Research article

Silting objects and recollements of extriangulated categories

  • Received: 03 April 2024 Revised: 15 August 2024 Accepted: 19 August 2024 Published: 23 August 2024
  • MSC : 16G10, 18E40, 18G80

  • In this paper, let (A,C,B) be a recollement of extriangulated categories. We construct a silting object and a tilting object from the two end terms of a recollement. We also show that the reverse direction holds under natural assumptions. Moreover, we show that our gluing preserves cotorsion pairs.

    Citation: Zhen Zhang, Shance Wang. Silting objects and recollements of extriangulated categories[J]. AIMS Mathematics, 2024, 9(9): 24796-24809. doi: 10.3934/math.20241207

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  • In this paper, let (A,C,B) be a recollement of extriangulated categories. We construct a silting object and a tilting object from the two end terms of a recollement. We also show that the reverse direction holds under natural assumptions. Moreover, we show that our gluing preserves cotorsion pairs.



    Belinson et al. [1] introduced the notions of recollements of abelian and triangulated categories, which were in connection with derived categories of sheaves on topological spaces, with the idea that one triangulated category may be glued together from two others. The recollements of abelian and triangulated categories are closely related, and they play an important role in algebraic geometry and representation theory; see, for instance, [2,3,4].

    Gluing techniques concerning the recollement of triangulated or abelian categories have been investigated by more and more experts and scholars. For example, the glued cotorsion pairs [5], the glued torsion pairs [6], and so on (e.g., [7,8]). In particular, Liu et al. [9] presented explicit constructions of the gluing of silting objects for a recollement of triangulated categories.

    Nakaoka and Palu [10] introduced an extriangulated category that extracts properties from triangulated categories and exact categories. In particular, exact categories and triangulated categories are extriangulated categories. There are a lot of examples of extriangulated categories, which are neither exact nor triangulated categories; see [10,11]. In the work of [12], Wang et al. defined the recollement of extriangulated categories, which generalized recollements of abelian categories and triangulated categories.

    Motivated by the applications of gluing techniques in recollements, we want to consider gluing a silting object and a tilting object from two other silting objects and tilting objects in a recollement of extriangulated categories.

    In this paper, let (A,C,B) be a recollement of extriangulated categories. Our first main result describes how to glue together silting objects σ in A and ω in B to obtain a silting object ρ in C; see Theorem 3.3. In the reverse direction, our second main result gives sufficient conditions for a silting object ρ of C, relative to the functors involved in the recollement, to induce silting objects in A and B; see Theorem 3.5, which partially extends the results of [9, Theorem 3.1], [13, Theorem 2.10] and [14, Theorem 2.5] in the framework of extriangulated categories. Our third main result constructs a glued tilting object from tilting objects in A and in B; see Theorems 4.2 and 4.3, which extend the result in [15, Theorem 3.5] to extriangulated categories.

    This section briefly recalls some definitions and basic properties of extriangulated categories from [10,11]. We omit some details here, but we refer the readers to [10].

    Let C be an additive category. All subcategories considered are full additive subcategories closed under isomorphisms. Let

    E:Cop×CAb

    be a biadditive functor, where Ab is an abelian category. For any pair of objects A,CC, an element δE(C,A) is called an E-extension.

    Let s be a correspondence that associates an equivalence class

    s(δ)=[AxByC]

    to any E-extension δE(C,A). This s is called a realization of E if it makes the diagrams in [10, Definition 2.9] commutative.

    We call a triplet (C,E,s) an extriangulated category if it satisfies:

    (1) E:Cop×CAb is a biadditive functor.

    (2) s is an additive realization of E.

    (3) E and s satisfy the compatibility conditions (ET3), (ET3)op, (ET4), and (ET4)op in [10, Definition 2.12].

    Remark 2.1. By [10, Example 2.13 and Proposition 3.22], triangulated categories and exact categories (with a condition concerning the smallness) are typical examples of extriangulated categories.

