Perverse sheaves and the Weil conjectures
Prof. Dr. Uwe JannsenSummer Term 16
Inhaltsverzeichnis
1 Triangulated categories 3
Appendix 1.A 19
2 Localization of categories and triangulated categories 22
3 t-structures 37
4 Derived functors 51
5 The six functors 53
6 Glueing of t-structures 57
7 60
8 61
9 t-exactness and the intermediate direct image 62
10 Laumon’s ℓ-adic Fourier transformation 65
11 Properties of perverse sheaves 71
1 Triangulated categories
Definition 1.1 An additive category C is a category whose Hom-sets are abelian groups,for which the composition of morphisms is bilinear, and for which there exist finite coproductsand products. (In particular, there is a zero object 0, which is final and cofinal). Then thecanonical morphism
X⨿Y → X
∏Y
is an isomorphism from the coproduct to the product.
Definition 1.2 A translation functor T on a category C is an automorphism
T : C → C .
For an object X in C, we also write X[n] for T nX, and for a morphism f in C we also writef [n] for T nf . A triangle in C (with respect to T ) is a diagram
X u // Y v // Z w // X[1]
with morphisms u, v and w. Write also
Z
+1
w
~~~~~~~~~~
X u // Y
v
__@@@@@@@@
where +1 indicates that w is a morphism Z → X[1]. A morphism of triangles is a commu-tative diagram
Xu //
f
��
Yv //
g
��
Zw //
h��
X[1]
f [1]
��X ′ u′ // Y ′ v′ // Z ′ w′
// X ′[1]
Definition 1.3 Let D be an additive category and let T be an additive automorphism ofD. For i ∈ Z the morphisms in
Homi(X,Y ) := Hom(X,T i(Y )) = Hom(X,Y [i])
are called morphisms of degree i. Define the composition
Homi(X,Y )×Hj(Y, Z) → Homi+j(X,Z)
by(f, g) 7→ T i(g) ◦ f .
We obtain a Z-graded category with homomorphism groups∏n∈Z
Homn(X,Y )).
D is called a triangulated category, if there is a class of triangles – called distinguishedtriangles – for which the following holds:
(TR1) (a) Every triangle which is isomorphic to a distinguished triangle is distinguished.
3
(b) Every morphism Xf→ Y can be extended to an distinguished triangle
Xf→ Y → Z → T (X) .
(c) The triangle Xid→ X → 0 → T (X) is distinguished.
(TR2) The triangleX
u→ Yv→ Z
w→ T (X)
is distinguished if and only if the triangle Yv→ Z
w→ T (X)−T (u)→ T (Y ) is distinguished.
(TR3) Every diagram with morphisms u, v, where the first square commutes and the rowsare distinguished triangles
X //
u
��
Y //
v
��
Z //
w
��
T (X)
T (u)
��X ′ // Y ′ // Z ′ // T (X ′)
can be extended to a morphism of triangles by a morphism w as indicated.
(TR4) (Octahedral axiom) Let
Xu // Y
i // Z ′ // TX
Yv // Z // X ′ j // TY
Xw // Z // Y ′ // TX
be three distinguished triangles with w = v ◦ u. Then the following equivalent propertieshold.
(i) There are morphisms f : Z ′ → Y ′, g : Y ′ → X ′, such that the diagram
X u //
id��
Y i //
v��
Z ′ //
f��
TX
id��
X w //
u��
Z //
id��
Y ′ //
g��
TX
Tu��
Yv // Z // X ′ j //
��
TY
T (i){{xxxxxxxx
TZ ′
commutes, and the third column is a distinguished triangle.
4
(ii) The diagram
Z ′
��666
6666
6666
6666
6
f
��
Y
i
::tttttttttt
v
��
X
u
<<xxxxxxxxx
w##F
FFFF
FFFF
TX
Z
$$JJJ
JJJJ
JJJ
��
Y ′
g��
DD����������������
X ′
j��
X ′
��T (Y )
T (i) // T (Z ′)
is commutative, and the column with the morphisms f and g is a distinguished triangle.
(iii) The diagram
Z ′ //
f
��
TX
Y
i
66nnnnnnnnnnnnnnn
v
��@@@
@@@@
@u
~~~~~~~~
Xw // Z //
!!CCC
CCCC
C Y ′ //
g��
TX
Tu��
X ′ j //
��
TY
T (i){{xxxxxxxx
TZ ′
is commutative, and the column with f and g is a distinguished triangle.
The equivalence of the conditions follows easily.
The octahedral axiom is an analogue of the second isomorphism theorem in abelian catego-ries. This is most obvious in the versions of (ii) and (iii): Z ′ is the analogue of Y/X and inthe category of complexes is given by Cone(u), Y ′ is the analogue of Z/X and is given byCone(w), and X ′ is the analogue of Z/Y and is given by Cone(v), and we obtain
Y ′/Z ′ =Z/X
Y/X∼=Z
Y= X ′ .
Additionally we get the information that
X ′ → T (Z ′)
is given by the composition T (i)j.
5
Lemma 1.4 The triangle X−id // X // 0 // TX is distinguished.
Proof : We have an isomorphism of triangles
X−idX // X // 0 // TX
X
idX
OO
idX // X //
−idX
OO
0
OO
// TX
T (idX)
OO
Definition 1.5 An additive functor F : D → D′ between triangulated categories is calledexact, or functor of triangulated categories, if it commutes with the translation functors Tand T ′ of D and D′:
FTX = T ′FX ,
and maps distinguished triangles in D to distinguished triangles in D′.
For working with triangulated categories, it is important to know that the above properties(TR1) to (TR4) hold. But, as observed by Peter May, the above system of conditions isredundant, and this is helpful for constructing triangulated categories.
Lemma 1.6 (TR3) follows from (TR4).
Proof Consider a diagram
(1) Xf //
i��
Yg //
j��
Z h // TX
T (i)��
X ′ f ′ // Y ′ g′ // Z ′ h′ // TX ′
where the first square is commutative. We are looking for a morphism k : Z → Z ′ makingthe diagram commutative.
Applying (TR1), we have a distinguished triangle
Xjf // Y ′ p // V
q // TX .
By (TR4) we get a commutative diagram
Xf //
id��
Yg //
j��
Zh //
s��
TX
id��
Xjf //
f��
Y ′ p //
id��
Vq //
t��
TX
T (f)��
Yj // Y ′ j′ // Y ′′ j′′ //
u��
TY
T (g)||yyyyyyyy
TZ
6
Similarly, for f ′i(= jf) we get the commutative diagram
X i //
id��
X ′ i′ //
f ′
��
X ′′ i′′ //
s′
��
TX
id��
Xf ′i //
i��
Y ′ p //
id��
Vq //
t′
��
TX
T (i)��
X ′ f ′ // Y ′ g′ // Z ′ h′ //
u′
��
TX ′
T (i′){{vvvvvvvvv
TX ′′
so thatpj = sg , tp = j′ , qs = h , j′′t = T (f)qpf ′ = s′i′ , t′p = g′ , qs′ = i′′ , h′t′ = T (i)q .
Now we let k = t′s : Z → Z ′. Then we havekg = t′sg = t′pj = g′j, andh′k = h′t′s = T (i)qs = T (i)h,so that k makes the diagram (1) commutative.
Lemma 1.7 Assume the following.
(TR2)(a) If the triangle
Xf // Y
g // Z h // X[1]
is distinguished, then
Yg // Z h // X[1]
−f [1] // Y [1]
is distinguished.
If (TR3) holds, then the converse of (TR2) (a) holds.
Proof Assume (g, h,−f [1]) is distinguished. By (TR2) (a) the triangle
X[1]−f [1] // Y [1]
−g[1] // Z[1]−h[1] // X[2]
is distinguished.
Now choose a distinguished triangle
Xf // Y
g′ // Z ′ h′ // X[1] .
Then, by (TR3) and exercise 1.16, we have an isomorphism of triangles
X[1]−f [1] //
id��
Y [1]−g′[1] //
id��
Z ′[1]−h′[1] //
α
��
X[2]
id��
X[1]−f [1] // Y [1]
−g[1] // Z[1]−h[1] // X[2] .
7
and hence an isomorphism of triangles
Xf //
id
��
Yg′ //
id
��
Z h′ //
α
��
X[1]
id��
Yf // Y
g // Zh // X[1] ,
so that the bottom triangle is distinguished by (TR1) (a).
We add some results of interest for working with triangulated categories.
Lemma 1.8 For a distinguished triangle
X u // Y v // Z w // X[1]
in a triangulated category D all compositions of consecutive morphisms are zero.
Proof : By (TR2) it suffices to show vu = 0. Now consider the following diagram
XidX //
id
��
X
u
��
// 0 //
g
��
X[1]
id��
Xu // Y
v // Z // X[1]
Here by (TR1) and assumption both rows are distinguished triangles, and the first squarecommutes. Hence by (TR3) the indicated dotted arrow exists making the diagram commu-tative. This implies vu = g0 = 0.
Lemma 1.9 IfX u // Y v // Z w // X[1]
is a distinguished triangle, then for any object A in D the sequence
HomD(A,X)u∗ // HomD(A, Y )
v∗ // HomD(A,Z)w∗ // HomD(A,X[1])
is exact.
Proof We already know that the composition of two consecutive morphisms is zero andagain it suffices to show the exactness at HomD(A, Y ). Consider f : A → Y with vf = 0.Then we have a morphism of triangles
AidA //
g
��
A //
f
��
0 //
��
A[1]
g[1]
��X
u // Yv // Z // A[1]
by (TR1), (TR2) and (TR3). Hence u∗(g) = f .
Lemma 1.10 LetX
u //
f
��
Yv //
g
��
Zw //
h��
X[1]
f(1)
��X ′ u′ // Y ′ v′ // Z ′ w′
// X ′[1]
8
be a morphism of distinguished triangles. If f and g are isomorphisms, then h is an isomor-phism as well.
Proof : By the previous lemma we get a commutative diagram with exact rows
Hom(A,X)u∗ //
fx��
Hom(A, Y )v∗ //
g∗��
Hom(A,Z)w∗ //
h∗��
Hom(A,X[1])u[1]∗ //
f [1]∗��
Hom(A, Y [1])
g[1]∗��
Hom(A,X ′)u′∗ // Hom(A, Y ′)
v′∗ / / Hom(A,Z ′)w′
∗ // Hom(A,X ′[1])u′[1]∗ // Hom(A, Y ′[1])
in which f∗, g∗, f [1]∗ and g[1]∗ are isomorphisms. Hence, by the five lemma, h∗ is an isomor-phism, and hence h is an isomorphism by the Yoneda lemma.
Now we discuss an important example of a triangulated category.
Let A be an additive category (not necessarily abelian!). Let C(A) be the category of the(unbounded) complexes in A: objects are diagrams
. . .→ Andn→ An+1 dn+1
→ An+2 → . . .
withAn ∈ Ob(A), dn ∈Mor(A), dn+1dn = 0 (n ∈ Z)
morphisms are commutative diagrams
. . . // An //
��
An+1 //
��
. . .
��. . . // Bn // Bn+1 // . . .
A complex A· is called bounded below, or bounded above, or bounded, respectively, if An = 0for n << 0, or for n >> 0, or for |n| >> 0, respectively. We obtain corresponding categoriesC+(A), C−(A), Cb(A).
Definition 1.11 For a complex A· and n ∈ Z let A·[n] be the n-th shifted complex:
(A·[n])i = Ai+n
diA·[n] = (−1)ndi+n
Definition 1.12 Let f : A· → B· be a morphism of complexes. The Cone Cone(f) of f isthe following complex:
Cone(f)n = Bn ⊕ An+1
dnCone(f) =
(dnB fn+1
−dn+1A
)In “elements” (b, a) 7→ (dnb + fn+1a,−dn+1a). Note that d(db + fa,−da) = (d2b + dfa −fda, d2b) = 0.
We obtain a sequence of complexes
0 → B· → Cone(f) → A·[1] → 0
9
Definition 1.13 A (naive) double complex in A is a diagram in A
......
// Am,n+1
OO
// Am+1,n+1
OO
//
. . . // Am,n
dII
OO
dI // Am+1,n
OO
// . . .
...
OO
...
OO
with d2I = 0 = d2II and dIdII = dIIdI, therefore a complex of complexes. If the consideredsums exist (i.e. Ai,j ∈ Cb(C(A)) or C(Cb(A) =) or C+(C+(A)) or in A there exist arbitrarysums), then the associated simple complex sA·,· is defined as
(sA·,·)n = ⊕p+q=n
Ap,q
d auf Ap,q = dI + (−1)pdII .
Other characterization: With dI and (−1)pdII on Ap,q, sA·,· is a non naive double complex(dIdII + dIIdI = 0) for which (dI + dII)
2 = 0.
Remark 1.14 With these definitions, Cone(f) is the simple complex associated to thedouble complex
A· f→ B·
first degree: −1 0
Definition 1.15 Two morphisms f, g : A· → B· of complexes are called homotopic (f ∼ g),if there is a homotopy between f and g, i.e., a family (hn)n∈Z of homomorphisms hn : An →Bn−1 in A with
f − g = dn−1hn + hn+1dn
And //
f−g��
hn
{{xxxxxxxx
An+1
hn+1||xxxxxxxxx
Bn−1d
// Bn
The class {f ∼ 0} of the zero homotopic morphisms form a two-sided ideal: If g ◦ f existsand g ∼ 0 or f ∼ 0, then g ◦ f ∼ 0; if f ∼ 0 and g ∼ 0, then f ⊕ g ∼ 0 in C(A).
Definition 1.16 The category K(A) of the complexes modulo homotopy (or homotopycategory of complexes) has the same objects as C(A) and as morphisms
HomK(A)(A·, B·) = HomC(A)(A
·, B·)/{f ∼ 0} .
Analogously one defines K+(A), K−(A), Kb(A). These are all additive categories, and A· p⇝A·[1] extends to a functor on K?(A).
10
Definition 1.17 For complexesA· andB· define the complex of abelian groupsHom·(A·, B·)by
Homn(A·, B·) = {families (fm : Am → Bm+n)} =∏m
HomA(Am, Bm+n) ,
where the differential d is defined as follows: for f = (fm) set df = (gm), with
gm = dBfm + (−1)m+1fm+1dA ∈ HomA(A
m, Bm+n+1), .
Obviously we haveZ0Hom·(A·, B·) = ker d0 = HomC(A)(A
·, B·)
B0Hom·(A·, B·) = im d−1 = Hom∼0C(A)(A
·, B·)
H0(Hom·(A·, B·)) = ker d0/imd−1 = HomK(A)(A·, B·)
Theorem 1.18 For an additive category A, K(A) becomes a triangulated category asfollows: Define the translation functor by T (A) := A[1] and call a triangle
X → Y → Z → X[1]
distinguished (or exact), if it is isomorphic to a triangle
(∗) A· f→ B· → Cone(f) → A·[1] .
Proof We give full proof, indicating the shortcuts due to Lemma 1.6 and 1.7. (TR1): We
only have to show that X · id→ X · → 0 → X[1] is exact, i.e., that C := Cone(idX) ∼= 0in K(A). This is (in an additive category) equivalent to idC = 0 in K(A), i.e. idC ∼ 0(homotopic to 0). This follows from the diagram
Xn ⊕Xn+1 (dpr1+pr2,−dpr2) //
id��
h=(0,pr1)
tthhhhhhhh
hhhhhhhh
hhXn+1 ⊕Xn+2
h=(0,pr1)ssgggggggggg
gggggggggg
ggg
Xn−1 ⊕Xn
(dpr1+pr2,−dpr2)// Xn ⊕Xn+1
(xn, xn+1)� d //
)
httiiiiiiii
iiiiiiii
iii(dxn + xn+1,−dxn+1)'
hssgggggggggg
gggggggggg
g
(0, xn)d7→ (d0 + xn,−dxn), (0, dxn + xn+1) ,
where (d0 + xn,−dxn) + (0, dxn + xn+1) = (xn, xn+1), so that id = dh+ hd.
(TR2): We can consider a distinguished triangle of the form
Xf // Y
i // Cone(f)p // X[1] .
