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Let $G$ be a locally compact abelian group and $f:G \to \mathbb{C}$ a function. Its Fourier transform (when it exists) is defined to be $$\widehat{f}(\chi) = \int_G f(g) \bar{\chi}(g) \mathrm{d} g,$$ Here $\chi \in \widehat{G} := \mathrm{Hom}(G, S^1)$ and $\mathrm{d} g$ is a choice of Haar measure on $G$.

Why does $\bar{\chi}$ appear and not $\chi$? Is this for historical reasons or is there a mathematical reason?

In practice one considers all characters $\chi$ at once (for example when performing Poisson summation), so it doesn't really make a difference including the complex conjugate in the definition.

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  • $\begingroup$ this definition ensures that the Fourier transform is zero if $\chi\neq f$, which is a more convenient statement than $\bar{\chi}\neq f$ $\endgroup$
    – Carlo Beenakker
    Commented Apr 12, 2023 at 10:45
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    $\begingroup$ I'd say probably mostly historical - for example note that this way round ensures that the Fourier inversion formula looks like $f(x)=\int_{\widehat{G}}\widehat{f}(\chi)\chi(x) \mathrm{d}\chi$, so it depends whether you view the decomposition of $f$ into characters or the decomposition of $\widehat{f}$ into physical space as more fundamental. Also a reason for this convention is that then you can write $\widehat{f}(\chi)=\langle f,\chi\rangle$ (where we conjugate the second argument of the inner product for e.g. positivity reasons so $\langle f,f\rangle=\| f\|^2$). $\endgroup$
    – Thomas Bloom
    Commented Apr 12, 2023 at 10:51
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    $\begingroup$ @CarloBeenakker, I do not think that that is true. Surely you meant the Fourier transform $\hat f(\chi)$ is non-$0$ if $\chi = f$? $\endgroup$
    – LSpice
    Commented Apr 12, 2023 at 13:01
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    $\begingroup$ The reason is that $\int f\overline{g}$ is an Hermitian dot product, while $\int fg$ is not. $\endgroup$
    – Alexandre Eremenko
    Commented Apr 12, 2023 at 13:07
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    $\begingroup$ @AlexandreEremenko, re, @‍ThomasBloom also suggested this, but also pointed out that, whichever way you do it, either the transform or its inverse is going to be complex-conjugate-less, so I think this just shifts the question to: why the transform? (To which the answer is surely, 🤷.) One might also consider modular, or even just $\overline{\mathbb Q_\ell}$, representation theory, in which setting there is no natural analogue of conjugation. $\endgroup$
    – LSpice
    Commented Apr 12, 2023 at 13:50

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This is an $L^2$ product: consider for instance the most classical Fourier expansion of $\mathbb Z$ periodic functions (or distributions) of one real variable. You have for $f$ locally square integrable $$ f(x)=\sum_{k\in \mathbb Z}\langle f, e_k\rangle_{L^2} e_k(x), $$ that is $ f(x)=\sum_{k\in \mathbb Z}\left(\int_0^1 f(y){e^{-2iπ ky}} dy\right) e^{2iπ kx}, $ simply since $ \langle f, e_k\rangle_{L^2}=\int_0^1 f(y)\overline{e_k(y)}dy. $

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  • $\begingroup$ Thanks for the answer. I think this nicely combines some of the points raised in the comments regarding the relation to inner products and the lack of minus sign present in the inverse Fourier transform. $\endgroup$
    – Daniel Loughran
    Commented Apr 13, 2023 at 19:38
  • $\begingroup$ Incidently a version of my question is also raised on signal processing stack exchange: dsp.stackexchange.com/questions/19004/… $\endgroup$
    – Daniel Loughran
    Commented Apr 13, 2023 at 19:39

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