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Let $C$ be a triangulated category, and $D \subset C$ a triangulated subcategory whose inclusion functor admits a right adjoint. Say that a map $x \to y$ in $C$ is a "D-equivalence" if its cone belongs to $D$.

Suppose $D_1$ and $D_2$ are two such subcategories. I want to consider a new subcategory, call it $D_1 \rtimes D_2$ in very disposable notation, given by those $x \in C$ that are $D_1$-equivalent to an object of $D_2$. I think the inclusion to $C$ again has a right adjoint, at least in a setting where this is equivalent to these subcategories being closed under infinite direct sums.

Is $D_1 \rtimes D_2$ a standard construction, or equivalent to a standard construction that makes its structure more transparent? For example is it the same as the join in the lattice of localizing subcategories of $C$? Right now I cannot even tell if $D_1 \rtimes D_2 = D_2 \rtimes D_1$

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  • $\begingroup$ Isn't it in both cases just the subcategory formed by objects which are extensions of $D_1$ by $D_2$ or of $D_2$ by $D_1$? Provided that the equivalence is symmetric: $x$ is $D$-equivalent to $y$ if there is map $x\to y$ or $y\to x$ whose cone is in $D$. Then I think $D_1\rtimes D_2$ would be just the triangulated subcategory generated by $D_1$ and $D_2$. $\endgroup$
    – Dragos Fratila
    Commented Mar 17, 2016 at 8:08
  • $\begingroup$ Hi Dragos. I think a $D$-equivalence $x \to y$ does not guarantee a $D$-equivalence $y \to x$. For example when $C$ is the derived category of abelian groups and $D$ is the subcategory of torsion groups, then $\mathbf{Z} \to \mathbf{Q}$ is a $D$-equivalence. I don't know if this disproves your conclusion, though. $\endgroup$
    – David Treumann
    Commented Mar 17, 2016 at 13:31
  • $\begingroup$ Oh, so when you say $x$ is $D$-equivalent to $y$ you mean precisely for a map $x\to y$? (i.e. exactly in this direction) I thought by equivalence you meant the symmetric closure of this. So I guess $D_1\rtimes D_2$ are extension of an object in $D_1$ by an object in $D_2$, and the other way around for $D_2\rtimes D_1$. Makes me think of torsion pairs in triang cats but they are not translation invariant in general... $\endgroup$
    – Dragos Fratila
    Commented Mar 17, 2016 at 15:11
  • $\begingroup$ I think I misunderstood your first comment, and also maybe now I'm not sure what you mean by "symmetric closure." By "D_1-equivalent" in the second paragraph I mean, "is connected by a chain of D_1-equivalences, $x \to x' \leftarrow \cdots \to y$," which I hope is the same as "$x$ and $y$ are isomorphic in a quotient category $C/D$." If that's enough to prove $D_1 \rtimes D_2$ is subcategory generated by $D_1$ and $D_2$, can you spoon feed me the explanation? $\endgroup$
    – David Treumann
    Commented Mar 17, 2016 at 17:43
  • $\begingroup$ Sorry for the late reply. So here's what I thought at first: $x$ is $D$-equivalent to $y$ if either there's a triangle $x\to y\to d\to $ or $y\to x\to d\to $ where $d\in D$. Now if $x$ is $D_1$-equivalent to an object of $D_2$ then there's a triangle $x\to d_2\to d_1\to $. But this implies also a triangle $d_1[-1]\to x\to d_2\to $ and so $x$ is $D_2$ equivalent to $d_1$. This seems to imply that the two equivalence relations "$D_1$ equivalent to an object of $D_2$" and "$D_2$ equivalent to an object of $D_1$" are the same... Or maybe I'm too naive and missing something? $\endgroup$
    – Dragos Fratila
    Commented Mar 19, 2016 at 20:25

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