DERIVED EQUIVALENCE IN SL_{2}(p^{2})

JOSEPH CHUANG

Abstract. We present a proof that Broué’s Abelian Defect Group Conjecture is true for the principal p-block of the group SL_{2}(p^{2}). Okuyama has independently obtained the same result using a different approach.

1. Introduction and preliminaries

Broué has posed a remarkable conjecture involving the derived categories of blocks of finite groups with abelian defect groups [2]. In the case of principal blocks this conjecture is particularly easy to state.

Conjecture 1 (Broué). Let K be an algebraically closed field of prime characteristic p, let G be a finite group with an abelian Sylow psubgroup P, and let H be the normalizer of P in G. Then the principal blocks of KG and KH have equivalent derived categories.

For good introductions to this conjecture we suggest [3] and [8]. The purpose of this paper is to show that Broué’s conjecture is true

for the group G = SL_{2}(p^{2}). Okuyama has independently obtained this result using an extension of the method he developed in [7]. We should also mention that the cases p = 2 and p = 3 were handled already in [13] and [7], respectively, and that the cases p = 5 and p = 7 have been settled independently by Holloway.

The proof given here is made possible by a new method for constructing derived equivalences due to Rickard (section 6 of [9]). We also rely heavily on calculations of cohomology in Carlson’s paper [4].

We will be dealing with the following categories associated to a finite-dimensional K-algebra Λ: mod(Λ), the category of finitely generated Λ-modules; stmod(Λ), the stable category of finitely generated Λ-modules, in which the objects are the same as in mod(Λ) and the morphisms are Λ-homomorphisms modulo those which factor through projective modules; and D^{b}(mod(Λ)), the derived category of bounded complexes of finitely generated Λ-modules. Viewing a module as a complex concentrated in degree zero defines a fully faithful functor from mod(Λ) to D^{b}(mod(Λ)); we will often identify a Λ-module with its image in D^{b}(mod(Λ)).

1991 Mathematics Subject Classification. 20C20.

If Λ is a symmetric algebra (e.g. a block of a finite group algebra) then stmod(Λ) is a triangulated category and may be identified with a quotient of D^{b}(mod(Λ)) in the following way [10]: the full subcategory P of D^{b}(mod(Λ)) consisting of objects isomorphic to bounded complexes of projective modules is a thick subcategory, and the composition of the embedding mod(Λ) → D^{b}(mod(Λ)) and the projection D^{b}(mod(Λ)) → D^{b}(mod(Λ))/P factors through the functor mod(Λ) → stmod(Λ); the resulting functor stmod(Λ) → D^{b}(mod(Λ))/P is an equivalence of triangulated categories. By ‘the’ image of an object X of D^{b}(mod(Λ)) in stmod(Λ) we will mean an object of stmod(Λ) which is associated via this equivalence to the image of X in D^{b}(mod(Λ))/P. So two corners of a distinguished triangle in D^{b}(mod(Λ)) have isomorphic images in stmod(Λ) if the third corner of the triangle lies in P.

If Λ and Γ are symmetric algebras we say they are Morita equivalent if their module categories are equivalent, stably equivalent if their stable categories are equivalent (as triangulated categories), and derived equivalent if their derived categories are equivalent (as triangulated categories). A stable equivalence stmod(Λ) → stmod(Γ) is of Morita type if it is induced by an exact functor between the corresponding module categories.

If we would like to prove that Λ and Γ are derived equivalent, one way to proceed is to find a tilting complex T for Λ (an object of P satisfying certain conditions) and show that Γ is isomorphic to the endomorphism ring of T; this is part of Rickard’s Morita theory for derived categories [11]. This approach was used for example by Rickard to prove that Conjecture 1 holds for groups with cyclic Sylow p-subgroups [10]. But this may not work well in more complicated cases: even if we have constructed an appropriate tilting complex T it may be very difficult to calculate its endomorphism ring. Furthermore we may not even know the structure of Γ explicitly. Okuyama used a theorem of Linckelmann [6] to develop a way around these problems in certain situations where one already has a stable equivalence of Morita type between the algebras Λ and Γ; he was then able to verify Conjecture 1 in a number of cases [7]. Partly in order to exploit Okuyama’s method, Rickard extended his theory, proving the following theorem.

Theorem 2 (Rickard [9]). Suppose Γ is a symmetric finite-dimensional K-algebra and let X_{1}, X_{2}, …, X_{r} be objects of D^{b}(mod(Γ)) which generate D^{b}(mod(Γ)) as a triangulated category and such that given n ≤ 0, the space

Hom(X_{i}, X_{j}[n])

is zero unless i = j and n = 0, in which case it is one-dimensional. Then there exists a K-algebra Γ′ and an equivalence

D^{b}(mod(Γ)) → D^{b}(mod(Γ′))

sending X_{1}, X_{2}, …, X_{r} to the simple Γ′-modules.

The application, suggested by Rickard in [9], which makes use of Okuyama’s method is stated here as a corollary.

Corollary (Rickard). Suppose Λ and Γ are finite-dimensional K-algebras, with Γ symmetric, and suppose there is a functor F : mod(Λ) → mod(Γ) inducing a stable equivalence. Let X_{1}, X_{2}, …, X_{r} be objects of D^{b}(mod(Γ)) satisfying the conditions of Theorem 2 and such that for every simple Λ-module S, the image of X_{i} in stmod(Γ) is isomorphic to F(S) for some i. Then Λ and Γ are derived equivalent.

Proof. By Theorem 2, there is a K-algebra Γ′ and an equivalence G : D^{b}(mod(Γ)) → D^{b}(mod(Γ′)) sending X_{1}, X_{2}, …, X_{r} to the simple Γ′- modules. Composing the stable equivalence stmod(Γ) → stmod(Γ′) induced by G (see Corollary 5.5 of [12]) with the equivalence stmod(Λ) → stmod(Γ) induced by F yields a stable equivalence of Morita type stmod(Λ) → stmod(Γ′) which sends simple Λ-modules to simple Γ′- modules. Thus by Theorem 2.1 of [6], Λ and Γ′ are Morita equivalent, and it follows that Λ and Γ are derived equivalent.

We now turn to the example under consideration: let G be the group SL_{2}(p^{2}), where we now assume that p is odd. We will follow closely the notation and methods of [4]. Let F be the field with p^{2} elements and let α be a generator of the multiplicative group F^{∗}. In G let P be the subgroup of upper unipotent matrices and let y be the diagonal matrix with diagonal entries α and α^{−1}. Then P is an elementary abelian p-group of order p^{2} and is a Sylow p-subgroup of G. The normalizer of P in G is H = ⟨y⟩P.