Here we haveCone(f)n = Y n ⊕X[1]
11
with differentiald(yn, xn+1) = (dyn + fxn+1, dxn+1)
andi(yn) = (yn, 0) , p(yn, xn+1) = xn+1 .
We construct a commutative diagram
(1) Y i // Cone(f)p // X[1]
−f [1] // Y [1]
Yi // Cone(f) α // Cone(i)
β //
π
OO
Y [1]
with a (homotopy-) isomorphism π, in which α and β are the canonical morphisms forCone(i) mit Cone(i)n = Y n ⊕Xn+1 ⊕ Y n+1:
α(yn, xn+1) = (yn, xn+1, 0) ,
β(yn, xn+1, yn+1) = yn+1 .
Then this shows that Y // Cone(f)p // X[1]
i[1] // Y [1] is a distinguished triangle.
We haveC = Cone(i) = Cone(f)⊕ Y [1] 0 Y ⊕X[1]⊕ Y [1]
with differential
dC(yn, xn+1, yn+1) = (dyn + fxn+1 + yn+1,−dxn+1,−dyn) .
Define π and a homotopic inverse ι of π by
π(yn, xn+1, yn+1) = xn+1 , ιxn+1 = (0, xn+1,−fxn+1) .
Then π and ι are morphisms of complexes:
πdC(yn, xn+1, yn) = π(dyn + fxn+1 + yn+1, dxn+1,−dyn)= −dxn+1 = dX[1]π(yn, xn+1, yn)
anddCι(xn+1) = d(0, xn+1,−fxn+1)
= (fxn+1 − fxn+1,−dxn+1, dfxn+1)
= (0,−dxn+1,−f(−dxn+1)) = ι(dX[1]xn+1) .
Moreover we have πι = idX[1]:
πι(xn+1) = π(0, xn+1,−fxn+1) = xn+1
Furthermore we have ιπ ∼ idCone(i): Consider the homotopy
h : Cone(i)n // Cone(i)n−1
(yn, xn+1, yn+1)� // (0, 0, yn)
12
Then we have
(idCone(i) − ιπ)(yn, xn+1, yn+1)
= (yn, xn+1, yn+1)− ιπ(yn, xn+1, yn)
= (yn, xn+1, yn+1)− ιxn+1
= (yn, xn+1, yn+1)− (0, xn+1,−fxn+1)
= (yn, 0, yn+1 + fxn+1)
and
(dh+ hd)(yn, xn+1, yn+1)
= d(0, 0, yn) + h(dyn + fxn+1 + yn+1,−dxn+1, dyn+1)
= (yn, 0, dyn) + (0, 0, dyn + fxn+1 + yn+1)
= (yn, 0, yn+1 + fxn+1)
Finally we show that the two right squares of (1) are commutative.
We have πα = p, so that the square in the middle is commutative:
πα(yn, xn+1) = π(yn, xn+1, 0) = xn+1 = p(yn, xn+1) .
Furthermore β ∼ −f [1]π, so that the square on the right hand side is commutative: Since πand ι are homotopy isomorphisms inverse to each other, it suffices to show that
βι ∼ −f [1]πι ∼ −f [1] .
But we have
βι(xn+1) = β(0, xn+1,−f(xn+1)) = −f(xn+1) = −f [1](xn+1) .
Now we show the backward direction in TR2. By Lemma 1.17 and (TR3) this is not necessary,but we get a short proof as follows. By applying the shown direction several times, by “rollingon” we obtain that
Z w // T (X)−T (u) // T (Y )
−T (v) // T (Z)
and then
T (X)−T (u) // T (Y )
−T (u)// T (Z)−T (w)// T 2(X)
are distinguished triangles. By rolling on twice again we obtain that
T 2(X)T 2(u) // T 2(Y )
T 2(u) // T 2(Z)T 2(w)// T 3(X)
is a distinguished triangle. This is up to renumbering n 7→ n− 2 the initial triangle
X u // Y v // Z w // X[1] .
(TR3): By Lemma 1.16 this follows from (TR4), but here is a direct proof. Without restrictionwe consider a diagram
(2) Xf //
u
��
Yα //
v
��
Cone(f)β //
w?��
X[1]
u
��X ′ f ′ // Y ′ α′
// Cone(f ′)β′
// X[1] ,
13
where the first square is commutative up to homotopy, and we need to find a morphismw which makes the diagram commutative (up to homotopy). By assumption there exists ahomotopy
hn : Xn → (Y n−1)′
withvnfn − (f ′)nun = sn+1dnX + dn−1
Y ′ sn .
Now definew : Cone(f)n = Y n ⊕Xn+1 → (Y ′)n ⊕ (X ′)n+1 = Cone(f ′)n
by
wn =
(vn sn+1
0 un+1
).
Then w is a morphism of complexes: We have
d
(vn sn+1
0 un+1
) (ynxn+1
)= d
(vnyn + dn+1xn+1
un+1xn+1
)
=
(dvnyn + dsn+1xn+1 + f ′un+1xn+1
−dun+1xn+1
)(vn sn+1
0 un+1
)d
(ynxn+1
)=
(vn sn+1
0 un+1
) (dyn + fxn+1
−dxn+1
)
=
(vndyn + vnfxn+1 − sn+1dxn+1
−un+1dxn+1
).
Since dvn = vnd, dun+1 = un+1d and vnf − (f ′)un+1 = sn+1dn + dn−1sn we have dw = wd.Furthermore both last squares commute: We have
wα(yn) =
(v sn+1
0 un+1
) (yn0
)=
(vyn0
)αv(yn) = α(vyn) =
(vyn0
)and
uβ
(ynxn+1
)= u(xn+1)
β′w
(ynxn+1
)= β′
(v sn+1
0 un+1
) (ynxn+1
)= β′
(vyn + sn+1xn+1
un+1xn+1
)= u(xn+1) .
(TR4): We consider three distinguished triangles
Xf // Y
i // Z ′ // X[1]
Yg // Z // X ′ j // Y [1]
Xgf // Z // Y ′ // X[1]
14
and look for morphisms Z ′ u // Y ′ and Y ′ v // X ′ , such that the diagram
(3) Xf //
id
��
Yi //
g
��
Z ′ //
u
��
X[1]
id��
Xgf //
f
��
Z //
id
��
Y ′ //
v
��
X[1]
u[1]
��Y
g // Z // X ′ j //
��
Y [1]
i[1]||xxxxxxxx
Z ′[1]
is commutative and the third column is a distinguished triangle. We can assume that
Z ′ = Cone(f) = Y ⊕X[1]Y ′ = Cone(gf) = Z ⊕X[1]X ′ = Cone(g) = Z ⊕ Y [1]
Define morphisms
u : Z ′ = Y ⊕X[1] → Y ′ = Z ⊕X[1]
(yn, xn+1) 7→ (gyn, xn+1)
v : Y ′ = Z ⊕X[1] → X ′ = Z ⊕ Y [1]
(zn, xn+1) 7→ (zn, fxn+1)
These are morphisms of complexes:
udZ′(yn, xn+1) = u(dyn + fxn+1,−dxn+1)
= (gdyn + gfxn+1,−dxn+1)
dY ′u(yn, xn+1) = dY ′(gyn, xn+1)
= (dgyn + gfxn+1,−dxn+1)
vdY ′(zn, xn+1) = v(dzn + gfxn+1,−dxn+1)
= (dzn + gfxn+1,−fdxn+1)
dX′v(zn, xn+1) = dX′(zn, fxn+1) = (dzn + gfxn+1, dfxn+1)
Now define X ′ → Z ′[1] as the composition
X ′ j // Y [1]i[1] // Z ′[1]
15
Then it follows easily that the diagram (3) above is commutative:
Y // Z ′ = Y ⊕X[1] // X[1]
yn� //
_
g
��
(yn, 0), (yn, xn+1)_
u
��
_
u
��
� // Xn+1_
id
��gyn
� // (gyn, 0), (gyn, xn+1)� // xn+1
Z // Y ′ = Z ⊕X[1] // X[1]
zn� //
_
id
��
(zn, 0), (zn, xn+1)_
v
��
_
v
��
� // Xn+1_
��zn
� // (zn, 0), (zn, fxn+1)� // fxn+1
Z // X ′ = Z ⊕ Y [1]j //
w
��
Y [1]
(zn, yn+1)_
��
yn+1-
vvmmmmmm
mmmmmm
(yn+1, 0) (yn+1, 0)
Z ′[1] = Y [1]⊕X[2]
Now we show that the triangle
(4) Z ′ u // Y ′ v // X ′ w // Z ′[1]
is distinguished.
For this we show that there are morphisms
ϕ : Cone(u) // X ′ , ψ : X ′ // Cone(u)
which are homotopy inverse to each other, such that we have a diagram
Z ′ u // Y ′ v //
��
X ′ w //
ψ��
Z ′[1]
Z ′ u // Y ′ α(u) // Cone(u)β(u) //
ϕ
OO
Z ′[1]
with ϕ ◦ α(u) = v and β(u) ◦ ψ = w.
We have
Z ′ = Cone(f) = Y ⊕X[1], with dZ′(yn, xn+1) = (dyn + fxn+1,−dxn+1)
Y ′ = Cone(gf) = Z ⊕X[1] with dY ′(zn, xn+1) = (dzn + gfxn+1,−dxn+1) ,
and
C = Cone(u) = Y ′ ⊕ Z ′[1], with
16
dC(zn, xn+1; yn+1, xn+2)
= (d(zn, xn+1) + u(yn+1, xn+2);−d(yn+1, xn+2))
= ((dzn + gfxn+1,−dxn+1) + (gyn+1, xn+2); (−dyn+1 − fxn+2, dxn+2))
= (dzn + gfxn+1 + gyn+1,−dxn+1 + xn+2; dyn+1 − fxn+2, dxn+2)
as well asX ′ = Cone(g) = Z ⊕ Y [1]
withdX′(zn, yn+1) = (dzn + gyn+1,−dyn+1) .
Define ϕ and ψ by
ϕn =
(idZn 0 0 00 fn+1 idY n+1 0
),
so that ϕn(zn, xn+1, yn+1, xn+2) = (zn, fxn+1 + yn+1), and
ψn =
idZn 00 00 idY n+1
0 0
,
so that ψn(zn, yn+1) = (zn, 0, yn+1, 0).
Then ϕ and ψ are morphisms of complexes: We have
dϕ(zn, xn+1, yn+1, xn+2)
= d(zn, fxn+1 + yn+1)
= (dzn + gfxn+1 + gyn+1,−dfxn+1 − dyn+1)
and
ϕd(zn, xn+1, yn+1, xn+2)
ϕ(dzn + gfxn+1 + gyn+1,−dxn+1 + xn+2,−dyn+1 − fxn+2,+dxn+2)
= (dzn + gfxn+1 + gyn+1,−fdxn+1 + fxn+2 − dyn+1 − fxn+2)
= (dzn + gfxn+1 + gyn+1,−fdxn+1 − dyn+1)
Furthermore we have
dψ(zn, yn+1)
= d(zn, 0, yn+1, 0)
= (dzn + gyn+1, 0,−dyn+1, 0)
and
ψd(zn, yn+1)
= ψ(dzn + gyn+1,−dyn+1)
= (dzn + gyn+1, 0,−dyn+1, 0)
Moreover we have ϕ ◦ ψ = idX′ :
ϕnψn(zn, yn+1) = ϕn(zn, 0, yn+1, 0) = (zn, yn+1) .
17
Finally we show that ψ ◦ ϕ ∼ idCone(u). Define the homotopy
sn : Cone(u)n // Cone(u)n−1
sn =
0 0 0 00 0 0 00 0 0 00 idX 0 0
,
i.e.,s(zn, xn+1, yn+1, xn+2) = (0, 0, 0, xn+1) .
Then idCone(u) − ψ ◦ ϕ = sd+ ds:
(idCone(u) − ψ ◦ ϕ)(zn, xn+1; yn+1, xn+2)
= (zn, xn+1; yn+1, xn+2)− ψ(ϕ(zn, xn+1; yn+1, xn+2))
= (zn, xn+1; yn+1, xn+2)− ψ(zn, fxn+1 + yn+1)
= (zn, xn+1; yn+1, xn+2)− (zn, 0, fxn+1 + yn+1, 0)
= (0, xn+1,−fxn+1, xn+2)
and
sd(zn, xn+1, yn+1, xn+2) + ds(zn, xn+1, yn+1, xn+2)
= s(dzn + gfxn+1 + gyn+1,−dxn+1 + xn+2,−dyn+1 − fxn+2, dxn+2) + d(0, 0 0, xn+1)
= (0, 0, 0,−dxn+1 + xn+2 + (0, xn+1,−fxn+1, dxn+1)
= (0, xn+1,−fxn+1, xn+2)
18
Appendix 1.A
The following result is due to J.L. Verdier.
Lemma 1.15 (The 3x3 lemma) Consider a diagram
Xf //
i
��
Yg //
j
��
Z h //
k
��
TX
T (i)
��
/.-,()*+1 /.-,()*+2X ′ f ′ //
i′
��
Y ′ g′ //
j′
��
Z ′ h′ //
k′
��
TX ′
T (i′)
��
/.-,()*+3 /.-,()*+4 /.-,()*+5X ′′ f ′′ //
i′′
��
Y ′′ g′′ //
j′′
��
Z ′′ h′′ //
k′′
��
TX ′′
−T (i′′)
��
/.-,()*+6 /.-,()*+7 /.-,()*+8TX
Tf // TYTg // TZ
−Th // T 2X ,
in which jf = f ′i, and the two first rows and the two left columns are distinguished triangles.Then there is an object Z ′′ and the dotted arrows f ′′, g′′, h′′, k, k′, k′′ such that the diagramis commutative except for the bottom right square which commutes up to the sign −1, andall four rows and columns are distinguished.
Proof The bottom row is distinguished by the isomorphism of triangles
TXTf // TY
Tg // TZ−Th // T 2X
TX
id
OO
−Tf // TY
−id
OO
−Tg // TZ
id
OO
−Th // T 2X
id
OO
where the bottom triangle is distinguished by (TR2). Similarly, the right column is distin-guished. Now we start as in the proof of Lemma 1.6:
Applying (TR1), we have a distinguished triangle
Xjf // Y ′ p // V
q // TX .
19
By (TR4) we get a commutative diagram of distinguished triangles
Xf //
id��
Yg //
j��
Z h //
s��
TX
id��
Xjf //
f��
Y ′ p //
id��
Vq //
t��
TX
T (f)��
Yj // Y ′ j′ // Y ′′ j′′ //
u��
TY
T (g)||yyyyyyyy
TZ
so that u = T (g)j′′. Similarly, for f ′i(= jf) we get the commutative diagram
Xi //
id��
X ′ i′ //
f ′
��
X ′′ i′′ //
s′
��
TX
id��
Xf ′i //
i��
Y ′ p //
id��
Vq //
t′
��
TX
T (i)��
X ′ f ′ // Y ′ g′ // Z ′ h′ //
u′
��
TX ′
T (i′){{vvvvvvvvv
TX ′′
so that u′ = T (i′)h′, and
pj = sg , tp = j′ , qs = h , j′′t = T (f)qpf ′ = s′i′ , t′p = g′ , qs′ = i′′ , h′t′ = T (i)q .
Now we let k = t′s : Z → Z ′. Then we havekg = t′sg = t′pj = g′j, andh′k = h′t′s = T (i)qs = T (i)h,so that k makes the diagram (1) commutative.Define k = t′s : Z → Z ′. Then we get the commutativity of the first two squares:/.-,()*+1 kg = t′sg = t′pj = g′j/.-,()*+2 h′k = h′t′s = T (i)qs = T (i)h.
Corollary (TR3) holds.
