## Semi-orthogonal decompositions and upper triangular algebras

There is a standard way to break a triangulated category in (hopefully simpler) pieces.

**Definition**. Let $\mathcal T$ be a triangulated category. A *semi-orthogonal decomposition* or *SOD* is pair of full triangulated subcategories $\mathcal A, \mathcal B$ of $\mathcal T$ such that

- $\mathcal A \subset \mathcal B^{\perp}$, ie \(\operatorname{Hom}(B,A) = 0\) for any $A \in \mathcal A$ and $B \in \mathcal B$.
- for any $X \in \mathcal T$ there exists a triangle \(B \to X \to A \to B[1]\) with $A \in \mathcal A$ and $B \in \mathcal B$.

We write \(\mathcal T = \langle \mathcal A, \mathcal B \rangle\) to denote an SOD.

The mental picture of a SOD is an upper triangular matrix.

**Example**. Let $A$ and $B$ be dg-algebras and let $P$ be an $A$-$B$ bimodule. We form a new dg-algebra $U$ whose elements are represented as \(\begin{pmatrix} a & p \\ 0 & b \end{pmatrix}\) since multiplication is given by what you would expect from the notation \(\begin{pmatrix} a_1 & p_1 \\ 0 & b_1 \end{pmatrix} \begin{pmatrix} a_2 & p_2 \\ 0 & b_2 \end{pmatrix} = \begin{pmatrix} a_1a_2 & a_1p_2 + p_1b_2 \\ 0 & b_1b_2 \end{pmatrix}\) Let $M$ be a $U$ dg-module. We have an inclusion of \(A \times B \to U\) so each $M$ has the structure of a $A \times B$ dg-module. We can take that two idempotents \(e_A := (1,0) \in A \times B \\ e_B := (0,1) \in A \times B\) and set \(M_A := e_A M \\ M_B := e_B M\) We have $M = M_A \oplus M_B$. Each $p \in K$ induces a $k$-linear map \(p : M \to M\) As $pe_A = 0$ and $e_Bp = 0$, we see that the only nonzero component in the decomposition $M = M_A \oplus M_B$ is the map \(p : M_B \to M_A.\) From the algebra structure we see that

- $P \otimes_k M_B \to M_A$ descends to a map \(\psi_M: P \otimes_B M_B \to M_A\) and
- the map \(\psi_M: P \otimes_B M_B \to M_A\) is a homomorphism of $A$ dg-modules.

A morphism of $U$ dg-modules $f : M \to N$ gives a pair $f_A : M_A \to N_A$ an $A$-linear map and $f_B : M_B \to N_B$ a $B$-linear map such that the diagram

commutes.

From the data of $(M,N,\psi)$ we can recover the structure of a $U$ dg-module on $M := M \oplus M^\prime$. Thus, we have an equivalence between $\operatorname{Mod} U$ and the category of triples $(M,N,\psi)$ with morphisms being pairs $(f,g)$ which make

commute.

Any $A$-module $M$ determines a $U$-module with $(M,0,0)$. Similarly, Any $B$-module $N$ determines a $U$-module with $(0,N,0)$. These given fully-faithful functors \(\imath_A : \operatorname{Mod} A \to \operatorname{Mod} U \\ \imath_B : \operatorname{Mod} B \to \operatorname{Mod} U\) Moreover, we have natural isomorphisms \(\operatorname{Hom}_U(\imath_A M, N) \cong \operatorname{Hom}_A(M,N_A)\) and \(\operatorname{Hom}_U(M,\imath_B N) \cong \operatorname{Hom}_B(M_B,N)\) Thus, \(\begin{aligned} \rho_A : \operatorname{Mod} U & \to \operatorname{Mod} A \\ M & \mapsto M_A \end{aligned}\) is right adjoint to $\imath_A$ and \(\begin{aligned} \lambda_A : \operatorname{Mod} U & \to \operatorname{Mod} B \\ M & \mapsto M_B \end{aligned}\) is left adjoint to $\imath_B$. Since both $\rho_A$ and $\lambda_B$ preserve acyclicity, we see that

**Lemma**. For any K-projective $A$-module $Q$, the $U$-module $\imath_A Q$ is also K-projective. For any K-injective $B$-module $I$, the $U$-module $\imath_B I$ is also K-injective.

**Corollary**. We have fully-faithful functors \(\imath_A : D(A) \to D(U) \\ \imath_B : D(B) \to D(U)\) The images are triangulated subcategories of $D(U)$.

We have \(\operatorname{Hom}_D(U)(\imath_A M, \imath_B N) = 0\) since \(\operatorname{Hom}_{D(U)}(\imath_A M, \imath_B N) \cong \operatorname{Hom}_{D(A)}(M, \rho_A \imath_B N) \cong \operatorname{Hom}_{D(A)}(M, 0) = 0\)

Finally, for any $M$ we have a the co-unit \(\epsilon_M : \imath_A \rho_A M \to M\) and the triangle \(\imath_A \rho_A M \to M \to C(\epsilon) \to \imath_A \rho_A M[1]\) Note that $\imath_A \rho_A M = M_A$ so we have a sequence \(0 \to M_A \to M \to M_B \to 0\) which is exact as graded $k$-modules. There is a natural map \(C(\epsilon) \to M_B\) which is a quasi-isomorphism. Thus, $C(\epsilon)$ lies in $D(B)$ (up to isomorphism) and we have a semi-orthogonal decomposition.

From considerations at the level of $\operatorname{Mod} U$, we might think this a completely orthogonal decomposition. However, that is rarely the case. To compute \(\operatorname{Hom}_{D(U)}(\imath_B N, \imath_A M)\) in terms of homotopy classes we need to replace $\imath_B N$ by a K-projective resolution or $\imath_A M$ by a K-injective resolution. Since $\imath_B$ does not necessarily preserve K-projectivity, we cannot simply replace $N$ in $K(B)$ and apply $\imath_B$. Similarly, we cannot resolve $M$ by a K-injective in $K(A)$ and call it a day.

Viewing $P$ as a functor \(\begin{aligned} \Phi_P : D(B) & \to D(A) \\ N & \mapsto P \overset{\mathbf{L}}{\otimes}_B N \end{aligned}\) we can think of $D(U)$ as gluing together a copy of $D(A)$ and $D(B)$ via $\Phi_P$.

We will see this and the ideas that popped in the discussion of $D(U)$ appearing when we discuss properties of semi-orthogonal decompositions in general.