The Bgg Category for Generalised Reductive Lie Algebras 3

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say g is a generalised reductive Lie algebra. It is easy to see that if J is the center of g, then g is a − reductive Lie algebra. Let g0 = n0 ⊕ h0 ⊕ n0 be the Cartan decomposition of g0. Let b0 = h0 ⊕ n0, b = b0 L J, n = n0 L J. It is easy to see that b is a maximal solvable subalgebra of g. + − Denote Φ to be the root system of g0, and let Φ be the positive root system, Φ is the negative

root system. We choose a simple root system ∆ = {α1, ..., αl} ⊂ Φ. Weyl group W is generated

by all the reﬂections according to Φ. The elements in W are denoted as sα(α ∈ Φ). Denote 1 ρ = 2 P α. α∈Φ+ According to the Cartan decomposition of g0 = h0 ⊕ L g0α, we can choose a set of generators α∈Φ

of g0 according to simple root system, denote as ei ∈ g0αi , fi ∈ g0−αi , [ei,fi]= hi ∈ h0(i = 1, ..., l).

Denote U(g) as the universal algebra of g. And U(g)0 is the universal algebra of g0.

Recall that a BGG category O is a subcategory of ModU(g0) whose objects satisfy following conditions (cf. [3]):

(O1) M is a ﬁnitely generated U(g0)-module.

(O2) M is h0-semisimple, that is, M is a weight module: M = Mµ, Mµ = {v ∈ M | h · v = L∗ µ∈h0 µ(h)v, ∀h ∈ h0}, Mµ is a weight space.

(O3) M is locally n0-finite: for each v ∈ M, the subspace U(n) · v of M is finite dimensional. (O4) All weight spaces of M are finite dimensional.

Definition 2.1. Category O′ is a subcategory of ModU(g) whose objects satisfy following conditions: (O′1) M is a finitely generated U(g)-module. ′ (O 2) M is h0-semisimple, that is, M is a weight module: M = Mµ, Mµ = {v ∈ M | h · v = L∗ µ∈h0 µ(h)v, ∀h ∈ h0}, Mµ is a weight space. (O′3) M is locally n-finite: for each v ∈ M, the subspace U(n) · v of M is finite dimensional. (O′4) All weight spaces of M are finite dimensional.

′ Proposition 2.2. Suppose M ∈ O , then M is a ﬁnitely generated U(g0)-module.

Proof. Suppose U(g)-module M has a set of generators v1, ..., vn. Denote V as a U(n)-module

generated by v1, ..., vn. Then V is ﬁnite dimensional. We can choose w1, .., wm as a basis of V . It

is obviously that M is generated by w1, .., wm as U(g0)-module. THE BGG CATEGORY FOR GENERALISED REDUCTIVE LIE ALGEBRAS 3

Corollary 2.3. Suppose M is an object of O′ as U(g)-module, then M is an object of BGG category

O as U(g0)-module.

Proposition 2.4. Suppose M is an object of O′, then M is an artinian module.

Proof. Suppose M is not an artinian module , then there is an inﬁnite U(g)-module chain

... ⊆ Mn ⊆ Mn−1 ⊆ ... ⊆ M0 = M.

It is also an inﬁnite U(g0)-module chain, but M is an artinian module in the BGG category O, contradictory.

Proposition 2.5. O′ is a noetherian category.

Proof. Similar to the proof above.

3. Verma modules in category O′

We can view J as a g0-module with adjoint action. Since g0 is a semisimple Lie algebra, then J can be decomposed into a direct sum of a ﬁnite number of irreducible submodules

J = L(λ1) ⊕ L(λ2) ⊕ ... ⊕ L(λn).

we denote J1 to be the direct sum of irreducible submodules of J with λi = 0 (i ∈ {1, 2..., n}), and

denote J2 to be the direct sum of irreducible submodules of J with λj 6= 0 (j ∈ {1, 2..., n}), then

J = J1 ⊕ J2. ′ To deﬁne Verma module in category O , we ﬁrst consider one-dimensional U(b)-module Cw(λ,g), ∗ ∗ λ ∈ h0, g ∈ J . h · w(λ,g) = λ(h)w(λ,g) for any h ∈ h0. u · w(λ,g) = g(u)w(λ,g) for any u ∈ J. The

action of n0 on w(λ,g) is follows by the proposition below.

