Gerstenhaber structure of the Hochschild cohomology of

Gerstenhaber structure of the Hochschild cohomology of gentlealgebras

Andrea Solotarjoint with Cristian Chaparro and Sibylle Schroll

Motivations to study gentle algebras

I Jacobian algebras arising from quivers and potentials,I algebras that are associated to triangulations of surfaces (Amiot (2016), Amiot and

Grimeland (2016), Assem et al. (2010), David-Roesler and Schiffler (2012),Labardini-Fragoso(2009)),

I they are an important element in “mirror symmetry” theory of 2-manifolds(Heiden-Katzarkov-Kontsevich, Lekili-Polishchuk),

I string theory (Cecotti).

DefinitionA finite dimensional algebra A is gentle if it is Morita equivalent to an algebra kQ/I such that:S1. Each vertex of Q is the source of at most two arrows and the target of at most two arrows.S2. For each arrow α in Q there is at most one arrow β in Q such that βα /∈ I and there is at

most one arrow γ such that αγ /∈ I.S3. For each arrow α in Q there is at most one arrow β in Q such that βα ∈ I and there is at

most one arrow γ such that αγ ∈ I.S4. I can be generated by paths of length 2.

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Notation

I k is an algebraically closed field,I Q is a finite connected quiver,I A = kQ/I is a k-algebra where I is an admissible ideal generated by a minimal set R of

paths of length 2,I B is the set of paths in Q which do not contain subpaths in R.

Aim

Study the representation category of a gentle algebra A. Notice that Db(modA) ↪→ Stmod(A),where A is the repetitive algebra of A, which is a self-injective algebra.

PropositionThe category Stmod(A) is equivalent to modZ(TA).

Theorem (Schroll)An algebra B is symmetric special biserial iff it is a Brauer graph algebra. Moreover, anyBrauer graph algebra with multiplicity 1 in every vertex is isomorphic to a trivial extension TAwith A a gentle algebra.

Tools

Hochschild cohomology of an algebra:I Encodes relevant information of the algebra.I Can be used to understand its structure.

Given a k-algebra A, HH∗(A)I is a graded vector space,I is a graded commutative algebra via the cup product,I is also equipped with the Gerstenhaber bracket:

[−,−] : HHm(A)⊗HHn(A)→ HHm+n−1(A), for all m,n ≥ 0

such that (HH∗(A), [−,−]) is a graded Lie algebra (shifted by 1).I all this structure is derived invariant.

Bracket

In particular HH1(A) ∼= Derk(A)Inder(A) is a graded Lie algebra, and the restriction of the bracket

coincides with the map induced by the conmutator of derivations.

The Gerstenhaber bracket is the algebraic analogue of the Schouten-Nijenhuis bracket in thecohomology of A = C∞(M).

Problem: how to compute it?

Bracket


coincides with the map induced by the conmutator of derivations.The Gerstenhaber bracket is the algebraic analogue of the Schouten-Nijenhuis bracket in thecohomology of A = C∞(M).


Bracket


coincides with the map induced by the conmutator of derivations.The Gerstenhaber bracket is the algebraic analogue of the Schouten-Nijenhuis bracket in thecohomology of A = C∞(M).


We fix some more notations.I A will always denote a gentle algebra.I B is the set of paths in Q that do not have elements of R as subpaths, so B is a basis of

the algebra A.I Bn is the set of paths of length n in B.

HH∗(A)

Some previous results.I Ladkani has computed the dimensions of the Hochschild cohomolgoy spaces of gentle

algebras and related them to the AAG-invariants.

I Redondo and Román have also computed these dimensions and provided a description interms of Bardzell’s resolution. They also proved that the cup product of two classes ofcocycles of odd degrees is zero, and that there are some non zero cup products ofelements of even degrees. Also, they proved that the brackets of two classes of cocycles ofeven positive degrees annihilate and that there are non zero brackets.

I Valdivieso has interpreted the dimensions of the Hochschild cohomology spacesgeometrically for Jacobian algebras.

HH∗(A)


algebras and related them to the AAG-invariants.I Redondo and Román have also computed these dimensions and provided a description in

terms of Bardzell’s resolution. They also proved that the cup product of two classes ofcocycles of odd degrees is zero, and that there are some non zero cup products ofelements of even degrees. Also, they proved that the brackets of two classes of cocycles ofeven positive degrees annihilate and that there are non zero brackets.


HH∗(A)


algebras and related them to the AAG-invariants.I Redondo and Román have also computed these dimensions and provided a description in

terms of Bardzell’s resolution. They also proved that the cup product of two classes ofcocycles of odd degrees is zero, and that there are some non zero cup products ofelements of even degrees. Also, they proved that the brackets of two classes of cocycles ofeven positive degrees annihilate and that there are non zero brackets.


