# The Geometry of the Slice Theorems

﻿﻿﻿When studying the Radon transform, we saw that we could reconstruct the Fourier transform of a function from its Radon Transform:

• The Fourier Transform integrates the product of a function with “waves” that are constant on hyperplanes.
• The Radon Transform computes the integral of a function over these hyperplanes.
• So the Fourier Transform of a function is a sort of phase-weighted sum of integrals over hyperplanes, i.e. a phase-weighted sum of the Radon transform.

When evaluating the Fourier Transform of a function $f$ at a point $\rho\omega$, the hyperplane $x\cdot\omega = s$ appears with “weight” $e^{-i\rho x\cdot\omega} = e^{-i\rho s}$. So we can write the Fourier transform of $f$ in terms of the Radon Transform as follows:

$\hat{f}(\rho\omega) = \int_{-\infty}^{\infty}e^{-i\rho s}Rf(s, \omega) ds = \mathfrak{F}_sRf(\rho, \omega)$

The same arguments apply to the X-Ray transform.  In fact, we can compute the Radon transform from the X-Ray transform at a hyperplane by integrating $Xf$ over a set of parallel lines that covers the hyperplane (there is more than one way to do this!).

To compute the Fourier Transform directly from the X-Ray transform, consider the following.  Say we want to compute the Fourier Transform of $f$ at a frequency vector $\eta$.  We can pick any orthogonal $\theta$, and have $\eta \in \theta^\perp$.  Now the “wave” with frequency $\eta$ will be constant in direction $\theta$, so to compute the Fourier Transform of $f$ at $\eta$ we just need to add up the integrals of $f$ along the lines in direction $\theta$$Xf(y, \theta)$ — weighted it by the value of the wave on these lines.  In other words

$\hat{f}(\eta) = \int_{\theta^\perp}e^{-iy\cdot\eta}Xf(y,\theta)dH_{\theta^\perp}(y) = \mathfrak{F}_{\theta^\perp}Xf(\eta, \theta)$

for $\eta \in \theta^\perp$.  Notice that our choice of $\theta$ was arbitrary, so this really gives us a continuum of formulas, one for each $\theta \in \eta^\perp$.