POINTED BRAUER TREES

Mary Schaps and Evelyne Zakay-Illouz*

Abstract. In this paper we establish a one-to-one correspondence between Brauer trees with an additional structure we call “pointing” and equivalence classes of tilting complexes of the corresponding Brauer star algebra which are of a special “combinatorial” nature. Under the correspondence, the identification of the indecomposable summands of the tilting complex with edges of the Brauer tree is completely natural, as is the multiplicity at the exceptional vertex. The procedure gives an algorithm for producing large numbers of distinct tilting complexes for the same Brauer tree.

§1. Introduction.

This paper is a sequel to [SZ1], but it is intended that they can be read independently. This work, [SZ1] and [SZ2] are all based on the dissertation of the second author [ZI2]. All the definitions and results needed from [SZ1] will be quoted in full. The basic subject is tilting theory for a particularly symmetrical class of algebras called the Brauer star algebras. The motivation for studying these algebras in depth lies in the modular representation theory of finite groups, and we will not repeat that here, but will take the importance of these algebras for granted.

Definition. Let e and 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. Let the cyclic group C_{e} = ⟨d⟩ act on the truncated polynomial ring A = K[x]/xn̂, d : x → ξx. The Brauer star algebra of type (e, m) is the skew group algebra

b = A[C_{e}]

Typeset by AMS-TEX

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

in which d and x commute via the relation

d^{−1}xd = ξx.

The algebra b has e distinct simple modules, corresponding to the idempotents

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

and satisfying

f_{i−1}x = xf_{i}.

The corresponding indecomposable projective left modules will be denoted by

P_{i} = bf_{i}, i = 1, …, e.

Each P_{i} is uniserial, and the projective cover of Rad(P_{i+1}) is P_{i}. 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}⟩. Any map ε_{j}^{ℓ}h̃_{ij}(= h̃_{ij}ε_{i}^{ℓ}) will be called normal homogeneous of degree ℓe + k, where k = j − i (mod e), 0 ≤ k < e.

Definition. A sequence of indices, i_{1} → ⋯ → i_{r} is called short if h̃_{i_{r−1}i_{r}} ∘ ⋯ ∘ h̃_{i_{2}i_{3}} ∘ h̃_{i_{1}i_{2}} = h̃_{i_{1}i_{r}} and it is called long if h̃_{i_{r−1}i_{r}} ∘ ⋯ ∘ h̃_{i_{1}i_{2}} = ε_{i_{r}}^{j} h̃_{i_{1}i_{r}}, j > 0.

We now consider complexes Q^{·} of projective modules. The notation Q^{·}[n] will indicate that the complex has been shifted n places to the left and the differential multiplied by (−1)^{n}.

POINTED BRAUER TREES 3

Definition. Let R be Noetherian ring. A bounded complex Q^{·} 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 Q^{·}.

A complex Q^{·} satisfying only (1) is called a partial tilting complex.

Definition. A partial tilting complex Q^{·} for the Brauer star algebra b is called two-restricted if it is a direct sum of shifts of the indecomposable complexes

S_{i} : 0 → P_{i} → 0

T_{ij} : 0 → P_{i} →^{h_{ij}} P_{j} → 0, i ≠ j

where the first non-zero term is in degree zero. The complexes S_{i}[n] and T_{ij}[n] will be called elementary. The map from T_{ij} to T_{ij} which is ε_{i}^{m} on P_{i} and zero on P_{j} will called the socle map. It is homotopic to the map which is zero on P_{i} and −ε_{j}^{m} on P_{j}.

Note. One can show that any indecomposable complex satisfying (1) above, which is nonzero in at most two degrees, is elementary.

§2. Two-restricted partial tilting complexes for the Brauer star algebra.

The main result of [SZ1] (from [ZI2]) was a complete description of the set of tworestricted partial tilting complexes, denoted by P T C_{2}, and their endomorphisms. We quote the results in full.

4 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

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

(3) r → s → u → t.

The second theorem summarizes the homomorphisms between the indecomposables.

Definition. Corresponding to the normal homogeneous maps ε_{j}^{ℓ}h̃_{ij} : P_{i} → P_{j}, we have induced chain maps ε_{j}^{ℓ}h̃_{ij} : S_{i} → S_{j}, which will be called normal homogeneous. As before, ε_{i} : S_{i} → S_{i} and h_{ij} = h̃_{ij} for i ≠ j.

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} + ℓe, 0 ≤ ℓ < 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 in the same

degree, there is a normal homogeneous map of degree zero in one direction and of degree em in the opposite direction.

POINTED BRAUER TREES 5

Finally, regarding distinct elementary complexes with a unique common index as in Theorem 2 (iii), we have more explicit information. Theorem 3. (Local Order Theorem) Let Q^{·} ∈ P T C_{2}.

(1) If {S_{i_{1}}, …, S_{i_{r}}} are a set of S_{i} in Q^{·}, such that i_{1} → i_{2} → ⋯ → i_{r} is short, and if we let i_{r+1} = i_{1}, 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}[n + 1]}_{i=1}^{k} ∪ {T_{rt_{j}}[n]}_{j=1}^{ℓ} are a set of elementary complexes 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}}[n] → T_{rt_{2}}[n] → ⋯ → T_{rt_{ℓ}}[n] → T_{s_{1}r}[n + 1] → ⋯ → T_{s_{k}r}[n + 1]

will be identity on P_{r}, whereas T_{s_{k}r}[n + 1] → T_{rt_{1}}[n] will be (−1)^{n}ε_{r}^{m} on P_{r}. If S_{r} is in Q^{·}, then n = 0 and S_{r} can be inserted between T_{rt_{ℓ}} and T_{s_{1}r}[1], with identity maps in P_{r}.

§3. Pointed Brauer trees and the statement of the main theorem.

The purpose of this paper is to give a direct connection between the combinatorics of a two-restricted tilting complex and the combinatorics of its endomorphism ring. Definition 3.1. A Brauer tree of type (e, m) is a connected acyclic graph with a set V of e + 1 vertices, a set E of e edges, and the following additional structures:

i) a cyclic ordering of the edges at each vertex, ii) the designation of one vertex, u_{0}, hereafter called the exceptional vertex,

which is assigned multiplicity m.

6 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Definition 3.2. Let B be a Brauer tree of type (e, m). A pointing on B is the choice, for each non-exceptional vertex u, of a pair of edges (i, j) which are adjacent in the cyclic ordering at u. If there is only one edge i at u, we take (i, i) as the required pair. The tree B together with a pointing will be called a pointed Brauer tree. We denote by B the set of pointed Brauer trees.

Remark 3.1. When a Brauer tree is represented by a planar embedding, the convention in the field is to represent the cyclic ordering at each vertex by the counterclockwise ordering of the edges in the plane. We will then represent the pointing (i, j) by placing a point in the sector between edge i and edge j in a small neighborhood of u, as in Fig. 1.

