REPRESENTATION THEORY. WEEK 4

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1 REPRESENTATION THEORY. WEEK 4 VERA SERANOVA 1. uced modules Let B A be rings and M be a B-module. Then one can construct induced module A B M = A B M as the quotient of a free abelian group with generators from A M by relations a 1 + a 2 ) m a 1 m a 2 m, a m 1 + m 2 ) a m 1 a m 2, ab m a bm, and A acts on A B M by left multiplication. Note that j : M A B M defined by is a homomorphism of B-modules. j m) = 1 m Lemma 1.1. Let N be an A-module, then for ϕ Hom B M, N) there exists a unique ψ Hom A A B M, N) such that ψ j = ϕ. Proof. Clearly, ψ must satisfy the relation ψ a m) = aψ 1 m) = aϕm). It is trivial to check that ψ is well defined. Exercise. Prove that for any B-module M there exists a unique A-module satisfying the conditions of Lemma 1.1. Corollary 1.2. Frobenius reciprocity.) For any B-module M and A-module N there is an isomorphism of abelian groups Hom B M, N) = Hom A A B M, N). Example. Let k F be a field extension. Then induction F k is an exact functor from the category of vector spaces over k to the category of vector spaces over F, in the sense that the short exact sequence becomes an exact sequence 0 V 1 V 2 V F k V 1 F k V 2 F k V 3 0. Date: September 27,

2 2 VERA SERANOVA In general, the latter property is not true. It is not difficult to see that induction is right exact, i.e. an exact sequence of B-modules induces an exact sequence of A-modules But an exact sequence M N 0 A B M A B N 0. 0 M N is not necessarily exact after induction. Later we discuss general properties of induction but now we are going to study induction for the case of groups. 2. uced representations for groups. Let H be a subgroup of and ρ : H L V ) be a representation. Then the induced representation H ρ is by definition a k )-module k ) kh) V. Lemma 2.1. The dimension of H ρ equals the product of dimρ and the index [ : H] of H. More precisely, let S is a set of representatives of left cosets, i.e. then = s S sh, 2.1) k ) kh) V = s S s V. For any t, s S there exist unique s S, h H such that ts = s h and the action of t is given by 2.2) t s v) = s ρ h v. Proof. Straightforward check. Lemma 2.2. Let χ = χ ρ and H χ denote the character of H ρ. Then 2.3) H χ t) = χ s 1 ts ) = 1 χ s 1 ts ). H Proof. 2.1) and 2.2) imply s S,s 1 ts H H χ t) = s S,s =s s,s 1 ts H χ h). Since s = s implies h = s 1 ts H, we obtain the formula for the induced character. Note also that χ s 1 ts) does not depend on a choice of s in a left coset.

3 REPRESENTATION THEORY. WEEK 4 3 Corollary 2.3. Let H be a normal subgroup in. Then H χ t) = 0 for any t / H. Theorem 2.4. For any ρ : L V ), σ: H L W), we have the identity 2.4) H χ σ, χ ρ ) = χ σ, Res H χ ρ ) H. Here a subindex indicates the group where we take inner product. Proof. It follows from Frobenius reciprocity, since dimhom H W, V ) = dimhom H W, V ). Note that 2.4) can be proved directly from 2.3). Define two maps Res H : C ) C H), H : C H) C ), the former is the restriction on a subgroup, the latter is defined by 2.3). Then for any f C ), g C H) 2.5) H g, f) = g, Res H f) H. Example 1. Let ρ be a trivial representation of H. Then H ρ is the permutation representation of obtained from the natural left action of on /H the set of left cosets). Example 2. Let = S 3, H = A 3, ρ be a non-trivial one dimensional representation of H one of two possible). Then H χ ρ 1) = 2, H χ ρ 12) = 0, H χ ρ 123) = 1. Thus, by induction we obtain an irreducible two-dimensional representation of. Now consider another subgroup K of = S 3 generated by the transposition 12), and let σ be the unique) non-trivial one-dimensional representation of K. Then K χ σ 1) = 3, K χ σ 12) = 1, H χ ρ 123) = Double cosets and restriction to a subgroup If K and H are subgroups of one can define the equivalence relation on : s t iff s KtH. The equivalence classes are called double cosets. We can choose a set of representative T such that = s T K th. We define the set of double cosets by K\/H. One can identify K\/H with K- orbits on S = /H in the obvious way and with -orbits on /K /H by the formula KtH K, th).

