On the conjugation action for quantum general linear groups

We consider the conjugation action of a quantum group over an arbitrary field. In particular we consider the coordinate algebra of a quantised general linear group G(n), at an arbitrary nonzero parameter q, and give analogues of results of Kostant and Richardson


Introduction
Domokos and Lenagan, [6], show that the algebra of polynomial invariants for the conjugation action of a quantum general linear group on quantum matrices of degree n, over a field of characteristic 0 at parameter q not a root of unity, is a commutative polynomial algebra in n variables.Further results are given in [7].
We consider this situation and also other aspects of the conjugation representation of a quantum linear group, as constructed in [5].We prove that the algebra of polynomial invariants, for arbitrary k and 0 = q ∈ k, is a commutative polynomial algebra.We consider the full algebra of invariants C(G) of the coordinate algebra of a quantum general linear group G and describe, Theorem 7.4, a (C(G), G)-module filtration of the coordinate algebra k[G].This is an analogue of the main result of [13] (which is itself an analogue of results of Richardson, [27, Theorem A] and Kostant, [24, Theorem 0.13]).We treat only the case in which G is a quantum general linear group, though it is clear that many of the arguments and results are valid also for the coordinate algebras of quantized enveloping algebras of semisimple complex Lie algebras.
One of the basic features of Steinberg's theory of conjugacy classes of a semisimple, simply connected group G (see, for example, [29]) is the description of the algebra of class functions C(G) as the subalgebra of the coordinate algebra of G spanned by the trace functions on rational G-modules.In [15], we extended this to give a description of the algebra C(G, H) of regular functions on G that are class functions relative to a subgroup H which enjoys a certain property which we call saturation.We give, as an investment for the future, the q analogue of the more general result concerning the algebra of relative invariants indicated above.The argument that we give, in order to give a description of C(G, H) in terms of trace functions, is a local one, using the structure of finite dimensional subcoalgebras of the coordinate algebra of G.With accessibility to a wider audience in mind, and for the sake of variety, we have chosen to present this by first dealing with the analogous properties for finite dimensional algebras and dualizing.Sections 2 and 3 are thus devoted to a version for finite dimensional algebras (satisfying certain conditions) of the general result on C(G, H), [15, Section 1, Theorem], with particular emphasis on quasi-hereditary algebras in Section 4. In Section 5 we transfer to the coalgebra setting.We apply the techniques developed so far to give, in Section 6, a description of the algebra of relative invariants C(G, H) in terms of shifted trace functions.In Section 7 we concentrate on the case G = H and give the results on (C(G), G)-module filtrations already mentioned.

Constrained algebras
Let k be a field.We write dim k V , or simply dim V , for the dimension of a finite dimensional vector space V over k.
Let S be a finite dimensional k-algebra which we assume to be split (or Schurian), i.e., that End S (L) = k, for every simple S-module L.
We write mod(S) for the category of finite dimensional left S-modules and write V ∈ mod(S) to indicate that V is a finite dimensional left S-module.An S-module will be assumed to be a left module and finite dimensional if no other indication is given.
We write J(S) for the Jacobson radical of S and write [S, S] for the k-span of all commutators [x, y] = xy − yx, x, y ∈ S. Definition 2.1.We say that S is (radically) constrained if J(S) ⊆ [S, S].
Note that if S is a constrained algebra and I is an ideal of S then J(S/I) = (J(S) + I)/I ⊆ ([S, S] + I)/I = [S/I, S/I] and so: (1) A factor algebra of a constrained algebra is constrained.
We write V * for the linear dual of a vector space V .For a subspaces H of V and L of V * we write H ⊥ and L ⊥ for the perpendicular subspaces of V * and V .Thus = 0 for all α ∈ H}.
For an S-module V and s ∈ S, we write trace(s, V ) for the trace of ρ(s), where ρ : S → End k (V ) is the representation afforded by V .We have the natural character χ V ∈ S * given by χ V (s) = trace(s, V ), s ∈ S.
We write C(S) for [S, S] ⊥ and C 0 (S) for the span of all natural characters χ V , V ∈ mod(S).Note that an element f of C 0 (S) vanishes on nilpotent elements (since the trace of a nilpotent endomorphism is 0) and hence f vanishes on J(S).Moreover, all elements of C 0 (S) vanish on [S, S] (since trace(xy, V ) = trace(yx, V ), for x, y ∈ S) and so C 0 (S) ⊆ C(S).The space C 0 (S) has k-basis χ 1 , . . ., χ n , where χ r = χ Vr , 1 ≤ r ≤ n, and V 1 , . . ., V n is a complete set of pairwise non-isomorphic simple left S-modules.In particular C 0 (S) has dimension l(S), the number of isomorphism types of simple left S-modules.Thus, writing Grot(S) for the Grothendieck group of mod(G), defined by short exact sequences: (3) From (2) we get that if S is constrained and V 1 , . . ., V n is a complete set of pairwise non-isomoprhic irreducible S-module then their natural characters χ 1 , . . ., χ n form a basis of C(S).
Given an S-bimodule M we write [S, M ] for the subspace of M spanned by all element sm − ms, with s ∈ S, m ∈ M .We write H i (S, M ) for the degree i Hochschild homology.Thus H 0 (S, M ) = M/[S, M ] and from Lemma 2.2 we have: (4) S is constrained if and only if H 0 (S, S) has dimension at most (equivalently equal to) l(S).
Moreover, from the Morita invariance property for Hochschild homology (see e.g., [30,Theorem 9.5.6])we have: (5) Suppose that S ′ is a finite dimensional k-algebra Morita equivalent to S. Then S ′ is split and S ′ is constrained if and only if S is constrained.
