3
$\begingroup$

I am reading these lecture notes By Daniel Bump about Character Theory on Abelian Groups. If $G$ is a group, then $G^*$ denotes its characters, the set of homomorphisms $\pi : G \to \mathbb C^{\times}$, where $\mathbb C^{\times}$ denotes the multiplicative group of the complex numbers. At the end he introduces Fourier Analysis on Abelian groups and mentiones with regard to the isomorphism of $G$ and $G^{\ast}$.

However $G$ and $G^{\ast}$ may or may not be isomorphic. We have seen that they are isomorphic if $G$ is finite; or if $G = \mathbb R$ (the additive group) or $\mathbb Q_p$ (the additive group of $p$-adic numbers), then $G \cong G^{\ast}$. But if $G = \mathbb R / \mathbb Z$ then $G^{\ast} = \mathbb Z$, and it is in this setting that most people first encounter Fourier analysis.

The functions $e^{inx}$ for $ \in \mathbb Z$ are characters, i.e. homomorphic functions from $\mathbb R / \mathbb Z$ to $\mathbb C$ (or form $\mathbb R$ with the restriction of being periodic), and in analysis we are interested in what funtion $f : \mathbb R \to \mathbb C$ that are periodic (i.e. which could be identified with function on $\mathbb R/\mathbb Z$) have a series representation in $\{ e^{inx} \}_{n \in \mathbb Z}$ (or the trigonometric functions as it is sometimes presented, but then I guess it should be isomorphic to $\mathbb N\times\mathbb N$, i.e. the positive coefficient for the $\sin$- and $\cos$-terms).

As written on Wikipedia:Convergence of Fourier Series, if a periodic function $f : \mathbb R \to \mathbb C$ is of bounded variation, then its Fourier series converges everywhere. But here is my question, as said above, the set of homomorphisms from $\mathbb R / \mathbb Z$ is isomorphic to $\mathbb Z$, but functions of bounded variation are not neccessariliy homomorphisms, i.e. they do not have to satisfy $f(x+y) = f(x)f(y)$, they are a more general class of functions. But when the Fourier Series is unique in this case, i.e. we have a bijection between the functions of bounded variation and elements from $\mathbb Z$ (i.e. the series), then we could not have an bijection between $\mathbb Z$ and the more restricted class of homomorphism, as some preimages of $\mathbb Z$ are not homomorphisms at all?

Hope someone could clarify in what sense every homomorphism $\mathbb R / \mathbb Z$ corresponds to an element from $\mathbb Z$, and what is their relation to the "ordinary" Fourier series everyone learns about in Analysis courses, which is not restricted to just homomorphic functions, but more general classes of periodic functions.

$\endgroup$
  • 1
    $\begingroup$ The homomorphisms are as you said the $x \mapsto e^{in x}$ for $n\in \mathbb{Z}$. But the Fourier series doesn't give you a bijection between the functions of bounded variation and $\mathbb{Z}$, it gives you an injection from the set of functions of bounded variation into the set of functions $\mathbb{Z}\to \mathbb{C}$, a function is mapped to the (double) sequence of its Fourier coefficients. $\endgroup$ – Daniel Fischer Oct 9 '15 at 12:50
  • $\begingroup$ Mhh, there seems something wrong. In the lecture notes it is written that if $G = \mathbb R /\mathbb Z$, then $G^{\ast} = \mathbb Z$, that means each homomorphism $G \to \mathbb C^{\times}$ should correspond to an element from $\mathbb Z$, should it be $G^{\ast} = \{ \varphi : \varphi : \mathbb Z \to \mathbb C \}$ instead? $\endgroup$ – StefanH Oct 9 '15 at 12:56
  • 1
    $\begingroup$ No, $G^{\ast}$ is the group of (continuous) homomorphisms $G \to \mathbb{C}^{\times}$, and that's isomorphic to $\mathbb{Z}$ just fine. What's wrong is "But when the Fourier Series is unique in this case, i.e. we have a bijection between the functions of bounded variation and elements from $\mathbb{Z}$", the Fourier series maps each function to the (double) sequence of its Fourier coefficients, not to an element of $\mathbb{Z}$. $\endgroup$ – Daniel Fischer Oct 9 '15 at 13:04
  • $\begingroup$ Okay, guess I have confused something here. But how does this isomorphism looks, in what sense is an element from $\mathbb Z$ a homomorphism $G \to \mathbb C^{\times}$? $\endgroup$ – StefanH Oct 9 '15 at 13:33
  • 1
    $\begingroup$ We get an isomorphism between $L^2(G)$ and $\ell^2(\mathbb{Z})$. But we have Fourier series for more general functions (for $f\in L^1(G)$, the Fourier series is directly defined, but as far as I know, there is no good characterisation of the sequences you get as Fourier coefficients of $L^1$ functions, nor which are the coefficients of continuous functions). $\endgroup$ – Daniel Fischer Oct 9 '15 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.