Lemma 3

Put $L = (\mathfrak{n}_-U)\cap U^0$.

(i) We have $L = (U\mathfrak{n}_+)\cap U^0$, and L is a two-sided ideal of $U^0$.

(ii) We have $U^0= V\oplus L$.

It is clear that $\mathfrak{n}_-U$ (resp. $U\mathfrak{n}_+$) is the set of linear combinations of the elements $u((q_i),(m_i),(p_i))$ such that $\sum q_i>0$ (resp. $\sum p_i>0$). On the other hand

This implies that $(\mathfrak{n}_-U)\cap U^0= (U\mathfrak{n}_+)\cap U^0$. Finally, $(\mathfrak{n}_-U)\cap U^0$ (resp. $(U\mathfrak{n}_+)\cap U^0$) is a right (resp. left) ideal of $U^0$, hence (i). Further, an element $u((q_i),(m_i),(p_i))$ that is in $U^0$ belongs to V (resp. to L) if and only if $p_1=\cdots =p_n=q_1=\cdots =q_n= 0$ (resp. $p_1+\cdots +p_n+q_1+\cdots +q_n>0$), hence (ii). Q.E.D.

In view of Lemma 3, the projection of $U^0$ onto V with kernel L is a homomorphism of algebras. It is called the Harish-Chandra homomorphism from $U^0$ to V (relative to B). Recall that V can be identified with the algebra of polynomial functions on $\mathfrak{h}^*$.