    For an extriangulated category C, we use the following notations in [10,11]:

    ● A sequence AaBbCδ is called an E-triangle.

    ● Let AaBbCδ be an E-triangle. A is called the CoCone of b and denoted by CoCone(B,C); C is called the Cone of a and denoted by Cone(A,B).

    ● An object P is called projective if, for any E-triangle AxByCδ and any morphism cC(P,C), there exists a morphism bC(P,B), satisfying yb=c. The class of all the projective objects in C is denoted by PC. An injective object can be dually defined, and the class of all the injective objects in C is denoted by IC.

    ● We say that C has enough projective objects if, for any object CC, there exists an E-triangle KxPyCδ satisfying PPC. Dually, we can define C as having enough injective objects.

    Let X, Y be subcategories of C.

    ● Denote Cone(X,Y) by the subcategory of C consisting of MC, which admits an E-triangle XYM in C, with XX and YY. We say that X is closed under Cones if Cone(X,X)X.

    ● Denote CoCone(X,Y) by the subcategory of C consisting of MC, which admits an E-triangle MXY in C, with XX and YY. We say that X is closed under CoCones if CoCone(X,X)X.

    ● We denote

    ΩX=CoCone(PC,X).

    And Ω is called the syzygy of X. Dually, we define the cosyzygy of X by

    ΣX=Cone(X,PC),

    see [16, Definition 4.2, and Proposition 4.3].

    For any subcategory X of C, put

    Ω0X=X,

    for k1, define ΩkX inductively by

    ΩkX=Ω(Ωk1X)=CoCone(PC,Ωk1X).

    ΩkX is called the k-th syzygy of X. Dually the k-th cosyzygy ΣkX by

    ΣkX=Cone(Σk1,IC)

    for k1 can be defined.

    For any k>0 and any objects X,Y, Liu and Nakaoka [16, Proposition 5.2] defined the higher extension groups in an extriangulated category having enough projective and injective objects as

    Ek+1(X,Y)=E(ΩkX,Y)=E(X,ΣkY).

    They showed the following result:

    Lemma 2.2. [16, Proposition 5.2] Let AfBgCδ be an E-triangle. For any object XC, there are long exact sequences

    C(X,A)C(X,f)C(X,B)C(X,g)C(X,C)(δ)XE(X,A)fE(X,B)gE(X,C)E2(X,A)

    and

    C(C,X)C(g,X)C(B,X)C(f,X)C(A,X)(δ)XE(C,X)gE(B,X)fE(A,X)E2(C,X).

    We define

    X={YCEk(X,Y)=0,k1}.

    Dually, we define

    X={YCEk(Y,X)=0,k1}.

    Definition 2.3. [17, Definition 3.1] Let C be an extriangulated category, X and Y subcategories of C. We call a pair (X,Y) a hereditary cotorsion pair if:

    (1) Both X and Y are closed under direct summands;

    (2) Ek(X,Y)=0 for any k1 and XX and YY;

    (3) C=Cone(Y,X);

    (4) C=CoCone(Y,X).

    We recall the concepts and basic properties of recollements of extriangulated categories from [12].

    Definition 2.4. [12, Definition 3.1] Let A, C, and B be three extriangulated categories. A recollement of C relative to A and B, denoted by (A,C,B), is a diagram

    (2.1)

    given by two exact functors i, j, two right exact functors i, j!, and two left exact functors i!, j, which satisfies the following conditions:

    (R1) (i,i,i!) and (j!,j,j) are adjoint triples.

    (R2) Im i=Ker j.

    (R3) i, j!, j are fully faithful.

    (R4) For each CC, there exists a left exact EC-triangle sequence

    ii!CCjjCiA

    with AA.

    (R5) For each CC, there exists a right exact EC-triangle sequence

    iA1j!jCCiiC

    with A1A.

    Remark 2.5. (1) If the categories A, B, and C are triangulated, then Definition 2.4 coincides with the definition of a recollement of triangulated categories [1].

