## Back to derived equivalences

Assume we have an abelian category $\mathcal A$ with $D(\mathcal A)$ compactly generated. Let $P$ be a compact object.

Let \(A := \mathbf{R}\operatorname{End}(P)\) be the dg-algebra of endomorphism. (Here we assume that $P$ is quasi-isomorphic to a K-projective or K-injective complex).

We have a exact functor \(\mathbf{R}\operatorname{Hom}(P,-) : D(\mathcal A) \to D(A)\)

**Proposition**. If \(\overline{\langle P \rangle} = D(A)^c\) then $\mathbf{R}\operatorname{Hom}(P,-)$ is an equivalence.

## **Proof**. (Expand to view)

Let $F := \mathbf{R}\operatorname{Hom}(P,-)$ for ease of notation. Let $\mathcal C$ be the full subcategory consisting of $X \in D(\mathcal A)$ such that $$ F: \operatorname{Hom}(P,X) \to \operatorname{Hom}(FP,FX) $$ is an isomorphism. By construction, $P[l] \in \mathcal C$ for all $l \in \mathbb{Z}$. Moreover, $\mathcal C$ is - triangulated and - closed under $\bigoplus$ as $P$ is compact. Thus, $\mathcal C = D(\mathcal A)$ as we assumed $D(\mathcal A)$ is compactly generated. Next, let $\mathcal D$ be the full subcategory $D(\mathcal A)$ consisting of objects $X$ such that $$ F : \operatorname{Hom}(X,Y) \to \operatorname{Hom}(FX,FY) $$ is an isomorphism for all $Y \in D(\mathcal A)$. We know that $P \in \mathcal D$ from the previous step. Moreover, $\mathcal D$ is - triangulated and - closed under coproducts since $$ \operatorname{Hom}(\bigoplus X_a, Y) \cong \prod \operatorname{Hom}(X_a,Y) $$ and $F$ commutes with coproducts. Thus $\mathcal D = D(\mathcal A)$. Finally, let $\mathcal E$ be the essential image of $F$. We have $A[l] \in \mathcal E$ for all $l \in \mathbb{Z}$. Moreover, $\mathcal E$, triangulated since we any $\phi: M \to N$ in $\mathcal E$ is of the form $F(\psi)$ by the previous step and the cone $\psi$ maps to the cone of $\phi$. $\mathcal E$ is closed under coproducts since $F$ commutes with them. Thus $\mathcal E = D(A)$ and we see that $F$ is fully-faithful and essentially surjective. ■

We can extract the proof technique of the previous statement.

**Proposition**. Let $F: \mathcal T \to \mathcal S$ be an exact functor between compactly generated triangulated categories. Assume there a set of compact generators $X_a$ for $\mathcal T$ such that

- $F(X_a)$ is a set of compact generators for $\mathcal S$ and
- \(F : \operatorname{Hom}(X_a,X_{a^\prime}) \to \operatorname{Hom}(FX_a,FX_{a^\prime})\) is fully-faithful for all $a,a^\prime$.

Then, $F$ is an equivalence.

The previous proposition gives a reasonable criteria to describe $D(\mathcal A)$ when you have a compact generator. If we know that \(H^\ast(A) = H^0(A)\) then we still donâ€™t know that $D(\mathcal A) \cong D(H^0(A))$ just yet. We need to say something more.

Let $A$ and $B$ be dg-modules and $U$ be a dg-$A$-$B$-bimodule. Then, we have a functor \(U \otimes_B - : \operatorname{Mod} B \to \operatorname{Mod} A\) where we grade via \((U \otimes_B N)^i = \sum_{p+q=i} U^p \otimes_B N^q\) and use the differential \(\delta(u \otimes n) = \delta_U(u) \otimes n + (-1)^{|u|} u \otimes \delta(n)\)

We can derive this, say using K-projective resolutions, to get \(U \overset{L}{\otimes}_B - : D(B) \to D(A)\)

As with the case of usual modules, we also have a right adjoint \(\operatorname{Hom}_A (U,-) : \operatorname{Mod} A \to \operatorname{Mod} B\) and its derived version \(\mathbb{R} \operatorname{Hom}_A (U,-) : D(A) \to D(B)\)

We can to consider these functor in a special case. Let $f: A \to B$ be quasi-isomorphism of dg-algebras. Then, $B$ is a $A$-$B$ dg-bimodule.

**Proposition**. If $f: A \to B$ is a quasi-isomorphism, then \(B \overset{\mathbf{L}}{\otimes}_A - : D(A) \to D(B)\) is an equivalence.

## **Proof**. (Expand to view)

We apply the previous proposition in the case $F = B \overset{\mathbf{L}}{\otimes}_A -$. Note that $$ B \overset{\mathbf{L}}{\otimes}_A A[l] \cong B[l] $$ for each $l \in \mathbb{Z}$. Thus $F$ takes a set of compact generators to a set of compact generators. Moreover, multiplication $$ a \mapsto a^\prime a $$ goes to multiplication by $f(a^\prime)$. Since $f$ is quasi-isomorphism, we see that $F$ is fully-faithful on subcategory consisting of the $A[l]$. ■

