The beautiful result to which I refer is that on a (locally) symmetric space $G/H$, $G$-invariant differential forms are automatically closed, hence represent cohomology. So you just have to use some Lie theory to find invariant forms.
More is, in fact, true. The cohomology is just given by the exterior algebra of invariant forms.
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This will tell you that certain structure constants $c^\gamma_{\alpha\beta} = 0$, where $\alpha,\beta,\gamma$ are in the $\mathfrak p$ range.
(Unless I'm messing up.)
Now dualize and think about the Maurer-Cartan forms $\omega^\alpha,\omega^\mu$. What we just said about structure constants says that $d\omega^\alpha \equiv 0\pmod {\omega^\mu}$.
Any form on $G/H$ is cohomologous to an invariant form (standard argument you've given here before). But any invariant form must be a polynomial with constant coefficients in the $\omega_\alpha$, and this computation shows it must be closed.
So, anyhow, any invariant form is closed. But that also tells you that the only exact invariant forms are $0$. So the complex of invariant forms gives the deRham cohomology.
You didn't make it into the paper acknowledgements since I ended up excising Gray, I think. But it would be a lie to not acknowledge you in the thesis, where acknowledgements (to me) representat people who made it possible.
@TedShifrin my source says that if I use a wrongly scaled $\mathrm dx$ then I get $\hat{\hat{f}}(x) = rf(-x)$ for $r>0$, but I'm wondering whether that should be $f(-rx)$ instead