For each B ∈ B, we consider the set of e possible pairs (B, v), where v is one of the non-exceptional vertices of B, hereafter called the marked vertex of the pair (B, v). The marked vertex will be used in the sequel to fix a numbering of the vertices of B. The set of all pairs (B, v) with B ∈ B will be denoted by B′.

Given a field K of characteristic zero or characteristic prime to em, we can associate to each Brauer tree an algebra which is basic, i.e., such that the algebra modulo its radical is a product of copies of the field. There are various ways to describe this algebra, depending upon one’s audience. In the sequel, [SZ2], which forms a part of the ongoing investigation of the Donald-Flanigan problem, we will describe it from the standpoint of algebra representation theory, in terms of quivers and relations. However, since this paper is addressed primarily to group theoreticians, we will follow the approach of Linckelmann [L]. This approach will be appropriately simplified and adjusted because we are working over a sufficiently large field K of arbitrary characteristic prime to em rather than over a complete valuation ring O with algebraically-closed residue field of characteristic p as Linckelmann does.

Definition 3.3. Let B be a Brauer tree, with vertex set V. The distance d(u) of any vertex u ∈ V from the exceptional vertex u_{0} is the number of edges in a

POINTED BRAUER TREES 7

minimal path from u to u_{0}. The distance parity of u is −1 or +1 as d(u) is odd or even.

Definition 3.4. Let B be a Brauer tree, and let I be a set in one-to-one correspondence with the edges of B. We define two permutations ρ, σ : I → I as follows: For each i ∈ I, let u_{σ}(i) be the vertex of positive distance parity belonging to the edge corresponding to i, let [i]_{σ} be the set of elements of i corresponding to edges adjacent to this vertex, and let σ : I → I be the permutation which maps every element of [i]_{σ} to the following element in the cyclic ordering at that vertex. We define u_{ρ}(i), [i]_{ρ} and ρ : I → I similarly for the vertexes of negative distance parity. Thus we get

⋃[i]_{σ} = ⋃[i]_{ρ} = I

and

[i]_{σ} ∩ [i]_{ρ} = {i}.

Remark 3.2. When the Brauer tree arises in the modular representation theory of blocks of cyclic defect group, the elements of i correspond to conjugacy classes of primitive idempotents, and the permutations σ and ρ describe the composition factors in descending order of uniserial submodules of the indecomposable left projective modules determined by these idempotents ([A] or [L], §2.1).

Definition 3.5. Let B be a Brauer tree of type (e, m), let I, σ, ρ be as in Definition 3.4, and let h(z) ∈ K[z] be a monic polynomial of degree m. The Brauer tree algebra for the data (B, h) is the K-algebra generated by the set I ∪ {r, s}, with the relations:

(a) i^{2} = i, ρ(i)r = ri, σ(i)s = si, for any i ∈ J

(b) ii′ = 0 for any pair i, i′ ∈ I, i ≠ i′

(c) rs = sr = 0 (d) If u_{σ}(i) is not exceptional,

is^{|[i]_{σ}|} = ir^{|[i]_{ρ}|}

(e) If u_{σ}(i) is exceptional and terminal,

ish(s^{|[i]_{σ}|}) = 0

(f) If u_{σ}(i) is exceptional and not terminal,

ih(s^{|[i]_{σ}|}) = ir^{|i_{ρ}]|}

Remark 3.3. The Brauer star algebra described in the introduction is the Brauer tree algebra of a star with e edges and the exceptional vertex at the center. The polynomial is h(z) = z^{m}, ρ is the identity permutation, and σ is the cyclic permuation (1 e e − 1 … 3 2), which describes the composition series in descending order. Note that the directions are reversed between the irreducible maps P_{i} → P_{i+1} and the composition series σ(i + 1) = i. Main Theorem. There is a one-to-one correspondence between the set of tworestricted tilting complexes for the Brauer star algebra b of type (e, m) and the set of pointed Brauer trees of type (e, m) with marked vertex. To a tilting complex Q^{·} in T C_{2} this correspondence associates a Brauer tree which is dual to the Brauer tree of End_{b}(Q^{·}).

The proof of this theorem will take the remainder of the paper. We first describe the maps in each direction. §4. Constructing a tilting complex from a pointed Brauer tree.

Let B be a pointed Brauer tree with marked vertex v. We recall that a Green’s walk for a planar tree is a counterclockwise circuit of the tree. In terms of the cyclic ordering of edges at a vertex, the Green’s walk can be obtained from an

POINTED BRAUER TREES 9

initial choice of vertex and adjacent edge by repeating the following recursion step 2e times:

Recursion step. Given a vertex u and edge i, one obtains a new vertex v as the other vertex of i, and a new edge j as the next edge at v in the cyclic ordering.

Definition 4.1. The vertex numbering of B induced by v is a vertex numbering obtained by taking a Green’s walk around the tree, starting at the point of vertex v, and numbering each vertex when its point is reached. We will refer to this as the numbering determined by a pointed Green walk. The exceptional vertex is numbered as 0. The vertex numbered by i will be denoted by u_{i} and thus will have u_{1} = v. The edge numbering of B induced by v is obtained from the vertex numbering by giving each edge the same number as the vertex farthest from the exceptional vertex. See Fig. 1 for an example of such a vertex numbering

Definition 4.2. The complex Q(B, v) determined by a pointed Brauer tree B with marked vertex v is the direct sum of e elementary 2-restricted complexes R_{i} chosen in one-to-one correspondence with the edges by recursion on d(u). We simultaneously define a displacement function s : V → ℤ.

(1) d(u_{0}) = 0: Set s(u_{0}) = 0. For each edge i connected to u_{0}, set s(u_{i}) = 0, and set R_{i} = S_{i}.

(2) Assume we have already chosen s(u_{i}) and an elementary complex R_{i′} for each edge i′ with d(u_{i}) < d. Let u_{i} be a vertex with d(u_{i}) = d.

(2a) For each edge j which occurs at u_{i} as one follows the cyclic ordering from i to the point at u_{i}, set s(u_{j}) = s(u_{i}) + 1, and set R_{j} = T_{ji}[s(u_{i}) + 1]. Note

that P_{i} occurs in degree −s(u_{i}) and P_{j} in degree −s(u_{j}).

(2b) For each edge k which occurs at u_{i} as one continues in the cyclic ordering from the point to i, set s(u_{k}) = s(u_{i}) − 1 and let R_{k} = T_{ik}[s(u_{i})]. Again P_{i}

occurs in degree −s(u_{i}) and P_{k} in degree −s(u_{k}).

10 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Define Q(B, v) = ⨁_{i=1}^{e} R_{i}.

