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Recollements & Hochschild Theory

Yang Han

KLMM, ISS, AMSS, Chinese Academy of Sciences

October 2 - 7, 2011, Shanghai.

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Plan of my talk :

1. (Perfect) recollements

2. Constructions of (perfect) recollements

3. Recollements and Hochschild dimensions

4. Recollements and derived functors

5. Recollements and Hochschild homology [Keller 1998]

6. Recollements and Hochschild cohomology

1 (Perfect) recollements

1.1. Recollements

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Definition 1. [Beilinson-Bernstein-Deligne 1982]

Let T1, T and T2 be triangulated categories. A recollement of T relative

to T1 and T2 is given by

T1 T T2

�

i∗�

j!

-

i∗ = i!-

j! = j∗

�

i!�

j∗

and denoted by 9-tuple (T1, T , T2, i∗, i∗ = i!, i

!, j!, j! = j∗, j∗) such that

(R1) (i∗, i∗), (i!, i!), (j!, j

!) and (j∗, j∗) are adjoint pairs of triangulated

functors;

(R2) i∗, j! and j∗ are full embeddings;

(R3) j!i∗ = 0 (and thus also i!j∗ = 0 and i∗j! = 0);

(R4) for each X ∈ T , there are triangles

j!j!X → X → i∗i

∗X →i!i

!X → X → j∗j∗X → .

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Viewpoint: Recollement ≈ Short exact sequence of 4-categories

Why recollements of triangulated categories?

1. Algebraic geometry : [Beilinsion-Bernstein-Deligne 1982] ...

2. Representation theory : [Cline-Parshall-Scott 1988] ...

Focus : recollements of derived categories of algebras.

Why recollements of derived categories of algebras?

1. Tilting theory : [Konig 1991] ...

2. Localization theory : [Miyachi 1991], [Krause 2008] ...

3. Homological invariants : global dimension [Wiedemann 1991] +

[Konig 1991] + [Angeleri Hugel-Konig-Liu-Yang 2012]; finitistic di-

mension [Happel 1993]; Hochschild homology + cyclic homology

[Keller 1998]; ...

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Notations & Terminologies :

k : a field

A : a k-algebra

ProjA (resp. projA) : the category of projective (resp. finitely gener-

ated projective) A-modules.

D(A) : the unbounded derived category of complexes of A-modules.

Kb(ProjA) (resp. K−(ProjA)) : the homotopy category of bounded

(resp. right bounded) complexes of projective A-modules.

Kb(projA) : the homotopy category of bounded complexes of finitely

generated projective A-modules.

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X : an object in D(A).

X⊥ = {Y ∈ D(A)|HomD(A)(X, Y [n]) = 0,∀ n ∈ Z}

TriaX : the smallest full triangulated subcategory of D(A) which con-

tains X and is closed under small coproducts.

X is exceptional if HomD(A)(X,X [n]) = 0 for all n ∈ Z\{0}.

X is compact if the functor HomD(A)(X,−) preserves small coproduct,

equivalently, X is perfect, i.e., isomorphic to an object in Kb(projA).

X is self-compact if HomD(A)(X,−) preserves small coproducts in

TriaX .

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An existness criterion of recollements:

Theorem 1. [Konig 1991]+[Jørgensen 2006]+ [Nicolas-Saorin 2009]

LetA1, A andA2 be algebras. ThenD(A) admits a recollement relative

toD(A1) andD(A2) if and only if there are objectsX1 andX2 inD(A)

such that

(1) EndD(A)(Xi) ∼= Ai as algebras for i = 1, 2;

(2) X2 (resp. X1) is exceptional and compact (resp. self-compact);

(3) X1 ∈ X⊥2 ;

(4) X⊥1 ∩X⊥

2 = {0}.

Remark 1. X1 = i∗A1 and X2 = j!A2.

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Example 1. Stratifying ideals. [Cline-Parshall-Scott 1996]

Let A be an algebra, e an idempotent of A, and AeA a stratifying ideal

ofA, i.e., the multiplication inA induces an isomorphismAe⊗eAeeA ∼=AeA and ToreAen (Ae, eA) = 0 for all n ≥ 1. Then there is a recollement

(D(A/AeA), D(A), D(eAe), i∗, i∗ = i!, i!, j!, j

! = j∗, j∗) where

i∗ = −⊗LA A/AeA, j! = −⊗L

eAe eA,

i∗ = i! = −⊗LA/AeA A/AeA, j! = j∗ = −⊗L

A Ae,

i! = RHomA(A/AeA,−), j∗ = RHomeAe(Ae,−).

