TITLE: Switching left and right adjoints in recollement situations
QUESTION [6 upvotes]: 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)

REPLY [4 votes]: 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.