HH∗(A)

We recall that we can interpret the Hochschild cohomology spaces HH∗(A) in terms of parallelpaths.

Def.Given a quiver Q and two paths α, β in Q, the pair (α, β) is a parallel pair if s(α) = s(β) andt(α) = t(β).

ei ej

α

β

Given two sets of paths X and Y in Q,

X||Y = {(α, β) ∈ X × Y : (α, β) is a parallel pair}.

The minimal resolution

Since a gentle algebra is in particular monomial, we will use Bardzell’s resolution -which is theminimal resolution- of A.

R : · · · → A|kΓm|A→ · · · → A|kΓ1|A→ A|kΓ0|Aµ−→ A

where kΓ0 = kQ0, kΓ1 = kQ1, for m > 1, kΓm is the k-vector space generated by the setΓm = {am...a1 | ai+1ai ∈ I, ai ∈ Q1, 1 ≤ i < m− 1} and | is the tensor product over kQ0.

The differential is given by

dm(1|am...a1|1) = am|am−1...a1|1 + (−1)m1|am...a2|a1

for each m ≥ 1 and µ is the augmentation induced by the multiplication of A.

HH0(A)

Proposition (CSS)Let A = kQ/I be a gentle algebra, there is a basis of HH0(A) which elements are:I (s(α), α) ∈ Q0||Bn with n ≥ 1 and val(s(α)) = 2,I∑e∈Q0

(e, e).

DefinitionLet (γ, α) ∈ Γn||(B \Q0) with γ 6= α and n ≥ 1. If γ and α do not share any arrow, we willsay that γ is a shortcut of α, if not, we will say that α is a deviation of γ.

ExampleIn the algebra determined by the following quiver, the parallel pair (f, dcba) in Q1||B4 is ashortcut, while (c, cb) ∈ Q1||B2 and cb is a deviation of c.

e1 e2 e3

e4

a c

df

b

Proposition (CSS)Given a gentle algebra A = kQ/I, the set of classes of the following cocycles is a basis ofHH1(A):I (a, α) where a is a shortcut of α and a does not appear in the elements of R,I (a, α) where α is a deviation of a, a does not appear in the elements of R and α = α2aα1

con α1, α2 /∈ Q0,I (a, a) for each arrow a in the complement of a maximal tree of Q,I in addition, in case char(k) = 2, the pairs (a, s(a)) with a a loop.

Example

Let Q be:

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d

with R = {a1b1, b3a2, b1b3, d2}, notice that A = kQ/I is a gentle algebra with

HH0(A) = k

( 5∑i=1

(ei, ei))

Shortcuts

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d

HH1(A) = k(b2, a2a1)⊕ k(c, dcb3b2b1)⊕ k(a1, a1)⊕ k(b1, b1)⊕ k(d, d)⊕k(d, e5)

Shortcuts

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

db2

a1 a2

HH1(A) = k(b2, a2a1)⊕k(c, dcb3b2b1)⊕ k(a1, a1)⊕ k(b1, b1)⊕ k(d, d)⊕k(d, e5)

Deviations

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d

HH1(A) = k(b2, a2a1)⊕k(c, dcb3b2b1)⊕ k(a1, a1)⊕ k(b1, b1)⊕ k(d, d)⊕k(d, e5)

Deviations

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d

b1

b2

b3

d

cc

HH1(A) = k(b2, a2a1)⊕ k(c, dcb3b2b1)⊕k(a1, a1)⊕ k(b1, b1)⊕ k(d, d)⊕k(d, e5)

Complement of a maximal tree

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d



e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d

a2

b3

b2

c



e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d

a2

b3

b2

c

b1

a1

d


Loops: if char(k) = 2

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

d


Loops: if char(k) = 2

e1

e2

e4

e3

e5

a1 a2

b1 b3

b2

c

dd


Example:Suppose char(k) 6= 2 and A = kQ/I with Q as follows:

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

and I the ideal generated by

R = {a2, bc3, c3d, df3, gf2, f1f5, f2j, h1h2, h2h1, jk1, k1k4, k3k2}.

