Ibrahim Assem

M Andrea Gatica

Ralf Schiffler

Ralf Schiffler 2011 The

The higher relation bimodule

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Introduction

The objective of this paper is to describe a new class of algebras, which we call higher relation extensions. Our motivation comes from the study of cluster-tilted algebras, introduced by Buan, Marsh and Reiten in [BMR], and in [CCS] for type A. Indeed, it was shown in [ABS] that an algebra A is cluster-tilted if and only if there exists a tilted algebra C such that A is isomorphic to the trivial extension of C by the C -C-bimodule Ext 2 C (DC, C). Moreover, a recipe for constructing the quiver of this trivial extension was given in [START_REF] Assem | Cluster-tilted algebras as trivial extensions[END_REF]Theorem 2.6]. The proof of the latter result rests on the fact that tilted algebras have global dimension 2.

Here, we consider the more general case of an algebra A having an arbitrary finite global dimension and consider its trivial extension by the bimodule i≥2 Ext i A (DA, A), which we call the higher relation bimodule. We believe that this class of algebras, which we call higher relation extensions, will be useful in the study of m-cluster-tilted algebras (see [FPT] [B]). Our first objective is to describe the ordinary quiver of the higher relation extension of A in the case where A is a string algebra in the sense of Butler and Ringel [BR]. We also assume that the quiver of A is a tree. This is no restriction, because the universal cover of a string algebra is a string tree [G]. Our theorem reads as follows.

Theorem 1.1 Let A = kQ/I be a string tree algebra. Then there exist two sequences (c ℓ ), (z ℓ ) of points of Q such that the arrows in the quiver of the higher relation extension are exactly those of Q plus one additional arrow from each z ℓ to c ℓ .

Our proof is constructive, in the sense that we give an algorithm allowing to construct explicitly the sequences (c ℓ ) and (z ℓ ) and thus the quiver of the higher relation extension.

We then consider the particular case where A is a gentle algebra. Gentle algebras form an important subclass of the class of string algebras. Part of their importance comes from the fact that this subclass is stable under derived equivalences [SZ]. While, as we show, the higher relation extension algebra of a gentle algebra is monomial but not necessarily gentle, we prove using our Theorem 1.1, that the tensor algebra of the higher relation bimodule is gentle.

Theorem 1.2 Let A = kQ/I be a gentle algebra, then the tensor algebra of the higher relation bimodule i≥2 Ext i A (DA, A) is gentle. The paper is organised as follows. In section 2, we fix the notation and recall some facts and results about string and gentle algebras. Section 3 is devoted to the computation of projective resolutions and injective coresolutions of uniserial modules over a string algebra. We study the top of the higher extension bimodule in section 4 and we prove Theorem 1.1 in section 5. Sections 6 and 7 are devoted to the case of gentle algebras.

Preliminaries

Notation

Throughout this paper, algebras are basic and connected finite dimensional algebras over an algebraically closed field k. Given an algebra A, there always exists a (unique) quiver Q = (Q 0 , Q 1 ) and (at least) an isomorphism A ∼ = kQ/I, where kQ is the path algebra of Q, and I is an admissible ideal of kQ, see, for instance, [ASS]. Such an isomorphism is called a presentation of the algebra. Given an algebra A, we denote by mod A the category of finitely generated right A-modules, and by D = Hom k (-, k) the standard duality between mod A and mod A op . For a point x in the quiver Q of A, we denote by P (x), I(x), S(x), e x respectively, the corresponding indecomposable projective module, injective module, simple module and primitive idempotent. We recall that a module M can be equivalently considered as a bound quiver representation M = (M i , M α ) i∈Q 0 ,α∈Q 1 . The projective, or injective, dimension of a module M is denoted by pd M, or id M, respectively. The global dimension of A is denoted by gldim A. For facts about the category mod A, we refer the reader to [ARS] or [ASS].

Trivial extensions

Let A be an algebra and M an A -A-bimodule . The trivial extension of A by M is the algebra A ⋉ M with underlying k-vector space

A ⊕ M = {(a, m)|a ∈ A, m ∈ M}
and the multiplication defined by

(a, m).(a ′ , m ′ ) = (a.a ′ , am ′ + ma ′ )
for a, a ′ ∈ A and m, m ′ ∈ M.

For instance, an algebra A is cluster-tilted if and only if there exists a tilted algebra C such that A is the trivial extension of C by the so-called relation bimodule Ext 2 C (DC, C), see [ABS].

The ordinary quiver of a trivial extension is computed as follows (see, for instance, [ABS]): let M be an A -A bimodule, then the quiver

Q A⋉M of A ⋉ M is given by 1) Q A⋉M 0 = (Q A ) 0 2) For z, c ∈ (Q A ) 0 ,

String algebras

Recall from [BR] (see also [WW]) that an algebra A is called a string algebra if there exists a presentation A = kQ/I (called a string presentation) such that: S1) I is generated by a set of paths (thus A is monomial). S2) Each point in Q is the source of at most two arrows and the target of at most two arrows.

S3

) For an arrow α, there is at most one arrow β and at most one arrow γ such that αβ / ∈ I and γα / ∈ I.

Whenever we deal with a string algebra A, we always assume that it is given by a string presentation A = kQ/I. We assume moreover that the relations (that is, the generators of I) are of minimal length.