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1.2. Perfect recollements

Definition 2. Let A1, A and A2 be algebras. A recollement (D(A1),

D(A), D(A2), i∗, i∗ = i!, i

!, j!, j! = j∗, j∗) is said to be perfect if i∗A1 is

perfect.

An existness criterion of perfect recollements:

Theorem 2. LetA1, A andA2 be algebras. ThenD(A) admits a perfect

recollement relative toD(A1) andD(A2) if and only if there are objects

Xi, i = 1, 2, in D(A) such that

(1) EndD(A)(Xi) ∼= Ai as algebras, ∀i = 1, 2;

(2) Xi is exceptional and perfect, ∀i = 1, 2;

(3) X1 ∈ X⊥2 ;

(4) X⊥1 ∩X⊥

2 = {0}.

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Examples.

(1) Derived equivalences. If the algebras A and B are derived equiva-

lent then D(A) admits a perfect recollement relative to 0 and D(B), or

to D(B) and 0.

(2) Triangular matrix algebras. Let A1 and A2 be algebras, M an A2-

A1-bimodule, and A =

[A1 0

M A2

]. Then X1 :=

[1A1

0

0 0

]A and

X2 :=

[0 0

0 1A2

]A satisfy all conditions in Criterion. Thus there is a

perfect recollement of D(A) relative to D(A1) and D(A2).

(3) Perfect stratifying ideals. Let A be an algebra, e an idempotent

of A, and AeA a perfect stratifying ideal of A, i.e., a stratifying ideal

which is perfect in D(A). Then there is a perfect recollement of D(A)

relative to D(A/AeA) and D(eAe).

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(4) Imperfect recollement. [Konig 1991, Example 9] Let A be the

infinite Kronecker algebra

[k 0

V k

], where V is an infinite-

dimensional k-vector space. Choose X2 to be the simple projective

A-module and X1 the other simple A-module. Then X1 and X2 satisfy

all conditions in Criterion. Thus D(A) admits a recollement relative to

D(k) and D(k), which is imperfect since X1 is isomorphic to an object

in Kb(ProjA) but not in Kb(projA).

Relations between examples :

Triangular matrix algebra Derived equivalence

⇓ ⇓Perfect stratifying ideal ⇒ Perfect recollement

⇓ ⇓Stratifying ideal ⇒ Recollement

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2 Constructions of (perfect) recollements

2.1. Tensor product algebras

Theorem 3. Let A1, A and A2 be algebras, and D(A) admit a (perfect)

recollement relative to D(A1) and D(A2). Then, for each algebra B,

D(B ⊗ A) admits a (perfect) recollement relative to D(B ⊗ A1) and

D(B ⊗ A2).

Proof.

• Criterion ⇒ ∃Xi ∈ D(A), i = 1, 2,3 · · ·

• Zi := B ⊗Xi, i = 1, 2.

• Criterion

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2.2. Opposite algebras

Theorem 4. Let A1, A and A2 be algebras, and D(A) admit a perfect

recollement relative to D(A1) and D(A2). Then D(Aop) admits a per-

fect recollement relative to D(Aop2 ) and D(Aop

1 ).

Proof.

• Criterion ⇒ ∃Xi ∈ D(A), i = 1, 2,3 · · ·

• Zi := RHomA(Xi, A), i = 1, 2.

• Criterion

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3 Recollements & Hochschild dimensions

3.1. Recollements & Global dimensions

Theorem 5. [Konig 1991] Let A1, A and A2 be algebras, and D(A)

admit a perfect recollement relative to D(A1) and D(A2). Then A is of

finite global dimension if and only if so are A1 and A2.

Proof.

• D−(A) : the derived category of complexes of A-modules with

right bounded cohomologies.

• ∃ a restricted recollement of D−(A) relative to D−(A1) and

D−(A2). [Konig 1991] or [Nicolas-Saorin 2011]

• [Konig 1991]

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Theorem 6. [Angeleri Hugel-Konig-Liu-Yang ≥2011] Let A1, A and

A2 be algebras, and (D(A1), D(A), D(A2), i∗, i∗ = i!, i

!, j!, j! = j∗, j∗)

a recollement. Then A is of finite global dimension if and only if so are

A1 and A2.