(One may also see this directly by considering R/[R, R] for an algebra R of the form End S (S ⊕ Se), where e ∈ S is an idempotent.)(6) If K is an extension field of k then the algebra S K = K ⊗ k S, obtained by base change, is constrained if and only if S is constrained.
This follows from the fact that l(S) = l(S K ) and the natural map We write Cent(R) for the centre of a ring R. Lemma 2.3.(i) If S is constrained and R is a symmetric, factor algebra of S then R is semisimple.(ii) If S is a symmetric algebra that is not semisimple then the dimension of the centre Cent(S) of S is greater than l(S).
Proof.(i) If I is an ideal of S then S/I is constrained, by (1).Thus it suffices to show that if S is symmetric (and constrained) then it is semisimple.Let ( , ) be a form on S that makes S a symmetric algebra.For a subspace V of S we now set gives Cent(S) ⊆ J(S) ⊥ .Thus J(S) ⊥ is an ideal containing the identity element of S and hence J(S) ⊥ = S and J(S) = 0. (ii) We have dim S/[S, S] > l(S), by Lemma 2.2, i.e., dim Cent(S) = dim [S, S] ⊥ > l(S).
. Remark 2.4.One of the main result of the paper [25], by H. Lenzing, is that in a finite dimensional algebra of finite global dimension, every nilpotent element is a sum of commutators, [25,Satz 5], or, expressed in our language, that such an algebra is constrained.In particular this gives that a quasihereditary algebra is constrained (we give a more direct argument for this in Remark 4.1 below).
As noted by Igusa, [22, Introduction], Lenzing's incisive result and the elementary observations above give a very short proof of the no loops conjecture for S a finite dimensional split algebra S. Suppose for a contradiction S is a split k-algebra of finite global dimension and Ext 1 S (L, L) = 0 for some non-zero simple module L. The algebra S is Morita equivalent to a basic algebra R over k.Now R also has finite global dimension and has a one dimensional module ).But we have J(A) = 0 and [A, A] = 0 so this algebra is not constrained.But then neither is R by, (1) above, and R has finite global dimension so we have a contradiction.
Remark 2.5.Suppose, for the moment, that k is an algebraically closed field of characteristic p > 0 and let g be the Lie algebra of a connected, semisimple, simply connected, algebraic group G over k.In [28], p370, the question is raised as to whether the dimension of the centre of the restricted enveloping algebra u(g) is equal to p r , where r is the rank of G, However, the algebra is symmetric, not semisimple and, by a theorem of Curtis (see e.g., [23,II,3.10 Proposition] ) the number of isomorphism classes of simple modules is p r and so, by Lemma 2.3 (ii) the centre has dimension greated than p r .A similar remark applies more generally to the algebra u n (g) of distributions of the nth Frobenius kernel of G.A more detailed, geometric, investigation into the centre, in the analogous case of the small quantum group, is to be found in [1].
Remark 2.6.We continue with the notation of Remark 2.5.Then G is split over the prime field via its structure as a Chevalley group.For n ≥ 1, let G(p n ) be the group of elements defined over the field of p n elements.The algebras kG(p n ) and u n are symmetric.In [4], the existence of interesting quasi-hereditary factor algebras of the group algebra kG(p n ) and of u n is demonstrated.By contrast the opposite problem of finding symmetric algebra factor algebras of a quasihereditary algebra has only trivial solutions.More precisely, a symmetric factor algebra of a quasihereditary algebra must be semisimple, by Lemma 2.3(i).In particular if H is a finite group such that kH occurs as a factor algebra of a quasihereditary algebra (or more generally a finite dimensional algebra of finite global dimension) then H has order prime to p.

Relative Constraint
We now consider a relative version of constraint.Let R be a subalgebra of the finite dimensional split k-algebra S. We define C(S, R) to be the subspace of the linear dual S * of S consisting of all elements f such that f (rs) = f (sr), for all r ∈ R, s ∈ S. Thus C(S, R) consists of all elements that vanish on the space [R, S], spanned by all commutators rs − sr, r ∈ R, s ∈ S, and Some of the elements of C(S, R) may be accounted for representation theoretically (analogously to the trace functions on S considered in Section 2) in the following way.Let V be a finite dimensional left S-module affording the representation ρ : S → End k (V ) and let θ be an element of End R (V ).We write χ θ or χ θ,V for the "twisted trace function", i.e., the element of S * defined by χ θ (s) = trace(θρ(s)), s ∈ S. It follows from the fact that θ commutes with the action of R that χ θ ∈ C(S, R).The set is a subspace of C(S, R).We define C 0 (S, R) ⊆ C(S, R) to be the sum of all spaces C 0 (S, R; V ), as V ranges over all finite dimensional S-modules.Note that C(S, R; V ) = C(S, R; W ) if V and W are isomorphic S-modules.Moreover, for finite dimensional S-modules V, W , we have It follows that C 0 (S, R) is the sum of the spaces C 0 (S, R; V ), as V ranges over all indecomposable S-modules (cf.[16,Section 2]).Definition 3.1.We say that R is a constrained subalgebra of S, or that the pair (S, R) is constrained, if C 0 (S, R) = C(S, R).Remark 3.2.Note that R is a constrained subalgebra if and only if dim H 0 (R, S) = dim C(S, R) (equivalently dim H 0 (R, S) ≤ dim C 0 (S, R)), by the argument of Lemma 2.2.Remark 3.3.We note that S is (radically) constrained if and only if the pair (S, S) is constrained.
If V ∈ mod(S) then χ V = χ θ , where θ : V → V is the identity map.Hence we have C 0 (S) ⊆ C 0 (S, S).If S is constrained then dim C 0 (S, S) ≥ dim C 0 (S) = dim H 0 (S, S) and so (S, S) is constrained.
