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Wrong way recollement for schemes

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WRONG WAY RECOLLEMENT FOR SCHEMES PETER J?RGENSEN The notion of a recollement of triangulated categories was introduced in [1,sec.1.4].The prototypical example of a recollement is D (Z )i ?i ?D (U ).j !Here X is a topological space equal to the union of the closed subset Z and the open complement U ,and D (Z ),D (X ),and D (U )are suitable derived categories of sheaves.The triangulated functors in a recolle-ment must satisfy various conditions,most importantly that (i ?,i ?),(i ?,i !),(j !,j ?),and (j ?,j ?)are adjoint pairs.The purpose of this note is to point out that,somewhat surpris-ingly,in the case of schemes,there is also a recollement which goes the other way.By way of notation,if X is a scheme then D (O X ),the derived category of sheaves of O X -modules,has the full subcategory D (X )consisting of complexes with quasi-coherent cohomology.If Z is a closed subscheme,there is also the full subcategory D Z (X )consisting of complexes with quasi-coherent cohomology supported on Z .Theorem.Let X be a quasi-compact quasi-separated scheme with a closed subscheme Z ,suppose that the open subscheme U =X ?Z is quasi-compact,and write U u ?→X for the inclusion.Then there is a recollement

D (U )R u ? L u ?D Z (X )

v

where v is the inclusion of the full subcategory D Z (X ).

2PETER J?RGENSEN

Before proving the Theorem,let me give the following Proposition which appears not to be stated in the literature,although it follows easily from known results and is well known to a number of people. Proposition.Let T be a compactly generated triangulated category with a set K which consists of compact objects and is closed under (de)suspension.Then there is a recollement

K⊥i?

i?

i!

K

j!

j?

where i?and j!are inclusions of full subcategories.

Here

K⊥={M∈T|Hom T(K,M)=0for each K in K},

while K is the smallest triangulated subcategory of T which contains K and is closed under set indexed coproducts;see[5,def.3.2.9]. Proof.It follows from the Thomason Localization Theorem,[4,thm.

2.1.1],that K is a compactly generated triangulated category which is generated by the objects of K.It is clear that the inclusion functor

T K

j!

is a triangulated functor respecting set indexed coproducts,so by the Brown Representability Theorem,[4,thm.4.1],there is a right adjoint j?to j!,

T j?

By[4,thm.2.1.3],the compact objects in the compactly generated category K are precisely the objects of K which are compact when viewed in T.Hence j!sends compact objects to compact objects,and so the right-adjoint j?respects set indexed coproducts by[4,thm.5.1]. And j?is triangulated by[5,lem.5.3.6],so by[4,thm.4.1]again,there is a right-adjoint j?to j?,

T j?

j?

WRONG WAY RECOLLEMENT3 As the inclusion j!is full and faithful,this is the situation considered in[3,prop.2.7]which gives a recollement

Ker j?i?

i?

i!

K j!

j?

where i?is the inclusion of the full subcategory Ker j?.

Finally,for N to be in Ker j?means j?N=0which holds precisely

if Hom K (K,j?N)=0for each K in K,because the objects of K

generate K .But

Hom K (K,j?N)~=Hom T(j!K,N)=Hom T(K,N),

so this again holds precisely if Hom T(K,N)=0for each K in K,that

is,if N is in K⊥.

So I can replace Ker j?by K⊥,and this gives the recollement of the Proposition. Proof(of Theorem).The category D Z(X)is a triangulated subcategory

of D(X)which is closed under set indexed coproducts.Hence D Z(X)

is a triangulated category with set indexed coproducts.Moreover,by [6,thm.6.8]there is an object E which is compact in D(X),sits in

D Z(X),and whose(de)suspensions generate D Z(X).Let K consist of all(de)suspensions of E.Then D Z(X)is a compactly generated

triangulated category with compact generators the objects of K.

By[4,thm.2.1.2],it follows that K D

Z(X),the smallest triangulated

subcategory of D Z(X)which contains K and is closed under set indexed

coproducts,is equal to D Z(X)itself.But K D

Z(X)is clearly equal to

K D(X),the smallest triangulated subcategory of D(X)which contains K and is closed under set indexed coproducts,so K D(X)is equal to D Z(X).

I shall simply denote K D(X)by K ,so

K =D Z(X).(1) By[7,prop.6.7]there is an adjoint pair of functors

D(O U)

R u?

