Wi Wi is inclusion into the jth summand. And because h is an isomorphism of representations, we will have hj commuting with . Since there are k maps

34、, they form a basis if they are linearly independent. If they were linearly dependent, then this would contradict that h is an isomorphism of vector spaces because we could construct a nontrivial vector in the kernel. Thus, every isomorphism arises in the way described in (c). 9. Exercise. Let Hi be

35、 the space of linear maps h: Wi ! V such that h s = s h. Show that the map h 7!h(e ) is an isomorphism of Hi onto Vi; .Solution. Suppose that h(e ) = 0 for some h. Then from (Ex. 2.8(b), we have h e = 0 inHi Wi. Since e 6= 0, we must have h = 0, so the map h 7!h(e ) is injective. By (Ex. 2.8(a) and

36、Proposition 8(a), dim Hi = dim Vi; , so we are done. 10. Exercise. Let x 2 Vi, and let V (x) be the smallest subrepresentation of V containing x. Let x1 be the image of x under p1 ; show that V (x) is the sum of the representations W (x1 ), = 1; : : : ; n. Deduce from this that V (x) is the direct s

37、um of at most n subrepresentations isomorphic to Wi.Solution. Since V (x) is invariant under s for all s 2 G, it is invariant under p1 since this is a linear combination of the s. So V (x) contains the x1 . Since W (x1 ) is generated by the images of x1 under 1, which V (x) is also invariant under,

38、we have W (x1 ) V (x). Their sum is a subrepresentation containing x, so is equal to V (x) by minimality. 3 SUBGROUPS, PRODUCTS, INDUCED REPRESENTATIONS63 Subgroups, products, induced representations1. Exercise. Show directly, using Schurs lemma, that each irreducible representation of an Abelian gr

39、oup, nite or not, has degree 1.Solution. Let : G ! GL(V ) be an irreducible representation of G, and pick s 2 G. Then for any other t 2 G, we have s t = t s because G is Abelian. By Schurs lemma (Proposition4(2), this means that s is a homothety. Since s was arbitrary, is irreducible if and only if

40、dim V = 1. 2. Exercise. Let be an irreducible representation of G of degree n and character ; let C be the center of G (i.e., the set of s 2 G such that st = ts for all t 2 G), and let c be its order.(a) Show that s is a homothety for each s 2 C. Deduce from this that j (s)j = n for all s 2 C.(b) Pr

41、ove the inequality n2 g=c.(c) Show that, if is faithful (i.e., s 6= 1 for s 6= 1), the group C is cyclic.Solution. For s 2 C, and any t 2 G, 4(2), s is a homothety. Since C has the identity, then must be a root ofFor (b), we havewe have s t = t s. By Schurs lemma (Proposition nite order, s has nite

42、order. If we write s as times unity, which means j (s)j = jn j = n, and nishes (a).1 = ( ) =1X(s) (s) = 1X(s) 2;jj jgs2Ggs2Gso this givesXXg =j (s)j2j (s)j2 = cn2;s2Gs2Cwhence the inequality g=cn2.By the classi cation of nitely generated Abelian groups, we can write C = Z=m1 Z=mk where every two mi

43、and mj have a common divisor 1. If k 1, then take a 2 Z=m1 0 0 and b 2 0 Z=m2 0 such that the orders of a and b have a common divisor and neither is the identity. Since a is an m1th root of unity and b is a m2th root of unity, we can nd some linear combination c1a + c2b 6= 0 such that its image unde

44、r is the identity. Hence, if is faithful, then k = 1, so C is cyclic. 3. Exercise. Let G be an Abelian group of order g, and let G be the set of irreducible charactersof G. If1; 2 belong to G, the same is true of their product 1 2. Show that this makes G anAbelian group of order g; the group G is ca

45、lled the dual of the group G. For x 2 G the mapping7! (x) is an irreducible character of G and so an element of the dual G of G. Show thatthe map of G into G thus obtained is an injective homomorphism; conclude (by comparing the orders of the two groups) that it is an isomorphism.Solution. That G is

46、 a group follows from the fact that every irreducible representation corre-sponds to a root of unity, and hence inverses exist. It is clear that G is Abelian because the product structure is just induced by multiplication of complex numbers. Also, is 1; : : : ; h are the irreducible representations

47、of G with multiplicity ni in the regular representation of G, thenPhn2 = g by Corollary 2 to Proposition 2.5. By (Ex. 3.1), each n1 = 1, so we must havei=1ih = g, which shows that G has order g.3 SUBGROUPS, PRODUCTS, INDUCED REPRESENTATIONS7For x 2 G, let a be the order of x. Then there is an irredu

48、cible representation that maps x toan ath root of unity. It follows that if x is nontrivial, the map7! (x) is nontrivial on G. Sox 7!( 7! (x) is an injective map G ! G. By the above argument, G has order g, so it is an isomorphism, which nishes (c). 4. Exercise. Show that each irreducible representa

49、tion of G is contained in a representation induced by an irreducible representation of H. Obtain from this another proof of the corollary to Theorem 9.Solution. Let V be the regular representation of G. If we take the trivial representation on H, then it is irreducible and induces the regular repres

50、entation and hence contains every irreducible representation of G. The last remark is a consequence of Theorem 12. 5. Exercise. Let (W; ) be a linear representation of H. Let V be the vector space of functions f : G ! W such that f(tu) = tf(u) for u 2 G, t 2 H. Let be the representation of G in V de

51、 ned by ( sf)(u) = f(us) for s; u 2 G. For w 2 W let fw 2 V be de ned by fw(t) = tw for t 2 H and fw(s) = 0 for s 2= H. Show that w 7!fw is an isomorphism of W onto the subspace W0 of V consisting of functions which vanish o H. Show that, if we identify W and W0 in this way, the representation (V; )

52、 is induced by the representation (W; ).Solution. The map w 7!fw is injective because fw(1) = w. For surjectivity, given a map f : G ! W such that f(tu) = tf(u) that vanishes o of H, take w = f(1). Then fw(t) =tf(1) = f(t) for all t2H. So w7!w0.fgives an isomorphism of vector spaces W =WLet R be a s

53、ystem of representatives of G=H. For each s 2 R and f 2 W0, sf vanishes o of sH. Any function G ! W can be de ned piecewise on the cosets of G, so any element of V canbe written uniquely as a sum of the formPs2Rffor some fs 2W. This shows that (V; )is induced by (W; ).ss06. Exercise. Suppose that G

54、is the direct product of two subgroups H and K. Let be a representation of G induced by a representation of H. Show that is isomorphic to rK , where rK denotes the regular representation of K.Solution. Write G = H K. In this case, H is a normal subgroup of G, and a system of representatives of G=H i

55、s given by (1; k) where 1 is the identity in H and k ranges over all elements of K. Let 1, 2, , and 0 be the characters of , rK , rK , and , respectively. Then for (h; k) 2 G, we have(h; k) = 1(h) 2(k) = (0jKj 1(h)k = 1k = 16where jKj denotes the order of K. By Theorem 12 and Proposition 1(iii) of Chapter 2,0(h; 1) = jH1j X 1(s 1hs) = jH1j X 1(h) = jjHGjj 1(h) = jKj1(h):s2Gs2GFor (h; k) with k 6= 1, the condition s 1(h; k)s 2 H never holds because conjugation is an automorphism and hence cannot map k to the identity. Thus, the sum in Theorem 12 is empty, so 0(h;