# Let $X$ be a projective variety. Is the bounded derived category of perfect complexes admissable in $D^b(X)$?

By an admissable subcategory $A$ in a triangulated category $B$, I mean a triangulated subcategory that has $A \oplus B \in A$, then $A$, $B \in A$, and so that there is either a right or left adjoint to the inclusion of $A$ into $B$.

Then, if $X$ is a projective variety (possibly with mild assumptions on the singularities), is $D^b_{perf}(X)$ admissible in $D^b(X)$? It clearly has the first condition, so I am curious about if an adjoint can be constructed.

I would like to know what is the best perfect approximation to a given complex.

Given a module, we can write $0 \to Tor(M) \to M \to M / Tor(M) \to 0$. For some rings (e.g. $\mathbb{Z}$), this is good enough, I think. But I don't know if there is or isn't something smarter one can do that works on any variety.

Follow up question: If so, then what is the (left or right) orthogonal to $D^b_{perf}(X)$ in $D^b(X)$?

I am trying to understand if $D^b(X)$ can be understood in general as a decomposition of some category of well behaved objects (perfect complexes), and some other category. What is that other category? Something torsion something?

• Perfect complexes are not closed under (homotopy) colimits nor (homotopy) limits, so there can be neither a left nor a right adjoint to the inclusion. The nearest thing I can think of is understanding Db(X) as "things assembled from perfect complexes", since every object is (canonically) a (homotopy) colimit of perfect complexes – Denis Nardin Oct 21 '16 at 18:24
• @DenisNardin: What is the CANONICAL way to represent a bounded complex as a homotopy colimit of perfect complexes? – Sasha Oct 21 '16 at 21:23
• @Sasha It is very silly: it is just the homotopy colimit indexed on the category of perfect complexes mapping to $C$. Since the category of perfect complexes is small, this makes sense. – Denis Nardin Oct 22 '16 at 0:39
• @DenisNardin: And how do you define a homotopy colimit over such a category in a triangulated category? All I know is how to define a homotopy colimit over naturals (as a cone of a morphism between infinite direct sum, as in the telescope construction), and my impression was that for more complicated categories one has to use enhancements. – Sasha Oct 22 '16 at 4:18
• @Sasha I am silently using the fact that $D^b(X)$ has a natural enhancement as a stable $\infty$-category. Sorry I didn't think it was worth mentioning (especially since the particular enhancement, you work with does not matter). – Denis Nardin Oct 22 '16 at 11:19

$D^b_{perf}(X)$ is not admissible in $D^b(X)$, unless $X$ is smooth (in which ase $D^b_{perf}(X) = D^b(X)$). Still one can consider the quotient category $D^b(X)/D^b_{perf}(X)$. It is called triangulated category of singularities, and was much studied.