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I have a question about the equivalence of derived categories. Let $\mathcal{A} = \mathcal{A}'\oplus \mathcal{A}''$ and $\mathcal{B} = \mathcal{B}' \oplus \mathcal{B}''$ are direct sum of abelian categories. Let $T$ be an complex of exact functors $$T: \text{ }\cdots \rightarrow T^{p-1} \rightarrow T^{p} \rightarrow T^{p+1} \rightarrow \cdots$$ (each $T^q$ is an functor from $\mathcal{A}$ to $\mathcal{B}$). Assume that for $T(X)$ is bounded for each $X\in \mathcal{A}$, then we can obtain a functor between their bounded derived categories, i.e. $T:$ $\mathcal{D}^b(\mathcal{A}) \rightarrow \mathcal{D}^b(\mathcal{B})$, by taking total complexes.

Assume that $T$ is an equivalence as a triangulated equivalence. Then it induce a isomorphism $[T]$ between their Grothendieck groups $[T]:\mathcal{K}(\mathcal{A}) \rightarrow \mathcal{K}(\mathcal{B})$. (Note that $\mathcal{K}(\mathcal{\mathcal{C}}) \cong \mathcal{K}(\mathcal{D}^b(\mathcal{C}) ) $ for every abelian category $\mathcal{C}$).

$\bf My$ $\bf Question:$ Is it true that $[T](\mathcal{K}(\mathcal{A}')) = \mathcal{K}(\mathcal{B}')$ implies the restriction $T|_{\mathcal{A}'}$ to $\mathcal{A}'$ is an equivalence between $\mathcal{D}^b(\mathcal{A'}) $ and $\mathcal{D}^b(\mathcal{B'})$? Furthermore, I was wondering when this is true. Thanks very much!

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It's certainly not true in general, as it could be that all the Grothendieck groups are zero, in which case the condition $[T](\mathcal{K}(\mathcal{A}')) = \mathcal{K}(\mathcal{B}')$ gives no information at all.

For example, let $\mathcal{V}$ be the category of (not necessarily finite dimensional) vector spaces over some field, so that $\mathcal{K}(\mathcal{V})=0$, and take $\mathcal{A}'=\mathcal{V}=\mathcal{B}''$ and $\mathcal{A}''=0=\mathcal{B}'$, with $T$ the identity functor.

As for when it is true, it's hard to know what kind of condition you would find useful without knowing more about the particular situation you're interested in. But if all the abelian categories involved have unique decompositions into indecomposable summands, and each summand has non-zero Grothendieck group then it will be true.

To see this, suppose $$\mathcal{A}=\bigoplus_{i\in I}\mathcal{A}_i$$ as a direct sum of indecomposable subcategories. Then each $\mathcal{D}^b(\mathcal{A}_i)$ is also indecomposable (as a triangulated category), since if $\mathcal{D}^b(\mathcal{A}_i)=\mathcal{D}\oplus\mathcal{D}'$, then $\mathcal{A}_i=(\mathcal{A}_i\cap\mathcal{D})\oplus(\mathcal{A}_i\cap\mathcal{D}')$, and so $\mathcal{A}_i$, and therefore $\mathcal{D}^b(\mathcal{A}_i)$, is contained in one of the summands.

Now $$\mathcal{D}^b(\mathcal{A})=\bigoplus_{i\in I}\mathcal{D}^b(\mathcal{A}_i)$$ and any direct summand $\mathcal{D}$ of $\mathcal{D}^b(\mathcal{A})$ is just the direct sum of those $\mathcal{D}^b(\mathcal{A}_i)$ that it contains, since if $\mathcal{D}^b(\mathcal{A})=\mathcal{D}\oplus\mathcal{D}'$ then for each $i$ $$\mathcal{D}^b(\mathcal{A}_i)=\left(\mathcal{D}^b(\mathcal{A}_i)\cap \mathcal{D}\right)\oplus\left(\mathcal{D}^b(\mathcal{A}_i)\cap\mathcal{D}'\right),$$ and so $\mathcal{D}^b(\mathcal{A}_i)$ is either contained in $\mathcal{D}$ or in $\mathcal{D}'$.

Since $$\mathcal{K}(\mathcal{A})=\bigoplus_{i\in I}\mathcal{K}(\mathcal{A}_i),$$ $\mathcal{D}$ is determined by which $\mathcal{K}(\mathcal{A}_i)$ are contained in $\mathcal{K}(\mathcal{D})$, so long as $\mathcal{K}(\mathcal{A}_i)\neq0$ for every $i$.

This is the case, for example, if the abelian categories are categories of finitely generated modules for finite dimensional algebras: if $\mathcal{A}=\text{mod-}A$ and $A=\prod_{i\in I}A_i$ as a product of indecomposable algebras, then $\mathcal{A}=\bigoplus_{i\in I}\text{mod-}A_i$ is the unique decomposition of $\mathcal{A}$ into indecomposable subcategories.

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  • $\begingroup$ Thank you very much! I was really wondering how can I prove your last statements. Could you give me some hints or some references? Thanks very much again. $\endgroup$ – GuNa Jul 16 '15 at 8:30
  • $\begingroup$ @GuNa It's pretty straightforward, but I've edited to add some details. $\endgroup$ – Jeremy Rickard Jul 16 '15 at 10:59
  • $\begingroup$ @GuNa I've added more details. $\endgroup$ – Jeremy Rickard Jul 17 '15 at 9:07
  • $\begingroup$ I'm very grateful to you for this. It is a very clear explanation. Thank you very much again! $\endgroup$ – GuNa Jul 17 '15 at 10:41

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