H_1(B, {{\mathbb C}})\oplus  H_1(B', {{\mathbb C}})\to  H_1(G, {{\mathbb C}}).$$ It is easy to see that $$\dim(H_1(B, {{\mathbb C}}) )=\dim (H_1(B',{{\mathbb C}}))=\dim (H_1(T,{{\mathbb C}}))=1$$ Since the quotient $G/[G,G]$ is finite we see that $H_1(G, {{\mathbb C}})=0$. Therefore $\dim(H_1(G,L))=2$.

On the other hand we have an exact sequence $$0 \to  {\mathcal K}_{\Omega}\to {\mathcal S}({\Omega}) \to L \to  0$$ and therefore an exact sequence $$H_1(G, {\mathcal S}({\Omega}))\to H_1(G, L )\to 
H_0 (G,  {\mathcal K}_{\Omega})\to H_0(G ,{\mathcal S}({\Omega})).$$ Since $H_1(G, {\mathcal S}({\Omega}))=H_1(T, {{\mathbb C}})$ we see $\dim(H_1(G, {\mathcal S}({\Omega}))=1$.

The map $a:H_1(G, {\mathcal S}({\Omega}))\to H_1(G, L )$ is an embedding.

It is sufficient to show that the map $H_1(G, {\mathcal S}({\Omega}))\to H_1(G, {\mathcal S}(G/B))$ induced by the composition $p_\star\circ a:L\to {\mathcal S}(G/B)$ is an embedding.

Since $H_1(G, {\mathcal S}({\Omega}))=H_1(T, {{\mathbb C}})$, $H_1(G, {\mathcal S}(G/B ))=H_1(B, {{\mathbb C}})$ and $H_{>0}(U,{{\mathbb C}})=0$, we see that this map is an isomorphism.

We can now finish the proof of Proposition \[regular\]. Since $G$ acts transitively on $G/B$ the map $\mu \to \int\limits _{G/B}\mu$ defines an isomorphism $H_0(G,{\mathcal S}(G/B))\to {{\mathbb C}}$. On the other hand, since $\int \limits_{{\Omega}}\nu =0$ for any $\nu \in {\mathcal K}_{\Omega}$ the map $H_0 (G, {\mathcal K}_{\Omega})\to H_0(G ,{\mathcal S}({\Omega}))$ equals zero. So $\dim  (H_0(G, 