HH0(A)

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

f1

f2

f3f4

f5

HH0(A)

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

f1

f2

f3f4

f5

HH1(A)

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

c1

c2

c3

HH1(A)

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

c1

c2

c3

HH1(A)

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

c1

c2

c3

HH1(A)

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

c1

f1

h1

k1

HH1(A)

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

c1

f1

h1

k1

HH0(A) = k(e5, f5f4f3f2f1)⊕ k(∑

(e, e))

HH1(A) = k(a, a)⊕ k(c1, c1)⊕ k(f1, f1)⊕ k(h1, h1)

⊕k(k1, k1)⊕ k(c1, c3c2)

Lie algebra structure of HH1(TA)

Theorem (Str)Given a finite dimensional monomial algebra A, the bracket defined by

[(a, α), (a′, α′)] = (a′, α′(a,α))− (a, α(a′,α′)),

for all (a, α), (a′, α′) ∈ Q1||B induces an isomorphism of Lie algebras between HH1(A) andker(d1)/ Im(d0).

Brackets in HH1(A)Non zero brackets:I (a, α) and (a′, α′) shortcuts:

[(a, α), (a′, α′)] ={

2(α, α) si a = α′ and α = a′,0 in any other case.

I (a, α) shortcut:

[(a, α), (a′, a′)] =

(a, α) if a = a′,−(a, α) if a′ appears in α,

0 in any other case.

I (a, α) deviation:

[(a, α), (a′, a′)] ={−(a, α) if a′ 6= α and a′ appears in α,


I a′ a loop:

[(a, a), (a′, s(a′))] ={−(a, ei) if a = a′,


Example

e1 e2

e3 e4 a

e8

e9

e5

e7

e6

e10 e11

e12

e15

e14

e13

c2

c3c1

b

d

f1

f2

f3f4

f5

g

h1

h2

j

k1 k2

k3k4

Figure: Non zero brackets: [(c1, c1), (c1, c3c2)] = −(c1, c3c2) and[(e5, f5f4f3f2f1), (f1, f1)] = −(e5, f5f4f3f2f1)

HHn(A) for gentle algebras, n ≥ 2

Proposition(Strametz) Let A be a monomial algebra. The Hochschild cohomology of A is the cohomologyof the cochain complex

0 −→ K(Γ0||B) d0

−→ K(Γ1||B) d1

−→ ...dn−1

−−−→ K(Γn||B) dn−→ ...

whered2m(γ, α) =

∑ρ=bγ∈Γ2m+1

χB(bα)(ρ, bα)−∑

ρ=γa∈Γ2m+1

χB(αa)(ρ, αa),

d2m+1(γ, α) =∑

ρ∈Γ2(m+1)

(ρ, ρ(γ,α)

).

DefinitionLet C = am...a1 be a cycle with m ≥ 1 and such that ai+1ai is a relation for all i,1 ≤ i ≤ m− 1.I We say that C is a complete cycle if a1am belongs to R.I C is a primitive cycle if it is not a non trivial power of another cycle.I A complete circuit is the set of all rotations of a complete cycle.

e1

e2

e3

e4

a1 a2

a3a4

DefinitionLet C = am...a1 be a cycle with m ≥ 1 and such that ai+1ai is a relation for all i,1 ≤ i ≤ m− 1.I We say that C is a complete cycle if a1am belongs to R.I C is a primitive cycle if it is not a non trivial power of another cycle.I A complete circuit is the set of all rotations of a complete cycle.

e1

e2

e3

e4

a1 a2

a3a4

A basis of HHn(A) for n ≥ 2

TheoremLet A = kQ/I be a gentle algebra with basis of paths B and n ≥ 2. The classes of thefollowing elements in k(Γn||B) give a basis of HHn(A).1. (γ, α) with maximal γ and, either α ∈ Q0 or α ∈ Bm>0 is such that α1 6= γ1 andαm 6= γn.

2. For each primitive complete circuit represented by a cycle C in Γm,a. the element

∑m

i=1

(Ck

i , s(ai))

with n = km even, and Ci = ai+m−1...ai.b. a pair

(a1C

k1 , a1

)with n odd, k such that km = n− 1 and with C1 = am...a1.

3. In addition to the previous elements, if char(k) = 2 for each complete primitive circuitrepresented by the cycle C in Γm,a. the element

∑m

i=1

(Ck

i , s(ai))

where n = mk odd, and Ci = ai+m−1...ai..b. the parallel pair

(a1C

k1 , a1

)with n even, k such that mk = n− 1 and C1 = am...a1.

Let’s take a closer look of the basis elements of each item:

(1) (γ, α) with maximal γ and, either α ∈ Q0 or α ∈ Bm>0 is such that α1 6= γ1 and αm 6= γn.

Notice that α ∈ Q0, then val(α) = 2 since γ is maximal and A is gentle.The corresponding configuration in the associated ribbon graph is:

××

×

× ×

×an

a1

a2

an−1

e

Figure: This local configuration corresponds to a maximal cycle γ = an...a1 ∈ Γn with a1an ∈ B,val(e) = 2 and e = s(α). The pair (γ, α) is in the basis of HHn(A).