Let σ be the Frobenius automorphism given by σ(β) = β^{p} for all β ∈ F. The simple KG-modules are described as follows. Let V_{1} be the standard two-dimensional KG-module and let V_{2} = σ(V_{1}). For i = 1, 2 and 0 ≤ t ≤ p − 1, let V_{i}^{(t)} be the t-th symmetric power of V_{i}. Let b = (b_{1}, b_{2}) be a pair of integers with 0 ≤ b_{i} ≤ p − 1 and let

M_{b} = V_{1}^{(b_{1})} ⊗ V_{2}^{(b_{2})}.

Each simple module is isomorphic to M_{b} for a unique b. For example M_{0,0} is the trivial module and M_{p−1,p−1} is the Steinberg module, which is simple and projective. M_{b} lies in the principal block if and only if b ∈ I where

I = {b : 0 ≤ b_{i} ≤ p − 1, b ≠ (p − 1, p − 1), and b_{1} + b_{2} is even}.

Note also that σ(M_{b_{1},b_{2}}) is isomorphic to M_{b_{2},b_{1}}.

For any integer j, let U_{j} be the one-dimensional KH-module on which y acts by multiplication by α^{j}. Each simple KH-module is isomorphic to U_{j} for some j, and U_{j} lies in the principal block if and only if j is even. The Frobenius automorphism σ preserves H so it acts on KH-modules as well: σ(U_{j}) is isomorphic to U_{pj}.

The composition factors of M_{b} regarded as a KH-module consist of the simple modules

{U_{m_{1}+pm_{2}} : −b_{i} ≤ m_{i} ≤ b_{i} and m_{1} + m_{2} ≡ b_{1} + b_{2} mod 2},

counted with multiplicities. Note also that U_{j} ⊗M_{b} lies in the principal block of KH if and only if j + b_{1} + b_{2} is even. In particular M_{b} lies in the principal block of KH for all b ∈ I.

Let A and B be the principal blocks of KG and KH, respectively. Because the Sylow subgroup P is a trivial intersection subgroup of G, restriction from A to B induces a stable equivalence (see [1]). Thus the following result, which will be proved in the course of this paper, together with the Corollary of Theorem 2 implies that Conjecture 1 is true for the group G = SL_{2}(p^{2}).

Theorem 3. There exist objects Z_{b} in D^{b}(mod(B)) indexed by b ∈ I such that

(1) for all b ∈ I, the image of Z_{b} in stmod(B) is isomorphic to M_{b};

(2) for all b, c ∈ I and all n ≤ 0, the space

Hom(Z_{b}, Z_{c}[n])

is zero unless b = c and n = 0, in which case it is one-dimensional;

(3) the objects Z_{b} generate D^{b}(mod(B)) as a triangulated category.

Corollary. A and B are derived equivalent.

Remark. For blocks of group algebras, there is an important strengthening of the notion of derived equivalence, which is now usually called splendid Rickard equivalence. This idea (at least for principal blocks) was introduced and argued for in [13], where it was conjectured that in the situation of Broué’s conjecture, there should be not only a derived equivalence but a splendid Rickard equivalence.

Carlson and Rouquier have proved that in certain situations one can deduce the existence of a splendid Rickard equivalence from that of a derived equivalence. In this paper we are in such a situation: as H/C_{G}(P) is a cyclic p′-group (of order (p^{2} − 1)/2) acting freely on P, Corollary 4.4 of [5] applies, and we conclude that A and B are splendid Rickard equivalent.

2. Construction of complexes

We begin by constructing the complexes Z_{b}; first, a few technical lemmas.

Lemma 4. Suppose i is 1 or 2, and suppose l and t are integers with 0 ≤ t < p−1. Let j = l+p^{i−1}(p+1+t) and j′ = l+p^{i−1}(p−1−t). Then any non-zero KH-homomorphism from U_{j} ⊗ V_{i}^{(p−1)} to U_{j′} ⊗ V_{i}^{(p−1)} has cokernel isomorphic to U_{l} ⊗ V_{i}^{(t)}.

Proof. By Lemma 2.1 of [4], as a KH-module V_{i}^{(p−1)} is uniserial with pairwise non-isomorphic composition factors. The same is true of the tensor product of any simple KH-module with V_{i}^{(p−1)}. Thus the space of KH-homomorphisms between any two such modules is at most onedimensional. By Lemma 2.2 of [4], U_{l}⊗V_{i}^{(t)} is isomorphic to the cokernel of some non-zero homomorphism from U_{j} ⊗ V_{i}^{(p−1)} to U_{j′} ⊗ V_{i}^{(p−1)}, and hence is isomorphic to the cokernel of any such homomorphism.

Lemma 5. Suppose t is an even non-negative integer less than p − 1 and suppose h is a positive integer. If h is even there is a distinguished triangle

Y ⟶ M_{p−1,t} ⟶ U_{h} ⊗ M_{p−1,t}[h] ⟶

in D^{b}(mod(B)), and if h is odd there is a distinguished triangle

Y ⟶ M_{p−1,t} ⟶ U_{h} ⊗ M_{p−1,p−2−t}[h] ⟶

in D^{b}(mod(B)), where in both cases Y is a bounded complex of projective modules and Y^{−i} = 0 for i ≥ h.

Proof. By Lemma 2.2 of [4] , there exists an exact sequence

… ⟶ X^{−2} ⟶^{δ^{−2}} X^{−1} ⟶^{δ^{−1}} X^{0} ⟶^{ϵ} V_{2}^{(t)},

where for s ≥ 0,

X^{−2s} = U_{p(p−1−t+2ps)} ⊗V_{2}^{(p−1)} and X^{−(2s+1)} = U_{p(p+1+t+2ps)} ⊗V_{2}^{(p−1)}.

Let C be the kernel of δ^{−(h−1)}. Then

[figure: ⋯; 0; X^{−(h−1)}; ⋯; X^{0}; 0; ⋯; ⋯; 0; C; X^{−(h−1)}; ⋯; X^{0}; 0; ⋯; ⋯; 0; C; 0; ⋯]

is a short exact sequence of bounded complexes of KH-modules, and the middle term is quasi-isomorphic to V_{2}^{(t)}. After tensoring this sequence with the module V_{1}^{(p−1)}, the middle term is quasi-isomorphic to M_{p−1,t}, so in the derived category we have a distinguished triangle

Y ⟶ M_{p−1,t} ⟶ V_{1}^{(p−1)} ⊗ C[h] ⟶,

where Y is a bounded complex of projective modules (because V_{1}^{(p−1)} ⊗ V_{2}^{(p−1)} is projective) and Y^{−i} = 0 for i ≥ h.