Now define f ′′ = t ◦ s′ : X ′′ → Y ′′, and apply (T1) to construct a distinguished triangle
X ′′ f ′′ // Y ′′ g′′ // Z ′′ h′′ // TX ′′
20
Applying (TR4), we get a commutative diagram
X ′′ s′ //
id��
Vt′ //
t��
Z ′ T (i′)h′//
k′
��
TX ′′
id��
X ′′ f ′′ //
s′
��
Y ′′ g′′ //
id��
Z ′′ h′′ //
k′′
��
TX ′′
T (s′)��
V t // Y ′′ T (g)j′′// TZ
−Ts //
��
TV
T (t′){{wwwwwwwww
TZ ′
Now we get the claims on the remaining squares:/.-,()*+3 f ′′i′ = ts′i′ = tpf ′ = j′f ′/.-,()*+4 g′′j′ = g′′tp = k′t′p = k′g′/.-,()*+5 h′′k′ = T (i′)h′/.-,()*+6 j′′f ′′ = j′′ts′ = T (f)qs′ = T (f)i′′/.-,()*+7 k′′g′′ = T (g)j′′/.-,()*+8 −T (h)k′′ = −T (qs)k′′ = −T (q)T (s)k′′ = T (q)T (s′)h′′ = T (i′′)h′′
(so that the squares /.-,()*+3 - /.-,()*+7 commute, and /.-,()*+8 anti-commutes).
21
2 Localization of categories and triangulated catego-
ries
Let C be a category and let S be a set of morphisms in C. The localization C[S−1] of C by Sshould have the following universal property:
There is a functor aS : C → C[S−1], so that we have: If F : C → C ′ is a functor in anothercategory, so that for every s ∈ S the morphism F (s) is invertible C ′ (therefore has aninverse), then there is a uniquely determined functor F : C[S−1] → C ′, which makes thefollowing diagram commutative:
C as //
F ��===
====
= C[S−1]
∃!F||xxxxxxxxx
C ′
Such a category “always exists, modulo set theoretic difficulties”. Existing constructions are(by proper conditions)
• a calculus of fractions (used here)
• closed model categories (Quillen)
Fix a set S of morphisms in C as above.
Definition 2.1 S allows calculus of fractions from the right if the following holds:
(FR1) For all X ∈ Ob(C), idX ∈ S, and S is closed under composition.
(FR2)r Every diagram in C as on the left hand side can be continued to a commutativediagram on the right hand side:
X ′
s��
Yf // X
⇝Y ′ f ′ //
s′
��
X ′
s��
Yf // X
where s and s′ are in S.
(FR3)r For every diagram
X ′ t // Xf //g
// Ys // Y ′
with s ∈ S and sf = sg, there is a t : X ′ → X in S with ft = gt.
Dually (by reversing the arrows) one defines the calculus of fractions from the left. If Sallows calculus of fractions from the right and the left, then one says that S allows calculusof fractions.
Theorem 2.2 If S allows calculus of fractions from the right, then C[S−1] can be defined asfollows. We have
ob(C[S−1]) = ob(C) .
22
For an object X in C, let S/X be the category of the morphisms in S over X, i.e., themorphisms
s : X ′ // X
in C with s ∈ S. The morphisms are commutative diagrams
X ′′ f //
s′′ !!CCC
CCCC
C X ′
s′~~||||||||
X
(s′, s′′ ∈ S)
The category S/X is cofiltered: It is
(i) non-empty,
(ii) for objects s : X ′ → X and s′ : X ′′ → X in S/X the diagram
X ′
s′
��X ′′ s′′ // X
can be continued by (FR2)r to a commutative diagram
X ′′′ f //
s′′′
��
X ′
s′
��X ′′ s′′ // X
with s′′′ ∈ S, so that one has morphisms
X ′
X ′′′
<<yyyyyyyy
""EEE
EEEE
E
X ′′
in S/X.
(iii) For morphisms
X ′′ f //g
//
s′′ !!BBB
BBBB
B X ′
s′~~}}}}}}}}
X ,
with s′f = s′′ = s′g, there is, by (FR3)r, a morphism s′′′ : X ′′′ → X ′′ with fs′′′ = gs′′′.
Therefore the dual category (S/X)0 is filtered, and we define
HomC[S−1](X,Y ) = lim→X′∈(S/X)0
Hom(X ′, Y ) .
23
This means: Every morphism in HomC[S−1](X,Y ) is given by a diagram
X ′
s��
f
AAA
AAAA
A
X Y
with f ∈ HomC(X,Y ) and s ∈ S. We also write fs−1 for this morphism. The compositionof morphisms
X ′
s��
f
!!BBB
BBBB
B Y ′
t��
g
AAA
AAAA
A
X Y Z
(s, t ∈ S)
is given by a commutative diagram
X ′′
t′
��
g
!!CCC
CCCC
C
X∗
s��
f
!!CCC
CCCC
C Y ′
t�� A
AAAA
AAA
X Y Z
where the commutative diagram ∗ with t′ ∈ S exists by (FR2)r. Two morphisms
X ′
s′
��
f
AAA
AAAA
A X ′′
s′′
��
g
!!BBB
BBBB
B
X Y X Y
with s′, s′′ ∈ S coincide in C[S−1], if and only if there is a commutative diagram
X ′
f
!!CCC
CCCC
Cs′
}}{{{{{{{{
X X ′′′ h //
��
s′′′oo
OO
Y
X ′′g
=={{{{{{{{s′′
aaCCCCCCCC
with s′′′ ∈ S.
Remark 2.3 If S allows calculus of fractions from the right and C is additive, then C[S−1] isadditive in a natural way and C → C[S−1] is an additive functor. This is implied by the factthat HomC[S−1](X,Y ) as a filtered limit of abelian groups is an additive group in a naturalway, and the bilinearity of compositions follows as well.
Remark 2.4 The concrete addition of morphisms can described as follows: Let fs−1 andgt−1 be morphisms from X to Y in C[S−1]. By (FR2)r we have morphisms f ′ and s′ with
24
s′ ∈ SX ′
f
!!BBB
BBBB
Bs
}}||||||||
X Z
s′��
uoo
f ′
OO
Y
X ′′g
==||||||||t
aaCCCCCCCC
which make the left hand triangle without u commutative, so that u = sf ′ = ts′ ∈ S. Then
fs−1 = ff ′u−1 and gt−1 = gs′u−1 ,
where ff ′ and gs′ are morphisms in C, so that one can add fs−1 and gt−1 in an obvious wayby the common denominator u.
Now we show the universal property of C[S−1]. Let
F : C → C ′
be an exact functor into another triangulated category, such that F (s) is invertible in C ′ foreach s ∈ S. Define a functor
F : C[S−1] → C ′
as follows. Since C[S−1] has the same objects as C, we necessarily have to define F on objectsby
F (X) = F (X) .
Next, for a morphism fs−1 : X → Y in C[S−1] with f a morphism in C and s a morphismin S, we necessarily have to define
F (fs−1) = F (f)F (s)−1 .
This is well-defined, since F (s) is invertible by assumption on C ′. For a composition ofmorphisms
X ′′
t′
��
h
!!CCC
CCCC
C
X ′
s��
f
!!CCC
CCCC
C Y ′
t��
g
AAA
AAAA
A
X Y Z
in C[S−1] (with s, t, t′ ∈ S) we have
F (gt−1)F (fs−1) = F (g)F (t)−1F (f)F (1)−1
= F (g)F (h)F (t′)−1F (s)−1
= F (gh)F (st′)−1
= F (gh(st′)−1) ,
which shows the compatibility with composition.
25
Finally we have F = aSF by construction.
Remark 2.5 Analogously one can define calculus of fractions from the left by reversing allarrows in the considerations above (thus passing to the dual category). In this case we have
HomC[S−1](X,Y ) = lim→Y ′∈S\Y
Hom(X,Y ′) ,
where S \ Y is the category of morphisms under Y :
Y
BBB
BBBB
B
��~~~~~~~~
Y1 // Y2 ,
which is filtered. In some sources (see i.e. SGA 4 12, [C.D.] of J.L. Verdier), calculus of
fractions is assumed from the left and from the right.
Now we apply the calculus of fractions to triangulated categories.
Definition 2.6 Let D be a triangulated category. A set S of morphisms with calculus offractions from the right in D is called compatible with the translation if we have
(FR4) s ∈ S ⇔ T (s) ∈ S.
Theorem 2.7 Let D be a triangulated category, and let S be a set of morphisms withcalculus of fractions from the right in D, which is compatible with the translation. Thenthere is a unique structure of a triangulated category on D[S−1] such that the followingholds:
(a) aS : D // D[S−1] is an exact functor.
(b) If F : D → D′ is an exact functor into a triangulated category D′, such that for all s ∈ Sthe image F (s) is an isomorphism in D′, then there is a uniquely determined exact functorF : D[S−1] → D′, which makes the diagram
D aS //
F ��???
????
? D[S−1]
F{{wwwwwwwww
D′
commutative.
Proof : (a) Define T on D[S−1] by
T (C) = T (C)
T (X X ′soo f // Y ) = T (X) T (X ′)T (s)oo T (f) // T (Y ) .
It follows from (FR4) that this is the unique functor with TaS = aST , and we define thedistinguished triangles in D[S−1] as those which are isomorphic to images of distinguishedtriangles under aS.
26
Now we show that D[S−1] is a triangulated category: The properties (TR1) (a), (TR1) (c)
and (TR2) are obviously fulfilled. We show (TR1) (b): Let fs−1 : X X ′s−1oo f // Y be a
morphism in D[S−1]. For f we have a distinguished triangle
X ′ f // Yg // Z
h // TX ′
in D. Then we obtain a commutative diagram
X ′ f //
s��
Yg //
id��
Z h //
id��
TX ′
T (s)��
Xfs−1
// Yg // Z
h // TX ,
such that the bottom triangle is isomorphic to the image of the distinguished triangle in Dat the top, and therefore is distinguished in D[S−1], by definition.
Now we first show (TR3). We use the following result.
Lemma 2.8 Consider a commutative diagram in D[S−1]
X ′ u′ // Y ′
X
m
OO
u // Y ,
n
OO
where u und u′ are morphisms in D. Then there is a commutative diagram
X ′ u′ // Y ′
/.-,()*+1X ′′
f
OO
u′′ //
s
��
Y ′′
g
OO
t
��
/.-,()*+2X
u // Y
in D, with s, t ∈ S, such that n = gt−1 and m = fs−1.
Proof (1) Assume that n = gt−1, with
Y Y ′′too g // Y ′ .
By (FR2)r we get a commutative diagram
X ′′1
u′′1 //
s1��
Y ′′
t��
Xu // Y
27
with s1 ∈ S.
(2) Let m = f0r−1 with r ∈ S. In the diagram in D[S−1]
X ′ u′ // Y ′′
X ′′2
f ′
OO
s2��
u′′2
AAA
AAAA
A
T
f0
II���������������������
r A
AAAA
AAA X ′′
1
u′′1 //
s1��
Y ′′
g
OO
t��
Xu // Y
we can write f0r−1s1 as a morphism f ′s−1
2 as indicated, and we let u′′2 = u′′1s2. The diagramin D
X ′ u′ // Y ′
X ′′2
u′′2 //
f ′
OO
Y ′′
g
OO
is only commutative in D[S−1], but this means that there is a morphism s3 : X′′ → X ′′
2 in Swith u′f ′s3 = gu′′2s3. Setting f = f ′s3 and u′′ = u′′2s3 and s = s1s2s3, we get a commutativediagram
X ′ u′ // Y ′
X ′′
faaDDDDDDDD
s3}}{{{{{{{{
u′′
��<<<
<<<<
<<<<
<<<<
<<<<
X ′′2
s2��
u′′2
))TTTTTTT
TTTTTTTT
TTTTTTT
f ′
OO
X ′′1
s1
��
u′′1 // Y ′′
g
OO
t
��X u // Y
and hence the claim.
By Lemma 2.8, we can prove (TR3) for D[S−1] as follows. Consider a diagram in D[S−1]
Xu //
m
��
Yv //
n
��
Zw // X[1]
m[1]
��X ′ u′ // Y ′ v′ // Z ′ w′
// X ′[1]
28
where the rows are distinguished triangles and the first square is commutative. We mayassume that the triangles are triangles in D, so that u, v, w, u′, v′, w′ ∈ D. By Lemma 2.8 wemay assume that we have a diagram in D
Xu // Y
v // Zw // X[1]
X ′′
s
OO
u′′ //
f
��
Y ′′ //
t
OO
g
��
Z ′′
u
OO
h��
// X ′′[1]
f [1]
��
s[1]
OO
X ′ // Y ′ // Z ′ // X ′[1]
with morphisms f and g in D, and s and t in S, where the first two squares are commutative,and where we have completed u′′ to a distinguished triangle. By (TR3) in D we have amorphism h in D as indicated giving a morphism (f, g, h) of triangles. On the other hand,we get a morphism u as indicated, such that (s, t, u) is a morphism of triangles. Now u mightnot be in S, but s and t become isomorphisms in D[S−1], and by Lemma 1.10, u becomesan isomorphism in D[S−1] as well, so that hu−1 is a well-defined morphism in D[S−1].
Finally we prove that (TR4) holds in D[S−1]. Let a = aS : D → D[S−1] be the canonicalfunctor, and consider a diagram in D[S−1]
Xf // Y //
g
��
U // X[1]
Xgf //
f
��
Z // V // X[1]
f [1]
��Y
g // Z //W // Y [1]
in which the rows are distinguished triangles and the two left squares are commutative.
By assumption there is an isomorphism of triangles in D[S−1]
Xf // Y // U // X[1]
aXaf //
≀
OO
aY //
≀
OO
aU //
≀
OO
aX[1]
≀
OO
for a distinguished triangle
Xf // Y // U // X[1]
in D, and an isomorphism of triangles in D[S−1]
Yg // Z //W // Y [1]
aY ′ //
≀
OO
aZ ′ //
≀
OO
aW ′ //
≀
OO
aY ′[1]
≀
OO
29
for a distinguished triangle
Y ′ // Z ′ //W ′ // Y ′[1]
in D. Now we use the following
Lemma 2.9 Given a diagram in D[S−1]
aY ′
aXaf // aY
≀
OO
where f : X → Y is a morphism in D and the vertical morphism is an isomorphism inD[S−1], then there exists a morphism f ′ : X ′ → Y ′ in D and an isomorphism aX → aX ′ inD[S−1] such that the diagram
aX ′ af ′ // aY ′
aX
≀
OO
af // aY
≀
OO
commutes.
Proof : The morphism aY → aY ′ is given by a morphism Y Ysoo g // Y ′ . Then we geta commutative diagram
X ′ f ′ // Y ′
X //
t��
Y
g
OO
s��
X // Y
with s, t ∈ S, where the lower square exists by (FR2)r.
We now conclude as follows. Since we have an isomorphism Y ′ ∼ // Y Y∼oo in D[S−1],
we get an isomorphism of distinguished triangles in D[S−1]
(1) aX ′ af ′ // aY ′ // aU ′
��
// aX ′[1]
≀
��
aXaf //
≀��
≀
OO
aY
≀��
≀
OO
X // Y // U // X[1]
where we have used the property (TR3) proved before.
Similarly we get an isomorphism of distinguished triangles
(2) aY ′ ag′ // aZ ′ // aW ′ // aY ′[1]
Y
≀
OO
g // Z
≀
OO
//W
≀
OO
// Y [1]
≀
OO
30
and an isomorphism of distinguished triangles
(3) aX ′ a(g′f ′)// aZ ′ // aV ′ // aX ′[1]
Y
≀
OO
gf // Z
≀
OO
// V
≀
OO
// X[1]
≀
OO
Since (TR4) holds for the three distinguished triangles in D, it also holds for the givendistinguished triangles in D[S−1].
I am indebted to Johannes Sprang for suggesting the above proof. As far as I know, this isthe first full proof of (TR4) for a localized triangulated category in the literature.
Now we consider the properties (a) and (b) in Theorem 2.7. Here (a) is fulfilled by definitionof the distinguished triangles in D[S−1]. For (b) let
D F // D′
be an exact functor into another triangulated category, such that all morphisms in S becomeinvertible in D′. Forgetting the triangulated structures we get a unique functor F : D[S−1] →D′ which makes the diagram
D aS //
F ��???
????
