Invariant bilinear form for a symmetrizable kac moody algebra

The prescribed Cartan form is symmetric on $H$: ${\alpha }_i(h_j)\mathrm{/}{d}_i=a_{ji}\mathrm{/}{d}_i=a_{ij}\mathrm{/}{d}_j$. Thus it defines a symmetric form on $H\oplus {\mathfrak{h}}^{\mathrm{\prime }\mathrm{\prime }}$. Put $p(h)=({\alpha }_i(h){)}_i$. Independence makes $p$ onto; $p{\mathrm{\mid }}_H$ has rank $r$, so both $\ker p$ and $\ker (p{\mathrm{\mid }}_H)$ have dimension $n-r$ and are equal. A vector annihilating the full Cartan form pairs to zero with all $h_i$, hence lies in $\ker p\subset H$. Write it $\sum c_i{h}_i$. Pairing with arbitrary $h$ gives $\sum (c_i\mathrm{/}{d}_i){\alpha }_i(h)=0$, whence all $c_i=0$. This proves Cartan nondegeneracy, even when $A$ is singular.

Use the principal height grading $\mathfrak{g}={⨁}_m{\mathfrak{g}}_m$. At heights $0,\pm 1$, set $(e_i,f_j)={\delta }_{ij}\mathrm{/}{d}_i$ and all unequal-total-height pairs to zero. Invariance with a Cartan element reduces to $(h,[e_i,f_j])={\delta }_{ij}{\alpha }_i(h)\mathrm{/}{d}_i=([h,e_i],f_j)$; all other possible nonzero triples at this stage are permutations of this equality. Suppose pairings up to height $N-1$ are defined and invariant whenever all relevant degrees have absolute value at most $N-1$. For $x\in {\mathfrak{g}}_N$ and $y={\sum }_a[u_a,v_a]\in {\mathfrak{g}}_{-N}$, with $u_a,v_a$ of strictly negative smaller heights, prescribe $(x,y)={\sum }_a([x,u_a],v_a)$ and extend symmetrically. Both arguments on the right have smaller absolute heights. Such bracket expressions exist since each half is generated in height one.

To check independence, write $x={\sum }_b[w_b,z_b]$ with positive smaller heights. For a single term on each side, the induction hypothesis, symmetry and Jacobi give $([[w,z],u],v)=([w,u],[z,v])-([z,u],[w,v])=(w,[u,[z,v]]-[v,[z,u]])=(w,[z,[u,v]])$. All inner mixed brackets have smaller absolute heights; each use of invariance therefore belongs to the induction hypothesis. Summing shows the proposed value equals ${\sum }_b(w_b,[z_b,y])$, which depends only on $y$ for a fixed expression of $x$. It was defined using actual $x$, so it is independent of both expressions. This argument also shows the value is zero when either sum of bracket expressions is zero.

Invariance for total height different from zero is automatic from orthogonality. If one of the three absolute heights is $N$ and the others are smaller, the equality follows from the definition when the extreme-height entry is first or last. For a middle entry $y=[u,v]$, expand $([x,[u,v]],z)=([v,x],[u,z])+([x,u],[v,z])=(x,[v,[z,u]]+[u,[v,z]])=(x,[[u,v],z])$ using the smaller-height invariance and Jacobi. If two heights are $N$, the third is zero. For $h\in \mathfrak{h}$, $y=[u,v]$ of height $N$ and $z$ of height $-N$, the same calculation gives $([h,[u,v]],z)=([h,u],[v,z])-([h,v],[u,z])=(h,[u,[v,z]]-[v,[u,z]])=(h,[[u,v],z])$. Symmetry and antisymmetry give all permutations, including a middle Cartan entry. Thus the induction extends full invariance at height $N$.

For homogeneous root weights, invariance implies $0=([h,x],y)+(x,[h,y])=(\alpha +\beta )(h)(x,y)$; hence unequal opposite weights pair to zero. Let $R$ be the radical. Invariance makes it an ideal, and step 1.1 gives $R\cap \mathfrak{h}=0$. If $R$ were nonzero, finite weight interpolation would give a nonzero homogeneous vector in it; take one of minimal positive absolute height. For a positive weight use the positive clause of F3; for a negative weight use its negative clause. Every opposite simple bracket is in $R$ of lower height, or in its zero Cartan intersection, so is zero. F3 then kills that vector, a contradiction. Hence the form is nondegenerate. Together with weight orthogonality and finite dimensions from F2 this proves perfect opposite-root pairings.

The Cartan map obeys $\nu (h_i)={\alpha }_i\mathrm{/}{d}_i$, so ${\nu }^{-1}({\alpha }_i)=d_i{h}_i$ and $({\alpha }_i,{\alpha }_j)=d_i{a}_{ij}$. For opposite vectors, $([x,y],h)=(x,[y,h])=\alpha (h)(x,y)$ for every $h$. The bracket lies in the Cartan by F2; step 1.1 identifies it as $(x,y){\nu }^{-1}(\alpha )$. Any invariant extension of the prescribed Cartan form has the height-one values of step 2.1, weight orthogonality, and the recursion of that step. Induction therefore proves uniqueness.