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Hello!

Given a triangulated category, one can look for semiorthogonal decompositions into (simpler?) triangulated subcategories.

I'd like to know if there's a way to attack the opposite problem, i.e. to classify the ways two given triangulated categories can be composed to give a big triangulated category which decomposes semiorthogonally into the given ones. Does anybody know about this?

I hope this question isn't too vague.

Thank you!

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4 Answers 4

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If $T = \langle A ,B\rangle$ is a semiorthogonal decomposition and $\alpha_\ast:A \to T$, $\beta_\ast:B \to T$ are the embedding functors then one can consider the functor $\beta^\ast\circ\alpha_\ast:A \to B$ (or its right adjoint $\alpha^!\circ\beta_\ast:B \to A$). This is called the gluing functor. Morally, the ways of gluing $A$ to $B$ are classified by gluing functors --- to each functor one should be able to associate a triangulated category $T$ with a s.o.d. into $A$ and $B$ for which the gluing functor is isomorphic to the given one. Because of the well-known problems with nonfunctoriality of the cone, one cannot hope to prove this precise statement. However, if everything has a DG-enhancement, one can.

Indeed, assume that $A = Hot(R)$, $B = Hot(S)$, where $R$ and $S$ are pretriangulated DG-algebras and assume that the functor $A \to B$ is realized by a $R-S$-bimodule $M$. Then one can consider a DG-algebra of the form $$ U = \left(\begin{array}{cc} R & 0 \cr M & S \end{array}\right). $$ Then $T := Hot(U)$ should give what you want.

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  • $\begingroup$ Sasha, that looks great, thanks alot. Is there a reference for the details? $\endgroup$
    – Hanno
    Oct 7, 2010 at 20:35
  • $\begingroup$ I am afraid there is no reference. This is a folklore. $\endgroup$
    – Sasha
    Oct 8, 2010 at 5:30
  • $\begingroup$ Still there is a key term - "recollement" $\endgroup$ Nov 17, 2014 at 8:05
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[Lidia Angeleri Hügel, Steffen Koenig, Qunhua Liu: On the uniqueness of stratifications of derived module categories] at http://arxiv.org/abs/1006.5301 (and other recent work by Koenig) should be relevant: this is Jordan-Hölder for (appropriate) triangulated categories.

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  • $\begingroup$ This is very interesting, thank you! $\endgroup$
    – Hanno
    Oct 7, 2010 at 20:34
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I'm not sure but Proposition 1.16 in the paper:

http://arxiv.org/pdf/0911.0172

by Iyama-Kato-Miyachi might be related to your question.

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There is an article by Kuznetsov and Lunts which lifts the glueing of triangulated categories to differential graded categories: http://arxiv.org/abs/1212.6170.

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