Kac Moody Algebras from Generalized Cartan Matrices — Examples

There are no distinct indices, so no Serre relations. The Cartan relations are $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Their right sides lie in the span of $e,h,f$, which is closed under brackets and contains all generators; thus the algebra has dimension at most three.

In traceless $2\times 2$ matrices put $E=E_{12}$, $F=E_{21}$ and $H=E_{11}-E_{22}$. Matrix multiplication gives $EF=E_{11}$, $FE=E_{22}$, $HE=E$, $EH=-E$, $HF=-F$, and $FH=F$. Thus $[E,F]=H$, $[H,E]=2E$ and $[H,F]=-2F$. These three matrices are independent and span every traceless matrix. The presentation map is onto this three-dimensional space and step 1.1 bounds its source dimension by three, so it is an isomorphism.

Set $z=[e_1,e_2]$. The two positive relations say $[e_1,z]=[e_2,z]=0$. Hence the span of $e_1,e_2,z$ is a Lie subalgebra containing the positive generators and equals the positive half. The same argument gives at most three dimensions for the negative half. Since $\det A=3$, the Cartan has dimension two, so F1 gives $\dim \mathfrak{g}\le 8$.

Take $e_1=E_{12}$, $e_2=E_{23}$, $f_1=E_{21}$, $f_2=E_{32}$, $h_1=E_{11}-E_{22}$ and $h_2=E_{22}-E_{33}$. The identity $[E_{ab},E_{cd}]={\delta }_{bc}{E}_{ad}-{\delta }_{da}{E}_{cb}$ gives $[e_1,e_2]=E_{13}$, $[f_2,f_1]=E_{31}$, $[e_i,f_j]={\delta }_{ij}{h}_i$, and zero second brackets $[E_{12},E_{13}]=[E_{23},E_{13}]=0$, with the analogous negative zeros. For a diagonal $h=\mathrm{diag}(t_1,t_2,t_3)$, $[h,E_{ab}]=(t_a-t_b)E_{ab}$. Thus ${\alpha }_1=t_1-t_2$, ${\alpha }_2=t_2-t_3$ give $({\alpha }_j(h_i){)}_{ij}=A$. All presentation relations hold, so F1 gives a homomorphism.

The six off-diagonal matrix units and the two displayed diagonal matrices are independent and span all traceless matrices. Step 1.2 therefore gives a surjection onto an eight-dimensional algebra; step 1.1 makes it an isomorphism. The diagonal commutator formula assigns the six weights stated in the example to these six units; the diagonal subspace has weight zero. Thus the root list is exhaustive, with each root multiplicity one.

The second matrix row is the negative of the first, and the first is nonzero; hence $\mathrm{rank}A=1$. Its diagonal is 2 and both off-diagonal entries are −2, so it is a GCM. Here $n=2$, and F1 gives minimal dimension $4-1=3$.

The coordinate values give ${\alpha }_0(h_0)=2$, ${\alpha }_1(h_0)=-2$, ${\alpha }_0(h_1)=-2$, ${\alpha }_1(h_1)=2$. If $a{\alpha }_0+b{\alpha }_1=0$, evaluation on $d$ gives $a=0$, then on $h_0$ gives $b=0$. Thus the two roots, as well as the supplied two coroots, are independent. Both roots annihilate the nonzero vector $h_0+h_1$, while $({\alpha }_0+{\alpha }_1)(d)=1$. Omitting $d$ would make the roots negatives of one another and destroy independence. The data are therefore a minimal realization with an essential complementary direction.

The matrix meets every GCM condition, is connected and symmetric, has determinant $4-9=-5$, and $A(1,1{)}^t=(-1,-1{)}^t$. Hence it is indefinite by F4.

The two positive Serre generators have degrees $(4,1)$ and $(1,4)$, each of total height five. Every element of their generated positive ideal is a linear combination of these and positive adjoints, so has no component below height five. The free bracket $[e_1,e_2]$ is nonzero: its tensor image is the difference of the distinct words $e_1{e}_2-e_2{e}_1$. F1 therefore ensures its degree-(1,1) class survives. It is a root vector of weight $\beta$.