? D[S−1]
F{{wwwwwwwww
D′
commutative. In addition, F is an exact functor, because it maps all distinguished trianglesin D[S−1], which by definition are isomorphic to images of distinguished triangles in Dunder aS, by construction to triangles in D′ which are isomorphic to images of distinguishedtriangles in D under aS:
More precisely, if we have an isomorphism of triangles in D[S−1]
(i) aSXf // aSY
g // aSZh // aSX[1]
X ′ f ′a−1//
αr−1
OO
Y ′ g′b−1//
βs−1
OO
Z ′ h′c−1//
γt−1
OO
X ′[1] ,
αs−1[1]
OO
with r, s, t, a, b, c ∈ S, where the upper triangle is the image under aS of a distinguishedtriangle in D, then we get an isomorphism of triangles in D′,
(ii) F (X)F (f) // F (Y )
F (g) // F (Z)F (h) // F (X)[1]
F (X ′)F (f ′)F (a)−1
//
F (α)F (r−1)
OO
F (Y ′)F (g′)F (b)−1
//
F (β)F (s−1)
OO
F (Z ′)F (h′)F (c)−1
//
F (γ)F (t−1)
OO
F (X ′)[1] ,
F (α)F (s)−1[1]
OO
so that the bottom triangle in (i) is mapped to a triangle isomorphic to the top triangle in(ii) and hence is distinguished in D′.
Remark 2.8 In the literature, the above result is sometimes shown under the followingadditional assumption
31
(FR5) If
X //
r
��
Y //
s
��
Z // T (X)
T (r)��
X ′ // Y ′ // Z ′ // T (X ′)
is a diagram with r, s ∈ S, where the first square is commutative and the rows are distinguis-hed triangles, we get a morphism t : Z → Z ′ in S, which makes the diagram commutative.
As we have seen, this assumption was not necessary. In fact, in the situation of (FR5), in anycase, by (TR3) we get a morphism t : Z → Z ′ which makes the diagram commutative, andby Lemma 1.10, t becomes an isomorphism in D[S−1], since r and s become isomorphisms.Therefore t is an isomorphism in D′ as well.
In a triangulated category, a set of morphisms with calculus of fractions from both sidesoften arises by the following construction.
Definition 2.9 Let (D) be a triangulated category. A triangulated subcategory N of D isa full subcategory which satisfies the following conditions:
N1: 0 ∈ N .
N2: X ∈ N ⇔ TX ∈ N .
N3: If X,Y ∈ N and X → Y → Z → X[1] is a distinguished triangle, then Z ∈ N .
Sometimes such a subcategory is also called a null system.
Theorem 2.10 Every triangulated subcategory N of D defines a multiplicative system Swith calculus of fractions which satisfies (FR4) and (FR5).
Proof : For a triangulated subcategory N of D define the set S = S(N ) by
s ∈ S(N ) ⇔ there is a distinguished triangle X s // Y // Z // X[1] with Z ∈ N .
We check the conditions (FR1) to (FR5).
(FR1): idX ∈ S by the distinguished triangle
XidX // X // 0 // X[1]
and the fact that 0 ∈ N by N1.
Next let s, t the morphisms in S(N ), such that the composition ts exists. Then, by assump-tion, there are distinguished triangles
Xs // Y // Z ′ // X[1]
Yt // Z // X ′ // Y [1]
with Z ′, X ′ in N . By the axiom (TR1) for triangulated categories there is a distinguishedtriangle
Xts // Z // Y ′ // X[1] .
32
By the axiom (TR4) there is a distinguished triangle
Z ′ // Y ′ // X ′ // Z ′[1] .
By applying (TR2) twice, we get a distinguished triangle
X ′ // Z ′[1] // Y ′[1] // X ′[1] .
Since X ′ and Z ′[1] are in N , we have Y ′[1] ∈ N by (N3), and hence Y ′ ∈ N by (N2). Thisshows that ts ∈ S(N ).
(FR2)r: We have to complete the left diagram to a commutative diagram as on the right:
X ′
s∈S��
Zf // Y
Wg //
t∈S��
X ′
s��
Zf // Y .
By assumption there is a distinguished triangle
X ′ s // Y k // Z ′ // X ′[1]
with Z ′ ∈ N . By (TR1) there is a distinguished triangle
Zkf // Z ′ //W // Z[1] ,
and by (TR2) a distinguished triangle
W [−1] // Zkf // Z ′ //W
Since the diagram
Zkf //
f��
Z ′
idZ′��
Yk // Z ′
is commutative, it can be extended to a morphism of distinguished triangles
Zkf //
f
��
Z ′ //
idZ′��
W [1] //
g
��
Z[1]
f [1]
��Y k // Z ′ // X[1] // Y [1]
by (TR3).
By (TR2) we obtain a morphism of distinguished triangles
Wt //
g
��
Zkf //
f
��
Z ′ //
idZ′��
W [1]
g[1]
��X
s // Z ′ k // Z ′ // X[1] .
33
Since Z ′ ∈ N , we have s ∈ S = S(N ). This gives the wanted diagram
Wg //
t��
X
s��
Zf // Y
with t ∈ S.
(FR2)ℓ: Here we have to complete the following left diagram to a commutative diagram ason the right:
X ′
Xg //
t∈S
OO
Y
X ′ f // Y ′
Zg //
t∈S
OO
Y
s∈S
OO
By assumption, there is a distinguished triangle
X t // X ′ k // Z ′ ℓ // X[1]
with Z ′ ∈ N . Furthermore, by (TR3), the commutative diagram
Z ′ ℓ //
idZ′��
X[1]
g[1]
��Z ′ g[1]ℓ // Y [1]
can be extended to a morphism of distinguished triangles
Z ′ ℓ //
idZ′��
X[1]−t[1] //
g[1]
��
X ′[1]−k[1] //
f [1]
��
Z ′[1]
idZ′[1]��
Z ′ g[1]ℓ // Y [1] // Y ′ // Z ′[1] .
By (TR2) we obtain a morphism of distinguished triangles
Xt //
g
��
X ′ k //
f
��
Z ′ //
idZ′��
X[1]
g[1]
��Y
s // Y ′ // Z ′ // Y [1]
Here t ∈ S, since Z ′ ∈ N . This gives the claim.
(FR3): In an additive category this means, that for a morphism f : X → Y we have:
sf = 0 for an s ∈ S ⇔ ft = 0 for an t ∈ S .
Let sf = 0. By definition we have a distinguished triangle
Zg // Y s // Y ′ // Z[1]
34
with Z ′ ∈ N , and by assumption we have a commutative square
X //
��
0
��
/.-,()*+1Y s // Y ′ .
By (TR1), (TR3) and (TR2) this can be extended to a morphism of distinguished triangles
(∗) Xidx //
h
��
X //
f
��
0 //
��
X[1]
��
/.-,()*+2Z ′[−1]
g // Y s // Z // Z ′ .
By (TR1) and (TR2) the morphism h can be extended to a distinguished triangle
Wt // X
h // Z ′[−1] //W [1] .
Here t ∈ S, since Z ′[−1] ∈ N . As above, the commutative diagram (gh = f)
Xh //
f
��
Z ′[−1]
g
��
/.-,()*+3Y
idY // Y
can be embedded into a morphism of distinguished triangles
(∗∗) Wt //
��
Xh //
f
��
Z ′[−1] //
g
��
W [1]
��
/.-,()*+4 /.-,()*+30 // Y
idY // Y // 0
This implies ft = 0, where t ∈ S.
The converse direction starts with /.-,()*+4 , obtains (∗∗) from it, since t ∈ S, and from this thecommutativity of /.-,()*+3 . This is equivalent to gh = f , hence to /.-,()*+2 , and hence to (∗), which gives/.-,()*+1 .(FR4): If s ∈ S, we have a distinguished triangle
(∗ ∗ ∗) Xs // Y
f // Zg // X[1]
35
with Z ∈ N . By (TR2) we have distinguished triangles
Yf // Z
g // X[1]−Ts // Y [1]
Zg // X[1]
−Ts // Y [1]−Tf // Z[1]
X[1]−Ts // Y [1]
−Tf // Z[1]−Tg // X[2] .
Since Z[1] ∈ N , we get −Ts ∈ S. By Lemma 1.4 and 0 ∈ N we have −idX[1] ∈ S and weconclude that (−Ts) ◦ (−idX[1]) = Ts ∈ S by (FR1) proved above. By going back, we getthe converse direction.
(FR5): Consider a diagram
Xf //
r
��
Y //
s
��
Z // X[1]
r[1]
��X ′ f ′ // Y ′ // Z ′ // X ′[1]
with r, s ∈ S, where the rows are distinguished triangles and the first square is commutative.By assumption we have distinguished triangles
Xr //
f
��
X ′ //
f ′
��
Z1// X[1]
Y s // Y ′ // Z2// Y [1]
with Z1, Z2 ∈ N , which are connected by the morphisms f and f ′. By Lemma 1.15 (3 × 3lemma), the diagram of solid arrows below can be completed by the dotted arrows
Xf //
r
��
Y //
s
��
Z //
t��
X[1]
��X ′ //
��
Y ′ //
��
Z ′ //
��
X ′[1]
��Z1
//
��
Z2
��
// Z3//
��
Z1[1]
��X[1]
f [1] // Y [1] // Z[1] // X[2]
such that all rows and all columns are distinguished triangles. By assumption we haveZ1, Z2 ∈ N , and hence Z3 ∈ N by axiom N3 of definition 2.9. Hence t ∈ S(N ) by defi-nition.
36
3 t-structures
Definition 3.1 Let D be a triangulated category.
(a) A t-structure on D consists of two strictly full subcategories D≤0 and D≥0 in D, suchthat with the definition D≤n := D≤0[−n] and D≥n = D≥0[−n] we have
(i) Hom(D≤0,D≥1) = 0.
(ii) D≤0 ⊂ D≤1, and D≥1 ⊆ D≥0.
(iii) For every object X in D there exists a distinguished triangle
A // X // B // A[1]
with A in D≤0 and B in D≥1.
(b) The heart of the t-structure is the full subcategory
C := D≤0 ∩ D≥0 .
Example 3.2 Let A be an abelian category. The natural t-structures on its derived categoryD(A) = K(A)/K∅(A) = K(A)[Quis−1] is given by
D(A)≤n = complexes C · with H i(C ·) = 0 for i > n .
D(A)≥n := complexes B· with H i(B·) = 0 for i < n
This is in fact a t-structure:
(iii): For a complex X · in D(A) define the short exact sequence
0 // τ≤0X· // X · // τ≥1X
· // 0
by
0 // 0 // X2 id // X2 // 0
0 //
OO
0 //
OO
X1 id //
OO
X1 //
OO
0
0 //
OO
ker d0 //
OO
X0 //
OO
im d0 //
OO
0
0 //
OO
X−1 id //
OO
X−1 //
OO
0 //
OO
0
0 //
OO
X−2 id //
OO
X−2 //
OO
0 //
OO
0
OO
37
Then one has H i(τ≤0X·) = 0 for i > 0, so that τ≤0X
· ∈ D≤0, and H i(τ≥1X·) = 0 for i ≤ 0,
so that τ≥1X· ∈ D≥1.
(i) If we have C · with H i(C ·) = 0 for i > 0 and B· with H i(B·) = 0 for i ≤ 0 then we have
HomD(A)(A·, B·) = 0 .
In fact, a morphism in D(A) is represented by a diagram
X ·
s��
f
!!CCC
CCCC
C
A· B·
with a quasi-isomorphism s and a morphism f in C(A). By assumption,
t : τ≤0X· // X ·
is a quasi-isomorphism, and so is
τ : B· // B·/τ≤0B· .
Then obviously rft = 0 in C(A), and this implies f = 0 in D(A), and hence fs−1 = 0.
(ii) is obvious.
The heart of this t-structure is given by the complexes which are concentrated in degreezero, and is equivalent to the category A itself, and hence an abelian category.
Remark 3.3 If (D≤0,D≥0) is a t-structure, then the shift by n, (D≤n,D≥n), is a t-structureas well.
Let D be t-category (i.e., a triangulated category with a t-structure).
Proposition 3.4 (i) The inclusion i≤n : D≤n ↪→ D has a right adjoint τ≤n : D → D≤n.
(ii) The inclusion j≥n : D≥n ↪→ D has a left adjoint τ≥n : D → D≥n.
(iii) For each X in D there is a unique morphism d ∈ Hom1(τ≥1X, τ≤0X) such that
τ≤0X // X // τ≥1Xd // τ≤0X[1]
is a distinguished triangle. Up to unique isomorphism this is the only distinguished triangle
A // X // B // A[1]
with A ∈ D≤0 and B ∈ D≥1.
We use the following lemma.
Lemma 3.5 Consider a diagram
X u //
f
��
Y v //
g
��
Z d //
h
��
X[1]
f [1]
��
/.-,()*+1 /.-,()*+2X ′ u′ // Y ′ v′ // Z ′ d′ // X[1]
38
where the rows are distinguished triangles and g : Y → Y ′ is a morphism. Then the followingconditions are equivalent
(a) v′gu = 0.
(b) There exists an f as indicated making the square /.-,()*+1 commutative.
(b’) There exists an h as indicated making the square /.-,()*+2 commutative.
(c) There exists a morphism (f, g, h) of triangles.
If these conditions hold, and if Hom−1(X,Z ′) = 0, then the morphism f of (b) (resp. h of(b’)) is unique.
Proof 1) The exactness of the sequence
Hom−1(X,Z ′) // Hom(X,X ′)(u′)∗ // Hom(X,Y ′)
(v′)∗ // Hom(X,Z ′)
f � // u′f, gu � // v′gu
shows that (a) ⇔ (b), and the unicity of f , if Hom−1(X,Z ′) = 0.
The implication (b) ⇒ (c) follows from (TR3): If f satisfies (b), then there exists h suchthat (f, g, h) is a morphism of triangles. The converse is trivial.
2) The exactness of the sequence
Hom(X[1], Z ′) / / Hom(Z,Z ′) v∗ // Hom(Y, Z ′) u∗ // Hom(X,Z ′)
h � // hv, v′g � // v′gu
shows that (a) ⇔ (b’), and the unicity of h if Hom−1(X,Z ′) = 0.
Proof of 3.4 By translation it suffices to show (i) for i≤0 : D≤0 → D.
Let
Af // X
g // Bd // A[1]
be as in 5.1 (iii), with A ∈ D≤0 and B ∈ D≥1. Then, for every T in D≤0 we have an exactsequence
Hom(T,B[−1]) //
(1)
Hom(T,A)f∗ // Hom(T,X) // Hom(T,B)
(2)
0 0
Here (1) holds, since B[−1] ∈ D≥2 ⊆ D≥1; and (2) holds, since B ⊆ D≥1. Hence f∗ is anisomorphism, and we can set τ≤0X := A, so that we get an adjunction isomorphism
Hom(T, τ≤0X) ∼ // Hom(i≤0T,X)
39
for all T in D≤0. This isomorphism is functorial in T and X: The functoriality in T is obvious,and if f : X → X ′ is a morphism, then we get a diagram
τ≤0Xi // X //
f
��
B // τ≤0X[1]
τ≤0X′ // X ′ j // B
in which jfi = 0, since Hom(D≤0,D≥1) = 0. By Lemma 3.5, and the fact that Hom(D≤0,D≥1) =0, we get a unique morphism of distinguished triangles
τ≤0X //
τ≤0f
��
X //
f
��
B //
��
τ≤0X[1]
τ≤0f [1]
��τ≤0X
′ // X ′ // B // τ≤0X′[1] .
(ii) For T ∈ D≥1 we have an exact sequence
Hom(τ≤0X,T ) //
(1)
Hom(X,T ) // Hom(B, T ) // Hom(τ≤0X[1], T )
(2)
0 0
Here (1) holds since τ≤0X ∈ D≤0 and T ∈ D≥1, and (2) holds, since τ≤0X[1] ∈ D≤0[1] = D≤−1
and T ∈ D≥1. Hence we get a functorial isomorphism
Hom(X, j≥1T )∼ // Hom(B, T ) ,
so that we can define τ≥1X := B. The functoriality of τ≥1 is shown as above.
Corollary 3.6 Let X u // Y // Z d // X[1] be a distinguished triangle. If Hom−1(X,Z) =0, then the following holds:
(i) The cone of u is unique up to unique isomorphism.
(ii) d is the unique morphism Z → X[1] making the above triangle distinguished.
