# Question about Fourier series

The Fourier series of a function $f: G \to \mathbb C$ where $G$ is a group is the representation of $f$ in terms of characters $\chi_g \in \mathrm{Hom}(G, S^1)$ of $G$.

I understand the case where $G$ is finite and discrete. Now I'm trying to generalise to periodic functions on $\mathbb R$ but I'm not so sure what I'm thinking is correct.

If $f: \mathbb R \to \mathbb C$ is a $[-R,R]$-periodic function, we set $G = \mathbb R / 2R\mathbb Z \cong [0, 2R)$. Then since $f$ is $2R$-periodic we can shift this by $R$ consider $G = [-R, R)$, makes no difference.

In the discrete case $|G|=n$ the characters are $e^{\frac{2 \pi i k}{n}}$ for $0 \leq k \leq n$.

Now for some reason, the characters are $e^{i kx}$ for all $k \in \mathbb Z$. How do I see that these are the characters of $[-R,R)$?

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I'm not sure what sort of equivalence you mean by $\mathbb R/2R\mathbb R\cong [0,2R)$. The first is a topological group (which is what allows us to find characters) while the second is not, and the two are not even homeomorphic or homotopy equivalent as topological spaces. – Alex Becker Jul 13 '12 at 15:31
@AlexBecker I meant "group isomorphism". – Matt N. Jul 13 '12 at 15:32

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But it's unclear to me what the group operation would look like on that. And since the standard topologies on the two are non-homeomorphic, it's not an isomorphism of topological groups, so for the purposes of Fourier analysis the two are different. – Alex Becker Jul 13 '12 at 15:36
@AlexBecker The group operation on $[0,2R)$ is addition mod $2R$. Ok, then my question remains : ) – Matt N. Jul 13 '12 at 15:42

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Not if we give $[0,1)$ the subspace topology from $\mathbb R$, as it is then noncompact and simply connected while the circle is compact and not simply connected. The difference is that in $[0,1)$ points in small neighborhoods of $1$ are not in small neighborhoods of $0$. – Alex Becker Jul 13 '12 at 18:06
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## 1 Answer

What you have is equivalent to the circle group $\mathbb R/\mathbb Z$, and its characters are indeed $e^{ikx}$ for $k\in \mathbb Z$. It is easy to verify that these are continuous group homomorphisms from $\mathbb R/\mathbb Z$ to itself (which is the definition of a character). To see that these are all characters of $\mathbb R/\mathbb Z$, suppose that $f:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is a continuous homomorphism. Assume first $f$ is increasing. Let $k$ be the degree of $f$ and note that we can lift $f$ to a map $g:\mathbb R\to\mathbb R$ with $g(x)=f(x)+k\lfloor x\rfloor$. The map $g$ is in fact a continuous homomorphism, as if $\{x\}+\{y\}<1$ ($\{\cdot\}$ denotes fractional part) then $$g(x+y)=f(x+y)+k\lfloor x+y\rfloor= f(x)+f(y) + k\lfloor x\rfloor + k\lfloor y\rfloor=g(x)+g(y)$$ and otherwise $$g(x+y)=f(x+y)+k\lfloor x+y\rfloor=f(x)+f(y)-\lim\limits_{x\to 1^-}g(x) + k\lfloor x\rfloor + k\lfloor y\rfloor+k=g(x)+g(y)$$ as $\lim\limits_{x\to 1^-} g(x)=k$, which also proves continuity. Restricting our attention to $\mathbb Q$ we see that $n\cdot g(m/n)=g(m)=km$ so $g(m/n)=km/n$, thus $g|_{\mathbb Q}(x)=kx$. By continuity we have that $g(x)=kx$ on $\mathbb R$, and it follows that $f(x)=kx$. Viewing the circle as the group of unit complex numbers, this is $e^{ikx}$. Similarly, if $f$ is decreasing we get $e^{-ikx}$. Thus the characters of $\mathbb R/\mathbb Z$ are $e^{ikx}$ for $k\in\mathbb Z$.

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