Definition 13.40.1. Let $\mathcal{D}$ be an additive category. Let $\mathcal{A} \subset \mathcal{D}$ be a full subcategory. The *right orthogonal* $\mathcal{A}^\perp $ of $\mathcal{A}$ is the full subcategory consisting of the objects $X$ of $\mathcal{D}$ such that $\mathop{\mathrm{Hom}}\nolimits (A, X) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. The *left orthogonal* ${}^\perp \mathcal{A}$ of $\mathcal{A}$ is the full subcategory consisting of the objects $X$ of $\mathcal{D}$ such that $\mathop{\mathrm{Hom}}\nolimits (X, A) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$.

## 13.40 Admissible subcategories

A reference for this section is [Section 1, Bondal-Kapranov].

Lemma 13.40.2. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ be a full subcategory invariant under all shifts. Consider a distinguished triangle

of $\mathcal{D}$. The following are equivalent

$Z$ is in $\mathcal{A}^\perp $, and

$\mathop{\mathrm{Hom}}\nolimits (A, X) = \mathop{\mathrm{Hom}}\nolimits (A, Y)$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$.

**Proof.**
By Lemma 13.4.2 the functor $\mathop{\mathrm{Hom}}\nolimits (A, -)$ is homological and hence we get a long exact sequence as in (13.3.5.1). Assume (1) and let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. Then we consider the exact sequence

Since $A[1] \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ we see that the first and last groups are zero. Thus we get (2). Assume (2) and let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. Then we consider the exact sequence

and we conclude that $\mathop{\mathrm{Hom}}\nolimits (A, Z) = 0$ as desired. $\square$

Lemma 13.40.3. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B} \subset \mathcal{D}$ be a full subcategory invariant under all shifts. Consider a distinguished triangle

of $\mathcal{D}$. The following are equivalent

$X$ is in ${}^\perp \mathcal{B}$, and

$\mathop{\mathrm{Hom}}\nolimits (Y, B) = \mathop{\mathrm{Hom}}\nolimits (Z, B)$ for all $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$.

**Proof.**
Dual to Lemma 13.40.2.
$\square$

Lemma 13.40.4. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ be a full subcategory invariant under all shifts. Then both the right orthogonal $\mathcal{A}^\perp $ and the left orthogonal ${}^\perp \mathcal{A}$ of $\mathcal{A}$ are strictly full, saturated^{1}, triangulated subcagories of $\mathcal{D}$.

**Proof.**
It is immediate from the definitions that the orthogonals are preserved under taking shifts, direct sums, and direct summands. Consider a distinguished triangle

of $\mathcal{D}$. By Lemma 13.4.16 it suffices to show that if $X$ and $Y$ are in $\mathcal{A}^\perp $, then $Z$ is in $\mathcal{A}^\perp $. This is immediate from Lemma 13.40.2. $\square$

Lemma 13.40.5. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ be a full triangulated subcategory of $\mathcal{D}$. For an object $X$ of $\mathcal{D}$ consider the property $P(X)$: there exists a distinguished triangle $A \to X \to B \to A[1]$ in $\mathcal{D}$ with $A$ in $\mathcal{A}$ and $B$ in $\mathcal{A}^\perp $.

If $X_1 \to X_2 \to X_3 \to X_1[1]$ is a distinguished triangle and $P$ holds for two out of three, then it holds for the third.

If $P$ holds for $X_1$ and $X_2$, then it holds for $X_1 \oplus X_2$.

**Proof.**
Let $X_1 \to X_2 \to X_3 \to X_1[1]$ be a distinguished triangle and assume $P$ holds for $X_1$ and $X_2$. Choose distinguished triangles

as in condition $P$. Since $\mathop{\mathrm{Hom}}\nolimits (A_1, A_2) = \mathop{\mathrm{Hom}}\nolimits (A_1, X_2)$ by Lemma 13.40.2 there is a unique morphism $A_1 \to A_2$ such that the diagram

commutes. Choose an extension of this to a diagram

as in Proposition 13.4.23. By TR3 we see that $Q_1 \cong B_1$ and $Q_2 \cong B_2$ and hence $Q_1, Q_2 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$. As $Q_1 \to Q_2 \to Q_3 \to Q_1[1]$ is a distinguished triangle we see that $Q_3 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$ by Lemma 13.40.4. Since $\mathcal{A}$ is a full triangulated subcategory, we see that $A_3$ is isomorphic to an object of $\mathcal{A}$. Thus $X_3$ satisfies $P$. The other cases of (1) follow from this case by translation. Part (2) is a special case of (1) via Lemma 13.4.11. $\square$

Lemma 13.40.6. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory of $\mathcal{D}$. For an object $X$ of $\mathcal{D}$ consider the property $P(X)$: there exists a distinguished triangle $A \to X \to B \to A[1]$ in $\mathcal{D}$ with $B$ in $\mathcal{B}$ and $A$ in ${}^\perp \mathcal{B}$.