Proposition 3.1. For any x ∈ n0, x · w(λ,g) = 0.

Proof. Without loss of generality, we may assume x ∈ n0 and [h, x] = α(h)x for any h ∈ h0, + α ∈ Φ . If x · w(λ,g) = cw(λ,g), c ∈ C, then h · x · w(λ,g) = cλ(h)w(λ,g) = [h, x] · w(λ,g) + x · h · w(λ,g) =

(cα(h)+ cλ(h))w(λ,g). So cα(h) = 0 for any h ∈ h0. There is a h0 ∈ h0 such that α(h0) 6= 0. Thus c must be zero. Finally we conclude x · w(λ,g) = 0 for any x ∈ n0. 4 YE REN

∗ To make one-dimensional space Cw(λ,g) to be a U(b)-module, there is no restriction to λ ∈ h0, but there are some restricted conditions about g ∈ J ∗.

Proposition 3.2. Suppose x ∈ n0, h ∈ h0, u1, u2, u ∈ J, then (1)g([h, u]) = 0. (2)g([x, u]) = 0.

(3)g([u1, u2]) = 0.

Proof. (1)h · u · w(λ,g) = λ(h)g(u)w(λ,g), [h, u] · w(λ,g) + u · h · w(λ,g) = g([h, u])w(λ,g) + λ(h)g(u)w(λ,g). Then we deduce that g([h, u]) = 0.

(2)x · u · w(λ,g) = 0. Since [x, u] ∈ J, [x, u] · w(λ,g) + u · x · w(λ,g) = g([x, u])w(λ,g). We deduce that g([e, u]) = 0.

(3)u1 · u2 · w(λ,g) = g(u1)g(u2)w(λ,g), [u1, u2] · w(λ,g) + u1 · u2 · w(λ,g) = g([u1, u2])w(λ,g) + g(u1)g(u2)w(λ,g). Then we deduce that g([u1, u2]) = 0.

Proposition 3.3. Suppose u ∈ J2, then g(u) = 0.

Proof. Without loss of generality, we may assume J2 = L(λ), λ 6= 0. We divided L(λ) as a direct sum of wight spaces, L(λ) = L(λ)µ. If u ∈ L(λ)µ, and µ 6= 0, then there is a h ∈ h Lµ makes µ(h) 6= 0 and g([h, u]) = µ(h)g(u) = 0, it follows that g(u)=0. If µ = 0, then u ∈ L(λ)0. − Suppose u0 is a lowest weight vector of L(λ), that is, [n0 , u0] = 0. Then u can be written as r1 r2 rl n + r1 r2 rl [ P kr1,...,rle1 e2 ...el , u0], e1, ..., el ∈ 0, r1, ..., rl ∈ Z , kr1,...,rl ∈ C. Actually, [e1 e2 ...el , u0] r1,...,rl ′ ′ ′ can be written as [e, u ] for some e ∈ n0 and for some u ∈ L(λ)µ, µ 6= 0. Since g([e, u ]) = 0, we get that g(u) = 0. It follows that for any u ∈ J2, g(u) = 0.

Proposition 3.4. Suppose u ∈ J1, x ∈ g0, then [x, u] = 0.

Proof. Without loss of generality, we may assume J1 = L(0), dimL(0)=1. It is a 1 dimensional − g0-module with adjoint action. If x ∈ h0, [x, u] = x · u = 0. If x ∈ n0 ⊕ n0 , h · x = α(h)x for any h ∈ h0, α ∈ Φ. Since dim L(0)=1, x · u = cu, c ∈ C. Then h · x · u =0=[h, x] · u + x · h · u = cα(h).

Thus for any h ∈ h0, cα(h) = 0. We can ﬁnd a h0 ∈ h0 such that α(h0) 6= 0. It follows that c = 0.

Thus [J1, g0] = 0.

∗ We deﬁne a set G = {g ∈ J | g(J2) = 0}. THE BGG CATEGORY FOR GENERALISED REDUCTIVE LIE ALGEBRAS 5

Proposition 3.5. An one-dimensional U(b)-module C(λ,g) is determined by a pair of functions ∗ λ ∈ h0 and g ∈ G. Conversely, suppose an one-dimensional U(b)-module C(λ,g) is determined by a ∗ ∗ pair of functions λ ∈ h0 and g ∈ J , then g ∈ G.