Let’s take a closer look of the basis elements of each item:(1) (γ, α) with maximal γ and, either α ∈ Q0 or α ∈ Bm>0 is such that α1 6= γ1 and αm 6= γn.

Notice that α ∈ Q0, then val(α) = 2 since γ is maximal and A is gentle.

The corresponding configuration in the associated ribbon graph is:

××

×

× ×

×an

a1

a2

an−1

e


Let’s take a closer look of the basis elements of each item:(1) (γ, α) with maximal γ and, either α ∈ Q0 or α ∈ Bm>0 is such that α1 6= γ1 and αm 6= γn.

Notice that α ∈ Q0, then val(α) = 2 since γ is maximal and A is gentle.The corresponding configuration in the associated ribbon graph is:

××

×

× ×

×an

a1

a2

an−1

e


α

×

×

×γ

a1a2

at−1at

Figure: Local configuration of a shortcut as in item (1) of the Theorem, where γ ∈ Γn andα = at...a1 ∈ Bt.

α

×

alii

a1i

×

b1i

×bl′ii

alkk

a1k

×

b1k

×

bl′kk

al1

a11

×

b11

×

bl′

1

a1i−1 a1

k−1

Figure: This local configuration corresponds to a deviation α = αk...α1 of γ = γk...γ1 as in item (1) ofthe Theorem and γi = b

l′ii ...b

1i , αi = ali

i ...a1i for any 1 ≤ i ≤ k. The parallel pair (γi, αi) is such that:

γi is a shortcut of αi, or αi is a cycle, or γi = αi is an arrow.

××

×

× ×

×

a1a2

a3 am

Figure: Local configuration of a primitive complete cycle α = am...a1 as in items (2)-(3).

The cup product

There are several different ways of computing the cup product. We do not want to use thecomparison morphisms between the minimal and the bar resolution.We will use the map ∆ : R → R⊗A R of complexes of A-bimodules defined by

∆(1|α|1) =n∑i=0

(1|an...ai+1|1)⊗A (1|ai...a1|1)

for any n ≥ 0 and any α ∈ Γn.

We will identify the parallel pair (γ, α) ∈ Γn||B with f(γ,α) ∈ HomA−A(A|KΓn|A,A) where

f(γ,α)(1|γ′|1) ={α if γ′ = γ,0 otherwise,

Let (γ, α) ∈ Γm||B and (γ′, α′) ∈ Γn||B, then

(γ, α) ^ (γ′, α′) ={

(γγ′, αα′) if γγ′ ∈ Γm+n and αα′ ∈ A,0 otherwise.

The cup product

There are several different ways of computing the cup product. We do not want to use thecomparison morphisms between the minimal and the bar resolution.We will use the map ∆ : R → R⊗A R of complexes of A-bimodules defined by

∆(1|α|1) =n∑i=0

(1|an...ai+1|1)⊗A (1|ai...a1|1)

for any n ≥ 0 and any α ∈ Γn.

We will identify the parallel pair (γ, α) ∈ Γn||B with f(γ,α) ∈ HomA−A(A|KΓn|A,A) where

f(γ,α)(1|γ′|1) ={α if γ′ = γ,0 otherwise,

Let (γ, α) ∈ Γm||B and (γ′, α′) ∈ Γn||B, then

(γ, α) ^ (γ′, α′) ={

(γγ′, αα′) if γγ′ ∈ Γm+n and αα′ ∈ A,0 otherwise.

DefinitionLet C = {C1, ..., Cm} be a primitive complete circuit of length m. The generating circuit Dassociated to C is the set of cycles {C2

1 , ..., C2m} if m is odd and char(k) 6= 2, and it is C

otherwise. The length of D is by definition the length of the cycles in D.

RemarkThe length of D is always even when char(k) 6= 2.

^∑

(D′j , ej) (γ′, α′) (a′, a′) (ej , α′)∑

(ej , ej)∑(ei, ei)

∑(D′j , ej) (γ′, α′) (a′, a′) (ej , α′)

∑(ej , ej)

(ei, α) (D′, D′) if D′ = α is a loopand char(k) = 2,

0 otherwise.0 0 0

(a, a) (aD′1, a) if a appears in D′1,0 otherwise. 0 0

(γ, α) 0 0∑(Di, ei)

∑(D2

i , ei) if D = D′,0 otherwise.

Table: Cup product in HH•(A).

