Helgason Fourier transform

Remarks

The quantity $A(x, b)$ is well defined: replacing $g$ by $g k'$ with $k' \in K$, or $k$ by $k m$ with $m \in M$, does not change $A(k^{-1} g)$, because $M$ normalizes $N$ and commutes with $A$. Geometrically, $A(x, b)$ is a vector-valued signed distance from the origin to the horocycle through $x$ with normal $b$.

If $f$ is bi-$K$-invariant, that is, radial on $X$, then $\tilde f(\lambda, b)$ does not depend on $b$ and equals the spherical transform $\hat f(\lambda)$, by Harish-Chandra's integral formula for spherical functions. With suitable normalizations of the measures, the inversion formula is $$f(x) = \frac{1}{\abs{W}} \int_{\mathfrak a^*} \int_B \tilde f(\lambda, b) \, e^{(i \lambda + \rho)(A(x, b))} \, \abs{c(\lambda)}^{-2} \, d\lambda \, db,$$ where $W$ is the Weyl group and $c$ the Harish-Chandra c-function.

Example: for $\mathrm{SL}(2, \R)$ acting on the upper half plane $\set{z : \operatorname{Im} z > 0}$, $B$ is the boundary circle. For the boundary point $b = \infty$, the function $e^{\rho(A(z, b))}$ is a power of $\operatorname{Im} z$, constant on the horocycles $\operatorname{Im} z = \text{const}$.