Proof. The second assertion is already proved, we now prove the ﬁrst assertion. We already know

[J, J] = 0, [b0, J1] = 0, [b0, J2] ⊂ J2, g ∈ G. Suppose x,y ∈ b, x = a1+b1+c1+d1,y = a2+b2+c2+d2, a1, a2 ∈ h0, b1, b2 ∈ b0, c1, c2 ∈ J1, d1, d2 ∈ J2, O`oˆ

x · y · w(λ,g) − y · x · w(λ,g) = (a1 + b1 + c1 + d1) · (a2 + b2 + c2 + d2) · w(λ,g)

−(a2 + b2 + c2 + d2) · (a1 + b1 + c1 + d1) · w(λ,g)

= (λ(a1)+ g(c1))(λ(a2)+ g(c2))w(λ,g)

−(λ(a2)+ g(c2))(λ(a1)+ g(c1))w(λ,g)

= 0.

[x,y] · w(λ,g) = [a1 + b1 + c1 + d1, a2 + b2 + c2 + d2] · w(λ,g) = [c1, c2] · w(λ,g) = 0. So x · y · w(λ,g) − y · x · w(λ,g) = [x,y] · w(λ,g). Then Cw(λ,g) becomes a U(b)-module.

Deﬁnition 3.6. We deﬁne Verma module in category O′£º

M(λ, g)= U(g) ⊗U(b) C(λ,g). w(λ,g) is a nonzero element of one-dimensional U(b)-module C(λ,g), and it satisﬁes: h · w(λ,g) = ∗ λ(h)w(λ,g) for any h ∈ h0. For any u ∈ J, u · w(λ,g) = g(u)w(λ,g). λ ∈ h0, g ∈ G.

4. The properties of category O′

′ + + + + Deﬁnition 4.1. Suppose M ∈ O , v ∈ M and n0 · v = 0. For any h ∈ h0, h · v = λ(h)v , ∗ + + + λ ∈ h0. For any u ∈ J, u · v = g(u)v , g ∈ G. We say v is a maximal vector of M with highest weight (λ, g). If M is generated by v+, we say M is a highest weight module.

Remark 4.2. The weights, weight modules and weight spaces mentioned in this paper are in the sense of semisimple Lie algebra. It is because they are based on the semisimple action of h0. The highest weights and highest weight modules mentioned in this paper are in the sense of generalised reductive Lie algebra, they are also highest weights and highest weight modules in the sense of semisimple Lie algebra. 6 YE REN

Proposition 4.3. Suppose M ∈ O′, and suppose there is a maximal vector in M with wight (λ, g). There is a U(g)-module homomorphism from M(λ, g) to M.

Proof. Suppose w(λ,g) is the maximal vector of M with weight (λ, g). According to the definition of Verma module, we can find a maximal vector w of M(λ, g), and the elements of M(λ, g) can be − expressed uniquely as u · w, u ∈ U(n0 ). We define a linear map

Φ : M(λ, g) → M.

− Φ(u.w) = u · w(λ,g), u ∈ U(n0 ). We need to show Φ is a U(g)-module homomorphism. That is to proof for any x ∈ U(g), Φ(xu · w) = xΦ(u · w). According to PBW basis theorem, we write − xu = P aibiciui, ai ∈ U(n0 ), bi ∈ U(h0), ci ∈ U(n0), ui ∈ U(J). biciui · w = ξw, ξ ∈ C, i

Φ(xu · w)=Φ(X ξaiw)= X ξaiu · w(λ,g). i i

xΦ(u · w)= xu · w(λ,g) = X ξaiu · w(λ,g). i Hence Φ(xu · w)= xΦ(u · w).

′ Proposition 4.4. Suppose M ∈ O , then the submodules of M are weight modules , that is h0- semisimple.

Proof. Refer to the proof of theorem 10.9 in [2]. We know that M = Mµi , i belongs to an L ∗ µi∈h0 index set I. Suppose N is a submodule of M, v ∈ N, v = vµi , vµi ∈ Mµi , there are only a P ∗ µi∈h0 ﬁnite number of sums that are not zero. We only need to prove every vµi ∈ N.