Suppose conversely that (S, S) is constrained.To show that S is constrained we may, by Section 2, (6), assume that k is algebraically closed.
Let V be an indecomposable finite dimensional left S-module affording the representation π : S → End k (V ).For θ ∈ End S (V ) we have θ = λ id V + ν, for some λ ∈ k and some nilpotent endomorphism ν (where id V : V → V is the identity map).But then, for s ∈ S, we have χ θ (s) = trace(π(s)θ) = trace(λπ(s)) + trace(π(s)ν).Since π(s)ν is nilpotent, its trace is zero and so we have χ θ = λχ V and χ θ ∈ C 0 (S).Now an element of C 0 (S, S) is a sum of elements χ θ , with θ ∈ End S (V ) and V indecomposable.Hence we have C 0 (S, S) = C 0 (S).Hence we have dim C 0 (S) = dim C 0 (S, S) ≥ dim H 0 (S, S) and S is constrained.Remark 3.4.Let M ∈ mod(R) and let π : R → End k (M ) be the representation afforded by M .Then End k (M ) is an R-bimodule, with rφ = π(r)φ, φr = φπ(r), for r ∈ R, φ ∈ End k (M ).Now End k (M ) has the non-singular symmetric bilinear form defined by (φ, ψ) = trace(φψ) and it is easy to check that End R (M ) = [R, End k (M )] ⊥ .We shall need this in the proof of the following result.Proposition 3.5.Let R be a subalgebra of S. (i) For M ∈ mod(S) we have C(S, R) = C 0 (S, R; M ) if and only if the representation ρ : S → End k (M ) induces an injection Proof.(i) For a subspace V of S * we now set Suppose now that R is a constrained subalgebra.We choose a basis Suppose, for a contradiction, we have.
Remark 3.6.Let K be a field extension of k.
Proof.We take M to be any faithful finite dimensional left S-module.The injection of R-bimodules S → End k (M ) is split (if S is semisimple because every injection of S-bimodules is split and if R is semisimple because every injection of R-bimodules is split) and so the induced map Remark 3.8.One may ask whether there is a relative version of Lenzing's result (described in Remark 2.4), i.e., whether one can give suitable homological conditions which imply that a subalgebra R is a constrained subalgebra of a finite dimensional algebra S.

Quasi-hereditary algebras
Let S be finite dimensional algebra over the field k.By abuse of notation we shall use the expression "S is a quasi-hereditary algebra" to indicate that, with respect to some labelling L(λ), λ ∈ Λ, of a complete set of pairwise inequivalent irreducible left S-modules, and some partial order ≤ on Λ, the category of finite dimensional left S-modules is a highest weight category (see for example [18,Appendix]).
We now assume now S is quasi-hereditary and that L(λ), λ ∈ Λ is a labelling of the irreducible left S-modules and ≤ a partial order with respect to which mod(S) is a highest weight category.For λ ∈ Λ let ∆(λ) (resp.∇(λ), resp.M (λ)) be the corresponding standard module (resp.costandard module, resp.indecomposable tilting module).
We write S op for the opposite algebra of S. For a finite dimensional left (resp.right) S-module X we regard the dual space X * = Hom k (X, k) as a right (resp.left) module in the natural way.Then L(λ) * , λ ∈ Λ is a complete set of pairwise non-isomorphic right S-modules.Moreover, S op is a quasihereditary for the labelling L(λ) * , λ ∈ Λ (with respect to the existing order on Λ).For λ ∈ Λ, the corresponding standard module (resp.costandard module, resp.indecomposable tilting module) is ∇(λ) * (resp.∆(λ) * , resp.M (λ) * ).
By a standard (resp.costandard) filtration of X ∈ mod(S) we mean a filtration 0 either 0 or a standard (resp.costandard) module.For X ∈ mod(S), we write X ∈ F(∆) (resp.X ∈ F(∇)) to indicate that X admits a standard filtration (resp.X admits a costandard filtration).
Remark 4.1.From this one quickly gets a sequence of ideals as in the usual definition of quasi-hereditary, and that S is constrained.Consider S as an S-bimodule.Then, for i ≥ 0, and λ, µ ∈ Λ, i ≥ 0, we have by [30, Lemma 9.1.3and Lemma 9.1.9],and this is k if λ = µ and i = 0 and 0 otherwise.Hence, by [18, Proposition A 2.2(iii)], the regular S bimodule has a ∆-filtration and is one if λ = µ and 0 otherwise.Thus, if Λ consists of the distinct elements λ 1 , . . ., λ n ordered so that λ i > λ j implies i < j, then S has a bimodule filtratiion (i.e.,a sequence of ideals) Let M be a finite dimensional S-bimodule.Recall the we have the Hochschild cohomology spaces H i (S, M ), i ≥ 0, and The dual space M * is naturally an S-bimodule with actions given by sα(m) = α(ms) and αs(m) = α(sm), α ∈ M * , s ∈ S, m ∈ M .Moreover the natural map M * × M → k induces a vector space isomorphism H 0 (S, M * ) → H 0 (S, M ) * .Taking M to be the dual S * of the regular bimodule we have C(S) = H 0 (S, S * ).Now S * has a bimodule filtration with sections ∆(λ, λ) * = ∇(λ) ⊗ ∆(λ) * , λ ∈ Λ and by [30, Lemma 91.3 and Lemma 9.1.9],and this is k for i = 0 and 0 for i = 1 so we get dim and S is constrained.
Definition 4.2.We shall say that a subalgebra R iof S is saturated if, for every λ ∈ Λ, the functor Hom R (∆(λ), −) is exact on short exact sequences of finite dimensional S-modules which admit a costandard filtration.
