Fix $g\in {\Omega}$ and let $T$ be the centralizer of $G$, thus ${\Omega}\cong G/T$. In view of Remark \[c\_ell\] it suffices to consider the case when the split rank of $T$ equals one; we also assume without loss of generality that $G$ is almost simple.

There are exactly two parabolic subgroups $P,\, P'\subsetneq G$ containing $T$. Let $U$, $U'$ be their unipotent radicals.

Consider the complex $$\label{compl_ae}
 0\to {\mathcal K}_{\Omega}\to {\mathcal S}(G/T)\to {\mathcal S}(G/TU)\oplus {\mathcal S}(G/TU')\to {{\mathbb C}}\to 0;$$ here ${\mathcal S}$ stands for the space of locally constant compactly supported measures as before, the third arrow send $\phi$ to $(pr_*(\phi), pr'_*(\phi))$, where $pr:G/T\to G/TU$, $pr':G/T\to G/TU'$ are the projections and the fourth arrow sends $(\phi, \phi')$ to $\int \phi-\int \phi'$.

\[compl\_ae\_lem\] The complex is exact.

Exactness at all the terms except for ${\mathcal S}(G/TU)\oplus {\mathcal S}(G/TU')$ is clear. Suppose that $(\psi,\psi'):{\mathcal S}(G/TU)\oplus {\mathcal S}(G/TU')\to {{\mathbb C}}$ is a linear functional vanishing on the image of ${\mathcal S}(G/T)$. Let $\tilde\psi:{\mathcal S}(G)\to {{\mathbb C}}$ be the composition of the direct image map ${\mathcal S}(G)\to {\mathcal S}(G/TU)$ with $\psi$. Then $\psi$ is a right $TU$ invariant generalized function on $G$. On the other hand, $-\psi$ is equal to the composition of the 