If $X_1 \to X_2 \to X_3 \to X_1[1]$ is a distinguished triangle and $P$ holds for two out of three, then it holds for the third.

If $P$ holds for $X_1$ and $X_2$, then it holds for $X_1 \oplus X_2$.

**Proof.**
Dual to Lemma 13.40.5.
$\square$

Lemma 13.40.7. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ be a full triangulated subcategory. The following are equivalent

the inclusion functor $\mathcal{A} \to \mathcal{D}$ has a right adjoint, and

for every $X$ in $\mathcal{D}$ there exists a distinguished triangle

\[ A \to X \to B \to A[1] \]in $\mathcal{D}$ with $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$.

If this holds, then $\mathcal{A}$ is saturated (Definition 13.6.1) and if $\mathcal{A}$ is strictly full in $\mathcal{D}$, then $\mathcal{A} = {}^\perp (\mathcal{A}^\perp )$.

**Proof.**
Assume (1) and denote $v : \mathcal{D} \to \mathcal{A}$ the right adjoint. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Set $A = v(X)$. We may extend the adjunction mapping $A \to X$ to a distinguished triangle $A \to X \to B \to A[1]$. Since

for $A' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$, we conclude that $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$ by Lemma 13.40.2.

Assume (2). We will contruct the adjoint $v$ explictly. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Choose $A \to X \to B \to A[1]$ as in (2). Set $v(X) = A$. Let $f : X \to Y$ be a morphism in $\mathcal{D}$. Choose $A' \to Y \to B' \to A'[1]$ as in (2). Since $\mathop{\mathrm{Hom}}\nolimits (A, A') = \mathop{\mathrm{Hom}}\nolimits (A, Y)$ by Lemma 13.40.2 there is a unique morphism $f' : A \to A'$ such that the diagram

commutes. Hence we can set $v(f) = f'$ to get a functor. To see that $v$ is adjoint to the inclusion morphism use Lemma 13.40.2 again.

Proof of the final statement. In order to prove that $\mathcal{A}$ is saturated we may replace $\mathcal{A}$ by the strictly full subcategory having the same isomorphism classes as $\mathcal{A}$; details omitted. Assume $\mathcal{A}$ is strictly full. If we show that $\mathcal{A} = {}^\perp (\mathcal{A}^\perp )$, then $\mathcal{A}$ will be saturated by Lemma 13.40.4. Since the incusion $\mathcal{A} \subset {}^\perp (\mathcal{A}^\perp )$ is clear it suffices to prove the other inclusion. Let $X$ be an object of ${}^\perp (\mathcal{A}^\perp )$. Choose a distinguished triangle $A \to X \to B \to A[1]$ as in (2). As $\mathop{\mathrm{Hom}}\nolimits (X, B) = 0$ by assumption we see that $A \cong X \oplus B[-1]$ by Lemma 13.4.11. Since $\mathop{\mathrm{Hom}}\nolimits (A, B[-1]) = 0$ as $B \in \mathcal{A}^\perp $ this implies $B[-1] = 0$ and $A \cong X$ as desired. $\square$

Lemma 13.40.8. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B} \subset \mathcal{D}$ be a full triangulated subcategory. The following are equivalent

the inclusion functor $\mathcal{B} \to \mathcal{D}$ has a left adjoint, and

for every $X$ in $\mathcal{D}$ there exists a distinguished triangle

\[ A \to X \to B \to A[1] \]in $\mathcal{D}$ with $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ and $A \in \mathop{\mathrm{Ob}}\nolimits ({}^\perp \mathcal{B})$.

If this holds, then $\mathcal{B}$ is saturated (Definition 13.6.1) and if $\mathcal{B}$ is strictly full in $\mathcal{D}$, then $\mathcal{B} = ({}^\perp \mathcal{B})^\perp $.

