CGP DERIVED SEMINAR
GABRIEL C. DRUMMOND-COLE
1. 6/20: Morimichi Kawasaki: on T -dg-modules
Today we consider category of functors and natural transformations. It was a
little confusing so if I make a mistake please point it out. Let T be a dg category.
Then F ∶ T → C(k) is a T -dg module if F is a dg-functor.
Remark 1.1. The category of chain complexes, C(k), has a dg-category structure
C
dg
(k) or C(k), so in this derived seminar, this was explained by Yoosik Kim. So
recall this first. We can
In other words, for any object x of T , there is a chain complex F
x
, and this
satisfies F
x
⊗ T (x, y) → F
y
and this satisfies the usual associativity and unit condi-
tions (these being that for any x and c ∈ F
x
that c ⊗ 1
x
→ c, where 1
x
is the unit of
T (x, x)).
The proof of the “in other words” is, take x, y, z in the objects of T and s in
T (x, y) and t in T (y, z). Then by the definition of a dg functor, we have the
following diagram:
T (x, y) ⊗ T (y, z)
//
T (x, z)
C(k)(F
x
, F
y
) ⊗ C(k)(F
y
, F
z
)
//
C(k)(F
x
, F
z
)
and this implies that F
(t○s)
= F
t
○ F
s
, which means that F
x
⊗ (t ○s) = (F
x
⊗ t)⊗s.
Here the bilinear map ⊗ is c ⊗ s ↦ F
s
(c).
Now the unit condition on dg functors, we have
k
e
x
//
e
F
x
%%
T (x, x)
F
x,x
C(k)(F
x
, F
x
)
and for any k in k, F
x,x
(e
x
(k)) = e
F
x
(k). Since e
F
x
(1
x
) = 1
F
x
in C(k)(F
x
, F
x
),
then c ⊗ 1
x
= c.
We want to consider the category T − mod of dg T -modules. First we, to
consider a category, we have to define morphisms. The set of morphisms between
T -dg modules F and F
′
from T to C(k) is the set of natural transformations
1
2 GABRIEL C. DRUMMOND-COLE
between dg functors. This means we have commutativity of the following diagram:
F
x
⊗ T (x, y)
//
F
y
F
′
x
⊗ T (x, y)
//
F
′
y
Then T − mod consists of the category of T -dg modules. Now we define a
model structure on T -dg modules. The computation here is simple. For f in
T − mod (F, F
′
), we say f is an equivalence if for all x in ob(T ), the component
f
x
∶ F
x
→ F
′
x
is an equivalence in C(k) (that is, a quasi-isomorphism).
Second, we define fibrations. A morphism f ∈ Mor(F, F
′
) is a fibration if for
all x, f
x
∶ F
x
→ F
′
x
is a fibration in C(k). Here fibration means surjection. This
induces the model structure on T − mod . To¨en’s pdf did not give a description of
cofibrations, but I think it’s hard to describe.
Today we will not check that this is a model structure.
Definition 1.1. The derived category of T -modules, D(T ) is Ho(T − mod )
∶
=
W
−1
(T − mod ), the localization with respect to the weak equivalences.
Next we’ll give a definition. I gave only the definition of T -dg modules. Let me
give examples.
(1) The trivial C(k)-dg module from C(k) to itself (the identity) is a trivial
module.
(2) The next examples come from To¨en, pages 15 and 16. Let T be a dg
category. Define f
∶
= h
x
∶ T → C(k) by y ↦ T (x, y), which is an object of
C(k). Moreover, for a morphism a, you get b ↦ a ○ b.
(3) For T any dg category, let h
x
∶ T
op
→ C(k), this takes y to T (y, x) and a
morphism a to b ↦ (b ○ a).
These three are the three main elementary examples. As an exercise, let T be a dg
category. We’ll prove exercise 2, that x ↦ h
x
defines a functor [T ] ↦ D(T
op
).
Proof. First, we construct the functor h ∶ T → T
op
− mod . If x is an object of T ,
then h
x
is a T
op
-module. So this was our example three. For a morphism, fix x
and y in the objects of T
op
. Then choose a in T
op
(y, x) = T (x, y). We can define
h
a
, a map in T
op
− mod (h
x
, h
y
) by the following. For z ∈ ob(T ) we define h
a
(z),
a map in C(k)(h
y
(z), h
x
(z)) by h
a
(z)(b) = b ○ a for any b ∈ T (y, z). In diagrams I
can write
a ↦ (z ↦ ((y
b
Ð→ z) ↦ (x
a
Ð→ y
b
Ð→ z))
Consider the restriction of h to Z
0
(T (x, y)) ∶ Z
0
(T (x, y)) → T
op
− mod (h
x
, h
y
).
Recall that H
0
(T (x, y)) = Z
0
T (x, y)/B
0
(T (x, y)). We still have to show the
well-definedness. Assume a, a
′
in Z
0
(T (x, y)), and assume that [a] = [a
′
] in
H
0
(T (x, y)), which means that a−a
′
∈ B
0
(T (x, y)). We have to prove that h
a
= h
a
′
in D(T
op
). Since a−a
′
is in B
0
(T (x, y)), there is an α in T
−1
(x, y) with a−a
′
= dα.
Then for any b in T (y, z), we have
b ○ a − b ○ a
′
= b ○ (a − a
′
) = b ○ dα = d(b ○ α) − db ○ α.
Therefore for all b in T (y, z) with db = 0, we have b ○ a − b ○ a
′
∈ I(d) ⊂ T (x, z).
Therefore h
a
= d
′
a
in D(T
op
).
I’ll finish here. Thank you.
