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Derived Morita equivalence

Definition. We say that two rings $R$ and $S$ are derived Morita equivalent if $D(R) \cong D(S)$.

The next goal will be to prove the following theorem of Rickard.

Theorem. If $R$ and $S$ are derived Morita equivalent, then there exists a $U \in D(R \otimes_{\mathbb{Z}} S)$ such that \(- \overset{\mathbf{L}}{\otimes}_R U : D(R) \to D(S)\) is an equivalence.

As a corollary of this, we know that \(\mathbf{R}\operatorname{Hom}_S(U,-) : D(S) \to D(R)\) is also an equivalence.

We start slowly.

Lemma. Assume that $R$ and $S$ are derived Morita equivalent. Then there exists a perfect complex of $S$- modules $P$ with \(\operatorname{Hom}_{D(S)}(P,P) \cong R, \\ \operatorname{Hom}_{D(S)}(P,P[i]) = 0 \text{ for } i \neq 0,\) and \(\overline{\langle P \rangle} = \operatorname{perf} S\)

Proof. (Expand to view)

Given an equivalence $F: D(R) \to D(S)$ we set $P := F(R)$. Then $$ F: \operatorname{Hom}_{D(R)}(R,R[i]) \overset{\sim}{\to} \operatorname{Hom}_{D(S)}(P,P[i]) $$ is an isomorphism. Also, $P$ generates as $\operatorname{perf} S$ as $R$ generates $\operatorname{perf} S$.

Definition. An object $P$ in a triangulated category $\mathcal T$ is called a generator if \(\overline{\langle P \rangle} = \mathcal T\) If $\mathcal S$ is compactly generated and \(\overline{\langle P \rangle} = \mathcal S^c\) we say $P$ is a compact generator.

$P$ is called tilting if \(\operatorname{Hom}(P,P[i]) = 0 \text{ for } i \neq 0\)

For a nice enough compactly generated triangulated, a compact generator captures everything. In particular, for derived categories of abelian categories.

But to precisely state how this works we need to take a digression into differential graded algebras and modules over them.

DG-algebras, DG-modules, and derived categories

Definition. Let $k$ be a commutative ring. A differential graded algebra or dg-algebra or DGA is $k$-algebra $A$ with a decomposition as $k$-modules \(A = \bigoplus_{i \in \mathbb{Z} } A^i\) and $k$-linear maps \(\delta^i : A^i \to A^{i+1}\) such that

  • multiplication is of degree $0$ \(A^i \cdot A^j \subseteq A^{i+j}\)
  • $\delta$ is a $k$-linear differential: $\delta^0(1) = 0$ and $\delta^{i+1} \delta^i = 0$ for all $i$
  • $\delta$ satisfies the graded Liebniz rule for multiplication in $A$ \(\delta(aa^\prime) = \delta(a)a^\prime + (-1)^{|a|} a \delta(a^\prime)\) where $a$ and $a^\prime$ are homogeneous of degrees $|a|$ and $|a^\prime|$.

A morphism $f: A \to B$ of dg-algebras is a $k$-algebra morphism with \(f(A^i) \subseteq B^i\) and \(\delta_B \circ f - f \circ \delta_A = 0\)

A dg-module $M$ over a dg-algebra $A$ is an $A$-module with a $k$-module decomposition \(M = \bigoplus_{i \in \mathbb{Z} } M^i\) and differential \(\delta^i : M^i \to M^{i+1}\) such that

  • \[A^i \cdot M^j \subseteq M^{i+j}\]
  • \[\delta_M(a \cdot m) = \delta_A(a) \cdot m + (-1)^{|a|} a \cdot \delta_M(m)\]

A morphism $\phi : M \to N$ of dg-modules over a dg-algebra $A$ is an $A$-module homomorphism with \(\phi(M^i) \subseteq N^i\) and \(\delta_N \circ \phi - \phi \circ \delta_M = 0\) Note that any $\phi$ induces a homomorphism of cohomology groups \(H^i(\phi) : H^i(M) \to H^i(N)\) so we can speak of quasi-isomorphisms.

We have an additive category $\operatorname{Mod} A$ of (left) dg-modules over a dg-algebra $A$. Inside there lies the full subcategory $\operatorname{Acyclic} A$ of dg-modules with \(H^\ast(M) = 0\)

The derived category $D(A)$ is the localization of $\operatorname{Mod} A$ at the class of quasi-isomorphisms with the same definition for the derived category of an abelian category, replacing $\operatorname{Ch} \mathcal A$ with $\operatorname{Mod} A$.

This is sensible thing to do because of the following lemma.

Lemma. If $A$ is a dg-algebra with $A^i = 0$ for $i \neq 0$, then there is an isomoprhism \(\operatorname{Ch}(\operatorname{Mod} A^0) \cong \operatorname{Mod} A\)

Proof. (Expand to view)

Any chain complex $(M^i,\delta^i)$ of $A^0$ modules gives a dg-module over $A$ with $$ M = \bigoplus M^i $$ and $\delta$ being the differential. And vice-versa.

Given two dg-modules $M$ and $N$, we have a chain complex whose degree $i$ term is \(\operatorname{Hom}^i(M,N) := \operatorname{Hom}_{\operatorname{gr} A}(M,N[i])\) where $\operatorname{gr} A$ is the underlying graded algebra of $A$ , ie just $A$ and toss out the differential. Morphisms are taken to be degree $0$.

