I am trying to solve a problem from the book "Fourier series and integrals" by McKean. :)


Prove that the group $\mathbb{Z}^\wedge$ (the dual to $\mathbb{Z}$) is isomorphic to $S^1$.

I have been given that $e_n\mapsto n$ gives an isomorphism between $(S^1)^{\wedge}$ and $\mathbb{Z}$. I tried to kinda mimic the idea and I ended up constructing the function:

$$\begin{cases}f:S^1\to\mathbb{Z}^{\wedge}\\ \alpha\mapsto \chi_{\alpha}. \end{cases}$$

In the book, they defined $S^1$ as "The circle $[0,1)$ under addition modulo $1$". When i read the sentence, I interpret it as the operation is addition on the unit circle, that is, if $\alpha,\beta\in S^1$ then $\alpha+\beta\in S^1$. But I don't think this is the correct interpretation of it, since $\alpha=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\beta=1$ are both on the unit circle, but $\alpha+\beta=\frac{3}{2}+\frac{\sqrt{3}}{2}$ and $|\alpha+\beta|=\sqrt{3}$ which shows us that $\alpha+\beta\not\in S^1$. Since I know multiplication of complex numbers on the unit circle gives us a group, I ended up thinking that's what they mean. Even though I don't know how I could interpret the above sentence as multiplication.

What we now have to do is to prove $f$ is a homomorphism, bijective, continuous and that its inverse is continuous.


For the homomorphism part, we have $f(\alpha\beta)=\chi_{\alpha\beta}(z)=(\alpha\beta)^z=\alpha^z\beta^z=f(\alpha)f(\beta)$.

$\text{ }$


I have two suggestions for the injectivity. Either we consider the equation $f(\alpha)=f(\beta)$ and try to conclude $\alpha=\beta$ or we try to prove kernel is trivial.

For the first suggestion, we have that $f(\alpha)=f(\beta)\iff\alpha^z=\beta^z\iff \alpha^z-\beta^z=0$ and I know a factor to $\alpha^z-\beta^z$ will be $\alpha-\beta$. Perhaps we can use it to conclude $\alpha=\beta$?

For my second suggestion, we have that $f(1)=\chi_1$ and $\chi_1(z)=1^z=1$, this shows us $1\in S^1$ gets mapped to the identity in $\mathbb{Z}^{\wedge}$. Perhaps it is trivial that this is the only element which gets mapped to $1\in\mathbb{Z}^{\wedge}$ and this is the better approach?

I have actually no suggestions or idea how to prove the map is surjective.

$\text{ }$


We have to check $f(\alpha)=\alpha^{z}$ is continuous for all $z$. But since $f(\alpha)=\alpha^z$ is a function from $S^1$ to $S^1$. Since it is a restriction of the map $\mathbb{C}\to\mathbb{C}$ defined by $\alpha\mapsto\alpha^z$, which is continuous, the map $f(\alpha)=\alpha^z$ is also continuous.

$\text{ }$

Continuity - Inverse function:

Unfortunately, I do not even know where to begin here. How Should I construct an inverse function? Can I "just" define it as $\chi_{\alpha}\mapsto\alpha?$ How do I prove it is continuous?

My questions are:

  • Is my interpretation of the group structure correct? If not, can you guide me in the right direction to help me understand this problem and be able to solve it?
  • Is my solution for the homomorphism part and the continuity correct?
  • What do you think about the proof of the injection? Do you think you can help me conclude it is, indeed, injective?
  • Can you help me with the surjection and the inverse map?

I appreciate any help, thanks!


"Addition modulo $1$" is not the same as addition of real numbers. Also, where you say $\frac 1 2 + \frac{\sqrt3} 2$ is on the unit circle, I suspect you meant $\frac 1 2 + i \frac{\sqrt3} 2.$ Not notice that $\frac 1 2 + i\frac{\sqrt3} 2 = \cos\frac \pi 3 + i\sin\frac\pi 3,$ and $\frac \pi 3 = \frac{\text{whole circle}} 6.$ The number within $[0,1)$ to which $\frac 1 2 + i\frac{\sqrt 3} 2$ corresponds is therefore $\frac 1 6 = \frac{\pi/3}{\text{whole circle}}. \vphantom{\dfrac{\displaystyle\sum}1}$ Suppose you have $\alpha,\beta\in[0,1).$ Then

\begin{align} & \Big(\cos(\alpha\times\text{whole circle)} + i\sin(\alpha\times \text{whole circle}) \Big) \\ \times {} & \Big( \cos(\beta\times\text{whole circle)} + i\sin(\beta\times \text{whole circle}) \Big) \\[10pt] = {} & \cos((\alpha+\beta)\times\text{whole circle)} + i \sin((\alpha+\beta)\times\text{whole circle}), \end{align} where $\alpha+\beta$ is the sum modulo $1$ of $\alpha$ and $\beta.$ For example, if $\alpha= 0.8$ and $\beta = 0.9,$ then the ordinary sum of $\alpha$ and $\beta$ is $1.7,$ so the sum modulo $1$ of $\alpha$ and $\beta$ is $0.7.$

  • $\begingroup$ Thank you for catching that, I meant $\frac{1}{2}+\frac{\sqrt{3}}{2}i$. :) Also, thank you for the clarification, what is meant by modulo. I have to think about what you wrote for a little while. But now, when you say it, to think about modulo in the way you described it in this answer should really have been the thing which first appeared in my mind. It's like when you work in, for instance, $\mathbb{Z}_3$.There we have $2+2=4\equiv 1$ (mod $3$). So perhaps I should have brought my inspiration from there! :) $\endgroup$ – user511893 Jan 21 '18 at 15:35

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