Timeline for Quick computation of the Pontryagin dual group of torus
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2015 at 14:18 | comment | added | LSpice | I'm not sure if I understand your response (particularly the second comment). I was suggesting that proving that $\widehat{\mathbb T}$ is a free Abelian group of rank 1 is not what @user21706 meant; rather, the goal is to show that the specific map $\mathbb Z \to \widehat{\mathbb T}$ given by $n \mapsto (z \mapsto z^n)$ is an isomorphism. | |
| Oct 18, 2015 at 19:15 | comment | added | Asaf | @LSpice, anyhow, the existence of roots you've mentioned is intimately related to the theory of divisible groups, whose structure theorem can be also achieved during the proof of the structure theorem of LCA groups. | |
| Oct 18, 2015 at 19:13 | comment | added | Asaf | @LSpice, on a rather abstract level, any homomorphisms between two Baire groups are automatically continuous. Moreover, even in the topological level, any rank $1$ subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$, I haven't claimed that by the so-called proof I've sketched you actually get the Pontryaign pairing (if you know that pairing beforehand, you can prove the isomorphism directly). | |
| Oct 16, 2015 at 3:44 | comment | added | LSpice | On the pure abstract-group level, even just knowing that the dual group is abstractly isomorphic to $\mathbb Z$ isn't enough, right? We've produced a rank-1 free subgroup of $\widehat{\mathbb T}$, but $\mathbb Z$ has proper such subgroups in abundance. (That is, we need the additional topological fact that the identity character $\mathbb T \to \mathbb C^\times$ given by $z \mapsto z$ has no proper root (that is a continuous homomorphism).) | |
| Feb 25, 2012 at 23:27 | history | answered | Asaf | CC BY-SA 3.0 |