The differential on \(\operatorname{Hom}(M,N) = \bigoplus_{i \in \mathbb{Z}} \operatorname{Hom}^i(M,N)\) is given by \(\delta \phi := \delta_N \circ \phi - (-1)^{|\phi|} \phi \circ \delta_M\)

The groups $H^0(\operatorname{Hom}(M,N))$ give the morphisms in the homotopy category of dg-modules over $A$, $K(A)$.

Following the course of $D(\mathcal A)$, we can also talk about K-projective and K-injective dg-modules.

A dg-module $P$ is K-projective if \(\operatorname{Hom}_{K(A)}(P,C) = 0\) for any $C \in \operatorname{Acyclic} A$. A dg-module $I$ is K-injective if \(\operatorname{Hom}_{K(A)}(C,I) = 0\) for any $C \in \operatorname{Acyclic} A$.

Proposition. Assuming there are enough K-projective dg-modules, then \(K-Proj(A) \to D(A)\) is an equivalence. Similarly, assuming there are enough K-injective dg-modules, \(K-Inj(A) \to D(A)\) is an equivalence.

Proof. (Expand to view)

The proof is completely analogous to [the one](/notes/2021_09_23/) for the derived category of an abelian category. We suppress it.

We need to determine if we have enough K-projectives and K-injectives. We have some obvious K-projective objects.

Lemma. There is a natural isomorphism \(\operatorname{Hom}_{K(A)}(A,M) \to H^0(M)\) for any dg-module $M$.

Proof. (Expand to view)

In the category $\operatorname{gr} A$, we have the map $$ \begin{aligned} \operatorname{Hom}_{\operatorname{gr} A}(A,M) & \to M^0 \\ \phi & \mapsto \phi(1) \end{aligned} $$ is an isomorphism. For the map $a \mapsto am$ to commute with differential we need $\delta_M(m) = 0$. The null-homotopic maps are those coming from $$ \operatorname{Hom}_{\operatorname{gr} A}(A,M[-1]) & \to M^{-1} $$ after applying $\delta_M$, ie the image of $\delta_M$. Thus, $$ \operatorname{Hom}_{K(A)}(A,M) \to H^0(M) $$

Consequently, we also have \(\operatorname{Hom}_{K(A)}(A[i],M) \to H^{-i}(M)\) Thus all $A[i]$ are K-projective.

To show we have enough K-projective dg-modules, we need to use the triangulated category structure on $K(A)$. It is completely analogous to $K(\mathcal A)$. Given a morphism $\phi : M \to N$, we can form the cone \(M[1] \oplus N, \begin{pmatrix} \delta_{M[1]} & \phi \\ 0 & \delta_N \end{pmatrix}\) The triangles for the structure are those isomorphic to \(M \overset{\phi}{\to} N \to C(\phi) \to M[1]\)

Proposition. This collection of triangles with the shift $[1]$ endow $K(A)$ with structure of a triangulated category.

Since the proof is very similar to that for $K(\mathcal A)$, we omit it. We turn to constructing enough K-projectives using a method similar to the “resolution” by compact objects in a compactly-generated triangulated category.

Let $M$ be a dg-module over $A$. Let $U_0$ be a generating set for the $k$-module $H^\ast(M)$ and choose a lift $\tilde{u} \in M$ for each $u \in U_0$. Set \(P_0 := \bigoplus_{u \in U_0} A[-|u|]\) This has a natural map \(\begin{aligned} P_0 & \to M \\ 1_{u} & \mapsto \tilde{u} \end{aligned}\) and we have exact sequence \(H^\ast(P_0) \to H^\ast(M) \to 0\) of graded $k$-modules. Let $N$ be the kernel of the map $H^\ast(P_0) \to H^\ast(M)$ and let $U_1$ be a generating set for it as a $k$-module and choose a lift $\tilde{u} \in P_0$ for each $u \in U_1$ and set \(Q_1 := \bigoplus_{u \in U_1} A[-|u|]\) The map $Q_1 \to P_0 \to M$ induces the zero map on cohomology. Since $Q_1$ is a direct sum of shifts of $A$, it is therefore the zero map in $K(A)$. Thus, we get an induced map \(P_1 := C(Q_1 \to P_0) \to M\)

Now, we proceed inductively to construct $P_i \to M$ and we set \(P := \operatorname{colim} P_i\)

We need to check that $P$ is

  • K-projective and
  • the map $P \to M$ is a quasi-isomorphism.

For K-projectivity, we have \(\operatorname{Hom}_{K(A)}(\operatorname{colim} P_i , C) = \operatorname{lim} \operatorname{Hom}_{K(A)}(P_i,C) = 0\) For a quasi-isomorphism, one can check that each \(H^\ast(P_i) \to H^\ast(M) \to 0\) is surjective. If $u \in H^\ast(P_i)$ maps to $0$, in $H^\ast(M)$, then by construction its image in $P_{i+1}$ is $0$ and $u$ is $0$ in the colimit.

Proposition. For any dg-algebra $A$ and any dg-module $M$, there is a K-projective dg-module $P$ and a quasi-isomorphism $P \to M$.

There are enough K-injective dg-modules but we outsource the proof.