A reduced walk ω in Q is called a string if it contains no zero relations.To each string ω in Q, we can associate a so-called string module [BR] in the following way. If ω is the stationary path at j, then M(ω) = S(j). Let ω = ω 1 ω 2 • • • ω t be a string, with each ω i an arrow or the inverse of an arrow. For each i such that 0 ≤ i ≤ t, let V i = k; and for 1 ≤ i ≤ t, let V ω i be the identity map sending x ∈ V i to x ∈ V i+1 if ω i is an arrow and otherwise the identity map sending x ∈ V i+1 to x ∈ V i . The string module M(ω) is then defined as follows: for each j ∈ Q 0 , M(ω) j is the direct sum of the vector spaces V i such that the source of ω i is j if j appears in ω, and otherwise M(ω) j = 0; for each α ∈ Q 1 , M(ω) α is the direct sum of the maps V ω i such that ω i = α or ω -1 i = α if α appears in ω, and otherwise M(ω) α = 0. A non-zero path ω in Q for a to b will sometimes be denoted by [a, b], whenever there is no ambiguity. Then, the corresponding string module is denoted by

M(ω) = M([a, b]).
We also recall that the endomorphism ring of a projective module over a string tree algebra A (a full subcategory of A) is also a string tree algebra.

Gentle algebras

Recall from [AS] that a string algebra A = kQ/I is called gentle if in addition to (S1), (S2), (S3), the bound quiver (Q, I) satisfies: G1) For an arrow α, there is at most one arrow β and at most one arrow γ such that αβ ∈ I and γα ∈ I.

G2) I is quadratic (that is, I is generated by paths of length 2).

For instance, cluster-tilted algebras of types A and à are gentle [ABCP].

Resolutions of uniserial modules

In this section, we compute minimal projective resolutions of an injective module, and dually minimal injective coresolutions of a projective module over a string algebra. Throughout, we let A = kQ/I be a string presentation.

Definition 3.1 Let [x 0 , y 0 ] be a non-zero path from x 0 to y 0 in Q. We define inductively the right maximal sequence of [x 0 , y 0 ] as follows. This is a finite sequence of non-zero paths

[x i 1 i 2 •••it , y i 1 i 2 •••it ] with i 1 = 0 and i j ∈ {0, 1} such that 1) Let [x 0 , y 00 ],
[x 0 , y 01 ] be the maximal non-zero paths starting at x 0 (where we agree that

[x 0 , y 0 ] is contained in [x 0 , y 00 ]).
Then we set

[x 00 , y 00 ] = [x 0 , y 00 ]\[x 0 , y 0 ] and [x 01 , y 01 ] = [x 0 , y 01 ]\[x 0 , y 0 ] = [x 0 , y 01 ]\{x 0 }. 2) Inductively, assume that [x 0i 2 •••i t-1 , y 0i 2 •••i t-1 ] has been defined. Let [x 0i 2 •••i t-1 , y 0i 2 •••i t-1 0 ] and [x 0i 2 •••i t-1 , y 0i 2 •••i t-1 1 ] be the maximal non-zero paths starting at x 0i 2 •••i t-1 , where we agree that [x 0i 2 •••i t-1 , y 0i 2 •••i t-1 ] is contained in [x 0i 2 •••i t-1 , y 0i 2 •••i t-1 0 ].
Then we set

[x 0i 2 •••i t-1 0 , y 0i 2 •••i t-1 0 ] = [x 0i 2 •••i t-1 , y 0i 2 •••i t-1 0 ]\[x 0i 2 •••i t-1 , y 0i 2 •••i t-1 ] and [x 0i 2 •••i t-1 1 , y 0i 2 •••i t-1 1 ] = [x 0i 2 •••i t-1 , y 0i 2 •••i t-1 1 ]\{x 0i 2 •••i t-1 }.
The left maximal sequence of a non-zero path is defined dually. However, we do it explicity for the convenience of the reader. Definition 3.2 Let [r 0 , s 0 ] be a non-zero path from r 0 to s 0 in Q. We define inductively the left maximal sequence of [r 0 , s 0 ] as follows. This is a finite sequence of non-zero paths

[r i 1 i 2 •••it , s i 1 i 2 •••it ] with i 1 = 0 and i j ∈ {0, 1} such that 1) Let [r 00 , s 0 ],[
r 01 , s 0 ] be the maximal non-zero paths ending at s 0 , where we agree that

[r 0 , s 0 ] is contained in [r 00 , s 0 ]. Then we set [r 00 , s 00 ] = [r 00 , y 0 ]\[r 0 , s 0 ] and [r 01 , s 01 ] = [r 01 , y 0 ]\[r 0 , s 0 ] = [r 01 , s 0 ]\{s 0 }. 2) Inductively, assume that [r 0i 2 •••i t-1 , s 0i 2 •••i t-1 ] has been defined. Let [r 0i 2 •••i t-1 , s 0i 2 •••i t-1 0 ] and [r 0i 2 •••i t-1 , s 0i 2 •••i t-1 1 ] be the maximal non-zero paths ending at s 0i 2 •••i t-1 , where we agree that [r 0i 2 •••i t-1 , s 0i 2 •••i t-1 ] is contained in [r 0i 2 •••i t-1 , s 0i 2 •••i t-1 0 ]. Then we set [r 0i 2 •••i t-1 0 , s 0i 2 •••i t-1 0 ] = [r 0i 2 •••i t-1 , s 0i 2 •••i t-1 0 ]\[r 0i 2 •••i t-1 , s 0i 2 •••i t-1 ] and [r 0i 2 •••i t-1 1 , s 0i 2 •••i t-1 1 ] = [r 0i 2 •••i t-1 , s 0i 2 •••i t-1 1 ]\{s 0i 2 •••i t-1 }.
Note that in both cases, some of the paths above might be empty and in this case, the points considered do not exist.