    (2) If the categories A, B, and C are abelian, then Definition 2.4 coincides with the definition of a recollement of abelian categories [3,4,6].

    Lemma 2.6. [12, Lemma 3.3] Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4.

    (1) All the natural transformations iiIdA, IdAi!i, IdBjj!, jjIdB are natural isomorphisms.

    (2) ij!=0=i!j.

    (3) i preserves projective objects, and i! preserves injective objects.

    (4) j! preserves projective objects, and j preserves injective objects.

    (5) If i! (resp. j) is exact, then i (resp. j) preserves projective objects.

    (6) If i (resp. j!) is exact, then i (resp. j) preserves injective objects.

    (7) If C has enough projectives, then A has enough projectives; if C has enough injectives, then A has enough injectives.

    (8) If C has enough projectives and j is exact, then B has enough projectives; if C has enough injectives and j! is exact, then B has enough injectives.

    (9) If i is exact, then j! is exact.

    (10) If i! is exact, then j is exact.

    The following results will be used frequently in this paper; their proof is very similar to [18, Proposition 2.8], we omit it.

    Lemma 2.7. Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4.

    (1) If A has enough projectives and i! is exact, then

    EkC(iX,Y)EkA(X,i!Y)

    for any k1, XA and YC.

    (2) If A has enough injectives and i is exact, then

    EkA(iX,Y)EkC(X,iY)

    for any k1, XC and YA.

    (3) If B has enough projectives and j! is exact, then

    EkC(j!Z,Y)EkB(Z,jY)

    for any k1, ZB and YC.

    In this section, we always assume A,C and B are three extriangulated categories with enough projective and injective objects.

    A subcategory X of C is called thick if it is closed under extensions, Cones, CoCones, and direct summands. Let thickX denote the smallest thick subcategory containing X. When X is a single object M, then thick(M)=thick (add M).

    Definition 3.1. An object σ in an extriangulated category C is silting if

    (1) Ek(σ,σ)=0 for all k1;

    (2) thick(σ)=C.

    For an object σ, add σ denotes the closure of σ under finite direct sums and summands. Clearly, when σ is a silting object, then add σ is a silting subcategory in the sense of [17, Definition 5.1]. When C is a well generated triangulated category, Definition 3.1 coincides with [13, Definition 1.11].

    Example 3.2. Let R be any ring, and let P(R) denote the category of right R-modules of finite projective dimension. Since P(R) is closed under extensions, it is an extriangulated category. We can easily check that R is a silting object of P(R).

    For any subcategory X of C, let

    X0=X0=X.

    We denote the full subcategories Xi and Xi for i>0 inductively by

    Xi=CoCone(X,Xi1),     Xi=Cone(Xi1,X).

    Put

    X:=i0Xi   and  X:=i0Xi.

    Proposition 3.3. Assume that C admits a recollement relative to A and B as (2.1) in Definition 2.4 and σ, ω are silting objects in A and B, respectively. If i is exact, then

    thick(iσj!ω)=C.

    Proof. Let CC. As i is exact, there exists an E-triangle by [12, Proposition 3.4],

    (1)  j!jCCiiC.

    Clearly, iCA and jCB. By [17, Theorem 5.7], there exist two hereditary cotorsion pairs ((add σ), (add σ)) in A and ((add ω), (add ω)) in B, respectively. So we get two E-triangles associated with the above two hereditary cotorsion pairs:

    (2)  NMiC   and    (3)  VUjC,

    where N(add σ), M(add σ) and V(add ω), U(add ω).

    Since i is exact, i(add σ) (add iσ) and i(add σ) (add iσ). So iiCthick(iσ) by the E-triangle (2). Similarly, since i is exact, j! is exact. So j!jCthick(j!ω) by the E-triangle (3). Hence Cthick(j!ωiσ) by the E-triangle (1). So

    C=thick(iσj!ω).

    This completes the proof.