DefinitionLet C = {C1, ..., Cm} be a primitive complete circuit of length m. The generating circuit Dassociated to C is the set of cycles {C2

1 , ..., C2m} if m is odd and char(k) 6= 2, and it is C

otherwise. The length of D is by definition the length of the cycles in D.

RemarkThe length of D is always even when char(k) 6= 2.

^∑

(D′j , ej) (γ′, α′) (a′, a′) (ej , α′)∑

(ej , ej)∑(ei, ei)

∑(D′j , ej) (γ′, α′) (a′, a′) (ej , α′)

∑(ej , ej)

(ei, α) (D′, D′) if D′ = α is a loopand char(k) = 2,

0 otherwise.0 0 0

(a, a) (aD′1, a) if a appears in D′1,0 otherwise. 0 0

(γ, α) 0 0∑(Di, ei)

∑(D2

i , ei) if D = D′,0 otherwise.

Table: Cup product in HH•(A).

PropositionLet A = kQ/I be a gentle algebra with basis of paths B. Suppose a ∈ Q1, α ∈ B and γ ∈ Γmwith m ≥ 1. The set consisting of the following elements generates (HH•(A),^) as gradedcommutative algebra.1.∑e∈Q(e, e),

2. (s(α), α) with α = an...a1 ∈ B \Q0 is such that val(s(α)) = 2 and a1an ∈ I,3. (a, a) where we choose exactly one arrow a in each fundamental cycle in Q,4. (γ, α) where γ is maximal and, either α ∈ Q0 with m ≥ 2 or α = an...a1 ∈ B \Q0 witha1 6= γ1 and an 6= γm,

5. for each generating circuit D = {D1, ..., Dl} of length m, the element∑li=1 (Di, ei) with

l the length of the primitive complete cycles in D and ei = s(Di).

How to compute the bracket using an alternative resolution

Some recent advances in the computation of the Gerstenhaber bracket using alternative resolutionshave been obtained by Witherspoon - Negron, Volkov and by Witherspoon-Negron-Volkov.

Another method, developed by Suárez-Alvarez, provides an easy way to compute the Lie actionof HH1(A) on HHn(A), for all n.Next I will give a brief description of this method.Let B be a k-algebra with a derivation δ : B → B. If M is a left B-module, a δ-operatoron M is a k-linear map f : M → M such that for all b ∈ B and all m ∈ M we havef(bm) = δ(b)m+ bf(m).


Some recent advances in the computation of the Gerstenhaber bracket using alternative resolutionshave been obtained by Witherspoon - Negron, Volkov and by Witherspoon-Negron-Volkov.Another method, developed by Suárez-Alvarez, provides an easy way to compute the Lie actionof HH1(A) on HHn(A), for all n.Next I will give a brief description of this method.

Let B be a k-algebra with a derivation δ : B → B. If M is a left B-module, a δ-operatoron M is a k-linear map f : M → M such that for all b ∈ B and all m ∈ M we havef(bm) = δ(b)m+ bf(m).


Some recent advances in the computation of the Gerstenhaber bracket using alternative resolutionshave been obtained by Witherspoon - Negron, Volkov and by Witherspoon-Negron-Volkov.Another method, developed by Suárez-Alvarez, provides an easy way to compute the Lie actionof HH1(A) on HHn(A), for all n.Next I will give a brief description of this method.Let B be a k-algebra with a derivation δ : B → B. If M is a left B-module, a δ-operatoron M is a k-linear map f : M → M such that for all b ∈ B and all m ∈ M we havef(bm) = δ(b)m+ bf(m).

If ε : P• →M is a projective resolution of M ,

· · · → P2 →d2→ P1 →d1 P0 →ε M → 0,

then a δ-lifting of f to P• is a sequence f• = (fi)i≥0 of δ-operators fi : Pi → Pi such that thediagram

· · · d3 // P2d2 //

f2

��

P1d1 //

f1

��

P0ε //

f0

��

M //

f

��

0

· · · d3 // P2d2 // P1

d1 // P0ε // M // 0

commutes.

Theorem [SA]Let M be a left B-module and suppose f is a δ-operator on M . There is a canonical morphism

∆•f : Ext•B(M,M)→ Ext•B(M,M)

of graded vector spaces, such that for every projective resolution ε : P• →M and each δ-liftingf• : P• → P• of f to P•, the following diagram commutes:

H(homB(P•,M))∆•f,P• //

∼=��

H(homB(P•,M))

∼=��

Ext•B(M,M)∆•f // Ext•B(M,M).

Given a k-derivation f , take B = A⊗Aop, M = A and δ = f ⊗ 1 + 1⊗ f . The map ∆•f givesthe bracket [f,−].