Our first result follows directly from the above definitions.

Theorem 3.3 Let A = kQ/I be a string algebra. a) If [x 0 , y 0 ] is a non-zero path in Q and • • • → P 3 → P 2 → P 1 → P 0 → M[x 0 , y 0 ] → 0 is a minimal projective resolution then, for l ≥ 1, P l-1 = P (x i 1 i 2 •••i l )
where the direct sum is taken over all l-tuples (0,

i 2 , • • • , i l ) such that i k ∈ {0, 1} for all k with 2 ≤ k ≤ l and the point x 0i 2 •••i l in definition 3.1 exists. b) If [r 0 .s 0 ] is a non-zero path in Q and 0 → M[r 0 , s 0 ] → I 0 → I 1 → I 2 → I 3 → • • • is a minimal injective coresolution then, for l ≥ 1, I l-1 = I(s i 1 i 2 •••i l )
where the direct sum is taken over all l-tuples (0,

i 2 , • • • , i l ) such that i k ∈ {0, 1} for all k with 2 ≤ k ≤ l and the point s 0i 2 •••i l in definition 3.2 exists.
Proof. We only prove a), since the proof of b) is dual. Clearly, the projective cover of the uniserial module M[x 0 , y 0 ] is P (x 0 ), whose support consists of the (at most two) maximal non-zero paths [x 0 , y 01 ] and [x 0 , y 01 ] starting at x 0 . Then,

Ω 1 M[x 0 .y 0 ] = M[x 00 , y 00 ] ⊕ M[x 01 , y 01 ]
where x 00 and x 01 are defined as above. The rest follows from an easy induction.

Example 3.4 Suppose that the string algebra is given by the following bound quiver. This sequence may be conveniently shown in the following diagram

[3, 9] K K K K K [4, 5] {6} [10, 15] M M M M M {11} {12} {13} [16, 17] .
The minimal projective resolution of M[3, 9] is the following (compare with the above diagram) 0 → P (13) → P (12) → → P (16) ⊕ P (11) ⊕ P (6) → P (10) ⊕ P (4)

→ P (3) → M[3, 9] → 0,
where the morphisms are induced by the corresponding paths.

Similarly, taking [r 0 , s 0 ] = [3, 9], the left maximal sequence is

{1} [3, 9]
from which we deduce the minimal injective coresolution

0 → M[3, 9] → I(9) → I(1) → 0.
We are interested in computing resolutions of injective and projective indecomposable modules. These modules are usually not uniserial, neither are in general their first syzygy or cosyzygy, respectively. In order to apply Theorem 3.3, the next lemma shows that we must start from the second. Lemma 3.5

a) The second syzygy of an indecomposable injective module is the direct sum of at most six uniserial modules.

b) The second cosyzygy of an indecomposable projective module is the direct sum of at most six uniserial modules.

Proof. Let I(c) be an indecomposable injective A-module. If I(c) is uniserial, then there is nothing to prove because of Theorem 3.3. Otherwise, let top I(c) = S(a 0 ) ⊕ S(a 1 ). Then the projective cover of I(c) is P (a 0 ) ⊕ P (a 1 ). Let [a i , b i ] and [a i , b ′ i ] be the two maximal non-zero paths starting at a i (with i = 0, 1), where we agree that [a

0 , c] is contained in [a 0 , b ′ 0 ] and [a 1 , c] is contained in [a 1 , b ′ 1 ]. Let d i be the direct successor of a i on the path [a i , b i ] then Ω 1 I(c) = M[d 0 , b 0 ] ⊕ M[d 1 , b 1 ] ⊕ M
where M is an indecomposable module, usually non-uniserial, such that top M = S(c) and soc M = S(b ′ 0 ) ⊕ S(b ′ 1 ). Hence, the projective cover of Ω 1 I(c) is P (d 0 ) ⊕ P (d 1 ) ⊕ P (c), and Ω 2 I(c) is the direct sum of at most six uniserial modules obtained as follows.

Let [d i , b i0 ], [d i , b i1
] be the maximal non-zero paths starting in d i (with i = 0, 1), where we agree that

[d i , b i ] is contained in [d i , b i0 ]. Then let [d i0 , b i0 ] = [d i , b i0 ]\[d i , b i ] [d i1 , b i1 ] = [d i , b i1 ]\[d i , b i ] = [d i , b i1 ]\{d i }.
Let also [c, c 0 ] and [c, c 1 ] be the maximal non-zero paths starting at c, where we agree that

[c, b ′ 0 ] is contained in [c, c 0 ] and [c, b ′ 1 ] is contained in [c, c 1 ]. We let [c ′ 0 , c 0 ] = [c, c 0 ]\[c, b ′ 0 ] and [c ′ 1 , c 1 ] = [c, c 1 ]\[c, b ′ 1 ]. It is then clear that Ω 2 I(c) = M[d 00 , b 00 ] ⊕ M[d 01 , b 01 ] ⊕ M[d 10 , b 10 ] ⊕ M[d 11 , b 11 ] ⊕ M[c ′ 0 , c 0 ] ⊕ M[c ′ 1 , c 1 ] which establishes a). Statement b) is dual.
Corollary 3.6 a) Let I(c) be an indecomposable injective module such that top(I(c)) = S(a 0 ) ⊕ S(a 1 ). Then I(c) has the following minimal projective resolution

