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On Maximal Diagonalizable Lie Subalgebras of the FirstHochschild CohomologyPatrick Le Meur aa CMLA, ENS Cachan, CNRS, UniverSud , Cachan, FrancePublished online: 07 Apr 2010.

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Communications in Algebra®, 38: 1325–1340, 2010Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870902915798

ON MAXIMAL DIAGONALIZABLE LIESUBALGEBRAS OF THE FIRST HOCHSCHILDCOHOMOLOGY

Patrick Le MeurCMLA, ENS Cachan, CNRS, UniverSud, Cachan, France

Let A be a basic connected finite dimensional algebra over an algebraically closedfield, with ordinary quiver without oriented cycles. Given a presentation of A by quiverand admissible relations, Assem and de la Peña have constructed an embedding of thespace of additive characters of the fundamental group of the presentation into the firstHochschild cohomology group of A. We compare the embeddings given by the differentpresentations of A. In some situations, we characterise the images of these embeddingsin terms of (maximal) diagonalizable subalgebras of the first Hochschild cohomologygroup (endowed with its Lie algebra structure).

Key Words: Finite dimensional algebra; Fundamental group; Hochschild cohomology; Representatetheory.

2000 Mathematics Subject Classification: Primary 16E40; Secondary 16G10.

INTRODUCTION

Let A be a finite dimensional algebra over an algebraically closed field k.The representation theory of A deals with the study of (right) A-modules. So weassume that A is basic and connected and it admits presentations A � kQ/I by its(unique) ordinary quiver Q and an ideal I of admissible relations. In the eighties,Martinez–Villa and de la Peña introduced the fundamental group �1�Q� I� of �Q� I�([17]). Like in topology, this group is defined using an equivalence relation ∼I (calledthe homotopy relation) on the set of unoriented paths in Q. This group is part ofthe so-called covering techniques initiated in [6, 18]. In particular, it has led to thedefinition of simple connectedness and strong simple connectedness for an algebra([2, 20]). Also, it has proved to be a very useful tool in representation theory.For example, it is proved in [19] that any domestic self-injective algebra admittinga Galois covering by a strongly simply connected locally bounded k-category isof quasi-tilted type. Note that in general, different presentations A � kQ/I andA� kQ/J may lead to non-isomorphic groups �1�Q� I� and �1�Q� J�.

The fundamental group �1�Q� I� behaves much like the fundamental group ofa topological space. For example, given a presentation � � kQ � A (with kernel I),

Received July 2, 2008; Revised February 26, 2009. Communicated by D. Zachaira.Address correspondence to Patrick Le Meur, CMLA, ENS Cachan, CNRS, UniverSud, 61

Avenue du President Wilson, Cachan F-94230, France; E-mail: Patrick.LeMeur@cmla.ens-cachan.fr

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Assem and de la Peña have defined an injective group homomorphism �� �Hom��1�Q� I�� k+� ↪→ HH1�A�. Here HH1�A� is the first Hochschild cohomologygroup Ext1Aop⊗A�A�A� ([12]). This result is to be compared with the classicalisomorphism Hom��1�X����

∼−→H1�X�� relating the singular cohomology to thefundamental group of a path connected topological space X. It is known from[10] that HH1�A� has a structure of Lie algebra, isomorphic to the Lie algebra ofderivations of A (with the commutator as Lie bracket) factored out by the idealof inner derivations. With this presentation of HH1�A�, the derivations that lie inthe image of ��, have been characterized in terms of diagonalizable derivations ([9],see also [7]).

The aim of this text is to characterise maximal diagonalisable Lie subalgebrasof HH1�A� using the subspaces lm���� associated to the different presentations �of A. Recall that, given a Lie algebra, the maximal diagonalizable (for the adjointrepresentation) subalgebras are related to Cartan subalgebras.

On the one hand, one can define a diagonalizability for elements in HH1�A�using the above notion of diagonalizable derivations. Also one can define thediagonalizability of a subset of HH1�A� (as the simultaneous diagonalizability ofits elements). It appears that lm���� is diagonalizable, and that any diagonalizablesubset of HH1�A� is contained in lm���� for some presentation � � kQ � A.

On the other hand, given two presentations � � kQ � A and � kQ � A withkernel I and J respectively, it is not easy to compare the groups �1�Q� I� and�1�Q� J� (and therefore �� and �). In some cases, this is possible, however. Forexample, assume that ��� u� is a bypass in Q (that is, � is an arrow and u isan oriented path which is parallel to � and distinct from �), that � ∈ k, and thatJ = ��u���I�. Here ��u�� � kQ

∼−→ kQ is the automorphism, called a transvection, whichmaps � to �+ �u, and which fixes any other arrow ([13]). In such a situation, if � ∼I

u (or � ∼J u), then there is a natural surjective group homomorphism �1�Q� J� ��1�Q� I� (or �1�Q� I� � �1�Q� J�, respectively); if � ∼I u and � ∼J u, then �1�Q� I�equals �1�Q� J�, and the natural homomorphisms are the identity maps; and if � �∼I

u and � �∼J u, then I = J and �1�Q� I� = �1�Q� J�. In each of these cases, we shallsee that there is a simple relation between �� and �.

