COMBINATORIAL PARTIAL TILTING COMPLEXES FOR THE BRAUER STAR ALGEBRAS

Mary Schaps and Evelyne Zakay-Illouz*

Abstract. In modular representation theory, the Brauer star algebras play a very special role, being the “local” blocks corresponding under the Brauer correspondence to the “global” blocks of a group ring of cyclic defect group. Every block of cyclic defect group can be obtained by tilting from a Brauer star algebra.

Among the various tilting complexes for the Brauer star algeras, we distinguish a subclass, which we will call two-restricted tilting complexes, of a special combinatorial character. In this paper we determine necessary conditions for a two-restricted complex to be a partial tilting complex for the Brauer star algebra.

§1. Introduction.

This paper is intended to be the first of three based on the doctoral dissertation of the second author. Since we cannot refer most readers to the original dissertation, which is in Hebrew, we intend to give a fairly detailed presentation in the published version.

The general subject is tilting complexes and deformations for blocks of cyclic defect group. The work was motivated by two problems: First, to find a Donald-Flanigan deformation of blocks of cyclic defect which would be more natural than that given in [S2]. Second, to try to understand the combinatorics of the Brauer tree in terms of the tilting theory. Rickard had already shown that a connection existed in [R2], where he constructed a tilting complex from a block of the “global” Brauer tree to the Brauer star. We chose to work in the opposite direction; to find tilting complexes for the Brauer star algebra from which one can “read off” the global Brauer tree directly.

Typeset by AMS-TEX 2 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

*Work done for a doctoral dissertation at Bar-Ilan University, and partially supported by the Bar-Ilan Research Authority.

Since not every tree is the Brauer tree of a block of a group ring, our results are not restricted to blocks of a group ring, and in fact, we will work with the larger class of algebras called Brauer tree algebras, over fields of arbitrary characteristic. However, the main application is to modular representation theory of finite groups. This paper corresponds to Chapters III and IV of the thesis, classifying two-restricted partial tilting complexes. The second paper, corresponding to Chapter V, will demonstrate a one-to-one correspondence between Brauer trees with an additional structure, called a pointing, and equivalence classes of two-restricted tilting complexes modulo cyclic rotation of the Brauer star. The third paper, Chapter VI of the thesis, will give a “natural” deformation based on tilting theory.

Let e, m be natural numbers, with e > 1. Let K be any field containing a primitive e-th root of unity ξ. Let n̂ = em + 1. Then the cyclic group C_{e} of order e acts on the truncated polynomial ring A = K[x] /xn̂ by letting the generator d of C_{e} send x to ξx.

We now construct the skew group ring

b = A[C_{e}],

where the algebra generation x of A and the group generator d of C_{e} commute according to the relation

xd = ξdx

or

d^{−1}xd = ξx.

The most important case of this construction occurs when K is a field of characteristic p, e is a divisor of p − 1, and m = (p^{c} − 1) /e. In this case

b →^{∼} K[C_{p^{c}} ⋊ C_{e}].

COMBINATORIAL PARTIAL TILTING COMPLEXES 3

The generator d of C_{e} induces an action on C_{p^{c}} = ⟨a⟩ given by d^{−1}ad = a^{r}, where r is an integer which is a primitive root of unity modulo e. The isomorphism between this case and the previous, more general case, is established by setting

x = \frac{1}{e} ∑_{j=0}^{e−1} r^{−j}a^{r^{j}}.

Then d^{−1}xd = rx, as desired.

This algebra b has been intensively studied in the context of modular representation theory. For a discussion of the quiver of b in a more general context, see [MS] or [SSS]. Alternatively, one can see b as a special case of the Brauer tree algebra, which we will discuss below.

Returning to the situation where K has an arbitrary characteristic, we set A = K[x]/xn̂ as before, and define idempotents of b = A[C_{e}] as

f_{i} = ∑_{j=0}^{e−1} ξ^{−ij}d^{j}, i = 1, ..., e.

This is a complete set of primitive orthogonal idempotents of b, and thus the indecomposable projectives are

P_{i} = bf_{i}, i = 1, ..., e.

We will denote the identity map on P_{i} by id_{i} . Lemma 1.1. The set {x^{s}f_{i} | s = 0, …, n̂ − 1; i = 1, …, e} is a basis of eigenvectors, with eigenvalues ξ^{s}, for the action of d. It satisfies

x^{s}f_{i} = f_{k}x^{s}, where k ≡ i − s (mod e).

Proof. Since {1, x, ..., x^{n−1}} is a basis for A, and {1, ..., d^{e−1}} are the elements of C_{e}, a basis for the skew group algebra A[C_{e}] is given by {x^{s}d^{i} | s = 0 …, n̂ − 1; i = 1, …, e}. We can make a change of basis for each s to {x^{s}f_{i} | s = 0, …, n̂ − 1, i = 1; …, e}.

4 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Now,

x^{s}f_{i} = x^{s} (\frac{1}{e} ∑_{j=0}^{e−1} ξ^{−ij}d^{j})

= (\frac{1}{e} ∑_{j=0}^{e−1} ξ^{−ij}ξ^{sj}d^{j}) x^{s}

= f_{k}x^{s}, where k ≡ i − s (mod e).

Definition. The basis {x^{s}f_{i}} will be called the normalized basis. The natural grading of K[x] /(x^{n}) induces a grading of A[C_{e}] in which the space of elements of degree s is spanned by x^{s}f_{1}, ..., x^{s}f_{e}. The unique b-homomorphism from P_{k} to P_{i} sending f_{k} to x^{s}f_{i} is said to be normal homogeneous of degree s. Corollary. If k = i, then there are m + 1 normal homogeneous maps q: P_{k} → P_{i}, for s = 0, e, 2e, ..., me. If k ≠ i, and {i − k}_{e} is the residue mod e, then there are m normal homogeneous maps, for s = {i − k}_{e} + ℓe, ℓ = 0, ..., m − 1.

§2. Brauer trees and tilting modules. Definition. Let T be any tree, together with a cyclic orientation on the edges adjacent to each vertex. Assume that T has e edges, and that there is one designated vertex, called the exceptional vertex which is assigned a multiplicity m ≥ 1. T is called a Brauer tree of type (e, m).

To any field k and any Brauer tree T, we can associate a k-algebra A_{T} called the Brauer tree algebra. We describe the algebra A_{T} in terms of composition series.

For each edge f_{r_{1}} and each vertex v of the edge which is not a non-exceptional terminal vertex, we define the standard circuit at v denoted by S(v, f_{r_{1}}) to be the sequence of edges in the cyclic ordering around v either one time or, if v is exceptional, m times.