We shall give a numerical criterion for saturation.For λ, µ ∈ Λ we set By left exactness of the Hom functor we have dim Hom R (X, Y ) ≤ a R (X, Y ) and we say that pair (X, Y ) ∈ F(∆)×F(∇) is well adapted for R (or just well adapted) if equality holds.Note (again by left exactness) that R is saturated if and only if (X, Y ) is well adapted for every and similarly (S * S : ∇(λ)) = dim ∆(λ).Putting all this together we have the following.
By a standard filtration of an S bimodule X we mean a standard filtration of X viewed as a left S ⊗ S op -module.Using Remark 4.1 is not difficult to deduce the following in a similar fashion.Lemma 4.4.R is saturated if and only if H 0 (R, −) is exact on short exact sequences of S-bimodules which admit a standard filtration and if R is saturated then H 1 (R, M ) = 0 for every S-bimodule which has a standard filtration.
Recall that a tilting module M for S is said to be full if M (λ) occurs as a component of M for all λ ∈ Λ.We shall need the following.The fact that ρ is faithful is well known.Indeed, if V is any module with a standard filtration then there exists a finite resolution where X 0 , X 1 , . . ., X n are all (partial) tilting modules (see e.g., the argument of [9, Section 1, Theorem]).In particular V embeds in a tilting module, and hence in a direct sum of copies of M .In the case V = S S we get that X 0 is faithful and since X 0 is a summand of a direct sum of copies of M , we get that M is faithful.
To treat the remaining part, we need a lemma.(ii) Follows from (i).
We now prove Proposition 4.5.As already noted, ρ is injective.Let λ ∈ Λ be a maximal element.Let π = Λ\{λ}.We write M = M 0 M 1 , where M 0 is a direct sum of modules of the form M (µ), µ ∈ π, and M 1 is a direct sum of copies of M (λ).Note that the image of ρ : , which has a standard S-bimodule filtration.Hence it suffices to prove that the cokernel of the restriction ρ ′ : S → End k (M 0 ) End k (M 1 ) has a standard filtration.
Recall, from [18,Appendix], that we have the functor O π : mod(S) → mod(S).For X ∈ mod(S), O π (X) is minimal among those submodules Y of X such that all composition factors of X/Y belong to {L(µ) | µ ∈ π}.If f : X → Y is a homomorphism of finite dimensional S-modules then O π (f ) is the restriction of f .We list the elements of Λ as λ 1 , λ 2 , . . ., λ n in such a way that i < j whenever λ i < λ j and λ n = λ.By Remark 4.1, there is a chain of ideals Note that all composition factors of S/I n come from {L(µ) | µ ∈ π} and L(λ) occurs in every quotient module of I n .Hence we must have I n = O π (S).We put Now S/I is a quasihereditary algebra, by [18, Appendix, Proposition A 3.7], and M 0 is a full tilting module for S/I.Hence we may assume, inductively, that the induced map ρ 0 : S/I → End k (M 0 ) is injective and has cokernel in F(∆).Hence by the Lemma 4.6(ii), it is enough to prove that the homomorphism σ : I → End k (M 1 ) (given by σ(s)m = sm, for s ∈ S, m ∈ M ) is injective with cokernel having a standard S ⊗ S op -module filtration.Certainly σ is injective (since M = M 0 M 1 is faithful and I acts as zero on M 0 ).We write Moreover, we have Hom k (Y i , Y j ) ∼ = M (λ) ⊗ M (λ) * , which has a standard S ⊗S op -module filtration, i = j.Thus it is enough to prove that the cokernel of the restriction By dimensions, we must have that the image of , which has a standard filtration, as an S-bimodule, so we are done.
By Proposition 3.5 we have the following.
Corollary 4.7.Let R be a saturated subalgebra.Then C(S, R) = C 0 (S, R, M ), for any full tilting module M and so R is constrained.In particular S is constrained.

Constrained coalgebras
We now consider the situation for coalgebras.Let (A, δ ǫ) be a coalgebra over k, which we assume split, i.e., End A (V ) = k for every irreducible comodule V .
Let S be an associative k-algebra.A left S-module V is called locally finite (dimensional) if every cyclic submodule (and hence every finitely generated submodule) is finite dimensional.If V is locally finite and {v i | i ∈ I} is a k-basis then the elements f ij ∈ S * , defined by the equations sv i = j∈I f ji (s)v j , for s ∈ S, i ∈ I, are called the coefficient functors of V (relative to this basis).The subspace of S * spanned by all f ij , i, j ∈ I, is independent of the choice of basis.We call it the coefficient space of V and denote it by cf(V ).
Let V be a right A-comodule V , with structure map τ : V → V ⊗ A. The elements f ij ∈ A, defined by the equations τ (v i ) = j∈J v j ⊗ f ji (for i ∈ I) are called the coefficient elements of A (relative to this basis).We write cf(V ) for the subspace of A spanned by all elements f ij , i, j ∈ I.It is independent of the choice of basis.
We now take S = A * , the dual algebra of A. We identify A with a subspace of S * via the natural map η : A → A * , η(a)(s) = s(a), for a ∈ A, s ∈ S. We write Comod(A) for the category of right A-comodules and comod(A) for the category of finite dimensional right A-comodules.
A locally finite left S-module V will be called.
In this way we obtain an equivalence of categories between the category of right A-comodules and A-admissible left S-modules.
Let R be a subalgebra of S. We define Viewing A as an S-bimodule, and hence R-bimodule, we have C(R, S) = H 0 (R, A).We write simply C(A) for C(S, A).Note that C(A) may be described directly in terms of the coalgebra structure as {f ∈ A | δ(f ) = T • δ(f )}, where T : A ⊗ A → A ⊗ A is the twist map, i.e., the linear map taking f ⊗ g to g ⊗ f , for f, g ∈ A.
