@tb But for a finite cyclic group $G$ it's not $G \leftrightarrow S^1$, it's $G \leftrightarrow \mu_n$ where $\mu_n$ is the group of $n$-th roots of unity, right?
Since I live in India, I forsee a danger of "iyen" bug biting me! Why the hell is this guy crazy trying to but together some words in logical fasion and succeeding almost every time luckily?
$$\partial_{x}(\cos(\theta)x)=\cos(\theta)-\left(\sin(\theta)(\partial_x\theta)\right)x,$$ or this?
@anon If dependent, then $\cos(\theta)$ is not constant. That is correct. So we have to have the other term -- but I am unsure here, well I will have the other term nevertheless or? (it is just a zero term in independent case)
If you don't see that $\theta$ varies with $x$ then I'm not sure you understand polar coordinates. If you rotate something about the origin then of course it's x-coordinate will vary along the angle.
@MattN I don't see a duality here. You can send a cyclic group $G$ of order $n$ to $S^1$ by choosing a primitive $n$-th root of unity $\mu_n$ and sending a generator of $G$ to $\mu_n$. But neither the choice of generator nor the choice of $\mu_n$ are canonical.
:,( But for $G = \mathbb Z / n \mathbb Z$ I can define an isomorphism from $G \to \widehat{\mu_n}$ as follows: $k \mapsto (\omega \mapsto \omega^k)$, no? Also, $\mu_n \to \widehat{G}$ via $\omega \mapsto (1 \mapsto \omega)$.
@anon Perhaps this q shows the idea I cannot understand here, I cannot understand the premise about the $z$ there with $\hat e_R$. I am uncertain whether it is a geometric interpretation or some other way?
@MattN yes, you can. The point is that there are choices involved in this isomorphism. There is no distinguished element $1$ in $G$ and there is no distinguished primitive $n$-th root of unity. You have to pick them. That's what is meant by "not canonical".
No, the duality is given by $\hat{G} \times G \to S^1$, $(\gamma,g) \mapsto \gamma(g)$.
In this duality there's no choice involved. By definition an element of $\hat{G}$ is a homomorphism $G \to S^1$, so you can evaluate that homomorphism on $g \in G$. *This* is the pairing (analogous to the pairing you have between a vector space and its dual space).
But you can show that there is an isomorphism between a finite abelian group $G$ and its dual group, but there is no distinguished such isomorphism. The point is that you can reduce to the case of a cyclic group and the map you wrote down can be interpreted as giving such an isomorphism. (by fixing a generator of $G$ and declaring that it is sent to $\omega^k$ - this determines a unique homomorphism $G \to S^1$ and every homomorphism $G \to S^1$ is of this form).
@MattN No, that's not the case. The dual group of $S^1$ is $\mathbb{Z}$ and the dual group of $\mathbb{Z}$ is $S^1$. You cannot relate those groups directly via an isomorhpism, but they are related by the duality.
There is an isomorphism from $\widehat{\mathbb Z}$ to $S^1$ and from $\widehat{S^1} $ to $\mathbb Z$. So I thought by giving an isomorphism from $\widehat{G}$ to $\mu_n$ and from $\widehat{\mu_n}$ to $G$ I get my duality.
Well, the dream of getting both silver and bronze specialist badges at once is dead. No way I can fill 15 answers to previously unanswered questions in the next few days.
@MattN I think you got it quite right, but you're putting it a bit confusingly. Fix a a primitive root of unity $\omega$ in order to identify $\mu_n = \langle \omega \rangle$. Now a homomorphism $\mathbb{Z}/n\mathbb{Z} \to S^1$ is determined by $1 \mapsto \omega^k$ and every such homomorphism is of this form.
Conversely, every homomorphism $\mu_n \to S^1$ is determined by what it does on $\omega$ and it must send $\omega$ to $\omega^k$. If $k \equiv l \pmod n$ then $\omega^k = \omega^l$, so such a homomorphism is determined by an element of $G$. This gives you an identification of $\hat{\mu}_n$ with $G$ (or $\hat{\hat{G}}$, rather).
@MattN that is the actual duality. By picking generators of $\hat{G} = \langle \omega \rangle$ and $G = \langle x \rangle$ (here $G$ is finite cyclic) you sort of "coordinatize" this duality.
@AsafKaragila Supersized font is also quite annoying. It makes my chat move around. Of course, now that I said this you'll have to reply to this message with a supersized font message.
@MattN what is really going on is that the duality gives you a map $G \to \hat{\hat{G}}$ and Pontryagin's theorem tells you that this is an isomorphism. You make this isomorphism explicit by fixing generators.
Remember what is going on for vector spaces: you have a pairing $V^\ast \times V \to \mathbb{R}$ given by $(\varphi,x) \mapsto \varphi(x)$. This gives you a map $V \to V^{\ast\ast}$ by sending $v$ to evaluation at $v$.