C is isomorphic to the cokernel of δ^{−(h+1)} : X^{−(h+1)} ⟶ X^{−h}. If h is even then

X^{−(h+1)} = U_{j} ⊗ V_{2}^{(p−1)} and X^{−h} = U_{j′} ⊗ V_{2}^{(p−1)},

where j = p(p + 1 + t + ph) and j′ = p(p − 1 − t + ph). Writing j = p^{2}h + p(p + 1 + t) and j′ = p^{2}h + p(p − 1 − t), and noting that U_{p^{2}h} = U_{h}, we see by Lemma 4 that C is isomorphic to U_{h} ⊗V_{2}^{(t)}. Thus the last term in the triangle above is isomorphic to U_{h} ⊗ M_{p−1,t}[h].

If on the other hand h is odd, then X^{−(h+1)} and X^{−h} are as above, but with j = p(p−1−t+p(h+1)) and j′ = p(p+1+t+p(h−1)). We may rewrite j = p^{2}h+p(p+1+(p−2−t)) and j′ = p^{2}h+p(p−1−(p−2−t)), so by Lemma 4, C is isomorphic to U_{h} ⊗ V_{2}^{(p−2−t)}. Thus the last term in the triangle above is isomorphic to U_{h} ⊗ M_{p−1,p−2−t}[h].

Lemma 6. Suppose b_{1} and b_{2} are non-negative integers less than p−1. Then there is an exact sequence of KH-modules

0 ⟶ U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶ U_{p−1−b_{1}} ⊗ M_{p−1,b_{2}} ⟶ M_{b} ⟶ 0.

Proof. By Lemma 2.1 of [4], there is an exact sequence

U_{j} ⊗ V_{1}^{(p−1)} ⟶ U_{j′} ⊗ V_{1}^{(p−1)} ⟶ U_{p−1−b_{1}} ⊗ V_{1}^{(p−1)} ⟶ V_{1}^{(b_{1})} ⟶ 0,

where j = p − 1 − b_{1} + 2p and j′ = p + 1 + b_{1}. Since j = p + (p + 1 + (p − 2 − b_{1})) and j′ = p + (p − 1 − (p − 2 − b_{1})), the cokernel of the first homomorphism in the sequence is isomorphic to U_{p} ⊗ V_{1}^{(p−2−b_{1})}, by Lemma 4. Thus there is an exact sequence

0 ⟶ U_{p} ⊗ V_{1}^{(p−2−b_{1})} ⟶ U_{p−1−b_{1}} ⊗ V_{1}^{(p−1)} ⟶ V_{1}^{(b_{1})} ⟶ 0.

Tensoring this with V_{2}^{(b_{2})} gives the desired exact sequence.

Now we construct the complexes Z_{b} and show that the first assertion of Theorem 3 holds. We begin by defining some subsets of the index set I. Let

I_{<} = {b ∈ I : b_{1} + b_{2} < p − 1},
I_{p−1} = {b ∈ I : b_{1} = p − 1 or b_{2} = p − 1},
I_{≥} = {b ∈ I : b_{1} + b_{2} ≥ p − 1, b_{1} ≠ p − 1, and b_{2} ≠ p − 1}.

I is the disjoint union of I_{<}, I_{p−1}, and I_{≥}. Given b ∈ I we define Z_{b} as follows:

• b ∈ I_{<}: In this case simply define Z_{b} = M_{b}. The condition that the image of Z_{b} in stmod(B) is isomorphic to M_{b} is trivially satisfied.

• b ∈ I_{p−1} and b_{1} = p − 1: If b_{2} < (p − 1)/2, define

Z_{p−1,b_{2}} = U_{b_{2}+2} ⊗ M_{p−1,b_{2}}[b_{2} + 2];

and if b_{2} ≥ (p − 1)/2, define

Z_{p−1,b_{2}} = U_{p−b_{2}} ⊗ M_{p−1,p−2−b_{2}}[p − b_{2}].

By lemma 5, the image of Z_{p−1,b_{2}} in stmod(B) is isomorphic to M_{p−1,b_{2}}. Note that the set of complexes constructed here may be written collectively as

{U_{s+2} ⊗ M_{p−1,s}[s + 2] : 0 ≤ s ≤ (p − 3)/2}.

• b ∈ I_{p−1} and b_{2} = p − 1: Define Z_{b_{1},p−1} = σ(Z_{p−1,b_{1}}), using the previous case. It is clear that the image of Z_{b_{1},p−1} in stmod(B) is isomorphic to M_{b_{1},p−1}. The set of complexes constructed here may be written collectively as

{U_{p(s+2)} ⊗ M_{s,p−1}[s + 2] : 0 ≤ s ≤ (p − 3)/2}.

• b ∈ I_{≥} and b_{2} ≤ b_{1}: By lemma 6, there is a distinguished triangle

U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶^{f} U_{p−1−b_{1}} ⊗ M_{p−1,b_{2}} ⟶ M_{b} ⟶

in D^{b}(mod(B)).

Let w = min{b_{2}, p − 2 − b_{2}}. Applying lemma 5 with t = b_{2} and h = w + b_{1} − p + 3 and then tensoring with U_{p−1−b_{1}}, we obtain a distinguished triangle

Y ⟶ U_{p−1−b_{1}} ⊗ M_{p−1,b_{2}} ⟶^{g} U_{w+2} ⊗ M_{p−1,w}[h] ⟶,

where Y is a bounded complex of projective modules and Y^{−i} = 0 for i ≥ h.

Define Z_{b} to be the third object in a distinguished triangle which contains the composite of f and g:

U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶^{g∘f} U_{w+2} ⊗ M_{p−1,w}[h] ⟶ Z_{b} ⟶ .

Then by the octahedral axiom there is a commutative diagram

[figure: U_{p} ⊗ M_{p−2−b_{1},b_{2}}; U_{p−1−b_{1}} ⊗ M_{p−1,b_{2}}; M_{b}; U_{p} ⊗ M_{p−2−b_{1},b_{2}}; U_{w+2} ⊗ M_{p−1,w}[h]; Z_{b}; Y [1]; Y [1]]

where all the rows and columns are distinguished triangles. Since Y [1] is a bounded complex of projective modules, the image of Z_{b} in stmod(B) is isomorphic to M_{b}.

