# $\varphi$ in $\operatorname{Hom}{(S^1, S^1)}$ are of the form $z^n$

I'd like to see a proof why $\varphi \in \operatorname{Hom}{(S^1, S^1)}$ looks like $z^n$ for an integer $n$.

At first I thought I could argue that if I have a homomorphism that maps $e^{ix}$ to some $e^{iy}$ for $x,y \in \mathbb R$ then $z = rx$ for some real number $r$. But on second thought I'm not sure why I can't have $\varphi (e^{ix} ) = e^{g(x)i}$ for a $g$ other than $g(x) = \lambda x$.

I'm also interested in seeing different proofs. I'm sure there are several ways to prove this.

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You need to invoke continuity somewhere. There is a proof in § 15 of these notes. – Zhen Lin Mar 23 '12 at 12:32
@ZhenLin Heh. "...we need a couple of lemmas..." -- I thought this proof was going to be a one liner. Thanks for the link. So that's one way of proving this. – Rudy the Reindeer Mar 23 '12 at 12:36
Yes, it feels like something that ought to be "easier" (assuming continuity is the added to the conditions.) – Thomas Andrews Mar 23 '12 at 13:25

Let $f : S^1 \to \mathbb{C}^*$ be a continuous(!) group homomorphism. I claim that $f(z)=z^n$ for some fixed $n \in \mathbb{N}$.

First of all, $f$ corresponds to a continuous group homomorphism $g : \mathbb{R} \to \mathbb{C}^*$ which is constant $1$ on $\mathbb{Z}$ (since $S^1 \cong \mathbb{R}/\mathbb{Z}$), via $g(t)=f\left(e^{2\pi i t}\right)$. There is some $a \in \mathbb{R}$ with $A:=\int_{0}^{a} g(t) dt \neq 0$ (otherwise the derivative $g(a)$ vanishes for all $a$, which is impossible since $g(0)=1$). For every $x \in \mathbb{R}$ it follows $A^{-1} \int_{x}^{x+a} g(t) dt = A^{-1} \int_{0}^{a} g(x) g(t) dt = g(x)$. In particular $g$ is differentiable and satisfies the differential equation $(Dg)(x)=A^{-1} (g(x+a)-g(x)) = A^{-1} (g(a)-1) g(x)$. Thus, there is some $b \in \mathbb{C}$ such that $g(x)=e^{bx}$ for all $x$. Since $1=g(1)=e^{b}$ it follows that $b=2\pi i n$ for some $n \in \mathbb{Z}$, meaning $f(z)=z^n$.

Remark: There are lots of non-continuous group homomorphisms $S^1 \to \mathbb{C}^*$. The reason is that some infinite-dimensional linear algebra and the theory of divisible abelian groups implies that there is an isomorphism of abelian groups $S^1 \cong \mathbb{Q}/\mathbb{Z} \oplus \mathbb{R}^{\oplus \mathbb{R}}$, and there are lots of group automorphisms of $\mathbb{R}^{\oplus \mathbb{R}}$.

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I don't understand your explanation why $\int_0^a g(t) dt$ can't be zero. Could you elaborate? Thanks! – Rudy the Reindeer Mar 23 '12 at 16:35
$a \mapsto \int_{0}^{a} g(t) dt$ is differentiable with derivative $a \mapsto g(a)$ (fundamental theorem of calculus). – Martin Brandenburg Mar 23 '12 at 17:13
I see. Thanks! And thanks for your answer! – Rudy the Reindeer Mar 24 '12 at 11:04

If you assume continuity then this follows fairly quickly from the fact that a continuous homomorphism between Lie groups is actually a Lie group homomorphism (i.e. it is automatically smooth). So if $\phi : S^1 \to S^1$ is a continuous group homomorphism then we can consider its differential on the Lie algebra $\phi_* : \mathbb R \to \mathbb R$. Being linear this has to have the form $x \mapsto cx$ for some $c \in \mathbb R$. But by properties of the exponential map of Lie groups, we have $$\phi(e^{ix}) = e^{i\phi_* x} = e^{icx}.$$ And now for this to be well-defined, we need $c \in \mathbb Z$.

