[kmill]

I think it's that exponentials simultaneously diagonalize time shifts, derivatives, and integrals. Convolution follows from these -- though there's something to be said for putting convolution above derivatives and integrals in importance.

(A deeper thing going on is that a periodic function can be thought of as a function whose domain is a circle. Circles have obvious rotational symmetry, and when you have symmetry you can use representation theory to decompose things into an (orthogonal) basis. In this case, rotations commute with each other so by some theory the decomposition is going to be entirely through eigenvectors of rotation, which happen to precisely be the exponential functions e^(n theta i) for n an integer. This decomposition is also an isomorphism that carries convolutions to point-wise products in both directions. Also: if you make it so the circle is the complex unit circle, a Fourier transform is the idea that you can create a Laurent polynomial that extends the function to the complex plane minus the origin.)

[bonoboTP]

Nice, indeed at the deeper level it's that if you want to make time shifts equivalent to rotation, you need to move in a circular fashion.

Minor point: to me derivatives are just one specific linear time invariant operator, a kind of convolution (with a generalized function) so I think LTI is the thing we really care about.

[bollu]

Does the representation theoretic perspective / harmonic analysis also explain why the Laplace transform works? I would be interested to know if it's possible to pick a different compact group from S1 to recover the Laplace transform

[kmill]

I've been thinking about this since my original comment actually, but there are complexities from using the noncompact group R. The exponential functions are still the simultaneous eigenvectors of time shifts.

The Laplace transform seems to be just using the fact that <f,g> = integrate(f(x) g(x), x from 0 to infinity) is an inner product for the space of square integrable functions (probably better would be <f,g> = integrate(f(x) conj(g(x)), x from 0 to infinity) as a Hermitian product). The various e^(ax) functions are linearly independent, so the functions g |-> <e^(ax), g> are linearly independent functionals. If the exponential functions are actually enough, then this means you can study a function by studying the vector consisting of its value through all the functionals, which is the Laplace transform.

The Laplace transform has a pretty bad inverse formula, partly because the exponential functions are not orthogonal with respect to the inner product.