Imaginary root spaces need not have multiplicity one

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.