Proof (i) For a diagram
Xu //
f
��
Yv //
g
��
Z //
h��
X[1]
f [1]
��X
u // Y // Z ′ // X[1]
with isomorphisms f, g the morphism h is uniquely determined by Lemma 3.5.
(ii) Apply Lemma 3.5 to
X u // Y v // Z d //
h��
X[1]
X u // Y v // Z ′ x // X[1] .
40
Necessarily we have h = idZ , so that x = d.
Remark 3.7 (i) The distinguished triangle
(3.7.1) τ≤0X // X // τ≥1X / / τ≤0X[1]
shows that the following conditions are equivalent:
(a) τ≤0X = 0
(a’) Hom(T,X) = 0 for every T ∈ D≤0.
(b) X∼ // τ≥1X is an isomorphism.
(b’) X ∈ D≥1.
The equivalence between (a’) and (b’) can be stated as follows
(c) D≥1 is the right orthogonal of D≤0 and D≤0 is the left orthogonal of D≥1.
(ii) Call a subcategory D′ of a triangulated category D stable by extension, if for any distin-guished triangle (X,Y, Z) in D with X,Z ∈ D′ we also have Y ∈ D′.
Then D≤0 and D≥1 are stable by extension. In fact, for a distinguished triangle
X // Y // Z // X[1]
with X,Z ∈ D≥1 and any T ∈ D≤0 we have an exact sequence
Hom(T,X) // Hom(T, Y ) // Hom(T, Z) ,
where Hom(T,X) = 0 = Hom(T, Z) by (i), so that Hom(T, Y ) = 0 and hence Y ∈ D≥1. Thecase of D≤0 is similar (dual). In particular, D≤0 and D≥1 are stable under finite direct sums.
(iii) For m ≤ n we have D≤m ⊂ D≤n, and therefore a commutative diagram for each X in D
τ≤mX //
im ""FFF
FFFF
Fτ≤nX
Xin
<<yyyyyyyy
which identifies τ≤mX with τ≤mτ≤nX, or more precisely with
τ≤m|D≤nτ≤nX
in the commutative diagramD
Dτ≤n //
τ≤m !!DDD
DDDD
D D≤n?�i≤n
OO
D≤m?�im,n
OO im
cc
Dually, we have a morphism τ≥nX → τ≥mX, identifying τ≥nτ≥mX with τ≥nX.
41
(iv) Let a ≤ b. Then the adjunction morphisms for 3.4 (i) and (ii) induce a commutativediagram
(∗) τ≤bXv //
��
Xv′ // τ≥aX
��τ≥aτ≤bX
φ // τ≤bτ≥aX
with a uniquely determined isomorphism φ. Here we have omitted the canonical inclusionsib : D≤b ↪→ D and ja : D≥a ↪→ D.
Proof : By the adjunction Hom(τ≤bX, jaτ≥aX) ∼= Hom(τ≥aτ≤bX, τ≥aX), the upper morphisminduces a unique morphism
φ1 : τ≥aτ≤bX // τ≥aX
makingτ≤bX //
��
τ≥aX
τ≥aτ≤bX
φ1
66mmmmmmmmmmmmmm
commutative. On the other hand, by adjunction again, there is a unique morphism
φ : τ≥aτ≤bX // τ≤bτ≥aX
making the diagramτ≥aX
��τ≥aτ≤bX
φ //
φ1
55lllllllllllllllτ≤bτ≥aX
commutative, and hence the diagram (∗) commutative.
To show that φ is an isomorphism, we apply (TR4) to the composition
τ<aXu // τ≤bX
v // X .
We get a commutative diagram
(∗) τ<aXu //
id
��
τ≤bX
v
��
α // Y //
β
��
/.-,()*+1τ<aX
w=vu //
u
��
Xv′ //
id
��
τ≥aX //
��τ≤bX
v // X // τ>bX //
42
where the rows are distinguished triangles, and where the column with Y is a distinguishedtriangle and makes the diagram commutative. Here the square /.-,()*+1 is commutative by 3.7(iii), and v and v′ are as in (∗) above.
Since τ<aX ∼= τ<aτ≤bX, we have a morphism of distinguished triangles by (TR3)
τ<aτ≤bX //
≀
τ≤bX // τ≥aτ≤bX //
φ1
��
τ<aτ≤bX[1]
≀
τ<aX // τ≤bX // Y // τ<aX[1]
and hence an isomorphism τ≥aτ≤bXφ1
∼// Y .
Similarly, since τ>bX = τ>bτ≥aX, we have an isomorphism of distinguished triangles
τ≤bτ≥aX // τ≥aX // τ>bτ≥aX // τ≤bτ≤aX[1]
φ2[1]
��Y
≀ φ2
OO
// τ≥aX // τ>bX // Y [1] ,
and hence an isomorphism Yφ2
∼// τ≤bτ≤aX .
Now, in the diagram (∗) we see that Y ∈ D≥a and Y ∈ D≤b. Hence the morphism α factorsthrough τ≥aτ≤bX, and the morphism β factors through τ≤bτ≥aX. This gives a commutativediagram
τ≤bXv //
��
α
##HHH
HHHH
HHH
X v′ // τ≥aX
��τ≥aτ≤bX
φ1
∼// Y
β::uuuuuuuuuu φ2
∼// τ≤bτ≥aX ,
so that the unique morphism at the bottom is an isomorphism.
Now we show
Theorem 3.8 (a) The heartC = D≤0 ∩ D≥0
of a t-structure (D≤0,D≥0) (see 3.1 (b)) is naturally an abelian category.
(b) The functorH0 : D → C
X 7→ H0(X) = τ≥0τ≤0X
is a cohomological functor.
Let D be a triangulated category, and let C be a full subcategory. Assume that the followingholds.
(3.8.1) Homi(X,Y ) := Hom(X,Y [i])
is zero for i < 0 and X,Y ∈ C.
43
Example 3.9 Let D = D(A) for an abelian category A, and let C be the image of thefunctor
A → DA 7→ A placed in degeree zero.
Proposition 3.10 Let f : X → Y be in C and extend it to a distinguished triangle
Xf // Y // S // X[1] .
Suppose that we have a commutative diagram
(3.10.1) K[2]
C
OO
Xf // Y
g //
β=={{{{{{{{{S
OO
// X[1]
K[1]
OO
α
<<xxxxxxxx
where K and C are in C, and the column is a distinguished triangle.
Then β : Y → C is a cokernel of f , and α[−1] : K → X is a kernel of f .
Proof For each object Z in C we have exact sequences
Hom(X[1], Z) // Hom(S, Z)g∗ // Hom(Y, Z)
f∗ // Hom(X,Z)
0
Hom(K[2], Z) // Hom(C,Z) // Hom(S, Z) // Hom(K[1], Z) ,
0 0
since Z,X,K ∈ C. Taking this together, we get an exact sequence
0 / / Hom(C,Z)β∗
// Hom(Y, Z)f∗ // Hom(X,Z)
for each Z in C, which shows the first claim.
44
On the other hand, we get exact sequences for each T ∈ C
Hom(T, Y [−1]) // Hom(T, S[−1]) // Hom(T,X)f∗ // Hom(T, Y )
0
Hom(T,C[−2]) // Hom(T,K) // Hom(T, S[−1]) // Hom(T,C[−1]) ,
0 0
since T,C,K ∈ C. Taking this together, we get an exact sequence
0 // Hom(T,K) // Hom(T,X)f∗ // Hom(T, Y )
for each T in C, which shows the second claim.
Example 3.11 In the case of example 3.9, where we consider A ⊆ D(A), by sending A ∈ Ato the complex . . . → 0 → A → 0 → . . . where A is placed in degree zero, the cone S of
f : X → Y is the complex . . .→ 0 → Xf→ Y → . . . withX in degree−1 and Y in degree 0. It
has the subcomplex ker(f)[1] = H−1(S)[1] and the quotient coker(f) = X/ ker(f)∼→ H0(S)
in degree zero. One gets a diagram as in Proposition 3.10.
Definition 3.12 Let C ⊂ D be as in (3.8.1). Call a morphism f : X → Y in C admissible(or C-admissible) if it is the basis of a diagram (3.10.1).
Remark 3.13 (a) If f is admissible and a monomorphism, then by 3.10 one has K = 0and hence S
∼→ C, so that S is in C, and is a cokernel of f in C by 3.10. Moreover (3.10.1)reduces to the distinguished triangle
Xf // Y // C //
with X,Y,C in C.
(b) If f is admissible and an epimorphism, then C = 0, and hence K[1]∼→ S, and (3.10.1)
reduces to a distinguished triangle
K // X // Y //
with K,X, Y ∈ C.
(c) Conversely, for every distinguished triangle
(∗) Xf // Y
g // Zd // X[1]
with X,Y, Z in C, f and g are admissible, f is a kernel of g and g is a cokernel of f . Bythe assumption that Hom(X,Y [i]) is zero for i < 0, so that Hom−1(X,Y ) = 0, d is uniquelydetermined by f and g, by Corollary 3.6 (ii).
45
In fact, by Lemma 1.9 we have an exact sequence
Hom(T, Z[−1]) // Hom(T,X)f∗ // Hom(T, Y )
g∗ / / Hom(T, Z)
0
for every T in C, which means that f is a kernel of g. On the other hand, we have an exactsequence
Hom(Z[−1], V ) // Hom(Z, V )g∗ // Hom(Y, V )
f∗ // Hom(X,V )
0
for every V in C, which means that g is a cokernel of f .
By this, the distinguished triangle
(∗) Xf // Y
g // Z // X[1]
shows that f is admissible. In fact, since g is a cokernel of f , in (3.10.1) S → C is anisomorphism, and (3.10.1) becomes (∗). On the other hand, since g is an epimorphism, thecokernel of g is zero, and the diagram (3.10.1) becomes
X[2]
0
OO
Yg // Z //
=={{{{{{{{{X[1]
OO
// Y [1]
X[1]
≀f [1]
<<yyyyyyyy
so that g is admissible.
The last claim is clear.
Definition 3.14 A sequence X → Y → Z in C is called admissible, if it is obtained by a
distinguished triangle in D by omitting the arrow Zd // X[1] .
Proposition 3.15 Assume that C is closed under finite direct sums in D. Then the followingconditions are equivalent:
(i) C is an abelian category, and the short exact sequences in C are admissible.
(ii) Every morphism in C is C-admissible.
46
Proof (ii) ⇒ (i): If f : X → Y is admissible, then by Proposition 3.10 it has a kernel anda cokernel. To show that C is an abelian category, we have to show that, in the diagram
Kα[−1] // X
f //
��
Yg // S
ker(f) coker(f)
coim(f) // im(f)
OO
coker(α[−1]) ker(g) ,
the natural morphism coim(f) → im(f) is an isomorphism.
But by (TR4) we obtain a commutative diagram with distinguished triangles as rows andthird column:
K // S[−1] //
��
C[−1] //
��
K[1]
Kα[−1] //
��
X // I //
��
K[1]
��S[−1] // X
f //
g
��
Y //
�
S
S // C
Since α[−1] is a monomorphism in C, and admissible by assumption, we have that I ∈ C, andis a cokernel of α[−1] by 3.13 (a), so that canonically coker(α[−1]) ∼ // I . On the otherhand, the exact sequence
Hom(T,C[−1]) // Hom(T, I) // Hom(T, Y )β∗ // Hom(T,C)
0
for any T ∈ C shows that canonically I∼ // ker(β) .
(i) ⇒ (ii): Let f : X → Y be a morphism in C. The two short exact sequences in C
0 // K // X // I // 0
ker(f) im(f)
0 // I // Y // C // 0
im(f) coker(f)
47
give, by (TR4), a commutative diagram of distinguished triangles
I[1] // K[2]
C
OO
C
OO
Xf // Y
OOβ
<<
// S
OO
// X[1]
X // I //
OO
K[1] //
α
;;OO
X[1] ,
so that f is admissible.
Definition 3.16 A full subcategory C of a triangulated category is called admissible, if itsatisfies (3.8.1) and the equivalent properties of Proposition 3.15.
Now we prove Theorem 3.8. Let D be a triangulated category, let (D≤0,D≥0) be a t-structureon D, and let C = D≥0 ∩ D≤0 be the heart of the t-structure. Then we have
(3.8.1) Hom(X,Y [i]) = 0 for i < 0 and X,Y ∈ C .
In fact, we have X ∈ D≤0 and Y ∈ D≥0 so that the claim follows from 3.1(i).
Therefore we can apply all results in this chapter.
By Proposition 3.15 ist suffices to show that every morphism f : X → Y in C is admissible.Consider a diagram
τ≤−1S[1]
τ≥0S
OO
Xf // Y // S
OO
// X[1]
τ≤−1S
OO
where the row and the column are distinguished triangles. By assumption, Y and X[1]are in D≥−1 ∩ D≤0, and by 3.7 (ii) this also holds for the extension S. Then in the verticaldistinguished triangle we must have τ≤−1S ∈ D≥−1∩D≤−1 = C[1], and τ≥0S ∈ D≥0∩D≤0 = C(compare Proposition 3.7 (iv)). Hence f is admissible.
3.17 Finally we show that for any distinguished triangle
X // Y // Z // X[1]
the sequence H0(X) → H0(Y ) → H0(Z) is exact, where H0(X) = τ≥oτ≤0X.
Case 1 If X,Y and Z are in D≤0, then the sequence H0(X) → H0(Y ) → H0(Z) → 0 isexact.
48
Proof For any U in D≤0 one has H0(U) = τ≥0U , and for any V in D≥0 one has H0(V ) =τ≤0V , and isomorphisms
(1) Hom(H0(U), H0(V )) ∼= Hom(U,H0(V )) ∼= Hom(U, V ) .
For T ∈ C (and hence in D≥0), the long exact Hom-sequence gives an exact sequence
0 = Hom(X[1], T ) // Hom(Z, T ) / / Hom(Y, T ) // Hom(X,T ) ,
where Hom(X[1], T ) ∼= Hom(X,T [−1]) = 0, because X ∈ D≤0 and T [−1] ∈ D≥1. By (1)this gives an exact sequence
0 // Hom(H0(Z), T ) // Hom(H0(Y ), T ) // Hom(H0(X), T ) .
Since this holds for all T ∈ D≥0, the wanted exactness follows.
Case 2 If X is in D≤0, the the sequence H0(X) → H0(Y ) → H0(Z) → 0 is exact.
Proof For all T in D≥1 the long exact Hom-sequence gives an isomorphism
0 = Hom(X[1], T ) // Hom(Z, T ) ∼ // Hom(Y, T ) // Hom(X,T ) = 0
since X ∈ D≤0, X[1] ∈ D≤−1, and T ∈ D≥1, and hence an isomorphism
τ≥1Y∼ // τ≥1Z .
By applying TR4 to the composition X → τ≤0Y → Y we obtain a commutative diagram
X // τ≤0Y //
��
Z ′ //
��
X[1]
X //
��
Y // Z //
β
��
X[1]
τ≤0Y // Y // τ≥1Y
Since the morphism Y → Z induces an isomorphism τ≥1Y∼→ τ≥1Z, we get a canonical
isomorphism Z ′ ∼= τ≤0Z and hence a distinguished triangle
X // τ≤0Y // τ≤0Z // ,
which can be treated by Case 1.
Case 3 If Z lies in D≥0, then
0 // H0(X) // H0(Y ) // H0(Z)
is exact. This case is dual to Case 2.
The general Case: Let the distinguished triangle
X // Y // Z // X[1]
49
be arbitrary. By (TR4) we get a commutative diagram
(1) τ≤0X // X //
��
τ≥1X //
��τ≤0X //
��
Y // U //
��X // Y // Z //
Case 2 and triangle (1) imply an exact sequence
H0(X) // H0(Y ) // H0(U) // 0
H0(τ≤0X)
Case 3 and the triangleU // Z // (τ≥1X)[1]
imply that 0 → H0(U) → H0(Z) is exact. Hence H0(X) → H0(Y ) → H0(X) is exact.
Remark 3.18 A t-structure is called non-degenerate, if∩n≥0
D≤n = 0 =∩n≥0
D≥n. In this case
the family of functors tH i is conservative and we have
X ∈ D≥0 ⇔ tH i(X) = 0 for all i < 0X ∈ D≤0 ⇔ tH i(X) = 0 for all i > 0 .
