# Gluing filtered object from associated graded pieces

So, I believe the following result is correct but do not know the exact reference (and not sure to what extent what I'm saying is true). If anyone could give a reference for this it would be great.

1) Consider three objects in some abelian category, $$F_1, F_2, F_3$$. Then objects with filtrations $$F^1 \subset F^2 \subset F^3$$ such that $$F^1 = F_1, F^2/F^1 = F_2, F^3/F^2 = F_3$$ are classified by triples $$(\alpha, \beta, \gamma)$$ of extension elements, $$\alpha \in Ext^1(F_2, F_1), \beta \in Ext^1(F_3, F_2), \gamma \in Ext^1(F_3, F_1)$$, such that the Yoneda product $$\alpha \beta \in Ext^2(F_3, F_1)$$ vanishes.

2) Similar result should hold for any amount of objects at least for the case of linear category in characteristic zero - the ways of gluing objects $$F_1, ..., F_n$$ into a filtered object $$F^1 \subset F^2 \subset ... \subset F^n$$ such that $$F^k / F^{k-1} = F_k$$ should be classified by Maurer-Cartan elements in the algebra $$\bigoplus_{j>i}RHom^{\bullet}(F_j, F_i)$$ (considered as either dgla or L-infinity algebra).

I also know that this type of questions frequently appear in the theory of mixed hodge structures (but unable to find any direct reference, too).

Edit: added forgotten $$\gamma$$

• It is not precisely correct. You need to specify how it is zero and this is an extra piece of information. For example you might have a case in which both the first pair and the second are glued trivially but the third object is glued no trivially to the first. In this case all the information lies in the trivialization – S. carmeli Jan 23 '20 at 17:33
• Excuse me, I've definitely forgot that I also need an element in Ext^1(F_3, F_1), but I think in case of three objects this is all data I need: you are talking about some kind of Massey product and for three objects there won't be any. – Lev Soukhanov Jan 24 '20 at 0:10
• One comment is that you can think of the equivalence you are defining as Koszul duality for representations of the $A_n$ quiver. The Koszul dual (linear) category to the $A_n$ quiver is the category $Ch_n$ whose representations are chain complexes. The structure you get on the associated gradeds is exactly a homotopy coherent chain complex $Ch_n \to D(A)$ which takes the $k$th object to $F_k [k]$. – Phil Tosteson Jan 24 '20 at 2:12
• @LevSoukhanov This "more data you need" is precisely there, from my perspective, to indicate the different options for trivializing $\alpha \beta$: every two such trivializations differ by a class in $Ext^1$. Note that it has still some ambiguity in this class of the form $\alpha \xi$ for $\xi:F_3\to F_2$, as in Kuhn's answer. – S. carmeli Jan 24 '20 at 7:44
• Lets choose some resolutions of our objects. Then I think we both agree that the data specified is an element $(\alpha, \beta, \gamma) \in \bigoplus_{j>i} RHom^{\bullet}(F_j, F_i)$, satisfying Maurer-Cartan condition $\alpha \beta = d \gamma$. The solutions of such equations are in $1-1$ correspondense to solutions of this equation in any quasiisomorphic dg-algebra. However, in this case this algebra is formal (quasiisomorphic to its cohomology), and the equation reduces to $\alpha \beta = 0$. This works only for char=0 however. – Lev Soukhanov Jan 24 '20 at 7:57

Here is how I would think about this: $$\alpha \in Ext^1(F_2,F_1)$$ corresponds to a short exact sequence $$0 \rightarrow F_1 \xrightarrow{i} F(1,2) \xrightarrow{p} F_2 \rightarrow 0.$$ Similarly $$\beta \in Ext^1(F_3,F_2)$$ corresponds to a short exact sequence $$0 \rightarrow F_2 \rightarrow F(2,3) \rightarrow F_3 \rightarrow 0.$$ The first short exact sequence induces a long exact sequence including $$Hom(F_3,F_2) \xrightarrow{\alpha \circ} Ext^1(F_3,F_1) \xrightarrow{i_*} Ext^1(F_3,F(1,2)) \xrightarrow{p_*} Ext^1(F_3,F_2) \xrightarrow{\alpha \circ} Ext^2(F_3,F_1),$$ and from this one sees that an $$F(1,2,3) \in Ext^1(F_3,F(1,2))$$ exists such that $$F(1,2,3)/F_1 = F(2,3)$$ if and only if $$\alpha \circ \beta = 0 \in Ext^2(F_3,F_1)$$. Furthermore, choices correspond to the image of $$i_*$$. Perhaps this is the classification you desire. I don't know of a reference, but the argument is just using basic triangulated category/homological algebra techniques.