Theorem [SA]Let M be a left B-module and suppose f is a δ-operator on M . There is a canonical morphism

∆•f : Ext•B(M,M)→ Ext•B(M,M)

of graded vector spaces, such that for every projective resolution ε : P• →M and each δ-liftingf• : P• → P• of f to P•, the following diagram commutes:

H(homB(P•,M))∆•f,P• //

∼=��

H(homB(P•,M))

∼=��

Ext•B(M,M)∆•f // Ext•B(M,M).

Given a k-derivation f , take B = A⊗Aop, M = A and δ = f ⊗ 1 + 1⊗ f . The map ∆•f givesthe bracket [f,−].

Jacobian algebras

Let’s consider the case of Jacobian algebras.

Recall: Let (S,M) be a Riemann surface with marked points and Π a triangulation. Supposefurthermore that (S,M) is unpunctured. The finite dimensional Jacobian algebra AΠ, whosequiver was defined by Formin, Shapiro and Thurston and the potential ωΠ was given by Labardini-Fragoso, is gentle -see [Assem, Brüstle, Charbonneau-Jodoin and Plamondon].The dimension of the Hochschild cohomology spaces of gentle algebras have been computed byRedondo and Román, and Valdivieso has given a geometric interpretation of these dimensions,proving that if the triangulation Π contains at least one internal triangle, then the cup productin HH∗(AΠ) is non trivial, and the same holds for the Gerstenhaber bracket if char(k) = 0.

Jacobian algebras

Let’s consider the case of Jacobian algebras.Recall: Let (S,M) be a Riemann surface with marked points and Π a triangulation. Supposefurthermore that (S,M) is unpunctured. The finite dimensional Jacobian algebra AΠ, whosequiver was defined by Formin, Shapiro and Thurston and the potential ωΠ was given by Labardini-Fragoso, is gentle -see [Assem, Brüstle, Charbonneau-Jodoin and Plamondon].

The dimension of the Hochschild cohomology spaces of gentle algebras have been computed byRedondo and Román, and Valdivieso has given a geometric interpretation of these dimensions,proving that if the triangulation Π contains at least one internal triangle, then the cup productin HH∗(AΠ) is non trivial, and the same holds for the Gerstenhaber bracket if char(k) = 0.

Jacobian algebras

Let’s consider the case of Jacobian algebras.Recall: Let (S,M) be a Riemann surface with marked points and Π a triangulation. Supposefurthermore that (S,M) is unpunctured. The finite dimensional Jacobian algebra AΠ, whosequiver was defined by Formin, Shapiro and Thurston and the potential ωΠ was given by Labardini-Fragoso, is gentle -see [Assem, Brüstle, Charbonneau-Jodoin and Plamondon].The dimension of the Hochschild cohomology spaces of gentle algebras have been computed byRedondo and Román, and Valdivieso has given a geometric interpretation of these dimensions,proving that if the triangulation Π contains at least one internal triangle, then the cup productin HH∗(AΠ) is non trivial, and the same holds for the Gerstenhaber bracket if char(k) = 0.

Valdivieso proved that in characteristic 0:I dimHH0(AΠ) = 1 + #{boundaries of type 0},I dimHH1(AΠ) = 1 + |Q1| − |Q0|+ #{B1}, where B1 is the set of boundaries of type 1,I dimHH6n(AΠ) = dimHH6n+1(AΠ) = #{internal triangles } for n > 0,I HHj(AΠ) = 0 in any other case.

Computing the cup product gives:I If char(k) 6= 2, then HH∗(A) is generated as an algebra by the basis of HH0(A), the

basis of HH1(A) and the basis of HH6(A).I If char(k) = 2, then HH∗(A) is generated as an algebra by the basis of HH0(A), the

basis of HH1(A) and the basis of HH3(A).

Valdivieso proved that in characteristic 0:I dimHH0(AΠ) = 1 + #{boundaries of type 0},I dimHH1(AΠ) = 1 + |Q1| − |Q0|+ #{B1}, where B1 is the set of boundaries of type 1,I dimHH6n(AΠ) = dimHH6n+1(AΠ) = #{internal triangles } for n > 0,I HHj(AΠ) = 0 in any other case.

Computing the cup product gives:I If char(k) 6= 2, then HH∗(A) is generated as an algebra by the basis of HH0(A), the

basis of HH1(A) and the basis of HH6(A).I If char(k) = 2, then HH∗(A) is generated as an algebra by the basis of HH0(A), the

basis of HH1(A) and the basis of HH3(A).

