Why do laplace and z and fourier transforms work?

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This may a little bit opaque, but maybe it's of some value.

All questions of "why the function E(x):=e^x" come down to
>because it's the group homomorphism between the two operations in the ring:
E(x+y) = E(x) · E[y]

Consider the collection of all complex numbers of norm 1.
This is e^{iu} for u in [0, 2pi].
It's a representation of U(1), which lives in the category of locally compact abelian groups.

For any group G, the set Hom(G,U(1)) inherits a group structure from U(1). (If you can't see what this means, ask.)
This is called the dual group and its elements are called "characters" X.

For example, consider R with addition. We'll see that this group is actually self-dual. Fix a real number p and then e.g. [math] x \mapsto e^{ipx} [/math] is an element [math] X_p [/math] of Hom(R,U(1)). It's a homomorphism, because [math] e^{ipa} · e^{ipb}= e^{ip(a+b)} [/math] . We can use the multiplication of phases in U(1) to define a group structure * on these X's by
[math]X_p*X_q \equiv (x \mapsto e^{ipx}) · (x\mapsto e^{iqx}) := x \mapsto e^{i(p+q)x} [/math].
We have a X_p for all real numbers p and so, in this case, as promised, we find Hom(R,U(1)) = R.
(That "new" R from constructing the dual is the domain of Fourier space!)

Now since we see Hom(-,U(1)) maps groups to groups, we can consider it as an auto-functor in the category of locally compact abelian groups.
It now turns out that the twice applied functor [math] Hom(Hom(-,U(1)),U(1)) [/math] is naturally isomorphic to the identity functor. Actually: If X in Hom(G,U(1)), then we can map g\in G to [math]g\mapsto (X\mapsto X(g))[/math].

Since locally compact groups have the Haar measure, one can form the space of integrable functions G -> C and carry it along with these functors. This procedure is the Fourier transform and it really works for all groups in this category! E.g. for U(1) itself, i.e. the periodic interval, we find that the dual group is Z and the associated transform is the Fourier series.

Here's a more workshop elaboration of the above:
en.wikipedia.org/wiki/Pontryagin_duality
en.wikipedia.org/wiki/Circle_group

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Now here a take on Laplace transforms that comes from a very different angle:

We're interested in series coefficients [math]a_n[/math] vs. inverse integral transforms [math]a(t)[/math].
We may compare the sum and integral of monomials [math] \sum_{n=0}^\infty z^n [/math] and [math]\int_0^\infty z^t\,{\mathrm d}t[/math].
We find they differ by Euler-MacLaurin sum terms 1/2, -1/12 etc:

[math]\sum_{n=0}^\infty z^n=\dfrac{1}{1-z}[/math]

[math]\int_0^\infty z^t\,{\mathrm d}t=\int_0^\infty {\mathrm e}^{t\log(z)}\,{\mathrm d}t=-\dfrac{1}{\log(z)}=-\dfrac{1}{1-z}-\dfrac{1}{2}+\dfrac{1}{12}(z-1)+{\mathcal O}\left((z-1)^2\right).[/math]

To match the integral computation to the sum, i.e. [math]-\log(z)\leftrightarrow 1-z[/math], we would first have to perform the conformal mapping [math]z\mapsto {\mathrm e}^{z-1}[/math]. Indeed,

[math]\int_0^\infty ({\mathrm e}^{z-1})^t\,{\mathrm d}t=\int_0^\infty {\mathrm e}^{(z-1)t}\,{\mathrm d}t=\dfrac{1}{1-z}[/math].

In other words, the smooth version of summands given by pairing [math]a_n z^n[/math] doesn't so much correspond to the pairing [math]a(t)\,z^t[/math] under the integral, but rather [math]a(t)\,{\mathrm e}^{(z-1)t}[/math].
Make that [math]a(t)\,{\mathrm e}^{-st}[/math] after a shift [math]z\mapsto 1-s[/math], so that [math]a(t)=1[/math] is connected to [math]\dfrac{1}{s}[/math].

If [math]a_0,a_1,a_2,\dots[/math] is a series, the function

[math]G[a](z)=\sum_{n=0}^\infty a_nz^n[/math]

is called it's generating function.
If [math]a(t)[/math] is a function, the corresponding function (shifted by one) is the Laplace transform

[math]L[a](s):=\int_0^\infty a(t)\,{\mathrm e}^{-st}\,{\mathrm d}t[/math].

For those two, the constant series/function [math]a_n=a(t)=1[/math] both lead to [math]\frac{1}{1-z}=\frac{1}{s}[/math].

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What about the Fourier transform for compact non-abelian groups as given via the Peter-Weyl theorem? How can you reconcile this with the locally compact abelian case? Is there a generalization encompassing both, and more definitions of FT?

en.wikipedia.org/wiki/Gelfand_representation

en.wikipedia.org/wiki/Wavelet_transform

en.wikipedia.org/wiki/Pontryagin_duality

en.wikipedia.org/wiki/Cartier_duality

en.wikipedia.org/wiki/Tannaka–Krein_duality

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*smiley face*

youtube.com/watch?v=r18Gi8lSkfM

Huly fug

Thanks bro. I always avoided C*-algebras. Looks like I have to overcome my dislike for algebra now.

Just look at this.

ncatlab.org/nlab/show/Fourier-Mukai transform

Wait wat