Let V be a finite dimensional right A-comodule which, viewed as a left S-module, affords the representation ρ : S → End k (V ).Then, for θ ∈ End R (V ), the element χ θ of S * , defined by χ θ (s) = trace(ρ(s)θ) belongs to A. If v 1 , . . ., v n is a k-basis of V with coefficient functions f ij and if θ : V → V is given by θ(v i ) = n j=1 θ ji v j then χ θ = n i,j=1 θ ij f ji .Moreover, from the familiar property of the trace function, we have χ θ ∈ C(A, R).
We define C 0 (R, A) to the subspace , where the sum is over all V ∈ comod(A).
For V ∈ comod(A) the element χ V of S * defined by χ V (s) = trace(s, V ) belongs to A. In fact if V has basis v 1 , . . ., v n and corresponding coefficient elements f ij , 1 ≤ i, j ≤ n, then we have χ V = f 11 + • • • + f nn .By arguments similar those of Section 2 we have that C 0 (A) is spanned by all χ V , V ∈ comod(A).We write K 0 (comod(A)) for the Grothendieck group of comod(A) and, for V ∈ comod(A) denote the class of V in K 0 (comod(A)) by [V ].
We have the following result analogous to Section 2, (2).
(ii) If A is a bialgebra then the above map is a k-algebra homomorphism.
We shall say that the coalgebra A is constrained if C 0 (A) = C(A).If A is finite dimensional we say that A is a quasi-hereditary coalgebra if the dual algebra A * is quasi-hereditary.We say that A is locally quasi-hereditary if each finite dimensional subcoalgebra is contained in a quasi-hereditary subcoalgebra.
Proposition 5.2.If A is locally quasi-hereditary then it is constrained.
Proof.Suppose first that A is finite dimensional and the dual algebra A * quasi-hereditary.We identify A with the double dual (A * ) * in the usual way.Then 6 Some remarks on relative invariants for quantum linear groups In this section we consider the algebra of invariants of the coordinate algebra of a quantum group G under the action by conjugation of a (quantum) subgroup H.We are especially interested in the case in which G is a product of general linear groups.In the final section we consider the case H = G and give more explicit results.
Let k be a field.
Let G be a quantum group over k.By a left G-module we mean a right k[G]-comodule.We write mod(G) for the category of finite dimensional left G-modules (i.e. the category of finite dimensional right k[G]-comodules).
Suppose now that A = (A, δ, ǫ) is a coalgebra.We extend the definition of C(A) to bicomodules.By an A-bicomodule we mean a triple (U, λ, ρ), where U is a k-space and λ : U → A ⊗ U and ρ : U → U ⊗ A are linear maps such that (U, λ) is a left A-comodule, (U, ρ) is a right A-comodule and Let Γ = A * be the dual algebra.Then an A-bicomodule U is a Γ-bimodule with actions γ • u = (1 ⊗ γ)ρ(u) and u • γ = (γ ⊗ 1)λ(u), for γ ∈ Γ, u ∈ U .
We say that an element u of an A-bicomodule U is stable if T λ(u) = ρ(u).Here T : A ⊗ U → U ⊗ A is the twist map, i.e. the k-linear map such that T (x ⊗ y) = y ⊗ x, for x ∈ A, y ∈ U .We write C(U ) for the space of all stable elements of U .
It is easy to check that C(U ) is the zeroth Hochschild cohomology space: Note that A = (A, δ, δ) is an A-bicomodule and that the more general definition of C(U ) agrees with the earlier definition of C(A), in case U = A.
By a G-bimodule we mean a k[G]-bicomodule.We now take A = k[G] and let S : k[G] → k[G] be the antipode.Let (U, λ, ρ) be an G-bimodule.We write C G (U ) for C(U ) is we wish to emphasise the role of G.We write U ad for the k-space U regarded as a left G-module via the adjoint action, i.e., via the structure map κ : U → U ⊗ A given by then we write V G for the space of fixed points, i.e., For V, W be finite dimensional G-modules we regard Hom k (V, W ) as a G-bimodule and identify V * ⊗ W with Hom k (V, W ) in the natural way.Lemma 6.1.Let U be a G-bimodule and let V, W be finite dimensional G-modules.
By a good filtration of a finite dimensional G(a)-module V we mean a filtration 0 is independent of the choice of good filtration, and will be denoted it is a linear isomorphism.It follows that, for a finite string of positive integers, the antipode of G(a), and hence of any quantum subgroup, is an isomorphism.In particular it is injective.Definition 6.3.A (quantum) subgroup H of G(a) will be called saturated if Hom H (∆(λ), −) is exact on short exact sequences of G(a)-modules with a good filtration.Remark 6.4.Let H be a quantum subgroup of G(a).Then H is saturated if and only (−) H is exact on short exact sequences of G(a)-modules with a good filtration.This may be seen as follows.For a G(a)-module V we have Hom H (k, V ) ∼ = V H so that if H is saturated then (−) H is exact on short exact sequences of G(a)-modules with a good filtration.Now suppose that (−) H is exact on short exact sequences of G(a)-modules with a good filtration.Let λ ∈ X + (a).If M ∈ mod(G(a)) has a good filtration then ∇(λ * ) ⊗ M has a good filtration, by [17,Section 4,(3) -modules with a good filtration then so is Hence we get that is exact and hence, by Lemma 6.1(ii), so is ).If λ ∈ π and V is an A-admissible Gmodule then we have Hom R (∆(λ), V ) = Hom H (∆(λ), V ).It follows that R is a saturated subalgebra of S, as in Section 4. Hence we have C 0 (S, R) = C(S, R).Identifying A with the dual of S * and using the fact that the action of Γ H on A factors though R we get Proposition 6.8.Let a be a finite sequence of positive integers.Suppose that H is a saturated subgroup of G(a).Let π be a saturated subset of X + (a).Then the algebra of relative class functions A(π) H is spanned by the shifted trace functions χ θ,M (λ) , with λ ∈ π, θ ∈ End H (V ).
