## 22.37 Equivalences of derived categories

Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. A natural question that arises in nature is what it means that $D(A, \text{d})$ is equivalent to $D(B, \text{d})$ as an $R$-linear triangulated category. This is a rather subtle question and it will turn out it isn't always the correct question to ask. Nonetheless, in this section we collection some conditions that guarantee this is the case.

We strongly urge the reader to take a look at the groundbreaking paper [Rickard] on this topic.

Lemma 22.37.1. Let $R$ be a ring. Let $(A, \text{d}) \to (B, \text{d})$ be a homomorphism of differential graded algebras over $R$, which induces an isomorphism on cohomology algebras. Then

$- \otimes _ A^\mathbf {L} B : D(A, \text{d}) \to D(B, \text{d})$

gives an $R$-linear equivalence of triangulated categories with quasi-inverse the restriction functor $N \mapsto N_ A$.

Proof. By Lemma 22.33.7 the functor $M \longmapsto M \otimes _ A^\mathbf {L} B$ is fully faithful. By Lemma 22.33.5 the functor $N \longmapsto R\mathop{\mathrm{Hom}}\nolimits (B, N) = N_ A$ is a right adjoint, see Example 22.33.6. It is clear that the kernel of $R\mathop{\mathrm{Hom}}\nolimits (B, -)$ is zero. Hence the result follows from Derived Categories, Lemma 13.7.2. $\square$

When we analyze the proof above we see that we obtain the following generalization for free.

Lemma 22.37.2. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras over $R$. Let $N$ be a differential graded $(A, B)$-bimodule. Assume that

1. $N$ defines a compact object of $D(B, \text{d})$,

2. if $N' \in D(B, \text{d})$ and $\mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(N, N'[n]) = 0$ for $n \in \mathbf{Z}$, then $N' = 0$, and

3. the map $H^ k(A) \to \mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(N, N[k])$ is an isomorphism for all $k \in \mathbf{Z}$.

Then

$- \otimes _ A^\mathbf {L} N : D(A, \text{d}) \to D(B, \text{d})$

gives an $R$-linear equivalence of triangulated categories.

Proof. By Lemma 22.33.7 the functor $M \longmapsto M \otimes _ A^\mathbf {L} N$ is fully faithful. By Lemma 22.33.5 the functor $N' \longmapsto R\mathop{\mathrm{Hom}}\nolimits (N, N')$ is a right adjoint. By assumption (3) the kernel of $R\mathop{\mathrm{Hom}}\nolimits (N, -)$ is zero. Hence the result follows from Derived Categories, Lemma 13.7.2. $\square$

Remark 22.37.3. In Lemma 22.37.2 we can replace condition (2) by the condition that $N$ is a classical generator for $D_{compact}(B, d)$, see Derived Categories, Proposition 13.37.6. Moreover, if we knew that $R\mathop{\mathrm{Hom}}\nolimits (N, B)$ is a compact object of $D(A, \text{d})$, then it suffices to check that $N$ is a weak generator for $D_{compact}(B, \text{d})$. We omit the proof; we will add it here if we ever need it in the Stacks project.

Sometimes the $B$-module $P$ in the lemma below is called an “$(A, B)$-tilting complex”.

Lemma 22.37.4. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Assume that $A = H^0(A)$. The following are equivalent

1. $D(A, \text{d})$ and $D(B, \text{d})$ are equivalent as $R$-linear triangulated categories, and

2. there exists an object $P$ of $D(B, \text{d})$ such that

1. $P$ is a compact object of $D(B, \text{d})$,

2. if $N \in D(B, \text{d})$ with $\mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(P, N[i]) = 0$ for $i \in \mathbf{Z}$, then $N = 0$,

3. $\mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(P, P[i]) = 0$ for $i \not= 0$ and equal to $A$ for $i = 0$.

The equivalence $D(A, \text{d}) \to D(B, \text{d})$ constructed in (2) sends $A$ to $P$.

Proof. Let $F : D(A, \text{d}) \to D(B, \text{d})$ be an equivalence. Then $F$ maps compact objects to compact objects. Hence $P = F(A)$ is compact, i.e., (2)(a) holds. Conditions (2)(b) and (2)(c) are immediate from the fact that $F$ is an equivalence.