• • • → j;(0,i 2 ,i 3 ) P (x j 0i 2 i 3 ) → j;(0,i 2 ) P (x j 0i 2 ) → j P (x j 0 ) → → P (d 0 ) ⊕ P (c) ⊕ P (d 1 ) → P (a 0 ) ⊕ P (a 1 ) → I(c) → 0
with the morphisms induced by the paths, where

{x j 0 | 1 ≤ j ≤ 6} = {d 00 , d 01 , d 10 , d 11 , c ′ 0 , c ′ 1 } and i j ∈ {0, 1}.
b) Let P (z) be an indecomposable projective module such that soc(P (z)) = S(w 0 ) ⊕ S(w 1 ).

Then P (z) has the following minimal injective coresolution

0 → P (z) → I(w 0 ) ⊕ I(w 1 ) → I(v 1 ) ⊕ I(z) ⊕ I(v 2 ) → j I(s j 0 ) → → j;(0,i 2 ) I(s j 0i 2 ) → j;(0,i 2 ,i 3 ) I(s j 0i 2 i 3 ) → . . .
with the morphisms induced by the paths, where {s j 0 | 1 ≤ j ≤ 6} are as above and i j ∈ {0, 1}.

Proof. This follows from Lemma 3.5 and Theorem 3.3.

Corollary 3.7 With the above notations a) All the points x j 0 , x j 0i 2 , • • • , x j 0i 2 ...i l are targets of relations.

b) All the points s j 0 , s j 0i 2 , • • • , s j 0i 2 ...i l are sources of relations.

Proof. This follows from the construction of these points.

4

The top of the higher relation bimodule of A by its higher relation bimodule is called the higher relation extension of A.

If gldim A ≤ 2, then the higher relation extension of A coincides with its relation extension, as defined in [ABS].

Our objective in this section is to construct the ordinary quiver of the higher relation extension of a string algera A of finite global dimension.

As mentioned in the introduction, we also assume that the ordinary quiver Q A of A is a tree.

Let thus A = kQ/I be a string algebra, with Q a tree and M be an

A -A-bimodule. We have rad M = M(rad A) + (rad A)M and then top M = M/[M(rad A) + (rad A)M]. If M = i≥2 Ext i A (DA, A), then, clearly, top M = i≥2 top Ext i A (DA, A).
In order to describe this top, we start by describing the modules top A Ext i A (I(c), A) and top Ext i A (DA, P (z)) A for all points c, z ∈ (Q A ) 0 .

In the following, we use the notation of section 3.

Proposition 4.2 Let A = kQ/I be a string tree algebra and l ≥ 0. Then Ext l+2 A (I(c), P (z)) = 0 if and only if one of the following two conditions hold: a) there exists a non-zero path ω : z

x j i 1 i 2 •••i l+1 not passing through x j i 1 i 2 •••i l and whose compositions with x j i 1 i 2 •••i l+1 x j i 1 i 2 •••i l+2 are both zero. b) z = x j i 1 i 2 •••i l and x j i 1 i 2 •••i l 0 , x j i 1 i 2 •••i l 1 both exist.
In this case, a non-zero element is induced from the difference of the two paths

x j i 1 i 2 •••i l x j i 1 i 2 •••i l 0 and x j i 1 i 2 •••i l x j i 1 i 2 •••i l 1 .
Remark 4.3 Observe that in case (b), we have the following situation

x j i 1 i 2 •••i l-1 v / / / o / o / o / o / o / o / o z=x j i 1 i 2 •••i l u 0 / / / o / o / o / o / o / o / o u 1 ( ( ( h ( h ( h ( h ( h ( h ( h ( h x j i 1 i 2 •••i l 0 x j i 1 i 2 •••i l 1
where vu 0 , vu 1 are zero paths.

Proof. Let

• • • → ⊕P (x j i 1 •••i l+2 ) d l+3 → ⊕P (x j i 1 •••i l+1 ) d l+2 → ⊕P (x j i 1 •••i l ) → • • • → P c → I(c)
P l+2 = ⊕P (x j i 1 •••i l+1 ) has two indecomposable summands P (x j i 1 •••i l 0 ), P (x j i 1 •••i l 1
), whose images d(P (x j i 1 •••i l 0 )) and d(P (x j i 1 •••i l 1 )) lie in the same indecomposable summand P (x j i 1 •••i l ) of P l+1 , together with two non-zero morphisms ν i : P (x j i 1 •••i l i l+1 ) → P (z) (i l+1 = 0, 1) such that there exist two morphisms γ i :

P (x j i 1 •••i l ) → P (z) with ν i = γ i d. P (x j i 1 •••i l 0 ) ⊕ P (x j i 1 •••i l 1 ) d / / ν 1 ν 2 P (x j i 1 •••i l ) [γ 1 ,γ 2 ]
u u l l l l l l l l l l l l l l l P (z)