We describe the composition series of the projective modules P_{i}, i = 1, …, e. The top and socle of P_{i} are the same simple module S_{i}. Rad(P_{i}) / Soc(P_{i}) is the

COMBINATORIAL PARTIAL TILTING COMPLEXES 5

direct sum of either one or two uniserial components, whose composition factors are the simples corresponding to the edges encountered in a standard circuit of the vertices at either end of the edge. If one of those vertices is a non-exceptional terminal vertex, then the corresponding factor is empty. A detailed description in the case of blocks of a group ring is given in [A].

In [R2], Rickard proved that any two blocks of a group ring with the same Brauer tree are derived equivalent. By the various reductions in [R1], this means that there is a complex Q^{·} of projective A-modules such that B = End_{A}(Q), where Q^{·} is a tilting complex according to the following definition. The notation Q^{·}[n] denotes shifting the complex n places to the left.

Definition. Let R be a Noetherian ring. A bounded complex of finitely-generated projective R-modules is called a tilting complex if

(1) Hom_{D^{b}(R)}(Q^{·}, Q^{·}[n]) = 0 whenever n ≠ 0.

(2) For any indecomposable projective P, define the stalk complex to be the

complex

P^{·} : 0 → P → 0.

Then every such P^{·} is in the triangulated category generated by the direct summands of direct sums of Q^{·}.

Q^{·} satisfying (1) and (2) is called a tilting complex for R. If it satisfies (1), it is called a partial tilting complex.

Rickard proved his result [R2] on derived equivalence of blocks of a group ring by showing that every block of a group ring which is a Brauer tree algebra can be tilted to an algebra Morita equivalent to the Brauer tree algebra of a star, i.e., to an algebra like the b defined above. Membrillo proved a similar result with regard to the generalized Brauer tree algebras, in which each vertex is allowed arbitrary multiplicity [M].

6 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Our interest in the subject of Brauer tree algebras came from deformation theory. We wanted to transfer the very easily described deformation of b given in [S2] to an arbitrary Brauer star algebra B, using the Rickard result [R3] on deformations of tilting complexes. Where Rickard and Membrillo were passing from B to b, we need to go from b to B. Furthermore, we wanted to get deformations which would be “homogeneous” with regard to the deformation parameters in a sense which will be made explicit in the sequel [SZ2]. Thus we restricted ourselves to tilting complexes of a particularly combinatorial form.

Definition. A complex of projectives over b = A[C_{e}] will be called two-restricted if it is a direct sum of shifts of complexes involving no more than two indecomposable projectives of the form

S_{r} :  0 → P_{r} → 0
T_{rt}:  0 → P_{r} →^{h} P_{t} → 0  r ≠ t.

where the map h is homogeneous of minimal degree. The complexes S_{r}[n], T_{rt}[n] will be called elementary.

Note. One can show that any indecomposable partial tilting complex containing at most two terms is elementary, and Theorem 2 below will establish that the elementary complexes are all indecomposable.

We recall some generalities about tilting complexes from [R1]. If A has e simple modules, and Q^{·} is a tilting complex over A, then Q^{·} is homotopic to a complex which is the direct sum of indecomposable complexes from exactly e distinct isomorphism classes. A partial tilting complex has no more than e distinct elementary complexes. If we take a tilting complex Q^{·} over A, and replace it by a complex Q^{′}˙ with the same distinct indecomposables but a different number of copies of each, we get an algebra B′ = End_{A}(Q^{′}˙ ) which is Morita equivalent to B = End_{A}(Q^{·}).

COMBINATORIAL PARTIAL TILTING COMPLEXES 7

We will generally work with tilting complexes in which there is a single copy of each indecomposable, so that the tilted algebra B will be basic.

An indecomposable complex can be recognized because its endomorphism ring is local. It will be a corollary of the results in this paper that the complexes S_{r} and T_{rt} are indecomposable. Thus if we can find a sum of e complexes S_{r} and T_{rt} which form a partial tilting complex, it will be a tilting complex. There are many other indecomposable complexes from which tilting complexes could be constructed; for example P_{r} →^{h} P_{t} →^{s} P_{t} →^{s} ... →^{s} P_{t}, where the map s takes the top of P_{t} to the socle of P_{t}. We concentrate on the complexes S_{r} and T_{rt} because they are particularly “combinatorial” in the following sense: they provide a minimal encoding of the information contained in an edge of the Brauer graph.

Notation. The set of all two-restricted partial tilting complexes will be denoted by P T C_{2}, and the subset of two-restricted tilting complexes by T C_{2}.

In this first paper, we will classify the elements of P T C_{2}, and determine the homomorphisms between their indecomposable components. In the second paper, we will show that there is a one-to-one correspondence between elements of T C_{2} modulo rotation of b and Brauer trees with an additional structure we will call a pointing. This will show that T C_{2} is adequate to produce all Brauer trees, but also will show that there are generally many different elements of T C_{2} which will produce a given Brauer tree. Finally, in the third paper, we will derive the existence of many different representations of A_{T} as a graded algebra, all equally “natural”, and all giving the same deformation.

§3. Statement of results.

In order to state the results in full, we need to introduce one more concept, that of a short sequence of indices.

Definition. Let b be the Brauer star algebra K[x]/(x^{em+1})[C_{e}], as above. We let {x^{s}f_{i}}_{s=0}^{em} be a basis for P_{i}, and define the following maps

ε_{i} : P_{i} → P_{i}  ε_{i}(f_{i}) = x^{e}f_{i}
h̃_{ij} : P_{i} → P_{j}  h̃_{ij}(f_{i}) = x^{k}f_{i},  k ≡ j − i (mod e) 0 ≤ k < e.

For i ≠ j, we will denote h̃_{ij} by h_{ij}, and for i = j by id_{i} . The homomorphism ε_{i}^{m} : P_{i} → P_{i} will be called the socle map, for the obvious reason that it maps the top of P_{i} into its socle ⟨x^{em}f_{i}⟩. Definition. Consider a sequence {r_{i}}_{i=1}^{ℓ} of elements of {1, ..., e}. Set

h = h̃_{r_{ℓ−1}r_{ℓ}} ∘ ... ∘ h̃_{r_{1}r_{2}} = ε_{r_{ℓ}}^{α} h̃_{r_{1}r_{ℓ}}.

Then the sequence is short if α = 0 and long if α > 0. We generally represent the sequence in the form r_{1} → r_{2} → ... → r_{ℓ}. Examples.

r → s → r is long
r → r → s is short
1 → 2 → 3 is short
1 → 3 → 2 is long.