Let $P$ be an object as in (2). Represent $P$ by a differential graded module with property (P). Set

$(E, \text{d}) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(P, P)$

Then $H^0(E) = A$ and $H^ k(E) = 0$ for $k \not= 0$ by Lemma 22.22.3 and assumption (2)(c). Viewing $P$ as a $(E, B)$-bimodule and using Lemma 22.37.2 and assumption (2)(b) we obtain an equivalence

$D(E, \text{d}) \to D(B, \text{d})$

sending $E$ to $P$. Let $E' \subset E$ be the differential graded $R$-subalgebra with

$(E')^ i = \left\{ \begin{matrix} E^ i & \text{if }i < 0 \\ \mathop{\mathrm{Ker}}(E^0 \to E^1) & \text{if }i = 0 \\ 0 & \text{if }i > 0 \end{matrix} \right.$

Then there are quasi-isomorphisms of differential graded algebras $(A, \text{d}) \leftarrow (E', \text{d}) \rightarrow (E, \text{d})$. Thus we obtain equivalences

$D(A, \text{d}) \leftarrow D(E', \text{d}) \rightarrow D(E, \text{d}) \rightarrow D(B, \text{d})$

by Lemma 22.37.1. $\square$

Remark 22.37.5. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Suppose given an $R$-linear equivalence

$F : D(A, \text{d}) \longrightarrow D(B, \text{d})$

of triangulated categories. Set $N = F(A)$. Then $N$ is a differential graded $B$-module. Since $F$ is an equivalence and $A$ is a compact object of $D(A, \text{d})$, we conclude that $N$ is a compact object of $D(B, \text{d})$. Since $A$ generates $D(A, \text{d})$ and $F$ is an equivalence, we see that $N$ generates $D(B, \text{d})$. Finally, $H^ k(A) = \mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(A, A[k])$ and as $F$ an equivalence we see that $F$ induces an isomorphism $H^ k(A) = \mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(N, N[k])$ for all $k$. In order to conclude that there is an equivalence $D(A, \text{d}) \longrightarrow D(B, \text{d})$ which arises from the construction in Lemma 22.37.2 all we need is a left $A$-module structure on $N$ compatible with derivation and commuting with the given right $B$-module structure. In fact, it suffices to do this after replacing $N$ by a quasi-isomorphic differential graded $B$-module. The module structure can be constructed in certain cases. For example, if we assume that $F$ can be lifted to a differential graded functor

$F^{dg} : \text{Mod}^{dg}_{(A, \text{d})} \longrightarrow \text{Mod}^{dg}_{(B, \text{d})}$

(for notation see Example 22.26.8) between the associated differential graded categories, then this holds. Another case is discussed in the proposition below.

Proposition 22.37.6. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $F : D(A, \text{d}) \to D(B, \text{d})$ be an $R$-linear equivalence of triangulated categories. Assume that

1. $A = H^0(A)$, and

2. $B$ is K-flat as a complex of $R$-modules.

Then there exists an $(A, B)$-bimodule $N$ as in Lemma 22.37.2.

Proof. As in Remark 22.37.5 above, we set $N = F(A)$ in $D(B, \text{d})$. We may assume that $N$ is a differential graded $B$-module with property (P). Set

$(E, \text{d}) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, N)$

Then $H^0(E) = A$ and $H^ k(E) = 0$ for $k \not= 0$ by Lemma 22.22.3. Moreover, by the discussion in Remark 22.37.5 and by Lemma 22.37.2 we see that $N$ as a $(E, B)$-bimodule induces an equivalence $- \otimes _ E^\mathbf {L} N : D(E, \text{d}) \to D(B, \text{d})$. Let $E' \subset E$ be the differential graded $R$-subalgebra with

$(E')^ i = \left\{ \begin{matrix} E^ i & \text{if }i < 0 \\ \mathop{\mathrm{Ker}}(E^0 \to E^1) & \text{if }i = 0 \\ 0 & \text{if }i > 0 \end{matrix} \right.$

Then there are quasi-isomorphisms of differential graded algebras $(A, \text{d}) \leftarrow (E', \text{d}) \rightarrow (E, \text{d})$. Thus we obtain equivalences

$D(A, \text{d}) \leftarrow D(E', \text{d}) \rightarrow D(E, \text{d}) \rightarrow D(B, \text{d})$

by Lemma 22.37.1. Note that the quasi-inverse $D(A, \text{d}) \to D(E', \text{d})$ of the left vertical arrow is given by $M \mapsto M \otimes _ A^\mathbf {L} A$ where $A$ is viewed as a $(A, E')$-bimodule, see Example 22.33.6. On the other hand the functor $D(E', \text{d}) \to D(B, \text{d})$ is given by $M \mapsto M \otimes _{E'}^\mathbf {L} N$ where $N$ is as above. We conclude by Lemma 22.34.3. $\square$