Proof.By a local finiteness argument we are reduced to the case in which π is finite, considered above.Remark 6.9.This applies in particular to the case in which G(a) = G(n) m and H = G(n),the diagonal subgroup.Taking π = Λ + (n) m , we get that is the q-analogue of the algebra of polynomial invariants for action of the general linear group on m-tuples of matrices, considered in [15], where an explicit set of generators is given.However, there is more work to be done to give an analogous set of generators in the quantum case.Another solution to this problem of describing generators (in the classical case) was given by Domokos and Zubkov, [8].Remark 6.10.One may replace the modules M (λ) in the proposition by any collection of modules which contain all modules M (λ), λ ∈ π, as summands.For r ≥ 0 we write Λ + (n, r) for the subset of Λ + (n) consisting of the partitions of r with at most n parts.For a partition λ we write λ ′ for the transpose partition.We have the natural G(n)-module E, as in [18], and, for r ≥ 0, the exterior power r E. Thus for a finite string of nonnegative integers α = (α 1 , . . ., α h ) the module For λ ∈ Λ + (n, r) the module λ ′ E is a tilting module for G(n) in which M (λ) occurs as component.More generally for a string of non-zero integers a = (a 1 , . . ., a m ) and λ = (λ(1), It is modules of this form that are used in the application of Proposition 6.8 in the classical case, [15].7 The coordinate algebra of a general linear group as a module over its algebra of class functions We here concentrate on the "absolute" case G = H = G(n).Proof.(i) Let ψ λ be the character of L(λ), for λ ∈ Λ + (n).For λ ∈ Λ + (n) the module λ ′ E has unique highest weight λ.Hence, for r ≥ 1, the characters of the modules λ ′ E, λ ∈ Λ + (n, r), and the characters of the modules L(λ), λ ∈ Λ + (n, r) are related by a unitriangular matrix.Hence, from Proposition 6.7, the characters of the λ ′ E, λ ∈ Λ + (n), form a basis for C(A(n)).The result follows.
For the case in which k has characteristic 0 and q is not a of unity, see [6].
We now consider the coordinate algebra k[G(n)] as a module over the algebra of invariants C(k[G(n)]) and the algebra A(n) as a module over C(A(n)).
We need to make a certain observation on flat modules.This is presumably well known but we include it since we do not know a precise reference (but cf [26, 4.E)]).Let A be a subring of a commutative ring B. Recall that if B is flat over A then Tor B i (B ⊗ A M, N ) is isomorphic to Tor A i (M, N ), for all i ≥ 0, all A-modules M and all B-modules N , see [30,Proposition 3.2.9].Lemma 7.2.Let B be a commutative ring and A a subring.Suppose A and B are regular Noetherian and that B is flat and finite over A. A B-module is flat if and only if it is flat as an A-module.
Proof.A flat B-module is also flat as an A-module, see [26, (3.B)].Now suppose M is a B-module which is flat as an A-module.To show that M is flat over B, it suffices, by [2, Chapter II, Section 3.4 Corollary], to show that, for each maximal ideal Q of B, the localization M Q is a flat as a module over the local ring B Q .Let P = Q ∩ A. The natural map M P → M Q is an isomorphism (since B is integral over A) so that M Q is flat as an A P -module.Thus we may replace A by A P and B by B Q and M by M Q .Thus we may assume that the inclusion of A in B is a homomorphism of regular local rings.
Suppose, for a contradiction that M is not flat.Then we have Tor B 1 (M, N ) = 0 for some finitely generated B-module N (see [ We set G = G(n), T = T (n), A = A(n) and J = C(A(n)).We shall consider A as a module for G (via the adjoint action) and J.For a Jmodule V and a k-space M we write |M | ⊗ V for the vector space M ⊗ V , viewed as a J-module via the action c(m ⊗ v) = m ⊗ cv (for c ∈ J, m ∈ M , v ∈ V ).By a (J, G)-module we mean a k-vector space V which is a Jmodule and a G-module in such a way that the action of J is by G-module endomorphisms of V .Morphisms of (J, G)-modules, (J, G)-submodules etc are defined in the usual way.Given a (J, G)-module V and a G-module M we regard the J-module |M | ⊗ V as a (J, G)-module, with G acting diagonally.We sometimes regard J as a (J, G)-module, with J acting via the regular action and G acting trivially.With Lemma 7.3 in place the argument of [13, 2.2 Theorem] now goes through and we obtain the following.(The second assertion follows from the first by localizing at the determinant.)Theorem 7.4.Suppose that q is a root of unity.Let λ(1), λ(2), . . .be a labelling of the elements of X + (n) such that i < j whenever λ(i) < λ(j).Then A has a filtration 0 = Y (0) ⊆ Y (1) ⊆ . .., as a (J, G)-module, with Corollary 7.5.Suppose that q is a root of unity.Then A is a free J-module.
We will give a strong version of Theorem 7.4 when q is not a root of unity.Our argument involves the A-regularity of the sequence φ 1 , . . ., φ n .
We have the coalgebra decomposition and algebra grading A = ∞ r=0 A r , where A r = A(n, r) is the span of the elements c i 1 j 1 . . .c irjr , with 1 ≤ i 1 , j 1 , . . ., i r , j r ≤ n.Note that J is a homogenous subspace of A. We set J + = r>0 J r .