50
4 Derived functors
Definition 4.1 For triangulated categories D and D′ let Homex(D,D′) be the group of exactfunctors F : D → D′.
Definition 4.2 Let F,G : D → D′ be exact functors between triangulated categories. Amorphism of exact functors η : F → G is a morphism of functors such that for every objectX of D the diagram
ηTX : F (TX) //
≀��
G(TX)
≀��
T ′(ηX) : T ′F (X) // T ′G(X)
commutes.
Definition 4.3 Let A,B be abelian categories, and let
F : K?(A) → K?′(B)
be an exact functor.
(a) Consider the canonical diagram
K?(A) F //
Q��
K?′(B)
Q′
��D?(A) D?′(B)
An exact functor RF : D?(A) → D?′(B) is called right derivative of F , if it represents thefunctor
Homex(D?(A),D?′(B)) → AbG 7→ Homex(Q
′F,GQ) .
This means that there is a morphism of exact functors
η0 : Q′F / / RFQ
K?(A) F //
Q
��
K?′(B)
Q′
��
η0
u} ttttttttt
ttttttttt
D?(A) RF // D?′(B) ,
and for each exact functor D?(A)G→ D?′(B) and each morphism of functors η : Q′F → GQ
there is a unique morphism of exact functors η : RF → G such that η equals the composition
Q′Fη0 // RFQ
η◦Q // GQ .
(b) Dually one defines left derived functors via ζ : LFQ⇒ Q′F .
51
Theorem 4.4 Let D′ ⊂ K?(A) be a full triangulated subcategory such that the followingholds:
(1) Each object X in K?(A) possesses a quasi-isomorphism
X → X ′ (resp. X ′ → X)
with X ′ in D′.
(2) If X ′ ∈ D′ and is acyclic (i.e., in D′ ∩K∅(A)), then F (X ′) is acyclic.
Then F has a right derivative (resp. left derivative), and with the notations of (1) one hasRF (X) = QF (X ′) (resp. LF (X) = Q(X ′)).
Proof (1) D′ = D′/D′ ∩K(A)ι→ D?(A) is an equivalence of categories.
(2) The restriction of F to D′ induces
D′/D ∩K∅(A) F // K?′(A)
(3) Define RF = F ρ, where ρ is a quasi-inverse of ι. (4) We have a commutative diagram
D′
F
66
ι∼
**
D′oo � � u // K?(A)
Q
��
// K?′(B)
Q′
��D?(A) G // D?′(B)
If X is an object in K?(A), take a quasi-isomorphism X → X ′ with X ′ ∈ D′. Then one hasa commutative diagram
Q′F (X) //
��
GQ(X)
≀��
Q′F (uX ′) // GQ(uX ′) ,
and
Hom(Q′F,GQ) = Hom(Q′Fu,GQu) = Hom(F v, Gιv) = Hom(F , Gι) = Hom(F ρ,G) .
Note: ρ(X) = X ′ implies that X ⋍ ιρ(X) = ιX ′ is a quasi-isomorphism.
Alternatively: use the Ind-object (Q′FX ′)X→X′Quis; this is essentially constant, isomorphicto (Q′FX ′)X→X′Quis
X′∈D′.
Lemma 4.5 Let A be an abelian category, and I ⊆ ob(A).
(a) If any object in A has a monomorphism A ↪→ I for an object in I, then for every complexA· in C≥n(A) there is a quasi-isomorphism
A· → I ·
into a complex I · in C≥n(A) with components in I.(b) In particular, if A has enough injectives, then right derivatives exist for any functor onD≥n(A).
(c) Dually, similar results hold. For example, one has left derivates if A has enough projec-tives.
52
5 The six functors
Following Grothendieck, these are the functors
Rf∗, Rf!, f∗, Rf !, RHom, ⊗L .
Here RHom is the right derivative of the sheaf-Hom, i.e., RHom(F,G) is the value at G ofthe right derivative of the functor G 7→ Hom(F,G). Furthermore ⊗L is the left derivative ofthe tensor product, i.e., the value at G the left derivative of the functor G 7→ F ⊗G. In bothcases one can also consider modules over a ring sheaf A.
We consider a fixed base scheme S, and the category CS of quasi-compact, quasi-separatedS-schemes with morphisms that are compactifiable over S:
Y ×S P
{{vvvvvvvvv
��
Xf //
��000000000000000
quasi-finite, separated 00
Y
��
P
properzzuuuuuuuuuu
S
This implies that f is separated, of finite type. If S = Spec(k) for a field k, one can considerall separated k-schemes of finite type.
Endow all sheaves in CS with the etale topology, and let A be an etale torsion ring-sheaf onS (for example A = constant sheaf Z/n). Let AX = a∗XA for aX : X → S.
By the previous consideration, for any morphism f : X → Y in CS we have a morphism
Rf∗ : D+(X,AX) → D+(Y,AY )
Theorem 5.1 (SGA 4, XVII 5.18) There is, up to canonical isomorphism, a unique way toassociate to each morphism f : X → Y in CS a functor
Rf! : D(X,AX) → D(Y,AX)
such that the following holds:
(i) If f is proper, then Rf! = Rf∗.
(ii) If j is an open immersion, then Rj! = j!.
(iii) f 7→ Rf! is “functorial”: One has canonical isomorphisms
R(gf)!∼ // Rg!Rf!
with “cocycle condition”.
53
(iv) For a commutative diagram
U � � j1 //
g��
X
f��
V � � j2 // Y
with open immersions j1 and j2 and proper morphisms g, f one has a canonical morphism
j2!Rg∗ // Rf∗j1! .
Theorem 5.2 (SGA 4 XVII 3.1.4) Rf! has a partial right adjoint Rf!, i.e., forK inD(X,AX)
and L ∈ D+(Y,AY ) one has a functorial isomorphism
Hom(Rf!K,L) ∼= Hom(K,Rf !L) .
Theorem 5.3 (Poincare duality, SGA XVIII 3.2.5) If f is smooth and purely of relativedimension d, then for n invertible on S and nA = 0 one has a canonical isomorphism
Rf !K ∼ // f ∗K(d)[2d] ,
where L(d) := L⊗Z/n Z/n(d). In particular, j! = j∗ for an open immersion.
Corollary 5.4 Let X be smooth, separated over a separably closed field k, of pure dimensiond, and F is a locally constant constructible Z/n-sheaf with n invertible in k. Then one hasa canonical isomorphism
H2d−i(X,F∨(d)) ∼ // H iC(X,F ) .
“Proof” One has
RHom(F,Z/n) = Hom(F,Z/n) (placed in degree 0) ,
since Z/n is an injective Z/n-sheaf. Then we have
RΓRHom(F,Z/n(d)[2d]) ∼ // RHom(RΓC(F ),Z/n)
RΓ(F∨(d)[2d])
Theorem 5.5 (Deligne’s finiteness theorem, SGA 4 12, “Theoremes de finitude on cohomo-
logie ℓ-adique”). Let S be a noetherian scheme of dimension 0 or 1, let f : X → Y be amorphism of S-schemes of finite type, and let F be an etale sheaf in the full subcategory ofconstructible Z/n-schemes on X. Then Rif∗F is constructible for all i ≥ 0, if n is invertibleon S.
Corollary 5.6 If X is of finite type over a separably closed field k, and n is invertible in k,then H i(X,F ) is finite for every constructible Z/n-sheaf F on X.
54
Definition 5.7 For a noetherian scheme X and n ∈ N let
D?c(Xet,Z/n) ⊆ D?(Xet,Z/n)
be the full subcategory of constructible complexes K ·, i.e., those complexes whose cohomo-logy sheaves are constructible.
Corollary 5.8 (a) Rf∗ maps D+c into D+
c .
(b) Rf∗ has finite cohomological dimension.
(c) Rf∗ maps Dbc to D
bc.
Corollary 5.9 The 6 functors respect
Dhctf =
{constructible complexes with finiteTor-dimension as Z/n-modules}
Lemma/Definition 5.10 (compare SGA 4 12[finitude] 4.7) Let X be separated of finite
type over a field k, and let n ∈ N be invertible in k. Let f : X → Spec(k) be the structuralmorphism.
(a) The dualizing complex of X in Dbc(Xet ,Z/n) is
KX := Rf !Z/n .
(b) The Verdier dual of a complex K in Dbc(Xet ,Z/n) is defined by
DXK := RHom(K,KX) .
(c) (Biduality) There is a canonical isomorphism in Dbc(Xet ,Z/n)
K∼ // DXDXK .
Corollary 5.11 Let h : X → Y be a morphism of separated algebraic k-schemes. Then Dexchanges Rf∗ and Rf!, and also f ∗ and Rf !, i.e., there are canonical functorial isomorphisms
Rh∗DXK ∼= DYRh!K
Rh!DXK ∼= DYRh∗K
DXh∗K ∼= Rh!DYK
DXRh!L ∼= h∗DYL
for K in Dbc(Xet ,Z/n) and L in Db
c(Yet ,Z/n).
Proof ForX //
f ##HHH
HHHH
HHY
g{{wwwwwwwww
Spec(k)
55
we haveDYRh!K = RHom(Rh!K,Rg
!Z/n)(1)= Rh∗RHom(K,Rh!Rg!Z/n)
= Rh∗RHom(K,Rf !Z/n)
= Rh∗DXK ,
where (1) comes from “local” Verdier duality.
Applied to DXK this givesDYRh!DXK ∼= Rh∗K
via biduality on X. Hence, by applying DY and biduality on Y we also get
Rh!DXK ∼= DYRh∗K .
The two remaining isomorphisms follow from the “induction formula” (SGA XVIII 3.1.12.2)
RHom(h∗L1, Rh!L2) ∼= Rh!RHom(L1, L2)
for L1, L2 in Dbc(Yet ,Z/n).
56
6 Glueing of t-structures
6.1 Let D,DU and DF be three triangulated categories and let
DFi∗ // D j∗ // DU
be exact functors such that the following holds.
(1) i∗ has an exact left adjoint i∗ and an exact right adjoint i!.
(2) j∗ has an exact left adjoint j! and an exact right adjoint j∗.
(3) One has j∗i∗ = 0, and hence i∗j! = 0 and i!j∗ = 0, by adjunction. For A in DF and B inDU we hence have
Hom(j!B, i∗A) = 0 and Hom(i∗A, j∗B) = 0 .
(4) For each object K in D there is a morphism
d : i∗i∗K // j!j
∗K, resp. d′ : j∗j∗K // i∗i
![K1]
such that
j!j∗K // K // i∗i
∗Kd // j!j
∗K[1] ,
resp.
i∗i!K // K // j∗j
∗Kd′ // i∗i!K[1]
are distinguished triangles.
(5) i∗, j! and j∗ are fully faithful, and the adjunction morphisms
i∗i∗ // id // i!i∗
andj∗j∗ // id // j∗j!
are isomorphisms.
Remark 6.2 The morphisms d and d′ in (4) are uniquely determined by (3): Consider
K // i∗i∗K
d //
h
��
j!j∗K[1]
K // i∗i∗K d // j!j
∗K[1]
where h exists by (TR3). By the exact sequence
Hom−1(j!j∗K, i∗i
∗K)
(3)
// Hom(i∗i∗K, i∗i
∗K) // Hom(K, i∗i∗K)
0
57
the morphism h is unique, hence the identity, which implies d = d.
Example 6.3 (a) Let X be a scheme, j : U ↪→ X an open immersion, and i : F ↪→ X aclosed immersion with X = i(F )
∪· j(U). Then the morphisms
D+(Fet ,Z/n)i∗ // D+(Xet ,Z/n)
j∗ // D+(Uet ,Z/n)
satisfy the conditions in 6.1:
i∗ = usual i∗
i! = usual Ri! (right derivative of i!)j! = usual j!j∗ = usual Rj∗
The adjunction properties hold by the formalism of the 6 functors.
(b) If X is of finite type over a field k, then one can replace D+ by D+c ,Dc or Db
c, by Deligne’sfiniteness theorem.
Theorem 6.4 (glueing theorem) In the situation of Theorem 6.1 let (D≤0U ,D≥0
U ) and (D≤0F ,D≥0
F )be t-structures on DU and DF , respectively. Then the categories
D≤0 = {K ∈ D | j∗K ∈ D≤0U and i∗K ∈ D≤0
F }
andD≥0 = {K ∈ D | j∗K ∈ D≥0
U and i!K ∈ D≥0F }
define t-structures on D (We say that these are obtained by glueing).
Proof We verify the conditions in 3.1 (a).
(i) Let K in D≤0 and L in D≥1. By 6.1 (4) we obtain an exact sequence
Hom(i∗i∗K,L) // Hom(K,L) // Hom(j!j
∗K,L)
Hom(i∗K, i!L) Hom(j∗K, j∗L)
0 0
and hence Hom(K,L) = 0.
(ii) We have D≤0 ⊆ D≤1 and D≥0 ⊇ D≥1 by the same properties for DU and DF .
(iii) Let K in D, and form the distinguished triangles
Y // Xα // j∗τ≥1j
∗X // ,
where α comes by adjunction from j∗X → τ≥1j∗X, and
A // Yβ // i∗τ≥i
∗Y // ,
58
where β comes by adjunction from i∗Y → τ≥1i∗Y .
By (TR4) we get a commutative diagram of distinguished triangles
A // Y //
��
i∗τ≥1i∗Y //
��A //
��
X // B //
��Y // X // j∗τ≥1j
∗X //
��i∗τ≥1i
∗Y [1]
Applying j∗ to the third column we get
j∗B ∼ // τ≥1j∗X ,
since j∗ is exact and j∗i∗ = 0 and j∗j∗ = id.
Applying j∗ to the second row we thus get
j∗A∼ // τ≤0j
∗X .
Applying i∗ to the first row we get
i∗A ∼ // τ≤0i∗Y ,
since i∗ is exact and i∗i∗ = id.
Applying i! to the third column we get
τ≥1i∗Y
∼ // i!B ,
since i!i∗ = id and i!j∗ = 0.
This implies A ∈ D≤0 and B ∈ D≥1, by definition, and hence 3.1 (iii).
Remark 6.5 The t-structure on D is non-degenerate if and only if this holds for the t-structures on DU and DF .
59
7
60
8
61
9 t-exactness and the intermediate direct image
Definition 9.1 Let f : D1 → D2 be an exact functor between triangulated categories whichare endowed with t-structures. Then f is called t-right exact if f(D≤0
1 ) ⊆ D≤02 , and t-left
exact if f(D≥01 ) ⊆ D≥0
2 , and f is called t-exact if f is t-right exact and t-left exact.
Proposition 9.2 Let C1 and C2 be the hearts of (the t-structures of) D1 and D2, respectively.
(a) If f is t-left exact (respectively, t-right exact), then the additive functor
tf := tH0f |C1 : C1 → C2
is left exact (respectively, right exact), and for K in D≤01 (respectively, in D≥0
1 ) one has anisomorphism
tf tH0K ∼ // tH0fK
(respectively, tH0fK ∼ // tf tH0K ).
(b) Let (f ∗, f∗) be a pair of adjoint exact functors f∗ : D1 → D2, f∗ : D2 → D1, where f
∗ isleft adjoint to f∗.
Then f ∗ is t-right exact if and only if f∗ is t-left exact, and in this case (tf ∗, tf∗) is a pair ofadjoint functors
C1tf∗
// C2tf∗oo
(tf ∗ left adjoint to tf∗).
(c) If f : D1 → D2 and g : D2 → D3 are t-left exact (respectively, t-right exact), then thesame holds for gf , and one has
t(gf) = tg ◦ tf .
Proof (a): Let 0 → X → Y → Z → 0 be an exact sequence in C1. Then there is adistinguished triangle
X // Y // Z // X[1]
in D1 (which is uniquely determined, see ), and hence we get a distinguished triangle
fX // fY // fZ // fX[1]
in D1. If f is t-right exact, then fX ∈ D≤02 , hence fX[1] ∈ D≤−1
2 , and therefore
tH0(fX[1]) = τ≤0τ≥0(fX[1]) = 0 .