In this case [ν 1 -ν 2 ] t : P (x j i 1 •••i l 0 ) ⊕ P (x j i 1 •••i l 1 ) → P (z) does not factor through P (x j i 1 •••i l ) because dim k Hom(P (x j i 1 •••i l 0 ) ⊕ P (x j i 1 •••i l 1 ), P (z)) = 2 while dim k Hom(P (x j i 1 •••i l ), P (z)) = 1
, since the algebra is a tree algebra. This shows that Ext l+2 A (I(c), P (z)) = 0. Conversely, suppose that Ext l+2 A (I(c),

P (z)) contains a non-zero element [f ]. Then [f ] is in the class of a morphism f : ⊕P (x j i 1 •••i l+1 ) → P (z) such that f d l+3 = 0.
Since A is a tree string algebra, there are at most two indecomposable summands on which f is non-zero, because otherwise there are non-zero paths from z to three points x j i 1 •••ı l+1 and these induce a full subcategory of type D 4 which contradicts the fact that A is string. Thus we get a morphism

f : P (x j i 1 •••i l+1 ) ⊕ P (x j ′ i ′ 1 •••i ′ l+1 ) → P (z) which does not factor through d l+2 . If z = x j i 1 •••i l then we must have j = j ′ , i = i ′ , • • • , i l = i ′ l and i l+1 = i ′ l+1 . Suppose z = x j i 1 •••i l . If both non-zero paths z x j i 1 •••i l+1 , z x j ′ i ′ 1 •••i ′ l+1
which induce f pass through x j i 1 •••i l then we have a contradiction to A being string. If one non-zero path z

x j i 1 •••i l+1 passes through x j i 1 •••i l then the other satisfies condition (a). Indeed, the composition with

x j ′ i ′ 1 i ′ 2 •••i ′ l+1 x j ′ i ′ 1 i ′ 2 •••i ′ l+2
vanishes because our original path corresponds to an element of Ext l+2 A (I(c), P (z)). Similarly, if z

x j i 1 •••i l+1
does not pass through neither

x j i 1 •••i l nor x j ′ i ′ 1 •••i ′ l+1
then both paths satisfy condition (a). The following example ilustrates condition (b).

Example 4.4 Let A be given by the quiver

3 1 / / 2 : : t t t t t t $ $ J J J J J J 4
bound by rad 2 A = 0. Then the minimal projective resolution of I( 1) is

0 → P (3) ⊕ P (4) → P (2) → P (1) → I(1) → 0.
Let j 1 : P (3) → P (2) and j 2 : P (4) → P (2) be the canonical inclusions, then it is easily seen that the morphism

[j 1 -j 2 ] t : P (3) ⊕ P (4) → P (2)
induces a non-zero element of Ext 2 A (I(1), P (2)).

Corollary 4.5 Assume A is a gentle tree algebra, then Ext l+2 A (I(c), P (z)) = 0 if and only if there exists a non-zero path ω : z

x j i 1 i 2 •••i l+1 not passing through x j i 1 i 2 •••i l and whose composi- tions with x j i 1 i 2 •••i l+1 x j i 1 i 2 •••i l+2 are both zero.
Proof. Indeed, if A is gentle, then condition (b) cannot occur as shown in the remark preceding the proof.

Corollary 4.6 a) Let ω : z x j i 1 i 2 •••i l+1 be a non-zero path as in Proposition 4.2 a). a1) Assume that a point x j i 1 i 2 •••i l+2 exists, then w induces an element of the top of A Ext l+2 A (I(c), A) if and only if z is the starting point of a relation of the form ωω ′ when ω ′ : x j i 1 i 2 •••i l+1 x j i 1 i 2 •••i l+2 a2) Assume that no point x j i 1 i 2 •••i l+2 exists, then w induces an element of the top of A Ext l+2 A (I(c), A) if and only if z = x j i 1 i 2 •••i l+1
and ω is the stationary path in z.

b) In the situation of Proposition 4.2 b), the class of the difference of the paths x j i 1 ...i l

x j i 1 ...i l 0 and x j i 1 ...i l

x j i 1 ...i l 1 in Ext l+2 A (I(c), P (z)) lies in the top of A Ext l+2 A (I(c), A) if and only if there are two minimal relations z x j i 1 ...i l 0

x j i 1 ...i l 0i l+2 and z x j i 1 ...i l 1

x j i 1 ...i l 1i l+2 .

Proof.

a1) The morphism f : P l+2 = ⊕P (x j i 1 ...i l i l+1 ) → P (z) induced by ω factors through P (s) where s is the source of a relation ending at x j i 1 i 2 •••i l+2 and such that s lies on the path ω. So, f induces an element on the top of A Ext l+2 A (I(c), A) if and only if s = z. a2) This follows from the fact that the morphism f :

P l+2 = ⊕P (x j i 1 ...i l i l+1 ) → P (z) factors through the identity of P (x j i 1 i 2 •••i l+1 ).
b) Let f be a representative of the class of the difference of paths x j i 1 ...i l 0

x j i 1 ...i l and

x j i 1 ...i l 1 x j i 1 ...i l in Ext l+2 A (I(c), P (x j i 1 ...i l )). Then f = [f 0 f 1 0] : P (x j i 1 ...i l 0 ) ⊕ P (x j i 1 ...i l 1 ) ⊕ P ℓ+2 -→ P (x j i 1 ...i l ).
Suppose first that there is no relation z x j i 1 ...i l 0

x j i 1 ...i l 0i l+2 . Then any relation ending at x j i 1 ...i l 0i l+2 must start at a successor y of z. Therefore there exists g : P (x j i 1 ...i l 0 ) → P (y) such that f 0 factors through g, whence f = [hg f 1 0], for some morphism h. So [f ] is not in the top of A Ext l+2 A (I(c), A). Conversely, if we have two minimal relations as in the statement, and [f ] is not in the top of A Ext l+2 A (I(c), A), then [f ] = [h][g] for some [g] ∈ Ext l+2 A (I(c), P (y)), which is represented by a morphism g : P (x j i 1 ...i l 0 ) ⊕ P (x j i 1 ...i l 1 ) -→ P (y).