Example 1. We have chosen the initial vertex v = u_{1} in the example of Fig. 1, so that each branch at the exceptional vertex determines a subinterval of [1, …, e].

The tilting complex that is determined from the Brauer tree in Fig. 1 is the direct sum of the following elementary complexes, where we omit the initial and final zeros in each elementary complex.

P_{7}  s(u_{7}) = 0
P_{1} →P_{7}  s(u_{1}) = 1
P_{1} →P_{3}  s(u_{3}) = 0
P_{2} →P_{3}  s(u_{2}) = 1
P_{3} → P_{4}  s(u_{4}) = −1
P_{5} →P_{7}  s(u_{5}) = 1
P_{5} →P_{6}  s(u_{6}) = 0
P_{7} → P_{8}  s(u_{8}) = −1
P_{11}  s(u_{11}) = 0
P_{9} →P_{11}  s(u_{9}) = 1
P_{9} →P_{10}  s(u_{10}) = 0

We wish to establish that the mapping Q(B, v) gives the desired correspondence. We first pause to collect some information about the tilting complexes produced by the algorithm.

12 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Definition 4.3. For any edge a, define the branch Y (a) of a to be the set consisting of a and of all edges and vertices such that the shortest path connecting them to the exceptional vertex u_{0} must include a. We call the vertex v of a nearest to u_{0} the root of Y (a), noting that by our definition it is not an element of Y (a), and we call the far vertex of a the source of Y (a). Assuming that we have fixed a pointing of the Brauer tree B, define st(a) to be the number of the first vertex assigned a number in the pointed Green’s walk, and end(a) to be the last vertex assigned a number. Lemma 4.1. Let a be any edge of the pointed Brauer tree B. Then

(1) For any vertex u_{i} ∈ Y (a),

st(a) → i → end(a)

is short, and if i ∉ Y (a), then i → st(a) → end(a) is short;

(2) if T_{ij}[m] is the elementary complex of some edge in Y (a), then

st(a) → i → j → end(a)

is short;

(3) If b_{1}, …, b_{k} are the edges in cyclic order from the entering edge a at vertex u_{i} to the point, and c_{1}, …, c_{ℓ} from the point around to a, then

st(a) =st(b_{1}) → end(b_{1}) → st(b_{2}) … end(b_{k}) → i →
st(c_{1}) → end(c_{1}) → st(c_{2}) → ⋯ → end(c_{ℓ}) = end(a)

is short.

Proof. (1) This is a consequence of the fact that the numbers are designed sequentially, and any sequential subsequence of the sequence

st(a) → st(a) + 1 → st(a) + 2 → ⋯ → e → 1 → ⋯ → st(a)

POINTED BRAUER TREES 13

is short.

(2) \underline{Case a}: If the edge b corresponding to T_{ij}[u] is attached before the point, then it is attached to u_{j}. The number for i is given before the number j, so st(a) → i → j → end(a) is short.

\underline{Case b}: If the edge b corresponds to T_{ij}[u] is after the point, then it is attached to u_{i} and the number i was given before j, so again

st(a) → i → j → end(a)

is short.

(3) This is immediate from the definition of the numbering process. Lemma 4.2. Q(B, v) is reduced and lies in P T C_{2}. Proof. Q(B, v) is by definition a two-restricted complex, being the direct sum of elementary two-restricted complexes.

We first claim that Q(B, v) is reduced. Each edge has a distinct number, so all the S_{i} are distinct. If i is an edge from u_{i′} to u_{i}, i ≠ 0, then R_{i} is the only elementary component of Q(B, v) which can be of the form T_{i′i}[n] or T_{i′i}[n′], so any such complex occurs only once.

We now verify that Q(B, v) satisfies all the conditions of Theorem 1.

(a) All elementary components S_{i} have their non-zero term in the same degree.

(b) If two elementary components have a common index, then the corresponding P_{i} appears at degree −s(u_{i}) in each of them.

(c) Suppose S_{i} and T_{jk}[n] are in Q(B, v), corresponding to edges a and b. If b ∈ Y (a) then by (2) and (3) of Lemma 4.1, we get

st(a) → j → k → i → end(a)

or

st(a) → i → j → k → end(a).

14 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

In either case j → i → k is long.

If b ∉ Y (a), then by (1) of Lemma 4.1 we have

i → st(b) → j → k → end(b),

and again j → i → k is long.

(d) Suppose T_{ij}[n], T_{kℓ}[n′] correspond to edges b and b′. If b ∉ Y (b′) and b′ ∉ Y (b), let a, a′ be the last distinct edges on the paths from b and b′ to u_{0}. Then by (3) of Lemma 4.1, we have

st(a) → i → j → end(a) → st(a′) → k → ℓ → end(a′),

which corresponds to possibility (1) in Theorem 1. If one of the edges b, b′ lies on the branch of the other, we may assume without loss of generality that b′ ∈ Y (b). One of the two vertices of b, u_{i} or u_{j} is the root.

\underline{Case 1}: u_{i} is the root. Then from Lemma 4.1, we get

i → st(b) → k → ℓ → j → end(b)

or

i → st(b) → j → k → ℓ → end(b).

\underline{Case 2}: u_{j} is the root. Then from Lemma 4.1, we get

j → st(b) → i → k → ℓ → end(b)

or

j → st(b) → k → ℓ → i → end(b).

The first possibility in each case corresponds to (2) of Theorem 1(d) and the second to (1) of Theorem 1(d).

Now that we have demonstrated that Q(B, v) ∈ P T C_{2}, it remains to show that Q(B, v) is actually a tilting complex, i.e., that we can generate stalk complexes

POINTED BRAUER TREES 15

0 → P_{i} → 0 for each projective P_{i}. This can be done inductively for each branch at the exceptional vertex, starting with the S_{i}. At each stage, given an edge from u_{i} to u_{j}, and given P_{i}^{·}[s(u_{i})], we get P_{j}^{·}[s(u_{j})] from the mapping cone of one of the following homomorphisms:

P_{i} → P_{j}  P_{i}
↓ id  ↓ id
P_{i}  P_{j} → P_{i}

In the first of the two cases, we must shift one place to the right, since then the mapping cone of a chain maps f : X → Y has modules X[1] ⊕ Y and differential (d_{X}^{i} + f^{i}, d_{Y}^{i−1}).

§5. Constructing a graph from a tilting complex.

We now give an algorithm for constructing a map in the opposite direction. We will begin with a two-restricted tilting complex Q^{·}, and construct a graph G(Q^{·}) in which the edges correspond to the elementary components of Q^{·}. Definition. Given a two-restricted tilting complex Q^{·} in which no elementary component appears with multiplicity greater than 1, let \overline{G}(Q^{·}) be the graph on e + 1 vertices (u_{0}, u_{1}, …, u_{e}), such that each of the e components Q^{·} represents an edge.