We gather together the following facts about short and long sequences.

(1) Cyclic permutation property: If the sequence r_{1} → r_{2} → ... → r_{ℓ} is short (or long), then the cyclic permutation

r_{i} → r_{i+1} → ... → r_{ℓ} → r_{1} → ... → r_{i−1}

is short (or long), respectively.

(2) Restriction property: If r_{1} → ... → r_{ℓ} is short, so is every subsequence.

COMBINATORIAL PARTIAL TILTING COMPLEXES 9

(3) Mixing property: If all the r_{i} are distinct, then up to cyclic permutation there is a unique permutation σ of indices such that r_{σ(1)} → r_{σ(2)} → ...r_{σ(ℓ)} is short.

(4) Refinement property: If r_{1} → ... → r_{ℓ} is short, and α^{1}, ..., α^{ℓ−1} are sequences such that for k = 1, ..., ℓ − 1

r_{k} → α^{k} → r_{k+1}

is short, then

r_{1} → α^{1} → r_{2} → α^{2} → ... → r_{ℓ}

is short.

Proof. The proofs of (1), (2) and (4) are simply repeated applications of the associability law for the maps h̃_{r_{i}r_{i+1}}. The proof of (3) is by induction, using the fact that if r → s → t is short, then r → t → s is long.

The first theorem will describe which pairs of indecomposable elements of P T C_{2} can occur together in a partial tilting complex.

Theorem 1. If Q^{·} is in P T C_{2}, then:

(a) All elementary components of Q^{·} of the form S_{r}[n] must have their non-zero term in the same degree, which we will assume to be zero.

(b) If two elementary components of Q^{·} have a common term P_{r}, it must occur with the same degree in each.

(c) If T_{rs}[n] and S_{k} occur together in Q^{·} then r → k → s is long.

(d) If T_{rs}[n] and T_{tu}[n′] occur in Q^{·}, with r, s, t, u all distinct, then one of the following sequences must be short.

(1) r → s → t → u

(2) r → t → u → s

10 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

(3) r → s → u → t. Remark. We divide the circle into e segments with end points numbered counterclockwise from 1 to e, and let \overline{rs} and \overline{tu} be directed counterclockwise segments. Then (1) represents the case where the ones \overline{rs} and \overline{tu} are disjoint with compatible orientations, (2) the case where I_{u} is included in \overline{rs}, and (3) the case \overline{rs} included in \overline{tu}.

The three excluded cases represent partial intersections

r → t → s → u
r → u → s → t
r → u → t → s.

This approach in terms of segments of a circle originated with [KZ1]. The actual problem treated by König and Zimmermann is different, and the two term sequences represent projective resolution in a hereditary order. Also, in their case all the tilting complexes are two-restricted, which is not the case for Brauer tree algebras. However, the combinatorics which were developed independently, turned out virtually equivalent. The case of Green orders treated by Zimmermann in [KZ2] presumably provides a link, since it uses methods similar to [KZ1] and reduces, modulo the prime, to the case of Brauer tree algebras.

The proof of Theorem 1 will depend on Propositions 1 - 5 in the next section. In these propositions, we will determine all homogeneous chain maps between indecomposable complexes which are not homotopic to zero. This information about the non-trivial chain maps will also be important in the sequel for constructing the endomorphism ring of a tilting complex Q^{·} in T C_{2}, so we will summarize in Theorem 2. However, since the statement of Theorem 2 will also include information about the degree of these chain maps, we first pause to show that this is a well-defined concept for chain maps in T C_{2}.

COMBINATORIAL PARTIAL TILTING COMPLEXES 11

Definition. A chain map ℓ^{·} between C^{·} and D^{·} is called normal homogeneous if each vertical map is normal homogeneous. Lemma 3.1. Any chain map ℓ^{·} between irreducible elements of P T C_{2} is a linear combination of normal homogeneous chain maps. Proof. If there is only one non-zero vertical arrow, this is obvious. Thus we need only consider the case

0 ⟶ P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
↓ ^{ℓ_{1}}  ↓ ^{ℓ_{2}}
0 ⟶ P_{t} ⟶^{h_{tu}} P_{u} ⟶ 0.

Set

ℓ_{1} = ( ∑_{i=0}^{m−1} a_{i} ε_{t}^{i} h̃_{rt}) + δ_{rt} a_{m} ε_{t}^{m} h̃_{rt}
ℓ_{2} = ( ∑_{i=0}^{m−1} b_{i} ε_{u}^{i} h̃_{su}) + δ_{su} b_{m} ε_{u}^{m} h̃_{su}.

Since ℓ^{·} is a chain map, the diagram is commutative, and thus

(∗)  ℓ_{2}h_{rs} = h_{tu}ℓ_{1}.

Define α and β by

h_{tu} h̃_{rt} = ε_{u}^{α} h̃_{ru}
h̃_{su} h_{rs} = ε_{u}^{β} h̃_{ru}.

We compute the two compositions in (∗):

h_{tu}ℓ_{1} = h_{tu} ( ∑_{i=0}^{m−1} a_{i} ε_{t}^{i} h̃_{rt} + δ_{rt} a_{m} ε_{t}^{m} h̃_{rt})

= ∑_{i=0}^{m−1} a_{i} ε_{u}^{i+α} h̃_{ru} + δ_{rt} δ_{ru} a_{m} ε_{u}^{m+α}

= ∑_{i=0}^{m−1} a_{i} ε_{u}^{i+α} h̃_{ru} since δ_{rt} δ_{ru} = 0 because t ≠ u.

12 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Similarly, ℓ_{2} h_{rs} = ∑_{j=0}^{m−1} b_{i} ε_{u}^{j+β} h̃_{ru}. Set a_{−1} = b_{−1} = 0. Then for 0 ≤ k ≤

m − 1 + δ_{ru}, we get

a_{k−α} = b_{k−β}.

