Glueing triangulated categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:45:11Z http://mathoverflow.net/feeds/question/41416 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41416/glueing-triangulated-categories Glueing triangulated categories Hanno Becker 2010-10-07T14:02:47Z 2010-10-07T20:27:40Z <p>Hello!</p> <p>Given a triangulated category, one can look for semiorthogonal decompositions into (simpler?) triangulated subcategories. </p> <p>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? </p> <p>I hope this question isn't too vague.</p> <p>Thank you!</p> http://mathoverflow.net/questions/41416/glueing-triangulated-categories/41419#41419 Answer by Yann Palu for Glueing triangulated categories Yann Palu 2010-10-07T14:20:58Z 2010-10-07T14:20:58Z <p>I'm not sure but Proposition 1.16 in the paper:</p> <p><a href="http://arxiv.org/pdf/0911.0172" rel="nofollow">http://arxiv.org/pdf/0911.0172</a></p> <p>by Iyama-Kato-Miyachi might be related to your question.</p> http://mathoverflow.net/questions/41416/glueing-triangulated-categories/41420#41420 Answer by Mariano Suárez-Alvarez for Glueing triangulated categories Mariano Suárez-Alvarez 2010-10-07T14:34:10Z 2010-10-07T14:34:10Z <p>[Lidia Angeleri Hügel, Steffen Koenig, Qunhua Liu: On the uniqueness of stratifications of derived module categories] at <a href="http://arxiv.org/abs/1006.5301" rel="nofollow">http://arxiv.org/abs/1006.5301</a> (and other recent work by Koenig) should be relevant: this is Jordan-Hölder for (appropriate) triangulated categories.</p> http://mathoverflow.net/questions/41416/glueing-triangulated-categories/41450#41450 Answer by Sasha for Glueing triangulated categories Sasha 2010-10-07T20:27:40Z 2010-10-07T20:27:40Z <p>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. </p> <p>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 $$U = \left(\begin{array}{cc} R &amp; 0 \cr M &amp; S \end{array}\right).$$ Then $T := Hot(U)$ should give what you want.</p>