Y(h − µj(h))v = Y(h − µj(h))vµi = Y((µi(h) − µj(h))vµi , h ∈ h0. j=6 i j=6 i j=6 i

We can ﬁnd a h ∈ h such that µi(h) 6= µj(h) for any j 6= i. For this h ∈ h we can get

Y((µi(h) − µj(h))vµi ∈ N. j=6 i Hence vµi ∈ N, and N = (Mµi N). L ∗ T µi∈h0

Proposition 4.5. M(λ, g) has a unique maximal submodule. THE BGG CATEGORY FOR GENERALISED REDUCTIVE LIE ALGEBRAS 7

Proof. Each proper submodule of M(λ, g) is a weight module. It cannot have λ as a wight, since

the one-dimensional space M(λ, g)λ generates M(λ, g), and the sum of all proper submodule is still proper. Thus M(λ, g) has a unique maximal submodule.

Deﬁnition 4.6. Suppose N is the maximal submodule of Verma module M(λ, g), We deﬁne L(λ, g) as M(λ, g)/N.

Proposition 4.7. Suppose M is an irreducible module in O′, then M is isomorphic to some L(λ, g), ∗ λ ∈ h0, g ∈ G.

Proof. Choose a nonzero vector v+ of M. Then v+ generates a ﬁnite dimensional U(n)-module V .

Since n is solvable, we can ﬁnd a 1 dimensional submodule of V, choose a nonzero vector w(λ,g) ∗ (λ ∈ h0, g ∈ G) of this submodule, h.w(λ,g) = λ(h)w(λ,g) for any h ∈ h0, u.w(λ,g) = g(u)w(λ,g) for any u ∈ J. x.w(λ,g) = 0 for any x ∈ n0. Since M is irreducible, dim V =1 and M is generated by w(λ,g). There is a surjective homomorphism from M(λ, g) to M, so M is isomorphic to L(λ, g).

Proposition 4.8. Suppose M is an object in O′, then M has a ﬁnite ﬁltration with nonzero quotients each of which is a highest weight module.

Proof. We may assume M is generated by v1, ..., vn. Let V be a U(n)-module generated by v1, ..., vn. Since M is locally n-finite, the dimension of V is finite. We use induction on dimV . If dimV = 1, it is clear that M itself is a highest weight module. Since n is solvable, we can find a 1 dimensional submodule of V , choose a nonzero vector v+ of this submodule, it is obvious that v+ is a maximal ′ vector. It generates a U(g)-submodule M1, M/M1 is still in the category O and generated by + + V/Cv . Since dimV/Cv < dimV , by induction M/M1 has a finite filtration with nonzero quotients each of which is a highest weight module. The preimages of submodules of M/M1 in the filtration with M1 form a desired filtration of M.

∗ Proposition 4.9. For any v ∈ M(λ, g), λ ∈ h0, g ∈ G, and for any u ∈ J, u · v = g(u)v.

Proof. Suppose w is a maximal vector of M(λ, g) with highest weight (λ, g), then every element − of M(λ, g) can be written as v = x · w for some x ∈ U(n0 ). If u ∈ J1, since u commutes with 8 YE REN

x, u · x · w = [u, x] · w + x · u · w = g(u)x · w. If u ∈ J2, since [J2, g0] ⊂ J2, g ∈ G then u · x · w = [u, x] · w + x · u · w = 0. Thus for any u ∈ J and any v ∈ M(λ, g), u · v = g(u)v.

′ From the proposition above, we can see that M(λ, g) ∈ O is h0 L J-semisimple.

Corollary 4.10. Suppose g0 6= g1, dimHomO′ (M(µ, g0), M(λ, g1)) = 0.

Corollary 4.11. Suppose M(µ, g′) is a submodule of M(λ, g), then g′ = g.

′ Proposition 4.12. Suppose M ∈ O , then the action of J2 on M is nilpotent.

Proof. Since M is an object of O′, there is a ﬁnite ﬁltration of M

0= M0 ⊂ M1 ⊂ ... ⊂ Mn = M.

∼ ∗ Mi/Mi−1=L(λi, gi),λi ∈ h0, gi ∈ G(i = 1, 2, ..., n). The action of J2 on L(λi, gi) is zero. Then J2 acts n times on M to be zero.

′ Proposition 4.13. Suppose M ∈ O , and M equals M(λ) as a U(g0)-module, then the action of

J2 is zero. It follows that U(g)-module M = M(λ, g) for some g ∈ G.