Dividing into cases according to α_{1} and β_{1}, we get the following decompositions of ℓ^{·} = (ℓ_{1}, ℓ_{2}) :

\underline{α = 0, β = 0}: (ℓ_{1}, ℓ_{2}) = ∑_{i=0}^{m−1} a_{i} (ε_{t}^{i}h̃_{rt}, ε_{u}^{i} h̃_{su}) + δ_{rt}a_{m} (ε_{r}^{m}, 0) + δ_{su}b_{m}(0, ε_{s}^{m})

\underline{α = 0, β = 1}: (ℓ_{1}, ℓ_{2}) = ∑_{i=0}^{m−1} a_{i} (ε_{t}^{i}h̃_{rt}, ε_{u}^{i−1}h̃_{su}) + δ_{rt}a_{m} (ε_{r}^{m}, 0) + b_{m−1} (0, ε_{u}^{m−1}h_{su})

\underline{α = 1, β = 0}: (ℓ_{1}, ℓ_{2}) = ∑_{j=0}^{m−1} b_{j} (ε_{t}^{j−1}h̃_{rt}, ε_{u}^{j}h̃_{su}) + a_{m−1} (ε_{t}^{m−1}h_{rt}, 0) + δ_{su}b_{m} (0, ε_{s}^{m}))

\underline{α = 1, β = 1}: In this case, if r = u, then a_{m−1} = b_{m−1}. Thus,

If r ≠ u, (ℓ_{1}, ℓ_{2}) = ∑_{i=1}^{m−2} b_{i} (ε_{t}^{i}h_{rt}, ε_{u}^{i}h_{su}) + a_{m−1} (ε_{t}^{m−1}h_{rt}, 0) + b_{m−1} (0, ε_{u}^{m−1}h_{su})

If r = u, (ℓ_{1}, ℓ_{2}) = ∑_{i=1}^{m−1} a_{i} (ε_{t}^{i}h_{rt}, ε_{u}^{i}h_{su}) .

□

This lemma means that if there is any non-zero homomorphism between indecomposable complexes, then there is a normal homogeneous non-zero homomorphism, so it suffices to study the normal homogeneous chain maps, of which there are a finite number.

Since we have reduced ourselves to the study of normal homogeneous chain maps, it will also be useful to know that if such a map is homotopic to zero, then homotopy can also be chosen with the same property.

Lemma 3.2. If a normal homogeneous chain map between two indecomposable elements of P T C_{2} is homotopic to zero, then we may choose the homotopy to be normal homogeneous.

COMBINATORIAL PARTIAL TILTING COMPLEXES 13

Proof. Let C^{·} and D^{·} be indecomposable elements of P T C_{2}. If there is a homogeneous non-zero chain map ℓ^{·} from C to D which is homotopic to zero with homotopy map T^{·}, we have

0 ⟶ C_{1} ⟶ C_{2} ⟶ C_{3} ⟶ 0
↓ ^{ℓ_{1}} ^{T_{1}} ↙ ↓ ^{ℓ_{2}} ^{T_{2}} ↙ ↓ ^{ℓ_{3}}
0 ⟶ D_{1} ⟶ D_{2} ⟶ D_{3} ⟶ 0

where T^{·} is a non-zero homotopy between ℓ^{·} and the zero map. We may assume that we have removed from T^{·} any homogeneous components whose composition with the relevant horizontal maps is zero.

Since each of C^{·} and D^{·} is of width no greater than two, we have four cases with non-zero homotopy : a square, a triangle with the base at the top or at the bottom and a parallelogram. In the first three cases it is not hard to show that the mild assumption made on T^{·}, that it not contain any irrelevant homogeneous factors, insures that it is normal homogeneous. Thus we will give the proof only for the more difficult case of a parallelogram. We have

0  ⟶ P_{r}  ⟶^{h_{rs}} P_{s}
↓ ^{0} ^{T_{1}} ↙ ↓ ^{ℓ_{2}}  ^{T_{2}} ↙ ↓ ^{0}
P_{t}  ⟶_{h_{tu}} P_{u} ⟶ 0

where h_{tu} T_{1} + T_{2}h_{rs} = ℓ_{2}. We have assumed that ℓ^{·} is normal homogeneous, so

ℓ_{2} = ε_{u}^{i_{0}} h̃_{ru} , 0 ≤ i_{0} ≤ m − 1 + δ_{ru}.

We have

T_{1} = ∑_{i=0}^{m−1} a_{i} ε_{t}^{i} h̃_{rt} + δ_{rt} a_{m} ε_{t}^{m}

T_{2} = ∑_{j=0}^{m−1} b_{j} ε_{u}^{j} h̃_{su} + δ_{su} a_{m} ε_{u}^{m}.

Choose α, β such that

h_{tu}h̃_{rt} = ε_{u}^{α}h̃_{ru}
h̃_{su}h_{rs} = ε_{u}^{β}h̃_{ru}.

Note that δ_{rt} = 1 ⇒ α = 0 and δ_{su} = 1 ⇒ β = 0

ℓ_{2} = ε_{u}^{i_{0}} h̃_{ru} = ( ∑_{i=0}^{m−1} a_{i} ε_{u}^{i+α} h̃_{ru}) + ( ∑_{j=0}^{m−1} b_{j} ε_{u}^{j+β} h̃_{ru}) .

Thus, for every k ≠ i_{0}, 0 ≤ k ≤ m − 1 + δ_{ru}, we have

a_{k−α} + b_{k−β} = 0.

Thus we may substitute for T^{·} a new homotopy T^{′·} in which a_{k−α} = b_{k−β} = 0 for k ≠ i_{0}. For k = i_{0}, we get

a_{i_{0}−α} + b_{i_{0}−β} = 1.

i_{0} − α and i_{0} − β cannot both be −1. Thus if we set one of a_{i_{0}−α} or b_{i_{0}−β} equal to one and the other equal to zero, we get the desired normal homogeneous homotopy T̃^{·}.

Definition. If ℓ^{·}: C^{·} → D^{·} is a normal homogeneous chain map with one non-zero map, then the degree of ℓ^{·} is the degree of its non-zero component. If ℓ^{·} has two non-zero maps, one of which is a map between two identical projectives, the degree of ℓ^{·} is the degree of the map between the identical projectives.

Remark. It will be a consequence of the main theorems of this paper that every normal homogeneous chain map between two indecomposable components of an element of P T C_{2} fulfills one of these two conditions and thus has a well-defined degree.

COMBINATORIAL PARTIAL TILTING COMPLEXES 15

Theorem 2. Between two elementary complexes in P T C_{2}, there are chain maps not homotopic to zero in the following cases:

(i) From S_{i} to S_{j} there are m + δ_{ij} normal homogeneous maps. If we let {j − i}_{e} represent the residue of j − i modulo e, then the maps have degree {j − i}_{e} + ke, 0 ≤ k ≤ m + δ_{ij}.

(ii) From T_{ij} to itself there are two normal homogeneous maps, the identity and

the socle map. (iii) Between two elementary components with one common index, there is a

normal homogeneous map of degree zero in one direction and of degree em in the opposite direction.

§4. Homomorphisms between indecomposable complexes.

