Timeline for Chain homotopy: Why du+ud and not du+vd?
Current License: CC BY-SA 2.5
18 events
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| Dec 19, 2009 at 6:47 | comment | added | Greg Kuperberg | Your exercise is related to the fact that every bounded chain complex over a field is a direct sum of complexes of the form $0 \to k \to 0$ and $0 \to k \to k \to 0$. | |
| Dec 18, 2009 at 19:16 | comment | added | darij grinberg | By the way, nice exercise for linear algebra classes (if someone reading this is holding any): Given a homomorphism between two chain complexes of vector spaces over a field $k$. (Both chain complexes are assumed to be bounded from below.) If this homomorphism is $0$ in homology, then show that it is a chain homotopy (i. e., $dw+wd$ for some $w$). But please check it first; I am not 100% sure about my proof. | |
| Dec 18, 2009 at 19:14 | comment | added | darij grinberg | I am not done thinking over this yet, but let me add that tensoring with $\mathbb{Z}\slash p$ and checking whether it's still homologically zero is not enough. For instance, if we alter David's second example such that the horizontal maps are multiplication by $p^2$ and the vertical map is multiplication by $p$. In this case, our map of chain complexes is homologically $0$ and remains so after tensoring with $\mathbb{Z}\slash p$ and $\mathbb{Z}\slash q$ for any other prime $q$, but it's not $du+vd$. So we should at least try $\mathbb{Z}\slash p^n$, if not even all $\mathbb{Z}\slash m$. | |
| Dec 18, 2009 at 17:08 | comment | added | Greg Kuperberg | (1) Don't worry what a spectral sequence is. It is a vast generalization of the Bockstein construction. It is good enough for you to just study the behavior of Bockstein. (2) I guess you are right that even the amended remark does not apply to the first example. Something is there, because both examples do have non-trivial Bockstein maps, but I'm not sure what. | |
| Dec 18, 2009 at 16:24 | comment | added | darij grinberg | So I assume that you are talking about the second example, and the question is whether any map between complexes of $\mathbb{Z}$-modules that is homologically $0$ even after tensoring with any arbitrary $\mathbb{Z}\slash p$ must be a map of the form $du+vd$. Well, this is an interesting question which I'll try to examie. But I fear that I don't really have the prerequisites for this, because I don't even know the definition of a spectral sequence. | |
| Dec 18, 2009 at 16:22 | comment | added | darij grinberg | "in David's examples": do you mean David's second example? As far as I understand, he made two examples: one of a $du+vd$ map which is not a $dw+wd$ map, and one of a homologically trivial map which is not a $du+vd$ map. The former can only become "better" (i. e., become $dw+wd$) after tensoring with $\mathbb{Z}\slash p$ (and it actually MUST become $dw+wd$, because after tensoring with $\mathbb{Z}\slash p$, every module becomes a vector space, and for vector spaces, $du+vd$ always rewrites as $dw+wd$ - at least, if our complex is bounded from one side). | |
| Dec 18, 2009 at 6:58 | comment | added | Greg Kuperberg | Darij: You're right that my remark and my thinking were both muddled. What I should have said is that, in David's examples, your generalized chain homotopies are no longer chain homotopies if you tensor the complexes with $\mathbb{Z}/p$; they create non-zero homology maps in the new coefficients. My intuition is that their behavior after tensoring can be expressed in terms of the Bockstein maps that become available. There is a "Bockstein spectral sequence", related to the universal coefficient theorem, that may help clarify the situation. | |
| Dec 16, 2009 at 19:50 | comment | added | darij grinberg | Still, even when the Bockstein homomorphisms are well-defined, what does "preserving" them mean? Aren't they canonical and therefore always preserved by a map of chain complexes? | |
| Dec 16, 2009 at 19:49 | comment | added | darij grinberg | On the other hand, the question is interesting even for flat modules. (Not even free modules of rank $1$, i. e. copies of the base ring, can ensure that every map of the form $du+vd$ that commutes with the boundaries has the form $dw+wd$ for some $w$.) | |
| Dec 16, 2009 at 18:47 | comment | added | darij grinberg | Greg, what do you mean by "preserving Bockstein homomorphisms"? At first, are they defined at all? The definition I know involves tensoring an exact sequence of abelian groups with $C_n$ for every $n$, hoping that this will produce another short exact sequence; however, this requires the $C_n$ to be flat. | |
| Dec 15, 2009 at 19:21 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 19:19 | comment | added | Reid Barton | In the last sentence, do you mean "0 mod p"? | |
| Dec 15, 2009 at 19:09 | comment | added | Greg Kuperberg | To expand on my previous remark, suppose that a pseudo-homotopy in this sense preserves all Bockstein maps. Then is it a true chain homotopy? My guess is, yes, at least if the chain complex is semibounded; and maybe also you need some tameness conditions on the abelian category in which you work. | |
| Dec 15, 2009 at 18:57 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 15:23 | comment | added | darij grinberg | Thanks a lot (particularly for the geometric interpretation of df + fd, which has nothing to do with my question but is a real gem). I'll ponder about your quotient category question a bit more. | |
| Dec 15, 2009 at 15:19 | comment | added | Greg Kuperberg | I think the point is that a true chain homotopy preserves not on homology, but also Bockstein homomorphisms. | |
| Dec 15, 2009 at 14:40 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 14:34 | history | answered | David E Speyer | CC BY-SA 2.5 |