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Let $\mathcal{T}' \subseteq \mathcal{T}$ be a full triangulated subcategory. Recall, $\mathcal{T}'$ is called $\textit{right admissible}$ if the inclusion $\mathcal{T}' \hookrightarrow \mathcal{T}$ admits a $\textit{right adjoint}$ $\pi : \mathcal{T} \rightarrow \mathcal{T}'$.

A left admissible subcategory is defined dually, and we call a subcategory $\textit{admissible}$ if it is both left and right admissible.

Equivalently, $\mathcal{T}' \subseteq \mathcal{T}$ is right admissible if and only if for every $A \in \mathcal{T}$, there exists a distinguished triangle

$B \rightarrow A \rightarrow C \rightarrow B[1]$

with $B \in \mathcal{T}'$ and $C \in \mathcal{T}'^{\perp}$, where $\mathcal{T}'^{\perp}$ denotes the $\textit{right orthogonal complement}$.

A similar equivalent condition exists for left admissible subcategories.

Admissible categories come up in many important contexts, for example, in semi-orthogonal decompositions.

My question is this: are there any examples where one can 'easily see' whether a given triangulated subcategory is admissible (or left admissible or right admissible) without having to check/prove the above abstract conditions?

For example, if $\mathcal{T} = D^{b}(X)$ is the bounded derived category of quasi-coherent sheaves on a scheme $X$, are there any conditions on $X$ that allows one to 'see' whether a triangulated subcategory is admissible or not?

I put 'easily see' and 'see' in quotes because I do not expect any non-trivial answer to make the above super easy, but the above two conditions, at least to me, seem quite abstract to check in most cases.

I am sure this is not a new question, and answers are available in the literature, but I am not found a source where such examples are 'collected'. Indeed, it would be useful to have such a collection to develop an intuition for these things.

Any help/pointing to references would be much appreciated.

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    $\begingroup$ The answer is much easier for unbounded derived categories of Grothendieck categories. In this case the category satisfies Brown representability, so full subcategories have left/right adjoint to the embedding functor iff they are closed under direct products/sums and direct summons. $\endgroup$
    – Denis T
    Commented Nov 18 at 15:13
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    $\begingroup$ The key word to ggogle is: "Bousfield localization". In the context of triangulated categories with small coproducts, Neemans book Triangulated Categories (Princeton Univ. Press) is all about this. If you are happy with stable $\infty$-categories, this is what the notion of presentable $\infty$-category is there to do for you. $\endgroup$
    – D.-C. Cisinski
    Commented Nov 18 at 19:23
  • $\begingroup$ If you want this kind of property for triangulated categories like the bounded derived category of coherent sheaves on a noetherian scheme, Neeman's notion of approximable triangulated category gives rather general answers; see arXiv:1806.06995 for a survey. Similar ideas may be found in this very nice paper of Ben-Zvi, Nadler and Preygel: arXiv:1312.7164 $\endgroup$
    – D.-C. Cisinski
    Commented Nov 18 at 19:23
  • $\begingroup$ DenisT and D.-C.Cisinski Thank you for your comments. This is much appreciated :) $\endgroup$
    – Sunny Sood
    Commented Nov 19 at 12:19

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One very useful example is the following: assume $\mathcal{T} = \mathrm{D}^b(\mathrm{coh}(X))$, and $\mathcal{T}' = \mathrm{D}^b(\mathrm{coh}(X'))$, where $X$ and $X'$ are smooth projective varieties. Then any fully faithful embedding $\mathcal{T}' \hookrightarrow \mathcal{T}$ is admissible, because in the geometric case any fully faithful functor $$ \mathrm{D}^b(\mathrm{coh}(X')) \hookrightarrow \mathrm{D}^b(\mathrm{coh}(X)) $$ is a Fourier--Mukai functor by Orlov's representability theorem, and any Fourier--Mukai functor between smooth projective varieties has both left and right adjoints.

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  • $\begingroup$ Suggested edit for clarity: "Then the image of every fully faithful functor $\mathcal{T}'\to \mathcal{T}$ is admissible." Otherwise there is no relationship between $\mathcal{T}$ and $\mathcal{T}'$. Or/similarly "in the geometric case the fully faithful functor" replace "the" with "every." $\endgroup$
    – Piotr Achinger
    Commented Nov 19 at 9:34
  • $\begingroup$ @PiotrAchinger: In the original question $\mathcal{T}'$ was assumed to be a subcategory of $\mathcal{T}$ (hence a fully faithful functor is fixed), so this is the assumption I am working with. $\endgroup$
    – Sasha
    Commented Nov 19 at 10:37
  • $\begingroup$ @Sasha: Just for the sake of clarity, can you please specify the relationship between $X$ and $X'$? $\endgroup$
    – Sunny Sood
    Commented Nov 19 at 12:17
  • $\begingroup$ @SunnySood You do not need any additional structure. You only need both the ambient category, and the full subcategory to be (abstractly) equivalent to $D^b$ of smooth protective variety; this is a property of separate categories, which is independent from choice of embedding. $\endgroup$
    – Denis T
    Commented Nov 19 at 14:01
  • $\begingroup$ @SunnySood: OK, I edited my answer following the suggestion of Piotr to make it more clear. So, indeed, as Denis explained the argument works for any fully faithful functor. $\endgroup$
    – Sasha
    Commented Nov 19 at 15:16

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