In order to formulate our main result, we use the quiver � of the homotopyrelations of the presentations of A ([13]). Its set of vertices is the set of thehomotopy relations ∼Ker��� associated to all the presentations � � kQ � A. Also, thereis an arrow ∼I → ∼J if there exists a transvection ��u�� such that J = ��u���I�such that � ∼J and such that the natural surjective group homomorphism is anonisomorphism �1�Q� I� � �1�Q� J�. The quiver � has been introduced in order tofind conditions under which an algebra admits a universal Galois covering. Thisexistence is related to the existence of a unique source (that is, a vertex which is thetarget of no arrow). Actually, under one of the two following conditions, � doeshave a unique source ([15, Prop. 2.11] and [16, Cor. 4.4]):

H1 Q has no double bypass and k has characteristic zero (a double bypass is a4-tuple ��� u� �� v� where ��� u� and ��� v� are bypasses such that the arrow �appears in the path u). In particular, Q has no oriented cycle.

H2 A is monomial (that is, A � kQ/I0 with I0 an ideal generated by a set of paths),and Q has neither oriented nor multiple arrows.

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Using these results, we prove the main theorem of the text.

Theorem 1. Assume that at least one of the two hypotheses (H1) or (H2) is satisfied.Then:

(i) The maximal diagonalizable subalgebras of HH1�A� are exactly the subalgebras ofthe form lm����, where � � kQ � A is a presentation such that ∼ker��� is the uniquesource of � .

(ii) If ���′ are two such subalgebras of HH1�A�, then there exists an algebraautomorphism � � A

∼−→A inducing a Lie algebra automorphism �∗ � HH1�A�∼−→

HH1�A� such that �′ = �∗���.

Note that the Lie algebra HH1�A� has already been studied (see [11, 21], forinstance).

The text is organised as follows. In Section 1 we recall all the definitions wewill need and prove some useful lemmas. In Section 2, we introduce the notionof diagonalizability in HH1�A�. In particular, we prove that a subset of HH1�A�is diagonalizable if and only if it is contained in lm���� for some presentation � �kQ � A. In Section 3 we compare the Lie algebra homomorphisms �� for differentpresentations � of A, using the quiver � . Finally, in Section 4 we prove Theorem 1.

This text is part of the author’s thesis ([14]) made at University Montpellier IIunder the supervision of Claude Cibils.

1. PRELIMINARIES

1.1. Terminology and Notations for Quivers

Let Q be a quiver. We write Q0 and Q1 for the set of vertices and of arrows,respectively. We read (oriented) paths from the right to the left, that is, we view apath u as a morphism and the concatenation vu of two paths u and v such that thesource of v equals the target of u as a composition of morphisms. Given x ∈ Q0, thetrivial path at x (of length 0, with source and target equal to x) is denoted by ex.Two paths are called parallel if they have the same source and the same target. Anoriented cycle in Q is a nontrivial path whose source and target are equal. If � ∈ Q1

we consider its formal inverse �−1 with source and target equal to the target andthe source of �, respectively. Hence, we get the double quiver Q such that Q0 = Q0

and Q1 = Q1 ∪ ��−1 � ∈ Q1�. Then, a walk in Q is exactly an oriented path in Q.Given a walk � = ��nn � � � �

�11 (with �i ∈ Q1, �i ∈ �±1�), its inverse �−1 is by definition

�−�11 � � � �−�n

n .

1.2. Presentations by Quiver and Admissible Relations

Let Q be a quiver. Its path algebra kQ is the k-algebra whose basis as a k-vector space is the set of paths in Q (including the trivial paths), and whose productis bilinearly induced by the concatenation of paths (if u� v are two paths such thatthe source of v is different from the target of u, then we set vu = 0). The unit of kQis∑

x∈Q0ex and kQ is finite dimensional if and only if Q is finite (that is, Q0 and Q1

are finite) and has no oriented cycles. We let kQ+ be the ideal of kQ generated bythe arrows.

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An admissible ideal of kQ is an ideal I such that �kQ+�N ⊆ I ⊆ �kQ+�2 forsome N � 2. In such a case, the elements of I are called relations and, following [17],a minimal relation of I is a relation

∑si=1 tiui �= 0 such that t1� � � � � ts ∈ k∗ and

u1� � � � � us are pairwise distinct paths in Q, and such that there is no nonemptyproper subset S ⊂ �1� � � � � s� satisfying

∑i∈S tiui ∈ I . In such a case, u1� � � � � us are

necessarily parallel. Note that I is generated by its minimal relations.Recall (see [4]) that any finite dimensional k-algerba A is Morita equivalent to

a basic one. If A is basic, then there exists a unique quiver Q, the ordinary quiverof A, and a surjective k-homomorphism � � kQ � A whose kernel is an admissibleideal of kQ. Also, ���ex� x ∈ Q0� is a complete set of primitive orthogonalidempotents of A. The homomorphism � is called a presentation (by quiver andadmissible relations). We have A � kQ/Ker���, and A is connected if and only if Qis connected.