Our eventual aim, summarized in Theorem 1 given in the previous section, is to describe all the partial tilting complexes Q^{·} ∈ P T C_{2}, as direct sums of shifts of the complexes S_{i} and T_{jk} of projective b modules. The condition for Q^{·} to be in P T C_{2} is that

Hom_{D^{b}(b)}(Q^{·}, Q^{·}[n]) = 0 for n ≠ 0.

By the standard properties of homomorphisms of direct sums, this is equivalent to showing that for any pair R and R′ of elementary indirect summands of Q^{·}, we have

Hom_{D^{b}(b)}(R, R^{·}[n]) = 0 for n ≠ 0.

Since our elementary complexes are all of a length less than or equal to two, there are at most three possible relative shifts for which a non-zero projective of R is positioned over a non-zero projective of R. Where a non-zero chain map exists we must decide whether or not it is homotopic to zero.

Since, for those partial tilting complexes which are actually tilting complexes, we will be interested in calculating the endomorphism ring, we want to determine

16 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

all maps in Hom_{b}(R, R′)[n]) which are not homotopic to zero. This information, which will be summarized in Theorem 2, will be determined from the same case-by-case study of homomorphisms between elementary components. We consider the following five cases,which will be treated in the subsequent five propositions.

Case 1: S with S.

Case 2: S with T, and no common indices. Case 3: S with T, and a common index. Case 4: T with T, and no common indices. Case 5: T with T, and common indices.

Once we have determined all non-trivial maps, we can combine the results into Theorem 1 by using the principle that if there are two or more relative positions which produce non-trivial maps, then the two elementary complexes cannot appear together at all in partial tilting complex. If there is exactly one, then that must be the relative position of the two elementary complexes in any element of P T C_{2}. Finally, if there are no non-trivial maps between the two elementary complexes, then they can appear together without any restrictions on their relative position.

Since in classifying all the maps we also show that they have a well-defined degree and determine what it is, we then get Theorem 2 more or less for free. We will also establish that the elementary complexes are in facat indecomposable, since their endomorphism rings will be local. Proposition 1. Two complexes, S_{r}[n] and S_{t}[n′], can appear together in a partial tilting complex in P T C_{2} if and only if n = n′. The normal homogeneous maps from S_{r}[n] to S_{t}[n] are the following

ε_{t}^{k}h̃_{rt} , 0 ≤ k ≤ m − 1 + δ_{rt}.

If {t − r}_{e} is the residue modulo e, the degrees are

{t − r}_{e} + ke , 0 ≤ k ≤ m − 1 + δ_{rt}.

COMBINATORIAL PARTIAL TILTING COMPLEXES 17

Proof. There are no non-zero chain maps from S_{r}[n] to S_{t}[n′] unless n = n′. If n ≠ n′, then there is a map after a shift of n′ − n, so S_{r}[n] and S_{r}[n′] cannot occur together. When n = n′, the normal homogeneous maps and their degree were calculated in Lemma 2.1. □

Henceforward we will assume that Q^{·} has been shifted so that if there is any indecomposable S_{r}, the non-zero projective module appears in degree zero.

Proposition 2. If S_{r} and T_{st}[n] have no common index, then they can appear together in a partial tilting complex in P T C_{2} if and only if r → s → t is short.

Proof. (⇐) Assume that r → s → t is short. We will show that every non-zero chain map in either direction is homotopic to zero.

Case 1. ℓ^{·}: S_{r} → T_{st}. If ℓ_{1} = ε_{s}^{i}h_{rs} is non-zero, so is h_{st} ∘ ℓ_{1} = ε_{t}^{i}h_{rt}, and thus the following diagram does not commute.

S_{r} :  0 ⟶ P_{r} ⟶ 0
↓ ^{ℓ_{1}}  ↓
T_{st}:  0 ⟶ P_{s} ⟶^{h_{st}} P_{t} ⟶ 0.

Case 2. ℓ^{·}: S_{r} → T_{st}[1]. If ℓ_{1} = ε_{t}^{i}h_{rt} is non-zero, then T_{1} = ε_{s}^{i}h_{rs} is a homotopy

0 ⟶ P_{r} ⟶ 0
↓ ^{T_{1}} ↙ ↓ ^{ℓ_{1}}
P_{s} ⟶^{h_{st}} P_{t} ⟶ 0.

Case 3. ℓ^{·}: T_{st} → S_{r}. As in Case 2, there is a non-trivial homotopy.

Case 4. ℓ^{·}: T_{st}[1] → S_{r}. As in Case 1, there are no well-defined chain maps.

18 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

(⇒) Now suppose that r → s → t is long, which implies, by the cyclic permutation property, that s → t → r is long. We obtain that these are two different non-trivial maps from S_{r} to T_{st}, with different shifts, so S_{r} and T_{st} cannot appear together in any relative position.

Case (1). ℓ′ : S_{r} → T_{st}.

0 ⟶ P_{r} ⟶ 0
↓ ^{ε_{s}^{m−1}h_{rs}}
0 ⟶ P_{s} ⟶ P_{t}

This map is well defined because h_{st} ∘ (ε_{s}^{m−1}h_{rs}) = 0, and no homotopy is possible.

Case (2). ℓ′ : S_{r} → T_{st}[1].

0 ⟶ P_{r} ⟶ 0
↓  ↓ ^{h_{rt}}
0 ⟶ P_{s} ⟶ P_{t} ⟶ 0

This map is clearly well defined, and no homotopy is possible because h_{st}∘(ε_{s}^{i}h_{rs}) = ε_{t}^{i+1}h_{rt}, and this cannot equal h_{rt}.

Proposition 3. If S_{r} and T_{st} or T_{st}[1] have a common index, then they can appear together in a partial tilting complex in P T C_{2} if and only if the common index appears in the same degree in each. The normal homogeneous homomorphisms are then the following:

(i) If r = s : ℓ^{·}: S_{r} → T_{rt} has ℓ_{1} = ε_{r}^{m} of degree em.

ℓ̄^{·}: T_{rt} → S_{r} has ℓ̄_{1} = id_{r} of degree 0.

(ii) If r = t: S_{r} → T_{sr}[1] has ℓ_{2} = id_{r} of degree 0. ℓ̄^{·}: T_{sr}[1] → S_{r} has ℓ̄_{2} = ε_{r}^{m} of degree em.

COMBINATORIAL PARTIAL TILTING COMPLEXES 19

Proof. We will first show that the four maps given in the proposition are welldefined and not homotopic to zero. This will establish that the common projectives must be in the same degree in any partial tilting module, since otherwise we would get a map to a shift.