Proof.

• If gl.dimA <∞ or gl.dimA2 <∞ then i∗A1 is isomorphic inD(A)

to a complex in Kb(ProjA).

• ∃ a restricted recollement of D−(A) relative to D−(A1) and

D−(A2). [Konig 1991] or [Nicolas-Saorin 2011]

• [Konig 1991]

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3.2. Recollements & Hochschild dimensions

A : an algebra

The enveloping algebra of A : Ae := Aop ⊗ A

The Hochschild dimension of A : h.dimA := pdAeA :

h.dimA = 0 ⇐⇒ Ae is semisimpleA f.g.⇐⇒ A is separable [Cartan-Eilenberg 1956]

h.dimA ≤ 1def⇐⇒ A is quasi-free [Cuntz-Quillen 1995], or

formally smooth [Kontsevich-Rosenberg 2000].

h.dimA <∞ def⇐⇒ A is smooth [Van den Bergh 2002]

[Cartan-Eilenberg 1956]+[Eilenberg-Rosenberg-Zelinsky 1957] ⇒A is smooth if and only if gl.dimAe <∞.

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Theorem 7. Let A1, A and A2 be algebras, and D(A) admit a perfect

recollement relative toD(A1) andD(A2). ThenA is smooth if and only

if so are A1 and A2.

Proof.

• ∃ a perfect recollement of D(Aop⊗A) relative to D(Aop⊗A1) and

D(Aop ⊗ A2).

• gl.dimAe <∞ iff gl.dimAop ⊗ Ai <∞,∀i = 1, 2.

• ∃ a perfect recollement of D(Aop ⊗ Aj) relative to D(Aop2 ⊗ Aj)

and D(Aop1 ⊗ Aj).

• gl.dimAop ⊗ Ai <∞ iff gl.dimAopi ⊗ Aj <∞,∀i, j = 1, 2.

• [Eilenberg-Rosenberg-Zelinsky 1957] ⇒ gl.dimAopi ⊗ Aj <∞ iff

gl.dimAei <∞,∀i = 1, 2.

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Corollary 1. Let A1 and A2 be algebras, M an A2-A1-bimodule, and

A =

[A1 0

M A2

]. Then A is smooth if and only if so are A1 and A2.

Question 1. Does “the middle algebra is smooth if and only if so are the

algebras on both sides” hold for any recollement of derived categories

of algebras?

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Theorem 8. Let A1, A and A2 be both left and right Noetherian al-

gebras, and D(A) admit a recollement relative to D(A1) and D(A2).

Then A is smooth if and only if so are A1 and A2.

Proof. ⇒ :

• h.dimA <∞⇒ gl.dimA <∞⇒ gl.dimAi <∞, i = 1, 2

• A1 and A2 are both left and right Noetherian ⇒ gl.dimAopi =

gl.dimAi <∞. [Auslander 1955]

• gl.dimAopi <∞ + h.dimA <∞⇒ gl.dimAop

i ⊗ A <∞, i = 1, 2

• ∃ a recollement (D(Aopi ⊗ A1), D(Aop

i ⊗ A), D(Aopi ⊗ A2), · · · ).

• gl.dimAopi ⊗ A <∞⇒ gl.dimAop

i ⊗ Aj <∞, i, j = 1, 2.

• [Eilenberg-Rosenberg-Zelinsky 1957] ⇒ gl.dimAei <∞, i = 1, 2.

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4 Recollements & Derived functors

Lemma 1. Let A1, A and A2 be algebras, and D(A) admit a recolle-

ment relative to D(A1) and D(A2). Then

(1) ∃ recollement (D(A1), D(A), D(A2), i∗, i∗ = i!, i

!, j!, j! = j∗, j∗), 3

j! = −⊗LA2Y2,

i∗ = i! = −⊗LA1Y1, j! = j∗ = RHomA(Y2,−),

i! = RHomA(Y1,−), j∗ = RHomA2(RHomA(Y2, A),−),

and

(2) ∀ algebra B, ∃ recollement (D(B ⊗ A1), D(B ⊗ A), D(B ⊗ A2),

I∗, I∗ = I!, I!, J!, J

! = J∗, J∗), 3

J! = −⊗LA2Y2,

I∗ = I! = −⊗LA1Y1, J ! = J∗ = RHomA(Y2,−),

I ! = RHomA(Y1,−), J∗ = RHomA2(RHomA(Y2, A),−),

where Yi ∈ D(Aopi ⊗ A), i = 1, 2.