1.3. Presentation of HH1�A�

Let A be a basic finite dimensional k-algebra, and let �e1� � � � � en� be a completeset of primitive orthogonal idempotents. A unitary derivation ([7]) is a k-linear mapd � A → A such that d�ab� = ad�b�+ d�a�b for any a� b ∈ A and such that d�ei� = 0for every i. Let Der0�A� be set of unitary derivations. It is a Lie algebra for thecommutator. In the sequel, all derivations will be unitary. So we shall call themderivations. Let E �= �

∑ni=1 tiei t1� � � � � tn ∈ k�. Then E is a semisimple subalgebra of

A and A = E ⊕ � where � is the radical of A. Let Int0�A� �= ��e � A → A� a ∈ A �→ea− ae e ∈ E�, this is an ideal of Der0�A�. Throughout this text, we shall use thefollowing presentation proved in [7].

Theorem 2 ([7]). HH1�A� � Der0�A�/Int0�A� as Lie algebras.

In the following lemma, we collect some useful properties on derivations.

Lemma 1.1. Let d ∈ Der0�A�, then d�ejAei� ⊆ ejAei. Assume that the ordinaryquiver Q of A has no oriented cycles, then d��� ⊆ � and d��2� ⊆ �2.

Proof. Since d is unitary, and since Q has no oriented cycles, we have d��� ⊆ �.So, d��2� ⊆ �2. �

If � � A∼−→A is a k-algebra automorphism such that ��ei� = ei for every i, then

the map d �→ � � d � �−1 induces a Lie algebra automorphism of HH1�A�, denotedby �∗ � HH1�A�

∼−→HH1�A�.

1.4. Fundamental Groups of Presentations

Let �Q� I� be a bound quiver (that is, Q is a finite quiver and I is an admissibleideal of kQ). The homotopy relation ∼I was defined in [17] as the equivalence classon the set of walks in Q generated by the following properties:

(1) ��−1 ∼I ey and �−1� ∼I ex for any arrow � with source x and target y;

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HOCHSCHILD COHOMOLOGY AND FUNDAMENTAL GROUPS 1329

(2) wvu ∼I wv′u if w� v� v′� u are walks such that the concatenations wvu and wv′u

are well defined and such that v ∼I v′;

(3) u ∼I v if u and v are paths in a same minimal relation (with a nonzero scalar).

Note that if r1� � � � � rt are minimal relations generating the ideal I , then the condition(3) above may be replaced by the following one ([5]):

(3′) u ∼I v if u and v are paths in Q appearing in ri (with a nonzero scalar) for somei ∈ �1� � � � � t�.

The ∼I -equivalence class of a walk � is be denoted by ���I . Let x0 ∈ Q0, following[17], the set of ∼I -equivalence classes of walks with source and target x0 is denotedby �1�Q� I� x0�. The concatenation of walks endows this set with a group structurewhose unit is �ex0 �I . This group is called the fundamental group of �Q� I� at x0. IfQ is connected, then the isomorphism class of �1�Q� I� x0� does not depend on thechoice of x0. In such a case, we write �1�Q� I� for �1�Q� I� x0�. If A is a basic finitedimensional k-algebra and if � � kQ � A is a presentation, the group �1�Q�Ker���� iscalled the fundamental group of the presentation �. The following well-known exampleshows that two presentations of A may have nonisomorphic fundamental groups.

Example 1.2. Let A = kQ/I , where Q is the quiver: and I = �ca�. Setx0 = 1. Then �1�Q� I� � � is generated by �b−1a�I . On the other hand, A � kQ/J ,where J = �ca− cb�, and �1�Q� J� is the trivial group.

In the sequel we shall use the following technical lemma.

Lemma 1.3. Let �Q� I� be a bound quiver where Q has no oriented cycles, and letd � kQ → kQ be a linear map such that d�I� ⊆ I , and d�u� = tuu for some tu ∈ k, forany path u. Let ≡I be the equivalence relation on the set of paths in Q generated by thecondition (3) defining ∼I . Then, the following implication holds for any paths u� v:

u ≡I v implies tu = tv�

Proof. We use a nonmultiplicative version of Gröbner bases ([1], see also [8]). Fixan arbitrary total order u1 < · · · < uN on the set of paths in Q, and let �u∗

1� � � � � u∗N �

be the basis of Homk�kQ� k� dual to �u1� � � � � uN �. Following [15, Sect. 1], theGröbner basis of the vector space I is the unique basis �r1� � � � � rt� defined by thethree following properties:

(i) rj ∈ uij+ Span�ui i < ij� for some ij , for every j;

(ii) u∗ij�rj′� = 0 unless j = j′;

(iii) i1 < · · · < it.

It follows from these properties that:

(iv) r = ∑tj=1 u

∗ij�r�rj for any r ∈ I .

Recall from [15, Sec. 1] that r1� � � � � rt are minimal relations of I so that ≡I isgenerated by the property (3′) defining ∼I . So we only need to prove that thatd�rj� ∈ k�rj for any j. We proceed by induction on j ∈ �1� � � � � t�. By assumption on

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d and thanks to (i), we have d�r1� ∈ I ∩ Span�ui i � i1�. Hence, (iii) and (iv) implythat d�r1� ∈ k�r1. Let j ∈ �1� � � � � d − 1� and assume that d�r1� ∈ k�r1� � � � � d�rj� ∈k�rj . By assumption on d and thanks to (i) and (ii), we have d�rj+1� ∈ I ∩Span�ui i � ij+1� and u∗

il�d�rj+1�� = 0 if l � j. So, (iii) and (iv) imply that d�rj+1� ∈

k�rj+1. This finishes the induction and proves the lemma. �

1.5. Comparison of Fundamental Groups

Let A be a basic finite dimensional k-algebra with ordinary quiver Q withoutoriented cycles. We defined the transvections in the introduction. A dilatation ([13])is an automorphism D � kQ

∼−→ kQ such that D�ei� = ei for any i and such thatD��� ∈ k�� for any � ∈ Q1. The following proposition will be useful in the sequel, itwas proved in [15].