Note that since u is the inclusion of an open subscheme,u?and hence also L u?is just restriction to U.It turns out that L u?and R u?restrict

4PETER J?RGENSEN

to the full subcategories D(U)and D(X),and so induce an adjoint pair

D(U)

R u?

This is obvious for L u?which is just restriction to U.For R u?it fol-lows from[2,thm.3.3.3]because u is a quasi-compact quasi-separated morphism.

I will now show

K⊥=Ess.Im R u?(2) where K⊥is taken inside D(X),while R u?is viewed as a functor D(U)?→D(X)and Ess.Im denotes essential image.To see?,let M be in Ess.Im R u?;that is,M~=R u?N for some N in D(U).Then Hom D(X)(K,M)~=Hom D(X)(K,R u?N)

~=Hom

( L u?K,N)

D(U)

(a)

=0

for each K in K,where(a)is because the cohomology of K is supported on Z whence L u?K~=0.Hence M is in K⊥.

To see?,note?rst that each M in D(X)determines a unit morphism

M?→R u? L u?M.

Let I be a K-injective resolution of M which exists by[7,thm.4.5]. Since L u?is just restriction, L u?I~=u?I.And u?has an exact left-adjoint u!so u?I is also K-injective whence R u?u?I~=u?u?I.Hence up to isomorphism,the unit morphism is just the canonical morphism

I?→u?u?I.

The restriction of this to U is an isomorphism because u is the inclu-sion of U into X,so completing the unit morphism to a distinguished triangle

M?→R u? L u?M?→C?→,

the cohomology of the cone C is supported on Z,that is,C is in D Z(X). Now let M be in K⊥;that is,

Hom D(X)(K,M)=0

for each K in K.Since also

Hom D(X)(K,R u? L u?M)~=Hom D(U)( L u?K, L u?M)=0

for each K in K,the distinguished triangle implies

Hom D(X)(K,C)=0

WRONG WAY RECOLLEMENT5 for each K in K.But C is in D Z(X)so this implies C=0.The distin-guished triangle thus shows that the unit morphism is an isomorphism,

M~=R u? L u?M,

so M~=R u?N for an N in D(U).Hence M is in Ess.Im R u?as desired. To conclude the proof,observe that each N in D(U)determines a counit morphism

L u?R u?N?→N.

Let J be a K-injective resolution of N.Then R u?J~=u?J and hence L u?R u?J~=u?u?J.Hence up to isomorphism,the counit morphism is just the canonical morphism

u?u?J?→J

which is an isomorphism because u is the inclusion of an open sub-scheme.

By adjoint functor theory this implies that

R u?is a full embedding of D(U)into D(X).(3) Let me now use the Proposition with T=D(X)and K from above, D(X)being compactly generated by[2,thm.3.1.1(2)].The Proposition

gives a recollement.Equation(1)says that I can replace K with D Z(X).Equation(2)says that I can replace K⊥with Ess.Im R u?. And equation(3)says that R u?is a full embedding of D(U)into D(X), so I can replace Ess.Im R u?with D(U).This gives the recollement of the Theorem.

Acknowledgement.I thank professor Leovigildo Alonso Tarr′?o for stimulating correspondence.The diagrams were typeset with XY-pic.

References

[1]A.A.Beilinson,J.Bernstein,and P.Deligne,Faisceaux pervers,Ast′e risque

100(1982),5–171(Vol.1of the proceedings of the conference“Analysis and topology on singular spaces”,Luminy,1981).

[2]A.Bondal and M.Van den Bergh,Generators and representability of functors

in commutative and noncommutative geometry,Mosc.Math.J.3(2003),1–36.

[3]J.-I.Miyachi,Localization of triangulated categories and derived categories,J.

Algebra141(1991),463–483.

[4]A.Neeman,The Grothendieck duality theorem via Bous?eld’s techniques and

Brown representability,J.Amer.Math.Soc.9(1996),205–236.

[5]

6PETER J?RGENSEN

[7]N.Spaltenstein,Resolutions of unbounded complexes,Compositio Math.65

(1988),121–154.

Department of Pure Mathematics,University of Leeds,Leeds LS2 9JT,United Kingdom

E-mail address:popjoerg@https://www.sodocs.net/doc/0816029271.html,

URL:https://www.sodocs.net/doc/0816029271.html,/~popjoerg

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