(i) If ℓ^{·}: S_{r} → T_{rt} is normal homogeneous, then ℓ_{1} must be ε_{r}^{i}. In order for the diagram to commute, we must have h_{rt} ε_{r}^{i} = 0, which will only be true if i = m. By the shape of the diagram, there is no possible homotopy, so ℓ_{1} = ε_{r}^{m} is a non-trivial homomorphism of degree em.

ℓ̄^{·}: T_{rt} → S_{r} is well-defined for any ℓ̄_{1} = ε_{r}^{i}, but if i > 0, then we have a homotopy T_{1} = ε_{r}^{i−1}h_{tr}. When i = 0, there is no normal homogeneous T such that

T ∘ h_{rt} = id_{r} .

(ii) is dual to (i).

It remains to show that if the corresponding indices are not positioned one over the other, then there is no non-trivial chain map.

(i′) \underline{r = s}: Consider ℓ^{·}: S_{r} → T_{rt}[1], with ℓ_{1} : P_{r} → P_{t} equal to ε^{j}h_{rt}. This has

a homotopy T = ε_{r}^{j} for any j.

Now take the opposite direction ℓ̄^{·}: T_{rt}[1] → S_{r}. If ℓ̄^{·} ≠ 0, then ℓ_{1} = ε_{r}^{j}h_{tr}, with j < m. Then, ℓ̄ cannot be well-defined because ε_{r}^{j}h_{tr} ∘ h_{rt} =

ε_{r}^{j+1} ≠ 0.
P_{r} ⟶^{h_{rt}} P_{t}
↓ ↘ ↓ ℓ_{1} = ε_{r}^{j}h_{tr}
0 ⟶ P_{r}.

(ii′) \underline{r = t}: Dual to (i’).

The following is the most difficult of the five propositions:

20 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Proposition 4. If T_{rs}[n] and T_{tu}[n′] appear together in a partial tilting complex in P T C_{2}, with all four indices distinct, then at least one of the following three sequences of indices must be short:

(1) r → s → t → u

(2) r → t → u → s

(3) r → s → u → t

This implies, by the mixing property in Lemma 3.1, that none of the following can be short.

(4) r → t → s → u

(5) r → u → s → t

(6) r → u → t → s

Proof. We will first show that in Cases (1), (2) and (3), there are no non-trivial homomorphisms, and thus these elementary complexes can appear together in some Q^{·} ∈ P T C_{s} with any desired shift. Since interchanging r, s with t, u produces the same three cases, it suffices to show that there are no non-trivial homomorphisms from T_{rs} to T_{tu}[n], for n = −1, 0, 1.

Case (1). r → s → t → u is short. n = −1 There is no well-defined map, as in case (i′) of Proposition 3. The first square does not commute.

0 ⟶ P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
↓ ↘^{ε_{t}^{i}h_{rt}} ↓^{ε_{t}^{i}h_{st}}
0 ⟶ P_{t}  ⟶_{h_{tu}} P_{u} ⟶ 0.

n = 0

0 ⟶  P_{r} ⟶^{h_{rs}} P_{s}  ⟶  0
ε_{t}^{i}h_{rt} ↓ ^{T} ↙ ↓^{ε_{u}^{i}h_{su}}
0 ⟶  P_{t} ⟶_{h_{tu}} P_{u}  ⟶ 0

This is well defined if i = j, but then T = ε_{t}^{i}h_{st} is a homotopy. n = 1

0 ⟶  P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
T_{1} ↙ ^{ε_{u}^{i}h_{ru}} ↓ ^{T_{2}} ↙
0 ⟶ P_{t} ⟶_{h_{tu}}  P_{u} ⟶ 0

T_{1} = ε_{t}^{i}h_{rt}, T_{2} = ε_{u}^{i}h_{su} is a homotopy. Case (2). r → t → u → s is short. n = −1 This is not well defined, as in Case 1. n = 0

0 ⟶  P_{r} ⟶^{h_{rs}} P_{s}  ⟶  0
ε_{t}^{i}h_{rt} ↓ ^{T} ↙ ↓^{ε_{u}^{j}h_{su}}
0 ⟶  P_{t} ⟶_{h_{tu}} P_{u}  ⟶ 0

This is well defined only if i = j + 1, but in that case T = ε. n = 1 There is a homotopy, as in n = 1 for Case 1. Case (3). r → s → n → t is short. If we rewrite this as t → r → s → u, it is clearly dual to Case 2, where r is interchanged with t, and s with n.

We now show that the Cases (4), (5), and (6) cannot occur in Q′ ∈ P T C_{2}, by exhibiting, in each case, two different shifts which produce non-trivial homomorphisms. Once again the cases are symmetric under the interchange of T_{rs} with

22 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

T_{tu}. We will in general leave to the reader the verification that the maps are not homomorphic to zero.

Case (4). r → t → s → u is short.

0 ⟶ P_{r} ⟶^{h_{rs}} P_{s}
↓ ^{ε_{t}^{m−1}h_{st}}
0 ⟶ P_{t} ⟶_{h_{tu}} P_{u} ⟶ 0

0 ⟶  P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
h_{rt} ↓  ↓ ^{h_{su}}
0 ⟶  P_{t} ⟶^{h_{tu}} P_{u} ⟶ 0

The second map is well defined because h_{su} ∘ h_{rs} = h_{ru} = h_{tu} ∘ h_{rt}. There is no homotopy because h_{st} ∘ h_{rs} = ε_{r}h_{rt}.

Case (5). t → r → u → s is short.

0 ⟶ P_{r} ⟶ P_{s}
↓ ↓ ^{h_{ru}} ↓
0 ⟶ P_{t} ⟶^{h_{tu}} P_{u} ⟶ 0
0 ⟶  P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
h_{rt} ↓  ↓ ^{h_{su}}
0 ⟶  P_{t} ⟶^{h_{tu}} P_{u} ⟶ 0

The second homomorphism is well defined because h_{su} ∘ h_{rs} = ε_{u}h_{ru} = h_{tu} ∘ h_{rt}.

Case (6). r → u → t → s is short.

0 ⟶ P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
↓  ↓ ^{ε_{t}^{m−1}h_{st}}
0 ⟶ P_{t} ⟶_{h_{tu}} P_{u} ⟶ 0
0 ⟶ P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
↓ ↓ ^{h_{ru}} ↓
0 ⟶ P_{t} ⟶^{h_{tu}} P_{u} ⟶ 0

The final proposition is parallel to Proposition 3, dealing with the case of common indices. Proposition 5. T_{rs} can occur in a two-restricted partial tilting complex, but not together with any shift of itself or of T_{sr}. If T_{rs} and T_{tu}[n] have unique common index, then they can occur together in a two-restricted partial tilting complex if and only if the common index occurs in the same degree and

(1) If n = −1, so that s = t, then r → s → u is short.