• b ∈ I_{≥} and b_{1} < b_{2}: Define Z_{b_{1},b_{2}} = σ(Z_{b_{2},b_{1}}). Again it is clear that the image of Z_{b} in stmod(B) is isomorphic to M_{b}.

We wish to record for future reference some details in the next-to-last construction above.

Lemma 7. Suppose b ∈ I_{≥} and b_{2} ≤ b_{1}. Then there exist distinguished triangles

U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶ U_{w+2} ⊗ M_{p−1,w}[h] ⟶ Z_{b} ⟶

and

Y ⟶ M_{b} ⟶ Z_{b} ⟶,

where w = min{b_{2}, p − 2 − b_{2}}, h = w + b_{1} − p + 3, and Y is a bounded complex of projective modules such that Y^{−i} = 0 for i ≥ h.

3. Homomorphisms

We prepare for a proof of the second assertion of Theorem 3 by calculating some Ext-groups involving B-modules. We will need a version of Theorem 2.6 of [4].

Theorem 8 (Carlson). Suppose b, c ∈ I, and suppose j, j′, and r are integers with r non-negative. Then the dimension of Ext_{KH}^{r}(U_{j} ⊗ M_{b}, U_{j′} ⊗ M_{c}) is equal to the number of triples (e, f, k) (where e = (e_{1}, e_{2}), f = (f_{1}, f_{2}), and k = (k_{1}, k_{2}) are pairs of non-negative integers) satisfying the following conditions:

(1) 2e_{1} + 2e_{2} + f_{1} + f_{2} = r;

(2) for every i, f_{i} is either 0 or 1;

(3) e_{i} = f_{i} = 0 whenever b_{i} or c_{i} is p − 1;

(4) if f_{i} = 0, then

max\{0, c_{i} − b_{i}\} ≤ k_{i} ≤ \begin{cases}c_{i} & if e_{i} = 0, \\ min\{c_{i}, p − 2 − b_{i}\} & if e_{i} > 0, \end{cases}

while if f_{i} = 1 then

max{0, b_{i} + c_{i} + 2 − p} ≤ k_{i} ≤ min{b_{i}, c_{i}}.

(5) a(e, f, k) ≡ 0 mod (p^{2} − 1) where

a(e, f, k) = j′ − j − 2(pe_{1} + e_{2}) + ∑_{i=1}^{2} p^{i−1}(b_{i} − c_{i} + 2k_{i} − f_{i}(2b_{i} + 2)).

Proof. The proof given by Carlson can be adapted with little alteration: The restrictions of U_{j} and U′_{j} to P are trivial, so Ext_{KP}^{r} (U_{j} ⊗ M_{b}, U_{j′} ⊗ M_{c}) has K-basis {θ(e, f, k)} indexed on triples (e, f, k) which satisfy conditions (1)-(4). A spectral sequence argument identifies Ext_{KH}^{r}(U_{j}⊗ M_{b}, U_{j′} ⊗ M_{c}) with the K(H/P)-fixed points of Ext_{KP}^{r} (U_{j} ⊗ M_{b}, U_{j′} ⊗ M_{c}). Because y acts on Ext_{KP}^{r} (U_{j} ⊗ M_{b}, U_{j′} ⊗ M_{c}) by scaling the basis vector θ(e, f, k) by α^{a(e,f,k)}, a basis for Ext_{KH}^{r}(U_{j} ⊗ M_{b}, U_{j′} ⊗ M_{c}) is indexed by triples (e, f, k) which in addition satisfy condition (5).

We will only provide proofs for a few parts of the following lemma, to give an idea of the type of arguments involved; similar reasoning can be used to prove the others.

Lemma 9. Suppose b, c ∈ I_{<} and b′, c′ ∈ I_{≥} with b′_{2} ≤ b′_{1} and c′_{2} ≤ c′_{1}; and suppose s and t are non-negative integers less than or equal to (p − 3)/2. Let w = min{b′_{2}, p − 2 − b′_{2}} and h = w + b′_{1} − p + 3. Then

1. Hom_{B}(M_{b}, M_{c}) = 0 if b ≠ c and is one-dimensional if b = c.

2. Ext_{B}^{r}(M_{b}, U_{p} ⊗ M_{p−2−c′_{1},c′_{2}}) = 0 for r = 0 or r = 1.

3. Ext_{B}^{r}(U_{s+2} ⊗ M_{p−1,s}, U_{t+2} ⊗ M_{p−1,t}) = 0 whenever r ≤ t − s unless s = t and r = 0 in which case it is one-dimensional.

4. Ext_{B}^{r}(U_{p(s+2)} ⊗ M_{s,p−1}, U_{t+2} ⊗ M_{p−1,t}) = 0 whenever r ≤ t − s.

5. Ext_{B}^{r}(M_{b}, U_{s+2} ⊗ M_{p−1,s}) = 0 whenever r ≤ s + 2.

6. Ext_{B}^{r}(M_{b′}, U_{s+2} ⊗ M_{p−1,s}) = 0 whenever h < r ≤ s + 2.

7. Ext_{B}^{r}(M_{b′}, U_{p(s+2)} ⊗ M_{s,p−1}) = 0 whenever h < r ≤ s + 2.

8. Ext_{B}^{r}(U_{p}⊗M_{p−2−b′_{1},b′_{2}}, U_{s+2}⊗M_{p−1,s}) = 0 whenever r ≤ min{h, s+ 2} unless s = w and r = h in which case it is one-dimensional.

9. Ext_{B}^{r}(U_{p}⊗M_{p−2−b′_{1},b′_{2}}, U_{p(s+2)}⊗M_{s,p−1}) = 0 whenever r ≤ min{h, s+ 2}

10. Hom_{B}(U_{s+2} ⊗ M_{p−1,s}, U_{p} ⊗ M_{p−2−c′_{1},c′_{2}}) = 0.

11. Hom_{B}(U_{1} ⊗ M_{b′_{1},p−2−b′_{2}}, U_{p} ⊗ M_{p−2−c′_{1},c′_{2}}) = 0.

12. Hom_{B}(U_{p} ⊗ M_{p−2−b′_{1},b′_{2}}, U_{p} ⊗ M_{p−2−c′_{1},c′_{2}}) = 0 if b′ ≠ c′ and is onedimensional if b′ = c′.