The bracket for Jacobian algebras

Suppose char(k) 6= 2. Since the generators of HH∗(A) as an algebra are in cohomologicaldegrees 0, 1 and 6, and the Gerstenhaber bracket is a graded biderivation with respect to thecup product, it is sufficient to compute the Lie algebra structure of HH1(A), together with itsaction on HH0(A) and HH6(A).

There is a basis of HH1(A) containing 1 + |Q1| − |Q0| elements of type [α||α], where α ∈ Q1,and we choose one of these elements for each fundamental cycle, and by #{B1} elements suchthat each of them can be represented by the class of a shortcut β||am...a1, where β ∈ Q1 andam...a1 is a path in A of length at least 2.

The bracket for Jacobian algebras

TheoremThe Lie structure of HH1(A) is as follows:1. [α1||α1, α2||α2] = 0 for all α1, α2.2. [β||am...a1, β

′||a′m...a′1] = 0 for all shortcuts β||am...a1 and β′||a′m...a′1.

3. [α||α, β||am...a1] ={β||am...a1 if exists j such that aj = α,0 if not.

Proof.It follows from the fact that the Gerstenhaber bracket in HH1(A) is just induced by thecommutator of derivations in A.

RemarkEach element of type [α||α] comes from a derivation of A given by α 7→ α and β 7→ 0 ifβ ∈ Q1 and β 6= α. This is the Eulerian derivation associated to the grading such thatdeg(α) = 1 and deg(β) = 0 for β ∈ Q1 and β 6= α. The brackets can be reinterpreted as

[α||α, β||am...a1] = deg(β||am...a1) · β||am...a1,

where the degree is taken with respect to the above mentioned grading.

We will compute the bracket [HH1(A), HH6(A)] with the method by Suárez-Álvarez. For this,given a derivation δ of A, we need to compute a family of δ-operators δi lifting δe = δ⊗1+⊗δ toour resolution of A. If δ comes from a grading deg, then δn can be chosen as δn(p) = degn(p)·p,where degn is the induced grading of Rn.

LemmaIf δ = β||an...a1 corresponds to a shortcut, then we can choose δn = 0 for n ≥ 2, andδ1(1⊗ v ⊗ 1) =

∑ni=1 an...ai+1 ⊗ ai ⊗ ai−1...a1 if v = β and 0 if not.

RemarkEach element of type [α||α] comes from a derivation of A given by α 7→ α and β 7→ 0 ifβ ∈ Q1 and β 6= α. This is the Eulerian derivation associated to the grading such thatdeg(α) = 1 and deg(β) = 0 for β ∈ Q1 and β 6= α. The brackets can be reinterpreted as

[α||α, β||am...a1] = deg(β||am...a1) · β||am...a1,

where the degree is taken with respect to the above mentioned grading.We will compute the bracket [HH1(A), HH6(A)] with the method by Suárez-Álvarez. For this,given a derivation δ of A, we need to compute a family of δ-operators δi lifting δe = δ⊗1+⊗δ toour resolution of A. If δ comes from a grading deg, then δn can be chosen as δn(p) = degn(p)·p,where degn is the induced grading of Rn.

LemmaIf δ = β||an...a1 corresponds to a shortcut, then we can choose δn = 0 for n ≥ 2, andδ1(1⊗ v ⊗ 1) =

∑ni=1 an...ai+1 ⊗ ai ⊗ ai−1...a1 if v = β and 0 if not.

PropositionGiven [α||α] in the basis of HH1(A) and [c2||e] in the basis of HH6(A), the bracket[α||α, c2||e] is as follows:

[α||α, c2||e] ={−2c2||e if α appears in c,

0 if not.

This coincides with the degree of c2||e with respect to the grading defined by α||α.

RemarkI [HH6(A), HH6(A)] = 0, since the result lives in HH11(A) = 0.I [HH0(A), HH6(A)] = 0, since the result lives in HH5(A) = 0.I [HH1(A), HH0(A)] is given by

I[α||α,

∑|Q0|i=1 ei||ei

]= 0 =

[β||an...a1,

∑|Q0|i=1 ei||ei

].

I [α||α, s(b1)||bm...b1] = #{i|α = bi} · s(b1)||bm...b1 for each cycle in Q arising from atriangle of type 0.

I the other brackets in [HH1(A), HH0(A)] are 0.

It is interesting to notice that the AAG invariant φA(1, 1) equals the dimension of the derived Liealgebra [HH1(A), HH1(A)], while φA(0, 3) equals the dimension ofHH6(A). As a consequence,we obtain an easy proof of the fact that φA(1, 0), φA(1, 1) and φA(0, 3) are derived invariants.