Proposition 1.4 ([15, Prop. 2.5]). Let I be an admissible ideal of kQ, let be anautomorphism of kQ, and let J = �I�. If is a dilatation, then ∼I and ∼J coincide.Assume that = ��u��:

a) If � ∼I u and � ∼J u, then ∼I and ∼J coincide;b) If � �∼I u and � ∼J u, then ∼J is generated by ∼I and � ∼J u;c) If � �∼I u and � �∼J u, then I = J and ∼I and ∼J coincide.

In particular, if � ∼J u, then the identity map on the set of walks in Q induces asurjective group homomorphism �1�Q� I� � �1�Q� J�.

Here generated means: generated as an equivalence relation on the set ofwalks in Q, and satisfying the conditions (1) and (2) in the definition of thehomotopy relation. If I� J are admissible ideals such that there exists ��u�� satisfyingJ = ��u���I�, � �∼I u and � ∼J u, then we say that ∼J is a direct successor of ∼I .Proposition 1.4 allows one to define a quiver � associated to A as follows ([13,Def. 4.1]):

i) �0 = �∼I I is an admissible ideal of kQ such that A � kQ/I�;ii) There is an arrow ∼I→∼′

U if ∼J is a direct successor of ∼I .

Example 1.5. Let A be as in Example 1.2, then J = ��cb�1�I�, and � is equal to∼I → ∼J .

The quiver � is finite, connected and has not oriented cycles ([13, Rem. 3,Prop. 4.2]). Moreover, if � has a unique source ∼I0

(that is, a vertex with no arrowending at it) then the fundamental group of any admissible presentation of A isa quotient of �1�Q� I0�. It was proved in [15] and [16] that � has a unique sourceunder one of the hypotheses (H1) or (H2) presented in the introduction. Moreover,the hypotheses (H1) and (H2) both ensure the following proposition which will beparticularly useful to prove Theorem 1.

Proposition 1.6 ([15, Lem. 4.3] and [16, Prop. 4.3]). Assume that at least one of thetwo hypotheses (H1) or (H2) is satisfied. Let ∼I0

�∼I ∈ � , where ∼I0is the unique source

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of � . Then there exist a dilatation D and a sequence of transvections �1�u1��1� � � � � �l�ul��l

such that:

a) I = D �l�ul��l� � � �1�u1��1

�I0�;b) If we set Ii �= �i�ui��i

� � � �1�u1��1�I0�, then �i ∼Ii

ui for every i.

If ∼I = ∼I0, then ∼I0

�∼I1� � � � �∼Il

�∼I coincide.

1.6. Comparison of the Fundamental Groupsand the Hochschild Cohomology

Let A be a basic finite dimensional k-algebra. Assume that the ordinary quiverQ of A has no oriented cycles. Let x0 ∈ Q0, and fix a maximal tree T of Q, that is, asubquiver of Q such that T0 = Q0 and such that the underlying graph of T is a tree.With these data, Assem and de la Peña have defined an injective homomorphism ofabelian groups �� � Hom��1�Q�Ker����� k+� ↪→ HH1�A� associated to any admissiblepresentation � � kQ � A ([3]). We recall the definition of �� and refer the readerto [3] for more details. For any x ∈ Q0, there exists a unique walk �x in T withsource x0, with target x, and of minimal length for these properties. Let � � kQ �

A be an admissible presentation, and let f ∈ Hom��1�Q�Ker����� k+� be a grouphomomorphism. Then, f defines a derivation f � A → A as follows: f ���u�� =f���−1

y u�x�∼Ker������u� for any path u with source x and target y. The following

proposition was proved in [3].

Proposition 1.7 ([3]). The map f �→ f induces an injective map of abelian groups

�� � Hom��1�Q�Ker����� k+� ↪→ HH1�A��

Note that �� is not surjective in general. Indeed, if A is the path algebraof the Kronecker quiver, then Ker��� = 0, dimk lm���� = 1, and dimk HH1�A� = 3.Note also that despite its definition, the homomorphism �� does not depend onthe choice of T . Indeed, let T ′ be another maximal tree, thus defining the walk �′xof minimal length in T ′ with source x0 and target x, for every vertex x. Given agroup homormorphism f � �1�Q�Ker���� → k+, there is a new derivation f � A →A (instead of f ) obtained by applying the previous construction to T ′ (instead ofto T ), that is, f ���u�� = f���′−1

y u�′x�Ker������u� for every path u in Q from x to y.Now let e = ∑

x∈Q0f���′−1

x �x�Ker����ex ∈ A. It is easily checked that f − f is the innerderivation associated to e. In particular, f and f have equal images in HH1�A�.So the construction of �� does not depend on the choice of the maximal tree T .

The product in k endows Hom��1�Q�Ker����� k+� with a commutativek-algebra structure. So it is also an abelian Lie algebra for the commutator. Thefollowing lemma proves that �� preserves this structure. The proof is just a directcomputation, so we omit it.