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Proof. (1)

• Criterion ⇒ ∃ homotopically projective Xi ∈ D(A), i = 1, 2,3· · ·

• Xi exceptional ⇒ ∃Yi ∈ D(Aopi ⊗ A),3 the fully faithful functor

−⊗LAiYi : D(Ai) → D(A) sends Ai to Xi. [Keller 1998, 1994]

• X2 perfect ⇒ ∃ recollement (X⊥2 , D(A), D(A2), i

′∗, i′∗ = i′!, i′!,

j!, j! = j∗, j∗). [Miyachi 2003]

• X1 self-compact ⇒ the essential image of − ⊗LA1Y1 : D(A1) →

D(A) is TriaX1.

• X1 ∈ X⊥2 and X⊥

1 ∩X⊥2 = {0} ⇒ TriaX1 = X⊥

2 .

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Proof. (2)

• Repeat the proof above, mutatis mutandis, ∃ recollement (D(B ⊗A1), D(B ⊗ A), D(B ⊗ A2), I

∗, I∗ = I!, I!, J!, J

! = J∗, J∗), 3

J! = −⊗LB⊗A2

(B ⊗ Y2),

I∗ = I! = −⊗LB⊗A1

(B ⊗ Y1), J! = J∗ = RHomB⊗A(B ⊗ Y2,−),

I ! = RHomB⊗A(B ⊗ Y1,−), J∗ = RHomB⊗A2(RHomB⊗A(B ⊗ Y2, B ⊗ A),−).

• All conditions on B ⊗Xi needed here have been checked already

in the proof of Construction.

• Cancel B in all derived functors:

I∗ = −⊗LB⊗A1

(B ⊗ Y1) ∼= −⊗LA1Y1

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Theorem 9. Let A1, A and A2 be algebras, and D(A) admit a recolle-

ment relative to D(A1) and D(A2). Then

(1) ∃ recollement (D(A1), D(A), D(A2), i∗, i∗ = i!, i

!, j!, j! = j∗, j∗), 3

i∗ = −⊗LA Y, j! = −⊗L

A2Y2,

i∗ = i! = −⊗LA1Y1, j! = j∗ = RHomA(Y2,−),

i! = RHomA(Y1,−), j∗ = RHomA2(RHomA(Y2, A),−),

and

(2) ∀ algebra B, ∃ recollement (D(B ⊗ A1), D(B ⊗ A), D(B ⊗ A2),

I∗, I∗ = I!, I!, J!, J

! = J∗, J∗), 3

I∗ = −⊗LA Y, J! = −⊗L

A2Y2,

I∗ = I! = −⊗LA1Y1, J ! = J∗ = RHomA(Y2,−),

I ! = RHomA(Y1,−), J∗ = RHomA2(RHomA(Y2, A),−),

where Y ∈ D(Aop ⊗ A1) and Yi ∈ D(Aopi ⊗ A), i = 1, 2.

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Proof. (1)

• Five Ok

• ∃ a recollement (D(Aop⊗A1), D(Aop⊗A), D(Aop⊗A2), I∗, I∗ =

I!, I!, J!, J

! = J∗, J∗), 3 Five = Five.

• Y := I∗A ∈ D(Aop ⊗ A1)

• i∗ ∼= −⊗LA Y !

(2) Repeat the proof above, mutatis mutandis, ...

Definition 3. Let A1, A and A2 be algebras. A recollement (D(A1),

D(A), D(A2), i∗, i∗ = i!, i

!, j!, j! = j∗, j∗) is said to be standard and

given by Y ∈ D(Aop ⊗A1) and Y2 ∈ D(Aop2 ⊗A) if i∗ ∼= −⊗L

A Y and

j! ∼= −⊗LA2Y2.