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Plus one. I can't verify your answer because I don't know enough about lie groups. But I'm grateful that someone posted an alternative proof. – Rudy the Reindeer Mar 24 '12 at 11:05

I'm going to post the proof given in the notes posted by Zhen rewritten in my own words:

First note that the proof needs the homomorphisms $S^1 \to S^1$ to be continuous.

(i) If $\alpha : \mathbb{R} \to \mathbb{R}$ is a continuous homomorphism then $\alpha$ is of the form $x \mapsto \lambda x$ for some $\lambda \in \mathbb{R}$. This follows directly from the fact that $\alpha$ is a linear map and one dimensional matrices are multiplication by scalars.

(ii) Continuous homomorphisms $\mathbb{R} \to S^1$ are of the form $e^{i \lambda x}$ for $\lambda \in \mathbb{R}$. To see this note that $(e^{ix}, \mathbb{R})$ is a covering space of $S^1$. Then by the unique lifting property we get that for a continuous homomorphism $f: \mathbb{R} \to S^1$ there is a unique continuous homomorphism $\alpha : \mathbb{R} \to \mathbb{R}$ such that $f = g \circ \alpha$ where $g (x) = e^{ix}$ is the covering map. By (i) we get that $f$ has to be of the form $x \mapsto e^{i\lambda x}$.

(iii) If $\varphi : S^1 \to S^1$ and $\psi : \mathbb{R} \to S^1$ are continuous homomorphisms then so is $\varphi \circ \psi : \mathbb{R} \to S^1$. So we know that $1 = \varphi (\psi (0))$. We also know $\psi$ has to map $0$ to $1$ hence $\psi (0) = e^{i 2 \pi k}$ for some $k \in \mathbb{Z}$. And we also know that $1 = e^{i 2 \pi n}$ for some $n \in \mathbb{Z}$. Hence $\varphi (z) = z^m$ for some $m \in \mathbb{Z}$.

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So this proof replaces the integral arguments in my proof with the help of covering spaces. – Martin Brandenburg Mar 24 '12 at 17:04

Other mechanisms: if one grants differentiability, and lift the source to the real line, then one finds that $\varphi$ satisfies a differential equatioon $\varphi'-c\varphi=0$... Another (not unrelated!) approach is to write a Fourier series for $\varphi$... which would include any merely-continuous $\varphi$, if one grants basic things about distributions and their Fourier expansions. Then the hom condition shows that the Fourier expansion has a single term, etc.

Edit: some details added. Again, continuity is assumed throughout. If one knows and/or proves differentiability, then $\varphi(x+y)=\varphi(x)\varphi(y)$ gives $\lim_{y\rightarrow 0} [\varphi(x+y)-\varphi(x)]/y= \varphi(x)\cdot \lim [\varphi(y)-1]/y$. This gives the differential equation for $\varphi$ on $\mathbb R$, parametrizing the circle by the exponential, for example. Thus, one finds all characters of $\mathbb R$. The ones that descend to the circle are the ones that are trivial on $2\pi\mathbb Z$.

Similarly, $\varphi(x)=\sum_n c_n e^{2\pi i nx}$ at least in an $L^2$ sense. It is not hard to determine that this is a group homomorphism only when there is a unique non-vanishing coefficient, and it is $1$.

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Thanks Paul, +1. Although I was hoping for slightly more detailed answers. : ) – Rudy the Reindeer Mar 25 '12 at 11:58
Nice, thank you. Another +1 from me for expanding on your answer even though I'd already accepted someone else's answer. – Rudy the Reindeer Mar 25 '12 at 20:46

$\newcommand{\Zobr}[3]{#1 \colon #2 \to #3}$I will try to add an "elementary" proof - not needing anything beyond the first course in general topology.

We will consider $(S^1,\cdot)$ as $([0,1),\oplus)$ where $\oplus$ is the addition modulo $1$.

Let $\Zobr f{[0,1)}{[0,1)}$ be any continuous homomorphism.