(1) A component S_{i} represents an edge connecting u_{0} and u_{i}.

(2) A component T_{jk}[m] represents an edge connecting u_{j} and u_{k}.

Lemma 5.1. If Q^{·} is a two-restricted tilting complex, then \overline{G}(Q^{·}) can be given the structure of a pointed Brauer tree B of type (e, m) with u_{0} as exceptional vertex. We will denote by G(Q^{·}) the pair (B, u_{1}) ∈ B′. Proof. Our first problem is to show that G(Q^{·}) is connected. Since there are no nontrivial maps between elementary complexes with no common indices, each stalk

16 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

complex S_{i} must be generated by the elementary complexes in the connected component containing u_{i}. By property (2) of tilting complexes, S_{i} can be generated recursively in a finite number of steps, each involving shifts, mapping cones, and cancellation of the factors

0 → P_{j} →^{id} P_{j} → 0

which generate all the complexes homotopic to zero. The total dimension of any T_{kℓ}[m] is 2n̂, as is the total dimension of each cancellable factor. Since the total dimension of S_{i} is n̂, we see that if the connected component contains only elementary complexes of the form T_{kℓ}[m], each mapping cone is of total dimension which is an even multiple of n̂, and it will never be possible to get a mapping cone which is homotopic to S_{i}. Thus for each i, 1 ≤ i ≤ e, the connected component of u_{i} must contain some S_{j}. Since all the S_{j} are connected at u_{0}, we conclude that G(Q^{·}) is connected.

Since \overline{G}(Q^{·}) has e + 1 vertices and e edges, if it is connected it must be a tree. Each edge can then be given the same number as the vertex farthest from the exceptional vertex.

By the Local Order Theorem, Theorem 3, the fact that Q^{·} is in P T C_{2} determines an ordering at each vertex, including the exceptional vertex. We denote the ordering by σ at vertices with distance parity +1, and by ρ at vertices of distance parity −1. Theorem 3 (2) determines the location of the point between the two edges whose corresponding homomorphism is the socle map. We choose u_{1} as marked vertex because it corresponds to the vertex numbered by 1 in \overline{G}(Q^{·}).

Naturally, the correspondence we are trying to establish will not be of mathematical interest unless there is some connection between G(Q^{·}) and the Brauer tree of the tilting of b given by Q^{·}. The following lemma establishes the last statement in the Main Theorem.

POINTED BRAUER TREES 17

Lemma 5.2. The Brauer tree of End_{D^{b}(b)}(Q^{·}) is G(Q^{·})^{∗}, where G^{∗} denotes taking the inverse cyclic ordering of the edges at each vertex of a graph G with an assigned cyclic ordering. Proof. It has already been established (cf. [R] or [M]), that any tilting of a Brauer tree algebra of type (e, m) is a Brauer tree algebra of type (e, m). However, since we must determine the particular choice of Brauer tree in this case, we will repeat the construction, basing ourselves on Theorem 2 and Theorem 3.

Let Q^{·} = ⨁_{i=1}^{e} R_{i}, where each R_{i} is the elementary complex corresponding to the edge numbered by i. Let π_{i} : Q^{·} → Q^{·} be the projection onto the factor R_{i}, and set I = {π_{i} | i = 1, …, e}. The permutations σ, ρ of the indices induce permutations of I, for which we will use the same names. Then, since the R_{i} are indecomposable and their sum is Q^{·}, the elements of I form a complete set of orthogonal primitive idempotents. We now define r to be the sum of the maps from the Local Order Theorem, at vertices of distance parity -1. We let s be the sum of the maps from the Local Order Theorem at the vertices of distance parity +1.

We now claim that the set I ∪ {r, s} satisfies the relations of the Brauer tree algebra for G(Q^{·})^{∗}, i.e., the relations given (a) – (f) in Definition 3.5, with ρ^{−1} and σ^{−1} in place of ρ and σ.

(a) Since the π_{i} are idempotents, they satisfy π_{i}^{2} = π_{i}. Since r is the sum of endomorphisms r_{i} given by extending to all of Q′ the homomorphism

\overline{r}_{i} : R_{i} → R_{ρ^{−1}(i)},

we get r ∘ π_{i} = π_{ρ^{−1}(i)} ∘ r. Similarly, since s is a sum of endomorphisms s_{i} gotten by extending to Q^{·} the homomorphism

\overline{S}_{i} : R_{i} → R_{σ^{−1}(i)},

we get

s ∘ π_{i} = π_{σ^{−1}(i)} ∘ s.

18 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

(b) Since Q^{·} is a direct sum, the projection π_{i} on the summands are orthogonal.

(c) The restriction of r ∘ s or s ∘ r to any summand R_{i} of Q′ will carry it to an indecomposable with no index in common with R_{i}, and thus the composition must be zero by Theorem 2(iii).

(d) Let R_{i} be an edge, such that neither vertex is exceptional. The composition of the maps from the Local Order Theorem is the socle map at the projective of positive distance parity and the negative of the socle map at the projective of negative distance parity. However, since the map from R_{i} to itself which is the socle map on each projective is homotopic to zero, these two maps coincide. Thus we get

π_{i} ∘ s^{|[π_{i}]_{σ^{−1}} |} = π_{i} ∘ r^{|[π_{i}]_{ρ^{−1}} |}.

(e) If the exceptional vertex is terminal, and R_{i} is the unique edge leading to

it, then

π_{i} ∘ s ∘ (s^{|[π_{i}]_{σ^{−1}} |})^{m} = ε_{i}^{m} ∘ s = 0.

(f) If R_{i} is an edge between the exceptional vertex and a non-exceptional vertex, then R_{i} is actually S_{i}. Then

π_{i} ∘ (s^{|[π_{i}]_{σ^{−1}} |})^{m} = ε_{i}^{m} = π_{i} ∘ r^{|[π_{i}]_{ρ^{−1}} |}.

By establishing (a) – (f), we have shown that End_{D^{b}(b)}(Q^{·}) is a quotient of the Bauer tree algebra with data G(Q^{·})^{∗} and h(z) = z^{m}. It remains to show that there are no additional relations. However, all the algebras involved are finite dimensional vector spaces over a field. End_{D^{b}(b)}(Q^{·}) has a basis consisting of the normal homogeneous maps given in Theorem 2. The Brauer tree algebra has a basis consisting of the elements i and products of the endomorphisms r_{i} and s_{i}. An edge-by-edge comparison based on the three cases in (d) – (f) shows that in both algebras the indecomposable projectives have the same dimension |[i]_{σ}| + |[i]_{ρ}| (or

POINTED BRAUER TREES 19

m|[i]_{σ}| + |[i]_{ρ}| if [i]_{σ} is exceptional). Letting v_{j} represent the valence of vertex u_{j}, then since there are v_{j} indecomposable projectives whose dimension involves v_{j} we get the total dimension

∑_{j≠i} v_{j}^{2} + m ∑ v_{0}^{2}.