13. Hom_{B}(U_{p(s+2)} ⊗ M_{s,p−1}, U_{p} ⊗ M_{p−2−c′_{1},c′_{2}}) = 0.

Proof. (1) Suppose that (e, f, k) is a triple satisfying the conditions in Theorem 8 with j = j′ = 0. Condition (1) implies that e_{1} = e_{2} = f_{1} = f_{2} = 0. Let γ_{i} = b_{i} − c_{i} + 2k_{i}. Then by condition (4) we have 0 = b_{i} − c_{i} + (0 + c_{i} − b_{i}) ≤ γ_{i} ≤ b_{i} − c_{i} + 2c_{i} = b_{i} + c_{i}, where the first inequality is an equality if and only if b_{i} = c_{i} and k_{i} = 0. We have γ_{1} + γ_{2} ≤ b_{1} + c_{1} + b_{2} + c_{2} ≤ 2(p − 2), so either γ_{1} ≤ p − 2 or γ_{2} ≤ p − 2. Suppose the latter. In order for condition (5) to be satisfied we need γ_{1} + pγ_{2} ≡ 0 mod (p^{2} − 1). But 0 ≤ γ_{1} + pγ_{2} ≤ 2(p − 2) + p(p − 2) < p^{2} − 1, so we must have γ_{1} +pγ_{2} = 0, which implies that b_{1} = c_{1}, b_{2} = c_{2}, and k_{1} = k_{2} = 0.

If γ_{1} ≤ p − 2, we note that γ_{2} + pγ_{1} ≡ p(γ_{1} + pγ_{2}) mod (p^{2} − 1) and that p(γ_{1} + pγ_{2}) ≡ 0 mod (p^{2} − 1) if and only if γ_{1} + pγ_{2} ≡ 0 mod (p^{2}−1). Thus we may use the argument above, interchanging the roles of γ_{1} and γ_{2}.

(3) Suppose that (e, f, k) is a triple satisfying the conditions in Theorem 8 with j = s + 2, b = (p − 1, s), j′ = t + 2, and c = (p − 1, t). Because b_{1} = p − 1, we have e_{1} = f_{1} = 0, by condition (3). We divide the argument into two cases according to the parity of r.

Suppose that r is even. Then by condition (1), we have r = 2e_{2} + f_{2} which implies that f_{2} is even and therefore zero, by condition (2). Condition (4) tells us that 0 ≤ k_{1} ≤ p − 1 and t − s ≤ k_{2} ≤ t, and that t−s = k_{2} if and only if s = t and k_{2} = 0. We have a = a(e, f, k) = (t−s−r)+2k_{1}+p(s−t+2k_{2}) ≡ 0 mod (p^{2}−1) by condition (5). Now a ≥ 0+2·0+p(s−t+2(t−s)) ≥ 0 with equalities throughout if and only if s = t and r = e_{2} = k_{1} = k_{2} = 0. This is in fact the only possibility, because a ≤ t + 2(p − 1) + p(s + t) ≤ (p − 3) + 2(p − 1) + p(p − 3) < p^{2} − 1.

Suppose on the other hand that r is odd. Then r = 2e_{2} + f_{2} implies that f_{2} is odd and therefore that f_{2} = 1. By condition (4), we have 0 ≤ k_{1} ≤ p − 1 and 0 ≤ k_{2} ≤ s. Let γ = (t − s − r) + 2k_{1} + p(p − 2 − s − t + 2k_{2}). Then γ ≡ a(e, f, k) mod (p^{2} − 1), so by condition (5) we have γ ≡ 0 mod (p^{2} −1). Now γ ≥ 0+2·0+ p(p−2−(p−3)+2·0) = p > 0 and because r is odd, s−t ≤ −1, so γ ≤ t+2(p−1)+p(p−2+s−t) ≤ (p−3)+2(p−1)+p(p−3) < p^{2}−1; we have arrived at a contradiction.

(5) Suppose that (e, f, k) is a triple satisfying the conditions in Theorem 8 with j = 0, j = s+2, and c = (p−1, s). Because c_{1} = p−1, we have e_{1} = f_{1} = 0, by condition (3). We divide the argument into two cases according to the parity of r.

Suppose r is even. Then r = 2e_{2} and f_{2} = 0 by conditions (1) and (2). Note that h ≥ 0. Thus r > 0 and e_{2} > 0 as well. By condition (4) we have p − 1 − b_{1} ≤ k_{1} ≤ p − 1 and max{0, s − b_{2}} ≤ k_{2} ≤ min{s, p − 2 − b_{2}}, and by condition (5) we have a = a(e, f, k) = (s+2−r)+b_{1}−(p−1)+2k_{1}+p(b_{2}−s+2k_{2}) ≡ 0 mod (p^{2} − 1). First note that a ≥ 0 + b_{1} − (p − 1) + 2(p − 1 − b_{1}) + p(b_{2} − s + 0 + (s − b_{2})) = (p − 1) − b_{1} > 0, so we must have a ≥ p^{2} − 1. Now suppose that either b_{2} + s ≤ p − 3 or b_{2} − s + 2(p − 2 − b_{2}) ≤ p − 3. Then b_{2} − s + 2k_{2} ≤ p − 3, so a ≤ (p − 3) + 2 + (p − 2) − (p − 1) + 2(p − 1) + p(p − 3) < p^{2} − 1, giving a contradiction. Hence we must have b_{2} + s ≥ p − 2 and b_{2} − s + 2(p − 2 − b_{2}) ≥ p − 2. These two inequalities together imply that b_{2} + s = p − 2. We then also have b_{1} ≤ p − 2 − b_{2} = s. Thus a ≤ s+2+s−(p−1)+2(p−1)+p(b_{2} +s) = 2s+1−p+p^{2} ≤ (p − 3) + 1 − p + p^{2} < p^{2} − 1, which is again a contradiction.

Suppose instead that r is odd. Then r = 2e_{2} + 1 and f_{2} = 1 by conditions (1) and (2). By condition (4) we have p − 1 − b_{1} ≤ k_{1} ≤ p − 1 and max{0, b_{2} + s + 2 − p} ≤ k_{2} ≤ min{b_{2}, s}. Let γ = (s+2−r)+b_{1} −(p−1)+2k_{1} +p(−b_{2} −s−2+p+2k_{2}). Then γ ≡ a(e, f, k) mod (p^{2} − 1), so by condition (5) we have γ ≡ 0 mod (p^{2} − 1). Now γ ≥ 0 + (p − 1) − b_{1} + p · 0 > 0 so we must have γ ≥ p^{2} − 1. If b_{2} = s, then b_{1} + s = b_{1} + b_{2} ≤ p − 2, so γ ≤ b_{1} + s + 2−r+(p−1)+p(p−2) ≤ (p−2)+2−1+(p−1)+p(p−2) < p^{2}−1, which gives a contradiction. So we may assume that b_{2} ≠ s. Then 2k_{2} ≤ b_{2}+s−1 and γ ≤ (p−3)+2+(p−2)+(p−1)+p(p−3) < p^{2}−1, which is also a contradiction.

