)=0 $ we see that the map $H_1(G, {\mathcal S}({\mathcal B}) )\to H_0 (G, {\mathcal K}_0)$ is an isomorphism. Since ${\mathcal B}=G/B$ we have: $$H_1(G, {\mathcal S}({\mathcal B}) )=H_1(B,{{\mathbb C}})=H_1(T,{{\mathbb C}})={{\mathbb C}}.$$

So $\dim( H_0 (G, {\mathcal K}_0))=1$.

To conclude the argument, recall the short exact sequence $0\to {\mathcal K}_0\to {\mathcal K}_u\to {{\mathbb C}}\to 0$.

Since $H_1(G,{{\mathbb C}})=0$ we have an exact sequence:

$$0\to H_0(G,{\mathcal K}_0) \to H_0(G,{\mathcal K}_u)\to {{\mathbb C}}\to 0$$ So $\dim(H_0(G,{\mathcal K}_u)=2$.

Let $r:{\mathcal K}_{cusp}\to {\mathcal K}_u$ be the restriction and $[r]:{\mathcal K}_{cusp}\to  H_0(G,{\mathcal K}_u)$ be the composition of $r$ and projection ${\mathcal K}_u\to H_0(G,{\mathcal K}_u) $.

\[r\_onto\] The map $[r]$ is onto.

Recall the map $\tau: H_0(G,{\mathcal K}) \to {\mathcal D}^G$, $[f]\to \hat f$. Let $\bar {\mathcal D}^G$ be the space of germs of invariant distributions at $e$ and $\bar{\tau}: H_0(G,{\mathcal K}) \to \bar {\mathcal D}^G$ be the composition of $\tau$ with the restriction map.

Corollary \[extensions\] implies that the map $\bar{\tau}$ vanishes on the kernel of the map $H_0(G,{\mathcal K})\to
H_0(G,{\mathcal K}_u )$. Thus it suffices to show that $\bar{\tau}|_{{\mathcal K}_{cusp}}$ has rank at least two, i.e. that there exist irreducible cuspidal representations $\rho_1$, $\rho_2$, such that their characters restricted to any $G$-invariant open neighborhood 