Timeline for How to lift a chain complex from $\mathbb{Z}/2\mathbb{Z}$ to $\mathbb{Z}$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2023 at 15:34 | vote | accept | unknown | ||
| Apr 30, 2023 at 20:56 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title; question in body
|
| Apr 30, 2023 at 20:37 | answer | added | Dmitri Pavlov | timeline score: 4 | |
| Apr 30, 2023 at 18:53 | comment | added | mme | This is "the structure theorem for chain complexes over F2". The hard part if you're trying to be concrete is to find such a splitting. | |
| Apr 30, 2023 at 18:51 | comment | added | mme | @unknown Change basis as suggested by Dave's comment. Split each $C_n$ as $H_n \oplus dC_{n+1} \oplus B_n$, where $d: B_n \to dC_n$ is a bijection. Choose a basis $B_n = \Bbb F_2^{b_n}$ and $H_n = \Bbb F_2^{h_n}$ and lift this to $C'_n = \Bbb Z^{h_n} \oplus \Bbb Z_{b_{n+1}} \oplus \Bbb Z_{b_n}$; the differential sends the last factor in $C'_n$ identically onto the second factor of $C'_{n-1}$, and is otherwise zero. All that remains is to show that you can lift the change of basis: every invertible matrix over $\Bbb F_2$ lifts to an invertible matrix over $\Bbb Z$ (lift elementary matrices) | |
| Apr 30, 2023 at 18:35 | comment | added | unknown | @DaveBenson excuse me if I'm being dense but I'm not sure how to use this. I have a list of matrices $\partial_n : \mathbb{F}_2^{a_n} \to \mathbb{F}_2^{a_{n+1}}$ with $\partial_n \partial_{n+1}=0$; how would I transform these into integer matrices? | |
| Apr 30, 2023 at 17:54 | comment | added | Dave Benson | A chain complex over $\mathbb{F}_2$ is just a direct sum of copies of $0\to\mathbb{F}_2\to 0$ and $0 \to \mathbb{F}_2\xrightarrow{\sim}\mathbb{F}_2 \to 0$ and shifts of these. Each of these has an obvious lift. | |
| Apr 30, 2023 at 15:36 | history | asked | unknown | CC BY-SA 4.0 |