Lemma 1.8. �� � Hom��1�Q�Ker����� k+� ↪→ HH1�A� is a Lie algebrahomomorphism. In particular, lm���� is an abelian Lie subalgebra of HH1�A�.

Throughout this text, A will be a basic connected finite dimensional k-algebrawith ordinary quiver Q without oriented cycles (Q0 = �1� � � � � n�). We fix a complete

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set �e1� � � � � en� of primitive orthogonal idempotents of A. So A = E ⊕ �, where E =k�e1 ⊕ · · · ⊕ k�en and � is the radical of A. Without loss of generality, we assume thatany presentation � � kQ � A is such that ��ei� = ei. Finally, in order to use the Liealgebra homomorphisms ��, we fix a maximal tree T in Q.

2. DIAGONALIZABILITY IN HH1�A�

The aim of this section is to prove some useful properties on the subspaceslm���� in terms of diagonalizability in HH1�A�. Note that diagonalizability wasintroduced for derivations of A in [9]. For short, a basis of A is a basis � of thek-vector space A such that: � ⊆ ⋃

i�j ejAei, such that �e1� � � � � en� ⊆ �, and such that�\�e1� � � � � en� ⊆ �. Note the following link between bases and presentations of A:

a) If � � kQ � A is a presentation of A, then there exists a basis � such that ���� ∈� for any � ∈ Q1 and such that any element of � is of the form ��u� with u apath in Q. We say that this basis � is adapted to �.

b) If � is a basis of A, then there exists a presentation � � kQ � A such that ���� ∈� for any � ∈ Q1. We say that the presentation � is adapted to �.

The property of being diagonalizable (as a linear map) is stable under the sumwith an inner derivation as the following lemma shows. The proof is immediate.

Lemma 2.1. Let u � A → A be a linear map, let e ∈ E, and let � be a basis of A.Then u is diagonal with respect to the basis � if and only if the same holds for u+ �e.

The preceding lemma justifies the following definition.

Definition 2.2. Let f ∈ HH1�A� and let d be a derivation representing f . Then fis called diagonalizable (and diagonal with respect to a basis � of A) if and only ifd is diagonalizable (and diagonal with respect to �, respectively).

The subset D ⊆ HH1�A� is called diagonalizable if and only if any there existsa basis � of A such that any f ∈ D is diagonal with respect to �.

The following proposition gives a criterion for a subset D ⊂ HH1�A� to bediagonalizable.

Proposition 2.3. Let D ⊆ HH1�A�. Then, D is diagonalizable if and only if everyelement of D is diagonalizable and �f� f ′� = 0 for any f� f ′ ∈ D.

Proof. Clearly, if D is diagonalizable, then so is every element of D and �f� f ′� = 0for every f� f ′ ∈ D. We prove the converse. For each f ∈ D, let df be a derivationrepresenting f . So df is diagonal with respect to some basis and it suffices to provethat this basis may be assumed to be the same for all f ∈ D. Note that df inducesa diagonalizable linear map df � ej�ei → ej�ei, for every i� j (see Lemma 1.1). Also,for every f� f ′ ∈ D, there exist scalars t�f�f

′�i ∈ k, for i ∈ �1� � � � � n�, such that �df � df ′ �

is the inner derivation �e�f�f ′� , where e�f�f′� = ∑n

i=1 t�f�f ′�i ei. Now, let i� j ∈ �1� � � � � n�.

Then, given f� f ′ ∈ D, we have two diagonalizable maps df � df ′ � ej�ei → ej�eiwhose commutator is equal to �t

�f�f ′�j − t

�f�f ′�i �Idej�ei

. So this commutator must be

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HOCHSCHILD COHOMOLOGY AND FUNDAMENTAL GROUPS 1333

zero. This shows that there exists a basis �i�j of ej�ei for which df � ej�ei →ej�ei has a diagonal matrix. So any f ∈ D is diagonal with respect to the basis� = �e1� � � � � en� ∪

⋃i�j �i�j which does not depend on f . This proves that D is

diagonalizable. �

Our main example of diagonalizable subspace of HH1�A� is lm����.

Proposition 2.4. Let � � kQ � A be a presentation. Then, lm���� is diagonalizable.

Proof. Let � be a basis of A adapted to �, and let I = Ker���. Then ���f� isdiagonal with respect to �, for every f ∈ Hom��1�Q� I�� k+�. �

In this section, we aim at proving that any diagonalizable subset of HH1�A�is contained in lm���� for some presentation �. It was proved in [9] that anydiagonalizable derivation (with suitable technical conditions) defines an element ofHH1�A� lying in lm���� for some �. We will use the following similar result.

Lemma 2.5. Let f ∈ HH1�A� be diagonalizable. Let � be a basis with respect towhich f is diagonal. Let � � kQ � A be a presentation adapted to �. Then f ∈ lm����.

Proof. Let I = Ker��� and let d � A → A be a derivation representing f . We setr �= ��r�, for any r ∈ kQ. Let � ∈ Q1. By assumption on �, there exists t� ∈ k suchthat d��� = t��. Let tu �= t�1 + · · · + t�n , for any path u = �n � � � �1 (with �i ∈ Q1). Sod�u� = tuu, because d is a derivation. More generally, if � = ��nn � � � �

�11 is a walk in

Q (with �i ∈ Q1), let us set t� �=∑n

i=1�−1��i t�i , with the convention that t� = 0 if �is trivial. We now prove that the map � �→ t� defines a group homomorphism g ��1�Q� I� → k+� ���I �→ t� and that f = ���g�.

