Is it possible to describe all integral transforms where the inverse transform is implemented by the same formula (with maybe a sign flipped somewhere). Fourier is such an example obviously. I am seeing this in the case for the explicit formulae of the Fourier-Harish-Chandra transform on some group manifold, but is this the case only in these examples or more generally?
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$\begingroup$ You get for free a lot of examples from (time reversible) PDEs, especially since you allow a "sign flip". For example, consider the free propagator for the linear Schrodinger equation $U(t) = e^{it \Delta}$ which can be expressed as an integral operator. You have that the backwards propagator is $U(-t) = \overline{U(t)}$. Similar things happen for the wave and Klein-Gordon equations. $\endgroup$– Willie WongCommented Jan 28, 2019 at 21:25
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$\begingroup$ Clever answer. The Fourier transform itself is used to obtain the Schroedinger propagator, by starting with $ U(t) = e^{-itk^2/2}$, then the Fourier gives the propagator $K = \frac{1}{{2\pi i t} e^{ix^2/t}$. And yes, $U(-t) * U(t) = 1$, but the Fourier transform itself not an example of this phenomenon as far as i can tell. $\endgroup$– KphysicsCommented Feb 2, 2019 at 9:01
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$\begingroup$ A self-dual integral transform is analogous (equivalent, even) to a unitary operator on a finite-dimensional space. These are 'coordinate transformations' in the sense that one basis of orthonormal functions is transformed to another. Therefore any transformation between function bases (e.g. wavelets are a basis, Fourier functions are a basis, dirac-delta distributions are a basis, probably many others) will be unitary. Measures between functions (which I think are analogous to symmetric postiive definite matrices) will have their own sets of self-dual operators I think. $\endgroup$– MyridiumCommented Jan 8, 2021 at 4:11
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