    Theorem 3.4. Let (A,C,B) be a recollement as (2.1) in Definition 2.4. Assume σ and ω are silting objects in A and B, respectively. If

    EkC(iσ,j!w)=0

    for any k1 and i is exact, then j!wiσ is a silting object in C.

    Proof. Since i is exact, j! is exact by Lemma 2.6(9). So

    EkC(j!ω,j!ω)EkB(ω,jj!ω)EkB(ω,ω)=0

    and

    EkC(j!ω,iσ)EkB(ω,jiσ)=0

    for any k1 by Lemmas 2.6(1) and 2.7(3). Since i is exact, i preserves injective objects. So

    EkC(iσ,iσ)EkA(iiσ,σ)EkA(σ,σ)=0

    by Lemmas 2.6(1) and 2.7(2). So for any k1, we get

    EkC(iσj!ω,iσj!ω)=EkC(iσ,iσ)EkC(iσ,j!ω)EkC(j!ω,iσ)EkC(j!ω,j!ω)=0.

    Hence j!wiσ is a silting object in C by Proposition 3.2 and Definition 3.1.

    Let σ and ω be silting objects; then there exist hereditary cotorsion pairs ((addσ), (addσ)) and ((add ω), (add ω)) in A and B by [17, Theorem 5.7], respectively. Define

    U:={CCiC(addσ),jC(addω)}

    and

    V:={CCi!C(addσ),jC(addω)}.

    By [12, Theorem 4.4], (U,V) is a glued hereditary cotorsion pair in C.

    On the other hand, by Theorem 3.3, we obtain a silting object

    ρ=j!ωiσ

    in C. So there exists a hereditary cotorsion pair ((add ρ), (add ρ)) in C.

    The following result indicates that our gluing preserves the cotorsion pair:

    Theorem 3.5. In the situation above, let

    ρ=j!ωiσ.

    Then

    (U,V)=((addρ),(addρ)).

    Proof. Since σ and ω are silting objects,

    (addσ)=σ

    and

    (addω)=ω

    by [17, Lemma 4.11]. By Theorem 3.3, ρ is a silting object in C. So

    thick(ρ)=C.

    Hence

    ρ=(addρ)=(addρ)

    by [17, Lemma 4.11]. So CV if and only if i!Cσ and jCω. By Lemma 2.7, the latter holds if and only if C(iσ) and C(j!ω) if and only if

    Cρ=(addρ).

    So

    (U,V)=((addρ),(addρ)).

    For the converse of Theorem 3.3, we have

    Theorem 3.6. Let (A,C,B) be a recollement as (2.1) in Definition 1.4. Assume that ρ is a silting object in C and i is exact,

    (1) If iiρ add ρ, then iρ is a silting object in A.

    (2) If j!jρ add ρ, then jρ is a silting object in B.

    In this case, ρ is a glued silting object with respect to two silting objects, σ in A and ω in B.

    Proof. (1) Since i is exact,

    EkA(iρ,iρ)EkC(ρ,iiρ)

    for any k1 by Lemma 2.11. Since iiρ add ρ,

    EkC(ρ,iiρ)=0

    by Definition 3.1. So

    EkA(iρ,iρ)=0.

    For any AA, then iAC. Since ρ is a silting object in C, there exists a hereditary cotorsion pair ((add ρ), (add ρ)) induced by ρ. So there exists an E-triangle V1U1iA with U1(add ρ) and V1(add ρ). Since i is exact, we get another E-triangle: iV1iU1iiA, Clearly, iV1(i(add ρ))thick(iρ) and iU1(i(add ρ))thick(iρ). Also

    iiAA,

    so Athick(iρ) and

    thick(iρ)=A.

    Hence, iρ is a silting object in A.

    (2) Since i is exact, j preserves injective objects by Lemma 2.10(6). So

    EkB(jρ,jρ)EkC(j!jρ,ρ)

    for any k1, by Lemma 2.7. Since j!jρ add ρ,

    EkB(jρ,jρ)=0.