Therefore we get an exact sequence
tH0X // tH0fY // tH0fZ // 0
tfX
62
If X is in D≤01 , then we have a distinguished triangle
τ<0X // X // tH0X // ,
τ≥0X
and hence a distinguished triangle
fτ<0X // fX // f(tH0X) //
with fτ<0X ∈ D<02 , so that we get an isomorphism
tH0(fX) ∼ // f(tH0X) .
The other case is dual.
(b): If f∗ is t-left exact, then for U in D>01 and V in D≤0
2 one has
Hom(f ∗V, U) = Hom(V, f∗U) = 0 ,
because f∗U ∈ D>02 . Since this holds for any U , we have τ>0f
∗V = 0, i.e., f ∗V ∈ D≤01 , so
that f ∗ is t-right exact.
If now A ∈ C1 and B ∈ C2, we have
H0(f ∗B) = τ≥0(f∗B) und H0(f∗A) = τ≤0(f∗B) ,
and hence a functorial isomorphism
Hom(H0(f ∗B), A) ∼ // Hom(f ∗B,A) = Hom(B, f∗A) Hom(A,H0(f∗B)) .∼oo
This, together with the dual arguments, proves (b).
(c): If f and g are t-left exact, and A ∈ C1, then fA ∈ D≥0, and t(gf)A = H0(g)H0(f)A bythe second part of (a). This, together with the dual arguments, proves (c).
Theorem 9.3 Let X be a scheme of finite type over a field k, and let ℓ be a prime invertiblein k.
Let i : Y ↪→ X be a closed immersion, and let j : U ↪→ X be the open immersion ofthe complement. Consider the triangulated categories Db
c(?,Qℓ) of Qℓ-constructible boundedcomplexes for ? = X,Y and U , endowed with the perverse t-structure p. Then the followingholds.
(a) j! and i∗ are p-right exact.
(b) j∗ and i∗ are p-exact.
(c) Rj∗ and Ri! are p-left exact.
(d) (pj!,pj∗ = j∗, pRj∗) and (pi∗,p i∗ = i∗,
pRi!) form triples of adjoint functors (where forf, g f is the left adjoint to g).
63
Proof By definition of the perverse t-structure p, j∗ is p-exact: For x ∈ U one has afactorization
ix : x� � ix // U � � j // X
so that i∗xj∗ = i∗x and Ri!xj
∗ = Ri!x. Furthermore, by definition i∗ is p-right exact (respects
pD≤0) and Ri! is p-left exact (respects pD≥0). The remaining claims follow from the adjunc-tions
(j!, j∗, Rj∗) , (i∗, i∗, Ri
!) .
In fact, j! is left adjoint to j∗ so that j! is p-right exact, and Rj∗ is right adjoint to j
∗ so thatRj∗ is p-left exact, by 9.2 (b).
Finally (d) follows from 9.2 (b).
We have a canonical morphism of functors from Dbc(U,Qℓ) to D
bc(X,Qℓ)
j! // Rj∗ ,
given by the canonical inclusionj!I
· // j∗I·
for any complex I · with injective components.
This induces a morphism of functors
Perv(Uet ,Qℓ) → Perv(Xet ,Qℓ)
pj! → pRj∗
by applying pH0 to the previous morphism of functors.
Definition 9.4 The intermediate direct image j!∗F of a perverse sheaf P in Perv(U,Qℓ) isby definition
j!∗F := im(pj!F → pRj∗F ) ∈ Perv(X,Qℓ) .
This gives a functor Perv(U,Qℓ) → Perv(X,Qℓ), and obviously one has j∗j!∗F = F .
64
10 Laumon’s ℓ-adic Fourier transformation
Let q = pn for a prime p, and let Fq be the finite field with q elements. Let V be a vectorspace of dimension d over Fq, and let
J := V(V ) := Spec(Sym(V )) ∼= AdFq.
Let Frobq : J → J be the q-Frobenius morphism: it is the identity on the underlyingtopological space and the map x 7→ xq on OJ . This is a morphism of algebraic groups.
We get an exact sequence
0 // J(Fq) // JFrobq−id// J // 0
(Artin-Schreier sequence). J → J is an etale covering, from which we get a continuousmorphism
π1(J) // J(Fq) .
Let
χ : J(Fq) // Q×ℓ
be a character, where ℓ is a prime = p. This gives a continuous homomorphism
π1(J) // J(Fq)χ−1
// Q×ℓ ,
and hence a smooth Qℓ-sheaf Lχ on J .
Remark 10.1 (1) One has a canonical isomorphism
Lχ|{0} ∼= Qℓ,0 .
(2) One has a canonical isomorphism
s∗LX ∼= p∗1Lχ ⊗ p∗2Lχ
where s : J × J → J is the addition morphism and pi : J × J → J are the projections to thei-th factor, i = 1, 2.
(3) One hasRΓc(J ⊗Fq k,Lχ) = 0 = RΓ(J ⊗Fq k,Lχ)
if k/Fq is an algebraically closed field extension and χ is a non-trivial character.
Let k be a perfect field of characteristic p, and let k be an algebraic closure of k. Fix a non-trivial character ψ : Fq → Qℓ, and let Lψ be the associated smooth Qℓ-sheaf on J = Ga,Fp .
Let S/k be of finite type, let E π // S be a vector bundle of rank r, and let E ′ π′// S be
the dual vector bundle.
Then we have a canonical pairing
⟨ , ⟩ : E ×S E′ // Ga,k .
65
Let L(⟨ , ⟩) be the pull-back of Lψ/Ga,k via ⟨ , ⟩.
Definition 10.2 The Fourier transformation associated to ψ on E π // S is the exactfunctor
Fψ : Dbc(E,Qℓ) // Db
c(E′,Qℓ)
defined byFψ(K) = Rpr′!(pr
∗K ⊗ Lψ(⟨ , ⟩))[r]
whereE × E ′
pr
{{wwwwwwwww
pr′
##HHH
HHHH
HH
E E ′
are the projections.
Let E ′′ π′′// S be the vector bundle which is bi-dual to E → S, and define the isomorphism
a : E ∼ // E ′′
by a(e) = −⟨e , ·⟩.
Theorem 10.3 Let F ′ be the Fourier transformation associated to ψ on E ′ π′/ / S . Then
there is a functorial isomorphism
(F ′ ◦ F)(K) ∼ // a∗K(r)
for K in Dbc(E,Qℓ).
Proof : We have a diagram
E ′ ×S E′′ β //
pr1
��
pr2
��
E ′′
E ×S E′ × E ′′
pr1,3
OO
α //
pr12
vvnnnnnn
nnnnnn pr23
((QQQQQ
QQQQQQ
Q E ′ ×S E′′
pr′′
OO
E ×S E′
pr′2 ((PPPPP
PPPPPP
PPPpr
zzvvvvvvvvvv
E ′ ×S E′′
pr′1vvmmmmmm
mmmmmm
mm pr′′
%%JJJJJ
JJJJJ
E E ′ E ′′
whereα(e, e′, e′′) = (e′, e′′ − a(e))β(e, e′′) = e′′ − a(e)
Claim 10.3.1 pr∗12L(⟨ , ⟩)⊗ pr∗23L(⟨ , ⟩) = α∗L(⟨ , ⟩)
66
Proof We have a commutative diagram
E × E ′ × E ′′
f
��
pr23
))SSSSSSS
SSSSSSS
pr12
uukkkkkkk
kkkkkk
k
E × E ′
⟨ , ⟩
��
E ′ × E ′′
⟨ , ⟩
��
Ga ×Ga
s
�� p2))SSS
SSSSSSSS
SSSSSS
p1uukkkk
kkkkkk
kkkkkk
Ga Ga Ga
with the obvious projections, f(e, e′, e′′) = (⟨ e, e′⟩, ⟨ e′, e′′⟩), and the sum map s.
By 9.1 (2) we havep∗1Lψ ⊗ p∗2Lψ = s∗Lψ
and hencep∗12Lψ(⟨ , ⟩)⊗ pr∗23Lψ(⟨ , ⟩) = f ∗s∗Lψ
Sinces ◦ f(e, e′, e′′) = ⟨ e, e′⟩+ ⟨ e′, e′′⟩ = ⟨ e′, e′′ − a(e)⟩ = ⟨ , ⟩ ◦ α(e, e′, e′′)
we get the claim.
Now by definition we have
(F ′ ◦ F)(K)
= Rpr′′! ((pr′1)
∗ ◦R(pr′2)!)(pr∗K ⊗ L(⟨ , ⟩))[r]⊗ L(⟨ , ⟩)[r]By proper base change (SGA 4, XVII 5.2.6) we have
(pr′1)∗ ◦R(pr′2)! = R(pr23)! ◦ pr∗12
Hence we get
F ′ ◦ F(K)
= Rpr′′! (R(pr23)! ◦ pr∗12(pr∗K ⊗ L(⟨ , ⟩))⊗ L(⟨ , ⟩)[2r](1)= R(pr′′ ◦ pr23)!(pr ◦ pr12)∗K ⊗ pr∗12L(⟨ , ⟩)⊗ pr∗23L(⟨ , ⟩)[2r]
= R(pr2 ◦ pr13)!(pr∗13 ◦ pr∗1K ⊗ α∗L(⟨ , ⟩)(2)= R(pr2)!(pr
∗1K ⊗R(pr13)!α
∗L(⟨ , ⟩)[2r](3)= R(pr2)!(pr
∗1K ⊗ β∗Rpr′′! L(⟨ , ⟩)[2r]
where (1) and (2) follow from the projection formula and (3) follows from base change.
Now we use
Proposition 10.4 For L in Db(S,Qℓ) one has a functorial isomorphism
F(π∗L[r]) ∼= σ′∗L(−r) ,
67
where π : E → S and π′ : E ′ → S are as above, and σ : S → E ′ is the zero section.
Proof : later.
With this we proceed as follows. By definition of F ′ we have Rpr′′! L(⟨ , ⟩)[r] = F ′(Qℓ,E′).
Applying 10.4 to π′ : E ′ // S , π′′ : E ′′ // Sσ′′
jj , and L = Qℓ we get
F ′(Qℓ,E′ [r]) = σ′′∗Qℓ,S(−r) .
Hence we getF ◦ F(K) ∼= R(pr2)!(pr
∗1K ⊗ β∗σ′′
∗Qℓ,S(−r)) .
Since
Eid×a //
π��
E × E ′′
�
S σ′′// E ′′
is cartesian, proper base change implies
β∗β′′∗ = (id× a)∗π
∗ .
Hence we getF ′ ◦ F(K) = R(pr2)!(pr
∗1K ⊗ (id× a)∗Qℓ,E(−r))
(1)= R(pr2)!((id× a)∗pr
∗1K(−r))
(2)= a∗K(−r) ,
where (1) holds by the projection formula, and (2) holds since pr2 ◦ (id × a) = a andpr1 ◦ (idE × a) = idE.
Proof of Proposition 10.4 We have a diagram
E ×S E′
pr
{{vvvvvvvvvv pr′
$$HHH
HHHH
HHH
⟨ , ⟩ // A1k
E
π$$II
IIIIII
IIIE ′
π′zzuuu
uuuuuu
uu
S σ′=0−section
JJ
and use the Fourier transform F for L = Lψ. By definition we have
(10.4.1)
F(π∗L[r])
= Rpr′!(pr∗π∗L[r]⊗ L(⟨ , ⟩))[r]
= Rpr′!((pr′)∗(π′)∗L[r]⊗ L(⟨ , ⟩))[r] (since πpr = π′pr′)
= (π′)∗L⊗Rpr′!L(⟨ , ⟩)[2r] (projection formula)
68
(10.4.2) On the other hand, by the cartesian diagram
E = E ×S Sid×σ′
//
π��
E × E ′
pr′
��S // E ′
and proper base change we have
(σ′)∗Rpr′!L(⟨ , ⟩) = Rπ!(id× σ′)∗L(⟨ , ⟩)= Rπ!(id× σ′)∗⟨ , ⟩∗L .
Since
EidE×σ′
//
π''PP
PPPPPP
PPPPPP
PP E × E ′ ⟨ , ⟩ // Ga
S0
66nnnnnnnnnnnnnnnn
commutes, we get(id× σ′)∗⟨ , ⟩∗Lψ = Qℓ,E
(trivial sheaf) by (10.1.1).
Finally we have Rpr′!L(⟨ , ⟩)|E′\σ(S) = 0, because for s′ ∈ E ′ we have, by proper base change,
(Ripr′!L(⟨ , ⟩))s′ = H ic(E ×S s′,L(x, s′))
= H ic(Ar
k × k(s′),Qℓ(χ)) ,
where χ : π1(Ark × k(s′)) → Qℓ
×is a character. Here χ is trivial if and only if s′ = 0.
In fact, let s′ = 0′. The cartesian diagram of Galois covers
X ′ ∼= Ar−1k(s′) ×k(s′) A1
k(s′)//
G′∼=Fp
��
A1k ×k k(s′)
G=Fp
��X := Ar ×k k(s′)
ψ(x,s′) // A1k ×k k(s′)
(where the bottom map is without restriction the projection to the last factor, since s′ = 0)shows that χ factors through G′. For the etale cohomology of affine space we have
H i(X ′,Qℓ) =
{Qℓ , i = 0 ,0 , i = 0 ,
and we obtain an isomorphism of G′-modules
H i(X ′,Qℓ(χ−1)) =
{Qℓ(χ
−1) , i = 0 ,0 , i = 0 .
Since H i(X,Qℓ(χ−1)) ∼= H i(X ′,Qℓ(χ
−1))G′(general property of Galois covers), we obtain
H i(X,Qℓ(χ−1) = 0 for all i = 0
69
Hence Poincare duality gives
H ic(X,Qℓ(χ)) =
{Qℓ(−r) , i = 2r, s′ = 0
0 , otherwise .
This impliesRpr′!L(⟨ , ⟩)|E′\σ(S) = 0
and henceRpr′!L(⟨ , ⟩) ∼= σ′
∗σ′∗L(⟨ , ⟩)
= σ′∗Qℓ(−r)[−2r]
as wanted.
70
11 Properties of perverse sheaves
In the following we consider perverse sheaves in D(X) = Dbc(X,Qℓ), where X is a variety
over the field k. Let i : Y ↪→ X be a closed subscheme, and let j : U ↪→ X be the opencomplement.We write Perv(X) for the category Perv(X,Qℓ) of perverse Qℓ-sheaves on X.
Remark 11.1 Obviously the perverse t-structure p on DX = Dbc(X,Qℓ) is obtained by
glueing the perverse t-structures on U and Y . The morphism of functors j! → j∗ is obtainedby the canonical distinguished triangle for K = j∗L :
j!j∗K
Adj // Kadi // i∗i
∗Kd // ,
where Adj and adi are the adjunction morphisms.
Lemma 11.2 (a) The inclusion of the full subcategory
D′ := {K ∈ DX | i∗K ∈ pD≤0Y } ↪→ DX
has a right adjoint τY≤0.
Proof (a) D′ = D≤′is obtained by glueing the perverse t-structure on Y and the degenerate
t-structure (DU , 0) (i.e., D≤′
U = DU , D≥′
U = 0) on U .
By 3.4 (i), the inclusion D′ ↪→ DX has a right adjoint.
(b) D′′ = D≥′′is obtained by glueing the perverse t-structure on DY and the degenerate
t-structure (0,DU), i.e. (D≤′= 0,D≥′
= DU)
By 3.4 (ii), we have a left adjoint of D′′ ↪→ DX .
Lemma 11.3 (a) By the constructions of Theorem 6.1 (glueing of t-structures), we obtainthe following description of τY≤0 =
pτY≤0: Form distinguished triangles
L // K // j∗τ′≥1j
∗K //
andA // L // i∗τ≥1i
∗K // .
Then we haveτY≤0K = A .
But we have τ ′≥1j∗K = 0, hence L = K, hence τY≤0K is given by a distinguished triangle
τY≤0K // K // i∗τ≥1i∗K // .
(b) For τY≥0 =pτY≥0 we form distinguished triangles
L // K // j∗τ′′≥1j
∗K //
A // L // i∗τ≥1i∗L //
A // K // B //
71
Then we have B = τY≥1. However, we have τ ′′≥1j∗K = j∗K, hence L = i∗i
!K; and from thedistinguished triangle
A // i∗i!K // i∗τ≥1i
!K //
we obtain A = i∗τ≤0i!K, and get a distinguished triangle
(11.3.1) i∗τ≤0i!K // K // τY≥1K // .