4 4 VERA SERANOVA Example. Let F q be a field of q elements and = L 2 F q ) def = L F 2 q). Let B be the subgroup of upper-triangular matrices in. Check that = q 2 1) q 2 q), B = q 1) 2 q and therefore [ : B] = q + 1. Identify /B with the set of lines P 1 in F 2 q and B\/B with -orbits on P1 P 1. Check that has only two orbits on P 1 P 1 : the diagonal and its complement. Thus, B\/B = 2 and where = B BsB, s = Theorem 3.1. Let T such that = s T KtH. Then where for any h shs 1. ) Res K H ρ = s T K K shs 1 ρs, ρ s h def = ρ s 1 hs, Proof. Let s T and W s = k K)s V ). Then by construction, W s is K-invariant and k ) kh) V = s T W s. Thus, we need to check that the representation of K in W s is isomorphic to K K shs 1 ρs. We define a homomorphism α : K K shs 1 V W s by α t v) = ts v for any t K, v V. It is well defined αth v t ρ s hv) = ths v ts ρ s 1 hsv = ts s 1 hs ) v ts ρ s 1 hsv = 0 and obviously surjective. Injectivity can be proved by counting dimensions. Example. Let us go back to our example B SL 2 F q ). Theorem 3.1 tells us that for any representation ρ of B B ρ = ρ H ρ, where H = B sbs 1 is a subgroup of diagonal matrices and ) ) a 0 b 0 ρ = ρ 0 b 0 a Corollary 3.2. If H is a normal subgroup of, then Res H H ρ = s /H ρ s.

5 REPRESENTATION THEORY. WEEK Mackey s criterion To find H χ, H χ) we can use Frobenius reciprocity and Theorem 3.1. H χ, H χ ) = Res H H χ, χ ) = H H H shs 1 χs, χ ) = H s T = χ s, Res H shs 1 χ) H shs 1 = χ, χ) H + χ s, Res H shs 1 χ) H shs 1. s T s T \{1} We call two representation disjoint if they do not have the same irreducible component, i.e. their characters are orthogonal. Theorem 4.1. Mackey s criterion) H ρ is irreducible iff ρ is irreducible and ρs and ρ are disjoint representations of H shs 1 for any s T \ {1}. Proof. Write the condition and use the above formula. H χ, H χ) = 1 Corollary 4.2. Let H be a normal subgroup of. Then H ρ is irreducible iff ρs is not isomorphic to ρ for any s /H, s / H. Remark 4.3. Note that if H is normal, then /H acts on the set of representations of H. In fact, this is a part of the action of the group AutH of automorphisms of H on the set of representation of H. eed, if ϕ AutH and ρ : H L V ) is a representation, then ρ ϕ : H L V ) defined by is a new representation of H. ρ ϕ t = ρ ϕt), 5. Some examples Let H be a subgroup of of index 2. Then H is normal and = H sh for some s \H. Suppose that ρ is an irreducible representation of H. There are two possibilities 1) ρ s is isomorphic to ρ; 2) ρ s is not isomorphic to ρ. Hence there are two possibilities for H ρ : 1) H ρ = σ σ, where σ and σ are two non-isomorphic irreducible representations of ; 2) H ρ is irreducible. For instance, let = S 5, H = A 5 and ρ 1,..., ρ 5 be irreducible representation of H, where the numeration is from lecture notes week 3. Then for i = 1, 2, 3 H ρ i = σ i σ i sgn),

6 6 VERA SERANOVA here sgn denotes the sign representation. Furthermore, H ρ 4 = H ρ 5 is irreducible. Thus S 5 has two 1, 5, 4-dimensional irreducible representations and one 6-dimensional. Now let be a subgroup of L 2 F q ) of matrices ) a b 0 1 We want to classify irreducible representations of over C. = q 2 q, has the following conjugacy classes ) ) ) a 0,,, in the last case a 1. Note that the subgroup H of matrices ) 1 b 0 1 is normal, /H = F q = Z q 1. Therefore has q 1 one-dimensional representations which can be lifted from /H. That leaves one more representation, its dimension must be q 1. We hope to obtain it by induction from H. Let σ be a non-trivial irreducible representation of H one-dimensional). Then dim H σ = q 1 as required. Note that for any previously constructed one-dimensional representation ρ of we have H σ, ρ) = σ, Res H ρ) H = 0, as Res H ρ is trivial. Therefore H σ is irreducible. The character takes values q 1, 1 and 0 on the corresponding conjugacy classes. Remark 5.1. To find all one-dimensional representation of a group, find its commutator, which is a subgroup generated by ghg 1 h 1 for all g, h. Onedimensional representations of are lifted from one-dimensional representations of /.