We work in an Abelian category. Consider Yoneda extensions, i.e., the Abelian groups Ext$^n(C,A)$ consisting (for $n \ge 1$) of equivalence classes of exact sequences starting at $A$ and ending at $C$ with $n+1$ maps in between, with the Baer sum as operation. See this Wikipedia page for reference.

The datum of a length $1$ extension, i.e., of an element of Ext$^1(C,A)$ (for some $A$ and $C$), corresponds to the datum of a filtration of length $2$, i.e., to the datum of an object $B$ and of a subobject $A \subseteq B$. Indeed, I am just saying that a short exact sequence $$ 0 \to A \to B \to C \to 0$$ gives the same information as the filtration $0 \subseteq A \subseteq B$, since $C \cong B/A$.

My question is if there is any relation between length $n$ extensions and length $n+1$ filtrations in general.

More specifically, if I have a nontrivial filtration of length $n+1$, can I obtain some nontrivial extension of length $n$? Or the converse?

This is already not so clear for $n=2$ (and we can stick to this case, if it makes things easier). An exact sequence

$$ 0 \to A \overset{\alpha}{\to} B \overset{\beta}{\to} C \overset{\gamma}{\to} D \to 0$$

is the same thing as the two short exact sequences

$$ 0 \to A \to B \to \ker(\gamma) \to 0 \quad \text{and} \quad 0 \to \ker(\gamma) \to C \to D \to 0,$$

but this is not the kind of thing that I am looking for, because this gives me two filtrations of length $2$ and not a single one of length $3$, which is what I would like to obtain somehow.

obstructionsto constructing length $n+1$ filtrations. $\endgroup$1more comment