On the Grothendieck Ring of a Hopf Algebra

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1 On the Grothendieck Ring of a Hopf Algebra H is a fin. dim l Hopf algebra over an alg. closed field k of char p 0, with augmentation ε, antipode S,.... (1) Grothendieck ring and character algebra The Grothendieck ring G 0 (H) is the abelian group generated by the iso. classes [V ] of finite dim l (right) H-modules V modulo the relations [V ] = [U] + [W ] for each s.e.s. 0 U V W 0. The iso. classes of simple modules form a Z-basis. Multiplication is given by = k. The character algebra R(H) is the k-subalgebra of H that is generated by the image of the character map χ : G 0 (H) H, [V ] χ V, where χ V (h) = trace V/k (v hv). Some basic relations: R(H) = G 0 (H) Z k and R(H) = ([H, H] + rad H) C(H) := [H, H] H. Here, C(H) = Cocom H is the algebra of trace forms on H. 1

2 Some examples: H = kg: (G a finite group) R(H) = k T (G) reg C(H) = k T (G) H = k G. Here T (G) is the set of conjugacy classes of G and T (G) reg the set of p-regular conj. classes. Furthermore, via Brauer characters, G 0 (H) G 0 (H) Z C = C T (G) reg. H = (kg) : G 0 (H) = ZG and R(H) = C(H) = H = kg. H = u(sl 2 ): (p > 2) where f p (X) = f p (X) = Let v G 0 (H) denote the class of the canonical 2-dim l H-module. Then G 0 (H) = Z[v] = Z[X]/ ( (X 2)f p (X) 2), (p 1)/2 k=0 (p 1)/2 k=0 ( 1) a p,k ( k + ap,k a p,k ) X k, a p,k = (p 1 2k)/4. ( 1) a p,k ( k+a p,k a p,k ) X k, a p,k = (p 1 2k)/4. is irreducible and satisfies f p (X) = (X 2) p 1 2 mod p. Thus: R(H) = k[x]/ ((X 2) p ) = kc p. 2

3 (2) On the class equation Assume H semisimple Some generalities: G 0 (H) is a semiprime ring with involution given by [V ] = [V ]. The form.,. : G 0 (H) G 0 (H) Z that is defined by [V ], [W ] = dim k Hom H (V, W ) is bilinear, associative, symmetric, and -invariant. A set of dual Z-bases of G 0 (H) is given by {[V 1 ],..., [V t ]} and {[V 1 ],..., [V t ] }, where V 1,..., V t is a complete set of simple H-modules. We will always take V 1 = k ε. By k-linear extension to R(H) = G 0 (H) Z k, one obtains a form β : R(H) R(H) k with analogous properties. Thus, R(H) is a symmetric algebra, with dual bases {χ 1,..., χ t } and {S (χ 1 ),..., S (χ t )}, where χ i = χ Vi are the irreducible characters of H. Note: χ 1 = ε = 1 H. If p = char k = 0 then R(H) is semisimple. The Class Equation (Kats-Zhu): Assume p = 0. Then, for every primitive idempotent e = e 2 R(H), one has dim k eh dim k H. Remark: For H = kg, this reduces to the fact that the size of each conjugacy class of G divides G, and for H = (kg), this says that the dimension of each simple kg-module divides G. 3

4 The proof: Wedderburn: Let ê Z(R(H)) be the centrally primitive idempotent with e = eê, let µ denote the character of the irreducible R(H)-module er(h), and let ω : Z(R(H)) k denote its central character. Then êr(h) = (er(h)) µ(1) and so êh = (eh ) µ(1). Hence, dim k êh = µ(1) dim k eh and the assertion becomes d := µ(1) dim H dim êh Z. Centrally primitive idempotents: Those can be expressed in terms of dual bases (cf. Curtis-Reiner I, p. 204). In our situation, ê = ω(ẽ) 1 ẽ with ẽ = i µ(s (χ i ))χ i. Writing L f (g) = fg (f, g H ), the element x = (f Tr H (L f )) H is an integral of H (Larson-Sweedler). Clearly, dim êh = Tr H (Lê) = ê(x). Furthermore, χ 1 (x) = x(1) = dim H and χ i (x) = 0 for i > 1, since V i x Vi H = {0}. Thus, ẽ(x) = µ(1) dim H and Consequently, d = ω(ẽ). dim êh = ê(x) = ω(ẽ) 1 µ(1) dim H. Integrality: If φ is any character of R(H) then φ(g 0 (H)) A := alg.int.(k). Indeed, any α G 0 (H) satisfies a monic polynomial over Z, and hence so does the image Φ(α) under the representation Φ belonging to φ, and all eigenvalues of Φ(α) as well. Thus, φ(α) = Tr(Φ(α)) A. In particular, ẽ = i µ(s (χ i ))χ i belongs to G 0 (H) Z A, and hence ẽ is integral over Z. Therefore, so is d = ω(ẽ) which implies d Z. 4