    For any BB, using the same proof method, we can also obtain

    thick(jρ)=B.

    Hence, jρ is a silting object in B.

    Denote iρ by σ and jρ by ω. Then σ is a silting object in A, and ω is a silting object in B. Moreover, iσ add ρ and j!ωadd ρ. So

    Ek1C(iσ,j!ω)=0.

    By Theorem 3.3, j!ωiσ is a silting object. Since i is exact, there exists an EC-triangle j!ωρiσ by [12, Proposition 3.4]. So

    ρj!ωiσ

    and ρ glued from σ and ω.

    By applying Theorem 3.5 to triangulated categories and using the fact that any triangulated category can be viewed as an extriangulated category, we get the following result:

    Corollary 3.7. When (A,C,B) is a recollement of triangulated categories, assume that i is exact. The silting object ρ glued from two silting objects σ in A and ω in B if and only if iiρ add ρ and j!jρ add ρ, which is a special case of [13, Theorem 2.26].

    In this section, we restrict the gluing procedures to tilting objects. We always assume that A,C,B are three extriangulated categories with enough projective and injective objects.

    The notion of tilting objects in an extriangulated category was introduced in [19, Definition 5.2], and we recall it.

    Definition 4.1. Let T be an object in an extriangulated category C. It is called tilting if

    (1) T(PC);

    (2) Ek(T,T)=0, for all k1;

    (3) PC (add T).

    Clearly, if T is a tilting object, then

    thick(T)=thick(PC).

    By [17, Proposition 4.10],

    thick(PC)=(PC)

    and by [19, Lemma 5.4], any tilting object of C is silting of (PC) and they coincide when every object in C has finite projective dimension by [19, Theorem 5.6].

    Theorem 4.2. Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4 and i, i! exact. Assume that T1 and T2 are tilting objects in A and B, respectively, such that

    EkC(iT1,j!T2)=0

    for all k1. Then there exists a glued tilting object ρ with respect to T1 and T2 in C.

    Proof. By [19, Lemma 5.4], T1 and T2 are silting objects of (PA) and (PB), respectively, and we have

    thick(T1)=thick(PA)=(PA)

    and

    thick(T2)=thick(PB)=(PB).

    Moreover, there exist two cotorsion pairs ((add T1), (add T1)) in (PA) and ((add T2), (add T2)) in (PB), respectively. Denote j!T2iT1 by ρ. Since i is exact, for any k1,

    EkC(iT1,iT1)EkA(T1,T1)=0,   EkC(j!T2,j!T2)EkB(T2,T2)=0

    and

    EkC(j!T2,iT1)EkB(T2,jiT1)=0

    by Lemmas 2.6 and 2.7.

    So

    EkC(ρ,ρ)=0,

    and ρ is self-orthogonal. Since i is exact, j! is exact. Since i! is exact, i preserves projective objects. So

    ρ=j!T2iT1(PC)

    by [19, Lemma 5.1].

    Since i is exact, for any C(PC), we have the following E-triangle by [12, Proposition 3.4],

    (1)  j!jCCiiC.

    Since both i and j preserves projective objects, iC(PA) and jC(PB). So we get two E-triangles associated with the above two cotorsion pairs:

    (2)  NMiC   and    (3)  VUjC,

    where N(add T1), M(add T1) and V(add T2), U(add T2).

    Since i is exact, iiCthick(iT1) by the E-triangle (2). Similarly, j!jCthick(j!T2) by the E-triangle (3). Hence Cthick(j!T2iT1) by the E-triangle (1). So

    thick(ρ)=(PC)

    and thus ρ is a silting object in (PC). Hence ((add ρ), (add ρ)) is a cotorsion pair in (PC). Since PC(PC), for any MPC, there exists an E-triangle VUM, where U (add ρ) and V (add ρ). Since M is a project object, the E-triangle splits. So M is a direct summand of U, and thus PC (add ρ). So ρ is a tilting object in C by Definition 4.1.