Definition 11.4 Let L ∈ DU . A complex K in DX with an isomorphism j∗K∼→ L is called
an extension of L (to X).
Proposition 11.5 Let L ∈ DU and n ∈ Z. Then up to unique isomorphism there is exactlyone extension K ∈ DX of L such that i∗K ∈ D≤n−1
Y and i!K ∈ D≥n+1Y .
In fact, the extension isK = L[n] := τY≤n−1j∗L = τ≥n+1j!L .
Proof Let K be an extension of L. By the canonical distinguished triangle (6.1(4), 6.3)
(11.5.1) i∗i!K // K // j∗j
∗Kd′ //
(where j∗K = L) and applying i∗ and shifting the triangle, we obtain the distinguishedtriangle
(11.5.2) i∗K // i∗j∗L // i!K[1] // .
This implies that the following properties are equivalent:
(a) i∗K ∈ D≤n+1Y and i!K ∈ D≥n+1
Y (⇔ i!K[1] ∈ D≥nY ).
(b) i!K[1] = τ≥ni∗j∗L.
(b’) i∗K = τ≤n−1i∗j∗L,
(c) K = τY≤n−1i∗j∗L,
(c’) K = τY≥n+1j!L.
In fact, the equivalences (a) ⇔ (b) ⇔ (b’) are obvious, and the distinguished triangle
(11.5.3) K // j∗L // i∗i!K[1] //
shows the equivalence between (b) and (c) , because of the distinguished triangle
(11.5.4) τY≤0j∗L // j∗L // i∗τ≥1i∗j∗L //
coming from 11.3 (a) for K = j∗L.
Exercise: Identify the morphisms
j∗L //
��
i∗i∗j∗L //
��
i∗τ≥ni∗j∗L
��i∗i
!K[1] //
id
55i∗i
∗i∗i!K[1] // i∗i
!K[1]
72
Dually note thati∗j∗L ∼= i!j!L[1] .
In fact, we have a distinguished triangle
(11.5.5) j!j∗K // K // i∗i
∗K // ,
hence for K = j∗L a distinguished triangle
(11.5.6) j!L // j∗L // i∗i∗j∗L // ,
and after shifting and applying i! a distinguished triangle
// i!j∗L // i!i∗i∗j∗L //
≀
i!j!L[1] // i!j∗L[1]
0 i∗j∗L 0
and hence the claim.
The distinguished triangle
(11.5.7) i∗i∗K[−1] // j!L // K //
then implies (b’) ⇔ (c’), by the distinguished triangle
(11.5.8) i∗τ≤0i!j!L // j!L // τY≥1j!L // 0
from 11.3 (b).
Uniqueness: (a) If i∗K ∈ D≤n−1Y , then we have isomorphisms
Hom(K, τY≤n−1j∗L)(1)∼= Hom(K, j∗L)
(2)∼= Hom(j∗K,L) ,
where (1) comes from 11.2 (a), and (2) is the adjunction map. The chosen isomorphismj∗K
∼→ L gives a unique morphism β : K → τY≤n−1j∗L, which is an isomorphism.
(b) The claims in (c) and (c’) follow similarly.
Proposition 11.6 For B in CU (= Perv(U,Qℓ)) we have:
(a) pj!B = τY≥0j!B = τY≤−2j∗B = B[−1]
(b) j!∗B = τY≥1j!B = τY≤−1j∗B = B[0]
(c) pj∗B = τY≥2j!B = τY≤0j∗B = B[1] .
Proof All nine terms are extensions of B toX (τY≤n and τY≥n only change on Y ). In particular,
one has j∗C = B for them. Further we have j!B ∈ pD≤0X (t-right-exactness of j!).
Consider the distinguished triangle
i∗τ≤−1i!j!B // j!B // τY≥0j!B //
73
(obtained in a similar way as (11.5.8)).
Here i∗τ≤−1i!j!B ∈ pD≤−1
X , since i∗ is t-exact. This shows that
pj!B = pH0(j!B) = pτ≥0j!B = τY≥0j!B ,
and by 11.5 this is equal to τY≤−2j∗B = B[−1]. The equalities in (c) follow in a similar way.
(b): By definition, i∗B[0] ∈ D≤−1, i!B ∈ D≥1Y , and by j∗B = B ∈ CU we have
B[0] ∈ CX .
(TR4) and 7.6. (b) give the following commutative diagram for K = j!B, with distinguishedtriangles in the two bottom rows
τY≥1K
OO
τY≥1K
OO
i∗τ≤−1i!K // K //
OO
τY≥1K //
OO
i∗τ≤−1i!K // i∗τ
Y≥0K //
OO
i∗pH0K //
OO
and hence a distinguished triangle
i∗pH0(i!K) // τY≥0K // τY≥1
// .
Here all three terms are in CX (note that i∗ is t-exact). Hence we obtain an exact sequencein CX
0 // i∗pH0(i!K) // B[−1] // B[0] // 0
pj!B
Correspondingly, 11.3 (a) gives a distinguished triangle and hence an exact sequence in CX
0 // τY≤−1j∗B // τY≤0j∗B // i∗pH0i∗j∗B // 0
B[0] B[1] = pj∗B
This shows thatj!∗B = im( pj!B → pj∗B) = B[0] .
Theorem 11.7 Let B ∈ CU = Perv(Uet ,Qℓ). Then j!∗B is the unique extension C of B
in CX = Perv(Xet ,Qℓ) with i∗C in D≤−1Y (i.e., in pD≤−1
c (Yet ,Qℓ)) and i!C in D≥1Y (i.e., in
pD≥1C (Yet ,Qℓ)).
74
Proof This follows immediately from 11.6 (b) and 11.8. Correspondingly, one obtains cha-racterizations of pj!B and pj∗B with 11.6 and 11.8.
Lemma 11.8 (a) The compositions pj∗ pi∗,pi∗ pj! and
pi! pj∗ are zero, and for A in CY andB in CU one has
Hom(pj!B,pi∗A) = 0 = Hom(pi∗A,
pj∗B) .
(b) For each C in CX the sequences
pj!pj∗C // C // pi∗
pi∗C // 0
and0 // pi∗
pi!C // C // pj∗pj∗C
are exact.
(c) pi∗,pj! and
pj∗ are fully faithful, and the adjunction morphisms
pi∗ pi∗ // id // pi! pi∗
pj∗ pj! // id // pj∗pj∗
are isomorphisms.
Proof This follows immediately from Theorem 9.3 and the corresponding properties for thefunctors j∗, j! etc. themselves. For example, we show
(a): The functors i! and j∗ are both t-left exact by 9.3 (c). Hence we have
pi! pj∗ =p(i!j∗) = 0 ,
since i!j∗ = 0.
(b): We have a distinguished triangle
j!j∗C // C // i∗i
∗C // ,
and j! and j∗ are t-right exact. Hence j!j
∗C ∈ pD≤0X , so that pH1(j!j
∗C) = 0. Now the claimfollows from the long exact pH ·-sequence.
(c): i∗i∗ = id⇒ pi∗ pi∗ =p(i∗i∗)
∼→ pid = id.
We are now ready to prove the following characterization of the intermediate direct image.
Theorem 11.9 Let B ∈ CU = Perv(Uet ,Qℓ). Then j!∗B is the uniquely determined extension
of B to CX = Perv(Xet ,Qℓ) which does not have any subobject or quotient in the essential
image CY of i∗ =pi∗ : CY → CX for CY = Perv(Yet ,Qℓ).
Proof By definition, j!∗B lies in CX . If C is an extension of B in X , then i∗C ∈ D≤0Y , and
hence i∗C ∈ D≤−1Y if and only if pi∗C = 0.
75
Dually we have i!C ∈ D≥0Y , and i!C ∈ D≥1
Y if and only if pi!C = 0.
Now C = j!∗B is just characterized by i∗C ∈ D≤−1Y and i!C ∈ D≥0
Y (see Theorem 11,7).
On the other hand, by the exact sequence from 11.8 (b),
pj!pj∗C // C // pi∗
pi∗C // 0 ,
and since Hom(pj!pj∗C,
pi∗pi∗C) = 0 by 11.8 (a), pi∗
pi∗C is the maximal quotient of Cwhich lies in CY .
Dually, pi∗pi!C is just the maximal sub-object of C which lies in CY .
11.10 Let again X be a scheme of finite type over a field k.
Let now j : U ↪→ X be the inclusion of a locally closed subscheme. Again we have the functorsj∗(= Rj∗), j
∗, j!, between the categories CX = Perv(Xet,Qℓ) and CU = Perv(Uet,Qℓ), and thecorresponding functors pj∗,
pj∗, pj! between them.
If k : V ↪→ U is a further such immersion, then one has similar transition morphisms, andtransivity isomorphisms (jk)∗ = j∗k∗, and similarly perverse variants pj, pk, with p(jk) =pj pk etc.
For an open immersion j the four functors are j∗, j∗, j!, j
∗, and for a closed immersion j = ithe four functors are i∗, i
∗, i∗, i!.
From this we get the following in the general case:
Lemma 11.11 (a) (j!, j!), (j∗, j∗), (
pj∗, pj∗)(pj!,
pj!) and (pj∗, pj∗) are pairs of adjoint func-tors.
(b) j! and j∗ are p-right exact, and j! and j∗ are t-left exact.
Proof: Obvious.
Definition 11.12 Let B ∈ DU . Then define
j!∗B = im(pj!B → pj∗B) ∈ DX .
Here the arrow is induced by the canonical morphism j!B → j∗B, which comes from id :B ∼= j∗j!B → B by adjunction.
Remark 11.13 We have a factorization
j!B //
∩p
pj!B = pτ≥0j!B // pj∗B = pτ≤0j∗B // j∗B
pD≤0
Theorem 11.14 (a) The category Perv(Xet ,Qℓ) is noetherian and artinian: every object
has finite length.
76
(b) Let j : V ↪→ X be the immersion of a locally closed irreducible subvariety of X forwhich (V ×k k)red is smooth, and let L be an irreducible smooth Qℓ-sheaf on V . Then j!∗(L[dimV ])‘ is a simple perverse sheaf on X.
(c) Every simple perverse sheaf on X is of this form.
Since the etale topology is not changed if we replace k by its perfect closure, we may assumethat k is perfect. Then we can replace “(V ×k k)red is smooth” by “V is smooth”. We startwith
Lemma 11.15 Let X be irreducible and smooth, and let L be a smooth Qℓ-sheaf on X. Ifj : U ↪→ X is a non-empty open subset of X, then for the perverse sheaf F := L[dimX] wehave
F = j!∗j∗F .
Proof: By Theorem 11.7 we have to show that for each x ∈ X ∖ U we have
Hi(i∗xF ) = 0 for all i ≥ − dim(x) ,
andHi(i!xF ) = 0 for all i ≤ − dim(x) .
Since dimx < dimX, the first claim holds, because we obviously have
Hi(i∗xF ) = Hi(i∗xL[dimX]) = Hi+dimX(i∗xL) = 0
for i+ dimX > i+ dim(x) ≥ 0.
On the other hand, by cohomological purity we have
Hi(i!xF ) = 0 for i < 2 codim(x)− dim(X) ,
and we have − dim(x) < − dim(x) + (dimX − dimx) = 2 codim(x)− dimX.
Lemma 11.16 If X is irreducible and smooth and L is an irreducible smooth Qℓ-sheaf onX, then the perverse sheaf G = L[dimX] is irreducible.
Proof Let G ⊆ F be a subobject. There is a non-empty Zariski open subset j : U ↪→ X suchthat j∗G is of the form M [dimX], for a smooth subsheaf M of L|U . Since X is normal,thehomomorphism of fundamental groups
π1(U) // // π1(X)
is surjective, because both groups are quotients of Gal(k(x)sep/k(x)). Hence L|U remainsirreducible (as a smooth sheaf). Hence we have M = 0 or M = L|U .
If M=0, then j∗G = 0, and henceG ∼= i∗
pi∗G
for the closed immersion i : Y = X ∖ U ↪→ X (compare 11.8 (b)).
77
If M = L|U , then j∗(F/G) = 0, so that F/G has support in Y , i.e., it lies in the essentialimage of
i∗ : Perv(Yet ,Qℓ) // Perv(Xet ,Qℓ) .
On the other hand, by Lemma 11.15 we have
F = j!∗j∗F ,
and by Theorem 11.9 every subobject or quotient of F in Perv(Xet ,Qℓ) with support in Y
is trivial, which gives a contradiction.
Now we prove Theorem 11.14: To show that the perverse sheaf j!∗(L[dimV ]) in 11.14(b) is simple, we use the factorization (where V is the closure of V in X)
j : V � � µ
open// V � � κ
closed// X
with an open immersion µ and a closed immersion κ. Then for F ∈ Perv(Vet ,Qℓ) we have
j!∗(F ) = Im(pj!F // pj∗F )
κ∗pµ!F κ∗
pµ∗F
= κ∗Im(pµ!F → pµ∗F )
= κ∗µ!∗F ,
since κ∗ =pκ∗ is exact.
Step 1: If F is simple (as a perverse sheaf on V ), then the same holds for µ!∗F : If G ⊆ µ!∗Fis a perverse subsheaf, then µ∗G ⊆ F is a perverse subsheaf; hence
µ∗G = 0 or µ∗(µ!∗F/G) = 0 .
Since F has no non-trivial subobjects or quotients with support in V ∖ V , we deduce thatG = 0 or G = µ!∗F .
Step 2: If F ∈ Perv(Vet ,Qℓ) is simple, then κ∗F is simple, too:
If G ⊆ κ∗F and ρ : X ∖ V ↪→ X is the open immersion, then ρ∗G ⊆ ρ∗κ∗F = 0, andhence G ∼= κ∗
pκ∗G for the perverse sheaf pκ∗G on V . By applying pκ∗ we have pκ∗G = 0 orpκ∗G = F , and hence also G = 0 or G = κ∗F .
Combining this with Lemma 11.16, we see that j!∗(L[dimV ]) is simple.
It remains to show that any perverse sheaf F on X has a finite composition series withquotients of the form j!∗(L[dimV ]) as in Theorem 11.14 (b).
In any case, there is a dense open Uj↪→ X such that U is smooth and equidimensional, and
F |U is of the form L[dimU ] for a smooth sheaf L on U . By noetherian induction we canassume that the claim holds on the closed complement i : Y ↪→ X of U .
78
Now we have the following
Lemma 11.17 Let C be a perverse sheaf on X.
(a) There are exact sequences of perverse sheaves
0 // pi∗pH−1i∗C // pj!
pj∗C // C // pi∗pi∗C // 0
0 // pi∗pi!C // C // pj∗
pj∗C // pi∗pH1i!C // 0
(b) C is in the essential image Perv(Y ) of
pi∗ : Perv(Y ) // Perv(X)
if and only if pj∗C = 0.
(c) Perv(Y ) is a Serre subcategory of Perv(X), and pj∗ induces an equivalence of categories
Perv(X)/Perv(Y ) ∼ // Perv(U) .
Proof (a) follows immediately from the long pH ·-sequences associated to the distinguishedtriangles in 6.1 (4).
(b) is obvious by the fact that pj∗ pi∗ = 0 and each of the sequences in (a).
(c) The first part follows from (b) and, the exactness of pj∗. For the second part see Asterisque100, Prop. 1.4.18.
Now we continue with the proof of 11.14. From Lemma 11.13 we see that there is a filtrationby perverse sheaves
F2 ⊆ F1 ⊆ F
such that F1/F2 = j!∗j∗F = j!∗L[dimU ], and such that F/F1 and F2 are in the essential
image of pi∗. If L is simple, then we are done.
If L↠ L′ is a simple quotient with kernel L′′ = 0, then the morphism
j!∗L[dimU ] // // j!∗L′[dimU ]
is surjective (since pj! is right exact), and the kernel F ′′ is an extension of L′′[dimU ]. Theclaim now follows by induction on the length of L. This finishes the proof of Theorem 11.14.
79