The ordering of the edges at a vertex u_{i} in the Brauer tree is determined by the composition factors in the corresponding uniserial submodule. However, the ordering of the composition factors in the endomorphism algebra is exactly dual to the ordering of the minimal maps between indecomposables in Q^{·}, and these describe the ordering of the edges in our graph G(Q^{·}). Thus the ordering in G(Q^{·})^{∗} is the ordering of the composition factors in End_{b}(Q^{·}). The vertex u_{0} acts as exceptional vertex, because there the sequence of minimal morphisms makes m circuits of u_{0}.

§6. The correspondence.

We now have well-defined maps

Q : B′ → T C_{2}
G : T C_{2} → B′,

and it remains to show that G ∘ Q = 1_{B′} and Q ∘ G = 1_{T C_{2}}. The first of these is relatively easy; the second is much harder. Proposition 6.1. If (B, v) is a pointed Brauer tree of type (e, m), then G(Q(B, v)) = (B, v). Proof. Let τ be the renumbering of the Brauer star b which renumbers 2 as 1, 3 as 2,…, i + 1 as i and 1 as e. This induces a permutation τ of T C_{2}; for any Q^{·} ∈ T C_{2}, τQ^{·} will be the tilting complex obtained by renumbering the projectives. Suppose that for some vertex v_{1} we can show that

G(Q(B, v_{1})) = (B, v_{1}).

20 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Let v_{2} be the vertex of B numbered as 2 in the numbering of B induced by v_{1}. Then Q(B_{,}v_{2}) = τ(Q(B, v_{1})). Now G(τQ^{·}) is identical as a pointed Brauer tree with G(Q^{·}), the only difference being a permutation in the numbering of the vertices. The vertex now numbered 1 is the vertex v_{2} which was once numbered 2, and then

G(τ(Q(B, v_{1}))) = (B, v_{2}).

Thus we get

G(Q(B, v_{2})) = (B, v_{2}).

We continute to apply cyclic permutations and conclude that for any v, G(Q(B, v)) = (B, v).

Take any branch at u_{0}, and choose v_{1} to be the vertex at which one first encounters a point in that branch while taking a Green’s walk around the branch. If we take v_{1} as the marked vertex, then the numbers in the branch will be an interval [1, 2, 3, …, s] in ℕ. Subsequent branches will give subsequent intervals. We will now prove that G(Q(B, v_{1})) = (B, v_{1}). This will then establish that G(Q(B, v)) = (B, v) for any vertex, by the argument given at the beginning of the proof. Lemma 6.1. In Q^{·} = Q(B, v_{1}), for any T_{kℓ}[n′] in Q_{1}, we have k < ℓ. Proof. We have already proven that for any branch a containing T_{kℓ}[n′], st(a) → k → ℓ → end(a) is short. Since all the numbers on the branch Y (a) lie in an interval [st(a), end(a)] of ℕ, we get

st(a) ≤ k ≤ ℓ ≤ end(a).

Since k ≠ ℓ, we get k < ℓ, as desired. Lemma 6.2. B and G(Q(B, v_{1})) are identical as pointed Brauer trees.

The underlying graph of G(Q(B, v_{1})) and of B is the same, because we have the same e + 1 vertices v_{1} = u_{1}, u_{2} …, u_{e}, and in both graphs there is an edge between

POINTED BRAUER TREES 21

u_{j} and u_{k}, j ≠ k ≠ 0, j < k iff there is an elementary complex T_{jk}[n′] in Q(B, v) and an edge between u_{0} and u_{i} iff S_{i} ∈ Q(B, v).

It remains to describe the order of the edges at each vertex and the position of the points.

1) The exceptional vertex, u_{0}.

Let a_{1}, …, a_{s} be the edges at u_{0} in counterclockwise order, beginning with the edge whose branch Y (a_{1}) contains the marked vertex v_{1}. Let S_{j_{i}} be the elementary complex corresponding to a_{i}. Then, by Lemma 4.1, we will have that

st(a_{1}) → j_{1} → end(a_{1}) → st(a_{2}) → j_{2} → end(a_{2}) → ⋯ → end(a_{s})

is a short sequence. This then implies, by the algorithm for G(Q^{·}), that the ordering of the edges in G(Q(B, v_{1})) will be S_{j_{1}}, …, S_{j_{s}}, just as in B.

2) Let u_{i} be any other vertex, let a be the entering edge, corresponding either to T_{ji}[s(i) + 1] or T_{ik}[s(i)]. Let b_{1}, …, b_{k} be the edges in order at u_{i} before the point and let c_{1}, …, c_{ℓ} be the edges after the point.

By combining (3) and (1) in Lemma 4.1, we have that

st(a) → b_{1} → b_{2} → ⋯ → b_{k} → i → c_{1} → ⋯ → c_{ℓ} → end(a)

is short. Since [st(a), end(a)] is an interval in ℕ, we have

b_{1} < b_{2} < ⋯ < b_{k} < i < c_{1} < ⋯ < c_{ℓ}.

We now consider the three possibilities for the elementary complex R_{i} corresponding to the entering edge a. We designate the point by “pt”.

(a) \underline{R_{i} = S_{i}} : In this case, by the Local Order Theorem, the above inequalities determine the ordering

S_{i}, T_{b_{1}i}[s(i) + 1], …, T_{b_{k}i}[s(i) + 1], pt, T_{ic_{1}}[s(i)], …, T_{ic_{ℓ}}[s(i)].

(b) \underline{R_{i} = T_{ji}[s(i) + 1]} : Let a′ be the branch at u_{0} containing u_{i}, then

st(a′) → j → st(i) → end(i) → end(a′) is short.

Thus

j < b_{1} < b_{2} < ⋯ < b_{k} < i < c_{1} < ⋯ < c_{ℓ},

so

j → b_{1} → b_{2} → ⋯ → b_{k} → i → c_{1} → ⋯ → c_{ℓ}

is short. Thus the ordering of the edges at u_{i}, as given by Theorem 3, is

T_{ji}[s(i) + 1], T_{b_{1}i}[s(i) + 1], pt, T_{ic_{1}}[s(i)] … T_{ic_{ℓ}}[s(i)].

(c) \underline{R_{i} = T_{ij}[s(i)]} : As before, let i′ be the source of the branch at u_{0} containing u_{i}. Then st(a′) → st(i) → end(i) → j → end(a′). We get

b_{1} < b_{2} < ⋯ < b_{k} < i < c_{1} < ⋯ < c_{ℓ} < j,

POINTED BRAUER TREES 23

and thus

b_{1} → b_{2} ⋯ → b_{k} → i → c_{1} → ⋯ → c_{ℓ} → j

is short. The corresponding sequence of edges is

T_{b_{1}i}[s(i) + 1], …, T_{b_{k}i}[s(i) + 1], pt, T_{ic_{1}}[s(i)], …, T_{ic_{ℓ}}[s(i)], T_{ij}[s(i)].

