# When may “summand of” be dropped from the definition of perfect dg module?

Let $\mathcal{A}$ be a small dg category. In Section 1 of Lunts-Orlov http://arxiv.org/pdf/0908.4187v5.pdf, $Perf(\mathcal{A})$ is defined to be the full DG subcategory of $\mathcal{S}\mathcal{F}(\mathcal{A})$ consisting of DG modules homotopy equivalent to a direct summand of a finitely generated semi-free DG module over $\mathcal{A}$. Here, $\mathcal{S}\mathcal{F}(\mathcal{A})$ is the DG category of semi-free DG modules over $\mathcal{A}$, and a module is considered free'' if it is a direct sum of shifts of representable modules.

Example 1.9 of Lunts-Orlov is the assertion that the bounded derived category of $\mathcal{A}$ is (the homotopy category of) $Perf(\mathcal{A})$.

Under what conditions on $\mathcal{A}$ are all objects of $Perf(\mathcal{A})$ actually homotopy equivalent (isomorphic in $H_0$) to finitely generated semi-free DG modules?

The motivation of this question is the case when $\mathcal{A}$ is a strands algebra from bordered Heegaard Floer homology (the structure as a small dg category comes from the distinguished idempotents of $\mathcal{A}$).The papers that decategorify such $\mathcal{A}$ typically identify the unbounded derived category of $\mathcal{A}$ with $H_0$ of the category of Type D structures homotopy equivalent to operationally bounded Type D structures. The bounded derived category is identified with the full subcategory on Type D structures homotopy equivalent to finitely generated operationally bounded Type D structures.

An operationally bounded Type D structure is (more or less) the same as a semi-free dg module $X$ with a choice of decomposition of $X$ as $\mathcal{A} \otimes_{\mathcal{I}} N$ with respect to the $\mathcal{A}$-action, where $\mathcal{I}$ is the idempotent ring of $\mathcal{A}$ and $N$ is a left $\mathcal{I}$-module. Type D structures are isomorphic or homotopy equivalent iff the corresponding dg modules are isomorphic or homotopy equivalent. The same is true for finite generation.

So, based on the usual definition of perfect dg modules, I'd expect at first to have to useType D structures which split inject in $H_0$ into a finitely generated operationally bounded Type D structure'' as the bounded derived category. But this is less convenient for computing $K_0$, and I'd rather be able to replace an object of the bounded derived category with an isomorphic object which literally is a finitely generated bounded Type D structure.

In other words, my question is: are there any sufficient conditions one can place on dg algebras $\mathcal{A}$ for this to work, which are general enough to cover some interesting examples of $\mathcal{A}$ with nonzero differential?

• You may have a look at section 1.5.6 in Beilinson's math.harvard.edu/~gaitsgde/grad_2009/motvo.pdf. – Mikhail Bondarko Apr 15 '16 at 20:44
• Thanks for the reference! If my dg algebra were negatively graded, then item 3 of the proposition in 1.5.6 would be exactly what I need. Unfortunately, the algebra I care about most is positively graded, and I think this case may be a bit harder. Do you have any results in your other papers which would cover positively-graded dg algebras? – Andy Manion Apr 21 '16 at 19:49
• Unfortunately I have never studied this setting. I beiieve that Orlov, Luntz, Efimov and others are the authors who have studied it in detail. – Mikhail Bondarko Apr 21 '16 at 20:06

For a DGA $A$, all perfect modules are equivalent to semi-free modules if and only if $K(A)$ is generated by the class of the rank 1 free module. Furthermore, a particular perfect module is equivalent to a semi-free module if it is in the subgroup generated by the rank 1 free module. $K$-theory is exactly the gap between semi-free and perfect modules, so this does not seem to me useful for computing $K$-theory. You may as well ask: which DGAs have trivial reduced $K$-theory; maybe you should, but I think that is a different question.

It is easy to see that this $K$-theory condition is necessary. The definition of semi-free can be rephrased as that the module is generated from shifts of $A$ by a finite sequence of cones; so its class in $K(A)$ is in the subgroup generated by $[A]$. For example, for the product of two fields $A=E\times F$, an $A$-module is just a pair of $E$- and $F$-vector spaces. They could have different dimensions, so $K(A)=\mathbb Z^2$. But a semi-free module must have equal Euler characteristics. For example, the cone of $(1,0)\colon A\to A$ has homology zero on the $E$ side and nontrivial on the $F$ side, but its Euler characteristics are both zero. So $(E,0)$ is a perfect module not equivalent to a semi-free one.

Thomason proved that this necessary condition is actually sufficient. Given a triangulated category $C$, a triangulated subcategory $D$ (ie, a full subcategory closed under cones (and isomorphisms)) is called dense if every object of $C$ is a summand of an object of $D$. Thomason proved that dense subcategories of $C$ are determined by their $K$-groups, which necessarily inject into $K(C)$. The dense subcategory $D\subset C$ corresponding to a subgroup $H\subset K(C)$ consists of the objects whose class is in $H$. The semi-free modules are dense in the perfect modules by definition of the perfect modules, so a perfect module is equivalent to a semi-free module if and only if its class is in the subgroup generated by the ring $\langle [A]\rangle$. In particular, for every perfect module $M$, the module $M\oplus M[1]$ has trivial class and thus has a semi-free model.

• I think I'm using a slightly different definition of semi-free than you are; I'll edit my question to clarify. I want $\mathcal{A}$ to be a small dg category, viewed as a dg algebra with set of distinguished idempotents. I'm using the definitions from Section 1 of Lunts-Orlov arxiv.org/pdf/0908.4187v5.pdf. For your example, $\mathcal{A}$ should have two objects and both $(E,0)$ and $(0,F)$ count as free modules. Sorry for the confusion. – Andy Manion Apr 7 '16 at 23:07
• What do you mean by "should"? The choice of objects is a choice and which objects are semi-free depends on that choice. In generalizing from an algebra to a category, we just replace the single free module with the many free modules. A module is equivalent to a semi-free object if it is in the subgroup of $K$ generated by the representable modules. Of course, there is a universal choice: if $A'=D_{parf}(A)^{op}$ then then $D_{parf}(A')=D_{parf}(A)$ but now every module is semi-free, indeed representable. – Ben Wieland Apr 8 '16 at 2:00
• You're right, sorry again for all the ambiguity. If idempotents in $\mathcal{A}$ don't split, then there'll be issues like the one you describe. But if $\mathcal{A}$ is idempotent-complete and has no differential, then you can write every indecomposable projective $\mathcal{A}$-module as a representable functor, and by (e.g.) Prop. 3.4 of eudml.org/doc/90218, anything split-injecting into a bounded complex of projective $\mathcal{A}$-modules is actually homotopy equivalent to a bounded complex of projective $\mathcal{A}$-modules. – Andy Manion Apr 9 '16 at 21:51
• What I'm really interested in is a general-enough condition on dg algebras $\mathcal{A}$ which ensures the same is still true even though $\mathcal{A}$ has nonzero differential. – Andy Manion Apr 9 '16 at 21:54
• Perhaps your question is: can one lift an idempotent from the homotopy category to a finitely generated model? . . . If the category has one object with endomorphisms the ring of functions on an affine elliptic curve, would you say that idempotents split? – Ben Wieland Apr 10 '16 at 15:34