Induced Characters That We Did Earlier and Look at What Frobenius Reciprocity Says About Them

Induced Characters That We Did Earlier and Look at What Frobenius Reciprocity Says About Them

32 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS I want to calculate IndZ/4 χ . By the proposition, the value of this in- Z/2 − duced character on 1,σ,σ2,σ3 is the index G : H = 2 times 1, 0, 1, 0 respectively. This gives 2, 0, 2, 0 as indicated| below| the character− ta- − ble for G = Z/4: G = Z/41σσ2 σ3 χ1 1 1 1 1 χ 1 11 1 2 − − χ3 1 i 1 i χ 1 i −1 −i 4 − − IndZ/4 χ 20 20 Z/2 − − By examination we see that Z/4 Ind χ = χ3 + χ4 Z/2 − 3.1.2. example 2. In the nonabelian case we have the following formula which is analogous to the one in the abelian case. Proposition 3.4. G 1 Ind χ(σ)= G : H (average value of χ(τστ − )) H | | Now let G = S3 and H = 1, (12) ∼= Z/2. Using the same notation as in the previous example, let{ χ be} the one dimensional character on H given by χ (1) = 1,χ (12) = − 1. We want to compute the induced −G − − character IndH χ . − G IndH χ (1) = G : H χ (1) = (3)(1) = 3 − | | − Since (12) has three conjugates only one of which lies in H, the average value of χ1 on these conjugates is 1 1 ( 1 + 0 + 0) = 3 − −3 So, G 1 3 IndH χ (12) = G : H − = − = 1 − | | 3 3 − ! " Since neither of the conjugates of (123) lie in H we have: G IndH χ (123) = 0 − MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 33 G So, IndH χ takes the values 3, 1, 0 on the conjugacy classes of G = − − S3. Put it below the character table of S3: G = S3 1 (12) (123) χ1 1 1 1 χ2 1 11 χ 20− 1 3 − G IndH χ 3 10 − − We can see that G IndH χ = χ2 + χ3 − 3.1.3. Frobenius reciprocity for characters. First I need some fancy no- tation for a very simple concept. If f : G C is any class function →G then the restriction of f to H, denoted ResH f, is the composition of f with the inclusion map j : H$ G: → G ResH f = f j : H C ◦ → Theorem 3.5 (Frobenius reciprocity). Suppose that g, h are class func- tions on G, H respectively. Then G G IndH h, g G = h, ResH g H Suppose for a moment# that$ this is# true. Then,$ letting h = χV and taking g to be the irreducible character g = χi, we get: G G IndH χV ,χi G = χV , ResH χi H = ni G Since ResH χi is# the character$ of# the G-module$ Si considered as an H-module, the number ni is a nonnegative integer, namely: ni = dimC HomH (V, Si) This implies that G IndH χV = χW where W is the G-module W = niSi. In other words, the induced character is an effective character (the character of some representa- tion). % Corollary 3.6. If h : H C is an effective character then so is G → IndH h : G C. → This is a rewording of the main theorem (Theorem 3.2). 34 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS Proof of Frobenius reciprocity for characters. Since G 1 1 Ind h(σ)= h(τστ − ) H H τ G | | &∈ the left hand side of our equation is 1 1 1 1 LHS = h(τστ − )g(σ− ) G H σ G τ G | | &∈ | | &∈ 1 1 1 1 Since g is a class function, g(σ− )=g(τσ− τ − ). Letting α = τστ − we get a sum of terms of the form 1 h(α)g(α− ) How many times does each such term occur? 1 Claim: The number of ways that α can be written as α = τστ − is exactly n = G . The proof| of| this claim is simple. For each τ G there is exactly 1 ∈ one σ which works, namely, σ = τ − ατ. This implies that 1 1 LHS = h(α)g(α− ) H α G | | &∈ Since h is a class function on H, h(α) = 0 if α/ H. Therefore, the sum can be restricted to α H and this expression∈ is equal to the RHS ∈ of the Frobenius reciprocity equation. ! 3.1.4. examples of Frobenius reciprocity. Let’s take the two example of induced characters that we did earlier and look at what Frobenius reciprocity says about them. In the case G = Z/4,H = Z/2, the restrictions of the four irreducible characters of G = Z/4 to H (given by the first and third columns) are: G ResH χ1 = χ+ G ResH χ2 = χ+ G ResH χ3 = χ − G ResH χ4 = χ − Frobenius reciprocity says that the number of times that χ appears G − in the decomposition of ResH χi is equal to the number of times that G χi appears in the decomposition of IndH χ . So, − G IndH χ = χ3 + χ4 − MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 35 In the case G = S ,H = 1, (12) , the restrictions of the three 3 { } irreducible characters of G = S3 to H, as given by the first two columns, are: G ResH χ1 = χ+ G ResH χ2 = χ − G ResH χ3 = (2, 0) = χ+ + χ − Since χ appears once in the restrictions of χ2,χ3 we have − G IndH χ = χ2 + χ3 − 3.1.5. induction-restriction tables. The results of the calculations in these two examples are summarized in the following tables which are called induction-restriction tables. For G = Z/4 and H = Z/2 the induction-restriction table is: χ+ χ − χ1 10 χ2 10 χ3 01 χ4 01 For G = S and H = 1, (12) the induction-restriction table is: 3 { } χ+ χ − χ1 10 χ2 01 χ3 11 G In both cases, the rows give the decompositions of ResH χi and the G columns give the decompositions of IndH χ . ± 36 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 3.2. Induced representations. Last time we proved that the in- G duced character IndH χV is the character of some representation which is uniquely determined up to isomorphism. Today I want to construct that representation explicitly. There are two methods, abstract and concrete. The abstract version is short and immediately implies Frobe- nius reciprocity. The concrete version is complicated but you can see what the representation actually looks like. Definition 3.7. If V is a representation of H and H G then the induced representation is defined to be ≤ G IndH V = C[G] [H] V ⊗C 3.2.1. Frobenius reciprocity. One of the main theorems follows imme- diately from basic properties of tensor product: Theorem 3.8 (Frobenius reciprocity). If V is a representation of H and W is a representation of G then G G HomG(IndH V, W ) ∼= HomH (V,ResH W ) This follows from: Theorem 3.9 (adjunction formula). Hom ( M V, N) = Hom ( V,Hom (M ,N)) R R S ⊗SS R ∼ S S R S And the easy formula: HomR(R, N) ∼= N Letting M = R and S R, we get the following. ⊆ Corollary 3.10. If S is a subring of R, V is an S-module and W is an R-module then Hom (R V, W ) = Hom (V, W ) R ⊗S ∼ S Putting R = C[G],S = C[H], this gives Frobenius reciprocity. Thus it suffices to prove the adjunction formula. Proof of adjunction formula. The first step follow from the definition of the tensor product. When S is a noncommutative ring, such as S = C[H], the tensor product M S V is sometimes called a “bal- anced product.” It is an abelian group⊗ characterized by the following universal property: (1) There is a mapping f : M V M S V which is (a) bilinear in the sense× that→ it is a⊗ homomorphism in each variable (f(x, ):V M S V and f( ,v):M M V are homomorphisms− → for⊗ all x, v) and− → ⊗S.

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