We now prove the second assertion of Theorem 3. Given b, c ∈ I and n ≤ 0, we aim to show that the space Hom(Z_{b}, Z_{c}[n]) is zero unless b = c and n = 0, in which case it is one-dimensional. Recall that if M and N are B-modules then

Hom_{D^{b}(mod(B))}(M[i], N[j]) = \begin{cases}0 & if j < i \\ Ext_{B}^{j−i}(M, N) & if j ≥ i. \end{cases}

If b ∈ I_{<} and c ∈ I_{<}, then Z_{b} is concentrated in degree zero and Z_{c}[n] is concentrated in degree −n. Hence if n < 0 then Hom(Z_{b}, Z_{c}[n]) = 0, and if n = 0 we may appeal to Lemma 9.1.

If b ∈ I_{p−1} and c ∈ I_{<}, then Z_{b} is concentrated in some negative degree, while Z_{c}[n] is concentrated in degree −n ≥ 0.

Suppose b ∈ I_{≥} and c ∈ I_{<}. We may assume that b_{2} ≤ b_{1} for if not we can apply the Frobenius automorphism σ. By Lemma 7, there is a distinguished triangle

U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶ U_{w+2} ⊗ M_{p−1,w}[h] ⟶ Z_{b} ⟶,

where h > 0. Applying the functor Hom(−, M_{c}) to this triangle gives rise to a long exact sequence, a segment of which is

Hom(U_{p} ⊗ M_{p−2−b_{1},b_{2}}, M_{c}[n − 1]) ⟶ Hom(Z_{b}, M_{c}[n])

⟶ Hom(U_{w+2} ⊗ M_{p−1,w}[h], M_{c}[n]).

Because n − 1 < 0 and n < h, the first and third terms of this segment are zero; thus the second term is zero as well.

If b ∈ I_{<} and c ∈ I_{p−1}, then applying σ if necessary, we may assume that c_{1} = p − 1 and then use Lemma 9.5.

Suppose b ∈ I_{p−1} and c ∈ I_{p−1}. We may assume that b_{1} = p − 1. If c_{1} = p − 1 we use Lemma 9.3 and if c_{2} = p − 1 we use Lemma 9.4.

Suppose b ∈ I_{≥} and c ∈ I_{p−1}. Applying σ if necessary, we may assume that b_{2} ≤ b_{1}. Suppose in addition that c_{1} = p − 1, so that Z_{c} = U_{s+2} ⊗ M_{p−1,s}[s + 2], for some 0 ≤ s ≤ (p − 3)/2. Let w = min{b_{2}, p−2−b_{2}} and h = w +b_{1} −p+3. Note that 0 ≤ w ≤ (p−3)/2. We divide the argument into three cases:

• s + 2 + n < h: By Lemma 7 we have a distinguished triangle

U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶ U_{w+2} ⊗ M_{p−1,w}[h] ⟶ Z_{b} ⟶ .

Applying the functor Hom(−, U_{s+2} ⊗M_{p−1,s}[s+2]) to this triangle gives rise to a long exact sequence, a segment of which is

Hom(U_{p} ⊗ M_{p−2−b_{1},b_{2}}, U_{s+2} ⊗ M_{p−1,s}[s + 2 + n − 1])

⟶ Hom(Z_{b}, U_{s+2} ⊗ M_{p−1,s}[s + 2 + n]) ⟶ Hom(U_{w+2} ⊗ M_{p−1,w}[h], U_{s+2} ⊗ M_{p−1,s}[s + 2 + n]).

The first term of this segment is zero by Lemma 9.8 because s+2+ n−1 ≤ min{h, s+2}, and the third term is zero since s+2+n < h; thus the second term is zero as well.

• s + 2 + n = h: Arguing as in the previous case we get an exact sequence

Hom(U_{p} ⊗ M_{p−2−b_{1},b_{2}}, U_{s+2} ⊗ M_{p−1,s}[h − 1])

⟶ Hom(Z_{b}, U_{s+2} ⊗ M_{p−1,s}[h])

⟶ Hom(U_{w+2} ⊗ M_{p−1,w}[h], U_{s+2} ⊗ M_{p−1,s}[h])
⟶ Hom(U_{p} ⊗ M_{p−2−b_{1},b_{2}}, U_{s+2} ⊗ M_{p−1,s}[h])

⟶ Hom(Z_{b}, U_{s+2} ⊗ M_{p−1,s}[h + 1]).

By Lemma 9.8, the first term is zero and the fourth term is zero unless s = w, in which case it is one-dimensional. By Lemma 9.3, the third term is zero unless s = w, in which case it is onedimensional. The fifth term is zero by the previous case. We conclude that the second term is zero, as desired.

• s + 2 + n > h: By Lemma 7 we have a distinguished triangle

Y ⟶ M_{b} ⟶ Z_{b} ⟶,

where Y is a bounded complex of projective modules such that Y^{−i} = 0 for i ≥ h. Applying the functor Hom(−, U_{s+2}⊗M_{p−1,s}[s+ 2]) to this triangle gives rise to a long exact sequence, a segment of which is

Hom(Y, U_{s+2} ⊗ M_{p−1,s}[s + 2 + n − 1])

⟶ Hom(Z_{b}, U_{s+2} ⊗ M_{p−1,s}[s + 2 + n]) ⟶ Hom(M_{b}, U_{s+2} ⊗ M_{p−1,s}[s + 2 + n]).

The first term is zero because Y is a bounded complex of projective modules, U_{s+2} ⊗ M_{p−1,s}[s + 2 + n − 1] is concentrated in degree −(s+2+n−1), and Y^{−(s+2+n−1)} = 0. Since h < s+2+n ≤ s+2, the third term is zero as well, by Lemma 9.6. Thus we conclude that the second term is zero.

If instead c_{2} = p − 1, then an analogous argument works, using parts 9, 4, and 7 of Lemma 9 in place of 8, 3, and 6.