First, we prove that the group homomorphism g � �1�Q� I� → k+ is welldefined. By definition of the scalar t�, we have the following:

(i) tex = 0 for any x ∈ Q0 and t�′� = t�′ + t� for any walks �� �′ such that the walk �′�is defined;

(ii) t�−1� = tex and t��−1 = tey for any arrow x�−→ y ∈ Q1;

(iii) twvu = twv′u for any walks w� v� v′� u such that tv = tv′ , and such that the walkswvu� wv′u are defined.

In order to prove that g is well defined, it only remains to prove that tu = tvwhenever u� v are paths in Q appearing in the same minimal relation of I (withnonzero scalars). For this purpose, let d′ � kQ → kQ be the linear map such thatd′�u� = tuu for any path u in k. Thus, d � � = � � d′. In particular, d′�I� ⊆ I . So wemay apply Lemma 1.3 to d′ and deduce that:

(iv) tu = tv if u� v are paths in Q lying in the support of a same minimal relationof I .

From (ii), (iii), and (iv) we deduce that we have a well-defined map g � �1�Q� I� →k� ���I �→ t�. Moreover, (i) proves that g is a group homomorphism.

Now we prove that f = ���g�. For any path u with source x and target y,we have g���−1

y u�x�I � = tu − t�y + t�x . Hence, ���g� ∈ HH1�A� is represented by the

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derivation g � A → A such that g�u� = �tu − t�y + t�x �u for any path u with source xand target y. Let us set e �= ∑

x∈Q0t�xex ∈ E. Therefore, g − �e = d. This proves that

f = ���g�. �

Now we can state the main result of this section. It is a direct consequence ofProposition 2.4 and of Lemma 2.5.

Proposition 2.6. Let D ⊆ HH1�A�. Then D is diagonalizable if and only if thereexists a presentation � � kQ � A such that D ⊆ lm����.

Remark that Lemma 2.5 also gives a sufficient condition for �� to be anisomorphism. Recall that A is called constricted if and only if dim eyAex = 1 for anyarrow x → y (this implies that Q has no multiple arrows). In [5] it was proved thatfor such an algebra, two different presentations have the same fundamental group.

Proposition 2.7. Assume that A is constricted. Let � � kQ � A be any presentation ofA. Then �� � Hom��1�Q� I�� k+� → HH1�A� is an isomorphism. In particular, HH1�A� isan abelian Lie algebra.

Proof. Since �� is one-to-one, we only need to prove that it is onto. Let � bea basis of A adapted to �, let f ∈ HH1�A�, and let d � A → A be a derivationrepresenting f . Let x

�−→ y be an arrow. Then eyAex = k����� so that there existst� ∈ k such that d������ = t�����. Let u = �n � � � �1 be any path in Q (with �i ∈ Q1).Since d is a derivation, we have d���u�� = �t�1 + · · · + t�n���u�. As a consequence, dis diagonal with respect to �. Moreover, � is adapted to �. So Lemma 2.5 provesthat f ∈ lm����. This proves that �� is an isomorphism. So HH1�A� is abelian. �

3. COMPARISON OF lm���� AND lm���� FOR DIFFERENTPRESENTATIONS � AND � OF A

If two presentations � and of A are related by a transvection or a dilatation,then there is a simple relation between the associated fundamental groups (seeProposition 1.4). In this section, we compare �� and �. We first compare �� and �when = � �D with D a dilatation. Recall that if J = D�I� with D a dilatation, then∼I and ∼J coincide, so that �1�Q� I� = �1�Q� J�.

Proposition 3.1. Let � � kQ � A be a presentation, let D � kQ∼−→ kQ be a dilatation.

Let �= � �D � kQ � A. Let I = Ker�� and J = Ker���, so that J = D�I�. Then� = ��.

Proof. Let f ∈ Hom��1�Q� I�� k+�. Then, ���f� and ��f� are represented by the

derivations d1 and d2, respectively, such that for any arrow x�−→ y

d1������ = f���−1y ��x�J � �����

d2����� = f���−1y ��x�I � ����

Therefore, d1������ = d2����� because D is a dilatation and because ∼I and ∼J

coincide. This implies that d1 = d2 and ���f� = ��f�. �

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The following example shows that Proposition 3.1 does not necessarily holdtrue if � and are two presentations of A such that ∼Ker��� and ∼Ker�� coincide.

Example 3.2. Assume that char�k� = 2, and let A = kQ/I , where Q is the quiver

and I = �da� fecb� fea+ dcb�. Let T be the maximal tree such that T1 = �b� c� e� f�.Let � � kQ � A = kQ/I be the natural projection. Let � �= a�cb�1 d�fe�1. Thus,I =��I�. Let �= � � � � kQ � A so that Ker�� = Ker��� = I . Observe that �1�Q� I�is the infinite cyclic group with generator �b−1c−1a�I . So let f � �1�Q� I�→ k+ be theunique group homomorphism such that f��b−1c−1a�I� = 1. Then ���f� is representedby the following derivation:

d1 � A −→ A��x� �−→ ��x� if x ∈ �a� d���x� �−→ 0 if x ∈ �b� c� e� f��

On the other hand, ��f� is represented by the derivation:

d2 � A −→ A��a� �−→ ��a�+ ��cb���d� �−→ ��d�+ ��fe���x� �−→ 0 if x ∈ �b� c� e� f��

It is easy to verify that d2 − d1 is not an inner derivation. Hence, �� �= �.