**Lemma**. If $A$ is a dg-algebra with $H^i(A) = 0$ for $i \neq 0$, then $A$ and $H^0(A)$ are quasi-isomorphic.

## **Proof**. (Expand to view)

Let $$ (\tau_{\geq l} A)^i = \begin{cases} A^i & i > l \\ A^l/\delta A^{l-1} & i = l \\ 0 & i < l \end{cases} $$ Then, we have a have a homomorphism $A \to \tau_{\geq l} A$ which is a quasi-isomorphism in degrees $\geq l$. If $H^i(A) = 0$ for $i < l$, then it is a quasi-isomorphism overall. Similarly, for $$ (\tau_{\leq l} A)^i = \begin{cases} A^i & i < l \\ \ker \delta^l & i = l \\ 0 & i > l \end{cases} $$ we have a homomorphism $\tau_{\leq l} A \to A$ which is a quasi-isomorphism in degrees $\leq l$ and overall if $H^i(A) = 0$ for $i > l$. If $H^i(A) = 0$ for $i \neq 0$, then

From our discussion before, if we assume that \(\operatorname{Hom}(P,P[l]) = 0 \text{ for } l \neq 0\) then we get

**Proposition**. $D(\mathcal A) \cong D(R)$ where $R$ is the ring $R = \operatorname{Hom}(P,P)$.

## **Proof**. (Expand to view)

We saw already that $D(\mathcal A) \cong D(A)$ where $$ A = \mathbf{R}\operatorname{Hom}(P,P) $$ The previous result shows that $$ D(A) \cong D(R). $$ ■

This gives us our recognition principle. Specializing to the case that $\mathcal A = \operatorname{Mod} S$ for a ring $S$, we have the following statement which is originally due to Rickard.

**Proposition**. Let $R$ and $S$ be rings. There is a equivalence $D(R) \cong D(S)$ if and only if there exists a perfect complex $P \in D(R)$ with

- \[\operatorname{Hom}_{D(R)}(P,P][l]) = \begin{cases} S & l = 0 \\ 0 & l \neq 0 \end{cases}\]
- the smallest thick triangulated subcategory generated by $P$ is all of $D(R)^c$.

## Kernels and integral transforms

We work here over a field $k$ for simplicity. These statements remain fine if we assume things involved are free (or flat) over $k$. In general, we would work the derived tensor product $B \overset{\mathbf{L}}{\otimes}_k A$ as a dg-algebra. Deriving this is beyond the scope of this course since dg-algebras do not form an additive category.

DG-bimodules give functors. We introduce some notation. Given a $B$-$A$ dg-module $U$, we write \(\Phi_U (M) := U \overset{\mathbf{L}}{\otimes}_A -\)

We can lift the identity functor to a bimodule using $A$ as a $A$-$A$ bimodule since \(A \overset{\mathbf{L}}{\otimes}_A M \cong M\) In general, this bimodule is denoted by $\Delta_A$ and called the *diagonal bimodule*.

Furthermore, we can lift composition to the level of bimodules. Let $A,B,C$ be dg-algebras. Assume we have a $B$-$A$ dg-module $U$ and a $C$-$A$ dg-module $V$. Then we can construct a $A$-$C$ bimodule via \(V \star U := V \overset{\mathbf{L}}{\otimes}_B U\)

**Proposition**. We have a natural isomorphism \(\Phi_{V \star U} \cong \Phi_V \circ \Phi_U\)

## **Proof**. (Expand to view)

We can assume that $U$ and $V$ are both K-projective. Then for $K$-projective $M$ we have $$ \Phi_V \circ \Phi_U (M) = V \otimes_B U \otimes_A M $$ which is just $\Phi_{V \star U}(M)$. ■

The bimodules of the form $N \otimes_k M$ act like projections as \((N \otimes_k M) \otimes_A M^\prime = N \otimes_k (M \otimes_A M^\prime)\) lies in the additive category generated by $N$.

If we have compact generators, then taking this exterior product gives another compact generator.

**Proposition**. Let $P$ be a compact generator of $D(A)$ and $Q$ a compact generator of $D(B)$. Then $P \otimes_k Q$ is a compact generator of $D(A \otimes_k B)$.

## **Proof**. (Expand to view)

We first check that $P \otimes_k Q$ is compact. We have $$ \operatorname{Hom}(P \otimes_k Q, \bigoplus M_a) \cong \operatorname{Hom}(P, \mathbf{R}\operatorname{Hom}(Q,\bigoplus M_a)) $$ Since $Q$ is compact, we have $$ \operatorname{Hom}(P, \bigoplus \mathbf{R}\operatorname{Hom}(Q,M_a)) \cong \operatorname{Hom}(P, \mathbf{R}\operatorname{Hom}(Q,\bigoplus M_a)) $$ Since $P$ is compact, $$ \bigoplus \operatorname{Hom}(P, \mathbf{R}\operatorname{Hom}(Q,M_a)) \cong \operatorname{Hom}(P, \bigoplus \mathbf{R}\operatorname{Hom}(Q,M_a)) $$ Next, we check generation. It turns out easier to show that $P \otimes_k Q$ generates a set of compact objects. Given this, the smallest thick triangulated category containing $P \otimes_k Q$ is all compact objects. Therefore, we just need to generate $A \otimes_k B$. We know that $$ A \in \overline{\langle P \rangle} $$ Since $\otimes_k Q$ is exact, we have $$ A \otimes_k Q \in \overline{\langle P \otimes_k Q \rangle} $$ We also have $$ B \in \overline{\langle Q \rangle} $$ so we have $$ A \otimes_k B \in \overline{\langle A \otimes_k Q \rangle} $$ Combining the two, we have $$ A \otimes_k B \in \overline{\langle P \otimes_k Q \rangle} $$ ■