Then y lies on the non-zero path z x j i 1 ...i l 0 or z x j i 1 ...i l 1 (or both) and y = z. But then g d l+3,0 : P (x j i 1 ...i l 0i l+2 ) → P (y) is non-zero, because it is given by the nonzero path y x j i 1 ...i l 0

x j i 1 ...i l 0i l+2 , and this contradicts the fact that [g] belongs to Ext l+2 A (I(c), P (y)). We summarise the results in the theorem below. For each point c in a string algebra A = kQ/I, we compute the minimal projective resolution of I(c) given in Corollary 3.6. Then for all l ≥ 0, the l + 2-nd term in the minimal projective resolution of I(c) is given by

P l+2 = j,(i 1 ,i 2 ,••• ,i l+1 ) P (x j i 1 i 2 •••i l+1 ). Whenever the point x j i 1 i 2 •••i l+2 exists, let z j i 1 i 2 •••i l+2 be the source of the relation ending in x j i 1 i 2 •••i l+2 and passing through x j i 1 i 2 •••i l+1 .
For each j ∈ {1, • • • , 6} and for each l ≥ 0, define

Z j i 1 i 2 •••i l+1 = z j i 1 i 2 •••i l+1 0 if x j i 1 i 2 •••i l+1 0 exists; x j i 1 i 2 •••i l+1 otherwise,
and let

ζ j i 1 ...i l+1 = [x j i 1 •••i l , Z j i 1 i 2 •••i l+1 ] if x j i 1 i 2 •••i l 0 , x j i 1 i 2 •••i l 1 both exist; x j i 1 •••i l , Z j i 1 i 2 •••i l+1 \ {x j i 1 •••i l } otherwise.
Dually, whenever the point s j i 1 i 2 •••i l+2 exists, let c j i 1 i 2 •••i l+2 be the target of the relation starting in s j i 1 i 2 •••i l+2 and passing through s j i 1 i 2 •••i l+1 . For each j ∈ {1, • • • , 6} and for each l ≥ 0, define

C j i 1 i 2 •••i l+1 = c j i 1 i 2 •••i l+1 0 if s j i 1 i 2 •••i l+1 0 exists; s j i 1 i 2 •••i l+1 otherwise,
and let

Θ j i 1 ...i l+1 = [C j i 1 i 2 •••i l+1 , s j i 1 •••i l ] if s j i 1 i 2 •••i l 0 , s j i 1 i 2 •••i l 1 both exist; C j i 1 i 2 •••i l+1 , s j i 1 •••i l \ {s j i 1 •••i l } otherwise.
Theorem 4.7 Let A = kQ/I be a string tree algebra and l ≥ 0. The following are equivalent

(a) Ext l+2 A (I(c), P (z)) = 0; (b) there exists j such that z ∈ ζ j i 1 i 2 •••i l+1 ; (c) there exists j such that c ∈ Θ j i 1 i 2 •••i l+1 .
Proof. The equivalence of (a) and (b) follows from Proposition 3.6 and from the definition of ζ j i 1 i 2 •••i l+1 , using the fact that if both points x j i 1 i 2 •••i l+2 exist then we have the following situation in the quiver

••• x j i 1 •••i l / / ••• / / z••• / / z j l,i 1 i 2 •••i l+1 0 / / •••• / / x j i 1 •••i l+1 ••• / / % % K K K K K K K K K K x j i 1 •••i l+1 0 x j i 1 •••i l+1 1 .
The equivalence of (a) and (c) follows from the dual argument.

Remark 4.8 One can easily compute the top of A Ext l+2 A (I(c), A) using Corollary 4.6.

The quiver of the higher relation extension

Knowing how to compute top A Ext i A (I(c), A) and top Ext i A (DA, P (z)) A allows us to find the new arrows of the higher relation extension of a string tree algebra A since they are in bijection with a basis of top Ext

i A (DA, A) = Ext i A (DA, A)/ rad(Ext i A (DA, A)) = Ext i A (DA, A) (rad A) Ext i A (DA, A) + Ext i A (DA, A)(rad A)
.

Note that Ext

P (y) h ; ; x x x x x x x x
where the map h is given by the left-multiplication by a path in Q from z to y, and the composition of f cy with P i+1 (I(c)) → P i (I(c)) is zero.

Dually, we can represent the elements of Ext

I(b)

Because A is gentle, at least one of the arrows α and β is new. Assume α is old and β is new. Then we have two such overlappings ω, ω ′ and no relation involving α and overlapping with ω or ω ′ , that is we have the following situation in the bound quiver of A.