We are now left with the case c ∈ I_{≥}. As before, we may assume that c_{2} ≤ c_{1}. By Lemma 7, we have a distinguished triangle

U_{p} ⊗ M_{p−2−c_{1},c_{2}} ⟶ U_{w+2} ⊗ M_{p−1,w}[h] ⟶ Z_{c} ⟶,

where w = min{c_{2}, p − 2 − c_{2}} and h = w + c_{1} − p + 3. Note that 0 ≤ w ≤ (p − 3)/2 and 0 < h ≤ w + 1.

Suppose first that b ∈ I_{<}. Applying the functor Hom(M_{b}, −) to the triangle above gives rise to a long exact sequence, a segment of which is

Hom(M_{b}, U_{w+2} ⊗ M_{p−1,w}[h + n]) ⟶ Hom(M_{b}, Z_{c}[n])

⟶ Hom(M_{b}, U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n + 1]).

The first term is zero by Lemma 9.5 and the third term is zero by Lemma 9.2; hence the second term is zero, as desired.

Now suppose that b ∈ I_{p−1}. Suppose further that b_{1} = p − 1, so we have that Z_{b} = U_{s+2} ⊗ M_{p−1,s}[s + 2] for some 0 ≤ s ≤ (p − 3)/2. Applying the functor Hom(U_{s+2}⊗M_{p−1,s}, −) to the triangle above gives rise to a long exact sequence, a segment of which is

Hom(U_{s+2} ⊗ M_{p−1,s}[s + 2], U_{w+2} ⊗ M_{p−1,w}[h + n])

⟶ Hom(U_{s+2} ⊗ M_{p−1,s}[s + 2], Z_{c}[n]) ⟶ Hom(U_{s+2} ⊗ M_{p−1,s}[s + 2], U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n + 1]).

The first term is zero by Lemma 9.3 because h+n−(s+2) ≤ (w +1)+ n−(s+2) ≤ w−s, and the third term is zero because n+1 < t+2; thus the second term is zero, as desired. If instead b_{2} = p − 1, an analogous argument which uses Lemma 9.4 works.

Finally suppose that b ∈ I_{≥}. Applying the functor Hom(Z_{b}, −) to the triangle above gives rise to a long exact sequence, a segment of which is

Hom(Z_{b}, U_{w+2} ⊗ M_{p−1,w}[h + n]) ⟶ Hom(Z_{b}, Z_{c}[n])

⟶ Hom(Z_{b}, U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n + 1]) ⟶ Hom(Z_{b}, U_{w+2} ⊗ M_{p−1,w}[h + n + 1]).

The last term may be rewritten as

Hom(Z_{b}, U_{w+2} ⊗ M_{p−1,w}[w + 2][h − (w + 1) + n]),

which we see is zero by noting that h − (w + 1) + n ≤ 0 and applying a previous case (b ∈ I_{≥} and c ∈ I_{p−1}). The first term is zero by a similar argument. Hence it suffices to show that Hom(Z_{b}, U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n + 1]) = 0 for all n ≤ 0 unless n = 0 and b = c, in which case it is one-dimensional. Suppose now that b_{2} ≤ b_{1}. By Lemma 7, there is a distinguished triangle

U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶ U_{w+2} ⊗ M_{p−1,w}[h] ⟶ Z_{b} ⟶

where w = min{b_{2}, p − 2 − b_{2}} and h = w + b_{1} − p + 3. Applying the functor Hom(−, U_{p} ⊗ M_{p−2−c_{1},c_{2}}) to this triangle gives rise to a long exact sequence, a segment of which is

Hom(U_{w+2} ⊗ M_{p−1,w}[h], U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n])

⟶ Hom(U_{p} ⊗ M_{p−2−b_{1},b_{2}}, U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n])

⟶ Hom(Z_{b}, U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n + 1])

⟶ Hom(U_{w+2} ⊗ M_{p−1,w}[h], U_{p} ⊗ M_{p−2−c_{1},c_{2}}[n + 1]).

If n < 0, then, remembering that h > 0, it is clear that the second and fourth terms are zero, and thus that the third term is zero, as desired. Finally if n = 0, then the first term is clearly zero, the fourth term is zero by Lemma 9.10, and by Lemma 9.12 the second term is zero unless b = c, in which case it is one-dimensional. It follows as desired that the third term is zero unless b = c, in which case it is one-dimensional.

If instead b_{1} < b_{2}, a similar argument using parts 11 and 13 of Lemma 9 works.

4. generation

Our final task is to show that the last statement of Theorem 3 holds. We take Z to be the full triangulated subcategory of D^{b}(mod(B)) generated by the complexes Z_{b}.

Lemma 10. Z contains the following modules:

1. M_{b}, whenever b ∈ I_{<};

2. U_{p} ⊗ M_{p−2−b_{1},b_{2}}, whenever b ∈ I_{≥} and b_{2} ≤ b_{1};

3. U_{1} ⊗ M_{b_{1},p−2−b_{2}}, whenever b ∈ I_{≥} and b_{1} ≤ b_{2};

4. U_{p+1} ⊗ M_{s,s}, whenever 0 ≤ s ≤ (p − 3)/2.

Proof. If b ∈ I_{<}, then M_{b} = Z_{b} is in Z, so part (1) is proved. If b ∈ I_{≥} and b_{2} ≤ b_{1}, then by Lemma 7 there is a distinguished triangle

U_{p} ⊗ M_{p−2−b_{1},b_{2}} ⟶ U_{w+2} ⊗ M_{p−1,w}[h] ⟶ Z_{b} ⟶,

where 0 ≤ w ≤ (p − 3)/2. The last term is in Z and the second term is as well, being a translate of Z_{c} for some c ∈ I_{p−1}; hence the first term is in Z, which proves part (2). Part (3) is proved similarly: apply σ to the triangle above.

Lastly we prove part (4). Applying part (2) with b_{1} = b_{2} = p − 2 − s, we have that U_{p} ⊗ M_{s,p−2−s} is in Z. Next, we use Lemma 6 with b_{1} = s and b_{2} = p − 2 − s. Applying σ to the resulting exact sequence and then tensoring with U_{p} we get a distinguished triangle

U_{p+1} ⊗ M_{s,s} ⟶ U_{p(s+2)} ⊗ M_{s,p−1} ⟶ U_{p} ⊗ M_{p−2−s,s} ⟶ .

We have just seen that the last term is in Z, and the second term is in Z because it is a translate of Z_{c} for some c ∈ I_{p−1}; hence the first term is in Z, as desired.