    For the converse direction,

    Theorem 4.3. Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4 and i, i! are exact. Assume that ρ is a tilting object in C.

    (1) If iiρ add ρ, then iρ is a tilting object in A.

    (2) If j!jρ add ρ, then jρ is a tilting object in B.

    In this case, ρ glued from two tilting objects in A and B.

    Proof. By Theorem 3.5, both iρ and jρ are self-orthogonal. Since i is exact and i preserves projective objects,

    iρ(iPC)(PA).

    Since i! is exact, j preserves projective objects. So

    jρ(jPC)(PB).

    On the other hand, since ρ is a tilting object in C, PC(addρ). For any MPA, then iMPC. So

    MiiM(addiρ).

    For any NPB, then j!NPC. So

    Njj!N(addjρ).

    Hence, iρ is a tilting object in A, and jρ is a tilting object in B, by Definition 4.1.

    Since i is exact, we have an E-triangle by [12, Proposition 3.4],

    (1)  j!jρρiiρ.

    Since iiρ add ρ and j!jρ add ρ,

    EkC(iiρ,j!jρ)=0.

    So

    ρiiρj!jρ

    and thus ρ glued from two tilting objects iρ in A and jρ in B.

    Corollary 4.4. (1) If we let C be a triangulated category that is K-linear over a field K, B a localizing subcategory of C, and

    A=ker(C(A,)).

    Then Theorem 4.2 is exactly [14, Theorem 2.5] when viewing triangulated categories as extriangulated categories.

    (2) Let Λ1, Λ, and Λ2 be artin algebras,

    A=modΛ1,   C=modΛ   andB=modΛ2.

    Then Theorem 4.2 is [15, Theorem 3.5].

    In this section, we present the conclusions of the paper.

    Let (A,C,B) be a recollement as (2.1) in Definition 2.4. Our first main result describes how to glue together silting objects σ in A and ω in B to obtain a silting object ρ in C; see Theorem 3.3.

    Theorem 5.1. Assume σ and ω are silting objects in A and B, respectively. If

    EkC(iσ,j!w)=0

    for any k1 and i is exact, then j!wiσ is a silting object in C.

    In the reverse direction, our second main result gives sufficient conditions for a silting object ρ of C, relative to the functors involved in the recollement, to induce silting objects in A and B; see Theorem 3.5, which partially extends the results of [9, Theorem 3.1], [13, Theorem 2.10] and [14, Theorem 2.5] in the framework of extriangulated categories.

    Theorem 5.2. Assume that ρ is a silting object in C and i is exact,

    (1) If iiρ add ρ, then iρ is a silting object in A.

    (2) If j!jρ add ρ, then jρ is a silting object in B.

    In this case, ρ glued from two silting objects, σ in A and ω in B.

    Our third main result constructs a glued tilting object from tilting objects in A and in B; see Theorems 4.2 and 4.3, which extend the result in [15, Theorem 3.5] to extriangulated categories.

    Theorem 5.3. Let i and i! be exact. Assume that T1 and T2 are tilting objects in A and B, respectively, such that

    EkC(iT1,j!T2)=0

    for all k1. Then there exists a glued tilting object ρ with respect to T1 and T2 in C.

    Theorem 5.4. Let i, i! be exact. Assume that ρ is a tilting object in C.

    (1) If iiρ add ρ, then iρ is a tilting object in A.

    (2) If j!jρ add ρ, then jρ is a tilting object in B.

    In this case, ρ glued from two tilting objects in A and B.

    Zhen Zhang: contributed the creative ideals and proof techniques for this paper; Shance Wang: consulted the relevant background of the paper and composed the article, encompassing the structure of the article and the modification of grammar. All authors have read and agreed to the published version of the manuscript.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This research was supported by the National Natural Science Foundation of China (No. 12101344) and Shan Dong Provincial Natural Science Foundation of China (No. ZR2015PA001).

    The authors declared that they have no conflicts of interest.



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