use of the term “admissible” is slightly more restrictive than the definition used
in [BBD].) If A T is admissible, then it is automatically an abelian category,
and every short exact sequence in A gives rise to a distinguished triangle in T.
The heart of any t-structure on T is admissible. For the following fact, see [BBD,
Remarque 3.1.17] or [BGS, Lemma 3.2.4].
Lemma 2.1. Let A be an admissible abelian subcategory of a triangulated cate-
gory T. The natural map
Y ) HomT(X,
Y )
is an isomorphism for i = 0, 1. If it is an isomorphism for i = 0, 1,...,k, then it
is injective for i = k + 1.
Next, we recall the “∗” operation for objects of a triangulated category D. If
X and Y are classes of objects in D, then we define
X Y = A D
there is a distinguished triangle
X A Y with X X , Y Y
By [BBD, Lemme 1.3.10], this operation is associative. In an abuse of notation,
when X is a singleton {X}, we will often write X Y rather than {X} Y. Note
that the zero object is a sort of “unit” for this operation. For instance, we have
X Y 0 = X Y. Given a class X , X 0 is the class of all objects isomorphic to
some object of X .
2.2. Tate twist. Many of our categories will be equipped with an automor-
phism known as a Tate twist, and denoted X X 1 . We will always assume that
Tate twists are “faithful,” meaning that for any nonzero object X, we have

X n if and only if n = 0.
A key example is the category Vectk of graded k-vector spaces, where the Tate
twist is the “shift of grading” functor. For X Vectk, let Xn denote its nth graded
component. Then X m is the graded vector space given by
(X m )n = Xn−m.
We regard k itself as an object of Vectk by placing it in degree 0.
If X and Y are objects of an additive category equipped with a Tate twist, we
let Hom(X, Y ) denote the graded vector space defined by
Hom(X, Y )n = Hom(X, Y −n ).
Notations like
−), and RHom(−, −) are defined similarly.
The following lemma is a graded analogue of [B2, Lemma 5].
Lemma 2.2. Let V be an object in
(1) If there are integers n1,...,nk such that 0 V n1 · · · V nk , then
V = 0.
(2) If there are integers n1,...,nk 0 such that k V V n1 · · · V nk ,
) = 0 for i 0, and

= k. For i 0,
) is concen-
trated in strictly positive degrees.
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