Thu s in all the cases we get back the same pointed Brauer graph.

We now turn to the difficult direction, showing that Q(G(Q^{·})) = Q^{·}. There are two essential difficulties: gaps and overlap. The first problem is that of gaps. Suppose we had a Brauer tree, and we lopped off some branches, leaving the old numbering. Then the corresponding sequence of vertex numbers as we took a Green’s walk around the small pointed tree would give a short sequence, but it would have gaps for the branches we pruned off, and there is nothing in the local information to prevent this.

The second problem is overlap. When we try to slice two short sequence together, we might get a long sequence unless we have some control on the start and end vertices in each branch. Because of this problem, we find it expedient to work outward by induction on the distance from the exceptional vertex, proving that as we take the Green’s walk on a pruned version of the tree we get a short sequence which possibly contains gaps. After a finite number of steps, we will have all e vertices in a short sequence. Then it is no longer possible that there be gaps.

We need one other key idea before we can proceed with our induction, the idea of a proper vertex. In the algorithm for constructing a complex from a tree, the entering edge at vertex i is always the first of the edges assigned a complex T_{∗i} or the last of the edges assigned a complex T_{i∗} as in (b) or (c), in the proof of Lemma 6.2. We must show that this is also true when we construct a tree from a tilting complex.

24 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

In a P T C_{2}, all occurrences of P_{i} occur with the same displacement s(i). Thus the shift of a T_{ij} is always s(i) = s(j) + 1. Rather than write this each time, we will write

T_{ij}[−] = T_{ij}[s(i)] = T_{ij}[s(j) + 1].

Definition. Let G(Q^{·}) be the tree constructed from a complex Q^{·} ∈ T C_{2}, with vertices u_{0}, …, u_{e}. Let us call a vertex u_{i} proper if it is u_{0}, or if it is adjacent to u_{0}, or if the entering edge is, in counterclockwise order, either the first edge of type T_{ji}[−] giving

T_{ji}[s(i) + 1], T_{b_{1}i}[s(i) + 1], pt, T_{ic_{1}}[s(i)] … T_{ic_{ℓ}}[s(i)].

or the last of type T_{ij}[−] giving

T_{b_{1}i}[s(i) + 1], …, T_{b_{k}i}[s(i) + 1], pt, T_{ic_{1}}[s(i)], …, T_{ic_{ℓ}}[s(i)], T_{ij}[s(i)].

We will call G(Q^{·}) proper if every vertex is proper. Definition. A sequence of edges a_{1}, …, a_{d} in a path outward from u_{0} will be called a nested zigzag if we can choose corner vertices with indices a_{d_{i}} where d_{1} = 1 < d_{2} < ⋯ < d_{r} = d such that

1) if J_{i} = a_{d_{i−1}} → a_{d_{i−1}+1} → ⋯ → a_{d_{i}}, then one of J_{i} or J_{i}^{∗} is short.

2) if i ≠ 1, r, then J_{i} → a_{d_{i−1}−1} is short if J_{i} is short and a_{d_{i−1}−1} → J_{i}^{∗} is short if J_{i}^{∗} is short. In the illustration, we have arranged the points so that a projection on the horizontal axis is in ascending order by index. We recall that u_{a_{i}} is the vertex of edge a_{i} which is farthest from the exceptional vertex.

Lemma 6.3. If a_{1}, a_{2} … a_{d} is a path from u_{0} to u_{a_{d}} in G(Q^{·}), and each of the vertices u_{a_{1}} … u_{a_{d−1}} is proper, then the sequence is a nested zigzag, and the corner vertices other than u_{a_{1}} are the vertices at which the point changes sides on the path.

Proof. Set d_{1} = 1 and let d_{2}, … d_{r−1} ≤ d − 1 to be the vertices at which the point changes sides.

1) Let J_{i} = a_{d_{i−1}} → a_{d_{i−1}+1} ⋯ → a_{d_{i}−1} → a_{d_{i}}. We claim that if the point is on the right side of u_{a_{d_{i−1}}} then J_{i} is short and if it is on the left of u_{a_{d_{i−1}}} then J_{i}^{∗} is short. Assume it is on the right (as occurs in Fig. 3 for i − 1 = 1 or i − 1 = 3). The complex corresponding to the edge a_{d_{i−1}+1} must have the form T_{a_{d_{i−1}} a_{d_{i−1}+1}}[−]. If the point is still on the right at the next vertex, then we get an edge with complex T_{a_{d_{i−1}+1}a_{d_{i−1}+2}}[−]. By the Local Order Theorem

a_{d_{i−1}} → a_{d_{i−1}+1} → a_{d_{i−1}+2}

is short, because the complex of the entering edge has the form of the complex of an edge before the point. Continuing until d_{i} we get J_{i} short.

If the point had been on the left side we would have gotten J_{i}^{∗} short by a dual argument.

2) We now assume that the point is on the left from u_{a_{d_{i}}} to u_{a_{d_{i+1}−1}}. At u_{a_{d_{i}}} we have an entering edge corresponding to a complex

T_{a_{d_{i}−1}a_{d_{i}}} [−],

and the next edge in the sequence corresponds to

T_{a_{d_{i}+1}a_{d_{i}}} [−].

Now we apply the fact that u_{a_{d_{i}}} is proper, to conclude that, since the entering edge is the first of its type, then by Theorem 3

a_{d_{i}−1} → a_{d_{i}+1} → a_{d_{i}}

is short. By 1) above we know that J_{i+1}^{∗} is short.

J_{i+1}^{∗} : a_{d_{i+1}} → a_{d_{i+1}−1} → ⋯ → a_{d_{i}+1} → a_{d_{i}}.

Since, by Theorem 1(d), we cannot have T_{a_{d_{i−1}} a_{d_{i}}} [−] appearing together in a tilting complex with T_{ℓℓ′}[−] where

ℓ′ → a_{d_{i}−1} → ℓ → a_{d_{i}}

is short, we conclude that

a_{d_{i−1}} → a_{k} → a_{k−1} → a_{d_{i}}

is short for all k = d_{i} + 1, …, d_{i+1}. Thus a_{d_{i−1}} → J_{i+1}^{∗} is short, as desired.The dual case when the point switches from left to right at d_{i} is exactly similar. Proposition 6.2. G(Q^{·}) is proper. Proof. We must prove that every vertex is proper. We work by induction on the length of the path leading outward from u_{0} to the vertex. Assume that our path is a_{1}, …, a_{d}, and that every vertex u_{a_{1}}, …, u_{a_{d−1}} is proper. We must show that u_{a_{d}} is proper.

