Contragredient algebra has a triangular decomposition

On $T({\mathbb{C}}^n)$ with letters $v_j$, let $F_j$ be left concatenation and let $H_h$ multiply a word of degree $-\beta$ by $(\lambda -\beta )(h)$, for any fixed $\lambda \in {\mathfrak{h}}^{\ast }$. Define $E_{i}1=0$ and $E_i(v_{j}a)=v_j{E}_i(a)+{\delta }_{ij}{H}_{h_i}(a)$. Thus $[E_i,F_j]={\delta }_{ij}{H}_{h_i}$ directly, the $H_h$ commute, and $[H_h,F_j]=-{\alpha }_j(h)F_j$. Every term of $E_i(a)$ has weight equal to the weight of $a$ plus ${\alpha }_i$, by induction on word length (the extra term occurs only when $i=j$). Hence $[H_h,E_i]={\alpha }_i(h)E_i$. All defining relations hold and give a representation for each $\lambda$.

A negative bracket word acts by left multiplication by the same tensor commutator word. Its value on 1 is that word. The composite of the free negative Lie algebra with this evaluation is the injection of F2, so its map into ${\tilde{\mathfrak{n}}}^{-}$ is injective as well as surjective. Also $H_{h}1=\lambda (h)$ for every $\lambda$, so a Cartan element killed by the presentation must be zero. The assignment $e_i\mapsto -f_i$, $f_i\mapsto -e_i$, $h\mapsto -h$ preserves each relation: for example $[-f_i,-e_j]=-{\delta }_{ij}{h}_i$. Its square is the identity. It proves freeness of the positive half too.

Jacobi gives $[f_i,[e_j,u]]=-{\delta }_{ij}[h_i,u]+[e_j,[f_i,u]]$. Induction on the length of the positive word $u$ therefore gives $[f_i,{\tilde{\mathfrak{n}}}^{+}]\subset \mathfrak{h}+{\tilde{\mathfrak{n}}}^{+}$. The analogous inclusion holds with signs reversed. The span $S={\tilde{\mathfrak{n}}}^{-}+\mathfrak{h}+{\tilde{\mathfrak{n}}}^{+}$ is consequently stable under brackets with every generator, so iterated brackets span only $S$ and $S=\tilde{\mathfrak{g}}$.

If $u_{-}+h+u_{+}=0$, evaluation on 1 in step 1.1 gives $u_{-}(1)+\lambda (h)1=0$. The first term has positive tensor length, so both terms vanish. Varying $\lambda$ kills $h$, and F2 kills $u_{-}$. Then $u_{+}=0$ as well. Homogeneity of the defining ideal gives a direct grading; Jacobi gives $[h,x]=\beta (h)x$ in degree $\beta$. Independence of the ${\alpha }_i$ identifies distinct degrees with distinct weights. Thus the displayed grading has exactly the asserted signs and zero part.