Kac Moody Algebras from Generalized Cartan Matrices

Let $V={\mathbb{C}}^n$ have basis $v_i$, and define $p_0:V\to {\mathbb{C}}^n$ by $p_0(v_i)=(a_{i1},\dots ,a_{in})$. Choose a complement $C$ of $\mathrm{im}{p}_0$ by finite elimination. On $H=V\oplus C$ put $p(v,c)=p_0(v)+c$, $h_i=(v_i,0)$ and ${\alpha }_j={\mathrm{pr}}_{j}p$. The map $p$ is onto, so its coordinate functionals are independent; the $h_i$ are independent and have the prescribed evaluations. Moreover $\dim H=n+(n-r)$.

In any realization with independent roots, $p:H\to {\mathbb{C}}^n$, $h\mapsto ({\alpha }_j(h){)}_j$, is onto. Its restriction to $V=\mathrm{span}(h_i)$ has rank $r$, so $\dim (H\mathrm{/}V)\ge n-r$. This proves the lower bound. At equality, $\ker p\subset V$, because $\dim \ker p=n-r=\dim \ker (p{\mathrm{\mid }}_V)$.

For two minimal realizations choose the same complement $C$ of the common row image in ${\mathbb{C}}^n$. Lift a basis of $C$ to each $H$ using surjectivity of $p$. The resulting linear sections $s:C\to H$ give $H=V\oplus s(C)$: an intersection vector has image both in $C$ and in the row image, hence zero; injectivity of $p$ on $s(C)$ then kills it. Dimensions give spanning. The map $h_i\mapsto h_i^{\mathrm{\prime }}$, $s(c)\mapsto s^{\mathrm{\prime }}(c)$ is invertible and commutes with $p$, so preserves every ${\alpha }_j$. All selections are finite Gaussian elimination.

Replace an adjacent inversion $x_j{x}_i$, $j>i$, by $x_i{x}_j+[x_j,x_i]$, expanding the bracket into basis vectors. On each term order the measure lexicographically by word length and number of inversions. The switched term decreases inversions; every bracket term decreases length. Each replacement has finitely many terms. The finitely branching reduction tree has finite depth: otherwise recursively selecting its first child with arbitrarily deep descendants gives an infinite descending sequence of measures. Thus reduction terminates in ordered words.

Two reductions on disjoint pairs commute, including their lower-length terms. The only overlapping pair is $zyx$ with $z>y>x$. Reducing the length-three terms in the two orders gives respectively $xyz+[y,x]z+y[z,x]+[z,y]x$ and $xyz+x[z,y]+[z,x]y+z[y,x]$. Their difference is $[[y,x],z]+[y,[z,x]]+[[z,y],x]$ after reducing length-two commutators. This is zero by Jacobi. Those length-two reductions are already unambiguous by induction on length, and this reasoning also holds inside a fixed word context.

Induct on the reduction measure to compare any two first reductions: the disjoint case joins exactly, and the overlap difference has zero normal form by step 2.1; all subsequent comparisons involve smaller measures. Linearity then gives a unique normal form $N$ for every finite polynomial. For any words $u,v$, $N(u(x_j{x}_i-x_i{x}_j-[x_j,x_i])v)=0$; the relations with the other order follow by antisymmetry and those with equal indices are zero. Hence $N$ kills the two-sided defining ideal. Conversely each reduction changes a polynomial by an element of that ideal. The ordered-word inclusion and $N$ are inverse maps after taking the quotient. In particular distinct length-one basis vectors remain independent.

Fix one homogeneous component $V$. Its spanning finite bracket words inherit a finite or countable enumeration; retaining each first word outside the span of its predecessors gives an ordered basis $b_0,b_1,\dots$ of $V$. For a specified subspace $W\le V$, put $E_r=\mathrm{span}(b_0,\dots ,b_r)$ and $W_r=W\cap E_r$. These finite-dimensional spaces exhaust $W$, and $\dim W_r-\dim W_{r-1}$ is zero or one. Starting with the empty basis, do nothing when the dimension is unchanged. When it rises at $r$, finite row reduction gives the unique $v_r\in W_r$ whose $b_r$-coefficient is $1$ and whose coefficients in the previous pivot columns are zero; append $v_r$. Indeed, existence comes from normalizing any element of $W_r\setminus W_{r-1}$ and eliminating its old pivots, while two such vectors differ by an element of $W_{r-1}$ with every pivot coefficient zero and hence are equal. Induction now shows that the vectors obtained through stage $r$ form a basis of $W_r$, so their union is a basis of $W$. Extend it to a basis of $V$ by scanning the $b_r$ and retaining the first vectors outside the span already obtained; the images of the added vectors form a quotient basis of $V\mathrm{/}W$. Applying this fixed construction to the supplied countable list of degrees gives compatible homogeneous bases without a family of choices. The same finite-coordinate exhaustion handles a countably spanned ungraded algebra. No basis for an arbitrary unbased vector space is asserted.

