# Invariants of objects in $D(\mathcal{A})$ for non-hereditary category $\mathcal{A}$

$$\newcommand\A{\mathcal{A}}$$Let $$\A$$ be an additive category, and $$D(\A)$$ be its derived category (i.e. the category of chain complexes of $$\A$$ localized at quasi-isomorphisms). It is easy to show that if $$X \simeq Y$$ in $$D(\A)$$, then such equivalence induces a graded isomorphism on their homologies $$H(X) \simeq H(Y)$$. Therefore, $$H(-)$$ is an invariant for objects in $$D(\A)$$.

If $$\A$$ is abelian and hereditary (i.e. $$\operatorname{Ext}_{\A}^2(-,-) \equiv 0$$; e.g. the category of abelian groups), then any bounded $$X \in D(\A)$$ is isomorphic to $$H(X)$$ in $$\A$$ (see [1][2]). (The converse seems to be true as well, according to [1].) Therefore, homology is a complete invariant in this case.

### Question

How about if $$\A$$ is abelian but non-hereditary (say those with $$\operatorname{Ext}_{\A}^{3}(-,-) \equiv 0$$ and so on)?

1. What are some examples of a pair of non-isomorphic objects that have graded-isomorphic homologies?
2. Are there other (complete) invariants?

### Remark

I asked a similar but perhaps harder question "Invariants of objects in $$\operatorname{Ch}(\mathrm{Ab})$$ up to chain homotopy" on MathOverflow before localizing at the quasi-isomorphisms. Turned out that may be too hard, so I try my luck here.

### Reference

• Although it would be fine with TeX's practice of ignoring whitespace, in MathJax you cannot separate your command definitions from the body by blank lines; doing so will force literal blank lines into the rendered post. Unfortunately, though it makes the source ugly, it is necessary to put the \$ that ends the command definitions on the same line as the beginning of the body. I have edited accordingly (also on MO). Nov 28, 2022 at 23:14

A simple way to construct an example for your first question is by truncating projective resolutions of objects, so in this answer I will assume that $$\mathcal{A}$$ is abelian and has enough projectives. As an example, you could take $$\mathcal{A}$$ to be the module category over an algebra.
Assume that $$\mathcal{A}$$ is not hereditary. Then there is an object in $$\mathcal{A}$$ of projective dimension at least $$2$$; let $$M$$ be such an object, and let $$\ldots \xrightarrow{f_3} P_2 \xrightarrow{f_2} P_1 \xrightarrow{f_1} P_0 \xrightarrow{f_0} M \to 0$$ be a projective resolution of $$M$$. Consider now the object $$X$$ in the derived category $$D(\mathcal{A})$$ given by the complex $$\ldots \to 0 \to P_1 \xrightarrow{f_1} P_0 \to 0 \to \ldots.$$ Then the homology $$H(X)$$ is non-zero in degree 0 (since $$H_0(X) \cong M$$) and in degree 1 (otherwise $$f_1$$ would be injective and $$M$$ would have projective dimension at most 1). Finally, consider the object $$X'$$ of $$D(\mathcal{A})$$ given by $$\ldots \to 0 \to H_1(X) \xrightarrow{0} M \to 0 \to \ldots.$$ Then $$X$$ and $$X'$$ have isomorphic graded homology, but they are not isomorphic objects of $$D(\mathcal{A})$$. To see that they are not isomorphic, one could note that, in $$D(\mathcal{A})$$, the functor $$\operatorname{Ext}^2(X',-)$$ does not vanish on $$\mathcal{A}$$ (since $$M$$ has projective dimension at least 2 and is a direct summand of $$X'$$) while $$\operatorname{Ext}^2(X,-)$$ does.