Proof. Let M decomposes to M = Mµ according to h0. Choose (0 6=) vλ ∈ Mλ. dimMµ 6= 0 L∗ µ∈h0 + if and only if (λ − µ) ∈ Z Φ. J2 decomposes to J2 = Jγ with the adjoint action of h0, choose L∗ γ∈h0 − ∗ (0 6=)u ∈ Jγ . then u · vλ ∈ Mλ+γ. If γ∈ / Z Φ, then u · vλ = 0. Denote Λ = {γ ∈ h0 | γ 6= 0, the − action of Jγ on vλ is not zero}, obviously Λ ⊂ Z Φ. We can deﬁne partial order on set Λ (α ≤ β if and only if (β − α) ∈ Z+Φ). Λ is a ﬁnite set, if set Λ is not empty, we can choose a minimal − element τ, let (0 6=) u ∈ Jτ . Since u · vλ 6= 0, there is a x ∈ U(n0 ) (x∈ / C) such that u · vλ = x · vλ.

[u, x] · vλ =0= u · x · vλ − x · u · vλ, so u · x · vλ = x · x · vλ, then the action of u is not nilpotent, a contradiction. So Λ is empty. Thus the action of Jγ (γ 6= 0) is zero. Let (0 6=) u ∈ J0, there is ′ ′ ′ ′ ′ a e ∈ n0, u ∈ Jτ (τ 6= 0) such that [e, u ] = u, so u · vλ = [e, u ] · vλ = e · u · vλ − u · e · vλ = 0.

Finitely we get the result that the action of J2 on vλ is zero. Since J2 is an ideal of g, so the action of J2 on M(λ) is zero.

′ Proposition 4.14. (1)Suppose U(g)-module M is an object in O , and M is irreducible as U(g0)- module, then M is irreducible as U(g)-module.

(2)Suppose U(g)-module N is a maximal submodule of M(λ, g) as U(g0)-module, then N is a maximal submodule of M(λ, g) as U(g)-module. THE BGG CATEGORY FOR GENERALISED REDUCTIVE LIE ALGEBRAS 9

Proof. (1)Suppose 0 $ N $ M as U(g)-modules, then 0 $ N $ M as U(g0)-modules, this contradicts with the fact that M is an irreducible U(g0)-module. ′ ′ (2)Suppose N $ N $ M as U(g)-modules, thenN $ N $ M as U(g0)-modules, this contradicts with the fact that N is a maximal U(g0)-submodule of M(λ, g).

Proposition 4.15. (1)Suppose N is a maximal submodule of M(λ, g) as U(g)-module, then N is a maximal submodule of M(λ) as U(g0)-module. ′ (2)Suppose M is an irreducible U(g)-module in category O , then M is also an irreducible U(g0)- module.

′ ′ Proof. (1)Suppose N $ N $ M as U(g0)-modules. Let N be the set of generators, we get a U(g)-module N ′′. Since for any u ∈ J and v ∈ N ′, u · v = g(u)v. So N ′ = N ′′. Then N $ N ′ $ M as U(g)-modules, this contradicts with the fact that N is a maximal U(g)-submodule of M(λ, g). (2)Since M is an irreducible U(g)-module, we can ﬁnd a maximal U(g)-submodule N of some

M(λ, g) such that M is U(g)-module isomorphic to M(λ, g)/N, N is also a maximal U(g0) submodule of M(λ, g), so M is U(g0)-module isomorphic to M(λ, g)/N. And M is an irreducible

U(g0)-module.

′ Proposition 4.16. Let M(λ, g) be a Verma module in O with maximal vector w(λ,g). Suppose (λ+ + (λ+ρ)(hi) ρ)(hi) ∈ Z \{0}, then fi w(λ,g) generates a proper submodule of M(λ, g). This submodule

is isomorphic to M(µ, g) where µ + ρ = sαi (λ + ρ).

n n n j n−j j Proof. Suppose n = (λ + ρ)(hi), ufi w(λ,g) = P j (−1) fi g((adfi) u)w(λ,g). If u ∈ J1, then j=0 j j j (adfi) u = 0, j = 1, 2, ..., n. If u ∈ J2, then (adfi) u ∈ J2 and g((adfi) u) = 0, j = 1, 2, ..., n. So n n n ufi w(λ,g) = g(u)fi w(λ,g). It is easy to check that for any x ∈ n0, xfi w(λ,g) = 0.

