Timeline for Quick computation of the Pontryagin dual group of torus
Current License: CC BY-SA 3.0
8 events
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| Feb 25, 2012 at 20:33 | comment | added | GH from MO | Personally I think that finding the characters of $\mathbb{T}^m$ is no easier than finding the continuous homomorphisms $\mathbb{R}^m\to\mathbb{T}^n$. But of course I am also open to see alternate proofs, some of which might be simpler than the one I outlined. | |
| Feb 25, 2012 at 20:24 | comment | added | user21706 | Yes, Bourbaki's General Topology Part 2 do too much things for me. I'm writing a paper and I only need to find the characters of $\mathbb{T}^d$ quickly. I can use Stone-Weierstrass theorem because I must use it in another part of the paper. | |
| Feb 25, 2012 at 20:18 | comment | added | GH from MO | What do you mean by "a more contained proof"? Bourbaki's General Topology Part 2 finds, with a detailed proof, all the continuous homomorphisms $\mathbb{R}^m\to\mathbb{T}^n$ (Proposition 3 on Page 81). From this statement (whose proof is straightforward) you can find all continuous homomorphisms $\mathbb{T}^m\to\mathbb{T}^n$ by composing them with the natural map $\mathbb{R}^m\to\mathbb{T}^m$. Finally, the special case $n=1$ gives the Pontryagin dual of $\mathbb{T}^m$ as given in your original post. It does not get simpler than that! | |
| Feb 25, 2012 at 20:07 | comment | added | user21706 | OK, thanks, anyway if someone find a more contained proof I'll be happy. | |
| Feb 25, 2012 at 19:58 | comment | added | GH from MO | $\mathbb{R}/\mathbb{Z}$ is clearly continuously isomorphic to the circle group (the isomorphism being $t\mapsto\exp(2\pi i t)$), so we talk about the same thing (specialize my post to $n=1$). For the details of the proof you should look up the given source, it takes only few pages and proves much more. Fancy tools like Stone--Weierstrass are not used in the proof. | |
| Feb 25, 2012 at 19:50 | comment | added | user21706 | In my setting the dual group of $G$ is the group of continuous homomorphisms $G \to \mathbb{T}^\times$, where $\mathbb{T}^\times = \{z \in \mathbb{C} : |z| = 1 \}$ is the circle group. So if I understand you say that if $\chi \in \widehat{\mathbb{T}^d}$ then I can extend $\chi$ to a continuous homomorphism $\chi^\prime : \mathbb{R}^d \to \mathbb{R}$. But I don't know how to do this, however I know how extend $\chi$ to a continuous homomorphism $\chi^\prime : \mathbb{R}^d \to \mathbb{T}^times$ by periodicity, I mean $\chi^\prime (x) = \chi(\mbox{fractional part of } x)$. | |
| Feb 25, 2012 at 19:27 | history | edited | GH from MO | CC BY-SA 3.0 |
added 4 characters in body
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| Feb 25, 2012 at 19:11 | history | answered | GH from MO | CC BY-SA 3.0 |