We consider the following set-up.Let R be a discrete valuation ring with field of fractions K and maximal ideal m generated by an element π, and let k = R/m be the residue field.Let q be an element of R\m and q be the image of q in k.We write A K for A K,q (n), write A for the R-subalgebra generated by the coefficient elements c ij , 1 ≤ i, j ≤ n.We write φ 1 , . . ., φ n for the characters of the exterior powers of the natural G(n)-module over K, as in Proposition 7.1.We write Ā for the algebra A k,q (n).We have the natural map A → Ā and, for a ∈ A, we write a for its image in Ā.Now A inherits a grading A = r≥0 A r , from A K = A K,q (n).Also Ā is graded and the natural map A → Ā takes A r to Ār , r ≥ 0. Lemma 7.6.If φ1 , . . ., φn is an Ā-regular sequence then φ 1 , . . ., φ n is an A-regular sequence.
Proof.Suppose φ 1 a = 0, for some a ∈ A. Then φ1 a = 0 and, since φ is not a zero divisor, a = 0, i.e., a ∈ πA.We write a = πb, for some b ∈ A. Then πφ 1 b = 0 so that φ 1 b = 0. Hence b ∈ πA and a ∈ π 2 A. Inductively, we get a ∈ π s A, for all s ≥ 1.But A is free over R so that s≥0 π s A = 0 and a = 0. Thus φ 1 is not a zero divisor.Now suppose that 1 ≤ m < n and a ∈ A. We set moreover, for each r, the R-module A r /Y r is finitely generated and hence (e.g. by Nakayama's Lemma) s≥0 π s (A r /Y r ) = 0. Hence we have s≥0 π s (A/Y ) = 0, i.e., s≥0 Let x ∈ X.Thus we have φ m+1 x = φ 1 x 1 + • • • + φ m x m , for some x 1 , . . ., x n ∈ A and hence φm+1 x = φ1 x 1 + • • • φm x m .But φ1 , . . ., φn is a regular sequence for Ā so there exist elements y 1 , . . ., y m of A such that x = φ1 y 1 + • • • + φm y m and we deduce that for any x ∈ X there exist y 1 , . . ., y m , z ∈ A such that In particular X ⊆ Y + πA.Now assume that s is a positive integer such that X ⊆ Y + π s A.
We fix a ∈ X and write for elements b 1 , . . ., b m , c of A.
Multiplying (3) by φ m+1 and using that φ m+1 a ∈ Y we get φ m+1 π s c ∈ Y .We claim that φ m+1 π s c ∈ π m i=1 φ i A. If not we choose 1 ≤ j ≤ m as small as possible such that We write φ m+1 π s c = 1≤i≤j for elements u i , 1 ≤ i ≤ j and v i , j + 1 ≤ i ≤ m of A. Taking (5) modulo π we get 1≤i≤j φi u i = 0. Since φ1 , . . ., φn is a regular sequence for Ā, we must have for some elements e i (with 1 ≤ i < j), and f of A. .By induction we have that a ∈ Y + π s A, for all s, i.e., a ∈ ∞ s=0 (Y + π s A), which, by (1), is Y .Thus a ∈ Y , i.e., we have a = φ 1 w 1 + • • • + φ m w m , for some w 1 , . . ., w m ∈ A. Thus φ 1 , . . ., φ n is an A-regular sequence.
Proof.We give the proof in several steps.
In the classical case q = 1 the elements φ 1 , . . ., φ n form a regular sequence (and if k is algebraically closed, the ideal generated by these elements is the defining ideal of the nullcone in the variety of n×n matrices).
Next assume q = 1 and q is a root of unity.By Corollary 7.5, A is free over J. Let b i , i ∈ I, be a J-basis of A. If a ∈ A and φ 1 a = 0 then writing a = i∈I f i b i , with f i ∈ J, we have i∈I φ 1 f i b i = 0. Hence for each i we have φ 1 f i = 0 and J is a domain so that f i = 0. Hence a = 0 and φ 1 is not a zero divisor.Now suppose that 1 ≤ m < n.Suppose a ∈ A and φ m+1 a = φ 1 a 1 + . . .+ φ m a m with a 1 , . . ., a m ∈ A. We write a = ∈I g i b i and a r = i∈I g ir b i , for 1 ≤ r ≤ m, with all g i , g ir ∈ J. Thus, for each i ∈ I, we have But J is the (commutative) polynomial algebra in φ 1 , . . ., φ n .In particular φ 1 , . . ., φ n is a J-regular sequence.Hence each Step 3.
By a similar argument to that of Step 2 one sees, for general 0 = q ∈ k and an extension field K of k the elements φ 1 , . . ., φ n ∈ A k,q (n) form a regular sequence if and only if the elements 1 ⊗ φ 1 , . . ., 1 ⊗ φ n of A K,q (n) = K ⊗ A k,q (n) form a regular sequence.
Now assume that q is algebraic over the prime field k 0 but not a root of unity.By Step 3, we may replace k by k 0 (q).If k has positive characteristic this implies that q is a root of unity.Hence k = Q(q) and that q satisfies an equation r 0 q d + r 1 q d−1 + • • • + r d = 0 for some positive integer d and integers r 0 , . . ., r d with r 0 = 0.
We choose a rational prime p not dividing r 0 .Let R 0 be the subring of k consisting of the elements u/v, with u, v integers and p not dividing v. Let R 1 = R 0 (q) and choose a maximal ideal m of R 1 not containing q.Let R 2 be the completion of R 1 at m and let K be the field of fractions.It is enough to show that the characters φ 1 , . . ., φ n form a regular sequence for A K = K ⊗ k A k,q (n).Thus we are reduced to the situation in which k is a p-adic number field and q ∈ R\m, where R is the ring of p-adic integers in k and m is its maximal ideal.