5 (3) Application: Projectives over smash products (with Loretta Tokoly) A sample result: THEOREM Let B = A#H be a smash product such that (1) A is commutative without idempotents 0, 1. (2) H is involutory, not semisimple ( p > 0), (3) 1 / [B, B]. Then p rank(p A ) holds for every f.g. projective B-module P. Recall that, for any f.g. projective A-module Q, rank(q) = dim k(m) Q A k(m), where m is any prime ideal of A and k(m) = Fract(A/m). By (1), rank(q) is independent of the choice of m. Remark: In the special case B = H (= k#h), the Theorem reduces to the fact that p dim k P for all projectives P over H. This needs (2). 5

6 Tools of proof: Frobenius reciprocity: Let A 1 and A 2 be H-module algebras and put B i = A i #H. If V i are B i -modules then the V 1 V 2 is a module over B 1 B 2 = (A1 A 2 )#(H H), and hence over B 1 by restriction along B 1 (A 1 A 2 )#(H H) a 1 #h (a 1 1 A2 )# h. Similarly for B 2. The following extends the classical Frobenius reciprocity isomorphism for group rings. PROPOSITION Let K be a Hopf subalgebra of H and put C i = A i #K B i (i = 1, 2). If V is a B 1 -module and W is a C 2 -module then, as B i -modules, (V C1 W ) Ci B i = V (W C2 B 2 ). Specializing to B 1 = B = A#H and B 2 = H we get: If V is a B- module and W is an H-module then V W is a B-module, and Frobenius reciprocity (with K = k) implies that if V is projective then V W is likewise, and V A B = V H. 6

7 Grothendieck groups: All this passes down to the level of Grothendieck groups: G 0 (B) and K 0 (B) are modules over the Grothendieck ring G 0 (H) via and the Cartan map K 0 (B) G 0 (B) is a module map. Further, Ind B A Res B A : G 0 (B) G 0 (A) G 0 (B) is multiplication by [H] G 0 (H) on G 0 (B), and similarly for K 0. In particular, [B] = [A B ] [H] holds in G 0 (B). The Hattori-Stallings trace map: If P is a f.g. projective module over the ring R then P = er n for some n and some idempotent matrix e = (e i,j ) M n (R). Putting ρ(p ) = e i,i + [R, R] R/[R, R], one obtains the Hattori-Stallings trace ρ = ρ R Some basic facts: : K 0 (R) R/[R, R]. ρ is functorial; If R is commutative without idempotents 0, 1, then χ factors as ρ = can. rank : K 0 (R) Z R. 7

8 The proof: Recall that the character map χ : G 0 (H) R(H) has kernel pg 0 (H). Condition (1) on H implies that χ H = 0 (c.f. Larson-Radford). Consequently, [H] = px for some x G 0 (H). Now let P be a f.g. projective B-module. Our goal is to show that Consider the commutative diagram (. ) A B K 0 (B) K 0 (A) p rank(p A ). rank Z ρ B B/[B, B] Since 1 / [B, B], by (3), the map Z B/[B, B] has kernel pz. Thus our assertion becomes ρ B (P A B) = 0. But as desired. ρ A can. A can. ρ B (P A B) = ρ B ([P ][H]) = pρ B ([P ]x) = 0, 8

9 COROLLARY (of proof) Assume (2) and (3) in the Theorem. In addition, assume that H is local and that K 0 (A) = Z[A]. Then p ρ(p ) holds for any ρ : G 0 (B) Z and any f.g. projective B- module P. In particular, B is not Morita equivalent to a domain. 9

Exercises on chapter 4

Exercises on chapter 4 Always R-algebra means associative, unital R-algebra. (There are other sorts of R-algebra but we won t meet them in this course.) 1. Let A and B be algebras over a field F. (i) Explain