Timeline for Invariants of objects in $\operatorname{Ch}(\mathrm{Ab})$ up to chain homotopy
Current License: CC BY-SA 4.0
26 events
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| Nov 29, 2022 at 23:44 | comment | added | Greg Friedman | @MartinBrandenburg I think mentioning possibly better notation is fine. I think telling a poster that their notation is a "bad idea" and focusing exclusively on the notation and not the question (when the poster's notation, at least in my opinion, is reasonably good enough and the question is pretty clear) risks being discouraging. It's possible I was being overly sensitive when I read your comment, but it struck me as being unnecessarily critical. In any case, I'm glad the original poster seems pretty undeterred! | |
| Nov 29, 2022 at 5:10 | comment | added | Fernando Muro | @Student Kom(ab) is the same as D(Ab). | |
| Nov 28, 2022 at 23:11 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 28, 2022 at 22:46 | history | edited | Student | CC BY-SA 4.0 |
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| Nov 28, 2022 at 21:52 | comment | added | Student | @FernandoMuro Thanks for hinting. I found literature on hereditary categories and how taking homologies in their derived category is a complete invariant (that includes $Ab$). I added this into the post. // For those who are following and got confused by Fernando's comment: Achim only talked about $Kom(ab)$, in which homology is also a complete invariant essentially because of Smith normal form (works for modules any PID). // Martin: I appreciate your suggestion. If there's anyway I can undo and fix all notations in all comments so far, I'm happy to do so. | |
| Nov 28, 2022 at 21:46 | history | edited | Student | CC BY-SA 4.0 |
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| Nov 28, 2022 at 21:42 | comment | added | Martin Brandenburg | @GregFriedman I do not understand why mentioning the standard notation should be a form of discouragement from asking. Changing the notation can be done easily in the current question via an edit or in future questions. Clearly, this doesn't affect if the question is being asked or not. What would you do if someone asks a question which starts "Let $\mathbf{Grp}$ be the category of finite abelian groups" or "Let $\pi := 5$? | |
| Nov 28, 2022 at 21:34 | history | edited | Student | CC BY-SA 4.0 |
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| Nov 28, 2022 at 17:21 | history | edited | Student | CC BY-SA 4.0 |
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| Nov 28, 2022 at 17:04 | history | edited | Student | CC BY-SA 4.0 |
Add a section on derived categories and hereditary categories
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| Nov 28, 2022 at 15:01 | comment | added | Tyler Lawson | The most straightforward generalization of homology that applies is the following. If $W$ is an object of $Ab$, then you get functors $Kom \to Ab$ given by $X \mapsto Hom_{Kom}(W[n],X).$ When $W = \Bbb Z$, these are the ordinary homology groups of $X$. These are essentially generated by both the ordinary homology functors $H_n$ and by the "$d$-torsion homology" functors $X \mapsto H_n (X_{d\text{-tors}})$, where $X_{d\text{-tors}} \subset X$ is the subcomplex of elements annihilated by a fixed integer $d$. | |
| Nov 28, 2022 at 14:38 | comment | added | Fernando Muro | @Student please have a look at Achim Krause’s comment. It’s a pretty standard argument. You’ll find it in many introductory texts on derived categories. Like Henning Krause’s “Derived categories, resolutions, and Brown representability”. | |
| Nov 28, 2022 at 13:07 | comment | added | Student | @FernandoMuro If it's hopeless as you said, perhaps the best I can ask for is the invariant for objects in the derived category? However, in your first comment, you seemed to suggest that $X \simeq Y$ in $D(Ab) \Leftrightarrow H(X) \simeq H(Y)$. By definition, I get how $\Rightarrow$ is done, but I don't see how you prove $\Leftarrow$. | |
| Nov 28, 2022 at 6:20 | comment | added | Mariano Suárez-Álvarez | Kom(A) is the notation for the category of complexes in Gelfand-Manin... | |
| Nov 28, 2022 at 4:39 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 28, 2022 at 4:20 | comment | added | Greg Friedman | @MartinBrandenburg Perhaps we should not be discouraging students from asking reasonable questions here by chastising them too much for not being familiar with our preferred notation, especially when they do such a clear job of establishing their notation for the questions. (And for what it's worth, I've certainly seen "Kom" before.) | |
| Nov 28, 2022 at 0:08 | comment | added | Fernando Muro | @Student There's no hope to classify objects in the homotopy category of chain complexes of abelian groups, I believe. I think this would contain the classification of pairs $(A,B)$ where $A$ is an abelian group and $B\subset A$ is a subgroup. This is a wild problem of representation theory. | |
| Nov 27, 2022 at 22:25 | comment | added | Martin Brandenburg | Please, do not change standard notation, since this is a source of confusion. $\mathbf{Ab}$ (or $Ab$) is the category of all abelian groups. If you want to consider the subcategory of f.g. abelian groups, you can name it $\mathbf{Ab}_{\text{fg}}$ for example. Small letters such as $\mathbf{ab}$ actually often indicate finite versions, so again it's a bad idea for this to be the category of free abelian groups. Better write $\mathbf{Ab}_{\text{free}}$ (for example). Also, $K(\mathcal{A})$ is the established notation for the homotopy category of $\mathcal{A}$. | |
| Nov 27, 2022 at 22:12 | comment | added | Achim Krause | It's a bit like asking for a canonical basis on an abelian group (note that the singular chain complex of a space comes with a preferred basis, which gets forgotten when passing into $\operatorname{Kom}(Ab)$) | |
| Nov 27, 2022 at 22:05 | comment | added | Student | @AchimKrause Thanks! Yes, for $Kom(ab)$ it's enough to look at their homologies (by Smith normal form, which works not only for $\mathbb{Z}$ but general PIDs). And for $Kom(Ab)$ in general, it is so hard that it remains open? -- I understand your second comment as follows (please correct me if I'm wrong): Given an isomorphism class $[X]$ in $Kom(Ab)$, there is no (why?) canonical construction of a space $\tilde{X}$ (up to space-level homotopy equivalence), so mentioning cup product does not make sense here. | |
| Nov 27, 2022 at 21:39 | comment | added | Achim Krause | The discussion of cup products seems unrelated: even if chain complexes come from spaces, any additional algebraic structure they might inherit from this is lost and doesn't matter anymore once you ask whether they are equivalent purely as objects in $\operatorname{Kom}(Ab)$. It's a bit like responding to the question "are finite sets bijective iff they have the same number of elements" with "for finite groups, we know that they are not characterized by their number of elements, and finite sets sometimes come from finite groups". | |
| Nov 27, 2022 at 21:33 | comment | added | Achim Krause | It is not hard to see that every chain complex of free abelian groups splits as a direct sum of $2$-term chain complexes. From this one sees that in your $\operatorname{Kom}(ab)$ two objects are isomorphic iff their homologies are isomorphic. | |
| Nov 27, 2022 at 20:25 | history | edited | Student | CC BY-SA 4.0 |
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| Nov 27, 2022 at 20:22 | comment | added | Student | @FernandoMuro Do you mean if I instead want invariant for objects in $D(Ab)$, then $H(-)$ is a complete invariant? - And yeah, I'm aware that the homotopy category is more complicated. Are there any partial results at least? | |
| Nov 27, 2022 at 20:18 | comment | added | Fernando Muro | If you considered the derived category rather than the homotopy category, homology would suffice. The homotopy category is more complicated. | |
| Nov 27, 2022 at 20:13 | history | asked | Student | CC BY-SA 4.0 |