Let F = R/m be the residue field and let q = q + m ∈ F .Let A R be the R-subalgebra of A K generated by all the coefficient elements c ij .We have the natural map A R → ĀR = A R /mA R = F ⊗ R A. For a ∈ A R we write a for the image in Ā.Now q is a root of unity, so by Step 2, φ1 , . . ., φn is an Ā-regular sequence.Thus by Lemma 7.6, φ 1 , . . ., φ n is a regular sequence in A R and hence in A = A K .
Step 5.There remains the case in which q is not algebraic over the prime field.Thus, by Step 3, we need only consider the case k = k 0 (t), the field of and so dim S/[S, S] = l(S).Thus dim S/[S, S] = dim C 0 (S) and [S, S] ⊆ C 0 (S) so that [S, S] = C 0 (S) ⊥ .(iv)⇒(i) We have J(S) ⊆ C 0 (S) ⊥ = [S, S].

Proposition 4 . 5 .
Let M be a full tilting module.Then the representation ρ : S → End k (M ) is injective and Coker(ρ) has a standard S-bimodule filtration.
has a standard S-bimodule filtration.Furthermore, by the above Lemma and induction on m, it suffices to prove that the cokernel of the map τ : I → End k (N ) has a standard S-bimodule filtration, where N = M (λ) and τ (x)n = xn, for x ∈ I, n ∈ N .Now N contains a submodule N 1 , say, isomorphic to ∆(λ), and N/N 1 belongs to π. Regarding End k (N ) = N ⊗ N * as a left S-module, we get O π (End k (N )) = N 1 ⊗ N * and hence τ takes I = O π (I) into N 1 ⊗ N * .Moreover, we have (N ⊗N * )/N 1 ⊗N * ∼ = (N/N 1 )⊗N * , which has a standard module filtration, as an S-bimodule, and hence it suffices to prove that the cokernel of the restriction τ ′ : I → N 1 ⊗ N * , i.e., τ ′ : I → Hom k (N, N 1 ), has a standard filtration.But now, we have a submodule N 2 of N such that N 2 ∈ F(∇) and N/N 2 ∼ = ∇(λ).Regarding Hom k (N 1 , N ) ∼ = N ⊗ N * 1 as a right S-module, we have that I maps into O

Proposition 7 . 1 .
(i) C(A(n)) is the free polynomial algebra on φ 1 , . . ., φ n , where φ r is the natural character of r E,1 ≤ r ≤ n. (ii) C(k[G(n)]) is the localization of C(A(n)) at the determinant d = φ n .
Thus φ(x) = 0 if and only if trace(ρ(s)θ) = 0 for all θ ∈ End R (M ), i.e., if and only if s ∈ C(S, R; M ) ′ .Hence φ is injective if and only if an element s ∈ S belongs to [R, S] if and only if it belongs to C 0 and (X, Y ) is well adapted then both pairs (X 1 , Y ) and (X 2 , Y ) are well adapted.Similarly if X ∈ F(∆) and 0 → Y 1 → Y → Y 2 → 0 is a short exact sequence in F(∇) and (X, Y ) is well adapted then both pairs (X, Y 1 ) and (X, Y 2 ) are well adapted.Now if X ∈ F(∆) and φ : P → X is a surjective homomorphism from a projective module then both P and the Ker(φ) belong to F(∆).Hence R is saturated provided that (P, Y ) is well adapted for every projective P ∈ mod(S).Similarly one deduces that indeed R is saturated provided that (P, I) is well adapted for all projective P ∈ mod(S) and injective I ∈ mod(S).It follows that R is saturated if and only if ( S S, S * S ) is well adapted.(Here S S denotes the left regular module, S S the right regular module, and S * S its dual).In other words, R is well adapted if and only if dim H 0 (R, S S ⊗ S S ) = dim Hom R ( S S, S * S ) is equal to a R ( S S, S * S ).Now we have By the statement "G is a quantum group" (over k) we indicate that we have in mind a Hopf algebra over k, denoted k[G] and called the coordinate algebra of G.By the statement "φ : G → H is a morphism of quantum groups" we indicate that G and H are quantum groups and we have in mind a morphism of Hopf algebras k[H] → k[G], called the comorphism of φ and denoted φ ♯ .We say that H is a (quantum) subgroup of the quantum group G to indicate that we have in mind a Hopf ideal I H , called the defining ideal of H and that H is the quantum group with coordinate algebra k[G]/I H .The inclusion map i : 26, Chapter 2, Section 3, Theorem 1]).Let i be maximal such that there exists a finitely a subalgebra of k[T (n)] via this embedding.Now k[t 1 , ..., t n ] is finite over the k[ t1 , ..., tn ] and hence over the algebra of invariants (under the symmetric group of degree n) Λ.Since Λ ≤ Λ ≤ k[t 1 , ..., t n ], we have that Λ is finite over Λ.Moreover, as is well known k[t 1 , ..., t n ] is free and hence flat over Λ.Moreover, k[t 1 , ..., t n ] is free over k[ t1 , ..., tn ] and k[ t1 , ..., tn ] is free over Λ. Hence k[t 1 , ..., t n ] is free, and hence flat over Λ.By integrality, we now have (see[26, (4.A) and (5.E)]) k[t 1 , . . ., t n ] is faithfully flat over Λ and k[t 1 , . . ., t n ] is faithfully flat over Λ.However, it follows from the definitions that if we have rings A ≤ B ≤ C with C faithfully flat over A and B faithfully flat over A then C if faithfully flat over B. Thus Λ is flat over Λ.The restriction map C