Evaluate a formal bracket recursively under any linear map $V\to L$. Bilinearity, antisymmetry and Jacobi vanish on evaluation because they hold in $L$. Thus evaluation factors uniquely through a Lie homomorphism $F(V)\to L$. The quotient itself has an antisymmetric bilinear bracket satisfying Jacobi by its defining relations. Bracket words form a countable spanning list, so first independent words supply a basis.

The linear inclusion $V\to U(F(V))$ gives an associative map $T(V)\to U(F(V))$. Conversely step 1.1 applied to the commutator Lie algebra of $T(V)$ gives $F(V)\to T(V)$, hence an associative map $U(F(V))\to T(V)$ by the tensor-quotient relations. Both composites fix $V$. The algebra $T(V)$ is generated by $V$; $U(F(V))$ is also generated by $V$ because each bracket word is an associative commutator polynomial. Therefore the composites are identities.

PBW for the basis selected in step 1.1 injects $F(V)$ into $U(F(V))$. Composing with the isomorphism of step 2.1 proves injectivity into $T(V)$. Every element of its image is a linear combination of bracket words, and every such word is in the image, proving the image description.

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.

Write $x\in J$ as ${\sum }_{\beta \in S}{x}_{\beta }$ with finite support. Choose $h\in \mathfrak{h}$ for which the distinct numbers $\beta (h)$ are pairwise different. Such an $h$ exists: the product of the finitely many nonzero linear polynomials $\beta -\gamma$ is nonzero over the infinite field $\mathbb{C}$, and a nonzero polynomial cannot vanish at all complex tuples (induct on the number of variables). Applying ${\prod }_{\gamma \mathrm{\ne }\beta }(\mathrm{ad}h-\gamma (h))\mathrm{/}(\beta (h)-\gamma (h))$ to $x$ extracts $x_{\beta }$ and keeps it in $J$. Thus $J$ is graded.

If $J\cap \mathfrak{h}=0$, every vector of $J$ has zero degree-zero component by step 1.1. The algebraic sum of all these ideals consists of finite sums of their vectors, so it also has zero degree-zero component. It is an ideal because bracketing distributes over a finite sum, and it contains every such ideal. This proves existence, maximality and uniqueness of $\mathfrak{r}$.

Cartan and positive generators preserve ${\mathfrak{r}}^{+}$. Bracketing a degree $\beta >0$ vector with $f_i$ gives degree $\beta -{\alpha }_i$. If this degree is zero, the result vanishes by step 2.1; if it has mixed signs it vanishes by F1; it cannot be strictly negative unless $\beta$ were zero or a forbidden fractional multiple of ${\alpha }_i$. The remaining degree is positive. Hence ${\mathfrak{r}}^{+}$ is stable under all generators and is an ideal. The sign-changing involution proves the negative assertion. The direct sum follows from F1.

Quotient each homogeneous component of F2 by its intersection with $\mathfrak{r}$. The split graded ideal has no zero component, so the resulting zero part is $\mathfrak{h}$ and all remaining parts have strictly positive or strictly negative degree. Every element still has finite support.

Every bracket of length $m$ is a linear combination of right-nested brackets of length $m$: apply $[[u,v],w]=[u,[v,w]]-[v,[u,w]]$ repeatedly to reduce the left bracket length. There are $n^m$ choices of letters for such a bracket. Thus the total height-$m$ subspace of either half, and hence each of its degree quotients, has dimension at most $n^m$.

Independence of the roots makes a lattice point on $\mathbb{C}{\alpha }_i$ an integer multiple of ${\alpha }_i$. A bracket word at positive degree $k{\alpha }_i$ uses only $e_i$; all words of length $k>1$ vanish since $[e_i,e_i]=0$. Degree ${\alpha }_i$ is spanned by $e_i$ and is nonzero since $[e_i,f_i]=h_i\mathrm{\ne }0$. The negative statement follows in the same way. If $\beta >0$ is not ${\alpha }_i$ and its simple reflection is a root, some coefficient at $j\mathrm{\ne }i$ is positive and remains unchanged by the reflection; the one-sign property forces the reflected root to stay positive.

Decompose $x$ into finitely many weights. For fixed $i$, the brackets of its distinct components with $f_i$ have distinct weights, so each bracket is zero. It suffices to treat a homogeneous $x$ of positive degree $\beta$. Let $J$ be the span of $x$ and all iterated $\mathrm{ad}{e}_i$ applied to $x$. Every such vector has degree $\beta +\gamma$ with $\gamma \in Q^{+}$, so $J\subset {\mathfrak{n}}^{+}$.