Let J be the set of pairs d = (d_{1}, d_{2}) of integers satisfying the following conditions:

1. −(p − 1) < d_{1} + d_{2} ≤ p − 1;

2. d_{1} + d_{2} is even;

3. d_{1} and d_{2} are either both non-negative or both non-positive.

It is easy to see that for any even integer j there exists d ∈ J such that j ≡ d_{1} + pd_{2} modulo p^{2} − 1; this implies that any simple B-module is isomorphic to U_{d_{1}+pd_{2}} for some d ∈ J.

For d ∈ J, let f(d) = min{|d_{1}|, |d_{2}|}. Define a partial order ⪯ on J as follows: if d = (d_{1}, d_{2}) and d′ = (d′_{1}, d′_{2}) are distinct elements of J, then set d ≺ d′ if and only if one of the following conditions is met:

1. f(d) < f(d′);

2. f(d) = f(d′) and |d_{1} + d_{2}| < |d′_{1} + d′_{2}|;

3. f(d) = f(d′) and |d_{1} + d_{2}| = |d′_{1} + d′_{2}| and d_{1} and d_{2} are both non-negative.

To prove that the complexes Z_{b} generate D^{b}(mod(B)) as a triangulated category it suffices to show that every simple B-module is in Z. We shall do this by proving that U_{d_{1}+pd_{2}} is in Z for each d ∈ J, inducting on the partial order ⪯. The only element of J minimal with respect to ⪯ is (0, 0), and U_{0+p·0} = U_{0} = M_{(0,0)} is in Z, by Lemma 10.1, so we may assume that d ∈ J and d ≠ (0, 0). The argument divides into four cases:

• d_{1}, d_{2} < 0: We have (−d_{1}, −d_{2}) ∈ I_{<} so by Lemma 10.1, Z contains M_{−d_{1},−d_{2}}. The composition factors of M_{−d_{1},−d_{2}} consist of the simple modules U_{m_{1}+pm_{2}}, where (m_{1}, m_{2}) runs over the set Ω of pairs (m_{1}, m_{2}) such that d_{1} ≤ m_{1} ≤ −d_{1} and d_{2} ≤ m_{2} ≤ −d_{2} and m_{1} + m_{2} is even. Now (d_{1}, d_{2}) ∈ Ω, so U_{d_{1}+pd_{2}} is a composition factor of M_{−d_{1},−d_{2}}. We will now show that every other composition factor of M_{−d_{1},−d_{2}} is in Z; it will follow that U_{d_{1}+pd_{2}} is in Z. To that end, let m ∈ Ω with m ≠ (d_{1}, d_{2}). We aim to find a m′ = (m′_{1}, m′_{2}) in J such that m′_{1} + pm′_{2} = m_{1} + pm_{2} and m′ ≺ d for then by induction we will have that U_{m_{1}+pm_{2}} is in Z.

If m_{1} and m_{2} are either both non-negative or both non-positive, then m ∈ J. Moreover f(m) ≤ f(d) and |m_{1} + m_{2}| ≤ |d_{1} + d_{2}| with equalities occurring only if m_{1} = −d_{1} and m_{2} = −d_{2}. Hence m ≺ d.

If m_{1} is positive, m_{2} is negative, and |m_{1}| > |m_{2}|, let m′_{1} = m_{1} −p and m′_{2} = m_{2} +1. Then m′_{1} +pm′_{2} = m_{1} +pm_{2}; in addition m′_{1} ≤ 0, m′_{2} ≤ 0, and m′_{1} + m′_{2} = m_{1} + m_{2} − (p − 1) > −(p − 1), so m′ = (m′_{1}, m′_{2}) ∈ J. Furthermore f(m′) ≤ |m_{2} + 1| < |m_{2}| ≤ f(d), so m′ ≺ d.

If m_{1} is positive, m_{2} is negative, and |m_{1}| ≤ |m_{2}|, let m′_{1} = m_{1} −1 and m′_{2} = m_{2} +p. Then m′_{1} +pm′_{2} = m_{1} +pm_{2}; in addition m′_{1} ≥ 0, m′_{2} ≥ 0, and m′_{1} + m′_{2} = m_{1} + m_{2} + p − 1 ≤ p − 1, so m′ = (m′_{1}, m′_{2}) ∈ J. Furthermore f(m′) ≤ |m_{1} − 1| < |m_{1}| ≤ f(d), so m′ ≺ d.

If m_{1} is negative and m_{2} is positive, we split the argument into the cases |m_{1}| ≥ |m_{2}| and |m_{1}| < |m_{2}|. These may be handled as above.

• d_{1}, d_{2} ≥ 0 and d_{1} = d_{2} > 0: Let s = d_{1} − 1. Then d_{1} + d_{2} ≤ p − 1 implies that 0 ≤ s ≤ (p − 3)/2. By Lemma 10.4, Z contains U_{p+1} ⊗ M_{s,s}. The composition factors of U_{p+1} ⊗ M_{s,s} consist of the simple modules U_{m_{1}+pm_{2}}, where (m_{1}, m_{2}) runs over the set Ω of pairs (m_{1}, m_{2}) such that −s + 1 ≤ m_{i} ≤ s + 1 and m_{1} + m_{2} is even. Thus U_{d_{1}+pd_{2}} is a composition factor of U_{p+1} ⊗ M_{s,s}, and by an argument similar to that for the previous case, one can show that every other composition factor is in Z.

• d_{1}, d_{2} ≥ 0 and d_{1} < d_{2}: By Lemma 10.2 with b_{1} = p − 2 − d_{1} and b_{2} = d_{2} − 1, we have that U_{p} ⊗ M_{d_{1},d_{2}−1} is in Z. One can show, as above, that U_{m_{1}+pm_{2}} is a composition factor of this module while any other composition factor is in Z.

• d_{1}, d_{2} ≥ 0 and d_{1} > d_{2}: Using Lemma 10.3 one can show that Z contains U_{1} ⊗ M_{d_{1}−1,d_{2}} and then argue as in previous cases.

acknowledgments

I am grateful to Jeremy Rickard for communicating to me his new theorem and to Raphaël Rouquier for explaining to me the relevance of Carlson’s paper. I would also like to thank the National Science Foundation for supporting my graduate studies at the University of Chicago in the summer of 1998, when a portion of this work was done.

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St. John’s College, Oxford OX1 3JP, UK

E-mail address: chuang@maths.ox.ac.uk