## Properties of SODs

The triangle required in a SOD is actually functorial.

Lemma. Let $\mathcal T = \langle \mathcal A, \mathcal B \rangle$ be a SOD. Let $f : X \to X^\prime$ be a morphism in $\mathcal T$. There are unique morphisms $f_A : A \to A^\prime$ and $f_B : B \to B^\prime$ so that the diagram

commutes.

Proof. (Expand to view)

We have a long exact sequence $$\cdots \to \operatorname{Hom}(B,A^\prime[-1]) \to \operatorname{Hom}(B,B^\prime) \to \operatorname{Hom}(B,X^\prime) \to \operatorname{Hom}(B,A^\prime) \to \cdots$$ with $$\operatorname{Hom}(B,A^\prime[-1]) = \operatorname{Hom}(B,A^\prime) = 0$$ so $$\operatorname{Hom}(B,B^\prime) \to \operatorname{Hom}(B,B^\prime)$$ is an isomorphism giving $f_B$ from $B \to X \overset{f}{\to} X^\prime$. Using the axioms of a triangulated category we can produce one $f_A$ making the diagram commute. Since $$\operatorname{Hom}(A,A^\prime) \to \operatorname{Hom}(X,A^\prime)$$ is also an isomorphism by similar considerations, we see that $f_A$ is unique.

From this we see that $A$ and $B$ occurring in a triangle $$B \to X \to A \to B[1]$$ depend on $X$ up to unique isomorphism. Thus, it makes sense to write $X_B$ and $X_A$ for (a choice of) them.

Corollary. The assignment \begin{aligned} \lambda_A : \mathcal T & \to \mathcal A \\ X & \mapsto X_A \\ f & \mapsto f_A \end{aligned} is a functor and left adjoint to $\imath_A : \mathcal A \to \mathcal T$.

The assignment \begin{aligned} \rho_B : \mathcal T & \to \mathcal B \\ X & \mapsto X_B \\ f & \mapsto f_B \end{aligned} is a functor and right adjoint to $\imath_B : \mathcal B \to \mathcal T$.

Proof. (Expand to view)

The previous lemma tells us that each are functors. For adjunction, we already saw that $$\operatorname{Hom}(B,B^\prime) \to \operatorname{Hom}(B,X^\prime)$$ and $$\operatorname{Hom}(A,A^\prime) \to \operatorname{Hom}(X,A^\prime)$$ are isomorphisms.

We have $\mathcal A \subset \mathcal B^{\perp}$ but more can be said.

Lemma. We have $$\mathcal A = \mathcal B^{\perp}$$ and $$\mathcal B = {}^{\perp} \mathcal A$$

Proof. (Expand to view)

Assume we have $X \in \mathcal T$ with $$\operatorname{Hom}(B,X) = 0$$ for all $B \in \mathcal B$. We have the triangle $$B \to X \to A \to B[1]$$ and the map $B \to X$ must be $0$. Therefore, $$A \cong X \oplus B$$ Now $B \in \mathcal B$ and is a summand of $\mathcal A$. Since $\mathcal A \subset \mathcal B^{\perp}$ and summand of $\mathcal A$ also lies in $\mathcal B^{\perp}$. Thus, $$\operatorname{Hom}(B,B) = 0$$ and $B = 0$. So $X \in \mathcal A$. The argument for the other equivalence is similar.

The previous results show that a single component in a SOD completely captures all information. We introduce some notation for this first.

Proposition. The following are equivalent:

• $\mathcal T = \langle \mathcal A, \mathcal B \rangle$
• $\mathcal A$ is left admissible.
• $\mathcal B$ is right admissible.

The previous proposition gives a bit more flexibility for verifying an SOD.

Example. Let $R$ be commutative Noetherian ring and $I \leq R$ an ideal. We say that $M \in \operatorname{Mod} R$ is $I$-torsion if for every $m \in M$ there exists $j > 0$ such that $I^jm = 0$.

Inside $D(R)$, we have the full triangulated subcategory $D_I(R)$ consisting of all complexes $C$ with $H^\ast(C)$ $I$-torsion.

The inclusion $D_I(R)$ admits a right adjoint $$M \mapsto \operatorname{colim}_n \operatorname{Hom}_R(R/I^n,J)$$ where $J$ is a K-injective replacement for $M$. The cohomology of this is often called the local cohomology of $M$ with respect to the ideal $I$.

From the proposition, there is a semi-orthogonal decomposition $$D(R) = \langle D_I(R)^{\perp}, D_I(R) \rangle$$

The objects of $D_I(R)^{\perp}$ are often $I$-local or torsion-free. There is an equivalence $$j^\ast : D_I(R)^{\perp} \to D(U)$$ where $U$ is the quasi-affine scheme $$U = \operatorname{Spec} R \setminus V(I)$$ and given by $j : U \to \operatorname{Spec} R$.

In general, $U$ is not going to be affine. But, here is a special case. Take $R = k[x]$ and $I = (x)$. Then, $$D_{(x)}(k[x])^{\perp} \cong D(k[x,x^{-1}])$$ and the map $$k[x] \to k[x,x^{-1}]$$ in the inclusion.

Given an admissible $\mathcal A \subseteq \mathcal T$, we get two SODs $$\mathcal T = \langle \mathcal A, {}^{\perp} \mathcal A \rangle$$ and $$\mathcal T = \langle \mathcal A^{\perp}, \mathcal A \rangle$$ and we can compare ${}^{\perp} \mathcal A$ and $\mathcal A^{\perp}$ \begin{aligned} L_{\mathcal A}: {}^{\perp} \mathcal A & \to \mathcal A^{\perp} \\ B & \mapsto \rho_{\mathcal A^{\perp} } \imath_{ {}^{\perp} \mathcal A} B \end{aligned} and \begin{aligned} R_{\mathcal A}: \mathcal A^{\perp} & \to {}^{\perp} \mathcal A \\ B & \mapsto \lambda_{ {}^{\perp} \mathcal A } \imath_{ \mathcal A^{\perp} } B \end{aligned} The functor $L_{\mathcal A}$ is called left mutation and $R_{\mathcal A}$ is right mutation.

Lemma. $L_{\mathcal A}$ and $R_{\mathcal A}$ are inverse.

Proof. (Expand to view)

Let $B \in {}^\perp \mathcal A$. To mutate it into $\mathcal A^\perp$, we fit it into a triangle $$A \to B \to B^\prime \to A_B[1]$$ To mutate it back we fit $B^\prime$ into a triangle of the form $$B^{\prime \prime} \to B^\prime A^\prime \to B^{\prime \prime}[1]$$ We notice that shifting the previous triangle gives $$B \to B^\prime \to A_B[1] \to B[1]$$ one such. Thus, $$R_{\mathcal A} L_{\mathcal A} \cong \operatorname{Id}$$ Similarly, $$L_{\mathcal A} R_{\mathcal A} \cong \operatorname{Id}$$

While the left and right orthogonals to an admissible $\mathcal A$ are equivalent, the actual subcategories in $\mathcal T$ can appear quite different.