The space $J$ is stable under all $e_i$ by construction and under $\mathfrak{h}$ by its homogeneous spanning vectors. Induct on the number of positive adjoints to prove stability under $f_i$. The base is $[f_i,x]=0$. For $y$ already treated, $[f_i,[e_j,y]]=-{\delta }_{ij}[h_i,y]+[e_j,[f_i,y]]$ lies in $J$. Thus $J$ is an ideal, and $J\cap \mathfrak{h}=0$ by its positive degrees. F1 forces $J=0$, hence $x=0$. The sign-changing involution interchanges the two conclusions.

The relations of F3 descend through the quotient of F2. For operators $E,F,H$ with $[E,F]=H$ and $[H,F]=-2F$, the identity $[E,F^m]=m{F}^{m-1}(H-m+1)$ follows from $[E,F^{m+1}]=[E,F^m]F+F^{m}H$ and $HF=F(H-2)$, starting at $m=1$. Take the adjoint operators of $e_i,f_i,h_i$. On $v=f_j$, $Ev=0$ and $Hv=-a_{ij}v$. For $m=1-a_{ij}$, the identity gives $E{F}^{m}v=m(-a_{ij}-m+1)F^{m-1}v=0$.

For $k\mathrm{\notin }\{i,j\}$, $\mathrm{ad}{e}_k$ commutes with $\mathrm{ad}{f}_i$ and kills $f_j$, so kills $F^{m}v$. For $k=j$, it commutes with $F$ and sends $f_j$ to $h_j$, so sends $F^{m}v$ to $F^m{h}_j$. Now $F{h}_j=a_{ji}{f}_i$ and $F^2{h}_j=0$. If $m\ge 2$ this vanishes; if $m=1$, $a_{ij}=0$ and the symmetric-zero axiom gives $a_{ji}=0$. Thus all $e_k$ kill the negative Serre vector.

F1 kills this vector in ${\mathfrak{n}}^{-}$. The sign-changing involution descending in F2 proves the positive relation with the same exponent. At no point has an ideal been asserted to be generated by these vectors.

For $D=\mathrm{ad}{e}_i$, F2 gives nilpotence on each $e_j$; the relations give $D{f}_j={\delta }_{ij}{h}_i$, $D^2{f}_i=-2e_i$, $D^3{f}_i=0$, and $D^{2}h=0$. For any derivation, $D^m[x,y]={\sum }_{k=0}^m\left(\genfrac{}{}{0px}{}{m}{k}\right)[D^{k}x,D^{m-k}y]$, proved by induction and Pascal addition. Hence nilpotence on generators propagates to every finite bracket word and every finite sum. The same holds for $\mathrm{ad}{f}_i$. Their pointwise finite exponentials preserve brackets by the binomial identity, with inverses obtained by negating the derivation.

Set $T_i=\exp (\mathrm{ad}{f}_i)\exp (-\mathrm{ad}{e}_i)\exp (\mathrm{ad}{f}_i)$. On the triple, the finite expansions use $\exp (\mathrm{ad}{f}_i)e_i=e_i-h_i-f_i$, $\exp (\mathrm{ad}{f}_i)h_i=h_i+2f_i$, and $\exp (-\mathrm{ad}{e}_i)f_i=f_i-h_i-e_i$. Substitution gives $T_i{e}_i=-f_i$, $T_i{f}_i=-e_i$, and $T_i{h}_i=-h_i$. It fixes $\ker {\alpha }_i$ in $\mathfrak{h}$. Writing $h=(h-{\alpha }_i(h)h_i\mathrm{/}2)+{\alpha }_i(h)h_i\mathrm{/}2$ gives $T_{i}h=h-{\alpha }_i(h)h_i$.

For $x\in {\mathfrak{g}}_{\beta }$, $[h,T_{i}x]=T_i[T_i^{-1}h,x]=\beta (s_{i}h)T_{i}x=(s_i\beta )(h)T_{i}x$. The inverse automorphism gives a bijection between these spaces. Products of $T_i$ implement each word in the generators of $W$, proving root and multiplicity invariance. Only the induced weight action, not independence of the lift from a word, is required.

If $DA$ is symmetric, extend $({\alpha }_i,{\alpha }_j)=d_i{a}_{ij}$ bilinearly. For $\lambda$ in the root span, $({\alpha }_i,\lambda )=d_i\lambda (h_i)$ and $({\alpha }_i,{\alpha }_i)=2d_i$. Expanding $(\lambda -\lambda (h_i){\alpha }_i,\mu -\mu (h_i){\alpha }_i)$ cancels the two cross terms against $2d_i\lambda (h_i)\mu (h_i)$, leaving $(\lambda ,\mu )$. This computation allows a degenerate form.

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.