Remark 22.37.7. Let $A, B, F, N$ be as in Proposition 22.37.6. It is not clear that $F$ and the functor $G(-) = - \otimes _ A^\mathbf {L} N$ are isomorphic. By construction there is an isomorphism $N = G(A) \to F(A)$ in $D(B, \text{d})$. It is straightforward to extend this to a functorial isomorphism $G(M) \to F(M)$ for $M$ is a differential graded $A$-module which is graded projective (e.g., a sum of shifts of $A$). Then one can conclude that $G(M) \cong F(M)$ when $M$ is a cone of a map between such modules. We don't know whether more is true in general.

Lemma 22.37.8. Let $R$ be a ring. Let $A$ and $B$ be $R$-algebras. The following are equivalent

1. there is an $R$-linear equivalence $D(A) \to D(B)$ of triangulated categories,

2. there exists an object $P$ of $D(B)$ such that

1. $P$ can be represented by a finite complex of finite projective $B$-modules,

2. if $K \in D(B)$ with $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(P, K) = 0$ for $i \in \mathbf{Z}$, then $K = 0$, and

3. $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(P, P) = 0$ for $i \not= 0$ and equal to $A$ for $i= 0$.

Moreover, if $B$ is flat as an $R$-module, then this is also equivalent to

1. there exists an $(A, B)$-bimodule $N$ such that $- \otimes _ A^\mathbf {L} N : D(A) \to D(B)$ is an equivalence.

Proof. The equivalence of (1) and (2) is a special case of Lemma 22.37.4 combined with the result of Lemma 22.36.6 characterizing compact objects of $D(B)$ (small detail omitted). The equivalence with (3) if $B$ is $R$-flat follows from Proposition 22.37.6. $\square$

Remark 22.37.9. Let $R$ be a ring. Let $A$ and $B$ be $R$-algebras. If $D(A)$ and $D(B)$ are equivalent as $R$-linear triangulated categories, then the centers of $A$ and $B$ are isomorphic as $R$-algebras. In particular, if $A$ and $B$ are commutative, then $A \cong B$. The rather tricky proof can be found in [Proposition 9.2, Rickard] or [Proposition 6.3.2, KZ]. Another approach might be to use Hochschild cohomology (see remark below).

Remark 22.37.10. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras which are derived equivalent, i.e., such that there exists an $R$-linear equivalence $D(A, \text{d}) \to D(B, \text{d})$ of triangulated categories. We would like to show that certain invariants of $(A, \text{d})$ and $(B, \text{d})$ coincide. In many situations one has more control of the situation. For example, it may happen that there is an equivalence of the form

$- \otimes _ A \Omega : D(A, \text{d}) \longrightarrow D(B, \text{d})$

for some differential graded $(A, B)$-bimodule $\Omega$ (this happens in the situation of Proposition 22.37.6 and is often true if the equivalence comes from a geometric construction). If also the quasi-inverse of our functor is given as

$- \otimes _ B^\mathbf {L} \Omega ' : D(B, \text{d}) \longrightarrow D(A, \text{d})$

for a differential graded $(B, A)$-bimodule $\Omega '$ (and as before such a module $\Omega '$ often exists in practice). In this case we can consider the functor

$D(A^{opp} \otimes _ R A, \text{d}) \longrightarrow D(B^{opp} \otimes _ R B, \text{d}),\quad M \longmapsto \Omega ' \otimes ^\mathbf {L}_ A M \otimes _ A^\mathbf {L} \Omega$

on derived categories of bimodules (use Lemma 22.28.3 to turn bimodules into right modules). Observe that this functor sends the $(A, A)$-bimodule $A$ to the $(B, B)$-bimodule $B$. Under suitable conditions (e.g., flatness of $A$, $B$, $\Omega$ over $R$, etc) this functor will be an equivalence as well. If this is the case, then it follows that we have isomorphisms of Hochschild cohomology groups

$HH^ i(A, \text{d}) = \mathop{\mathrm{Hom}}\nolimits _{D(A^{opp} \otimes _ R A, \text{d})}(A, A[i]) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(B^{opp} \otimes _ R B, \text{d})}(B, B[i]) = HH^ i(B, \text{d}).$

For example, if $A = H^0(A)$, then $HH^0(A, \text{d})$ is equal to the center of $A$, and this gives a conceptual proof of the result mentioned in Remark 22.37.9. If we ever need this remark we will provide a precise statement with a detailed proof here.

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