Timeline for Group cohomology of Q/Z
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 18, 2020 at 4:05 | vote | accept | qspt | ||
| Jul 18, 2020 at 4:04 | comment | added | qspt | I believe your answer shows that G cannot be $\mathbb{Q}/\mathbb{Z}$, because $\mathbb{Q}/\mathbb{Z}$ doesn't have any injections from $\mathbb{Z}$ whereas $A$ does. So this answers my question as far as the cohomology of $\mathbb{Q}/\mathbb{Z}$ is concerned. | |
| Jul 18, 2020 at 3:54 | comment | added | qspt | Perhaps I should submit a second question for this, but the original problem that led me to consider this was the following: For what G is $$ H^i(G) = \begin{cases} \mathbb{Z}, & i = 0,\\ 0, & i \equiv 1 \mod 2,\\ \mathbb{Q}/\mathbb{Z}, & i\equiv 0 \mod 2, i>1, \end{cases}?$$ | |
| Jul 18, 2020 at 3:32 | comment | added | qspt | I'm looking into the link you posted, but my conjecture before asking was that $A=\mathbb{Q}/\mathbb{Z}$. | |
| Jul 18, 2020 at 3:28 | comment | added | LSpice |
TeX note: rather than leaving math mode $H^2(\mathbb Q) \cong$ Ext$(\mathbb Q, \mathbb Z)$ $H^2(\mathbb Q) \cong$ Ext$(\mathbb Q, \mathbb Z)$, you can use \operatorname: $H^2(\mathbb Q) \cong \operatorname{Ext}(\mathbb Q, \mathbb Z)$ $H^2(\mathbb Q) \cong \operatorname{Ext}(\mathbb Q, \mathbb Z)$. Note the improved spacing and font matching. I have edited accordingly.
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| Jul 18, 2020 at 3:27 | history | edited | LSpice | CC BY-SA 4.0 |
Proofreading
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| Jul 18, 2020 at 3:11 | history | edited | Robert Kropholler | CC BY-SA 4.0 |
deleted 1659 characters in body
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| Jul 18, 2020 at 2:03 | review | First posts | |||
| Jul 18, 2020 at 7:46 | |||||
| Jul 18, 2020 at 2:03 | history | answered | Robert Kropholler | CC BY-SA 4.0 |