Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 2 / 4

Use the quantum plane $A=k⟨x,y⟩/(yx-qxy)$ and the quantum enveloping algebra of sl2 comultiplication from Question 2. The generator actions are $Kx=qx$, $Ky=q^{-1}y$, $Ex=0$, $Ey=x$, $Fx=y$, $Fy=0$. Extend them by the module algebra rulesThese descend to $A$: $E(yx)=q{x}^2=E(qxy)$ and $F(yx)=q{y}^2=F(qxy)$, while $K$ preserves $yx-qxy$. For the ordered monomial basis, induction with $[s+1]=q^{-1}[s]+q^s$ gives the quantum enveloping algebra action on the quantum planeTerms with a negative exponent have zero coefficients and are interpreted as zero.

We now verify the defining relations explicitly on every basis vector, rather than infer them merely from the generator actions. The $K$ weight changes by $q^2$ under $E$ and $q^{-2}$ under $F$, so $KE=q^{2}EK$ and $KF=q^{-2}FK$. AlsoThe middle identity follows by inserting $[j]=(q^j-q^{-j})/(q-q^{-1})$ and cancelling the terms $q^{r+s+1}$ and $q^{-r-s-1}$; the remaining numerator is $(q-q^{-1})(q^{r-s}-q^{s-r})$. Finally $K{K}^{-1}=1$. Thus all the rank-one defining relations hold. The phrase “Serre relations” here refers to these rank-one presentation relations; there are no additional relations involving distinct simple roots for ${\mathfrak{sl}}_2$.

Every ordered monomial in the degree-$n$ homogeneous component has the form $x^{n-j}{y}^j$, $0\le j\le n$. The formulas in part (a) showEach nonzero image still has total degree $n$; $K^{-1}$ does too. Since $E,F,K,K^{-1}$ generate the quantum enveloping algebra of sl2, the degree-$n$ component is a submodule, with basis $x^n,x^{n-1}y,\dots ,y^n$ and dimension $n+1$.

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

At a primitive cube root of unity, $[1]=1$, $[2]=-1$, $[3]=0$. In the degree-three quantum plane module, put $z_0=x^3$, $z_1=x^{2}y$, $z_2=x{y}^2$, $z_3=y^3$. The complete actions areHence $W=⟨z_0,z_3⟩$ is a two-dimensional trivial submodule; every line in $W$ is a simple module. The quotient by $W$ is a two-dimensional simple module, with distinct $K$ eigenvalues $q,q^{-1}$ and nonzero maps connecting its two weight spaces.

Let $S$ be a simple module that is a submodule of this cubic component. If $S\cap W\ne 0$, simplicity makes $S\subseteq W$, so $S$ is one of its lines. If $S\cap W=0$, the quotient map injects $S$ into the simple quotient. Its nonzero image is the entire quotient, so $\dim S=2$. Since $K$ is diagonalizable, such an $S$ must contain the unique $q$ weight space $\mathbb{C}{z}_1$. But $E{z}_1=z_0\ne 0$, which then belongs to $S\cap W$, a contradiction. Therefore the cubic-root quantum-plane socle gives the complete answer:All simple submodules lie in the two-dimensional $W$, so their sum cannot equal the four-dimensional cubic component. It is not a direct sum of simple modules. The obstruction is the nonzero extension to the simple two-dimensional quotient, not a failure of that quotient to be simple.