Timeline for morphism from a compact group to Z ?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2015 at 14:10 | comment | added | Todd Trimble | This is really a proof from the Book. Thanks very much. | |
| Jan 23, 2015 at 13:33 | history | edited | KConrad | CC BY-SA 3.0 |
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| Jan 23, 2015 at 12:54 | comment | added | Yemon Choi | I'm extremely happy to see an old question resurrected because of a new and informative answer, rather than merely bumped by reformatting or tweaking of existing answers | |
| Jan 23, 2015 at 11:28 | history | edited | Sean Eberhard | CC BY-SA 3.0 |
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| Jan 23, 2015 at 9:37 | comment | added | YCor | @LSpice I thought of a surjective homomorphism, which is no restriction... but $x\notin\phi^{-1}(\{0\})$ is fine. | |
| Jan 23, 2015 at 9:23 | comment | added | Sean Eberhard | @KConrad Thanks. It's clearer the way you suggest. | |
| Jan 23, 2015 at 9:23 | history | edited | Sean Eberhard | CC BY-SA 3.0 |
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| Jan 23, 2015 at 0:18 | comment | added | KConrad | You do not need $\phi(x) \not= 0$, as your argument directly shows every homomorphism $\phi \colon G \rightarrow \mathbf Z$ is identically $0$. Pick any $x \in G$ and your argument gives a homomorphism $\phi' \colon B\mathbf Z \rightarrow \mathbf Z$ where $\phi'(1) = \phi(x)$. You show $B\mathbf Z = B\mathbf R \times \widehat{\mathbf Z}$, and you give an argument that any homomorphism $B\mathbf R \rightarrow \mathbf Z$ is $0$ and any homomorphism $\widehat{\mathbf Z} \rightarrow \mathbf Z$ is $0$. Thus $\phi'$ is $0$, so $\phi(x) = \phi'(1) = 0$. Since $x \in G$ was arbitrary we're done. | |
| Jan 22, 2015 at 23:39 | comment | added | LSpice | @YCor, I think you mean $x \not\in \phi^{-1}(\{0\})$, right? | |
| Jan 22, 2015 at 22:43 | comment | added | YCor | Very nice. Let me slightly restate your argument: 1) replacing $G$ by the closure of some $x\in\phi^{-1}(\{1\})$, we can suppose that $G$ is abelian with a dense cyclic subgroup. 2) if $G$ is any connected compact abelian group then the result holds because $G$ is divisible (as a projective limit of tori) 3) now supposing $G$ has dense cyclic subgroup, $\phi$ vanishes on its unit connected component because of (2), hence we can suppose $G$ totally disconnected, hence a quotient of the profinite completion $\hat{Z}\simeq\prod_p\mathbf{Z}_p$, and your last argument finishes the job. | |
| Jan 22, 2015 at 22:20 | history | answered | Sean Eberhard | CC BY-SA 3.0 |