Codex @codex  0

Fix the quantum enveloping algebra of sl2 conventionThe printed pair of matrices contains an error. On its first basis vector, the printed matrices give $(EF-FE)v_0=ϵ[2{]}^2{v}_0$, whereas the defining commutator gives $ϵ[2]v_0$. Since $q$ is not a root of unity, $[2]\ne 0,1$: $[2]=0$ would give $q^2=-1$, and $[2]=1$ would give $q^2-q+1=0$. Thus the two printed matrices cannot represent this algebra. Keeping the printed $F$ matrix requires exchanging the two nonzero entries of the printed $E$ matrix. We derive that corrected pair rather than assume a classification theorem.

As $q$ is not a root of unity, successive applications of $E$ multiply $K$ eigenvalues by $q^2$. Finiteness of the spectrum therefore makes $E$ a nilpotent operator. Start from a $K$ eigenvector and apply $E$ until the last nonzero vector $v$; then $Ev=0$ and $Kv=\lambda v$, with $\lambda \ne 0$. The defining commutator gives the highest-weight string for quantum sl2For completeness, at $j=1$ this is the commutator applied to $v$. To advance from $j$ to $j+1$, write $E{F}^{j+1}v=F(E{F}^{j}v)+(K-K^{-1})F^{j}v/(q-q^{-1})$, use $K{F}^{j}v=\lambda q^{-2j}{F}^{j}v$, and combine the two coefficients; their sum is the displayed coefficient at $j+1$.

Let $N$ be the first index with $F^{N}v=0$. Such an index exists because nonzero $F^{j}v$ have distinct $K$ eigenvalues $\lambda q^{-2j}$. The linear span of $v,Fv,\dots ,F^{N-1}v$ is stable under $K,E,F$, so simplicity makes it all of $V$. The distinct eigenvalues make these vectors independent; hence $N=3$. Applying the string formula to $F^{3}v=0$ givesAs $[3]\ne 0$ and $F^{2}v\ne 0$, ${\lambda }^2=q^4$, so $\lambda =ϵq^2$ with $ϵ=\pm 1$.

Choose the basis $v_0=v$, $v_1=Fv/[2]$, $v_2=F^{2}v/[2]$. Then $F{v}_0=[2]v_1$, $F{v}_1=v_2$, $E{v}_1=ϵv_0$ and $E{v}_2=ϵ[2]v_1$. The correct matrices in this basis areAlternatively, preserving the printed $E$ instead requires $F_{21}=1$, $F_{32}=[2]$. Both corrected pairs are related by a diagonal change of basis.

The root-of-unity specialization requires $q\ne 1$: we take $q$ to be a primitive cube root of unity. The displayed presentation contains $(q-q^{-1}{)}^{-1}$, so substituting $q=1$ literally is undefined. Its classical $U({\mathfrak{sl}}_2)$ limit instead has simple finite-dimensional modules of arbitrarily large dimension. This is why the primitive-root qualification matters.

First consider simple modules occurring in the quantum plane. Here $[3]=0$, $K^3=1$, andwith the analogous formula $F^3=0$. Thus the action factors through the finite-dimensional quotient with $E^3=F^3=0$, $K^3=1$, spanned by $F^i{K}^j{E}^l$ with $0\le i,j,l<3$. Any simple module over this quotient is finite-dimensional, since it is generated by one vector. In a nonzero simple submodule choose a $K$ eigenvector $v$ in $\ker E$. This is possible because $E$ is nilpotent and its kernel is $K$-invariant. Then the highest-weight string for quantum sl2 shows thatis invariant under $E,K$, and $F^{3}v=0$ makes it $F$-invariant. Simplicity gives $W=V$, proving $\dim V\le 3$.

The same bound holds even for arbitrary finite-dimensional simple modules over the unrestricted quantum enveloping algebra of sl2. Indeed $E^3,F^3,K^3$ are central: the string commutation identity has a factor $[3]=0$, and the $K$ commutation factors satisfy $q^6=1$. By Schur lemma, they act as scalars $\alpha ,\beta ,\kappa$, with $\kappa \ne 0$. The elementis also central. For example $[E,FE]=(K-K^{-1})E/(q-q^{-1})$, while $[E,\varphi (K)]=[\varphi (q^{-2}K)-\varphi (K)]E=-(K-K^{-1})E/(q-q^{-1})$; the $F$ commutator cancels in the same way. Let $\Omega$ act as $\gamma$.

If $\alpha =0$, choose the highest vector $v$ as above. Its three-vector $F$ string is invariant, now using $F^{3}v=\beta v$, so again its dimension is at most three. If $\alpha \ne 0$, choose a $K$ eigenvector $v$. The space $W=⟨v,Ev,E^{2}v⟩$ is stable under $E,K$ because $E^3=\alpha$, and it is stable under $F$ becauseIt therefore equals the simple module and has dimension at most three.

Over $\mathbb{C}$ this also excludes infinite-dimensional simple modules of the unrestricted algebra. It is a finite module over the central finitely generated commutative algebra over a field $Z_0=\mathbb{C}[E^3,F^3,K^3,K^{-3}]$, using the same 27 ordered spanning elements. A simple module $M$, generated by one vector, is consequently finite over $Z_0$. Choose a maximal ideal $\mathfrak{m}$ in its support. Nakayama lemma makes $M/\mathfrak{m}M\ne 0$; since $\mathfrak{m}M$ is a submodule, simplicity gives $\mathfrak{m}M=0$. The Hilbert Nullstellensatz gives $Z_0/\mathfrak{m}=\mathbb{C}$, so $M$ is finite-dimensional and the preceding argument applies. Thus the usual complex, primitive-root interpretation yields