If d = 1, there is nothing to prove, for in that case the entering vertex corresponds to some S_{1} and by definition u_{a_{1}} is proper, being adjacent to u_{0}. Assume that d > 1. By Lemma 6.1, the path is a nested zigzag. Let d_{1} = 1, d_{2} … d_{r−1} be the corner vertices with d_{i} < d. Set d_{r} = d. Case 1. \underline{r = 1}. Consider the case that the point at a_{1} is on the right. We have a_{1} → a_{2} → ⋯ → a_{d}. Thus the entering edge corresponds to a complex T_{a_{d−1}a_{d}}[−].

Assume for the sake of contradiction that the vertex at a_{d} is not proper. Then T_{a_{d−1}a_{d}}[−] is not the first complex of the form T_{ℓa_{d}}[−]. There is a complex T_{ya_{d}}[−] with y → a_{d−1} → a_{d} short. However, by Theorem 1(d), we cannot have

a_{k−1} → y → a_{k} → a_{d},

for any k = 2, …, d − 1. Thus we get

y → a_{1} → a_{2} ⋯ → a_{d−1} → a_{d}

short, and in particular y → a_{1} → a_{d} short. This is a contradiction to Theorem 1(c).

The dual case follows in the same way.

28 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

Case 2. \underline{r > 1}. We now assume that the point is on the left at a_{d−1}. We now have that

a_{d_{r−1}−1} → a_{d} → a_{d−1} → ⋯ → a_{d_{r−1}}

is short. In order for u_{a_{d}} = u_{a_{dr}} not to be proper, we would have to have an edge corresponding to T_{a_{d}y} with

a_{d} → a_{d−1} → y short.

We cannot have a_{d} → a_{k} → y → a_{k−1} short by Theorem 1(d), and similarly we cannot have

a_{d_{r−1}−1} → a_{d} → a_{d_{r−1}} → y short,

again by Theorem 1(d) with T_{a_{d_{r−1}−1}a_{d_{r−1}}} [−] and T_{a_{d}y}[−]. The order a_{d_{r−1}−1} → y → a_{d} → a_{d_{r−1}} is also impossible by Theorem 1(d) for the same two complexes. Therefore, no such edge exists.

Definition. Let B be a pointed tree. Let d ≥ 0 be a natural number. Let B(0) = u_{0}. For any d > 0, define B(d) to be the subtree consisting of all vertices and edges of distance d′ ≤ d, and, for each vertex u with d(u) ≤ d, all vertices and edges in a forward and backward Green’s walk from u until the first point in each direction. B(d) will be called a truncated pointed tree.

Example 2.

POINTED BRAUER TREES 29

Our aim is to prove the following theorem. Continuity Theorem. Let Q^{·} ∈ T C_{2}, and let (B, v_{1}) = G(Q^{·}) be the corresponding Brauer tree. For any d ≥ 0, if we take a pointed Green’s walk around B(d) starting at u_{0}, then the vertices we encounter form a short sequence. Proof. Induction on d. For B(0) there is nothing to prove. We consider B(1). Let i_{1}, …, i_{s} be the edges at u_{0}. Let st(i_{j}) be the first point encountered on a Green’s walk at i_{j}, and end(i_{j}) the first point on the reverse Green’s walk.

Consider first the case that there is a single branch i at u_{0}. Letting a_{1}, … a_{d} be indices of the sequence of vertices on the forward walk and b_{1}, … b_{d′} the indices of the sequence on the backwards walk. Claim. st(i) = a_{d} → a_{d−1} → ⋯ → a_{1} → i → b_{1} → ⋯ → b_{d′} = end(i) is short. Since the points are all on the left on the a sequence and on the right in the b sequence, it is a consequence of Lemma 6.1 that each is monotonic. Since a_{d} → a_{d−1} → ⋯ → a_{1} is short and a_{k} → i → a_{k−1} is long for each k − 2, …, d, we get that a_{d} → a_{d−1} → ⋯ → a_{1} → i is short, and similarly i → b_{1} → ⋯ → b_{d′} is short. By the Local Order Theorem at i, we have a_{1} → i → b_{1} short. By the refinement property [SZ1], it will suffice to show that a_{d} → a_{1} → i → b_{1} → b_{d′} is short. By Lemma 4.1(1), a_{d} → i → b_{d′} is short. Thus the order given in the claim is the only arrangement of these five points satisfying all the conditions.

Now suppose that there are s > 1 edges at u_{0}.

Let α̃_{k} be the sequences of indices on the forward walk at vertex of index i_{k}, given by a_{kd_{k}} → a_{k(d_{k}−1)} → ⋯ → a_{k1}, and let β̃_{k} be the sequence of indices on the backward walk, b_{k1} → b_{k2} → ⋯ → b_{kd_{k}}. We must show that

(∗)  α̃_{1} → i_{1} → β̃_{1} → α̃_{2} → i_{2} → β̃_{2} ⋯ → α̃_{s} → i_{s} → β̃_{s}

is short. Some rearrangement is short. Each α̃ and β̃ sequence must preserve its internal order, and no α̃_{j} or β̃_{j} can overlap with any other α̃_{k} or β̃_{k} because of

30 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

contradiction with Theorem 1(d). Similarly, no i_{j} can enter into any α̃_{k} or β̃_{k} by Theorem 1(c). As we already proved in the claim above, each α̃_{j} → i_{j} → β̃_{j} must be short. Thus the problem remaining is to show that if j ≠ k, we have a short sequence

i_{j} → β̃_{j} → α̃_{k} → i_{k}.

Since the first edge in the left branch corresponds to T_{i_{j}b_{j1}}[−], by Theorem 1(c), we have i_{j} → b_{j1} → i_{k} short, and dually i_{j} → a_{k1} → i_{k} is short for any k ≠ j.

Thus the only possible rearrangement of all the i_{j}, α̃_{j}, β̃_{j} which is compatible is the arrangement (∗). This proves the theorem for B(1).

We now assume by induction that the theorem has been proven for all d′ < d, with d > 1. u_{j} at distance d − 1.

Let us start at a vertex u_{j} at distance d − 1 on B(d − 1). Suppose it has vertex u_{i} at distance d attached to u_{j} in B(d) by edge i. There are two dual situations depending on whether edge i is to the right of the point at u_{j} or to the left as we go outward from u_{0}. We will assume it is to the right, and thus that there are edges to the right, the other case being dual. Let i_{1} … i_{s} be the sequence of edges at u_{j} to the right of the point, with u_{i_{1}} ∈ B(d − 1) because it lies on the forward Green’s walk at u_{j}. From the Local Order Theorem, we have that

i_{1} → i_{2} → ⋯ → i_{s} → j.

is short. For each u_{i_{ℓ}}, let α̃_{ℓ} be the forward Green’s walk to the first point, and β̃_{ℓ} be the backward Green’s walk to the last point. We must prove that

(∗′)  α̃_{1} → i_{1} → β̃_{1} → α̃_{2} → ⋯ → i_{s} → β̃_{s} → j

is short. As was already established in Lemma 6.2, the α̃_{ℓ} and β̃_{k} are monotonic. They cannot overlap without violating Theorem (d). By Lemma 6.2, we have i_{k} → β̃_{k} → j since u_{i_{k}} is a corner vertex in the nested zigzag ending at the top of β̃_{k}, because the point lies to the left of u_{i_{k}}, and to the right at each a_{b_{kℓ}} in β̃_{k}.