Now we compare �� and � when = � � ��u�� and when the identity map onthe set of walks in Q induces a surjective group homomorphism �1�Q�Ker���� ��1�Q�Ker���.

Proposition 3.3. Let � � kQ � A be a presentation, let ��u�� � kQ∼−→ kQ be a

transvection, and let �= � � ��u�� � kQ � A. Set I = Ker��� and J = Ker��, so thatI = ��u���J�. Suppose that � ∼J u, and let p � �1�Q� I� � �1�Q� J� be the quotient map(see Proposition 1.4). Then, the following diagram commutes:

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where p∗ � Hom��1�Q� J�� k+� ↪→ Hom��1�Q� I�� k+� is the embedding induced by p. Inparticular, lm��� ⊆ lm����.

Proof. Recall that p is the map ���I �→ ���J . Let f ∈ Hom��1�Q� J�� k+�. So p∗�f�

is the composition �1�Q� I�p−→ �1�Q� J�

f−→ k. We know that ��f� and ���p∗�f�� are

represented by the derivations d1 and d2, respectively, such that for any arrow xa−→ y

d1��a�� = f���−1y a�x�J ��a� = p∗�f����−1

y a�x�I ��a�

d2���a�� = p∗�f����−1y a�x�I ���a��

Let us prove that d1 and d2 coincide on ��Q1�. Let xa−→ y be an arrow. If a �= �,

then �a� = ��a� and the above characterizations of d1 and d2 imply that d1���a�� =d1��a�� = d1���a��. Now assume that a = � so that: ��a� = �a�− ��u� and��−1

y a�x�J = ��−1y u�x�J (recall that a = � ∼J u). Thus:

d1���a�� = d1�����− �d1��u��

= f���−1y ��x�J ����− �f���−1

y u�x�J ��u�

= f���−1y ��x�J �����− ��u��

= p∗�f����−1y ��x�I �����

= d2������ = d2���a���

Hence, d1 and d2 are two derivations of A, and they coincide on ��Q1�. So d1 = d2

and ��f� = ���p∗�f�� for any f ∈ Hom��1�Q� J�� k+�. �

The following example shows that Proposition 3.3 does not necessarily holdtrue if � is a presentation of A and � � kQ → kQ is an automorphism such thatthe identity map on the walks in Q induces a surjective group homomorphism�1�Q�Ker���� � �1�Q�Ker�� � ���.

Example 3.4. Let A = kQ/I , where char�k� = 2, where Q is the quiver ofExample 3.2, and where I = �da� fea+ dcb�. Let � � kQ � A be the naturalprojection with kernel I , let � �= d�ef�1 a�cb�1, and let �= � � � � kQ � A. HenceKer�� = �da+ fecb� fea+ dcb�. Note that �1�Q�Ker���� � � is generated by�b−1c−1a�I and that �1�Q�Ker��� � �/2� is generated by �b−1c−1a�J . Note alsothat ∼Ker��� is weaker that ∼Ker��, so that the identity map on the set of walksin Q induces a surjective group homomorphism p � �1�Q�Ker���� � �1�Q�Ker���.Let T be the maximal tree such that T1 = �b� c� e� f�. Let f � �1�Q�Ker��� → k bethe group homomorphism such that f��b−1c−1a�J � = 1. On the one hand, ��f� ∈HH1�A� is represented by the derivation

d1 � A → A�x� �→ �x� if x ∈ �a� d��x� �→ 0 if x ∈ �b� c� e� f��

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HOCHSCHILD COHOMOLOGY AND FUNDAMENTAL GROUPS 1337

On the other hand, ���p∗�f�� ∈ HH1�A� is represented by the derivation

d2 � A → A�a� �→ �a�+ �cb��d� �→ �d�+ �fe��x� �→ 0 if x ∈ �b� c� e� f��

One checks easily that d2 − d1 is not inner, so that ��f� �= ���p∗�f��. Moreover,

lm���� and lm��� are one dimensional (because char�k� = 2, �1�Q�Ker���� � � and�1�Q�Ker��� � �/2�), and d1� d2 are not inner. Hence lm��� �⊆ lm����.

Actually, Proposition 3.3 does not work here because the automorphism� � �kQ� J� → �kQ� I� maps arrows to linear combination of paths which arenot homotopic for ∼I . For example, ��a� = a+ cb whereas a �∼I cb (recall that�1�Q� I� � � is generated by �b−1c−1a�I ).

Finally, we compare �� and � when = � � � with � � kQ∼−→ kQ an

automorphism such that Ker��� = Ker��.

Proposition 3.5. Let � � kQ � A be a presentation, and let I = Ker���. Let � �

kQ∼−→ kQ be an automorphism such that ��ei� = ei for every i and such that ��I� =

I . Let �= � � � � kQ � A so that Ker�� = I . Let � � A∼−→A be the k-algebra

automorphism such that � � = � �. Then, the following diagram commutes:

In particular, lm��� is equal to the image of lm���� under the Lie algebra

automorphism �∗ � HH1�A�∼−→HH1�A�, induced by � � A

∼−→A.