There is one normal homogeneous map in each direction

P_{r} ⟶  P_{s}
ε_{s}^{m} ↓↑ ^{id_{s}}
P_{s} ⟶ P_{u}

(2) If n = 0, and r = t, then s and u are symmetrical, and we may assume

r → s → u is short. There is one normal homogeneous map in each direction

P_{r} ⟶ P_{s}
id_{r} ↓ ↓^{h_{su}}
P_{r} ⟶ P_{u}

P_{r} ⟶ P_{s}
ε_{r}^{m}↑ ↑^{0}
P_{r} ⟶ P_{u}

If n = 0 and s = u, then r, t are symmetric so we may assume r → t → u is short.

(3) If n = 1, we get the case dual to n = −1, and must have t → r → s short.

P_{r} ⟶ P_{s}
id_{r} ↓↑ ^{ε_{r}^{m}}
P_{t} ⟶ P_{r}

There are two normal homogeneous maps from T_{rs} to itself, the identity and the socle maps.

Proof. Since we always have id : T_{rs} → T_{rs}, in order to show that T_{rs} can occur in Q′ ∈ P T C_{2}, it suffices to show that there are no non-trivial homomorphisms with a shift

P_{r} ⟶^{h_{rs}} P_{s}  P_{r}  ⟶^{h_{rs}} P_{s}
↓ ^{ε_{r}^{i}h_{sr}}  ^{ε_{r}^{i}} ↙ ↓ ^{ε_{s}^{i}h_{rs}} ↙ ^{ε_{s}^{i}}
P_{r} ⟶_{h_{rs}} P_{s}  P_{r} ⟶_{h_{rs}} P_{s}

The left hand map is never well defined for any i, and the right hand map always has the indicated homotopy.

T_{rs} cannot occur together with any shift of T_{sr} because the following two homomorphisms are non-trivial and thus whatever relative position is chosen, there is always a diagonal map

P_{r} ⟶ P_{s}  P_{r}  ⟶ P_{s}
↓ ^{ε_{s}^{m}}  ↓ ^{id^{r}}
P_{s} ⟶ P_{r}  P_{s} ⟶ P_{r}

COMBINATORIAL PARTIAL TILTING COMPLEXES 25

We are now prepared to consider a pair of elementary complexes, T_{rs} and T_{tu}[n], for n = −1, 0, 1, with one common index. The case n = 1 is dual to the case n = −1, and for n = 0, the case r = t is dual to s = u, so we will give the details only for n = −1 and n = 0 with s = u. We will first show that for the cases given in the proposition there are no non-trivial homomorphisms for other shifts. In addition, in the case n = −1, we must show that if r → s → u is long, there are two possible shifts which give non-trivial maps. n = −1 We first suppose that r → s → u is short, and show that the maps indicated in the proposition are non-trivial

0 ⟶ P_{r} ⟶  P_{s} ⟶ 0
ε_{s}^{m} ↓↑ ^{id^{s}}
P_{s} ⟶ P_{u} ⟶ 0

The map from top to bottom is well defined because ε_{s}^{m}h_{rs} = 0 = ε_{u}^{m}h_{su}, and no homotopy is possible. The map from top to bottom is automatically well defined, and there is no homotopy because id_{s} does not factor through h_{rs} or h_{su}.

Any homomorphism

0 ⟶  P_{r} ⟶^{h_{rs}} P_{s} ⟶ 0
ε_{s}^{i}h_{rs} ↓ ^{ε_{s}^{i}} ↙ ↓ ^{ε_{u}^{i}h_{su}}
0 ⟶  P_{s} ⟶^{h_{su}} P_{u} ⟶ 0

has the indicated homotopy and so does not give a non-trivial map

P_{r} ⟶^{h_{rs}} P_{s}
ε^{i}h_{rs} ↙ ↓ ↙ ^{ε_{s}^{j}h_{su}}
0 ⟶ P_{s} ⟶^{h_{su}} P_{u}

26 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

This is a homotopy with i = j if r → s → u is short. If r → s → u is long, it is a homotopy when i = j − 1 for j > 0, but the case j − 0 gives a non-trivial homomorphism. Thus, if r → s → u is long, T_{rs} and T_{su}[−1] cannot appear together because there is a non-trivial homomorphism with a different shift, and if r → s → u is short, they can. n = 0 We do the case s = u, and assume r → t → s is short.

P_{r} ⟶^{h_{rs}} P_{s}  P_{t}  ⟶^{h_{ts}} P_{s}
h_{rt} ↓ ↙ ↓ ^{id^{s}}  0 ↓ T ↙ ↓ ^{ε_{s}^{m}}
P_{t}  ⟶^{h_{ts}} P_{s}  P_{r} ⟶^{h_{rs}} P_{s}

In the left hand map there is no homotopy because id_{s} does not factor through h_{ts}. In the right hand case, if T = ε_{r}^{m−1}h_{sr}, then T ∘ h_{ts} ≠ 0.

It remains to show that there are no well-defined shifted maps

P_{r} ⟶ P_{s}  P_{r} ⟶^{h_{rs}} P_{s}
↓ ↘  0 ↙ ^{ε^{i}h_{rs}} ↓ ↙ ^{ε_{s}^{i}}
P_{t} ⟶ P_{s}  P_{t} ⟶  P_{s}

The left hand map is not well defined because the indicated diagonal is non-zero, and the right hand map has the indicated homotopy since we did not need the condition r → t → s is short. The same is true for maps from T_{ts} to T_{rs}. □

§5. Proof of theorems.

Now that we have Propositions 1 – 5, the proofs of Theorems 1 and 2 are simply a matter of organization. Proof of Theorem 1.

(a) By Proposition 1, all S_{i}[n] have their non-zero term in the same degree. Shifting Q′ if necessary we may assume that this is degree zero.

COMBINATORIAL PARTIAL TILTING COMPLEXES 27

(b) By Proposition 3, if S_{r} and T_{tu}[m] have a common projective, it occurs in the same degree in both. By Proposition 5, this is also true for T_{rs}[n′] and T_{tu}[n].

(c) By Proposition 4, if S_{r} and T_{st}[n] occur in Q, then r → s → t is short.

(d) This is Proposition 4. □ Proof of Theorem 2.

