Timeline for Quick computation of the Pontryagin dual group of torus
Current License: CC BY-SA 3.0
9 events
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| Feb 27, 2012 at 3:52 | comment | added | Noam D. Elkies | @michael-grade83: Use orthogonality of characters. Any character $\chi$ is uniformly approximable by a linear combination of the $e^{2\pi i k}$. If $\chi$ were not itself one of the $e^{2\pi i k}$ then $\chi$ would be orthogonal to functions arbitrarily close to itself, which is a contradiction. Likewise on the $d$-torus, with "$k$" interpreted as an element of ${\bf Z}^d$ rather than ${\bf Z}$. | |
| Feb 26, 2012 at 9:42 | comment | added | user21706 | @Noam D. Elkies I don't understand how conclude your argument. Yes, by Stone-Weierstrass theorem the "finite trigonometric polynomials" are dense in $C(\mathbb{T}, \mathbb{C})$ but how it mean that all characters of $\mathbb{T}$ are of the type $e^{2 \pi k}$? | |
| Feb 26, 2012 at 5:55 | comment | added | Kerry | @KConrad: Thanks, this is very clear. | |
| Feb 25, 2012 at 23:34 | comment | added | Noam D. Elkies | If you allow Stone-Weierstrass, all you need is that the "finite trigonometric polynomials" (= Laurent polynomials = finite linear combinations of $e^{2\pi i n x}$) constitute a $\bf C$-algebra of continuous functions on the circle that separates points and is closed under complex conjugations. "Separates points" is easy because $e^{2\pi i x}$ already separates points, and "closed under conjugation" is immediate. You can handle the $d$-torus the same way, using the fact that the $e^{2\pi i x_j}$ together already separate points. | |
| Feb 25, 2012 at 23:25 | history | edited | KConrad | CC BY-SA 3.0 |
added 171 characters in body
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| Feb 25, 2012 at 23:10 | comment | added | KConrad | @Changwei: you are not being too naive. The continuous homomorphisms from the unit circle to itself are the power maps $z \mapsto z^n$ for integers $n$, which compose with each other the same way that integers add (in the exponent), so the dual group of the unit circle is the integers. | |
| Feb 25, 2012 at 22:53 | vote | accept | user21706 | ||
| Feb 25, 2012 at 21:42 | comment | added | Kerry | Excuse me if I am being too naive: Is not this saying the representation of $S_{1}$ to $S_{1}$ is isomorphic to the integers? | |
| Feb 25, 2012 at 21:23 | history | answered | KConrad | CC BY-SA 3.0 |