The opposite Borels of a symmetrizable Kac–Moody algebra are root-degreewise dual Lie bialgebras

Put $H=\mathrm{span}\{h_i:i\in I\}$ and define $\rho :\mathfrak{h}\to {\mathbb{C}}^I$ by $\rho (x)=({\alpha }_i(x){)}_{i\in I}$. By [F1], $\rho$ is onto and $\rho (H)=\mathrm{im}(A^{\mathsf{T}})$ of dimension $r$. For the set $J$ in [F4], projection of this image to ${\mathbb{C}}^J$ is an isomorphism: it is surjective because its restriction to the $J$-coordinate subspace has matrix $A_J^{\mathsf{T}}$, and both spaces have dimension $r$. Hence $\mathrm{im}(A^{\mathsf{T}})\cap {\mathbb{C}}^{I\setminus J}=0$. For each $j\mathrm{\notin }J$, choose $D_j$ with $\rho (D_j)$ the $j$th coordinate vector. Their span intersects $H$ trivially, and its dimension $\mathrm{\mid }I\setminus J\mathrm{\mid }=\mathrm{\mid }I\mathrm{\mid }-r$ makes it a complement to $H$; the form normalization gives $(h_i\mathrm{\mid }{D}_j)={\delta }_{ij}\mathrm{/}{d}_i$. If $r=\mathrm{\mid }I\mathrm{\mid }$ this is the empty complementary family and $\mathfrak{h}=H$.

By [F3], the invariant form has perfect opposite-root pairings and is nondegenerate on the Cartan subalgebra. Also ${\mathfrak{n}}^{\pm }$ are positively and negatively root-graded, respectively.

For fixed $\beta ={\sum }_i{m}_i{\alpha }_i\in Q_{+}$, the degree-$\beta$ words in the finite simple-generator tensor algebra are finite in number, so the separate Serre presentations [F2] make $U({\mathfrak{n}}^{+})[\beta ]$ finite-dimensional; the same argument applies to $U({\mathfrak{n}}^{-})[-\beta ]$. The height-zero component is $\mathbb{C}1$ on both sides.

Each half is countably spanned by its finite bracket words. Enumerating those words by length and lexicographic order and retaining the first vectors outside the preceding span gives a countable ordered basis; applying [F5] provides the PBW basis and the symmetrization isomorphism for both halves. This construction uses no choice principle.

For $v_1,\dots ,v_m\in {\mathfrak{n}}^{+}$ and $w_1,\dots ,w_{\mathrm{\ell }}\in {\mathfrak{n}}^{-}$, define a pairing on the symmetric algebras to be zero when $m\mathrm{\ne }\mathrm{\ell }$, and when $m=\mathrm{\ell }$ put $⟨v_1\cdots v_m,w_1\cdots w_m{⟩}_S=\frac{1}{m!}{\sum }_{\sigma \in S_m}{\prod }_{t=1}^m(v_t\mathrm{\mid }{w}_{\sigma (t)})$, with the empty products paired as $1$. It is well defined under permutations of each list.

In a fixed root degree $\beta$ and symmetric length $m$, only finitely many tuples of positive roots sum to $\beta$. The tensor-product pairings on each such tuple are perfect by [F3]; averaging over $S_m$ identifies coinvariants with invariants because $m!\mathrm{\ne }0$ in $\mathbb{C}$, so the induced pairing on $S^m({\mathfrak{n}}^{+})[\beta ]\times S^m({\mathfrak{n}}^{-})[-\beta ]$ is perfect. Summing over the finitely many lengths $0\le m\le \mathrm{ht}(\beta )$ gives a perfect pairing on the full symmetric-algebra root components.

Let ${\mathfrak{h}}_0$ be a second copy of $\mathfrak{h}$ and define $B_{\mathfrak{d}}((x,a),(y,b))=\frac{1}{2}((x\mathrm{\mid }y)-(a\mathrm{\mid }b))$ on $\mathfrak{d}=\mathfrak{g}\oplus {\mathfrak{h}}_0$ with the componentwise bracket. Jacobi holds componentwise, and [F3] makes this form invariant, symmetric and nondegenerate. The maps ${\iota }_{+}$ and ${\iota }_{-}$ in (ii) are Lie homomorphisms because $\mathfrak{h}$ is abelian and the bracket of two Borel elements has zero Cartan component. Their images are complementary: the positive and negative root components split by the triangular decomposition, and the two Cartan copies split into diagonal and antidiagonal subspaces. Each image is isotropic, since the invariant form is orthogonal between the Cartan and nonzero root spaces and vanishes on pairs of positive roots or on pairs of negative roots. The cross pairing is $(h\mathrm{\mid }{h}^{\mathrm{\prime }})+\frac{1}{2}(x_{+}\mathrm{\mid }{x}_{-})$; it is degreewise perfect by [F3]. Within either Borel, a degree has only finitely many decompositions into degrees in its root cone, and all pieces are finite-dimensional. Thus these images satisfy the same-side finiteness required of the root-graded Manin triple; the whole double need not have finite decompositions.

Transport the invariant-form symmetric-power pairing of steps 1.5–1.6 through the PBW symmetrization isomorphisms of step 1.4. They preserve root degree, so they give a perfect pairing of $U({\mathfrak{n}}^{+})[\beta ]$ with $U({\mathfrak{n}}^{-})[-\beta ]$ for every $\beta \in Q_{+}$. Taking the direct sum of these finite-dimensional dualities gives the restricted graded-dual isomorphism in (i), in both directions; at $\beta =0$ it is the pairing $⟨1,1⟩=1$.

By [F6], the transposed brackets give dual Lie bialgebra structures on the two Borel copies. For $h\in \mathfrak{h}$, every bracket of two elements of ${\mathfrak{b}}^{-}$ has either negative root degree or is zero in degree zero, so its cross pairing with ${\iota }_{+}(h)$ vanishes; hence ${\delta }^{+}(h)=0$. The same argument gives ${\delta }^{-}(h)=0$. For a Cartan vector $h$, the determinant pairing gives $⟨e_i\wedge h_i,f_i\wedge h⟩={\alpha }_i(h)\mathrm{/}(2d_i^2)$, whereas $⟨e_i,[f_i,h]⟩={\alpha }_i(h)\mathrm{/}(2d_i)$. There are no other possible degree decompositions of the simple root except its simple-root and Cartan parts. Hence transposition gives ${\delta }^{+}(e_i)=d_i{e}_i\wedge h_i$, proving (ii) with the normalization used by the formal shuffle coproduct.