**Proof.**
Dual to Lemma 13.40.7.
$\square$

Definition 13.40.9. Let $\mathcal{D}$ be a triangulated category. A *right admissible* subcategory of $\mathcal{D}$ is a strictly full triangulated subcategory satisfying the equivalent conditions of Lemma 13.40.7. A *left admissible* subcategory of $\mathcal{D}$ is a strictly full triangulated subcategory satisfying the equivalent conditions of Lemma 13.40.8. A *two-sided admissible* subcategory is one which is both right and left admissible.

Let $\mathcal{A}$ be a right admissible subcategory of the triangulated category $\mathcal{D}$. Then we observe that for $X \in \mathcal{D}$ the distinguished triangle

with $A \in \mathcal{A}$ and $B \in \mathcal{A}^\perp $ is canonical in the following sense: for any other distinguished triangle $A' \to X \to B' \to A'[1]$ with $A' \in \mathcal{A}$ and $B' \in \mathcal{A}^\perp $ there is an isomorphism $(\alpha , \text{id}_ X, \beta ) : (A, X, B) \to (A', X, B')$ of triangles. The following proposition summarizes what was said above.

Proposition 13.40.10. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ and $\mathcal{B} \subset \mathcal{D}$ be subcategories. The following are equivalent

$\mathcal{A}$ is right admissible and $\mathcal{B} = \mathcal{A}^\perp $,

$\mathcal{B}$ is left admissible and $\mathcal{A} = {}^\perp \mathcal{B}$,

$\mathop{\mathrm{Hom}}\nolimits (A, B) = 0$ for all $A \in \mathcal{A}$ and $B \in \mathcal{B}$ and for every $X$ in $\mathcal{D}$ there exists a distinguished triangle $A \to X \to B \to A[1]$ in $\mathcal{D}$ with $A \in \mathcal{A}$ and $B \in \mathcal{B}$.

If this is true, then $\mathcal{A} \to \mathcal{D}/\mathcal{B}$ and $\mathcal{B} \to \mathcal{D}/\mathcal{A}$ are equivalences of triangulated categories, the right adjoint to the inclusion functor $\mathcal{A} \to \mathcal{D}$ is $\mathcal{D} \to \mathcal{D}/\mathcal{B} \to \mathcal{A}$, and the left adjoint to the inclusion functor $\mathcal{B} \to \mathcal{D}$ is $\mathcal{D} \to \mathcal{D}/\mathcal{A} \to \mathcal{B}$.

**Proof.**
The equivalence between (1), (2), and (3) follows in a straighforward manner from Lemmas 13.40.7 and 13.40.8 (small detail omitted). Denote $v : \mathcal{D} \to \mathcal{A}$ the right adjoint of the inclusion functor $i : \mathcal{A} \to \mathcal{D}$. It is immediate that $\mathop{\mathrm{Ker}}(v) = \mathcal{A}^\perp = \mathcal{B}$. Thus $v$ factors over a functor $\overline{v} : \mathcal{D}/\mathcal{B} \to \mathcal{A}$ by the universal property of the quotient. Since $v \circ i = \text{id}_\mathcal {A}$ by Categories, Lemma 4.24.4 we see that $\overline{v}$ is a left quasi-inverse to $\overline{i} : \mathcal{A} \to \mathcal{D}/\mathcal{B}$. We claim also the composition $\overline{i} \circ \overline{v}$ is isomorphic to $\text{id}_{\mathcal{D}/\mathcal{B}}$. Namely, suppose we have $X$ fitting into a distinguished triangle $A \to X \to B \to A[1]$ as in (3). Then $v(X) = A$ as was seen in the proof of Lemma 13.40.7. Viewing $X$ as an object of $\mathcal{D}/\mathcal{B}$ we have $\overline{i}(\overline{v}(X)) = A$ and there is a functorial isomorphism $\overline{i}(\overline{v}(X)) = A \to X$ in $\mathcal{D}/\mathcal{B}$. Thus we find that indeed $\overline{v} : \mathcal{D}/\mathcal{B} \to \mathcal{A}$ is an equivalence. To show that $\mathcal{B} \to \mathcal{D}/\mathcal{A}$ is an equivalence and the left adjoint to the inclusion functor $\mathcal{B} \to \mathcal{D}$ is $\mathcal{D} \to \mathcal{D}/\mathcal{A} \to \mathcal{B}$ is dual to what we just said.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #6021 by Sandeep on

Comment #6022 by Sandeep on

Comment #6172 by Johan on