F2 gives $(\beta ,\beta )=2+2-3-3=-2$. For each reflection, $({\alpha }_i,\lambda )=\lambda (h_i)$ and $({\alpha }_i,{\alpha }_i)=2$ make expansion of $(\lambda -\lambda (h_i){\alpha }_i{)}^2$ equal to $(\lambda ,\lambda )$. Thus every Weyl translate of a simple root has squared length 2. The root from step 1.2 has length −2, so cannot be such a translate and is imaginary by F3.

Take $A=\left(\begin{array}{cc}2& 0\\ -1& 2\end{array}\right)$. All four entries are integers, both diagonal entries are 2, and its off-diagonal entries 0 and −1 are nonpositive. Thus every hypothesis of the false claim holds.

Here $a_{12}=0$ but $a_{21}=-1\mathrm{\ne }0$. This violates the forward implication in the symmetric-zero condition of F1, so $A$ is not a GCM. It is the required counterexample.

Let $\mathfrak{s}$ be the traceless $3\times 3$ complex matrices and put $L=(\mathbb{C}[t,t^{-1}]\otimes \mathfrak{s})\oplus \mathbb{C}c\oplus \mathbb{C}d$. Define $c$ central, $[d,t^{m}x]=m{t}^{m}x$ and $[t^{m}x,t^{n}y]=t^{m+n}[x,y]+m{\delta }_{m,-n}\mathrm{tr}(xy)c$. The central term is antisymmetric since $m=-n$ on its support. Matrix trace satisfies $\mathrm{tr}([x,y]z)=\mathrm{tr}(x[y,z])$ by cyclic multiplication. In the Jacobi sum for three loop terms, the central coefficient is zero unless $m+n+k=0$; in that case it is $((m+n)+(n+k)+(k+m))\mathrm{tr}([x,y]z)=0$. The ordinary matrix part satisfies Jacobi by associativity. The Jacobi identity involving $d$ is the derivation rule: the central term on the right has factor $m+n=0$ on its support, and the loop terms have degree $m+n$. Thus L is a Lie algebra.

Let $H_1=E_{11}-E_{22}$ and $H_2=E_{22}-E_{33}$ and take $\mathfrak{h}=\mathbb{C}{H}_1\oplus \mathbb{C}{H}_2\oplus \mathbb{C}c\oplus \mathbb{C}d$. Put $h_1=H_1$, $h_2=H_2$, $h_0=c-H_1-H_2$. On a diagonal $H=\mathrm{diag}(z_1,z_2,z_3)$ set ${\alpha }_1(H)=z_1-z_2$, ${\alpha }_2(H)=z_2-z_3$, and ${\alpha }_0(H)=z_3-z_1$; all vanish on $c$, while ${\alpha }_0(d)=1$ and ${\alpha }_1(d)={\alpha }_2(d)=0$. Direct evaluation gives diagonal Cartan entries 2 and off-diagonal entries −1. The coroots are independent because $h_0$ has nonzero c-coordinate; the roots are independent because the d-coordinate separates ${\alpha }_0$ and the two finite diagonal differences are independent. The matrix has rank two: its row sum is zero and its upper 2×2 minor is 3. Thus $\dim \mathfrak{h}=4=2\cdot 3-2$ is minimal.

Take $e_0=t{E}_{31}$, $e_1=E_{12}$, $e_2=E_{23}$ and $f_0=t^{-1}{E}_{13}$, $f_1=E_{21}$, $f_2=E_{32}$. The matrix-unit bracket formula gives $[e_0,f_0]=E_{33}-E_{11}+c=h_0$ and $[e_i,f_j]={\delta }_{ij}{h}_i$ for every other pair. The diagonal and d-actions give the stated simple weights and their negatives. The positive pair brackets are $[e_0,e_1]=t{E}_{32}$, $[e_0,e_2]=-t{E}_{21}$, $[e_1,e_2]=E_{13}$; bracketing each again with either of its two participating e-generators gives zero by the matrix-unit formula. The negative pair brackets are $[f_0,f_1]=-t^{-1}{E}_{23}$, $[f_0,f_2]=t^{-1}{E}_{12}$ and $[f_1,f_2]=-E_{31}$, with the same double-bracket zeros. No central term occurs in these repeated brackets since the relevant matrix products have trace zero. Thus all Serre relations with exponent two hold, and F1 gives a homomorphism $\varphi :\mathfrak{g}(A)\to L$ preserving the Cartan.