Proof. Since � fixes the idempotents e1� � � � � en, we know that �∗ is well defined.Let f ∈ Hom��1�Q� I�� k+�. So ���f� and ��f� are represented by the derivations d1

and d2 respectively, such that for any arrow x�−→ y,

d1������ = f���−1y ��x�I ������

d2����� = f���−1y ��x�I �����

In order to prove that �∗����f�� = ��f� it suffices to prove that � � d1 = d2 � �. Letx

�−→ y be an arrow. Then:

d2 � ������� = d2����� because � = � �−1 and � � = � �= f���−1

y ��x�I �����

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On the other hand,

� � d1������ = f���−1y ��x�I ��������

= f���−1y ��x�I ���� because = � � � and � � = � ��

Hence, � � d1 and d2 � � are derivations of A which coincide on ��Q1�. So�∗����f�� = ��f� for any f ∈ Hom��1�Q� I� k+��. �

4. PROOF OF THEOREM 1

In this section, we prove Theorem 1. We begin with the following usefullemma.

Lemma 4.1. Assume that at least one of the two conditions (H1) or (H2) is satisfied.Let � � kQ � A be a presentation whose kernel I0 is such that ∼I0

is the unique source of� . Let � kQ � A be another presentation. Then, there exist �′ � kQ � A a presentationwith kernel I0 and a k-algebra automorphism � � A

∼−→A such that:

a) ��ei� = ei for any i,b) lm��� ⊆ lm���′� = �∗�lm�����.

If moreover ∼I and ∼I0coincide, then the above inclusion is an equality.

Proof. Let � � kQ/I0∼−→A and � kQ/I

∼−→A be the isomorphisms induced by �

and , respectively. Hence, −1 � � � kQ/I0∼−→ kQ/I is an isomorphism which maps

ei to ei for every i. Hence, there exists an automorphism � kQ∼−→ kQ which

maps ei to ei for every i and such that the following diagram commutes (see [14,Prop. 2.3.18], for instance):

where the vertical arrows are the natural projections. So, the following diagram iscommutative:

Let us apply Proposition 1.6 to I . We keep the notations �i� ui� �i� Ii of thatproposition. Let � �= −1D �l�ul��l

� � � �1�u1��1. Thus, ��I0� = I0, and �′ �= � � � �

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HOCHSCHILD COHOMOLOGY AND FUNDAMENTAL GROUPS 1339

kQ � A is a presentation with kernel I0. Thanks to Proposition 3.5 from which wekeep the notations, we know that

lm���′� = �∗�lm������ (1)

Let us show that lm��� ⊆ lm���′�. By construction, we have = � −1 =�′ −1

�1�u1��1� � � −1

�l�ul��lD−1. For simplicity, we use the following notations: 0 �= �′ and

i �= �′ −1�1�u1��1

� � � −1�i�ui��i

for i ∈ �1� � � � � l�. Note that Ii = Ker�i�. Since �i ∼Iiui and

i = i−1 � �i�ui�−�i, Proposition 3.3 implies that

lm��l � ⊆ lm��l−1� ⊆ · · · ⊆ lm��i � ⊆ lm��i−1

� ⊆ · · · ⊆ lm��0� = lm���′�� (2)

Moreover, = mD−1, where D−1 is a dilatation. Hence, (1), (2), and Proposition 3.1

imply that

lm��� = lm��l � ⊆ lm���′� = ���lm������ (3)

Now assume that ∼I is the unique source of � . Then Proposition 1.6 implies that thehomotopy relations ∼I0

�∼I1� � � � �∼Il

�∼I coincide. Therefore, for any i ∈ �1� � � � � l�,we have i−1 = i � �i�ui��i

, and �i ∼Ii−1ui. So Proposition 3.3 implies that lm��i−1

� ⊆lm��i �. This proves that all the inclusions in (2) are equalities, and so is the inclusionin (3). �

Now we can prove Theorem 1.

Proof of Theorem 1. (i) Let � be a maximal diagonalizable subalgebra ofHH1�A�. Thanks to Proposition 2.6, there exists a presentation � kQ � A such that� ⊆ lm���. On the other hand, Lemma 4.1, implies that there exists a presentation� � kQ � A such that ∼Ker��� is the unique source of � and such that lm��� ⊆ lm����.Hence, � ⊆ lm���� where lm���� is a diagonalizable subalgebra of HH1�A�, thanksto Proposition 2.4. The maximality of � forces � = lm����.

Conversely, let � kQ � A be a presentation such that ∼Ker�� is the uniquesource of � . Hence, lm��� is diagonalizable (thanks to Proposition 2.4) sothere exists a maximal diagonalizable subalgebra � of HH1�A� containing lm���.Thanks to the above description, we know that � = lm���� where � � kQ � A is apresentation such that ∼Ker��� is the unique source of � . Moreover, Lemma 4.1 gives

a k-algebra automorphism � � A∼−→A such that lm��� = ���lm�����. Since �∗ is a

Lie algebra automorphism of HH1�A�, the maximality of � = lm���� implies thatlm��� is maximal.

(ii) is a consequence of (i) and of Lemma 4.1. �

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