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Theorem 10. Let A1, A and A2 be algebras, and (D(A1), D(A),

D(A2), i∗, i∗ = i!, i

!, j!, j! = j∗, j∗) a standard recollement given by

Y ∈ D(Aop ⊗ A1) and Y2 ∈ D(Aop2 ⊗ A). Then

(1)

i∗ ∼= −⊗LA Y , j! ∼= −⊗L

A2Y2,

i∗ = i! ∼= RHomA1(Y ,−), j! = j∗ ∼= RHomA(Y2,−),

i! ∼= RHomA(RHomA1(Y ,A1),−), j∗ ∼= RHomA2

(RHomA(Y2, A),−),

and

(2) ∀ algebra B, ∃ a recollement (D(B ⊗ A1), D(B ⊗ A), D(B ⊗A2), I

∗, I∗ = I!, I!, J!, J

! = J∗, J∗), 3

I∗ ∼= −⊗LA Y , J!

∼= −⊗LA2Y2,

I∗ = I! ∼= RHomA1(Y ,−), J ! = J∗ ∼= RHomA(Y2,−),

I ! ∼= RHomA(RHomA1(Y ,A1),−), J∗ ∼= RHomA2

(RHomA(Y2, A),−).

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5 Recollements & Hochschild homology

The n-th Hochschild homology of an algebra A is HHn(A) :=

TorAe

n (A,A) ∼= H−n(A⊗LAe A).

Theorem 11. [Keller 1998] Let A,A1 and A2 be algebras, and D(A)

admit a recollement relative to D(A1) and D(A2). Then there is a

triangle in D(k):

A2 ⊗LAe

2A2 → A⊗L

Ae A→ A1 ⊗LAe

1A1 → .

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Proof.

• Realization of derived functors ⇒ ∃ a triangle in D(Ae):

RHomA(Y2, A)⊗LA2Y2 → A→ Y ⊗L

A1Y1 → .

• Triangulated functor −⊗LAe A⇒ ∃ a triangle in D(k):

(RHomA(Y2, A)⊗LA2Y2)⊗L

Ae A → A⊗LAe A→ (Y ⊗L

A1Y1)⊗L

Ae A →↓∼= ↓∼=

(Y2 ⊗LA RHomA(Y2, A))⊗L

Ae2A2 (Y1 ⊗L

A Y )⊗LAe

1A1

↓∼= ↓∼=A2 ⊗L

Ae2A2 A1 ⊗L

Ae1A1

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Theorem 12. [Keller 1998] Let A,A1 and A2 be algebras, and D(A)

admit a perfect recollement relative toD(A1) andD(A2). Then inD(k)

A⊗LAe A ∼= (A1 ⊗L

Ae1A1)⊕ (A2 ⊗L

Ae2A2).

Proof.

• Perfect recollements ≈ “derived triangular matrix algebras”!

• Denote by pX the homotopically projective resolution of X .

• A ∼ EndA(pj!j!A⊕pi∗i∗A) ∼

[EndA(pj!j!A) HomA(pi∗i∗A,pj!j!A)

0 EndA(pi∗i∗A)

]• A1 ∼ EndA1

(pi∗A) ∼ EndA(pi∗i∗A).

• A2 ∼ EndA2(pj!A) ∼ EndA(pj!j!A).

• Kadison’s method [Kadison 1989].

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6 Recollements & Hochschild cohomology

The n-th Hochschild cohomology of an algebra A is HHn(A) :=

ExtnAe(A,A) ∼= Hn(RHomAe(A,A)).

Lemma 2. Let A be an algebra and Xu→ Y

v→ Z → a triangle

in D(A) such that RHomA(X,Z) = 0 in D(k). Then there are three

triangles in D(k):

(1) RHomA(Y,X) → RHomA(Y, Y )φ→ RHomA(Z,Z) →

(2) RHomA(Z, Y ) → RHomA(Y, Y )ψ→ RHomA(X,X) →

(3) RHomA(Z,X) → RHomA(Y, Y )ϕ→ RHomA(X,X)⊕ RHomA(Z,Z) → .

Moreover, φ (resp. ψ, ϕ) induces a homomorphism of graded rings φ

(resp. ψ, ϕ) between cohomology rings.

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Proof. Triangles (1) and (2) :

• The bifunctor (−,−) := RHomA(−,−) ⇒ commutative diagram:

(X [1], Z[−1]) → (X [1], X) → (X [1], Y ) → (X [1], Z)

↓ ↓ ↓ ↓(Z,Z[−1]) → (Z,X) → (Z, Y ) → (Z,Z)

↓ ↓ ↓ ↓(Y, Z[−1]) → (Y,X) → (Y, Y ) → (Y, Z)

↓ ↓ ↓ ↓(X,Z[−1]) → (X,X) → (X,Y ) → (X,Z)

• RHomA(X,Z) = 0 in D(k) ⇒ four corners are zero.