First notice that $$f(x)=0 \qquad \Rightarrow \qquad f(n\times x)=0. \tag{1}$$

We will consider several cases:

• First suppose that $\operatorname{Ker} f=\{0\}$, which means that $f$ is an injective map.
Since $\frac12\oplus\frac12=0$, we see that $f(\frac12)\oplus f(\frac12)=0$, hence $f(\frac12)\in\{0,\frac12\}$. Since $f$ is injective, the only possibility is $$f\left(\frac12\right)=\frac12.$$
What about $\frac14$? We have $f(\frac14)\oplus f(\frac14)=\frac12$. We have again two possibilities: $f(\frac14)\in\{\frac14,\frac34\}$. Now the possibility $f(\frac14)=\frac34$ would contradict to injectivity, since by intermediate value theorem we would have $x\in(0,1/4)$ such that $f(x)=\frac12$. So we have $$f\left(\frac14\right)=\frac14.$$
By repeating of this argument we get $$f\left(\frac1{2^n}\right)=\frac1{2^n}$$ and, using (1), we get $$f\left(\frac{k}{2^n}\right)=\frac{k}{2^n}.$$ Since $\{\frac{k}{2^n}; k,n\in\mathbb N\}$ is dense in $[0,1)$ and $f$ is continuous, we get that $f$ is the identity map.

• Now suppose that $\operatorname{Ker} f\ne\{0\}$, i.e., there exists a non-zero $x$ with $f(x)=0$. Let us denote $$x_0=\inf \{x\in(0,1); f(x)=0\}.$$ By continuity $f(x_0)=0$.

• We show that $f(x_0)=0$ implies that $f\equiv0$. Suppose that there is a point such that $f(x)\ne 0$. Then there exists an $\varepsilon>0$ such that such that $f(y)\ne 0$ for each $y\in(x-\varepsilon,x+\varepsilon)$. Since $\inf \{x\in(0,1); f(x)=0\}=0$ there exists $x_1\in(0,\varepsilon)$ such that $f(x_1)=0$ and, consequently, $f(kx_1)=0$ for each integer $k$. Obviously, there exists $k$ such that $kx_1\in(x-\varepsilon,x+\varepsilon)$, which is a contradiction.

• So the only remaining case is that $x_0>0$. We claim that $x_0$ must be of the form $x_0=\frac1n$ for some $n\in\mathbb N$. To see this, consider the multiples $x_0, 2x_0,\dots,nx_0$, where $n$ is the smallest positive integer such that $nx_0\ge 1$. If the inequality $nx_0>1$ would be strict then $nx_0-1$ would belong to $\operatorname{Ker} f$ and it would be smaller than $x_0$, which contradicts the choice of $x_0$.
When we already know that $x_0=\frac1n$, it is obvious that is suffice to describe the map $f$ on the interval $[0,\frac1n)$. (Since $f$ is periodic with the period $\frac1n$.)
Let us consider interval $[0,\frac1n)$ with the operation $\oplus_n$, the addition modulo $\frac1n$. If we show that $\Zobr f{([0,\frac1n),\oplus_n)}{([0,1),\oplus)}$ is a homomorphism, then we have reduced this to the first case (since $[0,\frac1n),\oplus_n)$ is isomorphic to $S^1$, too). In this case the map $f$ will be $f \colon x\mapsto nx$.
So it only remains to check the definition of homeomorphisms. We get $$f(x\oplus_n y)=f(x\oplus y)=f(x)\oplus f(y).$$ The first equality holds since the difference between $x\oplus_n y$ a $x\oplus y$ is a multiple of $\frac1n$, and $f(\frac1n)=0$.

Whenever $\Zobr f{\mathbb R}{\mathbb R}$ is such that $$f(x+y)=f(x)+f(y)$$ and $f(1)\in\mathbb Z$, then it induces in a natural way a homomorphisms $\Zobr{\tilde f}{\mathbb R/\mathbb Z}{\mathbb R/\mathbb Z}$. From discontinuous solution of Cauchy equation with this property we can get many discontinuous homomorphisms from $S^1$ to $S^1$.

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I have the feeling that this approach reminds me strongly of some other question here at MSE, but I can't recall which question it was. – Martin Sleziak Nov 5 '12 at 19:23