Question: Is it possible to describe the other AAG invariants that are non zero for gentleJacobian algebras using HH∗ and the Gerstenhaber bracket?

Negron-Whiterspoon’s approach

They prove that given a resolution K ε−→ A of an algebra as a bimodule such thatI K admits an embedding K ι−→ B· of A-bimodules into the bar resolution such that µι = ε.I There is a section of ι which is a map of A-bimodules.I There is a diagonal ∆K : K → K⊗A K satisfying ∆Bι = (ι⊗A ι)∆K.

Under these hypotheses, consider the chain map

FK = ε⊗A idK − idK ⊗A ε : K ⊗A K → K.

They prove that it is a boundary in HomAe(K⊗AK,K) and there is a contracting homotopy φfor FK such that d(φ) = dKφ+ φdK⊗AK = FK.

Moreover, the bilinear operation

[f, g]φ = f ◦φ g − (−1)(|f |−1)(|g|−1)g ◦φ f

on HomAe(K, A) induces the Gerstenhaber bracket on HH•(A), where the φ-circle productf ◦φ g is defined by

(f ◦φ g)(ω) =∑

(−1)|ω1||g|f(φ(ω1 ⊗A g(ω2)⊗A ω3)),

with ω1, ω2 and ω3 given by evaluating the map

∆K(id⊗A ∆K) = ∆K(∆K ⊗A id) : K → K⊗A K ⊗A K

for ω.

This is very useful even if we do not need to use the map φ explicitly.

Moreover, the bilinear operation

[f, g]φ = f ◦φ g − (−1)(|f |−1)(|g|−1)g ◦φ f

on HomAe(K, A) induces the Gerstenhaber bracket on HH•(A), where the φ-circle productf ◦φ g is defined by

(f ◦φ g)(ω) =∑

(−1)|ω1||g|f(φ(ω1 ⊗A g(ω2)⊗A ω3)),

with ω1, ω2 and ω3 given by evaluating the map

∆K(id⊗A ∆K) = ∆K(∆K ⊗A id) : K → K⊗A K ⊗A K

for ω.This is very useful even if we do not need to use the map φ explicitly.

Steps:I We compute the brackets of generators in HH0 with generators in HHn, for n > 0,

which is easy.I We compute the brackets of generators in HH1 with generators in HHn, for n > 0, using

Suárez-Alvarez’s method.I We compute those brackets than can be deduced from the previous ones, taking into

account that the Gerstenhaber bracket is a graded biderivation with respect to bothvariables.

I For gentle algebras, the only brackets that we still need to compute are the bracketsamongst classes of cocycles coming from maximal paths, and brackets of classes of thesecocycles with the other families (brackets amongst generators coming from completecircuits can be deduced from the previous ones).

But for the last item, the only thing that we need to observe is that a maximal path in Γn willnever appear when applying ∆K(id⊗A ∆K) to an element in Γn+m−1 with m ≥ 2, so the lastfamily of brackets is zero.

Steps:I We compute the brackets of generators in HH0 with generators in HHn, for n > 0,

which is easy.I We compute the brackets of generators in HH1 with generators in HHn, for n > 0, using

Suárez-Alvarez’s method.I We compute those brackets than can be deduced from the previous ones, taking into

account that the Gerstenhaber bracket is a graded biderivation with respect to bothvariables.

I For gentle algebras, the only brackets that we still need to compute are the bracketsamongst classes of cocycles coming from maximal paths, and brackets of classes of thesecocycles with the other families (brackets amongst generators coming from completecircuits can be deduced from the previous ones).

But for the last item, the only thing that we need to observe is that a maximal path in Γn willnever appear when applying ∆K(id⊗A ∆K) to an element in Γn+m−1 with m ≥ 2, so the lastfamily of brackets is zero.

Gerstenhaber structure

Thus, we obtain the Gerstenhaber bracket in HH•(A) for A gentle.

[-,-]∑

(D′j , ej) (γ′, α′) (a′, a′) (ej , α′)∑

(ei, ei)∑(ei, ei) 0 0 0 0 0

(ei, α) (ei, ei)if D = α is a loopand char(K) = 2,

0 otherwise.0 −(ei, α) if a′ is an arrow in α,

0 otherwise. 0

(a, a) −k∑(

D′j , ej)

if a appears in D′,0 otherwise. dega(γ′, α′) · (γ′, α′) 0

(γ, α) 0 2(α, α) if γ = α′, α = γ′ ∈ Q1,0 otherwise.∑

(Di, ei) 0

Table: Brackets in HH•(A).

Thank you!

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Gerstenhaber structure of the Hochschild cohomology of