• α @ @ @ @ @ @ @ x / / Finally, if all three arrows α, β, γ are new, we get three overlappings starting at x. Because A is gentle, condition G1) implies that we have three arrows having x as a source, a contradiction. S1,G2) Suppose we have a minimal relation involving at least two paths in the sense of [MP].

Then, in the higher relation extension we have at least two paths c 1 , c 2 starting and ending at the same point with at least one new arrow in each of these paths. Let c i = c i1 α i c i2 where α i is a new arrow, i = 1, 2.

Assume first that there is exactly one new arrow on each path c i . Then each α i corresponds to an overlapping ω i in A starting at the target of α i and ending at its source, and this contradicts the assumption that A is a tree.

If c i contains several new arrows, the same argument as before applies.

This shows that the higher relation extension of A is monomial and hence that the tensor algebra is also monomial, and even has a quadratic relation ideal, because of Corollary 6.5.

S3) Suppose we have the following subquiver

• α & & M M M M M M x γ / / • • β
8 8 q q q q q q such that αγ and βγ are not in the ideal I of the tensor algebra T A ( i≥2 Ext i A (DA, A)). Then one of the three arrows is new. First assume that γ is a new arrow, then γ corresponds to an overlapping ω ending at x which implies that α or β must be a new arrow, say β which correspond to another overlapping ω ′ starting at x. But then the last arrow of ω and the first arrow of ω ′ are not bound by a relation and also α is not bound by a relation with the first arrow of ω ′ . This contradicts A being gentle.

Suppose now that α is a new arrow corresponding to an overlapping ω starting at x. Because of the first case, we may assume that γ is not new. Since β is not bound by

  the set of arrows in Q A⋉M from z to c equals the set of arrows in Q A from z to c plus dim k e z Me c e z M(rad A)e c + e z (rad A)Me c additional arrows from z to c. The latter arrows are called new arrows, while the former are the old arrows.

Definition 4. 1

 1 Let A be a finite dimensional algebra of finite global dimension. The A -Abimodule i≥2 Ext i A (DA, A) with the natural action is called the higher relation bimodule. The trivial extension A⋉ i≥2 Ext i A (DA, A)

  projective resolution of I(c). Recall that the morphisms d k are induced from the paths in Q. If condition (a) holds then it follows from the definition of Ext l+2 A (I(c), P (z)) that ω induces a non-zero element in Ext l+2 A (I(c), P (z)). If condition (b) holds, then

  i A (DA, A).e c = Ext i A (I(c), A) is a left A-module and that e z . Ext i A (DA, A) = Ext i A (DA, P (z)) is a right A-module. Given a right (left) A-module M, we denote by P i (M) the i-th term in a minimal projective resolution of M and by I i (M) the i-th term in a minimal injective coresolution of M. If we represent the elements of Ext i A (DA, A) as classes [f cz ] of morphisms f cz : P i (I(c)) → P (z) such that the composition of f cz with the map P i+1 (I(c)) → P i (I(c)) of the projective resolution is zero, then we are considering the left A-module structure of Ext i A (DA, A). Therefore, [f cz ] lies in (rad A) Ext i A (DA, A) if and only if [f cz ] ∈ Ext i A (I(c), A) lies in the radical of the left A-module Ext i A (I(c), A). In terms of morphisms, [f cz ] is in rad Ext i A (I(c), A) if and only if f cz factors non-trivially through another morphism f cy : P i (I(c)) → P (y) such that the following diagram is commutative P i (I(c))

  i A (DA, A) as classes [g cz ] of morphisms g cz : I(c) → I i (P (z)) such that the composition of g cz with the map I i (I(c)) → I i+1 (I(c)) of the injective coresolution is zero. This corresponds to the right A-module structure of Ext i A (DA, A). Therefore, [g cz ] lies in Ext i A (DA, A)(rad A) if and only if [g cz ] ∈ Ext i A (DA, P (z)) lies in the radical of the left A-module Ext i A (DA, P (z)). In terms of morphisms, [g cz ] is in rad Ext i A (DA, P (z)) if and only if g cz factors non-trivially through another morphism g bz : I(b) → I i (P (z)) such that the following diagram is commutative

where the map h ′ is given by the right-multiplication by a path in Q from b to c, and the composition of g bz with I i (P (c)) → P i+1 (P (c)) is zero.

Moreover there is an isomorphism of vector spaces LR :

such that LR([f cz ]) and f cz induce the same class in Ext i A (DA, A). Thus, [f cz ] is in rad Ext i A (DA, A) if and only if [f cz ] ∈ rad Ext i A (I(c), A) or LR([f cz ]) ∈ rad Ext i A (DA, P (z)).

Algorithm 5.1

• Compute top A Ext i (I(c), A) for all c ∈ Q 0 using Theorem 4.7 and Corollary 4.6. (For efficiency we can restrict to the points that are the source or the target of a relation because of Corollary 4.6 and Corollary 3.7).

• Let B i 0 = {ρ czj : c, z ∈ Q 0 , c the source or target of relations} be the set that spans the vector space top Ext i (DA, A).

• Compute top Ext i (DA, P (z)) A for each z such that ρ czj ∈ B i 0 using Theorem 4.7 and the dual statements of Corollary 4.6.

Each element of B i has a triple subcript czj, and each such element gives rise to exactly one new arrow z → c in the quiver of the higher relation extension.