∗ In BGG category, there is a theorem(cf. [3]): Let λ, µ ∈ h0

=a)If µ is strongly linked to λ, then U(g0)-module M(µ) ֒→ M(λ); In particular, [M(λ) : L(µ)] 6) 0. (b)If [M(λ) : L(µ)] 6= 0, then µ is strongly linked to λ. (M(λ) possesses a ﬁltration with simple quotients isomorphic to various L(µ), [M(λ) : L(µ)] is the multiplicity of L(µ).) In category O′, there is a similar theorem: 10 YE REN

∗ Theorem 4.17. Let λ, µ ∈ h0, g ∈ G : (a)If µ is strongly linked to λ, then U(g)-module M(µ, g) ֒→ M(λ, g); In particular, [M(λ, g) L(µ, g)] 6= 0. (b)If [M(λ, g) : L(µ, g)] 6= 0, then µ is strongly linked to λ.

→֒ (Proof. (a)If µ is strongly linked to λ, then there is a U(g0)-module homomorphism φ : M(µ, g M(λ, g). Let x ∈ M(µ, g), u ∈ J. Then φ(u · x) = φ(g(u)x) = g(u)φ(x) = u · φ(x). So φ can be extended to be a U(g)-module homomorphism.

(b)[M(λ, g) : L(µ, g)] 6= 0, then[M(λ, g) : L(µ, g)] 6=0 as U(g0)-modules, so µ is strongly linked to λ.

5. Projective modules in category O′

Deﬁnition 5.1. Let P ∈ O′, for any U(g)-module homomorphism φ : P → N, N ∈ O′, and for any U(g)-module epimorphism π : M → N, M ∈ O′, there is a U(g)-module homomorphism ψ : P → M, such that π ◦ ψ = φ, then P is a projective module in category O′.

′ ′ Remark 5.2. Let M ∈ O , if M is a projective U(g0)-module in BGG category O , then M may not be a projective U(g)-module. Indeed, There is no projective module in category O′.

Example 5.3. We ﬁrst give a example that is projective as U(g0)-module but not projective as

U(g)-module. Let g0 = sln(C), g = gln(C) = g0 ⊕ Cz. If λ is dominant, g(z) = 3, then M(λ, g)

is a projective U(g0)-module in BGG category(cf.[3] 3.8). We will show M(λ, g) is not projective U(g)-module in category O′.

Suppose w is a maximal vector of M(λ, g) with weight (λ, g). Denote U(g0)-module L1 =∼ L2 =∼

L(λ), let v1 be a maximal vector of L1 with weight λ, and v2 is a maximal vector of L2 with

weight λ. Let z · v1 = 3v1 + v2 and z · v2 = 3v2, then L1 ⊕ L2 becomes a U(g)-module. Let ϕ be

a U(g)-module homomorphism from M(λ, g) to L(λ, g), ϕ can also be viewed as a U(g0)-module homomorphism. Let v+ = ϕ(w) be a maximal vector of L(λ, g) with weight (λ, g). + Suppose π is a U(g0)-module epimorphism: k1v1 + k2v2 → k1v , k1, k2 ∈ C.ϕ ¯(w) is a weight

vector of L1⊕L2 with weight λ. Since M(λ, g) is a projective U(g0)-module, there is a U(g0)-module THE BGG CATEGORY FOR GENERALISED REDUCTIVE LIE ALGEBRAS 11

commutative diagram: M(λ, g) ss ϕ¯ sss sss ϕ yss π / / L1 ⊕ L2 L(λ, g) 0.

+ Since π(z · v1)= π(3v1 + v2) = 3v = z · π(v1), π(z · v2)= π(3v2)=0= z · π(v2), and z commutes with g0, then π is also a U(g)-module homomorphism. We will show that the diagram above is not a U(g)-module commutative diagram. If it is a U(g)-module commutative diagram, assume + − π(v1 + f2v2)= v , f2 ∈ U(n0 ), andϕ ¯(w) = (v1 + f2v2). But the mapϕ ¯ can not be a U(g)-module homomorphism. It is because

z · ϕ¯(w)= z · (v1 + f2v2) = 3v1 + v2 + 3f2v2.

ϕ¯(z · w)=3¯ϕ(w) = 3v1 + 3f2v2.

z · ϕ¯(w) 6=ϕ ¯(z · w).

Thus M(λ, g) is not a projective U(g)-module.

