In Fascieaux pervers, Beilinson, Bernstein and Deligne define a recollement situation as a triple of triangulated categories $\mathcal{D}_U,\mathcal{D}_F$ and $\mathcal{D}$, together with functors $$ i_*\colon \mathcal{D}_F\to \mathcal{D},\qquad j_*\colon \mathcal{D}\to \mathcal{D}_U $$ with suitable properties. In particular, $i_*$ has both a left adjoint $i^*$ and a right adjoint $i^!$. Beilinson, Bernstein and Deligne show that, in this situation, given $t$-structures $(\mathcal{D}^{\leq0}_U,\mathcal{D}^{\geq 0}_U)$ on $\mathcal{D}_U$ and $(\mathcal{D}^{\leq0}_F,\mathcal{D}^{\geq 0}_F)$ on $\mathcal{D}_F$ one obtains a $t$-structure $(\mathcal{D}^{\leq0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$ by setting $$ \mathcal{D}^{\leq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\leq 0} \text{ and } i^*K\in \mathcal{D}^{\leq0}_F\} $$ $$ \mathcal{D}^{\geq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\geq 0} \text{ and } i^!K\in \mathcal{D}^{\geq0}_F\} $$ It is not hard to check that this is indeed a $t$-structure on $\mathcal{D}$. One may wonder what happens by switching the role of the left and right adjoint of $i_*$ in the above definition, i.e., if one tries to define a $t$-structure by setting $$ \tilde{\mathcal{D}}^{\leq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\leq 0} \text{ and } i^!K\in \mathcal{D}^{\leq0}_F\} $$ $$ \tilde{\mathcal{D}}^{\geq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\geq 0} \text{ and } i^*K\in \mathcal{D}^{\geq0}_F\} $$ The argument that shows that $(\mathcal{D}^{\leq0},\mathcal{D}^{\geq 0})$ is a $t$-structure breaks down if one tries to adapt it to $(\tilde{\mathcal{D}}^{\leq0},\tilde{\mathcal{D}}^{\geq 0})$, precisely since the adjoints are not in the right place. However this is not a proof that $(\tilde{\mathcal{D}}^{\leq0},\tilde{\mathcal{D}}^{\geq 0})$ is not a $t$-structure, but only that it is not obviously so. And indeed, if one tries to set up a recollement situation in the much more symmetric setting of stable $\infty$-categories (where all constructions show a very nice symmetric behaviour due to the fact that every pullback is a pushout and vice versa), then it seems that not only also $(\tilde{\mathcal{D}}^{\leq0},\tilde{\mathcal{D}}^{\geq 0})$ is a $t$-structure, but that this actually coincides with $(\mathcal{D}^{\leq0},\mathcal{D}^{\geq 0})$ (this however does not imply that $i^*=i^!$). Or at least, this is what I believe we have shown with Fosco Loregian in http://arxiv.org/abs/1507.03913 Assuming our result is correct (which may well not be the case), the natural questions are:

  • is this actually true also for triangulated categories? (I have not been able to locate a statement like this in the literature)

  • it is this maybe false for a general triangulated category but true for a triangulated category which is the homotopy category of a stable $\infty$-category?

  • is this always manifestly false for triangulated categories? (something that would make me suspect of some mistake in my argument for stable $\infty$-categories, or a hint of the fact that giving a recollement situation in the stable setting is such a strong requirement that there are actually no natural examples of stable recollements)

  • 2
    $\begingroup$ Have you glanced at Appendix A in Higher Algebra? Lurie has a bunch of theorems on recollement of infty-cats and stable infty-cats etc. In any case, I would be very surprised if the stable infty-category story for t-structures is any different from the case of triangulated categories- after all, t-structures are defined on the homotopy category. $\endgroup$ Feb 25, 2016 at 0:40
  • 1
    $\begingroup$ Also I'm confused by the statement that the stable infty category case is nicer because 'push outs and pullbacks are the same'. The analog of this statement is that the triangles in a triangulated category behave simultaneously like cofiber and fiber sequences. But maybe I should go look at your proof and see what's used... $\endgroup$ Feb 25, 2016 at 0:42
  • $\begingroup$ Where exactly in appendix A? I can remember having performed a text search on Higher Algebra for recollements but that failed, so I missed the treatment in Appendix A. But now I've gone through it and I see it is actually relevant, thanks! Could you pint me to a precise point in the appendix that would be great. Me too would be surprised of a different behaviour in the triangulated case, the only explanations I can figure are a mistake in our proof or the fact that homotopy categories of stable categories are a bit special among triangulated categories. $\endgroup$ Feb 25, 2016 at 6:49
  • $\begingroup$ Concerning triangles in a triangulated category behaving simultaneusly as fiber and cofiber sequences, I'ma a bit confused: see Theorem 7.1.11 in Hovey's book. The point seems to be that in a general triangulated category $\Sigma$ and $\Omega$ are adjoint but not inverse each other. But I may be missing some point here. $\endgroup$ Feb 25, 2016 at 6:53
  • $\begingroup$ @domenicofiorenza $\Sigma$ is an equivalence by the definition of triangulated category... $\endgroup$
    – AAK
    Feb 25, 2016 at 9:53

1 Answer 1


The unexpected (and actually undue) symmetric behavior of stable recollements noticed in http://arxiv.org/abs/1507.03913 (Lemma 4.3 of version 1, therein called the Rorschach lemma) turned out to be the far reaching consequence of a typo in one of the commutative diagrams on page 9. This has now been corrected (i.e., Lemma 4.3, together with all its corollaries, has been removed). Luckily, this was only minimally affecting the remaining part of the article, which has now been revised accordingly.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.