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Proof. Triangle (3) :

• Octahedral axiom ⇒ commutative diagram:

(Z,Z[−1]) = (Z,Z[−1])

↓ ↓(Z,X) → (Y,X) → (X,X) → (Z[−1], X)

‖ ↓ ↓ ‖(Z,X) → (Y, Y ) → (X,X)⊕ (Z,Z) → (Z[−1], X)

↓ ↓(Z,Z) = (Z,Z)

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Theorem 13. Let A1, A and A2 be algebras, and (D(A1), D(A),

D(A2), i∗, i∗ = i!, i

!, j!, j! = j∗, j∗) a standard recollement given by

Y and Y2. Then there are three triangles in D(k):

(1) RHomAe(A,RHomA(Y2, A)⊗LA2Y2)

→ RHomAe(A,A)φ→ RHomAe

1(A1, A1) →

(2) RHomAe(RHomA1(Y, Y ), A)

→ RHomAe(A,A)ψ→ RHomAe

2(A2, A2) →

(3) RHomAe(RHomA1(Y, Y ),RHomA(Y2, A)⊗L

A2Y2)

→ RHomAe(A,A)ϕ→ RHomAe

1(A1, A1)⊕ RHomAe

2(A2, A2) → .

Moreover, φ (resp. ψ, ϕ) induces a homomorphism of graded rings φ

(resp. ψ, ϕ) between Hochschild cohomology rings.

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Proof.

• ∃ recollement (D(Aop ⊗ A1), D(Aop ⊗ A), D(Aop ⊗ A2), I∗, I∗ =

I!, I!, J!, J

! = J∗, J∗), 3

I∗ ∼= −⊗LA Y, J!

∼= −⊗LA2Y2,

I∗ = I! ∼= RHomA1(Y,−), J ! = J∗ ∼= RHomA(Y2,−),

I ! ∼= RHomA(RHomA1(Y,A1),−), J∗ ∼= RHomA2

(RHomA(Y2, A),−).

• ∃ a triangle J!J!A→ A→ I∗I

∗A→ in D(Aop ⊗ A).

• ∃ three triangles in D(k):

(1) RHomAe(A, J!J!A) → RHomAe(A,A) → RHomAe(I∗I

∗A, I∗I∗A) →

(2) RHomAe(I∗I∗A,A) → RHomAe(A,A) → RHomAe(J!J

!A, J!J!A) →

(3) RHomAe(I∗I∗A, J!J

!A) → RHomAe(A,A)

→ RHomAe(J!J!A, J!J

!A)⊕ RHomAe(I∗I∗A, I∗I

∗A) → .

• RHomAe(I∗I∗A, I∗I

∗A) ∼= RHomAe1(A1, A1) !

• RHomAe(J!J!A, J!J

!A) ∼= RHomAe2(A2, A2) !

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Corollary 2. LetA1, A andA2 be algebras, and (D(A1), D(A), D(A2),

i∗, i∗ = i!, i!, j!, j

! = j∗, j∗) a standard recollement given by Y and Y2.

Then there are three long exact sequences:

(1) · · · → ExtnAe(A,RHomA(Y2, A)⊗LA2Y2)

→ HHn(A)φn→ HHn(A1) → · · ·

(2) · · · → ExtnAe(RHomA1(Y, Y ), A)

→ HHn(A)ψn→ HHn(A2) → · · ·

(3) · · · → ExtnAe(RHomA1(Y, Y ),RHomA(Y2, A)⊗L

A2Y2)

→ HHn(A)ϕn→ HHn(A1)⊕HHn(A2) → · · · .

Moreover, ⊕n∈Nφn (resp. ⊕n∈Nψn, ⊕n∈Nϕn) is a homomorphism of

graded rings between Hochschild cohomology rings.

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Remark 2. Some of these long exact sequences had been studied:

• One-point extensions : [Happel 1989]

• Triangular matrix algebras : [Cibils 2000] + [Michelena-Platzeck

2000] + [Green-Solberg 2002] + [Cibils-Marcos-Redondo-Solotar

2003] + · · ·

• Stratified ideals : [Konig-Nagase 2009]

• Homological epimorphisms : [de la Pena-Xi 2006] + [Suarez-

Alvarez 2007]

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Thanks!