$f':=f-\sum _i ( f_i^{g_i}-f_i)$ satisfies the conditions of Corollary.

\[N’\] The space $H_0(G,{\mathcal K}_u)$ is two dimensional.

We first show that $\dim (H_0(G,{\mathcal K}_0))=1$.

Let ${\mathcal B}$ be the variety of Borel subgroups, $p:N\to {\mathcal B}$ the map which associates to $u\in N$ the Borel subgroup containing $u$. By definition we have an exact sequence $$0\to {\mathcal K}_0\to {\mathcal S}(N)\to {\mathcal S}({\mathcal B})\to 0$$ and therefore an exact sequence $$H_1(G, {\mathcal S}(N) )\to H_1(G, {\mathcal S}({\mathcal B}) )\to H_0 (G, {\mathcal K}_0)\to H_0(G ,{\mathcal S}(N))\to   H_0 (G ,{\mathcal S}({\mathcal B})).$$

$H_1(G, {\mathcal S}(N) )=0 $.

Fix a Borel subgroup $B=TU$. We can write $U$ as a union of open compact subgroup $U_1\subset ...U_n\subset ...$. Therefore ${\mathcal S}(N)={\mathcal S}(G/U)$ is the direct limit of ${\mathcal S}(G/U_n)$. Since the functor $M\mapsto H_1(G,M)$ commutes with direct limits it is sufficient to show that $H_1(G, {\mathcal S}(G/U_n) )=\{ 0\}$. Since $U_n\subset G$ is a compact subgroup the space $H_1(G, {\mathcal S}(G/U_n) )$ is a direct summand of ${\mathcal S}(G)$. Since $H_1(G, {\mathcal S}(G) )=0$ the Lemma is proven.

Since $G$ acts transitively on $N$ and on ${\mathcal B}$ we have $H_0 (G ,{\mathcal S}(N)){{\widetilde \longrightarrow}}H_0 (G ,{\mathcal S}({\mathcal B}))={{\mathbb C}}.$ Since $H_1(G, {\mathcal S}(N) 