Proposition 5.4. There is no projective module in category O′. In other words, the projective modules in U(g)-module category are not in category O′.

Proof. Let J be a g0-module with adjoint action, then J can be decomposed to J = L(λ1)⊕L(λ2)⊕

... ⊕ L(λn). Since L(λ1) is finite dimensional, we can find a u ∈ L(λ1) satisfies h · u = α(h)u for − − any h ∈ h0 and n0 · u = [n0 , u] = 0. Suppose P is a projective module in category O′, let N be a maximal submodule of P , and

P/N =∼ L(γ, g). Then there is a nature U(g)-module homomorphism φ1 from P to P/N =∼ L(γ, g).

Let v be a maximal vector of L(γ, g), w1 ∈ P , φ1(w1)= v.

We now construct a special U(g)-module. Let L(γ) and L(γ + α) be two irreducible U(g0)- modules. Let v1 be a maximal vector of L(γ) and v2 be a maximal vector of L(γ + α). We deﬁne u · v1 = g(u)v1 + v2 and u · v2 = g(u)v2. Each element of L(γ) can be written as f1v1 for some − − f1 ∈ U(n0 ). Since u commutes with n0 , so u · f1v1 = g(u)f1v1 + f1v2. Each element of L(γ + α) − can be written as f2v2 for some f2 ∈ U(n0 ). So u · f2v2 = g(u)f2v2. Thus we have deﬁned the

action of u on L(γ) ⊕ L(γ + α). Since u is a generator of L(λ1), then we can deﬁne the action of 12 YE REN

L(λ1) on L(γ) ⊕ L(γ + α). Let the action of L(λ2) ⊕ ... ⊕ L(λn) to be a scalar according to function g, then L(γ) ⊕ L(γ + α) becomes a U(g)-module.

Next, we deﬁne a U(g)-module epimorphism from L(γ)⊕L(γ+α) to L(γ, g). Let π1(f1v1+f2v2)= − f1v, f1,f2 ∈ U(n0 ). It is easy to check that π1 is a U(g)-module epimorphism. ′ Since P is a projective module in category O , there is a U(g)-module homomorphism φ2 satisﬁes

π1 ◦ φ2 = φ1. Then we have a commutative diagram

♥♥ P ♥♥♥ φ2♥♥♥ ♥♥♥ φ1 w♥♥ π1 L(γ) ⊕ L(γ + α) / L(γ, g) / 0.

−1 ′ Since π1 ◦ φ2(w1) = φ1(w1) = v, φ2(w1) ∈ π1 (v) = v1 + L(γ + α). Suppose φ2(w1) = v1 + v , ′ v ∈ L(γ + α). Let w2 = u · w1 − g(u)w1, we show w2 6=0. If u · w1 = g(u)w1, then

′ ′ u · φ2(w1)= u · (v1 + v )= g(u)v1 + v2 + g(u)v

′ φ2(u · w1)= φ2(g(u)w1)= g(u)v1 + g(u)v

u · φ2(w1) 6= φ2(u · w1)

So w2 6= 0. Next we show w1, w2 are linearly independent. Suppose k1w1 + k2w2 = 0, then

φ1(k1w1 + k2w2)= k1φ1(w1)= k1v = 0, so k1 = 0. Since w2 6= 0, we have k2 = 0.

Again, we construct a special U(g)-module. Consider U(g0)-module L(γ) ⊕ L(γ + α) ⊕ L(γ + α).

Let v1, v2, v3 be maximal vectors of three irreducible U(g0)-module respectively. We deﬁne u · v1 = g(u)v1 + v2 + v3, u · v2 = g(u)v2 + v3 and u · v3 = g(u)v3. Let the action of L(λ2) ⊕ ... ⊕ L(λn) to be scalars according to function g. Then L(γ) ⊕ L(γ + α) ⊕ L(γ + α) becomes a U(g)-module similarly.

Deﬁne a U(g)-module homomorphism π2 to be a projection from L(γ) ⊕ L(γ + α) ⊕ L(γ + α) to L(γ) ⊕ L(γ + α). If P is a projective module in category O′, there is a commutative diagram

❡❡❡ ❡❡❡❡❡❡ ♥♥ P φ3 ❡❡❡❡ ♥♥ ❡❡❡❡❡ φ2 ♥♥ ❡❡❡❡❡❡ ♥♥♥ ❡❡❡❡❡❡ ♥♥ φ1 r❡❡❡❡ w♥♥ π2 π1 L(γ) ⊕ L(γ + α) ⊕ L(γ + α) / L(γ) ⊕ L(γ + α) / L(γ, g) / 0.