The degree-zero generators produce all of $\mathfrak{s}$: their brackets yield $E_{13},E_{31}$ and $H_1,H_2$. From $t{E}_{31}$, brackets with $E_{23},E_{12}$ produce $t{E}_{21},-t{E}_{32}$, then brackets with $E_{12},E_{23}$ produce $t{H}_1,t{H}_2$. Bracketing these diagonals with the degree-zero matrix units produces every $t{E}_{ij}$, since for each $i\mathrm{\ne }j$ at least one of $H_1,H_2$ has distinct i,j entries. The same argument from $t^{-1}{E}_{13}$ starts with $[E_{21},t^{-1}{E}_{13}]=t^{-1}{E}_{23}$ and $[E_{32},t^{-1}{E}_{13}]=-t^{-1}{E}_{12}$ and gives all of $t^{-1}\mathfrak{s}$. Now $[\mathfrak{s},\mathfrak{s}]=\mathfrak{s}$, since diagonal differences are opposite-unit brackets and each off-diagonal unit is a nonzero scalar multiple of its bracket with a diagonal. Induction using $[t\mathfrak{s},t^{m-1}\mathfrak{s}]=t^m\mathfrak{s}$ for $m\ge 2$, and the negative counterpart, supplies every loop degree. The supplied Cartan contains $c,d$, so $\varphi$ is onto.

The map $\varphi$ is injective on the Cartan by step 2.1. Its kernel is an ideal of $\mathfrak{g}(A)$ disjoint from that Cartan, hence zero by [F4]. Thus $\varphi$ is an isomorphism. One can also check recognition directly in L. Its nonzero Cartan weight spaces are $\mathbb{C}{t}^m{E}_{ij}$, of weight ${\epsilon }_i-{\epsilon }_j+m\delta$, and $t^m{\mathfrak{d}}_0$, of weight $m\delta$ for $m\mathrm{\ne }0$, where ${\mathfrak{d}}_0$ is the traceless diagonal space and $\delta (d)=1$, $\delta ({\mathfrak{d}}_0\oplus \mathbb{C}c)=0$. Distinct listed weights are distinct functionals. Finite interpolation extracts a nonzero weight vector from any nonzero Cartan-disjoint ideal. A vector $t^m{E}_{ij}$ brackets with $t^{-m}{E}_{ji}$ to $E_{ii}-E_{jj}+mc\mathrm{\ne }0$ in the Cartan. A vector $t^{m}H$ in the diagonal space has $m\mathrm{\ne }0$; choose a traceless diagonal $K$ with $\mathrm{tr}(HK)\mathrm{\ne }0$. Such K exists since the trace matrix on $H_1,H_2$ is $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ of determinant 3. Its bracket with $t^{-m}K$ is the nonzero Cartan vector $m\mathrm{tr}(HK)c$. Both contradict disjointness.

The root coordinates give $\delta ={\alpha }_0+{\alpha }_1+{\alpha }_2$. By step 5.1 the space at $m\delta$, $m\mathrm{\ne }0$, has basis $t^m{H}_1,t^m{H}_2$, and so dimension two. F3 gives $(\delta ,\delta )=6-6=0$. A simple reflection preserves this root form by direct expansion using $({\alpha }_i,\lambda )=\lambda (h_i)$ and $({\alpha }_i,{\alpha }_i)=2$; hence every root in $W\mathrm{\Pi }$ has squared length 2. The nonzero root $m\delta$ has squared length zero, so is imaginary by F2. This proves the counterexample for every positive or negative nonzero integer m. At m=0 the four-dimensional space is the Cartan and is not a root space.