(i) From Proposition 1. (ii) From Proposition 5. (iii) From Proposition 3 and Proposition 5. In conclusion, having shown that in a two-restricted partial tilting complex all

occurrences of P_{r} are in the same degree, we want to summarize the information given by Propositions 1, 3, and 5 in a way that considers all S_{r} together, and all irreducibles with a common P_{r} together. As before, we let {t − s}_{e} be the residue modulo e of t − s between 0 and e − 1. Theorem 3. (Local order theorem)

(1) If {S_{i_{1}}, …, S_{i_{r}}} are a set of S_{i} in Q′ ∈ P T C_{2}, such that i_{1} → i_{2} → ⋯ → i_{r}

is short, then we have homomorphisms h_{i_{k}i_{k+1}} : S_{i_{k}} → S_{i_{k+1}} such that

h_{i_{k−1}i_{k}} ∘ ⋯ ∘ h_{i_{k+1}i_{k+2}} ∘ h_{i_{k}i_{k+1}} = ε_{i_{k}} for k = 1, …, r.

(2) If {T_{s_{i}r}[1]}_{i=1}^{k} ∪ {T_{rt_{j}}}_{j=1}^{ℓ} are a set of irreducibles in Q′, we can arrange

the indices so that

s_{1} → ⋯ → s_{k} → r → t_{1} → ⋯ → t_{ℓ}

is short. All of the maps

T_{rt_{1}} → T_{rt_{2}} → ⋯ → T_{rt_{ℓ}} → T_{s_{1}r}[1] → ⋯ → T_{s_{k}r}[1] will be identity on P_{r}, whereas T_{s_{k}r}[1] → T_{rt_{1}} will be ε^{m} on P_{r}. If S_{r} is in Q′, then S_{r} can be inserted between T_{rt_{ℓ}} and T_{s_{1}r}[1], with identity maps in P_{r}.

28 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Proof.

(1) This is direct from Proposition 1 and the definition of a short sequence.

(2) That each of the indicated maps is the identity on P_{r} is a consequence of Proposition 5, as is the fact that the map from T_{rt_{ℓ}} to T_{s_{1}r}[1] is ε_{r}^{m}. The possibility of inserting S_{i} comes from Proposition 1.

§6. Applications.

Our aim in studying these two-restricted partial tilting complexes was to give a classification to all T C_{2} in combinatorial terms, which will be done in the sequel, [SZ1]. It was, furthermore, important to show that the homomorphisms between the irreducible complexes could be taken to be homogeneous, for this allows a study of a class of homogeneous deformations of the Brauer tree algebras [SZ2] which is much more natural than that given in [S2], in the sense that it is local rather than global, and also that it is derived from a tilting complex.

Although the “two-restricted” condition means that we have considered only a subclass of all possible tilting complexes, it is a sufficiently large subclass to allow us to reach every possible Brauer tree algebra. In fact, it is too large for it is possible to reach each Brauer tree algebra in many possible ways. In the next paper we consider all the different possible “foldings” of the tilting complex. It had already been shown by Rouquier [R] that each tree can be reached by a “completely folded” complex with only two non-zero terms. In [RS] we show that Rickard’s combinatorial tree-to-star complex and the corresponding star-to-tree two-restricted complex give inverse equivalences, and reduce to the Rouquier two-term complex in the completely folded case. As Zimmermann pointed out in a very helpful conversation, if we go over to two-sided tilting complexes [Z] we can study the different possible tilting complexes which give the same Brauer tree algebra in terms of the group of two-sided selfequivalences of b. This group might be quite large; for the very simple case of the Brauer star algebra of type (2,1) Rouquier and Zimmermann [RoZ] calculated that, modulo shifts, it was the modular group, which is the free group on two generators

of order 2 and order 3. The entire subject merits further investigation.

We also hope that it might be possible to use this approach for some general-

ization of the Brauer star algebra with abelian but not cyclic defect group. This

might lead to some generalization of the Brauer tree. One such generalization has

already been made by Benson [B], for dihedral defect groups.

References

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[B] Benson D., Representations and Cohomology, Cambridge Studies in Advanced Mathematics, vol. 30, 1991.

[H] Happel D., On the derived category of a finite dimensional algebra, Comment. Math. Helvetici 62 (1987), 339-389.

[KZ1]König S. and Zimmermann A., Tilting hereditary orders, Comm. in Algebra 24 (1996),

1893-1913. [KZ2]König S. and Zimmermann A., Derived equivsalence for group rings, LNM, vol. 1685, 1998.

[MS] Mejer C. and Schaps M., Separable deformations of blocks with abelian defect group and of derived equivalent global blocks, Canadian Math. Soc. Conf. Proc. 18 (1996), 505-518.

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[R3] Rickard J., Lifting theorems for tilting complexes, J. of Algebra 142 (1991), 383-393.

[RS] Rickard J. and Schaps M., Folded titlting complexes for Brauer tree algebras, preprint.

[Ro] Rouquier R. and Zimmermann A., Picard groups for derived modular categories, preprint.

[RoZ] Rouquier R., From stable equivalence to Rickard equivalences for blocks with cyclic defect group, Proc. of Groups 93, Galway, St. Andrews, London Math. Soc. Lecture Notes Series 212, 512-523.

[S1] Schaps M., Deformations of finite dimensional algebras and their idempotents, Trans. AMS (1988), 843-856.

[S2] Schaps M., A modular version of Maschke’s theorem for groups with cyclic p-Sylow subgroup, J. of Algebra 163 (1994), 623-635.

[SSS] Schaps M., Shapira D. and Shlomo D., Quivers of blocks with normal defect group, preprint.

[SZ1] M. Schaps and Zakay-Illouz E., Pointed Brauer trees, preprint.

[SZ2] M. Schaps and Zakay-Illouz E., Homogeneous deformations of Brauer tree algebras, preprint.

[ZI1] Zakay-Illouz E., Basis-graphs and deformations for non-abelian groups of order 2^{j} · 3^{i} ≤ 24, i, j ≥ 0 with abelian p-Sylow subgroups, Master’s thesis, Bar-Ilan Univesity, 1993.

[ZI2] Zakay-Illouz E., The Green Correspondence between Separable Deformations, Ph.D. dissertation, Bar-Ilan University, 1999.

[Z] Zimmermann, A., A two-sided tilting complex for Green orders and Brauer tree algebras, J. of Alg. 187 (1997), no. 2, 446-473.

30 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ Department of Mathematics and Computer Science, Bar-Ilan University, 52900 Ramat-Gan, Israel Department of Mathematics, Jordan Valley College, Jordan Valley, Israel