POINTED BRAUER TREES 31

We next claim that for each k ≠ s, we have that

(∗∗)  i_{k} → β̃_{k} → α̃_{k+1} → i_{k+1} → j.

is short. To prove this, we first note that we have that

i_{k} → i_{k+1} → j

is short by the Local Order Theorem at vertex j and

i_{k} → β̃_{k} → j

is short because of Lemma 6.2 and the fact that u_{i_{k}} is a corner vertex in the nested zigzag leading to the top of β̃_{k}.

If the order were i_{k} → i_{k+1} → β̃_{k} → j, then we would have

i_{k} → i_{k+1} → b_{k1} → j.

Since T_{i_{k+1}j}[−] and T_{i_{k}b_{k1}}[−] are in Q^{·} this would contradict Theorem 1(d). Thus we have

i_{k} → β̃_{k} → i_{k+1} → β̃_{k+1} → j.

Since the sequence a_{k+1} → i_{k} → i_{k+1} → j is impossible for the same reason we see that α̃_{k+1} must also be between i_{k} and i_{k+1}. If either α̃_{k+1} or β̃_{k} is empty, we are finished. We cannot have overlaps by Theorem 1(d) and we cannot have

i_{k} → α̃_{k+1} → β̃_{k} → i_{k+1} → j

short because this would give i_{k} → a_{(k+1)1} → β_{k1} → i_{k+1} short, in contradiction to Theorem 1(d) and the presence of T_{i_{k}b_{k}1}[−] and T_{a_{(k+1)1}i_{k+1}}[−]. We have thus established that the sequence (∗∗) is short.

Combining all these results, we get that

i_{1} → β̃_{1} → α̃_{2} → i_{2} → ⋯ → i_{s} → β̃_{s} → j

32 MARY SCHAPS AND EVELYNE ZAKAY-ILLOUZ

is short. It remains only to add α_{1} to the beginning of the sequence. However, i_{1} lies on B(d − 1) as do the vertices of α̃_{1}, so we have α̃_{1} → i_{1} → j already from B(d − 1). This shows that (∗′) is short. Thus, for B(d) as well, the pointed Green’s walk produces a short sequence.

§7. The Main Theorem.

Main Theorem. There is a one-to-one correspondence between the set of tworestricted tilting complexes for the Brauer star algebra b of type (e, m) and the set of pointed Brauer trees of type (e, m) with a marked vertex. To a tilting complex Q^{·} in T C_{2}, this correspondence associates a Brauer tree which is dual to the Brauer tree of End_{b}(Q^{·}).

Proof. We have already defined the functions

G : T C_{2} → B′

and

Q : B′ → T C_{2}.

We have also shown that G ∘ Q(B, v) = (B, v) in Proposition 6.1.

We must now show that Q ∘ G(Q^{·}) = Q^{·}. By the Continuity Theorem, as we take a Green’s walk around G(Q^{·}) we get a short sequence of indices. We pass all the vertices 1, …, e, and the only short sequences with e vertices are cyclic permutations of

1 → 2 → ⋯ → e.

We conclude that the vertices form a complete sequence without gaps. The marked vertex is that numbered by 1. Thus as we take a pointed Green’s walk around the tree, the complex Q(B, v) produced by the algorithm Q is precisely the original tilting complex.

POINTED BRAUER TREES 33

§8. Applications.

The original motivation for this work was the study of deformations. The deformation of a block of cyclic defect given in [S2] was unnatural, in that it involved a number which had no particular connection to the structure of the block. We had hoped that an understanding of the tilting theory of the blocks would lead to a more natural definition, and that is the program carried out in the third paper of this series [SZ2], which represents the last section of the second author’s Ph.D. thesis.

We feel that the results have opened up other possible applications as well. One of the topics of research right now in tilting theory is the problem of self-equivalence of the derived category. The two-sided tilting complexes inducing a self-equivalence form a group, known as the derived Picard group. The two-restricted complexes form, in a sense, a complement to the self-equivalences of the Brauer star algebra. There are no two-restricted self-equivalences, yet the set of two-restricted complexes is large enough to reach every Brauer tree, usually in several ways. The first author and Rickard have shown that the complex inverse to a two-restricted complex Q^{·} is a version of the tree-to-star complexes in [R2], folded according to G(Q^{·}), [RS]. If we could construct two-sided complexes for these, then by taking the tilting complex for one pointing and the inverse complex for another, we would get a number of self-equivalences. In fact, by taking any tilting complex for the Brauer star, and multiplying it by the inverse of the complex for the corresponding Brauer tree, we should be able to factor a tilting as a self-equivalence followed by a two-restricted complex. It may be possible to generate all self-equivalences in this way.

A third possible application consists of considering special tilting complexes for other blocks whose Brauer correspondent has a simple structure. Two possibilities are: (1) to look at products of Brauer star algebras, or (2) groups of the form (C_{p^{n}} × C_{p^{n}}) ⋊ C_{e}, where the action of the generator on the two factors is inverse (as in the alternating group A4). Perhaps one can find some special set of tilting

complexes for these algebras which encode the structure of the block the way that

the Brauer tree and the two-restricted tilting complex do.

References

[L] Linckelmann M., Derived equivalence for cyclic blocks over a P -aic ring, Math. Z. 207 (1991), 293-304.

[M] Membrillo F.H., Homological Properties of Finite Dimensional Algebras, Ph.D. Thesis, Oxford University, 1993, pp. 6-13, 78.

[R1] Rickard J., Morita theory for derived categories, J. London Math. Soc. 39 (1989), no. 2, 436-456.

[R2] Rickard J., Derived categories and stable equivalence, J. of Pure and Applied Algebra 61 (1989), 303-317.

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

[RS] Rickard J. and Schaps M., Folded tilting complexes for the Brauer star algebra, preprint.

[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.

[SZ1] M. Schaps and Zakay-Illouz E., Combinatorial partial tilting complexes for the Brauer star algebras, 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.

Department of Mathematics and Computer Science, Bar-Ilan University, 52900 Ramat-Gan, Israel

Department of Mathematics, Jordan Valley College, Jordan Valley, Israel