Theorem 5.2 Let A = kQ/I be a string tree algebra. Then the algorithm 5.1 computes two sequences (c l ), (z l ) of vertices of Q A such that the arrows in the quiver of the higher relation extension are exactly those of Q A plus one additional arrow from each z l to c l .

Proof. This follows from the discussion preceding the algorithm.

Remark 5.3 The vertices (c l ), (z l ) are not necessarily distinct, there may be repetitions.

Example 5.4 Let A = kQ/I be the string algebra given by the following bound quiver:

Then there exists an element ρ 2,4 ∈ e 4 . top A Ext 2 A (I(2), A) which is not in top Ext 2 A (DA, P (4)) A .e 2 and therefore not in top A Ext 2 A (DA, A) A . Thus the quiver of the higher relation extension is

Example 5.5 In this example, the higher relation extension contains an Ext 2 -arrow 5 → 1 although there is no relation between the points 5 and 1. Let A = kQ/I be the string algebra given by the bound quiver:

Then the quiver of the higher relation extension of A is the following:

Example 5.6 Let A be the string algebra of Example 3.4. Then the quiver of the higher relation extension of A is the following: .

Then the quiver of the higher relation extension of A is the following:

Note the existence of a 2-cycle.

The higher relation bimodule for gentle algebras

Recall that a set of monomial relations {κ i } i=1,..,t is called an overlapping if the paths κ i and κ i+1 have a common subpath ϑ such that κ i = ϑ i ϑ and κ i+1 = ϑϑ i+1 , for all i = 1, .., t -1. A maximal t-overlapping is an overlapping {κ i } i=1,..,t such that there exists no monomial relation κ such that the sets {κ, κ i , i = 1, • • • , t} and {κ i , i = 1, • • • , t, κ} are an overlapping, see [GHZ, Gu].

Then, for the injective I(1) associated to the vertex 1, the sequence of x i 1 i 2 •••it is:

Proof. This follows from the construction of the points x i 1 i 2 •••it given in section 3. Remark 6.2 Observe that there may be other points x i 1 i 2 •••it where some i j = 0. In the Lemma we only consider one branch of the quiver which contains all the points x 00•••0 . Proposition 6.3 For every maximal t-overlapping κ = (κ 1 , • • • , κ t ) from c to z there is exactly one new arrow α(κ) : z → c in the higher relation extension which is induced by an element of Ext l+1 A (I(c), P (z)) and these are the only new arrows in the higher relation extension. Moreover, we have the following relations:

Proof. By Corollary 4.5, Ext l+2

A (I(c), P (z)) = 0 if and only if there is a non-zero path ω : z

and such that the compositions with the nonzero paths ω ′ i l+2 : But the previous Lemma implies that

contradicting that A is gentle. Therefore i l+2 does not exist, that is, pd I(c) = l + 2. Then we have the situation

Thus by Corollary 4.5, Ext l+2 A (I(c), P (z)) = 0 if and only if there is a non-zero path ω from z to

A (DA, A), we can apply the algorithm 5.1. Hence, ω induces an element of the top Ext l+2 A (DA, A) if and only if z is the starting point of the overlapping κ.

The result about the new arrows now follows from the algorithm.

Using the fact that Ext l+2 A (I(c), P (z)) = 0 if and only if there is a non-zero path ω from z to

Dually, one proves that α(κ)α 1 = 0, and the relations of the form ζρζ ′ occur since we are dealing with a trivial extension.

The following example shows that the higher relation extension of a gentle algebra is not necessarily gentle.

Example 6.4 Let A be given by the bound quiver

Then the higher relation extension coincides with the relation extension and has the quiver

bound by relations of length 2 and the relation ζρζ ′ , which is of length 3.

Corollary 6.5 The tensor algebra of the higher relation bimodule has the same quiver as the higher relation extension and has the relations in Proposition 6.3 (a). In particular its relation ideal is quadratic.

7 The tensor algebra of a gentle algebra Theorem 7.1 Let A be a gentle algebra.

(a) The tensor algebra T A ( i≥2 Ext i A (DA, A)) is gentle.

(b) The higher relation extension A ⋉ ( i≥2 Ext i A (DA, A)) is monomial. Proof. Since the universal cover of a gentle algebra is a gentle tree, we may assume that A is a tree. We prove the conditions S1), S2), S3), G1) and G2) of section 2. S2) At every point there are at most two incoming arrows (dually, outcoming arrows).

Suppose there are three arrows α, β, γ with target x (see figure

Then at least one, say γ, is a new arrow. Hence, γ corresponds to an overlapping ω = (ω 1 , ω 2 , • • • ) with source x and there is no relation involving α or β and overlapping with ω 1 , that is,

such that αγ and βγ are in the relation ideal of the tensor algebra T A ( i≥2 Ext i A (DA, A)). If γ is a new arrow corresponding to an overlapping ω γ ending at x then α or β must be new, say β, and corresponding to an overlapping ω β as above, which is bound by no relation with α. It follows from our description of the bound quiver that the new arrow β is not bound by a relation with γ, because γ is not in the overlapping ω β , and this is a contradiction. ACKNOWLEDGMENTS: The first author gratefully acknowledges support from the NSERC of Canada, the FQRNT of Québec and the Université de Sherbrooke, the second author gratefully acknowledges support from the NSERC of Canada and the CONICET of Argentina, and the third author gratefully acknowledges support from NSF grant DMS-1001637 and the University of Connecticut.