Let w3 = u · w2 − g(u)w2, we can prove that w3 6= 0 and w1, w2, w3 are linearly independent.

Proceeding above procedures, we can ﬁnd linearly independent elements w1, w2, ..., wn in U(n) · w1, THE BGG CATEGORY FOR GENERALISED REDUCTIVE LIE ALGEBRAS 13

n is suﬃciently large. It contradicts with the fact that P is locally n-ﬁnite. So P is not a projective module in category O′.

′ Proposition 5.5. Let M ∈ O , if M has a U(g0)-module standard ﬁltration, then M has a U(g)- module standard ﬁltration. Their length are the same.

− Proof. Since M has a U(g0)-module standard filtration, then M is U(n0 )-free (cf. [3] 3.7). Suppose v1, ..., vn generate U(g)-module M, let V be a U(n)-module generated by v1, ..., vn, then V is finite dimensional. Since n is solvable, we can find a suitable basis of V such that the action matrix of n is a upper triangular matrix, that is, there is a basis w1, w2, ..., wk of V such that U(g)-modules

Mi(i = 1, ..., k) are generated by wi, ..., wk. And there is a U(g)-module standard ﬁltration

0= M0 ⊂ M1 ⊂ ... ⊂ Mk = M.

∼ ∗ M0 = M(λ0, g0), Mi/Mi−1 = M(λi, gi)(λi ∈ h0, gi ∈ G, i = 1, 2, ..., k). It is also a U(g0)-module standard ﬁltration. The uniqueness of the length of standard ﬁltration shows their length are the same.

In the BGG category, P (λ) is a projective cover of L(λ). There is a theorem(BGG Reciprocity)(cf. ∗ [3]): Let λ, µ ∈ h0. Denote the multiplicity with which each Verma module M(µ) occurs in a standard ﬁltration of P (λ) by (P (λ) : M(µ)). Then (P (λ) : M(µ)) = [M(µ) : L(λ)]. According to the properties of projective cover, P (λ) is generated by one element. Assume P (λ) ∗ is generated by vλ, for any u ∈ J, let u · vλ = g(u)vλ, g ∈ J (indeed, g ∈ G), then U(g0)-module P (λ) become a U(g)-module, denoted as P (λ, g). According to the proposition above, we know that P (λ, g) has a U(g)-module standard ﬁltration. And there is a theorem in category O′:

∗ Theorem 5.6. Let λ, µ ∈ h0. then

(P (λ, g) : M(µ, g)) = [M(µ, g) : L(λ, g)].

Proof. There is a U(g)-module standard ﬁltration of P (λ, g):

0= M0 ⊂ M1 ⊂ ... ⊂ Mn = M. 14 YE REN

Mn = P (λ, g), Mi/Mi−1=∼M(λi, gi)(i=1, 2, ..., n). We ﬁrst show gi = g. Since Mn/Mn−1 =∼

M(λn, gn), there is a U(g)-module nature homomorphism:

φ : P (λ, g) → P (λ, g)/Mn−1 = M(λn, gn).

φ(vλ)= x · w(λn,gn). − vλ generates P (λ, g), w(λn,gn) is a maximal vector of M(λn, gn), x ∈ U(n0 ). For any u ∈ J, u · φ(vλ)= φ(u · vλ). On the one hand, u · φ(vλ)= u · x · w(λn,gn) = gn(u)x · w(λn,gn). On the other hand, φ(u · vλ)= φ(g(u)vλ)= g(u)x · w(λn ,gn). So gn = g and g ∈ G. Suppose y · vλ, y ∈ U(g0) is an element of P (λ, g), then for any u ∈ J, u · y · vλ = g(u)y · vλ. Hence gi = g, i = 1, ..., n. From this result we know that diﬀerent selected generators in P (λ) generate the same P (λ, g). Moreover,

(P (λ, g) : M(µ, g)) = (P (λ) : M(µ)) = [M(µ) : L(λ)] = [M(µ, g) : L(λ, g)].

Acknowledgements. I would like to thank professors Bin Shu, Lei Lin, Jun Hu for their helpful comments and suggestions.

References

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